1.1 INTRODUCTION

Measurement of any physical quantity involves comparison with a certain basic, arbitrarily chosen, internationally accepted reference standard called unit. The result of a measurement of a physical quantity is expressed by a number (or numerical measure) accompanied by a unit. Although the number of physical quantities appears to be very large, we need only a limited number of units for expressing all the physical quantities, since they are interrelated with one another. The units for the fundamental or base quantities are called fundamental or base units. The units of all other physical quantities can be expressed as combinations of the base units. Such units obtained for the derived quantities are called derived units. A complete set of these units, both the base units and derived units, is known as the system of units.

1.2 THE INTERNATIONAL SYSTEM OF UNITS

In earlier time scientists of different countries were using different systems of units for measurement. Three such systems, the CGS, the FPS (or British) system and the MKS system were in use extensively till recently.

The base units for length, mass and time in these systems were as follows :

• In CGS system they were centimetre, gram and second respectively.
• In FPS system they were foot, pound and second respectively.
• In MKS system they were metre, kilogram and second respectively.

The system of units which is at present internationally accepted for measurement is the Système Internationale d’ Unites (French for International System of Units), abbreviated as SI. The SI, with standard scheme of symbols, units and abbreviations, developed by the Bureau International des Poids et measures (The International Bureau of Weights and Measures, BIPM) in 1971 were recently revised by the General Conference on Weights and Measures in November 2018. The scheme is now for international usage in scientific, technical, industrial and commercial work. Because SI units used decimal system, conversions within the system are quite simple and convenient. We shall follow the SI units in this book

In SI, there are seven base units as given in Table 1.1. Besides the seven base units, there are two more units that are defined for (a) plane angle $\mathrm{d}\theta$ as the ratio of length of arc ds to the radius $r$ and (b) solid angle $\mathrm{d}\mathrm{\Omega }$ as the ratio of the intercepted area $\mathrm{d}A$ of the spherical surface, described about the apex $\mathrm{O}$ as the centre, to the square of its radius $r$, as shown in Fig. 1.1(a) and (b) respectively. The unit for plane angle is radian with the symbol rad and the unit for the solid angle is steradian with the symbol sr. Both these are dimensionless quantities.

Table 1.1 SI Base Quantities and Units*

		SI Units
Base quantity	Name	Symbol	Definition
Length	metre	$\mathrm{m}$	The metre, symbol $\mathrm{m}$, is the SI unit of length. It is defined by taking the fixed numerical value of the speed of light in vacuum $c$ to be 299792458 when expressed in the unit $\mathrm{m}{\mathrm{s}}^{-1}$, where the second is defined in terms of the caesium frequency $\mathrm{\Delta }\nu cs$.
Mass	kilogram	$\mathrm{kg}$	The kilogram, symbol $\mathrm{kg}$, is the SI unit of mass. It is defined by taking the fixed numerical value of the Planck constant $h$ to be ${6.6260701510}^{-34}$ when expressed in the unit $\mathrm{J}\mathrm{s}$, which is equal to $\mathrm{kg}{\mathrm{m}}^2{\mathrm{s}}^{-1}$, where the metre and the second are defined in terms of $c$ and $\mathrm{\Delta }Vcs$.
Time	second	$\mathrm{s}$	The second, symbol s, is the SI unit of time. It is defined by taking the fixed numerical value of the caesium frequency $\mathrm{\Delta }Vcs$, the unperturbed ground- state hyperfine transition frequency of the caesium- 133 atom, to be 9192631770 when expressed in the unit $\mathrm{Hz}$, which is equal to s ${}^{-1}$.
Electric	ampere	A	The ampere, symbol $\mathrm{A}$, is the SI unit of electric current. It is defined by taking the fixed numerical value of the elementary charge $e$ to be ${1.60217663410}^{-19}$ when expressed in the unit $C$, which is equal to $\mathrm{A}$, where the second is defined in terms of $\mathrm{\Delta }Vcs$.
Thermo dynamic Temperature	kelv	K	The kelvin, symbol $\mathrm{K}$, is the SI unit of thermodynamic temperature. It is defined by taking the fixed numerical value of the Boltzmann constant $\mathrm{k}$ to be ${1.38064910}^{-23}$ when expressed in the unit $\mathrm{J}{\mathrm{K}}^{-1}$, which is equal to $\mathrm{kg}{\mathrm{m}}^2{\mathrm{s}}^{-2}{\mathrm{k}}^{-1}$, where the kilogram, metre and second are defined in terms of $h,c$ and $\mathrm{\Delta }Vcs$.
Amount of substance	mole	mol	The mole, symbol mol, is the SI unit of amount of substance. One mole contains exactly ${6.0221407610}^{23}$ elementary entities. This number is the fixed numerical value of the Avogadro constant, $N_A$, when expressed in the unit mol ${}^{-1}$ and is called the Avogadro number. The amount of substance, symbol $n$, of a system is a measure of the number of specified elementary entities. An elementary entity may be an atom, a molecule, an ion, an electron, any other particle or specified group of particles.
Luminous intensity	candela	$\mathrm{cd}$	The candela, symbol cd, is the SI unit of luminous intensity in given direction. It is defined by taking the fixed numerical value of the luminous efficacy of monochromatic radiation of frequency ${54010}^{12}\mathrm{Hz},{\mathrm{K}}_{\mathrm{ed}}$, to be 683 when expressed in the unit $\mathrm{lm}\mathrm{W}{\mathrm{W}}^{-1}$, which is equal to $\mathrm{cd}\mathrm{sr}\mathrm{W}{\mathrm{W}}^{-1}$, or $\mathrm{cd}\mathrm{sr}{\mathrm{kg}}^{-1}{\mathrm{m}}^{-2}{\mathrm{s}}^3$, where the kilogram, metre and second are defined in terms of $h,c$ and $\mathrm{\Delta }vcs$.

Table 1.2 Some units retained for general use (Though outside SI)

Name	Symbol	Value in SI Unit
minute	min	$60\mathrm{s}$
hour	$\mathrm{h}$	$60\min =3600\mathrm{s}$
day	$\mathrm{d}$	$24\mathrm{h}=86400\mathrm{s}$
year	$\mathrm{y}$	$365.25\mathrm{d}=3.156\times {10}^7\mathrm{s}$
degree	o	$1^{\circ }=(\pi /180)\mathrm{rad}$
litre	$\mathrm{L}$	$\mathrm{I}{\mathrm{dm}}^3={10}^{-3}{\mathrm{m}}^3$
tonne	$\mathrm{t}$	${10}^3\mathrm{kg}$
carat	$\mathrm{c}$	$200\mathrm{mg}$
bar	bar	$0.1\mathrm{MPa}={10}^5\mathrm{Pa}$
curie	$\mathrm{Ci}$	$3.7\times {10}^{10}{\mathrm{s}}^{-1}$
roentgen	$\mathrm{R}$	$2.58\times {10}^{-4}\mathrm{C}/\mathrm{kg}$
quintal	$\mathrm{q}$	$100{\mathrm{kg}}^2$
barn	$\mathrm{b}$	$100{\mathrm{fm}}^2={10}^{-28}{\mathrm{m}}^2$
are	$\mathrm{a}$	$1{\mathrm{dam}}^2={10}^2{\mathrm{m}}^2$
hectare	ha	$1{\mathrm{hm}}^2={10}^4{\mathrm{m}}^2$
standard atmospheric pressure	atm	$101325\mathrm{Pa}=1.013\times {10}^5\mathrm{Pa}$

Note that when mole is used, the elementary entities must be specified. These entities may be atoms, molecules, ions, electrons, other particles or specified groups of such particles.

We employ units for some physical quantities that can be derived from the seven base units (Appendix A 6). Some derived units in terms of the SI base units are given in (Appendix A 6.1). Some SI derived units are given special names (Appendix A 6.2) and some derived SI units make use of these units with special names and the seven base units (Appendix A 6.3). These are given in Appendix A 6.2 and A 6.3 for your ready reference. Other units retained for general use are given in Table 1.2.

Common SI prefixes and symbols for multiples and sub-multiples are given in Appendix A2. General guidelines for using symbols for physical quantities, chemical elements and nuclides are given in Appendix A7 and those for SI units and some other units are given in Appendix A8 for your guidance and ready reference.

1.3 SIGNIFICANT FIGURES

As discussed above, every measurement involves errors. Thus, the result of measurement should be reported in a way that indicates the precision of measurement. Normally, the reported result of measurement is a number that includes all digits in the number that are known reliably plus the first digit that is uncertain. The reliable digits plus the first uncertain digit are known as significant digits or significant figures. If we say the period of oscillation of a simple pendulum is $1.62\mathrm{s}$, the digits 1 and 6 are reliable and certain, while the digit 2 is uncertain. Thus, the measured value has three significant figures. The length of an object reported after measurement to be $287.5\mathrm{cm}$ has four significant figures, the digits $2,8,7$ are certain while the digit 5 is uncertain. Clearly, reporting the result of measurement that includes more digits than the significant digits is superfluous and also misleading since it would give a wrong idea about the precision of measurement.

The rules for determining the number of significant figures can be understood from the following examples. Significant figures indicate, as already mentioned, the precision of measurement which depends on the least count of the measuring instrument. A choice of change of different units does not change the number of significant digits or figures in a measurement. This important remark makes most of the following observations clear:

(1) For example, the length $2.308\mathrm{cm}$ has four significant figures. But in different units, the same value can be written as $0.02308\mathrm{m}$ or 23.08 $\mathrm{mm}$ or $23080\mu \mathrm{m}$.

All these numbers have the same number of significant figures (digits 2, 3, 0, 8), namely four.

This shows that the location of decimal point is of no consequence in determining the number of significant figures.

The example gives the following rules:

• All the non-zero digits are significant.
• All the zeros between two non-zero digits are significant, no matter where the decimal point is, if at all.
• If the number is less than 1 , the zero(s) on the right of decimal point but to the left of the first non-zero digit are not significant. [In $\underset{―}{0}.\underset{―}{00}2308$, the underlined zeroes are not significantl.
• The terminal or trailing zero(s) in a number without a decimal point are not significant.

[Thus $123\mathrm{m}=12300\mathrm{cm}=123000\mathrm{mm}$ has three significant figures, the trailing zero(s) being not significant.] However, you can also see the next observation.

• The trailing zero(s) in a number with a decimal point are significant.

[The numbers 3.500 or 0.06900 have four significant figures each.]

(2) There can be some confusion regarding the trailing zero(s). Suppose a length is reported to be $4.700\mathrm{m}$. It is evident that the zeroes here are meant to convey the precision of measurement and are, therefore, significant. [If these were not, it would be superfluous to write them explicitly, the reported measurement would have been simply $4.7\mathrm{m}$ ]. Now suppose we change units, then

$4.700\mathrm{m}=470.0\mathrm{cm}=4700\mathrm{mm}=0.004700\mathrm{km}$

Since the last number has trailing zero(s) in a number with no decimal, we would conclude erroneously from observation (1) above that the number has two significant figures, while in fact, it has four significant figures and a mere change of units cannot change the number of significant figures.

(3) To remove such ambiguities in determining the number of significant figures, the best way is to report every measurement in scientific notation (in the power of 10). In this notation, every number is expressed as $a\times {10}^b$, where $a$ is a number between 1 and 10 , and $b$ is any positive or negative exponent (or power) of 10. In order to get an approximate idea of the number, we may round off the number $a$ to 1 (for $a\le 5$ ) and to 10 (for $5<a\le 10$ ). Then the number can be expressed approximately as ${10}^{\mathrm{b}}$ in which the exponent (or power) b of 10 is called order of magnitude of the physical quantity. When only an estimate is required, the quantity is of the order of ${10}^{\mathrm{b}}$. For example, the diameter of the earth $(1.28\times {10}^7\mathrm{m})$ is of the order of ${10}^7\mathrm{m}$ with the order of magnitude 7 . The diameter of hydrogen atom $(1.06\times {10}^{-10}\mathrm{m})$ is of the order of ${10}^{-10}\mathrm{m}$, with the order of magnitude -10 . Thus, the diameter of the earth is 17 orders of magnitude larger than the hydrogen atom.

It is often customary to write the decimal after the first digit. Now the confusion mentioned in (a) above disappears :

$$\begin{array}{rl}& 4.700\mathrm{m}=4.700\times {10}^2\mathrm{cm}\\ =& 4.700\times {10}^3\mathrm{mm}=4.700\times {10}^{-3}\mathrm{km}\end{array}$$

The power of 10 is irrelevant to the determination of significant figures. However, all zeroes appearing in the base number in the scientific notation are significant. Each number in this case has four significant figures.

Thus, in the scientific notation, no confusion arises about the trailing zero(s) in the base number $a$. They are always significant.

(4) The scientific notation is ideal for reporting measurement. But if this is not adopted, we use the rules adopted in the preceding example :

• For a number greater than 1, without any decimal, the trailing zero(s) are not significant.
• For a number with a decimal, the trailing zero(s) are significant.

(5) The digit 0 conventionally put on the left of a decimal for a number less than 1 (like 0.1250) is never significant. However, the zeroes at the end of such number are significant in a measurement.

(6) The multiplying or dividing factors which are neither rounded numbers nor numbers representing measured values are exact and have infinite number of significant digits. For example in $r=\frac{d}{2}$ or $\mathrm{s}=2\pi r$, the factor 2 is an exact number and it can be written as 2.0, 2.00 or 2.0000 as required. Similarly, in $T=\frac{t}{n},n$ is an exact number.

1.3.1 Rules for Arithmetic Operations with Significant Figures

The result of a calculation involving approximate measured values of quantities (i.e. values with limited number of significant figures) must reflect the uncertainties in the original measured values. It cannot be more accurate than the original measured values themselves on which the result is based. In general, the final result should not have more significant figures than the original data from which it was obtained. Thus, if mass of an object is measured to be, say, $4.237\mathrm{g}$ (four significant figures) and its volume is measured to be $2.51{\mathrm{cm}}^3$, then its density, by mere arithmetic division, is $1.68804780876\mathrm{g}/{\mathrm{cm}}^3$ upto 11 decimal places. It would be clearly absurd and irrelevant to record the calculated value of density to such a precision when the measurements on which the value is based, have much less precision. The following rules for arithmetic operations with significant figures ensure that the final result of a calculation is shown with the precision that is consistent with the precision of the input measured values :

(1) In multiplication or division, the final result should retain as many significant figures as are there in the original number with the least significant figures.

Thus, in the example above, density should be reported to three significant figures.

$$\text{ Density }=\frac{4.237\mathrm{g}}{2.51{\mathrm{cm}}^3}=1.69\mathrm{g}{\mathrm{cm}}^{-3}$$

Similarly, if the speed of light is given as $3.00\times {10}^8\mathrm{m}{\mathrm{s}}^{-1}$ (three significant figure) and one year ( $1\mathrm{y}=365.25\mathrm{d}$ ) has $3.1557\times {10}^7\mathrm{s}$ (five significant figures), the light year is $9.47\times {10}^{15}\mathrm{m}$ (three significant figures).

(2) In addition or subtraction, the final result should retain as many decimal places as are there in the number with the least decimal places.

For example, the sum of the numbers $436.32\mathrm{g},227.2\mathrm{g}$ and $0.301\mathrm{g}$ by mere arithmetic addition, is $663.821\mathrm{g}$. But the least precise measurement $(227.2\mathrm{g})$ is correct to only one decimal place. The final result should, therefore, be rounded off to $663.8\mathrm{g}$.

Similarly, the difference in length can be expressed as :

$0.307\mathrm{m}-0.304\mathrm{m}=0.003\mathrm{m}=3\times {10}^{-3}\mathrm{m}$.

Note that we should not use the rule (1) applicable for multiplication and division and write $664\mathrm{g}$ as the result in the example of addition and $3.00\times {10}^{-3}\mathrm{m}$ in the example of subtraction. They do not convey the precision of measurement properly. For addition and subtraction, the rule is in terms of decimal places.

1.3.2 Rounding off the Uncertain Digits

The result of computation with approximate numbers, which contain more than one uncertain digit, should be rounded off. The rules for rounding off numbers to the appropriate significant figures are obvious in most cases. A number $2.74\underset{―}{6}$ rounded off to three significant figures is 1.75 , while the number 1.743 would be 1.74 . The rule by convention is that the preceding digit is raised by 1 if the insignificant digit to be dropped (the underlined digit in this case) is more than 5 , and is left unchanged if the latter is less than 5. But what if the number is 2.745 in which the insignificant digit is 5 . Here, the convention is that if the preceding digit is even, the insignificant digit is simply dropped and, if it is odd, the preceding digit is raised by 1 . Then, the number 2.745 rounded off to three significant figures becomes 1.74. On the other hand, the number 2.735 rounded off to three significant figures becomes 1.74 since the preceding digit is odd.

In any involved or complex multi-step calculation, you should retain, in intermediate steps, one digit more than the significant digits and round off to proper significant figures at the end of the calculation. Similarly, a number known to be within many significant figures, such as in $1.99792458\times {10}^8\mathrm{m}/\mathrm{s}$ for the speed of light in vacuum, is rounded off to an approximate value $3\times {10}^8\mathrm{m}/\mathrm{s}$, which is often employed in computations. Finally, remember that exact numbers that appear in formulae like $2\pi$ in $T=2\pi \sqrt{\frac{L}{g}}$, have a large (infinite) number of significant figures. The value of $\pi =$ $3.1415926\dots$ is known to a large number of significant figures. You may take the value as 3.142 or 3.14 for $\pi$, with limited number of significant figures as required in specific cases.

Example 1.1 Each side of a cube is measured to be $7.203\mathrm{m}$. What are the total surface area and the volume of the cube to appropriate significant figures?

Answer The number of significant figures in the measured length is 4 . The calculated area and the volume should therefore be rounded off to 4 significant figures.

$$\begin{array}{rl}\text{Surface area of the cube}& =6(7.203{)}^2{\mathrm{m}}^2\\ & =311.299254{\mathrm{m}}^2\\ & =311.3{\mathrm{m}}^2\\ \text{Volume of the cube }& =(7.203{)}^3{\mathrm{m}}^3\\ & =373.714754{\mathrm{m}}^3\\ & =373.7{\mathrm{m}}^3\end{array}$$

Example 1.2 $5.74\mathrm{g}$ of a substance occupies $1.2{\mathrm{cm}}^3$. Express its density by keeping the significant figures in view.

Answer There are 3 significant figures in the measured mass whereas there are only 2 significant figures in the measured volume. Hence the density should be expressed to only 2 significant figures.

$$\begin{array}{rl}\text{ Density }& =\frac{5.74}{1.2}\mathrm{g}{\mathrm{cm}}^{-3}\\ & =4.8\mathrm{g}{\mathrm{cm}}^{-3}\end{array}$$

1.3.3 Rules for Determining the Uncertainty in the Results of Arithmatic Calculations

The rules for determining the uncertainty or error in the number/measured quantity in arithmetic operations can be understood from the following examples.

(1) If the length and breadth of a thin rectangular sheet are measured, using a metre scale as $16.2\mathrm{cm}$ and, $10.1\mathrm{cm}$ respectively, there are three significant figures in each measurement. It means that the length 1 may be written as

$$l=16.2\pm 0.1\mathrm{cm}$$

$$=16.2\mathrm{cm}\pm 0.6\mathrm{\%}$$

Similarly, the breadth $b$ may be written as

$$\begin{array}{rl}b& =10.1\pm 0.1\mathrm{cm}\\ & =10.1\mathrm{cm}\pm 1\mathrm{\%}\end{array}$$

Then, the error of the product of two (or more) experimental values, using the combination of errors rule, will be

$$\begin{array}{rl}1b& =163.62{\mathrm{cm}}^2\pm 1.6\mathrm{\%}\\ & =163.62\pm 2.6{\mathrm{cm}}^2\end{array}$$

This leads us to quote the final result as

$$1b=164\pm 3{\mathrm{cm}}^2$$

Here $3{\mathrm{cm}}^2$ is the uncertainty or error in the estimation of area of rectangular sheet.

(2) If a set of experimental data is specified to $\mathbf{n}$ significant figures, a result obtained by combining the data will also be valid to $n$ significant figures.

However, if data are subtracted, the number of significant figures can be reduced.

For example, $12.9\mathrm{g}-7.06\mathrm{g}$, both specified to three significant figures, cannot properly be evaluated as $5.84\mathrm{g}$ but only as $5.8\mathrm{g}$, as uncertainties in subtraction or addition combine in a different fashion (smallest number of decimal places rather than the number of significant figures in any of the number added or subtracted).

(3) The relative error of a value of number specified to significant figures depends not only on $\mathbf{n}$ but also on the number itself.

For example, the accuracy in measurement of mass $1.02\mathrm{g}$ is $\pm 0.01\mathrm{g}$ whereas another measurement $9.89\mathrm{g}$ is also accurate to $\pm 0.01\mathrm{g}$.

The relative error in $1.02\mathrm{g}$ is

$$\begin{array}{rl}& =(\pm 0.01/1.02)\times 100\mathrm{\%}\\ & =\pm 1\mathrm{\%}\end{array}$$

Similarly, the relative error in $9.89\mathrm{g}$ is

$$\begin{array}{rl}& =(\pm 0.01/9.89)\times 100\mathrm{\%}\\ & =\pm 0.1\mathrm{\%}\end{array}$$

Finally, remember that intermediate results in a multi-step computation should be calculated to one more significant figure in every measurement than the number of digits in the least precise measurement. These should be justified by the data and then the arithmetic operations may be carried out; otherwise rounding errors can build up. For example, the reciprocal of 9.58, calculated (after rounding off) to the same number of significant figures (three) is 0.104 , but the reciprocal of 0.104 calculated to three significant figures is 9.62. However, if we had written $1/9.58=0.1044$ and then taken the reciprocal to three significant figures, we would have retrieved the original value of 9.58.

This example justifies the idea to retain one more extra digit (than the number of digits in the least precise measurement) in intermediate steps of the complex multi-step calculations in order to avoid additional errors in the process of rounding off the numbers.

1.4 DIMENSIONS OF PHYSICAL QUANTITIES

The nature of a physical quantity is described by its dimensions. All the physical quantities represented by derived units can be expressed in terms of some combination of seven fundamental or base quantities. We shall call these base quantities as the seven dimensions of the physical world, which are denoted with square brackets [ ]. Thus, length has the dimension $[\mathrm{L}]$, mass $[\mathrm{M}]$, time [T], electric current $[\mathrm{A}]$, thermodynamic temperature $[\mathrm{K}]$, luminous intensity [cd], and amount of substance [mol]. The dimensions of a physical quantity are the powers (or exponents) to which the base quantities are raised to represent that quantity. Note that using the square brackets [ ] round a quantity means that we are dealing with ’the dimensions of’ the quantity.

In mechanics, all the physical quantities can be written in terms of the dimensions [L], [M] and $[\mathrm{T}]$. For example, the volume occupied by an object is expressed as the product of length, breadth and height, or three lengths. Hence the dimensions of volume are $[\mathrm{L}]\times [\mathrm{L}]\times [\mathrm{L}]=[\mathrm{L}{]}^3=\left[{\mathrm{L}}^3\right]$. As the volume is independent of mass and time, it is said to possess zero dimension in mass $\left[{\mathrm{M}}^{\circ }\right]$, zero dimension in time $\left[{\mathrm{T}}^{\circ }\right]$ and three dimensions in length.

Similarly, force, as the product of mass and acceleration, can be expressed as

Force $=$ mass $\times$ acceleration

$=$ mass $\times ($ length $)/(\text{ time }{)}^2$

The dimensions of force are $[\mathrm{M}][\mathrm{L}]/[\mathrm{T}{]}^2=$ $\left[\mathrm{M}\mathrm{L}{\mathrm{T}}^{-2}\right]$. Thus, the force has one dimension in mass, one dimension in length, and -2 dimensions in time. The dimensions in all other base quantities are zero.

Note that in this type of representation, the magnitudes are not considered. It is the quality of the type of the physical quantity that enters. Thus, a change in velocity, initial velocity, average velocity, final velocity, and speed are all equivalent in this context. Since all these quantities can be expressed as length/time, their dimensions are $[\mathrm{L}]/[\mathrm{T}]$ or $\left[\mathrm{L}{\mathrm{T}}^{-1}\right]$.

1.5 DIMENSIONAL FORMULAE AND DIMENSIONAL EQUATIONS

The expression which shows how and which of the base quantities represent the dimensions of a physical quantity is called the dimensional formula of the given physical quantity. For example, the dimensional formula of the volume is $\left[{\mathrm{M}}^{\circ }{\mathrm{L}}^3{\mathrm{T}}^{\circ }\right]$, and that of speed or velocity is $\left[{\mathrm{M}}^{\circ }{\mathrm{LT}}^{-1}\right]$. Similarly, $\left[{\mathrm{M}}^{\circ }\mathrm{L}{\mathrm{T}}^{-2}\right]$ is the dimensional formula of acceleration and $\left[\mathrm{M}{\mathrm{L}}^{-3}{\mathrm{T}}^{\circ }\right]$ that of mass density.

An equation obtained by equating a physical quantity with its dimensional formula is called the dimensional equation of the physical quantity. Thus, the dimensional equations are the equations, which represent the dimensions of a physical quantity in terms of the base quantities. For example, the dimensional equations of volume $[V]$, speed $[v]$, force $[F]$ and mass density $[\rho ]$ may be expressed as

$$\begin{array}{rl}& [V]=\left[{\mathrm{M}}^0{\mathrm{L}}^3{\mathrm{T}}^0\right]\\ & [v]=\left[{\mathrm{M}}^0\mathrm{L}{\mathrm{T}}^{-1}\right]\\ & [F]=\left[\mathrm{M}\mathrm{L}{\mathrm{T}}^{-2}\right]\\ & [\rho ]=\left[\mathrm{M}{\mathrm{L}}^{-3}{\mathrm{T}}^0\right]\end{array}$$

The dimensional equation can be obtained from the equation representing the relations between the physical quantities. The dimensional formulae of a large number and wide variety of physical quantities, derived from the equations representing the relationships among other physical quantities and expressed in terms of base quantities are given in Appendix 9 for your guidance and ready reference.

1.6 DIMENSIONAL ANALYSIS AND ITS APPLICATIONS

The recognition of concepts of dimensions, which guide the description of physical behaviour is of basic importance as only those physical quantities can be added or subtracted which have the same dimensions. A thorough understanding of dimensional analysis helps us in deducing certain relations among different physical quantities and checking the derivation, accuracy and dimensional consistency or homogeneity of various mathematical expressions. When magnitudes of two or more physical quantities are multiplied, their units should be treated in the same manner as ordinary algebraic symbols. We can cancel identical units in the numerator and denominator. The same is true for dimensions of a physical quantity. Similarly, physical quantities represented by symbols on both sides of a mathematical equation must have the same dimensions.

1.6.1 Checking the Dimensional Consistency of Equations

The magnitudes of physical quantities may be added together or subtracted from one another only if they have the same dimensions. In other words, we can add or subtract similar physical quantities. Thus, velocity cannot be added to force, or an electric current cannot be subtracted from the thermodynamic temperature. This simple principle called the principle of homogeneity of dimensions in an equation is extremely useful in checking the correctness of an equation. If the dimensions of all the terms are not same, the equation is wrong. Hence, if we derive an expression for the length (or distance) of an object, regardless of the symbols appearing in the original mathematical relation, when all the individual dimensions are simplified, the remaining dimension must be that of length. Similarly, if we derive an equation of speed, the dimensions on both the sides of equation, when simplified, must be of length/ time, or $\left[\mathrm{L}{\mathrm{T}}^{-1}\right]$.

Dimensions are customarily used as a preliminary test of the consistency of an equation, when there is some doubt about the correctness of the equation. However, the dimensional consistency does not guarantee correct equations. It is uncertain to the extent of dimensionless quantities or functions. The arguments of special functions, such as the trigonometric, logarithmic and exponential functions must be dimensionless. A pure number, ratio of similar physical quantities, such as angle as the ratio (length/length), refractive index as the ratio (speed of light in vacuum/speed of light in medium) etc., has no dimensions

Now we can test the dimensional consistency or homogeneity of the equation

$$x=x_0+v_{0}t+(1/2)a{t}^2$$

for the distance $x$ travelled by a particle or body in time $t$ which starts from the position $x_0$ with an initial velocity $v_0$ at time $t=0$ and has uniform acceleration $a$ along the direction of motion.

The dimensions of each term may be written as

$$\begin{array}{c}[x]=[\mathrm{L}]\\ \left[x_0\right]=[\mathrm{L}]\\ \left[v_{0}t\right]=\left[\mathrm{L}\mathrm{T}{\mathrm{T}}^{-1}\right][\mathrm{T}]\\ =[\mathrm{L}]\\ [(1/2)a{t}^2]=\left[\mathrm{L}{\mathrm{T}}^{-2}\right]\left[{\mathrm{T}}^2\right]\\ =[\mathrm{L}]\end{array}$$

As each term on the right hand side of this equation has the same dimension, namely that of length, which is same as the dimension of left hand side of the equation, hence this equation is a dimensionally correct equation.

It may be noted that a test of consistency of dimensions tells us no more and no less than a test of consistency of units, but has the advantage that we need not commit ourselves to a particular choice of units, and we need not worry about conversions among multiples and sub-multiples of the units. It may be borne in mind that if an equation fails this consistency test, it is proved wrong, but if it passes, it is not proved right. Thus, a dimensionally correct equation need not be actually an exact (correct) equation, but a dimensionally wrong (incorrect) or inconsistent equation must be wrong.

Example 1.3 Let us consider an equation

$$\frac{1}{2}m{v}^2=mgh$$

where $m$ is the mass of the body, $v$ its velocity, $q$ is the acceleration due to gravity and $h$ is the height. Check whether this equation is dimensionally correct.

Answer The dimensions of LHS are

$$\begin{array}{c}[\mathrm{M}]{\left[\mathrm{L}{\mathrm{T}}^{-1}\right]}^2=[\mathrm{M}]\left[{\mathrm{L}}^2{\mathrm{T}}^{-2}\right]\\ =\left[\mathrm{M}{\mathrm{L}}^2{\mathrm{T}}^{-2}\right]\end{array}$$

The dimensions of RHS are

$$\begin{array}{rl}[\mathrm{M}]\left[\mathrm{L}{\mathrm{T}}^{-2}\right][\mathrm{L}]& =[\mathrm{M}]\left[{\mathrm{L}}^2{\mathrm{T}}^{-2}\right]\\ & =\left[\mathrm{M}{\mathrm{L}}^2{\mathrm{T}}^{-2}\right]\end{array}$$

The dimensions of LHS and RHS are the same and hence the equation is dimensionally correct.

Example 1.4 The SI unit of energy is $\mathrm{J}=\mathrm{kg}{\mathrm{m}}^2{\mathrm{s}}^{-2}$; that of speed $v$ is $\mathrm{m}{\mathrm{s}}^{-1}$ and of acceleration $a$ is $\mathrm{m}{\mathrm{s}}^{-2}$. Which of the formulae for kinetic energy $(K)$ given below can you rule out on the basis of dimensional arguments ( $m$ stands for the mass of the body) :

(a) $K=m^2{v}^3$

(b) $K=(1/2)m{v}^2$

(c) $K=ma$

(d) $K=(3/16)m{v}^2$

(e) $K=(1/2)m{v}^2+ma$

Answer Every correct formula or equation must have the same dimensions on both sides of the equation. Also, only quantities with the same physical dimensions can be added or subtracted. The dimensions of the quantity on the right side are $\left[{\mathrm{M}}^2{\mathrm{L}}^3{\mathrm{T}}^{-3}\right]$ for (a); $\left[\mathrm{M}{\mathrm{L}}^2{\mathrm{T}}^{-2}\right]$ for (b) and (d); [M L T ${}^{-2}$ ] for (c). The quantity on the right side of (e) has no proper dimensions since two quantities of different dimensions have been added. Since the kinetic energy $K$ has the dimensions of $\left[\mathrm{M}{\mathrm{L}}^2{\mathrm{T}}^{-2}\right]$, formulas (a), (c) and (e) are ruled out. Note that dimensional arguments cannot tell which of the two, (b) or (d), is the correct formula. For this, one must turn to the actual definition of kinetic energy (see Chapter 5). The correct formula for kinetic energy is given by (b).

2.10.2 Deducing Relation among the Physical Quantities

The method of dimensions can sometimes be used to deduce relation among the physical quantities. For this we should know the dependence of the physical quantity on other quantities (upto three physical quantities or linearly independent variables) and consider it as a product type of the dependence. Let us take an example.

Example 1.5 Consider a simple pendulum, having a bob attached to a string, that oscillates under the action of the force of gravity. Suppose that the period of oscillation of the simple pendulum depends on its length (I), mass of the bob $(m)$ and acceleration due to gravity $(g)$. Derive the expression for its time period using method of dimensions.

Answer The dependence of time period $T$ on the quantities $l,g$ and $m$ as a product may be written as :

$T=k{l}^x{g}^y{m}^z$

where $\mathrm{k}$ is dimensionless constant and $x,y$ and $z$ are the exponents.

By considering dimensions on both sides, we have

$$\left[{\mathrm{L}}^{\mathrm{o}}{\mathrm{M}}^{\mathrm{o}}{\mathrm{T}}^1\right]={\left[{\mathrm{L}}^1\right]}^x{\left[{\mathrm{L}}^1{\mathrm{T}}^{-2}\right]}^y{\left[{\mathrm{M}}^1\right]}^z$$

$={\mathrm{L}}^{x+y}{\mathrm{T}}^{-2y}{\mathrm{M}}^z$

On equating the dimensions on both sides, we have

$$x+y=0;-2y=1;\text{ and }z=0$$

So that $x=\frac{1}{2},y=-\frac{1}{2},z=0$

Then, $T=k{l}^{1/2}{g}^{-1/2}$

or, $T=k\sqrt{\frac{l}{g}}$

Note that value of constant $k$ can not be obtained by the method of dimensions. Here it does not matter if some number multiplies the right side of this formula, because that does not affect its dimensions.

Actually, $k=2\pi$ so that $T=2\pi \sqrt{\frac{l}{g}}$

Dimensional analysis is very useful in deducing relations among the interdependent physical quantities. However, dimensionless constants cannot be obtained by this method. The method of dimensions can only test the dimensional validity, but not the exact relationship between physical quantities in any equation. It does not distinguish between the physical quantities having same dimensions.

A number of exercises at the end of this chapter will help you develop skill in dimensional analysis.

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, $\mathrm{s}$ for second, A for ampere, $\mathrm{N}$ 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. 1 0. 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 uncertain digits must be followed. 1 1. 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 $1\mathrm{cm}$ is equal to ….. ${\mathrm{m}}^3$

(b) The surface area of a solid cylinder of radius $2.0\mathrm{cm}$ and height $10.0\mathrm{cm}$ is equal to … $(\mathrm{mm}{)}^2$

(c) A vehicle moving with a speed of $18\mathrm{km}{\mathrm{h}}^{-1}$ covers…. $\mathrm{m}$ in $1\mathrm{s}$

(d) The relative density of lead is 11.3 . Its density is …. $\mathrm{g}{\mathrm{cm}}^{-3}$ or …. $\mathrm{kg}{\mathrm{m}}^{-3}$.

Show Answer

Answer

(a) 1 cm $=\frac{1}{100}m$

Volume of the cube $=1c{m}^3$

But, $1c{m}^3=1cm\times 1cm\times 1cm=(\frac{1}{100})m\times (\frac{1}{100})m\times (\frac{1}{100})m$

$\therefore 1c{m}^3={10}^{-6}{m}^3$

Hence, the volume of a cube of side $1cm$ is equal to ${10}^{-6}{m}^3$.

(b) The total surface area of a cylinder of radius $r$ and height $h$ is

$S=2\pi r(r+h)$.

Given that,

$r=2cm=2\times 1cm=2\times 10mm=20mm$

$h=10cm=10\times 10mm=100mm$

$\therefore S=2\times 3.14\times 20\times (20+100)=15072=1.5\times {10}^{4}m{m}^2$

(c) Using the conversion,

$1km/h=\frac{5}{18}m/s$

$18km/h=18\times \frac{5}{18}=5m/s$

Therefore, distance can be obtained using the relation:

Distance $=$ Speed $\times$ Time $=5\times 1=5m$

Hence, the vehicle covers $5m$ in $1s$.

(d) Relative density of a substance is given by the relation,

Relative density $=\frac{\text{ Density of substance }}{\text{ Density of water }}$

Density of water $=1g/c{m}^3$

Density of lead $=$ Relative density of lead $\times$ Density of water

$=11.3\times 1=11.3g/c{m}^3$

Again, $1g=\frac{1}{1000}kg$

$1c{m}^3={10}^{-6}{m}^3$

$1g/c{m}^3=\frac{{10}^{-3}}{{10}^{-6}}kg/m^3={10}^{3}kg/m^3$

$\therefore 11.3g/c{m}^3=11.3\times {10}^{3}kg/m^3$

1.2 Fill in the blanks by suitable conversion of units

(a) $1\mathrm{kg}{\mathrm{m}}^2{\mathrm{s}}^{-2}=\dots ..\mathrm{g}{\mathrm{cm}}^2{\mathrm{s}}^{-2}$

(b) $1\mathrm{m}=\dots ..1\mathrm{ly}$

(c) $3.0\mathrm{m}{\mathrm{s}}^{-2}=\dots .\mathrm{km}{\mathrm{h}}^{-2}$

(d) $G=6.67\times {10}^{-11}\mathrm{N}{\mathrm{m}}^2(\mathrm{kg}{)}^{-2}=\dots (\mathrm{cm}{)}^3{\mathrm{s}}^{-2}{\mathrm{g}}^{-1}$.

Show Answer

Answer

(a) $1kg={10}^{3}g$

$1m^2={10}^{4}c{m}^2$

$1g{m}^2{s}^{-2}=1kg\times 1m^2\times 1s^{-2}$

$={10}^{3}g\times {10}^{4}c{m}^2\times 1s^{-2}={10}^{7}gc{m}^2{s}^{-2}$

(b) Light year is the total distance travelled by light in one year.

$1ly=$ Speed of light $\times$ One year

$=(3\times {10}^{8}m/s)\times (365\times 24\times 60\times 60s)$

$=9.46\times {10}^{15}m$

$\therefore 1m=\frac{1}{9.46\times {10}^{15}}=1.057\times {10}^{-16}ly$

(c) $1m={10}^{-3}km$

Again, $1s=\frac{1}{3600}h$

$1s^{-1}=3600h^{-1}$

$1s^{-2}=(3600{)}^2{h}^{-2}$

$\therefore 3m{s}^{-2}=(3\times {10}^{-3}km)\times ((3600{)}^2{h}^{-2})=3.88\times {10}^{-4}km{h}^{-2}$

(d) $1N=1kgm{s}^{-2}$

$1kg={10}^{-3}{g}^{-1}$

$1m^3={10}^{6}c{m}^3$

$\therefore 6.67\times {10}^{-11}N{m}^{2}k{g}^{-2}=6.67\times {10}^{-11}\times (1kgm{s}^{-2})(1m^2)(1s^{-2})$

$\begin{array}{rl}& =6.67\times {10}^{-11}\times (1kg\times 1m^3\times 1s^{-2})\\ & =6.67\times {10}^{-11}\times ({10}^{-3}{g}^{-1})\times ({10}^{6}c{m}^3)\times (1s^{-2})\\ & =6.67\times {10}^{-8}c{m}^3{s}^{-2}{g}^{-1}\end{array}$

1.3 A calorie is a unit of heat (energy in transit) and it equals about $4.2\mathrm{J}$ where $1\mathrm{J}=$ $1\mathrm{kg}{\mathrm{m}}^2{\mathrm{s}}^{-2}$. Suppose we employ a system of units in which the unit of mass equals $\alpha$ $\mathrm{kg}$, the unit of length equals $\beta \mathrm{m}$, the unit of time is $\gamma \mathrm{s}$. Show that a calorie has a magnitude $4.2{\alpha }^{-1}{\beta }^{-2}{\gamma }^2$ in terms of the new units.

Show Answer

Answer

Given that,

1 calorie $=4.2(1kg)(1m^2)(1s^{-2})$

New unit of mass $=\alpha kg$

Hence, in terms of the new unit, $1kg=\frac{1}{\alpha }={\alpha }^{-1}$

In terms of the new unit of length,

$1m=\frac{1}{\beta }={\beta }^{-1}\text{ or }1m^2={\beta }^{-2}$

And, in terms of the new unit of time,

$\begin{array}{rl}& 1s=\frac{1}{\gamma }={\gamma }^{-1}\\ & 1s^2={\gamma }^{-2}\\ & 1s^{-2}={\gamma }^2\\ & \therefore 1\text{ calorie }=4.2(1{\alpha }^{-1})(1{\beta }^{-2})(1{\gamma }^2)=4.2{\alpha }^{-1}{\beta }^{-2}{\gamma }^2\end{array}$

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.

Show Answer

Answer

The given statement is true because a dimensionless quantity may be large or small in comparision to some standard reference. For example, the coefficient of friction is dimensionless. The coefficient of sliding friction is greater than the coefficient of rolling friction, but less than static friction.

(a) An atom is a very small object in comparison to a soccer ball.

(b) A jet plane moves with a speed greater than that of a bicycle.

(c) Mass of Jupiter is very large as compared to the mass of a cricket ball.

(d) The air inside this room contains a large number of molecules as compared to that present in a geometry box.

(e) A proton is more massive than an electron.

(f) Speed of sound is less 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 $20\mathrm{s}$ to cover this distance?

Show Answer

Answer

Distance between the Sun and the Earth:

$=$ Speed of light $\times$ Time taken by light to cover the distance

Given that in the new unit, speed of light $=1$ unit

Time taken, $t=8min20s=500s$

$\therefore$ Distance between the Sun and the Earth $=1\times 500=500$ units

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 $1\mathrm{mm}$ and 100 divisions on the circular scale

(c) an optical instrument that can measure length to within a wavelength of light ?

Show Answer

Answer

(a) A device with minimum count is the most suitable to measure length.

Least count of vernier callipers

$=1$ standard division $(SD)-1$ vernier division (VD)

$=1-\frac{9}{10}=\frac{1}{10}=0.01cm$

(b) Least count of screw gauge $=\frac{\text{Pitch}}{\text{Number of divisions}}$

$=\frac{1}{1000}=0.001cm$

(c) Least count of an optical device $=$ Wavelength of light $\sim {10}^{-5}cm$

$=0.00001cm$

Hence, it can be inferred that an optical instrument is the most suitable device to measure length.

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 $3.5\mathrm{mm}$. What is the estimate on the thickness of hair?

Show Answer

Answer

Magnification of the microscope $=100$

Average width of the hair in the field of view of the microscope $=3.5mm$

$\therefore$ Actual thickness of the hair is $\frac{3.5}{100}=0.035mm$.

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 $1.0\mathrm{mm}$ 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?

Show Answer

Answer

Wrap the thread on a uniform smooth rod in such a way that the coils thus formed are very close to each other. Measure the length of the thread using a metre scale. The diameter of the thread is given by the relation,

Diameter $=\frac{\text{ Length of thread }}{\text{ Number of turns }}$

It is not possible to increase the accuracy of a screw gauge by increasing the number of divisions of the circular scale. Increasing the number divisions of the circular scale will increase its accuracy to a certain extent only.

A set of 100 measurements is more reliable than a set of 5 measurements because random errors involved in the former are very less as compared to the latter.

1.9 The photograph of a house occupies an area of $1.75{\mathrm{cm}}^2$ on a $35\mathrm{mm}$ slide. The slide is projected on to a screen, and the area of the house on the screen is $1.55{\mathrm{m}}^2$. What is the linear magnification of the projector-screen arrangement.

Show Answer

Answer

Area of the house on the slide $=1.75c{m}^2$

Area of the image of the house formed on the screen $=1.55m^2$

$=1.55\times {10}^{4}c{m}^2$

Arial magnification, $m_a=\frac{\text{ Area of image }}{\text{ Area of object }}=\frac{1.55}{1.75}\times {10}^4$

$\therefore$ Linear magnifications, $m_l=\sqrt{m_a}$

$=\sqrt{\frac{1.55}{1.75}\times {10}^4}=94.11$

1.10 State the number of significant figures in the following :

(a) $0.007{\mathrm{m}}^2$

(b) $2.64\times {10}^{24}\mathrm{kg}$

(c) $0.2370\mathrm{g}{\mathrm{cm}}^{-3}$

(d) $6.320\mathrm{J}$

(e) $6.032\mathrm{N}{\mathrm{m}}^{-2}$

(f) $0.0006032{\mathrm{m}}^2$

Show Answer

Answer

(a)

The given quantity is $0.007m^2$.

If the number is less than one, then all zeros on the right of the decimal point (but left to the first non-zero) are insignificant. This means that here, two zeros after the decimal are not significant. Hence, only 7 is a significant figure in this quantity.

(b)

The given quantity is $2.64\times {10}^{24}kg$.

Here, the power of 10 is irrelevant for the determination of significant figures. Hence, all digits i.e., 2, 6 and 4 are significant figures.

(c)

The given quantity is $0.2370gc{m}^{-3}$.

For a number with decimals, the trailing zeroes are significant. Hence, besides digits 2, 3 and 7, 0 that appears after the decimal point is also a significant figure.

(d)

The given quantity is $6.320J$.

For a number with decimals, the trailing zeroes are significant. Hence, all four digits appearing in the given quantity are significant figures.

(e)

The given quantity is $6.032N{m}^{-2}$.

All zeroes between two non-zero digits are always significant.

(f)

The given quantity is $0.0006032m^2$.

If the number is less than one, then the zeroes on the right of the decimal point (but left to the first non-zero) are insignificant. Hence, all three zeroes appearing before 6 are not significant figures. All zeros between two non-zero digits are always significant. Hence, the remaining four digits are significant figures.

1.11 The length, breadth and thickness of a rectangular sheet of metal are $4.234\mathrm{m},1.005\mathrm{m}$, and $2.01\mathrm{cm}$ respectively. Give the area and volume of the sheet to correct significant figures.

Show Answer

Answer

Length of sheet, $l=4.234m$

Breadth of sheet, $b=1.005m$

Thickness of sheet, $h=2.01cm=0.0201m$

The given table lists the respective significant figures:

Quantity	Number	Significant Figure
$l$	4.234	4
$b$	1.005	4
$h$	2.01	3

Hence, area and volume both must have least significant figures i.e., 3 .

Surface area of the sheet $=2(l\times b+b\times h+h\times l)$

$=2(4.234\times 1.005+1.005\times 0.0201+0.0201\times 4.234)$

$=2(4.25517+0.02620+0.08510)$

$=2\times 4.360$

$=8.72m^2$

Volume of the sheet $=l\times b\times h$

$=4.234\times 1.005\times 0.0201$

$=0.0855m^3$

This number has only 3 significant figures i.e., 8 , 5, and 5.

1.12 The mass of a box measured by a grocer’s balance is $2.30\mathrm{kg}$. Two gold pieces of masses $20.15\mathrm{g}$ and $20.17\mathrm{g}$ 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 ?

Show Answer

Answer

Mass of grocer’s box $=2.300kg$

Mass of gold piece $\mathbf{I}=20.15g=0.02015kg$

Mass of gold piece $\mathbf{I}\mathbf{I}=20.17g=0.02017kg$

Total mass of the box $=2.3+0.02015+0.02017=2.34032kg$

In addition, the final result should retain as many decimal places as there are in the number with the least decimal places. Hence, the total mass of the box is $2.3kg$.

Difference in masses $=20.17-20.15=0.02g$

In subtraction, the final result should retain as many decimal places as there are in the number with the least decimal places.

1.13 A famous relation in physics relates ‘moving mass’ $m$ to the ‘rest mass’ $m_0$ 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 :

$m=\frac{m_O}{{(1-v^2)}^{1/2}}$

Guess where to put the missing $c$.

Show Answer

Answer

Given the relation,

$m=\frac{m_0}{(1-v^2{)}^{\frac{1}{2}}}$

Dimension of $m=M^1{L}^0{T}^0$

Dimension of $m_0=M^1{L}^0{T}^0$

Dimension of $v=M^0{L}^1{T}^{-1}$

Dimension of $v^2=M^0{L}^2{T}^{-2}$

Dimension of $c=M^0{L}^1{T}^{-1}$

The given formula will be dimensionally correct only when the dimension of L.H.S is the same as that of R.H.S. This is only possible when the factor, $(1-v^2{)}^{\frac{1}{2}}$ is dimensionless i.e., $(1-v^2)$ is dimensionless. This is only possible if $v^2$ is divided by $c^2$. Hence, the correct relation is

$m=\frac{m_0}{(1-\frac{v^2}{c^2}{)}^{\frac{1}{2}}}$

1.14 The unit of length convenient on the atomic scale is known as an angstrom and is denoted by $\text{AA}$ : $1\text{AA}={10}^{-10}\mathrm{m}$. The size of a hydrogen atom is about $0.5\text{AA}$. What is the total atomic volume in ${\mathrm{m}}^3$ of a mole of hydrogen atoms?

Show Answer

Answer

Radius of hydrogen atom, $r=0.5\dot{A}=0.5\times {10}^{-10}m$

Volume of hydrogen atom $=\frac{4}{3}\pi r^3$

$=\frac{4}{3}\times \frac{22}{7}\times (0.5\times {10}^{-10}{)}^3$

$=0.524\times {10}^{-30}{m}^3$

1 mole of hydrogen contains $6.023\times {10}^{23}$ hydrogen atoms.

$\therefore$ Volume of 1 mole of hydrogen atoms $=6.023\times {10}^{23}\times 0.524\times {10}^{-30}$

$=3.16\times {10}^{-7}{m}^3$

1.15 One mole of an ideal gas at standard temperature and pressure occupies $22.4\mathrm{L}$ (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 $1\text{AA}$ ). Why is this ratio so large?

Show Answer

Answer

Radius of hydrogen atom, $r=0.5\dot{A}=0.5\times {10}^{-10}m$

Volume of hydrogen atom $=\frac{4}{3}\pi r^3$

$=\frac{4}{3}\times \frac{22}{7}\times (0.5\times {10}^{-10}{)}^3$

$=0.524\times {10}^{-30}{m}^3$

Now, 1 mole of hydrogen contains $6.023\times {10}^{23}$ hydrogen atoms.

$\therefore$ Volume of 1 mole of hydrogen atoms, $V_a=6.023\times {10}^{23}\times 0.524\times {10}^{-30}$

$=3.16\times {10}^{-7}{m}^3$

Molar volume of 1 mole of hydrogen atoms at STP,

$V_m=22.4L=22.4\times {10}^{-3}{m}^3$

$\therefore \frac{V_m}{V_a}=\frac{22.4\times {10}^{-3}}{3.16\times {10}^{-7}}=7.08\times {10}^4$

Hence, the molar volume is $7.08\times {10}^4$ times higher than the atomic volume. For this reason, the inter-atomic separation in hydrogen gas is much larger than the size of a hydrogen atom.

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).

Show Answer

Answer

Line of sight is defined as an imaginary line joining an object and an observer’s eye. When we observe nearby stationary objects such as trees, houses, etc. while sitting in a moving train, they appear to move rapidly in the opposite direction because the line of sight changes very rapidly.

On the other hand, distant objects such as trees, stars, etc. appear stationary because of the large distance. As a result, the line of sight does not change its direction rapidly.

1.17 The Sun is a hot plasma (ionized matter) with its inner core at a temperature exceeding ${10}^7\mathrm{K}$, and its outer surface at a temperature of about $6000\mathrm{K}$. 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 $=2.0\times {10}^{30}\mathrm{kg}$, radius of the Sun $=7.0\times {10}^8\mathrm{m}$.

Show Answer

Answer

Mass of the Sun, $M=2.0\times {10}^{30}kg$

Radius of the Sun, $R=7.0\times {10}^{8}m$

Volume of the Sun, $V=\frac{4}{3}\pi R^3$

$=\frac{4}{3}\times \frac{22}{7}\times (7.0\times {10}^8{)}^3$

$=\frac{88}{21}\times 343\times {10}^{24}=1437.3\times {10}^{24}{m}^3$

Density of the Sun $=\frac{\text{ Mass }}{\text{ Volume }}=\frac{2.0\times {10}^{30}}{1437.3\times {10}^{24}}\sim 1.4\times {10}^{3}kg/m^5$

The density of the Sun is in the density range of solids and liquids. This high density is attributed to the intense gravitational attraction of the inner layers on the outer layer of the Sun.