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2.1 Two charges 5 × 10–8 C and –3 × 10–8 C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

2.2 A regular hexagon of side 10 cm has a charge 5 mC at each of its vertices. Calculate the potential at the centre of the hexagon.

2.2 INSTANTANEOUS VELOCITY AND SPEED The average velocity tells us how fast an object has been moving over a given time interval but does not tell us how fast it moves at different instants of time during that interval. For this, we define instantaneous velocity or simply velocity v at an instant t. The velocity at an instant is defined as the limit of the average velocity as the time interval ∆t becomes infinitesimally small. In other words, ∆ x v = lim (2.1a) ∆ t → 0 ∆ t Fig. 2.1 Determining velocity from position-time d x = (2.1b) graph. Velocity at t = 4 s is the slope of the dt tangent to the graph at that instant. lim where the symbol stands for the operation ∆→t 0 Now, we decrease the value of ∆t from 2 s to 1 of taking limit as ∆tg0 of the quantity on its s. Then line P1P2 becomes Q1Q2 and its slope right. In the language of calculus, the quantity gives the value of the average velocity over on the right hand side of Eq. (2.1a) is the the interval 3.5 s to 4.5 s. In the limit ∆t → 0, differential coefficient of x with respect to t and the line P1P2 becomes tangent to the position- d x time curve at the point P and the velocity at t is denoted by (see Appendix 2.1). It is the d t = 4 s is given by the slope of the tangent at rate of change of position with respect to time, that point. It is difficult to show this process graphically. But if we useat that instant. numerical method to obtain the value of We can use Eq. (2.1a) for obtaining the the velocity, the meaning of the limiting value of velocity at an instant either process becomes clear. For the graph shown graphically or numerically. Suppose that we in Fig. 2.1, x = 0.08 t3. Table 2.1 gives the want to obtain graphically the value of value of ∆x/∆t calculated for ∆t equal to 2.0 s, velocity at time t = 4 s (point P) for the motion 1.0 s, 0.5 s, 0.1 s and 0.01 s centred at t = of the car represented in Fig.2.1 calculation. 4.0 s. The second and third columns give theLet us take ∆t = 2 s centred at t = 4 s. Then, t  t    ∆ ∆by the definition of the average velocity, the t + t −     and t 2 = and the value of t1= 2 2slope of ( Fig. 2.1) gives the value of     line P1P2 average velocity over the interval 3 s to 5 s. fourth and the fifth columns give the ∆x Table 2.1 Limiting value of at t = 4 s ∆ t Reprint 2025-26 MOTION IN A STRAIGHT LINE 15 3 a + 16b – a – 4b corresponding values of x, i.e. x (t1) = 0.08 t1 = = 6.0 × b 2.0 -1 ⊳and x (t2) = 0.08 t23. The sixth column lists the = 6.0 × 2.5 =15 m s difference ∆x = x (t2) – x (t1) and the last column gives the ratio of ∆x and ∆t, i.e. the Note that for uniform motion, velocity is average velocity corresponding to the value the same as the average velocity at all of ∆t listed in the first column. instants. We see from Table 2.1 that as we decrease Instantaneous speed or simply speed is the the value of ∆t from 2.0 s to 0.010 s, the value of magnitude of velocity. For example, a velocity of the average velocity approaches the limiting + 24.0 m s–1 and a velocity of – 24.0 m s–1 — value 3.84 m s–1 which is the value of velocity at both have an associated speed of 24.0 m s-1. It should be noted that though average speed over dx t = 4.0 s, i.e. the value of at t = 4.0 s. In this a finite interval of time is greater or equal to the dt magnitude of the average velocity, manner, we can calculate velocity at each instantaneous speed at an instant is equal to instant for motion of the car. the magnitude of the instantaneous velocity at The graphical method for the determination that instant. Why so ? of the instantaneous velocity is always not a 2.3 ACCELERATIONconvenient method. For this, we must carefully plot the position–time graph and calculate the The velocity of an object, in general, changes value of average velocity as ∆t becomes smaller during its course of motion. How to describe and smaller. It is easier to calculate the value this change? Should it be described as the rate of velocity at different instants if we have data of change in velocity with distance or with of positions at different instants or exact time ? This was a problem even in Galileo’s expression for the position as a function of time. time. It was first thought that this change could Then, we calculate ∆x/∆t from the data for be described by the rate of change of velocity decreasing the value of ∆t and find the limiting with distance. But, through his studies of value as we have done in Table 2.1 or use motion of freely falling objects and motion of differential calculus for the given expression and objects on an inclined plane, Galileo concluded that the rate of change of velocity with time is dx calculate at different instants as done in a constant of motion for all objects in free fall. dt On the other hand, the change in velocity with the following example. distance is not constant – it decreases with the ⊳ increasing distance of fall. This led to the Example 2.1 The position of an object concept of acceleration as the rate of change moving along x-axis is given by x = a + bt2 of velocity with time. where a = 8.5 m, b = 2.5 m s–2 and t is The average acceleration a over a time interval measured in seconds. What is its velocity at is defined as the change of velocity divided by t = 0 s and t = 2.0 s. What is the average the time interval : velocity between t = 2.0 s and t = 4.0 s ? v 2 – v1 ∆v (2.2)Answer In notation of differential calculus, the a = = t 2 – t1 ∆tvelocity is where v2 and v1 are the instantaneous velocities dx d 2 -1 2b t = 5.0 t m s or simply velocities at time t2 and t1 . It is thev = = ( a + bt ) = dt dt average change of velocity per unit time. The SI At t = 0 s, v = 0 m s–1 and at t = 2.0 s, unit of acceleration is m s–2 . v = 10 m s-1 . On a plot of velocity versus time, the average acceleration is the slope of the straight line x ( 4.0 ) − x ( 2.0 )Average velocity = connecting the points corresponding to (v2, t2) 4.0 − 2.0 and (v1, t1). Reprint 2025-26 16 PHYSICS Instantaneous acceleration is defined in the (c) An object is moving in negative direction same way as the instantaneous velocity : with a negative acceleration. ∆v d v (d) An object is moving in positive direction lim a = = (2.3) till time t1, and then turns back with the d t ∆→ t 0 ∆ t same negative acceleration. The acceleration at an instant is the slope of the tangent to the v–t curve at that An interesting feature of a velocity-time instant. graph for any moving object is that the area Since velocity is a quantity having both under the curve represents the magnitude and direction, a change in displacement over a given time interval. A velocity may involve either or both of these general proof of this statement requires use of factors. Acceleration, therefore, may result calculus. We can, however, see that it is true from a change in speed (magnitude), a for the simple case of an object moving with change in direction or changes in both. Like constant velocity u. Its velocity-time graph is velocity, acceleration can also be positive, as shown in Fig. 2.4. negative or zero. Position-time graphs for motion with positive, negative and zero acceleration are shown in Figs. 2.4 (a), (b) and (c), respectively. Note that the graph curves upward for positive acceleration; downward for negative acceleration and it is a straight line for zero acceleration. Although acceleration can vary with time, our study in this chapter will be restricted to motion with constant acceleration. In this case, the average acceleration equals the constant value of acceleration during the interval. If the velocity of an object is vo at t = 0 and v at time t, we have v − v 0 a = or, v = v 0 + a t (2.4) t − 0 Fig. 2.3 Velocity–time graph for motions with Fig. 2.2 Position-time graph for motion with constant acceleration. (a) Motion in positive (a) positive acceleration; (b) negative direction with positive acceleration, acceleration, and (c) zero acceleration. (b) Motion in positive direction with Let us see how velocity-time graph looks like negative acceleration, (c) Motion in for some simple cases. Fig. 2.3 shows velocity- negative direction with negative acceleration, (d) Motion of an object withtime graph for motion with constant acceleration negative acceleration that changesfor the following cases : direction at time t1. Between times 0 to (a) An object is moving in a positive direction t1, it moves in positive x - direction with a positive acceleration. and between t1 and t2 it moves in the (b) An object is moving in positive direction opposite direction. with a negative acceleration. Reprint 2025-26 MOTION IN A STRAIGHT LINE 17 Fig. 2.4 Area under v–t curve equals displacement of the object over a given time interval. The v-t curve is a straight line parallel to the time axis and the area under it between t = 0 and t = T is the area of the rectangle of height u and base T. Therefore, area = u × T = uT which Fig. 2.5 Area under v-t curve for an object with is the displacement in this time interval. How uniform acceleration. come in this case an area is equal to a distance? Think! Note the dimensions of quantities on the two coordinate axes, and you will arrive at As explained in the previous section, the area the answer. under v-t curve represents the displacement. Therefore, the displacement x of the object is : Note that the x-t, v-t, and a-t graphs shown in several figures in this chapter have sharp 1 x = ( v – v 0 ) t + v 0 t (2.5)kinks at some points implying that the 2 functions are not differentiable at these But v − v 0 = a tpoints. In any realistic situation, the functions will be differentiable at all points 1 2 Therefore, x = a t + v 0 tand the graphs will be smooth. 2 What this means physically is that 1 2 or, x = v 0 t + at (2.6)acceleration and velocity cannot change 2 values abruptly at an instant. Changes are Equation (2.5) can also be written as always continuous. v + v 0 x = t = v t (2.7a)2.4 KINEMATIC EQUATIONS FOR 2 UNIFORMLY ACCELERATED MOTION where, For uniformly accelerated motion, we can derive some simple equations that relate displacement v + v 0 v = (constant acceleration only)(x), time taken (t), initial velocity (v0), final 2 velocity (v) and acceleration (a). Equation (2.4) (2.7b) already obtained gives a relation between final and initial velocities v and v0 of an object moving Equations (2.7a) and (2.7b) mean that the object with uniform acceleration a : has undergone displacement x with an average velocity equal to the arithmetic average of the v = v0 + at (2.4) initial and final velocities. From Eq. (2.4), t = (v – v0)/a. Substituting this in This relation is graphically represented in Fig. 2.5. Eq. (2.7a), we get The area under this curve is : Area between instants 0 and t = Area of triangle  v + v 0   v − v 0  v 2 − v 02 x = v t =ABC + Area of rectangle OACD  2   a = 2a 1 2 2 = (v – v 0 ) t + v 0 t v = v 0 + 2ax (2.8) 2 Reprint 2025-26 18 PHYSICS This equation can also be obtained by t v 0 + at ) d tsubstituting the value of t from Eq. (2.4) into Eq. = ∫ 0 ( (2.6). Thus, we have obtained three important equations : 1 2 x – x 0 = v 0 t + a t 2 v = v 0 + at 1 2 1 2 x = x 0 + v 0 t + a t x = v 0t + at 2 2 We can write v 2 = v 02 + 2ax (2.9a) d v d v d x d v a = = = v d t d x d t d x connecting five quantities v0, v, a, t and x. These or, v dv = a dxare kinematic equations of rectilinear motion for Integrating both sides,constant acceleration. The set of Eq. (2.9a) were obtained by v x v d v = a d xassuming that at t = 0, the position of the particle, ∫ v 0 ∫ x 0 x is 0. We can obtain a more general equation if we take the position coordinate at t = 0 as non- v 2 – v 02 = a ( x – x 0 ) zero, say x0. Then Eqs. (2.9a) are modified 2 (replacing x by x – x0 ) to : 2 2 v = v 0 + 2a ( x – x 0 ) v = v 0 + at The advantage of this method is that it can be used 1 2 for motion with non-uniform acceleration x = x 0 + v 0t + at (2.9b) also. 2 Now, we shall use these equations to some v 2 = v 02 + 2a ( x − x 0 ) (2.9c) important cases. ⊳ ⊳ ⊳ Example 2.3 A ball is thrown vertically Example 2.2 Obtain equations of motion upwards with a velocity of 20 m s–1 from for constant acceleration using method of the top of a multistorey building. The calculus. height of the point from where the ball is thrown is 25.0 m from the ground. (a) How high will the ball rise ? and (b) how longAnswer By definition will it be before the ball hits the ground? d v Take g = 10 m s–2. a = d t dv = a dt Answer (a) Let us take the y-axis in the Integrating both sides vertically upward direction with zero at the v t ∫ v 0 d v = ∫ 0 a d t ground, as shown in Fig. 2.6. Now vo = + 20 m s–1, t d t (a is a = – g = –10 m s–2, = a ∫ 0 v = 0 m s–1 constant) If the ball rises to height y from the point of v – v 0 = at launch, then using the equation 2 + 2 a 0 ( y – y 0 ) v = v 0 + at v 2 = v we get d x Further, v = 0 = (20)2 + 2(–10)(y – y0) d t Solving, we get, (y – y0) = 20 m. dx = v dt Integrating both sides (b) We can solve this part of the problem in two x t ways. Note carefully the methods used. ∫ x 0 dx = ∫v0 d t Reprint 2025-26 MOTION IN A STRAIGHT LINE 19 0 = 25 +20 t + (½) (-10) t2 Or, 5t2 – 20t – 25 = 0 Solving this quadratic equation for t, we get t = 5s Note that the second method is better since we do not have to worry about the path of the motion as the motion is under constant acceleration. ⊳ ⊳ Example 2.4 Free-fall : Discuss the motion of an object under free fall. Neglect air resistance. Answer An object released near the surface of the Earth is accelerated downward under the influence of the force of gravity. The magnitude of acceleration due to gravity is represented by g. If air resistance is neglected, the object is Fig. 2.6 said to be in free fall. If the height through which the object falls is small compared to the FIRST METHOD : In the first method, we split earth’s radius, g can be taken to be constant, the path in two parts : the upward motion (A to equal to 9.8 m s–2. Free fall is thus a case of B) and the downward motion (B to C) and motion with uniform acceleration. calculate the corresponding time taken t1 and We assume that the motion is in y-direction, t2. Since the velocity at B is zero, we have : more correctly in –y-direction because we v = vo + at choose upward direction as positive. Since the 0 = 20 – 10t1 acceleration due to gravity is always downward, Or, t1 = 2 s it is in the negative direction and we have This is the time in going from A to B. From B, or a = – g = – 9.8 m s–2 the point of the maximum height, the ball falls The object is released from rest at y = 0. Therefore, freely under the acceleration due to gravity. The v0 = 0 and the equations of motion become: ball is moving in negative y direction. We use v = 0 – g t = –9.8 t m s–1equation y = 0 – ½ g t2 = –4.9 t 2 m 1 2 y = y 0 + v 0t + at v2 = 0 – 2 g y = –19.6 y m2 s–2 2 These equations give the velocity and the We have, y0 = 45 m, y = 0, v0 = 0, a = – g = –10 m s–2 distance travelled as a function of time and also 0 = 45 + (½) (–10) t2 2 the variation of velocity with distance. The Solving, we get t2 = 3 s variation of acceleration, velocity, and distance, with time have been plotted in Fig. 2.7(a), (b)Therefore, the total time taken by the ball before and (c). it hits the ground = t1 + t2 = 2 s + 3 s = 5 s. SECOND METHOD : The total time taken can also be calculated by noting the coordinates of initial and final positions of the ball with respect to the origin chosen and using equation 1 2 y = y0 + v 0t + at 2 Now y0 = 25 m y = 0 m vo = 20 m s-1, a = –10m s–2, t = ? (a) Reprint 2025-26 20 PHYSICS traversed during successive intervals of time. Since initial velocity is zero, we have 2 y = −1 gt 2 Using this equation, we can calculate the position of the object after different time intervals, 0, τ, 2τ, 3τ… which are given in second column of Table 2.2. If we take (–1/ 2) gτ2 as y0 — the position coordinate after first time interval τ, then third column gives (b) the positions in the unit of yo. The fourth column gives the distances traversed in successive τs. We find that the distances are in the simple ratio 1: 3: 5: 7: 9: 11… as shown in the last column. This law was established by Galileo Galilei (1564-1642) who was the first to make quantitative studies of free fall. ⊳ ⊳ Example 2.6 Stopping distance of vehicles : When brakes are applied to a moving vehicle, the distance it travels before stopping is called stopping distance. It is (c) an important factor for road safety and depends on the initial velocity (v0) and theFig. 2.7 Motion of an object under free fall. braking capacity, or deceleration, –a that (a) Variation of acceleration with time. (b) Variation of velocity with time. is caused by the braking. Derive an (c) Variation of distance with time ⊳ expression for stopping distance of a vehicle in terms of vo and a. ⊳ Example 2.5 Galileo’s law of odd Answer Let the distance travelled by the vehicle numbers : “The distances traversed, during before it stops be ds. Then, using equation of equal intervals of time, by a body falling 2 motion v2 = vo + 2 ax, and noting that v = 0, we from rest, stand to one another in the same have the stopping distance ratio as the odd numbers beginning with unity [namely, 1: 3: 5: 7…...].” Prove it. – v 02 d s = 2aAnswer Let us divide the time interval of motion of an object under free fall into many Thus, the stopping distance is proportional to equal intervals τ and find out the distances the square of the initial velocity. Doubling the Table 2.2 Reprint 2025-26 MOTION IN A STRAIGHT LINE 21 initial velocity increases the stopping distance by a factor of 4 (for the same deceleration). For the car of a particular make, the braking distance was found to be 10 m, 20 m, 34 m and 50 m corresponding to velocities of 11, 15, 20 and 25 m/s which are nearly consistent with the above formula. Stopping distance is an important factor considered in setting speed limits, for example, in school zones. ⊳ ⊳ Example 2.7 Reaction time : When a situation demands our immediate action, it takes some time before we really respond. Reaction time is the time a person takes to observe, think Fig. 2.8 Measuring the reaction time. and act. For example, if a person is driving and suddenly a boy appears on the road, then the time elapsed before Answer The ruler drops under free fall. he slams the brakes of the car is the Therefore, vo = 0, and a = – g = –9.8 m s–2. The reaction time. Reaction time depends distance travelled d and the reaction time tr are on complexity of the situation and on related by an individual. You can measure your reaction time by a simple experiment. Take a ruler and ask your friend to drop it vertically through the gap between Or, your thumb and forefinger (Fig. 2.8). Given d = 21.0 cm and g = 9.8 m s–2 the reaction After you catch it, find the distance d time is travelled by the ruler. In a particular case, d was found to be 21.0 cm. ⊳ Estimate reaction time. SUMMARY 1. An object is said to be in motion if its position changes with time. The position of the object can be specified with reference to a conveniently chosen origin. For motion in a straight line, position to the right of the origin is taken as positive and to the left as negative. The average speed of an object is greater or equal to the magnitude of the average velocity over a given time interval. 2. Instantaneous velocity or simply velocity is defined as the limit of the average velocity as the time interval ∆t becomes infinitesimally small : ∆ x d x v = lim v = lim = ∆→t 0 ∆→t 0 ∆t d t The velocity at a particular instant is equal to the slope of the tangent drawn on position-time graph at that instant. Reprint 2025-26 22 PHYSICS 3. Average acceleration is the change in velocity divided by the time interval during which the change occurs : ∆ v a = ∆t 4. Instantaneous acceleration is defined as the limit of the average acceleration as the time interval ∆t goes to zero : ∆v d v a = lim a = lim = ∆→t 0 ∆→t 0 ∆ t d t The acceleration of an object at a particular time is the slope of the velocity-time graph at that instant of time. For uniform motion, acceleration is zero and the x-t graph is a straight line inclined to the time axis and the v-t graph is a straight line parallel to the time axis. For motion with uniform acceleration, x-t graph is a parabola while the v-t graph is a straight line inclined to the time axis. 5. The area under the velocity-time curve between times t1 and t2 is equal to the displacement of the object during that interval of time. 6. For objects in uniformly accelerated rectilinear motion, the five quantities, displacement x, time taken t, initial velocity v0, final velocity v and acceleration a are related by a set of simple equations called kinematic equations of motion : v = v0 + at 1 2 x = v0 t + at 2 v 2 = v 02 + 2ax if the position of the object at time t = 0 is 0. If the particle starts at x = x0 , x in above equations is replaced by (x – x0). Reprint 2025-26 MOTION IN A STRAIGHT LINE 23 POINTS TO PONDER 1. The origin and the positive direction of an axis are a matter of choice. You should first specify this choice before you assign signs to quantities like displacement, velocity and acceleration. 2. If a particle is speeding up, acceleration is in the direction of velocity; if its speed is decreasing, acceleration is in the direction opposite to that of the velocity. This statement is independent of the choice of the origin and the axis. 3. The sign of acceleration does not tell us whether the particle’s speed is increasing or decreasing. The sign of acceleration (as mentioned in point 3) depends on the choice of the positive direction of the axis. For example, if the vertically upward direction is chosen to be the positive direction of the axis, the acceleration due to gravity is negative. If a particle is falling under gravity, this acceleration, though negative, results in increase in speed. For a particle thrown upward, the same negative acceleration (of gravity) results in decrease in speed. 4. The zero velocity of a particle at any instant does not necessarily imply zero acceleration at that instant. A particle may be momentarily at rest and yet have non-zero acceleration. For example, a particle thrown up has zero velocity at its uppermost point but the acceleration at that instant continues to be the acceleration due to gravity. 5. In the kinematic equations of motion [Eq. (2.9)], the various quantities are algebraic, i.e. they may be positive or negative. The equations are applicable in all situations (for one dimensional motion with constant acceleration) provided the values of different quantities are substituted in the equations with proper signs. 6. The definitions of instantaneous velocity and acceleration (Eqs. (2.1) and (2.3)) are exact and are always correct while the kinematic equations (Eq. (2.9)) are true only for motion in which the magnitude and the direction of acceleration are constant during the course of motion. Reprint 2025-26 24 PHYSICS EXERCISES 2.1 In which of the following examples of motion, can the body be considered approximately a point object: (a) a railway carriage moving without jerks between two stations. (b) a monkey sitting on top of a man cycling smoothly on a circular track. (c) a spinning cricket ball that turns sharply on hitting the ground. (d) a tumbling beaker that has slipped off the edge of a table. 2.2 The position-time (x-t) graphs for two children A and B returning from their school O to their homes P and Q respectively are shown in Fig. 2.9. Choose the correct entries in the brackets below ; (a) (A/B) lives closer to the school than (B/A) (b) (A/B) starts from the school earlier than (B/A) (c) (A/B) walks faster than (B/A) (d) A and B reach home at the (same/different) time (e) (A/B) overtakes (B/A) on the road (once/twice). Fig. 2.9 2.3 A woman starts from her home at 9.00 am, walks with a speed of 5 km h–1 on a straight road up to her office 2.5 km away, stays at the office up to 5.00 pm, and returns home by an auto with a speed of 25 km h–1. Choose suitable scales and plot the x-t graph of her motion. 2.4 A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1 m long and requires 1 s. Plot the x-t graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 13 m away from the start. 2.5 A car moving along a straight highway with speed of 126 km h–1 is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop ? 2.6 A player throws a ball upwards with an initial speed of 29.4 m s–1. (a) What is the direction of acceleration during the upward motion of the ball ? (b) What are the velocity and acceleration of the ball at the highest point of its motion ? (c) Choose the x = 0 m and t = 0 s to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of x-axis, and give the signs of position, velocity and acceleration of the ball during its upward, and downward motion. (d) To what height does the ball rise and after how long does the ball return to the player’s hands ? (Take g = 9.8 m s–2 and neglect air resistance). 2.7 Read each statement below carefully and state with reasons and examples, if it is true or false ; A particle in one-dimensional motion (a) with zero speed at an instant may have non-zero acceleration at that instant (b) with zero speed may have non-zero velocity, (c) with constant speed must have zero acceleration, (d) with positive value of acceleration must be speeding up. Reprint 2025-26 MOTION IN A STRAIGHT LINE 25 2.8 A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one tenth of its speed. Plot the speed-time graph of its motion between t = 0 to 12 s. 2.9 Explain clearly, with examples, the distinction between : (a) magnitude of displacement (sometimes called distance) over an interval of time, and the total length of path covered by a particle over the same interval; (b) magnitude of average velocity over an interval of time, and the average speed over the same interval. [Average speed of a particle over an interval of time is defined as the total path length divided by the time interval]. Show in both (a) and (b) that the second quantity is either greater than or equal to the first. When is the equality sign true ? [For simplicity, consider one-dimensional motion only]. 2.10 A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h–1. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h–1. What is the (a) magnitude of average velocity, and (b) average speed of the man over the interval of time (i) 0 to 30 min, (ii) 0 to 50 min, (iii) 0 to 40 min ? [Note: You will appreciate from this exercise why it is better to define average speed as total path length divided by time, and not as magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero !] Fig. 2.10 2.11 In Exercises 2.9 and 2.10, we have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider instantaneous speed and magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why? 2.12 Look at the graphs (a) to (d) (Fig. 2.10) carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle. 2.13 Figure 2.11shows the x-t plot of one- dimensional motion of a particle. Is it correct to say from the graph that the particle moves Fig. 2.11 in a straight line for t < 0 and on a parabolic path for t >0 ? If not, suggest a suitable physical context for this graph. 2.14 A police van moving on a highway with a speed of 30 km h–1 fires a bullet at a thief’s car speeding away in the same direction with a speed of 192 km h–1. If the muzzle speed of the bullet is 150 m s–1, with what speed does the bullet hit the thief’s car ? (Note: Obtain that speed which is relevant for damaging the thief’s car). Reprint 2025-26 26 PHYSICS 2.15 Suggest a suitable physical situation for each of the following graphs (Fig 2.12): Fig. 2.12 2.16 Figure 2.13 gives the x-t plot of a particle executing one-dimensional simple harmonic motion. (You will learn about this motion in more detail in Chapter13). Give the signs of position, velocity and acceleration variables of the particle at t = 0.3 s, 1.2 s, – 1.2 s. Fig. 2.13 2.17 Figure 2.14 gives the x-t plot of a particle in one-dimensional motion. Three different equal intervals of time are shown. In which interval is the average speed greatest, and in which is it the least ? Give the sign of average velocity for each interval. Fig. 2.14 2.18 Figure 2.15 gives a speed-time graph of a particle in motion along a constant direction. Three equal intervals of time are shown. In which interval is the average acceleration greatest in magnitude? In which interval is the average speed greatest ? Choosing the positive direction as the constant direction of motion, give the signs of v and a in the three intervals. What are the accelerations at the points A, B, C and D ? Fig. 2.15 Reprint 2025-26 CHAPTER THREE MOTION IN A PLANE 3.1 INTRODUCTION In the last chapter we developed the concepts of position, displacement, velocity and acceleration that are needed to 3.1 Introduction describe the motion of an object along a straight line. We 3.2 Scalars and vectors found that the directional aspect of these quantities can be taken care of by + and – signs, as in one dimension only two3.3 Multiplication of vectors by real numbers directions are possible. But in order to describe motion of an 3.4 Addition and subtraction of object in two dimensions (a plane) or three dimensions vectors — graphical method (space), we need to use vectors to describe the above- 3.5 Resolution of vectors mentioned physical quantities. Therefore, it is first necessary to learn the language of vectors. What is a vector? How to3.6 Vector addition — analytical method add, subtract and multiply vectors ? What is the result of 3.7 Motion in a plane multiplying a vector by a real number ? We shall learn this to enable us to use vectors for defining velocity and3.8 Motion in a plane with constant acceleration acceleration in a plane. We then discuss motion of an object 3.9 Projectile motion in a plane. As a simple case of motion in a plane, we shall discuss motion with constant acceleration and treat in detail3.10 Uniform circular motion the projectile motion. Circular motion is a familiar class of Summary motion that has a special significance in daily-life situations. Points to ponder We shall discuss uniform circular motion in some detail. Exercises The equations developed in this chapter for motion in a plane can be easily extended to the case of three dimensions. 3.2 SCALARS AND VECTORS In physics, we can classify quantities as scalars or vectors. Basically, the difference is that a direction is associated with a vector but not with a scalar. A scalar quantity is a quantity with magnitude only. It is specified completely by a single number, along with the proper unit. Examples are : the distance between two points, mass of an object, the temperature of a body and the time at which a certain event happened. The rules for combining scalars are the rules of ordinary algebra. Scalars can be added, subtracted, multiplied and divided Reprint 2025-26 28 PHYSICS just as the ordinary numbers*. For example, represented by another position vector, OP′ if the length and breadth of a rectangle are denoted by r′. The length of the vector r 1.0 m and 0.5 m respectively, then its represents the magnitude of the vector and its perimeter is the sum of the lengths of the direction is the direction in which P lies as seen four sides, 1.0 m + 0.5 m +1.0 m + 0.5 m = from O. If the object moves from P to P′, the 3.0 m. The length of each side is a scalar vector PP′ (with tail at P and tip at P′) is called and the perimeter is also a scalar. Take the displacement vector corresponding to another example: the maximum and motion from point P (at time t) to point P′ (at time t′). minimum temperatures on a particular day are 35.6 °C and 24.2 °C respectively. Then, the difference between the two temperatures is 11.4 °C. Similarly, if a uniform solid cube of aluminium of side 10 cm has a mass of

2.5 A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = 10–12 F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?

2.7 kg, then its volume is 10–3 m3 (a scalar) and its density is 2.7×103 kg m–3 (a scalar). A vector quantity is a quantity that has both a magnitude and a direction and obeys the triangle law of addition or equivalently the Fig. 3.1 (a) Position and displacement vectors. parallelogram law of addition. So, a vector is (b) Displacement vector PQ and different specified by giving its magnitude by a number courses of motion. and its direction. Some physical quantities that It is important to note that displacement are represented by vectors are displacement, vector is the straight line joining the initial and velocity, acceleration and force. final positions and does not depend on the actual To represent a vector, we use a bold face type path undertaken by the object between the two in this book. Thus, a velocity vector can be positions. For example, in Fig. 3.1(b), given the represented by a symbol v. Since bold face is initial and final positions as P and Q, the difficult to produce, when written by hand, a displacement vector is the same PQ for different vector is often representedrv by an arrow placedrv paths of journey, say PABCQ, PDQ, and PBEFQ.over a letter, say . Thus, both v and Therefore, the magnitude of displacement is represent the velocity vector. The magnitude of either less or equal to the path length of an a vector is often called its absolute value, object between two points. This fact was indicated by |v| = v. Thus, a vector is emphasised in the previous chapter also whilerepresented by a bold face, e.g. by A, a, p, q, r, ... discussing motion along a straight line.x, y, with respective magnitudes denoted by light face A, a, p, q, r, ... x, y. 3.2.2 Equality of Vectors 3.2.1 Position and Displacement Vectors Two vectors A and B are said to be equal if, and only if, they have the same magnitude and theTo describe the position of an object moving in same direction.**a plane, we need to choose a convenient point, say O as origin. Let P and P′ be the positions of Figure 3.2(a) shows two equal vectors A and the object at time t and t′, respectively [Fig. 3.1(a)]. B. We can easily check their equality. Shift B We join O and P by a straight line. Then, OP is parallel to itself until its tail Q coincides with that the position vector of the object at time t. An of A, i.e. Q coincides with O. Then, since their arrow is marked at the head of this line. It is tips S and P also coincide, the two vectors are represented by a symbol r, i.e. OP = r. Point P′ is said to be equal. In general, equality is indicated * Addition and subtraction of scalars make sense only for quantities with same units. However, you can multiply and divide scalars of different units. ** In our study, vectors do not have fixed locations. So displacing a vector parallel to itself leaves the vector unchanged. Such vectors are called free vectors. However, in some physical applications, location or line of application of a vector is important. Such vectors are called localised vectors. Reprint 2025-26 MOTION IN A PLANE 29 The factor λ by which a vector A is multiplied could be a scalar having its own physical dimension. Then, the dimension of λ A is the product of the dimensions of λ and A. For example, if we multiply a constant velocity vector by duration (of time), we get a displacement vector. 3.4 ADDITION AND SUBTRACTION OF VECTORS — GRAPHICAL METHOD Fig. 3.2 (a) Two equal vectors A and B. (b) Two As mentioned in section 4.2, vectors, by vectors A′ and B′ are unequal though they definition, obey the triangle law or equivalently, are of the same length. the parallelogram law of addition. We shall now describe this law of addition using the graphical as A = B. Note that in Fig. 3.2(b), vectors A′ and method. Let us consider two vectors A and B that B′ have the same magnitude but they are not lie in a plane as shown in Fig. 3.4(a). The lengths equal because they have different directions. of the line segments representing these vectors Even if we shift B′ parallel to itself so that its tail are proportional to the magnitude of the vectors. Q′ coincides with the tail O′ of A′, the tip S′ of B′ To find the sum A + B, we place vector B so that does not coincide with the tip P′ of A′. its tail is at the head of the vector A, as in 3.3 MULTIPLICATION OF VECTORS BY REAL Fig. 3.4(b). Then, we join the tail of A to the head NUMBERS of B. This line OQ represents a vector R, that is, the sum of the vectors A and B. Since, in thisMultiplying a vector A with a positive number λ procedure of vector addition, vectors aregives a vector whose magnitude is changed by the factor λ but the direction is the same as that of A : λ A = λ A if λ > 0. For example, if A is multiplied by 2, the resultant vector 2A is in the same direction as A and has a magnitude twice of |A| as shown in Fig. 3.3(a). Multiplying a vector A by a negative number −λ gives another vector whose direction is opposite to the direction of A and whose magnitude is λ times |A|. Multiplying a given vector A by negative numbers, say –1 and –1.5, gives vectors as shown in Fig 3.3(b). (c) (d) Fig. 3.3 (a) Vector A and the resultant vector after multiplying A by a positive number 2. Fig. 3.4 (a) Vectors A and B. (b) Vectors A and B (b) Vector A and resultant vectors after added graphically. (c) Vectors B and A multiplying it by a negative number –1 added graphically. (d) Illustrating the and –1.5. associative law of vector addition. Reprint 2025-26 30 PHYSICS arranged head to tail, this graphical method is What is the physical meaning of a zero vector? called the head-to-tail method. The two vectors Consider the position and displacement vectors and their resultant form three sides of a triangle, in a plane as shown in Fig. 3.1(a). Now suppose so this method is also known as triangle method that an object which is at P at time t, moves to of vector addition. If we find the resultant of P′ and then comes back to P. Then, what is its B + A as in Fig. 3.4(c), the same vector R is displacement? Since the initial and final obtained. Thus, vector addition is commutative: positions coincide, the displacement is a “null vector”. A + B = B + A (3.1) Subtraction of vectors can be defined in termsThe addition of vectors also obeys the associative of addition of vectors. We define the differencelaw as illustrated in Fig. 3.4(d). The result of of two vectors A and B as the sum of two vectorsadding vectors A and B first and then adding A and –B :vector C is the same as the result of adding B and C first and then adding vector A : A – B = A + (–B) (3.5) (A + B) + C = A + (B + C) (3.2) It is shown in Fig 3.5. The vector –B is added to vector A to get R2 = (A – B). The vector R1 = A + BWhat is the result of adding two equal and is also shown in the same figure for comparison.opposite vectors ? Consider two vectors A and We can also use the parallelogram method to–A shown in Fig. 3.3(b). Their sum is A + (–A). find the sum of two vectors. Suppose we haveSince the magnitudes of the two vectors are the same, but the directions are opposite, the two vectors A and B. To add these vectors, we resultant vector has zero magnitude and is bring their tails to a common origin O as represented by 0 called a null vector or a zero shown in Fig. 3.6(a). Then we draw a line from vector : the head of A parallel to B and another line from the head of B parallel to A to complete a A – A = 0 |0|= 0 (3.3) parallelogram OQSP. Now we join the point of Since the magnitude of a null vector is zero, its the intersection of these two lines to the origin direction cannot be specified. O. The resultant vector R is directed from the The null vector also results when we multiply common origin O along the diagonal (OS) of the a vector A by the number zero. The main parallelogram [Fig. 3.6(b)]. In Fig.3.6(c), the properties of 0 are : triangle law is used to obtain the resultant of A A + 0 = A and B and we see that the two methods yield the λ 0 = 0 same result. Thus, the two methods are 0 A = 0 (3.4) equivalent. Fig. 3.5 (a) Two vectors A and B, – B is also shown. (b) Subtracting vector B from vector A – the result is R2. For comparison, addition of vectors A and B, i.e. R1 is also shown. Reprint 2025-26 MOTION IN A PLANE 31 Fig. 3.6 (a) Two vectors A and B with their tails brought to a common origin. (b) The sum A + B obtained using the parallelogram method. (c) The parallelogram method of vector addition is equivalent to the triangle method. ⊳ Example 3.1 Rain is falling vertically with 3.5 RESOLUTION OF VECTORS a speed of 35 m s–1. Winds starts blowing Let a and b be any two non-zero vectors in a after sometime with a speed of 12 m s–1 in plane with different directions and let A be east to west direction. In which direction another vector in the same plane (Fig. 3.8). A should a boy waiting at a bus stop hold can be expressed as a sum of two vectors — one his umbrella ? obtained by multiplying a by a real number and the other obtained by multiplying b by another real number. To see this, let O and P be the tail and head of the vector A. Then, through O, draw a straight line parallel to a, and through P, a straight line parallel to b. Let them intersect at Q. Then, we have A = OP = OQ + QP (3.6) But since OQ is parallel to a, and QP is parallel to b, we can write : OQ = λ a, and QP = µ b (3.7) Fig. 3.7 where λ and µ are real numbers. Answer The velocity of the rain and the wind Therefore, A = λ a + µ b (3.8)are represented by the vectors vr and vw in Fig.

2.4 A spherical conductor of radius 12 cm has a charge of 1.6 × 10–7C distributed uniformly on its surface. What is the electric field (a) inside the sphere (b) just outside the sphere (c) at a point 18 cm from the centre of the sphere?

2.5 POTENTIAL DUE TO A SYSTEM OF CHARGES Consider a system of charges q1, q2,…, qn with position vectors r1, r2,…, rn relative to some origin (Fig. 2.6). The potential V1 at P due to the charge q1 is 1 q1 V1 = 4 πε0 r1P where r1P is the distance between q1 and P. Similarly, the potential V2 at P due to q2 and V3 due to q3 are given by 1 q 2 1 q 3 V 2 = , V 3 = 4 πε0 r2P 4 πε0 r3P where r2P and r3P are the distances of P from charges q2 and q3, respectively; and so on for the potential due to other charges. By the FIGURE 2.6 Potential at a point due to a superposition principle, the potential V at P due system of charges is the sum of potentials to the total charge configuration is the algebraic due to individual charges. sum of the potentials due to the individual charges V = V1 + V2 + ... + Vn (2.17) 51 Reprint 2025-26 Physics 1  q1 q 2 q n  = + + ...... + (2.18) 4 πε0  r1P r2 P rnP  If we have a continuous charge distribution characterised by a charge density r (r), we divide it, as before, into small volume elements each of size Dv and carrying a charge rDv. We then determine the potential due to each volume element and sum (strictly speaking , integrate) over all such contributions, and thus determine the potential due to the entire distribution. We have seen in Chapter 1 that for a uniformly charged spherical shell, the electric field outside the shell is as if the entire charge is concentrated at the centre. Thus, the potential outside the shell is given by 1 q V = (r ≥ R ) [2.19(a)] 4 πε0 r where q is the total charge on the shell and R its radius. The electric field inside the shell is zero. This implies (Section 2.6) that potential is constant inside the shell (as no work is done in moving a charge inside the shell), and, therefore, equals its value at the surface, which is 1 q V = [2.19(b)] 4 πε0 R Example 2.2 Two charges 3 × 10–8 C and –2 × 10–8 C are located 15 cm apart. At what point on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero. Solution Let us take the origin O at the location of the positive charge. The line joining the two charges is taken to be the x-axis; the negative charge is taken to be on the right side of the origin (Fig. 2.7). FIGURE 2.7 Let P be the required point on the x-axis where the potential is zero. If x is the x-coordinate of P, obviously x must be positive. (There is no possibility of potentials due to the two charges adding up to zero for x < 0.) If x lies between O and A, we have 1  3 × 10 – 8 2 × 10 –8  − x × 10 –2 4 πε0  (15 − x ) × 10 –2  = 0 where x is in cm. That is, 3 2 − = 0 2.2 x 15 − x which gives x = 9 cm. If x lies on the extended line OA, the required condition is 3 2 − = 0 EXAMPLE x x − 15 Reprint 2025-26 Electrostatic Potential and Capacitance which gives x = 45 cm Thus, electric potential is zero at 9 cm and 45 cm away from the positive charge on the side of the negative charge. Note that the EXAMPLE formula for potential used in the calculation required choosing potential to be zero at infinity. 2.2 Example 2.3 Figures 2.8 (a) and (b) show the field lines of a positive and negative point charge respectively. Electric potential, equipotential-sufaces-12584/ FIGURE 2.8 equipotential (a) Give the signs of the potential difference VP – VQ; VB – VA. (b) Give the sign of the potential energy difference of a small negative charge between the points Q and P; A and B. surfaces: (c) Give the sign of the work done by the field in moving a small positive charge from Q to P. (d) Give the sign of the work done by the external agency in moving a small negative charge from B to A. (e) Does the kinetic energy of a small negative charge increase or decrease in going from B to A? Solution 1 (a) As V ∝ , VP > VQ. Thus, (VP – VQ) is positive. Also VB is less negative r than VA . Thus, VB > VA or (VB – VA) is positive. (b) A small negative charge will be attracted towards positive charge. The negative charge moves from higher potential energy to lower potential energy. Therefore the sign of potential energy difference of a small negative charge between Q and P is positive. Similarly, (P.E.)A > (P.E.)B and hence sign of potential energy differences is positive. http://video.mit.edu/watch/4-electrostatic-potential-elctric-energy-ev-conservative-field- (c) In moving a small positive charge from Q to P, work has to be done by an external agency against the electric field. Therefore, work done by the field is negative. (d) In moving a small negative charge from B to A work has to be done by the external agency. It is positive. EXAMPLE (e) Due to force of repulsion on the negative charge, velocity decreases and hence the kinetic energy decreases in going from B to A. 2.3 53 Reprint 2025-26 Physics 2.6 EQUIPOTENTIAL SURFACES An equipotential surface is a surface with a constant value of potential at all points on the surface. For a single charge q, the potential is given by Eq. (2.8): 1 q V = 4 πεo r This shows that V is a constant if r is constant. Thus, equipotential surfaces of a single point charge are concentric spherical surfaces centred at the charge. Now the electric field lines for a single charge q are radial lines starting from or ending at the charge, depending on whether q is positive or negative. Clearly, the electric field at every point is normal to the equipotential surface passing through that point. This is true in general: for any charge configuration, equipotential surface through a point is normal to the electric field at that point. The proof of this statement is simple. If the field were not normal to the equipotential surface, it would have non-zero component along the surface. To move a unit test charge against the direction of the component of the field, work would have to be done. But this is in contradiction to the definition of an equipotential FIGURE 2.9 For a surface: there is no potential difference between any two points on the single charge q surface and no work is required to move a test charge on the surface. (a) equipotential The electric field must, therefore, be normal to the equipotential surface surfaces are at every point. Equipotential surfaces offer an alternative visual picture spherical surfaces in addition to the picture of electric field lines around a charge centred at the configuration. charge, and (b) electric field lines are radial, starting from the charge if q > 0. FIGURE 2.10 Equipotential surfaces for a uniform electric field. For a uniform electric field E, say, along the x-axis, the equipotential surfaces are planes normal to the x-axis, i.e., planes parallel to the y-z plane (Fig. 2.10). Equipotential surfaces for (a) a dipole and (b) two identical positive charges are shown in Fig. 2.11. FIGURE 2.11 Some equipotential surfaces for (a) a dipole, 54 (b) two identical positive charges. Reprint 2025-26 Electrostatic Potential and Capacitance 2.6.1 Relation between field and potential Consider two closely spaced equipotential surfaces A and B (Fig. 2.12) with potential values V and V + dV, where dV is the change in V in the direction of the electric field E. Let P be a point on the surface B. d l is the perpendicular distance of the surface A from P. Imagine that a unit positive charge is moved along this perpendicular from the surface B to surface A against the electric field. The work done in this process is |E|dl. This work equals the potential difference VA–VB. Thus, |E|d l = V – (V + dV)= – dV V i.e., |E|= −δ (2.20) δl Since dV is negative, dV = – |dV|. we can rewrite FIGURE 2.12 From the Eq (2.20) as potential to the field. δV δV E = − = + (2.21) δl δl We thus arrive at two important conclusions concerning the relation between electric field and potential: (i) Electric field is in the direction in which the potential decreases steepest. (ii) Its magnitude is given by the change in the magnitude of potential per unit displacement normal to the equipotential surface at the point.

2.3 Two charges 2 mC and –2 mC are placed at points A and B 6 cm apart. (a) Identify an equipotential surface of the system. (b) What is the direction of the electric field at every point on this surface?

2.8 In a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10–3 m2 and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor? 79 Reprint 2025-26 Physics 2.9 Explain what would happen if in the capacitor given in Exercise 2.8, a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates, (a) while the voltage supply remained connected. (b) after the supply was disconnected. 2.10 A 12pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor? 2.11 A 600pF capacitor is charged by a 200V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process? Reprint 2025-26 Chapter Three CURRENT ELECTRICITY 3.1 INTRODUCTION In Chapter 1, all charges whether free or bound, were considered to be at rest. Charges in motion constitute an electric current. Such currents occur naturally in many situations. Lightning is one such phenomenon in which charges flow from the clouds to the earth through the atmosphere, sometimes with disastrous results. The flow of charges in lightning is not steady, but in our everyday life we see many devices where charges flow in a steady manner, like water flowing smoothly in a river. A torch and a cell-driven clock are examples of such devices. In the present chapter, we shall study some of the basic laws concerning steady electric currents. 3.2 ELECTRIC CURRENT Imagine a small area held normal to the direction of flow of charges. Both the positive and the negative charges may flow forward and backward across the area. In a given time interval t, let q+ be the net amount (i.e., forward minus backward) of positive charge that flows in the forward direction across the area. Similarly, let q– be the net amount of negative charge flowing across the area in the forward direction. The net amount of charge flowing across the area in the forward direction in the time interval t, then, is q = q+– q–. This is proportional to t for steady current Reprint 2025-26 Physics and the quotient q I = (3.1) t is defined to be the current across the area in the forward direction. (If it turn out to be a negative number, it implies a current in the backward direction.) Currents are not always steady and hence more generally, we define the current as follows. Let DQ be the net charge flowing across a cross- section of a conductor during the time interval Dt [i.e., between times t and (t + Dt)]. Then, the current at time t across the cross-section of the conductor is defined as the value of the ratio of DQ to Dt in the limit of Dt tending to zero, ∆Q lim (3.2) I (t ) ≡ t 0 ∆→ ∆t In SI units, the unit of current is ampere. An ampere is defined through magnetic effects of currents that we will study in the following chapter. An ampere is typically the order of magnitude of currents in domestic appliances. An average lightning carries currents of the order of tens of thousands of amperes and at the other extreme, currents in our nerves are in microamperes. 3.3 ELECTRIC CURRENTS IN CONDUCTORS An electric charge will experience a force if an electric field is applied. If it is free to move, it will thus move contributing to a current. In nature, free charged particles do exist like in upper strata of atmosphere called the ionosphere. However, in atoms and molecules, the negatively charged electrons and the positively charged nuclei are bound to each other and are thus not free to move. Bulk matter is made up of many molecules, a gram of water, for example, contains approximately 1022 molecules. These molecules are so closely packed that the electrons are no longer attached to individual nuclei. In some materials, the electrons will still be bound, i.e., they will not accelerate even if an electric field is applied. In other materials, notably metals, some of the electrons are practically free to move within the bulk material. These materials, generally called conductors, develop electric currents in them when an electric field is applied. If we consider solid conductors, then of course the atoms are tightly bound to each other so that the current is carried by the negatively charged electrons. There are, however, other types of conductors like electrolytic solutions where positive and negative charges both can move. In our discussions, we will focus only on solid conductors so that the current is carried by the negatively charged electrons in the background of fixed positive ions. Consider first the case when no electric field is present. The electrons will be moving due to thermal motion during which they collide with the fixed ions. An electron colliding with an ion emerges with the same speed as before the collision. However, the direction of its velocity after the collision is completely random. At a given time, there is no preferential 82 direction for the velocities of the electrons. Thus on the average, the Reprint 2025-26 Current Electricity number of electrons travelling in any direction will be equal to the number of electrons travelling in the opposite direction. So, there will be no net electric current. Let us now see what happens to such a piece of conductor if an electric field is applied. To focus our thoughts, imagine the conductor in the shape of a cylinder of radius R (Fig. 3.1). Suppose we now take two thin circular discs FIGURE 3.1 Charges +Q and –Q put at the ends of a dielectric of the same radius and put of a metallic cylinder. The electrons will drift positive charge +Q distributed over one disc because of the electric field created to and similarly –Q at the other disc. We attach neutralise the charges. The current thus the two discs on the two flat surfaces of the will stop after a while unless the charges +Q cylinder. An electric field will be created and and –Q are continuously replenished. is directed from the positive towards the negative charge. The electrons will be accelerated due to this field towards +Q. They will thus move to neutralise the charges. The electrons, as long as they are moving, will constitute an electric current. Hence in the situation considered, there will be a current for a very short while and no current thereafter. We can also imagine a mechanism where the ends of the cylinder are supplied with fresh charges to make up for any charges neutralised by electrons moving inside the conductor. In that case, there will be a steady electric field in the body of the conductor. This will result in a continuous current rather than a current for a short period of time. Mechanisms, which maintain a steady electric field are cells or batteries that we shall study later in this chapter. In the next sections, we shall study the steady current that results from a steady electric field in conductors.

12.5 A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n = 4 level. Determine the wavelength and frequency of photon.

12.7 The radius of the innermost electron orbit of a hydrogen atom is 5.3×10–11 m. What are the radii of the n = 2 and n =3 orbits?

12.3 BEHAVIOUR OF GASES where M is the mass of the gas containing N Properties of gases are easier to understand than molecules, M0 is the molar mass and NA the those of solids and liquids. This is mainly Avogadro’s number. Using Eqs. (12.4) and (12.3) because in a gas, molecules are far from each can also be written as other and their mutual interactions are PV = kB NT or P = kB nT negligible except when two molecules collide. Gases at low pressures and high temperatures much above that at which they liquefy (or solidify) approximately satisfy a simple relation ) between their pressure, temperature and volume –1 given by (see Chapter 10) K –1 PV = KT (12.1) mol Jfor a given sample of the gas. Here T is the ( temperature in kelvin or (absolute) scale. K is a T pV µ constant for the given sample but varies with the volume of the gas. If we now bring in the idea of atoms or molecules, then K is proportional to the number of molecules, (say) N in the sample. We can write K = N k . Observation tells P (atm) us that this k is same for all gases. It is called Fig.12.1 Real gases approach ideal gas behaviour at Boltzmann constant and is denoted by k B. low pressures and high temperatures. P1V1 P2 V2 where n is the number density, i.e. number ofAs = = constant = kB (12.2) N 1 T1 N 2 T2 molecules per unit volume. kB is the Boltzmann constant introduced above. Its value in SI unitsif P, V and T are same, then N is also same for all is 1.38 × 10–23 J K–1.gases. This is Avogadro’s hypothesis, that the Another useful form of Eq. (12.3) isnumber of molecules per unit volume is ρRT the same for all gases at a fixed temperature and P = (12.5) pressure. The number in 22.4 litres of any gas M 0 Reprint 2025-26 KINETIC THEORY 247 where ρ is the mass density of the gas. etc. in a vessel of volume V at temperature T and A gas that satisfies Eq. (12.3) exactly at all pressure P. It is then found that the equation of pressures and temperatures is defined to be an state of the mixture is : ideal gas. An ideal gas is a simple theoretical PV = ( µ1 + µ2 +… ) RT (12.7)model of a gas. No real gas is truly ideal. Fig. 12.1 shows departures from ideal gas RT RT i.e. P = µ1 + µ2 + ... (12.8)behaviour for a real gas at three different V V temperatures. Notice that all curves approach = P1 + P2 + … (12.9)the ideal gas behaviour for low pressures and high temperatures. Clearly P1 = µ1 R T/V is the pressure that At low pressures or high temperatures the gas 1 would exert at the same conditions of molecules are far apart and molecular volume and temperature if no other gases were interactions are negligible. Without interactions present. This is called the partial pressure of the the gas behaves like an ideal one. gas. Thus, the total pressure of a mixture of ideal If we fix µ and T in Eq. (12.3), we get gases is the sum of partial pressures. This is Dalton’s law of partial pressures. PV = constant (12.6) i.e., keeping temperature constant, pressure of a given mass of gas varies inversely with volume. This is the famous Boyle’s law. Fig. 12.2 shows comparison between experimental P-V curves and the theoretical curves predicted by Boyle’s law. Once again you see that the agreement is good at high temperatures and low pressures. Next, if you fix P, Eq. (12.1) shows that V ∝ T i.e., for a fixed pressure, the volume of a gas is proportional to its absolute temperature T (Charles’ law). See Fig. 12.3. Fig. 12.3 Experimental T-V curves (solid lines) for CO2 at three pressures compared with Charles’ law (dotted lines). T is in units of 300 K and V in units of 0.13 litres. We next consider some examples which give us information about the volume occupied by the molecules and the volume of a single molecule. ⊳ Example 12.1 The density of water is 1000 kg m–3. The density of water vapour at 100 °C and 1 atm pressure is 0.6 kg m–3. The volume of a molecule multiplied by the total Fig.12.2 Experimental P-V curves (solid lines) for number gives ,what is called, molecular steam at three temperatures compared with volume. Estimate the ratio (or fraction) of Boyle’s law (dotted lines). P is in units of 22 the molecular volume to the total volume atm and V in units of 0.09 litres. occupied by the water vapour under the Finally, consider a mixture of non-interacting above conditions of temperature and pressure.ideal gases: µ moles of gas 1, µ moles of gas 2, 1 2 Reprint 2025-26 248 PHYSICS Answer For a given mass of water molecules, number of molecules and (ii) mass density the density is less if volume is large. So the of neon and oxygen in the vessel. Atomic volume of the vapour is 1000/0.6 = 1/(6 × 10 -4 ) mass of Ne = 20.2 u, molecular mass of O2times larger. If densities of bulk water and water = 32.0 u. molecules are same, then the fraction of molecular volume to the total volume in liquid Answer Partial pressure of a gas in a mixture is state is 1. As volume in vapour state has the pressure it would have for the same volume increased, the fractional volume is less by the and temperature if it alone occupied the vessel. same amount, i.e. 6×10-4. ⊳ (The total pressure of a mixture of non-reactive ⊳ gases is the sum of partial pressures due to its Example 12.2 Estimate the volume of a constituent gases.) Each gas (assumed ideal) water molecule using the data in Example obeys the gas law. Since V and T are common to 12.1. the two gases, we have P1V = µ 1 RT and P2V = Answer In the liquid (or solid) phase, the µ2 RT, i.e. (P1/P2) = (µ1 / µ2). Here 1 and 2 refer to neon and oxygen respectively. Since (P1/P2) =molecules of water are quite closely packed. The (3/2) (given), (µ1/ µ2) = 3/2.density of water molecule may therefore, be (i) By definition µ1 = (N1/NA ) and µ2 = (N2/NA)regarded as roughly equal to the density of bulk where N1 and N2 are the number of moleculeswater = 1000 kg m–3. To estimate the volume of of 1 and 2, and NA is the Avogadro’s number.a water molecule, we need to know the mass of Therefore, (N1/N2) = (µ1 / µ2) = 3/2.a single water molecule. We know that 1 mole (ii) We can also write µ1 = (m1/M1) and µ2 =of water has a mass approximately equal to (2 + 16)g = 18 g = 0.018 kg. (m2/M2) where m1 and m2 are the masses of 1 and 2; and M1 and M2 are their molecular Since 1 mole contains about 6 × 1023 masses. (Both m1 and M1; as well as m2 andmolecules (Avogadro’s number), the mass of a molecule of water is (0.018)/(6 × 1023) kg = M2 should be expressed in the same units). If ρ1 and ρ2 are the mass densities of 1 and3 × 10–26 kg. Therefore, a rough estimate of the 2 respectively, we havevolume of a water molecule is as follows : Volume of a water molecule ρ1 m 1 / V m 1 µ1  M 1  = (3 × 10–26 kg)/ (1000 kg m–3) = = = × ρ2 m 2 / V m 2 µ2  M 2  = 3 × 10–29 m3 = (4/3) π (Radius)3 3 20.2 Hence, Radius ≈ 2 ×10-10 m = 2 Å ⊳ = × = 0.947 2 32.0 ⊳ ⊳ Example 12.3 What is the average distance between atoms (interatomic distance) in water? Use the data given in 12.4 KINETIC THEORY OF AN IDEAL GAS Examples 12.1 and 12.2. Kinetic theory of gases is based on the molecular picture of matter. A given amount of gas is a Answer : A given mass of water in vapour state collection of a large number of molecules has 1.67×103 times the volume of the same mass (typically of the order of Avogadro’s number) that of water in liquid state (Ex. 12.1). This is also are in incessant random motion. At ordinary the increase in the amount of volume available pressure and temperature, the average distance for each molecule of water. When volume between molecules is a factor of 10 or more than increases by 103 times the radius increases by the typical size of a molecule (2 Å). Thus, V1/3 or 10 times, i.e., 10 × 2 Å = 20 Å. So the interaction between molecules is negligible and average distance is 2 × 20 = 40 Å. ⊳ we can assume that they move freely in straight lines according to Newton’s first law. However,⊳ Example 12.4 A vessel contains two non- occasionally, they come close to each other, reactive gases : neon (monatomic) and experience intermolecular forces and their oxygen (diatomic). The ratio of their partial velocities change. These interactions are called pressures is 3:2. Estimate the ratio of (i) collisions. The molecules collide incessantly against each other or with the walls and change Reprint 2025-26 KINETIC THEORY 249 their velocities. The collisions are considered to the wall. Thus, the number of molecules with be elastic. We can derive an expression for the velocity (vx, vy, vz ) hitting the wall in time ∆t is pressure of a gas based on the kinetic theory. ½A vx ∆t n, where n is the number of molecules We begin with the idea that molecules of a per unit volume. The total momentum gas are in incessant random motion, colliding transferred to the wall by these molecules in against one another and with the walls of the time ∆t is: container. All collisions between molecules Q = (2mvx) (½ n A vx ∆t ) (12.10) among themselves or between molecules and the The force on the wall is the rate of momentum walls are elastic. This implies that total kinetic transfer Q/∆t and pressure is force per unit energy is conserved. The total momentum is area : conserved as usual. P = Q /(A ∆t) = n m vx 2 (12.11) Actually, all molecules in a gas do not have 12.4.1 Pressure of an Ideal Gas the same velocity; there is a distribution in velocities. The above equation, therefore, standsConsider a gas enclosed in a cube of side l. Take for pressure due to the group of molecules withthe axes to be parallel to the sides of the cube, speed vx in the x-direction and n stands for theas shown in Fig. 12.4. A molecule with velocity number density of that group of molecules. The (vx, vy, vz ) hits the planar wall parallel to yz- total pressure is obtained by summing over theplane of area A (= l2). Since the collision is elastic, contribution due to all groups:the molecule rebounds with the same velocity; its y and z components of velocity do not change P = n m v x2 (12.12) in the collision but the x-component reverses where v 2x is the average of vx 2 . Now the gas sign. That is, the velocity after collision is is isotropic, i.e. there is no preferred direction (-vx, vy, vz ) . The change in momentum of the of velocity of the molecules in the vessel. molecule is: –mvx – (mvx) = – 2mvx . By the Therefore, by symmetry, principle of conservation of momentum, the momentum imparted to the wall in the collision v 2x = v y2 = v z2 = 2mvx . 2 2 2 2 = (1/3) [ v x + v y + v z ] = (1/3) v (12.13) where v is the speed and v 2 denotes the mean of the squared speed. Thus P = (1/3) n m v 2 (12.14) Some remarks on this derivation. First, though we choose the container to be a cube, the shape of the vessel really is immaterial. For a vessel of arbitrary shape, we can always choose a small infinitesimal (planar) area and carry through the steps above. Notice that both A and ∆t do not appear in the final result. By Pascal’s law, given in Ch. 9, pressure in one portion of Fig. 12.4 Elastic collision of a gas molecule with the the gas in equilibrium is the same as anywhere wall of the container. else. Second, we have ignored any collisions in To calculate the force (and pressure) on the the derivation. Though this assumption is wall, we need to calculate momentum imparted difficult to justify rigorously, we can qualitatively to the wall per unit time. In a small time interval see that it will not lead to erroneous results. The ∆t, a molecule with x-component of velocity vx number of molecules hitting the wall in time ∆t will hit the wall if it is within the distance vx ∆t was found to be ½ n Avx ∆t. Now the collisions from the wall. That is, all molecules within the are random and the gas is in a steady state. volume Avx ∆t only can hit the wall in time ∆t. Thus, if a molecule with velocity (vx, vy, vz ) But, on the average, half of these are moving acquires a different velocity due to collision with towards the wall and the other half away from some molecule, there will always be some other Reprint 2025-26 250 PHYSICS molecule with a different initial velocity which P = (1/3) [n1m1 v1 2 + n2 m2 v 22 +… ] (12.20) after a collision acquires the velocity (vx, vy, vz ). In equilibrium, the average kinetic energy of If this were not so, the distribution of velocities the molecules of different gases will be equal. would not remain steady. In any case we are That is, finding v x2 . Thus, on the whole, molecular ½ m1 v1 2 = ½ m2 v 22 = (3/2) kB Tcollisions (if they are not too frequent and the so thattime spent in a collision is negligible compared to time between collisions) will not affect the P = (n1 + n2 +… ) kB T (12.21) calculation above. which is Dalton’s law of partial pressures. From Eq. (12.19), we can get an idea of the12.4.2 Kinetic Interpretation of Temperature typical speed of molecules in a gas. At a Equation (13.14) can be written as temperature T = 300 K, the mean square speed PV = (1/3) nV m v 2 (12.15a) of a molecule in nitrogen gas is : PV = (2/3) N x ½ m v 2 (12.15b) M N 2 28 –26 where N (= nV) is the number of molecules in m = = 26 = 4.65 × 10 kg. N A 6.02 × 10the sample. The quantity in the bracket is the average v 2 = 3 kB T / m = (516)2 m2s-2 translational kinetic energy of the molecules in 2 The square root of v is known as root mean the gas. Since the internal energy E of an ideal square (rms) speed and is denoted by vrms,gas is purely kinetic*, 2 ( We can also write v 2 as < v2 >.) E = N × (1/2) m v (12.16) vrms = 516 m s-1 Equation (12.15) then gives : The speed is of the order of the speed of sound PV = (2/3) E (12.17) in air. It follows from Eq. (12.19) that at the same We are now ready for a kinetic interpretation temperature, lighter molecules have greater rms of temperature. Combining Eq. (12.17) with the speed. ⊳ideal gas Eq. (12.3), we get Example 12.5 A flask contains argon and E = (3/2) kB NT (12.18) chlorine in the ratio of 2:1 by mass. The or E/ N = ½ m v 2 = (3/2) kBT (12.19) temperature of the mixture is 27 °C. Obtain i.e., the average kinetic energy of a molecule is the ratio of (i) average kinetic energy per proportional to the absolute temperature of the molecule, and (ii) root mean square speed gas; it is independent of pressure, volume or vrms of the molecules of the two gases. the nature of the ideal gas. This is a fundamental Atomic mass of argon = 39.9 u; Molecular result relating temperature, a macroscopic mass of chlorine = 70.9 u. measurable parameter of a gas (a thermodynamic variable as it is called) to a Answer The important point to remember is thatmolecular quantity, namely the average kinetic the average kinetic energy (per molecule) of anyenergy of a molecule. The two domains are connected by the Boltzmann constant. We note (ideal) gas (be it monatomic like argon, diatomic in passing that Eq. (12.18) tells us that internal like chlorine or polyatomic) is always equal to energy of an ideal gas depends only on (3/2) kBT. It depends only on temperature, and temperature, not on pressure or volume. With is independent of the nature of the gas. this interpretation of temperature, kinetic theory (i) Since argon and chlorine both have the same of an ideal gas is completely consistent with the temperature in the flask, the ratio of average ideal gas equation and the various gas laws kinetic energy (per molecule) of the two gasesbased on it. is 1:1. For a mixture of non-reactive ideal gases, the total pressure gets contribution from each gas (ii) Now ½ m vrms2 = average kinetic energy per in the mixture. Equation (12.14) becomes molecule = (3/2) ) kBT where m is the mass * E denotes the translational part of the internal energy U that may include energies due to other degrees of freedom also. See section 12.5. Reprint 2025-26 KINETIC THEORY 251 of a molecule of the gas. Therefore, v 2 ( rms ) Ar (m )Cl ( M )Cl 70.9 = = 2 v rms (m ) Ar ( M ) Ar = 39.9 =1.77 ( )Cl where M denotes the molecular mass of the gas. (For argon, a molecule is just an atom of argon.) Taking square root of both sides, v ( rms ) Ar ( vrms )Cl = 1.33 You should note that the composition of the mixture by mass is quite irrelevant to the above calculation. Any other proportion by mass of argon and chlorine would give the same answers to (i) and (ii), provided the temperature remains Fig. 12.5 Molecules going through a porous wall. unaltered. ⊳ ⊳ ⊳ Example 12.7 (a) When a molecule (or an Example 12.6 Uranium has two isotopes elastic ball) hits a ( massive) wall, it of masses 235 and 238 units. If both are rebounds with the same speed. When a ball present in Uranium hexafluoride gas which hits a massive bat held firmly, the same would have the larger average speed ? If thing happens. However, when the bat is atomic mass of fluorine is 19 units, moving towards the ball, the ball rebounds estimate the percentage difference in with a different speed. Does the ball move speeds at any temperature. faster or slower? (Ch.5 will refresh your Answer At a fixed temperature the average memory on elastic collisions.) energy = ½ m <v2 > is constant. So smaller the (b) When gas in a cylinder is compressed mass of the molecule, faster will be the speed. by pushing in a piston, its temperature The ratio of speeds is inversely proportional to rises. Guess at an explanation of this in the square root of the ratio of the masses. The terms of kinetic theory using (a) above. masses are 349 and 352 units. So (c) What happens when a compressed gas v349 / v352 = ( 352/ 349)1/2 = 1.0044 . pushes a piston out and expands. What ∆ V would you observe ? Hence difference = 0.44 %. (d) Sachin Tendulkar used a heavy cricket V bat while playing. Did it help him in [235U is the isotope needed for nuclear fission. anyway ?To separate it from the more abundant isotope 238U, the mixture is surrounded by a porous cylinder. The porous cylinder must be thick and Answer (a) Let the speed of the ball be u relative narrow, so that the molecule wanders through to the wicket behind the bat. If the bat is moving individually, colliding with the walls of the long towards the ball with a speed V relative to the pore. The faster molecule will leak out more than wicket, then the relative speed of the ball to bat the slower one and so there is more of the lighter is V + u towards the bat. When the ball rebounds molecule (enrichment) outside the porous (after hitting the massive bat) its speed, relative cylinder (Fig. 12.5). The method is not very to bat, is V + u moving away from the bat. So efficient and has to be repeated several times relative to the wicket the speed of the rebounding for sufficient enrichment.]. ⊳ ball is V + (V + u) = 2V + u, moving away from the When gases diffuse, their rate of diffusion is wicket. So the ball speeds up after the collision inversely proportional to square root of the with the bat. The rebound speed will be less than masses (see Exercise 12.12 ). Can you guess the u if the bat is not massive. For a molecule this explanation from the above answer? would imply an increase in temperature. Reprint 2025-26 252 PHYSICS You should be able to answer (b) (c) and (d) to the axis joining the two oxygen atoms about based on the answer to (a). which the molecule can rotate*. The molecule (Hint: Note the correspondence, pistonà bat, thus has two rotational degrees of freedom, each of which contributes a term to the total energycylinder à wicket, molecule à ball.) ⊳ consisting of translational energy tε and rotational energy εr.12.5 LAW OF EQUIPARTITION OF ENERGY The kinetic energy of a single molecule is εt + εr = 1 mv x2 + 1 mv y2 + 1 mv z2 + 1 I 1ω12 + 1 I 2ω22 (12.25) 2 2 2 2 2 1 2 1 2 1 2 εt = mv x + mv y + mv z (12.22) 2 2 2 For a gas in thermal equilibrium at temperature T the average value of energy denoted by < tε > is 1 2 1 2 1 2 3 εt = mv x + mv y + mv z = k B T (12.23) 2 2 2 2 Since there is no preferred direction, Eq. (12.23) implies 1 2 1 1 2 1 mv x = k B T , mv y = k B T , 2 2 2 2 Fig. 12.6 The two independent axes of rotation of a diatomic molecule 1 2 1 mv z = k B T (12.24) 2 2 where ω1 and ω2 are the angular speeds about A molecule free to move in space needs three the axes 1 and 2 and I1, I2 are the corresponding coordinates to specify its location. If it is moments of inertia. Note that each rotational constrained to move in a plane it needs two; and degree of freedom contributes a term to the if constrained to move along a line, it needs just energy that contains square of a rotational one coordinate to locate it. This can also be variable of motion. expressed in another way. We say that it has We have assumed above that the O2 molecule one degree of freedom for motion in a line, two is a ‘rigid rotator’, i.e., the molecule does not for motion in a plane and three for motion in vibrate. This assumption, though found to be space. Motion of a body as a whole from one true (at moderate temperatures) for O2, is notpoint to another is called translation. Thus, a always valid. Molecules, like CO, even at molecule free to move in space has three moderate temperatures have a mode of translational degrees of freedom. Each vibration, i.e., its atoms oscillate along the translational degree of freedom contributes a interatomic axis like a one-dimensional term that contains square of some variable of 2 oscillator, and contribute a vibrational energymotion, e.g., ½ mvx and similar terms in term εv to the total energy:vy and vz. In, Eq. (12.24) we see that in thermal equilibrium, the average of each such term is 1  d y  2 1 2 εv = m + ky½ kBT . 2  d t  2 Molecules of a monatomic gas like argon have only translational degrees of freedom. But what ε = εt + εr + ε v (12.26) about a diatomic gas such as O2 or N2? A where k is the force constant of the oscillator molecule of O2 has three translational degrees and y the vibrational co-ordinate. of freedom. But in addition it can also rotate Once again the vibrational energy terms in about its centre of mass. Figure 12.6 shows the Eq. (12.26) contain squared terms of vibrational two independent axes of rotation 1 and 2, normal variables of motion y and dy/dt . * Rotation along the line joining the atoms has very small moment of inertia and does not come into play for quantum mechanical reasons. See end of section 12.6. Reprint 2025-26 KINETIC THEORY 253 At this point, notice an important feature in where Cp is the molar specific heat at constant Eq.(12.26). While each translational and pressure. Thus, rotational degree of freedom has contributed only 5 one ‘squared term’ in Eq.(12.26), one vibrational Cp = R (12.30) mode contributes two ‘squared terms’ : kinetic 2 and potential energies. C p 5 The ratio of specific heats γ = = (12.31) Each quadratic term occurring in the C v 3 expression for energy is a mode of absorption of energy by the molecule. We have seen that in 12.6.2 Diatomic Gases thermal equilibrium at absolute temperature T, As explained earlier, a diatomic molecule treated for each translational mode of motion, the as a rigid rotator, like a dumbbell, has 5 degrees average energy is ½ kBT. The most elegant of freedom: 3 translational and 2 rotational. principle of classical statistical mechanics (first Using the law of equipartition of energy, the total proved by Maxwell) states that this is so for each internal energy of a mole of such a gas is mode of energy: translational, rotational and 5 5 vibrational. That is, in equilibrium, the total U = k B T × N A = RT (12.32) 2 2 energy is equally distributed in all possible The molar specific heats are then given by energy modes, with each mode having an average energy equal to ½ kBT. This is known as the law 5 7 Cv (rigid diatomic) = R, Cp = R (12.33)of equipartition of energy. Accordingly, each 2 2 translational and rotational degree of freedom 7 of a molecule contributes ½ kBT to the energy, γ (rigid diatomic) = (12.34) while each vibrational frequency contributes 5 If the diatomic molecule is not rigid but has 2 × ½ kBT = kBT , since a vibrational mode has in addition a vibrational mode both kinetic and potential energy modes.  5  7 The proof of the law of equipartition of energy U =  k B T + k B T  N A = RT is beyond the scope of this book. Here, we shall  2  2 apply the law to predict the specific heats of gases 7 9 9 theoretically. Later, we shall also discuss briefly, C v = R , C p = R , γ = R (12.35) 2 2 7 the application to specific heat of solids. 12.6.3 Polyatomic Gases 12.6 SPECIFIC HEAT CAPACITY In general a polyatomic molecule has 3 12.6.1 Monatomic Gases translational, 3 rotational degrees of freedom and a certain number ( f ) of vibrational modes.The molecule of a monatomic gas has only three According to the law of equipartition of energy,translational degrees of freedom. Thus, the it is easily seen that one mole of such a gas hasaverage energy of a molecule at temperature 3T is (3/2)kBT . The total internal energy of a mole  3 U = kBT + kBT + f kBT  NA of such a gas is  2 2 3 3 U = k B T × N A = RT (12.27) i.e.,Cv = (3 + f ) R, Cp = (4 + f ) R, 2 2 ( 4 + f ) γ = (12.36) The molar specific heat at constant volume, ( 3 + f ) Cv, is Note that Cp – Cv = R is true for any ideal d U 3 gas, whether mono, di or polyatomic. Cv (monatomic gas) = = RT (12.28) d T 2 Table 12.1 summarises the theoretical For an ideal gas, predictions for specific heats of gases ignoring Cp – Cv = R (12.29) any vibrational modes of motion. The values are Reprint 2025-26 254 PHYSICS in good agreement with experimental values of Answer Using the gas law PV = µRT, you can specific heats of several gases given in Table 12.2. easily show that 1 mol of any (ideal) gas at Of course, there are discrepancies between standard temperature (273 K) and pressure predicted and actual values of specific heats of (1 atm = 1.01 × 105 Pa) occupies a volume of 22.4 several other gases (not shown in the table), such litres. This universal volume is called molar volume. Thus the cylinder in this exampleas Cl2, C2H6 and many other polyatomic gases. contains 2 mol of helium. Further, since heliumUsually, the experimental values for specific is monatomic, its predicted (and observed) molar heats of these gases are greater than the specific heat at constant volume, Cv = (3/2) R,predicted values as given in Table12.1 suggesting and molar specific heat at constant pressure, that the agreement can be improved by including Cp = (3/2) R + R = (5/2) R. Since the volume of vibrational modes of motion in the calculation. the cylinder is fixed, the heat required is The law of equipartition of energy is, thus, well determined by Cv. Therefore, verified experimentally at ordinary temperatures. Heat required = no. of moles × molar specific heat × rise in temperature Table 12.1 Predicted values of specific heat = 2 × 1.5 R × 15.0 = 45 R capacities of gases (ignoring vibrational modes) = 45 × 8.31 = 374 J. ⊳ 12.6.4 Specific Heat Capacity of Solids Nature of Cv Cp Cp - Cv g Gas (J mol-1 K-1) (J mol-1 K-1) (J mol-1 K-1) We can use the law of equipartition of energy to determine specific heats of solids. Consider a Monatomic 12.5 20.8 8.31 1.67 solid of N atoms, each vibrating about its mean Diatomic 20.8 29.1 8.31 1.40 position. An oscillation in one dimension has average energy of 2 × ½ kBT = kBT . In three Triatomic 24.93 33.24 8.31 1.33 dimensions, the average energy is 3 kBT. For a mole of solid, N = NA, and the total energy is Table12.2 Measured values of specific heat U = 3 kBT × NA = 3 RT capacities of some gases Now at constant pressure ∆Q = ∆U + P∆V = ∆U, since for a solid ∆V is negligible. Hence, ∆Q ∆ U C = = = 3 R (12.37) ∆T ∆T Table 12.3 Specific Heat Capacity of some solids at room temperature and atmospheric pressure As Table 12.3 shows the prediction generally ⊳ Example 12.8 A cylinder of fixed capacity agrees with experimental values at ordinary 44.8 litres contains helium gas at standard temperature (Carbon is an exception). temperature and pressure. What is the amount of heat needed to raise the 12.7 MEAN FREE PATH temperature of the gas in the cylinder by Molecules in a gas have rather large speeds of 15.0 °C ? (R = 8.31 J mo1–1 K–1). the order of the speed of sound. Yet a gas leaking Reprint 2025-26 KINETIC THEORY 255 from a cylinder in a kitchen takes considerable are moving and the collision rate is determined time to diffuse to the other corners of the room. by the average relative velocity of the molecules. The top of a cloud of smoke holds together for Thus we need to replace <v> by <v r> in Eq. hours. This happens because molecules in a gas (12.38). A more exact treatment gives have a finite though small size, so they are bound 2 2 nπd (12.40)to undergo collisions. As a result, they cannot l = 1/ ( ) move straight unhindered; their paths keep Let us estimate l and τ for air molecules with getting incessantly deflected. average speeds <v> = ( 485m/s). At STP 0.02 × 1023 ( ) n = –3 22.4 × 10 ( ) = 2.7 × 10 25 m -3. Taking, d = 2 × 10–10 m, τ = 6.1 × 10–10 s t and l = 2.9 × 10–7 m ≈ 1500 d (12.41) v As expected, the mean free path given by d Eq. (12.40) depends inversely on the number density and the size of the molecules. In a highly evacuated tube n is rather small and the mean d free path can be as large as the length of the tube. ⊳ Example 12.9 Estimate the mean free path for a water molecule in water vapour at 373 K. Use information from Exercises 12.1 and Eq. Fig. 12.7 The volume swept by a molecule in time ∆t (12.41) above. in which any molecule will collide with it. Answer The d for water vapour is same as that Suppose the molecules of a gas are spheres of of air. The number density is inverselydiameter d. Focus on a single molecule with the proportional to absolute temperature.average speed <v>. It will suffer collision with any molecule that comes within a distance d 25 273 25 –3 So n = 2.7 × 10 × = 2 × 10 mbetween the centres. In time ∆t, it sweeps a 373 volume πd2 <v> ∆t wherein any other molecule –7 Hence, mean free path l = 4 × 10 m ⊳will collide with it (see Fig. 12.7). If n is the number of molecules per unit volume, the Note that the mean free path is 100 times the molecule suffers nπd2 <v> ∆t collisions in time interatomic distance ~ 40 Å = 4 × 10-9 m calculated ∆t. Thus the rate of collisions is nπd2 <v> or the earlier. It is this large value of mean free path that time between two successive collisions is on the leads to the typical gaseous behaviour. Gases can average, not be confined without a container. τ = 1/(nπ <v> d2 ) (12.38) Using, the kinetic theory of gases, the bulk The average distance between two successive measurable properties like viscosity, heat collisions, called the mean free path l, is : conductivity and diffusion can be related to the l = <v> τ = 1/(nπd2) (12.39) microscopic parameters like molecular size. It is In this derivation, we imagined the other through such relations that the molecular sizes molecules to be at rest. But actually all molecules were first estimated. Reprint 2025-26 256 PHYSICS SUMMARY 1. The ideal gas equation connecting pressure (P), volume (V) and absolute temperature (T ) is PV = µ RT = kB NT where µ is the number of moles and N is the number of molecules. R and kB are universal constants. R R = 8.314 J mol–1 K–1, kB = N A = 1.38 × 10–23 J K–1 Real gases satisfy the ideal gas equation only approximately, more so at low pressures and high temperatures. 2. Kinetic theory of an ideal gas gives the relation 1 2 P = n m v 3 where n is number density of molecules, m the mass of the molecule and v 2 is the mean of squared speed. Combined with the ideal gas equation it yields a kinetic interpretation of temperature. 1 2 3 2 1/2 3k B T m v = k B T , v rms = v = 2 2 ( ) m This tells us that the temperature of a gas is a measure of the average kinetic energy of a molecule, independent of the nature of the gas or molecule. In a mixture of gases at a fixed temperature the heavier molecule has the lower average speed. 3. The translational kinetic energy 3 E = kB NT. 2 This leads to a relation 2 PV = E 3 4. The law of equipartition of energy states that if a system is in equilibrium at absolute temperature T, the total energy is distributed equally in different energy modes of absorption, the energy in each mode being equal to ½ kB T. Each translational and rotational degree of freedom corresponds to one energy mode of absorption and has energy ½ kB T. Each vibrational frequency has two modes of energy (kinetic and potential) with corresponding energy equal to 2 × ½ kB T = kB T. 5. Using the law of equipartition of energy, the molar specific heats of gases can be determined and the values are in agreement with the experimental values of specific heats of several gases. The agreement can be improved by including vibrational modes of motion. 6. The mean free path l is the average distance covered by a molecule between two successive collisions : 1 l = 2 2 n πd where n is the number density and d the diameter of the molecule. Reprint 2025-26 KINETIC THEORY 257 POINTS TO PONDER 1. Pressure of a fluid is not only exerted on the wall. Pressure exists everywhere in a fluid. Any layer of gas inside the volume of a container is in equilibrium because the pressure is the same on both sides of the layer. 2. We should not have an exaggerated idea of the intermolecular distance in a gas. At ordinary pressures and temperatures, this is only 10 times or so the interatomic distance in solids and liquids. What is different is the mean free path which in a gas is 100 times the interatomic distance and 1000 times the size of the molecule. 3. The law of equipartition of energy is stated thus: the energy for each degree of freedom in thermal equilibrium is ½ k T. Each quadratic term in the total energy expression of a molecule is to be counted asB a degree of freedom. Thus, each vibrational mode gives 2 (not 1) degrees of freedom (kinetic and potential energy modes), corresponding to the energy 2 × ½ k T = k T. B B 4. Molecules of air in a room do not all fall and settle on the ground (due to gravity) because of their high speeds and incessant collisions. In equilibrium, there is a very slight increase in density at lower heights (like in the atmosphere). The effect is small since the potential energy (mgh) for ordinary heights is much less than the average kinetic energy ½ mv2 of the molecules. 5. < v2 > is not always equal to ( < v >)2. The average of a squared quantity is not necessarily the square of the average. Can you find examples for this statement. EXERCISESEXERCISESEXERCISESEXERCISESEXERCISES 12.112.112.112.112.1 Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP. Take the diameter of an oxygen molecule to be 3 Å. 12.212.212.212.212.2 Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP : 1 atmospheric pressure, 0 °C). Show that it is 22.4 litres. 12.312.312.312.312.3 Figure 12.8 shows plot of PV/T versus P for 1.00×10–3 kg of oxygen gas at two different temperatures. y T1 PV –1 T2 (J K ) T x P Fig.Fig.Fig.Fig.Fig. 12.812.812.812.812.8 (a) What does the dotted plot signify? (b) Which is true: T1 > T2 or T1 < T2? (c) What is the value of PV/T where the curves meet on the y-axis? Reprint 2025-26 258 PHYSICS (d) If we obtained similar plots for 1.00×10–3 kg of hydrogen, would we get the same value of PV/T at the point where the curves meet on the y-axis? If not, what mass of hydrogen yields the same value of PV/T (for low pressure high temperature region of the plot) ? (Molecular mass of H2 = 2.02 u, of O2 = 32.0 u, R = 8.31 J mo1–1 K–1.) 12.412.412.412.412.4 An oxygen cylinder of volume 30 litre has an initial gauge pressure of 15 atm and a temperature of 27 °C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17 °C. Estimate the mass of oxygen taken out of the cylinder (R = 8.31 J mol–1 K–1, molecular mass of O2 = 32 u). 12.512.512.512.512.5 An air bubble of volume 1.0 cm3 rises from the bottom of a lake 40 m deep at a temperature of 12 °C. To what volume does it grow when it reaches the surface, which is at a temperature of 35 °C ? 12.612.612.612.612.6 Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 25.0 m3 at a temperature of 27 °C and 1 atm pressure. 12.712.712.712.712.7 Estimate the average thermal energy of a helium atom at (i) room temperature (27 °C), (ii) the temperature on the surface of the Sun (6000 K), (iii) the temperature of 10 million kelvin (the typical core temperature in the case of a star). 12.812.812.812.812.8 Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains neon (monatomic), the second contains chlorine (diatomic), and the third contains uranium hexafluoride (polyatomic). Do the vessels contain equal number of respective molecules ? Is the root mean square speed of molecules the same in the three cases? If not, in which case is vrms the largest ? 12.912.912.912.912.9 At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at – 20 °C ? (atomic mass of Ar = 39.9 u, of He = 4.0 u). 12.1012.1012.1012.1012.10 Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 17 0C. Take the radius of a nitrogen molecule to be roughly 1.0 Å. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of N2 = 28.0 u). Reprint 2025-26 CHAPTER THIRTEEN OSCILLATIONS 13.1 INTRODUCTION In our daily life we come across various kinds of motions. You have already learnt about some of them, e.g., rectilinear 13.1 Introduction motion and motion of a projectile. Both these motions are non-repetitive. We have also learnt about uniform circular13.2 Periodic and oscillatory motions motion and orbital motion of planets in the solar system. In 13.3 Simple harmonic motion these cases, the motion is repeated after a certain interval of 13.4 Simple harmonic motion time, that is, it is periodic. In your childhood, you must have and uniform circular enjoyed rocking in a cradle or swinging on a swing. Both motion these motions are repetitive in nature but different from the 13.5 Velocity and acceleration periodic motion of a planet. Here, the object moves to and fro in simple harmonic motion about a mean position. The pendulum of a wall clock executes 13.6 Force law for simple a similar motion. Examples of such periodic to and fro harmonic motion motion abound: a boat tossing up and down in a river, the

12.4 The ground state energy of hydrogen atom is –13.6 eV. What are the kinetic and potential energies of the electron in this state?

12.3 A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom make a transition from the upper level to the lower level?

12.6 (a) Using the Bohr’s model calculate the speed of the electron in a hydrogen atom in the n = 1, 2, and 3 levels. (b) Calculate the orbital period in each of these levels.

12.1 (a) No different from (b) Thomson’s model; Rutherford’s model (c) Rutherford’s model (d) Thomson’s model; Rutherford’s model (e) Both the models

3.6 VECTOR ADDITION – ANALYTICAL then, a vector T = a + b – c has components : METHOD T x = a x + b x − c x Although the graphical method of adding vectors Ty = a y + b y − c y (3.23b) helps us in visualising the vectors and the T z = a z + b z − c z .resultant vector, it is sometimes tedious and has limited accuracy. It is much easier to add vectors ⊳ Example 3.2 Find the magnitude andby combining their respective components. direction of the resultant of two vectors AConsider two vectors A and B in x-y plane with and B in terms of their magnitudes and components Ax, Ay and Bx, By : angle θ between them. ɵ ɵ A = A x i + A y j (3.18) * Note that angles α, β, and γ are angles in space. They are between pairs of lines, which are not coplanar. Reprint 2025-26 34 PHYSICS ⊳ Example 3.3 A motorboat is racing towards north at 25 km/h and the water current in that region is 10 km/h in the direction of 60° east of south. Find the resultant velocity of the boat. Answer The vector vb representing the velocity of the motorboat and the vector vc representing Fig. 3.10 the water current are shown in Fig. 3.11 in Answer Let OP and OQ represent the two vectors directions specified by the problem. Using the A and B making an angle θ (Fig. 3.10). Then, parallelogram method of addition, the resultant using the parallelogram method of vector R is obtained in the direction shown in the addition, OS represents the resultant vector R : figure. R = A + B SN is normal to OP and PM is normal to OS. From the geometry of the figure, OS2 = ON2 + SN2 but ON = OP + PN = A + B cos θ SN = B sin θ OS2 = (A + B cos θ)2 + (B sin θ)2 or, R2 = A2 + B2 + 2AB cos θ R = A 2 + B 2 + 2AB cosθ (3.24a) In ∆ OSN, SN = OS sinα = R sinα, and in ∆ PSN, SN = PS sin θ = B sin θ Therefore, R sin α = B sin θ R B or, = (3.24b) sin θ sin α Fig. 3.11 Similarly, PM = A sin α = B sin β We can obtain the magnitude of R using the Law A B or, = (3.24c) of cosine : sin β sin α Combining Eqs. (3.24b) and (3.24c), we get R = v 2b + v c2 + 2v bv c cos120 o R A B = = (3.24d) = 25 2 + 10 2 + 2 × 25 × 10 ( -1/2 ) ≅ 22 km/h sin θ sin β sin α To obtain the direction, we apply the Law of sines Using Eq. (3.24d), we get: R vc v c B = sin θ or, sin φ = sin α = sin θ (3.24e) sin θ sin φ R R where R is given by Eq. (3.24a). 10 × sin120 10 3 = = ≅ 0.397 SN B sin θ 21.8 2 × 21.8or, tan α= = (3.24f) OP + PN A + B cos θ φ ≅ 23.4 ⊳ Equation (3.24a) gives the magnitude of the 3.7 MOTION IN A PLANEresultant and Eqs. (3.24e) and (3.24f) its direction. Equation (3.24a) is known as the Law of cosines In this section we shall see how to describe and Eq. (3.24d) as the Law of sines. ⊳ motion in two dimensions using vectors. Reprint 2025-26 MOTION IN A PLANE 35 3.7.1 Position Vector and Displacement Suppose a particle moves along the curve shown The position vector r of a particle P located in a by the thick line and is at P at time t and P′ at plane with reference to the origin of an x-y time t′ [Fig. 3.12(b)]. Then, the displacement is : reference frame (Fig. 3.12) is given by ∆r = r′ – r (3.25) ɵ ɵ and is directed from P to P′. r = x i + y j We can write Eq. (3.25) in a component form: where x and y are components of r along x-, and y- axes or simply they are the coordinates of ɵ ɵ ɵ ɵ ∆r = x' i + y' j − x i + y j ( ) ( )the object. ɵ ɵ = i ∆ x + j ∆ y where ∆x = x ′ – x, ∆y = y′ – y (3.26) Velocity v The average velocity ( ) of an object is the ratio of the displacement and the corresponding time interval : ɵ ɵ ∆ r ∆ x i + ∆y j ɵ ∆ x ɵ ∆ y v = = = i + j (3.27) ∆ t ∆ t ∆ t ∆ t Or, v = v x ˆi + v y j (a) ∆ r Since v = , the direction of the average velocity ∆t is the same as that of ∆r (Fig. 3.12). The velocity (instantaneous velocity) is given by the limiting value of the average velocity as the time interval approaches zero : ∆ r d r v = lim = (3.28) ∆ t → 0 ∆ t d t The meaning of the limiting process can be easily understood with the help of Fig 3.13(a) to (d). In these figures, the thick line represents the path of an object, which is at P at time t. P1, P2 and (b) P3 represent the positions of the object after Fig. 3.12 (a) Position vector r. (b) Displacement ∆r and times ∆t1,∆t2, and ∆t3. ∆r1, ∆r2, and ∆r3 are the average velocity v of a particle. displacements of the object in times ∆t1, ∆t2, and Fig. 3.13 As the time interval ∆t approaches zero, the average velocity approaches the velocity v. The direction of v is parallel to the line tangent to the path. Reprint 2025-26 36 PHYSICS ∆t3, respectively. The direction of the average velocity v is shown in figures (a), (b) and (c) for three decreasing values of ∆t, i.e. ∆t1,∆t2, and ∆t3, (∆t1 > ∆t2 > ∆t3). As ∆t → 0, ∆r → 0 and is along the tangent to the path [Fig. 3.13(d)]. Therefore, the direction of velocity at any point on the path of an object is tangential to the path at that point and is in the direction of motion. We can express v in a component form : Fig. 3.14 The components vx and vy of velocity v and the angle θ it makes with x-axis. Note that dr vx = v cos θ, vy = v sin θ. v = dt The acceleration (instantaneous acceleration)  ∆x ɵ ∆y ɵ  is the limiting value of the average acceleration = lim  i + j  (3.29) ∆t → 0 ∆t ∆t  as the time interval approaches zero : ∆ v ∆x ∆y ɵ ɵ a = lim (3. 32a) + = i lim j lim ∆t → 0 ∆t ∆t → 0 ∆t ∆ t → 0 ∆ t ɵ ɵ ɵ dx ɵ dy ɵ ɵ Since ∆ v = ∆ v x i + ∆ v y j, we have + jOr, v = i = v x i + v y j . dt dt ɵ ∆ v x ɵ ∆ v y dx dy a = i lim + j lim ∆ t → 0 ∆ t ∆ t → 0 ∆ twhere v x = dt , v y = dt (3.30a) ɵ ɵ So, if the expressions for the coordinates x and Or, a = a x i + a y j (3.32b) y are known as functions of time, we can use d v y d v xthese equations to find vx and vy. (3.32c)* where, a x = , a y = The magnitude of v is then d t d t 2 2 As in the case of velocity, we can understand v = v x + v y (3.30b) graphically the limiting process used in defining and the direction of v is given by the angle θ : acceleration on a graph showing the path of the object’s motion. This is shown in Figs. 3.15(a) to  v y  (d). P represents the position of the object at v y −1 tanθ = , θ = tan     (3.30c) time t and P1, P2, P3 positions after time ∆t1, ∆t2,  v x  v x ∆t3, respectively (∆t 1> ∆t2>∆t3). The velocity vectors vx, vy and angle θ are shown in Fig. 3.14 for a at points P, P1, P2, P3 are also shown in Figs. 3.15 velocity vector v at point p. (a), (b) and (c). In each case of ∆t, ∆v is obtained using the triangle law of vector addition. ByAcceleration definition, the direction of average acceleration The average acceleration a of an object for a is the same as that of ∆v. We see that as ∆t time interval ∆t moving in x-y plane is the change decreases, the direction of ∆v changes and in velocity divided by the time interval : consequently, the direction of the acceleration ɵ ɵ changes. Finally, in the limit ∆t g0 [Fig. 3.15(d)], j v x i + v y ∆ v ∆ ( ∆ v y ɵ ) ∆ v x ɵ a = = = i + j (3.31a) the average acceleration becomes the ∆ t ∆ t ∆ t ∆ t instantaneous acceleration and has the direction ɵ ɵ as shown. Or, a = a x i + a y j . (3.31b) * In terms of x and y, ax and ay can be expressed as Reprint 2025-26 MOTION IN A PLANE 37 x (m) Fig. 3.15 The average acceleration for three time intervals (a) ∆t1, (b) ∆t2, and (c) ∆t3, (∆t1> ∆t2> ∆t3). (d) In the limit ∆t g0, the average acceleration becomes the acceleration. Note that in one dimension, the velocity and the acceleration of an object are always along -1  v y  − 1  4  ° θ = tan   = tan   ≅ 53 with x-axis. the same straight line (either in the same  v x   3  direction or in the opposite direction). ⊳ However, for motion in two or three dimensions, velocity and acceleration vectors may have any angle between 0° and 180° 3.8 MOTION IN A PLANE WITH CONSTANT between them. ACCELERATION ⊳ Suppose that an object is moving in x-y plane Example 3.4 The position of a particle is and its acceleration a is constant. Over an given by interval of time, the average acceleration will r = 3.0t ˆi + 2 .0t 2 ˆj + 5 .0 kˆ equal this constant value. Now, let the velocity where t is in seconds and the of the object be v0 at time t = 0 and v at time t. coefficients have the proper units for r to Then, by definition be in metres. (a) Find v(t) and a(t) of the v − v 0 v − v 0 particle. (b) Find the magnitude and a = = t − 0 t direction of v(t) at t = 1.0 s. Or, v = v 0 + a t (3.33a) Answer In terms of components : v x = v ox + a x t dr d ɵ ɵ ɵ v ( t ) = = 3.0 t i + 2.0t2 j + 5.0 k ( ) v y = v oy + a y t (3.33b) d t dt ɵ ɵ = 3.0 i + 4.0t j Let us now find how the position r changes with d v time. We follow the method used in the one- a ( t ) = = +4.0ɵj dt dimensional case. Let ro and r be the position a = 4.0 m s–2 along y- direction vectors of the particle at time 0 and t and let the velocities at these instants be vo and v. Then,At t = 1.0 s, v = 3.0ˆi + 4.0 ˆj over this time interval t, the average velocity is 2 2 -1 (vo + v)/2. The displacement is the average It’s magnitude is v = 3 + 4 = 5.0 m s velocity multiplied by the time interval : and direction is Reprint 2025-26 38 PHYSICS ˆ  v + v 0   ( v 0 + at ) + v 0  = 5.0 t + 1.5 t 2 i + 1.0 t 2 ˆj ( ) t t = r − r0 = 2  2    2 Therefore, x (t ) = 5.0 t + 1.5 t 1 2 y (t ) = +1.0 t 2 at = v 0 t + 2 Given x (t) = 84 m, t = ? 1 2 5.0 t + 1.5 t 2 = 84 ⇒ t = 6 s Or, r = r0 + v 0t + at (3.34a) At t = 6 s, y = 1.0 (6)2 = 36.0 m 2 d r 2.0 t ˆjIt can be easily verified that the derivative of Now, the velocity v = = ( 5.0 + 3.0 t ) ˆi + d t d r ɵ ɵEq. (3.34a), i.e. gives Eq.(3.33a) and it also At t = 6 s, v = 23.0 i + 12.0 j d t satisfies the condition that at t=0, r = ro. speed = v = 232 + 122 ≅ 26 m s−1 . ⊳Equation (3.34a) can be written in component form as 3.9 PROJECTILE MOTION 1 2 x = x 0 + v ox t + a x t As an application of the ideas developed in the 2 previous sections, we consider the motion of a 1 2 projectile. An object that is in flight after being y = y 0 + v oy t + a y t (3.34b) thrown or projected is called a projectile. Such 2 a projectile might be a football, a cricket ball, a One immediate interpretation of Eq.(3.34b) is that baseball or any other object. The motion of a the motions in x- and y-directions can be treated projectile may be thought of as the result of two independently of each other. That is, motion in separate, simultaneously occurring components a plane (two-dimensions) can be treated as two of motions. One component is along a horizontal separate simultaneous one-dimensional direction without any acceleration and the other motions with constant acceleration along two along the vertical direction with constant perpendicular directions. This is an important acceleration due to the force of gravity. It was result and is useful in analysing motion of objects Galileo who first stated this independency of the in two dimensions. A similar result holds for three horizontal and the vertical components ofdimensions. The choice of perpendicular projectile motion in his Dialogue on the greatdirections is convenient in many physical world systems (1632).situations, as we shall see in section 3.9 for In our discussion, we shall assume that theprojectile motion. air resistance has negligible effect on the motion ⊳ Example 3.5 A particle starts from origin of the projectile. Suppose that the projectile is launched with velocity vo that makes an angle at t = 0 with a velocity 5.0 î m/s and moves in x-y plane under action of a force which θo with the x-axis as shown in Fig. 3.16. produces a constant acceleration of (3.0iɵ+2.0jɵ) m/s 2. (a) What is the After the object has been projected, the acceleration acting on it is that due to gravity y-coordinate of the particle at the instant which is directed vertically downward: its x-coordinate is 84 m ? (b) What is the ɵ speed of the particle at this time ? a = −g j Or, ax = 0, ay = – g (3.35) Answer From Eq. (3.34a) for r0 = 0, the position The components of initial velocity vo are : of the particle is given by 1 at 2 vox = vo cos θo r (t ) = v 0 t + 2 2 voy= vo sin θo (3.36) t 3.0ˆi + 2.0 ˆj = 5.0 ˆi t + (1/2 )( ) Reprint 2025-26 MOTION IN A PLANE 39 Now, since g, θo and vo are constants, Eq. (3.39) is of the form y = a x + b x2, in which a and b are constants. This is the equation of a parabola, i.e. the path of the projectile is a parabola (Fig. 3.17). Fig 3.16 Motion of an object projected with velocity vo at angle θ0. If we take the initial position to be the origin of the reference frame as shown in Fig. 3.16, we have : xo = 0, yo = 0 Then, Eq.(3.34b) becomes : x = vox t = (vo cos θo ) t and y = (vo sin θo ) t – ( ½ )g t2 (3.37) Fig. 3.17 The path of a projectile is a parabola. The components of velocity at time t can be Time of maximum height obtained using Eq.(3.33b) : How much time does the projectile take to reach the vx = vox = vo cos θo maximum height ? Let this time be denoted by tm. Since at this point, vy= 0, we have from Eq. (3.38): vy = vo sin θo – g t (3.38) vy = vo sinθo – g tm = 0 Equation (3.37) gives the x-, and y-coordinates Or, tm = vo sinθo /g (3.40a) of the position of a projectile at time t in terms of The total time Tf during which the projectile is two parameters — initial speed vo and projection in flight can be obtained by putting y = 0 in angle θo. Notice that the choice of mutually Eq. (3.37). We get : perpendicular x-, and y-directions for the analysis of the projectile motion has resulted in Tf = 2 (vo sin θo )/g (3.40b) a simplification. One of the components of Tf is known as the time of flight of the projectile. velocity, i.e. x-component remains constant We note that Tf = 2 tm , which is expected throughout the motion and only the because of the symmetry of the parabolic path. y- component changes, like an object in free fall Maximum height of a projectile in vertical direction. This is shown graphically The maximum height hm reached by theat few instants in Fig. 3.17. Note that at the point projectile can be calculated by substituting of maximum height, vy= 0 and therefore, t = tm in Eq. (3.37) : -1 v y 2 = oθ = tan    g  v 0 sinθ0 v 0 sinθ0 v x y = h m = ( v 0 sinθ0 )  −   Equation of path of a projectile  g  2  g  What is the shape of the path followed by the ( v 0 sinθ0 ) 2 projectile? This can be seen by eliminating the Or, h m = (3.41) 2gtime between the expressions for x and y as given in Eq. (3.37). We obtain: Horizontal range of a projectile g 2 The horizontal distance travelled by a projectile from x y = ( tan θo ) x − (3.39) its initial position (x = y = 0) to the position where it 2 (v o cosθo )2 passes y = 0 during its fall is called the horizontal Reprint 2025-26 40 PHYSICS range, R. It is the distance travelled during the time y (t) = yo + voy t +(1/2) ay t2 of flight Tf . Therefore, the range R is Here, xo = yo = 0, voy = 0, ay = –g = –9.8 m s-2, R = (vo cos θo) (Tf ) vox = 15 m s-1. The stone hits the ground when y(t) = – 490 m. =(vo cos θo) (2 vo sin θo)/g – 490 m = –(1/2)(9.8) t2. 2 sin 2θ0 This gives t =10 s. v 0 Or, R = (3.42a) The velocity components are vx = vox and g vy = voy – g t Equation (3.42a) shows that for a given so that when the stone hits the ground : projection velocity vo , R is maximum when sin vox = 15 m s–1 2θ0 is maximum, i.e., when θ0 = 450. voy = 0 – 9.8 × 10 = – 98 m s–1 The maximum horizontal range is, therefore, Therefore, the speed of the stone is 2 v 0 v 2x + v y2 = 15 2 + 98 2 = 99 m s −1 ⊳ R m = (3.42b) g ⊳ ⊳ Example 3.6 Galileo, in his book Two new Example 3.8 A cricket ball is thrown at a sciences, stated that “for elevations which speed of 28 m s–1 in a direction 30° above exceed or fall short of 45° by equal amounts, the horizontal. Calculate (a) the maximum the ranges are equal”. Prove this statement. height, (b) the time taken by the ball to return to the same level, and (c) the distance from the thrower to the point Answer For a projectile launched with velocity where the ball returns to the same level. vo at an angle θo , the range is given by v 02 sin2θ0 R = Answer (a) The maximum height is given by g 2 2 sin sin 30 ( 28 ( v 0 θo ) ° ) m = m =Now, for angles, (45° + α) and ( 45° – α), 2θo is h 2 2 g ( 9.8 )(90° + 2α) and ( 90° – 2α) , respectively. The values of sin (90° + 2α) and sin (90° – 2α) are 14 × 14 = = 10.0 mthe same, equal to that of cos 2α. Therefore, 2 × 9.8 ranges are equal for elevations which exceed or fall short of 45° by equal amounts α. ⊳ (b) The time taken to return to the same level is Tf = (2 vo sin θo )/g = (2× 28 × sin 30° )/9.8 ⊳ = 28/9.8 s = 2.9 s Example 3.7 A hiker stands on the edge (c) The distance from the thrower to the point of a cliff 490 m above the ground and where the ball returns to the same level is throws a stone horizontally with an initial speed 2 of 15 m s-1. Neglecting air resistance, v o sin2θo 28 × 28 × sin60o ( ) = = 69 m ⊳ find the time taken by the stone to reach R = g 9.8 the ground, and the speed with which it hits the ground. (Take g = 9.8 m s-2 ). 3.10 UNIFORM CIRCULAR MOTION Answer We choose the origin of the x-,and y- When an object follows a circular path at a axis at the edge of the cliff and t = 0 s at the constant speed, the motion of the object is called instant the stone is thrown. Choose the positive uniform circular motion. The word “uniform” direction of x-axis to be along the initial velocity refers to the speed, which is uniform (constant) and the positive direction of y-axis to be the throughout the motion. Suppose an object is vertically upward direction. The x-, and y- moving with uniform speed v in a circle of radius components of the motion can be treated R as shown in Fig. 3.18. Since the velocity of the independently. The equations of motion are : object is changing continuously in direction, the x (t) = xo + vox t object undergoes acceleration. Let us find the magnitude and the direction of this acceleration. Reprint 2025-26 MOTION IN A PLANE 41 Fig. 3.18 Velocity and acceleration of an object in uniform circular motion. The time interval ∆t decreases from (a) to (c) where it is zero. The acceleration is directed, at each point of the path, towards the centre of the circle. Let r and r′ be the position vectors and v and r′ be ∆θ. Since the velocity vectors v and v′ are v′ the velocities of the object when it is at point P always perpendicular to the position vectors, the and P′ as shown in Fig. 3.18(a). By definition, angle between them is also ∆θ . Therefore, the velocity at a point is along the tangent at that triangle CPP′ formed by the position vectors and point in the direction of motion. The velocity the triangle GHI formed by the velocity vectors vectors v and v′ are as shown in Fig. 3.18(a1). v, v′ and ∆v are similar (Fig. 3.18a). Therefore, ∆v is obtained in Fig. 3.18 (a2) using the triangle the ratio of the base-length to side-length for law of vector addition. Since the path is circular, one of the triangles is equal to that of the other v is perpendicular to r and so is v′ to r′. triangle. That is : Therefore, ∆v is perpendicular to ∆r. Since  ∆v  ∆ v ∆ r average acceleration is along ∆v  a =  , the =  ∆t  v R average acceleration a is perpendicular to ∆r. If we place ∆v on the line that bisects the angle ∆ r Or, ∆ v = v between r and r′, we see that it is directed towards R the centre of the circle. Figure 3.18(b) shows the Therefore, same quantities for smaller time interval. ∆v and ∆ v v ∆ r v ∆r hence a is again directed towards the centre. a = lim = lim = lim t ∆ t → 0 R ∆ t R ∆ t → 0 ∆ tIn Fig. 3.18(c), ∆tŽ 0 and the average ∆ t → 0 ∆ acceleration becomes the instantaneous If ∆t is small, ∆θ will also be small and then arc acceleration. It is directed towards the centre*. PP′ can be approximately taken to be|∆r|: Thus, we find that the acceleration of an object ∆ r ≅v∆ t in uniform circular motion is always directed ∆r towards the centre of the circle. Let us now find ≅ v the magnitude of the acceleration. ∆ t The magnitude of a is, by definition, given by ∆ r lim = v ∆v Or, ∆ t → 0 ∆ t a = lim ∆ t → 0 ∆t Let the angle between position vectors r and Therefore, the centripetal acceleration ac is : * In the limit ∆tŽ0, ∆r becomes perpendicular to r. In this limit ∆v→ 0 and is consequently also perpendicular to V. Therefore, the acceleration is directed towards the centre, at each point of the circular path. Reprint 2025-26 42 PHYSICS 2 2 2  v  v R ω 2  R ac = = = ω a c =  R v = v2/R (3.43) R R Thus, the acceleration of an object moving with 2 a c = ω R (3.46) speed v in a circle of radius R has a magnitude v 2/R and is always directed towards the centre. The time taken by an object to make one revolution This is why this acceleration is called centripetal is known as its time period T and the number ofacceleration (a term proposed by Newton). A revolution made in one second is called itsthorough analysis of centripetal acceleration was frequency ν (=1/T). However, during this timefirst published in 1673 by the Dutch scientist the distance moved by the object is s = 2πR.Christiaan Huygens (1629-1695) but it was probably known to Newton also some years earlier. Therefore, v = 2πR/T =2πRν (3.47) “Centripetal” comes from a Greek term which means In terms of frequency ν, we have ‘centre-seeking’. Since v and R are constant, the ω = 2πν magnitude of the centripetal acceleration is also v = 2πRν constant. However, the direction changes — ac = 4π2 ν2R (3.48)pointing always towards the centre. Therefore, a ⊳centripetal acceleration is not a constant vector. Example 3.9 An insect trapped in a We have another way of describing the circular groove of radius 12 cm moves along velocity and the acceleration of an object in the groove steadily and completes 7 uniform circular motion. As the object moves revolutions in 100 s. (a) What is the from P to P′ in time ∆t (= t′ – t), the line CP angular speed, and the linear speed of the (Fig. 3.18) turns through an angle ∆θ as shown motion? (b) Is the acceleration vector a in the figure. ∆θ is called angular distance. We constant vector ? What is its magnitude ? define the angular speed ω (Greek letter omega) as the time rate of change of angular Answer This is an example of uniform circular displacement : motion. Here R = 12 cm. The angular speed ω is ∆θ given by ω = ∆t (3.44) ω = 2π/T = 2π × 7/100 = 0.44 rad/s The linear speed v is :Now, if the distance travelled by the object during the time ∆t is ∆s, i.e. PP′ is ∆s, then : v =ω R = 0.44 s-1 × 12 cm = 5.3 cm s-1 ∆ s The direction of velocity v is along the tangent v = ∆t to the circle at every point. The acceleration is directed towards the centre of the circle. Sincebut ∆s = R ∆θ. Therefore : this direction changes continuously, ∆θ v = R = R ω acceleration here is not a constant vector. ∆ t However, the magnitude of acceleration is v = R ω (3.45) constant: a = ω2 R = (0.44 s–1)2 (12 cm)We can express centripetal acceleration ac in terms of angular speed : = 2.3 cm s-2 ⊳ Reprint 2025-26 MOTION IN A PLANE 43 SUMMARY 1. Scalar quantities are quantities with magnitudes only. Examples are distance, speed, mass and temperature. 2. Vector quantities are quantities with magnitude and direction both. Examples are displacement, velocity and acceleration. They obey special rules of vector algebra. 3. A vector A multiplied by a real number λ is also a vector, whose magnitude is λ times the magnitude of the vector A and whose direction is the same or opposite depending upon whether λ is positive or negative. 4. Two vectors A and B may be added graphically using head-to-tail method or parallelogram method. 5. Vector addition is commutative : A + B = B + A It also obeys the associative law : (A + B) + C = A + (B + C) 6. A null or zero vector is a vector with zero magnitude. Since the magnitude is zero, we don’t have to specify its direction. It has the properties : A + 0 = A λ0 = 0 0 A = 0 7. The subtraction of vector B from A is defined as the sum of A and –B : A – B = A+ (–B) 8. A vector A can be resolved into component along two given vectors a and b lying in the same plane : A = λ a + µ b where λ and µ are real numbers. 9. A unit vector associated with a vector A has magnitude 1 and is along the vector A: A ˆn = A ɵ ɵ ɵ The unit vectors i, j, k are vectors of unit magnitude and point in the direction of the x-, y-, and z-axes, respectively in a right-handed coordinate system. 10. A vector A can be expressed as ɵ ɵ A = A x i + Ay j where Ax, Ay are its components along x-, and y -axes. If vector A makes an angle θ A y 2 2 with the x-axis, then Ax = A cos θ, Ay=A sin θ and A = A = A x + A y , tanθ = . A x 11. Vectors can be conveniently added using analytical method. If sum of two vectors A and B, that lie in x-y plane, is R, then : ɵ R = R x i + Ry ɵ,j where, Rx = Ax + Bx, and Ry = Ay + By ɵ ɵ 12. The position vector of an object in x-y plane is given by r = x i + y j and the displacement from position r to position r’ is given by ∆r = r′− r ɵ ɵ = ( x ′ − x ) i + (y ′ − y ) j ɵ ɵ = ∆x i + ∆y j 13. If an object undergoes a displacement ∆r in time ∆t, its average velocity is given by ∆ r v = . The velocity of an object at time t is the limiting value of the average velocity ∆t as ∆t tends to zero : Reprint 2025-26 44 PHYSICS ∆ r d r v = lim = . It can be written in unit vector notation as : ∆ t →0 ∆ t dt ɵ ɵ ɵ dx dy dz = y v = v x i + v y j + v z k where v x = dt , v dt , v z = dt When position of an object is plotted on a coordinate system, v is always tangent to the curve representing the path of the object. 14. If the velocity of an object changes from v to v′in time ∆t, then its average acceleration v − v' ∆ v is given by: a = = ∆ t ∆t The acceleration a at any time t is the limiting value of a as ∆t Ž0 : ∆ v dv a = lim = ∆ t → 0 ∆ t dt ɵ ɵ ɵ In component form, we have : a = a x i + a y j + a z k dvy dv x dv z where, a x = , a y = , a z = dt dt dt 15. If an object is moving in a plane with constant acceleration a = a = a x2 + a y2 and its position vector at time t = 0 is ro, then at any other time t, it will be at a point given by: 1 2 r = ro + v o t + at 2 and its velocity is given by : v = vo + a t where vo is the velocity at time t = 0 In component form : 1 2 x = x o + v ox t + a x t 2 1 2 y = yo + v oy t + a y t 2 v x = v ox + a x t v y = v oy + a y t Motion in a plane can be treated as superposition of two separate simultaneous one- dimensional motions along two perpendicular directions 16. An object that is in flight after being projected is called a projectile. If an object is projected with initial velocity vo making an angle θo with x-axis and if we assume its initial position to coincide with the origin of the coordinate system, then the position and velocity of the projectile at time t are given by : x = (vo cos θo) t y = (vo sin θo) t − (1/2) g t2 vx = vox = vo cos θo vy = vo sin θo − g t The path of a projectile is parabolic and is given by : 2 gx y = ( tanθ0 ) x – 2 2 (v o cosθo ) The maximum height that a projectile attains is : Reprint 2025-26 MOTION IN A PLANE 45 (v o sinq o )2 h m = 2g The time taken to reach this height is : v o sinθo t m = g The horizontal distance travelled by a projectile from its initial position to the position it passes y = 0 during its fall is called the range, R of the projectile. It is : v o2 R = sin2θo g 17. When an object follows a circular path at constant speed, the motion of the object is called uniform circular motion. The magnitude of its acceleration is ac = v2 /R. The direction of ac is always towards the centre of the circle. The angular speed ω, is the rate of change of angular distance. It is related to velocity v by v = ω R. The acceleration is ac = ω 2R. If T is the time period of revolution of the object in circular motion and ν is its frequency, we have ω = 2π ν, v = 2πνR, ac = 4π2ν2R Reprint 2025-26 46 PHYSICS POINTS TO PONDER 1. The path length traversed by an object between two points is, in general, not the same as the magnitude of displacement. The displacement depends only on the end points; the path length (as the name implies) depends on the actual path. The two quantities are equal only if the object does not change its direction during the course of motion. In all other cases, the path length is greater than the magnitude of displacement. 2. In view of point 1 above, the average speed of an object is greater than or equal to the magnitude of the average velocity over a given time interval. The two are equal only if the path length is equal to the magnitude of displacement. 3. The vector equations (3.33a) and (3.34a) do not involve any choice of axes. Of course, you can always resolve them along any two independent axes. 4. The kinematic equations for uniform acceleration do not apply to the case of uniform circular motion since in this case the magnitude of acceleration is constant but its direction is changing. 5. An object subjected to two velocities v1 and v2 has a resultant velocity v = v1 + v2. Take care to distinguish it from velocity of object 1 relative to velocity of object 2 : v12= v1 − v2. Here v1 and v2 are velocities with reference to some common reference frame. 6. The resultant acceleration of an object in circular motion is towards the centre only if the speed is constant. 7. The shape of the trajectory of the motion of an object is not determined by the acceleration alone but also depends on the initial conditions of motion ( initial position and initial velocity). For example, the trajectory of an object moving under the same acceleration due to gravity can be a straight line or a parabola depending on the initial conditions. EXERCISES 3.1 State, for each of the following physical quantities, if it is a scalar or a vector : volume, mass, speed, acceleration, density, number of moles, velocity, angular frequency, displacement, angular velocity. 3.2 Pick out the two scalar quantities in the following list : force, angular momentum, work, current, linear momentum, electric field, average velocity, magnetic moment, relative velocity. 3.3 Pick out the only vector quantity in the following list : Temperature, pressure, impulse, time, power, total path length, energy, gravitational potential, coefficient of friction, charge. 3.4 State with reasons, whether the following algebraic operations with scalar and vector physical quantities are meaningful : (a) adding any two scalars, (b) adding a scalar to a vector of the same dimensions , (c) multiplying any vector by any scalar, (d) multiplying any two scalars, (e) adding any two vectors, (f) adding a component of a vector to the same vector. 3.5 Read each statement below carefully and state with reasons, if it is true or false : (a) The magnitude of a vector is always a scalar, (b) each component of a vector is always a scalar, (c) the total path length is always equal to the magnitude of the displacement vector of a particle. (d) the average speed of a particle (defined as total path length divided by the time taken to cover the path) is either greater or equal to the magnitude of average velocity of the particle over the same interval of time, (e) Three vectors not lying in a plane can never add up to give a null vector. 3.6 Establish the following vector inequalities geometrically or otherwise : (a) |a+b| < |a| + |b| (b) |a+b| > ||a| −|b|| Reprint 2025-26 MOTION IN A PLANE 47 (c) |a−b| < |a| + |b| (d) |a−b| > ||a| − |b|| When does the equality sign above apply? 3.7 Given a + b + c + d = 0, which of the following statements are correct : Q (a) a, b, c, and d must each be a null vector, (b) The magnitude of (a + c) equals the magnitude of ( b + d), (c) The magnitude of a can never be greater than the sum of the magnitudes of b, c, and d, (d) b + c must lie in the plane of a and d if a and d are not collinear, and in the line of a and d, if they are collinear ? 3.8 Three girls skating on a circular ice ground of radius 200 m start from a point P on the edge of the ground and reach a point Q diametrically opposite to P following different paths as shown in Fig. 3.19. What is the magnitude of the displacement vector for each ? For Fig. 3.19 which girl is this equal to the actual length of path skate ? 3.9 A cyclist starts from the centre O of a circular park of radius 1 km, reaches the edge P of the park, then cycles along the circumference, and returns to the centre along QO as shown in Fig. 3.20. If the round trip takes 10 min, what is the (a) net displacement, (b) average velocity, and (c) average speed of the cyclist ? Fig. 3.20 3.10 On an open ground, a motorist follows a track that turns to his left by an angle of 600 after every 500 m. Starting from a given turn, specify the displacement of the motorist at the third, sixth and eighth turn. Compare the magnitude of the displacement with the total path length covered by the motorist in each case. 3.11 A passenger arriving in a new town wishes to go from the station to a hotel located 10 km away on a straight road from the station. A dishonest cabman takes him along a circuitous path 23 km long and reaches the hotel in 28 min. What is (a) the average speed of the taxi, (b) the magnitude of average velocity ? Are the two equal ? 3.12 The ceiling of a long hall is 25 m high. What is the maximum horizontal distance that a ball thrown with a speed of 40 m s-1 can go without hitting the ceiling of the hall ? 3.13 A cricketer can throw a ball to a maximum horizontal distance of 100 m. How much high above the ground can the cricketer throw the same ball ? Reprint 2025-26 48 PHYSICS 3.14 A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and direction of acceleration of the stone ? 3.15 An aircraft executes a horizontal loop of radius 1.00 km with a steady speed of 900 km/h. Compare its centripetal acceleration with the acceleration due to gravity. 3.16 Read each statement below carefully and state, with reasons, if it is true or false : (a) The net acceleration of a particle in circular motion is always along the radius of the circle towards the centre (b) The velocity vector of a particle at a point is always along the tangent to the path of the particle at that point (c) The acceleration vector of a particle in uniform circular motion averaged over one cycle is a null vector 3.17 The position of a particle is given by r = 3.0t ˆi − 2.0t 2 ˆj + 4.0 kˆ m where t is in seconds and the coefficients have the proper units for r to be in metres. (a) Find the v and a of the particle? (b) What is the magnitude and direction of velocity of the particle at t = 2.0 s ? 3.18 A particle starts from the origin at t = 0 s with a velocity of 10.0 jɵ m/s and moves in ɵ ɵ the x-y plane with a constant acceleration of 8.0 i + 2.0 j m s-2. (a) At what time is ( ) the x- coordinate of the particle 16 m? What is the y-coordinate of the particle at that time? (b) What is the speed of the particle at the time ? 3.19 ɵi and ɵj are unit vectors along x- and y- axis respectively. What is the magnitude ɵ ɵ ɵ ɵ and direction of the vectors i + j , and i − j ? What are the components of a vector ɵ ɵ ɵ ɵ ɵ ɵ A= 2 i + 3 j along the directions of i + j and i −?j [You may use graphical method] 3.20 For any arbitrary motion in space, which of the following relations are true : (a) vaverage = (1/2) (v (t1) + v (t2)) (b) v average = [r(t2) - r(t1) ] /(t2 – t1) (c) v (t) = v (0) + a t (d) r (t) = r (0) + v (0) t + (1/2) a t2 (e) a average =[ v (t2) - v (t1 )] /( t2 – t1) (The ‘average’ stands for average of the quantity over the time interval t1 to t2) 3.21 Read each statement below carefully and state, with reasons and examples, if it is true or false : A scalar quantity is one that (a) is conserved in a process (b) can never take negative values (c) must be dimensionless (d) does not vary from one point to another in space (e) has the same value for observers with different orientations of axes. 3.22 An aircraft is flying at a height of 3400 m above the ground. If the angle subtended at a ground observation point by the aircraft positions 10.0 s a part is 30°, wat is the speed of the aircraft ? Reprint 2025-26 CHAPTER FOUR LAWS OF MOTION 4.1 INTRODUCTION In the preceding Chapter, our concern was to describe the motion of a particle in space quantitatively. We saw that 4.1 Introduction uniform motion needs the concept of velocity alone whereas 4.2 Aristotle’s fallacy non-uniform motion requires the concept of acceleration in 4.3 The law of inertia addition. So far, we have not asked the question as to what governs the motion of bodies. In this chapter, we turn to this4.4 Newton’s first law of motion basic question.4.5 Newton’s second law of Let us first guess the answer based on our common motion experience. To move a football at rest, someone must kick it.4.6 Newton’s third law of motion To throw a stone upwards, one has to give it an upward

3.2 A battery of emf 10 V and internal resistance 3 Ω is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?

3.7 and are in the direction specified by the problem. Using the rule of vector addition, we see that the resultant of vr and vw is R as shown in the figure. The magnitude of R is 2 2 2 2 −1 −1 R = v r + v w = 35 + 12 m s = 37 m s The direction θ that R makes with the vertical is given by v w 12 tan θ = = = 0.343 Fig. 3.8 (a) Two non-colinear vectors a and b. v r 35 (b) Resolving a vector A in terms of vectors Or, θ = tan-1 ( 0.343 ) = 19° a and b. Therefore, the boy should hold his umbrella We say that A has been resolved into two in the vertical plane at an angle of about 19o component vectors λ a and µ b along a and b with the vertical towards the east. ⊳ Reprint 2025-26 32 PHYSICS respectively. Using this method one can resolve and A2 is parallel to ɵj, we have :a given vector into two component vectors along a set of two vectors – all the three lie in the same A1= Ax ɵi , A2 = Ay ɵj (3.11) plane. It is convenient to resolve a general vector where Ax and Ay are real numbers.along the axes of a rectangular coordinate system using vectors of unit magnitude. These Thus, A = Ax ɵi + Ay ɵj (3.12) are called unit vectors that we discuss now. A unit vector is a vector of unit magnitude and This is represented in Fig. 3.9(c). The quantities points in a particular direction. It has no Ax and Ay are called x-, and y- components of the dimension and unit. It is used to specify a vector A. Note that Ax is itself not a vector, but direction only. Unit vectors along the x-, y- and ɵi is a vector, and so is Ay ɵj. Using simplez-axes of a rectangular coordinate system are Ax trigonometry, we can express Ax and Ay in terms denoted by ɵi , ɵj and ˆk , respectively, as shown of the magnitude of A and the angle θ it makes in Fig. 3.9(a). with the x-axis : Since these are unit vectors, we have Ax = A cos θ Ay = A sin θ (3.13) ˆi  = ˆj  = ˆk =1 (3.9) As is clear from Eq. (3.13), a component of a These unit vectors are perpendicular to each vector can be positive, negative or zero other. In this text, they are printed in bold face depending on the value of θ. with a cap (^) to distinguish them from other Now, we have two ways to specify a vector A vectors. Since we are dealing with motion in two in a plane. It can be specified by : dimensions in this chapter, we require use of (i) its magnitude A and the direction θ it makes only two unit vectors. If we multiply a unit vector, with the x-axis; or say ˆn by a scalar, the result is a vector (ii) its components Ax and Ay λ = λ ˆn. In general, a vector A can be written as If A and θ are given, Ax and Ay can be obtained using Eq. (3.13). If Ax and Ay are given, A and θ A = |A| ˆn (3.10) can be obtained as follows : where ˆn is a unit vector along A. 2 2 2 2 2 2 A x + A y = A cos θ + A sin θ We can now resolve a vector A in terms = A2 of component vectors that lie along unit vectors iˆ and ɵj. Consider a vector A that lies in x-y Or, A = A 2x + Ay2 (3.14) plane as shown in Fig. 3.9(b). We draw lines from the head of A perpendicular to the coordinate Ay A y tan θ = , θ = tan− 1 axes as in Fig. 3.9(b), and get vectors A1 and A2 And A x A x (3.15) such that A1 + A2 = A. Since A1 is parallel to ɵi Fig. 3.9 (a) Unit vectors ɵi , ɵj and ɵk lie along the x-, y-, and z-axes. (b) A vector A is resolved into its components Ax and Ay along x-, and y- axes. (c) A1 and A2 expressed in terms of ɵi and ɵj . Reprint 2025-26 MOTION IN A PLANE 33 So far we have considered a vector lying in ɵ ɵ B = B x i + B y jan x-y plane. The same procedure can be used Let R be their sum. We haveto resolve a general vector A into three components along x-, y-, and z-axes in three R = A + B dimensions. If α, β, and γ are the angles* ɵ ɵ ɵ ɵ = + (3.19a) ( A x i + A y j ) ( B x i + B y j )between A and the x-, y-, and z-axes, respectively [Fig. 3.9(d)], we have Since vectors obey the commutative and associative laws, we can arrange and regroup the vectors in Eq. (3.19a) as convenient to us : ɵ ɵ j (3.19b) R = ( A x + B x ) i + ( A y + B y ) ɵ ɵ SinceR = R x i + R y j (3.20) we have, R x = A x + B x , R y = A y + B y (3.21) Thus, each component of the resultant vector R is the sum of the corresponding components of A and B. In three dimensions, we have ɵ ɵ ɵ A = A x i + Ay j + A z k ɵ ɵ ɵ B = B x i + B y j + B z k (d) ɵ ɵ ɵ R = A + B = R x i + R y j + R z kFig. 3.9 (d) A vector A resolved into components along x-, y-, and z-axes with R x = A x + B x y = A y + B yA x = A cos α, A y = A cos β, A z = A cos γ (3.16a) R In general, we have R z = A z + B z (3.22) A = Ax ˆi + Ay ˆj + Az kˆ (3.16b) This method can be extended to addition and The magnitude of vector A is subtraction of any number of vectors. For A = A x2 + Ay2 + A z2 (3.16c) example, if vectors a, b and c are given as ɵ ɵ ɵ A position vector r can be expressed as a = a x i + a y j + a z k ɵ ɵ ɵ r = x i + y j + z k (3.17) ɵ ɵ ɵ b = b x i + b y j + b z k where x, y, and z are the components of r along ɵ ɵ ɵ x-, y-, z-axes, respectively. c = c x i + c y j + c z k (3.23a)

3.1 The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4 Ω, what is the maximum current that can be drawn from the battery?

3.3 At room temperature (27.0 °C) the resistance of a heating element is 100 Ω. What is the temperature of the element if the resistance is found to be 117 Ω, given that the temperature coefficient of the material of the resistor is 1.70 × 10–4 °C–1. 105 Reprint 2025-26 Physics 3.4 A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6.0 × 10–7 m2, and its resistance is measured to be 5.0 W. What is the resistivity of the material at the temperature of the experiment? 3.5 A silver wire has a resistance of 2.1 W at 27.5 °C, and a resistance of 2.7 W at 100 °C. Determine the temperature coefficient of resistivity of silver. 3.6 A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds to a steady value of 2.8 A. What is the steady temperature of the heating element if the room temperature is 27.0 °C? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is 1.70 × 10–4 °C–1. 3.7 Determine the current in each branch of the network shown in Fig. 3.20: FIGURE 3.20 3.8 A storage battery of emf 8.0 V and internal resistance 0.5 W is being charged by a 120 V dc supply using a series resistor of 15.5 W. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit? 3.9 The number density of free electrons in a copper conductor estimated in Example 3.1 is 8.5 × 1028 m–3. How long does an electron take to drift from one end of a wire 3.0 m long to its other end? The area of cross-section of the wire is 2.0 × 10–6 m2 and it is carrying a current of 3.0 A. Reprint 2025-26 Chapter Four MOVING CHARGES AND MAGNETISM 4.1 INTRODUCTION Both Electricity and Magnetism have been known for more than 2000 years. However, it was only about 200 years ago, in 1820, that it was realised that they were intimately related. During a lecture demonstration in the summer of 1820, Danish physicist Hans Christian Oersted noticed that a current in a straight wire caused a noticeable deflection in a nearby magnetic compass needle. He investigated this phenomenon. He found that the alignment of the needle is tangential to an imaginary circle which has the straight wire as its centre and has its plane perpendicular to the wire. This situation is depicted in Fig.4.1(a). It is noticeable when the current is large and the needle sufficiently close to the wire so that the earth’s magnetic field may be ignored. Reversing the direction of the current reverses the orientation of the needle [Fig. 4.1(b)]. The deflection increases on increasing the current or bringing the needle closer to the wire. Iron filings sprinkled around the wire arrange themselves in concentric circles with the wire as the centre [Fig. 4.1(c)]. Oersted concluded that moving charges or currents produced a magnetic field in the surrounding space. Following this, there was intense experimentation. In 1864, the laws obeyed by electricity and magnetism were unified and formulated by Reprint 2025-26 Physics James Maxwell who then realised that light was electromagnetic waves. Radio waves were discovered by Hertz, and produced by J.C.Bose and G. Marconi by the end of the 19th century. A remarkable scientific and technological progress took place in the 20th century. This was due to our increased understanding of electromagnetism and the invention of devices for production, amplification, transmission and detection of electromagnetic waves. FIGURE 4.1 The magnetic field due to a straight long current-carrying wire. The wire is perpendicular to the plane of the paper. A ring of compass needles surrounds the wire. The orientation of the needles is shown when (a) the current emerges out of the plane of the paper, (b) the current moves into the plane of the paper. (c) The arrangement of iron filings around the wire. The darkened ends of the needle represent north poles. The effect of the earth’s magnetic field is neglected. In this chapter, we will see how magnetic field exerts forces on moving charged particles, like electrons, protons, and current-carrying wires. We shall also learn how currents produce magnetic fields. We shall see how particles can be accelerated to very high energies in a cyclotron. We shall study how currents and voltages are detected by a galvanometer.(1777–1851) In this and subsequent Chapter on magnetism, we adopt the following convention: A current or a field (electric or magnetic) emerging out of the plane of the paper is depicted by a dot (¤). A current or a field going into the plane of the paper is depicted by a cross ()*. Hans Christian Oersted Figures. 4.1(a) and 4.1(b) correspond to these twoOERSTED (1777–1851) Danish situations, respectively. physicist and chemist, professor at Copenhagen. 4.2 MAGNETIC FORCE He observed that a compass needle suffers a 4.2.1 Sources and fields deflection when placed Before we introduce the concept of a magnetic field B, weCHRISTIAN near a wire carrying an electric current. This shall recapitulate what we have learnt in Chapter 1 about discovery gave the first the electric field E. We have seen that the interaction empirical evidence of a between two charges can be considered in two stages.HANS connection between electric The charge Q, the source of the field, produces an electric and magnetic phenomena. field E, where * A dot appears like the tip of an arrow pointed at you, a cross is like the feathered 108 tail of an arrow moving away from you. Reprint 2025-26 Moving Charges and Magnetism E = Q ˆr / (4pe0)r2 (4.1) where ˆr is unit vector along r, and the field E is a vector field. A charge q interacts with this field and experiences a force F given by F = q E = q Q ˆr / (4pe0) r 2 (4.2) As pointed out in the Chapter 1, the field E is not just an artefact but has a physical role. It can convey energy and momentum and is not established instantaneously but takes finite time to propagate. The concept of a field was specially stressed by Faraday and was incorporated by Maxwell in his unification of electricity and magnetism. In addition to depending on each point in space, it can also vary with time, i.e., be a function of time. In our Hendrik Antoon Lorentz discussions in this chapter, we will assume that the fields (1853 – 1928) Dutch do not change with time. theoretical physicist, The field at a particular point can be due to one or professor at Leiden. He investigated themore charges. If there are more charges the fields add HENDRIK relationship between vectorially. You have already learnt in Chapter 1 that this electricity, magnetism, and is called the principle of superposition. Once the field is mechanics. In order to known, the force on a test charge is given by Eq. (4.2). explain the observed effect Just as static charges produce an electric field, the of magnetic fields on ANTOONcurrents or moving charges produce (in addition) a emitters of light (Zeeman magnetic field, denoted by B (r), again a vector field. It effect), he postulated the existence of electric chargeshas several basic properties identical to the electric field. in the atom, for which he It is defined at each point in space (and can in addition was awarded the Nobel Prize depend on time). Experimentally, it is found to obey the in 1902. He derived a set of LORENTZprinciple of superposition: the magnetic field of several transformation equations sources is the vector addition of magnetic field of each (known after him, as individual source. Lorentz transformation equations) by some tangled (1853 4.2.2 Magnetic Field, Lorentz Force mathematical arguments, – but he was not aware thatLet us suppose that there is a point charge q (moving these equations hinge on a with a velocity v and, located at r at a given time t) in new concept of space andpresence of both the electric field E (r) and the magnetic 1928) time. field B (r). The force on an electric charge q due to both of them can be written as F = q [ E (r) + v × B (r)] º Felectric +Fmagnetic (4.3) This force was given first by H.A. Lorentz based on the extensive experiments of Ampere and others. It is called the Lorentz force. You have already studied in detail the force due to the electric field. If we look at the interaction with the magnetic field, we find the following features. (i) It depends on q, v and B (charge of the particle, the velocity and the magnetic field). Force on a negative charge is opposite to that on a positive charge. (ii) The magnetic force q [ v × B ] includes a vector product of velocity 109 and magnetic field. The vector product makes the force due to magnetic Reprint 2025-26 Physics field vanish (become zero) if velocity and magnetic field are parallel or anti-parallel. The force acts in a (sideways) direction perpendicular to both the velocity and the magnetic field. Its direction is given by the screw rule or right hand rule for vector (or cross) product as illustrated in Fig. 4.2. (iii) The magnetic force is zero if charge is not moving (as then |v|= 0). Only a moving charge feels the magnetic force. The expression for the magnetic force helps us to define the unit of the magnetic field, if one FIGURE 4.2 The direction of the magnetic takes q, F and v, all to be unity in the force force acting on a charged particle. (a) The equation F = q [ v × B] =q v B sin q ˆn , where q is force on a positively charged particle with the angle between v and B [see Fig. 4.2 (a)]. The velocity v and making an angle q with the magnitude of magnetic field B is 1 SI unit, when magnetic field B is given by the right-hand the force acting on a unit charge (1 C), moving rule. (b) A moving charged particle q is perpendicular to B with a speed 1m/s, is one deflected in an opposite sense to –q in the newton. presence of magnetic field. Dimensionally, we have [B] = [F/qv] and the unit of B are Newton second / (coulomb metre). This unit is called tesla (T) named after Nikola Tesla (1856 – 1943). Tesla is a rather large unit. A smaller unit (non-SI) called gauss (=10–4 tesla) is also often used. The earth’s magnetic field is about 3.6 × 10–5 T. 4.2.3 Magnetic force on a current-carrying conductor We can extend the analysis for force due to magnetic field on a single moving charge to a straight rod carrying current. Consider a rod of a uniform cross-sectional area A and length l. We shall assume one kind of mobile carriers as in a conductor (here electrons). Let the number density of these mobile charge carriers in it be n. Then the total number of mobile charge carriers in it is nlA. For a steady current I in this conducting rod, we may assume that each mobile carrier has an average drift velocity vd (see Chapter 3). In the presence of an external magnetic field B, the force on these carriers is: F = (nlA)q vd ´ B where q is the value of the charge on a carrier. Now nq vd is the current density j and |(nq vd)|A is the current I (see Chapter 3 for the discussion of current and current density). Thus, F = [(nq vd )lA] × B = [ jAl ] ´ B = Il ´ B (4.4) where l is a vector of magnitude l, the length of the rod, and with a direction identical to the current I. Note that the current I is not a vector. In the last step leading to Eq. (4.4), we have transferred the vector sign from j to l. Equation (4.4) holds for a straight rod. In this equation, B is the external magnetic field. It is not the field produced by the current-carrying rod. If the wire has an arbitrary shape we can calculate the Lorentz force on it by considering it as a collection of linear strips dlj and summing F  Idl j × B j This summation can be converted to an integral in most cases. Reprint 2025-26 Moving Charges and Magnetism Example 4.1 A straight wire of mass 200 g and length 1.5 m carries a current of 2 A. It is suspended in mid-air by a uniform horizontal magnetic field B (Fig. 4.3). What is the magnitude of the magnetic field? FIGURE 4.3 Solution From Eq. (4.4), we find that there is an upward force F, of magnitude IlB,. For mid-air suspension, this must be balanced by the force due to gravity: m g = I lB m g B = I l Interactive Charged 0.2 × 9.8 = = 0.65 T 2 × 1.5 Note that it would have been sufficient to specify m/l, the mass per EXAMPLE particles unit length of the wire. The earth’s 4 × 10–5 T and we have ignored it. magnetic field is approximately 4.1 moving demonstration: in a Example 4.2 If the magnetic field is parallel to the positive y-axis and the charged particle is moving along the positive x-axis (Fig. 4.4), which way would the Lorentz force be for (a) an electron (negative magnetic charge), (b) a proton (positive charge). field. http://www.phys.hawaii.edu/~teb/optics/java/partmagn/index.html FIGURE 4.4 Solution The velocity v of particle is along the x-axis, while B, the magnetic field is along the y-axis, so v × B is along the z-axis (screw rule or right-hand thumb rule). So, (a) for electron it will be along –z EXAMPLE axis. (b) for a positive charge (proton) the force is along +z axis. 4.2 111 Reprint 2025-26 Physics 4.3 MOTION IN A MAGNETIC FIELD We will now consider, in greater detail, the motion of a charge moving in a magnetic field. We have learnt in Mechanics (see Class XI book, Chapter 5) that a force on a particle does work if the force has a component along (or opposed to) the direction of motion of the particle. In the case of motion of a charge in a magnetic field, the magnetic force is perpendicular to the velocity of the particle. So no work is done and no change in the magnitude of the velocity is produced (though the direction of momentum may be changed). [Notice that this is unlike the force due to an electric field, qE, which can have a component parallel (or antiparallel) to motion and thus can transfer energy in addition to momentum.] We shall consider motion of a charged particle in a uniform magnetic field. First consider the case of v perpendicular to B. The perpendicular force, q v × B, acts as a centripetal force and produces a circular motion perpendicular to the magnetic field. The particle will describe a circle if v and B are perpendicular to each other (Fig. 4.5). If velocity has a component along B, this component remains unchanged as the motion along the magnetic field will not be affected by the magnetic field. The motion in a plane perpendicular to B is as before a circular one, thereby producing a helical motion (Fig. 4.6). You have already learnt in earlier classes (See Class XI, Chapter 3) that if r is the radius of the circular path of a particle, then a force of m v2 / r, acts perpendicular to the path towards the centre of the circle, and is called the centripetal force. If the FIGURE 4.5 Circular motion velocity v is perpendicular to the magnetic field B, the magnetic force is perpendicular to both v and B and acts like a centripetal force. It has a magnitude q v B. Equating the two expressions for centripetal force, m v 2/r = q v B, which gives r = m v / qB (4.5) for the radius of the circle described by the charged particle. The larger the momentum, the larger is the radius and bigger the circle described. If w is the angular frequency, then v = w r. So, w = 2p n = q B/ m [4.6(a)] which is independent of the velocity or energy . Here n is the frequency of rotation. The independence of n from energy has important application in the design of a cyclotron. The time taken for one revolution is T= 2p/ w º 1/n. If there is a component of the velocity FIGURE 4.6 Helical motion parallel to the magnetic field (denoted by v||), 112 it will make the particle move along the field and the path of the Reprint 2025-26 Moving Charges and Magnetism particle would be a helical one (Fig. 4.6). The distance moved along the magnetic field in one rotation is called pitch p. Using Eq. [4.6 (a)], we have p = v||T = 2pm v|| / q B [4.6(b)] The radius of the circular component of motion is called the radius of the helix. Example 4.3 What is the radius of the path of an electron (mass 9 × 10-31 kg and charge 1.6 × 10–19 C) moving at a speed of 3 ×107 m/s in a magnetic field of 6 × 10–4 T perpendicular to it? What is its frequency? Calculate its energy in keV. ( 1 eV = 1.6 × 10–19 J). Solution Using Eq. (4.5) we find r = m v / (qB) = 9 ×10–31 kg × 3 × 107 m s–1 / ( 1.6 × 10–19 C × 6 × 10–4 T) = 28 × 10–2 m = 28 cm n = v / (2 pr) = 17×106 s–1 = 17×106 Hz =17 MHz. 2 EXAMPLE E = (½ )mv = (½ ) 9 × 10–31 kg × 9 × 1014 m2/s2 = 40.5 ×10–17 J ≈ 4×10–16 J = 2.5 keV. 4.3

9.15 Apply mirror equation and the condition: (a) f < 0 (concave mirror); u < 0 (object on left) (b) f > 0; u < 0 (c) f > 0 (convex mirror) and u < 0 (d) f < 0 (concave mirror); f < u < 0 to deduce the desired result.

9.17 (a) sin i¢c = 1.44/1.68 which gives i¢c = 59°. Total internal reflection takes place when i > 59° or when r < rmax = 31°. Now, (sin i /sin r max max ) = 1.68 , which gives imax ~ 60°. Thus, all incident rays of angles in the range 0 < i < 60° will suffer total internal reflections in the pipe. (If the length of the pipe is finite, which it is in practice, there will be a lower limit on i determined by the ratio of the diameter to the length of the pipe.) (b) If there is no outer coating, i¢c = sin–1(1/1.68) = 36.5°. Now, i = 90° will have r = 36.5° and i¢ = 53.5° which is greater than i¢c. Thus, all incident rays (in the range 53.5° < i < 90°) will suffer total internal reflections.

9.8 A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12cm from P. At what point does the beam converge if the lens is (a) a convex lens of focal length 20cm, and (b) a concave lens of focal length 16cm?

9.5 A small bulb is placed at the bottom of a tank containing water to a depth of 80cm. What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is 1.33. (Consider the bulb to be a point source.)

9.20 (a) (i) Let a parallel beam be the incident from the left on the convex lens first. f1 = 30 cm and u1 = – , give v1 = + 30 cm. This image becomes a virtual object for the second lens. f2 = –20 cm, u 2 = + (30 – 8) cm = + 22 cm which gives, v2 = – 220 cm. The parallel incident beam appears to diverge from a point 216 cm from the centre of the two-lens system. (ii) Let the parallel beam be incident from the left on the concave lens first: f1 = – 20 cm, u1 = – ¥, give v1 = – 20 cm. This image becomes a real object for the second lens: f2 = + 30 cm, u2 = – (20 + 8) cm = – 28 cm which gives, v2 = – 420 cm. The parallel incident beam appears to diverge from a point 416 cm on the left of the centre of the two-lens system. Clearly, the answer depends on which side of the lens system the parallel beam is incident. Further we do not have a simple lens equation true for all u (and v) in terms of a definite constant of the system (the constant being determined by f1 and f2, and the separation between the lenses). The notion of effective focal length, therefore, does not seem to be meaningful for this system. (b) u1 = – 40 cm, f1 = 30 cm, gives v1= 120 cm. Magnitude of magnification due to the first (convex) lens is 3. u 2 = + (120 – 8) cm = +112 cm (object virtual); 112 × 20 f2 = – 20 cm which gives v2 = − cm 92 Magnitude of magnification due to the second (concave) 347 Reprint 2025-26 Physics lens = 20/92. Net magnitude of magnification = 0.652 Size of the image = 0.98 cm 9.21 If the refracted ray in the prism is incident on the second face at the critical angle ic, the angle of refraction r at the first face is (60°–ic). Now, ic = sin–1 (1/1.524) ~ 41° Therefore, r = 19° sin i = 0.4962; i ~ 30° 1 1 1 9.22 (a) + = v 9 10 i.e., v = – 90 cm, Magnitude of magnification = 90/9 = 10. Each square in the virtual image has an area 10 × 10 × 1 mm2 = 100 mm2 = 1 cm2 (b) Magnifying power = 25/9 = 2.8 (c) No, magnification of an image by a lens and angular magnification (or magnifying power) of an optical instrument are two separate things. The latter is the ratio of the angular size of the object (which is equal to the angular size of the image even if the image is magnified) to the angular size of the object if placed at the near point (25 cm). Thus, magnification magnitude is |(v/u)| and magnifying power is (25/ |u|). Only when the image is located at the near point |v| = 25 cm, are the two quantities equal. 9.23 (a) Maximum magnifying power is obtained when the image is at the near point (25 cm) u = – 7.14 cm. (b) Magnitude of magnification = (25/ |u|) = 3.5. (c) Magnifying power = 3.5 Yes, the magnifying power (when the image is produced at 25 cm) is equal to the magnitude of magnification. 9.24 Magnification = ( 6.25 / 1) = 2.5 v = +2.5u 1 1 1    2.5u u 10 i.e.,u = – 6 cm |v| = 15 cm The virtual image is closer than the normal near point (25 cm) and cannot be seen by the eye distinctly. 9.25 (a) Even though the absolute image size is bigger than the object size, the angular size of the image is equal to the angular size of the object. The magnifier helps in the following way: without it object would be placed no closer than 25 cm; with it the object can be placed much closer. The closer object has larger angular size than the same object at 25 cm. It is in this sense that angular magnification is achieved. (b) Yes, it decreases a little because the angle subtended at the eye is then slightly less than the angle subtended at the lens. The Reprint 2025-26 Answers effect is negligible if the image is at a very large distance away. [Note: When the eye is separated from the lens, the angles subtended at the eye by the first object and its image are not equal.] (c) First, grinding lens of very small focal length is not easy. More important, if you decrease focal length, aberrations (both spherical and chromatic) become more pronounced. So, in practice, you cannot get a magnifying power of more than 3 or so with a simple convex lens. However, using an aberration corrected lens system, one can increase this limit by a factor of 10 or so. (d) Angular magnification of eye-piece is [(25/fe) + 1] ( fe in cm) which increases if fe is smaller. Further, magnification of the objective v O 1 = is given by | u O | (| u O |/ f O ) − 1 which is large when |u O | is slightly greater than fO. The micro- scope is used for viewing very close object. So |u O | is small, and so is fO. (e) The image of the objective in the eye-piece is known as ‘eye-ring’. All the rays from the object refracted by objective go through the eye-ring. Therefore, it is an ideal position for our eyes for viewing. If we place our eyes too close to the eye-piece, we shall not collect much of the light and also reduce our field of view. If we position our eyes on the eye-ring and the area of the pupil of our eye is greater or equal to the area of the eye-ring, our eyes will collect all the light refracted by the objective. The precise location of the eye-ring naturally depends on the separation between the objective and the eye-piece. When you view through a microscope by placing your eyes on one end,the ideal distance between the eyes and eye-piece is usually built-in the design of the instrument.

9.3 A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again?

9.27 (a) m = ( fO/fe) = 28 f O  f O  (b) m = 1 + = 33.6 f e  25  349 Reprint 2025-26 Physics 9.28 (a) fO + fe = 145 cm (b) Angle subtended by the tower = (100/3000) = (1/30) rad. Angle subtended by the image produced by the objective h h = = f O 140 Equating the two, h = 4.7 cm. (c) Magnification (magnitude) of the eye-piece = 6. Height of the final image (magnitude) = 28 cm. 9.29 The image formed by the larger (concave) mirror acts as virtual object for the smaller (convex) mirror. Parallel rays coming from the object at infinity will focus at a distance of 110 mm from the larger mirror. The distance of virtual object for the smaller mirror = (110 –20) = 90 mm. The focal length of smaller mirror is 70 mm. Using the mirror formula, image is formed at 315 mm from the smaller mirror. 9.30 The reflected rays get deflected by twice the angle of rotation of the mirror. Therefore, d/1.5 = tan 7°. Hence d = 18.4 cm. 9.31 n = 1.33 CHAPTER 10 10.1 (a) Reflected light: (wavelength, frequency, speed same as incident light) l = 589 nm, n = 5.09 ´ 1014 Hz, c = 3.00 ´ 108 m s–1 (b) Refracted light: (frequency same as the incident frequency) n = 5.09 ´ 1014Hz v = (c/n) = 2.26 × 108 m s–1, l = (v/n) = 444 nm 10.2 (a) Spherical (b) Plane (c) Plane (a small area on the surface of a large sphere is nearly planar). 10.3 (a) 2.0 × 108 m s–1 (b) No. The refractive index, and hence the speed of light in a medium, depends on wavelength. [When no particular wavelength or colour of light is specified, we may take the given refractive index to refer to yellow colour.] Now we know violet colour deviates more than red in a glass prism, i.e. nv > nr. Therefore, the violet component of white light travels slower than the red component. 1.2 10 – 2  0.28 10 – 3 10.4  m = 600 nm 4 14. 10.5 K/4 10.6 (a) 1.17 mm (b) 1.56 mm 10.7 0.15° 350 10.8 tan–1(1.5) ~ 56.3o Reprint 2025-26 Answers

9.18 For fixed distance s between object and screen, the lens equation does not give a real solution for u or v if f is greater than s/4. Therefore, fmax = 0.75 m.

14.2 Which of the statements given in Exercise 14.1 is true for p-type semiconductos.

14.1 In an n-type silicon, which of the following statement is true: (a) Electrons are majority carriers and trivalent atoms are the dopants. (b) Electrons are minority carriers and pentavalent atoms are the dopants. (c) Holes are minority carriers and pentavalent atoms are the dopants. (d) Holes are majority carriers and trivalent atoms are the dopants.

14.3 Carbon, silicon and germanium have four valence electrons each. These are characterised by valence and conduction bands separated 341 Reprint 2025-26 Physics by energy band gap respectively equal to (Eg)C, (Eg)Si and (Eg)Ge. Which of the following statements is true? (a) (Eg)Si < (Eg)Ge < (Eg)C (b) (Eg)C < (Eg)Ge > (Eg)Si (c) (Eg)C > (Eg)Si > (Eg)Ge (d) (Eg)C = (Eg)Si = (Eg)Ge 14.4 In an unbiased p-n junction, holes diffuse from the p-region to n-region because (a) free electrons in the n-region attract them. (b) they move across the junction by the potential difference. (c) hole concentration in p-region is more as compared to n-region. (d) All the above. 14.5 When a forward bias is applied to a p-n junction, it (a) raises the potential barrier. (b) reduces the majority carrier current to zero. (c) lowers the potential barrier. (d) None of the above. 14.6 In half-wave rectification, what is the output frequency if the input frequency is 50 Hz. What is the output frequency of a full-wave rectifier for the same input frequency. Reprint 2025-26 Notes Reprint 2025-26 Physics APPENDICES APPENDIX A 1 THE GREEK ALPHABET APPENDIX A 2 COMMON SI PREFIXES AND SYMBOLS FOR MULTIPLES AND SUB-MULTIPLES Reprint 2025-26 AppendicesAnswers APPENDIX A 3 SOME IMPORTANT CONSTANTS OTHER USEFUL CONSTANTS 345 Reprint 2025-26 Physics ANSWERS CHAPTER 9 9.1 v = –54 cm. The image is real, inverted and magnified. The size of the image is 5.0 cm. As u ® f, v ® ¥; for u < f, image is virtual. 9.2 v = 6.7 cm. Magnification = 5/9, i.e., the size of the image is 2.5 cm. As u ® ¥; v ® f (but never beyond) while m ® 0. 9.3 1.33; 1.7 cm 9.4 nga = 1.51; nwa = 1.32; ngw = 1.144; which gives sin r = 0.6181 i.e., r ~ 38°. 9.5 r = 0.8 × tan ic and sin ci = 1/1.33 ≅ 0.75 , where r is the radius (in m) of the largest circle from which light comes out and ic is the critical angle for water-air interface, Area = 2.6 m2 9.6 n ≅ 1.53 and Dm for prism in water ≅ 10° 9.7 R = 22 cm 9.8 Here the object is virtual and the image is real. u = +12 cm (object on right; virtual) (a) f = +20 cm. Image is real and at 7.5 cm from the lens on its right side. (b) f = –16 cm. Image is real and at 48 cm from the lens on its right side. 9.9 v = 8.4 cm, image is erect and virtual. It is diminished to a size 1.8 cm. As u ® ¥, v ® f (but never beyond f while m ® 0). Note that when the object is placed at the focus of the concave lens (21 cm), the image is located at 10.5 cm (not at infinity as one might wrongly think). 9.10 A diverging lens of focal length 60 cm 9.11 (a) ve = –25 cm and fe = 6.25 cm give ue = –5 cm; vO = (15 – 5) cm = 10 cm, fO = uO = – 2.5 cm; Magnifying power = 20 (b) uO = – 2.59 cm. Magnifying power = 13.5. 9.12 Angular magnification of the eye-piece for image at 25 cm 25 25   1  11; | u e |= cm = 2 .27cm ; vO = 7.2 cm 2.5 11 Separation = 9.47 cm; Magnifying power = 88 9.13 24; 150 cm 9.14 (a) Angular magnification = 1500 346 (b) Diameter of the image = 13.7 cm. Reprint 2025-26 Answers

14.5 THE PRINCIPLE OF SUPERPOSITION the medium. If the waves arrive in a region OF WAVES simultaneously, and therefore, overlap, the net displacement y (x,t) is given by What happens when two wave pulses travelling in opposite directions cross each other y (x, t) = y1(x, t) + y2(x, t) (14.25) (Fig. 14.9)? It turns out that wave pulses If we have two or more waves moving in the continue to retain their identities after they have medium the resultant waveform is the sum of crossed. However, during the time they overlap, wave functions of individual waves. That is, if the wave pattern is different from either of the the wave functions of the moving waves are Reprint 2025-26 288 PHYSICS y1 = f1(x–vt), y2 = f2(x–vt), .......... .......... yn = fn (x–vt) then the wave function describing the disturbance in the medium is y = f1(x – vt)+ f2(x – vt)+ ...+ fn(x – vt) n = ∑ f ( x − vt ) (14.26) i i =1 The principle of superposition is basic to the phenomenon of interference. For simplicity, consider two harmonic travelling waves on a stretched string, both with the same ω (angular frequency) and k (wave number), and, therefore, the same wavelength Fig. 14.10 The resultant of two harmonic waves of λ. Their wave speed will be identical. Let us equal amplitude and wavelength further assume that their amplitudes are equal according to the principle of superposition. and they are both travelling in the positive The amplitude of the resultant wave depends on the phase difference φ, whichdirection of x-axis. The waves only differ in their is zero for (a) and π for (b)initial phase. According to Eq. (14.2), the two waves are described by the functions: φ between the constituent two waves: y1(x, t) = a sin (kx – ωt) (14.27) A(φ) = 2a cos ½φ (14.32) For φ = 0, when the waves are in phase, and y2(x, t) = a sin (kx – ωt + φ ) (14.28) y ( x , t ) = 2a sin ( kx − ωt ) (14.33) The net displacement is then, by the principle i.e., the resultant wave has amplitude 2a, theof superposition, given by largest possible value for A. For φ = π , the y (x, t) = a sin (kx – ωt) + a sin (kx – ωt + φ) waves are completely, out of phase and the (14.29) resultant wave has zero displacement   ( kx − ωt ) + ( kx − ωt + φ)  φ everywhere at all times = a  2sin   cos  y (x, t) = 0 (14.34)   2  2  Eq. (14.33) refers to the so-called constructive (14.30) interference of the two waves where the where we have used the familiar trignometric amplitudes add up in the resultant wave. Eq. identity for sin A + sin B . We then have (14.34) is the case of destructive intereference where the amplitudes subtract out in the φ  φ resultant wave. Fig. 14.10 shows these two cases y ( x , t ) = 2 a cos sin  kx − ωt +  (14.31) 2  2  of interference of waves arising from the principle of superposition.Eq. (14.31) is also a harmonic travelling wave in the positive direction of x-axis, with the same 14.6 REFLECTION OF WAVES frequency and wavelength. However, its initial So far we considered waves propagating in an φ unbounded medium. What happens if a pulse phase angle is . The significant thing is that 2 or a wave meets a boundary? If the boundary is its amplitude is a function of the phase difference rigid, the pulse or wave gets reflected. The Reprint 2025-26 WAVES 289 phenomenon of echo is an example of reflection If on the other hand, the boundary point is by a rigid boundary. If the boundary is not not rigid but completely free to move (such as in completely rigid or is an interface between two the case of a string tied to a freely moving ring different elastic media, the situation is some on a rod), the reflected pulse has the same phase what complicated. A part of the incident wave is and amplitude (assuming no energy dissipation) reflected and a part is transmitted into the as the incident pulse. The net maximum second medium. If a wave is incident obliquely displacement at the boundary is then twice the on the boundary between two different media amplitude of each pulse. An example of non- rigid the transmitted wave is called the refracted boundary is the open end of an organ pipe. wave. The incident and refracted waves obey To summarise, a travelling wave or pulse Snell’s law of refraction, and the incident and suffers a phase change of π on reflection at a reflected waves obey the usual laws of rigid boundary and no phase change on reflection. reflection at an open boundary. To put this Fig. 14.11 shows a pulse travelling along a mathematically, let the incident travelling stretched string and being reflected by the wave be boundary. Assuming there is no absorption of y 2 ( x , t ) = a sin ( kx − ωt )energy by the boundary, the reflected wave has the same shape as the incident pulse but it At a rigid boundary, the reflected wave is given suffers a phase change of π or 1800 on reflection. by This is because the boundary is rigid and the yr(x, t) = a sin (kx – ωt + π). disturbance must have zero displacement at all = – a sin (kx – ωt) (14.35) times at the boundary. By the principle of At an open boundary, the reflected wave is given superposition, this is possible only if the reflected by and incident waves differ by a phase of π, so that yr(x, t) = a sin (kx – ωt + 0). the resultant displacement is zero. This = a sin (kx – ωt) (14.36) reasoning is based on boundary condition on a Clearly, at the rigid boundary, y = y 2 + y r = 0rigid wall. We can arrive at the same conclusion dynamically also. As the pulse arrives at the wall, at all times. it exerts a force on the wall. By Newton’s Third 14.6.1 Standing Waves and Normal Modes Law, the wall exerts an equal and opposite force We considered above reflection at one boundary. on the string generating a reflected pulse that But there are familiar situations (a string fixed differs by a phase of π. at either end or an air column in a pipe with either end closed) in which reflection takes place at two or more boundaries. In a string, for example, a wave travelling in one direction will get reflected at one end, which in turn will travel and get reflected from the other end. This will go on until there is a steady wave pattern set up on the string. Such wave patterns are called standing waves or stationary waves. To see this mathematically, consider a wave travelling along the positive direction of x-axis and a reflected wave of the same amplitude and wavelength in the negative direction of x-axis. From Eqs. (14.2) and (14.4), with φ = 0, we get: y1(x, t) = a sin (kx – ωt) y2(x, t) = a sin (kx + ωt) The resultant wave on the string is, according to the principle of superposition: Fig. 14.11 Reflection of a pulse meeting a rigid boundary. y (x, t) = y1(x, t) + y2(x, t) Reprint 2025-26 290 PHYSICS = a [sin (kx – ωt) + sin (kx + ωt)] nodes; the points at which the amplitude is the largest are called antinodes. Fig. 14.12 showsUsing the familiar trignometric identity a stationary wave pattern resulting fromSin (A+B) + Sin (A–B) = 2 sin A cosB we get, superposition of two travelling waves in y (x, t) = 2a sin kx cos ωt (14.37) opposite directions. Note the important difference in the wave The most significant feature of stationary pattern described by Eq. (14.37) from that waves is that the boundary conditions constrain described by Eq. (14.2) or Eq. (14.4). The terms the possible wavelengths or frequencies of kx and ωt appear separately, not in the vibration of the system. The system cannot combination kx - ωt. The amplitude of this wave oscillate with any arbitrary frequency (contrast is 2a sin kx. Thus, in this wave pattern, the this with a harmonic travelling wave), but is amplitude varies from point-to-point, but each characterised by a set of natural frequencies or element of the string oscillates with the same normal modes of oscillation. Let us determine angular frequency ω or time period. There is no these normal modes for a stretched string fixed phase difference between oscillations of different at both ends. elements of the wave. The string as a whole First, from Eq. (14.37), the positions of nodes vibrates in phase with differing amplitudes at (where the amplitude is zero) are given by different points. The wave pattern is neither sin kx = 0 . moving to the right nor to the left. Hence, they which implies are called standing or stationary waves. The kx = nπ; n = 0, 1, 2, 3, ... amplitude is fixed at a given location but, as Since, k = 2π/λ , we get remarked earlier, it is different at different locations. The points at which the amplitude is nλ zero (i.e., where there is no motion at all) are x = ; n = 0, 1, 2, 3, ... (14.38) 2 Fig. 14.12 Stationary waves arising from superposition of two harmonic waves travelling in opposite directions. Note that the positions of zero displacement (nodes) remain fixed at all times. Reprint 2025-26 WAVES 291 Clearly, the distance between any two speed of wave determined by the properties of λ the medium. The n = 2 frequency is called the successive nodes is . In the same way, the second harmonic; n = 3 is the third harmonic 2 and so on. We can label the various harmonics bypositions of antinodes (where the amplitude is the symbol νn ( n = 1, 2, ...).the largest) are given by the largest value of sin Fig. 14.13 shows the first six harmonics of akx : sin k x = 1 stretched string fixed at either end. A string need not vibrate in one of these modes only.which implies Generally, the vibration of a string will be a kx = (n + ½) π ; n = 0, 1, 2, 3, ... superposition of different modes; some modes With k = 2π/λ, we get may be more strongly excited and some less. Musical instruments like sitar or violin are λ based on this principle. Where the string is x = (n + ½) ; n = 0, 1, 2, 3, ... (14.39) 2 plucked or bowed, determines which modes are Again the distance between any two consecutive more prominent than others. Let us next consider normal modes of λ antinodes is . Eq. (14.38) can be applied to oscillation of an air column with one end closed 2 the case of a stretched string of length L fixed at both ends. Taking one end to be at x = 0, the boundary conditions are that x = 0 and x = L are positions of nodes. The x = 0 condition is already satisfied. The x = L node condition requires that the length L is related to λ by λ L = n ; n = 1, 2, 3, ... (14.40) 2 Thus, the possible wavelengths of stationary waves are constrained by the relation 2L λ = ; n = 1, 2, 3, … (14.41) n with corresponding frequencies nv v = , for n = 1, 2, 3, (14.42) 2L We have thus obtained the natural frequencies - the normal modes of oscillation of the system. The lowest possible natural frequency of a system is called its fundamental mode or the first harmonic. For the stretched string fixed at either end v it is given by v = , corresponding 2 L Fig. 14.13 The first six harmonics of vibrations of a stretched to n = 1 of Eq. (14.42). Here v is the string fixed at both ends. Reprint 2025-26 292 PHYSICS and the other open. A glass tube partially filled modes of this system is more complex. This with water illustrates this system. The end in problem involves wave propagation in two contact with water is a node, while the open end dimensions. However, the underlying physics is is an antinode. At the node the pressure the same. changes are the largest, while the displacement is minimum (zero). At the open end - the u Example 14.5 A pipe, 30.0 cm long, is open antinode, it is just the other way - least pressure at both ends. Which harmonic mode of the change and maximum amplitude of pipe resonates a 1.1 kHz source? Will displacement. Taking the end in contact with resonance with the same source be water to be x = 0, the node condition (Eq. 14.38) observed if one end of the pipe is closed ? is already satisfied. If the other end x = L is an Take the speed of sound in air as antinode, Eq. (14.39) gives 330 m s–1.  1  λ n +L = , for n = 0, 1, 2, 3, …  2  2 Answer The first harmonic frequency is given by The possible wavelengths are then restricted by v v the relation : ν1 = λ1 = 2 L (open pipe) where L is the length of the pipe. The frequency 2 L λ = , for n = 0, 1, 2, 3,... (14.43) of its nth harmonic is: ( n + 1 / 2 ) nv νn = 2L , for n = 1, 2, 3, ... (open pipe) The normal modes – the natural frequencies – of the system are First few modes of an open pipe are shown in Fig. 14.15.  1  v For L = 30.0 cm, v = 330 m s–1, ; n = 0, 1, 2, 3, ... (14.44) ν =  n + 2  2 L n 330 (m s − 1 ) νn = = 550 n s–1 The fundamental frequency corresponds to n = 0, 0.6 (m) v Clearly, a source of frequency 1.1 kHz will and is given by . The higher frequencies resonate at v2, i.e. the second harmonic. 4 L are odd harmonics, i.e., odd multiples of the v v fundamental frequency : 3 , 5 , etc. 4 L 4 L Fig. 14.14 shows the first six odd harmonics of air column with one end closed and the other open. For a pipe open at both ends, each end is an antinode. It is then easily seen that an open air column at both ends generates all harmonics (See Fig. 14.15). The systems above, strings and air columns, can also undergo forced oscillations (Chapter 13). If the external frequency is close to one of the natural frequencies, the system shows resonance. Normal modes of a circular membrane rigidly clamped to the circumference as in a tabla are determined by the boundary condition that no Fundamental point on the circumference of the membrane or third fifth vibrates. Estimation of the frequencies of normal first harmonic harmonic harmonic Reprint 2025-26 WAVES 293 Fig. 14.15 Standing waves in an open pipe, first four harmonics are depicted. while tuning their instruments with each other. They go on tuning until their sensitive ears do seventh ninth eleventh not detect any beats. harmonic harmonic harmonic To see this mathematically, let us consider two harmonic sound waves of nearly equal Fig. 14.14 Normal modes of an air column open at angular frequency ω1 and ω2 and fix the location one end and closed at the other end. Only to be x = 0 for convenience. Eq. (14.2) with a the odd harmonics are seen to be possible. suitable choice of phase (φ = π/2 for each) and, assuming equal amplitudes, gives Now if one end of the pipe is closed (Fig. 14.15), it follows from Eq. (14.15) that the fundamental s1 = a cos ω1t and s2 = a cos ω2t (14.45) frequency is Here we have replaced the symbol y by s, v v since we are referring to longitudinal not transverse displacement. Let ω1 be the (slightly) ν1 = λ1 = 4 L (pipe closed at one end) greater of the two frequencies. The resultant and only the odd numbered harmonics are displacement is, by the principle of present : superposition, s = s1 + s2 = a (cos ω1 t + cos ω2 t) 3v 5v ν3 = , ν5 = , and so on. Using the familiar trignometric identity for 4 L 4 L cos A + cosB, we get For L = 30 cm and v = 330 m s–1, the (ω1 - ω2 ) t (ω1 + ω2 ) t fundamental frequency of the pipe closed at one = 2 a cos cos (14.46) end is 275 Hz and the source frequency 2 2 corresponds to its fourth harmonic. Since this which may be written as : harmonic is not a possible mode, no resonance s = [2 a cos ωb t ] cos ωat (14.47) will be observed with the source, the moment If |ω1 – ω2| <<ω1, ω2, ωa >> ωb, th one end is closed. ⊳ where 14.7 BEATS (ω1 − ω2 ) (ω1 + ω2 ) ωb = and ωa = ‘Beats’ is an interesting phenomenon arising 2 2 from interference of waves. When two harmonic Now if we assume |ω1 – ω2| <<ω1, which means sound waves of close (but not equal) frequencies ωa >> ωb, we can interpret Eq. (14.47) as follows. are heard at the same time, we hear a sound of The resultant wave is oscillating with the average similar frequency (the average of two close angular frequency ωa; however its amplitude is frequencies), but we hear something else also. not constant in time, unlike a pure harmonic We hear audibly distinct waxing and waning of wave. The amplitude is the largest when the the intensity of the sound, with a frequency term cos ωb t takes its limit +1 or –1. In other equal to the difference in the two close words, the intensity of the resultant wave waxes frequencies. Artists use this phenomenon often and wanes with a frequency which is 2ωb = ω1 – Reprint 2025-26 294 PHYSICS ω2. Since ω = 2πν, the beat frequency νbeat, is given by νbeat = ν1 – ν2 (14.48) Fig. 14.16 illustrates the phenomenon of beats for two harmonic waves of frequencies 11 Hz and 9 Hz. The amplitude of the resultant wave shows beats at a frequency of 2 Hz. Musical Pillars Temples often have some pillars portraying human figures playing musical instru- ments, but seldom do these pillars themselves produce music. At the Nellaiappar temple in Tamil Nadu, gentle taps on a cluster of pillars carved out of a single piece of rock produce the basic notes of Indian classical music, viz. Sa, Re, Ga, Ma, Pa, Dha, Ni, Sa. Vibrations of these pillars depend on elasticity of the stone used, Fig. 14.16 Superposition of two harmonic waves, one its density and shape. of frequency 11 Hz (a), and the other of Musical pillars are categorised into three frequency 9Hz (b), giving rise to beats of frequency 2 Hz, as shown in (c). types: The first is called the Shruti Pillar, as it can produce the basic notes — the “swaras”. The second type is the Gana u Example 14.6 Two sitar strings A and B Thoongal, which generates the basic tunes playing the note ‘Dha’ are slightly out of that make up the “ragas”. The third variety tune and produce beats of frequency 5 Hz. is the Laya Thoongal pillars that produce The tension of the string B is slightly “taal” (beats) when tapped. The pillars at the increased and the beat frequency is found Nellaiappar temple are a combination of the to decrease to 3 Hz. What is the original Shruti and Laya types. frequency of B if the frequency of A is Archaeologists date the Nelliappar 427 Hz ? temple to the 7th century and claim it was built by successive rulers of the Pandyan Answer Increase in the tension of a string dynasty. increases its frequency. If the original frequency The musical pillars of Nelliappar and of B (νB) were greater than that of A (νA ), further several other temples in southern India like increase in νB should have resulted in an those at Hampi (picture), Kanyakumari, and increase in the beat frequency. But the beat Thiruvananthapuram are unique to the frequency is found to decrease. This shows that country and have no parallel in any other νB < νA. Since νA – νB = 5 Hz, and νA = 427 Hz, we part of the world. get νB = 422 Hz. ⊳ Reprint 2025-26 WAVES 295 SUMMARY 1. Mechanical waves can exist in material media and are governed by Newton’s Laws. 2. Transverse waves are waves in which the particles of the medium oscillate perpendicular to the direction of wave propagation. 3. Longitudinal waves are waves in which the particles of the medium oscillate along the direction of wave propagation. 4. Progressive wave is a wave that moves from one point of medium to another. 5. The displacement in a sinusoidal wave propagating in the positive x direction is given by y (x, t) = a sin (kx – ωt + φ) where a is the amplitude of the wave, k is the angular wave number, ω is the angular frequency, (kx – ωt + φ) is the phase, and φ is the phase constant or phase angle. 6. Wavelength λ of a progressive wave is the distance between two consecutive points of the same phase at a given time. In a stationary wave, it is twice the distance between two consecutive nodes or antinodes. 7. Period T of oscillation of a wave is defined as the time any element of the medium takes to move through one complete oscillation. It is related to the angular frequency ω through the relation 2π T = ω 8. Frequency v of a wave is defined as 1/T and is related to angular frequency by ω ν = 2 π ω λ 9. Speed of a progressive wave is given by v = = = λν k T 10. The speed of a transverse wave on a stretched string is set by the properties of the string. The speed on a string with tension T and linear mass density µ is T v = µ 11. Sound waves are longitudinal mechanical waves that can travel through solids, liquids, or gases. The speed v of sound wave in a fluid having bulk modulus B and density ρ is B v = ρ The speed of longitudinal waves in a metallic bar is Y v = ρ For gases, since B = γP, the speed of sound is γP v = ρ Reprint 2025-26 296 PHYSICS 12. When two or more waves traverse simultaneously in the same medium, the displacement of any element of the medium is the algebraic sum of the displacements due to each wave. This is known as the principle of superposition of waves n f i ( x − vt ) y = ∑ i =1 13. Two sinusoidal waves on the same string exhibit interference, adding or cancelling according to the principle of superposition. If the two are travelling in the same direction and have the same amplitude a and frequency but differ in phase by a phase constant φ, the result is a single wave with the same frequency ω :  1   1  y (x, t) =  2a cos 2 φ sin  kx − ωt + 2 φ If φ = 0 or an integral multiple of 2π, the waves are exactly in phase and the interference is constructive; if φ= π, they are exactly out of phase and the interference is destructive. 14. A travelling wave, at a rigid boundary or a closed end, is reflected with a phase reversal but the reflection at an open boundary takes place without any phase change. For an incident wave yi (x, t) = a sin (kx – ωt ) the reflected wave at a rigid boundary is yr (x, t) = – a sin (kx + ωt ) For reflection at an open boundary yr (x,t ) = a sin (kx + ωt) 15. The interference of two identical waves moving in opposite directions produces standing waves. For a string with fixed ends, the standing wave is given by y (x, t) = [2a sin kx ] cos ωt Standing waves are characterised by fixed locations of zero displacement called nodes and fixed locations of maximum displacements called antinodes. The separation between two consecutive nodes or antinodes is λ/2. A stretched string of length L fixed at both the ends vibrates with frequencies given by n v v = , n = 1, 2, 3, ... 2 L The set of frequencies given by the above relation are called the normal modes of oscillation of the system. The oscillation mode with lowest frequency is called the fundamental mode or the first harmonic. The second harmonic is the oscillation mode with n = 2 and so on. A pipe of length L with one end closed and other end open (such as air columns) vibrates with frequencies given by v v = ( n + ½) , n = 0, 1, 2, 3, ... 2L The set of frequencies represented by the above relation are the normal modes of oscillation of such a system. The lowest frequency given by v/4L is the fundamental mode or the first harmonic. 16. A string of length L fixed at both ends or an air column closed at one end and open at the other end or open at both the ends, vibrates with certain frequencies called their normal modes. Each of these frequencies is a resonant frequency of the system. 17. Beats arise when two waves having slightly different frequencies, ν1 and ν2 and comparable amplitudes, are superposed. The beat frequency is νbeat = ν1 ~ ν2 Reprint 2025-26 WAVES 297 POINTS TO PONDER 1. A wave is not motion of matter as a whole in a medium. A wind is different from the sound wave in air. The former involves motion of air from one place to the other. The latter involves compressions and rarefactions of layers of air. 2. In a wave, energy and not the matter is transferred from one point to the other. 3. In a mechanical wave, energy transfer takes place because of the coupling through elastic forces between neighbouring oscillating parts of the medium. 4. Transverse waves can propagate only in medium with shear modulus of elasticity, Longitudinal waves need bulk modulus of elasticity and are therefore, possible in all media, solids, liquids and gases. 5. In a harmonic progressive wave of a given frequency, all particles have the same amplitude but different phases at a given instant of time. In a stationary wave, all particles between two nodes have the same phase at a given instant but have different amplitudes. 6. Relative to an observer at rest in a medium the speed of a mechanical wave in that medium (v) depends only on elastic and other properties (such as mass density) of the medium. It does not depend on the velocity of the source. EXERCISES 14.1 A string of mass 2.50 kg is under a tension of 200 N. The length of the stretched string is 20.0 m. If the transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end? 14.2 A stone dropped from the top of a tower of height 300 m splashes into the water of a pond near the base of the tower. When is the splash heard at the top given that the speed of sound in air is 340 m s–1 ? (g = 9.8 m s–2) 14.3 A steel wire has a length of 12.0 m and a mass of 2.10 kg. What should be the tension in the wire so that speed of a transverse wave on the wire equals the speed of sound in dry air at 20 °C = 343 m s–1. γP 14.4 Use the formula v = to explain why the speed of sound in air ρ (a) is independent of pressure, (b) increases with temperature, (c) increases with humidity. Reprint 2025-26 298 PHYSICS 14.5 You have learnt that a travelling wave in one dimension is represented by a function y = f (x, t) where x and t must appear in the combination x – v t or x + v t, i.e. y = f (x ± v t). Is the converse true? Examine if the following functions for y can possibly represent a travelling wave : (a) (x – vt )2 (b) log [(x + vt)/x0] (c) 1/(x + vt) 14.6 A bat emits ultrasonic sound of frequency 1000 kHz in air. If the sound meets a water surface, what is the wavelength of (a) the reflected sound, (b) the transmitted sound? Speed of sound in air is 340 m s –1 and in water 1486 m s–1. 14.7 A hospital uses an ultrasonic scanner to locate tumours in a tissue. What is the wavelength of sound in the tissue in which the speed of sound is 1.7 km s–1 ? The operating frequency of the scanner is 4.2 MHz. 14.8 A transverse harmonic wave on a string is described by y(x, t) = 3.0 sin (36 t + 0.018 x + π/4) where x and y are in cm and t in s. The positive direction of x is from left to right. (a) Is this a travelling wave or a stationary wave ? If it is travelling, what are the speed and direction of its propagation ? (b) What are its amplitude and frequency ? (c) What is the initial phase at the origin ? (d) What is the least distance between two successive crests in the wave ? 14.9 For the wave described in Exercise 14.8, plot the displacement (y) versus (t) graphs for x = 0, 2 and 4 cm. What are the shapes of these graphs? In which aspects does the oscillatory motion in travelling wave differ from one point to another: amplitude, frequency or phase ? 14.10 For the travelling harmonic wave y(x, t) = 2.0 cos 2π (10t – 0.0080 x + 0.35) where x and y are in cm and t in s. Calculate the phase difference between oscillatory motion of two points separated by a distance of (a) 4 m, (b) 0.5 m, (c) λ/2, (d) 3λ/4 14.11 The transverse displacement of a string (clamped at its both ends) is given by  2π  y(x, t) = 0.06 sin  3 x  cos (120 πt) where x and y are in m and t in s. The length of the string is 1.5 m and its mass is 3.0 ×10–2 kg. Answer the following : (a) Does the function represent a travelling wave or a stationary wave? (b) Interpret the wave as a superposition of two waves travelling in opposite directions. What is the wavelength, frequency, and speed of each wave ? Reprint 2025-26 WAVES 299 (c) Determine the tension in the string. 14.12 (i) For the wave on a string described in Exercise 15.11, do all the points on the string oscillate with the same (a) frequency, (b) phase, (c) amplitude? Explain your answers. (ii) What is the amplitude of a point 0.375 m away from one end? 14.13 Given below are some functions of x and t to represent the displacement (transverse or longitudinal) of an elastic wave. State which of these represent (i) a travelling wave, (ii) a stationary wave or (iii) none at all: (a) y = 2 cos (3x) sin (10t) (b) y = 2 x − vt (c) y = 3 sin (5x – 0.5t) + 4 cos (5x – 0.5t) (d) y = cos x sin t + cos 2x sin 2t 14.14 A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of 45 Hz. The mass of the wire is 3.5 × 10–2 kg and its linear mass density is 4.0 × 10–2 kg m–1. What is (a) the speed of a transverse wave on the string, and (b) the tension in the string? 14.15 A metre-long tube open at one end, with a movable piston at the other end, shows resonance with a fixed frequency source (a tuning fork of frequency 340 Hz) when the tube length is 25.5 cm or 79.3 cm. Estimate the speed of sound in air at the temperature of the experiment. The edge effects may be neglected. 14.16 A steel rod 100 cm long is clamped at its middle. The fundamental frequency of longitudinal vibrations of the rod are given to be 2.53 kHz. What is the speed of sound in steel? 14.17 A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a 430 Hz source ? Will the same source be in resonance with the pipe if both ends are open? (speed of sound in air is 340 m s–1). 14.18 Two sitar strings A and B playing the note ‘Ga’ are slightly out of tune and produce beats of frequency 6 Hz. The tension in the string A is slightly reduced and the beat frequency is found to reduce to 3 Hz. If the original frequency of A is 324 Hz, what is the frequency of B? 14.19 Explain why (or how): (a) in a sound wave, a displacement node is a pressure antinode and vice versa, (b) bats can ascertain distances, directions, nature, and sizes of the obstacles without any “eyes”, (c) a violin note and sitar note may have the same frequency, yet we can distinguish between the two notes, (d) solids can support both longitudinal and transverse waves, but only longitudinal waves can propagate in gases, and (e) the shape of a pulse gets distorted during propagation in a dispersive medium. Reprint 2025-26

13.5 VELOCITY AND ACCELERATION IN SIMPLE HARMONIC MOTION The speed of a particle v in uniform circular motion is its angular speed ω times the radius of the circle A. v = ω A (13.8) The direction of velocity v at a time t is along the tangent to the circle at the point where the particle is located at that instant. From the Fig. 13.12 The acceleration, a(t), of the particle P′ isgeometry of Fig. 13.11, it is clear that the velocity the projection of the acceleration a of theof the projection particle P′ at time t is reference particle P. v(t) = –ωA sin (ωt + φ ) (13.9) Eq. (13.11) gives the acceleration of a particle in SHM. The same equation can again be obtained directly by differentiating velocity v(t) given by Eq. (13.9) with respect to time: d a (t ) = v (t ) (13.12) d t We note from Eq. (13.11) the important Fig. 13.11 The velocity, v (t), of the particle P′ is property that acceleration of a particle in SHM the projection of the velocity v of the is proportional to displacement. For x(t) > 0, reference particle, P. a(t) < 0 and for x(t) < 0, a(t) > 0. Thus, whatever Reprint 2025-26 OSCILLATIONS 267 the value of x between –A and A, the acceleration (b) Using Eq. (13.9), the speed of the body a(t) is always directed towards the centre. = – (5.0 m)(2π s–1) sin [(2π s–1) ×1.5 s For simplicity, let us put φ = 0 and write the + π/4] expression for x (t), v (t) and a(t) = – (5.0 m)(2π s–1) sin [(3π + π/4)] x(t) = A cos ωt, v(t) = – ω Asin ωt, a(t)=–ω2 A cos ωt = 10π × 0.707 m s–1 The corresponding plots are shown in Fig. 13.13. = 22 m s–1 All quantities vary sinusoidally with time; only (c) Using Eq.(13.10), the acceleration of the their maxima differ and the different plots differ body in phase. x varies between –A to A; v(t) varies = –(2π s–1)2 × displacement from –ωA to ωA and a(t) from –ω2A to ω2A. With = – (2π s–1)2 × (–3.535 m) respect to displacement plot, velocity plot has a = 140 m s–2 ⊳ phase difference of π/2 and acceleration plot has a phase difference of π. 13.6 FORCE LAW FOR SIMPLE HARMONIC MOTION Using Newton’s second law of motion, and the expression for acceleration of a particle undergoing SHM (Eq. 13.11), the force acting on a particle of mass m in SHM is F (t ) = ma = –mω2 x (t) i.e., F (t) = –k x (t) (13.13) where k = mω2 (13.14a) k or ω = (13.14b) m Like acceleration, force is always directed towards the mean position—hence it is sometimes called the restoring force in SHM. To summarise the discussion so far, simple harmonic motion can be defined in two equivalent ways, either by Eq. (13.4) for displacement or by Eq. (13.13) that gives Fig. 13.13 Displacement, velocity and acceleration of its force law. Going from Eq. (13.4) to Eq. (13.13) a particle in simple harmonic motion have required us to differentiate two times. Likewise, the same period T, but they differ in phase by integrating the force law Eq. (13.13) two times, we can get back Eq. (13.4). u Example 13.5 A body oscillates with SHM Note that the force in Eq. (13.13) is linearly according to the equation (in SI units), proportional to x(t). A particle oscillating under x = 5 cos [2π t + π/4]. such a force is, therefore, calling a linear harmonic oscillator. In the real world, the force At t = 1.5 s, calculate the (a) displacement, may contain small additional terms proportional (b) speed and (c) acceleration of the body. to x2, x3, etc. These then are called non-linear oscillators. Answer The angular frequency ω of the body = 2π s–1 and its time period T = 1 s. u Example 13.6 Two identical springs of At t = 1.5 s spring constant k are attached to a block (a) displacement = (5.0 m) cos [(2π s–1)× of mass m and to fixed supports as shown 1.5 s + π/4] in Fig. 13.14. Show that when the mass is = (5.0 m) cos [(3π + π/4)] displaced from its equilibrium position on = –5.0 × 0.707 m either side, it executes a simple harmonic = –3.535 m motion. Find the period of oscillations. Reprint 2025-26 268 PHYSICS 13.7 ENERGY IN SIMPLE HARMONIC MOTION Both kinetic and potential energies of a particle in SHM vary between zero and their maximum values. In section 13.5 we have seen that the velocity of a particle executing SHM, is a periodic Fig. 13.14 function of time. It is zero at the extreme positions of displacement. Therefore, the kinetic energy (K) Answer Let the mass be displaced by a small of such a particle, which is defined as distance x to the right side of the equilibrium position, as shown in Fig. 13.15. Under this 1 2 K = mv situation the spring on the left side gets 2 1 2 2 2 = m ω A sin (ωt + φ) 2 1 2 2 = k A sin (ωt + φ) (13.15) 2 is also a periodic function of time, being zero when the displacement is maximum and maximum when the particle is at the mean Fig. 13.15 position. Note, since the sign of v is immaterial in K, the period of K is T/2. elongated by a length equal to x and that on What is the potential energy (U) of a particle the right side gets compressed by the same executing simple harmonic motion? In length. The forces acting on the mass are Chapter 6, we have seen that the concept of then, potential energy is possible only for conservative forces. The spring force F = –kx is a conservative F1 = –k x (force exerted by the spring on force, with associated potential energy the left side, trying to pull the mass towards the mean 1 2 U = k x position) (13.16) 2 F2 = –k x (force exerted by the spring on Hence the potential energy of a particle the right side, trying to push the executing simple harmonic motion is, mass towards the mean position) 1 2 The net force, F, acting on the mass is then U(x) = k x given by, 2 F = –2kx 1 2 2 Hence the force acting on the mass is = k A cos (ωt + φ) (13.17) 2proportional to the displacement and is directed towards the mean position; therefore, the motion Thus, the potential energy of a particle executed by the mass is simple harmonic. The executing simple harmonic motion is also time period of oscillations is, periodic, with period T/2, being zero at the mean m position and maximum at the extreme T = 2 π displacements. 2k ⊳ Reprint 2025-26 OSCILLATIONS 269 It follows from Eqs. (13.15) and (13.17) that Observe that both kinetic energy and the total energy, E, of the system is, potential energy in SHM are seen to be always positive in Fig. 13.16. Kinetic energy can, of E = U + K course, be never negative, since it is proportional to the square of speed. Potential 1 1 = k A 2 cos 2 (ωt + φ) + k A 2 sin 2 (ωt + φ) energy is positive by choice of the undermined 2 2 constant in potential energy. Both kinetic energy and potential energy peak twice during each period of SHM. For x = 0, the energy is 1 2 2 2 = k A cos (ωt + φ) + sin (ωt + φ)  kinetic; at the extremes x = ±A, it is all   2 potential energy. In the course of motion Using the familiar trigonometric identity, the between these limits, kinetic energy increases value of the expression in the brackets is unity. at the expense of potential energy or Thus, vice-versa. 1 2 E = k A (13.18) u Example 13.7 A block whose mass is 1 kg 2 is fastened to a spring. The spring has a The total mechanical energy of a harmonic spring constant of 50 N m–1. The block is pulled to a distance x = 10 cm from itsoscillator is thus independent of time as equilibrium position at x = 0 on a frictionlessexpected for motion under any conservative surface from rest at t = 0. Calculate the force. The time and displacement dependence kinetic, potential and total energies of the of the potential and kinetic energies of a block when it is 5 cm away from the mean linear simple harmonic oscillator are shown position. in Fig. 13.16. Answer The block executes SHM, its angular frequency, as given by Eq. (13.14b), is k ω = m 1 50 N m– = 1kg = 7.07 rad s–1 Its displacement at any time t is then given by, x(t) = 0.1 cos (7.07t) Fig. 13.16 Kinetic energy, potential energy and total energy as a function of time [shown in (a)] Therefore, when the particle is 5 cm away from and displacement [shown in (b)] of a particle in SHM. The kinetic energy and potential the mean position, we have energy both repeat after a period T/2. The total energy remains constant at all t or x. 0.05 = 0.1 cos (7.07t) Reprint 2025-26 270 PHYSICS let it go. The stone executes a to and fro motion,Or cos (7.07t) = 0.5 and hence it is periodic with a period of about two seconds. 3 We shall show that this periodic motion is sin (7.07t) = = 0.866 2 simple harmonic for small displacements from Then, the velocity of the block at x = 5 cm is = 0.1 × 7.07 × 0.866 m s–1 = 0.61 m s–1 Hence the K.E. of the block, 1 2 = m v 2 = ½[1kg × (0.6123 m s–1 )2 ] = 0.19 J (a) The P.E. of the block, 1 2 = k x 2 = ½(50 N m–1 × 0.05 m × 0.05 m) = 0.0625 J The total energy of the block at x = 5 cm, = K.E. + P.E. = 0.25 J we also know that at maximum displacement, K.E. is zero and hence the total energy of the (b) system is equal to the P.E. Therefore, the total Fig. 13.17 (a) A bob oscillating about its mean position. (b) The radial force T-mg cosθ energy of the system, provides centripetal force but no torque = ½(50 N m–1 × 0.1 m × 0.1 m ) about the support. The tangential force mg sinθ provides the restoring torque. = 0.25 J which is same as the sum of the two energies at the mean position. Consider simple pendulum a displacement of 5 cm. This is in conformity — a small bob of mass m tied to an inextensible with the principle of conservation of energy. ⊳ massless string of length L. The other end of the string is fixed to a rigid support. The bob13.8 The Simple Pendulum oscillates in a plane about the vertical lineIt is said that Galileo measured the periods of a through the support. Fig. 13.17(a) shows thisswinging chandelier in a church by his pulse system. Fig. 13.17(b) is a kind of ‘free-body’beats. He observed that the motion of the chandelier was periodic. The system is a kind diagram of the simple pendulum showing the of pendulum. You can also make your own forces acting on the bob. pendulum by tying a piece of stone to a long Let θ be the angle made by the string with unstretchable thread, approximately 100 cm the vertical. When the bob is at the mean long. Suspend your pendulum from a suitable position, θ = 0 support so that it is free to oscillate. Displace There are only two forces acting on the bob; the stone to one side by a small distance and the tension T along the string and the vertical Reprint 2025-26 OSCILLATIONS 271 force due to gravity (=mg). The force mg can be Table 13.1 sin θ as ma function of angle θ resolved into the component mg cosθ along the string and mg sinθ perpendicular to it. Since (degrees) (radians) sin the motion of the bob is along a circle of length L and centre at the support point, the bob has a radial acceleration (ω2L) and also a tangental acceleration; the latter arises since motion along the arc of the circle is not uniform. The radial acceleration is provided by the net radial force T –mg cosθ, while the tangential acceleration is provided by mg sinθ. It is more convenient to Equation (13.24) is mathematically, identical towork with torque about the support since the radial force gives zero torque. Torque τ about Eq. (13.11) except that the variable is angular displacement. Hence we have proved that forthe support is entirely provided by the tangental small q, the motion of the bob is simple harmonic.component of force From Eqs. (13.24) and (13.11), τ = –L (mg sinθ ) (13.19) This is the restoring torque that tends to reduce mgL ω =angular displacement — hence the negative Isign. By Newton’s law of rotational motion, τ = I α (13.20) and where I is the moment of inertia of the system about the support and α is the angular I T = 2π (13.25)acceleration. Thus, mgL I α = –m g sin θ L (13.21) Now since the string of the simple pendulum is massless, the moment of inertia I is simply mL2. Eq. (13.25) then gives the well-known Or, formula for time period of a simple pendulum. m g L α = − sin θ (13.22) L I T = 2π (13.26) g We can simplify Eq. (13.22) if we assume that the displacement θ is small. We know that sin θ u Example 13.8 What is the length of a can be expressed as, simple pendulum, which ticks seconds ? θ3 θ5 Answer From Eq. (13.26), the time period of a sin θ = θ− + ± ... (13.23) 3! 5! simple pendulum is given by, L where θ is in radians. T = 2π Now if θ is small, sin θ can be approximated g by θ and Eq. (13.22) can then be written as, From this relation one gets, gT 2 mgL L = 2 θ α = − (13.24) 4π I The time period of a simple pendulum, which In Table 13.1, we have listed the angle θ in ticks seconds, is 2 s. Therefore, for g = 9.8 m s–2 degrees, its equivalent in radians, and the value of the function sin θ . From this table it and T = 2 s, L is can be seen that for θ as large as 20 degrees, 9.8(m s –2 ) × 4(s 2 ) = sin θ is nearly the same as θ expressed 2 4π in radians. = 1 m ⊳ Reprint 2025-26 272 PHYSICS SUMMARY 1. The motion that repeats itself is called periodlic motion. 2. The period T is the time reequired for one complete oscillation, or cycle. It is related to the frequency v by, 1 T = v The frequency ν of periodic or oscillatory motion is the number of oscillations per unit time. In the SI, it is measured in hertz : 1 hertz = 1 Hz = 1 oscillation per second = 1s–1 3. In simple harmonic motion (SHM), the displacement x (t) of a particle from its equilibrium position is given by, x (t) = A cos (ωt + φ ) (displacement), in which A is the amplitude of the displacement, the quantity (ωt + φ ) is the phase of the motion, and φ is the phase constant. The angular frequency ω is related to the period and frequency of the motion by, 2π ω= = 2πν (angular frequency). T 4. Simple harmonic motion can also be viewed as the projection of uniform circular motion on the diameter of the circle in which the latter motion occurs. 5. The particle velocity and acceleration during SHM as functions of time are given by, v (t) = –ωA sin (ωt + φ ) (velocity), a (t) = –ω2A cos (ωt + φ ) = –ω2x (t) (acceleration), Thus we see that both velocity and acceleration of a body executing simple harmonic motion are periodic functions, having the velocity amplitude vm=ω A and acceleration amplitude am =ω 2A, respectively. 6. The force acting in a simple harmonic motion is proportional to the displacement and is always directed towards the centre of motion. 7. A particle executing simple harmonic motion has, at any time, kinetic energy K = ½ mv2 and potential energy U = ½ kx2. If no friction is present the mechanical energy of the system, E = K + U always remains constant even though K and U change with time. 8. A particle of mass m oscillating under the influence of Hooke’s law restoring force given by F = – k x exhibits simple harmonic motion with k ω = (angular frequency) m m T = 2π (period) k Such a system is also called a linear oscillator. 9. The motion of a simple pendulum swinging through small angles is approximately simple harmonic. The period of oscillation is given by, L T = 2π g Reprint 2025-26 OSCILLATIONS 273 POINTS TO PONDER 1. The period T is the least time after which motion repeats itself. Thus, motion repeats itself after nT where n is an integer. 2. Every periodic motion is not simple harmonic motion. Only that periodic motion governed by the force law F = – k x is simple harmonic. 3. Circular motion can arise due to an inverse-square law force (as in planetary motion) as well as due to simple harmonic force in two dimensions equal to: –mω2r. In the latter case, the phases of motion, in two perpendicular directions (x and y) must differ by π/2. Thus, for example, a particle subject to a force –mω2r with initial position (0, A) and velocity (ωA, 0) will move uniformly in a circle of radius A. 4. For linear simple harmonic motion with a given ω, two initial conditions are necessary and sufficient to determine the motion completely. The initial conditions may be (i) initial position and initial velocity or (ii) amplitude and phase or (iii) energy and phase. 5. From point 4 above, given amplitude or energy, phase of motion is determined by the initial position or initial velocity. 6. A combination of two simple harmonic motions with arbitrary amplitudes and phases is not necessarily periodic. It is periodic only if frequency of one motion is an integral multiple of the other’s frequency. However, a periodic motion can always be expressed as a sum of infinite number of harmonic motions with appropriate amplitudes. 7. The period of SHM does not depend on amplitude or energy or the phase constant. Contrast this with the periods of planetary orbits under gravitation (Kepler’s third law). 8. The motion of a simple pendulum is simple harmonic for small angular displacement. 9. For motion of a particle to be simple harmonic, its displacement x must be expressible in either of the following forms : x = A cos ωt + B sin ωt x = A cos (ωt + α ), x = B sin (ωt + β ) The three forms are completely equivalent (any one can be expressed in terms of any other two forms). Thus, damped simple harmonic motion is not strictly simple harmonic. It is approximately so only for time intervals much less than 2m/b where b is the damping constant. Reprint 2025-26 274 PHYSICS Exercises 13.1 Which of the following examples represent periodic motion? (a) A swimmer completing one (return) trip from one bank of a river to the other and back. (b) A freely suspended bar magnet displaced from its N-S direction and released. (c) A hydrogen molecule rotating about its centre of mass. (d) An arrow released from a bow. 13.2 Which of the following examples represent (nearly) simple harmonic motion and which represent periodic but not simple harmonic motion? (a) the rotation of earth about its axis. (b) motion of an oscillating mercury column in a U-tube. (c) motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lower most point. (d) general vibrations of a polyatomic molecule about its equilibrium position. 13.3 Fig. 13.18 depicts four x-t plots for linear motion of a particle. Which of the plots represent periodic motion? What is the period of motion (in case of periodic motion) ? Fig. 18.18 Reprint 2025-26 OSCILLATIONS 275 13.4 Which of the following functions of time represent (a) simple harmonic, (b) periodic but not simple harmonic, and (c) non-periodic motion? Give period for each case of periodic motion (ω is any positive constant): (a) sin ωt – cos ωt (b) sin3 ωt (c) 3 cos (π/4 – 2ωt) (d) cos ωt + cos 3ωt + cos 5ωt (e) exp (–ω2t2) (f) 1 + ωt + ω2t2 13.5 A particle is in linear simple harmonic motion between two points, A and B, 10 cm apart. Take the direction from A to B as the positive direction and give the signs of velocity, acceleration and force on the particle when it is (a) at the end A, (b) at the end B, (c) at the mid-point of AB going towards A, (d) at 2 cm away from B going towards A, (e) at 3 cm away from A going towards B, and (f) at 4 cm away from B going towards A. 13.6 Which of the following relationships between the acceleration a and the displacement x of a particle involve simple harmonic motion? (a) a = 0.7x (b) a = –200x2 (c) a = –10x (d) a = 100x3 13.7 The motion of a particle executing simple harmonic motion is described by the displacement function, x(t) = A cos (ωt + φ ). If the initial (t = 0) position of the particle is 1 cm and its initial velocity is ω cm/s, what are its amplitude and initial phase angle ? The angular frequency of the particle is π s–1. If instead of the cosine function, we choose the sine function to describe the SHM : x = B sin (ωt + α), what are the amplitude and initial phase of the particle with the above initial conditions. 13.8 A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20 cm. A body suspended from this balance, when displaced and released, oscillates with a period of 0.6 s. What is the weight of the body ? 13.9 A spring having with a spring constant 1200 N m–1 is mounted on a horizontal table as shown in Fig. 13.19. A mass of 3 kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0 cm and released. Fig. 13.19 Determine (i) the frequency of oscillations, (ii) maximum acceleration of the mass, and (iii) the maximum speed of the mass. Reprint 2025-26 276 PHYSICS 13.10 In Exercise 13.9, let us take the position of mass when the spring is unstreched as x = 0, and the direction from left to right as the positive direction of x-axis. Give x as a function of time t for the oscillating mass if at the moment we start the stopwatch (t = 0), the mass is (a) at the mean position, (b) at the maximum stretched position, and (c) at the maximum compressed position. In what way do these functions for SHM differ from each other, in frequency, in amplitude or the initial phase? 13.11 Figures 13.20 correspond to two circular motions. The radius of the circle, the period of revolution, the initial position, and the sense of revolution (i.e. clockwise or anti-clockwise) are indicated on each figure. Fig. 13.20 Obtain the corresponding simple harmonic motions of the x-projection of the radius vector of the revolving particle P, in each case. 13.12 Plot the corresponding reference circle for each of the following simple harmonic motions. Indicate the initial (t =0) position of the particle, the radius of the circle, and the angular speed of the rotating particle. For simplicity, the sense of rotation may be fixed to be anticlockwise in every case: (x is in cm and t is in s). (a) x = –2 sin (3t + π/3) (b) x = cos (π/6 – t) (c) x = 3 sin (2πt + π/4) (d) x = 2 cos πt 13.13 Figure 13.21(a) shows a spring of force constant k clamped rigidly at one end and a mass m attached to its free end. A force F applied at the free end stretches the spring. Figure 13.21 (b) shows the same spring with both ends free and attached to a mass m at either end. Each end of the spring in Fig. 13.21(b) is stretched by the same force F. Fig. 13.21 (a) What is the maximum extension of the spring in the two cases ? (b) If the mass in Fig. (a) and the two masses in Fig. (b) are released, what is the period of oscillation in each case ? Reprint 2025-26 OSCILLATIONS 277 13.14 The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 1.0 m. If the piston moves with simple harmonic motion with an angular frequency of 200 rad/min, what is its maximum speed ? 13.15 The acceleration due to gravity on the surface of moon is 1.7 m s–2. What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is 3.5 s ? (g on the surface of earth is 9.8 m s–2) 13.16 A simple pendulum of length l and having a bob of mass M is suspended in a car. The car is moving on a circular track of radius R with a uniform speed v. If the pendulum makes small oscillations in a radial direction about its equilibrium position, what will be its time period ? 13.17 A cylindrical piece of cork of density of base area A and height h floats in a liquid of density ρl. The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a period hρ T = 2π ρ1 g where ρ is the density of cork. (Ignore damping due to viscosity of the liquid). 13.18 One end of a U-tube containing mercury is connected to a suction pump and the other end to atmosphere. A small pressure difference is maintained between the two columns. Show that, when the suction pump is removed, the column of mercury in the U-tube executes simple harmonic motion. Reprint 2025-26 CHAPTER FOURTEEN WAVES 14.1 INTRODUCTION In the previous Chapter, we studied the motion of objects oscillating in isolation. What happens in a system, which is a collection of such objects? A material medium provides

13.5 The Q value of a nuclear reaction A + b ® C + d is defined by Q = [ mA + mb – mC – md]c2 where the masses refer to the respective nuclei. Determine from the given data the Q-value of the following reactions and state whether the reactions are exothermic or endothermic. (i) 11 H+13 H →12 H+12 H (ii) 126 C+126 C →1020 Ne+ 24 He Atomic masses are given to be m ( 12 H ) = 2.014102 u m ( 13 H) = 3.016049 u m ( 126 C ) = 12.000000 u m ( 1020 Ne ) = 19.992439 u

13.2 Obtain the binding energy of the nuclei 5626Fe and 20983 Bi in units of MeV from the following data: m ( 5626Fe ) = 55.934939 u m ( 20983 Bi ) = 208.980388 u

13.2 PERIODIC AND OSCILLATORY MOTIONS Very often, the body undergoing periodic motion has an equilibrium position somewhereFig. 13.1 shows some periodic motions. Suppose inside its path. When the body is at this positionan insect climbs up a ramp and falls down, it no net external force acts on it. Therefore, if it iscomes back to the initial point and repeats the left there at rest, it remains there forever. If the process identically. If you draw a graph of its body is given a small displacement from the height above the ground versus time, it would position, a force comes into play which tries to look something like Fig. 13.1 (a). If a child climbs bring the body back to the equilibrium point, up a step, comes down, and repeats the process giving rise to oscillations or vibrations. For identically, its height above the ground would example, a ball placed in a bowl will be in look like that in Fig. 13.1 (b). When you play the equilibrium at the bottom. If displaced a little game of bouncing a ball off the ground, between from the point, it will perform oscillations in the your palm and the ground, its height versus time bowl. Every oscillatory motion is periodic, but graph would look like the one in Fig. 13.1 (c). every periodic motion need not be oscillatory. Note that both the curved parts in Fig. 13.1 (c) Circular motion is a periodic motion, but it is are sections of a parabola given by the Newton’s not oscillatory. equation of motion (see section 2.6), There is no significant difference between 1 2 oscillations and vibrations. It seems that when h = ut + gt for downward motion, and 2 the frequency is small, we call it oscillation (like, the oscillation of a branch of a tree), while when 1 2 h = ut – gt for upward motion, the frequency is high, we call it vibration (like, 2 the vibration of a string of a musical instrument). with different values of u in each case. These Simple harmonic motion is the simplest form are examples of periodic motion. Thus, a motion of oscillatory motion. This motion arises when that repeats itself at regular intervals of time is the force on the oscillating body is directly called periodic motion. proportional to its displacement from the mean position, which is also the equilibrium position. Further, at any point in its oscillation, this force is directed towards the mean position. In practice, oscillating bodies eventually (a) come to rest at their equilibrium positions because of the damping due to friction and other dissipative causes. However, they can be forced to remain oscillating by means of some external periodic agency. We discuss the phenomena of damped and forced oscillations later in the chapter. Any material medium can be pictured as a (b) collection of a large number of coupled oscillators. The collective oscillations of the constituents of a medium manifest themselves as waves. Examples of waves include water waves, seismic waves, electromagnetic waves. We shall study the wave phenomenon in the next chapter. (c) 13.2.1 Period and frequency We have seen that any motion that repeats itself at regular intervals of time is called periodic motion. The smallest interval of time after which the motion is repeated is called its Fig. 13.1 Examples of periodic motion. The period T period. Let us denote the period by the symbol is shown in each case. T. Its SI unit is second. For periodic motions, Reprint 2025-26 OSCILLATIONS 261 which are either too fast or too slow on the scale as a displacement variable [see Fig.13.2(b)]. The of seconds, other convenient units of time are term displacement is not always to be referred used. The period of vibrations of a quartz crystal is expressed in units of microseconds (10–6 s) abbreviated as µs. On the other hand, the orbital period of the planet Mercury is 88 earth days. The Halley’s comet appears after every 76 years. The reciprocal of T gives the number of repetitions that occur per unit time. This quantity is called the frequency of the periodic motion. It is represented by the symbol ν. The relation between ν and T is Fig. 13.2(a) A block attached to a spring, the other ν = 1/T (13.1) end of which is fixed to a rigid wall. The block moves on a frictionless surface. The The unit of ν is thus s–1. After the discoverer of motion of the block can be described in radio waves, Heinrich Rudolph Hertz (1857–1894), terms of its distance or displacement x a special name has been given to the unit of from the equilibrium position. frequency. It is called hertz (abbreviated as Hz). Thus, 1 hertz = 1 Hz =1 oscillation per second =1 s–1 (13.2) Note, that the frequency, ν, is not necessarily an integer. u Example 13.1 On an average, a human heart is found to beat 75 times in a minute. Calculate its frequency and period. Fig.13.2(b) An oscillating simple pendulum; its Answer The beat frequency of heart = 75/(1 min) motion can be described in terms of = 75/(60 s) angular displacement θ from the vertical. = 1.25 s–1 = 1.25 Hz in the context of position only. There can be The time period T = 1/(1.25 s–1) many other kinds of displacement variables. The = 0.8 s ⊳ voltage across a capacitor, changing with time in an AC circuit, is also a displacement variable.13.2.2 Displacement In the same way, pressure variations in time inIn section 3.2, we defined displacement of a the propagation of sound wave, the changingparticle as the change in its position vector. In electric and magnetic fields in a light wave arethis chapter, we use the term displacement examples of displacement in different contexts.in a more general sense. It refers to change The displacement variable may take bothwith time of any physical property under positive and negative values. In experiments onconsideration. For example, in case of rectilinear oscillations, the displacement is measured formotion of a steel ball on a surface, the distance different times.from the starting point as a function of time is The displacement can be represented by a its position displacement. The choice of origin mathematical function of time. In case of periodic is a matter of convenience. Consider a block motion, this function is periodic in time. One of attached to a spring, the other end of the spring the simplest periodic functions is given by is fixed to a rigid wall [see Fig.13.2(a)]. Generally, it is convenient to measure displacement of the f (t) = A cos ωt (13.3a) body from its equilibrium position. For an If the argument of this function, ωt, is oscillating simple pendulum, the angle from the increased by an integral multiple of 2π radians, vertical as a function of time may be regarded the value of the function remains the same. The Reprint 2025-26 262 PHYSICS function f (t) is then periodic and its period, T, (ii) This is an example of a periodic motion. It is given by can be noted that each term represents a 2 π periodic function with a different angular T = (13.3b) frequency. Since period is the least interval ω of time after which a function repeats its Thus, the function f (t) is periodic with period T, value, sin ωt has a period T0= 2π/ω ; cos 2 ωt f (t) = f (t+T ) has a period π/ω =T0/2; and sin 4 ωt has a period 2π/4ω = T0/4. The period of the firstThe same result is obviously correct if we term is a multiple of the periods of the lastconsider a sine function, f (t ) = A sin ωt. Further, two terms. Therefore, the smallest intervala linear combination of sine and cosine functions of time after which the sum of the threelike, terms repeats is T0, and thus, the sum is a f (t) = A sin ωt + B cos ωt (13.3c) periodic function with a period 2π/ω. is also a periodic function with the same period (iii) The function e–ωt is not periodic, itT. Taking, decreases monotonically with increasing A = D cos φ and B = D sin φ time and tends to zero as t → ∞ and thus, Eq. (13.3c) can be written as, never repeats its value. (iv) The function log(ωt) increases f (t) = D sin (ωt + φ ) , (13.3d) monotonically with time t. It, therefore, Here D and φ are constant given by never repeats its value and is a non- periodic function. It may be noted that as  B  t → ∞, log(ωt) diverges to ∞. It, therefore, 2 2 – 1 D = A + B and φ= tan  A  cannot represent any kind of physical displacement. ⊳ The great importance of periodic sine and cosine functions is due to a remarkable result 13.3 SIMPLE HARMONIC MOTION proved by the French mathematician, Jean Consider a particle oscillating back and forth Baptiste Joseph Fourier (1768–1830): Any about the origin of an x-axis between the limits periodic function can be expressed as a +A and –A as shown in Fig. 13.3. This oscillatory superposition of sine and cosine functions motion is said to be simple harmonic if the of different time periods with suitable displacement x of the particle from the origin coefficients. varies with time as : x (t) = A cos (ω t + φ) (13.4) u Example 13.2 Which of the following functions of time represent (a) periodic and (b) non-periodic motion? Give the period for each case of periodic motion [ω is any positive constant]. (i) sin ωt + cos ωt Fig. 13.3 A particle vibrating back and forth about the origin of x-axis, between the limits +A (ii) sin ωt + cos 2 ωt + sin 4 ωt and –A. (iii) e–ωt (iv) log (ωt) where A, ω and φ are constants. Thus, simple harmonic motion (SHM) is not Answer any periodic motion but one in which displacement is a sinusoidal function of time.(i) sin ωt + cos ωt is a periodic function, it can Fig. 13.4 shows the positions of a particle also be written as 2 sin (ωt + π/4). executing SHM at discrete value of time, each Now 2 sin (ωt + π/4)= 2 sin (ωt + π/4+2π) interval of time being T/4, where T is the period of motion. Fig. 13.5 plots the graph of x versus t, = 2 sin [ω (t + 2π/ω) + π/4] which gives the values of displacement as a The periodic time of the function is 2π/ω. continuous function of time. The quantities A, Reprint 2025-26 OSCILLATIONS 263 any loss of generality]. As the cosine function of time varies from +1 to –1, the displacement varies between the extremes A and – A. Two simple harmonic motions may have same ω and φ but different amplitudes A and B, as shown in Fig. 13.7 (a). While the amplitude A is fixed for a given SHM, the state of motion (position and velocity) of the particle at any time t is determined by the Fig. 13.4 The location of the particle in SHM at the discrete values t = 0, T/4, T/2, 3T/4, T, 5T/4. The time after which motion repeats itself is T. T will remain fixed, no matter what location you choose as the initial (t = Fig. 13.7 (a) A plot of displacement as a function of 0) location. The speed is maximum for zero time as obtained from Eq. (14.4) with displacement (at x = 0) and zero at the φ = 0. The curves 1 and 2 are for two extremes of motion. different amplitudes A and B. ω and φ which characterize a given SHM have standard names, as summarised in Fig. 13.6. argument (ωt + φ) in the cosine function. This Let us understand these quantities. time-dependent quantity, (ωt + φ) is called the The amplitutde A of SHM is the magnitude phase of the motion. The value of plase at t = 0 of maximum displacement of the particle. is φ and is called the phase constant (or phase [Note, A can be taken to be positive without angle). If the amplitude is known, φ can be determined from the displacement at t = 0. Two simple harmonic motions may have the same A and ω but different phase angle φ, as shown in Fig. 13.7 (b). Finally, the quantity ω can be seen to be related to the period of motion T. Taking, for simplicity, φ = 0 in Eq. (13.4), we have Fig. 13.5 Displacement as a continuous function of time for simple harmonic motion. x (t) : displacement x as a function of time t A : amplitude ω : angular frequency ωt + φ : phase (time-dependent) φ : phase constant Fig. 13.7 (b) A plot obtained from Eq. (13.4). The curves 3 and 4 are for φ = 0 and -π/4 respectively. The amplitude A is same for Fig. 13.6 The meaning of standard symbols both the plots. in Eq. (13.4) Reprint 2025-26 264 PHYSICS x(t) = A cos ωt (13.5) This function represents a simple harmonic motion having a period T = 2π/ω and a Since the motion has a period T, x (t) is equal to phase angle (–π/4) or (7π/4) x (t + T). That is, (b) sin2 ωt = ½ – ½ cos 2 ωt A cos ωt = A cos ω (t + T ) (13.6) The function is periodic having a period Now the cosine function is periodic with period T = π/ω. It also represents a harmonic 2π, i.e., it first repeats itself when the argument motion with the point of equilibrium ½ instead of zero. ⊳changes by 2π. Therefore, occurring at ω(t + T ) = ωt + 2π 13.4 SIMPLE HARMONIC MOTION AND UNIFORM CIRCULAR MOTION that is ω = 2π/ T (13.7) In this section, we show that the projection of uniform circular motion on a diameter of the ω is called the angular frequency of SHM. Its circle follows simple harmonic motion. A S.I. unit is radians per second. Since the simple experiment (Fig. 13.9) helps us visualise frequency of oscillations is simply 1/T, ω is 2π this connection. Tie a ball to the end of a string times the frequency of oscillation. Two simple and make it move in a horizontal plane about harmonic motions may have the same A and φ, a fixed point with a constant angular speed. but different ω, as seen in Fig. 13.8. In this plot The ball would then perform a uniform circular the curve (b) has half the period and twice the motion in the horizontal plane. Observe the frequency of the curve (a). ball sideways or from the front, fixing your attention in the plane of motion. The ball will appear to execute to and fro motion along a horizontal line with the point of rotation as the midpoint. You could alternatively observe the shadow of the ball on a wall which is perpendicular to the plane of the circle. In this process what we are observing is the motion of the ball on a diameter of the circle normal to the direction of viewing. Fig. 13.8 Plots of Eq. (13.4) for φ = 0 for two different periods. u Example 13.3 Which of the following functions of time represent (a) simple Fig. 13.9 Circular motion of a ball in a plane viewed harmonic motion and (b) periodic but not edge-on is SHM. simple harmonic? Give the period for each case. Fig. 13.10 describes the same situation (1) sin ωt – cos ωt mathematically. Suppose a particle P is moving (2) sin2 ωt uniformly on a circle of radius A with angular Answer speed ω. The sense of rotation is anticlockwise. (a) sin ωt – cos ωt The initial position vector of the particle, i.e., = sin ωt – sin (π/2 – ωt) the vector OP at t = 0 makes an angle of φ with = 2 cos (π/4) sin (ωt – π/4) the positive direction of x-axis. In time t, it will = √2 sin (ωt – π/4) cover a further angle ωt and its position vector Reprint 2025-26 OSCILLATIONS 265 u Example 13.4 The figure given below depicts two circular motions. The radius of the circle, the period of revolution, the initial position and the sense of revolution are indicated in the figures. Obtain the simple harmonic motions of the x-projection of the radius vector of the rotating particle P in each case. Fig. 13.10 will make an angle of ωt + φ with the +ve x-axis. Next, consider the projection of the position vector OP on the x-axis. This will be Answer OP′. The position of P′ on the x-axis, as the (a) At t = 0, OP makes an angle of 45o = π/4 rad particle P moves on the circle, is given by with the (positive direction of) x-axis. After x(t) = A cos (ωt + φ ) 2 π time t, it covers an angle t in thewhich is the defining equation of SHM. This T shows that if P moves uniformly on a circle, anticlockwise sense, and makes an angle its projection P′ on a diameter of the circle executes SHM. The particle P and the circle of 2 πt + π with the x-axis. on which it moves are sometimes referred to T 4 as the reference particle and the reference circle, The projection of OP on the x-axis at time t respectively. is given by, We can take projection of the motion of P on any diameter, say the y-axis. In that case, the  2π π  x (t) = A cos t +displacement y(t) of P′ on the y-axis is given by  T 4  y = A sin (ωt + φ) For T = 4 s, which is also an SHM of the same amplitude as that of the projection on x-axis, but differing  2π π  x(t) = A cos t +by a phase of π/2.  4 4  In spite of this connection between circular motion and SHM, the force acting on a particle which is a SHM of amplitude A, period 4 s, in linear simple harmonic motion is very πdifferent from the centripetal force needed to and an initial phase* = . keep a particle in uniform circular motion. 4 * The natural unit of angle is radian, defined through the ratio of arc to radius. Angle is a dimensionless quantity. Therefore it is not always necessary to mention the unit ‘radian’ when we use π, its multiples or submultiples. The conversion between radian and degree is not similar to that between metre and centimetre or mile. If the argument of a trigonometric function is stated without units, it is understood that the unit is radian. On the other hand, if degree is to be used as the unit of angle, then it must be shown explicitly. For example, sin(150) means sine of 15 degree, but sin(15) means sine of 15 radians. Hereafter, we will often drop ‘rad’ as the unit, and it should be understood that whenever angle is mentioned as a numerical value, without units, it is to be taken as radians. Reprint 2025-26 266 PHYSICS (b) In this case at t = 0, OP makes an angle of where the negative sign shows that v (t) has a π direction opposite to the positive direction of 90o = with the x-axis. After a time t, it x-axis. Eq. (13.9) gives the instantaneous 2 2π velocity of a particle executing SHM, where covers an angle of t in the clockwise T displacement is given by Eq. (13.4). We can, of  π 2π  course, obtain this equation without using sense and makes an angle of  2 − T t  geometrical argument, directly by differentiating (Eq. 13.4) with respect of t: with the x-axis. The projection of OP on the x-axis at time t is given by d v(t) = x (t ) (13.10)  π 2π  d t x(t) = B cos  2 − T t  The method of reference circle can be similarly used for obtaining instantaneous acceleration  2π  of a particle undergoing SHM. We know that the = B sin  T t  centripetal acceleration of a particle P in uniform For T = 30 s, circular motion has a magnitude v2/A or ω2A, and it is directed towards the centre i.e., the  π  direction is along PO. The instantaneous x(t) = B sin  15 t  acceleration of the projection particle P′ is then (See Fig. 13.12)  π π  a (t) = –ω2A cos (ωt + φ) Writing this as x (t) = B cos  15 t − 2 , and comparing with Eq. (13.4). We find that this = –ω2x (t) (13.11) represents a SHM of amplitude B, period 30 s, π and an initial phase of − . ⊳ 2

13.4 1.23 351 Reprint 2025-26 Physics 13.5 (i) Q = –4.03 MeV; endothermic (ii) Q = 4.62 MeV; exothermic 56 – 2m 28 Al = 26.90 MeV; not possible. 13.6 Q = m ( 26 Fe ) ( 13 ) 13.7 4.536 × 1026 MeV 13.8 About 4.9 × 104 y 13.9 360 KeV CHAPTER 14 14.1 (c) 14.2 (d) 14.3 (c) 14.4 (c) 14.5 (c) 14.6 50 Hz for half-wave, 100 Hz for full-wave Reprint 2025-26 Bibligraphy BIBLIOGRAPHY TEXTBOOKS For additional reading on the topics covered in this book, you may like to consult one or more of the following books. Some of these books however are more advanced and contain many more topics than this book. 1 Ordinary Level Physics, A.F. Abbott, Arnold-Heinemann (1984). 2 Advanced Level Physics, M. Nelkon and P. Parker, 6th Edition, Arnold-Heinemann (1987). 3 Advanced Physics, Tom Duncan, John Murray (2000). 4 Fundamentals of Physics, David Halliday, Robert Resnick and Jearl Walker, 7th Edition John Wily (2004). 5 University Physics (Sears and Zemansky’s), H.D. Young and R.A. Freedman, 11th Edition, Addison—Wesley (2004). 6 Problems in Elementary Physics, B. Bukhovtsa, V. Krivchenkov, G. Myakishev and V. Shalnov, MIR Publishers, (1971). 7 Lectures on Physics (3 volumes), R.P. Feynman, Addision – Wesley (1965). 8 Berkeley Physics Course (5 volumes) McGraw Hill (1965). a. Vol. 1 – Mechanics: (Kittel, Knight and Ruderman) b. Vol. 2 – Electricity and Magnetism (E.M. Purcell) c. Vol. 3 – Waves and Oscillations (Frank S. Crawford) d. Vol. 4 – Quantum Physics (Wichmann) e. Vol. 5 – Statistical Physics (F. Reif ) 9 Fundamental University Physics, M. Alonso and E. J. Finn, Addison – Wesley (1967). 10 College Physics, R.L. Weber, K.V. Manning, M.W. White and G.A. Weygand, Tata McGraw Hill (1977). 11 Physics: Foundations and Frontiers, G. Gamow and J.M. Cleveland, Tata McGraw Hill (1978). 12 Physics for the Inquiring Mind, E.M. Rogers, Princeton University Press (1960). 13 PSSC Physics Course, DC Heath and Co. (1965) Indian Edition, 14 Physics Advanced Level, Jim Breithampt, Stanley Thornes Publishers (2000). 15 Physics, Patrick Fullick, Heinemann (2000). 16 Conceptual Physics, Paul G. Hewitt, Addision—Wesley (1998). 17 College Physics, Raymond A. Serway and Jerry S. Faughn, Harcourt Brace and Co. (1999). 18 University Physics, Harris Benson, John Wiley (1996). 19 University Physics, William P. Crummet and Arthur B. Western, Wm.C. Brown (1994). 20 General Physics, Morton M. Sternheim and Joseph W. Kane, John Wiley (1988). 21 Physics, Hans C. Ohanian, W.W. Norton (1989). Reprint 2025-26 Physics 22 Advanced Physics, Keith Gibbs, Cambridge University Press (1996). 23 Understanding Basic Mechanics, F. Reif, John Wiley (1995). 24 College Physics, Jerry D. Wilson and Anthony J. Buffa, Prentice Hall (1997). 25 Senior Physics, Part – I, I.K. Kikoin and A.K. Kikoin, MIR Publishers (1987). 26 Senior Physics, Part – II, B. Bekhovtsev, MIR Publishers (1988). 27 Understanding Physics, K. Cummings, Patrick J. Cooney, Priscilla W. Laws and Edward F. Redish, John Wiley (2005). 28 Essentials of Physics, John D. Cutnell and Kenneth W. Johnson, John Wiley (2005). GENERAL BOOKS For instructive and entertaining general reading on science, you may like to read some of the following books. Remember however, that many of these books are written at a level far beyond the level of the present book. 1 Mr. Tompkins in paperback, G. Gamow, Cambridge University Press (1967). 2 The Universe and Dr. Einstein, C. Barnett, Time Inc. New York (1962). 3 Thirty years that Shook Physics, G. Gamow, Double Day, New York (1966). 4 Surely You’re Joking, Mr. Feynman, R.P. Feynman, Bantam books (1986). 5 One, Two, Three… Infinity, G. Gamow, Viking Inc. (1961). 6 The Meaning of Relativity, A. Einstein, (Indian Edition) Oxford and IBH Pub. Co. (1965). 7 Atomic Theory and the Description of Nature, Niels Bohr, Cambridge (1934). 8 The Physical Principles of Quantum Theory, W. Heisenberg, University of Chicago Press (1930). 9 The Physics—Astronomy Frontier, F. Hoyle and J.V. Narlikar, W.H. Freeman (1980). 10 The Flying Circus of Physics with Answer, J. Walker, John Wiley and Sons (1977). 11 Physics for Everyone (series), L.D. Landau and A.I. Kitaigorodski, MIR Publisher (1978). Book 1: Physical Bodies Book 2: Molecules Book 3: Electrons Book 4: Photons and Nuclei. 12 Physics can be Fun, Y. Perelman, MIR Publishers (1986). 13 Power of Ten, Philip Morrison and Eames, W.H. Freeman (1985). 14 Physics in your Kitchen Lab., I.K. Kikoin, MIR Publishers (1985). 15 How Things Work: The Physics of Everyday Life, Louis A. Bloomfield, John Wiley (2005). 16 Physics Matters: An Introduction to Conceptual Physics, James Trefil and Robert M. Hazen, John Wiley (2004). 354 Reprint 2025-26

13.4 Obtain approximately the ratio of the nuclear radii of the gold isotope 197 79 Au and the silver isotope 10747 Ag .

6.11 Dynamics of rotational the motion of extended bodies. motion about a fixed axis A large class of problems with extended bodies can be

6.12 ANGULAR MOMENTUM IN CASE OF For computing the total angular momentum ROTATION ABOUT A FIXED AXIS of the whole rigid body, we add up the contribution of each particle of the body. We have studied in section 6.7, the angular momentum of a system of particles. We already Thus know from there that the time rate of total We denote by L ⊥ and L z the components of angular momentum of a system of particles L respectively perpendicular to the z-axis andabout a point is equal to the total external torque along the z-axis;on the system taken about the same point. When OC i × m i v i (6.42a)the total external torque is zero, the total angular L ⊥= ∑ momentum of the system is conserved. where mi and vi are respectively the mass and We now wish to study the angular momentum the velocity of the ith particle and Ci is the centrein the special case of rotation about a fixed axis. of the circle described by the particle; The general expression for the total angular momentum of the system of n particles is N and ˆ (6.42b) L = =∑i 1 ri × p i (6.25b) or L z = Iωk We first consider the angular momentum of The last step follows since the perpendicular a typical particle of the rotating rigid body. We distance of the ith particle from the axis is ri; andthen sum up the contributions of individual by definition the moment of inertia of the body particles to get L of the whole body. m i ri2 . For a typical particle l = r × p. As seen in the about the axis of rotation is I =∑ last section r = OP = OC + CP [Fig. 6.17(b)]. With Note L = L z + L ⊥ (6.42c)p = m v , l = ( OC × m v ) + ( CP × m v ) The rigid bodies which we have mainly considered in this chapter are symmetric about The magnitude of the linear velocity v of the the axis of rotation, i.e. the axis of rotation is particle at P is given by v = ωr⊥ where r⊥ is the one of their symmetry axes. For such bodies, for length of CP or the perpendicular distance of P a given OCi, for every particle which has a from the axis of rotation. Further, v is tangential velocity vi , there is another particle of velocity at P to the circle which the particle describes. –vi located diametrically opposite on the circle Using the right-hand rule one can check that with centre Ci described by the particle. TogetherCP × v is parallel to the fixed axis. The unit vector along the fixed axis (chosen as the z-axis) such pairs will contribute zero to L ⊥ and as a is ˆk . Hence result for symmetric bodies L ⊥ is zero, and CP × m v = r⊥ (mv ) kˆ hence z = Iωkˆ (6.42d) = mr⊥2ω kˆ (since υ = ωr⊥ ) L = L Similarly, we can check that OC × v is For bodies, which are not symmetric about perpendicular to the fixed axis. Let us denote the axis of rotation, L is not equal to Lz and hence the part of l along the fixed axis (i.e. the z-axis) L does not lie along the axis of rotation. by lz, then Referring to Table 6.1, can you tell in which l z = CP × m v = mr⊥2ωkˆ cases L = Lz will not apply? Let us differentiate Eq. (6.42b). Since ˆk is a and l = l z + OC × m v fixed (constant) vector, we get We note that lz is parallel to the fixed axis, ˆbut l is not. In general, for a particle, the angular I ω) k d ( L z ) =  d (  d t momentum l is not along the axis of rotation, d t i.e. for a particle, l and ω are not necessarily Now, Eq. (6.28b) states parallel. Compare this with the corresponding dL fact in translation. For a particle, p and v are = τ dtalways parallel to each other. Reprint 2025-26 122 PHYSICS As we have seen in the last section, only We have already derived this equation using those components of the external torques which the work - kinetic energy route. are along the axis of rotation, need to be taken into account, when we discuss rotation about a 6.12.1 Conservation of angular momentum fixed axis. This means we can take τ = τkˆ . We are now in a position to revisit the principle of conservation of angular momentum in the Since L = L z + L ⊥ and the direction of Lz (vector context of rotation about a fixed axis. From Eq. ˆk ) is fixed, it follows that for rotation about a (6.43c), if the external torque is zero, fixed axis, Lz = Iω = constant (6.44) For symmetric bodies, from Eq. (6.42d), Lz d L z = τkˆ (6.43a) may be replaced by L .(L and Lz are respectively d t the magnitudes of L and Lz.) This then is the required form, for fixed axis d L rotation, of Eq. (6.29a), which expresses theand ⊥= 0 (6.43b) dt general law of conservation of angular momentum Thus, for rotation about a fixed axis, the of a system of particles. Eq. (6.44) applies to many component of angular momentum perpendicular situations that we come across in daily life. You may do this experiment with your friend. Sit on a to the fixed axis is constant. As L z = Iωkˆ , we swivel chair (a chair with a seat, free to rotate get from Eq. (6.43a), about a pivot) with your arms folded and feet not resting on, i.e., away from, the ground. Ask your d ( Iω) = τ (6.43c) friend to rotate the chair rapidly. While the chair d t is rotating with considerable angular speed If the moment of inertia I does not change with stretch your arms horizontally. What happens? time, Your angular speed is reduced. If you bring back d dω your arms closer to your body, the angular speed ( Iω) = I = Iα increases again. This is a situation where thed t d t principle of conservation of angular momentumand we get from Eq. (6.43c), is applicable. If friction in the rotational τ = I α (6.41) Fig 6.32 (a) A demonstration of conservation of Fig 6.32 (b) An acrobat employing the principle of angular momentum. A girl sits on a conservation of angular momentum in swivel chair and stretches her arms/ her performance. brings her arms closer to the body. Reprint 2025-26 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 123 mechanism is neglected, there is no external A circus acrobat and a diver take advantage torque about the axis of rotation of the chair and of this principle. Also, skaters and classical, hence Iω is constant. Stretching the arms Indian or western, dancers performing a increases I about the axis of rotation, resulting in pirouette (a spinning about a tip–top) on the toes decreasing the angular speed ω. Bringing the of one foot display ‘mastery’ over this principle. arms closer to the body has the opposite effect. Can you explain? SUMMARY 1. Ideally, a rigid body is one for which the distances between different particles of the body do not change, even though there are forces on them. 2. A rigid body fixed at one point or along a line can have only rotational motion. A rigid body not fixed in some way can have either pure translational motion or a combination of translational and rotational motions. 3. In rotation about a fixed axis, every particle of the rigid body moves in a circle which lies in a plane perpendicular to the axis and has its centre on the axis. Every Point in the rotating rigid body has the same angular velocity at any instant of time. 4. In pure translation, every particle of the body moves with the same velocity at any instant of time. 5. Angular velocity is a vector. Its magnitude is ω = dθ/dt and it is directed along the axis of rotation. For rotation about a fixed axis, this vector ω has a fixed direction. 6. The vector or cross product of two vector a and b is a vector written as a×b. The magnitude of this vector is absinθ and its direction is given by the right handed screw or the right hand rule. 7. The linear velocity of a particle of a rigid body rotating about a fixed axis is given by v = ω × r, where r is the position vector of the particle with respect to an origin along the fixed axis. The relation applies even to more general rotation of a rigid body with one point fixed. In that case r is the position vector of the particle with respect to the fixed point taken as the origin. 8. The centre of mass of a system of n particles is defined as the point whose position vector is ri ∑m i R = M 9. Velocity of the centre of mass of a system of particles is given by V = P/M, where P is the linear momentum of the system. The centre of mass moves as if all the mass of the system is concentrated at this point and all the external forces act at it. If the total external force on the system is zero, then the total linear momentum of the system is constant. 10. The angular momentum of a system of n particles about the origin is n L = ri × pi i =∑1 The torque or moment of force on a system of n particles about the origin is τ = ∑ri × Fi 1 The force Fi acting on the ith particle includes the external as well as internal forces. Assuming Newton’s third law of motion and that forces between any two particles act along the line joining the particles, we can show τint = 0 and Reprint 2025-26 124 PHYSICS dL = τ ext dt 11. A rigid body is in mechanical equilibrium if (1) it is in translational equilibrium, i.e., the total external force on it is zero : Fi = 0 , ∑ and (2) it is in rotational equilibrium, i.e. the total external torque on it is zero : Fi = 0 . ∑ τi = ∑ri × 12. The centre of gravity of an extended body is that point where the total gravitational torque on the body is zero. 13. The moment of intertia of a rigid body about an axis is defined by the formula I m i ri2 =∑ where ri is the perpendicular distance of the ith point of the body from the axis. The 1 2 kinetic energy of rotation is K = Iω . 2 POINTS TO PONDER 1. To determine the motion of the centre of mass of a system no knowledge of internal forces of the system is required. For this purpose we need to know only the external forces on the body. 2. Separating the motion of a system of particles as the motion of the centre of mass, (i.e., the translational motion of the system) and motion about (i.e. relative to) the centre of mass of the system is a useful technique in dynamics of a system of particles. One example of this technique is separating the kinetic energy of a system of particles K as the kinetic energy of the system about its centre of mass K′ and the kinetic energy of the centre of mass MV2/2, K = K′ + MV2/2 3. Newton’s Second Law for finite sized bodies (or systems of particles) is based in Newton’s Second Law and also Newton’s Third Law for particles. 4. To establish that the time rate of change of the total angular momentum of a system of particles is the total external torque in the system, we need not only Newton’s second law for particles, but also Newton’s third law with the provision that the forces between any two particles act along the line joining the particles. 5. The vanishing of the total external force and the vanishing of the total external torque are independent conditions. We can have one without the other. In a couple, total external force is zero, but total torque is non-zero. 6. The total torque on a system is independent of the origin if the total external force is zero. 7. The centre of gravity of a body coincides with its centre of mass only if the gravitational field does not vary from one part of the body to the other. Reprint 2025-26 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 125 8. The angular momentum L and the angular velocity ω are not necessarily parallel vectors. However, for the simpler situations discussed in this chapter when rotation is about a fixed axis which is an axis of symmetry of the rigid body, the relation L = Iω holds good, where I is the moment of the inertia of the body about the rotation axis. EXERCISES 6.1 Give the location of the centre of mass of a (i) sphere, (ii) cylinder, (iii) ring, and (iv) cube, each of uniform mass density. Does the centre of mass of a body necessarily lie inside the body ? 6.2 In the HCl molecule, the separation between the nuclei of the two atoms is about 1.27 Å (1 Å = 10-10 m). Find the approximate location of the CM of the molecule, given that a chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus. 6.3 A child sits stationary at one end of a long trolley moving uniformly with a speed V on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, what is the speed of the CM of the (trolley + child) system ? 6.4 Show that the area of the triangle contained between the vectors a and b is one half of the magnitude of a × b. 6.5 Show that a.(b × c) is equal in magnitude to the volume of the parallelepiped formed on the three vectors , a, b and c. 6.6 Find the components along the x, y, z axes of the angular momentum l of a particle, whose position vector is r with components x, y, z and momentum is p with components px, py and pz. Show that if the particle moves only in the x-y plane the angular momentum has only a z-component. 6.7 Two particles, each of mass m and speed v, travel in opposite directions along parallel lines separated by a distance d. Show that the angular momentum vector of the two particle system is the same whatever be the point about which the angular momentum is taken. 6.8 A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in Fig.6.33. The angles made by the strings with the vertical are 36.9° and 53.1° respectively. The bar is 2 m long. Calculate the distance d of the centre of gravity of the bar from its left end. Fig. 6.33 6.9 A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its centre of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel. Reprint 2025-26 126 PHYSICS 6.10 Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time. 6.11 A solid cylinder of mass 20 kg rotates about its axis with angular speed 100 rad s-1. The radius of the cylinder is 0.25 m. What is the kinetic energy associated with the rotation of the cylinder? What is the magnitude of angular momentum of the cylinder about its axis? 6.12 (a) A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of 40 rev/min. How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to 2/5 times the initial value ? Assume that the turntable rotates without friction. (b) Show that the child’s new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy? 6.13 A rope of negligible mass is wound round a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N ? What is the linear acceleration of the rope ? Assume that there is no slipping. 6.14 To maintain a rotor at a uniform angular speed of 200 rad s-1, an engine needs to transmit a torque of 180 N m. What is the power required by the engine ? (Note: uniform angular velocity in the absence of friction implies zero torque. In practice, applied torque is needed to counter frictional torque). Assume that the engine is 100% efficient. 6.15 From a uniform disk of radius R, a circular hole of radius R/2 is cut out. The centre of the hole is at R/2 from the centre of the original disc. Locate the centre of gravity of the resulting flat body. 6.16 A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5 g are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm. What is the mass of the metre stick? 6.17 The oxygen molecule has a mass of 5.30 × 10-26 kg and a moment of inertia of 1.94 ×10-46 kg m2 about an axis through its centre perpendicular to the lines joining the two atoms. Suppose the mean speed of such a molecule in a gas is 500 m/s and that its kinetic energy of rotation is two thirds of its kinetic energy of translation. Find the average angular velocity of the molecule. Reprint 2025-26 CHAPTER SEVEN GRAVITATION 7.1 INTRODUCTION Early in our lives, we become aware of the tendency of all material objects to be attracted towards the earth. Anything

6.4 LINEAR MOMENTUM OF A SYSTEM OF Eq. (6.15), this also means that when the PARTICLES total external force on the system is zero the velocity of the centre of mass remainsLet us recall that the linear momentum of a constant. (We assume throughout the particle is defined as discussion on systems of particles in this p = m v (6.12) chapter that the total mass of the system Let us also recall that Newton’s second law remains constant.) written in symbolic form for a single particle is Note that on account of the internal forces, dp i.e. the forces exerted by the particles on one F = (6.13) another, the individual particles may have dt Reprint 2025-26 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 101 complicated trajectories. Yet, if the total external force acting on the system is zero, the centre of mass moves with a constant velocity, i.e., moves uniformly in a straight line like a free particle. The vector Eq. (6.18a) is equivalent to three scalar equations, Px = c1, Py = c2 and Pz = c3 (6.18 b) (a) (b) Here Px, Py and Pz are the components of the total linear momentum vector P along the x–, y– Fig. 6.14 (a) Trajectories of two stars, S1 (dotted line) and z–axes respectively; c1, c2 and c3 are and S2 (solid line) forming a binary constants. system with their centre of mass C in uniform motion. (b) The same binary system, with the centre of mass C at rest. move back to back with their centre of mass remaining at rest as shown in Fig.6.13 (b). In many problems on the system of particles, as in the above radioactive decay problem, it is convenient to work in the centre of mass frame rather than in the laboratory frame of reference. (a) (b) In astronomy, binary (double) stars is a common occurrence. If there are no external forces, the centre of mass of a double star Fig. 6.13 (a) A heavy nucleus radium (Ra) splits into moves like a free particle, as shown in Fig.6.14 a lighter nucleus radon (Rn) and an alpha (a). The trajectories of the two stars of equal particle (nucleus of helium atom). The CM mass are also shown in the figure; they look of the system is in uniform motion. complicated. If we go to the centre of mass (b) The same spliting of the heavy nucleus radium (Ra) with the centre of mass at frame, then we find that there the two stars rest. The two product particles fly back are moving in a circle, about the centre of to back. mass, which is at rest. Note that the position of the stars have to be diametrically opposite As an example, let us consider the to each other [Fig. 6.14(b)]. Thus in our frame radioactive decay of a moving unstable particle, of reference, the trajectories of the stars are a combination of (i) uniform motion in a straightlike the nucleus of radium. A radium nucleus line of the centre of mass and (ii) circulardisintegrates into a nucleus of radon and an orbits of the stars about the centre of mass.alpha particle. The forces leading to the decay As can be seen from the two examples,are internal to the system and the external separating the motion of different parts of aforces on the system are negligible. So the total system into motion of the centre of mass andlinear momentum of the system is the same motion about the centre of mass is a verybefore and after decay. The two particles useful technique that helps in understanding produced in the decay, the radon nucleus and the motion of the system. the alpha particle, move in different directions in such a way that their centre of mass moves 6.5 VECTOR PRODUCT OF TWO VECTORS along the same path along which the original decaying radium nucleus was moving We are already familiar with vectors and their [Fig. 6.13(a)]. use in physics. In chapter 5 (Work, Energy, Power) If we observe the decay from the frame of we defined the scalar product of two vectors. An reference in which the centre of mass is at rest, important physical quantity, work, is defined as the motion of the particles involved in the decay a scalar product of two vector quantities, force looks particularly simple; the product particles and displacement. Reprint 2025-26 102 PHYSICS We shall now define another product of two A simpler version of the right hand rule is vectors. This product is a vector. Two important the following : Open up your right hand palm quantities in the study of rotational motion, and curl the fingers pointing from a to b. Your namely, moment of a force and angular stretched thumb points in the direction of c. momentum, are defined as vector products. It should be remembered that there are two angles between any two vectors a and b . In Definition of Vector Product Fig. 6.15 (a) or (b) they correspond to θ(as shown) A vector product of two vectors a and b is a and (3600– θ). While applying either of the above vector c such that rules, the rotation should be taken through the (i) magnitude of c = c = ab sinθ where a and b smaller angle (<1800) between a and b. It is θ are magnitudes of a and b and θ is the here. angle between the two vectors. Because of the cross (×) used to denote the (ii) c is perpendicular to the plane containing vector product, it is also referred to as cross product. a and b. • Note that scalar product of two vectors is (iii) if we take a right handed screw with its head commutative as said earlier, a.b = b.a lying in the plane of a and b and the screw The vector product, however, is not perpendicular to this plane, and if we turn commutative, i.e. a × b ≠ b × a the head in the direction from a to b, then The magnitude of both a × b and b × a is the the tip of the screw advances in the direction same ( ab sin θ ); also, both of them are of c. This right handed screw rule is perpendicular to the plane of a and b. But the illustrated in Fig. 6.15a. rotation of the right-handed screw in case of Alternately, if one curls up the fingers of a × b is from a to b, whereas in case of b × a it right hand around a line perpendicular to the is from b to a. This means the two vectors are plane of the vectors a and b and if the fingers in opposite directions. We have are curled up in the direction from a to b, then a × b = − b × a the stretched thumb points in the direction of • Another interesting property of a vector c, as shown in Fig. 6.15b. product is its behaviour under reflection. Under reflection (i.e. on taking the plane mirror image) we have x →− x , y →−y and z →− z . As a result all the components of a vector change sign and thus a →−a , b →−b . What happens to a × b under reflection? a × b →−( a ) × ( − b ) = a × b Thus, a × b does not change sign under reflection. • Both scalar and vector products are distributive with respect to vector addition. Thus, a.( b + c ) = a.b + a.c a × ( b + c ) = a × b + a × c (a) (b) • We may write c = a × b in the component form. For this we first need to obtain some elementary cross products: Fig. 6.15 (a) Rule of the right handed screw for (i) a × a = 0 (0 is a null vector, i.e. a vector defining the direction of the vector with zero magnitude) product of two vectors. This follows since magnitude of a × a is (b) Rule of the right hand for defining the direction of the vector product. a 2 sin0° = 0 . Reprint 2025-26 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 103 From this follow the results ˆ ˆ ˆ i j k (i) ˆi × ˆi = 0, ˆj × ˆj = 0, kˆ × kˆ = 0 a × b = 3 − 4 5 = 7 ˆi − ˆj − 5 kˆ (ii) ˆi × ˆj = kˆ − 2 1 − 3 Note that the magnitude of ˆi × ˆj is sin900 Note b × a = −7ˆi + ˆj + 5 kˆ ⊳ or 1, since ˆi and ˆj both have unit magnitude and the angle between them is 900. 6.6 ANGULAR VELOCITY AND ITS RELATION WITH LINEAR VELOCITY Thus, ˆi × ˆj is a unit vector. A unit vector In this section we shall study what is angular perpendicular to the plane of ˆi and ˆj and velocity and its role in rotational motion. We related to them by the right hand screw rule is have seen that every particle of a rotating body moves in a circle. The linear velocity of the ˆk . Hence, the above result. You may verify particle is related to the angular velocity. The similarly, relation between these two quantities involves ˆ j × kˆ = ˆi and kˆ × ˆi = ˆj a vector product which we learnt about in the last section. From the rule for commutation of the cross Let us go back to Fig. 6.4. As said above, inproduct, it follows: rotational motion of a rigid body about a fixed ˆ j × ˆi = − kˆ , kˆ × ˆj = − ˆi, ˆi × kˆ = − ˆj axis, every particle of the body moves in a circle, Note if ˆi, ˆj, kˆ occur cyclically in the above vector product relation, the vector product is positive. If ˆi, ˆj, kˆ do not occur in cyclic order, the vector product is negative. Now, a × b = (a x ˆi + a y ˆj + a z kˆ ) × (b x ˆi + b y ˆj + b z kˆ ) = a x b y kˆ − a x b z ˆj − a y b x kˆ + a y b z ˆi + a z b x ˆj − a z b y ˆi = (a y b z − a z b y )i + (a z b x − a x b z ) j + (a x b y − a y b x )k We have used the elementary cross products in obtaining the above relation. The expression for a × b can be put in a determinant form which is easy to remember. ˆ ˆ ˆ i j k a × b = a x a y a z b x b y b z u Example 6.4 Find the scalar and vector Fig. 6.16 Rotation about a fixed axis. (A particle (P) of the rigid body rotating about the fixed products of two vectors. a = (z-) axis moves in a circle with centre (C) and b = on the axis.) Answer which lies in a plane perpendicular to the axis a i b = (3ˆi − 4 ˆj + 5 kˆ )i( − 2ˆi + ˆj − 3 kˆ ) and has its centre on the axis. In Fig. 6.16 we redraw Fig. 6.4, showing a typical particle (at a = −6 − 4 − 15 point P) of the rigid body rotating about a fixed = −25 axis (taken as the z-axis). The particle describes Reprint 2025-26 104 PHYSICS a circle with a centre C on the axis. The radius and points out in the direction in which a right of the circle is r, the perpendicular distance of handed screw would advance, if the head of the the point P from the axis. We also show the screw is rotated with the body. (See Fig. 6.17a). linear velocity vector v of the particle at P. It is The magnitude of this vector is ω = d θ dt along the tangent at P to the circle. referred as above. Let P′ be the position of the particle after an interval of time ∆t (Fig. 6.16). The angle PCP′ describes the angular displacement ∆θ of the particle in time ∆t. The average angular velocity of the particle over the interval ∆t is ∆θ/∆t. As ∆t tends to zero (i.e. takes smaller and smaller values), the ratio ∆θ/∆t approaches a limit which is the instantaneous angular velocity dθ/dt of the particle at the position P. We denote the instantaneous angular velocity by ω (the Greek letter omega). We know from our study Fig. 6.17 (a) If the head of a right handed screw of circular motion that the magnitude of linear rotates with the body, the screw velocity v of a particle moving in a circle is advances in the direction of the angular related to the angular velocity of the particle ω velocity ω. If the sense (clockwise or by the simple relation υ = ωr , where r is the anticlockwise) of rotation of the body changes, so does the direction of ω.radius of the circle. We observe that at any given instant the relation v = ωr applies to all particles of the rigid body. Thus for a particle at a perpendicular distance ri from the fixed axis, the linear velocity at a given instant vi is given by v i = ωri (6.19) The index i runs from 1 to n, where n is the total number of particles of the body. For particles on the axis, r = 0 , and hence v = ω r = 0. Thus, particles on the axis are stationary. This verifies that the axis is fixed. Note that we use the same angular velocity ω for all the particles. We therefore, refer to ω as the angular velocity of the whole body. We have characterised pure translation of a body by all parts of the body having the same Fig. 6.17 (b) The angular velocity vector ω is directed velocity at any instant of time. Similarly, we along the fixed axis as shown. The linear may characterise pure rotation by all parts of velocity of the particle at P is v = ω × r. the body having the same angular velocity at It is perpendicular to both ωωωωω and r and any instant of time. Note that this is directed along the tangent to the circle described by the particle. characterisation of the rotation of a rigid body about a fixed axis is just another way of saying We shall now look at what the vector as in Sec. 6.1 that each particle of the body moves product ω × r corresponds to. Refer to Fig. in a circle, which lies in a plane perpendicular 6.17(b) which is a part of Fig. 6.16 reproduced to the axis and has the centre on the axis. to show the path of the particle P. The figure In our discussion so far the angular velocity shows the vector ω directed along the fixed (z–) appears to be a scalar. In fact, it is a vector. We axis and also the position vector r = OP of the shall not justify this fact, but we shall accept particle at P of the rigid body with respect to it. For rotation about a fixed axis, the angular the origin O. Note that the origin is chosen to velocity vector lies along the axis of rotation, be on the axis of rotation. Reprint 2025-26 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 105 Now ω × r = ω × OP = ω × (OC + CP) If the axis of rotation is fixed, the direction But ω × OC = 00000 as ωωωωω is along OC of ωωωωω and hence, that of α is fixed. In this case Hence ω × r = ω × CP the vector equation reduces to a scalar equation dω α = (6.22) The vector ω × CP is perpendicular to ω, i.e. dt to the z-axis and also to CP, the radius of the circle described by the particle at P. It is 6.7 TORQUE AND ANGULAR MOMENTUM therefore, along the tangent to the circle at P. In this section, we shall acquaint ourselves with Also, the magnitude of ω × CP is ω (CP) since two physical quantities (torque and angular ω and CP are perpendicular to each other. We momentum) which are defined as vector products shall denote CP by ⊥r and not by r, as we did of two vectors. These as we shall see, are earlier. especially important in the discussion of motion Thus, ω × r is a vector of magnitude ωr⊥ of systems of particles, particularly rigid bodies. and is along the tangent to the circle described by the particle at P. The linear velocity vector v 6.7.1 Moment of force (Torque) at P has the same magnitude and direction. We have learnt that the motion of a rigid body, Thus, in general, is a combination of rotation and v = ωωωωω × r (6.20) translation. If the body is fixed at a point or along In fact, the relation, Eq. (6.20), holds good a line, it has only rotational motion. We know even for rotation of a rigid body with one point that force is needed to change the translationalfixed, such as the rotation of the top [Fig. 6.6(a)]. In this case r represents the position vector of state of a body, i.e. to produce linear the particle with respect to the fixed point taken acceleration. We may then ask, what is the as the origin. analogue of force in the case of rotational We note that for rotation about a fixed motion? To look into the question in a concrete axis, the direction of the vector ω does not situation let us take the example of opening or change with time. Its magnitude may, closing of a door. A door is a rigid body which however, change from instant to instant. For can rotate about a fixed vertical axis passing the more general rotation, both the magnitude and the direction of ωωωωω may change through the hinges. What makes the door from instant to instant. rotate? It is clear that unless a force is applied the door does not rotate. But any force does not 6.6.1 Angular acceleration do the job. A force applied to the hinge line You may have noticed that we are developing cannot produce any rotation at all, whereas a the study of rotational motion along the lines force of given magnitude applied at right angles of the study of translational motion with which to the door at its outer edge is most effective in we are already familiar. Analogous to the kinetic producing rotation. It is not the force alone, but variables of linear displacement (s) and velocity how and where the force is applied is important (v) in translational motion, we have angular in rotational motion. displacement (θ) and angular velocity (ω) in The rotational analogue of force in linear rotational motion. It is then natural to define motion is moment of force. It is also referred to in rotational motion the concept of angular as torque or couple. (We shall use the words acceleration in analogy with linear acceleration moment of force and torque interchangeably.) defined as the time rate of change of velocity in We shall first define the moment of force for the translational motion. We define angular special case of a single particle. Later on we acceleration α as the time rate of change of shall extend the concept to systems of particles angular velocity. Thus, including rigid bodies. We shall also relate it to d ω a change in the state of rotational motion, i.e. is α = (6.21) dt angular acceleration of a rigid body. Reprint 2025-26 106 PHYSICS of the line of action of F from the origin and F⊥=( F sin θ) is the component of F in the direction perpendicular to r. Note that τ = 0 if r = 0, F = 0 or θ = 00 or 1800 . Thus, the moment of a force vanishes if either the magnitude of the force is zero, or if the line of action of the force passes through the origin. One may note that since r × F is a vector product, properties of a vector product of two vectors apply to it. If the direction of F is reversed, the direction of the moment of force is reversed. If directions of both r and F are reversed, the direction of the moment of force remains the same. 6.7.2 Angular momentum of a particle Just as the moment of a force is the rotational analogue of force in linear motion, the quantity angular momentum is the rotational analogue Fig. 6.18 τττττ ===== r × F, τττττ is perpendicular to the plane of linear momentum. We shall first define containing r and F, and its direction is angular momentum for the special case of a given by the right handed screw rule. single particle and look at its usefulness in the context of single particle motion. We shall then If a force acts on a single particle at a point extend the definition of angular momentum to P whose position with respect to the origin O is systems of particles including rigid bodies. given by the position vector r (Fig. 6.18), the Like moment of a force, angular momentum moment of the force acting on the particle with is also a vector product. It could also be referred respect to the origin O is defined as the vector to as moment of (linear) momentum. From this product term one could guess how angular momentum τ = r × F (6.23) is defined. The moment of force (or torque) is a vector Consider a particle of mass m and linear quantity. The symbol τττττ stands for the Greek momentum p at a position r relative to the origin letter tau. The magnitude of τττττ is O. The angular momentum l of the particle with τ = r F sinθ (6.24a) respect to the origin O is defined to be l = r × p (6.25a)where r is the magnitude of the position vector r, i.e. the length OP, F is the magnitude of force The magnitude of the angular momentum F and θ is the angle between r and F as vector is shown. l = r p sinθ (6.26a) Moment of force has dimensions M L2 T -2. where p is the magnitude of p and θ is the angle Its dimensions are the same as those of work between r and p. We may write or energy. It is, however, a very different physical l = r p⊥ or r ⊥ p (6.26b)quantity than work. Moment of a force is a vector, while work is a scalar. The SI unit of where r⊥ (= r sinθ) is the perpendicular distance moment of force is newton metre (N m). The of the directional line of p from the origin and magnitude of the moment of force may be p ⊥=( p sin θ) is the component of p in a directionwritten perpendicular to r. We expect the angular τ = (r sin θ)F = r⊥ F (6.24b) momentum to be zero (l = 0), if the linear or τ = r F sin θ = rF ⊥ (6.24c) momentum vanishes (p = 0), if the particle is at the origin (r = 0), or if the directional line of p where r⊥ = r sinθ is the perpendicular distance passes through the origin θ = 00 or 1800. Reprint 2025-26 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 107 The physical quantities, moment of a force and angular momentum, have an important An experiment with the bicycle rim relation between them. It is the rotational Take a analogue of the relation between force and linear bicycle rim momentum. For deriving the relation in the and extend context of a single particle, we differentiate its axle on l = r × p with respect to time, both sides. Tie two d l d = ( r × p ) s t r i n g s d t d t at both ends Applying the product rule for differentiation A and B, to the right hand side, as shown in the d d r d p ( r × p ) = × p + r × a d j o i n i n g d t d t d t figure. Hold Now, the velocity of the particle is v = dr/dt both the and p = m v Initially After s t r i n g s together in dr one hand such that the rim is vertical. If you Because of this × p = v × m v = 0, dt leave one string, the rim will tilt. Now keeping the rim in vertical position with both the stringsas the vector product of two parallel vectors in one hand, put the wheel in fast rotation vanishes. Further, since dp / dt = F, around the axle with the other hand. Then leave d p one string, say B, from your hand, and observe r × = r × F = t dt what happens. The rim keeps rotating in a vertical plane d and the plane of rotation turns around the string Hence ( r × p ) = τ A which you are holding. We say that the axis dt of rotation of the rim or equivalently or (6.27) its angular momentum precesses about the string A. Thus, the time rate of change of the angular The rotating rim gives rise to an angular momentum of a particle is equal to the torque momentum. Determine the direction of this acting on it. This is the rotational analogue of angular momentum. When you are holding the the equation F = dp/dt, which expresses rotating rim with string A, a torque is generated. (We leave it to you to find out how the torque isNewton’s second law for the translational motion generated and what its direction is.) The effect of a single particle. of the torque on the angular momentum is to make it precess around an axis perpendicular Torque and angular momentum for a system to both the angular momentum and the torque. of particles Verify all these statements. To get the total angular momentum of a system of particles about a given point we need to add vectorially the angular momenta of individual particles. Thus, for a system of n particles, particle has mass mi and velocity vi) We may write the total angular momentum of a system of particles as (6.25b) The angular momentum of the ith particle is given by li = ri × pi This is a generalisation of the definition of angular momentum (Eq. 6.25a) for a singlewhere ri is the position vector of the ith particle particle to a system of particles.with respect to a given origin and p = (mivi) is Using Eqs. (6.23) and (6.25b), we getthe linear momentum of the particle. (The Reprint 2025-26 108 PHYSICS d L d d l Note that like Eq.(6.17), Eq.(6.28b) holds = ( l ) = ∑ = ∑ τ (6.28a) good for any system of particles, whether it is a d t d t i d t i rigid body or its individual particles have all where τi is the torque acting on the ith particle; kinds of internal motion. τi = ri × Fi Conservation of angular momentum The force Fi on the ith particle is the vector ext If τext = 0, Eq. (6.28b) reduces to Fi sum of external forces acting on the particle d L = 0 and the internal forces iFint exerted on it by the dt other particles of the system. We may therefore or L = constant. (6.29a) separate the contribution of the external and Thus, if the total external torque on a system the internal forces to the total torque of particles is zero, then the total angular momentum of the system is conserved, i.e. τ = ∑ τ i = ∑ ri × Fi as remains constant. Eq. (6.29a) is equivalent to i i three scalar equations, τ = τext + τ int , Lx = K1, Ly = K2 and Lz = K3 (6.29 b) Here K1, K2 and K3 are constants; Lx, Ly and τ ext = ∑ri × Fi ext Lz are the components of the total angular where i momentum vector L along the x,y and z axes respectively. The statement that the total i × Fiint τ int = ∑r and angular momentum is conserved means that i each of these three components is conserved. We shall assume not only Newton’s third law Eq. (6.29a) is the rotational analogue of of motion, i.e. the forces between any two particles Eq. (6.18a), i.e. the conservation law of the total of the system are equal and opposite, but also that linear momentum for a system of particles. these forces are directed along the line joining the Like Eq. (6.18a), it has applications in many two particles. In this case the contribution of the practical situations. We shall look at a few of internal forces to the total torque on the system is the interesting applications later on in zero, since the torque resulting from each action- this chapter. reaction pair of forces is zero. We thus have, τint = 0 and therefore τ = τττext.ττ u Example 6.5 Find the torque of a force Since τ = ∑ τ i , it follows from Eq. (6.28a) + – about the origin. The force acts on a particle whose position vector is .that d L = τ ext (6.28 b) Answer Here r = ˆi − ˆj + kˆ d t and F = 7 ˆi + 3 ˆj − 5 kˆ . Thus, the time rate of the total angular We shall use the determinant rule to find themomentum of a system of particles about a τ = r × Fpoint (taken as the origin of our frame of torque reference) is equal to the sum of the external torques (i.e. the torques due to external forces) acting on the system taken about the same point. Eq. (6.28 b) is the generalisation of the single particle case of Eq. (6.23) to a system of particles. Note that when we have only one or ⊳ particle, there are no internal forces or torques. Eq.(6.28 b) is the rotational analogue of Example 6.6 Show that the angular u momentum about any point of a single d P = Fext (6.17) particle moving with constant velocity d t remains constant throughout the motion. Reprint 2025-26 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 109 Answer Let the particle with velocity v be at acceleration nor angular acceleration. This means point P at some instant t. We want to calculate (1) the total force, i.e. the vector sum of the the angular momentum of the particle about an forces, on the rigid body is zero; arbitrary point O. n F1 + F2 + ... + Fn = =∑i 1 Fi = 0 (6.30a) If the total force on the body is zero, then the total linear momentum of the body does not change with time. Eq. (6.30a) gives the condition for the translational equilibrium of the body. (2) The total torque, i.e. the vector sum of the torques on the rigid body is zero, n τ + τ i = 0 (6.30b) 1 2 + ... + τ n = =∑i 1 τ Fig 6.19 If the total torque on the rigid body is zero, The angular momentum is l = r × mv. Its the total angular momentum of the body does magnitude is mvr sinθ, where θ is the angle not change with time. Eq. (6.30 b) gives the between r and v as shown in Fig. 6.19. Although condition for the rotational equilibrium of the the particle changes position with time, the line body. of direction of v remains the same and hence One may raise a question, whether theOM = r sin θ. is a constant. rotational equilibrium condition [Eq. 6.30(b)] Further, the direction of l is perpendicular remains valid, if the origin with respect to whichto the plane of r and v. It is into the page of the the torques are taken is shifted. One can showfigure.This direction does not change with time. Thus, l remains the same in magnitude and that if the translational equilibrium condition direction and is therefore conserved. Is there [Eq. 6.30(a)] holds for a rigid body, then such a any external torque on the particle? ⊳ shift of origin does not matter, i.e. the rotational equilibrium condition is independent of the 6.8 EQUILIBRIUM OF A RIGID BODY location of the origin about which the torques are taken. Example 6.7 gives a proof of this result We are now going to concentrate on the motion in a special case of a couple, i.e. two forcesof rigid bodies rather than on the motion of acting on a rigid body in translationalgeneral systems of particles. equilibrium. The generalisation of this result to We shall recapitulate what effect the external forces have on a rigid body. (Henceforth n forces is left as an exercise. we shall omit the adjective ‘external’ because Eq. (6.30a) and Eq. (6.30b), both, are vector unless stated otherwise, we shall deal with only equations. They are equivalent to three scalar external forces and torques.) The forces change equations each. Eq. (6.30a) corresponds to the translational state of the motion of the rigid n n n body, i.e. they change its total linear =∑i 1 Fix = 0 , =∑i 1 Fiy = 0 and =∑i 1 Fiz = 0 (6.31a)momentum in accordance with Eq. (6.17). But this is not the only effect the forces have. The where Fix, Fiy and Fiz are respectively the x, y and total torque on the body may not vanish. Such z components of the forces Fi. Similarly, Eq. a torque changes the rotational state of motion (6.30b) is equivalent to three scalar equations of the rigid body, i.e. it changes the total angular n n 0momentum of the body in accordance with τix = 0 , τiy = and (6.31b) =∑i 1 =∑i 1 Eq. (6.28 b). where τix, τiy and τiz are respectively the x, y and A rigid body is said to be in mechanical z components of the torque τi .equilibrium, if both its linear momentum and Eq. (6.31a) and (6.31b) give six independentangular momentum are not changing with time, conditions to be satisfied for mechanicalor equivalently, the body has neither linear Reprint 2025-26 110 PHYSICS equilibrium of a rigid body. In a number of problems all the forces acting on the body are coplanar. Then we need only three conditions to be satisfied for mechanical equilibrium. Two of these conditions correspond to translational equilibrium; the sum of the components of the forces along any two perpendicular axes in the plane must be zero. The third condition corresponds to rotational equilibrium. The sum of the components of the torques along any axis Fig. 6.20 (b) perpendicular to the plane of the forces must be zero. The force at B in Fig. 6.20(a) is reversed in Fig. 6.20(b). Thus, we have the same rod with The conditions of equilibrium of a rigid body two forces of equal magnitude but acting inmay be compared with those for a particle, which opposite diretions applied perpendicular to the we considered in earlier chapters. Since rod, one at end A and the other at end B. Here consideration of rotational motion does not apply the moments of both the forces are equal, but to a particle, only the conditions for translational they are not opposite; they act in the same sense equilibrium (Eq. 6.30 a) apply to a particle. Thus, and cause anticlockwise rotation of the rod. The for equilibrium of a particle the vector sum of total force on the body is zero; so the body is in all the forces on it must be zero. Since all these translational equilibrium; but it is not in forces act on the single particle, they must be rotational equilibrium. Although the rod is not fixed in any way, it undergoes pure rotation (i.e.concurrent. Equilibrium under concurrent rotation without translation).forces was discussed in the earlier chapters. A pair of forces of equal magnitude but acting A body may be in partial equilibrium, i.e., it in opposite directions with different lines of may be in translational equilibrium and not in action is known as a couple or torque. A couple rotational equilibrium, or it may be in rotational produces rotation without translation. equilibrium and not in translational When we open the lid of a bottle by turning equilibrium. it, our fingers are applying a couple to the lid Consider a light (i.e. of negligible mass) rod [Fig. 6.21(a)]. Another known example is a compass needle in the earth’s magnetic field as(AB) as shown in Fig. 6.20(a). At the two ends (A shown in the Fig. 6.21(b). The earth’s magneticand B) of which two parallel forces, both equal field exerts equal forces on the north and southin magnitude and acting along same direction poles. The force on the North Pole is towards are applied perpendicular to the rod. the north, and the force on the South Pole is toward the south. Except when the needle points in the north-south direction; the two forces do not have the same line of action. Thus there is a couple acting on the needle due to the earth’s magnetic field. Fig. 6.20 (a) Let C be the midpoint of AB, CA = CB = a. the moment of the forces at A and B will both be equal in magnitude (aF ), but opposite in sense as shown. The net moment on the rod will be zero. The system will be in rotational equilibrium, but it will not be in translational fingers apply a couple to turnequilibrium; F ≠ 0 Fig. 6.21(a) Our ∑ the lid. Reprint 2025-26 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 111 length. This point is called the fulcrum. A see- saw on the children’s playground is a typical example of a lever. Two forces F1 and F2, parallel to each other and usually perpendicular to the lever, as shown here, act on the lever at distances d1 and d2 respectively from the fulcrum as shown in Fig. 6.23. Fig. 6.21(b) The Earth’s magnetic field exerts equal and opposite forces on the poles of a Fig. 6.23 compass needle. These two forces form a couple. The lever is a system in mechanical equilibrium. Let R be the reaction of the supportu Example 6.7 Show that moment of a at the fulcrum; R is directed opposite to the couple does not depend on the point about forces F1 and F2. For translational equilibrium, which you take the moments. R – F1 – F2 = 0 (i) Answer For considering rotational equilibrium we take the moments about the fulcrum; the sum of moments must be zero, d1F1 – d2F2 = 0 (ii) Normally the anticlockwise (clockwise) moments are taken to be positive (negative). Note R acts at the fulcrum itself and has zero moment about the fulcrum. Fig. 6.22 In the case of the lever force F1 is usually Consider a couple as shown in Fig. 6.22 some weight to be lifted. It is called the load and acting on a rigid body. The forces F and -F act its distance from the fulcrum d1 is called the respectively at points B and A. These points have load arm. Force F2 is the effort applied to lift the position vectors r1 and r2 with respect to origin load; distance d2 of the effort from the fulcrum O. Let us take the moments of the forces about is the effort arm. the origin. Eq. (ii) can be written as The moment of the couple = sum of the d1F1 = d2 F2 (6.32a) moments of the two forces making the couple or load arm × load = effort arm × effort = r1 × (–F) + r2 × F The above equation expresses the principle = r2 × F – r1 × F of moments for a lever. Incidentally the ratio = (r2–r1) × F F1/F2 is called the Mechanical Advantage (M.A.); But r1 + AB = r2, and hence AB = r2 – r1. F1 d 2 The moment of the couple, therefore, is M.A. = = (6.32b) F2 d1AB × F. Clearly this is independent of the origin, the If the effort arm d2 is larger than the load point about which we took the moments of the arm, the mechanical advantage is greater than forces. ⊳ one. Mechanical advantage greater than one means that a small effort can be used to lift a 6.8.1 Principle of moments large load. There are several examples of a lever An ideal lever is essentially a light (i.e. of around you besides the see-saw. The beam of a negligible mass) rod pivoted at a point along its balance is a lever. Try to find more such Reprint 2025-26 112 PHYSICS examples and identify the fulcrum, the effort and The CG of the cardboard is so located that effort arm, and the load and the load arm of the the total torque on it due to the forces m1g, m2g lever in each case. …. etc. is zero. You may easily show that the principle of If ri is the position vector of the ith particle moment holds even when the parallel forces F1 of an extended body with respect to its CG, then and F2 are not perpendicular, but act at some the torque about the CG, due to the force of angle, to the lever. gravity on the particle is τi = ri × mi g. The total gravitational torque about the CG is zero, i.e. 6.8.2 Centre of gravity i × m i g = 0 (6.33) τ g = ∑ τ i = ∑r Many of you may have the experience of We may therefore, define the CG of a body balancing your notebook on the tip of a finger. as that point where the total gravitational torque Figure 6.24 illustrates a similar experiment that on the body is zero. you can easily perform. Take an irregular- We notice that in Eq. (6.33), g is the same shaped cardboard having mass M and a narrow for all particles, and hence it comes out of the tipped object like a pencil. You can locate by trial summation. This gives, since g is non-zero, and error a point G on the cardboard where it ir = 0. Remember that the position vectorscan be balanced on the tip of the pencil. (The ∑mi cardboard remains horizontal in this position.) (ri) are taken with respect to the CG. Now, in This point of balance is the centre of gravity (CG) accordance with the reasoning given below of the cardboard. The tip of the pencil provides Eq. (6.4a) in Sec. 6.2, if the sum is zero, the origin a vertically upward force due to which the must be the centre of mass of the body. Thus, cardboard is in mechanical equilibrium. As the centre of gravity of the body coincides with shown in the Fig. 6.24, the reaction of the tip is the centre of mass in uniform gravity or gravity- equal and opposite to Mg and hence the cardboard is in translational equilibrium. It is also in rotational equilibrium; if it were not so, due to the unbalanced torque it would tilt and fall. There are torques on the card board due to the forces of gravity like m1g, m2g …. etc, acting on the individual particles that make up the cardboard. Fig. 6.25 Determining the centre of gravity of a body Fig. 6.24 Balancing a cardboard on the tip of a of irregular shape. The centre of gravity G pencil. The point of support, G, is the lies on the vertical AA1 through the point centre of gravity. of suspension of the body A. Reprint 2025-26 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 113 free space. We note that this is true because = 30 cm, PG = 5 cm, AK1= BK2 = 10 cm and K1G = the body being small, g does not K2G = 25 cm. Also, W= weight of the rod = 4.00 vary from one point of the body to the other. If kg and W 1= suspended load = 6.00 kg; the body is so extended that g varies from part R1 and R2 are the normal reactions of the to part of the body, then the centre of gravity support at the knife edges. and centre of mass will not coincide. Basically, For translational equilibrium of the rod, the two are different concepts. The centre of R1+R2 –W1 –W = 0 (i) mass has nothing to do with gravity. It depends Note W1 and W act vertically down and R1 only on the distribution of mass of the body. and R2 act vertically up. In Sec. 6.2 we found out the position of the For considering rotational equilibrium, we centre of mass of several regular, homogeneous take moments of the forces. A convenient point objects. Obviously the method used there gives to take moments about is G. The moments of us also the centre of gravity of these bodies, if R2 and W1 are anticlockwise (+ve), whereas the they are small enough. moment of R1 is clockwise (-ve). Figure 6.25 illustrates another way of For rotational equilibrium, determining the CG of an irregular shaped body –R1 (K1G) + W1 (PG) + R2 (K2G) = 0 (ii) like a cardboard. If you suspend the body from It is given that W = 4.00g N and W1 = 6.00g some point like A, the vertical line through A N, where g = acceleration due to gravity. We passes through the CG. We mark the vertical take g = 9.8 m/s2. AA1. We then suspend the body through other With numerical values inserted, from (i) points like B and C. The intersection of the R1 + R2 – 4.00g – 6.00g = 0 verticals gives the CG. Explain why the method or R1 + R2 = 10.00g N (iii) works. Since the body is small enough, the = 98.00 N method allows us to determine also its centre From (ii), – 0.25 R1 + 0.05 W1 + 0.25 R2 = 0 of mass. or R1 – R2 = 1.2g N = 11.76 N (iv) From (iii) and (iv), R1 = 54.88 N, u Example 6.8 A metal bar 70 cm long and R2 = 43.12 N 4.00 kg in mass supported on two knife- Thus the reactions of the support are about edges placed 10 cm from each end. A 6.00 55 N at K1 and 43 N at K2. ⊳ kg load is suspended at 30 cm from one end. Find the reactions at the knife-edges. u Example 6.9 A 3m long ladder weighing (Assume the bar to be of uniform cross 20 kg leans on a frictionless wall. Its feet section and homogeneous.) rest on the floor 1 m from the wall as shown in Fig.6.27. Find the reaction forces of the Answer wall and the floor. Answer Fig. 6.26 Figure 6.26 shows the rod AB, the positions of the knife edges K1 and K2 , the centre of gravity of the rod at G and the suspended load at P. Note the weight of the rod W acts at its centre of gravity G. The rod is uniform in cross section and homogeneous; hence G is at the centre of the rod; AB = 70 cm. AG = 35 cm, AP Fig. 6.27 Reprint 2025-26 114 PHYSICS The ladder AB is 3 m long, its foot A is at from the axis, the linear velocity is υi = ir ω. The distance AC = 1 m from the wall. From kinetic energy of motion of this particle is Pythagoras theorem, BC = 2 2 m. The forces 1 2 1 2 2 on the ladder are its weight W acting at its centre k i = m i υi = m i ri ω 2 2 of gravity D, reaction forces F1 and F2 of the wall where mi is the mass of the particle. The totaland the floor respectively. Force F1 is kinetic energy K of the body is then given byperpendicular to the wall, since the wall is the sum of the kinetic energies of individualfrictionless. Force F2 is resolved into two particles,components, the normal reaction N and the force of friction F. Note that F prevents the ladder n 1 n 2 2 from sliding away from the wall and is therefore K = ∑ k i = ∑ (m i ri ω ) i =1 2 i =1 directed toward the wall. For translational equilibrium, taking the Here n is the number of particles in the body. forces in the vertical direction, Note ωis the same for all particles. Hence, taking N – W = 0 (i) ω out of the sum, Taking the forces in the horizontal direction, n 1 2 2 i ri ) F – F1 = 0 (ii) K = 2 ω ( ∑i =1 m For rotational equilibrium, taking the We define a new parameter characterisingmoments of the forces about A, the rigid body, called the moment of inertia I , 2 2 F1 −(1/2) W = 0 (iii) given by Now W = 20 g = 20 × 9.8 N = 196.0 N n 2 I = ∑ m i ri (6.34)From (i) N = 196.0 N i =1 With this definition,From (iii) F1 = W 4 2 = 196.0/4 2 = 34.6 N 1 2 From (ii) F = F1 = 34.6 N K = Iω (6.35) 2 2 2 Note that the parameter I is independent of F2 = F + N = 199.0 N the magnitude of the angular velocity. It is a The force F2 makes an angle α with the characteristic of the rigid body and the axis horizontal, about which it rotates. −1 Compare Eq. (6.35) for the kinetic energy oftan α = N F = 4 2 , α = tan (4 2) ≈ 80 ⊳ a rotating body with the expression for the kinetic energy of a body in linear (translational)6.9 MOMENT OF INERTIA motion, We have already mentioned that we are 1 2developing the study of rotational motion parallel K = m υ 2to the study of translational motion with which Here, m is the mass of the body and v is itswe are familiar. We have yet to answer one major velocity. We have already noted the analogy question in this connection. What is the between angular velocity ω (in respect of analogue of mass in rotational motion? We shall rotational motion about a fixed axis) and linear attempt to answer this question in the present velocity v (in respect of linear motion). It is then section. To keep the discussion simple, we shall evident that the parameter, moment of inertia consider rotation about a fixed axis only. Let us I, is the desired rotational analogue of mass in try to get an expression for the kinetic energy of linear motion. In rotation (about a fixed axis), a rotating body. We know that for a body rotating the moment of inertia plays a similar role as about a fixed axis, each particle of the body moves mass does in linear motion. We now apply the definition Eq. (6.34), toin a circle with linear velocity given by Eq. (6.19). calculate the moment of inertia in two simple cases.(Refer to Fig. 6.16). For a particle at a distance Reprint 2025-26 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 115 (a) Consider a thin ring of radius R and mass change in its rotational motion, it can be M, rotating in its own plane around its centre regarded as a measure of rotational inertia of with angular velocity ω. Each mass element the body; it is a measure of the way in which of the ring is at a distance R from the axis, different parts of the body are distributed at and moves with a speed Rω. The kinetic different distances from the axis. Unlike the energy is therefore, mass of a body, the moment of inertia is not a fixed quantity but depends on distribution of 1 2 1 2 2 K = M υ = MR ω mass about the axis of rotation, and the 2 2 orientation and position of the axis of rotation Comparing with Eq. (6.35) we get I = MR 2 with respect to the body as a whole. As a for the ring. measure of the way in which the mass of a rotating rigid body is distributed with respect to the axis of rotation, we can define a new parameter, the radius of gyration. It is related to the moment of inertia and the total mass of the body. Notice from the Table 6.1 that in all cases, we can write I = Mk2, where k has the dimension of length. For a rod, about the perpendicular axis at its midpoint, k 2 = L2 12, i.e. k = L 12 . Similarly, k = R/2 for the circular disc about its diameter. The length k is a geometric property of the body and axis of rotation. It is called the radius of Fig. 6.28 A light rod of length l with a pair of gyration. The radius of gyration of a body masses rotating about an axis through about an axis may be defined as the distance the centre of mass of the system and perpendicular to the rod. The total mass from the axis of a mass point whose mass is of the system is M. equal to the mass of the whole body and whose moment of inertia is equal to the moment of (b) Next, take a rigid rod of negligible mass of inertia of the body about the axis. length of length l with a pair of small masses, Thus, the moment of inertia of a rigid body rotating about an axis through the centre of depends on the mass of the body, its shape and mass perpendicular to the rod (Fig. 6.28). size; distribution of mass about the axis of Each mass M/2 is at a distance l/2 from rotation, and the position and orientation of the the axis. The moment of inertia of the masses axis of rotation. is therefore given by From the definition, Eq. (6.34), we can infer (M/2) (l/2)2 + (M/2)(l/2)2 that the dimensions of moments of inertia are Thus, for the pair of masses, rotating about ML2 and its SI units are kg m2. the axis through the centre of mass The property of this extremely important perpendicular to the rod 2 quantity I, as a measure of rotational inertia of I = Ml / 4 the body, has been put to a great practical use. Table 6.1 simply gives the moment of inertia of The machines, such as steam engine and thevarious familiar regular shaped bodies about automobile engine, etc., that produce rotationalspecific axes. (The derivations of these motion have a disc with a large moment ofexpressions are beyond the scope of this inertia, called a flywheel. Because of its largetextbook and you will study them in higher classes.) moment of inertia, the flywheel resists the As the mass of a body resists a change in its sudden increase or decrease of the speed of the state of linear motion, it is a measure of its inertia vehicle. It allows a gradual change in the speed in linear motion. Similarly, as the moment of and prevents jerky motions, thereby ensuring inertia about a given axis of rotation resists a a smooth ride for the passengers on the vehicle. Reprint 2025-26 116 PHYSICS Table 6.1 Moments of inertia of some regular shaped bodies about specific axes Z Body Axis Figure I (1) Thin circular Perpendicular to M R 2 ring, radius R plane, at centre (2) Thin circular Diameter M R2/2 ring, radius R (3) Thin rod, Perpendicular to M L2/12 length L rod, at mid point (4) Circular disc, Perpendicular to M R2/2 radius R disc at centre (5) Circular disc, Diameter M R2/4 radius R (6) Hollow cylinder, Axis of cylinder M R2 radius R (7) Solid cylinder, Axis of cylinder M R2/2 radius R (8) Solid sphere, Diameter 2 M R2/5 radius R 6.10 KINEMATICS OF ROTATIONAL MOTION translation. We wish to take this analogy further. ABOUT A FIXED AXIS In doing so we shall restrict the discussion only We have already indicated the analogy between to rotation about fixed axis. This case of motion rotational motion and translational motion. For involves only one degree of freedom, i.e., needs example, the angular velocity ω plays the same only one independent variable to describe the role in rotation as the linear velocity v in motion. This in translation corresponds to linear Reprint 2025-26 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 117 motion. This section is limited only to kinematics. We shall turn to dynamics in later sections. We recall that for specifying the angular displacement of the rotating body we take any particle like P (Fig.6.29) of the body. Its angular displacement θ in the plane it moves is the angular displacement of the whole body; θ is measured from a fixed direction in the plane of motion of P, which we take to be the x′-axis, chosen parallel to the x-axis. Note, as shown, the axis of rotation is the z – axis and the plane of the motion of the particle is the x - y plane. Fig. 6.29 also shows θ0, the angular displacement at t = 0. We also recall that the angular velocity is the time rate of change of angular displacement, ω = dθ/dt. Note since the axis of rotation is fixed, there is no need to treat angular velocity as a Fig.6.29 Specifying the angular position of a rigid vector. Further, the angular acceleration, α = body. dω/dt. u Example 6.10 Obtain Eq. (6.36) from first The kinematical quantities in rotational principles. motion, angular displacement (θ), angular velocity (ω) and angular acceleration (α) Answer The angular acceleration is uniform, respectively are analogous to kinematic hence quantities in linear motion, displacement (x), dω velocity (v) and acceleration (a). We know the = α = constant (i) kinematical equations of linear motion with d t uniform (i.e. constant) acceleration: Integrating this equation, α dt + c v = v0 + at (a) ω = ∫ 1 2 x = x 0 + υ0t + at (b) = αt + c (as α is constant) 2 At t = 0, ω = ω0 (given) 2 2 υ = υ0 + 2ax (c) From (i) we get at t = 0, ω = c = ω0 Thus, ω = αt + ω0 as required. where x0 = initial displacement and v0= initial With the definition of ω = dθ/dt we may velocity. The word ‘initial’ refers to values of the integrate Eq. (6.36) to get Eq. (6.37). This quantities at t = 0 derivation and the derivation of Eq. (6.38) is left The corresponding kinematic equations for as an exercise. rotational motion with uniform angular acceleration are: u Example 6.11 The angular speed of a motor wheel is increased from 1200 rpm to ω= ω0 + αt (6.36) 3120 rpm in 16 seconds. (i) What is its angular acceleration, assuming the 1 2 θ = θ0 + ω0t + αt (6.37) acceleration to be uniform? (ii) How many 2 revolutions does the engine make during and ω2 = ω0 2 + 2α(θ– θ0 ) (6.38) this time? Answer where θ0= initial angular displacement of the (i) We shall use ω = ω0 + αt rotating body, and ω0 = initial angular velocity ω0 = initial angular speed in rad/s of the body. Reprint 2025-26 118 PHYSICS = 2π × angular speed in rev/s It is, however, necessary that these correspondences are established on sound 2π × angular speed in rev/min dynamical considerations. This is what we now = 60 s/min turn to. Before we begin, we note a simplification 2π × 1200 that arises in the case of rotational motion = rad/s 60 about a fixed axis. Since the axis is fixed, only those components of torques, which are along = 40π rad/s the direction of the fixed axis need to be Similarly ω = final angular speed in rad/s considered in our discussion. Only these 2π × 3120 components can cause the body to rotate about = rad/s the axis. A component of the torque 60 perpendicular to the axis of rotation will tend to = 2π × 52 rad/s turn the axis from its position. We specifically = 104 π rad/s assume that there will arise necessary forces of constraint to cancel the effect of the ∴Angular acceleration perpendicular components of the (external) torques, so that the fixed position of the axis ω − ω will be maintained. The perpendicular α = 0 = 4 π rad/s2 t components of the torques, therefore need not be taken into account. This means that for our The angular acceleration of the engine calculation of torques on a rigid body: = 4π rad/s2 (1) We need to consider only those forces that (ii) The angular displacement in time t is lie in planes perpendicular to the axis. given by Forces which are parallel to the axis will give torques perpendicular to the axis and need 1 2 θ = ω0 t + αt not be taken into account. 2 (2) We need to consider only those components 1 2 of the position vectors which are = (40π × 16 + × 4π × 16 ) rad 2 perpendicular to the axis. Components of position vectors along the axis will result in = (640π + 512π) rad torques perpendicular to the axis and need = 1152π rad not be taken into account. 1152π = 576 ⊳ Work done by a torqueNumber of revolutions = 2π

6.12 Angular momentum in case solved by considering them to be rigid bodies. Ideally a of rotation about a fixed axis rigid body is a body with a perfectly definite and unchanging shape. The distances between all pairs of Summary particles of such a body do not change. It is evident from Points to Ponder this definition of a rigid body that no real body is truly rigid, Exercises since real bodies deform under the influence of forces. But in many situations the deformations are negligible. In a number of situations involving bodies such as wheels, tops, steel beams, molecules and planets on the other hand, we can ignore that they warp (twist out of shape), bend or vibrate and treat them as rigid. 6.1.1 What kind of motion can a rigid body have? Let us try to explore this question by taking some examples of the motion of rigid bodies. Let us begin with a rectangular Reprint 2025-26 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 93 most common way to constrain a rigid body so that it does not have translational motion is to fix it along a straight line. The only possible motion of such a rigid body is rotation. The line or fixed axis about which the body is rotating is its axis of rotation. If you look around, you will come across many examples of rotation about an axis, a ceiling fan, a potter’s wheel, a giant wheel in a fair, a merry-go-round and so on (Fig Fig 6.1 Translational (sliding) motion of a block down 6.3(a) and (b)). an inclined plane. (Any point like P1 or P2 of the block moves with the same velocity at any instant of time.) block sliding down an inclined plane without any sidewise movement. The block is taken as a rigid body. Its motion down the plane is such that all the particles of the body are moving together, i.e. they have the same velocity at any instant of time. The rigid body here is in pure translational motion (Fig. 6.1). In pure translational motion at any instant of time, all particles of the body have the same velocity. Consider now the rolling motion of a solid metallic or wooden cylinder down the same (a) inclined plane (Fig. 6.2). The rigid body in this problem, namely the cylinder, shifts from the top to the bottom of the inclined plane, and thus, seems to have translational motion. But as Fig.

6.2 Use Lenz’s law to determine the direction of induced current in the situations described by Fig. 6.16: (a) A wire of irregular shape turning into a circular shape; 175 Reprint 2025-26 Physics (b) A circular loop being deformed into a narrow straight wire. FIGURE 6.16 6.3 A long solenoid with 15 turns per cm has a small loop of area 2.0 cm2 placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, what is the induced emf in the loop while the current is changing? 6.4 A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of uniform magnetic field of magnitude 0.3 T directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is 1 cm s–1 in a direction normal to the (a) longer side, (b) shorter side of the loop? For how long does the induced voltage last in each case? 6.5 A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad s–1 about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5 T parallel to the axis exists everywhere. Calculate the emf developed between the centre and the ring. 6.6 A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5.0 m s–1, at right angles to the horizontal component of the earth’s magnetic field, 0.30 ´ 10–4 Wb m–2. (a) What is the instantaneous value of the emf induced in the wire? (b) What is the direction of the emf? (c) Which end of the wire is at the higher electrical potential? 6.7 Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V induced, give an estimate of the self-inductance of the circuit. 6.8 A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil? Reprint 2025-26 Chapter Seven ALTERNATING CURRENT 7.1 INTRODUCTION We have so far considered direct current (dc) sources and circuits with dc sources. These currents do not change direction with time. But voltages and currents that vary with time are very common. The electric mains supply in our homes and offices is a voltage that varies like a sine function with time. Such a voltage is called alternating voltage (ac voltage) and the current driven by it in a circuit is called the alternating current (ac current)*. Today, most of the electrical devices we use require ac voltage. This is mainly because most of the electrical energy sold by power companies is transmitted and distributed as alternating current. The main reason for preferring use of ac voltage over dc voltage is that ac voltages can be easily and efficiently converted from one voltage to the other by means of transformers. Further, electrical energy can also be transmitted economically over long distances. AC circuits exhibit characteristics which are exploited in many devices of daily use. For example, whenever we tune our radio to a favourite station, we are taking advantage of a special property of ac circuits – one of many that you will study in this chapter. * The phrases ac voltage and ac current are contradictory and redundant, respectively, since they mean, literally, alternating current voltage and alternating current current. Still, the abbreviation ac to designate an electrical quantity displaying simple harmonic time dependance has become so universally accepted that we follow others in its use. Further, voltage – another phrase commonly used means potential difference between two points. Reprint 2025-26 Physics 7.2 AC VOLTAGE APPLIED TO A RESISTOR Figure 7.1 shows a resistor connected to a source ε of ac voltage. The symbol for an ac source in a circuit diagram is . We consider a source which produces sinusoidally varying potential difference across its terminals. Let this potential difference, also called ac voltage, be given by v = vm sin ωt (7.1) where vm is the amplitude of the oscillating potential difference and ω is its angular frequency. Nicola Tesla (1856 – 1943) Serbian-American scientist, inventor and genius. He conceived the idea of the rotating1943) magnetic field, which is the – basis of practically all alternating current machinery, and which(1856 helped usher in the age of FIGURE 7.1 AC voltage applied to a resistor. electric power. He also invented among other To find the value of current through the resistor, we things the induction motor, ε()t = 0 (refer to Section the polyphase system of ac apply Kirchhoff’s loop rule ∑TESLA power, and the high 3.12), to the circuit shown in Fig. 7.1 to get frequency induction coil v m sin ωt = i R (the Tesla coil) used in radio and television sets and v m i = sin ωtNICOLA other electronic equipment. or R The SI unit of magnetic field is named in his honour. Since R is a constant, we can write this equation as i = i m sin ωt (7.2) where the current amplitude im is given by v m i m = (7.3) R Equation (7.3) is Ohm’s law, which for resistors, works equally well for both ac and dc voltages. The voltage across a pure resistor and the current through it, given by Eqs. (7.1) and (7.2) are plotted as a function of time in Fig. 7.2. Note, in particular that both v and i reach zero, minimum and maximum values at the same time. Clearly, the voltage and current are in phase with FIGURE 7.2 In a pure resistor, the voltage and each other. current are in phase. The We see that, like the applied voltage, the current varies minima, zero and maxima sinusoidally and has corresponding positive and negative values occur at the same during each cycle. Thus, the sum of the instantaneous current respective times. values over one complete cycle is zero, and the average current 178 is zero. The fact that the average current is zero, however, does Reprint 2025-26 Alternating Current not mean that the average power consumed is zero and that there is no dissipation of electrical energy. As you know, Joule heating is given by i2R and depends on i2 (which is always positive whether i is positive or negative) and not on i. Thus, there is Joule heating and dissipation of electrical energy when an ac current passes through a resistor. The instantaneous power dissipated in the resistor is p = i 2 R = i m2 R sin 2 ωt (7.4) The average value of p over a cycle is* p = < i 2 R > = < i m2 R sin 2 ωt > [7.5(a)] where the bar over a letter (here, p) denotes its average George Westinghouse value and <......> denotes taking average of the quantity (1846 – 1914) A leading inside the bracket. Since, i2m and R are constants, proponent of the use of p = i m2 R < sin 2 ωt > [7.5(b)] alternating current overUsing the trigonometric identity, sin2 wt = direct current. Thus, GEORGE he came into conflict 1/2 (1– cos 2wt), we have < sin2 wt > = (1/2) (1– < cos 2wt >) with Thomas Alva Edison, and since < cos2wt > = 0**, we have, an advocate of direct 2 1 current. Westinghouse < sin ωt > = was convinced that the 2 technology of alternating Thus, current was the key to 1 2 the electrical future. p = i m R [7.5(c)] He founded the famous 2 Company named after him WESTINGHOUSE To express ac power in the same form as dc power and enlisted the services (P = I2R), a special value of current is defined and used. of Nicola Tesla and It is called, root mean square (rms) or effective current other inventors in the (1846(Fig. 7.3) and is denoted by Irms or I. development of alternating current motors and – apparatus for the transmission of high tension current, pioneering 1914) in large scale lighting. FIGURE 7.3 The rms current I is related to the peak current im by I = mi / 2 = 0.707 im. 1 T F (t ) d t* The average value of a function F (t) over a period T is given by F (t ) = T ∫0 1 T 1  sin 2ωt  T 1 < cos 2ωt > = ∫ cos 2ω t dt = = [ sin 2ω T − 0 ] = 0** T 0 T  2ω  0 2ωT 179 Reprint 2025-26 Physics It is defined by 2 1 2 i m I = i = i m = 2 2 = 0.707 im (7.6) In terms of I, the average power, denoted by P is 1 2 2 P = p = i m R = I R (7.7) 2 Similarly, we define the rms voltage or effective voltage by v m V = = 0.707 vm (7.8) 2 From Eq. (7.3), we have vm = imR v m i m or, = R 2 2 or, V = IR (7.9) Equation (7.9) gives the relation between ac current and ac voltage and is similar to that in the dc case. This shows the advantage of introducing the concept of rms values. In terms of rms values, the equation for power [Eq. (7.7)] and relation between current and voltage in ac circuits are essentially the same as those for the dc case. It is customary to measure and specify rms values for ac quantities. For example, the household line voltage of 220 V is an rms value with a peak voltage of vm = 2 V = (1.414)(220 V) = 311 V In fact, the I or rms current is the equivalent dc current that would produce the same average power loss as the alternating current. Equation (7.7) can also be written as P = V2 / R = I V (since V = I R) Example 7.1 A light bulb is rated at 100W for a 220 V supply. Find (a) the resistance of the bulb; (b) the peak voltage of the source; and (c) the rms current through the bulb. Solution (a) We are given P = 100 W and V = 220 V. The resistance of the bulb is 2 V 2 ( 220 V ) R = = = 484 Ω P 100 W (b) The peak voltage of the source is V 7.1 v m = 2V = 311 (c) Since, P = I V P 100 W I 0.454A EXAMPLE V 220 V Reprint 2025-26 Alternating Current

11.6 SPECIFIC HEAT CAPACITY average energy of 2 × ½ kBT = kBT. In three dimensions, the average energy is 3 kBT.Suppose an amount of heat ∆Q supplied to a For a mole of a solid, the total energy is substance changes its temperature from T to T + ∆T. We define heat capacity of a substance U = 3 kBT × NA = 3 RT (∵kBT × NA = R) (see Chapter 10) to be Now, at constant pressure, ∆Q = ∆U + P ∆V ≅ ∆ Q ∆U, since for a solid ∆V is negligible. Therefore, S = (11.4) ∆ T ∆ Q ∆U C = = = 3 R (11.7) We expect ∆Q and, therefore, heat capacity S ∆ T ∆ T to be proportional to the mass of the substance. Table 11.1 Specific and molar heat capacities Further, it could also depend on the of some solids at room temperature, i.e., a different amount of heat may temperature and atmospheric be needed for a unit rise in temperature at pressure different temperatures. To define a constant –v Speci"c heat Molar speci"c characteristic of the substance and Substance –1 –1 –1 –1 (J kg K ) heat (J mol K ) independent of its amount, we divide S by the mass of the substance m in kg : S  1  ∆Q s = (11.5) m =  m  ∆T s is known as the specific heat capacity of the substance. It depends on the nature of the As Table 11.1 shows, the experimentally substance and its temperature. The unit of measured values which generally agrees withspecific heat capacity is J kg–1 K–1. Reprint 2025-26 232 PHYSICS predicted value 3R at ordinary temperatures. ideal gas, we have a simple relation. (Carbon is an exception.) The agreement is known to break down at low temperatures. Cp – Cv = R (11.8) Specific heat capacity of water where Cp and Cv are molar specific heat capacities of an ideal gas at constant pressure The old unit of heat was calorie. One calorie and volume respectively and R is the universal was earlier defined to be the amount of heat gas constant. To prove the relation, we begin required to raise the temperature of 1g of water with Eq. (11.3) for 1 mole of the gas : by 1°C. With more precise measurements, it was found that the specific heat of water varies ∆Q = ∆U + P ∆V slightly with temperature. Figure 11.5 shows If ∆Q is absorbed at constant volume, ∆V = 0this variation in the temperature range 0 to 100 °C.  ∆Q   ∆ U   ∆ U  (11.9) C v =  ∆T  v =  ∆T  v =  ∆T  where the subscript v is dropped in the last step, since U of an ideal gas depends only on temperature. (The subscript denotes the quantity kept fixed.) If, on the other hand, ∆Q is absorbed at constant pressure,  ∆ Q   ∆ U   ∆V  (11.10) C p =  ∆T  p =  ∆T  p + P  ∆ T  p The subscript p can be dropped from the Fig. 11.5 Variation of specific heat capacity of first term since U of an ideal gas depends only water with temperature. on T. Now, for a mole of an ideal gas For a precise definition of calorie, it was, PV = RTtherefore, necessary to specify the unit temperature interval. One calorie is defined which gives to be the amount of heat required to raise the temperature of 1g of water from 14.5 °C to  ∆ V  P = R (11.11)15.5 °C. Since heat is just a form of energy,  ∆ T  p it is preferable to use the unit joule, J. In SI units, the specific heat capacity of water Equations (11.9) to (11.11) give the desired is 4186 J kg–1 K–1 i.e. 4.186 J g–1 K–1. The so relation, Eq. (11.8). called mechanical equivalent of heat defined 11.7THERMODYNAMIC STATE VARIABLES as the amount of work needed to produce AND EQUATION OF STATE 1 cal of heat is in fact just a conversion factor between two different units of energy : calorie Every equilibrium state of a thermodynamic to joule. Since in SI units, we use the unit joule system is completely described by specific for heat, work or any other form of energy, the values of some macroscopic variables, also term mechanical equivalent is now called state variables. For example, an superfluous and need not be used. equilibrium state of a gas is completely As already remarked, the specific heat specified by the values of pressure, volume, capacity depends on the process or the temperature, and mass (and composition if conditions under which heat capacity transfer there is a mixture of gases). A thermodynamic takes place. For gases, for example, we can system is not always in equilibrium. For example, define two specific heats : specific heat a gas allowed to expand freely against vacuum capacity at constant volume and specific is not an equilibrium state [Fig. 11.6(a)]. During heat capacity at constant pressure. For an the rapid expansion, pressure of the gas may Reprint 2025-26 THERMODYNAMICS 233 not be uniform throughout. Similarly, a mixture temperature do not. To decide which variable is of gases undergoing an explosive chemical extensive and which intensive, think of a reaction (e.g. a mixture of petrol vapour and relevant system in equilibrium, and imagine that air when ignited by a spark) is not an it is divided into two equal parts. The variables that remain unchanged for each part areequilibrium state; again its temperature and intensive. The variables whose values get halvedpressure are not uniform [Fig. 11.6(b)]. in each part are extensive. It is easily seen, forEventually, the gas attains a uniform example, that internal energy U, volume V, total temperature and pressure and comes to mass M are extensive variables. Pressure P, thermal and mechanical equilibrium with its temperature T, and density ρ are intensive surroundings. variables. It is a good practice to check the consistency of thermodynamic equations using this classification of variables. For example, in the equation ∆Q = ∆U + P ∆V quantities on both sides are extensive*. (The product of an intensive variable like P and an extensive quantity ∆V is extensive.) 11.8 THERMODYNAMIC PROCESSES 11.8.1 Quasi-static process Consider a gas in thermal and mechanical equilibrium with its surroundings. The pressure of the gas in that case equals the external Fig. 11.6 (a) The partition in the box is suddenly pressure and its temperature is the same as removed leading to free expansion of the that of its surroundings. Suppose that the gas. (b) A mixture of gases undergoing an external pressure is suddenly reduced (say by explosive chemical reaction. In both lifting the weight on the movable piston in the situations, the gas is not in equilibrium and container). The piston will accelerate outward. cannot be described by state variables. During the process, the gas passes through In short, thermodynamic state variables states that are not equilibrium states. The non- describe equilibrium states of systems. The equilibrium states do not have well-defined various state variables are not necessarily pressure and temperature. In the same way, if independent. The connection between the state a finite temperature difference exists between variables is called the equation of state. For the gas and its surroundings, there will be a example, for an ideal gas, the equation of state rapid exchange of heat during which the gas is the ideal gas relation will pass through non-equilibrium states. In due course, the gas will settle to an equilibrium P V = µ R T state with well-defined temperature and For a fixed amount of the gas i.e. given µ, there pressure equal to those of the surroundings. The are thus, only two independent variables, say P free expansion of a gas in vacuum and a mixture and V or T and V. The pressure-volume curve of gases undergoing an explosive chemical for a fixed temperature is called an isotherm. reaction, mentioned in section 11.7 are also Real gases may have more complicated examples where the system goes through non- equations of state. equilibrium states. The thermodynamic state variables are of two Non-equilibrium states of a system are difficult kinds: extensive and intensive. Extensive to deal with. It is, therefore, convenient to variables indicate the ‘size’ of the system. imagine an idealised process in which at every Intensive variables such as pressure and stage the system is an equilibrium state. Such a * As emphasised earlier, Q is not a state variable. However, ∆Q is clearly proportional to the total mass of system and hence is extensive. Reprint 2025-26 234 PHYSICS process is, in principle, infinitely slow, hence the A process in which the temperature of the name quasi-static (meaning nearly static). The system is kept fixed throughout is called an system changes its variables (P, T, V ) so slowly isothermal process. The expansion of a gas in that it remains in thermal and mechanical a metallic cylinder placed in a large reservoir of equilibrium with its surroundings throughout. fixed temperature is an example of an isothermal In a quasi-static process, at every stage, the process. (Heat transferred from the reservoir to difference in the pressure of the system and the the system does not materially affect the external pressure is infinitesimally small. The temperature of the reservoir, because of its very large heat capacity.) In isobaric processes thesame is true of the temperature difference pressure is constant while in isochoricbetween the system and its surroundings processes the volume is constant. Finally, if the(Fig.11.7). To take a gas from the state (P, T ) to another state (P ′, T ′ ) via a quasi-static process, system is insulated from the surroundings and no heat flows between the system and thewe change the external pressure by a very small surroundings, the process is adiabatic. The amount, allow the system to equalise its pressure definitions of these special processes are with that of the surroundings and continue the summarised in Table. 11.2 process infinitely slowly until the system achieves the pressure P ′. Similarly, to change Table 11.2 Some special thermodynamic the temperature, we introduce an infinitesimal processes temperature difference between the system and the surrounding reservoirs and by choosing reservoirs of progressively different temperatures T to T ′, the system achieves the temperature T ′. We now consider these processes in some detail : 11.8.2 Isothermal process For an isothermal process (T fixed), the ideal gas equation gives PV = constant i.e., pressure of a given mass of gas varies inversely as its volume. This is nothing but Boyle’s Law. Suppose an ideal gas goes isothermally (at temperature T ) from its initial state (P1, V1) to Fig. 11.7 In a quasi-static process, the temperature the final state (P2, V 2). At any intermediate stage of the surrounding reservoir and the with pressure P and volume change from V to external pressure differ only infinitesimally V + ∆V (∆V small) from the temperature and pressure of the system. ∆W = P ∆ V A quasi-static process is obviously a Taking (∆V → 0) and summing the quantity hypothetical construct. In practice, processes ∆W over the entire process, that are sufficiently slow and do not involve V2 accelerated motion of the piston, large W = ∫ P d V temperature gradient, etc., are reasonably V1 approximation to an ideal quasi-static process. V2 d V V2We shall from now on deal with quasi-static = µ RT = µRT In ∫ (11.12) Vprocesses only, except when stated otherwise. V1 V 1 Reprint 2025-26 THERMODYNAMICS 235 where in the second step we have made use of We can calculate, as before, the work done in the ideal gas equation PV = µ RT and taken the an adiabatic change of an ideal gas from the constants out of the integral. For an ideal gas, state (P1, V1, T1) to the state (P2, V2, T2). internal energy depends only on temperature. V 2Thus, there is no change in the internal energy W = ∫ P d Vof an ideal gas in an isothermal process. The V1First Law of Thermodynamics then implies that heat supplied to the gas equals the work done by the gas : Q = W. Note from Eq. (11.12) that for V2 > V1, W > 0; and for V2 < V1, W < 0. That is, in an isothermal expansion, the gas absorbs heat and does work while in an isothermal (11.15) compression, work is done on the gas by the environment and heat is released. From Eq. (11.14), the constant is P1V1γ or P2V2γ 11.8.3 Adiabatic process γ  γ  P2 V 2 P1V 1In an adiabatic process, the system is insulated W = 1 − 1   −1 γ γ − from the surroundings and heat absorbed or 1 − γ  V2 V1  released is zero. From Eq. (11.1), we see that work done by the gas results in decrease in its 1 µR(T1 − T2 ) = [ P2 V2 − P1V1 ] = (11.16)internal energy (and hence its temperature for 1 − γ γ − 1 an ideal gas). We quote without proof (the result that you will learn in higher courses) that for As expected, if work is done by the gas in an an adiabatic process of an ideal gas. adiabatic process (W > 0), from Eq. (11.16), P V γ = const (11.13) T2 < T1. On the other hand, if work is done on the gas (W < 0), we get T2 > T1 i.e., thewhere γ is the ratio of specific heats (ordinary temperature of the gas rises. or molar) at constant pressure and at constant volume. 11.8.4 Isochoric process Cp In an isochoric process, V is constant. No work γ = Cv is done on or by the gas. From Eq. (11.1), the heat absorbed by the gas goes entirely to change Thus if an ideal gas undergoes a change in its internal energy and its temperature. The its state adiabatically from (P1, V1) to (P2, V2) : change in temperature for a given amount of γ = P2 V2γ (11.14) heat is determined by the specific heat of the P1 V1 gas at constant volume. Figure11.8 shows the P-V curves of an ideal gas for two adiabatic processes connecting two 11.8.5 Isobaric process isotherms. In an isobaric process, P is fixed. Work done by the gas is W = P (V2 – V1) = µ R (T2 – T1) (11.17) Since temperature changes, so does internal energy. The heat absorbed goes partly to increase internal energy and partly to do work. The change in temperature for a given amount of heat is determined by the specific heat of the gas at constant pressure. 11.8.6 Cyclic process In a cyclic process, the system returns to its initial state. Since internal energy is a stateFig. 11.8 P-V curves for isothermal and adiabatic variable, ∆U = 0 for a cyclic process. From processes of an ideal gas. Reprint 2025-26 236 PHYSICS Eq. (11.1), the total heat absorbed equals the 11.10 REVERSIBLE AND IRREVERSIBLE work done by the system. PROCESSES Imagine some process in which a thermodynamic11.9 SECOND LAW OF THERMODYNAMICS system goes from an initial state i to a final state The First Law of Thermodynamics is the f. During the process the system absorbs heat Q principle of conservation of energy. Common from the surroundings and performs work W on experience shows that there are many it. Can we reverse this process and bring both conceivable processes that are perfectly the system and surroundings to their initial allowed by the First Law and yet are never states with no other effect anywhere ? Experience observed. For example, nobody has ever seen suggests that for most processes in nature this a book lying on a table jumping to a height by is not possible. The spontaneous processes ofitself. But such a thing would be possible if nature are irreversible. Several examples can bethe principle of conservation of energy were cited. The base of a vessel on an oven is hotterthe only restriction. The table could cool than its other parts. When the vessel is removed,spontaneously, converting some of its internal heat is transferred from the base to the otherenergy into an equal amount of mechanical parts, bringing the vessel to a uniformenergy of the book, which would then hop to a height with potential energy equal to the temperature (which in due course cools to the mechanical energy it acquired. But this never temperature of the surroundings). The process happens. Clearly, some additional basic cannot be reversed; a part of the vessel will not principle of nature forbids the above, even get cooler spontaneously and warm up the base. though it satisfies the energy conservation It will violate the Second Law of Thermodynamics, principle. This principle, which disallows if it did. The free expansion of a gas is irreversible. many phenomena consistent with the First The combustion reaction of a mixture of petrol Law of Thermodynamics is known as the and air ignited by a spark cannot be reversed. Second Law of Thermodynamics. Cooking gas leaking from a gas cylinder in the The Second Law of Thermodynamics gives kitchen diffuses to the entire room. The diffusion a fundamental limitation to the efficiency of a process will not spontaneously reverse and bring heat engine and the co-efficient of the gas back to the cylinder. The stirring of a performance of a refrigerator. In simple terms, liquid in thermal contact with a reservoir will it says that efficiency of a heat engine can convert the work done into heat, increasing the never be unity. For a refrigerator, the Second internal energy of the reservoir. The process Law says that the co-efficient of performance cannot be reversed exactly; otherwise it would can never be infinite. The following two amount to conversion of heat entirely into work, statements, one due to Kelvin and Planck violating the Second Law of Thermodynamics. denying the possibility of a perfect heat engine, Irreversibility is a rule rather an exception and another due to Clausius denying the in nature. possibility of a perfect refrigerator or heat Irreversibility arises mainly from two causes: pump, are a concise summary of these one, many processes (like a free expansion, or observations. an explosive chemical reaction) take the system Kelvin-Planck statement to non-equilibrium states; two, most processes No process is possible whose sole result is the involve friction, viscosity and other dissipative absorption of heat from a reservoir and the effects (e.g., a moving body coming to a stop and complete conversion of the heat into work. losing its mechanical energy as heat to the floor and the body; a rotating blade in a liquid coming Clausius statement to a stop due to viscosity and losing its No process is possible whose sole result is the mechanical energy with corresponding gain in transfer of heat from a colder object to a the internal energy of the liquid). Since hotter object. dissipative effects are present everywhere and It can be proved that the two statements can be minimised but not fully eliminated, most above are completely equivalent. processes that we deal with are irreversible. Reprint 2025-26 THERMODYNAMICS 237 A thermodynamic process (state i → state f ) in a reversible heat engine operating between is reversible if the process can be turned back two temperatures, heat should be absorbed such that both the system and the surroundings (from the hot reservoir) isothermally and return to their original states, with no other released (to the cold reservoir) isothermally. We change anywhere else in the universe. From the thus have identified two steps of the reversible preceding discussion, a reversible process is an heat engine : isothermal process at temperature idealised notion. A process is reversible only if T1 absorbing heat Q1 from the hot reservoir, and another isothermal process at temperature T2it is quasi-static (system in equilibrium with the releasing heat Q2 to the cold reservoir. Tosurroundings at every stage) and there are no complete a cycle, we need to take the systemdissipative effects. For example, a quasi-static from temperature T1 to T2 and then back fromisothermal expansion of an ideal gas in a temperature T2 to T1. Which processes shouldcylinder fitted with a frictionless movable piston we employ for this purpose that are reversible?is a reversible process. A little reflection shows that we can only adopt Why is reversibility such a basic concept in reversible adiabatic processes for these thermodynamics ? As we have seen, one of the purposes, which involve no heat flow from any concerns of thermodynamics is the efficiency reservoir. If we employ any other process that is with which heat can be converted into work. not adiabatic, say an isochoric process, to take The Second Law of Thermodynamics rules out the system from one temperature to another, we the possibility of a perfect heat engine with 100% shall need a series of reservoirs in the efficiency. But what is the highest efficiency temperature range T2 to T1 to ensure that at each possible for a heat engine working between two stage the process is quasi-static. (Remember reservoirs at temperatures T1 and T2? It turns again that for a process to be quasi-static and out that a heat engine based on idealised reversible, there should be no finite temperature reversible processes achieves the highest difference between the system and the reservoir.) efficiency possible. All other engines involving But we are considering a reversible engine that irreversibility in any way (as would be the case operates between only two temperatures. Thus for practical engines) have lower than this adiabatic processes must bring about the limiting efficiency. temperature change in the system from T1 to T2 and T2 to T1 in this engine.11.11 CARNOT ENGINE Suppose we have a hot reservoir at temperature T1 and a cold reservoir at temperature T2. What is the maximum efficiency possible for a heat engine operating between the two reservoirs and what cycle of processes should be adopted to achieve the maximum efficiency ? Sadi Carnot, a French engineer, first considered this question in 1824. Interestingly, Carnot arrived at the correct answer, even though the basic concepts of heat and thermodynamics had yet to be firmly established. We expect the ideal engine operating between two temperatures to be a reversible engine. Irreversibility is associated with dissipative Fig. 11.9 Carnot cycle for a heat engine with an effects, as remarked in the preceding section, ideal gas as the working substance. and lowers efficiency. A process is reversible if it is quasi-static and non-dissipative. We have A reversible heat engine operating between seen that a process is not quasi-static if it two temperatures is called a Carnot engine. We involves finite temperature difference between have just argued that such an engine must have the system and the reservoir. This implies that the following sequence of steps constituting one Reprint 2025-26 238 PHYSICS cycle, called the Carnot cycle, shown in Fig. 11.9. We have taken the working substance In  V 3  of the Carnot engine to be an ideal gas. T2   V4  = 1 −  T1   V2  (11.23)(a) Step 1 → 2 Isothermal expansion of the gas In  V1  taking its state from (P1, V1, T1) to (P2, V2, T1). Now since step 2 → 3 is an adiabatic process, The heat absorbed by the gas (Q1) from the γ −1 γ −1 reservoir at temperature T1 is given by T1 V 2 = T2 V3 Eq. (11.12). This is also the work done (W1 → 2) by the gas on the environment. 1 /( γ −1 ) V2  T2   V2  i.e. = (11.24) 3  T1  W1 → 2 = Q1 = µ R T1 ln  V1  (11.18) V (b) Step 2 → 3 Adiabatic expansion of the gas Similarly, since step 4 → 1 is an adiabatic from (P2, V2, T1) to (P3, V3, T2) process Work done by the gas, using γ −1 γ −1 Eq. (11.16), is T2 V 4 = T1 V1 µR ( T1 − T2 ) 1 /γ −1 W2 →=3 (11.19) V1  T2  γ − 1 i.e. = (11.25) V4  T1  (c) Step 3 → 4 Isothermal compression of the gas from (P3, V3, T2) to (P4, V4, T2). From Eqs. (11.24) and (11.25), 2 Heat released (Q2) by the gas to the reservoir V3 = V (11.26) at temperature T2 is given by Eq. (11.12). This V4 V1 is also the work done (W3 → 4) on the gas by the environment. Using Eq. (11.26) in Eq. (11.23), we get  V 3  2 W 3 → 4 = Q2 = µRT2 ln η = 1 −T (Carnot engine) (11.27)  V 4  (11.20) T1 (d) Step 4 → 1 Adiabatic compression of the gas from (P4, V4, T2) to (P1,V1, T1). We have already seen that a Carnot engine is a reversible engine. Indeed it is the only Work done on the gas, [using Eq.(11.16), is reversible engine possible that works between two reservoirs at different temperatures. Each  T1 − T2  W4 → 1 = µR (11.21) step of the Carnot cycle given in Fig. 11.9 can  γ -1  be reversed. This will amount to taking heat Q2 from the cold reservoir at T2, doing work W on From Eqs. (11.18) to (11.21) total work done by the gas in one complete cycle is the system, and transferring heat Q1 to the hot reservoir. This will be a reversible refrigerator. W = W1 → 2 + W2 → 3 – W3 → 4 – W4 → 1 We next establish the important result  V2   V3  (sometimes called Carnot’s theorem) that = µ RT1 ln  V1  – µ RT2 ln  V4  (11.22) (a) working between two given temperatures T1 and T2 of the hot and cold reservoirs respectively, The efficiency η of the Carnot engine is no engine can have efficiency more than that of the Carnot engine and (b) the efficiency of the W Q 2 η = = 1 − Carnot engine is independent of the nature of Q1 Q1 the working substance. Reprint 2025-26 THERMODYNAMICS 239 To prove the result (a), imagine a reversible reservoir and delivers the same amount of work (Carnot) engine R and an irreversible engine I in one cycle, without any change in the source working between the same source (hot reservoir) or anywhere else. This is clearly against the and sink (cold reservoir). Let us couple the Kelvin-Planck statement of the Second Law of engines, I and R, in such a way so that I acts Thermodynamics. Hence the assertion ηI > ηR like a heat engine and R acts as a refrigerator. is wrong. No engine can have efficiency greater Let I absorb heat Q1 from the source, deliver than that of the Carnot engine. A similar work W ′ and release the heat Q1- W′ to the sink. argument can be constructed to show that a We arrange so that R returns the same heat Q1 reversible engine with one particular substance to the source, taking heat Q2 from the sink and cannot be more efficient than the one using requiring work W = Q1 – Q2 to be done on it. another substance. The maximum efficiency of Now suppose ηR < ηI i.e. if R were to act a Carnot engine given by Eq. (11.27) is as an engine it would give less work output independent of the nature of the system performing the Carnot cycle of operations. Thus we are justified in using an ideal gas as a system in the calculation of efficiency η of a Carnot engine. The ideal gas has a simple equation of I state, which allows us to readily calculate η, but the final result for η, [Eq. (11.27)], is true for any Carnot engine. R This final remark shows that in a Carnot cycle, Q1 T1 = W (11.28) T2 Q 2 is a universal relation independent of the natureFig. 11.10 An irreversible engine (I) coupled to a reversible refrigerator (R). If W ′ > W, this of the system. Here Q1 and Q2 are respectively, would amount to extraction of heat the heat absorbed and released isothermally W′ – W from the sink and its full (from the hot and to the cold reservoirs) in a conversion to work, in contradiction with Carnot engine. Equation (11.28), can, therefore, the Second Law of Thermodynamics. be used as a relation to define a truly universal thermodynamic temperature scale that is than that of I i.e. W < W ′ for a given Q1. With R independent of any particular properties of the acting like a refrigerator, this would mean system used in the Carnot cycle. Of course, for Q2 = Q1 – W > Q1 – W ′. Thus, on the whole, an ideal gas as a working substance, this the coupled I-R system extracts heat universal temperature is the same as the ideal (Q1 – W) – (Q1 – W ′) = (W ′ – W ) from the cold gas temperature introduced in section 11.9. SUMMARY 1. The zeroth law of thermodynamics states that ‘two systems in thermal equilibrium with a third system separately are in thermal equilibrium with each other’. The Zeroth Law leads to the concept of temperature. 2. Internal energy of a system is the sum of kinetic energies and potential energies of the molecular constituents of the system. It does not include the over-all kinetic energy of the system. Heat and work are two modes of energy transfer to the system. Heat is the energy transfer arising due to temperature difference between the system and the surroundings. Work is energy transfer brought about by other means, such as moving the piston of a cylinder containing the gas, by raising or lowering some weight connected to it. Reprint 2025-26 240 PHYSICS 3. The first law of thermodynamics is the general law of conservation of energy applied to any system in which energy transfer from or to the surroundings (through heat and work) is taken into account. It states that ∆Q = ∆U + ∆W where ∆Q is the heat supplied to the system, ∆W is the work done by the system and ∆U is the change in internal energy of the system. 4. The specific heat capacity of a substance is defined by 1 ∆Q s = m ∆T where m is the mass of the substance and ∆Q is the heat required to change its temperature by ∆T. The molar specific heat capacity of a substance is defined by 1 ∆ Q C = µ ∆T where µ is the number of moles of the substance. For a solid, the law of equipartition of energy gives C = 3 R which generally agrees with experiment at ordinary temperatures. Calorie is the old unit of heat. 1 calorie is the amount of heat required to raise the temperature of 1 g of water from 14.5 °C to 15.5 °C. 1 cal = 4.186 J. 5. For an ideal gas, the molar specific heat capacities at constant pressure and volume satisfy the relation Cp – Cv = R where R is the universal gas constant. 6. Equilibrium states of a thermodynamic system are described by state variables. The value of a state variable depends only on the particular state, not on the path used to arrive at that state. Examples of state variables are pressure (P ), volume (V ), temperature (T ), and mass (m ). Heat and work are not state variables. An Equation of State (like the ideal gas equation PV = µ RT ) is a relation connecting different state variables. 7. A quasi-static process is an infinitely slow process such that the system remains in thermal and mechanical equilibrium with the surroundings throughout. In a quasi-static process, the pressure and temperature of the environment can differ from those of the system only infinitesimally. 8. In an isothermal expansion of an ideal gas from volume V1 to V2 at temperature T the heat absorbed (Q) equals the work done (W ) by the gas, each given by  V 2   Q = W = µ R T ln   V1  9. In an adiabatic process of an ideal gas γ PV = constant C p where γ = C v Work done by an ideal gas in an adiabatic change of state from (P1, V1, T1) to (P2, V2, T2) is µR ( T1 − T2 ) W = γ – 1 Reprint 2025-26 THERMODYNAMICS 241 10. The second law of thermodynamics disallows some processes consistent with the First Law of Thermodynamics. It states Kelvin-Planck statement No process is possible whose sole result is the absorption of heat from a reservoir and complete conversion of the heat into work. Clausius statement No process is possible whose sole result is the transfer of heat from a colder object to a hotter object. Put simply, the Second Law implies that no heat engine can have efficiency η equal to 1 or no refrigerator can have co-efficient of performance α equal to infinity. 11. A process is reversible if it can be reversed such that both the system and the surroundings return to their original states, with no other change anywhere else in the universe. Spontaneous processes of nature are irreversible. The idealised reversible process is a quasi-static process with no dissipative factors such as friction, viscosity, etc. 12. Carnot engine is a reversible engine operating between two temperatures T1 (source) and T2 (sink). The Carnot cycle consists of two isothermal processes connected by two adiabatic processes. The efficiency of a Carnot engine is given by T 2 η= 1 − (Carnot engine) T1 No engine operating between two temperatures can have efficiency greater than that of the Carnot engine. 13. If Q > 0, heat is added to the system If Q < 0, heat is removed to the system If W > 0, Work is done by the system If W < 0, Work is done on the system Quantity Symbol Dimensions Unit Remark Co-efficienty of volume αv [K–1] K–1 αv = 3 α1 expansion Heat supplied to a system ∆Q [ML2 T–2] J Q is not a state variable Specific heat capacity s [L2 T–2 K–1] J kg–1 K–1 dt Thermal Conductivity K [MLT–3 K–1] J s–1 K–1 H = – KA dx POINTS TO PONDER 1. Temperature of a body is related to its average internal energy, not to the kinetic energy of motion of its centre of mass. A bullet fired from a gun is not at a higher temperature because of its high speed. 2. Equilibrium in thermodynamics refers to the situation when macroscopic variables describing the thermodynamic state of a system do not depend on time. Equilibrium of a system in mechanics means the net external force and torque on the system are zero. Reprint 2025-26 242 PHYSICS 3. In a state of thermodynamic equilibrium, the microscopic constituents of a system are not in equilibrium (in the sense of mechanics). 4. Heat capacity, in general, depends on the process the system goes through when heat is supplied. 5. In isothermal quasi-static processes, heat is absorbed or given out by the system even though at every stage the gas has the same temperature as that of the surrounding reservoir. This is possible because of the infinitesimal difference in temperature between the system and the reservoir. EXERCISES 11.1 A geyser heats water flowing at the rate of 3.0 litres per minute from 27 °C to 77 °C. If the geyser operates on a gas burner, what is the rate of consumption of the fuel if its heat of combustion is 4.0 × 104 J/g ? 11.2 What amount of heat must be supplied to 2.0 × 10–2 kg of nitrogen (at room temperature) to raise its temperature by 45 °C at constant pressure ? (Molecular mass of N2 = 28; R = 8.3 J mol–1 K–1.) 11.3 Explain why (a) Two bodies at different temperatures T1 and T2 if brought in thermal contact do not necessarily settle to the mean temperature (T1 + T2 )/2. (b) The coolant in a chemical or a nuclear plant (i.e., the liquid used to prevent the different parts of a plant from getting too hot) should have high specific heat. (c) Air pressure in a car tyre increases during driving. (d) The climate of a harbour town is more temperate than that of a town in a desert at the same latitude. 11.4 A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the pressure of the gas increase if the gas is compressed to half its original volume ? 11.5 In changing the state of a gas adiabatically from an equilibrium state A to another equilibrium state B, an amount of work equal to 22.3 J is done on the system. If the gas is taken from state A to B via a process in which the net heat absorbed by the system is 9.35 cal, how much is the net work done by the system in the latter case ? (Take 1 cal = 4.19 J) 11.6 Two cylinders A and B of equal capacity are connected to each other via a stopcock. A contains a gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally insulated. The stopcock is suddenly opened. Answer the following : (a) What is the final pressure of the gas in A and B ? (b) What is the change in internal energy of the gas ? (c) What is the change in the temperature of the gas ? (d) Do the intermediate states of the system (before settling to the final equilibrium state) lie on its P-V-T surface ? 11.7 An electric heater supplies heat to a system at a rate of 100W. If system performs work at a rate of 75 joules per second. At what rate is the internal energy increasing? Reprint 2025-26 THERMODYNAMICS 243 11.8 A thermodynamic system is taken from an original state to an intermediate state by the linear process shown in Fig. (11.13) Fig. 11.11 Its volume is then reduced to the original value from E to F by an isobaric process. Calculate the total work done by the gas from D to E to F Reprint 2025-26 CHAPTER TWELVE KINETIC THEORY 12.1 INTRODUCTION Boyle discovered the law named after him in 1661. Boyle, Newton and several others tried to explain the behaviour of 12.1 Introduction gases by considering that gases are made up of tiny atomic 12.2 Molecular nature of matter particles. The actual atomic theory got established more than 12.3 Behaviour of gases 150 years later. Kinetic theory explains the behaviour of gases 12.4 Kinetic theory of an ideal gas based on the idea that the gas consists of rapidly moving 12.5 Law of equipartition of energy atoms or molecules. This is possible as the inter-atomic forces, 12.6 Specific heat capacity which are short range forces that are important for solids and liquids, can be neglected for gases. The kinetic theory12.7 Mean free path was developed in the nineteenth century by Maxwell, Summary Boltzmann and others. It has been remarkably successful. It Points to ponder gives a molecular interpretation of pressure and temperature Exercises of a gas, and is consistent with gas laws and Avogadro’s hypothesis. It correctly explains specific heat capacities of many gases. It also relates measurable properties of gases such as viscosity, conduction and diffusion with molecular parameters, yielding estimates of molecular sizes and masses. This chapter gives an introduction to kinetic theory. 12.2 MOLECULAR NATURE OF MATTER Richard Feynman, one of the great physicists of 20th century considers the discovery that “Matter is made up of atoms” to be a very significant one. Humanity may suffer annihilation (due to nuclear catastrophe) or extinction (due to environmental disasters) if we do not act wisely. If that happens, and all of scientific knowledge were to be destroyed then Feynman would like the ‘Atomic Hypothesis’ to be communicated to the next generation of creatures in the universe. Atomic Hypothesis: All things are made of atoms - little particles that move around in perpetual motion, attracting each other when they are a little distance apart, but repelling upon being squeezed into one another. Speculation that matter may not be continuous, existed in many places and cultures. Kanada in India and Democritus Reprint 2025-26 KINETIC THEORY 245 Atomic Hypothesis in Ancient India and Greece Though John Dalton is credited with the introduction of atomic viewpoint in modern science, scholars in ancient India and Greece conjectured long before the existence of atoms and molecules. In the Vaiseshika school of thought in India founded by Kanada (Sixth century B.C.) the atomic picture was developed in considerable detail. Atoms were thought to be eternal, indivisible, infinitesimal and ultimate parts of matter. It was argued that if matter could be subdivided without an end, there would be no difference between a mustard seed and the Meru mountain. The four kinds of atoms (Paramanu — Sanskrit word for the smallest particle) postulated were Bhoomi (Earth), Ap (water), Tejas (fire) and Vayu (air) that have characteristic mass and other attributes, were propounded. Akasa (space) was thought to have no atomic structure and was continuous and inert. Atoms combine to form different molecules (e.g. two atoms combine to form a diatomic molecule dvyanuka, three atoms form a tryanuka or a triatomic molecule), their properties depending upon the nature and ratio of the constituent atoms. The size of the atoms was also estimated, by conjecture or by methods that are not known to us. The estimates vary. In Lalitavistara, a famous biography of the Buddha written mainly in the second century B.C., the estimate is close to the modern estimate of atomic size, of the order of 10 –10 m. In ancient Greece, Democritus (Fourth century B.C.) is best known for his atomic hypothesis. The word ‘atom’ means ‘indivisible’ in Greek. According to him, atoms differ from each other physically, in shape, size and other properties and this resulted in the different properties of the substances formed by their combination. The atoms of water were smooth and round and unable to ‘hook’ on to each other, which is why liquid /water flows easily. The atoms of earth were rough and jagged, so they held together to form hard substances. The atoms of fire were thorny which is why it caused painful burns. These fascinating ideas, despite their ingenuity, could not evolve much further, perhaps because they were intuitive conjectures and speculations not tested and modified by quantitative experiments - the hallmark of modern science. in Greece had suggested that matter may consist of matter. The theory is now well accepted by of indivisible constituents. The scientific ‘Atomic scientists. However even at the end of the Theory’ is usually credited to John Dalton. He nineteenth century there were famous scientists proposed the atomic theory to explain the laws who did not believe in atomic theory ! of definite and multiple proportions obeyed by From many observations, in recent times we elements when they combine into compounds. now know that molecules (made up of one or The first law says that any given compound has, more atoms) constitute matter. Electron a fixed proportion by mass of its constituents. microscopes and scanning tunnelling The second law says that when two elements microscopes enable us to even see them. The form more than one compound, for a fixed mass size of an atom is about an angstrom (10 -10 m). of one element, the masses of the other elements In solids, which are tightly packed, atoms are are in ratio of small integers. spaced about a few angstroms (2 Å) apart. In To explain the laws Dalton suggested, about liquids the separation between atoms is also 200 years ago, that the smallest constituents about the same. In liquids the atoms are not of an element are atoms. Atoms of one element as rigidly fixed as in solids, and can move are identical but differ from those of other around. This enables a liquid to flow. In gases elements. A small number of atoms of each the interatomic distances are in tens of element combine to form a molecule of the angstroms. The average distance a molecule compound. Gay Lussac’s law, also given in early can travel without colliding is called the mean 19th century, states: When gases combine free path. The mean free path, in gases, is of chemically to yield another gas, their volumes the order of thousands of angstroms. The atoms are in the ratios of small integers. Avogadro’s are much freer in gases and can travel long law (or hypothesis) says: Equal volumes of all distances without colliding. If they are not gases at equal temperature and pressure have enclosed, gases disperse away. In solids and the same number of molecules. Avogadro’s law, liquids the closeness makes the interatomic force when combined with Dalton’s theory explains important. The force has a long range attraction Gay Lussac’s law. Since the elements are often and a short range repulsion. The atoms attract in the form of molecules, Dalton’s atomic theory when they are at a few angstroms but repel when can also be referred to as the molecular theory they come closer. The static appearance of a gas Reprint 2025-26 246 PHYSICS is misleading. The gas is full of activity and the is 6.02 × 1023. This is known as Avogadro number equilibrium is a dynamic one. In dynamic and is denoted by NA. The mass of 22.4 litres of equilibrium, molecules collide and change their any gas is equal to its molecular weight in grams speeds during the collision. Only the average at S.T.P (standard temperature 273 K and properties are constant. pressure 1 atm). This amount of substance is Atomic theory is not the end of our quest, but called a mole (see Chapter 1 for a more precise the beginning. We now know that atoms are not definition). Avogadro had guessed the equality of indivisible or elementary. They consist of a numbers in equal volumes of gas at a fixed nucleus and electrons. The nucleus itself is made temperature and pressure from chemical up of protons and neutrons. The protons and reactions. Kinetic theory justifies this hypothesis. neutrons are again made up of quarks. Even The perfect gas equation can be written as quarks may not be the end of the story. There PV = µ RT (12.3)may be string like elementary entities. Nature always has surprises for us, but the search for where µ is the number of moles and R = NA truth is often enjoyable and the discoveries kB is a universal constant. The temperature T is absolute temperature. Choosing kelvin scale forbeautiful. In this chapter, we shall limit ourselves absolute temperature, R = 8.314 J mol–1K–1.to understanding the behaviour of gases (and a Herelittle bit of solids), as a collection of moving molecules in incessant motion. M N µ = = (12.4) M 0 N A

11.3 The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted?

11.4 HEAT, INTERNAL ENERGY AND WORK associated with various random motions of its The Zeroth Law of Thermodynamics led us to molecules. We will see in the next chapter that the concept of temperature that agrees with our in a gas this motion is not only translational commonsense notion. Temperature is a marker (i.e. motion from one point to another in the of the ‘hotness’ of a body. It determines the volume of the container); it also includes rotational and vibrational motion of thedirection of flow of heat when two bodies are molecules (Fig. 11.3).placed in thermal contact. Heat flows from the body at a higher temperature to the one at lower temperature. The flow stops when the temperatures equalise; the two bodies are then in thermal equilibrium. We saw in some detail how to construct temperature scales to assign temperatures to different bodies. We now describe the concepts of heat and other relevant quantities like internal energy and work. Fig. 11.3 (a) Internal energy U of a gas is the sum of the kinetic and potential energies of its The concept of internal energy of a system is molecules when the box is at rest. Kinetic not difficult to understand. We know that every energy due to various types of motion bulk system consists of a large number of (translational, rotational, vibrational) is to molecules. Internal energy is simply the sum of be included in U. (b) If the same box is the kinetic energies and potential energies of moving as a whole with some velocity, these molecules. We remarked earlier that in the kinetic energy of the box is not to be thermodynamics, the kinetic energy of the included in U. system, as a whole, is not relevant. Internal energy is thus, the sum of molecular kinetic and potential energies in the frame of reference relative to which the centre of mass of the system is at rest. Thus, it includes only the (disordered) energy associated with the random motion of molecules of the system. We denote the internal energy of a system by U. Though we have invoked the molecular picture to understand the meaning of internal energy, as far as thermodynamics is concerned, U is simply a macroscopic variable of the system. The important thing about internal energy is that it depends only on the state of the system, not on how that state was achieved. Internal energy U of a system is an example of a thermodynamic ‘state variable’ – its value depends only on the given state of the system, not on history i.e. not on the ‘path’ taken to arrive at that state. Thus, the internal energy of a given mass of gas depends on its state described by Fig. 11.4 Heat and work are two distinct modes of specific values of pressure, volume and energy transfer to a system that results in temperature. It does not depend on how this change in its internal energy. (a) Heat is energy transfer due to temperaturestate of the gas came about. Pressure, volume, difference between the system and the temperature, and internal energy are surroundings. (b) Work is energy transfer thermodynamic state variables of the system brought about by means (e.g. moving the (gas) (see section 11.7). If we neglect the small piston by raising or lowering some weight intermolecular forces in a gas, the internal connected to it) that do not involve such a energy of a gas is just the sum of kinetic energies temperature difference. Reprint 2025-26 230 PHYSICS What are the ways of changing internal 11.5 FIRST LAW OF THERMODYNAMICS energy of a system ? Consider again, for We have seen that the internal energy U of a simplicity, the system to be a certain mass of system can change through two modes of energy gas contained in a cylinder with a movable transfer : heat and work. Let piston as shown in Fig. 11.4. Experience shows ∆Q = Heat supplied to the system by thethere are two ways of changing the state of the surroundingsgas (and hence its internal energy). One way is to put the cylinder in contact with a body at a ∆W = Work done by the system on the higher temperature than that of the gas. The surroundings temperature difference will cause a flow of ∆U = Change in internal energy of the system energy (heat) from the hotter body to the gas, The general principle of conservation of thus increasing the internal energy of the gas. energy then implies that The other way is to push the piston down i.e. to ∆Q = ∆U + ∆W (11.1) do work on the system, which again results in i.e. the energy (∆Q) supplied to the system goesincreasing the internal energy of the gas. Of in partly to increase the internal energy of thecourse, both these things could happen in the system (∆U) and the rest in work on thereverse direction. With surroundings at a lower environment (∆W). Equation (11.1) is known astemperature, heat would flow from the gas to the First Law of Thermodynamics. It is simplythe surroundings. Likewise, the gas could push the general law of conservation of energy applied the piston up and do work on the surroundings. to any system in which the energy transfer from In short, heat and work are two different modes or to the surroundings is taken into account. of altering the state of a thermodynamic system Let us put Eq. (11.1) in the alternative form and changing its internal energy. The notion of heat should be carefully ∆Q – ∆W = ∆U (11.2) distinguished from the notion of internal energy. Heat is certainly energy, but it is the energy in Now, the system may go from an initial state transit. This is not just a play of words. The to the final state in a number of ways. For distinction is of basic significance. The state of example, to change the state of a gas from a thermodynamic system is characterised by its (P1, V1) to (P2, V2), we can first change the volume of the gas from V1 to V2, keeping itsinternal energy, not heat. A statement like ‘a pressure constant i.e. we can first go the stategas in a given state has a certain amount of heat’ is as meaningless as the statement that (P1, V2) and then change the pressure of the gas from P1 to P2, keeping volume constant, to‘a gas in a given state has a certain amount take the gas to (P2, V2). Alternatively, we canof work’. In contrast, ‘a gas in a given state first keep the volume constant and then keep has a certain amount of internal energy’ is a the pressure constant. Since U is a state perfectly meaningful statement. Similarly, the variable, ∆U depends only on the initial and statements ‘a certain amount of heat is final states and not on the path taken by the supplied to the system’ or ‘a certain amount gas to go from one to the other. However, ∆Q of work was done by the system’ are perfectly and ∆W will, in general, depend on the path meaningful. taken to go from the initial to final states. From To summarise, heat and work in the First Law of Thermodynamics, Eq. (11.2), thermodynamics are not state variables. They it is clear that the combination ∆Q – ∆W, is are modes of energy transfer to a system however, path independent. This shows that resulting in change in its internal energy, if a system is taken through a process in which ∆U = 0 (for example, isothermal expansion ofwhich, as already mentioned, is a state variable. an ideal gas, see section 11.8), In ordinary language, we often confuse heat with internal energy. The distinction between ∆Q = ∆W them is sometimes ignored in elementary physics books. For proper understanding of i.e., heat supplied to the system is used up thermodynamics, however, the distinction is entirely by the system in doing work on the crucial. environment. Reprint 2025-26 THERMODYNAMICS 231 If the system is a gas in a cylinder with a If the amount of substance is specified in movable piston, the gas in moving the piston does terms of moles µ (instead of mass m in kg ), we work. Since force is pressure times area, and can define heat capacity per mole of the area times displacement is volume, work done substance by by the system against a constant pressure P is S 1 ∆Q C = = (11.6) ∆W = P ∆V µ µ ∆T C is known as molar specific heat capacity of where ∆V is the change in volume of the gas. the substance. Like s, C is independent of the Thus, for this case, Eq. (11.1) gives amount of substance. C depends on the nature ∆Q = ∆U + P ∆V (11.3) of the substance, its temperature and the conditions under which heat is supplied. The As an application of Eq. (11.3), consider the unit of C is J mo1–1 K–1. As we shall see later (in change in internal energy for 1 g of water when connection with specific heat capacity of gases), we go from its liquid to vapour phase. The additional conditions may be needed to define measured latent heat of water is 2256 J/g. i.e., C or s. The idea in defining C is that simple for 1 g of water ∆Q = 2256 J. At atmospheric predictions can be made in regard to molar pressure, 1 g of water has a volume 1 cm3 in specific heat capacities. liquid phase and 1671 cm3 in vapour phase. Table 11.1 lists measured specific and molar heat capacities of solids at atmospheric pressure Therefore, and ordinary room temperature. ∆W =P (Vg –Vl ) = 1.013 ×105 × (1671 × 10–6) =169.2 J We will see in Chapter 12 that predictions of specific heats of gases generally agree with Equation (11.3) then gives experiment. We can use the same law of equipartition of energy that we use there to ∆U = 2256 – 169.2 = 2086.8 J predict molar specific heat capacities of solids We see that most of the heat goes to increase (See Section 12.5 and 12.6). Consider a solid of the internal energy of water in transition from N atoms, each vibrating about its mean the liquid to the vapour phase. position. An oscillator in one dimension has

11.3 ZEROTH LAW OF THERMODYNAMICS (a) Imagine two systems A and B, separated by an adiabatic wall, while each is in contact with a third system C, via a conducting wall [Fig. 11.2(a)]. The states of the systems (i.e., their macroscopic variables) will change until both A and B come to thermal equilibrium with C. After this is achieved, suppose that the adiabatic wall between A and B is replaced by a conducting wall and C is insulated from A and B by an adiabatic wall [Fig.11.2(b)]. It is found that the states of A and B change no (b) further i.e. they are found to be in thermal Fig. 11.2 (a) Systems A and B are separated by an equilibrium with each other. This observation adiabatic wall, while each is in contact forms the basis of the Zeroth Law of with a third system C via a conducting Thermodynamics, which states that ‘two wall. (b) The adiabatic wall between A systems in thermal equilibrium with a third and B is replaced by a conducting wall, system separately are in thermal equilibrium while C is insulated from A and B by an adiabatic wall.with each other’. R.H. Fowler formulated this * Both the variables need not change. It depends on the constraints. For instance, if the gases are in containers of fixed volume, only the pressures of the gases would change to achieve thermal equilibrium. Reprint 2025-26 THERMODYNAMICS 229

11.2 THERMAL EQUILIBRIUM independent variables. Let the pressure and Equilibrium in mechanics means that the net volume of the gases be (PA, VA) and (PB, VB) external force and torque on a system are zero. respectively. Suppose first that the two systems The term ‘equilibrium’ in thermodynamics appears are put in proximity but are separated by an * Thermodynamics may also involve other variables that are not so obvious to our senses e.g. entropy, enthalpy, etc., and they are all macroscopic variables. However, a thermodynamic state is specified by five state variables viz., pressure, volume, temperature, internal energy and entropy. Entropy is a measure of disorderness in the system. Enthalpy is a measure of total heat content of the system. Reprint 2025-26 228 PHYSICS adiabatic wall – an insulating wall (can be law in 1931 long after the first and second Laws movable) that does not allow flow of energy (heat) of thermodynamics were stated and so numbered. from one to another. The systems are insulated The Zeroth Law clearly suggests that when two from the rest of the surroundings also by similar systems A and B, are in thermal equilibrium, adiabatic walls. The situation is shown there must be a physical quantity that has the schematically in Fig. 11.1 (a). In this case, it is same value for both. This thermodynamic found that any possible pair of values (PA, VA) will variable whose value is equal for two systems in be in equilibrium with any possible pair of values thermal equilibrium is called temperature (T ). (PB, VB). Next, suppose that the adiabatic wall is Thus, if A and B are separately in equilibrium replaced by a diathermic wall – a conducting wall with C, TA = TC and TB = TC. This implies that that allows energy flow (heat) from one to another. TA = TB i.e. the systems A and B are also in It is then found that the macroscopic variables of thermal equilibrium. the systems A and B change spontaneously until We have arrived at the concept of temperature both the systems attain equilibrium states. After formally via the Zeroth Law. The next question that there is no change in their states. The is : how to assign numerical values to situation is shown in Fig. 11.1(b). The pressure temperatures of different bodies ? In other words, and volume variables of the two gases change to how do we construct a scale of temperature ? (PB ′, VB ′) and (PA ′, VA ′) such that the new states Thermometry deals with this basic question to of A and B are in equilibrium with each other*. which we turn in the next section. There is no more energy flow from one to another. We then say that the system A is in thermal equilibrium with the system B. What characterises the situation of thermal equilibrium between two systems ? You can guess the answer from your experience. In thermal equilibrium, the temperatures of the two systems are equal. We shall see how does one arrive at the concept of temperature in thermodynamics? The Zeroth law of thermodynamics provides the clue.

11.8 Light of frequency 7.21 × 1014 Hz is incident on a metal surface. Electrons with a maximum speed of 6.0 × 105 m/s are ejected from the surface. What is the threshold frequency for photoemission of electrons?

11.10 What is the de Broglie wavelength of (a) a bullet of mass 0.040 kg travelling at the speed of 1.0 km/s, (b) a ball of mass 0.060 kg moving at a speed of 1.0 m/s, and (c) a dust particle of mass 1.0 × 10–9 kg drifting with a speed of 2.2 m/s ?

4.4 A horizontal overhead power line carries a current of 90 A in east to west direction. What is the magnitude and direction of the magnetic field due to the current 1.5 m below the line?

4.10 Two moving coil meters, M1 and M2 have the following particulars: R1 = 10 Ω, N1 = 30, A1 = 3.6 × 10–3 m2, B1 = 0.25 T R2 = 14 Ω, N2 = 42, A2 = 1.8 × 10–3 m2, B2 = 0.50 T (The spring constants are identical for the two meters). Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M2 and M1.

4.7 Two long and parallel straight wires A and B carrying currents of 8.0 A and 5.0 A in the same direction are separated by a distance of 4.0 cm. Estimate the force on a 10 cm section of wire A.

4.6 A 3.0 cm wire carrying a current of 10 A is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be 0.27 T. What is the magnetic force on the wire?

4.8 A closely wound solenoid 80 cm long has 5 layers of windings of 400 turns each. The diameter of the solenoid is 1.8 cm. If the current carried is 8.0 A, estimate the magnitude of B inside the solenoid near its centre.

4.9 A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30º with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?

4.7 Conservation of momentum push. A breeze causes the branches of a tree to swing; a 4.8 Equilibrium of a particle strong wind can even move heavy objects. A boat moves in a 4.9 Common forces in mechanics flowing river without anyone rowing it. Clearly, some external 4.10 Circular motion agency is needed to provide force to move a body from rest. 4.11 Solving problems in Likewise, an external force is needed also to retard or stop mechanics motion. You can stop a ball rolling down an inclined plane by applying a force against the direction of its motion. Summary In these examples, the external agency of force (hands, Points to ponder wind, stream, etc) is in contact with the object. This is not Exercises always necessary. A stone released from the top of a building accelerates downward due to the gravitational pull of the earth. A bar magnet can attract an iron nail from a distance. This shows that external agencies (e.g. gravitational and magnetic forces ) can exert force on a body even from a distance. In short, a force is required to put a stationary body in motion or stop a moving body, and some external agency is needed to provide this force. The external agency may or may not be in contact with the body. So far so good. But what if a body is moving uniformly (e.g. a skater moving straight with constant speed on a horizontal ice slab) ? Is an external force required to keep a body in uniform motion? Reprint 2025-26 50 PHYSICS 4.2 ARISTOTLE’S FALLACY true law of nature for forces and motion, one has to imagine a world in which uniform motion isThe question posed above appears to be simple. possible with no frictional forces opposing. ThisHowever, it took ages to answer it. Indeed, the is what Galileo did.correct answer to this question given by Galileo in the seventeenth century was the foundation 4.3 THE LAW OF INERTIA of Newtonian mechanics, which signalled the Galileo studied motion of objects on an inclined birth of modern science. plane. Objects (i) moving down an inclined plane The Greek thinker, Aristotle (384 B.C– 322 accelerate, while those (ii) moving up retard. B.C.), held the view that if a body is moving, (iii) Motion on a horizontal plane is an interme- something external is required to keep it moving. diate situation. Galileo concluded that an object According to this view, for example, an arrow moving on a frictionless horizontal plane must shot from a bow keeps flying since the air behind neither have acceleration nor retardation, i.e. it the arrow keeps pushing it. The view was part of should move with constant velocity (Fig. 4.1(a)).an elaborate framework of ideas developed by Aristotle on the motion of bodies in the universe. Most of the Aristotelian ideas on motion are now known to be wrong and need not concern us. For our purpose here, the Aristotelian law of motion may be phrased thus: An external force (i) (ii) (iii)is required to keep a body in motion. Fig. 4.1(a) Aristotelian law of motion is flawed, as we shall Another experiment by Galileo leading to thesee. However, it is a natural view that anyone same conclusion involves a double inclined plane.would hold from common experience. Even a A ball released from rest on one of the planes rollssmall child playing with a simple (non-electric) down and climbs up the other. If the planes are toy-car on a floor knows intuitively that it needs smooth, the final height of the ball is nearly the to constantly drag the string attached to the toy- same as the initial height (a little less but never car with some force to keep it going. If it releases greater). In the ideal situation, when friction is the string, it comes to rest. This experience is absent, the final height of the ball is the same common to most terrestrial motion. External as its initial height. forces seem to be needed to keep bodies in If the slope of the second plane is decreased motion. Left to themselves, all bodies eventually and the experiment repeated, the ball will still come to rest. reach the same height, but in doing so, it will What is the flaw in Aristotle’s argument? The travel a longer distance. In the limiting case, when answer is: a moving toy car comes to rest because the slope of the second plane is zero (i.e. is a the external force of friction on the car by the floor horizontal) the ball travels an infinite distance. opposes its motion. To counter this force, the child In other words, its motion never ceases. This is, has to apply an external force on the car in the of course, an idealised situation (Fig. 4.1(b)). direction of motion. When the car is in uniform motion, there is no net external force acting on it: the force by the child cancels the force ( friction) by the floor. The corollary is: if there were no friction, the child would not be required to apply any force to keep the toy car in uniform motion. The opposing forces such as friction (solids) and viscous forces (for fluids) are always present in the natural world. This explains why forces by external agencies are necessary to overcome the frictional forces to keep bodies in uniform motion. Now we understand where Aristotle Fig. 4.1(b) The law of inertia was inferred by Galileo went wrong. He coded this practical experience from observations of motion of a ball on a in the form of a basic argument. To get at the double inclined plane. Reprint 2025-26 LAWS OF MOTION 51 In practice, the ball does come to a stop after accomplished almost single-handedly by Isaac moving a finite distance on the horizontal plane, Newton, one of the greatest scientists of all times. because of the opposing force of friction which Newton built on Galileo’s ideas and laid the can never be totally eliminated. However, if there foundation of mechanics in terms of three laws were no friction, the ball would continue to move of motion that go by his name. Galileo’s law of with a constant velocity on the horizontal plane. inertia was his starting point which he formu- Galileo thus, arrived at a new insight on lated as the first law of motion: motion that had eluded Aristotle and those who Every body continues to be in its state followed him. The state of rest and the state of of rest or of uniform motion in a straight uniform linear motion (motion with constant line unless compelled by some external velocity) are equivalent. In both cases, there is force to act otherwise. Ideas on Motion in Ancient Indian Science Ancient Indian thinkers had arrived at an elaborate system of ideas on motion. Force, the cause of motion, was thought to be of different kinds : force due to continuous pressure (nodan), as the force of wind on a sailing vessel; impact (abhighat), as when a potter’s rod strikes the wheel; persistent tendency (sanskara) to move in a straight line(vega) or restoration of shape in an elastic body; transmitted force by a string, rod, etc. The notion of (vega) in the Vaisesika theory of motion perhaps comes closest to the concept of inertia. Vega, the tendency to move in a straight line, was thought to be opposed by contact with objects including atmosphere, a parallel to the ideas of friction and air resistance. It was correctly summarised that the different kinds of motion (translational, rotational and vibrational) of an extended body arise from only the translational motion of its constituent particles. A falling leaf in the wind may have downward motion as a whole (patan) and also rotational and vibrational motion (bhraman, spandan), but each particle of the leaf at an instant only has a definite (small) displacement. There was considerable focus in Indian thought on measurement of motion and units of length and time. It was known that the position of a particle in space can be indicated by distance measured along three axes. Bhaskara (1150 A.D.) had introduced the concept of ‘instantaneous motion’ (tatkaliki gati), which anticipated the modern notion of instantaneous velocity using Differential Calculus. The difference between a wave and a current (of water) was clearly understood; a current is a motion of particles of water under gravity and fluidity while a wave results from the transmission of vibrations of water particles. no net force acting on the body. It is incorrect to The state of rest or uniform linear motion both assume that a net force is needed to keep a body imply zero acceleration. The first law of motion can, in uniform motion. To maintain a body in therefore, be simply expressed as: uniform motion, we need to apply an external If the net external force on a body is zero, its force to ecounter the frictional force, so that acceleration is zero. Acceleration can be non the two forces sum up to zero net external zero only if there is a net external force on force. the body. To summarise, if the net external force is zero, Two kinds of situations are encountered in thea body at rest continues to remain at rest and a application of this law in practice. In somebody in motion continues to move with a uniform examples, we know that the net external forcevelocity. This property of the body is called on the object is zero. In that case we caninertia. Inertia means ‘resistance to change’. A body does not change its state of rest or conclude that the acceleration of the object is uniform motion, unless an external force zero. For example, a spaceship out in compels it to change that state. interstellar space, far from all other objects and with all its rockets turned off, has no net 4.4 NEWTON’S FIRST LAW OF MOTION external force acting on it. Its acceleration, Galileo’s simple, but revolutionary ideas according to the first law, must be zero. If it is dethroned Aristotelian mechanics. A new in motion, it must continue to move with a mechanics had to be developed. This task was uniform velocity. Reprint 2025-26 52 PHYSICS More often, however, we do not know all the The acceleration of the car cannot be accounted forces to begin with. In that case, if we know for by any internal force. This might sound that an object is unaccelerated (i.e. it is either surprising, but it is true. The only conceivable at rest or in uniform linear motion), we can infer external force along the road is the force of from the first law that the net external force on friction. It is the frictional force that accelerates the object must be zero. Gravity is everywhere. the car as a whole. (You will learn about friction For terrestrial phenomena, in particular, every in section 4.9). When the car moves with object experiences gravitational force due to the constant velocity, there is no net external force. earth. Also objects in motion generally experience The property of inertia contained in the First friction, viscous drag, etc. If then, on earth, an law is evident in many situations. Suppose we object is at rest or in uniform linear motion, it is are standing in a stationary bus and the driver not because there are no forces acting on it, but starts the bus suddenly. We get thrown because the various external forces cancel out backward with a jerk. Why ? Our feet are in touch i.e. add up to zero net external force. with the floor. If there were no friction, we would Consider a book at rest on a horizontal surface remain where we were, while the floor of the bus Fig. (4.2(a)). It is subject to two external forces : would simply slip forward under our feet and the the force due to gravity (i.e. its weight W) acting back of the bus would hit us. However, downward and the upward force on the book by fortunately, there is some friction between the the table, the normal force R . R is a self-adjusting feet and the floor. If the start is not too sudden, force. This is an example of the kind of situation i.e. if the acceleration is moderate, the frictional mentioned above. The forces are not quite known force would be enough to accelerate our feet fully but the state of motion is known. We observe along with the bus. But our body is not strictly the book to be at rest. Therefore, we conclude a rigid body. It is deformable, i.e. it allows some from the first law that the magnitude of R equals relative displacement between different parts. that of W. A statement often encountered is : What this means is that while our feet go with “Since W = R, forces cancel and, therefore, the book the bus, the rest of the body remains where it is is at rest”. This is incorrect reasoning. The correct due to inertia. Relative to the bus, therefore, we statement is : “Since the book is observed to be at are thrown backward. As soon as that happens, rest, the net external force on it must be zero, however, the muscular forces on the rest of the according to the first law. This implies that the body (by the feet) come into play to move the body normal force R must be equal and opposite to the along with the bus. A similar thing happens weight W ”. when the bus suddenly stops. Our feet stop due to the friction which does not allow relative motion between the feet and the floor of the bus. But the rest of the body continues to move forward due to inertia. We are thrown forward. The restoring muscular forces again come into play and bring the body to rest. ⊳ Example 4.1 An astronaut accidentally gets separated out of his small spaceship accelerating in inter stellar space at a Fig. 4.2 (a) a book at rest on the table, and (b) a car constant rate of 100 m s–2. What is the moving with uniform velocity. The net force acceleration of the astronaut the instant after is zero in each case. he is outside the spaceship ? (Assume that Consider the motion of a car starting from there are no nearby stars to exert rest, picking up speed and then moving on a gravitational force on him.) smooth straight road with uniform speed (Fig. Answer Since there are no nearby stars to exert(4.2(b)). When the car is stationary, there is no gravitational force on him and the smallnet force acting on it. During pick-up, it spaceship exerts negligible gravitationalaccelerates. This must happen due to a net attraction on him, the net force acting on theexternal force. Note, it has to be an external force. Reprint 2025-26 LAWS OF MOTION 53 astronaut, once he is out of the spaceship, is act. One reason is that the cricketer allows a zero. By the first law of motion the acceleration longer time for his hands to stop the ball. As of the astronaut is zero. ⊳ you may have noticed, he draws in the hands backward in the act of catching the ball4.5 NEWTON’S SECOND LAW OF MOTION (Fig. 4.3). The novice, on the other hand, The first law refers to the simple case when the keeps his hands fixed and tries to catch the net external force on a body is zero. The second ball almost instantly. He needs to provide a law of motion refers to the general situation when much greater force to stop the ball instantly, there is a net external force acting on the body. and this hurts. The conclusion is clear: force It relates the net external force to the not only depends on the change in momentum, acceleration of the body. but also on how fast the change is brought Momentum about. The same change in momentum Momentum of a body is defined to be the product brought about in a shorter time needs a of its mass m and velocity v, and is denoted greater applied force. In short, the greater the by p: rate of change of momentum, the greater is p = m v (4.1) the force. Momentum is clearly a vector quantity. The following common experiences indicate the importance of this quantity for considering the effect of force on motion. • Suppose a light-weight vehicle (say a small car) and a heavy weight vehicle (say a loaded truck) are parked on a horizontal road. We all know that a much greater force is needed to push the truck than the car to bring them to the same speed in same time. Similarly, a greater opposing force is needed to stop a heavy body than a light body in the same time, if they are moving with the same speed. • If two stones, one light and the other heavy, are dropped from the top of a building, a Fig. 4.3 Force not only depends on the change in person on the ground will find it easier to catch momentum but also on how fast the change is brought about. A seasoned cricketer draws the light stone than the heavy stone. The in his hands during a catch, allowing greater mass of a body is thus an important time for the ball to stop and hence requires a parameter that determines the effect of force smaller force. on its motion. • Speed is another important parameter to consider. A bullet fired by a gun can easily • Observations confirm that the product of pierce human tissue before it stops, resulting mass and velocity (i.e. momentum) is basic to in casualty. The same bullet fired with the effect of force on motion. Suppose a fixed moderate speed will not cause much damage. force is applied for a certain interval of time Thus for a given mass, the greater the speed, on two bodies of different masses, initially at the greater is the opposing force needed to stop rest, the lighter body picks up a greater speed the body in a certain time. Taken together, than the heavier body. However, at the end of the product of mass and velocity, that is the time interval, observations show that each momentum, is evidently a relevant variable body acquires the same momentum. Thus of motion. The greater the change in the the same force for the same time causes momentum in a given time, the greater is the the same change in momentum for force that needs to be applied. • A seasoned cricketer catches a cricket ball different bodies. This is a crucial clue to the second law of motion. coming in with great speed far more easily • In the preceding observations, the vector than a novice, who can hurt his hands in the Reprint 2025-26 54 PHYSICS character of momentum has not been evident. ∆p ∆p In the examples so far, momentum and change F ∝ or F = k ∆t ∆ t in momentum both have the same direction. But this is not always the case. Suppose a where k is a constant of proportionality. Taking stone is rotated with uniform speed in a ∆p the limit ∆t → 0, the term becomes the horizontal plane by means of a string, the ∆t magnitude of momentum is fixed, but its derivative or differential co-efficient of p with direction changes (Fig. 4.4). A force is needed d p to cause this change in momentum vector. respect to t, denoted by . Thus dt This force is provided by our hand through the string. Experience suggests that our hand d p F = k (4.2) needs to exert a greater force if the stone is d t rotated at greater speed or in a circle of For a body of fixed mass m, smaller radius, or both. This corresponds to greater acceleration or equivalently a greater d p d d v = (m v ) = m = m a (4.3) rate of change in momentum vector. This d t d t d t suggests that the greater the rate of change i.e the Second Law can also be written as in momentum vector the greater is the force F = k m a (4.4) applied. which shows that force is proportional to the product of mass m and acceleration a. The unit of force has not been defined so far. In fact, we use Eq. (4.4) to define the unit of force. We, therefore, have the liberty to choose any constant value for k. For simplicity, we choose k = 1. The second law then is dp F = = m a (4.5) dt In SI unit force is one that causes an acceleration of 1 m s-2 to a mass of 1 kg. This unit is known as newton : 1 N = 1 kg m s-2. Let us note at this stage some important points Fig. 4.4 Force is necessary for changing the direction about the second law : of momentum, even if its magnitude is constant. We can feel this while rotating a 1. In the second law, F = 0 implies a = 0. The second stone in a horizontal circle with uniform speed law is obviously consistent with the first law. by means of a string. 2. The second law of motion is a vector law. It is These qualitative observations lead to the equivalent to three equations, one for each second law of motion expressed by Newton as component of the vectors : follows : d p x F x = = ma xThe rate of change of momentum of a body is d t directly proportional to the applied force and d p ytakes place in the direction in which the force F y = = ma y acts. d t dp zThus, if under the action of a force F for time F z = =m a z (4.6) dtinterval ∆t, the velocity of a body of mass m changes from v to v + ∆v i.e. its initial momentum This means that if a force is not parallel to the velocity of the body, but makes some anglep = m v changes by ∆ p = m ∆v . According to the with it, it changes only the component of Second Law, velocity along the direction of force. The Reprint 2025-26 LAWS OF MOTION 55 component of velocity normal to the force Answer The retardation ‘a’ of the bullet remains unchanged. For example, in the (assumed constant) is given by motion of a projectile under the vertical – 90 × 90 – u 2 gravitational force, the horizontal component m s −2 = – 6750 m s −2 a = = of velocity remains unchanged (Fig. 4.5). 2s 2 × 0.6 3. The second law of motion given by Eq. (4.5) is The retarding force, by the second law of applicable to a single point particle. The force motion, is F in the law stands for the net external force = 0.04 kg × 6750 m s-2 = 270 N on the particle and a stands for acceleration of the particle. It turns out, however, that the The actual resistive force, and therefore, law in the same form applies to a rigid body or, retardation of the bullet may not be uniform. The answer therefore, only indicates the average even more generally, to a system of particles. resistive force. ⊳ In that case, F refers to the total external force ⊳ on the system and a refers to the acceleration Example 4.3 The motion of a particle of of the system as a whole. More precisely, a is 1 2 the acceleration of the centre of mass of the mass m is described by y = ut + gt . Find 2 system about which we shall study in detail in the force acting on the particle. Chapter 6. Any internal forces in the system are not to be included in F. Answer We know 1 2 y = ut + gt 2 Now, d y v = = u + gt d t dv acceleration, a = = g d t Fig. 4.5 Acceleration at an instant is determined by Then the force is given by Eq. (4.5) the force at that instant. The moment after a F = ma = mg stone is dropped out of an accelerated train, Thus the given equation describes the motion it has no horizontal acceleration or force, if of a particle under acceleration due to gravity air resistance is neglected. The stone carries no memory of its acceleration with the train and y is the position coordinate in the direction a moment ago. of g. ⊳ 4. The second law of motion is a local relation Impulse which means that force F at a point in space We sometimes encounter examples where a large (location of the particle) at a certain instant force acts for a very short duration producing a of time is related to a at that point at that finite change in momentum of the body. For instant. Acceleration here and now is example, when a ball hits a wall and bounces determined by the force here and now, not by back, the force on the ball by the wall acts for a any history of the motion of the particle very short time when the two are in contact, yet the force is large enough to reverse the momentum (See Fig. 4.5). of the ball. Often, in these situations, the force ⊳ and the time duration are difficult to ascertain Example 4.2 A bullet of mass 0.04 kg separately. However, the product of force and time, moving with a speed of 90 m s–1 enters a which is the change in momentum of the body heavy wooden block and is stopped after a remains a measurable quantity. This product is distance of 60 cm. What is the average called impulse: resistive force exerted by the block on the bullet? Impulse = Force × time duration = Change in momentum (4.7) Reprint 2025-26 56 PHYSICS A large force acting for a short time to produce a Thus, according to Newtonian mechanics, finite change in momentum is called an impulsive force never occurs singly in nature. Force is the force. In the history of science, impulsive forces were mutual interaction between two bodies. Forces put in a conceptually different category from always occur in pairs. Further, the mutual forces ordinary forces. Newtonian mechanics has no such between two bodies are always equal and distinction. Impulsive force is like any other force – opposite. This idea was expressed by Newton in except that it is large and acts for a short time. the form of the third law of motion. ⊳ To every action, there is always an equal and Example 4.4 A batsman hits back a ball opposite reaction. straight in the direction of the bowler without changing its initial speed of 12 m s–1. Newton’s wording of the third law is so crisp and If the mass of the ball is 0.15 kg, determine beautiful that it has become a part of common the impulse imparted to the ball. (Assume language. For the same reason perhaps, linear motion of the ball) misconceptions about the third law abound. Let us note some important points about the third law, particularly in regard to the usage of theAnswer Change in momentum terms : action and reaction. = 0.15 × 12–(–0.15×12) 1. The terms action and reaction in the third law = 3.6 N s, mean nothing else but ‘force’. Using different Impulse = 3.6 N s, terms for the same physical concept in the direction from the batsman to the bowler. can sometimes be confusing. A simple and clear way of stating the third law is as This is an example where the force on the ball follows :by the batsman and the time of contact of the ball and the bat are difficult to know, but the Forces always occur in pairs. Force on a impulse is readily calculated. ⊳ body A by B is equal and opposite to the force on the body B by A.

5.2 NOTIONS OF WORK AND KINETIC to be proportional to the speed of the drop ENERGY: THE WORK-ENERGY THEOREM but is otherwise undetermined. Consider The following relation for rectilinear motion under a drop of mass 1.00 g falling from a height constant acceleration a has been encountered 1.00 km. It hits the ground with a speed of in Chapter 3, 50.0 m s-1. (a) What is the work done by the v2 − u2 = 2 as (5.2) gravitational force ? What is the work done where u and v are the initial and final speeds by the unknown resistive force? and s the distance traversed. Multiplying both Answer (a) The change in kinetic energy of the sides by m/2, we have drop is 1 2 1 2 1 2 mv − mu = mas = Fs (5.2a) ∆ K = m v − 0 2 2 2 where the last step follows from Newton’s Second 1 -3 = × 10 × 50 × 50 Law. We can generalise Eq. (5.2) to three 2 dimensions by employing vectors = 1.25 J v2 − u2 = 2 a.d where we have assumed that the drop is initially at rest. Here a and d are acceleration and displacement Assuming that g is a constant with a value vectors of the object respectively. 10 m/s2, the work done by the gravitational force Once again multiplying both sides by m/2 , we obtain is, 1 1 mv 2 − mu 2 = m a.d = F.d (5.2b) Wg = mgh 2 2 = 10-3 ×10 ×103 The above equation provides a motivation for = 10.0 J the definitions of work and kinetic energy. The (b) From the work-energy theorem left side of the equation is the difference in the quantity ‘half the mass times the square of the ∆ K = W g + W r speed’ from its initial value to its final value. We where Wr is the work done by the resistive force call each of these quantities the ‘kinetic energy’, on the raindrop. Thus denoted by K. The right side is a product of the Wr = ∆K − Wg displacement and the component of the force = 1.25 −10 along the displacement. This quantity is called = − 8.75 J ‘work’ and is denoted by W. Eq. (5.2b) is then is negative. ⊳ Kf − Ki = W (5.3) 5.3 WORK where Ki and Kf are respectively the initial and As seen earlier, work is related to force and the final kinetic energies of the object. Work refers displacement over which it acts. Consider a to the force and the displacement over which it constant force F acting on an object of mass m. acts. Work is done by a force on the body over The object undergoes a displacement d in the a certain displacement. positive x-direction as shown in Fig. 5.2. Equation (5.2) is also a special case of the work-energy (WE) theorem : The change in kinetic energy of a particle is equal to the work done on it by the net force. We shall generalise the above derivation to a varying force in a later section. ⊳ Example 5.2 It is well known that a raindrop falls under the influence of the Fig. 5.2 An object undergoes a displacement d downward gravitational force and the under the influence of the force F. opposing resistive force. The latter is known Reprint 2025-26 74 PHYSICS The work done by the force is defined to be Table 5.1 Alternative Units of Work/Energy in J the product of component of the force in the direction of the displacement and the magnitude of this displacement. Thus W = (F cos θ)d = F.d (5.4) We see that if there is no displacement, there is no work done even if the force is large. Thus, when you push hard against a rigid brick wall, ⊳ the force you exert on the wall does no work. Yet Example 5.3 A cyclist comes to a skidding your muscles are alternatively contracting and stop in 10 m. During this process, the force relaxing and internal energy is being used up on the cycle due to the road is 200 N and and you do get tired. Thus, the meaning of work is directly opposed to the motion. (a) How in physics is different from its usage in everyday much work does the road do on the cycle ? language. (b) How much work does the cycle do on the road ? No work is done if : (i) the displacement is zero as seen in the example above. A weightlifter holding a 150 Answer Work done on the cycle by the road is kg mass steadily on his shoulder for 30 s the work done by the stopping (frictional) force does no work on the load during this time. on the cycle due to the road. (ii) the force is zero. A block moving on a smooth (a) The stopping force and the displacement make horizontal table is not acted upon by a an angle of 180o (π rad) with each other. horizontal force (since there is no friction), but Thus, work done by the road, may undergo a large displacement. Wr = Fd cosθ (iii) the force and displacement are mutually = 200 × 10 × cos π perpendicular. This is so since, for θ= π/2 rad = – 2000 J (= 90o), cos (π/2) = 0. For the block moving on It is this negative work that brings the cycle a smooth horizontal table, the gravitational to a halt in accordance with WE theorem. force mg does no work since it acts at right (b) From Newton’s Third Law an equal and angles to the displacement. If we assume that opposite force acts on the road due to the the moon’s orbits around the earth is cycle. Its magnitude is 200 N. However, the perfectly circular then the earth’s road undergoes no displacement. Thus, gravitational force does no work. The moon’s work done by cycle on the road is zero. ⊳ instantaneous displacement is tangential while the earth’s force is radially inwards and The lesson of Example 5.3 is that though the θ = π/2. force on a body A exerted by the body B is always Work can be both positive and negative. If θ is equal and opposite to that on B by A (Newton’s between 0o and 90o, cos θ in Eq. (5.4) is positive. Third Law); the work done on A by B is not If θ is between 90o and 180o, cos θ is negative. necessarily equal and opposite to the work done In many examples the frictional force opposes on B by A. displacement and θ = 180o. Then the work done 5.4 KINETIC ENERGY by friction is negative (cos 180o = –1). As noted earlier, if an object of mass m has From Eq. (5.4) it is clear that work and energy velocity v, its kinetic energy K ishave the same dimensions, [ML2T–2]. The SI unit of these is joule (J), named after the famous British 1 1 2physicist James Prescott Joule (1811-1869). Since K = m v. v = mv (5.5) 2 2work and energy are so widely used as physical concepts, alternative units abound and some of Kinetic energy is a scalar quantity. The kinetic these are listed in Table 5.1. energy of an object is a measure of the work an Reprint 2025-26 WORK, ENERGY AND POWER 75 Table 5.2 Typical kinetic energies (K) object can do by the virtue of its motion. This This is illustrated in Fig. 5.3(a). Adding notion has been intuitively known for a long time. successive rectangular areas in Fig. 5.3(a) we The kinetic energy of a fast flowing stream get the total work done as has been used to grind corn. Sailing x f ships employ the kinetic energy of the wind. Table W ≅ F (x )∆x (5.6) ∑

5.2 lists the kinetic energies for various x i objects. where the summation is from the initial position ⊳ xi to the final position xf. Example 5.4 In a ballistics demonstration a police officer fires a bullet of mass 50.0 g If the displacements are allowed to approach with speed 200 m s-1 (see Table 5.2) on soft zero, then the number of terms in the sum plywood of thickness 2.00 cm. The bullet increases without limit, but the sum approaches emerges with only 10% of its initial kinetic a definite value equal to the area under the curve energy. What is the emergent speed of the in Fig. 5.3(b). Then the work done is bullet ? xf W = lim F (x )∆xAnswer The initial kinetic energy of the bullet ∆ x → 0 ∑ x i is mv2/2 = 1000 J. It has a final kinetic energy xfof 0.1×1000 = 100 J. If vf is the emergent speed x ) d x (5.7)of the bullet, = ∫F ( i 1 2 x mv f = 100 J where ‘lim’ stands for the limit of the sum when 2 ∆x tends to zero. Thus, for a varying force 2 × 100 J the work done can be expressed as a definite v f = 0. 05 kg integral of force over displacement (see also Appendix 3.1). = 63.2 m s–1 The speed is reduced by approximately 68% (not 90%). ⊳

5.8 THE CONSERVATION OF MECHANICAL a ball of mass m being dropped from a cliff of ENERGY height H. For simplicity we demonstrate this important principle for one-dimensional motion. Suppose that a body undergoes displacement ∆x under the action of a conservative force F. Then from the WE theorem we have, ∆K = F(x) ∆x If the force is conservative, the potential energy function V(x) can be defined such that − ∆V = F(x) ∆x The above equations imply that Fig. 5.5 The conversion of potential energy to kinetic ∆K + ∆V = 0 energy for a ball of mass m dropped from a ∆(K + V ) = 0 (5.10) height H. Reprint 2025-26 WORK, ENERGY AND POWER 79 The total mechanical energies E0, Eh, and EH Answer (i) There are two external forces on of the ball at the indicated heights zero (ground the bob : gravity and the tension (T ) in the level), h and H, are string. The latter does no work since the displacement of the bob is always normal to the EH = mgH (5.11 a) string. The potential energy of the bob is thus 1 2 E h = mgh + mv h (5.11 b) associated with the gravitational force only. The 2 2 total mechanical energy E of the system is E0 = (1/2) mvf (5.11 c) conserved. We take the potential energy of the The constant force is a special case of a spatially system to be zero at the lowest point A. Thus, dependent force F(x). Hence, the mechanical at A : energy is conserved. Thus EH = E0 1 2 1 2 E = mv0 (5.12) or, mgH = mv f 2 2 v f = 2 gH [Newton’s Second Law] a result that was obtained in section 5.7 for a where TA is the tension in the string at A. At thefreely falling body. highest point C, the string slackens, as the Further, tension in the string (TC) becomes zero. EH = Eh Thus, at Cwhich implies, 2 1 2 v h = 2 g(H − h) (5.11 d) E = mv c + 2mgL (5.13) 2 and is a familiar result from kinematics. At the height H, the energy is purely potential. mvc2It is partially converted to kinetic at height h and mg = [Newton’s Second Law] (5.14) L is fully kinetic at ground level. This illustrates the conservation of mechanical energy. where vC is the speed at C. From Eqs. (5.13) ⊳ and (5.14) Example 5.7 A bob of mass m is suspended 5 by a light string of length L . It is imparted a E = mgL horizontal velocity vo at the lowest point A 2 such that it completes a semi-circular Equating this to the energy at A trajectory in the vertical plane with the string 5 m 2 becoming slack only on reaching the topmost mgL = v 0 point, C. This is shown in Fig. 5.6. Obtain an 2 2 expression for (i) vo; (ii) the speeds at points or, v 0 = 5 gL B and C; (iii) the ratio of the kinetic energies (ii) It is clear from Eq. (5.14) (KB/KC) at B and C. Comment on the nature of the trajectory of the bob after it reaches vC = gL the point C. At B, the energy is 1 2 E = mv B + mgL 2 Equating this to the energy at A and employing the result from (i), namely v 02 = 5 gL , 1 2 1 2 mv B + mgL = mv 0 2 2 5 = m g L Fig. 5.6 2 Reprint 2025-26 80 PHYSICS ∴ vB = 3 gL k x m2 W = + (5.16) 2 (iii) The ratio of the kinetic energies at B and C is : 1 2 mv B K B 2 3 = = 2 1 K C 1 mvC 2 At point C, the string becomes slack and the velocity of the bob is horizontal and to the left. If the connecting string is cut at this instant, the bob will execute a projectile motion with horizontal projection akin to a rock kicked horizontally from the edge of a cliff. Otherwise the bob will continue on its circular path and complete the revolution. ⊳

5.5 WORK DONE BY A VARIABLE FORCE A constant force is rare. It is the variable force, which is more commonly encountered. Fig. 5.3 is a plot of a varying force in one dimension. If the displacement ∆x is small, we can take the force F (x) as approximately constant and the work done is then ∆W =F (x) ∆x Fig. 5.3(a) Reprint 2025-26 76 PHYSICS The work done by the frictional force is Wf → area of the rectangle AGHI Wf = (−50) × 20 = − 1000 J The area on the negative side of the force axis has a negative sign. ⊳ 5.6 THE WORK-ENERGY THEOREM FOR A VARIABLE FORCE We are now familiar with the concepts of workFig. 5.3 (a) The shaded rectangle represents the work done by the varying force F(x), over and kinetic energy to prove the work-energy the small displacement ∆x, ∆W = F(x) ∆x. theorem for a variable force. We confine (b) adding the areas of all the rectangles we ourselves to one dimension. The time rate of find that for ∆x →0, the area under the curve change of kinetic energy is is exactly equal to the work done by F(x). d K d  1 2  =⊳ d t  2 m v  Example 5.5 A woman pushes a trunk on d t a railway platform which has a rough d v surface. She applies a force of 100 N over a = m v d t distance of 10 m. Thereafter, she gets progressively tired and her applied force = F v (from Newton’s Second Law) reduces linearly with distance to 50 N. The d x total distance through which the trunk has = F d t been moved is 20 m. Plot the force applied Thus by the woman and the frictional force, which dK = Fdx is 50 N versus displacement. Calculate the Integrating from the initial position (x i ) to final work done by the two forces over 20 m. position ( x f ), we have Answer K f x f F dx ∫ d K = ∫ K i x i where, Ki and K f are the initial and final kinetic energies corresponding to x i and x f. x f F d x or (5.8a) K f − K i = ∫ Fig. 5.4 Plot of the force F applied by the woman and x i the opposing frictional force f versus From Eq. (5.7), it follows that displacement. Kf − Ki = W (5.8b) The plot of the applied force is shown in Fig. 5.4. At x = 20 m, F = 50 N (≠ 0). We are given Thus, the WE theorem is proved for a variable that the frictional force f is |f|= 50 N. It opposes force. motion and acts in a direction opposite to F. It While the WE theorem is useful in a variety of is therefore, shown on the negative side of the problems, it does not, in general, incorporate the force axis. complete dynamical information of Newton’s The work done by the woman is second law. It is an integral form of Newton’s WF → area of the rectangle ABCD + area of second law. Newton’s second law is a relation the trapezium CEID between acceleration and force at any instant of 1 time. Work-energy theorem involves an integral WF = 100 × 10 + (100 + 50) × 10 over an interval of time. In this sense, the temporal 2 = 1000 + 750 (time) information contained in the statement of = 1750 J Newton’s second law is ‘integrated over’ and is Reprint 2025-26 WORK, ENERGY AND POWER 77 not available explicitly. Another observation is that are like ‘compressed springs’. They possess a Newton’s second law for two or three dimensions large amount of potential energy. An earthquake is in vector form whereas the work-energy results when these fault lines readjust. Thus, theorem is in scalar form. In the scalar form, potential energy is the ‘stored energy’ by virtue information with respect to directions contained of the position or configuration of a body. The in Newton’s second law is not present. body left to itself releases this stored energy in ⊳ the form of kinetic energy. Let us make our notion Example 5.6 A block of mass m = 1 kg, of potential energy more concrete. moving on a horizontal surface with speed The gravitational force on a ball of mass m is vi = 2 m s–1 enters a rough patch ranging mg . g may be treated as a constant near the earth from x = 0.10 m to x = 2.01 m. The retarding surface. By ‘near’ we imply that the height h of force Fr on the block in this range is inversely the ball above the earth’s surface is very small proportional to x over this range, compared to the earth’s radius RE (h <<RE) so that −k we can ignore the variation of g near the earth’s Fr = for 0.1 < x < 2.01 m surface*. In what follows we have taken the x upward direction to be positive. Let us raise the = 0 for x < 0.1m and x > 2.01 m ball up to a height h. The work done by the external where k = 0.5 J. What is the final kinetic agency against the gravitational force is mgh. This energy and speed vf of the block as it work gets stored as potential energy. crosses this patch ? Gravitational potential energy of an object, as a function of the height h, is denoted by V(h) and it Answer From Eq. (5.8a) is the negative of work done by the gravitational 2.01 ( −k ) force in raising the object to that height. d x V (h) = mgh K f = K i + ∫ x 0.1 If h is taken as a variable, it is easily seen that the gravitational force F equals the negative of 1 2 2.01 = mv i − k ln ( x ) 0.1 the derivative of V(h) with respect to h. Thus, 2 d F = − V(h) = −m g 1 2 d h = mv i − k ln (2.01/0.1) 2 The negative sign indicates that the = 2 − 0.5 ln (20.1) gravitational force is downward. When released, the ball comes down with an increasing speed. = 2 − 1.5 = 0.5 J Just before it hits the ground, its speed is given v f = 2K f / m = 1 m s−1 by the kinematic relation, v2 = 2gh This equation can be written as Here, note that ln is a symbol for the natural 1logarithm to the base e and not the logarithm to the base 10 [ln X = loge X = 2.303 log10 X]. ⊳ 2 m v2 = m g h which shows that the gravitational potential5.7 THE CONCEPT OF POTENTIAL ENERGY energy of the object at height h, when the object The word potential suggests possibility or is released, manifests itself as kinetic energy of capacity for action. The term potential energy the object on reaching the ground. brings to one’s mind ‘stored’ energy. A stretched Physically, the notion of potential energy is bow-string possesses potential energy. When it applicable only to the class of forces where work is released, the arrow flies off at a great speed. done against the force gets ‘stored up’ as energy. The earth’s crust is not uniform, but has When external constraints are removed, it discontinuities and dislocations that are called manifests itself as kinetic energy. Mathematically, fault lines. These fault lines in the earth’s crust (for simplicity, in one dimension) the potential * The variation of g with height is discussed in Chapter 7 on Gravitation. Reprint 2025-26 78 PHYSICS energy V(x) is defined if the force F(x) can be which means that K + V, the sum of the kinetic written as and potential energies of the body is a constant. Over the whole path, xi to xf, this means that d V F ( x ) = − d x Ki + V(xi ) = Kf + V(xf) (5.11) The quantity K +V(x), is called the totalThis implies that mechanical energy of the system. Individually xf Vf the kinetic energy K and the potential energy ∫ F(x) d x = − ∫ d V = Vi − V f V(x) may vary from point to point, but the sum x i Vi is a constant. The aptness of the term The work done by a conservative force such as ‘conservative force’ is now clear. gravity depends on the initial and final positions Let us consider some of the definitions of a only. In the previous chapter we have worked conservative force. on examples dealing with inclined planes. If an l A force F(x) is conservative if it can be derived object of mass m is released from rest, from the from a scalar quantity V(x) by the relation top of a smooth (frictionless) inclined plane of given by Eq. (5.9). The three-dimensional height h, its speed at the bottom generalisation requires the use of a vector is 2 gh irrespective of the angle of inclination. derivative, which is outside the scope of this book.Thus, at the bottom of the inclined plane it l The work done by the conservative forceacquires a kinetic energy, mgh. If the work done depends only on the end points. This can be or the kinetic energy did depend on other factors seen from the relation, such as the velocity or the particular path taken W = Kf – Ki = V (xi) – V(xf)by the object, the force would be called non- which depends on the end points. conservative. l A third definition states that the work done The dimensions of potential energy are by this force in a closed path is zero. This is [ML2T –2] and the unit is joule (J), the same as once again apparent from Eq. (5.11) since kinetic energy or work. To reiterate, the change xi = xf .in potential energy, for a conservative force, ∆V is equal to the negative of the work done by Thus, the principle of conservation of total mechanical energy can be stated asthe force ∆V = − F(x) ∆x (5.9) The total mechanical energy of a system is In the example of the falling ball considered in conserved if the forces, doing work on it, are this section we saw how potential energy was conservative. The above discussion can be made moreconverted to kinetic energy. This hints at an concrete by considering the example of theimportant principle of conservation in mechanics, gravitational force once again and that of thewhich we now proceed to examine. spring force in the next section. Fig. 5.5 depicts

5.10 POWER Often it is interesting to know not only the work u Example 5.10 An elevator can carry a done on an object, but also the rate at which maximum load of 1800 kg (elevator + this work is done. We say a person is physically passengers) is moving up with a constant fit if he not only climbs four floors of a building speed of 2 m s–1. The frictional force opposing but climbs them fast. Power is defined as the the motion is 4000 N. Determine the time rate at which work is done or energy is minimum power delivered by the motor to transferred. the elevator in watts as well as in horse The average power of a force is defined as the power. ratio of the work, W, to the total time t taken Answer The downward force on the elevator is W Pav = F = m g + = (1800 × 10) + 4000 = 22000 N t Ff The motor must supply enough power to balanceThe instantaneous power is defined as the this force. Hence,limiting value of the average power as time interval approaches zero, P = F. v = 22000 × 2 = 44000 W = 59 hp ⊳ d W 5.11 COLLISIONS P = (5.20) d t In physics we study motion (change in position). The work dW done by a force F for a displacement At the same time, we try to discover physical dr is dW = F.dr. The instantaneous power can quantities, which do not change in a physical also be expressed as process. The laws of momentum and energy conservation are typical examples. In this d r P = F. section we shall apply these laws to a commonly d t encountered phenomena, namely collisions. Several games such as billiards, marbles or = F.v (5.21) carrom involve collisions.We shall study the where v is the instantaneous velocity when the collision of two masses in an idealised form. force is F. Consider two masses m1 and m2. The particle Power, like work and energy, is a scalar m1 is moving with speed v1i , the subscript ‘i’ quantity. Its dimensions are [ML2T–3]. In the SI, implying initial. We can cosider m2 to be at rest. its unit is called a watt (W). The watt is 1 J s–1. No loss of generality is involved in making such The unit of power is named after James Watt, a selection. In this situation the mass m1 one of the innovators of the steam engine in the collides with the stationary mass m2 and this eighteenth century. is depicted in Fig. 5.10. There is another unit of power, namely the horse-power (hp) 1 hp = 746 W This unit is still used to describe the output of automobiles, motorbikes, etc. We encounter the unit watt when we buy electrical goods such as bulbs, heaters and refrigerators. A 100 watt bulb which is on for 10 hours uses 1 kilowatt hour (kWh) of energy. 100 (watt) × 10 (hour) Fig. 5.10 Collision of mass m1, with a stationary mass m2. = 1000 watt hour The masses m1 and m2 fly-off in different =1 kilowatt hour (kWh) directions. We shall see that there are = 103 (W) × 3600 (s) relationships, which connect the masses, the = 3.6 × 106 J velocities and the angles. Reprint 2025-26 84 PHYSICS 5.11.1 Elastic and Inelastic Collisions The loss in kinetic energy on collision is In all collisions the total linear momentum is 1 2 1 2 conserved; the initial momentum of the system ∆ K = m 1 v1i − (m 1 + m 2 )v f 2 2 is equal to the final momentum of the system. One can argue this as follows. When two objects 2 1 2 1 m 1 2 collide, the mutual impulsive forces acting over = m 1v1i − v1i [using Eq. (5.22)] the collision time ∆t cause a change in their 2 2 m 1 + m 2 respective momenta : 1 2  m 1  ∆p1 = F12 ∆t = 2 m 1v1i 1 − m 1 + m 2  ∆p2 = F21 ∆t where F12 is the force exerted on the first particle 1 m 1m 2 2by the second particle. F21 is likewise the force = v1i 2 m 1 + m 2exerted on the second particle by the first particle. Now from Newton’s third law, F12 = − F21. This which is a positive quantity as expected.implies ∆p1 + ∆p2 = 0 Consider next an elastic collision. Using the above nomenclature with θ1 = θ2 = 0, the The above conclusion is true even though the momentum and kinetic energy conservation forces vary in a complex fashion during the equations are collision time ∆t. Since the third law is true at every instant, the total impulse on the first object m1v1i = m1v1f + m2v2f (5.23) is equal and opposite to that on the second. 2 2 2 m 1v1i = m 1v1 f + m 2 v 2 f (5.24) On the other hand, the total kinetic energy of the system is not necessarily conserved. The From Eqs. (5.23) and (5.24) it follows that, impact and deformation during collision may generate heat and sound. Part of the initial kinetic m 1v1i (v 2 f − v1i ) = m 1v1 f (v 2 f − v1 f ) energy is transformed into other forms of energy. A useful way to visualise the deformation during or, v 2 f (v1i − v1 f ) = v12i − v12f collision is in terms of a ‘compressed spring’. If the ‘spring’ connecting the two masses regains = (v1i − v1 f )(v 1i + v 1 f ) its original shape without loss in energy, then the initial kinetic energy is equal to the final Hence, ∴ v 2 f = v1i + v1 f (5.25) kinetic energy but the kinetic energy during the Substituting this in Eq. (5.23), we obtaincollision time ∆t is not constant. Such a collision is called an elastic collision. On the other hand (m 1 − m 2 ) v1 f = v1i (5.26)the deformation may not be relieved and the two m 1 + m 2 bodies could move together after the collision. A 2m 1v1icollision in which the two particles move together and v 2 f = (5.27) m 1 + m 2after the collision is called a completely inelastic collision. The intermediate case where the Thus, the ‘unknowns’ {v1f, v2f} are obtained in deformation is partly relieved and some of the terms of the ‘knowns’ {m1, m2, v1i}. Special cases initial kinetic energy is lost is more common and of our analysis are interesting. is appropriately called an inelastic collision. Case I : If the two masses are equal 5.11.2 Collisions in One Dimension v1f = 0 Consider first a completely inelastic collision v2f = v1i in one dimension. Then, in Fig. 5.10, The first mass comes to rest and pushes off the θ 1 = θ 2 = 0 second mass with its initial speed on collision. Case II : If one mass dominates, e.g. m2 > > m1 m1v1i = (m1+m2)vf (momentum conservation) v1f ~ − v1i v2f ~ 0 m 1 The heavier mass is undisturbed while the v f = v1i (5.22) m 1 + m 2 lighter mass reverses its velocity. Reprint 2025-26 WORK, ENERGY AND POWER 85 ⊳ dimensional, where the initial velocities and the Example 5.11 Slowing down of neutrons: final velocities lie in a plane. In a nuclear reactor a neutron of high speed (typically 107 m s–1) must be slowed 5.11.3 Collisions in Two Dimensions to 103 m s–1 so that it can have a high Fig. 5.10 also depicts the collision of a moving probability of interacting with isotope 235U92 mass m1 with the stationary mass m2. Linear momentum is conserved in such a collision. and causing it to fission. Show that a neutron can lose most of its kinetic energy Since momentum is a vector this implies three in an elastic collision with a light nuclei equations for the three directions {x, y, z}. like deuterium or carbon which has a mass Consider the plane determined by the final of only a few times the neutron mass. The velocity directions of m1 and m2 and choose it to material making up the light nuclei, usually be the x-y plane. The conservation of the heavy water (D2O) or graphite, is called a z-component of the linear momentum implies moderator. that the entire collision is in the x-y plane. The x- and y-component equations are Answer The initial kinetic energy of the neutron m1v1i = m1v1f cos θ1 + m2v2f cos θ2 (5.28)is 0 = m1v1f sin θ1 − m2v2f sin θ2 (5.29) 1 2 K 1i = m 1v1i 2 One knows {m1, m2, v1i} in most situations. There are thus four unknowns {v1f, v2f , θ1 and θ2}, andwhile its final kinetic energy from Eq. (5.26) only two equations. If θ 1 = θ 2 = 0, we regain 1 2 1  m 1 − m 2  2 2 Eq. (5.23) for one dimensional collision. K 1 f = m 1v1 f = m 1 v1i 2 2  m 1 + m 2  If, further the collision is elastic, 1 2 1 2 1 2 m1v1i = m1v1 f + m 2v 2 f (5.30) The fractional kinetic energy lost is 2 2 2 2 We obtain an additional equation. That still K 1 f  m 1 − m 2  f 1 = = leaves us one equation short. At least one of K 1i  m 1 + m 2  the four unknowns, say θ1, must be made known while the fractional kinetic energy gained by the for the problem to be solvable. For example, θ1 moderating nuclei K2f /K1i is can be determined by moving a detector in an angular fashion from the x to the y axis. Given f2 = 1 − f1 (elastic collision) {m1, m2, v1i , θ1} we can determine {v1f , v2f, θ2} 4m 1m 2 from Eqs. (5.28)-(5.30). = 2 ⊳ (m1 + m 2 ) Example 5.12 Consider the collision depicted in Fig. 5.10 to be between two One can also verify this result by substituting billiard balls with equal masses m1 = m2.from Eq. (5.27). The first ball is called the cue while the For deuterium m2 = 2m1 and we obtain second ball is called the target. The f1 = 1/9 while f2 = 8/9. Almost 90% of the billiard player wants to ‘sink’ the target neutron’s energy is transferred to deuterium. For ball in a corner pocket, which is at an carbon f1 = 71.6% and f2 = 28.4%. In practice, angle θ2 = 37°. Assume that the collision ishowever, this number is smaller since head-on elastic and that friction and rotational collisions are rare. ⊳ motion are not important. Obtain θ1. If the initial velocities and final velocities of Answer From momentum conservation, since both the bodies are along the same straight line, the masses are equal then it is called a one-dimensional collision, or head-on collision. In the case of small spherical v1i = v 1f + v 2f bodies, this is possible if the direction of travel of body 1 passes through the centre of body 2 ⋅ or v 1i 2 = ( v1 f + v 2 f ) ( v1 f + v 2 f ) which is at rest. In general, the collision is two- = v1 f 2 + v 2 f 2 + 2 v1 f .v 2 f Reprint 2025-26 86 PHYSICS 2 2 The matter simplifies greatly if we consider= (5.31) { v1 f + v 2 f + 2v1 f v 2 f cos (θ1 + 37 ° ) } spherical masses with smooth surfaces, and assume that collision takes place only when the Since the collision is elastic and m1 = m2 it follows bodies touch each other. This is what happensfrom conservation of kinetic energy that in the games of marbles, carrom and billiards. v1i 2 = v1 f 2 + v 2 f 2 (5.32) In our everyday world, collisions take place only when two bodies touch each other. But considerComparing Eqs. (5.31) and (5.32), we get a comet coming from far distances to the sun, or cos (θ1 + 37°) = 0 alpha particle coming towards a nucleus and or θ1 + 37° = 90° going away in some direction. Here we have to deal with forces involving action at a distance. Thus, θ1 = 53° Such an event is called scattering. The velocities This proves the following result : when two equal and directions in which the two particles go away masses undergo a glancing elastic collision with depend on their initial velocities as well as the one of them at rest, after the collision, they will type of interaction between them, their masses, move at right angles to each other. ⊳ shapes and sizes. SUMMARY 1. The work-energy theorem states that the change in kinetic energy of a body is the work done by the net force on the body. Kf - Ki = Wnet 2. A force is conservative if (i) work done by it on an object is path independent and depends only on the end points {xi, xj}, or (ii) the work done by the force is zero for an arbitrary closed path taken by the object such that it returns to its initial position. 3. For a conservative force in one dimension, we may define a potential energy function V(x) such that d V ( x ) F ( x ) = − d x x f F ( x ) d x or Vi − V f = ∫ x i 4. The principle of conservation of mechanical energy states that the total mechanical energy of a body remains constant if the only forces that act on the body are conservative. 5. The gravitational potential energy of a particle of mass m at a height x about the earth’s surface is V(x) = m g x where the variation of g with height is ignored. 5. The elastic potential energy of a spring of force constant k and extension x is 1 2 V ( x ) = k x 2 7. The scalar or dot product of two vectors A and B is written as A.B and is a scalar quantity given by :A.B = AB cos θ, where θ is the angle between A and B. It can be positive, negative or zero depending upon the value of θ. The scalar product of two vectors can be interpreted as the product of magnitude of one vector and component of the other vector along the first vector. For unit vectors : ˆi ⋅ ˆi = ˆj ⋅ ˆj = kˆ ⋅ kˆ = 1 and ˆi ⋅ ˆj = ˆj ⋅ kˆ = kˆ ⋅ ˆi = 0 Scalar products obey the commutative and the distributive laws. Reprint 2025-26 WORK, ENERGY AND POWER 87 POINTS TO PONDER 1. The phrase ‘calculate the work done’ is incomplete. We should refer (or imply clearly by context) to the work done by a specific force or a group of forces on a given body over a certain displacement. 2. Work done is a scalar quantity. It can be positive or negative unlike mass and kinetic energy which are positive scalar quantities. The work done by the friction or viscous force on a moving body is negative. 3. For two bodies, the sum of the mutual forces exerted between them is zero from Newton’s Third Law, F12 + F21 = 0 But the sum of the work done by the two forces need not always cancel, i.e. W12 + W21 ≠ 0 However, it may sometimes be true. 4. The work done by a force can be calculated sometimes even if the exact nature of the force is not known. This is clear from Example 5.2 where the WE theorem is used in such a situation. 5. The WE theorem is not independent of Newton’s Second Law. The WE theorem may be viewed as a scalar form of the Second Law. The principle of conservation of mechanical energy may be viewed as a consequence of the WE theorem for conservative forces. 5. The WE theorem holds in all inertial frames. It can also be extended to non- inertial frames provided we include the pseudoforces in the calculation of the net force acting on the body under consideration. 7. The potential energy of a body subjected to a conservative force is always undetermined upto a constant. For example, the point where the potential energy is zero is a matter of choice. For the gravitational potential energy mgh, the zero of the potential energy is chosen to be the ground. For the spring potential energy kx2/2 , the zero of the potential energy is the equilibrium position of the oscillating mass. 8. Every force encountered in mechanics does not have an associated potential energy. For example, work done by friction over a closed path is not zero and no potential energy can be associated with friction. 9. During a collision : (a) the total linear momentum is conserved at each instant of the collision ; (b) the kinetic energy conservation (even if the collision is elastic) applies after the collision is over and does not hold at every instant of the collision. In fact the two colliding objects are deformed and may be momentarily at rest with respect to each other. Reprint 2025-26 88 PHYSICS EXERCISES 5.1 The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative: (a) work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket. (b) work done by gravitational force in the above case, (c) work done by friction on a body sliding down an inclined plane, (d) work done by an applied force on a body moving on a rough horizontal plane with uniform velocity, (e) work done by the resistive force of air on a vibrating pendulum in bringing it to rest. 5.2 A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1. Compute the (a) work done by the applied force in 10 s, (b) work done by friction in 10 s, (c) work done by the net force on the body in 10 s, (d) change in kinetic energy of the body in 10 s, and interpret your results. 5.3 Given in Fig. 5.11 are examples of some potential energy functions in one dimension. The total energy of the particle is indicated by a cross on the ordinate axis. In each case, specify the regions, if any, in which the particle cannot be found for the given energy. Also, indicate the minimum total energy the particle must have in each case. Think of simple physical contexts for which these potential energy shapes are relevant. Fig. 5.11 Reprint 2025-26 WORK, ENERGY AND POWER 89 5.4 The potential energy function for a particle executing linear simple harmonic motion is given by V(x) = kx2/2, where k is the force constant of the oscillator. For k = 0.5 N m-1, the graph of V(x) versus x is shown in Fig. 5.12. Show that a particle of total energy 1 J moving under this Fig. 5.12 potential must ‘turn back’ when it reaches x = ± 2 m. 5.5 Answer the following : (a) The casing of a rocket in flight burns up due to friction. At whose expense is the heat energy required for burning obtained? The rocket or the atmosphere? (b) Comets move around the sun in highly elliptical orbits. The gravitational force on the comet due to the sun is not Fig. 5.13 normal to the comet’s velocity in general. Yet the work done by the gravitational force over every complete orbit of the comet is zero. Why ? (c) An artificial satellite orbiting the earth in very thin atmosphere loses its energy gradually due to dissipation against atmospheric resistance, however small. Why then does its speed increase progressively as it comes closer and closer to the earth ? (d) In Fig. 5.13(i) the man walks 2 m carrying a mass of 15 kg on his hands. In Fig. 5.13(ii), he walks the same distance pulling the rope behind him. The rope goes over a pulley, and a mass of 15 kg hangs at its other end. In which case is the work done greater ? 5.6 Underline the correct alternative : (a) When a conservative force does positive work on a body, the potential energy of the body increases/decreases/remains unaltered. (b) Work done by a body against friction always results in a loss of its kinetic/potential energy. (c) The rate of change of total momentum of a many-particle system is proportional to the external force/sum of the internal forces on the system. (d) In an inelastic collision of two bodies, the quantities which do not change after the collision are the total kinetic energy/total linear momentum/total energy of the system of two bodies. 5.7 State if each of the following statements is true or false. Give reasons for your answer. (a) In an elastic collision of two bodies, the momentum and energy of each body is conserved. (b) Total energy of a system is always conserved, no matter what internal and external forces on the body are present. (c) Work done in the motion of a body over a closed loop is zero for every force in nature. (d) In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system. 5.8 Answer carefully, with reasons : (a) In an elastic collision of two billiard balls, is the total kinetic energy conserved during the short time of collision of the balls (i.e. when they are in contact) ? (b) Is the total linear momentum conserved during the short time of an elastic collision of two balls ? Reprint 2025-26 90 PHYSICS (c) What are the answers to (a) and (b) for an inelastic collision ? (d) If the potential energy of two billiard balls depends only on the separation distance between their centres, is the collision elastic or inelastic ? (Note, we are talking here of potential energy corresponding to the force during collision, not gravitational potential energy). 5.9 A body is initially at rest. It undergoes one-dimensional motion with constant acceleration. The power delivered to it at time t is proportional to (i) t1/2 (ii) t (iii) t3/2 (iv) t2 5.10 A body is moving unidirectionally under the influence of a source of constant power. Its displacement in time t is proportional to (i) t1/2 (ii) t (iii) t3/2 (iv) t2 5.11 A body constrained to move along the z-axis of a coordinate system is subject to a constant force F given by F = −ˆi + 2 ˆj + 3 kˆ N where ˆ,i ˆ,j kˆ are unit vectors along the x-, y- and z-axis of the system respectively. What is the work done by this force in moving the body a distance of 4 m along the z-axis ? 5.12 An electron and a proton are detected in a cosmic ray experiment, the first with kinetic energy 10 keV, and the second with 100 keV. Which is faster, the electron or the proton ? Obtain the ratio of their speeds. (electron mass = 9.11×10-31 kg, proton mass = 1.67×10–27 kg, 1 eV = 1.60 ×10–19 J). 5.13 A rain drop of radius 2 mm falls from a height of 500 m above the ground. It falls with decreasing acceleration (due to viscous resistance of the air) until at half its original height, it attains its maximum (terminal) speed, and moves with uniform speed thereafter. What is the work done by the gravitational force on the drop in the first and second half of its journey ? What is the work done by the resistive force in the entire journey if its speed on reaching the ground is 10 m s–1 ? 5.14 A molecule in a gas container hits a horizontal wall with speed 200 m s–1 and angle 30° with the normal, and rebounds with the same speed. Is momentum conserved in the collision ? Is the collision elastic or inelastic ? 5.15 A pump on the ground floor of a building can pump up water to fill a tank of volume 30 m3 in 15 min. If the tank is 40 m above the ground, and the efficiency of the pump is 30%, how much electric power is consumed by the pump ? 5.16 Two identical ball bearings in contact with each other and resting on a frictionless table are hit head-on by another ball bearing of the same mass moving initially with a speed V. If the collision is elastic, which of the following (Fig. 5.14) is a possible result after collision ? Fig. 5.14 Reprint 2025-26 WORK, ENERGY AND POWER 91 5.17 The bob A of a pendulum released from 30o to the vertical hits another bob B of the same mass at rest on a table as shown in Fig. 5.15. How high does the bob A rise after the collision ? Neglect the size of the bobs and assume the collision to be elastic. 5.18 The bob of a pendulum is released from a horizontal position. If the length of the pendulum is 1.5 m, what is the speed with which the bob arrives at the lowermost point, given that it dissipated 5% of its initial energy against air resistance ? 5.19 A trolley of mass 300 kg carrying a sandbag of 25 kg is moving uniformly with a speed of 27 km/h on a Fig. 5.15 frictionless track. After a while, sand starts leaking out of a hole on the floor of the trolley at the rate of 0.05 kg s–1. What is the speed of the trolley after the entire sand bag is empty ? 5.20 A body of mass 0.5 kg travels in a straight line with velocity v =a x3/2 where a = 5 m–1/2 s–1. What is the work done by the net force during its displacement from x = 0 to x = 2 m ? 5.21 The blades of a windmill sweep out a circle of area A. (a) If the wind flows at a velocity v perpendicular to the circle, what is the mass of the air passing through it in time t ? (b) What is the kinetic energy of the air ? (c) Assume that the windmill converts 25% of the wind’s energy into electrical energy, and that A = 30 m2, v = 36 km/h and the density of air is 1.2 kg m–3. What is the electrical power produced ? 5.22 A person trying to lose weight (dieter) lifts a 10 kg mass, one thousand times, to a height of 0.5 m each time. Assume that the potential energy lost each time she lowers the mass is dissipated. (a) How much work does she do against the gravitational force ? (b) Fat supplies 3.8 × 107J of energy per kilogram which is converted to mechanical energy with a 20% efficiency rate. How much fat will the dieter use up? 5.23 A family uses 8 kW of power. (a) Direct solar energy is incident on the horizontal surface at an average rate of 200 W per square meter. If 20% of this energy can be converted to useful electrical energy, how large an area is needed to supply 8 kW? (b) Compare this area to that of the roof of a typical house. Reprint 2025-26 CHAPTER SIX SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 6.1 INTRODUCTION In the earlier chapters we primarily considered the motion of a single particle. (A particle is ideally represented as a 6.1 Introduction point mass having no size.) We applied the results of our 6.2 Centre of mass study even to the motion of bodies of finite size, assuming 6.3 Motion of centre of mass that motion of such bodies can be described in terms of the

1.3 SIGNIFICANT FIGURES figures in a measurement. This important remark makes most of the followingAs discussed above, every measurement observations clear:involves errors. Thus, the result of (1) For example, the length 2.308 cm has fourmeasurement should be reported in a way that significant figures. But in different units, theindicates the precision of measurement. same value can be written as 0.02308 m or 23.08Normally, the reported result of measurement is a number that includes all digits in the mm or 23080 µm. number that are known reliably plus the first All these numbers have the same number of digit that is u Reprint 2025-26 4 PHYSICS This shows that the location of decimal point is negative exponent (or power) of 10. In order to of no consequence in determining the number get an approximate idea of the number, we may of significant figures. round off the number a to 1 (for a ≤5) and to 10 The example gives the following rules : (for 5<a ≤10). Then the number can be • All the non-zero digits are significant. expressed approximately as 10b in which the • All the zeros between two non-zero digits exponent (or power) b of 10 is called order of magnitude of the physical quantity. When only are significant, no matter where the an estimate is required, the quantity is of the decimal point is, if at all. order of 10b. For example, the diameter of the • If the number is less than 1, the zero(s) earth (1.28×107m) is of the order of 107m with on the right of decimal point but to the the order of magnitude 7. The diameter of left of the first non-zero digit are not hydrogen atom (1.06 ×10–10m) is of the order of significant. [In 0.00 2308, the underlined 10–10m, with the order of magnitude zeroes are not significant]. –10. Thus, the diameter of the earth is 17 orders • The terminal or trailing zero(s) in a of magnitude larger than the hydrogen atom. It is often customary to write the decimal after number without a decimal point are not the first digit. Now the confusion mentioned in significant. (a) above disappears : [Thus 123 m = 12300 cm = 123000 mm has 4.700 m = 4.700 × 102 cm three significant figures, the trailing zero(s) = 4.700 × 103 mm = 4.700 × 10–3 km being not significant.] However, you can also The power of 10 is irrelevant to the see the next observation. determination of significant figures. However, all • The trailing zero(s) in a number with a zeroes appearing in the base number in the decimal point are significant. scientific notation are significant. Each number [The numbers 3.500 or 0.06900 have four in this case has four significant figures. significant figures each.] Thus, in the scientific notation, no confusion (2) There can be some confusion regarding the arises about the trailing zero(s) in the base trailing zero(s). Suppose a length is reported to number a. They are always significant. be 4.700 m. It is evident that the zeroes here (4) The scientific notation is ideal for reporting are meant to convey the precision of measurement. But if this is not adopted, we use measurement and are, therefore, significant. [If the rules adopted in the preceding example : these were not, it would be superfluous to write • For a number greater than 1, without anythem explicitly, the reported measurement decimal, the trailing zero(s) are notwould have been simply 4.7 m]. Now suppose we change units, then significant. • For a number with a decimal, the trailing4.700 m = 470.0 cm = 4700 mm = 0.004700 km zero(s) are significant. Since the last number has trailing zero(s) in a number with no decimal, we would conclude (5) The digit 0 conventionally put on the left of a erroneously from observation (1) above that the decimal for a number less than 1 (like 0.1250) number has two significant figures, while in is never significant. However, the zeroes at the fact, it has four significant figures and a mere end of such number are significant in a change of units cannot change the number of measurement. significant figures. (6) The multiplying or dividing factors which are (3) To remove such ambiguities in neither rounded numbers nor numbers determining the number of significant representing measured values are exact and figures, the best way is to report every have infinite number of significant digits. For measurement in scientific notation (in the dpower of 10). In this notation, every number is example in r = or s = 2πr, the factor 2 is an expressed as a × 10b, where a is a number 2 between 1 and 10, and b is any positive or exact number and it can be written as 2.0, 2.00 Reprint 2025-26 UNITS AND MEASUREMENT 5 decimal place. The final result should, therefore, t or 2.0000 as required. Similarly, in T = , n is be rounded off to 663.8 g. n Similarly, the difference in length can be an exact number. expressed as : 1.3.1 Rules for Arithmetic Operations with 0.307 m – 0.304 m = 0.003 m = 3 ×10–3 m. Significant Figures Note that we should not use the rule (1) applicable The result of a calculation involving approximate for multiplication and division and write 664 g as measured values of quantities (i.e. values with the result in the example of addition and limited number of significant figures) must 3.00 × 10–3 m in the example of subtraction. They reflect the umeasured values. It cannot be more accurate properly. For addition and subtraction, the rule than the original measured values themselves is in terms of decimal places. on which the result is based. In general, the final result should not have more significant 1.3.2 Rounding off the Ufigures than the original data from which it was The result of computation with approximate obtained. Thus, if mass of an object is measured numbers, which contain more than one to be, say, 4.237 g (four significant figures) and uits volume is measured to be 2.51 cm3, then its for rounding off numbers to the appropriate density, by mere arithmetic division, is significant figures are obvious in most cases. A 1.68804780876 g/cm3 upto 11 decimal places. number 2.746 rounded off to three significant It would be clearly absurd and irrelevant to figures is 2.75, while the number 1.743 would record the calculated value of density to such a be 1.74. The rule by convention is that the precision when the measurements on which the preceding digit is raised by 1 if the value is based, have much less precision. The insignificant digit to be dropped (the following rules for arithmetic operations with underlined digit in this case) is more than significant figures ensure that the final result 5, and is left unchanged if the latter is less of a calculation is shown with the precision that than 5. But what if the number is 2.745 in is consistent with the precision of the input which the insignificant digit is 5. Here, themeasured values : convention is that if the preceding digit is(1) In multiplication or division, the final even, the insignificant digit is simplyresult should retain as many significant dropped and, if it is odd, the preceding digitfigures as are there in the original number with the least significant figures. is raised by 1. Then, the number 2.745 rounded Thus, in the example above, density should off to three significant figures becomes 1.74. On be reported to three significant figures. the other hand, the number 2.735 rounded off to three significant figures becomes 1.74 since 4.237g -3 Density = = 1.69 g cm the preceding digit is odd. 3 2.51 cm In any involved or complex multi-step Similarly, if the speed of light is given as calculation, you should retain, in intermediate 3.00 × 108 m s-1 (three significant figure) and steps, one digit more than the significant digits one year (1y = 365.25 d) has 3.1557 × 107 s (five and round off to proper significant figures at the significant figures), the light year is 9.47 × 1015 m end of the calculation. Similarly, a number (three significant figures). known to be within many significant figures, such as in 2.99792458 × 108 m/s for the speed (2) In addition or subtraction, the final result of light in vacuum, is rounded off to anshould retain as many decimal places as are approximate value 3 × 108 m/s , which is oftenthere in the number with the least employed in computations. Finally, rememberdecimal places. that exact numbers that appear in formulae like For example, the sum of the numbers 436.32 g, 227.2 g and 0.301 g by mere arithmetic L addition, is 663.821 g. But the least precise 2 π in T = 2π , have a large (infinite) number measurement (227.2 g) is correct to only one g Reprint 2025-26 6 PHYSICS of significant figures. The value of π = = 16.2 cm ± 0.6 %. 3.1415926.... is known to a large number of significant figures. You may take the value as Similarly, the breadth b may be written as 3.142 or 3.14 for π, with limited number of b = 10.1 ± 0.1 cm significant figures as required in specific = 10.1 cm ± 1 % cases. Then, the error of the product of two (or more)⊳ Example 1.1 Each side of a cube is experimental values, using the combination of measured to be 7.203 m. What are the errors rule, will be total surface area and the volume of the l b = 163.62 cm2 + 1.6% cube to appropriate significant figures? = 163.62 + 2.6 cm2 Answer The number of significant figures in the measured length is 4. The calculated area This leads us to quote the final result as and the volume should therefore be rounded off l b = 164 + 3 cm2 to 4 significant figures. Here 3 cm2 is the uSurface area of the cube = 6(7.203)2 m2 estimation of area of rectangular sheet. = 311.299254 m2 (2) If a set of experimental data is specified = 311.3 m2 to n significant figures, a result obtained by combining the data will also be valid to n Volume of the cube = (7.203)3 m3 significant figures. = 373.714754 m3 However, if data are subtracted, the number of = 373.7 m3 ⊳ significant figures can be reduced. ⊳ Example 1.2 5.74 g of a substance For example, 12.9 g – 7.06 g, both specified to occupies 1.2 cm3. Express its density by three significant figures, cannot properly be keeping the significant figures in view. evaluated as 5.84 g but only as 5.8 g, as u in a different fashion (smallest number ofmeasured mass whereas there are only 2 decimal places rather than the number of significant figures in the measured volume. significant figures in any of the number added Hence the density should be expressed to only or subtracted). 2 significant figures. (3) The relative error of a value of number 5.74 −3 specified to significant figures depends not Density = g cm 1.2 only on n but also on the number itself. = 4.8 g cm--3 . ⊳ For example, the accuracy in measurement of mass 1.02 g is ± 0.01 g whereas another 1.3.3 Rules for Determining the U in the Results of Arithmetic The relative error in 1.02 g is Calculations = (± 0.01/1.02) × 100 % = ± 1%The rules for determining the u Similarly, the relative error in 9.89 g iserror in the number/measured quantity in = (± 0.01/9.89) × 100 %arithmetic operations can be understood from = ± 0.1 %the following examples. Finally, remember that intermediate results in(1) If the length and breadth of a thin rectangular sheet are measured, using a metre a multi-step computation should be scale as 16.2 cm and, 10.1 cm respectively, there calculated to one more significant figure in are three significant figures in each every measurement than the number of measurement. It means that the length l may digits in the least precise measurement. be written as These should be justified by the data and then l = 16.2 ± 0.1 cm the arithmetic operations may be carried out; Reprint 2025-26 UNITS AND MEASUREMENT 7 otherwise rounding errors can build up. For mass, one dimension in length, and –2 example, the reciprocal of 9.58, calculated (after dimensions in time. The dimensions in all other rounding off) to the same number of significant base quantities are zero. figures (three) is 0.104, but the reciprocal of Note that in this type of representation, the magnitudes are not considered. It is the quality0.104 calculated to three significant figures is of the type of the physical quantity that enters. 9.62. However, if we had written 1/9.58 = 0.1044 Thus, a change in velocity, initial velocity, and then taken the reciprocal to three significant average velocity, final velocity, and speed are figures, we would have retrieved the original all equivalent in this context. Since all these value of 9.58. quantities can be expressed as length/time, This example justifies the idea to retain one their dimensions are [L]/[T] or [L T–1]. more extra digit (than the number of digits in the least precise measurement) in intermediate 1.5 DIMENSIONAL FORMULAE AND steps of the complex multi-step calculations in DIMENSIONAL EQUATIONS order to avoid additional errors in the process The expression which shows how and which of of rounding off the numbers. the base quantities represent the dimensions of a physical quantity is called the dimensional

1.14 Consider a uniform electric field E = 3 × 103 î N/C. (a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane? (b) What is the flux through the same square if the normal to its plane makes a 60° angle with the x-axis?

1.18 A point charge of 2.0 mC is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?

1.00 nF for two values of R: (i) R = 100 W and (ii) R = 200 W. For the source applied vm = 1 100 V. w0 for this case is = 1.00×106 LC FIGURE 7.14 Variation of im with w for tworad/s. cases: (i) R = 100 W, (ii) R = 200 W, We see that the current amplitude is L = 1.00 mH. maximum at the resonant frequency. Since im = vm / R at resonance, the current amplitude for case (i) is twice to that for case (ii). Resonant circuits have a variety of applications, for example, in the tuning mechanism of a radio or a TV set. The antenna of a radio accepts signals from many broadcasting stations. The signals picked up in the antenna acts as a source in the tuning circuit of the radio, so the circuit can be driven at many frequencies. But to hear one particular radio station, we tune the radio. In tuning, we vary the capacitance of a capacitor in the tuning circuit such that the resonant frequency of the circuit becomes nearly equal to the frequency of the radio signal received. When this happens, the amplitude of the current with the frequency of the signal of the particular radio station in the circuit is maximum. It is important to note that resonance phenomenon is exhibited by a circuit only if both L and C are present in the circuit. Only then do the voltages across L and C cancel each other (both being out of phase) and the current amplitude is vm/R, the total source voltage appearing across R. This means that we cannot have resonance in a RL or RC circuit. 189 Reprint 2025-26 Physics Example 7.6 A resistor of 200 W and a capacitor of 15.0 mF are connected in series to a 220 V, 50 Hz ac source. (a) Calculate the current in the circuit; (b) Calculate the voltage (rms) across the resistor and the capacitor. Is the algebraic sum of these voltages more than the source voltage? If yes, resolve the paradox. Solution Given R = 200 Ω, C = 15.0 µF = 15.0 × 10 −6 F V = 220 V, ν = 50Hz (a) In order to calculate the current, we need the impedance of the circuit. It is Z = R 2 + X C2 = R 2 + (2πνC )− 2 = (200 Ω )2 + (2 × 3.14 × 50 × 15.0 × 10 −6 F )−2 = (200 Ω )2 + (212.3 Ω )2 = 291.67 Ω Therefore, the current in the circuit is V 220 V I = = = 0.755 A Z 291.5 Ω (b) Since the current is the same throughout the circuit, we have V R = I R = (0.755 A)(200 Ω=) 151V VC = I X C = (0.755 A)(212.3 Ω=) 160.3 V The algebraic sum of the two voltages, VR and VC is 311.3 V which is more than the source voltage of 220 V. How to resolve this paradox? As you have learnt in the text, the two voltages are not in the same phase. Therefore, they cannot be added like ordinary numbers. The two voltages are out of phase by ninety degrees. Therefore, the total of these voltages must be obtained using the Pythagorean theorem: V R +C = V R2 + VC2 7.6 = 220 V Thus, if the phase difference between two voltages is properly taken into account, the total voltage across the resistor and the capacitor EXAMPLE is equal to the voltage of the source. 7.7 POWER IN AC CIRCUIT: THE POWER FACTOR We have seen that a voltage v = vm sinwt applied to a series RLC circuit drives a current in the circuit given by i = im sin(wt + f) where v m − 1  X C − X L  i m = Z and φ = tan  R  190 Therefore, the instantaneous power p supplied by the source is Reprint 2025-26 Alternating Current p = v i = (v m sin ωt ) × [ i m sin(ωt + φ)] v m i m = [ cosφ− cos(2ωt + φ)] (7.29) 2 The average power over a cycle is given by the average of the two terms in R.H.S. of Eq. (7.29). It is only the second term which is time-dependent. Its average is zero (the positive half of the cosine cancels the negative half). Therefore, v m i m v m i m P = cos φ = cos φ 2 2 2 = V I cos φ [7.30(a)] This can also be written as, P = I 2 Z cos φ [7.30(b)] So, the average power dissipated depends not only on the voltage and current but also on the cosine of the phase angle φ between them. The quantity cosφ is called the power factor. Let us discuss the following cases: CaseCaseCaseCaseCase (i)(i)(i)(i)(i) Resistive circuit: If the circuit contains only pure R, it is called resistive. In that case φ = 0, cos φ = 1. There is maximum power dissipation. CaseCaseCaseCaseCase (ii)(ii)(ii)(ii)(ii) Purely inductive or capacitive circuit: If the circuit contains only an inductor or capacitor, we know that the phase difference between voltage and current is π/2. Therefore, cos φ = 0, and no power is dissipated even though a current is flowing in the circuit. This current is sometimes referred to as wattless current. CaseCaseCaseCaseCase (iii)(iii)(iii)(iii)(iii) LCR series circuit: In an LCR series circuit, power dissipated is given by Eq. (7.30) where φ = tan–1 (Xc – XL )/ R. So, φ may be non-zero in a RL or RC or RCL circuit. Even in such cases, power is dissipated only in the resistor. CaseCaseCaseCaseCase (iv)(iv)(iv)(iv)(iv) Power dissipated at resonance in LCR circuit: At resonance Xc – XL= 0, and φ = 0. Therefore, cosφ = 1 and P = I2Z = I2 R. That is, maximum power is dissipated in a circuit (through R) at resonance. ExampleExampleExampleExampleExample 7.77.77.77.77.7 (a) For circuits used for transporting electric power, a low power factor implies large power loss in transmission. Explain. (b) Power factor can often be improved by the use of a capacitor of appropriate capacitance in the circuit. Explain. SolutionSolutionSolutionSolutionSolution (a) We know that P = I V cosφ where cosφ is the power factor. To supply a given power at a given voltage, if cosφ is small, we have to increase current accordingly. But this will lead to large power loss (I2R) in transmission. EEEE (b)Suppose in a circuit, current I lags the voltage by an angle φ. Then power factor cosφ =R/Z. EXAMPLEXAMPLEXAMPLEXAMPLEXAMPLE We can improve the power factor (tending to 1) by making Z tend to 7.77.77.77.77.7 us understand, with the help of a phasor diagram (Fig. 7.15) R. Let 191 Reprint 2025-26 Physics FIGURE 7.15 how this can be achieved. Let us resolve I into two components. Ip along the applied voltage V and Iq perpendicular to the applied voltage. Iq as you have learnt in Section 7.7, is called the wattless component since corresponding to this component of current, there is no power loss. IP is known as the power component because it is in phase with the voltage and corresponds to power loss in the circuit. It’s clear from this analysis that if we want to improve power factor, 7.7 we must completely neutralize the lagging wattless current Iq by an equal leading wattless current I¢q. This can be done by connecting a capacitor of appropriate value in parallel so that Iq and I¢q cancel EXAMPLE each other and P is effectively Ip V. Example 7.8 A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which R = 3 W, L = 25.48 mH, and C = 796 mF. Find (a) the impedance of the circuit; (b) the phase difference between the voltage across the source and the current; (c) the power dissipated in the circuit; and (d) the power factor. Solution (a) To find the impedance of the circuit, we first calculate XL and XC. XL = 2 pnL = 2 × 3.14 × 50 × 25.48 × 10–3 W = 8 W 1 X C = 2 π νC 1 = 6 = 4 Ω 2 × 3.14 × 50 × 796 × 10− Therefore, Z = R 2 + ( X L − X C )2 = 3 2 + (8 − 4)2 = 5 W X C − X L 7.8 (b) Phase difference, f = tan–1 R −1  4 − 8  EXAMPLE = tan  3 = −53. 1°192 Reprint 2025-26 Alternating Current Since f is negative, the current in the circuit lags the voltage across the source. (c) The power dissipated in the circuit is P = I 2 R im 1  283  = Now, I = 2 2  5 = 40A EXAMPLE Therefore, P = (40 A )2 × 3 Ω= 4800 W (d) Power factor = cos cos –53.1 0.6 7.8 Example 7.9 Suppose the frequency of the source in the previous example can be varied. (a) What is the frequency of the source at which resonance occurs? (b) Calculate the impedance, the current, and the power dissipated at the resonant condition. Solution (a) The frequency at which the resonance occurs is 1 1 ω0 = = LC 25.48 × 10 −3 × 796 × 10 −6 = 222.1rad/s ω0 221.1 νr = = Hz = 35.4Hz 2 π 2 × 3.14 (b) The impedance Z at resonant condition is equal to the resistance: Z = R = 3 Ω The rms current at resonance is V V  283  1 = = = = 66.7 A Z R  2  3 The power dissipated at resonance is P = I 2 × R = (66.7)2 × 3 = 13.35 kW EXAMPLE You can see that in the present case, power dissipated at resonance is more than the power dissipated in Example 7.8. 7.9 Example 7.10 At an airport, a person is made to walk through the doorway of a metal detector, for security reasons. If she/he is carrying anything made of metal, the metal detector emits a sound. On what principle does this detector work? Solution The metal detector works on the principle of resonance in ac circuits. When you walk through a metal detector, you are, in fact, walking through a coil of many turns. The coil is connected to a capacitor tuned so that the circuit is in resonance. When you walk through with metal in your pocket, the impedance of the EXAMPLE circuit changes – resulting in significant change in current in the circuit. This change in current is detected and the electronic circuitry causes a sound to be emitted as an alarm. 7.10 193 Reprint 2025-26 Physics 7.8 TRANSFORMERS For many purposes, it is necessary to change (or transform) an alternating voltage from one to another of greater or smaller value. This is done with a device called transformer using the principle of mutual induction. A transformer consists of two sets of coils, insulated from each other. They are wound on a soft-iron core, either one on top of the other as in Fig. 7.16(a) or on separate limbs of the core as in Fig. 7.16(b). One of the coils called the primary coil has Np turns. The other coil is called the secondary coil; it has Ns turns. Often the primary coil is the input coil and the secondary coil is the output coil of the transformer. FIGURE 7.16 Two arrangements for winding of primary and secondary coil in a transformer: (a) two coils on top of each other, (b) two coils on separate limbs of the core. When an alternating voltage is applied to the primary, the resulting current produces an alternating magnetic flux which links the secondary and induces an emf in it. The value of this emf depends on the number of turns in the secondary. We consider an ideal transformer in which the primary has negligible resistance and all the flux in the core links both primary and secondary windings. Let f be the flux in each turn in the core at time t due to current in the primary when a voltage vp is applied to it. Then the induced emf or voltage es, in the secondary with Ns turns is dφ εs = − N s (7.31) d t The alternating flux f also induces an emf, called back emf in the primary. This is dφ εp = − N p (7.32) d t But ep = vp. If this were not so, the primary current would be infinite since the primary has zero resistance (as assumed). If the secondary is an open circuit or the current taken from it is small, then to a good approximation es = vs Reprint 2025-26 Alternating Current where vs is the voltage across the secondary. Therefore, Eqs. (7.31) and (7.32) can be written as dφ v s = − N s [7.31(a)] d t dφ v p = − N p [7.32(a)] d t From Eqs. [7.31 (a)] and [7.32 (a)], we have v s N s = (7.33) v p N p Note that the above relation has been obtained using three assumptions: (i) the primary resistance and current are small; (ii) the same flux links both the primary and the secondary as very little flux escapes from the core, and (iii) the secondary current is small. If the transformer is assumed to be 100% efficient (no energy losses), the power input is equal to the power output, and since p = i v, ipvp = isvs (7.34) Although some energy is always lost, this is a good approximation, since a well designed transformer may have an efficiency of more than 95%. Combining Eqs. (7.33) and (7.34), we have i p v s N s = = (7.35) i s v p N p Since i and v both oscillate with the same frequency as the ac source, Eq. (7.35) also gives the ratio of the amplitudes or rms values of corresponding quantities. Now, we can see how a transformer affects the voltage and current. We have:  N s   N p  Vs = V p and I s = I p (7.36)  N p   N s  That is, if the secondary coil has a greater number of turns than the primary (Ns > Np), the voltage is stepped up (Vs > Vp). This type of arrangement is called a step-up transformer. However, in this arrangement, there is less current in the secondary than in the primary (Np/Ns < 1 and Is < Ip). For example, if the primary coil of a transformer has 100 turns and the secondary has 200 turns, Ns/Np = 2 and Np/Ns=1/2. Thus, a 220V input at 10A will step-up to 440 V output at 5.0 A. If the secondary coil has less turns than the primary (Ns < Np), we have a step-down transformer. In this case, Vs < Vp and Is > Ip. That is, the voltage is stepped down, or reduced, and the current is increased. The equations obtained above apply to ideal transformers (without any energy losses). But in actual transformers, small energy losses do occur due to the following reasons: (i) Flux Leakage: There is always some flux leakage; that is, not all of the flux due to primary passes through the secondary due to poor 195 Reprint 2025-26 Physics design of the core or the air gaps in the core. It can be reduced by winding the primary and secondary coils one over the other. (ii) Resistance of the windings: The wire used for the windings has some resistance and so, energy is lost due to heat produced in the wire (I 2R). In high current, low voltage windings, these are minimised by using thick wire. (iii) Eddy currents: The alternating magnetic flux induces eddy currents in the iron core and causes heating. The effect is reduced by using a laminated core. (iv) Hysteresis: The magnetisation of the core is repeatedly reversed by the alternating magnetic field. The resulting expenditure of energy in the core appears as heat and is kept to a minimum by using a magnetic material which has a low hysteresis loss. The large scale transmission and distribution of electrical energy over long distances is done with the use of transformers. The voltage output of the generator is stepped-up (so that current is reduced and consequently, the I 2R loss is cut down). It is then transmitted over long distances to an area sub-station near the consumers. There the voltage is stepped down. It is further stepped down at distributing sub-stations and utility poles before a power supply of 240 V reaches our homes. SUMMARY 1. An alternating voltage v = v m sin ω t applied to a resistor R drives a v m current i = im sinwt in the resistor, i m = . The current is in phase with R the applied voltage. 2. For an alternating current i = im sin wt passing through a resistor R, the average power loss P (averaged over a cycle) due to joule heating is ( 1/2 )i 2mR. To express it in the same form as the dc power (P = I 2R), a special value of current is used. It is called root mean square (rms) current and is donoted by I: i m I = = 0.707 i m 2 Similarly, the rms voltage is defined by vm V = = 0.707 v m 2 We have P = IV = I 2R 3. An ac voltage v = vm sin wt applied to a pure inductor L, drives a current in the inductor i = im sin (wt – p/2), where im = vm/XL. XL = wL is called inductive reactance. The current in the inductor lags the voltage by p/2. The average power supplied to an inductor over one complete cycle is zero. Reprint 2025-26 Alternating Current 4. An ac voltage v = vm sinwt applied to a capacitor drives a current in the capacitor: i = im sin (wt + p/2). Here, v m 1 i m = , X C = is called capacitive reactance. X C ωC The current through the capacitor is p/2 ahead of the applied voltage. As in the case of inductor, the average power supplied to a capacitor over one complete cycle is zero. 5. For a series RLC circuit driven by voltage v = vm sin wt, the current is given by i = im sin (wt + f) v m where i m = R 2 + ( X C − X L )2 −1 X C − X L and φ = tan R is called the impedance of the circuit. Z = R 2 + ( X C − X L )2 The average power loss over a complete cycle is given by P = V I cosf The term cosf is called the power factor. 6. In a purely inductive or capacitive circuit, cosf = 0 and no power is dissipated even though a current is flowing in the circuit. In such cases, current is referred to as a wattless current. 7. The phase relationship between current and voltage in an ac circuit can be shown conveniently by representing voltage and current by rotating vectors called phasors. A phasor is a vector which rotates about the origin with angular speed w. The magnitude of a phasor represents the amplitude or peak value of the quantity (voltage or current) represented by the phasor. The analysis of an ac circuit is facilitated by the use of a phasor diagram. 8. A transformer consists of an iron core on which are bound a primary coil of Np turns and a secondary coil of Ns turns. If the primary coil is connected to an ac source, the primary and secondary voltages are related by  N s  Vs =  N p  V p and the currents are related by  N p  Is = I p  N s  If the secondary coil has a greater number of turns than the primary, the voltage is stepped-up (Vs > Vp). This type of arrangement is called a step- up transformer. If the secondary coil has turns less than the primary, we have a step-down transformer. 197 Reprint 2025-26 Physics Physical quantity Symbol Dimensions Unit Remarks 2 –3 v m rms voltage V [M L T A –1] V V = , v m is the 2 amplitude of the ac voltage. i m rms current I [ A] A I = , im is the amplitude of 2 the ac current. Reactance: Inductive XL [M L2 T –3 A–2] XL = L Capacitive XC [M L2 T –3 A–2] XC = 1/C Impedance Z [M L2 T –3 A–2] Depends on elements present in the circuit. 1 Resonant wr or w0 [T –1] Hz w0  for a frequency LC series RLC circuit ω0 L 1 Quality factor Q Dimensionless Q = = for a series R ω0 C R RLC circuit. Power factor Dimensionless = cosf, f is the phase difference between voltage applied and current in the circuit. POINTS TO PONDER 1. When a value is given for ac voltage or current, it is ordinarily the rms value. The voltage across the terminals of an outlet in your room is normally 240 V. This refers to the rms value of the voltage. The amplitude of this voltage is v m = 2V = 2(240) = 340 V 2. The power rating of an element used in ac circuits refers to its average power rating. 3. The power consumed in an ac circuit is never negative. 4. Both alternating current and direct current are measured in amperes. But how is the ampere defined for an alternating current? It cannot be derived from the mutual attraction of two parallel wires carrying ac 198 currents, as the dc ampere is derived. An ac current changes direction Reprint 2025-26 Alternating Current with the source frequency and the attractive force would average to zero. Thus, the ac ampere must be defined in terms of some property that is independent of the direction of the current. Joule heating is such a property, and there is one ampere of rms value of alternating current in a circuit if the current produces the same average heating effect as one ampere of dc current would produce under the same conditions. 5. In an ac circuit, while adding voltages across different elements, one should take care of their phases properly. For example, if VR and VC are voltages across R and C, respectively in an RC circuit, then the total voltage across RC combination is V RC = VR2 + VC2 and not VR + VC since VC is p/2 out of phase of VR. 6. Though in a phasor diagram, voltage and current are represented by vectors, these quantities are not really vectors themselves. They are scalar quantities. It so happens that the amplitudes and phases of harmonically varying scalars combine mathematically in the same way as do the projections of rotating vectors of corresponding magnitudes and directions. The ‘rotating vectors’ that represent harmonically varying scalar quantities are introduced only to provide us with a simple way of adding these quantities using a rule that we already know as the law of vector addition. 7. There are no power losses associated with pure capacitances and pure inductances in an ac circuit. The only element that dissipates energy in an ac circuit is the resistive element. 8. In a RLC circuit, resonance phenomenon occur when XL = XC or 1 ω0 = . For resonance to occur, the presence of both L and C LC elements in the circuit is a must. With only one of these (L or C ) elements, there is no possibility of voltage cancellation and hence, no resonance is possible. 9. The power factor in a RLC circuit is a measure of how close the circuit is to expending the maximum power. 10. In generators and motors, the roles of input and output are reversed. In a motor, electric energy is the input and mechanical energy is the output. In a generator, mechanical energy is the input and electric energy is the output. Both devices simply transform energy from one form to another. 11. A transformer (step-up) changes a low-voltage into a high-voltage. This does not violate the law of conservation of energy. The current is reduced by the same proportion. 199 Reprint 2025-26 Physics EXERCISES 7.1 A 100 W resistor is connected to a 220 V, 50 Hz ac supply. (a) What is the rms value of current in the circuit? (b) What is the net power consumed over a full cycle? 7.2 (a) The peak voltage of an ac supply is 300 V. What is the rms voltage? (b) The rms value of current in an ac circuit is 10 A. What is the peak current? 7.3 A 44 mH inductor is connected to 220 V, 50 Hz ac supply. Determine the rms value of the current in the circuit. 7.4 A 60 mF capacitor is connected to a 110 V, 60 Hz ac supply. Determine the rms value of the current in the circuit. 7.5 In Exercises 7.3 and 7.4, what is the net power absorbed by each circuit over a complete cycle. Explain your answer. 7.6 A charged 30 mF capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit? 7.7 A series LCR circuit with R = 20 W, L = 1.5 H and C = 35 mF is connected to a variable-frequency 200 V ac supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle? 7.8 Figure 7.17 shows a series LCR circuit connected to a variable frequency 230 V source. L = 5.0 H, C = 80mF, R = 40 W. FIGURE 7.17 (a) Determine the source frequency which drives the circuit in resonance. (b) Obtain the impedance of the circuit and the amplitude of current at the resonating frequency. (c) Determine the rms potential drops across the three elements of the circuit. Show that the potential drop across the LC combination is zero at the resonating frequency. Reprint 2025-26 Chapter Eight ELECTROMAGNETIC WAVES 8.1 INTRODUCTION In Chapter 4, we learnt that an electric current produces magnetic field and that two current-carrying wires exert a magnetic force on each other. Further, in Chapter 6, we have seen that a magnetic field changing with time gives rise to an electric field. Is the converse also true? Does an electric field changing with time give rise to a magnetic field? James Clerk Maxwell (1831-1879), argued that this was indeed the case – not only an electric current but also a time-varying electric field generates magnetic field. While applying the Ampere’s circuital law to find magnetic field at a point outside a capacitor connected to a time-varying current, Maxwell noticed an inconsistency in the Ampere’s circuital law. He suggested the existence of an additional current, called by him, the displacement current to remove this inconsistency. Maxwell formulated a set of equations involving electric and magnetic fields, and their sources, the charge and current densities. These equations are known as Maxwell’s equations. Together with the Lorentz force formula (Chapter 4), they mathematically express all the basic laws of electromagnetism. The most important prediction to emerge from Maxwell’s equations is the existence of electromagnetic waves, which are (coupled) time- varying electric and magnetic fields that propagate in space. The speed of the waves, according to these equations, turned out to be very close to Reprint 2025-26 Physics the speed of light( 3 ×108 m/s), obtained from optical measurements. This led to the remarkable conclusion that light is an electromagnetic wave. Maxwell’s work thus unified the domain of electricity, magnetism and light. Hertz, in 1885, experimentally demonstrated the existence of electromagnetic waves. Its technological use by Marconi and others led in due course to the revolution in communication that we are witnessing today. In this chapter, we first discuss the need for displacement current and its consequences. Then we present a descriptive account of electromagnetic waves. James Clerk Maxwell The broad spectrum of electromagnetic waves, (1831 – 1879) Born in stretching from g rays (wavelength ~10–12 m) to long Edinburgh, Scotland, radio waves (wavelength ~106 m) is described. was among the greatest physicists of the nineteenth century. He 8.2 DISPLACEMENT CURRENT derived the thermal We have seen in Chapter 4 that an electrical current velocity distribution of produces a magnetic field around it. Maxwell showed molecules in a gas and that for logical consistency, a changing electric field must was among the first to obtain reliable also produce a magnetic field. This effect is of great estimates of molecular importance because it explains the existence of radio parameters from waves, gamma rays and visible light, as well as all other measurable quantities forms of electromagnetic waves. like viscosity, etc. To see how a changing electric field gives rise to Maxwell’s greatest a magnetic field, let us consider the process of acheivement was the unification of the laws of charging of a capacitor and apply Ampere’s circuital electricity and law given by (Chapter 4) magnetism (discovered by Coulomb, Oersted, “B.dl = m0 i (t) (8.1)(1831–1879) Ampere and Faraday) to find magnetic field at a point outside the capacitor. into a consistent set of Figure 8.1(a) shows a parallel plate capacitor C which equations now called Maxwell’s equations. is a part of circuit through which a time-dependent From these he arrived at current i (t) flows . Let us find the magnetic field at a the most important point such as P, in a region outside the parallel plateMAXWELL conclusion that light is capacitor. For this, we consider a plane circular loop of an electromagnetic radius r whose plane is perpendicular to the direction wave. Interestingly, of the current-carrying wire, and which is centred Maxwell did not agree symmetrically with respect to the wire [Fig. 8.1(a)]. FromCLERK with the idea (strongly suggested by the symmetry, the magnetic field is directed along the Faraday’s laws of circumference of the circular loop and is the same in electrolysis) that magnitude at all points on the loop so that if B is theJAMES electricity was magnitude of the field, the left side of Eq. (8.1) is B (2p r). particulate in nature. So we have B (2pr) = m0i (t) (8 .2) 202 Reprint 2025-26 Electromagnetic Waves Now, consider a different surface, which has the same boundary. This is a pot like surface [Fig. 8.1(b)] which nowhere touches the current, but has its bottom between the capacitor plates; its mouth is the circular loop mentioned above. Another such surface is shaped like a tiffin box (without the lid) [Fig. 8.1(c)]. On applying Ampere’s circuital law to such surfaces with the same perimeter, we find that the left hand side of Eq. (8.1) has not changed but the right hand side is zero and not m0i, since no current passes through the surface of Fig. 8.1(b) and (c). So we have a contradiction; calculated one way, there is a magnetic field at a point P; calculated another way, the magnetic field at P is zero. Since the contradiction arises from our use of Ampere’s circuital law, this law must be missing something. The missing term must be such that one gets the same magnetic field at point P, no matter what surface is used. We can actually guess the missing term by looking carefully at Fig. 8.1(c). Is there anything passing through the surface S between the plates of the capacitor? Yes, of course, the electric field! If the plates of the capacitor have an area A, and a total charge Q, the magnitude of the electric field E between the plates is (Q/A)/e0 (see Eq. 2.41). The field is perpendicular to the surface S of Fig. 8.1(c). It has the same magnitude over the area A of the capacitor plates, and vanishes outside it. So what is the electric flux FE through the surface S ? Using Gauss’s law, it is 1 Q Q ΦE = E A = A = (8.3) ε0 A ε0 Now if the charge Q on the capacitor plates changes with time, there is a current i = (dQ/dt), so that using Eq. (8.3), we have dΦE d  Q  1 d Q = FIGURE 8.1 A d t d t  ε0 = ε0 d t parallel plate capacitor C, as part of This implies that for consistency, a circuit through which a time  dΦE  dependent current i (t) flows, (a) a loop of ε0  d t  = i (8.4) radius r, to determine This is the missing term in Ampere’s circuital law. If we generalise magnetic field at a this law by adding to the total current carried by conductors through point P on the loop; the surface, another term which is e0 times the rate of change of electric (b) a pot-shaped surface passingflux through the same surface, the total has the same value of current i through the interior for all surfaces. If this is done, there is no contradiction in the value of B between the capacitor obtained anywhere using the generalised Ampere’s law. B at the point P plates with the loop shown in (a) as itsis non-zero no matter which surface is used for calculating it. B at a rim; (c) a tiffin- point P outside the plates [Fig. 8.1(a)] is the same as at a point M just shaped surface with inside, as it should be. The current carried by conductors due to flow of the circular loop as charges is called conduction current. The current, given by Eq. (8.4), is a its rim and a flat circular bottom S new term, and is due to changing electric field (or electric displacement, between the capacitor an old term still used sometimes). It is, therefore, called displacement plates. The arrows show uniform electriccurrent or Maxwell’s displacement current. Figure 8.2 shows the electric field between the and magnetic fields inside the parallel plate capacitor discussed above. capacitor plates. The generalisation made by Maxwell then is the following. The source of a magnetic field is not just the conduction electric current due to flowing 203 Reprint 2025-26 Physics charges, but also the time rate of change of electric field. More precisely, the total current i is the sum of the conduction current denoted by ic, and the displacement current denoted by id (= ✒0 (d✂E/dt)). So we have dΦE i = i c + i d = i c + ε0 (8.5) d t In explicit terms, this means that outside the capacitor plates, we have only conduction current ic = i, and no displacement current, i.e., id = 0. On the other hand, inside the capacitor, there is no conduction current, i.e., ic = 0, and there is only displacement current, so that id = i. The generalised (and correct) Ampere’s circuital law has the same form as Eq. (8.1), with one difference: “the total current passing through any surface of which the closed loop is the perimeter” is the sum of the conduction current and the displacement current. The generalised law is dΦE B idl = µ0 i c + µ0 ε0 (8.6) Ñ∫ dt and is known as Ampere-Maxwell law. In all respects, the displacement current has the same physical effects as the conduction current. In some cases, for example, steady electric fields in a conducting wire, the displacement current may be zero since the electric field E does not change with time. In other FIGURE 8.2 (a) The cases, for example, the charging capacitor above, both conduction electric and magnetic and displacement currents may be present in different regions of fields E and B between space. In most of the cases, they both may be present in the same the capacitor plates, at region of space, as there exist no perfectly conducting or perfectly the point M. (b) A cross insulating medium. Most interestingly, there may be large regions sectional view of Fig. (a). of space where there is no conduction current, but there is only a displacement current due to time-varying electric fields. In such a region, we expect a magnetic field, though there is no (conduction) current source nearby! The prediction of such a displacement current can be verified experimentally. For example, a magnetic field (say at point M) between the plates of the capacitor in Fig. 8.2(a) can be measured and is seen to be the same as that just outside (at P). The displacement current has (literally) far reaching consequences. One thing we immediately notice is that the laws of electricity and magnetism are now more symmetrical*. Faraday’s law of induction states that there is an induced emf equal to the rate of change of magnetic flux. Now, since the emf between two points 1 and 2 is the work done per unit charge in taking it from 1 to 2, the existence of an emf implies the existence of an electric field. So, we can rephrase Faraday’s law of electromagnetic induction by saying that a magnetic field, changing with time, gives rise to an electric field. Then, the fact that an electric field changing with time gives rise to a magnetic field, is the symmetrical counterpart, and is * They are still not perfectly symmetrical; there are no known sources of magnetic field (magnetic monopoles) analogous to electric charges which are sources of 204 electric field. Reprint 2025-26 Electromagnetic Waves a consequence of the displacement current being a source of a magnetic field. Thus, time- dependent electric and magnetic fields give rise to each other! Faraday’s law of electromagnetic induction and Ampere-Maxwell law give a quantitative expression of this statement, with the current being the total current, as in Eq. (8.5). One very important consequence of this symmetry is the existence of electromagnetic waves, which we discuss qualitatively in the next section. MAXWELL’S EQUATIONS IN VACUUM 1. “E.dA = Q/✒0 (Gauss’s Law for electricity) 2. “B.dA = 0 (Gauss’s Law for magnetism) –dΦB 3. “E.dl == (Faraday’s Law) d t d ΦE + µ0 ε0 (Ampere – Maxwell Law) 4. “B.dl = µ0 i c d t

1.4 DIMENSIONS OF PHYSICAL QUANTITIES formula of the given physical quantity. For example, the dimensional formula of the volume The nature of a physical quantity is described is [M° L3 T°], and that of speed or velocity is by its dimensions. All the physical quantities [M° L T-1]. Similarly, [M° L T–2] is the dimensional represented by derived units can be expressed formula of acceleration and [M L–3 T°] that of in terms of some combination of seven mass density. fundamental or base quantities. We shall call An equation obtained by equating a physical these base quantities as the seven dimensions quantity with its dimensional formula is called of the physical world, which are denoted with the dimensional equation of the physical square brackets [ ]. Thus, length has the quantity. Thus, the dimensional equations are dimension [L], mass [M], time [T], electric current the equations, which represent the dimensions [A], thermodynamic temperature [K], luminous of a physical quantity in terms of the base intensity [cd], and amount of substance [mol]. quantities. For example, the dimensional The dimensions of a physical quantity are the equations of volume [V], speed [v], force [F ] and powers (or exponents) to which the base mass density [ρ] may be expressed as quantities are raised to represent that [V] = [M0 L3 T0] quantity. Note that using the square brackets [v] = [M0 L T–1] [ ] round a quantity means that we are dealing [F] = [M L T–2] with ‘the dimensions of’ the quantity. [ρ] = [M L–3 T0] In mechanics, all the physical quantities can be written in terms of the dimensions [L], [M] The dimensional equation can be obtained and [T]. For example, the volume occupied by from the equation representing the relations between the physical quantities. Thean object is expressed as the product of length, dimensional formulae of a large number andbreadth and height, or three lengths. Hence the wide variety of physical quantities, derived fromdimensions of volume are [L] × [L] × [L] = [L]3 = [L3]. the equations representing the relationships As the volume is independent of mass and time, among other physical quantities and expressed it is said to possess zero dimension in mass [M°], in terms of base quantities are given in zero dimension in time [T°] and three Appendix 9 for your guidance and ready dimensions in length. reference. Similarly, force, as the product of mass and acceleration, can be expressed as 1.6 DIMENSIONAL ANALYSIS AND ITS Force = mass × acceleration APPLICATIONS = mass × (length)/(time)2 The recognition of concepts of dimensions, which The dimensions of force are [M] [L]/[T]2 = guide the description of physical behaviour is [M L T–2]. Thus, the force has one dimension in of basic importance as only those physical Reprint 2025-26 8 PHYSICS quantities can be added or subtracted which such as angle as the ratio (length/length), have the same dimensions. A thorough refractive index as the ratio (speed of light in understanding of dimensional analysis helps us vacuum/speed of light in medium) etc., has no in deducing certain relations among different dimensions. physical quantities and checking the derivation, Now we can test the dimensional consistency accuracy and dimensional consistency or or homogeneity of the equation homogeneity of various mathematical 2expressions. When magnitudes of two or more x = x 0 + v 0 t + (1/2 ) a t physical quantities are multiplied, their units for the distance x travelled by a particle or body should be treated in the same manner as in time t which starts from the position x0 with ordinary algebraic symbols. We can cancel an initial velocity v0 at time t = 0 and has uniformidentical units in the numerator and acceleration a along the direction of motion. denominator. The same is true for dimensions The dimensions of each term may be written as of a physical quantity. Similarly, physical [x] = [L] quantities represented by symbols on both sides of a mathematical equation must have the same [x0 ] = [L] dimensions. [v0 t] = [L T–1] [T] = [L] [(1/2) a t2] = [L T–2] [T2]1.6.1 Checking the Dimensional Consistency of Equations = [L] As each term on the right hand side of this The magnitudes of physical quantities may be equation has the same dimension, namely that added together or subtracted from one another of length, which is same as the dimension of only if they have the same dimensions. In other left hand side of the equation, hence this words, we can add or subtract similar physical equation is a dimensionally correct equation. quantities. Thus, velocity cannot be added to It may be noted that a test of consistency of force, or an electric current cannot be subtracted dimensions tells us no more and no less than a from the thermodynamic temperature. This test of consistency of units, but has the simple principle called the principle of advantage that we need not commit ourselves homogeneity of dimensions in an equation is to a particular choice of units, and we need not extremely useful in checking the correctness of worry about conversions among multiples and an equation. If the dimensions of all the terms sub-multiples of the units. It may be borne in are not same, the equation is wrong. Hence, if mind that if an equation fails this consistency we derive an expression for the length (or test, it is proved wrong, but if it passes, it is distance) of an object, regardless of the symbols not proved right. Thus, a dimensionally correct appearing in the original mathematical relation, equation need not be actually an exact when all the individual dimensions are (correct) equation, but a dimensionally wrong simplified, the remaining dimension must be (incorrect) or inconsistent equation must be that of length. Similarly, if we derive an equation wrong. of speed, the dimensions on both the sides of ⊳equation, when simplified, must be of length/ Example 1.3 Let us consider an equation time, or [L T–1]. Dimensions are customarily used as a 1 2 m v = m g h preliminary test of the consistency of an 2 equation, when there is some doubt about the where m is the mass of the body, v its correctness of the equation. However, the velocity, g is the acceleration due to dimensional consistency does not guarantee gravity and h is the height. Check correct equations. It is uof dimensionless quantities or functions. The correct. arguments of special functions, such as the trigonometric, logarithmic and exponential Answer The dimensions of LHS are functions must be dimensionless. A pure [M] [L T–1 ]2 = [M] [ L2 T–2] number, ratio of similar physical quantities, = [M L2 T–2] Reprint 2025-26 UNITS AND MEASUREMENT 9 The dimensions of RHS are string, that oscillates under the action of [M][L T–2] [L] = [M][L2 T–2] the force of gravity. Suppose that the period = [M L2 T–2] of oscillation of the simple pendulum depends on its length (l), mass of the bobThe dimensions of LHS and RHS are the same and hence the equation is dimensionally correct. ⊳ (m) and acceleration due to gravity (g). Derive the expression for its time period ⊳ using method of dimensions. Example 1.4 The SI unit of energy is J = kg m2 s–2; that of speed v is m s–1 and Answer The dependence of time period T on of acceleration a is m s–2. Which of the the quantities l, g and m as a product may be formulae for kinetic energy (K) given below written as : can you rule out on the basis of T = k lx gy mz dimensional arguments (m stands for the mass of the body) : where k is dimensionless constant and x, y (a) K = m2 v3 and z are the exponents. (b) K = (1/2)mv2 By considering dimensions on both sides, we (c) K = ma have (d) K = (3/16)mv2 o o 1 1 x 1 –2 y 1 z [L M T ]=[L ] [L T ] [M ] (e) K = (1/2)mv2 + ma = Lx+y T–2y Mz Answer Every correct formula or equation must On equating the dimensions on both sides, have the same dimensions on both sides of the we have equation. Also, only quantities with the same x + y = 0; –2y = 1; and z = 0 physical dimensions can be added or 1 1 subtracted. The dimensions of the quantity on So that x = , y = – , z = 0 2 2the right side are [M2 L3 T–3] for (a); [M L2 T–2] for (b) and (d); [M L T–2] for (c). The quantity on the Then, T = k l½ g–½ right side of (e) has no proper dimensions since two quantities of different dimensions have been l or, T = kadded. Since the kinetic energy K has the g dimensions of [M L2 T–2], formulas (a), (c) and (e) Note that value of constant k can not be obtainedare ruled out. Note that dimensional arguments by the method of dimensions. Here it does notcannot tell which of the two, (b) or (d), is the matter if some number multiplies the right sidecorrect formula. For this, one must turn to the of this formula, because that does not affect itsactual definition of kinetic energy (see dimensions.Chapter 5). The correct formula for kinetic energy is given by (b). ⊳ l Actually, k = 2π so that T = 2π ⊳ 1.6.2 Deducing Relation among the g Physical Quantities The method of dimensions can sometimes be Dimensional analysis is very useful in deducing used to deduce relation among the physical relations among the interdependent physical quantities. For this we should know the quantities. However, dimensionless constants dependence of the physical quantity on other cannot be obtained by this method. The method quantities (upto three physical quantities or of dimensions can only test the dimensional linearly independent variables) and consider it validity, but not the exact relationship between as a product type of the dependence. Let us take physical quantities in any equation. It does not an example. distinguish between the physical quantities having same dimensions.⊳ Example 1.5 Consider a simple A number of exercises at the end of this pendulum, having a bob attached to a chapter will help you develop skill in dimensional analysis. Reprint 2025-26 10 PHYSICS SUMMARY 1. Physics is a quantitative science, based on measurement of physical quantities. Certain physical quantities have been chosen as fundamental or base quantities (such as length, mass, time, electric current, thermodynamic temperature, amount of substance, and luminous intensity). 2. Each base quantity is defined in terms of a certain basic, arbitrarily chosen but properly standardised reference standard called unit (such as metre, kilogram, second, ampere, kelvin, mole and candela). The units for the fundamental or base quantities are called fundamental or base units. 3. Other physical quantities, derived from the base quantities, can be expressed as a combination of the base units and are called derived units. A complete set of units, both fundamental and derived, is called a system of units. 4. The International System of Units (SI) based on seven base units is at present internationally accepted unit system and is widely used throughout the world. 5. The SI units are used in all physical measurements, for both the base quantities and the derived quantities obtained from them. Certain derived units are expressed by means of SI units with special names (such as joule, newton, watt, etc). 6. The SI units have well defined and internationally accepted unit symbols (such as m for metre, kg for kilogram, s for second, A for ampere, N for newton etc.). 7. Physical measurements are usually expressed for small and large quantities in scientific notation, with powers of 10. Scientific notation and the prefixes are used to simplify measurement notation and numerical computation, giving indication to the precision of the numbers. 8. Certain general rules and guidelines must be followed for using notations for physical quantities and standard symbols for SI units, some other units and SI prefixes for expressing properly the physical quantities and measurements. 9. In computing any physical quantity, the units for derived quantities involved in the relationship(s) are treated as though they were algebraic quantities till the desired units are obtained. 10. In measured and computed quantities proper significant figures only should be retained. Rules for determining the number of significant figures, carrying out arithmetic operations with them, and ‘rounding off ‘ the u 11. The dimensions of base quantities and combination of these dimensions describe the nature of physical quantities. Dimensional analysis can be used to check the dimensional consistency of equations, deducing relations among the physical quantities, etc. A dimensionally consistent equation need not be actually an exact (correct) equation, but a dimensionally wrong or inconsistent equation must be wrong. EXERCISES Note : In stating numerical answers, take care of significant figures. 1.1 Fill in the blanks (a) The volume of a cube of side 1 cm is equal to .....m3 (b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to ...(mm)2 (c) A vehicle moving with a speed of 18 km h–1 covers....m in 1 s (d) The relative density of lead is 11.3. Its density is ....g cm–3 or ....kg m–3. 1.2 Fill in the blanks by suitable conversion of units (a) 1 kg m2 s–2 = ....g cm2 s–2 (b) 1 m = ..... ly (c) 3.0 m s–2 = .... km h–2 (d) G = 6.67 × 10–11 N m2 (kg)–2 = .... (cm)3 s–2 g–1. Reprint 2025-26 UNITS AND MEASUREMENT 11 1.3 A calorie is a unit of heat (energy in transit) and it equals about 4.2 J where 1J = 1 kg m2 s–2. Suppose we employ a system of units in which the unit of mass equals α kg, the unit of length equals β m, the unit of time is γ s. Show that a calorie has a magnitude 4.2 α–1 β–2 γ 2 in terms of the new units. 1.4 Explain this statement clearly : “To call a dimensional quantity ‘large’ or ‘small’ is meaningless without specifying a standard for comparison”. In view of this, reframe the following statements wherever necessary : (a) atoms are very small objects (b) a jet plane moves with great speed (c) the mass of Jupiter is very large (d) the air inside this room contains a large number of molecules (e) a proton is much more massive than an electron (f) the speed of sound is much smaller than the speed of light. 1.5 A new unit of length is chosen such that the speed of light in vacuum is unity. What is the distance between the Sun and the Earth in terms of the new unit if light takes 8 min and 20 s to cover this distance ? 1.6 Which of the following is the most precise device for measuring length : (a) a vernier callipers with 20 divisions on the sliding scale (b) a screw gauge of pitch 1 mm and 100 divisions on the circular scale (c) an optical instrument that can measure length to within a wavelength of light ? 1.7 A student measures the thickness of a human hair by looking at it through a microscope of magnification 100. He makes 20 observations and finds that the average width of the hair in the field of view of the microscope is 3.5 mm. What is the estimate on the thickness of hair ? 1.8 Answer the following : (a)You are given a thread and a metre scale. How will you estimate the diameter of the thread ? (b)A screw gauge has a pitch of 1.0 mm and 200 divisions on the circular scale. Do you think it is possible to increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular scale ? (c) The mean diameter of a thin brass rod is to be measured by vernier callipers. Why is a set of 100 measurements of the diameter expected to yield a more reliable estimate than a set of 5 measurements only ? 1.9 The photograph of a house occupies an area of 1.75 cm2 on a 35 mm slide. The slide is projected on to a screen, and the area of the house on the screen is 1.55 m2. What is the linear magnification of the projector-screen arrangement. 1.10 State the number of significant figures in the following : (a) 0.007 m2 (b) 2.64 × 1024 kg (c) 0.2370 g cm–3 (d) 6.320 J (e) 6.032 N m–2 (f) 0.0006032 m2 1.11 The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m, and 2.01 cm respectively. Give the area and volume of the sheet to correct significant figures. 1.12 The mass of a box measured by a grocer’s balance is 2.30 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box. What is (a) the total mass of the box, (b) the difference in the masses of the pieces to correct significant figures ? 1.13 A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ mo of a particle in terms of its speed v and the speed of light, c. (This relation first arose as a consequence of special relativity due to Albert Einstein). A boy recalls the relation almost correctly but forgets where to put the constant c. He writes : m 0 m = 2 1/2 . 1 − v ( ) Guess where to put the missing c. Reprint 2025-26 12 PHYSICS 1.14 The unit of length convenient on the atomic scale is known as an angstrom and is denoted by Å: 1 Å = 10–10 m. The size of a hydrogen atom is about 0.5 Å. What is the total atomic volume in m3 of a mole of hydrogen atoms ? 1.15 One mole of an ideal gas at standard temperature and pressure occupies 22.4 L (molar volume). What is the ratio of molar volume to the atomic volume of a mole of hydrogen ? (Take the size of hydrogen molecule to be about 1 Å). Why is this ratio so large ? 1.16 Explain this common observation clearly : If you look out of the window of a fast moving train, the nearby trees, houses etc. seem to move rapidly in a direction opposite to the train’s motion, but the distant objects (hill tops, the Moon, the stars etc.) seem to be stationary. (In fact, since you are aware that you are moving, these distant objects seem to move with you). 1.17 The Sun is a hot plasma (ionized matter) with its inner core at a temperature exceeding 107 K, and its outer surface at a temperature of about 6000 K. At these high temperatures, no substance remains in a solid or liquid phase. In what range do you expect the mass density of the Sun to be, in the range of densities of solids and liquids or gases ? Check if your guess is correct from the following data : mass of the Sun = 2.0 ×1030 kg, radius of the Sun = 7.0 × 108 m. Reprint 2025-26 CHAPTER TWO MOTION IN A STRAIGHT LINE 2.1 INTRODUCTION Motion is common to everything in the universe. We walk, run and ride a bicycle. Even when we are sleeping, air moves 2.1 Introduction into and out of our lungs and blood flows in arteries and 2.2 Instantaneous velocity and veins. We see leaves falling from trees and water flowing speed down a dam. Automobiles and planes carry people from one 2.3 Acceleration place to the other. The earth rotates once every twenty-four 2.4 Kinematic equations for hours and revolves round the sun once in a year. The sun uniformly accelerated motion itself is in motion in the Milky Way, which is again moving

1.20 A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm from the centre of the sphere is 1.5 × 103 N/C and points radially inward, what is the net charge on the sphere? 43 Reprint 2025-26 Physics 1.21 A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of 80.0 mC/m2. (a) Find the charge on the sphere. (b) What is the total electric flux leaving the surface of the sphere? 1.22 An infinite line charge produces a field of 9 × 104 N/C at a distance of 2 cm. Calculate the linear charge density. 1.23 Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude 17.0 × 10–22 C/m2. What is E: (a) in the outer region of the first plate, (b) in the outer region of the second plate, and (c) between the plates? Reprint 2025-26 Chapter Two ELECTROSTATIC POTENTIAL AND CAPACITANCE 2.12.12.12.12.1 IIINTRODUCTIONIINTRODUCTIONNTRODUCTIONNTRODUCTIONNTRODUCTION In Chapters 5 and 7 (Class XI), the notion of potential energy was introduced. When an external force does work in taking a body from a point to another against a force like spring force or gravitational force, that work gets stored as potential energy of the body. When the external force is removed, the body moves, gaining kinetic energy and losing an equal amount of potential energy. The sum of kinetic and potential energies is thus conserved. Forces of this kind are called conservative forces. Spring force and gravitational force are examples of conservative forces. Coulomb force between two (stationary) charges is also a conservative force. This is not surprising, since both have inverse-square dependence on distance and differ mainly in the proportionality constants – the masses in the gravitational law are replaced by charges in Coulomb’s law. Thus, like the potential energy of a mass in a gravitational field, we can define electrostatic potential energy of a charge in an electrostatic field. Consider an electrostatic field EEEEE due to some charge configuration. First, for simplicity, consider the field E due to a charge Q placed at the origin. Now, imagine that we bring a test charge q from a point R to a point P against the repulsive force on it due to the charge Q. With reference Reprint 2025-26 Physics to Fig. 2.1, this will happen if Q and q are both positive or both negative. For definiteness, let us take Q, q > 0. Two remarks may be made here. First, we assume that the test charge q is so small that it does not disturb the original configuration, namely the charge Q at the origin (or else, we keep Q fixed at the origin by some unspecified force). Second, in bringing the charge q fromFIGURE 2.1 A test charge q (> 0) is moved from the point R to the R to P, we apply an external force Fext just enough to point P against the repulsive counter the repulsive electric force FE (i.e, Fext= –FE). force on it by the charge Q (> 0) This means there is no net force on or acceleration of placed at the origin. the charge q when it is brought from R to P, i.e., it is brought with infinitesimally slow constant speed. In this situation, work done by the external force is the negative of the work done by the electric force, and gets fully stored in the form of potential energy of the charge q. If the external force is removed on reaching P, the electric force will take the charge away from Q – the stored energy (potential energy) at P is used to provide kinetic energy to the charge q in such a way that the sum of the kinetic and potential energies is conserved. Thus, work done by external forces in moving a charge q from R to P is WRP = – = (2.1) This work done is against electrostatic repulsive force and gets stored as potential energy. At every point in electric field, a particle with charge q possesses a certain electrostatic potential energy, this work done increases its potential energy by an amount equal to potential energy difference between points R and P. Thus, potential energy difference ∆U = U P − U R = W RP (2.2) (Note here that this displacement is in an opposite sense to the electric force and hence work done by electric field is negative, i.e., –WRP .) Therefore, we can define electric potential energy difference between two points as the work required to be done by an external force in moving (without accelerating) charge q from one point to another for electric field of any arbitrary charge configuration. Two important comments may be made at this stage: (i) The right side of Eq. (2.2) depends only on the initial and final positions of the charge. It means that the work done by an electrostatic field in moving a charge from one point to another depends only on the initial and the final points and is independent of the path taken to go from one point to the other. This is the fundamental characteristic of a conservative force. The concept of the potential energy would not be meaningful if the work depended on the path. The path-independence of work done by an electrostatic field can be proved using the 46 Coulomb’s law. We omit this proof here. Reprint 2025-26 Electrostatic Potential and Capacitance (ii) Equation (2.2) defines potential energy difference in terms of the physically meaningful quantity work. Clearly, potential energy so defined is undetermined to within an additive constant.What this means is that the actual value of potential energy is not physically significant; it is only the difference of potential energy that is significant. We can always add an arbitrary constant a to potential energy at every point, since this will not change the potential energy difference: (U P + α) − (U R + α) = U P − U R Put it differently, there is a freedom in choosing the point where potential energy is zero. A convenient choice is to have electrostatic potential energy zero at infinity. With this choice, if we take the point R at infinity, we get from Eq. (2.2) Count Alessandro Volta (1745 – 1827) Italian W ∞ P = U P − U ∞ = U P (2.3) physicist, professor at Since the point P is arbitrary, Eq. (2.3) provides us with a Pavia. Volta established that the animal electri- COUNTdefinition of potential energy of a charge q at any point. city observed by LuigiPotential energy of charge q at a point (in the presence of field Galvani, 1737–1798, indue to any charge configuration) is the work done by the experiments with frog external force (equal and opposite to the electric force) in muscle tissue placed in bringing the charge q from infinity to that point. contact with dissimilar metals, was not due to 2.2 ELECTROSTATIC POTENTIAL any exceptional property of animal tissues but ALESSANDROConsider any general static charge configuration. We define was also generated potential energy of a test charge q in terms of the work done whenever any wet body on the charge q. This work is obviously proportional to q, since was sandwiched between the force at any point is qE, where E is the electric field at that dissimilar metals. This VOLTA point due to the given charge configuration. It is, therefore, led him to develop the convenient to divide the work by the amount of charge q, so first voltaic pile, orthat the resulting quantity is independent of q. In other words, battery, consisting of a (1745 work done per unit test charge is characteristic of the electric large stack of moist disks of cardboard (electro-field associated with the charge configuration. This leads to lyte) sandwiched the idea of electrostatic potential V due to a given charge between disks of metal –1827) configuration. From Eq. (2.1), we get: (electrodes). Work done by external force in bringing a unit positive charge from point R to P  U P − U R  = VP – VR = (2.4)  q  where VP and VR are the electrostatic potentials at P and R, respectively. Note, as before, that it is not the actual value of potential but the potential difference that is physically significant. If, as before, we choose the potential to be zero at infinity, Eq. (2.4) implies: Work done by an external force in bringing a unit positive charge from infinity to a point = electrostatic potential (V ) at that point. 47 Reprint 2025-26 Physics In other words, the electrostatic potential (V ) at any point in a region with electrostatic field is the work done in bringing a unit positive charge (without acceleration) from infinity to that point. The qualifying remarks made earlier regarding potential energy also apply to the definition of potential. To obtain the work done per unit test charge, we should take an infinitesimal test charge FIGURE 2.2 Work done on a test charge q dq, obtain the work done dW in bringing it from by the electrostatic field due to any given infinity to the point and determine the ratio charge configuration is independent dW/dq. Also, the external force at every point of the of the path, and depends only on path is to be equal and opposite to the electrostatic its initial and final positions. force on the test charge at that point. 2.3 POTENTIAL DUE TO A POINT CHARGE Consider a point charge Q at the origin (Fig. 2.3). For definiteness, take Q to be positive. We wish to determine the potential at any point P with position vector r from the origin. For that we must calculate the work done in bringing a unit positive test charge from infinity to the point P. For Q > 0, the work done against the repulsive force on the test charge is positive. Since work done is independent of the path, we choose a convenient path – along the radial direction from infinity to the point P. At some intermediate point P¢ on the path, the electrostatic force on a unit positive charge is FIGURE 2.3 Work done in bringing a unit positive test charge from infinity to the Q × 1 rˆ ′ (2.5) point P, against the repulsive force of 2 4 πε0r ' charge Q (Q > 0), is the potential at P due to the charge Q. where ˆ′r is the unit vector along OP¢. Work done against this force from r¢ to r¢ + Dr¢ is Q ∆W = − 2 ∆′r (2.6) 4 πε0r ' The negative sign appears because for Dr¢ < 0, DW is positive. Total work done (W) by the external force is obtained by integrating Eq. (2.6) from r¢ = ¥ to r¢ = r, r Q Q r Q = dr ′ = ε 0r ′ 2 4 πε0r ′ ∞ 4 πε0r (2.7) W = − ∫4∞ π This, by definition is the potential at P due to the charge Q Q V (r ) = (2.8) 48 4 πε0r Reprint 2025-26 Electrostatic Potential and Capacitance Equation (2.8) is true for any sign of the charge Q, though we considered Q > 0 in its derivation. For Q < 0, V < 0, i.e., work done (by the external force) per unit positive test charge in bringing it from infinity to the point is negative. This is equivalent to saying that work done by the electrostatic force in bringing the unit positive charge form infinity to the point P is positive. [This is as it should be, since for Q < 0, the force on a unit positive test charge is attractive, so that the electrostatic force and the displacement (from infinity to P) are FIGURE 2.4 Variation of potential V with r [in units of in the same direction.] Finally, we (Q/4pe0) m-1] (blue curve) and field with r [in units of (Q/4pe0) m-2] (black curve) for a point charge Q.note that Eq. (2.8) is consistent with the choice that potential at infinity be zero. Figure (2.4) shows how the electrostatic potential ( 1/r) and the electrostatic field (1/r 2 ) varies with r. Example 2.1 (a) Calculate the potential at a point P due to a charge of 4 × 10–7C located 9 cm away. (b) Hence obtain the work done in bringing a charge of 2 × 10–9 C from infinity to the point P. Does the answer depend on the path along which the charge is brought? Solution (a) = 4 × 104 V (b) W = qV = 2 × 10–9C × 4 × 104V = 8 × 10–5 J No, work done will be path independent. Any arbitrary infinitesimal path can be resolved into two perpendicular displacements: One along EXAMPLE r and another perpendicular to r. The work done corresponding to the later will be zero. 2.1

10.5 In Young’s double-slit experiment using monochromatic light of wavelength l, the intensity of light at a point on the screen where path difference is l, is K units. What is the intensity of light at a point where path difference is l/3?

10.4 In a Young’s double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment.

10.6 SPECIFIC HEAT CAPACITY heat absorbed or given off to change the temperature of unit mass of it by one unit. ThisTake some water in a vessel and start heating it quantity is referred to as the specific heaton a burner. Soon you will notice that bubbles capacity of the substance.begin to move upward. As the temperature is If ∆Q stands for the amount of heat absorbedraised the motion of water particles increases or given off by a substance of mass m when ittill it becomes turbulent as water starts boiling. undergoes a temperature change ∆T, then theWhat are the factors on which the quantity of specific heat capacity, of that substance is givenheat required to raise the temperature of a bysubstance depend? In order to answer this question in the first step, heat a given quantity S 1 ∆ Q s = = (10.11)of water to raise its temperature by, say 20 °C m m ∆ T and note the time taken. Again take the same The specific heat capacity is the property of amount of water and raise its temperature by the substance which determines the change in 40 °C using the same source of heat. Note the the temperature of the substance (undergoing time taken by using a stopwatch. You will find it no phase change) when a given quantity of heat takes about twice the time and therefore, double is absorbed (or given off) by it. It is defined as the the quantity of heat required raising twice the amount of heat per unit mass absorbed or given temperature of same amount of water. off by the substance to change its temperature In the second step, now suppose you take by one unit. It depends on the nature of the double the amount of water and heat it, using substance and its temperature. The SI unit of the same heating arrangement, to raise the specific heat capacity is J kg–1 K–1. temperature by 20 °C, you will find the time taken If the amount of substance is specified in is again twice that required in the first step. terms of moles µ, instead of mass m in kg, we In the third step, in place of water, now heat can define heat capacity per mole of the the same quantity of some oil, say mustard oil, substance by and raise the temperature again by 20 °C. Now note the time by the same stopwatch. You will (10.12) find the time taken will be shorter and therefore, where C is known as molar specific heat the quantity of heat required would be less than capacity of the substance. Like S, C also that required by the same amount of water for depends on the nature of the substance and its the same rise in temperature. temperature. The SI unit of molar specific heat The above observations show that the quantity capacity is J mol–1 K–1. of heat required to warm a given substance However, in connection with specific heat depends on its mass, m, the change in capacity of gases, additional conditions may betemperature, ∆T and the nature of substance. needed to define C. In this case, heat transferThe change in temperature of a substance, when can be achieved by keeping either pressure ora given quantity of heat is absorbed or rejected volume constant. If the gas is held underby it, is characterised by a quantity called the constant pressure during the heat transfer, thenheat capacity of that substance. We define heat it is called the molar specific heat capacity atcapacity, S of a substance as constant pressure and is denoted by Cp. On ∆Q the other hand, if the volume of the gas is S = (10.10) ∆T maintained during the heat transfer, then the where ∆Q is the amount of heat supplied to corresponding molar specific heat capacity is the substance to change its temperature from T called molar specific heat capacity at constant to T + ∆T. volume and is denoted by Cv. For details see You have observed that if equal amount of Chapter 11. Table 10.3 lists measured specific heat is added to equal masses of different heat capacity of some substances at atmospheric substances, the resulting temperature changes pressure and ordinary temperature while Table will not be the same. It implies that every 10.4 lists molar specific heat capacities of some substance has a unique value for the amount of gases. From Table 10.3 you can note that water Reprint 2025-26 THERMAL PROPERTIES OF MATTER 209 Table 10.3 Specific heat capacity of some substances at room temperature and atmospheric pressure Substance Specific heat capacity Substance Specific heat capacity (J kg–1 K–1) (J kg–1 K–1) Aluminium 900.0 Ice 2060 Carbon 506.5 Glass 840 Copper 386.4 Iron 450 Lead 127.7 Kerosene 2118 Silver 236.1 Edible oil 1965 Tungesten 134.4 Mercury 140 Water 4186.0 has the highest specific heat capacity compared equal to the heat gained by the colder body, to other substances. For this reason water is also provided no heat is allowed to escape to the used as a coolant in automobile radiators, as surroundings. A device in which heat well as, a heater in hot water bags. Owing to its measurement can be done is called a high specific heat capacity, water warms up calorimeter. It consists of a metallic vessel and more slowly than land during summer, and stirrer of the same material, like copper or consequently wind from the sea has a cooling aluminium. The vessel is kept inside a wooden effect. Now, you can tell why in desert areas, jacket, which contains heat insulating material, the earth surface warms up quickly during the like glass wool etc. The outer jacket acts as a day and cools quickly at night. heat shield and reduces the heat loss from the Table 10.4 Molar specific heat capacities of inner vessel. There is an opening in the outer some gases jacket through which a mercury thermometer can be inserted into the calorimeter (Fig. 10.20). Gas Cp (J mol–1K–1) Cv(J mol–1K–1) The following example provides a method by He 20.8 12.5 which the specific heat capacity of a given solid can be determinated by using the principle, heat H2 28.8 20.4 gained is equal to the heat lost. N2 29.1 20.8 ⊳ Example 10.3 A sphere of 0.047 kg O2 29.4 21.1 aluminium is placed for sufficient time in a CO2 37.0 28.5 vessel containing boiling water, so that the sphere is at 100 °C. It is then immediately transfered to 0.14 kg copper calorimeter 10.7 CALORIMETRY containing 0.25 kg water at 20 °C. The A system is said to be isolated if no exchange or temperature of water rises and attains a transfer of heat occurs between the system and steady state at 23 °C. Calculate the specific its surroundings. When different parts of an heat capacity of aluminium. isolated system are at different temperature, a quantity of heat transfers from the part at higher Answer In solving this example, we shall use temperature to the part at lower temperature. the fact that at a steady state, heat given by an The heat lost by the part at higher temperature aluminium sphere will be equal to the heat is equal to the heat gained by the part at lower absorbed by the water and calorimeter. temperature. Mass of aluminium sphere (m1) = 0.047 kg Calorimetry means measurement of heat. Initial temperature of aluminium sphere =100 °C When a body at higher temperature is brought Final temperature = 23 °C in contact with another body at lower Change in temperature (∆T)=(100 °C-23°C)= 77 °C temperature, the heat lost by the hot body is Let specific heat capacity of aluminium be sAl. Reprint 2025-26 210 PHYSICS The amount of heat lost by the aluminium sphere = m 1s Al ∆ T = 0.047kg × s Al × 77 °C Mass of water (m2) = 0.25 kg Mass of calorimeter (m3) = 0.14 kg Initial temperature of water and calorimeter=20 °C Final temperature of the mixture = 23 °C Change in temperature (∆T2) = 23 °C – 20 °C = 3 °C Specific heat capacity of water (sw) = 4.18 × 103 J kg–1 K–1 Specific heat capacity of copper calorimeter = 0.386 × 103 J kg–1 K–1 The amount of heat gained by water and calorimeter = m2 sw ∆T2 + m3scu∆T2 Fig. 10.9 A plot of temperature versus time showing = (m2sw + m3scu) (∆T2) the changes in the state of ice on heating = (0.25 kg × 4.18 × 103 J kg–1 K–1 + 0.14 kg × (not to scale). 0.386 × 103 J kg–1 K–1) (23 °C – 20 °C) The change of state from solid to liquid is called In the steady state heat lost by the aluminium melting or fusion and from liquid to solid is calledsphere = heat gained by water + heat gained by freezing. It is observed that the temperaturecalorimeter. remains constant until the entire amount of the So, 0.047 kg × sAl × 77 °C solid substance melts. That is, both the solid and = (0.25 kg × 4.18 × 103 J kg–1 K–1+ 0.14 kg × the liquid states of the substance coexist in 0.386 × 103 J kg–1 K–1)(3 °C) thermal equilibrium during the change of sAl = 0.911 kJ kg –1 K–1 ⊳ states from solid to liquid. The temperature at which the solid and the liquid states of the substance is in thermal equilibrium with each10.8 CHANGE OF STATE other is called its melting point. It is Matter normally exists in three states: solid, characteristic of the substance. It also depends liquid and gas. A transition from one of these on pressure. The melting point of a substance states to another is called a change of state. Two at standard atomspheric pressure is called its common changes of states are solid to liquid normal melting point. Let us do the following and liquid to gas (and, vice versa). These changes activity to understand the process of melting can occur when the exchange of heat takes place of ice. between the substance and its surroundings. Take a slab of ice. Take a metallic wire and To study the change of state on heating or fix two blocks, say 5 kg each, at its ends. Put cooling, let us perform the following activity. the wire over the slab as shown in Fig. 10.10. Take some cubes of ice in a beaker. Note the You will observe that the wire passes through temperature of ice. Start heating it slowly on a the ice slab. This happens due to the fact that constant heat source. Note the temperature after just below the wire, ice melts at lower every minute. Continuously stir the mixture of temperature due to increase in pressure. When water and ice. Draw a graph between the wire has passed, water above the wire freezes temperature and time (Fig. 10.9). You will observe again. Thus, the wire passes through the slab no change in the temperature as long as there and the slab does not split. This phenomenon is ice in the beaker. In the above process, the of refreezing is called regelation. Skating is temperature of the system does not change even possible on snow due to the formation of water though heat is being continuously supplied. The under the skates. Water is formed due to the heat supplied is being utilised in changing the increase of pressure and it acts as a state from solid (ice) to liquid (water). lubricant. Reprint 2025-26 THERMAL PROPERTIES OF MATTER 211 100 °C when it again becomes steady. The heat supplied is now being utilised to change water from liquid state to vapour or gaseous state. The change of state from liquid to vapour (or gas) is called vaporisation. It is observed that the temperature remains constant until the entire amount of the liquid is converted into vapour. That is, both the liquid and vapour states of the substance coexist in thermal equilibrium, during the change of state from liquid to vapour. The temperature at which the liquid and the vapour states of the substance coexist is called Fig. 10.10 its boiling point. Let us do the following activity After the whole of ice gets converted into water to understand the process of boiling of water. and as we continue further heating, we shall see Take a round-bottom flask, more than halfthat temperature begins to rise (Fig.10.9). The temperature keeps on rising till it reaches nearly filled with water. Keep it over a burner and fix a Triple Point The temperature of a substance remains constant during its change of state (phase change). A graph between the temperature T and the Pressure P of the substance is called a phase diagram or P – T diagram. The following figure shows the phase diagram of water and CO2. Such a phase diagram divides the P – T plane into a solid-region, the vapour-region and the liquid-region. The regions are separated by the curves such as sublimation curve (BO), fusion curve (AO) and vaporisation curve (CO). The points on sublimation curve represent states in which solid and vapour phases coexist. The point on the sublimation curve BO represent states in which the solid and vapour phases co-exist. Points on the fusion curve AO represent states in which solid and liquid phase coexist. Points on the vapourisation curve CO represent states in which the liquid and vapour phases coexist. The temperature and pressure at which the fusion curve, the vaporisation curve and the sublimation curve meet and all the three phases of a substance coexist is called the triple point of the substance. For example the triple point of water is represented by the temperature 273.16 K and pressure 6.11×10–3 Pa. (a) (b) Figure : Pressure-temperature phase diagrams for (a) water and (b) CO2 (not to the scale). Reprint 2025-26 212 PHYSICS thermometer and steam outlet through the cork cork. Keep the flask turned upside down on the of the flask (Fig. 10.11). As water gets heated in stand. Pour ice-cold water on the flask. Water the flask, note first that the air, which was vapours in the flask condense reducing the dissolved in the water, will come out as small pressure on the water surface inside the flask. bubbles. Later, bubbles of steam will form at Water begins to boil again, now at a lower the bottom but as they rise to the cooler water temperature. Thus boiling point decreases with near the top, they condense and disappear. decrease in pressure. Finally, as the temperature of the entire mass This explains why cooking is difficult on hills. of the water reaches 100 °C, bubbles of steam At high altitudes, atmospheric pressure is lower, reach the surface and boiling is said to occur. reducing the boiling point of water as compared The steam in the flask may not be visible but as to that at sea level. On the other hand, boiling it comes out of the flask, it condenses as tiny point is increased inside a pressure cooker by droplets of water, giving a foggy appearance. increasing the pressure. Hence cooking is faster. The boiling point of a substance at standard atmospheric pressure is called its normal boiling point. However, all substances do not pass through the three states: solid-liquid-gas. There are certain substances which normally pass from the solid to the vapour state directly and vice versa. The change from solid state to vapour state without passing through the liquid state is called sublimation, and the substance is said to sublime. Dry ice (solid CO2) sublimes, so also iodine. During the sublimation process both the solid and vapour states of a substance coexist in thermal equilibrium. 10.8.1 Latent Heat In Section 10.8, we have learnt that certain amount of heat energy is transferred between a substance and its surroundings when it undergoes a change of state. The amount of heat per unit mass transferred during change of state of the substance is called latent heat of the substance for the process. For example, if heat Fig. 10.11 Boiling process. is added to a given quantity of ice at –10 °C, the temperature of ice increases until it reaches its If now the steam outlet is closed for a few melting point (0 °C). At this temperature, the seconds to increase the pressure in the flask, addition of more heat does not increase the you will notice that boiling stops. More heat temperature but causes the ice to melt, or would be required to raise the temperature changes its state. Once the entire ice melts, (depending on the increase in pressure) before adding more heat will cause the temperature of boiling begins again. Thus boiling point increases the water to rise. A similar situation with increase in pressure. occurs during liquid gas change of state at the Let us now remove the burner. Allow water to boiling point. Adding more heat to boiling water cool to about 80 °C. Remove the thermometer and causes vaporisation, without increase in steam outlet. Close the flask with the airtight temperature. Reprint 2025-26 THERMAL PROPERTIES OF MATTER 213 Table 10.5 Temperatures of the change of state and latent heats for various substances at 1 atm pressure Substance Melting Lf Boiling Lv Point (°C) (105J kg–1) Point (°C) (105J kg–1) Ethanol –114 1.0 78 8.5 Gold 1063 0.645 2660 15.8 Lead 328 0.25 1744 8.67 Mercury –39 0.12 357 2.7 Nitrogen –210 0.26 –196 2.0 Oxygen –219 0.14 –183 2.1 Water 0 3.33 100 22.6 The heat required during a change of state Note that when heat is added (or removed) depends upon the heat of transformation and during a change of state, the temperature the mass of the substance undergoing a change remains constant. Note in Fig. 10.12 that the of state. Thus, if mass m of a substance slopes of the phase lines are not all the same, undergoes a change from one state to the other, which indicate that specific heats of the various then the quantity of heat required is given by states are not equal. For water, the latent heat of Q = m L fusion and vaporisation are Lf = 3.33 × 105 J kg–1 or L = Q/m (10.13) and Lv = 22.6 × 105 J kg–1, respectively. That is, where L is known as latent heat and is a 3.33 × 105 J of heat is needed to melt 1 kg ice at characteristic of the substance. Its SI unit is 0 °C, and 22.6 × 105 J of heat is needed to convert J kg–1. The value of L also depends on the 1 kg water into steam at 100 °C. So, steam at pressure. Its value is usually quoted at standard 100 °C carries 22.6 × 105 J kg–1 more heat than atmospheric pressure. The latent heat for a solid- water at 100 °C. This is why burns from steam liquid state change is called the latent heat of are usually more serious than those from fusion (Lf), and that for a liquid-gas state change boiling water. is called the latent heat of vaporisation (Lv). ⊳ These are often referred to as the heat of fusion Example 10.4 When 0.15 kg of ice at 0 °C and the heat of vaporisation. A plot of is mixed with 0.30 kg of water at 50 °C in a temperature versus heat for a quantity of water container, the resulting temperature is is shown in Fig. 10.12. The latent heats of some 6.7 °C. Calculate the heat of fusion of ice. substances, their freezing and boiling points, are (swater = 4186 J kg–1 K–1) given in Table 10.5. Answer Heat lost by water = msw (θf–θi)w = (0.30 kg) (4186 J kg–1 K–1) (50.0 °C – 6.7 °C) = 54376.14 J Heat required to melt ice = m2Lf = (0.15 kg) Lf Heat required to raise temperature of ice water to final temperature = mIsw (θf–θi)I = (0.15 kg) (4186 J kg–1 K –1) (6.7 °C – 0 °C) = 4206.93 J Heat lost = heat gained Fig. 10.12 Temperature versus heat for water at 54376.14 J = (0.15 kg) Lf + 4206.93 J 1 atm pressure (not to scale). Lf = 3.34×105 J kg–1. ⊳ Reprint 2025-26 214 PHYSICS ⊳ temperature difference. What are the different Example 10.5 Calculate the heat required ways by which this energy transfer takes to convert 3 kg of ice at –12 °C kept in a place? There are three distinct modes of heat calorimeter to steam at 100 °C at transfer: conduction, convection and radiation atmospheric pressure. Given specific heat (Fig. 10.13). capacity of ice = 2100 J kg–1 K–1, specific heat capacity of water = 4186 J kg– 1 K–1, latent heat of fusion of ice = 3.35 × 105 J kg–1 and latent heat of steam = 2.256 ×106 J kg–1. Answer We have Mass of the ice, m = 3 kg specific heat capacity of ice, sice = 2100 J kg–1 K–1 specific heat capacity of water, swater = 4186 J kg–1 K–1 latent heat of fusion of ice, Lf ice = 3.35 × 105 J kg–1 latent heat of steam, Lsteam Fig. 10.13 Heating by conduction, convection and = 2.256 × 106 J kg–1 radiation. Now, Q = heat required to convert 3 kg of 10.9.1 Conduction ice at –12 °C to steam at 100 °C, Conduction is the mechanism of transfer of heat Q1 = heat required to convert ice at between two adjacent parts of a body because –12 °C to ice at 0 °C. of their temperature difference. Suppose, one end = m sice ∆T1 = (3 kg) (2100 J kg–1. of a metallic rod is put in a flame, the other end K–1) [0–(–12)]°C = 75600 J of the rod will soon be so hot that you cannot Q2 = heat required to melt ice at hold it by your bare hands. Here, heat transfer 0 °C to water at 0 °C takes place by conduction from the hot end of = m Lf ice = (3 kg) (3.35 × 105 J kg–1) the rod through its different parts to the other = 1005000 J end. Gases are poor thermal conductors, while Q3 = heat required to convert water liquids have conductivities intermediate between at 0 °C to water at 100 °C. solids and gases. = msw ∆T2 = (3kg) (4186J kg–1 K–1) Heat conduction may be described (100 °C) quantitatively as the time rate of heat flow in a = 1255800 J material for a given temperature difference. Q4 = heat required to convert water Consider a metallic bar of length L and uniform cross-section A with its two ends maintained at at 100 °C to steam at 100 °C. different temperatures. This can be done, for = m Lsteam = (3 kg) (2.256 ×106 example, by putting the ends in thermal contact J kg–1) with large reservoirs at temperatures, say, TC and = 6768000 J TD, respectively (Fig. 10.14). Let us assume theSo, Q = Q1 + Q2 + Q3 + Q4 ideal condition that the sides of the bar are fully = 75600J + 1005000 J insulated so that no heat is exchanged between + 1255800 J + 6768000 J the sides and the surroundings. = 9.1×106 J ⊳ After sometime, a steady state is reached; the temperature of the bar decreases uniformly with 10.9 HEAT TRANSFER distance from TC to TD; (TC>TD). The reservoir at C supplies heat at a constant rate, whichWe have seen that heat is energy transfer transfers through the bar and is given out atfrom one system to another or from one part the same rate to the reservoir at D. It is foundof a system to another part, arising due to Reprint 2025-26 THERMAL PROPERTIES OF MATTER 215 prohibited and keeps the room cooler. In some situations, heat transfer is critical. In a nuclear reactor, for example, elaborate heat transfer systems need to be installed so that the enormous energy produced by nuclear fission in the core transits out sufficiently fast, thus preventing the core from overheating. Table 10.6 Thermal conductivities of some material Fig. 10.14 Steady state heat flow by conduction in a bar with its two ends maintained at Material Thermal conductivity temperatures TC and TD; (TC > TD). (J s–1 m–1 K–1 ) Metals experimentally that in this steady state, the rate of flow of heat (or heat current) H is proportional Silver 406 to the temperature difference (TC – TD) and the Copper 385 area of cross-section A and is inversely Aluminium 205 proportional to the length L : Brass 109 Steel 50.2 TC – TD Lead 34.7 H = KA (10.14) L Mercury 8.3 The constant of proportionality K is called the thermal conductivity of the material. The Non-metals greater the value of K for a material, the more rapidly will it conduct heat. The SI unit of K is Insulating brick 0.15 J s–1 m –1 K–1 or W m –1 K–1. The thermal Concrete 0.8 conductivities of various substances are listed Body fat 0.20 in Table 10.6. These values vary slightly with Felt 0.04 temperature, but can be considered to be Glass 0.8 constant over a normal temperature range. Ice 1.6 Compare the relatively large thermal Glass wool 0.04 conductivities of good thermal conductors and, Wood 0.12 metals, with the relatively small thermal Water 0.8 conductivities of some good thermal insulators, such as wood and glass wool. You may have Gases noticed that some cooking pots have copper coating on the bottom. Being a good conductor Air 0.024 of heat, copper promotes the distribution of heat Argon 0.016 over the bottom of a pot for uniform cooking. Hydrogen 0.14 Plastic foams, on the other hand, are good ⊳ insulators, mainly because they contain pockets Example 10.6 What is the temperature of of air. Recall that gases are poor conductors, the steel-copper junction in the steady and note the low thermal conductivity of air in state of the system shown in Fig. 10.15. the Table 10.5. Heat retention and transfer are Length of the steel rod = 15.0 cm, length important in many other applications. Houses of the copper rod = 10.0 cm, temperature made of concrete roofs get very hot during of the furnace = 300 °C, temperature of the summer days because thermal conductivity of other end = 0 °C. The area of cross section concrete (though much smaller than that of a of the steel rod is twice that of the copper metal) is still not small enough. Therefore, people, rod. (Thermal conductivity of steel usually, prefer to give a layer of earth or foam = 50.2 J s –1 m –1 K –1; and of copper insulation on the ceiling so that heat transfer is = 385 J s–1m–1 K–1). Reprint 2025-26 216 PHYSICS Answer Given, L1 = L2= L = 0.1 m, A1 = A2= A= 0.02 m2 K1 = 79 W m–1 K –1, K2 = 109 W m–1 K–1, T1 = 373 K, and T2 = 273 K. Under steady state condition, the heat current (H1) through iron bar is equal to the Fig. 10.15 heat current (H2) through brass bar. Answer The insulating material around the rods So, H = H1 = H2 reduces heat loss from the sides of the rods. T1 – T 0 ) K 2 A 2 ( T 0 – T 2 )Therefore, heat flows only along the length of K 1 A1 ( = = the rods. Consider any cross section of the rod. L 1 L 2 In the steady state, heat flowing into the element For A1 = A2 = A and L1 = L2 = L, this equation must equal the heat flowing out of it; otherwise leads to there would be a net gain or loss of heat by the K1 (T1 – T0) = K2 (T0 – T2) element and its temperature would not be Thus, the junction temperature T0 of the two steady. Thus in the steady state, rate of heat bars is flowing across a cross section of the rod is the same at every point along the length of the ( K1T1 + K 2 T2 ) combined steel-copper rod. Let T be the T0 = ( K1 + K 2 ) temperature of the steel-copper junction in the Using this equation, the heat current H through steady state. Then, either bar is K1 A1 (300 − T ) K 2 A 2 ( T – 0 ) K1 A ( T1 – T0 ) K 2 A ( T0 – T2 ) = H = = L 1 L 2 L L where 1 and 2 refer to the steel and copper rod respectively. For A1 = 2 A2, L1 = 15.0 cm, L2 = 10.0 cm, K1 = 50.2 J s–1 m–1 K –1, K2 = 385 J s–1 m–1 K –1, we have Using these equations, the heat current H′ 50.2 × 2 ( 300 − T ) 385 T = through the compound bar of length L1 + L2 = 2L 15 10 and the equivalent thermal conductivity K′, of which gives T = 44.4 °C ⊳ the compound bar are given by ⊳ K ′ A ( T1 – T2 ) Example 10.7 An iron bar (L1 = 0.1 m, A1 = H ′ = = H 0.02 m 2, K 1 = 79 W m –1 K –1) and a 2 L brass bar (L2 = 0.1 m, A2 = 0.02 m2, 2 K1 K 2 K2 = 109 W m–1K–1) are soldered end to end K ′ = K1 + K 2 as shown in Fig. 10.16. The free ends of the iron bar and brass bar are maintained ( K1T1 + K 2 T2 ) at 373 K and 273 K respectively. Obtain (i) T0 = ( K 1 + K 2 ) expressions for and hence compute (i) the temperature of the junction of the two bars, 79 W m –1 K –1 109 W m –1 K –1 273 K ) ( )( 373 K ) + ( )( (ii) the equivalent thermal conductivity of = –1 –1 –1 –1 79 W m K + 109 W m K the compound bar, and (iii) the heat current through the compound bar. = 315 K 2 K 1 K 2 (ii) K ′ = K 1 + K 2 2×(79 W m –1 K –1 ) ×(109 W m –1 K –1 ) = –1 –1 –1 –1 79 W m K +109 W m K Fig 10.16 = 91.6 W m–1 K–1 Reprint 2025-26 THERMAL PROPERTIES OF MATTER 217 of water do. This occurs both because water has K ′ A ( T1 – T2 )(iii) H ′ = H = a greater specific heat capacity and because 2 L mixing currents disperse the absorbed heat 91.6 W m K × 0.02 m ( –1 –1 2 throughout the great volume of water. The air ) ( ) × ( 373 K–273 K ) in contact with the warm ground is heated by = 2× (0.1 m ) conduction. It expands, becoming less dense = 916.1 W ⊳ than the surrounding cooler air. As a result, the warm air rises (air currents) and the other air 10.9.2 Convection moves (winds) to fill the space-creating a sea breeze near a large body of water. Cooler air Convection is a mode of heat transfer by actual descends, and a thermal convection cycle is set motion of matter. It is possible only in fluids. up, which transfers heat away from the land.Convection can be natural or forced. In natural At night, the ground loses its heat more quickly,convection, gravity plays an important part. and the water surface is warmer than the land.When a fluid is heated from below, the hot part As a result, the cycle is reveresed (Fig. 10.17).expands and, therefore, becomes less dense. The other example of natural convection isBecause of buoyancy, it rises and the upper the steady surface wind on the earth blowingcolder part replaces it. This again gets heated, in from north-east towards the equator, therises up and is replaced by the relatively colder so-called trade wind. A resonable explanationpart of the fluid. The process goes on. This mode is as follows: the equatorial and polar regions ofof heat transfer is evidently different from the earth receive unequal solar heat. Air at theconduction. Convection involves bulk transport of different parts of the fluid. earth’s surface near the equator is hot, while In forced convection, material is forced to move the air in the upper atmosphere of the poles is by a pump or by some other physical means. The cool. In the absence of any other factor, a common examples of forced convection systems convection current would be set up, with the are forced-air heating systems in home, the air at the equatorial surface rising and moving human circulatory system, and the cooling out towards the poles, descending and system of an automobile engine. In the human streaming in towards the equator. The rotation body, the heart acts as the pump that circulates of the earth, however, modifies this convection blood through different parts of the body, current. Because of this, air close to the equator transferring heat by forced convection and has an eastward speed of 1600 km/h, while it maintaining it at a uniform temperature. is zero close to the poles. As a result, the air Natural convection is responsible for many descends not at the poles but at 30° N (North) familiar phenomena. During the day, the latitude and returns to the equator. This is ground heats up more quickly than large bodies called trade wind. Fig. 10.17 Convection cycles. Reprint 2025-26 218 PHYSICS 10.9.3 Radiation contents of the bottle. The outer wall similarly reflects back any incoming radiation. The spaceConduction and convection require some between the walls is evacuted to reduce material as a transport medium. These modes conduction and convection losses and the flask of heat transfer cannot operate between bodies is supported on an insulator, like cork. The separated by a distance in vacuum. But the device is, therefore, useful for preventing hot earth does receive heat from the Sun across a contents (like, milk) from getting cold, or huge distance. Similarly, we quickly feel the alternatively, to store cold contents (like, ice). warmth of the fire nearby even though air conducts poorly and before convection takes 10.9.4 Blackbody Radiation some time to set in. The third mechanism for We have so far not mentioned the wavelength heat transfer needs no medium; it is called content of thermal radiation. The important radiation and the energy so transferred by thing about thermal radiation at any electromagnetic waves is called radiant energy. temperature is that it is not of one (or a few) In an electromagnetic wave, electric and wavelength(s) but has a continuous spectrum magnetic fields oscillate in space and time. Like from the small to the long wavelengths. The any wave, electromagnetic waves can have energy content of radiation, however, varies for different wavelengths and can travel in vacuum different wavelengths. Figure 10.18 gives the with the same speed, namely the speed of light experimental curves for radiation energy per unit i.e., 3 × 108 m s–1 . You will learn these matters in area per unit wavelength emitted by a blackbody more detail later, but you now know why heat versus wavelength for different temperatures. transfer by radiation does not need any medium and why it is so fast. This is how heat is transferred to the earth from the Sun through empty space. All bodies emit radiant energy, whether they are solid, liquid or gas. The electromagnetic radiation emitted by a body by virtue of its temperature, like radiation by a red hot iron or light from a filament lamp is called thermal radiation. When this thermal radiation falls on other bodies, it is partly reflected and partly absorbed. The amount of heat that a body can absorb by radiation depends on the colour of the body. We find that black bodies absorb and emit radiant energy better than bodies of lighter Fig. 10.18: Energy emitted versus wavelength colours. This fact finds many applications in our for a blackbody at different daily life. We wear white or light coloured clothes temperatures in summer, so that they absorb the least heat Notice that the wavelength λm for which energy from the Sun. However, during winter, we use is the maximum decreases with increasing dark coloured clothes, which absorb heat from temperature. The relation between λm and T is the sun and keep our body warm. The bottoms of given by what is known as Wien’s Displacement utensils for cooking food are blackened so that Law: they absorb maximum heat from fire and transfer λm T = constant (10.15) it to the vegetables to be cooked. Similarly, a Dewar flask or thermos bottle is The value of the constant (Wien’s constant) a device to minimise heat transfer between the is 2.9 × 10–3 m K. This law explains why the contents of the bottle and outside. It consists colour of a piece of iron heated in a hot flame of a double-walled glass vessel with the inner first becomes dull red, then reddish yellow, and and outer walls coated with silver. Radiation finally white hot. Wien’s law is useful for from the inner wall is reflected back to the estimating the surface temperatures of celestial Reprint 2025-26 THERMAL PROPERTIES OF MATTER 219 bodies like, the moon, Sun and other stars. Light For a body with emissivity e, the relation from the moon is found to have a maximum modifies to intensity near the wavelength 14 µm. By Wien’s H = eσ A (T4 – Ts 4) (10.18)law, the surface of the moon is estimated to have a temperature of 200 K. Solar radiation has a As an example, let us estimate the heat maximum at λm = 4753 Å. This corresponds to radiated by our bodies. Suppose the surface area T = 6060 K. Remember, this is the temperature of a person’s body is about 1.9 m2 and the room of the surface of the sun, not its interior. temperature is 22 °C. The internal body The most significant feature of the temperature, as we know, is about 37 °°C. The blackbody radiation curves in Fig. 10.18 is that skin temperature may be 28°°C (say). Thethey are universal. They depend only on the emissivity of the skin is about 0.97 for thetemperature and not on the size, shape or relevant region of electromagnetic radiation. Thematerial of the blackbody. Attempts to explain rate of heat loss is:blackbody radiation theoretically, at the beginning of the twentieth century, spurred the H = 5.67 × 10–8 × 1.9 × 0.97 × {(301)4 – (295)4} quantum revolution in physics, as you will learn in later courses. = 66.4 W Energy can be transferred by radiation over which is more than half the rate of energy large distances, without a medium (i.e., in production by the body at rest (120 W). To vacuum). The total electromagnetic energy prevent this heat loss effectively (better than radiated by a body at absolute temperature T ordinary clothing), modern arctic clothing has is proportional to its size, its ability to radiate an additional thin shiny metallic layer next to (called emissivity) and most importantly to its the skin, which reflects the body’s radiation. temperature. For a body, which is a perfect radiator, the energy emitted per unit time (H) 10.10 NEWTON’S LAW OF COOLING is given by We all know that hot water or milk when left on H = AσT 4 (10.16) a table begins to cool, gradually. Ultimately it attains the temperature of the surroundings. To where A is the area and T is the absolute study how slow or fast a given body can cool on temperature of the body. This relation obtained exchanging heat with its surroundings, let us experimentally by Stefan and later proved perform the following activity. theoretically by Boltzmann is known as Stefan- Take some water, say 300 mL, in a Boltzmann law and the constant σ is called calorimeter with a stirrer and cover it with a Stefan-Boltzmann constant. Its value in SI units two-holed lid. Fix the stirrer through one hole is 5.67 × 10–8 W m–2 K–4. Most bodies emit only a and fix a thermometer through another hole fraction of the rate given by Eq. 10.16. A substance in the lid and make sure that the bulb of like lamp black comes close to the limit. One, thermometer is immersed in the water. Note therefore, defines a dimensionless fraction e the reading of the thermometer. This reading called emissivity and writes, T1 is the temperature of the surroundings. H = AeσT 4 (10.17) Heat the water kept in the calorimeter till it Here, e = 1 for a perfect radiator. For a tungsten attains a temperature, say 40 °C above room lamp, for example, e is about 0.4. Thus, a tungsten temperature (i.e., temperature of the lamp at a temperature of 3000 K and a surface surroundings). Then, stop heating the water area of 0.3 cm2 radiates at the rate H = 0.3 × by removing the heat source. Start the 10–4 × 0.4 × 5.67 × 10–8 × (3000)4 = 60 W. stop-watch and note the reading of the A body at temperature T, with surroundings thermometer after a fixed interval of time, say at temperatures Ts, emits, as well as, receives after every one minute of stirring gently with energy. For a perfect radiator, the net rate of the stirrer. Continue to note the temperature loss of radiant energy is (T2) of water till it attains a temperature about 5 °C above that of the surroundings. Then, plot H = σA (T 4 – Ts4) Reprint 2025-26 220 PHYSICS a graph by taking each value of temperature From Eqs. (10.15) and (10.16) we have ∆T = T2 – T1 along y-axis and the coresponding dT 2value of t along x-axis (Fig. 10.19). – m s = k ( T2 – T1 ) dt dT2 k = – dt = – K dt (10.21) T 2 – T1 ms where K = k/m s On integrating, ∆ loge (T2 – T1) = – K t + c (10.22) or T2 = T1 + C′ e–Kt; where C′ = ec (10.23) Equation (10.23) enables you to calculate the time of cooling of a body through a particular Fig. 10.19 Curve showing cooling of hot water range of temperature. with time. For small temperature differences, the rate From the graph you can infer how the cooling of cooling, due to conduction, convection, and of hot water depends on the difference of its radiation combined, is proportional to the temperature from that of the surroundings. You difference in temperature. It is a valid will also notice that initially the rate of cooling approximation in the transfer of heat from a is higher and decreases as the temperature of radiator to a room, the loss of heat through the the body falls. wall of a room, or the cooling of a cup of tea on The above activity shows that a hot body loses the table. heat to its surroundings in the form of heat radiation. The rate of loss of heat depends on the difference in temperature between the body and its surroundings. Newton was the first to study, in a systematic manner, the relation between the heat lost by a body in a given enclosure and its temperature. According to Newton’s law of cooling, the rate of loss of heat, – dQ/dt of the body is directly proportional to the difference of temperature ∆T = (T2–T1) of the body and the surroundings. The law holds good only for small difference of temperature. Also, the loss of heat by radiation depends upon the nature of the surface of the body and the area of the exposed surface. We Fig. 10.20 Verification of Newton’s Law of cooling. can write Newton’s law of cooling can be verified with the help of the experimental set-up shown in – (10.19) Fig. 10.20(a). The set-up consists of a double- walled vessel (V) containing water between where k is a positive constant depending upon the two walls. A copper calorimeter (C)the area and nature of the surface of the body. containing hot water is placed inside theSuppose a body of mass m and specific heat double-walled vessel. Two thermometerscapacity s is at temperature T2. Let T1 be the through the corks are used to note thetemperature of the surroundings. If the temperatures T2 of water in calorimeter andtemperature falls by a small amount dT2 in time dt, then the amount of heat lost is T1 of hot water in between the double walls, respectively. Temperature of hot water in the dQ = ms dT2 calorimeter is noted after equal intervals of ∴ Rate of loss of heat is given by time. A graph is plotted between log e (T2–T1) [or ln(T2–T1)] and time (t). The nature of the dQ dT2 = ms (10.20) dt dt Reprint 2025-26 THERMAL PROPERTIES OF MATTER 221 graph is observed to be a straight line having 8 °C a negative slope as shown in Fig. 10.20(b). This = K ( 70 °C ) is in support of Eq. 10.22. 2 min ⊳ The average of 69 °C and 71 °C is 70 °C, which Example 10.8 A pan filled with hot food is 50 °C above room temperature. K is the same cools from 94 °C to 86 °C in 2 minutes when for this situation as for the original. the room temperature is at 20 °C. How long will it take to cool from 71 °C to 69 °C? 2 °C = K (50 °C) Time Answer The average temperature of 94 °C and When we divide above two equations, we 86 °C is 90 °C, which is 70 °C above the room have temperature. Under these conditions the pan cools 8 °C in 2 minutes. 8 °C/2 min K (70 °C) Using Eq. (10.21), we have = 2 °C/time K (50 °C) Change in temperature = K ∆ T Time = 0.7 min Time = 42 s ⊳ SUMMARY 1. Heat is a form of energy that flows between a body and its surrounding medium by virtue of temperature difference between them. The degree of hotness of the body is quantitatively represented by temperature. 2. A temperature-measuring device (thermometer) makes use of some measurable property (called thermometric property) that changes with temperature. Different thermometers lead to different temperature scales. To construct a temperature scale, two fixed points are chosen and assigned some arbitrary values of temperature. The two numbers fix the origin of the scale and the size of its unit. 3. The Celsius temperature (tC) and the Farenheit temperare (tF)are related by tF = (9/5) tC + 32 4. The ideal gas equation connecting pressure (P), volume (V) and absolute temperature (T) is : PV = µRT where µ is the number of moles and R is the universal gas constant. 5. In the absolute temperature scale, the zero of the scale corresponds to the temperature where every substance in nature has the least possible molecular activity. The Kelvin absolute temperature scale (T ) has the same unit size as the Celsius scale (Tc ), but differs in the origin : TC = T – 273.15 6. The coefficient of linear expansion (αl ) and volume expansion (αv ) are defined by the relations : ∆=l αl ∆ T l ∆V = αV ∆ T V Reprint 2025-26 222 PHYSICS where ∆l and ∆V denote the change in length l and volume V for a change of temperature ∆T. The relation between them is : αv = 3 αl 7. The specific heat capacity of a substance is defined by 1 ∆Q s = m ∆T where m is the mass of the substance and ∆Q is the heat required to change its temperature by ∆T. The molar specific heat capacity of a substance is defined by 1 ∆ Q C = µ ∆ T where µ is the number of moles of the substance. 8. The latent heat of fusion (Lf) is the heat per unit mass required to change a substance from solid into liquid at the same temperature and pressure. The latent heat of vaporisation (Lv) is the heat per unit mass required to change a substance from liquid to the vapour state without change in the temperature and pressure. 9. The three modes of heat transfer are conduction, convection and radiation. 10. In conduction, heat is transferred between neighbouring parts of a body through molecular collisions, without any flow of matter. For a bar of length L and uniform cross section A with its ends maintained at temperatures TC and TD, the rate of flow of heat H is : T − T H = K A C D L where K is the thermal conductivity of the material of the bar. 11. Newton’s Law of Cooling says that the rate of cooling of a body is proportional to the excess temperature of the body over the surroundings : d Q = – k ( T2 – T1 ) d t Where T1 is the temperature of the surrounding medium and T2 is the temperature of the body. Reprint 2025-26 THERMAL PROPERTIES OF MATTER 223 POINTS TO PONDER 1. The relation connecting Kelvin temperature (T ) and the Celsius temperature tc T = tc + 273.15 and the assignment T = 273.16 K for the triple point of water are exact relations (by choice). With this choice, the Celsius temperature of the melting point of water and boiling point of water (both at 1 atm pressure) are very close to, but not exactly equal to 0 °C and 100 °C respectively. In the original Celsius scale, these latter fixed points were exactly at 0 °C and 100 °C (by choice), but now the triple point of water is the preferred choice for the fixed point, because it has a unique temperature. 2. A liquid in equilibrium with vapour has the same pressure and temperature throughout the system; the two phases in equilibrium differ in their molar volume (i.e. density). This is true for a system with any number of phases in equilibrium. 3. Heat transfer always involves temperature difference between two systems or two parts of the same system. Any energy transfer that does not involve temperature difference in some way is not heat. 4. Convection involves flow of matter within a fluid due to unequal temperatures of its parts. A hot bar placed under a running tap loses heat by conduction between the surface of the bar and water and not by convection within water. EXERCISES 10.1 The triple points of neon and carbon dioxide are 24.57 K and 216.55 K respectively. Express these temperatures on the Celsius and Fahrenheit scales. 10.2 Two absolute scales A and B have triple points of water defined to be 200 A and 350 B. What is the relation between TA and TB ? 10.3 The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law : R = Ro [1 + α (T – To )] The resistance is 101.6 Ω at the triple-point of water 273.16 K, and 165.5 Ω at the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4 Ω ? 10.4 Answer the following : (a) The triple-point of water is a standard fixed point in modern thermometry. Why ? What is wrong in taking the melting point of ice and the boiling point of water as standard fixed points (as was originally done in the Celsius scale) ? (b) There were two fixed points in the original Celsius scale as mentioned above which were assigned the number 0 °C and 100 °C respectively. On the absolute scale, one of the fixed points is the triple-point of water, which on the Kelvin absolute scale is assigned the number 273.16 K. What is the other fixed point on this (Kelvin) scale ? (c) The absolute temperature (Kelvin scale) T is related to the temperature tc on the Celsius scale by tc = T – 273.15 Why do we have 273.15 in this relation, and not 273.16 ? (d) What is the temperature of the triple-point of water on an absolute scale whose unit interval size is equal to that of the Fahrenheit scale ? 10.5 Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The following observations are made : Reprint 2025-26 224 PHYSICS Temperature Pressure Pressure thermometer A thermometer B Triple-point of water 1.250 × 105 Pa 0.200 × 105 Pa Normal melting point 1.797 × 105 Pa 0.287 × 105 Pa of sulphur (a) What is the absolute temperature of normal melting point of sulphur as read by thermometers A and B ? (b) What do you think is the reason behind the slight difference in answers of thermometers A and B ? (The thermometers are not faulty). What further procedure is needed in the experiment to reduce the discrepancy between the two readings ? 10.6 A steel tape 1m long is correctly calibrated for a temperature of 27.0 °C. The length of a steel rod measured by this tape is found to be 63.0 cm on a hot day when the temperature is 45.0 °C. What is the actual length of the steel rod on that day ? What is the length of the same steel rod on a day when the temperature is 27.0 °C ? Coefficient of linear expansion of steel = 1.20 × 10–5 K–1 . 10.7 A large steel wheel is to be fitted on to a shaft of the same material. At 27 °C, the outer diameter of the shaft is 8.70 cm and the diameter of the central hole in the wheel is 8.69 cm. The shaft is cooled using ‘dry ice’. At what temperature of the shaft does the wheel slip on the shaft? Assume coefficient of linear expansion of the steel to be constant over the required temperature range : αsteel = 1.20 × 10–5 K–1. 10.8 A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27.0 °C. What is the change in the diameter of the hole when the sheet is heated to 227 °C? Coefficient of linear expansion of copper = 1.70 × 10–5 K–1. 10.9 A brass wire 1.8 m long at 27 °C is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of –39 °C, what is the tension developed in the wire, if its diameter is 2.0 mm ? Co-efficient of linear expansion of brass = 2.0 × 10–5 K–1; Young’s modulus of brass = 0.91 × 1011 Pa. 10.10 A brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at 250 °C, if the original lengths are at 40.0 °C? Is there a ‘thermal stress’ developed at the junction ? The ends of the rod are free to expand (Co-efficient of linear expansion of brass = 2.0 × 10–5 K–1, steel = 1.2 × 10–5 K–1 ). 10.11 The coefficient of volume expansion of glycerine is 49 × 10–5 K–1. What is the fractional change in its density for a 30 °C rise in temperature ? 10.12 A 10 kW drilling machine is used to drill a bore in a small aluminium block of mass 8.0 kg. How much is the rise in temperature of the block in 2.5 minutes, assuming 50% of power is used up in heating the machine itself or lost to the surroundings. Specific heat of aluminium = 0.91 J g–1 K–1. 10.13 A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500 °C and then placed on a large ice block. What is the maximum amount of ice that can melt? (Specific heat of copper = 0.39 J g–1 K–1; heat of fusion of water = 335 J g–1 ). 10.14 In an experiment on the specific heat of a metal, a 0.20 kg block of the metal at 150 °C is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 150 cm3 of water at 27 °C. The final temperature is 40 °C. Compute the specific heat of the metal. If heat losses to the surroundings are not negligible, is your answer greater or smaller than the actual value for specific heat of the metal ? 10.15 Given below are observations on molar specific heats at room temperature of some common gases. Reprint 2025-26 THERMAL PROPERTIES OF MATTER 225 Gas Molar specific heat (Cv ) (cal mo1–1 K–1) Hydrogen 4.87 Nitrogen 4.97 Oxygen 5.02 Nitric oxide 4.99 Carbon monoxide 5.01 Chlorine 6.17 The measured molar specific heats of these gases are markedly different from those for monatomic gases. Typically, molar specific heat of a monatomic gas is 2.92 cal/mol K. Explain this difference. What can you infer from the somewhat larger (than the rest) value for chlorine ? 10.16 A child running a temperature of 101°F is given an antipyrin (i.e. a medicine that lowers fever) which causes an increase in the rate of evaporation of sweat from his body. If the fever is brought down to 98 °F in 20 minutes, what is the average rate of extra evaporation caused, by the drug. Assume the evaporation mechanism to be the only way by which heat is lost. The mass of the child is 30 kg. The specific heat of human body is approximately the same as that of water, and latent heat of evaporation of water at that temperature is about 580 cal g–1. 10.17 A ‘thermacole’ icebox is a cheap and an efficient method for storing small quantities of cooked food in summer in particular. A cubical icebox of side 30 cm has a thickness of 5.0 cm. If 4.0 kg of ice is put in the box, estimate the amount of ice remaining after 6 h. The outside temperature is 45 °C, and co-efficient of thermal conductivity of thermacole is 0.01 J s–1 m–1 K–1. [Heat of fusion of water = 335 × 103 J kg–1] 10.18 A brass boiler has a base area of 0.15 m2 and thickness 1.0 cm. It boils water at the rate of 6.0 kg/min when placed on a gas stove. Estimate the temperature of the part of the flame in contact with the boiler. Thermal conductivity of brass = 109 J s–1 m–1 K–1 ; Heat of vaporisation of water = 2256 × 103 J kg–1. 10.19 Explain why : (a) a body with large reflectivity is a poor emitter (b) a brass tumbler feels much colder than a wooden tray on a chilly day (c) an optical pyrometer (for measuring high temperatures) calibrated for an ideal black body radiation gives too low a value for the temperature of a red hot iron piece in the open, but gives a correct value for the temperature when the same piece is in the furnace (d) the earth without its atmosphere would be inhospitably cold (e) heating systems based on circulation of steam are more efficient in warming a building than those based on circulation of hot water 10.20 A body cools from 80 °C to 50 °C in 5 minutes. Calculate the time it takes to cool from 60 °C to 30 °C. The temperature of the surroundings is 20 °C. Reprint 2025-26 CHAPTER ELEVEN THERMODYNAMICS 11.1 INTRODUCTION In previous chapter we have studied thermal properties of matter. In this chapter we shall study laws that govern thermal energy. We shall study the processes where work is 11.1 Introduction converted into heat and vice versa. In winter, when we rub 11.2 Thermal equilibrium our palms together, we feel warmer; here work done in rubbing 11.3 Zeroth law of produces the ‘heat’. Conversely, in a steam engine, the ‘heat’ Thermodynamics of the steam is used to do useful work in moving the pistons,

10.3 (a) The refractive index of glass is 1.5. What is the speed of light in glass? (Speed of light in vacuum is 3.0 × 108 m s–1) (b) Is the speed of light in glass independent of the colour of light? If not, which of the two colours red and violet travels slower in a glass prism?

7.5 ACCELERATION DUE TO GRAVITY OF THE EARTH 4π 3 density and hence its mass is M E = R E ρ 3 The earth can be imagined to be a sphere made where ME is the mass of the earth RE is its radiusof a large number of concentric spherical shells and ρ is the density. On the other hand thewith the smallest one at the centre and the largest one at its surface. A point outside the 4π 3 ρr and mass of the sphere Mr of radius r isearth is obviously outside all the shells. Thus, 3 Reprint 2025-26 GRAVITATION 133 hence its distance from the centre of the earth is (RE + h ). If F (h) denoted the magnitude of the force on the point mass m , we get from G m M E Eq. (7.5) : = 3 r (7.10) R E If the mass m is situated on the surface of GM E m F (h ) = earth, then r = RE and the gravitational force on ( R E + h )2 (7.13) it is, from Eq. (7.10) The acceleration experienced by the point M E m F = G 2 (7.11) mass is F (h )/ m ≡ g (h ) and we get R E The acceleration experienced by the mass F (h ) GM E . g (h ) = = (7.14)m, which is usually denoted by the symbol g is m ( R E + h )2 related to F by Newton’s 2nd law by relation This is clearly less than the value of g on the F = mg. Thus GM E . g = surface of earth : GM F For h << R E , we can E R E2 g = = 2 (7.12) m R E expand the RHS of Eq. (7.14) : E Acceleration g is readily measurable. RE is a g (h ) = 2 GM 2 = g (1 + h / R E )−2known quantity. The measurement of G by R E (1 + h / R E ) Cavendish’s experiment (or otherwise), combined h << 1 , using binomial expression,with knowledge of g and RE enables one to For R E estimate ME from Eq. (7.12). This is the reason  2h why there is a popular statement regarding g (h ) ≅ g 1 − . (7.15)Cavendish : “Cavendish weighed the earth”.  RE  7.6 ACCELERATION DUE TO GRAVITY BELOW Equation (7.15) thus tells us that for small AND ABOVE THE SURFACE OF EARTH heights h above the value of g decreases by a factor (1 − 2h / R E ).Consider a point mass m at a height h above the Now, consider a point mass m at a depthsurface of the earth as shown in Fig. 7.8(a). The d below the surface of the earth (Fig. 7.8(b)),radius of the earth is denoted by RE . Since this so that its distance from the centre of thepoint is outside the earth, earth is ( R E − d ) as shown in the figure. The earth can be thought of as being composed of a smaller sphere of radius (RE – d ) and a spherical shell of thickness d. The force on m due to the outer shell of thickness d is zero because the result quoted in the previous section. As far as the smaller sphere of radius ( RE – d ) is concerned, the point mass is outside it and hence according to the result quoted earlier, the force due to this smaller sphere is just as if the entire mass of the smaller sphere is concentrated at the centre. If Ms is the mass of the smaller sphere, then, Ms/ME = ( RE – d)3 / RE3 ( 7.16) Since mass of a sphere is proportional to be Fig. 7.8 (a) g at a height h above the surface of the cube of its radius. earth. Reprint 2025-26 134 PHYSICS close to the surface of earth, at distances from the surface much smaller than the radius of the earth. In such cases, the force of gravity is practically a constant equal to mg, directed towards the centre of the earth. If we consider a Ms ME point at a height h1 from the surface of the earth and another point vertically above it at a height h2 from the surface, the work done in lifting the particle of mass m from the first to the second position is denoted by W12 Fig. 7.8 (b) g at a depth d. In this case only the smaller W12 = Force × displacement = mg (h2 – h1) (7.20) sphere of radius (RE–d) contributes to g. Thus the force on the point mass is If we associate a potential energy W(h) at a point at a height h above the surface such that F (d) = G Ms m / (RE – d ) 2 (7.17) W(h) = mgh + Wo (7.21) Substituting for Ms from above , we get (where Wo = constant) ; F (d) = G ME m ( RE – d ) / RE 3 (7.18) then it is clear that and hence the acceleration due to gravity at W12 = W(h2) – W(h1) (7.22) a depth d, The work done in moving the particle is just the difference of potential energy between its F ( d ) final and initial positions.Observe that the g(d) = is m constant Wo cancels out in Eq. (7.22). Setting h = 0 in the last equation, we get W ( h = 0 ) = Wo. F (d ) GM E g ( d ) = = 3 ( R E − d ) . h = 0 means points on the surface of the earth. m R E Thus, Wo is the potential energy on the surface of the earth. R E − d = g = g (1 − d / R E ) (7.19) If we consider points at arbitrary distance R E from the surface of the earth, the result just Thus, as we go down below earth’s surface, derived is not valid since the assumption that the acceleration due gravity decreases by a factor the gravitational force mg is a constant is no (1 − d / R E ). The remarkable thing about longer valid. However, from our discussion we know that a point outside the earth, the force of acceleration due to earth’s gravity is that it is gravitation on a particle directed towards the maximum on its surface decreasing whether you centre of the earth is go up or down. G ME m F = 2 (7.23)7.7 GRAVITATIONAL POTENTIAL ENERGY r where ME = mass of earth, m = mass of theWe had discussed earlier the notion of potential particle and r its distance from the centre of theenergy as being the energy stored in the body at earth. If we now calculate the work done inits given position. If the position of the particle lifting a particle from r = r1 to r = r2 (r2 > r1) alongchanges on account of forces acting on it, then a vertical path, we get instead of Eq. (7.20) the change in its potential energy is just the amount of work done on the body by the force. r2 G M m W12 2 d rAs we had discussed earlier, forces for which the =∫r1 r work done is independent of the path are the conservative forces.  1 1  = − G M E m − (7.24) The force of gravity is a conservative force  r2 r1  and we can calculate the potential energy of a In place of Eq. (7.21), we can thus associate body arising out of this force, called the a potential energy W(r) at a distance r, such that gravitational potential energy. Consider points Reprint 2025-26 GRAVITATION 135 G M E m W (r ) =− + W1 , (7.25) r valid for r > R , so that once again W12 = W(r2) – W(r1). Setting r = infinity in the last equation, we get W ( r = infinity ) = W1 . Thus, W1 is the potential energy at infinity. One should note that only the difference of potential energy between two points has a definite meaning from Eqs. (7.22) and (7.24). One conventionally sets W1 equal to zero, so that the potential energy at a point is just the amount of work done in displacing the particle from infinity to that point. We have calculated the potential energy at a point of a particle due to gravitational forces Fig. 7.9 on it due to the earth and it is proportional to the mass of the particle. The gravitational The gravitational potential at the centre of potential due to the gravitational force of the the square r = 2 l/2 is ( )earth is defined as the potential energy of a particle of unit mass at that point. From the G m U (r ) = − 4 2 . ⊳earlier discussion, we learn that the gravitational l potential energy associated with two particles of masses m1 and m2 separated by distance by a 7.8 ESCAPE SPEED distance r is given by If a stone is thrown by hand, we see it falls back Gm 1m 2 to the earth. Of course using machines we can V = – (if we choose V = 0 as r →∞) r shoot an object with much greater speeds and It should be noted that an isolated system of with greater and greater initial speed, the object particles will have the total potential energy that scales higher and higher heights. A natural equals the sum of energies (given by the above query that arises in our mind is the following: equation) for all possible pairs of its constituent ‘can we throw an object with such high initial particles. This is an example of the application speeds that it does not fall back to the earth?’ of the superposition principle. The principle of conservation of energy helps us to answer this question. Suppose the object ⊳ Example 7.3 Find the potential energy of did reach infinity and that its speed there was a system of four particles placed at the Vf. The energy of an object is the sum of potential vertices of a square of side l. Also obtain and kinetic energy. As before W1 denotes that the potential at the centre of the square. gravitational potential energy of the object at infinity. The total energy of the projectile at infinity then is Answer Consider four masses each of mass m at the corners of a square of side l; See Fig. 7.9. 2 mV fWe have four mass pairs at distance l and two E ( ∞=) W1 + (7.26) 2 diagonal pairs at distance 2 l If the object was thrown initially with a speed Hence, Vi from a point at a distance (h+RE) from the G m 2 G m 2 centre of the earth (RE = radius of the earth), its W (r ) = − 4 − 2 l 2 l energy initially was 2 2 1 2 GmM E 2 G m  1  G m E (h + R E ) = mVi – + W1 (7.27) 5.41 = − 2 (h + R E ) 2 +  = − l  l  2  Reprint 2025-26 136 PHYSICS By the principle of energy conservation ⊳ Eqs. (7.26) and (7.27) must be equal. Hence Example 7.4 Two uniform solid spheres 2 2 of equal radii R, but mass M and 4 M have mV f mVi GmM E a centre to centre separation 6 R, as shown – = (7.28) 2 (h + R E ) 2 in Fig. 7.10. The two spheres are held fixed. The R.H.S. is a positive quantity with a A projectile of mass m is projected from the minimum value zero hence so must be the L.H.S. surface of the sphere of mass M directly Thus, an object can reach infinity as long as Vi towards the centre of the second sphere. is such that Obtain an expression for the minimum speed v of the projectile so that it reaches mVi 2 GmM E the surface of the second sphere. – ≥ 0 (7.29) 2 (h + R E ) The minimum value of Vi corresponds to the case when the L.H.S. of Eq. (7.29) equals zero. Thus, the minimum speed required for an object to reach infinity (i.e. escape from the earth) Fig. 7.10corresponds to 1 2 GmM E Answer The projectile is acted upon by two m = (7.30) mutually opposing gravitational forces of the two ( Vi )min 2 h + R E spheres. The neutral point N (see Fig. 7.10) is defined as the position where the two forces If the object is thrown from the surface of cancel each other exactly. If ON = r, we have the earth, h = 0, and we get G M m 4 G M m = 2GM E r 2 (6 R −r )2 (Vi )min = (7.31) (6R – r)2 = 4r2 R E 6R – r = ±2r r = 2R or – 6R. 2 The neutral point r = – 6R does not concern Using the relation g = GM E / R E , we get us in this example. Thus ON = r = 2R. It is sufficient to project the particle with a speed (Vi )min = 2 gR E (7.32) which would enable it to reach N. Thereafter, the greater gravitational pull of 4M would suffice. The mechanical energy at the surface Using the value of g and RE, numerically of M is (Vi)min≈11.2 km/s. This is called the escape 1 2 G M m 4 G M mspeed, sometimes loosely called the escape E i = m v − − . velocity. 2 R 5 R Equation (7.32) applies equally well to an At the neutral point N, the speed approaches object thrown from the surface of the moon with zero. The mechanical energy at N is purely g replaced by the acceleration due to Moon’s potential. gravity on its surface and rE replaced by the G M m 4 G M m − EN = − .radius of the moon. Both are smaller than their 2 R 4 R values on earth and the escape speed for the From the principle of conservation of moon turns out to be 2.3 km/s, about five times mechanical energy smaller. This is the reason that moon has no atmosphere. Gas molecules if formed on the 1 2 GM 4GM GM GMsurface of the moon having velocities larger than v − − = − − this will escape the gravitational pull of the 2 R 5 R 2R R moon. or Reprint 2025-26 GRAVITATION 137 + h) with speed V. Its 2 2 G M  4 1  traverses a distance 2π(RE v = − time period T therefore is R  5 2  2π( R E + h ) 2π( R E + h )3 / 2 T = = (7.37)  3 G M 1/2 V ⊳ G M E v =  5 R  on substitution of value of V from Eq. (7.35). A point to note is that the speed of the projectile Squaring both sides of Eq. (7.37), we get is zero at N, but is nonzero when it strikes the 2 T = k ( RE + h)3 (where k = 4 π2 / GME) (7.38)heavier sphere 4 M. The calculation of this speed is left as an exercise to the students. which is Kepler’s law of periods, as applied to motion of satellites around the earth. For a 7.9 EARTH SATELLITES satellite very close to the surface of earth h can be neglected in comparison to RE in Eq. (7.38). Earth satellites are objects which revolve around Hence, for such satellites, T is To, where the earth. Their motion is very similar to the motion of planets around the Sun and hence T 0 = 2π R E / g (7.39) Kepler’s laws of planetary motion are equally If we substitute the numerical values applicable to them. In particular, their orbits g ≃ 9.8 m s-2 and RE = 6400 km., we get around the earth are circular or elliptic. Moon is the only natural satellite of the earth with a near 6.4 × 10 6 T 0 = 2π s circular orbit with a time period of approximately 9.8

7.4 AC VOLTAGE APPLIED TO AN INDUCTOR Figure 7.5 shows an ac source connected to an inductor. Usually, inductors have appreciable resistance in their windings, but we shall assume that this inductor has negligible resistance. Thus, the circuit is a purely inductive ac circuit. Let the voltage across the source be v = vm sinw t. Using the Kirchhoff’s loop rule, ε ()t = 0 , and since there ∑ is no resistor in the circuit, d i v − L = 0 (7.10) d t where the second term is the self-induced Faraday FIGURE 7.5 An ac source emf in the inductor; and L is the self-inductance of connected to an inductor. * Though voltage and current in ac circuit are represented by phasors – rotating vectors, they are not vectors themselves. They are scalar quantities. It so happens that the amplitudes and phases of harmonically varying scalars combine mathematically in the same way as do the projections of rotating vectors of corresponding magnitudes and directions. The rotating vectors that represent harmonically varying scalar quantities are introduced only to provide us with a simple way of adding these quantities using a rule that we already know. 181 Reprint 2025-26 Physics the inductor. The negative sign follows from Lenz’s law (Chapter 6). Combining Eqs. (7.1) and (7.10), we have d i v v m = = sin ωt (7.11) d t L L Equation (7.11) implies that the equation for i(t), the current as a function of time, must be such that its slope di/dt is a sinusoidally varying quantity, with the same phase as the source voltage and an amplitude given by vm/L. To obtain the current, we integrate di/dt with respect to circuits: time: di v m series d t = sin(ωt )d t ∫ d t L ∫ RLC and get, and v m i = − cos( ω t ) + constant C ω L L, The integration constant has the dimension of current and is time- R, independent. Since the source has an emf which oscillates symmetrically about zero, the current it sustains also oscillates symmetrically about zero, so that no constant or time-independent component of the current exists. Therefore, the integration constant is zero. containing, Using  π  circuits − cos(ωt ) = sin ω t − , we have  2  ac of  π  i = i m sin  ωt − 2  diagrams v m (7.12) where i m = is the amplitude of the current. The quantity w L is ω L analogous to the resistance and is called inductive reactance, denoted Phasor by XL: on XL = w L (7.13) The amplitude of the current is, then v m animation i m = (7.14) X L The dimension of inductive reactance is the same as that of resistance and its SI unit is ohm (W). The inductive reactance limits the current in a Interactive http://www.animations.physics.unsw.edu.au//jw/AC.html purely inductive circuit in the same way as the resistance limits the current in a purely resistive circuit. The inductive reactance is directly proportional to the inductance and to the frequency of the current. A comparison of Eqs. (7.1) and (7.12) for the source voltage and the current in an inductor shows that the current lags the voltage by p/2 or one-quarter (1/4) cycle. Figure 7.6 (a) shows the voltage and the current phasors in the present case at instant t1. The current phasor I is p/2 behind the voltage phasor V. When rotated with frequency w counter- clockwise, they generate the voltage and current given by Eqs. (7.1) and 182 (7.12), respectively and as shown in Fig. 7.6(b). Reprint 2025-26 Alternating Current FIGURE 7.6 (a) A Phasor diagram for the circuit in Fig. 7.5. (b) Graph of v and i versus wt. We see that the current reaches its maximum value later than the  T π/ 2 voltage by one-fourth of a period = . You have seen that an  4 ω  inductor has reactance that limits current similar to resistance in a dc circuit. Does it also consume power like a resistance? Let us try to find out. The instantaneous power supplied to the inductor is  π  p L = i v = i m sin ω t − ×v m sin (ωt )  2  = −i m vm cos (ωt ) sin (ωt ) i m v m = − sin ( 2ωt ) 2 So, the average power over a complete cycle is i m v m PL = − sin ( 2ωt ) 2 i m v m = − sin ( 2ωt ) = 0, 2 since the average of sin (2wt) over a complete cycle is zero. Thus, the average power supplied to an inductor over one complete cycle is zero. Example 7.2 A pure inductor of 25.0 mH is connected to a source of 220 V. Find the inductive reactance and rms current in the circuit if the frequency of the source is 50 Hz. Solution The inductive reactance, X L = 2 πνL = 2 × 3 .14 × 50 × 25 × 10 –3 Ω = 7.85W The rms current in the circuit is EXAMPLE V 220 V I = = = 28 A 7.2 7.85 Ω X L 183 Reprint 2025-26 Physics 7.5 AC VOLTAGE APPLIED TO A CAPACITOR Figure 7.7 shows an ac source e generating ac voltage v = vm sin wt connected to a capacitor only, a purely capacitive ac circuit. When a capacitor is connected to a voltage source in a dc circuit, current will flow for the short time required to charge the capacitor. As charge accumulates on the capacitor plates, the voltage across them increases, opposing the current. That is, a capacitor in a dc circuit will limit or oppose the current as it charges. When the capacitor is fully charged, the current in the circuit falls to zero. When the capacitor is connected to an ac source, as in Fig. 7.7, it limits or regulates the current, but FIGURE 7.7 An ac source does not completely prevent the flow of charge. The connected to a capacitor. capacitor is alternately charged and discharged as the current reverses each half cycle. Let q be the charge on the capacitor at any time t. The instantaneous voltage v across the capacitor is q v = (7.15) C From the Kirchhoff’s loop rule, the voltage across the source and the capacitor are equal, q v m sin ωt = C d q To find the current, we use the relation i = dt d i = (vm C sin ωt ) = ωC v m cos(ωt ) d t  π  Using the relation, cos(ωt ) = sin  ω t + 2  , we have  π  (7.16) i = i m sin  ωt + 2  where the amplitude of the oscillating current is im = wCvm. We can rewrite it as vm i m = (1/ωC ) Comparing it to im= vm/R for a purely resistive circuit, we find that (1/wC) plays the role of resistance. It is called capacitive reactance and is denoted by Xc, Xc= 1/wC (7.17) so that the amplitude of the current is vm i m = (7.18)184 X C Reprint 2025-26 Alternating Current The dimension of capacitive reactance is the same as that of resistance and its SI unit is ohm (Ω). The capacitive reactance limits the amplitude of the current in a purely capacitive circuit in the same way as the resistance limits the current in a purely resistive circuit. But it is inversely proportional to the frequency and the capacitance. A comparison of Eq. (7.16) with the FIGURE 7.8 (a) A Phasor diagram for the circuit equation of source voltage, Eq. (7.1) shows that in Fig. 7.7. (b) Graph of v and i versus ωt. the current is π/2 ahead of voltage. Figure 7.8(a) shows the phasor diagram at an instant t1. Here the current phasor I is π/2 ahead of the voltage phasor V as they rotate counterclockwise. Figure 7.8(b) shows the variation of voltage and current with time. We see that the current reaches its maximum value earlier than the voltage by one-fourth of a period. The instantaneous power supplied to the capacitor is pc = i v = im cos(ωt)vm sin(ωt) = imvm cos(ωt) sin(ωt) i m vm = sin(2ωt ) (7.19) 2 So, as in the case of an inductor, the average power i m v m i m v m PC = sin(2ωt ) = sin(2ωt ) = 0 2 2 since <sin (2ωt)> = 0 over a complete cycle. Thus, we see that in the case of an inductor, the current lags the voltage by π/2 and in the case of a capacitor, the current leads the voltage by π/2. Example 7.3 A lamp is connected in series with a capacitor. Predict your observations for dc and ac connections. What happens in each case if the capacitance of the capacitor is reduced? Solution When a dc source is connected to a capacitor, the capacitor gets charged and after charging no current flows in the circuit and the lamp will not glow. There will be no change even if C is reduced. With ac source, the capacitor offers capacitative reactance (1/ωC) and the current flows in the circuit. Consequently, the lamp will shine. EXAMPLE Reducing C will increase reactance and the lamp will shine less brightly than before. 7.3 Example 7.4 A 15.0 µF capacitor is connected to a 220 V, 50 Hz source. Find the capacitive reactance and the current (rms and peak) in the circuit. If the frequency is doubled, what happens to the capacitive reactance and the current? Solution The capacitive reactance is 1 1 = = 212 Ω X C = EXAMPLE 2 π (50Hz)(15.0 × 10 −6 F ) 2 π νC The rms current is 7.4 185 Reprint 2025-26 Physics V 220 V I = = = 1.04 A X C 212 Ω The peak current is m = 2 I = (1.41)(1.04 A ) = 1.47 A 7.4 i This current oscillates between +1.47A and –1.47 A, and is ahead of the voltage by p/2. If the frequency is doubled, the capacitive reactance is halved and EXAMPLE consequently, the current is doubled. Example 7.5 A light bulb and an open coil inductor are connected to an ac source through a key as shown in Fig. 7.9. FIGURE 7.9 The switch is closed and after sometime, an iron rod is inserted into the interior of the inductor. The glow of the light bulb (a) increases; (b) decreases; (c) is unchanged, as the iron rod is inserted. Give your answer with reasons. 7.5 SolutionmagnetizesAsthetheironiron increasingrod is inserted,the magneticthe magneticfieldfieldinsideinsideit. theHence,coil the inductance of the coil increases. Consequently, the inductive reactance of the coil increases. As a result, a larger fraction of the applied ac voltage appears across the inductor, leaving less voltage EXAMPLE across the bulb. Therefore, the glow of the light bulb decreases. 7.6 AC VOLTAGE APPLIED TO A SERIES LCR CIRCUIT Figure 7.10 shows a series LCR circuit connected to an ac source e. As usual, we take the voltage of the source to be v = vm sin wt. If q is the charge on the capacitor and i the current, at time t, we have, from Kirchhoff’s loop rule: d i q L + i R + = v (7.20) d t C We want to determine the instantaneous current i and its phase relationship to the applied alternating voltage v. We shall solve this problem by two methods. First, we use the technique of FIGURE 7.10 A series LCR circuit phasors and in the second method, we solve connected to an ac source. Eq. (7.20) analytically to obtain the time– 186 dependence of i. Reprint 2025-26 Alternating Current 7.6.1 Phasor-diagram solution From the circuit shown in Fig. 7.10, we see that the resistor, inductor and capacitor are in series. Therefore, the ac current in each element is the same at any time, having the same amplitude and phase. Let it be i = im sin(wt+f) (7.21) where fis the phase difference between the voltage across the source and the current in the circuit. On the basis of what we have learnt in the previous sections, we shall construct a phasor diagram for the present case. Let I be the phasor representing the current in the circuit as given by Eq. (7.21). Further, let VL, VR, VC, and V represent the voltage across the inductor, resistor, capacitor and the source, respectively. From previous section, we know that VR is parallel to I, VC is p/2 behind I and VL is p/2 ahead of I. VL, VR, VC and I are shown in Fig. 7.11(a) with apppropriate phase- relations. The length of these phasors or the amplitude of VR, VC and VL are: vRm = im R, vCm = im XC, vLm = im XL (7.22) The voltage Equation (7.20) for the circuit can be written as vL + vR + vC = v (7.23) The phasor relation whose vertical component gives the above equation is FIGURE 7.11 (a) Relation between the phasors VL, VR, VC, and I, (b) Relation VL + VR + VC = V (7.24) between the phasors VL, VR, and (VL + VC) This relation is represented in Fig. 7.11(b). Since for the circuit in Fig. 7.10. VC and VL are always along the same line and in opposite directions, they can be combined into a single phasor (VC + VL) which has a magnitude ½vCm – vLm½. Since V is represented as the hypotenuse of a right-triangle whose sides are VR and (VC + VL), the pythagorean theorem gives: 2 2 2 vm = v Rm + (v Cm − v Lm ) Substituting the values of vRm, vCm, and vLm from Eq. (7.22) into the above equation, we have v m2 = (i m R )2 + (i m X C − i m X L )2 = i m2  R 2 + ( X C − X L )2  v m or, i m = 2 2 [7.25(a)] R + ( X C − X L ) By analogy to the resistance in a circuit, we introduce the impedance Z in an ac circuit: v m i m = [7.25(b)] Z where Z = R 2 + ( X C − X L )2 (7.26) 187 Reprint 2025-26 Physics Since phasor I is always parallel to phasor VR, the phase angle f is the angle between VR and V and can be determined from Fig. 7.12: vCm − v Lm tan φ = v Rm Using Eq. (7.22), we have X C − X L tan φ = (7.27) R Equations (7.26) and (7.27) are graphically shown in Fig. (7.12). FIGURE 7.12 Impedance This is called Impedance diagram which is a right-triangle with diagram. Z as its hypotenuse. Equation 7.25(a) gives the amplitude of the current and Eq. (7.27) gives the phase angle. With these, Eq. (7.21) is completely specified. If XC > XL, f is positive and the circuit is predominantly capacitive. Consequently, the current in the circuit leads the source voltage. If XC < XL, f is negative and the circuit is predominantly inductive. Consequently, the current in the circuit lags the source voltage. Figure 7.13 shows the phasor diagram and variation of v and i with wt for the case XC > XL. Thus, we have obtained the amplitude and phase of current for an LCR series circuit using the technique of phasors. But this method of analysing ac circuits suffers from certain disadvantages. First, the phasor diagram say nothing about the initial condition. One can take any arbitrary value of t (say, t1, as done throughout this chapter) and draw different phasors which show the relative angle between different phasors. The solution so obtained is called the steady-state solution. This is not a general FIGURE 7.13 (a) Phasor diagram of V and I. solution. Additionally, we do have a (b) Graphs of v and i versus w t for a series LCR transient solution which exists even for circuit where XC > XL. v = 0. The general solution is the sum of the transient solution and the steady-state solution. After a sufficiently long time, the effects of the transient solution die out and the behaviour of the circuit is described by the steady-state solution. 7.6.2 Resonance An interesting characteristic of the series RLC circuit is the phenomenon of resonance. The phenomenon of resonance is common among systems that have a tendency to oscillate at a particular frequency. This frequency is called the system’s natural frequency. If such a system is driven by an energy source at a frequency that is near the natural frequency, the amplitude of oscillation is found to be large. A familiar example of this phenomenon is a child on a swing. The swing has a natural frequency for swinging back and forth like a pendulum. If the child pulls on the Reprint 2025-26 Alternating Current rope at regular intervals and the frequency of the pulls is almost the same as the frequency of swinging, the amplitude of the swinging will be large. For an RLC circuit driven with voltage of amplitude vm and frequency w, we found that the current amplitude is given by v m v m i m = = 2 2 Z R + ( X C − X L ) with Xc = 1/wC and XL = wL. So if w is varied, then at a particular frequency Z = R 2 + 0 2 = R . Thisw0, Xc = XL, and the impedance is minimum ( ) frequency is called the resonant frequency: 1 X c = X L or = ω0 L ω0 C 1 or ω0 = (7.28) LC At resonant frequency, the current amplitude is maximum; im = vm/R. Figure 7.16 shows the variation of im with w in a RLC series circuit with L = 1.00 mH, C =

8.2 A parallel plate capacitor (Fig. 8.6) made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to 213 a 230 V ac supply with a (angular) frequency of 300 rad s–1. Reprint 2025-26 Physics (a) What is the rms value of the conduction current? (b) Is the conduction current equal to the displacement current? (c) Determine the amplitude of B at a point 3.0 cm from the axis between the plates. FIGURE 8.6 8.3 What physical quantity is the same for X-rays of wavelength 10–10 m, red light of wavelength 6800 Å and radiowaves of wavelength 500m? 8.4 A plane electromagnetic wave travels in vacuum along z-direction. What can you say about the directions of its electric and magnetic field vectors? If the frequency of the wave is 30 MHz, what is its wavelength? 8.5 A radio can tune in to any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band? 8.6 A charged particle oscillates about its mean equilibrium position with a frequency of 10 9 Hz. What is the frequency of the electromagnetic waves produced by the oscillator? 8.7 The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is B0 = 510 nT. What is the amplitude of the electric field part of the wave? 8.8 Suppose that the electric field amplitude of an electromagnetic wave is E0 = 120 N/C and that its frequency is n = 50.0 MHz. (a) Determine, B0,w, k, and l. (b) Find expressions for E and B. 8.9 The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula E = hn (for energy of a quantum of radiation: photon) and obtain the photon energy in units of eV for different parts of the electromagnetic spectrum. In what way are the different scales of photon energies that you obtain related to the sources of electromagnetic radiation? 8.10 In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 × 1010 Hz and amplitude 48 V m–1. (a) What is the wavelength of the wave? (b) What is the amplitude of the oscillating magnetic field? (c) Show that the average energy density of the E field equals the average energy density of the B field. [c = 3 × 108 m s–1.] Reprint 2025-26

8.3 ELECTROMAGNETIC WAVES 8.3.1 Sources of electromagnetic waves How are electromagnetic waves produced? Neither stationary charges nor charges in uniform motion (steady currents) can be sources of electromagnetic waves. The former produces only electrostatic fields, while the latter produces magnetic fields that, however, do not vary with time. It is an important result of Maxwell’s theory that accelerated charges radiate electromagnetic waves. The proof of this basic result is beyond the scope of this book, but we can accept it on the basis of rough, qualitative reasoning. Consider a charge oscillating with some frequency. (An oscillating charge is an example of accelerating charge.) This produces an oscillating electric field in space, which produces an oscillating magnetic field, which in turn, is a source of oscillating electric field, and so on. The oscillating electric and magnetic fields thus regenerate each other, so to speak, as the wave propagates through the space. The frequency of the electromagnetic wave naturally equals the frequency of oscillation of the charge. The energy associated with the propagating wave comes at the expense of the energy of the source – the accelerated charge. From the preceding discussion, it might appear easy to test the prediction that light is an electromagnetic wave. We might think that all we needed to do was to set up an ac circuit in which the current oscillate at the frequency of visible light, say, yellow light. But, alas, that is not possible. The frequency of yellow light is about 6 × 1014 Hz, while the frequency that we get even with modern electronic circuits is hardly about 1011 Hz. This is why the experimental demonstration of electromagnetic 205 Reprint 2025-26 Physics wave had to come in the low frequency region (the radio wave region), as in the Hertz’s experiment (1887). Hertz’s successful experimental test of Maxwell’s theory created a sensation and sparked off other important works in this field. Two important achievements in this connection deserve mention. Seven years after Hertz, Jagdish Chandra Bose, working at Calcutta (now Kolkata), succeeded in producing and observing electromagnetic waves of much shorter 8.1 wavelength (25 mm to 5 mm). His experiment, like that of Hertz’s, was confined to the laboratory. At around the same time, Guglielmo Marconi in Italy followed Hertz’s work and succeeded in transmitting EXAMPLE electromagnetic waves over distances of many kilometres. Heinrich Rudolf Hertz Marconi’s experiment marks the beginning of the field of (1857 – 1894) German communication using electromagnetic waves. physicist who was the first to broadcast and 8.3.2 Nature of electromagnetic wavesHEINRICH receive radio waves. He It can be shown from Maxwell’s equations that electric produced electro- and magnetic fields in an electromagnetic wave are magnetic waves, sent them through space, and perpendicular to each other, and to the direction of measured their wave- propagation. It appears reasonable, say from ourRUDOLF length and speed. He discussion of the displacement current. Consider showed that the nature Fig. 8.2. The electric field inside the plates of the capacitor of their vibration, is directed perpendicular to the plates. The magnetic reflection and refraction field this gives rise to via the displacement current is was the same as that ofHERTZ along the perimeter of a circle parallel to the capacitor light and heat waves, plates. So B and E are perpendicular in this case. This establishing their identity for the first time. is a general feature. He also pioneered In Fig. 8.3, we show a typical example of a plane research on discharge of electromagnetic wave propagating along the z direction electricity through gases, (the fields are shown as a function of the z coordinate, at and discovered the(1857–1894) a given time t). The electric field Ex is along the x-axis, photoelectric effect. and varies sinusoidally with z, at a given time. The magnetic field By is along the y-axis, and again varies sinusoidally with z. The electric and magnetic fields Ex and By are perpendicular to each other, and to the direction z of propagation. We can write Ex and By as follows: Ex= E0 sin (kz–wt) [8.7(a)] By= B0 sin (kz–wt) [8.7(b)] Here k is related to the wave length FIGURE 8.3 A linearly polarised electromagnetic wave, l of the wave by the usual propagating in the z-direction with the oscillating electric field E equation along the x-direction and the oscillating magnetic field B along the y-direction. 2 π k = (8.8) 206 λ Reprint 2025-26 Electromagnetic Waves and ω is the angular frequency. k is the magnitude of the wave vector (or propagation vector) k and its direction describes the direction of propagation of the wave. The speed of propagation of the wave is (ω/k). Using Eqs. [8.7(a) and (b)] for Ex and By and Maxwell’s equations, one finds that ω = ck, where, c = 1/ µ0ε0 [8.9(a)] The relation ω = ck is the standard one for waves (see for example, Section 14.4 of class XI Physics textbook). This relation is often written in terms of frequency, ν (=ω/2π) and wavelength, λ (=2π/k) as  2π  2 πν = c  λ  or νλ = c [8.9(b)] It is also seen from Maxwell’s equations that the magnitude of the electric and the magnetic fields in an electromagnetic wave are related as B0 = (E0/c) (8.10) We here make remarks on some features of electromagnetic waves. They are self-sustaining oscillations of electric and magnetic fields in free space, or vacuum. They differ from all the other waves we have studied so far, in respect that no material medium is involved in the vibrations of the electric and magnetic fields. But what if a material medium is actually there? We know that light, an electromagnetic wave, does propagate through glass, for example. We have seen earlier that the total electric and magnetic fields inside a medium are described in terms of a permittivity ε and a magnetic permeability µ (these describe the factors by which the total fields differ from the external fields). These replace ε0 and µ0 in the description to electric and magnetic fields in Maxwell’s equations with the result that in a material medium of permittivity ε and magnetic permeability µ, the velocity of light becomes, 1 v = µε (8.11) Thus, the velocity of light depends on electric and magnetic properties of the medium. We shall see in the next chapter that the refractive index of one medium with respect to the other is equal to the ratio of velocities of light in the two media. The velocity of electromagnetic waves in free space or vacuum is an important fundamental constant. It has been shown by experiments on electromagnetic waves of different wavelengths that this velocity is the same (independent of wavelength) to within a few metres per second, out of a value of 3×108 m/s. The constancy of the velocity of em waves in vacuum is so strongly supported by experiments and the actual value is so well known now that this is used to define a standard of length. The great technological importance of electromagnetic waves stems from their capability to carry energy from one place to another. The radio and TV signals from broadcasting stations carry energy. Light carries energy from the sun to the earth, thus making life possible on the earth. 207 Reprint 2025-26 Physics Example 8.1 A plane electromagnetic wave of frequency 25 MHz travels in free space along the x-direction. At a particular point in space and time, E = 6.3 ˆj V/m. What is B at this point? Solution Using Eq. (8.10), the magnitude of B is E B = c 6.3 V/m –8 = 8 = 2.1 × 10 T 3 × 10 m/s 8.1 To find the direction, we note that E is along y-direction and the wave propagates along x-axis. Therefore, B should be in a direction perpendicular to both x- and y-axes. Using vector algebra, E × B should be along x-direction. Since, (+ ˆj ) × (+ ˆk ) = ˆi , B is along the z-direction. EXAMPLE Thus, B = 2.1 × 10–8 ˆk T Example 8.2 The magnetic field in a plane electromagnetic wave is given by By = (2 × 10–7) T sin (0.5×103x+1.5×1011t). (a) What is the wavelength and frequency of the wave? (b) Write an expression for the electric field. Solution (a) Comparing the given equation with   x t   By=B0 Sin 2p +  λ T  spectrum  2π We get, λ = 3 m = 1.26 cm, 0.5 × 10 1 11 and = ν= 1.5 × 10 /2 π = 23.9 GHz T ( ) 8.2 (b) E0 = B0c = 2×10–7 T × 3 × 108 m/s = 6 × 101 V/mElectromagnetic http://www.fnal.gov/pub/inquiring/more/light http://imagine.gsfc.nasa.gov/docs/science/ The electric field component is perpendicular to the direction of propagation and the direction of magnetic field. Therefore, the electric field component along the z-axis is obtained as EXAMPLE Ez = 60 sin (0.5 × 103x + 1.5 × 1011 t) V/m 8.4 ELECTROMAGNETIC SPECTRUM At the time Maxwell predicted the existence of electromagnetic waves, the only familiar electromagnetic waves were the visible light waves. The existence of ultraviolet and infrared waves was barely established. By the end of the nineteenth century, X-rays and gamma rays had also been discovered. We now know that, electromagnetic waves include visible light waves, X-rays, gamma rays, radio waves, microwaves, ultraviolet and infrared waves. The classification of em waves according to frequency is the electromagnetic spectrum (Fig. 8.4). There is no sharp division between one kind of wave and the next. The classification is based roughly on how the waves are produced and/or detected. We briefly describe these different types of electromagnetic waves, in 208 order of decreasing wavelengths. Reprint 2025-26 Electromagnetic Waves FIGURE 8.4 The electromagnetic spectrum, with common names for various part of it. The various regions do not have sharply defined boundaries. 8.4.1 Radio waves Radio waves are produced by the accelerated motion of charges in conducting wires. They are used in radio and television communication systems. They are generally in the frequency range from 500 kHz to about 1000 MHz. The AM (amplitude modulated) band is from 530 kHz to 1710 kHz. Higher frequencies upto 54 MHz are used for short wave bands. TV waves range from 54 MHz to 890 MHz. The FM (frequency modulated) radio band extends from 88 MHz to 108 MHz. Cellular phones use radio waves to transmit voice communication in the ultrahigh frequency (UHF) band. How these waves are transmitted and received is described in Chapter 15. 8.4.2 Microwaves Microwaves (short-wavelength radio waves), with frequencies in the gigahertz (GHz) range, are produced by special vacuum tubes (called klystrons, magnetrons and Gunn diodes). Due to their short wavelengths, they are suitable for the radar systems used in aircraft navigation. Radar also provides the basis for the speed guns used to time fast balls, tennis- serves, and automobiles. Microwave ovens are an interesting domestic application of these waves. In such ovens, the frequency of the microwaves is selected to match the resonant frequency of water molecules so that energy from the waves is transferred efficiently to the kinetic energy of 209the molecules. This raises the temperature of any food containing water. Reprint 2025-26 Physics 8.4.3 Infrared waves Infrared waves are produced by hot bodies and molecules. This band lies adjacent to the low-frequency or long-wave length end of the visible spectrum. Infrared waves are sometimes referred to as heat waves. This is because water molecules present in most materials readily absorb infrared waves (many other molecules, for example, CO2, NH3, also absorb infrared waves). After absorption, their thermal motion increases, that is, they heat up and heat their surroundings. Infrared lamps are used in physical therapy. Infrared radiation also plays an important role in maintaining the earth’s warmth or average temperature through the greenhouse effect. Incoming visible light (which passes relatively easily through the atmosphere) is absorbed by the earth’s surface and re- radiated as infrared (longer wavelength) radiations. This radiation is trapped by greenhouse gases such as carbon dioxide and water vapour. Infrared detectors are used in Earth satellites, both for military purposes and to observe growth of crops. Electronic devices (for example semiconductor light emitting diodes) also emit infrared and are widely used in the remote switches of household electronic systems such as TV sets, video recorders and hi-fi systems. 8.4.4 Visible rays It is the most familiar form of electromagnetic waves. It is the part of the spectrum that is detected by the human eye. It runs from about 4 × 1014 Hz to about 7 × 1014 Hz or a wavelength range of about 700 – 400 nm. Visible light emitted or reflected from objects around us provides us information about the world. Our eyes are sensitive to this range of wavelengths. Different animals are sensitive to different range of wavelengths. For example, snakes can detect infrared waves, and the ‘visible’ range of many insects extends well into the utraviolet. 8.4.5 Ultraviolet rays It covers wavelengths ranging from about 4 × 10–7 m (400 nm) down to 6 × 10–10m (0.6 nm). Ultraviolet (UV) radiation is produced by special lamps and very hot bodies. The sun is an important source of ultraviolet light. But fortunately, most of it is absorbed in the ozone layer in the atmosphere at an altitude of about 40 – 50 km. UV light in large quantities has harmful effects on humans. Exposure to UV radiation induces the production of more melanin, causing tanning of the skin. UV radiation is absorbed by ordinary glass. Hence, one cannot get tanned or sunburn through glass windows. Welders wear special glass goggles or face masks with glass windows to protect their eyes from large amount of UV produced by welding arcs. Due to its shorter wavelengths, UV radiations can be focussed into very narrow beams for high precision applications such as LASIK (Laser- assisted in situ keratomileusis) eye surgery. UV lamps are used to kill germs in water purifiers. Ozone layer in the atmosphere plays a protective role, and hence its depletion by chlorofluorocarbons (CFCs) gas (such as freon) is a matter 210 of international concern. Reprint 2025-26 Electromagnetic Waves 8.4.6 X-rays Beyond the UV region of the electromagnetic spectrum lies the X-ray region. We are familiar with X-rays because of its medical applications. It covers wavelengths from about 10–8 m (10 nm) down to 10–13 m (10–4 nm). One common way to generate X-rays is to bombard a metal target by high energy electrons. X-rays are used as a diagnostic tool in medicine and as a treatment for certain forms of cancer. Because X-rays damage or destroy living tissues and organisms, care must be taken to avoid unnecessary or over exposure. 8.4.7 Gamma rays They lie in the upper frequency range of the electromagnetic spectrum and have wavelengths of from about 10–10m to less than 10–14m. This high frequency radiation is produced in nuclear reactions and also emitted by radioactive nuclei. They are used in medicine to destroy cancer cells. Table 8.1 summarises different types of electromagnetic waves, their production and detections. As mentioned earlier, the demarcation between different regions is not sharp and there are overlaps. TABLE 8.1 DIFFERENT TYPES OF ELECTROMAGNETIC WAVES Type Wavelength range Production Detection Radio > 0.1 m Rapid acceleration and Receiver’s aerials decelerations of electrons in aerials Microwave 0.1m to 1 mm Klystron valve or Point contact diodes magnetron valve Infra-red 1mm to 700 nm Vibration of atoms Thermopiles and molecules Bolometer, Infrared photographic film Light 700 nm to 400 nm Electrons in atoms emit The eye light when they move from Photocells one energy level to a Photographic film lower energy level Ultraviolet 400 nm to 1nm Inner shell electrons in Photocells atoms moving from one Photographic film energy level to a lower level X-rays 1nm to 10–3 nm X-ray tubes or inner shell Photographic film electrons Geiger tubes Ionisation chamber Gamma rays <10–3 nm Radioactive decay of the -do- nucleus 211 Reprint 2025-26 Physics SUMMARY 1. Maxwell found an inconsistency in the Ampere’s law and suggested the existence of an additional current, called displacement current, to remove this inconsistency. This displacement current is due to time-varying electric field and is given by dΦΕ di = ε0 dt and acts as a source of magnetic field in exactly the same way as conduction current. 2. An accelerating charge produces electromagnetic waves. An electric charge oscillating harmonically with frequency n, produces electromagnetic waves of the same frequency n. An electric dipole is a basic source of electromagnetic waves. 3. Electromagnetic waves with wavelength of the order of a few metres were first produced and detected in the laboratory by Hertz in 1887. He thus verified a basic prediction of Maxwell’s equations. 4. Electric and magnetic fields oscillate sinusoidally in space and time in an electromagnetic wave. The oscillating electric and magnetic fields, E and B are perpendicular to each other, and to the direction of propagation of the electromagnetic wave. For a wave of frequency n, wavelength l, propagating along z-direction, we have E = Ex (t) = E0 sin (kz – w t )   z     z t   = E0 sin 2 π  λ − νt  = E 0 sin 2 π  λ − T   B = By(t) = B0 sin (kz – w t)   z     z t   = B 0 sin 2 π  λ − νt  = B 0 sin  2 π  λ − T   They are related by E0/B0 = c. 5. The speed c of electromagnetic wave in vacuum is related to m0 and e0 (the free space permeability and permittivity constants) as follows: c = 1/ µ0 ε0 . The value of c equals the speed of light obtained from optical measurements. Light is an electromagnetic wave; c is, therefore, also the speed of light. Electromagnetic waves other than light also have the same velocity c in free space. The speed of light, or of electromagnetic waves in a material medium is given by v = 1/ µε where m is the permeability of the medium and e its permittivity. 6. The spectrum of electromagnetic waves stretches, in principle, over an infinite range of wavelengths. Different regions are known by different names; g-rays, X-rays, ultraviolet rays, visible rays, infrared rays, microwaves and radio waves in order of increasing wavelength from 10–2 Å or 10–12 m to 106 m. They interact with matter via their electric and magnetic fields which set in oscillation charges present in all matter. The detailed interaction and so the mechanism of absorption, scattering, etc., depend on the wavelength of the electromagnetic wave, and the nature of the atoms and molecules 212 in the medium. Reprint 2025-26 Electromagnetic Waves POINTS TO PONDER 1. The basic difference between various types of electromagnetic waves lies in their wavelengths or frequencies since all of them travel through vacuum with the same speed. Consequently, the waves differ considerably in their mode of interaction with matter. 2. Accelerated charged particles radiate electromagnetic waves. The wavelength of the electromagnetic wave is often correlated with the characteristic size of the system that radiates. Thus, gamma radiation, having wavelength of 10–14 m to 10–15 m, typically originate from an atomic nucleus. X-rays are emitted from heavy atoms. Radio waves are produced by accelerating electrons in a circuit. A transmitting antenna can most efficiently radiate waves having a wavelength of about the same size as the antenna. Visible radiation emitted by atoms is, however, much longer in wavelength than atomic size. 3. Infrared waves, with frequencies lower than those of visible light, vibrate not only the electrons, but entire atoms or molecules of a substance. This vibration increases the internal energy and consequently, the temperature of the substance. This is why infrared waves are often called heat waves. 4. The centre of sensitivity of our eyes coincides with the centre of the wavelength distribution of the sun. It is because humans have evolved with visions most sensitive to the strongest wavelengths from the sun. EXERCISES

8.2 STRESS AND STRAIN forces are applied parallel to the cross-sectional When forces are applied on a body in such a area of the cylinder, as shown in Fig. 8.1(b), manner that the body is still in static equilibrium, there is relative displacement between the it is deformed to a small or large extent depending opposite faces of the cylinder. The restoring force upon the nature of the material of the body and per unit area developed due to the applied the magnitude of the deforming force. The tangential force is known as tangential or deformation may not be noticeable visually in shearing stress. many materials but it is there. When a body is As a result of applied tangential force, there subjected to a deforming force, a restoring force is a relative displacement ∆x between opposite is developed in the body. This restoring force is faces of the cylinder as shown in the Fig. 8.1(b). equal in magnitude but opposite in direction to The strain so produced is known as shearing the applied force. The restoring force per unit area strain and it is defined as the ratio of relative is known as stress. If F is the force applied normal displacement of the faces ∆x to the length of the to the cross–section and A is the area of cross cylinder L. section of the body, ∆x Magnitude of the stress = F/A (8.1) Shearing strain = = tan θ (8.3) L The SI unit of stress is N m–2 or pascal (Pa) and its dimensional formula is [ ML–1T–2 ]. where θ is the angular displacement of the There are three ways in which a solid may cylinder from the vertical (original position of the change its dimensions when an external force cylinder). Usually θ is very small, tan θ acts on it. These are shown in Fig. 8.1. In is nearly equal to angle θ, (if θ = 10°, for Fig.8.1(a), a cylinder is stretched by two equal example, there is only 1% difference between θ forces applied normal to its cross-sectional area. and tan θ). The restoring force per unit area in this case is It can also be visualised, when a book is called tensile stress. If the cylinder is pressed with the hand and pushed horizontally, compressed under the action of applied forces, as shown in Fig. 8.2 (c). the restoring force per unit area is known as Thus, shearing strain = tan θ ≈ θ (8.4) compressive stress. Tensile or compressive In Fig. 8.1 (d), a solid sphere placed in the fluid stress can also be termed as longitudinal stress. under high pressure is compressed uniformly on In both the cases, there is a change in the all sides. The force applied by the fluid acts in length of the cylinder. The change in the length perpendicular direction at each point of the ∆L to the original length L of the body (cylinder surface and the body is said to be under in this case) is known as longitudinal strain. hydraulic compression. This leads to decrease (a) (b) (c) (d) Fig. 8.1 (a) A cylindrical body under tensile stress elongates by ∆L (b) Shearing stress on a cylinder deforming it by an angle θ(c) A body subjected to shearing stress (d) A solid body under a stress normal to the surface at every point (hydraulic stress). The volumetric strain is ∆V/V, but there is no change in shape. Reprint 2025-26 MECHANICAL PROPERTIES OF SOLIDS 169 in its volume without any change of its compression and shear stress may also be geometrical shape. obtained. The stress-strain curves vary from The body develops internal restoring forces material to material. These curves help us to that are equal and opposite to the forces applied understand how a given material deforms with by the fluid (the body restores its original shape increasing loads. From the graph, we can see and size when taken out from the fluid). The that in the region between O to A, the curve is internal restoring force per unit area in this case linear. In this region, Hooke’s law is obeyed. is known as hydraulic stress and in magnitude The body regains its original dimensions when is equal to the hydraulic pressure (applied force the applied force is removed. In this region, the per unit area). solid behaves as an elastic body. The strain produced by a hydraulic pressure is called volume strain and is defined as the ratio of change in volume (∆V) to the original volume (V). ∆V Volume strain = (8.5) V Since the strain is a ratio of change in dimension to the original dimension, it has no units or dimensional formula.

8.3 HOOKE’S LAW Stress and strain take different forms in the situations depicted in the Fig. (8.1). For small deformations the stress and strain are proportional to each other. This is known as Fig. 8.2 A typical stress-strain curve for a metal. Hooke’s law. Thus, In the region from A to B, stress and strain stress ∝ strain are not proportional. Nevertheless, the body still stress = k × strain (8.6) returns to its original dimension when the load where k is the proportionality constant and is is removed. The point B in the curve is known known as modulus of elasticity. as yield point (also known as elastic limit) and Hooke’s law is an empirical law and is found the corresponding stress is known as yield to be valid for most materials. However, there strength (σy) of the material. are some materials which do not exhibit this If the load is increased further, the stress linear relationship. developed exceeds the yield strength and strain increases rapidly even for a small change in the 8.4 STRESS-STRAIN CURVE stress. The portion of the curve between B and D shows this. When the load is removed, say at The relation between the stress and the strain some point C between B and D, the body does for a given material under tensile stress can be not regain its original dimension. In this case, found experimentally. In a standard test of even when the stress is zero, the strain is not tensile properties, a test cylinder or a wire is zero. The material is said to have a permanent stretched by an applied force. The fractional set. The deformation is said to be plastic change in length (the strain) and the applied deformation. The point D on the graph is the force needed to cause the strain are recorded. ultimate tensile strength (σu) of the material. The applied force is gradually increased in steps Beyond this point, additional strain is produced and the change in length is noted. A graph is even by a reduced applied force and fracture plotted between the stress (which is equal in occurs at point E. If the ultimate strength and magnitude to the applied force per unit area) and fracture points D and E are close, the material the strain produced. A typical graph for a metal is said to be brittle. If they are far apart, the is shown in Fig. 8.2. Analogous graphs for material is said to be ductile. Reprint 2025-26 170 PHYSICS 8.5 ELASTIC MODULI The proportional region within the elastic limit of the stress-strain curve (region OA in Fig. 8.2) is of great importance for structural and manufacturing engineering designs. The ratio of stress and strain, called modulus of elasticity, is found to be a characteristic of the material. 8.5.1 Young’s Modulus Experimental observation show that for a given material, the magnitude of the strain produced is same whether the stress is tensile or compressive. The ratio of tensile (or compressive) stress (σ) to the longitudinal strain (ε) is defined as Fig. 8.3 Stress-strain curve for the elastic tissue of Young’s modulus and is denoted by the symbol Y. Aorta, the large tube (vessel) carrying blood from the heart. σ Y = (8.7) As stated earlier, the stress-strain behaviour ε varies from material to material. For example, From Eqs. (8.1) and (8.2), we have rubber can be pulled to several times its original length and still returns to its original shape. Fig. Y = (F/A)/(∆L/L) 8.3 shows stress-strain curve for the elastic = (F × L) /(A × ∆L) (8.8) tissue of aorta, present in the heart. Note that Since strain is a dimensionless quantity, the although elastic region is very large, the material unit of Young’s modulus is the same as that of does not obey Hooke’s law over most of the region. stress i.e., N m–2 or Pascal (Pa). Table 8.1 gives Secondly, there is no well defined plastic region. the values of Young’s moduli and yield strengths Substances like tissue of aorta, rubber etc. of some material. which can be stretched to cause large strains From the data given in Table 8.1, it is noticed are called elastomers. that for metals Young’s moduli are large. Table 8.1 Young’s moduli and yield strenghs of some material # Substance tested under compression Reprint 2025-26 MECHANICAL PROPERTIES OF SOLIDS 171 Therefore, these materials require a large force Answer The copper and steel wires are under to produce small change in length. To increase a tensile stress because they have the same the length of a thin steel wire of 0.1 cm2 cross- tension (equal to the load W) and the same area sectional area by 0.1%, a force of 2000 N is of cross-section A. From Eq. (8.7) we have stress required. The force required to produce the same = strain × Young’s modulus. Therefore strain in aluminium, brass and copper wires W/A = Yc × (∆Lc/Lc) = Ys × (∆Ls/Ls) having the same cross-sectional area are 690 N, where the subscripts c and s refer to copper 900 N and 1100 N respectively. It means that and stainless steel respectively. Or, steel is more elastic than copper, brass and ∆Lc/∆Ls = (Ys/Yc) × (Lc/Ls) aluminium. It is for this reason that steel is Given Lc = 2.2 m, Ls = 1.6 m, preferred in heavy-duty machines and in From Table 9.1 Yc = 1.1 × 1011 N.m–2, and structural designs. Wood, bone, concrete and Ys = 2.0 × 1011 N.m–2. glass have rather small Young’s moduli. ∆Lc/∆Ls = (2.0 × 1011/1.1 × 1011) × (2.2/1.6) = 2.5. The total elongation is given to be u Example 8.1 A structural steel rod has a ∆Lc + ∆Ls = 7.0 × 10-4 m radius of 10 mm and a length of 1.0 m. A Solving the above equations, 100 kN force stretches it along its length. ∆Lc = 5.0 × 10-4 m, and ∆Ls = 2.0 × 10-4 m. Calculate (a) stress, (b) elongation, and (c) Therefore strain on the rod. Young’s modulus, of W = (A × Yc × ∆Lc)/Lc structural steel is 2.0 × 1011 N m-2. = π (1.5 × 10-3)2 × [(5.0 × 10-4 × 1.1 × 1011)/2.2] = 1.8 × 102 N ⊳ Answer We assume that the rod is held by a clamp at one end, and the force F is applied at uExample 8.3 In a human pyramid in a the other end, parallel to the length of the rod. circus, the entire weight of the balanced Then the stress on the rod is given by group is supported by the legs of a performer F F who is lying on his back (as shown in Fig. Stress = = 2 8.4). The combined mass of all the persons A πr 3 performing the act, and the tables, plaques 100 × 10 N = etc. involved is 280 kg. The mass of the −2 2 3.14 × 10 m performer lying on his back at the bottom of ( ) = 3.18 × 108 N m–2 the pyramid is 60 kg. Each thighbone (femur) of this performer has a length of 50 cm andThe elongation, an effective radius of 2.0 cm. Determine the ( F/A ) L ∆L = amount by which each thighbone gets Y compressed under the extra load. 8 –2 3.18 × 10 N m 1m ) ( )( = 11 –2 2 × 10 N m = 1.59 × 10–3 m = 1.59 mm The strain is given by Strain = ∆L/L = (1.59 × 10–3 m)/(1m) = 1.59 × 10–3 = 0.16 % ⊳ u Example 8.2 A copper wire of length 2.2 m and a steel wire of length 1.6 m, both of diameter 3.0 mm, are connected end to end. When stretched by a load, the net elongation is found to be 0.70 mm. Obtain the load applied. Fig. 8.4 Human pyramid in a circus. Reprint 2025-26 172 PHYSICS Answer Total mass of all the performers, tables, Table 8.2 Shear moduli (G) of some common materialsplaques etc. = 280 kg Mass of the performer = 60 kg Material G (109 Nm–2 Mass supported by the legs of the performer or GPa) at the bottom of the pyramid Aluminium 25 = 280 – 60 = 220 kg Brass 36 Weight of this supported mass Copper 42 = 220 kg wt. = 220 × 9.8 N = 2156 N. Glass 23 Weight supported by each thighbone of the Iron 70 performer = ½ (2156) N = 1078 N. Lead 5.6 From Table 9.1, the Young’s modulus for bone Nickel 77 is given by Steel 84 Y = 9.4 × 109 N m–2. Tungsten 150 Wood 10 Length of each thighbone L = 0.5 m the radius of thighbone = 2.0 cm u Example 8.4 A square lead slab of side 50 Thus the cross-sectional area of the thighbone A = π × (2 × 10-2)2 m2 = 1.26 × 10-3 m2. cm and thickness 10 cm is subject to a shearing force (on its narrow face) of 9.0 × Using Eq. (9.8), the compression in each 104 N. The lower edge is riveted to the floor. thighbone (∆L) can be computed as How much will the upper edge be displaced? ∆L = [(F × L)/(Y × A)] = [(1078 × 0.5)/(9.4 × 109 × 1.26 × 10-3)] Answer The lead slab is fixed and the force is = 4.55 × 10-5 m or 4.55 × 10-3 cm. applied parallel to the narrow face as shown in This is a very small change! The fractional Fig. 8.6. The area of the face parallel to which decrease in the thighbone is ∆L/L = 0.000091 or this force is applied is 0.0091%. ⊳ A = 50 cm × 10 cm = 0.5 m × 0.1 m8.5.2 Shear Modulus = 0.05 m2 The ratio of shearing stress to the corresponding Therefore, the stress applied is shearing strain is called the shear modulus of = (9.4 × 104 N/0.05 m2) the material and is represented by G. It is also = 1.80 × 106 N.m–2 called the modulus of rigidity. G = shearing stress (σs)/shearing strain G = (F/A)/(∆x/L) = (F × L)/(A × ∆x) (8.10) Similarly, from Eq. (9.4) G = (F/A)/θ = F/(A × θ) (8.11) The shearing stress σs can also be expressed as σs = G × θ (8.12) aaaaaaaaaaaaaaaaaaaaa aaaaaaaaaaaaaaaaaaaaa SI unit of shear modulus is N m–2 or Pa. The Fig. 8.5 shear moduli of a few common materials are given in Table 9.2. It can be seen that shear We know that shearing strain = (∆x/L)= Stress /G. modulus (or modulus of rigidity) is generally less Therefore the displacement ∆x = (Stress × L)/G than Young’s modulus (from Table 9.1). For most = (1.8 × 106 N m–2 × 0.5m)/(5.6 × 109 N m–2) materials G ≈ Y/3. = 1.6 × 10–4 m = 0.16 mm ⊳ Reprint 2025-26 MECHANICAL PROPERTIES OF SOLIDS 173 8.5.3 Bulk Modulus Table 8.3 Bulk moduli (B) of some common Materials In Section (8.3), we have seen that when a body is submerged in a fluid, it undergoes a hydraulic Material B (109 N m–2 or GPa) stress (equal in magnitude to the hydraulic Solids pressure). This leads to the decrease in the Aluminium 72volume of the body thus producing a strain called volume strain [Eq. (8.5)]. The ratio of hydraulic Brass 61 stress to the corresponding hydraulic strain is Copper 140called bulk modulus. It is denoted by symbol B. B = – p/(∆V/V) (8.12) Glass 37 The negative sign indicates the fact that with Iron 100 an increase in pressure, a decrease in volume occurs. That is, if p is positive, ∆V is negative. Nickel 260 Thus for a system in equilibrium, the value of Steel 160 bulk modulus B is always positive. SI unit of bulk modulus is the same as that of pressure Liquids i.e., N m–2 or Pa. The bulk moduli of a few common Water 2.2 materials are given in Table 8.3. The reciprocal of the bulk modulus is called Ethanol 0.9 compressibility and is denoted by k. It is defined Carbon disulphide 1.56 as the fractional change in volume per unit increase in pressure. Glycerine 4.76 k = (1/B) = – (1/∆p) × (∆V/V) (8.13) Mercury 25 It can be seen from the data given in Table Gases8.3 that the bulk moduli for solids are much larger than for liquids, which are again much Air (at STP) 1.0 × 10–4 larger than the bulk modulus for gases (air). Table 8.4 Stress, strain and various elastic moduli Type of Stress Strain Change in Elastic Name of State of stress shape volume Modulus Modulus Matter Tensile Two equal and Elongation or Yes No Y = (F×L)/ Young’s Solid or opposite forces compression (A×∆L) modulus compressive perpendicular to parallel to force (σ = F/A) opposite faces direction (∆L/L) (longitudinal strain) Shearing Two equal and Pure shear, θ Yes No G = F/(A×θ) Shear Solid (σs = F/A) opposite forces modulus parallel to oppoiste or modulus surfaces forces of rigidity in each case such that total force and total torque on the body vanishes Hydraulic Forces perpendicular Volume change No Yes B = –p/(∆V/V) Bulk Solid, liquid everywhere to the (compression or modulus and gas surface, force per elongation) unit area (pressure) (∆V/V) same everywhere. Reprint 2025-26 174 PHYSICS Thus, solids are the least compressible, whereas, 8.5.5 Elastic Potential Energy gases are the most compressible. Gases are about in a Stretched Wire a million times more compressible than solids! When a wire is put under a tensile stress, work Gases have large compressibilities, which vary is done against the inter-atomic forces. This with pressure and temperature. The work is stored in the wire in the form of elastic incompressibility of the solids is primarily due potential energy. When a wire of original length to the tight coupling between the neighbouring L and area of cross-section A is subjected to a atoms. The molecules in liquids are also bound deforming force F along the length of the wire, with their neighbours but not as strong as in let the length of the wire be elongated by l. Then solids. Molecules in gases are very poorly from Eq. (8.8), we have F = YA × (l/L). Here Y is coupled to their neighbours. the Young’s modulus of the material of the wire. Table 8.4 shows the various types of stress, Now for a further elongation of infinitesimal strain, elastic moduli, and the applicable state small length dl, work done dW is F × dl or YAldl/ of matter at a glance. L. Therefore, the amount of work done (W) in increasing the length of the wire from L to L + l, u Example 8.5 The average depth of Indian that is from l = 0 to l = l is Ocean is about 3000 m. Calculate the l YAl YA l 2 dl = × fractional compression, ∆V/V, of water at W = ∫0 2 L L the bottom of the ocean, given that the bulk 2 modulus of water is 2.2 × 109 N m–2. (Take 1  l  W = × Y ×   × AL g = 10 m s–2) 2  L  1 Answer The pressure exerted by a 3000 m = × Young’s modulus × strain2 × column of water on the bottom layer 2 p = hρ g = 3000 m × 1000 kg m–3 × 10 m s–2 volume of the wire = 3 × 107 kg m–1 s-2 1 × stress × strain × volume of the = 3 × 107 N m–2 = 2 Fractional compression ∆V/V, is wire ∆V/V = stress/B = (3 × 107 N m-2)/(2.2 × 109 N m–2) This work is stored in the wire in the form of = 1.36 × 10-2 or 1.36 % ⊳ elastic potential energy (U). Therefore the elastic potential energy per unit volume of the wire (u) is 1 8.5.4 POISSON’S RATIO u = ×σε (8.14) 2 The strain perpendicular to the applied force is called lateral strain. Simon Poisson pointed out 8.6 APPLICATIONS OF ELASTIC that within the elastic limit, lateral strain is BEHAVIOUR OF MATERIALS directly proportional to the longitudinal strain. The elastic behaviour of materials plays an The ratio of the lateral strain to the longitudinal important role in everyday life. All engineering strain in a stretched wire is called Poisson’s designs require precise knowledge of the elastic ratio. If the original diameter of the wire is d behaviour of materials. For example while and the contraction of the diameter under stress designing a building, the structural design of is ∆d, the lateral strain is ∆d/d. If the original the columns, beams and supports require length of the wire is L and the elongation under knowledge of strength of materials used. Have stress is ∆L, the longitudinal strain is ∆L/L. you ever thought why the beams used in Poisson’s ratio is then (∆d/d)/(∆L/L) or (∆d/∆L) construction of bridges, as supports etc. have × (L/d). Poisson’s ratio is a ratio of two strains; a cross-section of the type I? Why does a heap it is a pure number and has no dimensions or of sand or a hill have a pyramidal shape? units. Its value depends only on the nature of Answers to these questions can be obtained material. For steels the value is between 0.28 and from the study of structural engineering which 0.30, and for aluminium alloys it is about 0.33. is based on concepts developed here. Reprint 2025-26 MECHANICAL PROPERTIES OF SOLIDS 175 Cranes used for lifting and moving heavy loads This relation can be derived using what you from one place to another have a thick metal have already learnt and a little calculus. From rope to which the load is attached. The rope is Eq. (8.16), we see that to reduce the bending pulled up using pulleys and motors. Suppose we for a given load, one should use a material with want to make a crane, which has a lifting a large Young’s modulus Y. For a given material, capacity of 10 tonnes or metric tons (1 metric increasing the depth d rather than the breadth ton = 1000 kg). How thick should the steel rope b is more effective in reducing the bending, since be? We obviously want that the load does not δ is proportional to d -3 and only to b-1(of course deform the rope permanently. Therefore, the the length l of the span should be as small as extension should not exceed the elastic limit. possible). But on increasing the depth, unlessFrom Table 8.1, we find that mild steel has a yield strength (σy) of about 300 × 106 N m–2. Thus, the load is exactly at the right place (difficult to the area of cross-section (A) of the rope should arrange in a bridge with moving traffic), the at least be deep bar may bend as shown in Fig. 8.7(b). This A ≥ W/σy = Mg/σy (8.15) is called buckling. To avoid this, a common = (104 kg × 9.8 m s-2)/(300 × 106 N m-2) compromise is the cross-sectional shape shown = 3.3 × 10-4 m2 in Fig. 8.7(c). This section provides a large load- corresponding to a radius of about 1 cm for a bearing surface and enough depth to prevent rope of circular cross-section. Generally a bending. This shape reduces the weight of the large margin of safety (of about a factor of ten beam without sacrificing the strength and in the load) is provided. Thus a thicker rope of hence reduces the cost. radius about 3 cm is recommended. A single wire of this radius would practically be a rigid rod. So the ropes are always made of a number of thin wires braided together, like in pigtails, for ease in manufacture, flexibility and strength. A bridge has to be designed such that it can withstand the load of the flowing traffic, the force of winds and its own weight. Similarly, in the design of buildings the use of beams and columns is very common. In both the cases, the overcoming of the problem of bending of beam under a load is of prime importance. The beam should not bend too much or break. Let us consider the case of a beam loaded at the centre and supported near its ends as shown in Fig. 8.6. A bar of length l, breadth b, and depth d when loaded at the centre by a load W sags by (a) (b) (c) an amount given by Fig. 8.7 Different cross-sectional shapes of a δ = W l 3/(4bd 3Y) (8.16) beam. (a) Rectangular section of a bar; (b) A thin bar and how it can buckle; (c) Commonly used section for a load bearing bar. The use of pillars or columns is also very common in buildings and bridges. A pillar with rounded ends as shown in Fig. 8.9(a) supports less load than that with a distributed shape at the ends [Fig. 8.9(b)]. The precise design of a bridge or a building has to take into account the conditions under which it will function, the Fig. 8.6 A beam supported at the ends and loaded cost and long period, reliability of usable at the centre. material, etc. Reprint 2025-26 176 PHYSICS shearing stress to the rocks under which they can flow. The stress due to all the material on the top should be less than the critical shearing stress at which the rocks flow. At the bottom of a mountain of height h, the force per unit area due to the weight of the mountain is hρg where ρ is the density of the material of the mountain and g is the acceleration due to gravity. The material at the bottom experiences this force in the vertical direction, and the sides of the mountain are free. Therefore, this is not a case of pressure or bulk compression. (a) (b) There is a shear component, approximately hρg Fig. 8.8 Pillars or columns: (a) a pillar with rounded itself. Now the elastic limit for a typical rock is ends, (b) Pillar with distributed ends. 30 × 107 N m-2. Equating this to hρg, with The answer to the question why the maximum ρ = 3 × 103 kg m-3 gives height of a mountain on earth is ~10 km can hρg = 30 × 107 N m-2 . also be provided by considering the elastic h = 30 × 107 N m-2/(3 × 103 kg m-3 × 10 m s-2) properties of rocks. A mountain base is not under = 10 km uniform compression and this provides some which is more than the height of Mt. Everest! SUMMARY 1. Stress is the restoring force per unit area and strain is the fractional change in dimension. In general there are three types of stresses (a) tensile stress — longitudinal stress (associated with stretching) or compressive stress (associated with compression), (b) shearing stress, and (c) hydraulic stress. 2. For small deformations, stress is directly proportional to the strain for many materials. This is known as Hooke’s law. The constant of proportionality is called modulus of elasticity. Three elastic moduli viz., Young’s modulus, shear modulus and bulk modulus are used to describe the elastic behaviour of objects as they respond to deforming forces that act on them. A class of solids called elastomers does not obey Hooke’s law. 3. When an object is under tension or compression, the Hooke’s law takes the form F/A = Y∆L/L where ∆L/L is the tensile or compressive strain of the object, F is the magnitude of the applied force causing the strain, A is the cross-sectional area over which F is applied (perpendicular to A) and Y is the Young’s modulus for the object. The stress is F/A. 4. A pair of forces when applied parallel to the upper and lower faces, the solid deforms so that the upper face moves sideways with respect to the lower. The horizontal displacement ∆L of the upper face is perpendicular to the vertical height L. This type of deformation is called shear and the corresponding stress is the shearing stress. This type of stress is possible only in solids. In this kind of deformation the Hooke’s law takes the form F/A = G × ∆L/L where ∆L is the displacement of one end of object in the direction of the applied force F, and G is the shear modulus. 5. When an object undergoes hydraulic compression due to a stress exerted by a surrounding fluid, the Hooke’s law takes the form p = B (∆V/V), where p is the pressure (hydraulic stress) on the object due to the fluid, ∆V/V (the volume strain) is the absolute fractional change in the object’s volume due to that pressure and B is the bulk modulus of the object. Reprint 2025-26 MECHANICAL PROPERTIES OF SOLIDS 177 POINTS TO PONDER 1. In the case of a wire, suspended from celing and stretched under the action of a weight (F) suspended from its other end, the force exerted by the ceiling on it is equal and opposite to the weight. However, the tension at any cross-section A of the wire is just F and not 2F. Hence, tensile stress which is equal to the tension per unit area is equal to F/A. 2. Hooke’s law is valid only in the linear part of stress-strain curve. 3. The Young’s modulus and shear modulus are relevant only for solids since only solids have lengths and shapes. 4. Bulk modulus is relevant for solids, liquid and gases. It refers to the change in volume when every part of the body is under the uniform stress so that the shape of the body remains unchanged. 5. Metals have larger values of Young’s modulus than alloys and elastomers. A material with large value of Young’s modulus requires a large force to produce small changes in its length. 6. In daily life, we feel that a material which stretches more is more elastic, but it a is misnomer. In fact material which stretches to a lesser extent for a given load is considered to be more elastic. 7. In general, a deforming force in one direction can produce strains in other directions also. The proportionality between stress and strain in such situations cannot be described by just one elastic constant. For example, for a wire under longitudinal strain, the lateral dimensions (radius of cross section) will undergo a small change, which is described by another elastic constant of the material (called Poisson ratio). 8. Stress is not a vector quantity since, unlike a force, the stress cannot be assigned a specific direction. Force acting on the portion of a body on a specified side of a section has a definite direction. EXERCISES 8.1 A steel wire of length 4.7 m and cross-sectional area 3.0 × 10-5 m2 stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 × 10–5 m2 under a given load. What is the ratio of the Young’s modulus of steel to that of copper? 8.2 Figure 8.9 shows the strain-stress curve for a given material. What are (a) Young’s modulus and (b) approximate yield strength for this material? Fig. 8.9 Reprint 2025-26 178 PHYSICS 8.3 The stress-strain graphs for materials A and B are shown in Fig. 8.10. Fig. 8.10 The graphs are drawn to the same scale. (a) Which of the materials has the greater Young’s modulus? (b) Which of the two is the stronger material? 8.4 Read the following two statements below carefully and state, with reasons, if it is true or false. (a) The Young’s modulus of rubber is greater than that of steel; (b) The stretching of a coil is determined by its shear modulus. 8.5 Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in Fig. 8.11. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m. Compute the elongations of the steel and the brass wires. Fig. 8.11 8.6 The edge of an aluminium cube is 10 cm long. One face of the cube is firmly fixed to a vertical wall. A mass of 100 kg is then attached to the opposite face of the cube. The shear modulus of aluminium is 25 GPa. What is the vertical deflection of this face? 8.7 Four identical hollow cylindrical columns of mild steel support a big structure of mass 50,000 kg. The inner and outer radii of each column are 30 and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column. 8.8 A piece of copper having a rectangular cross-section of 15.2 mm × 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain? 8.9 A steel cable with a radius of 1.5 cm supports a chairlift at a ski area. If the maximum stress is not to exceed 108 N m–2, what is the maximum load the cable can support ? 8.10 A rigid bar of mass 15 kg is supported symmetrically by three wires each 2.0 m long. Those at each end are of copper and the middle one is of iron. Determine the ratios of their diameters if each is to have the same tension. 8.11 A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev/s at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm2. Calculate the elongation of the wire when the mass is at the lowest point of its path. Reprint 2025-26 MECHANICAL PROPERTIES OF SOLIDS 179 8.12 Compute the bulk modulus of water from the following data: Initial volume = 100.0 litre, Pressure increase = 100.0 atm (1 atm = 1.013 × 105 Pa), Final volume = 100.5 litre. Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large. 8.13 What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is 1.03 × 103 kg m–3? 8.14 Compute the fractional change in volume of a glass slab, when subjected to a hydraulic pressure of 10 atm. 8.15 Determine the volume contraction of a solid copper cube, 10 cm on an edge, when subjected to a hydraulic pressure of 7.0 × 106 Pa. 8.16 How much should the pressure on a litre of water be changed to compress it by 0.10%? carry one quarter of the load. Reprint 2025-26 CHAPTER NINE MECHANICAL PROPERTIES OF FLUIDS 9.1 INTRODUCTION In this chapter, we shall study some common physical properties of liquids and gases. Liquids and gases can flow 9.1 Introduction and are therefore, called fluids. It is this property that 9.2 Pressure distinguishes liquids and gases from solids in a basic way. 9.3 Streamline flow Fluids are everywhere around us. Earth has an envelop of 9.4 Bernoulli’s principle air and two-thirds of its surface is covered with water. Water is not only necessary for our existence; every mammalian9.5 Viscosity body constitute mostly of water. All the processes occurring9.6 Surface tension in living beings including plants are mediated by fluids. Thus Summary understanding the behaviour and properties of fluids is Points to ponder important. Exercises How are fluids different from solids? What is common in Additional exercises liquids and gases? Unlike a solid, a fluid has no definite Appendix shape of its own. Solids and liquids have a fixed volume, whereas a gas fills the entire volume of its container. We have learnt in the previous chapter that the volume of solids can be changed by stress. The volume of solid, liquid or gas depends on the stress or pressure acting on it. When we talk about fixed volume of solid or liquid, we mean its volume under atmospheric pressure. The difference between gases and solids or liquids is that for solids or liquids the change in volume due to change of external pressure is rather small. In other words solids and liquids have much lower compressibility as compared to gases. Shear stress can change the shape of a solid keeping its volume fixed. The key property of fluids is that they offer very little resistance to shear stress; their shape changes by application of very small shear stress. The shearing stress of fluids is about million times smaller than that of solids. 9.2 PRESSURE A sharp needle when pressed against our skin pierces it. Our skin, however, remains intact when a blunt object with a wider contact area (say the back of a spoon) is pressed against it with the same force. If an elephant were to step on a man’s chest, his ribs would crack. A circus performer across whose Reprint 2025-26 MECHANICAL PROPERTIES OF FLUIDS 181 chest a large, light but strong wooden plank is In principle, the piston area can be made placed first, is saved from this accident. Such arbitrarily small. The pressure is then defined everyday experiences convince us that both the in a limiting sense as force and its coverage area are important. Smaller lim ∆F the area on which the force acts, greater is the P = ∆A → 0 (9.2) ∆Aimpact. This impact is known as pressure. Pressure is a scalar quantity. We remind the When an object is submerged in a fluid at reader that it is the component of the force rest, the fluid exerts a force on its surface. This normal to the area under consideration and not force is always normal to the object’s surface. the (vector) force that appears in the numerator This is so because if there were a component of in Eqs. (9.1) and (9.2). Its dimensions are force parallel to the surface, the object will also [ML–1T–2]. The SI unit of pressure is N m–2. It has exert a force on the fluid parallel to it; as a been named as pascal (Pa) in honour of the consequence of Newton’s third law. This force French scientist Blaise Pascal (1623-1662) who will cause the fluid to flow parallel to the surface. carried out pioneering studies on fluid pressure. Since the fluid is at rest, this cannot happen. A common unit of pressure is the atmosphere Hence, the force exerted by the fluid at rest has (atm), i.e. the pressure exerted by the to be perpendicular to the surface in contact atmosphere at sea level (1 atm = 1.013 × 105 Pa). with it. This is shown in Fig.9.1(a). Another quantity, that is indispensable in The normal force exerted by the fluid at a point describing fluids, is the density ρ. For a fluid of may be measured. An idealised form of one such mass m occupying volume V, pressure-measuring device is shown in Fig. m ρ = (9.3) 9.1(b). It consists of an evacuated chamber with V a spring that is calibrated to measure the force The dimensions of density are [ML–3]. Its SI acting on the piston. This device is placed at a unit is kg m–3. It is a positive scalar quantity. A point inside the fluid. The inward force exerted liquid is largely incompressible and its density by the fluid on the piston is balanced by the is therefore, nearly constant at all pressures. outward spring force and is thereby measured. Gases, on the other hand exhibit a large variation in densities with pressure. The density of water at 4oC (277 K) is 1.0 × 103 kg m–3. The relative density of a substance is the ratio of its density to the density of water at 4oC. It is a dimensionless positive scalar quantity. For example the relative density of aluminium is 2.7. Its density is 2.7 × 103 kg m–3. The densities of some common fluids are displayed in Table 9.1. Table 9.1 Densities of some common fluids (a) (b) at STP* Fig. 9.1 (a) The force exerted by the liquid in the beaker on the submerged object or on the walls is normal (perpendicular) to the surface at all points. (b) An idealised device for measuring pressure. If F is the magnitude of this normal force on the piston of area A then the average pressure Pav is defined as the normal force acting per unit area. F Pav = (9.1) A * STP means standard temperature (00C) and 1 atm pressure. Reprint 2025-26 182 PHYSICS ⊳ this element of area corresponding to the normal Example 9.1 The two thigh bones (femurs), forces Fa, Fb and Fc as shown in Fig. 9.2 on the each of cross-sectional area10 cm2 support faces BEFC, ADFC and ADEB denoted by Aa, Ab the upper part of a human body of mass 40 and Ac respectively. Then kg. Estimate the average pressure Fb sinθ = Fc, Fb cosθ = Fa (by equilibrium) sustained by the femurs. Ab sinθ = Ac, Ab cosθ = Aa (by geometry) Thus, Answer Total cross-sectional area of the Fb Fc Fafemurs is A = 2 × 10 cm2 = 20 × 10–4 m2. The = = ; Pb = Pc = Pa (9.4) force acting on them is F = 40 kg wt = 400 N Ab A c A a (taking g = 10 m s–2). This force is acting Hence, pressure exerted is same in all vertically down and hence, normally on the directions in a fluid at rest. It again reminds us femurs. Thus, the average pressure is that like other types of stress, pressure is not a F 5 −2 vector quantity. No direction can be assigned Pav = = 2 × 10 N m ⊳ A to it. The force against any area within (or bounding) a fluid at rest and under pressure is 9.2.1 Pascal’s Law normal to the area, regardless of the orientation of the area. The French scientist Blaise Pascal observed that Now consider a fluid element in the form of a the pressure in a fluid at rest is the same at all horizontal bar of uniform cross-section. The bar points if they are at the same height. This fact is in equilibrium. The horizontal forces exerted may be demonstrated in a simple way. at its two ends must be balanced or the pressure at the two ends should be equal. This proves that for a liquid in equilibrium the pressure is same at all points in a horizontal plane. Suppose the pressure were not equal in different parts of the fluid, then there would be a flow as the fluid will have some net force acting on it. Hence in the absence of flow the pressure in the fluid must be same everywhere in a horizontal plane. 9.2.2 Variation of Pressure with Depth Fig. 9.2 Proof of Pascal’s law. ABC-DEF is an Consider a fluid at rest in a container. In element of the interior of a fluid at rest. Fig. 9.3 point 1 is at height h above a point 2. This element is in the form of a right- The pressures at points 1 and 2 are P1 and P2 angled prism. The element is small so that respectively. Consider a cylindrical element of the effect of gravity can be ignored, but it fluid having area of base A and height h. As the has been enlarged for the sake of clarity. fluid is at rest the resultant horizontal forces Fig. 9.2 shows an element in the interior of a should be zero and the resultant vertical forces fluid at rest. This element ABC-DEF is in the should balance the weight of the element. The form of a right-angled prism. In principle, this forces acting in the vertical direction are due to prismatic element is very small so that every the fluid pressure at the top (P1A) acting part of it can be considered at the same depth downward, at the bottom (P2A) acting upward. from the liquid surface and therefore, the effect If mg is weight of the fluid in the cylinder we of the gravity is the same at all these points. have But for clarity we have enlarged this element. (P2 − P1) A = mg (9.5) The forces on this element are those exerted by Now, if ρ is the mass density of the fluid, we the rest of the fluid and they must be normal to have the mass of fluid to be m = ρV= ρhA so the surfaces of the element as discussed above. that Thus, the fluid exerts pressures Pa, Pb and Pc on P2 − P1= ρgh (9.6) Reprint 2025-26 MECHANICAL PROPERTIES OF FLUIDS 183 Fig 9.4 Illustration of hydrostatic paradox. The three vessels A, B and C contain different amounts of liquids, all upto the same height. ⊳ Example 9.2 What is the pressure on a swimmer 10 m below the surface of a lake? Answer HereFig.9.3 Fluid under gravity. The effect of gravity is illustrated through pressure on a vertical h = 10 m and ρ = 1000 kg m-3. Take g = 10 m s–2 cylindrical column. From Eq. (9.7) P = Pa + ρgh Pressure difference depends on the vertical = 1.01 × 105 Pa + 1000 kg m–3 × 10 m s–2 × 10 m = 2.01 × 105 Padistance h between the points (1 and 2), mass ≈ 2 atmdensity of the fluid ρ and acceleration due to This is a 100% increase in pressure from gravity g. If the point 1 under discussion is surface level. At a depth of 1 km, the increase shifted to the top of the fluid (say, water), which in pressure is 100 atm! Submarines are designed is open to the atmosphere, P1 may be replaced to withstand such enormous pressures. ⊳ by atmospheric pressure (Pa) and we replace P2 by P. Then Eq. (9.6) gives 9.2.3 Atmospheric Pressure and Gauge Pressure P = Pa + ρgh (9.7) Thus, the pressure P, at depth below the The pressure of the atmosphere at any point is equal to the weight of a column of air of unitsurface of a liquid open to the atmosphere is cross-sectional area extending from that pointgreater than atmospheric pressure by an to the top of the atmosphere. At sea level, it is amount ρgh. The excess of pressure, P − Pa, at 1.013 × 105 Pa (1 atm). Italian scientist depth h is called a gauge pressure at that point. Evangelista Torricelli (1608–1647) devised for The area of the cylinder is not appearing in the first time a method for measuring the expression of absolute pressure in Eq. (9.7). atmospheric pressure. A long glass tube closed Thus, the height of the fluid column is important at one end and filled with mercury is inverted and not cross-sectional or base area or the shape into a trough of mercury as shown in Fig.9.5 (a). of the container. The liquid pressure is the same This device is known as ‘mercury barometer’. at all points at the same horizontal level (same The space above the mercury column in the tube depth). The result is appreciated through the contains only mercury vapour whose pressure example of hydrostatic paradox. Consider three P is so small that it may be neglected. Thus, vessels A, B and C [Fig.9.4] of different shapes. the pressure at Point A=0. The pressure inside They are connected at the bottom by a horizontal the coloumn at Point B must be the same as the pipe. On filling with water, the level in the three pressure at Point C, which is atmospheric vessels is the same, though they hold different pressure, Pa. amounts of water. This is so because water at Pa = ρgh (9.8) where ρ is the density of mercury and h is thethe bottom has the same pressure below each height of the mercury column in the tube.section of the vessel. Reprint 2025-26 184 PHYSICS In the experiment it is found that the mercury column in the barometer has a height of about 76 cm at sea level equivalent to one atmosphere (1 atm). This can also be obtained using the value of ρ in Eq. (9.8). A common way of stating pressure is in terms of cm or mm of mercury (Hg). A pressure equivalent of 1 mm is called a torr (after Torricelli). 1 torr = 133 Pa. The mm of Hg and torr are used in medicine and physiology. In meteorology, a common unit is the bar and millibar. 1 bar = 105 Pa An open tube manometer is a useful (b) The open tube manometer instrument for measuring pressure differences. Fig 9.5 Two pressure measuring devices. It consists of a U-tube containing a suitable Pressure is same at the same level on both liquid i.e., a low density liquid (such as oil) for sides of the U-tube containing a fluid. For measuring small pressure differences and a liquids, the density varies very little over wide high density liquid (such as mercury) for large ranges in pressure and temperature and we can pressure differences. One end of the tube is open treat it safely as a constant for our present to the atmosphere and the other end is purposes. Gases on the other hand, exhibits connected to the system whose pressure we want large variations of densities with changes in to measure [see Fig. 9.5 (b)]. The pressure P at A pressure and temperature. Unlike gases, liquids is equal to pressure at point B. What we are, therefore, largely treated as incompressible. normally measure is the gauge pressure, which ⊳ Example 9.3 The density of the is P − Pa, given by Eq. (9.8) and is proportional to atmosphere at sea level is 1.29 kg/m3. manometer height h. Assume that it does not change with altitude. Then how high would the atmosphere extend? Answer We use Eq. (9.7) ρgh = 1.29 kg m–3 × 9.8 m s2 × h m = 1.01 × 105 Pa ∴ h = 7989 m ≈ 8 km In reality the density of air decreases with height. So does the value of g. The atmospheric cover extends with decreasing pressure over 100 km. We should also note that the sea level atmospheric pressure is not always 760 mm of Hg. A drop in the Hg level by 10 mm or more is a sign of an approaching storm. ⊳ ⊳ Example 9.4 At a depth of 1000 m in an ocean (a) what is the absolute pressure? (b) What is the gauge pressure? (c) Find the force acting on the window of area 20 cm × 20 cm of a submarine at this depth, the interior of which is maintained at sea- level atmospheric pressure. (The density of sea water is 1.03 × 103 kg m-3, g = 10 m s–2.) Fig 9.5 (a) The mercury barometer. Reprint 2025-26 MECHANICAL PROPERTIES OF FLUIDS 185 Answer Here h = 1000 m and ρ = 1.03 × 103 kg m-3. law. In these devices, fluids are used for (a) From Eq. (9.6), absolute pressure transmitting pressure. In a hydraulic lift, as P = Pa + ρgh shown in Fig. 9.6 (b), two pistons are separated = 1.01 × 105 Pa by the space filled with a liquid. A piston of small + 1.03 × 103 kg m–3 × 10 m s–2 × 1000 m cross-section A1 is used to exert a force F1 directly F1 = 104.01 × 105 Pa ≈ 104 atm on the liquid. The pressure P = A1 is (b) Gauge pressure is P − Pa = ρgh = Pg transmitted throughout the liquid to the larger Pg = 1.03 × 103 kg m–3 × 10 ms2 × 1000 m cylinder attached with a larger piston of area A2, = 103 × 105 Pa which results in an upward force of P × A2. ≈ 103 atm Therefore, the piston is capable of supporting a (c) The pressure outside the submarine is large force (large weight of, say a car, or a truck, P = Pa + ρgh and the pressure inside it is Pa. F1 A 2 Hence, the net pressure acting on the placed on the platform) F2 = PA2 = A1 . By window is gauge pressure, Pg = ρgh. Since the area of the window is A = 0.04 m2, the changing the force at A1, the platform can be force acting on it is moved up or down. Thus, the applied force has F = Pg A = 103 × 105 Pa × 0.04 m2 = 4.12 × 105 N A 2 ⊳ been increased by a factor of A1 and this factor is the mechanical advantage of the device. The9.2.4 Hydraulic Machines example below clarifies it. Let us now consider what happens when we change the pressure on a fluid contained in a vessel. Consider a horizontal cylinder with a piston and three vertical tubes at different points [Fig. 9.6 (a)]. The pressure in the horizontal cylinder is indicated by the height of liquid column in the vertical tubes. It is necessarily the same in all. If we push the piston, the fluid level rises in all the tubes, again reaching the same level in each one of them. Fig 9.6 (b) Schematic diagram illustrating the principle behind the hydraulic lift, a device used to lift heavy loads. Fig 9.6 (a) Whenever external pressure is applied ⊳ Example 9.5 Two syringes of different on any part of a fluid in a vessel, it is cross-sections (without needles) filled with equally transmitted in all directions. water are connected with a tightly fitted rubber tube filled with water. Diameters of This indicates that when the pressure on the the smaller piston and larger piston are cylinder was increased, it was distributed 1.0 cm and 3.0 cm respectively. (a) Find uniformly throughout. We can say whenever the force exerted on the larger piston when external pressure is applied on any part of a a force of 10 N is applied to the smaller fluid contained in a vessel, it is transmitted piston. (b) If the smaller piston is pushed undiminished and equally in all directions. in through 6.0 cm, how much does the This is another form of the Pascal’s law and it larger piston move out? has many applications in daily life. A number of devices, such as hydraulic lift Answer (a) Since pressure is transmitted and hydraulic brakes, are based on the Pascal’s undiminished throughout the fluid, Reprint 2025-26 186 PHYSICS –2 2 important advantage of the system is that the 3/2 10 m × π A ( ) set up by pressing pedal is transmitted F2 = 2 F1 = 2 × 10 N pressure A1 equally to all cylinders attached to the four 1/2 10 –2 m × π ( ) wheels so that the braking effort is equal on = 90 N all wheels.(b) Water is considered to be perfectly incompressible. Volume covered by the 9.3 STREAMLINE FLOWmovement of smaller piston inwards is equal to volume moved outwards due to the larger piston. So far we have studied fluids at rest. The study L1 A1 = L 2 A2 of the fluids in motion is known as fluid dynamics. When a water tap is turned on slowly, the water flow is smooth initially, but loses its smoothness when the speed of the outflow is increased. In studying the motion of fluids, we j 0.67 × 10-2 m = 0.67 cm focus our attention on what is happening to Note, atmospheric pressure is common to both various fluid particles at a particular point in pistons and has been ignored. ⊳ space at a particular time. The flow of the fluid ⊳ is said to be steady if at any given point, the Example 9.6 In a car lift compressed air velocity of each passing fluid particle remains exerts a force F1 on a small piston having constant in time. This does not mean that the a radius of 5.0 cm. This pressure is velocity at different points in space is same. The transmitted to a second piston of radius velocity of a particular particle may change as it 15 cm (Fig 9.7). If the mass of the car to be moves from one point to another. That is, at some lifted is 1350 kg, calculate F1. What is the other point the particle may have a different pressure necessary to accomplish this velocity, but every other particle which passes task? (g = 9.8 ms-2). the second point behaves exactly as the previous particle that has just passed that point. Each Answer Since pressure is transmitted particle follows a smooth path, and the paths of undiminished throughout the fluid, the particles do not cross each other. = 1470 N ≈ 1.5 × 103 N The air pressure that will produce this force is This is almost double the atmospheric pressure. ⊳ Fig. 9.7 The meaning of streamlines. (a) A typical Hydraulic brakes in automobiles also work on trajectory of a fluid particle. the same principle. When we apply a little force (b) A region of streamline flow. on the pedal with our foot the master piston moves inside the master cylinder, and the The path taken by a fluid particle under a pressure caused is transmitted through the steady flow is a streamline. It is defined as a brake oil to act on a piston of larger area. A large curve whose tangent at any point is in the force acts on the piston and is pushed down direction of the fluid velocity at that point. expanding the brake shoes against brake lining. Consider the path of a particle as shown in In this way, a small force on the pedal produces Fig.9.7 (a), the curve describes how a fluid a large retarding force on the wheel. An particle moves with time. The curve PQ is like a Reprint 2025-26 MECHANICAL PROPERTIES OF FLUIDS 187 permanent map of fluid flow, indicating how the but their directions are parallel. Figure 9.8 (b) fluid streams. No two streamlines can cross, for gives a sketch of turbulent flow. if they do, an oncoming fluid particle can go either one way or the other and the flow would not be steady. Hence, in steady flow, the map of flow is stationary in time. How do we draw closely spaced streamlines ? If we intend to show streamline of every flowing particle, we would end up with a continuum of lines. Consider planes perpendicular to the direction of fluid flow e.g., at three points P, R and Q in Fig.9.7 (b). The plane Fig. 9.8 (a) Some streamlines for fluid flow. pieces are so chosen that their boundaries be (b) A jet of air striking a flat plate placed determined by the same set of streamlines. This perpendicular to it. This is an example means that number of fluid particles crossing of turbulent flow. the surfaces as indicated at P, R and Q is the same. If area of cross-sections at these points 9.4 BERNOULLI’S PRINCIPLE are AP,AR and AQ and speeds of fluid particles are vP, vR and vQ, then mass of fluid ∆mP crossing Fluid flow is a complex phenomenon. But we at AP in a small interval of time ∆t is ρPAPvP ∆t. can obtain some useful properties for steady Similarly mass of fluid ∆mR flowing or crossing or streamline flows using the conservation at AR in a small interval of time ∆t is ρRARvR ∆t of energy. and mass of fluid ∆mQ is ρQAQvQ ∆t crossing at Consider a fluid moving in a pipe of varying AQ. The mass of liquid flowing out equals the cross-sectional area. Let the pipe be at varying mass flowing in, holds in all cases. Therefore, heights as shown in Fig. 9.9. We now suppose ρPAPvP∆t = ρRARvR∆t = ρQAQvQ∆t (9.9) that an incompressible fluid is flowing through For flow of incompressible fluids the pipe in a steady flow. Its velocity must ρP = ρR = ρQ change as a consequence of equation of Equation (9.9) reduces to continuity. A force is required to produce this APvP = ARvR = AQvQ (9.10) acceleration, which is caused by the fluid which is called the equation of continuity and surrounding it, the pressure must be different it is a statement of conservation of mass in flow in different regions. Bernoulli’s equation is a of incompressible fluids. In general general expression that relates the pressure Av = constant (9.11) difference between two points in a pipe to both Av gives the volume flux or flow rate and velocity changes (kinetic energy change) and remains constant throughout the pipe of flow. elevation (height) changes (potential energy Thus, at narrower portions where the change). The Swiss Physicist Daniel Bernoulli streamlines are closely spaced, velocity developed this relationship in 1738. increases and its vice versa. From (Fig 9.7b) it Consider the flow at two regions 1 (i.e., BC) is clear that AR > AQ or vR < vQ, the fluid is and 2 (i.e., DE). Consider the fluid initially lying accelerated while passing from R to Q. This is between B and D. In an infinitesimal time associated with a change in pressure in fluid interval ∆t, this fluid would have moved. Suppose flow in horizontal pipes. v1 is the speed at B and v2 at D, then fluid initially Steady flow is achieved at low flow speeds. at B has moved a distance v1∆t to C (v1∆t is small Beyond a limiting value, called critical speed, enough to assume constant cross-section along this flow loses steadiness and becomes BC). In the same interval ∆t the fluid initially at turbulent. One sees this when a fast flowing D moves to E, a distance equal to v2∆t. Pressures stream encounters rocks, small foamy P1 and P2 act as shown on the plane faces of whirlpool-like regions called ‘white water areas A1 and A2 binding the two regions. The rapids are formed. work done on the fluid at left end (BC) is W1 = Figure 9.8 displays streamlines for some P1A1(v1∆t) = P1∆V. Since the same volume ∆V typical flows. For example, Fig. 9.8(a) describes passes through both the regions (from the a laminar flow where the velocities at different equation of continuity) the work done by the fluid points in the fluid may have different magnitudes at the other end (DE) is W2 = P2A2(v2∆t) = P2∆V or, Reprint 2025-26 188 PHYSICS the work done on the fluid is –P2∆V. So the total In words, the Bernoulli’s relation may be work done on the fluid is stated as follows: As we move along a streamline W1 – W2 = (P1− P2) ∆V the sum of the pressure (P), the kinetic energy Part of this work goes into changing the kinetic  ρv2 energy of the fluid, and part goes into changing per unit volume and the potential energythe gravitational potential energy. If the density  2  of the fluid is ρ and ∆m = ρA1v1∆t = ρ∆V is the per unit volume (ρgh) remains a constant. mass passing through the pipe in time ∆t, then Note that in applying the energy conservation change in gravitational potential energy is principle, there is an assumption that no energy ∆U = ρg∆V (h2 − h1) is lost due to friction. But in fact, when fluids The change in its kinetic energy is flow, some energy does get lost due to internal 1 friction. This arises due to the fact that in a fluid ∆K = ρ ∆V (v2 2 − v1 2) flow, the different layers of the fluid flow with 2 different velocities. These layers exert frictional We can employ the work – energy theorem forces on each other resulting in a loss of energy. (Chapter 6) to this volume of the fluid and This property of the fluid is called viscosity and this yields is discussed in more detail in a later section. The 1 lost kinetic energy of the fluid gets converted into (P1− P2) ∆V = ρ ∆V (v2 2 − v1 2) + ρg∆V (h2 − h1) heat energy. Thus, Bernoulli’s equation ideally 2 applies to fluids with zero viscosity or non- We now divide each term by ∆V to obtain viscous fluids. Another restriction on application of Bernoulli theorem is that the fluids must be 1 incompressible, as the elastic energy of the fluid (P1− P2) = ρ (v2 2 − v1 2) + ρg (h2 − h1) is also not taken into consideration. In practice, 2 it has a large number of useful applications and We can rearrange the above terms to obtain can help explain a wide variety of phenomena 1 1 2 2 for low viscosity incompressible fluids. P1 + ρv1 + ρgh1 = P2+ ρv2 + ρgh2 2 2 Bernoulli’s equation also does not hold for non- (9.12) steady or turbulent flows, because in that This is Bernoulli’s equation. Since 1 and 2 situation velocity and pressure are constantly refer to any two locations along the pipeline, we fluctuating in time. may write the expression in general as When a fluid is at rest i.e., its velocity is zero everywhere, Bernoulli’s equation becomes 1 P + ρv2 + ρgh = constant (9.13) P1 + ρgh1 = P2 + ρgh2 2 (P1− P2) = ρg (h2 − h1) which is same as Eq. (9.6). 9.4.1 Speed of Efflux: Torricelli’s Law The word efflux means fluid outflow. Torricelli discovered that the speed of efflux from an open tank is given by a formula identical to that of a freely falling body. Consider a tank containing a liquid of density ρ with a small hole in its side at a height y1 from the bottom (see Fig. 9.10). The air above the liquid, whose surface is at height y2, is at pressure P. From the equation of continuity [Eq. (9.10)] we have Fig. 9.9 The flow of an ideal fluid in a pipe of varying v1 A1 = v2 A2 cross section. The fluid in a section of length v1∆t moves to the section of length v2∆t in A1 v 2 = v1 time ∆t. A 2 Reprint 2025-26 MECHANICAL PROPERTIES OF FLUIDS 189 from its parabolic trajectory as it moves through air. This deviation can be partly explained on the basis of Bernoulli’s principle. (i) Ball moving without spin: Fig. 9.11(a) shows the streamlines around a non-spinning ball moving relative to a fluid. From the symmetry of streamlines it is clear that the velocity of fluid (air) above and below the ball at corresponding points is the same resulting in zero pressure difference. The air therefore, exerts no upward or downward force on the ball. Fig. 9.10 Torricelli’s law. The speed of efflux, v1, (ii) Ball moving with spin: A ball which is from the side of the container is given by the application of Bernoulli’s equation. spinning drags air along with it. If the If the container is open at the top to the surface is rough more air will be dragged. atmosphere then v1 = 2 g h . Fig 9.11(b) shows the streamlines of air for a ball which is moving and spinning at the same time. The ball is moving If the cross-sectional area of the tank A2 is forward and relative to it the air is moving much larger than that of the hole (A2 >>A1), then backwards. Therefore, the velocity of air we may take the fluid to be approximately at rest above the ball relative to the ball is larger at the top, i.e., v2 = 0. Now, applying the Bernoulli and below it is smaller (see Section 9.3).equation at points 1 and 2 and noting that at the hole P1 = Pa, the atmospheric pressure, we The stream lines, thus, get crowded above have from Eq. (9.12) and rarified below. This difference in the velocities of air results 1 2 Pa + ρ v1 + ρ g y1 = P + ρ g y 2 in the pressure difference between the lower and 2 upper faces and there is a net upward force on Taking y2 – y1 = h we have the ball. This dynamic lift due to spining is called 2 ( P − Pa ) Magnus effect. v1 = 2 g h + (9.14) ρ Aerofoil or lift on aircraft wing: Figure 9.11 When P >>Pa and 2 g h may be ignored, the (c) shows an aerofoil, which is a solid piece speed of efflux is determined by the container shaped to provide an upward dynamic lift pressure. Such a situation occurs in rocket when it moves horizontally through air. The propulsion. On the other hand, if the tank is cross-section of the wings of an aeroplane open to the atmosphere, then P = Pa and looks somewhat like the aerofoil shown in Fig. v1 = 2g h (9.15) 9.11 (c) with streamlines around it. When the This is also the speed of a freely falling body. aerofoil moves against the wind, the Equation (9.15) represents Torricelli’s law. orientation of the wing relative to flow direction causes the streamlines to crowd together 9.4.2 Dynamic Lift above the wing more than those below it. The Dynamic lift is the force that acts on a body, flow speed on top is higher than that below it. such as airplane wing, a hydrofoil or a spinning There is an upward force resulting in a ball, by virtue of its motion through a fluid. In dynamic lift of the wings and this balances many games such as cricket, tennis, baseball, the weight of the plane. The following example or golf, we notice that a spinning ball deviates illustrates this. Reprint 2025-26 190 PHYSICS (a) (b) (c) Fig 9.11 (a) Fluid streaming past a static sphere. (b) Streamlines for a fluid around a sphere spinning clockwise. (c) Air flowing past an aerofoil. ⊳ Example 9.7 A fully loaded Boeing aircraft vav = (v2 + v1)/2 = 960 km/h = 267 m s-1, has a mass of 3.3 × 105 kg. Its total wing we have area is 500 m2. It is in level flight with a ∆ P speed of 960 km/h. (a) Estimate the (v 2 – v1 ) / v av = 2 ≈ 0.08 ρ v av pressure difference between the lower and The speed above the wing needs to be only 8 upper surfaces of the wings (b) Estimate % higher than that below. ⊳ the fractional increase in the speed of the air on the upper surface of the wing relative to the lower surface. [The density of air is ρ 9.5 VISCOSITY = 1.2 kg m-3] Most of the fluids are not ideal ones and offer some resistance to motion. This resistance to fluid motion is like an internal friction analogous to friction when Answer (a) The weight of the Boeing aircraft is a solid moves on a surface. It is called viscosity. balanced by the upward force due to the This force exists when there is relative motion pressure difference between layers of the liquid. Suppose we consider ∆P × A = 3.3 × 105 kg × 9.8 a fluid like oil enclosed between two glass plates ∆P = (3.3 × 105 kg × 9.8 m s–2) / 500 m2 as shown in Fig. 9.12 (a). The bottom plate is fixed while the top plate is moved with a constant = 6.5 ×103 Nm-2 velocity v relative to the fixed plate. If oil is (b) We ignore the small height difference replaced by honey, a greater force is required to between the top and bottom sides in Eq. (9.12). move the plate with the same velocity. Hence The pressure difference between them is we say that honey is more viscous than oil. The then fluid in contact with a surface has the same ρ 2 2 velocity as that of the surfaces. Hence, the layer ∆P = ( v 2 – v1 ) of the liquid in contact with top surface moves 2 where v2 is the speed of air over the upper with a velocity v and the layer of the liquid in surface and v1 is the speed under the bottom contact with the fixed surface is stationary. The surface. velocities of layers increase uniformly from bottom (zero velocity) to the top layer (velocity 2 ∆ P (v 2 – v1 )= v). For any layer of liquid, its upper layer pulls ρ ( v 2 + v1 ) it forward while lower layer pulls it backward. Taking the average speed This results in force between the layers. This Reprint 2025-26 MECHANICAL PROPERTIES OF FLUIDS 191 type of flow is known as laminar. The layers of change of strain’ or ‘strain rate’ i.e. ∆x/(l ∆t) or liquid slide over one another as the pages of a v/l instead of strain itself. The coefficient of book do when it is placed flat on a table and a viscosity (pronounced ‘eta’) for a fluid is defined horizontal force is applied to the top cover. When as the ratio of shearing stress to the strain rate. a fluid is flowing in a pipe or a tube, then velocity (9.16)of the liquid layer along the axis of the tube is maximum and decreases gradually as we move The SI unit of viscosity is poiseiulle (Pl). Its towards the walls where it becomes zero, other units are N s m-2 or Pa s. The dimensions Fig. 9.12 (b). The velocity on a cylindrical surface of viscosity are [ML-1T-1]. Generally, thin liquids, in a tube is constant. like water, alcohol, etc., are less viscous than thick liquids, like coal tar, blood, glycerine, etc. The coefficients of viscosity for some common fluids are listed in Table 9.2. We point out two facts about blood and water that you may find interesting. As Table 9.2 indicates, blood is ‘thicker’ (more viscous) than water. Further, the relative viscosity (η/ηwater) of blood remains constant between 0 oC and 37 oC. (a) Fig. 9.13 Measurement of the coefficient of viscosity of a liquid. (b) Fig 9.12 (a) A layer of liquid sandwiched between The viscosity of liquids decreases with two parallel glass plates, in which the temperature, while it increases in the case of gases. lower plate is fixed and the upper one is moving to the right with velocity v ⊳ (b) velocity distribution for viscous flow in a pipe. Example 9.8 A metal block of area 0.10 m2 is connected to a 0.010 kg mass via a string On account of this motion, a portion of liquid, that passes over an ideal pulley (considered which at some instant has the shape ABCD, massless and frictionless), as in Fig. 9.13. take the shape of AEFD after short interval of A liquid with a film thickness of 0.30 mm time (∆t). During this time interval the liquid has is placed between the block and the table. undergone a shear strain of ∆x/l. Since, the When released the block moves to the right strain in a flowing fluid increases with time with a constant speed of 0.085 m s-1. Find continuously. Unlike a solid, here the stress is the coefficient of viscosity of the liquid. found experimentally to depend on ‘rate of Reprint 2025-26 192 PHYSICS Answer The metal block moves to the right This is known as Stokes’ law. We shall not because of the tension in the string. The tension derive Stokes’ law. T is equal in magnitude to the weight of the This law is an interesting example of retarding suspended mass m. Thus, the shear force F is force, which is proportional to velocity. We can study its consequences on an object falling F = T = mg = 0.010 kg × 9.8 m s–2 = 9.8 × 10-2 N through a viscous medium. We consider a Shear stress on the fluid = F/A = N/m2 raindrop in air. It accelerates initially due to gravity. As the velocity increases, the retarding Strain rate = force also increases. Finally, when viscous force plus buoyant force becomes equal to the force due to gravity, the net force becomes zero and so does the acceleration. The sphere (raindrop) then descends with a constant velocity. Thus, in = equilibrium, this terminal velocity vt is given by 6πηavt = (4π/3) a3 (ρ-σ)g = 3.46 ×10-3 Pa s where ρ and σ are mass densities of sphere and ⊳ the fluid, respectively. We obtain Table 9.2 The viscosities of some fluids vt = 2a2 (ρ-σ)g / (9η) (9.18) Fluid T(oC) Viscosity (mPl) So the terminal velocity vt depends on the Water 20 1.0 square of the radius of the sphere and inversely 100 0.3 on the viscosity of the medium. Blood 37 2.7 You may like to refer back to Example 6.2 in Machine Oil 16 113 this context. 38 34 ⊳ Glycerine 20 830 Example 9.9 The terminal velocity of a Honey – 200 copper ball of radius 2.0 mm falling through Air 0 0.017 a tank of oil at 20oC is 6.5 cm s-1. Compute the viscosity of the oil at 20oC. Density of 40 0.019 oil is 1.5 ×103 kg m-3, density of copper is 8.9 × 103 kg m-3. 9.5.1 Stokes’ Law When a body falls through a fluid it drags the Answer We have vt = 6.5 × 10-2 ms-1, a = 2 × 10-3 m,layer of the fluid in contact with it. A relative g = 9.8 ms-2, ρ = 8.9 × 103 kg m-3,motion between the different layers of the fluid is set and, as a result, the body experiences a σ =1.5 ×103 kg m-3. From Eq. (9.18) retarding force. Falling of a raindrop and swinging of a pendulum bob are some common examples of such motion. It is seen that the viscous force is proportional to the velocity of = 9.9 × 10-1 kg m–1 s–1 ⊳ the object and is opposite to the direction of motion. The other quantities on which the force F depends are viscosity η of the fluid and radius 9.6 SURFACE TENSION a of the sphere. Sir George G. Stokes (1819– You must have noticed that, oil and water do 1903), an English scientist enunciated clearly not mix; water wets you and me but not ducks; the viscous drag force F as mercury does not wet glass but water sticks to F = 6 π ηav (9.17) it, oil rises up a cotton wick, inspite of gravity, Reprint 2025-26 MECHANICAL PROPERTIES OF FLUIDS 193 Sap and water rise up to the top of the leaves of Let us consider a molecule near the surface the tree, hair of a paint brush do not cling Fig. 9.14(b). Only lower half side of it is together when dry and even when dipped in surrounded by liquid molecules. There is some water but form a fine tip when taken out of it. negative potential energy due to these, but All these and many more such experiences are obviously it is less than that of a molecule in related with the free surfaces of liquids. As bulk, i.e., the one fully inside. Approximately liquids have no definite shape but have a it is half of the latter. Thus, molecules on a definite volume, they acquire a free surface when liquid surface have some extra energy in poured in a container. These surfaces possess comparison to molecules in the interior. A some additional energy. This phenomenon is liquid, thus, tends to have the least surface known as surface tension and it is concerned area which external conditions permit. with only liquid as gases do not have free Increasing surface area requires energy. Most surfaces. Let us now understand this surface phenomenon can be understood in phenomena. terms of this fact. What is the energy required for having a molecule at the surface? As9.6.1 Surface Energy mentioned above, roughly it is half the energy A liquid stays together because of attraction required to remove it entirely from the liquid between molecules. Consider a molecule well i.e., half the heat of evaporation. inside a liquid. The intermolecular distances are Finally, what is a surface? Since a liquid such that it is attracted to all the surrounding consists of molecules moving about, there cannot molecules [Fig. 9.14(a)]. This attraction results be a perfectly sharp surface. The density of the in a negative potential energy for the molecule, liquid molecules drops rapidly to zero around which depends on the number and distribution z = 0 as we move along the direction indicated of molecules around the chosen one. But the Fig 9.14 (c) in a distance of the order of a few average potential energy of all the molecules is molecular sizes. the same. This is supported by the fact that to take a collection of such molecules (the liquid) 9.6.2 Surface Energy and Surface Tension and to disperse them far away from each other As we have discussed that an extra energy is in order to evaporate or vaporise, the heat of associated with surface of liquids, the creation evaporation required is quite large. For water it of more surface (spreading of surface) keeping is of the order of 40 kJ/mol. other things like volume fixed requires a Fig. 9.14 Schematic picture of molecules in a liquid, at the surface and balance of forces. (a) Molecule inside a liquid. Forces on a molecule due to others are shown. Direction of arrows indicates attraction of repulsion. (b) Same, for a molecule at a surface. (c) Balance of attractive (AI and repulsive (R) forces. Reprint 2025-26 194 PHYSICS horizontal liquid film ending in bar free to slide We make the following observations from over parallel guides Fig (9.15). above: (i) Surface tension is a force per unit length (or surface energy per unit area) acting in the plane of the interface between the plane of the liquid and any other substance; it also is the extra energy that the molecules at the interface have as compared to molecules in the interior. (ii) At any point on the interface besides the Fig. 9.15 Stretching a film. (a) A film in equilibrium; boundary, we can draw a line and imagine (b) The film stretched an extra distance. equal and opposite surface tension forces Suppose that we move the bar by a small S per unit length of the line acting distance d as shown. Since the area of the perpendicular to the line, in the plane of surface increases, the system now has more the interface. The line is in equilibrium. To energy, this means that some work has been be more specific, imagine a line of atoms or done against an internal force. Let this internal molecules at the surface. The atoms to the force be F, the work done by the applied force is left pull the line towards them; those to the F.d = Fd. From conservation of energy, this is right pull it towards them! This line of stored as additional energy in the film. If the atoms is in equilibrium under tension. If surface energy of the film is S per unit area, the the line really marks the end of the extra area is 2dl. A film has two sides and the interface, as in Figure 9.14 (a) and (b) there liquid in between, so there are two surfaces and is only the force S per unit length the extra energy is acting inwards. Table 9.3 gives the surface tension of various S (2dl) = Fd (9.19) liquids. The value of surface tension depends Or, S=Fd/2dl = F/2l (9.20) on temperature. Like viscosity, the surface This quantity S is the magnitude of surface tension of a liquid usually falls with tension. It is equal to the surface energy per unit temperature. area of the liquid interface and is also equal to Table 9.3 Surface tension of some liquids at thethe force per unit length exerted by the fluid on temperatures indicated with the the movable bar. heats of the vaporisation So far we have talked about the surface of one liquid. More generally, we need to consider Liquid Temp (oC) Surface Heat of fluid surface in contact with other fluids or solid Tension vaporisation (N/m) (kJ/mol)surfaces. The surface energy in that case depends on the materials on both sides of the surface. For example, if the molecules of the Helium –270 0.000239 0.115 materials attract each other, surface energy is Oxygen –183 0.0132 7.1 reduced while if they repel each other the surface energy is increased. Thus, more Ethanol 20 0.0227 40.6 appropriately, the surface energy is the energy Water 20 0.0727 44.16 of the interface between two materials and Mercury 20 0.4355 63.2 depends on both of them. Reprint 2025-26 MECHANICAL PROPERTIES OF FLUIDS 195 A fluid will stick to a solid surface if the by θ. It is different at interfaces of different pairs surface energy between fluid and the solid is of liquids and solids. The value of θ determines smaller than the sum of surface energies whether a liquid will spread on the surface of a between solid-air, and fluid-air. Now there is solid or it will form droplets on it. For example, attraction between the solid surface and the water forms droplets on lotus leaf as shown in liquid. It can be directly measured Fig. 9.17 (a) while spreads over a clean plastic experimentaly as schematically shown in Fig. plate as shown in Fig. 9.17(b). 9.16. A flat vertical glass plate, below which a vessel of some liquid is kept, forms one arm of the balance. The plate is balanced by weights on the other side, with its horizontal edge just over water. The vessel is raised slightly till the liquid just touches the glass plate and pulls it down a little because of surface tension. Weights (a)are added till the plate just clears water. (b) Fig. 9.17 Different shapes of water drops with interfacial tensions (a) on a lotus leaf (b) on a clean plastic plate. Fig. 9.16 Measuring Surface Tension. We consider the three interfacial tensions at all the three interfaces, liquid-air, solid-air and Suppose the additional weight required is W. solid-liquid denoted by Sla, Ssa and Ssl , respectivelyThen from Eq. 9.20 and the discussion given as given in Fig. 9.17 (a) and (b). At the line of there, the surface tension of the liquid-air contact, the surface forces between the three media interface is must be in equilibrium. From the Fig. 9.17(b) the Sla = (W/2l) = (mg/2l ) (9.21) following relation is easily derived. where m is the extra mass and l is the length of Sla cos θ + Ssl = Ssa (9.22) the plate edge. The subscript (la) emphasises The angle of contact is an obtuse angle if the fact that the liquid-air interface tension is involved. Ssl > Sla as in the case of water-leaf interface while it is an acute angle if Ssl < Sla as in the case of water-plastic interface. When θ is an 9.6.3 Angle of Contact obtuse angle then molecules of liquids are The surface of liquid near the plane of contact, attracted strongly to themselves and weakly to with another medium is in general curved. The those of solid, it costs a lot of energy to create a angle between tangent to the liquid surface at liquid-solid surface, and liquid then does not the point of contact and solid surface inside the wet the solid. This is what happens with water liquid is termed as angle of contact. It is denoted on a waxy or oily surface, and with mercury on Reprint 2025-26 196 PHYSICS any surface. On the other hand, if the molecules so that of the liquid are strongly attracted to those of (Pi – Po) = (2 Sla/ r) (9.25) the solid, this will reduce Ssl and therefore, In general, for a liquid-gas interface, the cos θ may increase or θ may decrease. In this convex side has a higher pressure than the case θ is an acute angle. This is what happens concave side. For example, an air bubble in a for water on glass or on plastic and for kerosene liquid, would have higher pressure inside it. oil on virtually anything (it just spreads). Soaps, See Fig 9.18 (b). detergents and dying substances are wetting agents. When they are added the angle of contact becomes small so that these may penetrate well and become effective. Water proofing agents on the other hand are added to create a large angle of contact between the water and fibres. 9.6.4 Drops and Bubbles Fig. 9.18 Drop, cavity and bubble of radius r. One consequence of surface tension is that free A bubble Fig 9.18 (c) differs from a drop liquid drops and bubbles are spherical if effects and a cavity; in this it has two interfaces. Applying of gravity can be neglected. You must have seen the above argument we have for a bubble this especially clearly in small drops just formed (Pi – Po) = (4 Sla/ r) (9.26) in a high-speed spray or jet, and in soap bubbles blown by most of us in childhood. Why are drops This is probably why you have to blow hard, and bubbles spherical? What keeps soap but not too hard, to form a soap bubble. A little bubbles stable? extra air pressure is needed inside! As we have been saying repeatedly, a liquid- air interface has energy, so for a given volume 9.6.5 Capillary Rise the surface with minimum energy is the one with One consequence of the pressure difference the least area. The sphere has this property. across a curved liquid-air interface is the well- Though it is out of the scope of this book, but known effect that water rises up in a narrow you can check that a sphere is better than at tube in spite of gravity. The word capilla means least a cube in this respect! So, if gravity and hair in Latin; if the tube were hair thin, the rise other forces (e.g. air resistance) were ineffective, would be very large. To see this, consider a liquid drops would be spherical. vertical capillary tube of circular cross section Another interesting consequence of surface (radius a) inserted into an open vessel of water tension is that the pressure inside a spherical (Fig. 9.19). The contact angle between water and drop Fig. 9.18(a) is more than the pressure outside. Suppose a spherical drop of radius r is in equilibrium. If its radius increase by ∆r. The extra surface energy is [4π(r + ∆r) 2- 4πr2] Sla = 8πr ∆r Sla (9.23) If the drop is in equilibrium this energy cost is balanced by the energy gain due to expansion under the pressure difference (Pi – Po) between the inside of the bubble and the outside. The work done is Fig. 9.19 Capillary rise, (a) Schematic picture of a narrow tube immersed water. W = (Pi – Po) 4πr2∆r (9.24) (b) Enlarged picture near interface. Reprint 2025-26 MECHANICAL PROPERTIES OF FLUIDS 197 glass is acute. Thus the surface of water in the ⊳ Example 9.10 The lower end of a capillary capillary is concave. This means that there is tube of diameter 2.00 mm is dipped 8.00 a pressure difference between the two sides cm below the surface of water in a beaker. of the top surface. This is given by What is the pressure required in the tube in order to blow a hemispherical bubble at (Pi – Po) =(2S/r) = 2S/(a sec θ ) its end in water? The surface tension of = (2S/a) cos θ (9.27) water at temperature of the experiments is Thus the pressure of the water inside the 7.30 × 10-2 Nm-1. 1 atmospheric pressure = 1.01 × 105 Pa, density of water = 1000 kg/m3,tube, just at the meniscus (air-water interface) g = 9.80 m s-2. Also calculate the excess is less than the atmospheric pressure. Consider pressure. the two points A and B in Fig. 9.19(a). They must be at the same pressure, namely Answer The excess pressure in a bubble of gas P0 + h ρ g = Pi = PA (9.28) in a liquid is given by 2S/r, where S is the where ρ is the density of water and h is called surface tension of the liquid-gas interface. You the capillary rise [Fig. 9.19(a)]. Using should note there is only one liquid surface in Eq. (9.27) and (9.28) we have this case. (For a bubble of liquid in a gas, there are two liquid surfaces, so the formula for h ρ g = (Pi – P0) = (2S cos θ )/a (9.29) excess pressure in that case is 4S/r.) The The discussion here, and the Eqs. (9.24) and radius of the bubble is r. Now the pressure (9.25) make it clear that the capillary rise is due outside the bubble Po equals atmospheric pressure plus the pressure due to 8.00 cm ofto surface tension. It is larger, for a smaller a. water column. That is Typically it is of the order of a few cm for fine Po = (1.01 × 105 Pa + 0.08 m × 1000 kg m–3 capillaries. For example, if a = 0.05 cm, using × 9.80 m s–2) the value of surface tension for water (Table 9.3), = 1.01784 × 105 Pa we find that Therefore, the pressure inside the bubble is h = 2S/(ρ g a) Pi = Po + 2S/r -1 = 1.01784 × 105 Pa + (2 × 7.3 × 10-2 Pa m/10-3 m) 2×(0.073 N m ) = (1.01784 + 0.00146) × 105 Pa = 3 -3 -2 -4 (10 kg m ) (9.8 m s )(5 × 10 m) = 1.02 × 105 Pa where the radius of the bubble is taken = 2.98 × 10–2 m = 2.98 cm to be equal to the radius of the capillary tube, Notice that if the liquid meniscus is convex, since the bubble is hemispherical ! (The answer as for mercury, i.e., if cos θ is negative then from has been rounded off to three significant figures.) Eq. (9.28) for example, it is clear that the liquid The excess pressure in the bubble is 146 Pa. will be lower in the capillary ! ⊳ SUMMARY 1. The basic property of a fluid is that it can flow. The fluid does not have any resistance to change of its shape. Thus, the shape of a fluid is governed by the shape of its container. 2. A liquid is incompressible and has a free surface of its own. A gas is compressible and it expands to occupy all the space available to it. 3. If F is the normal force exerted by a fluid on an area A then the average pressure Pav is defined as the ratio of the force to area F Pav = A Reprint 2025-26 198 PHYSICS 4. The unit of the pressure is the pascal (Pa). It is the same as N m-2. Other common units of pressure are 1 atm = 1.01×105 Pa 1 bar = 105 Pa 1 torr = 133 Pa = 0.133 kPa 1 mm of Hg = 1 torr = 133 Pa 5. Pascal’s law states that: Pressure in a fluid at rest is same at all points which are at the same height. A change in pressure applied to an enclosed fluid is transmitted undiminished to every point of the fluid and the walls of the containing vessel. 6. The pressure in a fluid varies with depth h according to the expression P = Pa + ρgh where ρ is the density of the fluid, assumed uniform. 7. The volume of an incompressible fluid passing any point every second in a pipe of non uniform crossection is the same in the steady flow. v A = constant ( v is the velocity and A is the area of crossection) The equation is due to mass conservation in incompressible fluid flow. 8. Bernoulli’s principle states that as we move along a streamline, the sum of the pressure (P), the kinetic energy per unit volume (ρv2/2) and the potential energy per unit volume (ρgy) remains a constant. P + ρv2/2 + ρgy = constant The equation is basically the conservation of energy applied to non viscuss fluid motion in steady state. There is no fluid which have zero viscosity, so the above statement is true only approximately. The viscosity is like friction and converts the kinetic energy to heat energy. 9. Though shear strain in a fluid does not require shear stress, when a shear stress is applied to a fluid, the motion is generated which causes a shear strain growing with time. The ratio of the shear stress to the time rate of shearing strain is known as coefficient of viscosity, η. where symbols have their usual meaning and are defined in the text. 10. Stokes’ law states that the viscous drag force F on a sphere of radius a moving with velocity v through a fluid of viscosity is, F = 6πηav. 11. Surface tension is a force per unit length (or surface energy per unit area) acting in the plane of interface between the liquid and the bounding surface. It is the extra energy that the molecules at the interface have as compared to the interior. POINTS TO PONDER 1. Pressure is a scalar quantity. The definition of the pressure as “force per unit area” may give one false impression that pressure is a vector. The “force” in the numerator of the definition is the component of the force normal to the area upon which it is impressed. While describing fluids as a concept, shift from particle and rigid body mechanics is required. We are concerned with properties that vary from point to point in the fluid. 2. One should not think of pressure of a fluid as being exerted only on a solid like the walls of a container or a piece of solid matter immersed in the fluid. Pressure exists at all points in a fluid. An element of a fluid (such as the one shown in Fig. 9.4) is in equilibrium because the pressures exerted on the various faces are equal. Reprint 2025-26 MECHANICAL PROPERTIES OF FLUIDS 199 3. The expression for pressure P = Pa + ρgh holds true if fluid is incompressible. Practically speaking it holds for liquids, which are largely incompressible and hence is a constant with height. 4. The gauge pressure is the difference of the actual pressure and the atmospheric pressure. P – Pa = Pg Many pressure-measuring devices measure the gauge pressure. These include the tyre pressure gauge and the blood pressure gauge (sphygmomanometer). 5. A streamline is a map of fluid flow. In a steady flow two streamlines do not intersect as it means that the fluid particle will have two possible velocities at the point. 6. Bernoulli’s principle does not hold in presence of viscous drag on the fluid. The work done by this dissipative viscous force must be taken into account in this case, and P2 [Fig. 9.9] will be lower than the value given by Eq. (9.12). 7. As the temperature rises the atoms of the liquid become more mobile and the coefficient of viscosity, η falls. In a gas the temperature rise increases the random motion of atoms and η increases. 8. Surface tension arises due to excess potential energy of the molecules on the surface in comparison to their potential energy in the interior. Such a surface energy is present at the interface separating two substances at least one of which is a fluid. It is not the property of a single fluid alone. EXERCISES 9.1 Explain why (a) The blood pressure in humans is greater at the feet than at the brain (b) Atmospheric pressure at a height of about 6 km decreases to nearly half of its value at the sea level, though the height of the atmosphere is more than 100 km (c) Hydrostatic pressure is a scalar quantity even though pressure is force divided by area. 9.2 Explain why (a) The angle of contact of mercury with glass is obtuse, while that of water with glass is acute. (b) Water on a clean glass surface tends to spread out while mercury on the same surface tends to form drops. (Put differently, water wets glass while mercury does not.) Reprint 2025-26 200 PHYSICS (c) Surface tension of a liquid is independent of the area of the surface (d) Water with detergent disolved in it should have small angles of contact. (e) A drop of liquid under no external forces is always spherical in shape 9.3 Fill in the blanks using the word(s) from the list appended with each statement: (a) Surface tension of liquids generally ... with temperatures (increases / decreases) (b) Viscosity of gases ... with temperature, whereas viscosity of liquids ... with temperature (increases / decreases) (c) For solids with elastic modulus of rigidity, the shearing force is proportional to ... , while for fluids it is proportional to ... (shear strain / rate of shear strain) (d) For a fluid in a steady flow, the increase in flow speed at a constriction follows (conservation of mass / Bernoulli’s principle) (e) For the model of a plane in a wind tunnel, turbulence occurs at a ... speed for turbulence for an actual plane (greater / smaller) 9.4 Explain why (a) To keep a piece of paper horizontal, you should blow over, not under, it (b) When we try to close a water tap with our fingers, fast jets of water gush through the openings between our fingers (c) The size of the needle of a syringe controls flow rate better than the thumb pressure exerted by a doctor while administering an injection (d) A fluid flowing out of a small hole in a vessel results in a backward thrust on the vessel (e) A spinning cricket ball in air does not follow a parabolic trajectory 9.5 A 50 kg girl wearing high heel shoes balances on a single heel. The heel is circular with a diameter 1.0 cm. What is the pressure exerted by the heel on the horizontal floor ? 9.6 Toricelli’s barometer used mercury. Pascal duplicated it using French wine of density 984 kg m–3. Determine the height of the wine column for normal atmospheric pressure. 9.7 A vertical off-shore structure is built to withstand a maximum stress of 109 Pa. Is the structure suitable for putting up on top of an oil well in the ocean ? Take the depth of the ocean to be roughly 3 km, and ignore ocean currents. 9.8 A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000 kg. The area of cross-section of the piston carrying the load is 425 cm2. What maximum pressure would the smaller piston have to bear ? 9.9 A U-tube contains water and methylated spirit separated by mercury. The mercury columns in the two arms are in level with 10.0 cm of water in one arm and 12.5 cm of spirit in the other. What is the specific gravity of spirit ? 9.10 In the previous problem, if 15.0 cm of water and spirit each are further poured into the respective arms of the tube, what is the difference in the levels of mercury in the two arms ? (Specific gravity of mercury = 13.6) 9.11 Can Bernoulli’s equation be used to describe the flow of water through a rapid in a river ? Explain. 9.12 Does it matter if one uses gauge instead of absolute pressures in applying Bernoulli’s equation ? Explain. 9.13 Glycerine flows steadily through a horizontal tube of length 1.5 m and radius 1.0 cm. If the amount of glycerine collected per second at one end is 4.0 × 10–3 kg s–1, what is the pressure difference between the two ends of the tube ? (Density of glycerine = 1.3 × 103 kg m–3 and viscosity of glycerine = 0.83 Pa s). [You may also like to check if the assumption of laminar flow in the tube is correct]. 9.14 In a test experiment on a model aeroplane in a wind tunnel, the flow speeds on the upper and lower surfaces of the wing are 70 m s–1and 63 m s-1 respectively. What is the lift on the wing if its area is 2.5 m2 ? Take the density of air to be 1.3 kg m–3. 9.15 Figures 9.20(a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is incorrect ? Why ? Reprint 2025-26 MECHANICAL PROPERTIES OF FLUIDS 201 Fig. 9.20 9.16 The cylindrical tube of a spray pump has a cross-section of 8.0 cm2 one end of which has 40 fine holes each of diameter 1.0 mm. If the liquid flow inside the tube is 1.5 m min–1, what is the speed of ejection of the liquid through the holes ? 9.17 A U-shaped wire is dipped in a soap solution, and removed. The thin soap film formed between the wire and the light slider supports a weight of 1.5 × 10–2 N (which includes the small weight of the slider). The length of the slider is 30 cm. What is the surface tension of the film ? 9.18 Figure 9.21 (a) shows a thin liquid film supporting a small weight = 4.5 × 10–2 N. What is the weight supported by a film of the same liquid at the same temperature in Fig. (b) and (c) ? Explain your answer physically. Fig. 9.21 9.19 What is the pressure inside the drop of mercury of radius 3.00 mm at room temperature ? Surface tension of mercury at that temperature (20 °C) is 4.65 × 10–1 N m–1. The atmospheric pressure is 1.01 × 105 Pa. Also give the excess pressure inside the drop. 9.20 What is the excess pressure inside a bubble of soap solution of radius 5.00 mm, given that the surface tension of soap solution at the temperature (20 °C) is 2.50 × 10–2 N m–1 ? If an air bubble of the same dimension were formed at depth of 40.0 cm inside a container containing the soap solution (of relative density 1.20), what would be the pressure inside the bubble ? (1 atmospheric pressure is 1.01 × 105 Pa). Reprint 2025-26 CHAPTER TEN THERMAL PROPERTIES OF MATTER 10.1 INTRODUCTION We all have common sense notions of heat and temperature. Temperature is a measure of ‘hotness’ of a body. A kettle 10.1 Introduction with boiling water is hotter than a box containing ice. In 10.2 Temperature and heat physics, we need to define the notion of heat, temperature, 10.3 Measurement of etc., more carefully. In this chapter, you will learn what heat temperature is and how it is measured, and study the various proceses by 10.4 Ideal-gas equation and which heat flows from one body to another. Along the way, absolute temperature you will find out why blacksmiths heat the iron ring before 10.5 Thermal expansion fitting on the rim of a wooden wheel of a horse cart and why 10.6 Specific heat capacity the wind at the beach often reverses direction after the sun 10.7 Calorimetry goes down. You will also learn what happens when water boils or freezes, and its temperature does not change during these10.8 Change of state processes even though a great deal of heat is flowing into or10.9 Heat transfer out of it.10.10 Newton’s law of cooling Summary 10.2 TEMPERATURE AND HEAT Points to ponder We can begin studying thermal properties of matter with Exercises definitions of temperature and heat. Temperature is a relative Additional Exercises measure, or indication of hotness or coldness. A hot utensil is said to have a high temperature, and ice cube to have a low temperature. An object that has a higher temperature than another object is said to be hotter. Note that hot and cold are relative terms, like tall and short. We can perceive temperature by touch. However, this temperature sense is somewhat unreliable and its range is too limited to be useful for scientific purposes. We know from experience that a glass of ice-cold water left on a table on a hot summer day eventually warms up whereas a cup of hot tea on the same table cools down. It means that when the temperature of body, ice-cold water or hot tea in this case, and its surrounding medium are different, heat transfer takes place between the system and the surrounding medium, until the body and the surrounding medium are at the same temperature. We also know that in the case of glass tumbler of ice-cold water, heat flows from the environment to Reprint 2025-26 THERMAL PROPERTIES OF MATTER 203 the glass tumbler, whereas in the case of hot tea, it flows from the cup of hot tea to the environment. So, we can say that heat is the form of energy transferred between two (or more) systems or a system and its surroundings by virtue of temperature difference. The SI unit of heat energy transferred is expressed in joule (J) while SI unit of temperature is Kelvin (K), and degree Celsius (oC) is a commonly used unit of temperature. When an object is heated, many changes may take place. Its temperature may rise, it may expand or change state. We will study the effect of heat on different bodies in later sections. Fig. 10.1 A plot of Fahrenheit temperature (tF) versus10.3 MEASUREMENT OF TEMPERATURE Celsius temperature (tc). A measure of temperature is obtained using a thermometer. Many physical properties of A relationship for converting between the two materials change sufficiently with temperature. scales may be obtained from a graph of Some such properties are used as the basis for Fahrenheit temperature (tF) versus celsius constructing thermometers. The commonly used temperature (tC) in a straight line (Fig. 10.1), property is variation of the volume of a liquid whose equation is with temperature. For example, in common t F – 32 t C = (10.1)liquid–in–glass thermometers, mercury, alcohol 180 100 etc., are used whose volume varies linearly with temperature over a wide range. 10.4 IDEAL-GAS EQUATION AND Thermometers are calibrated so that a ABSOLUTE TEMPERATURE numerical value may be assigned to a given temperature in an appropriate scale. For the Liquid-in-glass thermometers show different definition of any standard scale, two fixed readings for temperatures other than the fixed reference points are needed. Since all points because of differing expansion properties. substances change dimensions with A thermometer that uses a gas, however, gives temperature, an absolute reference for the same readings regardless of which gas is expansion is not available. However, the used. Experiments show that all gases at low necessary fixed points may be correlated to the densities exhibit same expansion behaviour. The physical phenomena that always occur at the variables that describe the behaviour of a given same temperature. The ice point and the steam quantity (mass) of gas are pressure, volume, and point of water are two convenient fixed points temperature (P, V, and T )(where T = t + 273.15; and are known as the freezing and boiling t is the temperature in °C). When temperature points, respectively. These two points are the is held constant, the pressure and volume of a temperatures at which pure water freezes and quantity of gas are related as PV = constant. boils under standard pressure. The two familiar This relationship is known as Boyle’s law, after temperature scales are the Fahrenheit Robert Boyle (1627–1691), the English Chemist temperature scale and the Celsius temperature who discovered it. When the pressure is held scale. The ice and steam point have values constant, the volume of a quantity of the gas is 32 °F and 212 °F, respectively, on the Fahrenheit related to the temperature as V/T = constant. scale and 0 °C and 100 °C on the Celsius scale. This relationship is known as Charles’ law, On the Fahrenheit scale, there are 180 equal after French scientist Jacques Charles (1747– intervals between two reference points, and on 1823). Low-density gases obey these the Celsius scale, there are 100. laws, which may be combined into a single Reprint 2025-26 204 PHYSICS Fig. 10.3 A plot of pressure versus temperature and Fig. 10.2 Pressure versus temperature of a low extrapolation of lines for low density gases density gas kept at constant volume. indicates the same absolute zero temperature. relationship. Notice that since PV = constant named after the British scientist Lord Kelvin. On and V/T = constant for a given quantity of gas, this scale, – 273.15 °C is taken as the zero point, then PV/T should also be a constant. This that is 0 K (Fig. 10.4). relationship is known as ideal gas law. It can be written in a more general form that applies not just to a given quantity of a single gas but to any quantity of any low-density gas and is known as ideal-gas equation: or PV = µRT (10.2) where, µ is the number of moles in the sample of gas and R is called universal gas constant: R = 8.31 J mol–1 K–1 In Eq. 10.2, we have learnt that the pressure and volume are directly proportional to temperature : PV ∝ T. This relationship allows a gas to be used to measure temperature in a constant volume gas thermometer. Holding the volume of a gas constant, it gives P ∝T. Thus, with a constant-volume gas thermometer, Fig. 10.4 Comparision of the Kelvin, Celsius and temperature is read in terms of pressure. A plot Fahrenheit temperature scales. The size of unit in Kelvin and Celsius of pressure versus temperature gives a straight temperature scales is the same. So, temperature line in this case, as shown in Fig. 10.2. on these scales are related by However, measurements on real gases deviate from the values predicted by the ideal gas law T = tC + 273.15 (10.3) at low temperature. But the relationship is linear over a large temperature range, and it looks as 10.5 THERMAL EXPANSION though the pressure might reach zero with You may have observed that sometimes sealed decreasing temperature if the gas continued to bottles with metallic lids are so tightly screwed be a gas. The absolute minimum temperature that one has to put the lid in hot water for some for an ideal gas, therefore, inferred by time to open it. This would allow the metallic lid extrapolating the straight line to the axis, as in to expand, thereby loosening it to unscrew Fig. 10.3. This temperature is found to be easily. In case of liquids, you may have observed – 273.15 °C and is designated as absolute zero. that mercury in a thermometer rises, when the Absolute zero is the foundation of the Kelvin thermometer is put in slightly warm water. If temperature scale or absolute scale temperature we take out the thermometer from the warm Reprint 2025-26 THERMAL PROPERTIES OF MATTER 205 water the level of mercury falls again. Similarly, Table 10.1 Values of coefficient of linear in case of gases, a balloon partially inflated in a expansion for some material cool room may expand to full size when placed in warm water. On the other hand, a fully Material αl (10–5 K–1) inflated balloon when immersed in cold water Aluminium 2.5 would start shrinking due to contraction of the Brass 1.8 air inside. Iron 1.2 It is our common experience that most Copper 1.7 substances expand on heating and contract on Silver 1.9 cooling. A change in the temperature of a body Gold 1.4 Glass (pyrex) 0.32causes change in its dimensions. The increase Lead 0.29in the dimensions of a body due to the increase in its temperature is called thermal expansion. The expansion in length is called linear Similarly, we consider the fractional change expansion. The expansion in area is called area expansion. The expansion in volume is called ∆V in volume, , of a substance for temperature volume expansion (Fig. 10.5). V change ∆T and define the coefficient of volume expansion (or volume expansivity), as (10.5) Here αV is also a characteristic of the substance but is not strictly a constant. It depends in general on temperature (Fig 10.6). It ∆l ∆A ∆V = a l ∆ T = 2a l ∆T = 3a l ∆ T is seen that αV becomes constant only at a high l A V temperature. (a) Linear expansion (b) Area expansion (c) Volume expansion Fig. 10.5 Thermal Expansion. If the substance is in the form of a long rod, then for small change in temperature, ∆T, the fractional change in length, ∆l/l, is directly proportional to ∆T. (10.4) where α1 is known as the coefficient of linear expansion (or linear expansivity) and is characteristic of the material of the rod. In Table Fig. 10.6 Coefficient of volume expansion of copper 10.1, typical average values of the coefficient of as a function of temperature. linear expansion for some material in the temperature range 0 °C to 100 °C are given. From Table 10.2 gives the values of coefficient of this Table, compare the value of αl for glass and volume expansion of some common substances copper. We find that copper expands about five in the temperature range 0–100 °C. You can see times more than glass for the same rise in that thermal expansion of these substances temperature. Normally, metals expand more and (solids and liquids) is rather small, with material, have relatively high values of αl. Reprint 2025-26 206 PHYSICS like pyrex glass and invar (a special iron-nickel lakes and ponds, freeze at the top first. As a lake alloy) having particularly low values of αV. From cools toward 4 °C, water near the surface loses this Table we find that the value of αv for energy to the atmosphere, becomes denser, and alcohol (ethanol) is more than mercury and sinks; the warmer, less dense water near the expands more than mercury for the same rise bottom rises. However, once the colder water on in temperature. top reaches temperature below 4 °C, it becomes less dense and remains at the surface, where it Table 10.2 Values of coefficient of volume freezes. If water did not have this property, lakes expansion for some substances and ponds would freeze from the bottom up, which would destroy much of their animal and Material αv ( K–1) plant life. Aluminium 7 × 10–5 Gases, at ordinary temperature, expand more Brass 6 × 10–5 than solids and liquids. For liquids, the Iron 3.55 × 10–5 coefficient of volume expansion is relatively Paraffin 58.8 × 10–5 independent of the temperature. However, for Glass (ordinary) 2.5 × 10–5 gases it is dependent on temperature. For an Glass (pyrex) 1 × 10–5 ideal gas, the coefficient of volume expansion at Hard rubber 2.4 × 10–4 constant pressure can be found from the ideal Invar 2 × 10–6 gas equation: Mercury 18.2 × 10–5 PV = µRT Water 20.7 × 10–5 At constant pressure Alcohol (ethanol) 110 × 10–5 P∆V = µR ∆T ∆V ∆ T Water exhibits an anomalous behaviour; it = V T contracts on heating between 0 °C and 4 °C. The volume of a given amount of water decreases as 1 i.e., αv = for ideal gas (10.6) it is cooled from room temperature, until its T temperature reaches 4 °C, [Fig. 10.7(a)]. Below At 0 °C, αv = 3.7 × 10–3 K–1, which is much 4 °C, the volume increases, and therefore, the larger than that for solids and liquids. density decreases [Fig. 10.7(b)]. Equation (10.6) shows the temperature This means that water has the maximum dependence of αv; it decreases with increasing density at 4 °C. This property has an important temperature. For a gas at room temperature and environmental effect: bodies of water, such as constant pressure, αv is about 3300 × 10–6 K–1, as Temperature (°C) Temperature (°C) (a) (b) Fig. 10.7 Thermal expansion of water. Reprint 2025-26 THERMAL PROPERTIES OF MATTER 207 much as order(s) of magnitude larger than the Answer coefficient of volume expansion of typical liquids. ∆A3 = (∆a) (∆b) a (∆b) There is a simple relation between the ∆Al =coefficient of volume expansion (αv) and coefficient of linear expansion (αl). Imagine a ∆b cube of length, l, that expands equally in all directions, when its temperature increases by b ∆T. We have ∆a a ∆l = αl l ∆T so, ∆V = (l+∆l)3 – l3 ≃ 3l2 ∆l (10.7) In Equation (10.7), terms in (∆l)2 and (∆l)3 have been neglected since ∆l is small compared to l. So ∆A2 = b (∆a) 3V ∆ l Fig. 10.8 ∆ V = = 3V α l ∆T (10.8) l Consider a rectangular sheet of the solid which gives material of length a and breadth b (Fig. 10.8 ). αv = 3αl (10.9) When the temperature increases by ∆T, a increases by ∆a = αl a∆T and b increases by ∆b What happens by preventing the thermal = αlb ∆T. From Fig. 10.8, the increase in area expansion of a rod by fixing its ends rigidly? ∆A = ∆A1 +∆A2 + ∆A3 Clearly, the rod acquires a compressive strain ∆A = a ∆b + b ∆a + (∆a) (∆b) due to the external forces provided by the rigid = a αlb ∆T + b αl a ∆T + (αl)2 ab (∆T)2 support at the ends. The corresponding stress = αl ab ∆T (2 + αl ∆T) = αl A ∆T (2 + αl ∆T) set up in the rod is called thermal stress. For Since αl ≃ 10–5 K–1, from Table 10.1, theexample, consider a steel rail of length 5 m and product αl ∆T for fractional temperature is smallarea of cross-section 40 cm2 that is prevented in comparision with 2 and may be neglected. from expanding while the temperature rises by Hence, 10 °C. The coefficient of linear expansion of steel is αl(steel) = 1.2 × 10–5 K–1. Thus, the compressive ⊳ ∆l strain is = αl(steel) ∆T = 1.2 × 10–5 × 10=1.2 × 10–4. l ⊳ Youngs modulus of steel is Y (steel) = 2 × 1011 N m–2. Example 10.2 A blacksmith fixes iron ring on the rim of the wooden wheel of a horseTherefore, the thermal stress developed is cart. The diameter of the rim and the iron ∆F  ∆l  = Y steel 2.4 × 107 N m –2, which ring are 5.243 m and 5.231 m, respectively A  l = at 27 °C. To what temperature should the corresponds to an external force of ring be heated so as to fit the rim of the wheel?  ∆l  ∆F = AYsteel  l  = 2.4 × 107 × 40 × 10–4 j 105N. If Answer two such steel rails, fixed at their outer ends, Given, T1 = 27 °Care in contact at their inner ends, a force of this magnitude can easily bend the rails. LT1 = 5.231 m ⊳ LT2 = 5.243 m Example 10.1 Show that the coefficient of So, area expansion, (∆A/A)/∆T, of a LT2 =LT1 [1+αl (T2–T1)] rectangular sheet of the solid is twice its linear expansivity, αl. 5.243 m = 5.231 m [1 + 1.20×10–5 K–1 (T2–27 °C)] or T2 = 218 °C. ⊳ Reprint 2025-26 208 PHYSICS

27.3 days which is also roughly equal to the Which is approximately 85 minutes. rotational period of the moon about its own axis. ⊳ Example 7.5 The planet Mars has twoSince, 1957, advances in technology have enabled moons, phobos and delmos. (i) phobos hasmany countries including India to launch artificial a period 7 hours, 39 minutes and an orbitalearth satellites for practical use in fields like radius of 9.4 ×103 km. Calculate the masstelecommunication, geophysics and meteorology. of mars. (ii) Assume that earth and mars We will consider a satellite in a circular orbit move in circular orbits around the sun,of a distance (RE + h) from the centre of the earth, with the martian orbit being 1.52 timeswhere RE = radius of the earth. If m is the mass the orbital radius of the earth. What isof the satellite and V its speed, the centripetal the length of the martian year in days ?force required for this orbit is mV 2 Answer (i) We employ Eq. (7.38) with the sun’s F(centripetal) = (7.33) ( R E + h ) mass replaced by the martian mass Mm directed towards the centre. This centripetal force 2 4 π 2 3 T = Ris provided by the gravitational force, which is GM m G m M E 4 π 2 R 3 F(gravitation) = 2 (7.34) Mm = 2 ( R E + h ) G T where ME is the mass of the earth. 2 3 18 Equating R.H.S of Eqs. (7.33) and (7.34) and 4 × ( 3.14 ) × ( 9.4 ) × 10 = -11 2cancelling out m, we get 6.67 × 10 × ( 459 × 60 ) 2 G M E 2 3 18 V = (7.35) 4 × ( 3.14 ) × ( 9.4 ) × 10 ( R E + h ) M m = 2 -5 6.67 × ( 4.59 × 6 ) × 10 Thus V decreases as h increases. From = 6.48 × 1023 kg. equation (7.35),the speed V for h = 0 is (ii) Once again Kepler’s third law comes to our V 2 (h = 0) = GM / R E = gR E (7.36) aid, where we have used the relation T M2 R MS3 2 2 = 3 g = GM / R E . In every orbit, the satellite T E R ES Reprint 2025-26 138 PHYSICS where RMS is the mars -sun distance and RES is − 13  1 2   1    d the earth-sun distance. = 10    ( 24 × 60 × 60 ) 2  ( 1 / 1000 ) 3 km 3  ∴ TM = (1.52)3/2 × 365 = 1.33 ×10–14 d2 km–3 = 684 days Using Eq. (7.38) and the given value of k, We note that the orbits of all planets except the time period of the moon is Mercury and Mars are very close to being 2 T = (1.33 × 10-14)(3.84 × 105)3 circular. For example, the ratio of the semi- T = 27.3 d ⊳ minor to semi-major axis for our Earth is, Note that Eq. (7.38) also holds for elliptical b/a = 0.99986. ⊳ orbits if we replace (RE+h) by the semi-major axis ⊳ of the ellipse. The earth will then be at one of Example 7.6 Weighing the Earth : You the foci of this ellipse. are given the following data: g = 9.81 ms–2, RE = 6.37×106 m, the distance to the moon R 7.10 ENERGY OF AN ORBITING SATELLITE = 3.84×108 m and the time period of the moon’s revolution is 27.3 days. Obtain the Using Eq. (7.35), the kinetic energy of the satellite mass of the Earth ME in two different ways. in a circular orbit with speed v is 1 m v 2Answer From Eq. (7.12) we have K i E = 2 g R E2 M E = Gm M E G = , (7.40) 2( R E + h ) 6 2 Considering gravitational potential energy at 9.81 × ( 6.37 × 10 ) = -11 infinity to be zero, the potential energy at distance 6.67 × 10 (Re+h) from the centre of the earth is = 5.97× 1024 kg. The moon is a satellite of the Earth. From G m M E P .E = − (7.41)the derivation of Kepler’s third law [see Eq. ( R E + h ) (7.38)] The K.E is positive whereas the P.E is 2 4 π2R 3 negative. However, in magnitude the K.E. is half T = G M E the P.E, so that the total E is E 4 π2R 3 E = K .E + P .E = − G m M ME = G T 2 2( R E + h ) (7.42) 4 × 3.14 × 3.14 × ( 3.84 ) 3 × 10 24 The total energy of an circularly orbiting = -11 2 satellite is thus negative, with the potential 6.67 × 10 × ( 27.3 × 24 × 60 × 60 ) energy being negative but twice is magnitude of = 6.02 × 1024 kg the positive kinetic energy. Both methods yield almost the same answer, When the orbit of a satellite becomes the difference between them being less than 1%. elliptic, both the K.E. and P.E. vary from point ⊳ to point. The total energy which remains constant is negative as in the circular orbit case. ⊳ Example 7.7 Express the constant k of Eq. This is what we expect, since as we have (7.38) in days and kilometres. Given discussed before if the total energy is positive or k = 10–13 s2 m–3. The moon is at a distance zero, the object escapes to infinity. Satellites of 3.84 × 105 km from the earth. Obtain its are always at finite distance from the earth and time-period of revolution in days. hence their energies cannot be positive or zero. Answer Given k = 10–13 s2 m–3 Reprint 2025-26 GRAVITATION 139 The change in the total energy is⊳ Example 7.8 A 400 kg satellite is in a circular ∆E = Ef – Ei orbit of radius 2RE about the Earth. How much energy is required to transfer it to a circular orbit of radius 4RE? What are the changes in  E E = G M E m =  the kinetic and potential energies ?  G M2  m R 8 R E  R E  8 Answer Initially, g m R E = 9.81 × 400 × 6. 37 × 106 = 3.13 × 10 9 J ∆ E = G M E m 8 8 E i = − 4 R E The kinetic energy is reduced and it mimics While finally ∆E, namely, ∆K = Kf – Ki = – 3.13 × 109 J. The change in potential energy is twice the G M E m E f = − change in the total energy, namely 8 R E ∆V = Vf – Vi = – 6.25 × 109 J ⊳ SUMMARY 1. Newton’s law of universal gravitation states that the gravitational force of attraction between any two particles of masses m1 and m2 separated by a distance r has the magnitude m 1m 2 F = G 2 r where G is the universal gravitational constant, which has the value 6.672 ×10–11 N m2 kg–2. 2. If we have to find the resultant gravitational force acting on the particle m due to a number of masses M1, M2, ….Mn etc. we use the principle of superposition. Let F1, F2, ….Fn be the individual forces due to M1, M2, ….Mn, each given by the law of gravitation. From the principle of superposition each force acts independently and uninfluenced by the other bodies. The resultant force FR is then found by vector addition n FR = F1 + F2 + ……+ Fn = ∑ Fi i = 1 where the symbol ‘Σ’ stands for summation. 3. Kepler’s laws of planetary motion state that (a) All planets move in elliptical orbits with the Sun at one of the focal points (b) The radius vector drawn from the Sun to a planet sweeps out equal areas in equal time intervals. This follows from the fact that the force of gravitation on the planet is central and hence angular momentum is conserved. (c) The square of the orbital period of a planet is proportional to the cube of the semi-major axis of the elliptical orbit of the planet The period T and radius R of the circular orbit of a planet about the Sun are related by 2  4 π 2  3 T =   R  G M s  where Ms is the mass of the Sun. Most planets have nearly circular orbits about the Sun. For elliptical orbits, the above equation is valid if R is replaced by the semi-major axis, a. 4. The acceleration due to gravity. (a) at a height h above the earth’s surface G M E g ( h ) = 2 ( R E + h ) G M E  2 h  ≈ 2 1 − for h << RE R E  R E  Reprint 2025-26 140 PHYSICS  2 h  G M E g (h ) = g ( 0 ) 1 − where g ( 0 ) = 2  R E  R E (b) at depth d below the earth’s surface is d   1 − g (d ) = G M2 E 1 − d  = g ( 0 ) R E  R E   R E  5. The gravitational force is a conservative force, and therefore a potential energy function can be defined. The gravitational potential energy associated with two particles separated by a distance r is given by G m1 m 2 V = − r where V is taken to be zero at r → ∞. The total potential energy for a system of particles is the sum of energies for all pairs of particles, with each pair represented by a term of the form given by above equation. This prescription follows from the principle of superposition. 6. If an isolated system consists of a particle of mass m moving with a speed v in the vicinity of a massive body of mass M, the total mechanical energy of the particle is given by 1 G M m E = m v 2− 2 r That is, the total mechanical energy is the sum of the kinetic and potential energies. The total energy is a constant of motion. 7. If m moves in a circular orbit of radius a about M, where M >> m, the total energy of the system is G M m E = − 2a with the choice of the arbitrary constant in the potential energy given in the point 5., above. The total energy is negative for any bound system, that is, one in which the orbit is closed, such as an elliptical orbit. The kinetic and potential energies are G M m K = 2a G M m V = − a 8. The escape speed from the surface of the earth is 2 G M E ve = = 2 gR E R E and has a value of 11.2 km s–1. 9. If a particle is outside a uniform spherical shell or solid sphere with a spherically symmetric internal mass distribution, the sphere attracts the particle as though the mass of the sphere or shell were concentrated at the centre of the sphere. 10. If a particle is inside a uniform spherical shell, the gravitational force on the particle is zero. If a particle is inside a homogeneous solid sphere, the force on the particle acts toward the centre of the sphere. This force is exerted by the spherical mass interior to the particle. Reprint 2025-26 GRAVITATION 141 POINTS TO PONDER 1. In considering motion of an object under the gravitational influence of another object the following quantities are conserved: (a) Angular momentum (b) Total mechanical energy Linear momentum is not conserved 2. Angular momentum conservation leads to Kepler’s second law. However, it is not special to the inverse square law of gravitation. It holds for any central force. 3. In Kepler’s third law (see Eq. (7.1) and T2 = KS R3. The constant KS is the same for all planets in circular orbits. This applies to satellites orbiting the Earth [(Eq. (7.38)]. 4. An astronaut experiences weightlessness in a space satellite. This is not because the gravitational force is small at that location in space. It is because both the astronaut and the satellite are in “free fall” towards the Earth. 5. The gravitational potential energy associated with two particles separated by a distance r is given by G m 1 m 2 V = – + constant r The constant can be given any value. The simplest choice is to take it to be zero. With this choice G m 1 m 2 V = – r This choice implies that V → 0 as r → ∞. Choosing location of zero of the gravitational energy is the same as choosing the arbitrary constant in the potential energy. Note that the gravitational force is not altered by the choice of this constant. 6. The total mechanical energy of an object is the sum of its kinetic energy (which is always positive) and the potential energy. Relative to infinity (i.e. if we presume that the potential energy of the object at infinity is zero), the gravitational potential energy of an object is negative. The total energy of a satellite is negative. 7. The commonly encountered expression m g h for the potential energy is actually an approximation to the difference in the gravitational potential energy discussed in the point 6, above. 8. Although the gravitational force between two particles is central, the force between two finite rigid bodies is not necessarily along the line joining their centre of mass. For a spherically symmetric body however the force on a particle external to the body is as if the mass is concentrated at the centre and this force is therefore central. 9. The gravitational force on a particle inside a spherical shell is zero. However, (unlike a metallic shell which shields electrical forces) the shell does not shield other bodies outside it from exerting gravitational forces on a particle inside. Gravitational shielding is not possible. EXERCISES 7.1 Answer the following : (a) You can shield a charge from electrical forces by putting it inside a hollow conductor. Can you shield a body from the gravitational influence of nearby matter by putting it inside a hollow sphere or by some other means ? (b) An astronaut inside a small space ship orbiting around the earth cannot detect gravity. If the space station orbiting around the earth has a large size, can he hope to detect gravity ? (c) If you compare the gravitational force on the earth due to the sun to that due to the moon, you would find that the Sun’s pull is greater than the moon’s pull. (you can check this yourself using the data available in the succeeding exercises). However, the tidal effect of the moon’s pull is greater than the tidal effect of sun. Why ? Reprint 2025-26 142 PHYSICS 7.2 Choose the correct alternative : (a) Acceleration due to gravity increases/decreases with increasing altitude. (b) Acceleration due to gravity increases/decreases with increasing depth (assume the earth to be a sphere of uniform density). (c) Acceleration due to gravity is independent of mass of the earth/mass of the body. (d) The formula –G Mm(1/r2 – 1/r1) is more/less accurate than the formula mg(r2 – r1) for the difference of potential energy between two points r2 and r1 distance away from the centre of the earth. 7.3 Suppose there existed a planet that went around the Sun twice as fast as the earth. What would be its orbital size as compared to that of the earth ? 7.4 Io, one of the satellites of Jupiter, has an orbital period of 1.769 days and the radius of the orbit is 4.22 × 108 m. Show that the mass of Jupiter is about one-thousandth that of the sun. 7.5 Let us assume that our galaxy consists of 2.5 × 1011 stars each of one solar mass. How long will a star at a distance of 50,000 ly from the galactic centre take to complete one revolution ? Take the diameter of the Milky Way to be 105 ly. 7.6 Choose the correct alternative: (a) If the zero of potential energy is at infinity, the total energy of an orbiting satellite is negative of its kinetic/potential energy. (b) The energy required to launch an orbiting satellite out of earth’s gravitational influence is more/less than the energy required to project a stationary object at the same height (as the satellite) out of earth’s influence. 7.7 Does the escape speed of a body from the earth depend on (a) the mass of the body, (b) the location from where it is projected, (c) the direction of projection, (d) the height of the location from where the body is launched? 7.8 A comet orbits the sun in a highly elliptical orbit. Does the comet have a constant (a) linear speed, (b) angular speed, (c) angular momentum, (d) kinetic energy, (e) potential energy, (f) total energy throughout its orbit? Neglect any mass loss of the comet when it comes very close to the Sun. 7.9 Which of the following symptoms is likely to afflict an astronaut in space (a) swollen feet, (b) swollen face, (c) headache, (d) orientational problem. 7.10 In the following two exercises, choose the correct answer from among the given ones: The gravitational intensity at the centre of a hemispherical shell of uniform mass density has the direction indicated by the arrow (see Fig 7.11) (i) a, (ii) b, (iii) c, (iv) 0. Fig. 7.11 7.11 For the above problem, the direction of the gravitational intensity at an arbitrary point P is indicated by the arrow (i) d, (ii) e, (iii) f, (iv) g. 7.12 A rocket is fired from the earth towards the sun. At what distance from the earth’s centre is the gravitational force on the rocket zero ? Mass of the sun = 2×1030 kg, mass of the earth = 6×1024 kg. Neglect the effect of other planets etc. (orbital radius = 1.5 × 1011 m). 7.13 How will you ‘weigh the sun’, that is estimate its mass? The mean orbital radius of the earth around the sun is 1.5 × 108 km. 7.14 A saturn year is 29.5 times the earth year. How far is the saturn from the sun if the earth is 1.50 × 108 km away from the sun ? 7.15 A body weighs 63 N on the surface of the earth. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth ? Reprint 2025-26 GRAVITATION 143 7.16 Assuming the earth to be a sphere of uniform mass density, how much would a body weigh half way down to the centre of the earth if it weighed 250 N on the surface ? 7.17 A rocket is fired vertically with a speed of 5 km s-1 from the earth’s surface. How far from the earth does the rocket go before returning to the earth ? Mass of the earth = 6.0 × 1024 kg; mean radius of the earth = 6.4 × 106 m; G = 6.67 × 10–11 N m2 kg–2. 7.18 The escape speed of a projectile on the earth’s surface is 11.2 km s–1. A body is projected out with thrice this speed. What is the speed of the body far away from the earth? Ignore the presence of the sun and other planets. 7.19 A satellite orbits the earth at a height of 400 km above the surface. How much energy must be expended to rocket the satellite out of the earth’s gravitational influence? Mass of the satellite = 200 kg; mass of the earth = 6.0×1024 kg; radius of the earth = 6.4 × 106 m; G = 6.67 × 10–11 N m2 kg–2. 7.20 Two stars each of one solar mass (= 2×1030 kg) are approaching each other for a head on collision. When they are a distance 109 km, their speeds are negligible. What is the speed with which they collide ? The radius of each star is 104 km. Assume the stars to remain undistorted until they collide. (Use the known value of G). 7.21 Two heavy spheres each of mass 100 kg and radius 0.10 m are placed 1.0 m apart on a horizontal table. What is the gravitational force and potential at the mid point of the line joining the centres of the spheres ? Is an object placed at that point in equilibrium? If so, is the equilibrium stable or unstable ? Reprint 2025-26