Book Insights

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

12.8 A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?

12.9 In accordance with the Bohr’s model, find the quantum number that characterises the earth’s revolution around the sun in an orbit of radius 1.5 × 1011 m with orbital speed 3 × 104 m/s. (Mass of earth 305 = 6.0 × 1024 kg.) Reprint 2025-26 Physics Chapter Thirteen NUCLEI 13.1 INTRODUCTION In the previous chapter, we have learnt that in every atom, the positive charge and mass are densely concentrated at the centre of the atom forming its nucleus. The overall dimensions of a nucleus are much smaller than those of an atom. Experiments on scattering of a-particles demonstrated that the radius of a nucleus was smaller than the radius of an atom by a factor of about 104. This means the volume of a nucleus is about 10–12 times the volume of the atom. In other words, an atom is almost empty. If an atom is enlarged to the size of a classroom, the nucleus would be of the size of pinhead. Nevertheless, the nucleus contains most (more than 99.9%) of the mass of an atom. Does the nucleus have a structure, just as the atom does? If so, what are the constituents of the nucleus? How are these held together? In this chapter, we shall look for answers to such questions. We shall discuss various properties of nuclei such as their size, mass and stability, and also associated nuclear phenomena such as radioactivity, fission and fusion. 13.2 ATOMIC MASSES AND COMPOSITION OF NUCLEUS The mass of an atom is very small, compared to a kilogram; for example, the mass of a carbon atom, 12C, is 1.992647 × 10–26 kg. Kilogram is not 306 a very convenient unit to measure such small quantities. Therefore, a Reprint 2025-26 Nuclei different mass unit is used for expressing atomic masses. This unit is the atomic mass unit (u), defined as 1/12th of the mass of the carbon (12C) atom. According to this definition mass of one 12 C atom 1u = 12 1.992647 × 10 −26 kg = 12 = 1.660539 × 10 −27 kg (13.1) The atomic masses of various elements expressed in atomic mass unit (u) are close to being integral multiples of the mass of a hydrogen atom. There are, however, many striking exceptions to this rule. For example, the atomic mass of chlorine atom is 35.46 u. Accurate measurement of atomic masses is carried out with a mass spectrometer, The measurement of atomic masses reveals the existence of different types of atoms of the same element, which exhibit the same chemical properties, but differ in mass. Such atomic species of the same element differing in mass are called isotopes. (In Greek, isotope means the same place, i.e. they occur in the same place in the periodic table of elements.) It was found that practically every element consists of a mixture of several isotopes. The relative abundance of different isotopes differs from element to element. Chlorine, for example, has two isotopes having masses 34.98 u and 36.98 u, which are nearly integral multiples of the mass of a hydrogen atom. The relative abundances of these isotopes are

12.6 (a) 2.18 × 106 m/s; 1.09 × 106 m/s; 7.27 × 105 m/s (b) 1.52 × 10–16 s; 1.22 × 10–15 s; 4.11 × 10–15 s.

12.6 DE BROGLIE’S EXPLANATION OF BOHR’S SECOND POSTULATE OF QUANTISATION Of all the postulates, Bohr made in his model of the atom, perhaps the most puzzling is his second postulate. It states that the angular momentum of the electron orbiting around the nucleus is quantised (that is, Ln = nh/2p; n = 1, 2, 3 …). Why should the angular momentum have only those values that are integral multiples of h/2p? The French physicist Louis de Broglie explained this puzzle in 1923, ten years after Bohr proposed his model. We studied, in Chapter 11, about the de Broglie’s hypothesis that material particles, such as electrons, also have a wave nature. C. J. Davisson and L. H. Germer later experimentally verified the wave nature of electrons in 1927. Louis de Broglie argued that the electron in its circular orbit, as proposed by Bohr, must be seen as a particle wave. In analogy to waves travelling on a string, particle waves too can lead to standing waves under resonant conditions. From FIGURE 12.8 A standing wave Chapter 14 of Class XI Physics textbook, we know that when is shown on a circular orbit a string is plucked, a vast number of wavelengths are excited. where four de Broglie wavelengths fit into theHowever only those wavelengths survive which have nodes circumference of the orbit. at the ends and form the standing wave in the string. It means that in a string, standing waves are formed when the total distance travelled by a wave down the string and back is one wavelength, two wavelengths, or any integral number of wavelengths. Waves with other wavelengths interfere with themselves upon reflection and their amplitudes quickly drop to zero. For an electron moving in nth circular orbit of radius rn, the total distance is the circumference of the orbit, 3012prn. Thus Reprint 2025-26 Physics 2p rn = nl, n = 1, 2, 3... (12.12) Figure 12.8 illustrates a standing particle wave on a circular orbit for n = 4, i.e., 2prn = 4l, where l is the de Broglie wavelength of the electron moving in nth orbit. From Chapter 11, we have l = h/p, where p is the magnitude of the electron’s momentum. If the speed of the electron is much less than the speed of light, the momentum is mvn. Thus, l = h/ mvn. From Eq. (12.12), we have 2p rn = n h/mvn or m vn rn = nh/2p This is the quantum condition proposed by Bohr for the angular momentum of the electron [Eq. (12.15)]. In Section 12.5, we saw that this equation is the basis of explaining the discrete orbits and energy levels in hydrogen atom. Thus de Broglie hypothesis provided an explanation for Bohr’s second postulate for the quantisation of angular momentum of the orbiting electron. The quantised electron orbits and energy states are due to the wave nature of the electron and only resonant standing waves can persist. Bohr’s model, involving classical trajectory picture (planet-like electron orbiting the nucleus), correctly predicts the gross features of the hydrogenic atoms*, in particular, the frequencies of the radiation emitted or selectively absorbed. This model however has many limitations. Some are: (i) The Bohr model is applicable to hydrogenic atoms. It cannot be extended even to mere two electron atoms such as helium. The analysis of atoms with more than one electron was attempted on the lines of Bohr’s model for hydrogenic atoms but did not meet with any success. Difficulty lies in the fact that each electron interacts not only with the positively charged nucleus but also with all other electrons. The formulation of Bohr model involves electrical force between positively charged nucleus and electron. It does not include the electrical forces between electrons which necessarily appear in multi-electron atoms. (ii) While the Bohr’s model correctly predicts the frequencies of the light emitted by hydrogenic atoms, the model is unable to explain the relative intensities of the frequencies in the spectrum. In emission spectrum of hydrogen, some of the visible frequencies have weak intensity, others strong. Why? Experimental observations depict that some transitions are more favoured than others. Bohr’s model is unable to account for the intensity variations. Bohr’s model presents an elegant picture of an atom and cannot be generalised to complex atoms. For complex atoms we have to use a new and radical theory based on Quantum Mechanics, which provides a more complete picture of the atomic structure. * Hydrogenic atoms are the atoms consisting of a nucleus with positive charge +Ze and a single electron, where Z is the proton number. Examples are hydrogen atom, singly ionised helium, doubly ionised lithium, and so forth. In these 302 atoms more complex electron-electron interactions are nonexistent. Reprint 2025-26 Atoms SUMMARY 1. Atom, as a whole, is electrically neutral and therefore contains equal amount of positive and negative charges. 2. In Thomson’s model, an atom is a spherical cloud of positive charges with electrons embedded in it. 3. In Rutherford’s model, most of the mass of the atom and all its positive charge are concentrated in a tiny nucleus (typically one by ten thousand the size of an atom), and the electrons revolve around it. 4. Rutherford nuclear model has two main difficulties in explaining the structure of atom: (a) It predicts that atoms are unstable because the accelerated electrons revolving around the nucleus must spiral into the nucleus. This contradicts the stability of matter. (b) It cannot explain the characteristic line spectra of atoms of different elements. 5. Atoms of most of the elements are stable and emit characteristic spectrum. The spectrum consists of a set of isolated parallel lines termed as line spectrum. It provides useful information about the atomic structure. 6. To explain the line spectra emitted by atoms, as well as the stability of atoms, Niel’s Bohr proposed a model for hydrogenic (single elctron) atoms. He introduced three postulates and laid the foundations of quantum mechanics: (a) In a hydrogen atom, an electron revolves in certain stable orbits (called stationary orbits) without the emission of radiant energy. (b) The stationary orbits are those for which the angular momentum is some integral multiple of h/2p. (Bohr’s quantisation condition.) That is L = nh/2p, where n is an integer called the principal quantum number. (c) The third postulate states that an electron might make a transition from one of its specified non-radiating orbits to another of lower energy. When it does so, a photon is emitted having energy equal to the energy difference between the initial and final states. The frequency (n) of the emitted photon is then given by hn = Ei – Ef An atom absorbs radiation of the same frequency the atom emits, in which case the electron is transferred to an orbit with a higher value of n. Ei + hn = Ef 7. As a result of the quantisation condition of angular momentum, the electron orbits the nucleus at only specific radii. For a hydrogen atom it is given by  n 2   h  2 4 πε0 rn = 2  m   2 π  e The total energy is also quantised: me 4 E n = − 2 2 2 8n ε0 h = –13.6 eV/n2 The n = 1 state is called ground state. In hydrogen atom the ground state energy is –13.6 eV. Higher values of n correspond to excited states (n > 1). Atoms are excited to these higher states by collisions with other atoms or electrons or by absorption of a photon of right frequency. 303 Reprint 2025-26 Physics 8. de Broglie’s hypothesis that electrons have a wavelength λ = h/mv gave an explanation for Bohr’s quantised orbits by bringing in the wave- particle duality. The orbits correspond to circular standing waves in which the circumference of the orbit equals a whole number of wavelengths. 9. Bohr’s model is applicable only to hydrogenic (single electron) atoms. It cannot be extended to even two electron atoms such as helium. This model is also unable to explain for the relative intensities of the frequencies emitted even by hydrogenic atoms. POINTSPOINTSPOINTSPOINTSPOINTS TOTOTOTOTO PONDERPONDERPONDERPONDERPONDER 1. Both the Thomson’s as well as the Rutherford’s models constitute an unstable system. Thomson’s model is unstable electrostatically, while Rutherford’s model is unstable because of electromagnetic radiation of orbiting electrons. 2. What made Bohr quantise angular momentum (second postulate) and not some other quantity? Note, h has dimensions of angular momentum, and for circular orbits, angular momentum is a very relevant quantity. The second postulate is then so natural! 3. The orbital picture in Bohr’s model of the hydrogen atom was inconsistent with the u quantum mechanics in which Bohr’s orbits are regions where the electron may be found with large probability. 4. Unlike the situation in the solar system, where planet-planet gravitational forces are very small as compared to the gravitational force of the sun on each planet (because the mass of the sun is so much greater than the mass of any of the planets), the electron-electron electric force interaction is comparable in magnitude to the electron- nucleus electrical force, because the charges and distances are of the same order of magnitude. This is the reason why the Bohr’s model with its planet-like electron is not applicable to many electron atoms. 5. Bohr laid the foundation of the quantum theory by postulating specific orbits in which electrons do not radiate. Bohr’s model include only one quantum number n. The new theory called quantum mechanics supports Bohr’s postulate. However in quantum mechanics (more generally accepted), a given energy level may not correspond to just one quantum state. For example, a state is characterised by four quantum numbers (n, l, m, and s), but for a pure Coulomb potential (as in hydrogen atom) the energy depends only on n. 6. In Bohr model, contrary to ordinary classical expectation, the frequency of revolution of an electron in its orbit is not connected to the frequency of spectral line. The later is the difference between two orbital energies divided by h. For transitions between large quantum numbers (n to n – 1, n very large), however, the two coincide as expected. 7. Bohr’s semiclassical model based on some aspects of classical physics and some aspects of modern physics also does not provide a true picture of the simplest hydrogenic atoms. The true picture is quantum mechanical affair which differs from Bohr model in a number of fundamental ways. But then if the Bohr model is not strictly correct, why do we bother about it? The reasons which make Bohr’s model still useful are: Reprint 2025-26 Atoms (i) The model is based on just three postulates but accounts for almost all the general features of the hydrogen spectrum. (ii) The model incorporates many of the concepts we have learnt in classical physics. (iii) The model demonstrates how a theoretical physicist occasionally must quite literally ignore certain problems of approach in hopes of being able to make some predictions. If the predictions of the theory or model agree with experiment, a theoretician then must somehow hope to explain away or rationalise the problems that were ignored along the way. EXERCISES

12.4 BOHR MODEL OF THE HYDROGEN ATOM The model of the atom proposed by Rutherford assumes that the atom, consisting of a central nucleus and revolving electron is stable much like sun-planet system which the model imitates. However, there are some fundamental differences between the two situations. While the planetary system is held by gravitationalforce, the nucleus-electron system being charged NIELS objects, interact by Coulomb’s Law of force. We know that an object which moves in a circle is being constantly accelerated – the acceleration being centripetal in nature. According to classical HENRIK electromagnetic theory, an accelerating charged particle emits radiation in the form of electromagnetic waves. The energy of an accelerating electron should therefore, Niels Henrik David Bohrcontinuously decrease. The electron would spiral (1885 – 1962) Danish DAVID inward and eventually fall into the nucleus (Fig. 12.6). physicist who explained the Thus, such an atom can not be stable. Further, spectrum of hydrogen atom according to the classical electromagnetic theory, the based on quantum ideas. BOHR He gave a theory of nuclearfrequency of the electromagnetic waves emitted by the fission based on the liquid- revolving electrons is equal to the frequency of drop model of nucleus. revolution. As the electrons spiral inwards, their angular Bohr contributed to the (1885 velocities and hence their frequencies would change clarification of conceptual – continuously, and so will the frequency of the light problems in quantum emitted. Thus, they would emit a continuous spectrum, mechanics, in particular by in contradiction to the line spectrum actually observed. proposing the comple- 1962) Clearly Rutherford model tells only a part of the story mentary principle. implying that the classical ideas are not sufficient to explain the atomic structure. 297 Reprint 2025-26 Physics FIGURE 12.6 An accelerated atomic electron must spiral into the nucleus as it loses energy. Example 12.4 According to the classical electromagnetic theory, calculate the initial frequency of the light emitted by the electron revolving around a proton in hydrogen atom. Solution From Example 12.3 we know that velocity of electron moving around a proton in hydrogen atom in an orbit of radius 5.3 × 10–11 m is 2.2 × 10–6 m/s. Thus, the frequency of the electron moving around the proton is v 2.2 × 10 6 m s −1 ν = = −11 2 π r 2 π 5.3 × 10 m ( ) 12.4 » 6.6 × 1015 Hz. According to the classical electromagnetic theory we know that the frequency of the electromagnetic waves emitted by the revolving electrons is equal to the frequency of its revolution around the nucleus. EXAMPLE Thus the initial frequency of the light emitted is 6.6 × 1015 Hz. It was Niels Bohr (1885 – 1962) who made certain modifications in this model by adding the ideas of the newly developing quantum hypothesis. Niels Bohr studied in Rutherford’s laboratory for several months in 1912 and he was convinced about the validity of Rutherford nuclear model. Faced with the dilemma as discussed above, Bohr, in 1913, concluded that in spite of the success of electromagnetic theory in explaining large-scale phenomena, it could not be applied to the processes at the atomic scale. It became clear that a fairly radical departure from the established principles of classical mechanics and electromagnetism would be needed to understand the structure of atoms and the relation of atomic structure to atomic spectra. Bohr combined classical and early quantum concepts and gave his theory in the form of three postulates. These are : (i) Bohr’s first postulate was that an electron in an atom could revolve in certain stable orbits without the emission of radiant energy, contrary to the predictions of electromagnetic theory. According to 298 this postulate, each atom has certain definite stable states in which it Reprint 2025-26 Atoms can exist, and each possible state has definite total energy. These are called the stationary states of the atom. (ii) Bohr’s second postulate defines these stable orbits. This postulate states that the electron revolves around the nucleus only in those orbits for which the angular momentum is some integral multiple of h/2π where h is the Planck’s constant (= 6.6 × 10–34 J s). Thus the angular momentum (L) of the orbiting electron is quantised. That is L = nh/2π (12.5) (iii) Bohr’s third postulate incorporated into atomic theory the early quantum concepts that had been developed by Planck and Einstein. It states that an electron might make a transition from one of its specified non-radiating orbits to another of lower energy. When it does so, a photon is emitted having energy equal to the energy difference between the initial and final states. The frequency of the emitted photon is then given by hν = Ei – Ef (12.6) where Ei and Ef are the energies of the initial and final states and Ei > Ef. For a hydrogen atom, Eq. (12.4) gives the expression to determine the energies of different energy states. But then this equation requires the radius r of the electron orbit. To calculate r, Bohr’s second postulate about the angular momentum of the electron–the quantisation condition – is used. The radius of nth possible orbit thus found is  n 2   h  2 4 πε0 rn =    2 (12.7)  m   2 π  e The total energy of the electron in the stationary states of the hydrogen atom can be obtained by substituting the value of orbital radius in Eq. (12.4) as  e 2   m  2 π  2  e 2  E n = −   2      8 πε0   n  h   4 πε0  me 4 or E n = − 2 2 2 (12.8) 8n ε0 h Substituting values, Eq. (12.8) yields −18 2.18 × 10 E n = − 2 J (12.9) n Atomic energies are often expressed in electron volts (eV) rather than joules. Since 1 eV = 1.6 × 10–19 J, Eq. (12.9) can be rewritten as 13.6 E n = − 2 eV (12.10) n The negative sign of the total energy of an electron moving in an orbit means that the electron is bound with the nucleus. Energy will thus be required to remove the electron from the hydrogen atom to a distance 299 infinitely far away from its nucleus (or proton in hydrogen atom). Reprint 2025-26 Physics 12.4.1 Energy levels The energy of an atom is the least (largest negative value) when its electron is revolving in an orbit closest to the nucleus i.e., the one for which n = 1. For n = 2, 3, ... the absolute value of the energy E is smaller, hence the energy is progressively larger in the outer orbits. The lowest state of the atom, called the ground state, is that of the lowest energy, with the electron revolving in the orbit of smallest radius, the Bohr radius, a 0. The energy of this state (n = 1), E1 is –13.6 eV. Therefore, the minimum energy required to free the electron from the ground state of the hydrogen atom is 13.6 eV. It is called the ionisation energy of the hydrogen atom. This prediction of the Bohr’s model is in excellent agreement with the experimental value of ionisation energy. At room temperature, most of the hydrogen atoms are in ground state. When a hydrogen atom receives energy by processes such as electron collisions, the atom may acquire sufficient energy to raise the electron to higher energy states. The atom is then said to be in an excited state. From Eq. (12.10), for n = 2; the energy E2 is –3.40 eV. It means that the energy required to excite an electron in hydrogen atom to its first excited state, is an FIGURE 12.7 The energy level energy equal to E2 – E1 = –3.40 eV – (–13.6) eV = 10.2 eV. diagram for the hydrogen atom. Similarly, E3 = –1.51 eV and E3 – E1 = 12.09 eV, or to exciteThe electron in a hydrogen atom the hydrogen atom from its ground state (n = 1) to second at room temperature spends most of its time in the ground excited state (n = 3), 12.09 eV energy is required, and so state. To ionise a hydrogen on. From these excited states the electron can then fall back atom an electron from the to a state of lower energy, emitting a photon in the process. ground state, 13.6 eV of energy Thus, as the excitation of hydrogen atom increases (that is must be supplied. (The horizontal as n increases) the value of minimum energy required to lines specify the presence of free the electron from the excited atom decreases. allowed energy states.) The energy level diagram* for the stationary states of a hydrogen atom, computed from Eq. (12.10), is given in Fig. 12.7. The principal quantum number n labels the stationary states in the ascending order of energy. In this diagram, the highest energy state corresponds to n =¥ in Eq, (12.10) and has an energy of 0 eV. This is the energy of the atom when the electron is completely removed (r = ¥) from the nucleus and is at rest. Observe how the energies of the excited states come closer and closer together as n increases. 12.5 THE LINE SPECTRA OF THE HYDROGEN ATOM According to the third postulate of Bohr’s model, when an atom makes a transition from the higher energy state with quantum number ni to the lower energy state with quantum number nf (nf < ni), the difference of energy is carried away by a photon of frequency nif such that * An electron can have any total energy above E = 0 eV. In such situations the 300 electron is free. Thus there is a continuum of energy states above E = 0 eV, as shown in Fig. 12.7. Reprint 2025-26 Atoms hvif = Eni – Enf (12.11) Since both nf and ni are integers, this immediately shows that in transitions between different atomic levels, light is radiated in various discrete frequencies. The various lines in the atomic spectra are produced when electrons jump from higher energy state to a lower energy state and photons are emitted. These spectral lines are called emission lines. But when an atom absorbs a photon that has precisely the same energy needed by the electron in a lower energy state to make transitions to a higher energy state, the process is called absorption. Thus if photons with a continuous range of frequencies pass through a rarefied gas and then are analysed with a spectrometer, a series of dark spectral absorption lines appear in the continuous spectrum. The dark lines indicate the frequencies that have been absorbed by the atoms of the gas. The explanation of the hydrogen atom spectrum provided by Bohr’s model was a brilliant achievement, which greatly stimulated progress towards the modern quantum theory. In 1922, Bohr was awarded Nobel Prize in Physics.

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.

9.19 A screen is placed 90cm from an object. The image of the object on the screen is formed by a convex lens at two different locations separated by 20cm. Determine the focal length of the lens.

9.6 A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. The angle of minimum deviation is measured to be 40°. What is the refractive index of the material of the prism? The refracting angle of the prism is 60°. If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of light.

9.21 At what angle should a ray of light be incident on the face of a prism of refracting angle 60° so that it just suffers total internal reflection at the other face? The refractive index of the material of the prism is 1.524.

9.7 Double-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 20cm?

9.6 REFRACTION THROUGH A PRISM Figure 9.21 shows the passage of light through a triangular prism ABC. The angles of incidence and refraction at the first face AB are i and r1, while the angle of incidence (from glass to air) at the second face AC is r2 and the angle of refraction or emergence e. The angle between the emergent ray RS and the direction of the incident ray PQ is called the angle of deviation, d. In the quadrilateral AQNR, two of the angles (at the vertices Q and R) are right angles. Therefore, the sum of the other angles of the quadrilateral is 180°. FIGURE 9.21 A ray of light passing through a triangular glass prism. ÐA + ÐQNR = 180° From the triangle QNR, r1 + r2 + ÐQNR = 180° Comparing these two equations, we get r1 + r2 = A (9.34) The total deviation d is the sum of deviations at the two faces, d = (i – r1 ) + (e – r2 ) that is, d = i + e – A (9.35) Thus, the angle of deviation depends on the angle of incidence. A plot between the angle of deviation and angle of incidence is shown in Fig. 9.22. You can see that, in general, any given value of d, except for i = e, corresponds to two values i and hence of e. This, in fact, is expected from the symmetry of i and e in Eq. (9.35), i.e., d remains the same if i 239 Reprint 2025-26 Physics and e are interchanged. Physically, this is related to the fact that the path of ray in Fig. 9.21 can be traced back, resulting in the same angle of deviation. At the minimum deviation Dm, the refracted ray inside the prism becomes parallel to its base. We have d = Dm, i = e which implies r1 = r2. Equation (9.34) gives A 2r = A or r = (9.36) 2 In the same way, Eq. (9.35) gives Dm = 2i – A, or i = (A + Dm)/2 (9.37) The refractive index of the prism is FIGURE 9.22 Plot of angle of deviation (d) n 2 sin[( A + D m )/2] versus angle of incidence (i) for a n 21 = = (9.38) triangular prism. n1 sin[ A /2] The angles A and Dm can be measured experimentally. Equation (9.38) thus provides a method of determining refractive index of the material of the prism. For a small angle prism, i.e., a thin prism, Dm is also very small, and we get sin[( A + Dm )/2] A + Dm ) /2 ≃ ( n 21 = sin[ A /2] A /2 Dm = (n21–1)A It implies that, thin prisms do not deviate light much. 9.7 OPTICAL INSTRUMENTS A number of optical devices and instruments have been designed utilising reflecting and refracting properties of mirrors, lenses and prisms. Periscope, kaleidoscope, binoculars, telescopes, microscopes are some examples of optical devices and instruments that are in common use. Our eye is, of course, one of the most important optical device the nature has endowed us with. We have already studied about the human eye in Class X. We now go on to describe the principles of working of the microscope and the telescope. 9.7.1 The microscope A simple magnifier or microscope is a converging lens of small focal length (Fig. 9.23). In order to use such a lens as a microscope, the lens is held near the object, one focal length away or less, and the eye is positioned close to the lens on the other side. The idea is to get an erect, magnified and virtual image of the object at a distance so that it can be viewed comfortably, i.e., at 25 cm or more. If the object is at a distance f, the 240 image is at infinity. However, if the object is at a distance slightly less Reprint 2025-26 Ray Optics and Optical Instruments FIGURE 9.23 A simple microscope; (a) the magnifying lens is located such that the image is at the near point, (b) the angle subtanded by the object, is the same as that at the near point, and (c) the object near the focal point of the lens; the image is far off but closer than infinity. than the focal length of the lens, the image is virtual and closer than infinity. Although the closest comfortable distance for viewing the image is when it is at the near point (distance D @ 25 cm), it causes some strain on the eye. Therefore, the image formed at infinity is often considered most suitable for viewing by the relaxed eye. We show both cases, the first in Fig. 9.23(a), and the second in Fig. 9.23(b) and (c). The linear magnification m, for the image formed at the near point D, by a simple microscope can be obtained by using the relation 241 Reprint 2025-26 Physics v  1 1   v  m = = v – 1 – u  v f =  f  Now according to our sign convention, v is negative, and is equal in magnitude to D. Thus, the magnification is  D  m = 1 + (9.39)  f  Since D is about 25 cm, to have a magnification of six, one needs a convex lens of focal length, f = 5 cm. Note that m = h¢/h where h is the size of the object and h¢ the size of the image. This is also the ratio of the angle subtended by the image to that subtended by the object, if placed at D for comfortable viewing. (Note that this is not the angle actually subtended by the object at the eye, which is h/u.) What a single-lens simple magnifier achieves is that it allows the object to be brought closer to the eye than D. We will now find the magnification when the image is at infinity. In this case we will have to obtained the angular magnification. Suppose the object has a height h. The maximum angle it can subtend, and be clearly visible (without a lens), is when it is at the near point, i.e., a distance D. The angle subtended is then given by  h  tan θo =  D » qo (9.40) We now find the angle subtended at the eye by the image when the object is at u. From the relations h ′ v = m = h u we have the angle subtended by the image h ′ h v h tan θi = = ⋅ = »q. The angle subtended by the object, when it −v −v u −u is at u = –f.  h  θi = (9.41)  f  as is clear from Fig. 9.23(c). The angular magnification is, therefore  θi  D m = (9.42)  θo = f This is one less than the magnification when the image is at the near point, Eq. (9.39), but the viewing is more comfortable and the difference in magnification is usually small. In subsequent discussions of optical instruments (microscope and telescope) we shall assume the image to be 242 at infinity. Reprint 2025-26 Ray Optics and Optical Instruments FIGURE 9.24 Ray diagram for the formation of image by a compound microscope. The A simple microscope has a limited maximum magnification (£ 9) for realistic focal lengths. For much larger magnifications, one uses two lenses, one compounding the effect of the other. This is known as a compound world’s microscope. A schematic diagram of a compound microscope is shown in Fig. 9.24. The lens nearest the object, called the objective, forms a largestreal, inverted, magnified image of the object. This serves as the object for the second lens, the eyepiece, which functions essentially like a simple microscope or magnifier, produces the final image, which is enlarged optical and virtual. The first inverted image is thus near (at or within) the focal plane of the eyepiece, at a distance appropriate for final image formation at infinity, or a little closer for image formation at the near point. Clearly,the final image is inverted with respect to the original object. http://astro.nineplanets.org/bigeyes.html telescopes We now obtain the magnification due to a compound microscope. The ray diagram of Fig. 9.24 shows that the (linear) magnification due to the objective, namely h¢/h, equals h ′ L m O = = (9.43) h f o where we have used the result  h   h ′  tanβ =  f o =  L  Here h¢ is the size of the first image, the object size being h and fo being the focal length of the objective. The first image is formed near the focal point of the eyepiece. The distance L, i.e., the distance between the second focal point of the objective and the first focal point of the eyepiece (focal length fe) is called the tube length of the compound microscope. 243 Reprint 2025-26 Physics As the first inverted image is near the focal point of the eyepiece, we use the result from the discussion above for the simple microscope to obtain the (angular) magnification me due to it [Eq. (9.39)], when the final image is formed at the near point, is  D  m e =  1 +  [9.44(a)]  f e  When the final image is formed at infinity, the angular magnification due to the eyepiece [Eq. (9.42)] is me = (D/fe ) [9.44(b)] Thus, the total magnification [(according to Eq. (9.33)], when the image is formed at infinity, is L   D  m = m om e =    (9.45)  f o   f e  Clearly, to achieve a large magnification of a small object (hence the name microscope), the objective and eyepiece should have small focal lengths. In practice, it is difficult to make the focal length much smaller than 1 cm. Also large lenses are required to make L large. For example, with an objective with fo = 1.0 cm, and an eyepiece with focal length fe = 2.0 cm, and a tube length of 20 cm, the magnification is L   D  m = m o m e =     f o   f e  20 25 250 1 2 Various other factors such as illumination of the object, contribute to the quality and visibility of the image. In modern microscopes, multi- component lenses are used for both the objective and the eyepiece to improve image quality by minimising various optical aberrations (defects) in lenses. 9.7.2 Telescope The telescope is used to provide angular magnification of distant objects (Fig. 9.25). It also has an objective and an eyepiece. But here, the objective has a large focal length and a much larger aperture than the eyepiece. Light from a distant object enters the objective and a real image is formed in the tube at its second focal point. The eyepiece magnifies this image producing a final inverted image. The magnifying power m is the ratio of the angle b subtended at the eye by the final image to the angle a which the object subtends at the lens or the eye. Hence h f o f o m . (9.46) 244 f e h f e Reprint 2025-26 Ray Optics and Optical Instruments In this case, the length of the telescope tube is fo + fe. Terrestrial telescopes have, in addition, a pair of inverting lenses to make the final image erect. Refracting telescopes can be used both for terrestrial and astronomical observations. For example, consider a telescope whose objective has a focal length of 100 cm and the eyepiece a focal length of 1 cm. The magnifying power of this telescope is m = 100/1 = 100. Let us consider a pair of stars of actual separation 1¢ (one minute of arc). The stars appear as though they are separated by an angle of 100 × 1¢ = 100¢ =1.67°. FIGURE 9.25 A refracting telescope. The main considerations with an astronomical telescope are its light gathering power and its resolution or resolving power. The former clearly depends on the area of the objective. With larger diameters, fainter objects can be observed. The resolving power, or the ability to observe two objects distinctly, which are in very nearly the same direction, also depends on the diameter of the objective. So, the desirable aim in optical telescopes is to make them with objective of large diameter. The largest lens objective in use has a diameter of 40 inch (~1.02 m). It is at the Yerkes Observatory in Wisconsin, USA. Such big lenses tend to be very heavy and therefore, difficult to make and support by their edges. Further, it is rather difficult and expensive to make such large sized lenses which form images that are free from any kind of chromatic aberration and distortions. For these reasons, modern telescopes use a concave mirror rather than a lens for the objective. Telescopes with mirror objectives are called reflecting telescopes. There is no chromatic aberration in a mirror. Mechanical support is much less of a problem since a mirror weighs much less than a lens of equivalent optical quality, and can be supported over its entire back surface, not just over its rim. One obvious problem with a reflecting telescope is that the objective mirror focusses light inside 245 Reprint 2025-26 Physics FIGURE 9.26 Schematic diagram of a reflecting telescope (Cassegrain). the telescope tube. One must have an eyepiece and the observer right there, obstructing some light (depending on the size of the observer cage). This is what is done in the very large 200 inch (~5.08 m) diameters, Mt. Palomar telescope, California. The viewer sits near the focal point of the mirror, in a small cage. Another solution to the problem is to deflect the light being focussed by another mirror. One such arrangement using a convex secondary mirror to focus the incident light, which now passes through a hole in the objective primary mirror, is shown in Fig. 9.26. This is known as a Cassegrain telescope, after its inventor. It has the advantages of a large focal length in a short telescope. The largest telescope in India is in Kavalur, Tamil Nadu. It is a 2.34 m diameter reflecting telescope (Cassegrain). It was ground, polished, set up, and is being used by the Indian Institute of Astrophysics, Bangalore. The largest reflecting telescopes in the world are the pair of Keck telescopes in Hawaii, USA, with a reflector of 10 metre in diameter. SUMMARY 1. Reflection is governed by the equation Ði = Ðr¢ and refraction by the Snell’s law, sini/sinr = n, where the incident ray, reflected ray, refracted ray and normal lie in the same plane. Angles of incidence, reflection and refraction are i, r ¢ and r, respectively. 2. The critical angle of incidence ic for a ray incident from a denser to rarer medium, is that angle for which the angle of refraction is 90°. For i > ic, total internal reflection occurs. Multiple internal reflections in diamond (ic @ 24.4°), totally reflecting prisms and mirage, are some examples of total internal reflection. Optical fibres consist of glass fibres coated with a thin layer of material of lower refractive index. Light incident at an angle at one end comes out at the other, after multiple internal reflections, even if the fibre is bent. Reprint 2025-26 Ray Optics and Optical Instruments 3. Cartesian sign convention: Distances measured in the same direction as the incident light are positive; those measured in the opposite direction are negative. All distances are measured from the pole/optic centre of the mirror/lens on the principal axis. The heights measured upwards above x-axis and normal to the principal axis of the mirror/ lens are taken as positive. The heights measured downwards are taken as negative. 4. Mirror equation: 1 1 1 + = v u f where u and v are object and image distances, respectively and f is the focal length of the mirror. f is (approximately) half the radius of curvature R. f is negative for concave mirror; f is positive for a convex mirror. 5. For a prism of the angle A, of refractive index n 2 placed in a medium of refractive index n1, n 2 sin ( A + D m ) / 2  n 21 = = n 1 sin ( A / 2 ) where Dm is the angle of minimum deviation. 6. For refraction through a spherical interface (from medium 1 to 2 of refractive index n1 and n 2, respectively) n 2 n 1 n 2 − n 1 − = v u R Thin lens formula 1 1 1 − = v u f Lens maker’s formula 1 ( n 2 − n1 )  1 1  = − f n1  R1 R 2  R1 and R2 are the radii of curvature of the lens surfaces. f is positive for a converging lens; f is negative for a diverging lens. The power of a lens P = 1/f. The SI unit for power of a lens is dioptre (D): 1 D = 1 m–1. If several thin lenses of focal length f1, f2, f3,.. are in contact, the effective focal length of their combination, is given by 1 1 1 1 = + + + … f f 1 f 2 f 3 The total power of a combination of several lenses is P = P1 + P2 + P3 + … 7. Dispersion is the splitting of light into its constituent colour. 247 Reprint 2025-26 Physics 8. Magnifying power m of a simple microscope is given by m = 1 + (D/f), where D = 25 cm is the least distance of distinct vision and f is the focal length of the convex lens. If the image is at infinity, m = D/f. For a compound microscope, the magnifying power is given by m = me × m0 where me = 1 + (D/fe), is the magnification due to the eyepiece and mo is the magnification produced by the objective. Approximately, L D m = × f o f e where fo and fe are the focal lengths of the objective and eyepiece, respectively, and L is the distance between their focal points. 9. Magnifying power m of a telescope is the ratio of the angle b subtended at the eye by the image to the angle a subtended at the eye by the object. β f o m = = α f e where f0 and fe are the focal lengths of the objective and eyepiece, respectively. POINTS TO PONDER 1. The laws of reflection and refraction are true for all surfaces and pairs of media at the point of the incidence. 2. The real image of an object placed between f and 2f from a convex lens can be seen on a screen placed at the image location. If the screen is removed, is the image still there? This question puzzles many, because it is difficult to reconcile ourselves with an image suspended in air without a screen. But the image does exist. Rays from a given point on the object are converging to an image point in space and diverging away. The screen simply diffuses these rays, some of which reach our eye and we see the image. This can be seen by the images formed in air during a laser show. 3. Image formation needs regular reflection/refraction. In principle, all rays from a given point should reach the same image point. This is why you do not see your image by an irregular reflecting object, say the page of a book. 4. Thick lenses give coloured images due to dispersion. The variety in colour of objects we see around us is due to the constituent colours of the light incident on them. A monochromatic light may produce an entirely different perception about the colours on an object as seen in white light. 5. For a simple microscope, the angular size of the object equals the angular size of the image. Yet it offers magnification because we can keep the small object much closer to the eye than 25 cm and hence have it subtend a large angle. The image is at 25 cm which we can see. Without the microscope, you would need to keep the small object at 25 cm which would subtend a very small angle. Reprint 2025-26 Ray Optics and Optical Instruments EXERCISES

9.4 Figures 9.27(a) and (b) show refraction of a ray in air incident at 60° with the normal to a glass-air and water-air interface, respectively. Predict the angle of refraction in glass when the angle of incidence in water is 45° with the normal to a water-glass interface [Fig. 9.27(c)]. FIGURE 9.27

9.1 A small candle, 2.5 cm in size is placed at 27 cm in front of a concave mirror of radius of curvature 36 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved?

9.31 Figure 9.30 shows an equiconvex lens (of refractive index 1.50) in contact with a liquid layer on top of a plane mirror. A small needle with its tip on the principal axis is moved along the axis until its inverted image is found at the position of the needle. The distance of the needle from the lens is measured to be 45.0cm. The liquid is removed and the experiment is repeated. The new distance is measured to be 30.0cm. What is the refractive index of the liquid? FIGURE 9.30 253 Reprint 2025-26 Physics Notes Reprint 2025-26 Wave Optics Chapter Ten WAVE OPTICS 10.1 INTRODUCTION In 1637 Descartes gave the corpuscular model of light and derived Snell’s law. It explained the laws of reflection and refraction of light at an interface. The corpuscular model predicted that if the ray of light (on refraction) bends towards the normal then the speed of light would be greater in the second medium. This corpuscular model of light was further developed by Isaac Newton in his famous book entitled OPTICKS and because of the tremendous popularity of this book, the corpuscular model is very often attributed to Newton. In 1678, the Dutch physicist Christiaan Huygens put forward the wave theory of light – it is this wave model of light that we will discuss in this chapter. As we will see, the wave model could satisfactorily explain the phenomena of reflection and refraction; however, it predicted that on refraction if the wave bends towards the normal then the speed of light would be less in the second medium. This is in contradiction to the prediction made by using the corpuscular model of light. It was much later confirmed by experiments where it was shown that the speed of light in water is less than the speed in air confirming the prediction of the wave model; Foucault carried out this experiment in 1850. The wave theory was not readily accepted primarily because of Newton’s authority and also because light could travel through vacuum255 Reprint 2025-26 Physics and it was felt that a wave would always require a medium to propagate from one point to the other. However, when Thomas Young performed his famous interference experiment in 1801, it was firmly established that light is indeed a wave phenomenon. The wavelength of visible light was measured and found to be extremely small; for example, the wavelength of yellow light is about 0.6 mm. Because of the smallness of the wavelength of visible light (in comparison to the dimensions of typical mirrors and lenses), light can be assumed to approximately travel in straight lines. This is the field of geometrical optics, which we had discussed in the previous chapter. Indeed, the branch of optics in which one completely neglects the finiteness of the wavelength is called geometrical optics and a ray is defined as the path of energy propagation in the limit of wavelength tending to zero. After the interference experiment of Young in 1801, for the next 40 years or so, many experiments were carried out involving the interference and diffraction of lightwaves; these experiments could only be satisfactorily explained by assuming a wave model of light. Thus, around the middle of the nineteenth century, the wave theory seemed to be very well established. The only major difficulty was that since it was thought that a wave required a medium for its propagation, how could light waves propagate through vacuum. This was explained when Maxwell put forward his famous electromagnetic theory of light. Maxwell had developed a set of equations describing the laws of electricity and magnetism and using these equations he derived what is known as the wave equation from which he predicted the existence of electromagnetic waves*. From the wave equation, Maxwell could calculate the speed of electromagnetic waves in free space and he found that the theoretical value was very close to the measured value of speed of light. From this, he propounded that light must be an electromagnetic wave. Thus, according to Maxwell, light waves are associated with changing electric and magnetic fields; changing electric field produces a time and space varying magnetic field and a changing magnetic field produces a time and space varying electric field. The changing electric and magnetic fields result in the propagation of electromagnetic waves (or light waves) even in vacuum. In this chapter we will first discuss the original formulation of the Huygens principle and derive the laws of reflection and refraction. In Sections 10.4 and 10.5, we will discuss the phenomenon of interference which is based on the principle of superposition. In Section 10.6 we will discuss the phenomenon of diffraction which is based on Huygens- Fresnel principle. Finally in Section 10.7 we will discuss the phenomenon of polarisation which is based on the fact that the light waves are transverse electromagnetic waves. * Maxwell had predicted the existence of electromagnetic waves around 1855; it was much later (around 1890) that Heinrich Hertz produced radiowaves in the laboratory. J.C. Bose and G. Marconi made practical applications of the Hertzian 256 waves Reprint 2025-26 Wave Optics

9.23 (a) At what distance should the lens be held from the card sheet in Exercise 9.22 in order to view the squares distinctly with the maximum possible magnifying power? (b) What is the magnification in this case? (c) Is the magnification equal to the magnifying power in this case? Explain.

9.26 Assume microscope in normal use i.e., image at 25 cm. Angular magnification of the eye-piece 25 =  1  6 5 Magnification of the objective 30 =  5 6 1 1 1 − = 5u O u O 1.25 which gives uO= –1.5 cm; v0= 7.5 cm. |ue| (25/6) cm = 4.17 cm. The separation between the objective and the eye-piece should be (7.5 + 4.17) cm = 11.67 cm. Further the object should be placed 1.5 cm from the objective to obtain the desired magnification.

9.9 An object of size 3.0cm is placed 14cm in front of a concave lens of focal length 21cm. Describe the image produced by the lens. What happens if the object is moved further away from the lens? 249 Reprint 2025-26 Physics 9.10 What is the focal length of a convex lens of focal length 30cm in contact with a concave lens of focal length 20cm? Is the system a converging or a diverging lens? Ignore thickness of the lenses. 9.11 A compound microscope consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15cm. How far from the objective should an object be placed in order to obtain the final image at (a) the least distance of distinct vision (25cm), and (b) at infinity? What is the magnifying power of the microscope in each case? 9.12 A person with a normal near point (25 cm) using a compound microscope with objective of focal length 8.0 mm and an eyepiece of focal length 2.5cm can bring an object placed at 9.0mm from the objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope, 9.13 A small telescope has an objective lens of focal length 144cm and an eyepiece of focal length 6.0cm. What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece? 9.14 (a) A giant refracting telescope at an observatory has an objective lens of focal length 15m. If an eyepiece of focal length 1.0cm is used, what is the angular magnification of the telescope? (b) If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is 3.48 × 106m, and the radius of lunar orbit is 3.8 × 108m. 9.15 Use the mirror equation to deduce that: (a) an object placed between f and 2f of a concave mirror produces a real image beyond 2f. (b) a convex mirror always produces a virtual image independent of the location of the object. (c) the virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole. (d) an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image. [Note: This exercise helps you deduce algebraically properties of images that one obtains from explicit ray diagrams.] 9.16 A small pin fixed on a table top is viewed from above from a distance of 50cm. By what distance would the pin appear to be raised if it is viewed from the same point through a 15cm thick glass slab held parallel to the table? Refractive index of glass = 1.5. Does the answer depend on the location of the slab? 9.17 (a) Figure 9.28 shows a cross-section of a ‘light pipe’ made of a glass fibre of refractive index 1.68. The outer covering of the pipe is made of a material of refractive index 1.44. What is the range of the angles of the incident rays with the axis of the pipe for which total reflections inside the pipe take place, as shown in the figure. FIGURE 9.28 Reprint 2025-26 Ray Optics and Optical Instruments (b) What is the answer if there is no outer covering of the pipe?

9.20 (a) Determine the ‘effective focal length’ of the combination of the two lenses in Exercise 9.10, if they are placed 8.0cm apart with their principal axes coincident. Does the answer depend on which side of the combination a beam of parallel light is incident? Is the notion of effective focal length of this system useful at all? (b) An object 1.5 cm in size is placed on the side of the convex lens in the arrangement (a) above. The distance between the object and the convex lens is 40 cm. Determine the magnification produced by the two-lens system, and the size of the image.

9.4 TOTAL INTERNAL REFLECTION When light travels from an optically denser medium to a rarer medium at the interface, it is partly reflected back into the same medium and partly refracted to the second medium. This reflection is called the internal reflection. When a ray of light enters from a denser medium to a rarer medium, it bends away from the normal, for example, the ray AO1 B in Fig. 9.11. The incident ray AO1 is partially reflected (O1C) and partially transmitted (O1B) or refracted, the angle of refraction (r) being larger than the angle of 229incidence (i). As the angle of incidence increases, so does the angle of Reprint 2025-26 Physics refraction, till for the ray AO3, the angle of refraction is p/2. The refracted ray is bent so much away from the normal that it grazes the surface at the interface between the two media. This is shown by the ray AO3 D in Fig. 9.11. If the angle of incidence is increased still further (e.g., the ray AO4), refraction is not possible, and the incident ray is totally reflected. This is called total internal reflection. When light gets reflected by a surface, normally some FIGURE 9.11 Refraction and internal reflection fraction of it gets transmitted. The of rays from a point A in the denser medium reflected ray, therefore, is always less (water) incident at different angles at the interface intense than the incident ray, howsoever with a rarer medium (air). smooth the reflecting surface may be. In total internal reflection, on the other hand, no transmission of light takes place. The angle of incidence corresponding to an angle of refraction 90°, say ÐAO3N, is called the critical angle (ic ) for the given pair of media. We see from Snell’s law [Eq. (9.10)] that if the relative refractive index of the refracting medium is less than one then, since the maximum value of sin r is unity, there is an upper limit to the value of sin i for which the law can be satisfied, that is, i = ic such that sin ic = n 21 (9.12) For values of i larger than ic, Snell’s law of refraction cannot be satisfied, and hence no refraction is possible. The refractive index of denser medium 1 with respect to rarer medium 2 will be n12 = 1/sinic. Some typical critical angles are listed in Table 9.1. TABLE 9.1 CRITICAL ANGLE OF SOME TRANSPARENT MEDIA WITH RESPECT TO AIR Substance medium Refractive index Critical angle Water 1.33 48.75 Crown glass 1.52 41.14 Dense flint glass 1.62 37.31 Diamond 2.42 24.41 A demonstration for total internal reflection All optical phenomena can be demonstrated very easily with the use of a laser torch or pointer, which is easily available nowadays. Take a glass beaker with clear water in it. Add a few drops of milk or any other suspension to water and stir so that water becomes a little turbid. Take a laser pointer and shine its beam through the turbid water. You will find that the path of the beam inside the water shines brightly. 230 Reprint 2025-26 Ray Optics and Optical Instruments Shine the beam from below the beaker such that it strikes at the upper water surface at the other end. Do you find that it undergoes partial reflection (which is seen as a spot on the table below) and partial refraction [which comes out in the air and is seen as a spot on the roof; Fig. 9.12(a)]? Now direct the laser beam from one side of the beaker such that it strikes the upper surface of water more obliquely [Fig. 9.12(b)]. Adjust the direction of laser beam until you find the angle for which the refraction above the water surface is totally absent and the beam is totally reflected back to water. This is total internal reflection at its simplest. Pour this water in a long test tube and shine the laser light from top, as shown in Fig. 9.12(c). Adjust the direction of the laser beam such that it is totally internally reflected every time it strikes the walls of the tube. This is similar to what happens in optical fibres. Take care not to look into the laser beam directly and not to point it at anybody’s face. 9.4.19.4.19.4.19.4.19.4.1 TotalTotalTotalTotalTotal internalinternalinternalinternalinternal reflectionreflectionreflectionreflectionreflection ininininin naturenaturenaturenaturenature andandandandand itsitsitsitsits technologicaltechnologicaltechnologicaltechnologicaltechnological applicationsapplicationsapplicationsapplicationsapplications (i) Prism: Prisms designed to bend light by 90° or by 180° make use of total internal reflection [Fig. 9.13(a) and (b)]. Such a prism is also used to invert images without changing their size [Fig. 9.13(c)]. In the first two cases, the critical angle ic for the material of the prism FIGUREFIGUREFIGUREFIGUREFIGURE 9.129.129.129.129.12 must be less than 45°. We see from Table 9.1 that this is true for both Observing total internal crown glass and dense flint glass. reflection in water with (ii) Optical fibres: Nowadays optical fibres are extensively used for a laser beam (refraction transmitting audio and video signals through long distances. Optical due to glass of beaker fibres too make use of the phenomenon of total internal reflection. neglected being very Optical fibres are fabricated with high quality composite glass/quartz thin). fibres. Each fibre consists of a core and cladding. The refractive index of the material of the core is higher than that of the cladding. When a signal in the form of light is directed at one end of the fibre at a suitable angle, it undergoes repeated total internal reflections along the length of the fibre and finally comes out at the other end (Fig. 9.14). Since light undergoes total internal reflection at each stage, there is no appreciable loss in the intensity of the light signal. Optical fibres are fabricated such that light reflected at one side of inner surface strikes the other at an angle larger than the critical angle. Even if the fibre is bent, light can easily travel along its length. Thus, an optical fibre can be used to act as an optical pipe. FIGUREFIGUREFIGUREFIGUREFIGURE 9.139.139.139.139.13 Prisms designed to bend rays by A bundle of optical fibres can be put to 90° and 180° or to invert image without changing several uses. Optical fibres are extensively its size make use of total internal reflection. used for transmitting and receiving 231 Reprint 2025-26 Physics electrical signals which are converted to light by suitable transducers. Obviously, optical fibres can also be used for transmission of optical signals. For example, these are used as a ‘light pipe’ to facilitate visual examination of internal organs like esophagus, stomach and intestines. You might have seen a commonly available decorative lamp with fine FIGURE 9.14 Light undergoes successive total plastic fibres with their free ends forming a internal reflections as it moves through an fountain like structure. The other end of the optical fibre. fibres is fixed over an electric lamp. When the lamp is switched on, the light travels from the bottom of each fibre and appears at the tip of its free end as a dot of light. The fibres in such decorative lamps are optical fibres. The main requirement in fabricating optical fibres is that there should be very little absorption of light as it travels for long distances inside them. This has been achieved by purification and special preparation of materials such as quartz. In silica glass fibres, it is possible to transmit more than 95% of the light over a fibre length of 1 km. (Compare with what you expect for a block of ordinary window glass 1 km thick.) 9.5 REFRACTION AT SPHERICAL SURFACES AND BY LENSES We have so far considered refraction at a plane interface. We shall now consider refraction at a spherical interface between two transparent media. An infinitesimal part of a spherical surface can be regarded as planar and the same laws of refraction can be applied at every point on the surface. Just as for reflection by a spherical mirror, the normal at the point of incidence is perpendicular to the tangent plane to the spherical surface at that point and, therefore, passes through its centre of curvature. We first consider refraction by a single spherical surface and follow it by thin lenses. A thin lens is a transparent optical medium bounded by two surfaces; at least one of which should be spherical. Applying the formula for image formation by a single spherical surface successively at the two surfaces of a lens, we shall obtain the lens maker’s formula and then the lens formula. 9.5.1 Refraction at a spherical surface Figure 9.15 shows the geometry of formation of image I of an object O on the principal axis of a spherical surface with centre of curvature C, and radius of curvature R. The rays are incident from a medium of refractive index n1, to another of refractive index n 2. As before, we take the aperture (or the lateral size) of the surface to be small compared to other distances involved, so that small angle approximation can be made. In particular, NM will be taken to be nearly equal to the length of the perpendicular from the point N on the principal axis. We have, for small angles, MN 232 tan ÐNOM = OM Reprint 2025-26 Ray Optics and Optical Instruments MN tan ÐNCM = MC MN tan ÐNIM = MI Now, for DNOC, i is the exterior angle. Therefore, i = ÐNOM + ÐNCM MN MN i = + (9.13) OM MC Similarly, FIGURE 9.15 Refraction at a spherical r = ÐNCM – ÐNIM surface separating two media. MN MN i.e., r = − (9.14) MC MI Now, by Snell’s law n1 sin i = n 2 sin r or for small angles n1i = n 2r Substituting i and r from Eqs. (9.13) and (9.14), we get n 1 n 2 n 2 − n 1 + = (9.15) OM MI MC Here, OM, MI and MC represent magnitudes of distances. Applying the Cartesian sign convention, OM = –u, MI = +v, MC = +R Substituting these in Eq. (9.15), we get n 2 n 1 n 2 − n 1 − = (9.16) v u R Equation (9.16) gives us a relation between object and image distance in terms of refractive index of the medium and the radius of curvature of the curved spherical surface. It holds for any curved spherical surface. Example 9.5 Light from a point source in air falls on a spherical glass surface (n = 1.5 and radius of curvature = 20 cm). The distance of the light source from the glass surface is 100 cm. At what position the image is formed? Solution We use the relation given by Eq. (9.16). Here u = – 100 cm, v = ?, R = + 20 cm, n1 = 1, and n2 = 1.5. We then have 1.5 1 0.5 + = v 100 20 or v = +100 cm The image is formed at a distance of 100 cm from the glass surface, EXAMPLE in the direction of incident light. 9.5 233 Reprint 2025-26 Physics 9.5.2 Refraction by a lens Figure 9.16(a) shows the geometry of image formation by a double convex lens. The image formation can be seen in terms of two steps: (i) The first refracting surface forms the image I1 of the object O [Fig. 9.16(b)]. The image I1 acts as a virtual object for the second surface that forms the image at I [Fig. 9.16(c)]. Applying Eq. (9.15) to the first interface ABC, we get n 1  n 2  n 2  n 1 (9.17) OB BI1 BC1 A similar procedure applied to the second interface* ADC gives,  n 2  n 1  n 2  n 1 (9.18) DI1 DI DC 2 For a thin lens, BI1 = DI1. Adding Eqs. (9.17) and (9.18), we get n1 n1  1 1  + = (n 2 − n1 ) + (9.19) OB DI  BC1 DC 2  Suppose the object is at infinity, i.e., OB ® ¥ and DI = f, Eq. (9.19) gives n1  1 1  = (n 2 − n1 ) + (9.20) f  BC1 DC 2  The point where image of an object placed at infinity is formed is called the focus F, of the lens and the distance f gives its focal length. A lens has two foci, F and F¢, on either side of it (Fig. 9.17). By the sign convention, BC1 = + R1, DC2 = –R2 So Eq. (9.20) can be written as (9.21) Equation (9.21) is known as the lens maker’s formula. It is useful to design lenses of desired focal length using surfaces FIGURE 9.16 (a) The position of object, and the of suitable radii of curvature. Note that the image formed by a double convex lens, formula is true for a concave lens also. In (b) Refraction at the first spherical surface and that case R1is negative, R2 positive and (c) Refraction at the second spherical surface. therefore, f is negative. * Note that now the refractive index of the medium on the right side of ADC is n1 while on its left it is n2. Further DI1 is negative as the distance is measured234 against the direction of incident light. Reprint 2025-26 Ray Optics and Optical Instruments From Eqs. (9.19) and (9.20), we get n1 n1 n1 + = (9.22) OB DI f Again, in the thin lens approximation, B and D are both close to the optical centre of the lens. Applying the sign convention, BO = – u, DI = +v, we get 1 1 1 − = (9.23) v u f Equation (9.23) is the familiar thin lens formula. Though we derived it for a real image formed by a convex lens, the formula is valid for both convex as well as concave lenses and for both real and virtual images. It is worth mentioning that the two foci, F and F¢, of a double convex or concave lens are equidistant from the optical centre. The focus on the side of the (original) source of light is called the first focal point, whereas the other is called the second focal point. To find the image of an object by a lens, we can, in principle, take any two rays emanating from a point on an object; trace their paths using the laws of refraction and find the point where the refracted rays meet (or appear to meet). In practice, however, it is convenient to choose any two of the following rays: (i) A ray emanating from the object parallel to the principal axis of the lens after refraction passes through the second principal focus F¢ (in a convex lens) or appears to diverge (in a concave lens) from the first principal focus F. (ii) A ray of light, passing through the optical centre of the lens, emerges without any deviation after refraction. (iii) (a) A ray of light passing through the first principal focus of a convex lens [Fig. 9.17(a)] emerges parallel to the principal axis after refraction. (b) A ray of light incident on a concave lens appearing to meet the principal axis at second focus point emerges parallel to the principal axis after refraction [Fig. 9.17(b)]. Figures 9.17(a) and (b) illustrate these rules for a convex and a concave lens, respectively. You should practice drawing similar ray diagrams for different positions of the object with respect to the lens and also verify that the lens formula, Eq. (9.23), holds good for all cases. Here again it must be remembered that each point on an object gives out infinite number of rays. All these rays will pass through the same image point after refraction at the lens. FIGURE 9.17 Tracing rays through (a) Magnification (m) produced by a lens is convex lens (b) concave lens. defined, like that for a mirror, as the ratio of the 235 size of the image to that of the object. Proceeding Reprint 2025-26 Physics in the same way as for spherical mirrors, it is easily seen that for a lens h ′ v m = = (9.24) h u When we apply the sign convention, we see that, for erect (and virtual) image formed by a convex or concave lens, m is positive, while for an inverted (and real) image, m is negative. Example 9.6 A magician during a show makes a glass lens with n = 1.47 disappear in a trough of liquid. What is the refractive index of the liquid? Could the liquid be water? Solution 9.6 The refractive index of the liquid must be equal to 1.47 in order to make the lens disappear. This means n1 = n2. This gives 1/f =0 or f ® ¥. The lens in the liquid will act like a plane sheet of glass. No, EXAMPLE the liquid is not water. It could be glycerine. 9.5.3 Power of a lens Power of a lens is a measure of the convergence or divergence, which a lens introduces in the light falling on it. Clearly, a lens of shorter focal length bends the incident light more, while converging it in case of a convex lens and diverging it in case of a concave lens. The power P of a lens is defined as the tangent of the angle by which it converges or diverges a beam of light parallel to the principal axis falling at unit distance from the optical centre (Fig. 9.18). FIGURE 9.18 Power of a lens. h 1 1 tan δ = ; if h = 1, tan δ = or δ = for small f f f value of d. Thus, 1 P = (9.25) f The SI unit for power of a lens is dioptre (D): 1D = 1m–1. The power of a lens of focal length of 1 metre is one dioptre. Power of a lens is positive for a converging lens and negative for a diverging lens. Thus, when an optician prescribes a corrective lens of power + 2.5 D, the required lens is a convex lens of focal length + 40 cm. A lens of power of – 4.0 D means a concave lens of focal length – 25 cm. Example 9.7 (i) If f = 0.5 m for a glass lens, what is the power of the lens? (ii) The radii of curvature of the faces of a double convex lens 9.7 are 10 cm and 15 cm. Its focal length is 12 cm. What is the refractive index of glass? (iii) A convex lens has 20 cm focal length in air. What is focal length in water? (Refractive index of air-water = 1.33, refractive EXAMPLE index for air-glass = 1.5.) Reprint 2025-26 Ray Optics and Optical Instruments Solution (i) Power = +2 dioptre. (ii) Here, we have f = +12 cm, R1 = +10 cm, R2 = –15 cm. Refractive index of air is taken as unity. We use the lens formula of Eq. (9.22). The sign convention has to be applied for f, R1 and R2. Substituting the values, we have 1  1 1  12 = (n − 1)  10 −−15  This gives n = 1.5. (iii) For a glass lens in air, n2 = 1.5, n1 = 1, f = +20 cm. Hence, the lens formula gives 1  1 1  = 0.5 − 20  R1 R 2  For the same glass lens in water, n2 = 1.5, n1 = 1.33. Therefore, 1 .33  1 1  = (1 .5 − 1 .33 ) − (9.26) EXAMPLE f  R1 R 2  Combining these two equations, we find f = + 78.2 cm. 9.7 9.5.4 Combination of thin lenses in contact Consider two lenses A and B of focal length f1 and f2 placed in contact with each other. Let the object be placed at a point O beyond the focus of the first lens A (Fig. 9.19). The first lens produces an image at I1. Since image I1 is real, it serves as a virtual object for the second lens B, producing the final image at I. It must, however, be borne in mind that formation of image by the first lens is presumed only to facilitate determination of the position of the FIGURE 9.19 Image formation by a final image. In fact, the direction of rays emerging combination of two thin lenses in contact. from the first lens gets modified in accordance with the angle at which they strike the second lens. Since the lenses are thin, we assume the optical centres of the lenses to be coincident. Let this central point be denoted by P. For the image formed by the first lens A, we get 1 1 1 − = (9.27) v1 u f 1 For the image formed by the second lens B, we get 1 1 1 − = (9.28) v v1 f 2 Adding Eqs. (9.27) and (9.28), we get 1 1 1 1 − = + (9.29) v u f 1 f 2 If the two lens-system is regarded as equivalent to a single lens of focal length f, we have 237 Reprint 2025-26 Physics 1 1 1 − = v u f so that we get 1 1 1 = + (9.30) f f 1 f 2 The derivation is valid for any number of thin lenses in contact. If several thin lenses of focal length f1, f2, f3,... are in contact, the effective focal length of their combination is given by 1 1 1 1 = + + + … (9.31) f f 1 f 2 f 3 In terms of power, Eq. (9.31) can be written as P = P1 + P2 + P3 + … (9.32) where P is the net power of the lens combination. Note that the sum in Eq. (9.32) is an algebraic sum of individual powers, so some of the terms on the right side may be positive (for convex lenses) and some negative (for concave lenses). Combination of lenses helps to obtain diverging or converging lenses of desired magnification. It also enhances sharpness of the image. Since the image formed by the first lens becomes the object for the second, Eq. (9.25) implies that the total magnification m of the combination is a product of magnification (m1, m 2, m 3,...) of individual lenses m = m1 m2 m3 ... (9.33) Such a system of combination of lenses is commonly used in designing lenses for cameras, microscopes, telescopes and other optical instruments. Example 9.8 Find the position of the image formed by the lens combination given in the Fig. 9.20. FIGURE 9.20 Solution Image formed by the first lens 1 1 1 − = 1 u 1 f 1 9.8 v 1 1 1 − = v1 −30 10 or v1 = 15 cm EXAMPLE Reprint 2025-26 Ray Optics and Optical Instruments The image formed by the first lens serves as the object for the second. This is at a distance of (15 – 5) cm = 10 cm to the right of the second lens. Though the image is real, it serves as a virtual object for the second lens, which means that the rays appear to come from it for the second lens. 1 1 1 − = v 2 10 − 10 or v2 = ¥ The virtual image is formed at an infinite distance to the left of the second lens. This acts as an object for the third lens. 1 1 1 − = v 3 u 3 f 3 1 1 1 or = + v 3 ∞ 30 EXAMPLE or v3 = 30 cm The final image is formed 30 cm to the right of the third lens. 9.8

9.2 A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror.

9.24 What should be the distance between the object in Exercise 9.23 and the magnifying glass if the virtual image of each square in the figure is to have an area of 6.25 mm2. Would you be able to see the squares distinctly with your eyes very close to the magnifier? [Note: Exercises 9.22 to 9.24 will help you clearly understand the difference between magnification in absolute size and the angular magnification (or magnifying power) of an instrument.] 251 Reprint 2025-26 Physics 9.25 Answer the following questions: (a) The angle subtended at the eye by an object is equal to the angle subtended at the eye by the virtual image produced by a magnifying glass. In what sense then does a magnifying glass provide angular magnification? (b) In viewing through a magnifying glass, one usually positions one’s eyes very close to the lens. Does angular magnification change if the eye is moved back? (c) Magnifying power of a simple microscope is inversely proportional to the focal length of the lens. What then stops us from using a convex lens of smaller and smaller focal length and achieving greater and greater magnifying power? (d) Why must both the objective and the eyepiece of a compound microscope have short focal lengths? (e) When viewing through a compound microscope, our eyes should be positioned not on the eyepiece but a short distance away from it for best viewing. Why? How much should be that short distance between the eye and eyepiece? 9.26 An angular magnification (magnifying power) of 30X is desired using an objective of focal length 1.25cm and an eyepiece of focal length 5cm. How will you set up the compound microscope? 9.27 A small telescope has an objective lens of focal length 140cm and an eyepiece of focal length 5.0cm. What is the magnifying power of the telescope for viewing distant objects when (a) the telescope is in normal adjustment (i.e., when the final image is at infinity)? (b) the final image is formed at the least distance of distinct vision (25cm)? 9.28 (a) For the telescope described in Exercise 9.27 (a), what is the separation between the objective lens and the eyepiece? (b) If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the image of the tower formed by the objective lens? (c) What is the height of the final image of the tower if it is formed at 25cm? 9.29 A Cassegrain telescope uses two mirrors as shown in Fig. 9.26. Such a telescope is built with the mirrors 20mm apart. If the radius of curvature of the large mirror is 220mm and the small mirror is 140mm, where will the final image of an object at infinity be? 9.30 Light incident normally on a plane mirror attached to a galvanometer coil retraces backwards as shown in Fig. 9.29. A current in the coil produces a deflection of 3.5o of the mirror. What is the displacement of the reflected spot of light on a screen placed 1.5 m away? Reprint 2025-26 Ray Optics and Optical Instruments FIGURE 9.29

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

14.7 Beats the cork pieces move up and down but do not move away Summary from the centre of disturbance. This shows that the water Points to ponder mass does not flow outward with the circles, but rather a Exercises moving disturbance is created. Similarly, when we speak, the sound moves outward from us, without any flow of air from one part of the medium to another. The disturbances produced in air are much less obvious and only our ears or a microphone can detect them. These patterns, which move without the actual physical transfer or flow of matter as a whole, are called waves. In this Chapter, we will study such waves. Waves transport energy and the pattern of disturbance has information that propagate from one point to another. All our communications essentially depend on transmission of sig- nals through waves. Speech means production of sound waves in air and hearing amounts to their detection. Often, communication involves different kinds of waves. For exam- ple, sound waves may be first converted into an electric cur- rent signal which in turn may generate an electromagnetic wave that may be transmitted by an optical cable or via a Reprint 2025-26 WAVES 279 satellite. Detection of the original signal will usu- We shall illustrate this connection through ally involve these steps in reverse order. simple examples. Not all waves require a medium for their Consider a collection of springs connected to propagation. We know that light waves can one another as shown in Fig. 14.1. If the spring travel through vacuum. The light emitted by at one end is pulled suddenly and released, the stars, which are hundreds of light years away, disturbance travels to the other end. What has reaches us through inter-stellar space, which is practically a vacuum. The most familiar type of waves such as waves on a string, water waves, sound waves, seismic waves, etc. is the so-called mechanical waves. Fig. 14.1 A collection of springs connected to each These waves require a medium for propagation, other. The end A is pulled suddenly they cannot propagate through vacuum. They generating a disturbance, which then involve oscillations of constituent particles and propagates to the other end. depend on the elastic properties of the medium. The electromagnetic waves that you will learn happened? The first spring is disturbed from its in Class XII are a different type of wave. equilibrium length. Since the second spring is Electromagnetic waves do not necessarily require connected to the first, it is also stretched or a medium - they can travel through vacuum. compressed, and so on. The disturbance moves Light, radiowaves, X-rays, are all electromagnetic from one end to the other; but each spring only waves. In vacuum, all electromagnetic waves executes small oscillations about its equilibrium have the same speed c, whose value is : position. As a practical example of this situation, consider a stationary train at a railway station. c = 299, 792, 458 ms–1. (14.1) Different bogies of the train are coupled to each A third kind of wave is the so-called Matter other through a spring coupling. When an waves. They are associated with constituents of engine is attached at one end, it gives a push to matter : electrons, protons, neutrons, atoms and the bogie next to it; this push is transmitted from molecules. They arise in quantum mechanical one bogie to another without the entire train description of nature that you will learn in your being bodily displaced. later studies. Though conceptually more abstract Now let us consider the propagation of sound than mechanical or electro-magnetic waves, they waves in air. As the wave passes through air, it have already found applications in several compresses or expands a small region of air. This devices basic to modern technology; matter causes a change in the density of that region, waves associated with electrons are employed say δρ, this change induces a change in pressure, in electron microscopes. δp, in that region. Pressure is force per unit area, In this chapter we will study mechanical so there is a restoring force proportional to waves, which require a material medium for the disturbance, just like in a spring. In this their propagation. case, the quantity similar to extension or The aesthetic influence of waves on art and compression of the spring is the change in literature is seen from very early times; yet the density. If a region is compressed, the molecules first scientific analysis of wave motion dates back in that region are packed together, and they tend to the seventeenth century. Some of the famous to move out to the adjoining region, thereby scientists associated with the physics of wave increasing the density or creating compression motion are Christiaan Huygens (1629-1695), in the adjoining region. Consequently, the air Robert Hooke and Isaac Newton. The in the first region undergoes rarefaction. If a understanding of physics of waves followed the region is comparatively rarefied the surrounding physics of oscillations of masses tied to springs air will rush in making the rarefaction move to and physics of the simple pendulum. Waves in the adjoining region. Thus, the compression or elastic media are intimately connected with rarefaction moves from one region to another, harmonic oscillations. (Stretched strings, coiled making the propagation of a disturbance springs, air, etc., are examples of elastic media). possible in air. Reprint 2025-26 280 PHYSICS In solids, similar arguments can be made. In a crystalline solid, atoms or group of atoms are arranged in a periodic lattice. In these, each atom or group of atoms is in equilibrium, due to forces from the surrounding atoms. Displacing one atom, keeping the others fixed, leads to restoring forces, exactly as in a spring. So we can think of atoms in a lattice as end points, with springs between pairs of them. In the subsequent sections of this chapter we are going to discuss various characteristic Fig. 14.3 A harmonic (sinusoidal) wave travelling properties of waves. along a stretched string is an example of a transverse wave. An element of the string 14.2 TRANSVERSE AND LONGITUDINAL in the region of the wave oscillates about WAVES its equilibrium position perpendicular to the direction of wave propagation. We have seen that motion of mechanical waves involves oscillations of constituents of the position as the pulse or wave passes through medium. If the constituents of the medium them. The oscillations are normal to the oscillate perpendicular to the direction of wave direction of wave motion along the string, so this propagation, we call the wave a transverse wave. is an example of transverse wave. If they oscillate along the direction of wave We can look at a wave in two ways. We can fix propagation, we call the wave a longitudinal an instant of time and picture the wave in space. wave. This will give us the shape of the wave as a Fig.14.2 shows the propagation of a single whole in space at a given instant. Another way pulse along a string, resulting from a single up is to fix a location i.e. fix our attention on a and down jerk. If the string is very long compared particular element of string and see its oscillatory motion in time. Fig. 14.4 describes the situation for longitudinal waves in the most familiar example of the propagation of sound waves. A long pipe filled with air has a piston at one end. A single sudden push forward and pull back of the piston will generate a pulse of condensations (higher density) and rarefactions (lower density) in the medium (air). If the push-pull of the piston is continuous and periodic (sinusoidal), a Fig. 14.2 When a pulse travels along the length of a stretched string (x-direction), the elements of the string oscillate up and down (y- direction) to the size of the pulse, the pulse will damp out before it reaches the other end and reflection from that end may be ignored. Fig. 14.3 shows a similar situation, but this time the external agent gives a continuous periodic sinusoidal up Fig. 14.4 Longitudinal waves (sound) generated in a and down jerk to one end of the string. The pipe filled with air by moving the piston up resulting disturbance on the string is then a and down. A volume element of air oscillates sinusoidal wave. In either case the elements of in the direction parallel to the direction of the string oscillate about their equilibrium mean wave propagation. Reprint 2025-26 WAVES 281 sinusoidal wave will be generated propagating u Example 14.1 Given below are some in air along the length of the pipe. This is clearly examples of wave motion. State in each case an example of longitudinal waves. if the wave motion is transverse, longitudinal The waves considered above, transverse or or a combination of both: longitudinal, are travelling or progressive waves (a) Motion of a kink in a longitudinal spring since they travel from one part of the medium produced by displacing one end of the to another. The material medium as a whole spring sideways. does not move, as already noted. A stream, for (b) Waves produced in a cylinder example, constitutes motion of water as a whole. containing a liquid by moving its piston back and forth.In a water wave, it is the disturbance that moves, (c) Waves produced by a motorboat sailing not water as a whole. Likewise a wind (motion in water. of air as a whole) should not be confused with a (d) Ultrasonic waves in air produced by a sound wave which is a propagation of vibrating quartz crystal. disturbance (in pressure density) in air, without the motion of air medium as a whole. Answer In transverse waves, the particle motion is (a) Transverse and longitudinal normal to the direction of propagation of the (b) Longitudinal wave. Therefore, as the wave propagates, each (c) Transverse and longitudinal element of the medium undergoes a shearing (d) Longitudinal ⊳ strain. Transverse waves can, therefore, be 14.3 DISPLACEMENT RELATION INpropagated only in those media, which can A PROGRESSIVE WAVE sustain shearing stress, such as solids and not in fluids. Fluids, as well as, solids can sustain For mathematical description of a travelling compressive strain; therefore, longitudinal wave, we need a function of both position x and time t. Such a function at every instant shouldwaves can be propagated in all elastic media. give the shape of the wave at that instant. Also,For example, in medium like steel, both at every given location, it should describe thetransverse and longitudinal waves can motion of the constituent of the medium at thatpropagate, while air can sustain only location. If we wish to describe a sinusoidal longitudinal waves. The waves on the surface travelling wave (such as the one shown in Fig. of water are of two kinds: capillary waves and 14.3) the corresponding function must also be gravity waves. The former are ripples of fairly sinusoidal. For convenience, we shall take the short wavelength—not more than a few wave to be transverse so that if the position of centimetre—and the restoring force that the constituents of the medium is denoted by x, produces them is the surface tension of water. the displacement from the equilibrium position Gravity waves have wavelengths typically may be denoted by y. A sinusoidal travelling ranging from several metres to several hundred wave is then described by: meters. The restoring force that produces these y ( x , t ) = a sin( kx −ωt + φ) (14.2)waves is the pull of gravity, which tends to keep The term φ in the argument of sine functionthe water surface at its lowest level. The means equivalently that we are considering aoscillations of the particles in these waves are linear combination of sine and cosine functions:not confined to the surface only, but extend with diminishing amplitude to the very bottom. The y ( x , t ) = A sin(kx − ωt ) + B cos(kx − ωt ) (14.3) particle motion in water waves involves a From Equations (14.2) and (14.3), complicated motion—they not only move up and  B down but also back and forth. The waves in an a = A 2 + B 2 and φ= tan −1ocean are the combination of both longitudinal  A  and transverse waves. To understand why Equation (14.2) It is found that, generally, transverse and represents a sinusoidal travelling wave, take a longitudinal waves travel with different speed fixed instant, say t = t0. Then, the argument of in the same medium. the sine function in Equation (14.2) is simply Reprint 2025-26 282 PHYSICS kx + constant. Thus, the shape of the wave (at any fixed instant) as a function of x is a sine wave. Similarly, take a fixed location, say x = x0. Then, the argument of the sine function in Equation (14.2) is constant -ωt. The displacement y, at a fixed location, thus, varies sinusoidally with time. That is, the constituents of the medium at different positions execute simple harmonic motion. Finally, as t increases, x must increase in the positive direction to keep kx – ωt + φ constant. Thus, Eq. (14.2) represents a sinusiodal (harmonic) wave travelling along the positive direction of the x-axis. On the other hand, a function y ( x , t ) = a sin( kx + ω t + φ ) (14.4) represents a wave travelling in the negative direction of x-axis. Fig. (14.5) gives the names of the various physical quantities appearing in Eq. (14.2) that we now interpret. y(x,t) : displacement as a function of position x and time t a : amplitude of a wave ω : angular frequency of the wave Fig. 14.6 A harmonic wave progressing along the k : angular wave number positive direction of x-axis at different times. kx–ωt+φ : initial phase angle (a+x = 0, t = 0) Using the plots of Fig. 14.6, we now define Fig. 14.5 The meaning of standard symbols in the various quantities of Eq. (14.2). Eq. (14.2) 14.3.1 Amplitude and Phase Fig. 14.6 shows the plots of Eq. (14.2) for In Eq. (14.2), since the sine function varies different values of time differing by equal between 1 and –1, the displacement y (x,t) varies intervals of time. In a wave, the crest is the between a and –a. We can take a to be a positive point of maximum positive displacement, the constant, without any loss of generality. Then, trough is the point of maximum negative a represents the maximum displacement of the displacement. To see how a wave travels, we constituents of the medium from their can fix attention on a crest and see how it equilibrium position. Note that the displacement progresses with time. In the figure, this is y may be positive or negative, but a is positive. shown by a cross (×) on the crest. In the same It is called the amplitude of the wave. manner, we can see the motion of a particular The quantity (kx – ωt + φ) appearing as the constituent of the medium at a fixed location, argument of the sine function in Eq. (14.2) issay at the origin of the x-axis. This is shown called the phase of the wave. Given theby a solid dot (•). The plots of Fig. 14.6 show amplitude a, the phase determines thethat with time, the solid dot (•) at the origin displacement of the wave at any position andmoves periodically, i.e., the particle at the origin oscillates about its mean position as at any instant. Clearly φ is the phase at x = 0 the wave progresses. This is true for any other and t = 0. Hence, φ is called the initial phase location also. We also see that during the time angle. By suitable choice of origin on the x-axis the solid dot (•) has completed one full and the intial time, it is possible to have φ = 0. oscillation, the crest has moved further by a Thus there is no loss of generality in dropping certain distance. φ, i.e., in taking Eq. (14.2) with φ = 0. Reprint 2025-26 WAVES 283 14.3.2 Wavelength and Angular Wave Number The minimum distance between two points having the same phase is called the wavelength of the wave, usually denoted by λ. For simplicity, we can choose points of the same phase to be crests or troughs. The wavelength is then the distance between two consecutive crests or troughs in a wave. Taking φ = 0 in Eq. (14.2), Fig. 14.7 An element of a string at a fixed location the displacement at t = 0 is given by oscillates in time with amplitude a and period T, as the wave passes over it. y ( x ,0) = a sin kx (14.5) Since the sine function repeats its value after Now, the period of oscillation of the wave is the every 2π change in angle, time it takes for an element to complete one full oscillation. That is − a sin ωt = −a sin ω(t + T) = − a sin(ωt + ωT) That is the displacements at points x and at Since sine function repeats after every 2π, 2nπx + 2π k ωT = 2π or ω = (14.7) T are the same, where n=1,2,3,... The 1east is called the angular frequency of the wave.distance between points with the same ω displacement (at any given instant of time) is Its SI unit is rad s –1. The frequency ν is the obtained by taking n = 1. λ is then given by number of oscillations per second. Therefore, 1 ω 2π 2π ν= = (14.8) λ= or k = (14.6) T 2π k λ ν is usually measured in hertz. k is the angular wave number or propagation In the discussion above, reference has always constant; its SI unit is radian per metre or been made to a wave travelling along a string or rad m−*1 a transverse wave. In a longitudinal wave, the displacement of an element of the medium is 14.3.3 Period, Angular Frequency and parallel to the direction of propagation of the Frequency wave. In Eq. (14.2), the displacement function Fig. 14.7 shows again a sinusoidal plot. It for a longitudinal wave is written as, describes not the shape of the wave at a certain s(x, t) = a sin (kx – ωt + φ) (14.9) instant but the displacement of an element (at any fixed location) of the medium as a function where s(x, t) is the displacement of an element of time. We may for, simplicity, take Eq. (14.2) of the medium in the direction of propagation with φ = 0 and monitor the motion of the element of the wave at position x and time t. In Eq. (14.9), a is the displacement amplitude; othersay at x = 0 . We then get quantities have the same meaning as in case y (0, t ) = a sin( −ωt ) of a transverse wave except that the displacement function y (x, t) is to be replaced = −a sin ωt by the function s (x, t). * Here again, ‘radian’ could be dropped and the units could be written merely as m–1. Thus, k represents 2π times the number of waves (or the total phase difference) that can be accommodated per unit length, with SI units m–1. Reprint 2025-26 284 PHYSICS the shape of the wave at two instants of time, u Example 14.2 A wave travelling along a which differ by a small time internal ∆t. The string is described by, entire wave pattern is seen to shift to the right (positive direction of x-axis) by a distance ∆x. In y(x, t) = 0.005 sin (80.0 x – 3.0 t), particular, the crest shown by a dot (• ) moves a in which the numerical constants are in SI units (0.005 m, 80.0 rad m–1, and 3.0 rad s–1). Calculate (a) the amplitude, (b) the wavelength, and (c) the period and frequency of the wave. Also, calculate the displacement y of the wave at a distance x = 30.0 cm and time t = 20 s ? Answer On comparing this displacement equation with Eq. (14.2), Fig. 14.8 Progression of a harmonic wave from time y (x, t) = a sin (kx – ωt), t to t + ∆t. where ∆t is a small interval. The wave pattern as a whole shifts to the we find right. The crest of the wave (or a point with (a) the amplitude of the wave is 0.005 m = 5 mm. any fixed phase) moves right by the distance (b) the angular wave number k and angular ∆x in time ∆t. frequency ω are distance ∆x in time ∆t. The speed of the wave is k = 80.0 m–1 and ω = 3.0 s–1 then ∆x/∆t. We can put the dot (• ) on a point We, then, relate the wavelength λ to k through with any other phase. It will move with the same Eq. (14.6), speed v (otherwise the wave pattern will not λ = 2π/k remain fixed). The motion of a fixed phase point on the wave is given by 2π = −1 kx – ωt = constant (14.10) 80.0 m Thus, as time t changes, the position x of the = 7.85 cm fixed phase point must change so that the phase (c) Now, we relate T to ω by the relation remains constant. Thus, T = 2π/ω kx – ωt = k(x+∆x) – ω(t+∆t) 2π or k ∆x – ω∆t =0 = −1 3.0 s Taking ∆x, ∆t vanishingly small, this gives = 2.09 s dx ω and frequency, v = 1/T = 0.48 Hz = = v (14.11) dt k The displacement y at x = 30.0 cm and Relating ω to T and k to λ, we get time t = 20 s is given by 2πν λ y = (0.005 m) sin (80.0 × 0.3 – 3.0 × 20) v = = λν= (14.12) 2π/λ T = (0.005 m) sin (–36 + 12π) = (0.005 m) sin (1.699) Eq. (14.12), a general relation for all progressive = (0.005 m) sin (970) j 5 mm ⊳ waves, shows that in the time required for one full oscillation by any constituent of the medium, the 14.4 THE SPEED OF A TRAVELLING WAVE wave pattern travels a distance equal to the To determine the speed of propagation of a wavelength of the wave. It should be noted that travelling wave, we can fix our attention on any the speed of a mechanical wave is determined by particular point on the wave (characterised by the inertial (linear mass density for strings, mass some value of the phase) and see how that point density in general) and elastic properties (Young’s moves in time. It is convenient to look at the modulus for linear media/ shear modulus, bulk motion of the crest of the wave. Fig. 14.8 gives modulus) of the medium. The medium determines Reprint 2025-26 WAVES 285 the speed; Eq. (14.12) then relates wavelength to arising due to an external force). It does not frequency for the given speed. Of course, as depend on wavelength or frequency of the wave remarked earlier, the medium can support both itself. In higher studies, you will come across transverse and longitudinal waves, which will have waves whose speed is not independent of different speeds in the same medium. Later in this frequency of the wave. Of the two parameters λ chapter, we shall obtain specific expressions for and ν the source of disturbance determines the the speed of mechanical waves in some media. frequency of the wave generated. Given the speed of the wave in the medium and the 14.4.1 Speed of a Transverse Wave on frequency Eq. (14.12) then fixes the wavelength Stretched String v The speed of a mechanical wave is determined λ = (14.15) by the restoring force setup in the medium when ν it is disturbed and the inertial properties (mass u Example 14.3 A steel wire 0.72 m longdensity) of the medium. The speed is expected to has a mass of 5.0 ×10–3 kg. If the wire isbe directly related to the former and inversely to under a tension of 60 N, what is the speedthe latter. For waves on a string, the restoring of transverse waves on the wire ?force is provided by the tension T in the string. The inertial property will in this case be linear Answer Mass per unit length of the wire,mass density µ, which is mass m of the string divided by its length L. Using Newton’s Laws of −3 5.0 × 10 kgMotion, an exact formula for the wave speed on µ = 0. 72 m a string can be derived, but this derivation is outside the scope of this book. We shall, = 6.9 ×10–3 kg m–1 therefore, use dimensional analysis. We already know that dimensional analysis alone can never Tension, T = 60 N yield the exact formula. The overall The speed of wave on the wire is given by dimensionless constant is always left T 60 N −1undetermined by dimensional analysis. = v = = 93 m s ⊳ µ 6.9 × 10− 3 kg m −1 The dimension of µ is [ML–1] and that of T is like force, namely [MLT–2]. We need to combine 14.4.2 Speed of a Longitudinal Wavethese dimensions to get the dimension of speed (Speed of Sound)v [LT–1]. Simple inspection shows that the quantity T/µ has the relevant dimension In a longitudinal wave, the constituents of the − 2 medium oscillate forward and backward in the MLT   2 − 2   =  L T  direction of propagation of the wave. We have [ ML ]   already seen that the sound waves travel in the Thus if T and µ are assumed to be the only form of compressions and rarefactions of small volume elements of air. The elastic property thatrelevant physical quantities, determines the stress under compressional strain is the bulk modulus of the medium defined T v = C (14.13) by (see Chapter 8) µ ∆P where C is the undetermined constant of B = (14.16)dimensional analysis. In the exact formula, it −∆V/V turms out, C=1. The speed of transverse waves Here, the change in pressure ∆P produces a on a stretched string is given by ∆V volumetric strain . B has the same dimension V T v = (14.14) as pressure and given in SI units in terms of µ pascal (Pa). The inertial property relevant for the Note the important point that the speed v propagation of wave is the mass density ρ, with depends only on the properties of the medium T dimensions [ML–3]. Simple inspection reveals and µ (T is a property of the stretched string that quantity B/ρ has the relevant dimension: Reprint 2025-26 286 PHYSICS ML − 2 T − 2   2 − 2 Liquids and solids generally have higher speed    L T =  (14.17) of sound than gases. [Note for solids, the speed − 3    ML    being referred to is the speed of longitudinal Thus, if B and ρ are considered to be the only waves in the solid]. This happens because they are much more difficult to compress than gasesrelevant physical quantities, and so have much higher values of bulk modulus. B Now, see Eq. (14.19). Solids and liquids have v = C (14.18) higher mass densities ( ρ) than gases. But the ρ corresponding increase in both the modulus (B)where, as before, C is the undetermined constant of solids and liquids is much higher. This is thefrom dimensional analysis. The exact derivation reason why the sound waves travel faster inshows that C=1. Thus, the general formula for solids and liquids.longitudinal waves in a medium is: We can estimate the speed of sound in a gas B in the ideal gas approximation. For an ideal gas, v = (14.19) ρ the pressure P, volume V and temperature T are related by (see Chapter 10). For a linear medium, like a solid bar, the lateral expansion of the bar is negligible and we PV = NkBT (14.21) may consider it to be only under longitudinal where N is the number of molecules in volumestrain. In that case, the relevant modulus of V, kB is the Boltzmann constant and T theelasticity is Young’s modulus, which has the temperature of the gas (in Kelvin). Therefore, forsame dimension as the Bulk modulus. an isothermal change it follows from Eq.(14.21) Dimensional analysis for this case is the same that as before and yields a relation like Eq. (14.18), V∆P + P∆V = 0 with an undetermined C, which the exact derivation shows to be unity. Thus, the speed of ∆ P or − = Plongitudinal waves in a solid bar is given by ∆V/V Hence, substituting in Eq. (14.16), we have Y v = (14.20) B = P ρ Therefore, from Eq. (14.19) the speed of a where Y is the Young’s modulus of the material longitudinal wave in an ideal gas is given by, of the bar. Table 14.1 gives the speed of sound in some media. v = P (14.22) ρ Table 14.1 Speed of Sound in some Media This relation was first given by Newton and is known as Newton’s formula. u Example 14.4 Estimate the speed of sound in air at standard temperature and pressure. The mass of 1 mole of air is 29.0 ×10–3 kg. Answer We know that 1 mole of any gas occupies 22.4 litres at STP. Therefore, density of air at STP is: ρo = (mass of one mole of air)/ (volume of one mole of air at STP) −3 29.0 × 10 kg = −3 3 22.4 × 10 m = 1.29 kg m–3 Reprint 2025-26 WAVES 287 According to Newton’s formula for the speed of sound in a medium, we get for the speed of sound in air at STP, = 280 m s–1 (14.23) ⊳ The result shown in Eq.(14.23) is about 15% smaller as compared to the experimental value of 331 m s–1 as given in Table 14.1. Where did we go wrong ? If we examine the basic assumption made by Newton that the pressure variations in a medium during propagation of sound are isothermal, we find that this is not correct. It was pointed out by Laplace that the pressure variations in the propagation of sound waves are so fast that there is little time for the heat flow to maintain constant temperature. These variations, therefore, are adiabatic and not isothermal. For adiabatic processes the ideal Fig. 14.9 Two pulses having equal and opposite gas satisfies the relation (see Section 11.8), γ displacements moving in opposite PV = constant directions. The overlapping pulses add up i.e. ∆(PVγ ) = 0 to zero displacement in curve (c). or P γ V γ –1 ∆V + V γ ∆P = 0 pulses. Figure 14.9 shows the situation when where γ is the ratio of two specific heats, two pulses of equal and opposite shapes move Cp/Cv. towards each other. When the pulses overlap, Thus, for an ideal gas the adiabatic bulk the resultant displacement is the algebraic summodulus is given by, of the displacement due to each pulse. This is ∆P known as the principle of superposition of waves. Bad = − ∆V/V According to this principle, each pulse moves = γP as if others are not present. The constituents of The speed of sound is, therefore, from Eq. the medium, therefore, suffer displacments due (14.19), given by, to both and since the displacements can be positive and negative, the net displacement is γ P an algebraic sum of the two. Fig. 14.9 gives v = (14.24) ρ graphs of the wave shape at different times. Note the dramatic effect in the graph (c); the This modification of Newton’s formula is referred displacements due to the two pulses have exactly to as the Laplace correction. For air cancelled each other and there is zero γ = 7/5. Now using Eq. (14.24) to estimate the speed displacement throughout. of sound in air at STP, we get a value 331.3 m s–1, To put the principle of superposition which agrees with the measured speed. mathematically, let y1 (x,t) and y2 (x,t) be the displacements due to two wave disturbances in

14.6 Reflection of waves some cork pieces on the disturbed surface, it is seen that

14.5 p-n JUNCTION A p-n junction is the basic building block of many semiconductor devices like diodes, transistor, etc. A clear understanding of the junction behaviour is important to analyse the working of other semiconductor devices. We will now try to understand how a junction is formed and how the junction behaves under the influence of external applied voltage (also called bias). 14.5.1 p-n junction formation Consider a thin p-type silicon (p-Si) semiconductor wafer. By adding Formationprecisely a small quantity of pentavelent impurity, part of the p-Si wafer can be converted into n-Si. There are several processes by which a and semiconductor can be formed. The wafer now contains p-region and n-region and a metallurgical junction between p-, and n- region. working Two important processes occur during the formation of a p-n junction: of diffusion and drift. We know that in an n-type semiconductor, the p-nconcentration of electrons (number of electrons per unit volume) is more compared to the concentration of holes. Similarly, in a p-type semiconductor, the concentration of holes is more than the concentration junction of electrons. During the formation of p-n junction, and due to the concentration gradient across p-, and n- sides, holes diffuse from p-side diode to n-side (p ® n) and electrons diffuse from n-side to p-side (n ® p). This motion of charge carries gives rise to diffusion current across the junction. When an electron diffuses from n ® p, it leaves behind an ionised donor on n-side. This ionised donor (positive charge) is immobile as it is bonded to the surrounding atoms. As the electrons continue to diffuse http://hyperphysics.phy-astr.gsu.edu/hbase/solids/pnjun.html from n ® p, a layer of positive charge (or positive space-charge region) on n-side of the junction is developed. Similarly, when a hole diffuses from p ® n due to the concentration gradient, it leaves behind an ionised acceptor (negative charge) which is immobile. As the holes continue to diffuse, a layer of negative charge (or negative space-charge region) on the p-side of the junction is developed. This space-charge region on either side of the junction together is known as depletion region as the electrons and holes taking part in the initial movement across the junction depleted the region of its free charges (Fig. 14.10). The thickness of depletion region is of the order of one-tenth of a micrometre. Due to the positive space-charge region on n-side of the junction and negative space charge region on p-side of the junction, an electric field directed from positive charge towards negative charge develops. Due to this field, an electron on p-side of the junction moves to n-side and a hole on n-side of the junction moves to p- FIGURE 14.10 p-n junction side. The motion of charge carriers due to the electric field formation process. is called drift. Thus a drift current, which is opposite in direction to the diffusion current (Fig. 14.10) starts. 333 Reprint 2025-26 Physics Initially, diffusion current is large and drift current is small. As the diffusion process continues, the space-charge regions on either side of the junction extend, thus increasing the electric field strength and hence drift current. This process continues until the diffusion current equals the drift current. Thus a p-n junction is formed. In a p-n junction under equilibrium there is no net current. The loss of electrons from the n-region and the gain of electron by the p-region causes a difference of potential across the junction of the two regions. The polarity of this potential is such as to oppose further flow of carriers so that a condition of equilibrium exists. Figure 14.11 shows the p-n junction at equilibrium and the potential across the junction. The n-material has lost electrons, and p material has acquired electrons. The n material is thus positive relative to the p FIGURE 14.11 (a) Diode under material. Since this potential tends to prevent the movement of equilibrium (V = 0), (b) Barrier electron from the n region into the p region, it is often called a potential under no bias. barrier potential. Example 14.3 Can we take one slab of p-type semiconductor and physically join it to another n-type semiconductor to get p-n junction? 14.3 Solution No! Any slab, howsoever flat, will have roughness much larger than the inter-atomic crystal spacing (~2 to 3 Å) and hence continuous contact at the atomic level will not be possible. The junction EXAMPLE will behave as a discontinuity for the flowing charge carriers. 14.6 SEMICONDUCTOR DIODE A semiconductor diode [Fig. 14.12(a)] is basically a p-n junction with metallic contacts provided at the ends for the application of an external voltage. It is a two terminal device. A p-n junction diode is symbolically represented as shown in Fig. 14.12(b). The direction of arrow indicates the conventional direction of current (when the diode is under forward p n bias). The equilibrium barrier potential can be altered by applying an external voltage V across the diode. The FIGURE 14.12 (a) Semiconductor diode, situation of p-n junction diode under equilibrium (b) Symbol for p-n junction diode. (without bias) is shown in Fig. 14.11(a) and (b). 14.6.1 p-n junction diode under forward bias When an external voltage V is applied across a semiconductor diode such that p-side is connected to the positive terminal of the battery and n-side to the negative terminal [Fig. 14.13(a)], it is said to be forward biased. The applied voltage mostly drops across the depletion region and the voltage drop across the p-side and n-side of the junction is negligible. (This is because the resistance of the depletion region – a region where there are no charges – is very high compared to the resistance of n-side 334 and p-side.) The direction of the applied voltage (V ) is opposite to the Reprint 2025-26 Semiconductor Electronics: Materials, Devices and Simple Circuits built-in potential V0. As a result, the depletion layer width decreases and the barrier height is reduced [Fig. 14.13(b)]. The effective barrier height under forward bias is (V0 – V ). If the applied voltage is small, the barrier potential will be reduced only slightly below the equilibrium value, and only a small number of carriers in the material—those that happen to be in the uppermost energy levels—will possess enough energy to cross the junction. So the current will be small. If we increase the applied voltage significantly, the barrier height will be reduced and more number of carriers will have the required energy. Thus the current increases. Due to the applied voltage, electrons from n-side cross the depletion region and reach p-side (where they are minority carries). Similarly, holes from p-side cross the junction and reach the n-side (where they are minority carries). This process under forward bias is known as minority carrier injection. At the FIGURE 14.13 (a) p-njunction boundary, on each side, the minority carrier junction diode under forwardconcentration increases significantly compared to the locations bias, (b) Barrier potentialfar from the junction. (1) without battery, (2) Low Due to this concentration gradient, the injected electrons on battery voltage, and (3) High p-side diffuse from the junction edge of p-side to the other end voltage battery. of p-side. Likewise, the injected holes on n-side diffuse from the junction edge of n-side to the other end of n-side (Fig. 14.14). This motion of charged carriers on either side gives rise to current. The total diode forward current is sum of hole diffusion current and conventional current due to electron diffusion. The magnitude of this current is usually in mA. 14.6.2 p-n junction diode under reverse bias When an external voltage (V ) is applied across the diode such FIGURE 14.14 Forward bias that n-side is positive and p-side is negative, it is said to be minority carrier injection. reverse biased [Fig.14.15(a)]. The applied voltage mostly drops across the depletion region. The direction of applied voltage is same as the direction of barrier potential. As a result, the barrier height increases and the depletion region widens due to the change in the electric field. The effective barrier height under reverse bias is (V0 + V ), [Fig. 14.15(b)]. This suppresses the flow of electrons from n ® p and holes from p ® n. Thus, diffusion current, decreases enormously compared to the diode under forward bias. The electric field direction of the junction is such that if electrons on p-side or holes on n-side in their random motion come close to the junction, they will be swept to its majority zone. This drift of carriers gives rise to current. The drift current is of the order of a few mA. This is quite low because it is due to the motion of carriers from their minority side to their majority side across the junction. The drift current is also there under forward bias but it is negligible (mA) when compared with current due to injected carriers which is usually in mA. The diode reverse current is not very much dependent on the applied voltage. Even a small voltage is sufficient to sweep the minority carriers 335from one side of the junction to the other side of the junction. The current Reprint 2025-26 Physics is not limited by the magnitude of the applied voltage but is limited due to the concentration of the minority carrier on either side of the junction. The current under reverse bias is essentially voltage independent upto a critical reverse bias voltage, known as breakdown voltage (Vbr ). When V = Vbr, the diode reverse current increases sharply. Even a slight increase in the bias voltage causes large change in the current. If the reverse current is not limited by an external circuit below the rated value (specified by the manufacturer) the p-n junction will get destroyed. Once it exceeds the rated value, the diode gets destroyed due to overheating. This can happen even for the diode under forward bias, if the forward current exceeds the rated value. The circuit arrangement for studying the V-I characteristics of a diode, (i.e., the variation of current as a function of applied FIGURE 14.15 (a) Diode voltage) are shown in Fig. 14.16(a) and (b). The battery is connected under reverse bias, to the diode through a potentiometer (or reheostat) so that the(b) Barrier potential under applied voltage to the diode can be changed. For different values reverse bias. of voltages, the value of the current is noted. A graph between V and I is obtained as in Fig. 14.16(c). Note that in forward bias measurement, we use a milliammeter since the expected current is large (as explained in the earlier section) while a micrometer is used in reverse bias to measure the current. You can see in Fig. 14.16(c) that in forward FIGURE 14.16 Experimental circuit arrangement for studying V-I characteristics of a p-n junction diode (a) in forward bias, (b) in reverse bias. (c) Typical V-I 336 characteristics of a silicon diode. Reprint 2025-26 Semiconductor Electronics: Materials, Devices and Simple Circuits bias, the current first increases very slowly, almost negligibly, till the voltage across the diode crosses a certain value. After the characteristic voltage, the diode current increases significantly (exponentially), even for a very small increase in the diode bias voltage. This voltage is called the threshold voltage or cut-in voltage (~0.2V for germanium diode and ~0.7 V for silicon diode). For the diode in reverse bias, the current is very small (~mA) and almost remains constant with change in bias. It is called reverse saturation current. However, for special cases, at very high reverse bias (break down voltage), the current suddenly increases. This special action of the diode is discussed later in Section 14.8. The general purpose diode are not used beyond the reverse saturation current region. The above discussion shows that the p-n junction diode primerly allows the flow of current only in one direction (forward bias). The forward bias resistance is low as compared to the reverse bias resistance. This property is used for rectification of ac voltages as discussed in the next section. For diodes, we define a quantity called dynamic resistance as the ratio of small change in voltage DV to a small change in current DI: ∆ V rd = (14.6) ∆I Example 14.4 The V-I characteristic of a silicon diode is shown in the Fig. 14.17. Calculate the resistance of the diode at (a) ID = 15 mA and (b) VD = –10 V. FIGURE 14.17 Solution Considering the diode characteristics as a straight line between I = 10 mA to I = 20 mA passing through the origin, we can calculate the resistance using Ohm’s law. (a) From the curve, at I = 20 mA, V = 0.8 V; I = 10 mA, V = 0.7 V rfb = DV/DI = 0.1V/10 mA = 10 W (b) From the curve at V = –10 V, I = –1 mA, EXAMPLE Therefore, rrb = 10 V/1mA= 1.0 × 107 W 14.4 337 Reprint 2025-26 Physics 14.7 APPLICATION OF JUNCTION DIODE AS A RECTIFIER From the V-I characteristic of a junction diode we see that it allows current to pass only when it is forward biased. So if an alternating voltage is applied across a diode the current flows only in that part of the cycle when the diode is forward biased. This property is used to rectify alternating voltages and the circuit used for this purpose is called a rectifier. If an alternating voltage is applied across a diode in series with a load, a pulsating voltage will appear across the load only during the half cycles of the ac input during which the diode is forward biased. Such rectifier circuit, as shown in Fig. 14.18, is called a half-wave rectifier. The secondary of a transformer supplies the desired ac voltage across terminals A and B. When the voltage at A is positive, the diode is forward biased and it conducts. When A is negative, the diode is reverse-biased and it does not conduct. The reverse saturation current of a diode is negligible and can be considered equal to zero for practical purposes. (The reverse breakdown voltage of the diode must be sufficiently higher than the peak ac voltage at the secondary of the transformer to protect the diode from reverse breakdown.) Therefore, in the positive half-cycle of ac there FIGURE 14.18 (a) Half-wave rectifier is a current through the load resistor RL and we circuit, (b) Input ac voltage and output get an output voltage, as shown in Fig. 14.18(b), voltage waveforms from the rectifier circuit. whereas there is no current in the negative half- cycle. In the next positive half-cycle, again we get the output voltage. Thus, the output voltage, though still varying, is restricted to only one direction and is said to be rectified. Since the rectified output of this circuit is only for half of the input ac wave it is called as half-wave rectifier. The circuit using two diodes, shown in Fig. 14.19(a), gives output rectified voltage corresponding to both the positive as well as negative half of the ac cycle. Hence, it is known as full-wave rectifier. Here the p-side of the two diodes are connected to the ends of the secondary of the transformer. The n-side of the diodes are connected together and the output is taken between this common point of diodes and the midpoint of the secondary of the transformer. So for a full-wave rectifier the secondary of the transformer is provided with a centre tapping and so it is called centre-tap transformer. As can be seen from Fig.14.19(c) the voltage rectified by each diode is only half the total secondary voltage. Each diode rectifies only for half the cycle, but the two do so for alternate cycles. Thus, the output between their common terminals and the centre- tap of the transformer becomes a full-wave rectifier output. (Note that there is another circuit of full wave rectifier which does not need a centre- 338 tap transformer but needs four diodes.) Suppose the input voltage to A Reprint 2025-26 Semiconductor Electronics: Materials, Devices and Simple Circuits with respect to the centre tap at any instant is positive. It is clear that, at that instant, voltage at B being out of phase will be negative as shown in Fig.14.19(b). So, diode D1 gets forward biased and conducts (while D2 being reverse biased is not conducting). Hence, during this positive half cycle we get an output current (and a output voltage across the load resistor RL) as shown in Fig.14.19(c). In the course of the ac cycle when the voltage at A becomes negative with respect to centre tap, the voltage at B would be positive. In this part of the cycle diode D1 would not conduct but diode D2 would, giving an output current and output voltage (across RL) during the negative half cycle of the input ac. Thus, we get output voltage during both the positive as well as the negative half of the cycle. Obviously, this is a more efficient circuit for getting rectified voltage or current than the half- wave rectifier. The rectified voltage is in the form of pulses of the shape of half sinusoids. Though it is unidirectional it does not have a steady value. To get steady dc output from the pulsating voltage normally a capacitor is connected across the output terminals (parallel to the load RL). One can also use an inductor in series with RL for FIGURE 14.19 (a) A Full-wave rectifierthe same purpose. Since these additional circuit; (b) Input wave forms given to thecircuits appear to filter out the ac ripple diode D1 at A and to the diode D2 at B;and give a pure dc voltage, so they are (c) Output waveform across the called filters. load RL connected in the full-wave Now we shall discuss the role of rectifier circuit. capacitor in filtering. When the voltage across the capacitor is rising, it gets charged. If there is no external load, it remains charged to the peak voltage of the rectified output. When there is a load, it gets discharged through the load and the voltage across it begins to fall. In the next half-cycle of rectified output it again gets charged to the peak value (Fig. 14.20). The rate of fall of the voltage across the capacitor depends inversely upon the product of capacitance C and the effective resistance RL used in the circuit and is called the time constant. To make the time constant large value of C should be large. So capacitor input filters use large capacitors. The output voltage obtained by using capacitor input filter is nearer to the peak voltage of the rectified voltage. This type of filter is most widely used in power supplies. 339 Reprint 2025-26 Physics FIGURE 14.20 (a) A full-wave rectifier with capacitor filter, (b) Input and output voltage of rectifier in (a). SUMMARY 1. Semiconductors are the basic materials used in the present solid state electronic devices like diode, transistor, ICs, etc. 2. Lattice structure and the atomic structure of constituent elements decide whether a particular material will be insulator, metal or semiconductor. 3. Metals have low resistivity (10–2 to 10–8 Wm), insulators have very high resistivity (>108 W m–1), while semiconductors have intermediate values of resistivity. 4. Semiconductors are elemental (Si, Ge) as well as compound (GaAs, CdS, etc.). 5. Pure semiconductors are called ‘intrinsic semiconductors’. The presence of charge carriers (electrons and holes) is an ‘intrinsic’ property of the material and these are obtained as a result of thermal excitation. The number of electrons (ne) is equal to the number of holes (nh ) in intrinsic conductors. Holes are essentially electron vacancies with an effective positive charge. 6. The number of charge carriers can be changed by ‘doping’ of a suitable impurity in pure semiconductors. Such semiconductors are known as extrinsic semiconductors. These are of two types (n-type and p-type). 7. In n-type semiconductors, ne >> nh while in p-type semiconductors nh >> ne. 8. n-type semiconducting Si or Ge is obtained by doping with pentavalent atoms (donors) like As, Sb, P, etc., while p-type Si or Ge can be obtained by doping with trivalent atom (acceptors) like B, Al, In etc. 9. nenh = ni 2 in all cases. Further, the material possesses an overall charge neutrality. 10. There are two distinct band of energies (called valence band and conduction band) in which the electrons in a material lie. Valence band energies are low as compared to conduction band energies. All energy levels in the valence band are filled while energy levels in the conduction band may be fully empty or partially filled. The electrons in the conduction band are free to move in a solid and are responsible for the conductivity. The extent of conductivity depends upon the energy gap (Eg) between the top of valence band (EV ) and the bottom of the conduction band EC. The electrons from valence band can be excited by Reprint 2025-26 Semiconductor Electronics: Materials, Devices and Simple Circuits heat, light or electrical energy to the conduction band and thus, produce a change in the current flowing in a semiconductor. 11. For insulators Eg > 3 eV, for semiconductors Eg is 0.2 eV to 3 eV, while for metals Eg » 0. 12. p-n junction is the ‘key’ to all semiconductor devices. When such a junction is made, a ‘depletion layer’ is formed consisting of immobile ion-cores devoid of their electrons or holes. This is responsible for a junction potential barrier. 13. By changing the external applied voltage, junction barriers can be changed. In forward bias (n-side is connected to negative terminal of the battery and p-side is connected to the positive), the barrier is decreased while the barrier increases in reverse bias. Hence, forward bias current is more (mA) while it is very small (mA) in a p-n junction diode. 14. Diodes can be used for rectifying an ac voltage (restricting the ac voltage to one direction). With the help of a capacitor or a suitable filter, a dc voltage can be obtained. POINTS TO PONDER 1. The energy bands (EC or EV) in the semiconductors are space delocalised which means that these are not located in any specific place inside the solid. The energies are the overall averages. When you see a picture in which EC or EV are drawn as straight lines, then they should be respectively taken simply as the bottom of conduction band energy levels and top of valence band energy levels. 2. In elemental semiconductors (Si or Ge), the n-type or p-type semiconductors are obtained by introducing ‘dopants’ as defects. In compound semiconductors, the change in relative stoichiometric ratio can also change the type of semiconductor. For example, in ideal GaAs the ratio of Ga:As is 1:1 but in Ga-rich or As-rich GaAs it could respectively be Ga1.1 As0.9 or Ga0.9 As1.1. In general, the presence of defects control the properties of semiconductors in many ways. EXERCISES

14.4 The speed of a travelling a circle. If you continue dropping pebbles in the pond, you wave see circles rapidly moving outward from the point where the

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 .

13.7 Energy in simple harmonic piston in a steam engine going back and forth, etc. Such a motion motion is termed as oscillatory motion. In this chapter we

13.3 A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin is entirely made of 6329Cu atoms (of mass 62.92960 u).

13.8 The simple pendulum study this motion. Summary The study of oscillatory motion is basic to physics; its Points to ponder concepts are required for the understanding of many physical Exercises phenomena. In musical instruments, like the sitar, the guitar or the violin, we come across vibrating strings that produce pleasing sounds. The membranes in drums and diaphragms in telephone and speaker systems vibrate to and fro about their mean positions. The vibrations of air molecules make the propagation of sound possible. In a solid, the atoms vibrate about their equilibrium positions, the average energy of vibrations being proportional to temperature. AC power supply give voltage that oscillates alternately going positive and negative about the mean value (zero). The description of a periodic motion, in general, and oscillatory motion, in particular, requires some fundamental concepts, like period, frequency, displacement, amplitude and phase. These concepts are developed in the next section. Reprint 2025-26 260 PHYSICS

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 ?

11.6 The threshold frequency for a certain metal is 3.3 × 1014 Hz. If light of frequency 8.2 × 1014 Hz is incident on the metal, predict the cut- off voltage for the photoelectric emission.

11.4 Heat, internal energy and which in turn rotate the wheels of the train. work In physics, we need to define the notions of heat, 11.5 First law of temperature, work, etc. more carefully. Historically, it took a thermodynamics long time to arrive at the proper concept of ‘heat’. Before the 11.6 Specific heat capacity modern picture, heat was regarded as a fine invisible fluid 11.7 Thermodynamic state filling in the pores of a substance. On contact between a hot variables and equation of body and a cold body, the fluid (called caloric) flowed from state the colder to the hotter body ! This is similar to what happens 11.8 Thermodynamic processes when a horizontal pipe connects two tanks containing water 11.9 Second law of up to different heights. The flow continues until the levels of thermodynamics water in the two tanks are the same. Likewise, in the ‘caloric’ 11.10 Reversible and irreversible picture of heat, heat flows until the ‘caloric levels’ (i.e., the processes temperatures) equalise. 11.11 Carnot engine In time, the picture of heat as a fluid was discarded in favour of the modern concept of heat as a form of energy. An Summary important experiment in this connection was due to Benjamin Points to ponder Thomson (also known as Count Rumford) in 1798. He Exercises observed that boring of a brass cannon generated a lot of heat, indeed enough to boil water. More significantly, the amount of heat produced depended on the work done (by the horses employed for turning the drill) but not on the sharpness of the drill. In the caloric picture, a sharper drill would scoop out more heat fluid from the pores; but this was not observed. A most natural explanation of the observations was that heat was a form of energy and the experiment demonstrated conversion of energy from one form to another–from work to heat. Reprint 2025-26 THERMODYNAMICS 227 Thermodynamics is the branch of physics that in a different context : we say the state of a system deals with the concepts of heat and temperature is an equilibrium state if the macroscopic and the inter-conversion of heat and other forms variables that characterise the system do not of energy. Thermodynamics is a macroscopic change in time. For example, a gas inside a closed science. It deals with bulk systems and does not rigid container, completely insulated from its go into the molecular constitution of matter. In surroundings, with fixed values of pressure, fact, its concepts and laws were formulated in the volume, temperature, mass and composition that nineteenth century before the molecular picture do not change with time, is in a state of of matter was firmly established. Thermodynamic thermodynamic equilibrium. description involves relatively few macroscopic variables of the system, which are suggested by common sense and can be usually measured directly. A microscopic description of a gas, for example, would involve specifying the co-ordinates and velocities of the huge number of molecules constituting the gas. The description in kinetic theory of gases is not so detailed but it does involve molecular distribution of velocities. Thermodynamic description of a gas, on the other (a) hand, avoids the molecular description altogether. Instead, the state of a gas in thermodynamics is specified by macroscopic variables such as pressure, volume, temperature, mass and composition that are felt by our sense perceptions and are measurable*. The distinction between mechanics and thermodynamics is worth bearing in mind. In mechanics, our interest is in the motion of particles (b) or bodies under the action of forces and torques. Fig. 11.1 (a) Systems A and B (two gases) separated Thermodynamics is not concerned with the by an adiabatic wall – an insulating wall motion of the system as a whole. It is concerned that does not allow flow of heat. (b) The with the internal macroscopic state of the body. same systems A and B separated by a When a bullet is fired from a gun, what changes diathermic wall – a conducting wall that is the mechanical state of the bullet (its kinetic allows heat to flow from one to another. In this case, thermal equilibrium is attainedenergy, in particular), not its temperature. When in due course. the bullet pierces a wood and stops, the kinetic energy of the bullet gets converted into heat, In general, whether or not a system is in a state changing the temperature of the bullet and the of equilibrium depends on the surroundings and surrounding layers of wood. Temperature is the nature of the wall that separates the system related to the energy of the internal (disordered) from the surroundings. Consider two gases A and motion of the bullet, not to the motion of the bullet B occupying two different containers. We know as a whole. experimentally that pressure and volume of a given mass of gas can be chosen to be its two

11.7 The work function for a certain metal is 4.2 eV. Will this metal give hotoelectric emission for incident radiation of wavelength 330 nm?

11.4 Monochromatic light of wavelength 632.8 nm is produced by a helium-neon laser. The power emitted is 9.42 mW. (a) Find the energy and momentum of each photon in the light beam, (b) How many photons per second, on the average, arrive at a target irradiated by this beam? (Assume the beam to have uniform cross-section which is less than the target area), and (c) How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon?

11.2 The work function of caesium metal is 2.14 eV. When light of frequency 6 ×1014Hz is incident on the metal surface, photoemission of electrons occurs. What is the (a) maximum kinetic energy of the emitted electrons, (b) Stopping potential, and (c) maximum speed of the emitted photoelectrons?

11.9 Light of wavelength 488 nm is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the stopping (cut-off) potential of photoelectrons is 0.38 V. Find the work function of the material from which the emitter is made.

11.5 In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be 4.12 × 10–15 V s. Calculate the value of Planck’s constant.

11.1 to study the variation of photocurrent with (a) intensity of radiation, (b) frequency of incident radiation, FIGURE 11.1 Experimental (c) the potential difference between the plates A and C, arrangement for study of and (d) the nature of the material of plate C. Light of photoelectric effect. different frequencies can be used by putting appropriate coloured filter or coloured glass in the path of light falling 277 Reprint 2025-26 Physics on the emitter C. The intensity of light is varied by changing the distance of the light source from the emitter. 11.4.1 Effect of intensity of light on photocurrent The collector A is maintained at a positive potential with respect to emitter C so that electrons ejected from C are attracted towards collector A. Keeping the frequency of the incident radiation and the potential fixed, the intensity of light is varied and the resulting photoelectric current is measured each time. It is found that the photocurrent increases linearly with intensity of incident light as shown graphically in Fig. 11.2. The photocurrent is directly proportional to the number of photoelectrons emitted per FIGURE 11.2 Variation of second. This implies that the number of photoelectrons Photoelectric current with emitted per second is directly proportional to the intensity intensity of light. of incident radiation. 11.4.2 Effect of potential on photoelectric current We first keep the plate A at some positive potential with respect to the plate C and illuminate the plate C with light of fixed frequency n and fixed intensity I1. We next vary the positive potential of plate A gradually and measure the resulting photocurrent each time. It is found that the photoelectric current increases with increase in positive (accelerating) potential. At some stage, for a certain positive potential of plate A, all the emitted electrons are collected by the plate A and the photoelectric current becomes maximum or saturates. If we increase the accelerating potential of plate A further, the photocurrent does not increase. This maximum value of the photoelectric current is called saturation current. Saturation current corresponds to the case when all the photoelectrons emitted by the emitter plate C reach the collector plate A. We now apply a negative (retarding) potential to the plate A with respect to the plate C and make it increasingly negative gradually. When the polarity is reversed, the electrons are repelled and only the sufficiently energetic electrons are able to reach the collector A. The photocurrent is found to decrease rapidly until it drops to zero at a certain sharply defined, critical value of the negative potential V0 on the plate A. For a particular frequency of incident radiation, the minimum negative (retarding) potential V0 given to the plate A for which the photocurrent stops or becomes zero is called the cut- off or stopping potential. The interpretation of the observation FIGURE 11.3 Variation of photocurrent with in terms of photoelectrons is collector plate potential for different straightforward. All the photoelectrons 278 intensity of incident radiation. emitted from the metal do not have the Reprint 2025-26 Dual Nature of Radiation and Matter same energy. Photoelectric current is zero when the stopping potential is sufficient to repel even the most energetic photoelectrons, with the maximum kinetic energy (Kmax), so that Kmax = e V0 (11.1) We can now repeat this experiment with incident radiation of the same frequency but of higher intensity I2 and I3 (I3 > I2 > I1). We note that the saturation currents are now found to be at higher values. This shows that more electrons are being emitted per second, proportional to the intensity of incident radiation. But the stopping potential remains the same as that for the incident radiation of intensity I1, as shown graphically in Fig. 11.3. Thus, for a given frequency of the incident radiation, the stopping potential is independent of its intensity. In other words, the maximum kinetic energy of photoelectrons depends on the light source and the emitter plate material, but is independent of intensity of incident radiation. 11.4.3 Effect of frequency of incident radiation on stopping potential We now study the relation between the frequency n of the incident radiation and the stopping potential V0. We suitably adjust the same intensity of light radiation at various frequencies and study the variation of photocurrent with collector plate potential. The resulting variation is shown in Fig. 11.4. We obtain different values of stopping potential but the same value of the saturation current for incident radiation of different frequencies. The energy of the emitted electrons depends on the frequency of the incident radiations. The stopping potential is FIGURE 11.4 Variation of photoelectric current more negative for higher frequencies of incident with collector plate potential for different radiation. Note from Fig. 11.4 that the stopping frequencies of incident radiation. potentials are in the order V03 > V02 > V01 if the frequencies are in the order n3 > n2 > n1 . This implies that greater the frequency of incident light, greater is the maximum kinetic energy of the photoelectrons. Consequently, we need greater retarding potential to stop them completely. If we plot a graph between the frequency of incident radiation and the corresponding stopping potential for different metals we get a straight line, as shown in Fig. 11.5. The graph shows that (i) the stopping potential V0 varies linearly with the frequency of incident radiation for a given photosensitive material. FIGURE 11.5 Variation of stopping potential V0 (ii) there exists a certain minimum cut-off with frequency n of incident radiation for a given photosensitive material. frequency n0 for which the stopping potential 279 is zero. Reprint 2025-26 Physics These observations have two implications: (i) The maximum kinetic energy of the photoelectrons varies linearly with the frequency of incident radiation, but is independent of its intensity. (ii) For a frequency n of incident radiation, lower than the cut-off frequency n0, no photoelectric emission is possible even if the intensity is large. This minimum, cut-off frequency n0, is called the threshold frequency. It is different for different metals. Different photosensitive materials respond differently to light. Selenium is more sensitive than zinc or copper. The same photosensitive substance gives different response to light of different wavelengths. For example, ultraviolet light gives rise to photoelectric effect in copper while green or red light does not. Note that in all the above experiments, it is found that, if frequency of the incident radiation exceeds the threshold frequency, the photoelectric emission starts instantaneously without any apparent time lag, even if the incident radiation is very dim. It is now known that emission starts in a time of the order of 10– 9 s or less. We now summarise the experimental features and observations described in this section. (i) For a given photosensitive material and frequency of incident radiation (above the threshold frequency), the photoelectric current is directly proportional to the intensity of incident light (Fig. 11.2). (ii) For a given photosensitive material and frequency of incident radiation, saturation current is found to be proportional to the intensity of incident radiation whereas the stopping potential is independent of its intensity (Fig. 11.3). (iii) For a given photosensitive material, there exists a certain minimum cut-off frequency of the incident radiation, called the threshold frequency, below which no emission of photoelectrons takes place, no matter how intense the incident light is. Above the threshold frequency, the stopping potential or equivalently the maximum kinetic energy of the emitted photoelectrons increases linearly with the frequency of the incident radiation, but is independent of its intensity (Fig. 11.5). (iv) The photoelectric emission is an instantaneous process without any apparent time lag (~10– 9s or less), even when the incident radiation is made exceedingly dim. 11.5 PHOTOELECTRIC EFFECT AND WAVE THEORY OF LIGHT The wave nature of light was well established by the end of the nineteenth century. The phenomena of interference, diffraction and polarisation were explained in a natural and satisfactory way by the wave picture of light. According to this picture, light is an electromagnetic wave consisting of electric and magnetic fields with continuous distribution of energy over 280 the region of space over which the wave is extended. Let us now see if this Reprint 2025-26 Dual Nature of Radiation and Matter wave picture of light can explain the observations on photoelectric emission given in the previous section. According to the wave picture of light, the free electrons at the surface of the metal (over which the beam of radiation falls) absorb the radiant energy continuously. The greater the intensity of radiation, the greater are the amplitude of electric and magnetic fields. Consequently, the greater the intensity, the greater should be the energy absorbed by each electron. In this picture, the maximum kinetic energy of the photoelectrons on the surface is then expected to increase with increase in intensity. Also, no matter what the frequency of radiation is, a sufficiently intense beam of radiation (over sufficient time) should be able to impart enough energy to the electrons, so that they exceed the minimum energy needed to escape from the metal surface . A threshold frequency, therefore, should not exist. These expectations of the wave theory directly contradict observations (i), (ii) and (iii) given at the end of sub-section 11.4.3. Further, we should note that in the wave picture, the absorption of energy by electron takes place continuously over the entire wavefront of the radiation. Since a large number of electrons absorb energy, the energy absorbed per electron per unit time turns out to be small. Explicit calculations estimate that it can take hours or more for a single electron to pick up sufficient energy to overcome the work function and come out of the metal. This conclusion is again in striking contrast to observation (iv) that the photoelectric emission is instantaneous. In short, the wave picture is unable to explain the most basic features of photoelectric emission.

11.6 EINSTEIN’S PHOTOELECTRIC EQUATION: ENERGY QUANTUM OF RADIATION In 1905, Albert Einstein (1879-1955) proposed a radically new picture of electromagnetic radiation to explain photoelectric effect. In this picture, photoelectric emission does not take place by continuous absorption of energy from radiation. Radiation energy is built up of discrete units – the so called quanta of energy of radiation. Each quantum of radiant energy has energy hn, where h is Planck’s constant and n the frequency of light. In photoelectric effect, an electron absorbs a quantum of energy (hn ) of radiation. If this quantum of energy absorbed exceeds the minimum energy needed for the electron to escape from the metal surface (work function f0), the electron is emitted with maximum kinetic energy Kmax = hn – f0 (11.2) More tightly bound electrons will emerge with kinetic energies less than the maximum value. Note that the intensity of light of a given frequency is determined by the number of photons incident per second. Increasing the intensity will increase the number of emitted electrons per second. However, the maximum kinetic energy of the emitted photoelectrons is determined by the energy of each photon. Equation (11.2) is known as Einstein’s photoelectric equation. We now see how this equation accounts in a simple and elegant manner all the observations on photoelectric effect given at the end of sub-section 281 11.4.3. Reprint 2025-26 Physics · According to Eq. (11.2), Kmax depends linearly on n, and is independent of intensity of radiation, in agreement with observation. This has happened because in Einstein’s picture, photoelectric effect arises from the absorption of a single quantum of radiation by a single electron. The intensity of radiation (that is proportional to the number of energy quanta per unit area per unit time) is irrelevant to this basic process. · Since Kmax must be non-negative, Eq. (11.2 ) implies that photoelectric emission is possible only if h n > f0 or n > n0 , where φ0 Albert Einstein (1879 – n0 = (11.3) 1955) Einstein, one of the h greatest physicists of all Equation (11.3) shows that the greater the work time, was born in Ulm, function f0, the higher the minimum or threshold Germany. In 1905, he frequency n0 needed to emit photoelectrons. Thus, published three path- breaking papers. In the there exists a threshold frequency n0 (= f0/h) for the first paper, he introduced metal surface, below which no photoelectric emission the notion of light quanta is possible, no matter how intense the incident (now called photons) and radiation may be or how long it falls on the surface. used it to explain the · In this picture, intensity of radiation as noted above, features of photoelectric effect. In the second paper, is proportional to the number of energy quanta per he developed a theory of unit area per unit time. The greater the number of Brownian motion, energy quanta available, the greater is the number of confirmed experimentally a electrons absorbing the energy quanta and greater, few years later and provided therefore, is the number of electrons coming out of a convincing evidence of the atomic picture of matter. the metal (for n > n0). This explains why, for n > n0, The third paper gave birth photoelectric current is proportional to intensity. to the special theory of · In Einstein’s picture, the basic elementary process relativity. In 1916, he involved in photoelectric effect is the absorption of a published the general light quantum by an electron. This process is1955) theory of relativity. Some of – Einstein’s most significant instantaneous. Thus, whatever may be the intensity later contributions are: the i.e., the number of quanta of radiation per unit area notion of stimulated per unit time, photoelectric emission is instantaneous. emission introduced in an Low intensity does not mean delay in emission, since(1879 alternative derivation of the basic elementary process is the same. Intensity Planck’s blackbody radiation law, static model only determines how many electrons are able to of the universe which participate in the elementary process (absorption of a started modern cosmology, light quantum by a single electron) and, therefore, the quantum statistics of a gas photoelectric current.EINSTEIN of massive bosons, and a Using Eq. (11.1), the photoelectric equation, Eq. (11.2), critical analysis of the foundations of quantum can be written as mechanics. In 1921, he was e V0 = h n – f0; for ν≥ ν0 awarded the Nobel Prize in physics for his contribution h φ0 (11.4) ν −ALBERT to theoretical physics and or V0 = the photoelectric effect. e e This is an important result. It predicts that the V0 282 versus n curve is a straight line with slope = (h/e), Reprint 2025-26 Dual Nature of Radiation and Matter independent of the nature of the material. During 1906-1916, Millikan performed a series of experiments on photoelectric effect, aimed at disproving Einstein’s photoelectric equation. He measured the slope of the straight line obtained for sodium, similar to that shown in Fig. 11.5. Using the known value of e, he determined the value of Planck’s constant h. This value was close to the value of Planck’s contant (= 6.626 × 10–34J s) determined in an entirely different context. In this way, in 1916, Millikan proved the validity of Einstein’s photoelectric equation, instead of disproving it. The successful explanation of photoelectric effect using the hypothesis of light quanta and the experimental determination of values of h and φ0, in agreement with values obtained from other experiments, led to the acceptance of Einstein’s picture of photoelectric effect. Millikan verified photoelectric equation with great precision, for a number of alkali metals over a wide range of radiation frequencies.

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?

10.2 What is the shape of the wavefront in each of the following cases: (a) Light diverging from a point source. (b) Light emerging out of a convex lens when a point source is placed at its focus. (c) The portion of the wavefront of light from a distant star intercepted by the Earth.

10.7 POLARISATION Consider holding a long string that is held horizontally, the other end of which is assumed to be fixed. If we move the end of the string up and down in a periodic manner, we will generate a wave propagating in the +x direction (Fig. 10.17). Such a wave could be described by the following equation FIGURE 10.17 (a) The curves represent the displacement of a string at t = 0 and at t = Dt, respectively when a sinusoidal wave is propagating in the +x-direction. (b) The curve represents the time variation of the displacement at x = 0 when a sinusoidal wave is propagating in the +x-direction. At x = Dx, the time variation of the displacement will be slightly displaced to the right. y (x,t) = a sin (kx – wt) (10.15) where a and w(= 2pn) represent the amplitude and the angular frequency of the wave, respectively; further, 2 π λ = (10.16) k represents the wavelength associated with the wave. We had discussed propagation of such waves in Chapter 14 of Class XI textbook. Since the displacement (which is along the y direction) is at right angles to the direction of propagation of the wave, we have what is known as a transverse wave. Also, since the displacement is in the y direction, it is often referred to as a y-polarised wave. Since each point on the string moves on a straight line, the wave is also referred to as a linearly polarised 269 Reprint 2025-26 Physics wave. Further, the string always remains confined to the x-y plane and therefore it is also referred to as a plane polarised wave. In a similar manner we can consider the vibration of the string in the x-z plane generating a z-polarised wave whose displacement will be given by z (x,t) = a sin (kx – wt) (10.17) It should be mentioned that the linearly polarised waves [described by Eqs. (10.15) and (10.17)] are all transverse waves; i.e., the displacement of each point of the string is always at right angles to the direction of propagation of the wave. Finally, if the plane of vibration of the string is changed randomly in very short intervals of time, then we have what is known as an unpolarised wave. Thus, for an unpolarised wave the displacement will be randomly changing with time though it will always be perpendicular to the direction of propagation. Light waves are transverse in nature; i.e., the electric field associated with a propagating light wave is always at right angles to the direction of propagation of the wave. This can be easily demonstrated using a simple polaroid. You must have seen thin plastic like sheets, which are called polaroids. A polaroid consists of long chain molecules aligned in a particular direction. The electric vectors (associated with the propagating light wave) along the direction of the aligned molecules get absorbed. Thus, if an unpolarised light wave is incident on such a polaroid then the light wave will get linearly polarised with the electric vector oscillating along a direction perpendicular to the aligned molecules; this direction is known as the pass-axis of the polaroid. Thus, if the light from an ordinary source (like a sodium lamp) passes through a polaroid sheet P1, it is observed that its intensity is reduced by half. Rotating P1 has no effect on the transmitted beam and transmitted intensity remains constant. Now, let an identical piece of polaroid P2 be placed before P1. As expected, the light from the lamp is reduced in intensity on passing through P2 alone. But now rotating P1 has a dramatic effect on the light coming from P2. In one position, the intensity transmitted by P2 followed by P1 is nearly zero. When turned by 90° from this position, P1 transmits nearly the full intensity emerging from P2 (Fig. 10.18). The experiment at figure 10.18 can be easily understood by assuming that light passing through the polaroid P2 gets polarised along the pass- axis of P2. If the pass-axis of P2 makes an angle q with the pass-axis of P1, then when the polarised beam passes through the polaroid P2, the component E cos q (along the pass-axis of P2) will pass through P2. Thus, as we rotate the polaroid P1 (or P2), the intensity will vary as: I = I0 cos2q (10.18) where I0 is the intensity of the polarized light after passing through P1. This is known as Malus’ law. The above discussion shows that the Reprint 2025-26 Wave Optics FIGURE 10.18 (a) Passage of light through two polaroids P2 and P1. The transmitted fraction falls from 1 to 0 as the angle between them varies from 0° to 90°. Notice that the light seen through a single polaroid P1 does not vary with angle. (b) Behaviour of the electric vector when light passes through two polaroids. The transmitted polarisation is the component parallel to the polaroid axis. The double arrows show the oscillations of the electric vector. intensity coming out of a single polaroid is half of the incident intensity. By putting a second polaroid, the intensity can be further controlled from 50% to zero of the incident intensity by adjusting the angle between the pass-axes of two polaroids. Polaroids can be used to control the intensity, in sunglasses, windowpanes, etc. Polaroids are also used in photographic cameras and 3D movie cameras. Example 10.2 Discuss the intensity of transmitted light when a polaroid sheet is rotated between two crossed polaroids? Solution Let I0 be the intensity of polarised light after passing through the first polariser P1. Then the intensity of light after passing through second polariser P2 will be I = I 0cos 2θ, where q is the angle between pass axes of P1 and P2. Since P1 and P3 are crossed the angle between the pass axes of P2 and P3 will be (p/2–q). Hence the intensity of light emerging from P3 will be  π  – θ I = I 0 cos 2θ cos 2  2  EXAMPLE = I0 cos2q sin2q =(I0/4) sin22q Therefore, the transmitted intensity will be maximum when q = p/4. 10.2 271 Reprint 2025-26 Physics SUMMARY 1. Huygens’ principle tells us that each point on a wavefront is a source of secondary waves, which add up to give the wavefront at a later time. 2. Huygens’ construction tells us that the new wavefront is the forward envelope of the secondary waves. When the speed of light is independent of direction, the secondary waves are spherical. The rays are then perpendicular to both the wavefronts and the time of travel is the same measured along any ray. This principle leads to the well known laws of reflection and refraction. 3. The principle of superposition of waves applies whenever two or more sources of light illuminate the same point. When we consider the intensity of light due to these sources at the given point, there is an interference term in addition to the sum of the individual intensities. But this term is important only if it has a non-zero average, which occurs only if the sources have the same frequency and a stable phase difference. 4. Young’s double slit of separation d gives equally spaced interference fringes. 5. A single slit of width a gives a diffraction pattern with a central λ 2λ maximum. The intensity falls to zero at angles of ± , ± , etc., a a with successively weaker secondary maxima in between. 6. Natural light, e.g., from the sun is unpolarised. This means the electric vector takes all possible directions in the transverse plane, rapidly and randomly, during a measurement. A polaroid transmits only one component (parallel to a special axis). The resulting light is called linearly polarised or plane polarised. When this kind of light is viewed through a second polaroid whose axis turns through 2p, two maxima and minima of intensity are seen. POINTS TO PONDER 1. Waves from a point source spread out in all directions, while light was seen to travel along narrow rays. It required the insight and experiment of Huygens, Young and Fresnel to understand how a wave theory could explain all aspects of the behaviour of light. 2. The crucial new feature of waves is interference of amplitudes from different sources which can be both constructive and destructive, as shown in Young’s experiment. 3. Diffraction phenomena define the limits of ray optics. The limit of the ability of microscopes and telescopes to distinguish very close objects is set by the wavelength of light. 4. Most interference and diffraction effects exist even for longitudinal waves like sound in air. But polarisation phenomena are special to transverse waves like light waves. Reprint 2025-26 Wave Optics EXERCISES

10.6 A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a Young’s double-slit experiment. (a) Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm. (b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide? 273 Reprint 2025-26 Physics Chapter Eleven DUAL NATURE OF RADIATION AND MATTER 11.1 INTRODUCTION The Maxwell’s equations of electromagnetism and Hertz experiments on the generation and detection of electromagnetic waves in 1887 strongly established the wave nature of light. Towards the same period at the end of 19th century, experimental investigations on conduction of electricity (electric discharge) through gases at low pressure in a discharge tube led to many historic discoveries. The discovery of X-rays by Roentgen in 1895, and of electron by J. J. Thomson in 1897, were important milestones in the understanding of atomic structure. It was found that at sufficiently low pressure of about 0.001 mm of mercury column, a discharge took place between the two electrodes on applying the electric field to the gas in the discharge tube. A fluorescent glow appeared on the glass opposite to cathode. The colour of glow of the glass depended on the type of glass, it being yellowish-green for soda glass. The cause of this fluorescence was attributed to the radiation which appeared to be coming from the cathode. These cathode rays were discovered, in 1870, by William Crookes who later, in 1879, suggested that these rays consisted of streams of fast moving negatively charged particles. The British physicist J. J. Thomson (1856-1940) confirmed this hypothesis. By applying mutually perpendicular electric and magnetic fields across the discharge 274 tube, J. J. Thomson was the first to determine experimentally the speed Reprint 2025-26 Dual Nature of Radiation and Matter and the specific charge [charge to mass ratio (e/m)] of the cathode ray particles. They were found to travel with speeds ranging from about 0.1 to 0.2 times the speed of light (3 ×108 m/s). The presently accepted value of e/m is 1.76 × 1011 C/kg. Further, the value of e/m was found to be independent of the nature of the material/metal used as the cathode (emitter), or the gas introduced in the discharge tube. This observation suggested the universality of the cathode ray particles. Around the same time, in 1887, it was found that certain metals, when irradiated by ultraviolet light, emitted negatively charged particles having small speeds. Also, certain metals when heated to a high temperature were found to emit negatively charged particles. The value of e/m of these particles was found to be the same as that for cathode ray particles. These observations thus established that all these particles, although produced under different conditions, were identical in nature. J. J. Thomson, in 1897, named these particles as electrons, and suggested that they were fundamental, universal constituents of matter. For his epoch-making discovery of electron, through his theoretical and experimental investigations on conduction of electricity by gasses, he was awarded the Nobel Prize in Physics in 1906. In 1913, the American physicist R. A. Millikan (1868-1953) performed the pioneering oil-drop experiment for the precise measurement of the charge on an electron. He found that the charge on an oil-droplet was always an integral multiple of an elementary charge, 1.602 × 10–19 C. Millikan’s experiment established that electric charge is quantised. From the values of charge (e) and specific charge (e/m), the mass (m) of the electron could be determined.

10.1 Monochromatic light of wavelength 589 nm is incident from air on a water surface. What are the wavelength, frequency and speed of (a) reflected, and (b) refracted light? Refractive index of water is 1.33.

10.5 INTERFERENCE OF LIGHT WAVES AND YOUNG’S EXPERIMENT We will now discuss interference using light waves. If we use two sodium lamps illuminating two pinholes (Fig. 10.11) we will not observe any interference fringes. This is because of the fact that the light wave emitted from an ordinary source (like a sodium lamp) undergoes abrupt phase changes in times of the order of 10–10 seconds. Thus the light waves coming out from two independent sources of light will not have any fixed phase relationship and would be incoherent, when this FIGURE 10.11 If two sodium happens, as discussed in the previous section, the lamps illuminate two pinholes intensities on the screen will add up. S1 and S2, the intensities will add The British physicist Thomas Young used an up and no interference fringes will ingenious technique to “lock” the phases of the waves be observed on the screen. emanating from S1 and S2. He made two pinholes S1 and S2 (very close to each other) on an opaque screen [Fig. 10.12(a)]. These were illuminated by another pinholes that was in turn, lit by a bright source. Light waves spread out from S and fall on both S1 and S2. S1 and S2 then behave like two coherent sources because light waves coming out from S1 and S2 are derived from the same original source and any abrupt phase change in S will manifest in exactly similar phase changes in the light coming out from S1 and S2. Thus, the two sources S1 and S2 will be locked in phase; i.e., they will be coherent like the two vibrating needle in our water wave example [Fig. 10.8(a)]. The spherical waves emanating from S1 and S2 will produce interference fringes on the screen GG¢, as shown in Fig. 10.12(b). The positions of maximum and minimum intensities can be calculated by using the analysis given in Section 10.4. (a) (b) FIGURE 10.12 Young’s arrangement to produce interference pattern. 265 Reprint 2025-26 Physics We will have constructive interference resulting in a bright xd region when = nl. That is, D n λD x = xn = ; n = 0, ± 1, ± 2, ... (10.13) d On the other hand, we will have destructive xd 1 interference resulting in a dark region when = (n+ ) l D 2 that is  1 D1829) ) ; n  0,  1,  2 (10.14) x = xn = (n+– 2 d Thomas Young Thus dark and bright bands appear on the screen, (1773 – 1829) English as shown in Fig. 10.13. Such bands are called fringes. physicist, physician and Equations (10.13) and (10.14) show that dark and(1773 Egyptologist. Young worked bright fringes are equally spaced. on a wide variety of scientific problems, ranging from the structure of the eye and the mechanism ofYOUNG vision to the decipherment of the Rosetta stone. He revived the wave theory of light and recognised that interference phenomenaTHOMAS provide proof of the wave properties of light. FIGURE 10.13 Computer generated fringe pattern produced by two point source S1 and S2 on the screen GG¢ (Fig. 10.12); d = 0.025 mm, D = 5 cm and l = 5 × 10–5 cm.) (Adopted from OPTICS by A. Ghatak, Tata McGraw Hill Publishing Co. Ltd., New Delhi, 2000.) 10.6 DIFFRACTION If we look clearly at the shadow cast by an opaque object, close to the region of geometrical shadow, there are alternate dark and bright regions just like in interference. This happens due to the phenomenon of diffraction. Diffraction is a general characteristic exhibited by all types of waves, be it sound waves, light waves, water waves or matter waves. Since the wavelength of light is much smaller than the dimensions of most 266 obstacles; we do not encounter diffraction effects of light in everyday Reprint 2025-26 Wave Optics observations. However, the finite resolution of our eye or of optical instruments such as telescopes or microscopes is limited due to the phenomenon of diffraction. Indeed the colours that you see when a CD is viewed is due to diffraction effects. We will now discuss the phenomenon of diffraction. 10.6.1 The single slit In the discussion of Young’s experiment, we stated that a single narrow slit acts as a new source from which light spreads out. Even before Young, early experimenters – including Newton – had noticed that light spreads out from narrow holes and slits. It seems to turn around corners and enter regions where we would expect a shadow. These effects, known as diffraction, can only be properly understood using wave ideas. After all, you are hardly surprised to hear sound waves from someone talking around a corner! When the double slit in Young’s experiment is replaced by a single narrow slit (illuminated by a monochromatic source), a broad pattern with a central bright region is seen. On both sides, there are alternate dark and bright regions, the intensity becoming weaker away from the centre (Fig. 10.15). To understand this, go to Fig. 10.14, which shows a parallel beam of light falling normally on a single slit LN of width a. The diffracted light goes on to meet FIGURE 10.14 The geometry of path a screen. The midpoint of the slit is M. differences for diffraction by a single slit. A straight line through M perpendicular to the slit plane meets the screen at C. We want the intensity at any point P on the screen. As before, straight lines joining P to the different points L,M,N, etc., can be treated as parallel, making an angle q with the normal MC. The basic idea is to divide the slit into much smaller parts, and add their contributions at P with the proper phase differences. We are treating different parts of the wavefront at the slit as secondary sources. Because the incoming wavefront is parallel to the plane of the slit, these sources are in phase. It is observed that the intensity has a central maximum at q = 0 and other secondary maxima at q l (n+1/2) l/a, which go on becoming weaker and weaker with increasing n. The minima (zero intensity) are at q l nl/a, n = ±1, ±2, ±3, .... FIGURE 10.15 Intensity The photograph and intensity pattern corresponding distribution and photograph of to it is shown in Fig. 10.15. fringes due to diffraction There has been prolonged discussion about at single slit. difference between intereference and diffraction among 267 Reprint 2025-26 Physics scientists since the discovery of these phenomena. In this context, it is interesting to note what Richard Feynman* has said in his famous Feynman Lectures on Physics: No one has ever been able to define the difference between interference and diffraction satisfactorily. It is just a question of usage, and there is no specific, important physical difference between them. The best we can do is, roughly speaking, is to say that when there are only a few sources, say two interfering sources, then the result is usually called interference, but if there is a large number of them, it seems that the word diffraction is more often used. In the double-slit experiment, we must note that the pattern on the screen is actually a superposition of single-slit diffraction from each slit or hole, and the double-slit interference pattern. 10.6.210.6.210.6.210.6.210.6.2 SeeingSeeingSeeingSeeingSeeing thethethethethe singlesinglesinglesinglesingle slitslitslitslitslit diffractiondiffractiondiffractiondiffractiondiffraction patternpatternpatternpatternpattern It is surprisingly easy to see the single-slit diffraction pattern for oneself. The equipment needed can be found in most homes –– two razor blades and one clear glass electric bulb preferably with a straight filament. One has to hold the two blades so that the edges are parallel and have a narrow slit in between. This is easily done with the thumb and forefingers (Fig. 10.16). Keep the slit parallel to the filament, right in front of the eye. Use spectacles if you normally do. With slight adjustment of the width of the slit and the parallelism of the edges, the pattern should be seen with its bright and dark bands. Since the position of all the bands (except the central one) depends on wavelength, they will show some colours. Using a filter for red or blue will make the fringes clearer. With both filters available, the wider fringes for red compared to blue FIGUREFIGUREFIGUREFIGUREFIGURE 10.1610.1610.1610.1610.16 can be seen. Holding two blades to In this experiment, the filament plays the role of the first slit S in form a single slit. A bulb filament viewed Fig. 10.15. The lens of the eye focuses the pattern on the screen (the through this shows retina of the eye). clear diffraction With some effort, one can cut a double slit in an aluminium foil with bands. a blade. The bulb filament can be viewed as before to repeat Young’s experiment. In daytime, there is another suitable bright source subtending a small angle at the eye. This is the reflection of the Sun in any shiny convex surface (e.g., a cycle bell). Do not try direct sunlight – it can damage the eye and will not give fringes anyway as the Sun subtends an angle of (1/2)°. In interference and diffraction, light energy is redistributed. If it reduces in one region, producing a dark fringe, it increases in another region, producing a bright fringe. There is no gain or loss of energy, which is consistent with the principle of conservation of energy. * Richard Feynman was one of the recipients of the 1965 Nobel Prize in Physics 268 for his fundamental work in quantum electrodynamics. Reprint 2025-26 Wave Optics

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

75.4 and 24.6 per cent, respectively. Thus, the average mass of a chlorine atom is obtained by the weighted average of the masses of the two isotopes, which works out to be 75.4 × 34.98 + 24.6 × 36.98 = 100 = 35.47 u which agrees with the atomic mass of chlorine. Even the lightest element, hydrogen has three isotopes having masses 1.0078 u, 2.0141 u, and 3.0160 u. The nucleus of the lightest atom of hydrogen, which has a relative abundance of 99.985%, is called the proton. The mass of a proton is −27 m p = 1.00727 u = 1.67262 × 10 kg (13.2) This is equal to the mass of the hydrogen atom (= 1.00783u), minus the mass of a single electron (me = 0.00055 u). The other two isotopes of hydrogen are called deuterium and tritium. Tritium nuclei, being unstable, do not occur naturally and are produced artificially in laboratories. The positive charge in the nucleus is that of the protons. A proton carries one unit of fundamental charge and is stable. It was earlier thought that the nucleus may contain electrons, but this was ruled out later using arguments based on quantum theory. All the electrons of an atom are outside the nucleus. We know that the number of these electrons outside 307the nucleus of the atom is Z, the atomic number. The total charge of the Reprint 2025-26 Physics atomic electrons is thus (–Ze), and since the atom is neutral, the charge of the nucleus is (+Ze). The number of protons in the nucleus of the atom is, therefore, exactly Z, the atomic number. Discovery of Neutron Since the nuclei of deuterium and tritium are isotopes of hydrogen, they must contain only one proton each. But the masses of the nuclei of hydrogen, deuterium and tritium are in the ratio of 1:2:3. Therefore, the nuclei of deuterium and tritium must contain, in addition to a proton, some neutral matter. The amount of neutral matter present in the nuclei of these isotopes, expressed in units of mass of a proton, is approximately equal to one and two, respectively. This fact indicates that the nuclei of atoms contain, in addition to protons, neutral matter in multiples of a basic unit. This hypothesis was verified in 1932 by James Chadwick who observed emission of neutral radiation when beryllium nuclei were bombarded with alpha-particles (a-particles are helium nuclei, to be discussed in a later section). It was found that this neutral radiation could knock out protons from light nuclei such as those of helium, carbon and nitrogen. The only neutral radiation known at that time was photons (electromagnetic radiation). Application of the principles of conservation of energy and momentum showed that if the neutral radiation consisted of photons, the energy of photons would have to be much higher than is available from the bombardment of beryllium nuclei with a-particles. The clue to this puzzle, which Chadwick satisfactorily solved, was to assume that the neutral radiation consists of a new type of neutral particles called neutrons. From conservation of energy and momentum, he was able to determine the mass of new particle ‘as very nearly the same as mass of proton’. The mass of a neutron is now known to a high degree of accuracy. It is m n = 1.00866 u = 1.6749×10–27 kg (13.3) Chadwick was awarded the 1935 Nobel Prize in Physics for his discovery of the neutron. A free neutron, unlike a free proton, is unstable. It decays into a proton, an electron and a antineutrino (another elementary particle), and has a mean life of about 1000s. It is, however, stable inside the nucleus. The composition of a nucleus can now be described using the following terms and symbols: Z - atomic number = number of protons [13.4(a)] N - neutron number = number of neutrons [13.4(b)] A - mass number = Z + N = total number of protons and neutrons [13.4(c)] One also uses the term nucleon for a proton or a neutron. Thus the number of nucleons in an atom is its mass number A. Nuclear species or nuclides are shown by the notation ZA X where X is the chemical symbol of the species. For example, the nucleus of gold is denoted by 19779 Au . It contains 197 nucleons, of which 79 are protons 308 and the rest118 are neutrons. Reprint 2025-26 Nuclei The composition of isotopes of an element can now be readily explained. The nuclei of isotopes of a given element contain the same number of protons, but differ from each other in their number of neutrons. Deuterium, 12 H, which is an isotope of hydrogen, contains one proton and one neutron. Its other isotope tritium, 13 H, contains one proton and two neutrons. The element gold has 32 isotopes, ranging from A =173 to A = 204. We have already mentioned that chemical properties of elements depend on their electronic structure. As the atoms of isotopes have identical electronic structure they have identical chemical behaviour and are placed in the same location in the periodic table. All nuclides with same mass number A are called isobars. For example, the nuclides 13 H and 32He are isobars. Nuclides with same neutron number N but different atomic number Z, for example 19880 Hg and 19779 Au , are called isotones.