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

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

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