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

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.

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.