## Chapter 01 Units and Measurements

### Multiple Choice Questions (MCQs)

**1.** The number of significant figures in 0.06900 is

(a) 5

(b) 4

(c) 2

(d) 3

## Show Answer

**Thinking Process**

If the number is less than 1, the zero(s) on the right of decimal point and before the first non-zero digit are not significant.

**Answer**
(b) In $\underline{0}$. $\underline{6} 6900$, the underlined zeroes are not significant. Hence, number of significant figures are four (6900).

**2.** The sum of the numbers 436.32, 227.2 and 0.301 in appropriate significant figures is

(a) 663.821

(b) 664

(c) 663.8

(d) 663.82

## Show Answer

**Answer**
(b) The sum of the numbers can be calculated as 663.821 arithmetically. The number with least decimal places is 227.2 is correct to only one decimal place.

The final result should, therefore be rounded off to one decimal place i.e., 664 .

Note In calculating the sum, we should not confuse with the number of decimal places and significant figures. The result should have least number of decimal places.

**3.** The mass and volume of a body are $4.237 g$ and $2.5 cm^{3}$, respectively. The density of the material of the body in correct significant figures is

(a) $1.6048 g cm^{-3}$

(b) $1.69 g cm^{-3}$

(c) $1.7 g cm^{-3}$

(d) $1.695 g cm^{-3}$

## Show Answer

**Thinking Process**

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.

**Answer**
(c) In this question, density should be reported to two significant figures.

$ \text { Density }=\frac{4.237 g}{2.5 cm^{3}}=1.6948 $

As rounding off the number, we get density $=1.7$

**4.** The numbers 2.745 and 2.735 on rounding off to 3 significant figures will give

(a) 2.75 and 2.74

(b) 2.74 and 2.73

(c) 2.75 and 2.73

(d) 2.74 and 2.74

## Show Answer

**Answer**(d) Rounding off 2.745 to 3 significant figures it would be 2.74 . Rounding off 2.735 to 3 significant figures it would be 2.74 .

**5.** The length and breadth of a rectangular sheet are $16.2 cm$ and $10.1 cm$, respectively. The area of the sheet in appropriate significant figures and error is

(a) $164 \pm 3 cm^{2}$

(b) $163.62 \pm 2.6 cm^{2}$

(c) $163.6 \pm 2.6 cm^{2}$

(d) $163.62 \pm 3 cm^{2}$

## Show Answer

**Thinking Process**

If $\Delta x$ is error in a physical quantity, then relative error is calculated as $\frac{\Delta x}{x}$.

**Answer**
(a) Given,

$ \begin{aligned} \text { length } l & =(16.2 \pm 0.1) cm \\ \text { Breadth } b & =(10.1 \pm 0.1) cm \\ \text { Area } A & =l \times b \\ & =(16.2 cm) \times(10.1 cm)=163.62 cm^{2} \end{aligned} $

Rounding off to three significant digits, area $A=164 cm^{2}$

$ \begin{aligned} \frac{\Delta A}{A} & =\frac{\Delta l}{l}+\frac{\Delta b}{b}=\frac{0.1}{16.2}+\frac{0.1}{10.1} \\ & =\frac{1.01+1.62}{16.2 \times 10.1}=\frac{2.63}{163.62} \\ \Rightarrow \quad \Delta A & =A \times \frac{2.63}{163.62}=163.62 \times \frac{2.63}{163.62}=2.63 cm^{2} \\ \Delta A & =3 cm^{2} \quad \text { (By rounding off to one significant figure) } \\ \therefore \quad \text { Area, } A & =A \pm \Delta A=(164 \pm 3) cm^{2} . \end{aligned} $

**6.** Which of the following pairs of physical quantities does not have same dimensional formula?

(a) Work and torque

(b) Angular momentum and Planck’s constant

(c) Tension and surface tension

(d) Impulse and linear momentum

## Show Answer

**Answer**
(c) (a) Work $=$ force $\times$ distance $=[MLT^{-2}][L][ML^{2} T^{-2}]$ Torque $=$ force $\times$ distance $=[ML^{2} T^{-2}]$

(b) Angular momentum $=mvr=[M][LT^{-1}][L]=[ML^{2} T^{-1}]$ Planck’s constant $=\frac{E}{V}=\frac{[ML^{2} T^{-2}]}{[T^{-1}]}=[ML^{2} T^{-1}]$

(c) Tension $=$ force $=[MLT^{-2}]$

Surface tension $=\frac{\text { force }}{\text { length }}=\frac{[MLT^{-2}]}{[L]}=[ML^{0} T^{-2}]$

(d) Impulse $=$ force $\times$ time $=[MLT^{-2}][T]=[MLT^{-1}]$

Momentum $=$ mass $\times$ velocity $=[M][LT^{-1}]=[MLT^{-1}]$

**Note One should not be confused with the similar form tension in both the physical quantities-surface tension and tension. Dimensional formula for both of them is not same**.

**7.** Measure of two quantities along with the precision of respective measuring instrument is

$A=2.5 ms^{-1} \pm 0.5 ms^{-1}, B=0.10 s \pm 0.01 s$. The value of $A B$ will be

(a) $(0.25 \pm 0.08) m$

(b) $(0.25 \pm 0.5) m$

(c) $(0.25 \pm 0.05) m$

(d) $(0.25 \pm 0.135) m$

## Show Answer

**Answer**
(a) Given,

$ \begin{aligned} A & =2.5 ms^{-1} \pm 0.5 ms^{-1}, B=0.10 s \pm 0.01 s \\ x & =A B=(2.5)(0.10)=0.25 m \\ \frac{\Delta x}{x} & =\frac{\Delta A}{A}+\frac{\Delta B}{B} \\ & =\frac{0.5}{2.5}+\frac{0.01}{0.10}=\frac{0.05+0.025}{0.25}=\frac{0.075}{0.25} \end{aligned} $

$\Delta x=0.075=0.08 m$, rounding off to two significant figures.

$A B=(0.25 \pm 0.08) m$

**8.** You measure two quantities as $A=1.0 m \pm 0.2 m, B=2.0 m \pm 0.2 m$. We should report correct value for $\sqrt{A B}$ as

(a) $1.4 m \pm 0.4 m$

(b) $1.41 m \pm 0.15 m$

(c) $1.4 m \pm 0.3 m$

(d) $1.4 m \pm 0.2 m$

## Show Answer

**Answer**
(d) Given,

$ \begin{aligned} & A=1.0 m \pm 0.2 m, B=2.0 m \pm 0.2 m \\ & Y=\sqrt{A B}=\sqrt{(1.0)(2.0)}=1.414 m \end{aligned} $

Let,

Rounding off to two significant digit $Y=1.4 m$

$\frac{\Delta Y}{Y}=\frac{1}{2}[\frac{\Delta A}{A}+\frac{\Delta B}{B}]=\frac{1}{2}[\frac{0.2}{1.0}+\frac{0.2}{2.0}]=\frac{0.6}{2 \times 2.0}$

$\Rightarrow \Delta Y=\frac{0.6 Y}{2 \times 2.0}=\frac{0.6 \times 1.4}{2 \times 2.0}=0.212$

Rounding off to one significant digit $\Delta Y=0.2 m$

Thus, correct value for $\sqrt{A B}=r+\Delta r=1.4 \pm 0.2 m$

**9.** Which of the following measurement is most precise?

(a) $5.00 mm$

(b) $5.00 cm$

(c) $5.00 m$

(d) $5.00 km$

## Show Answer

**Answer**
(a) All given measurements are correct upto two decimal places. As here $5.00 mm$ has the smallest unit and the error in $5.00 mm$ is least (commonly taken as $0.01 mm$ if not specified), hence, $5.00 mm$ is most precise.

Note In solving these type of questions, we should be careful about units although their magnitude is same.

**10.** The mean length of an object is $5 cm$. Which of the following measurements is most accurate?

(a) $4.9 cm$

(b) $4.805 cm$

(c) $5.25 cm$

(d) $5.4 cm$

## Show Answer

**Answer**
(a) Given length

$ l=5 cm $

Now, checking the errors with each options one by one, we get

Error $\Delta l_1$ is least.

$ \begin{aligned} & \Delta l_1=5-4.9=0.1 cm \\ & \Delta l_2=5-4.805=0.195 cm \\ & \Delta l_3=5.25-5=0.25 cm \\ & \Delta l_4=5.4-5=0.4 cm \end{aligned} $

Hence, $4.9 cm$ is most precise.

**11.** Young’s modulus of steel is $1.9 \times 10^{11} N / m^{2}$. When expressed in CGS units of dyne $/ cm^{2}$, it will be equal to $(1 N=10^{5}.$ dyne, $.1 m^{2}=10^{4} cm^{2})$

(a) $1.9 \times 10^{10}$

(b) $1.9 \times 10^{11}$

(c) $1.9 \times 10^{12}$

(d) $1.9 \times 10^{13}$

## Show Answer

**Answer**
(c) Given, Young’s modulus

$ \begin{aligned} Y & =1.9 \times 10^{11} N / m^{2} \\ 1 N & =10^{5} \text { dyne } \\ Y & =1.9 \times 10^{11} \times 10^{5} \text { dyne } / m^{2} \\ 1 m & =100 cm \\ Y & =1.9 \times 10^{11} \times 10^{5} \text { dyne } /(100)^{2} cm^{2} \\ & =1.9 \times 10^{16-4} \text { dyne } / cm^{2} \\ Y & =1.9 \times 10^{12} \text { dyne } / cm^{2} \end{aligned} $

Hence,

We know that

Note While we are going through units conversion, we should keep in mind that proper relation between units are mentioned.

**12.** If momentum $(p)$, area $(A)$ and time $(T)$ are taken to be fundamental quantities, then energy has the dimensional formula

(a) $[pA^{-1} T^{1}]$

(b) $[p^{2} AT]$

(c) $[p A^{-1 / 2} T]$

(d) $[pA^{1 / 2} T]$

## Show Answer

**Answer**
(d) Given, fundamental quantities are momentum $(p)$, area $(A)$ and time $(T)$. We can write energy $E$ as

$ \begin{aligned} & E \propto p^{a} A^{b} T^{c} \\ & E=k p^{a} A^{A} T^{c} \end{aligned} $

where $k$ is dimensionless constant of proportionality.

Dimensions of

$ \begin{aligned} E & =[E]=[ML^{2} T^{-2}] \text { and }[p]=[MLT^{-1}] \\ {[A] } & =[L^{2}] \\ {[T] } & =[T] \\ {[E] } & =[K][p]^{a}[A]^{b}[T]^{c} \end{aligned} $

Putting all the dimensions, we get

$ \begin{aligned} ML^{2} T^{-2} & =[MLT^{-1}]^{a}[L^{2}]^{b}[T]^{c} \\ & =M^{a} L^{2 b+a} T^{-a+c} \end{aligned} $

By principle of homogeneity of dimensions,

$ a =1,2 b+a=2 $

$\Rightarrow 2 b+1 =2 $

$\Rightarrow b =1 / 2-a+c=-2 $

$\Rightarrow c =-2+a=-2+1=-1 $

$\text{ Hence, } E =p A^{1 / 2} T^{-1}$

### Multiple Choice Questions (More Than One Options)

**13.** On the basis of dimensions, decide which of the following relations for the displacement of a particle undergoing simple harmonic motion is not correct?

(a) $y=a \sin 2 \pi t / T$

(b) $y=a \sin v t$

(c) $y=\frac{a}{T} \sin (\frac{t}{a})$

(d) $y=a \sqrt{2}(\sin \frac{2 \pi t}{T}-\cos \frac{2 \pi t}{T})$

## Show Answer

**Thinking Process**

We know that angle is dimensionless. Here, $a$ is displacement and $y$ is also displacement, hence they are having same dimensions.

**Answer**
$(b, c)$

Now, by principle of homogeneity of dimensions LHS and RHS of (a) and (d) will be same and is $L$.

For (c)

$ \begin{aligned} & {[LHS]=L} \\ & {[RHS]=\frac{L}{T}=LT^{-1}} \\ & {[LHS] \neq[RHS]} \end{aligned} $

Hence, (c) is not correct option.

In option (b) dimension of angle is [vt ]i.e., $L$

$ \begin{matrix} \Rightarrow & R H S=L . L=L^{2} \text { and } L H S=L \\ \Rightarrow & L H S \neq R H S . \end{matrix} $

So, option (b) is also not correct.

**14.** If $P, Q, R$ are physical quantities, having different dimensions, which of the following combinations can never be a meaningful quantity?

(a) $(P-Q) / R$

(b) $P Q-R$

(c) $P Q / R$

(d) $(P R-Q^{2}) / R$

(e) $(R+Q) / P$

## Show Answer

**Thinking Process**

We should keep in mind that when two physical quantities are added or subtracted they should have same dimensions.

**Answer**
$(a, e)$

In this question, it is given that $P, Q$ and $R$ are having different dimensions, hence they cannot be added or subtracted, so we can say that (a) and (e) are not meaningful. We cannot say about the dimension of product of these quantities, hence (b), (c) and (d) may be meaningful.

Note In this question, we are certain about the quantity which is never meaningful but we should keep in mind that others may or may not be meaningful.

**15.** Photon is quantum of radiation with energy $E=h v$, where $v$ is frequency and $h$ is Planck’s constant. The dimensions of $h$ are the same as that of

(a) linear impulse

(b) angular impulse

(c) linear momentum

(d) angular momentum

## Show Answer

**Answer**

( $b, d)$

We know that energy of radiation, $E=h v$

$ [h]=\frac{[E]}{[v]}=\frac{[ML^{2} T^{-2}]}{[T^{-1}]}=[ML^{2} T^{-1}] $

Dimension of linear impulse $=$ Dimension of momentum $=[MLT^{-1}]$

As we know that linear impulse $J=\Delta P$

$\Rightarrow \quad$ Angular impulse $=\tau d t=\Delta L=$ Change in angular momentum

Hence, dimension of angular impulse

$ \begin{aligned} & =\text { Dimension of angular momentum } \\ & =[ML^{2} T^{-1}] . \end{aligned} $

This is similar to the dimension of Planck’s constant $h$.

**16.** If Planck’s constant $(h)$ and speed of light in vacuum (c) are taken as two fundamental quantities, which one of the following can, in addition, be taken to express length, mass and time in terms of the three chosen fundamental quantities?

(a) Mass of electron $(m_{e})$

(b) Universal gravitational constant (G)

(c) Charge of electron (e)

(d) Mass of proton $(m_{p})$

## Show Answer

**Answer**

$(a, b, d)$

We know that dimension of $h=[h]=[ML^{2} T^{-1}]$

$[c]=[LT^{-1}],[m_{e}]=M$

$[G]=[M^{-1} L^{3} T^{-2}]$

$[e]=[A T],[m_{p}]=[M]$

$[\frac{h c}{G}]=\frac{[ML^{2} T^{-1}][LT^{-1}]}{[M^{-1} L^{3} T^{-2}]}=[M^{2}]$

$M=\sqrt{\frac{h c}{G}}$

Similarly,

$\frac{h}{c}=\frac{[ML^{2} T^{-1}]}{[LT^{-1}]}=[ML]$

$L=\frac{h}{c M}=\frac{h}{c} \sqrt{\frac{G}{h c}}=\frac{\sqrt{G h}}{c^{3 / 2}}$

As,

$ c=L T^{-1} $

$\Rightarrow \quad[T]=\frac{[L]}{[c]}=\frac{\sqrt{G h}}{c^{3 / 2} \cdot c}=\frac{\sqrt{G h}}{c^{5 / 2}}$

Hence, (a), (b) or (d) any can be used to express $L, M$ and $T$ in terms of three chosen fundamental quantities.

**17.** Which of the following ratios express pressure?

(a) Force/Area

(b) Energy/Volume

(c) Energy/Area

(d) Force/Volume

## Show Answer

**Thinking Process**

While solving this type of questions, we should first write an expression and try to express it in terms of quantities given in the option.

**Answer**
$(a, b)$

We know that pressure $=\frac{\text { Force }}{\text { Area }}$

$ \text { Pressure }=\frac{\text { Force } \times \text { Distance }}{\text { Area } \times \text { Distance }}=\frac{\text { Work }}{\text { Volume }}=\frac{\text { Energy }}{\text { Volume }} $

Note Here, we should keep it in mind that above values are not exactly equal but these are equivalent with respect to their units.

**18.** Which of the following are not a unit of time?

(a) Second

(b) Parsec

(c) Year

(d) Light year

## Show Answer

**Answer**
$(b, d)$

We know that 1 light year $=9.46 \times 10^{11} m$

$=$ distance that light travels in 1 year with speed $3 \times 10^{8} m / s$.

1 parsec $=3.08 \times 10^{16} m$

$=$ Distance at which average radius of earth’s orbit subtends an angle of 1 parsecond

Here, second and year represent time.

### Very Short Answer Type Questions

**19.** Why do we have different units for the same physical quantity?

## Show Answer

**Answer**
The value of any given physical quantity may vary over a wide range, therefore, different units of same physical quantity are required.

e.g., The length of a pen can be easily measured in $cm$, the height of a tree can be measured in metres, the distance between two cities can be measured in kilometres and distance between two heavenly bodies can be measured in light year.

**20.** The radius of atom is of the order of $1 \AA$ and radius of nucleus is of the order of fermi. How many magnitudes higher is the volume of atom as compared to the volume of nucleus?

## Show Answer

**Answer**

$ \begin{aligned} \text { Radius of atom } & =1 \AA=10^{-10} m \\ \text { Radius of nucleus } & =1 \text { fermi }=10^{-15} m \\ \text { Volume of atom } & =V_{A}=\frac{4}{3} \pi R_A^{3} \\ \text { Volume of nucleus } V_{N} & =\frac{4}{3} \pi R_N^{3} \\ \frac{V_{A}}{V_{N}} & =\frac{\frac{4}{3} \pi R_A^{3}}{\frac{4}{3} \pi R_N^{3}}=(\frac{R_{A}}{R_{N}})^{3}=(\frac{10^{-10}}{10^{-15}})^{3}=10^{15} \end{aligned} $

Note In such type of questions, always change the value in same unit.

**21.** Name the device used for measuring the mass of atoms and molecules.

## Show Answer

**Answer**A mass spectrograph is used for measuring the mass of atoms and molecules.

**22.** Express unified atomic mass unit in $kg$.

## Show Answer

**Answer**
One atomic mass unit is the $\frac{1}{12}$ of the mass of $a_6 C^{12}$ atom.

Mass of one mole of $ _6 C^{12}$ atom $=12 g$

Number of atoms in one mole $=$ Avogadro’s number

$ =6.023 \times 10^{23} $

$ \begin{aligned} \therefore \quad \text { Mass of one } _6 C^{12} \text { atom } & =\frac{12}{6.023 \times 10^{23}} g \\ 1 amu & =\frac{1}{12} \times \text { mass of one } _6 C^{12} \text { atom } \\ \therefore \quad 1 amu & =(\frac{1}{12} \times \frac{12}{6.023 \times 10^{23}}) g=1.67 \times 10^{-24} g \\ & =1.67 \times 10^{-27} kg \quad(\because 1 g=10^{-3} kg) \end{aligned} $

**23.** A function $f(\theta)$ is defined as $f(\theta)=1-\theta+\frac{\theta^{2}}{2 !}-\frac{\theta^{3}}{3 !}+\frac{\theta^{4}}{4 !}+\ldots$

Why is it necessary for $f(\theta)$ to be a dimensionless quantity?

## Show Answer

**Answer**Since, $f(\theta)$ is a sum of different powers of $\theta$ and it is a dimensionless quantity. By principle of homogeneity as RHS is dimensionless, hence LHS should also be dimensionless.

**24.** Why length, mass and time are chosen as base quantities in mechanics?

## Show Answer

**Answer**
Length, mass and time are chosen as base quantities in mechanics because

(i) Length, mass and time cannot be derived from one another, that is these quantities are independent.

(ii) All other quantities in mechanics can be expressed in terms of length, mass and time.

### Short Answer Type Questions

**25.** (a) The earth-moon distance is about 60 earth radius. What will be the diameter of the earth (approximately in degrees) as seen from the moon?

(b) Moon is seen to be of $(1 / 2)^{\circ}$ diameter from the earth. What must be the relative size compared to the earth?

(c) From parallax measurement, the sun is found to be at a distance of about 400 times the earth-moon distance. Estimate the ratio of sun-earth diameters.

## Show Answer

**Thinking Process**

To solve this question, we have to treat radius of earth as an arc as seen from the moon.

**Answer**
(a) Angle subtended at distance $r$ due to an arc of length $l$ is

Given,

$ \begin{aligned} l & =R_{E} ; r=60 R_{E} \\ \theta & =\frac{R_{E}}{60 R_{E}}=\frac{1}{60} rad=\frac{1}{60} \times \frac{180}{\pi} \text { degree } \\ & =\frac{3}{\pi} \approx 1^{\circ} \end{aligned} $

Hence, angle subtended by diameter of the earth $=2 \theta=2^{\circ}$.

(b) Given that moon is seen as $(\frac{1}{2})^{\circ}$ diameter and earth is seen as $2^{\circ}$ diameter.

Hence,

$ \frac{\text { Diameter of earth }}{\text { Diameter of moon }}=\frac{(2 / \pi) rad}{(\frac{1}{2 \pi}) rad}=4 $

(c) From parallax measurement given that sun is at a distance of about 400 times the earth-moon distance, hence, $\frac{r_{\text {sun }}}{r_{\text {moon }}}=400$

(Suppose, here $r$ stands for distance and $D$ for diameter) sun and moon both appear to be of the same angular diameter as seen from the earth.

$ \begin{matrix} \therefore & \frac{D_{\text {sun }}}{r_{\text {sun }}}=\frac{D_{\text {moon }}}{r_{\text {moon }}} \\ \therefore & \frac{D_{\text {sun }}}{D_{\text {moon }}}=400 \\ \text { But } & \frac{D_{\text {earth }}}{D_{\text {moon }}}=4 \\ \therefore & \frac{D_{\text {sun }}}{D_{\text {earth }}}=100 \end{matrix} $

**26.** Which of the following time measuring devices is most precise?

(a) A wall clock

(b) A stop watch

(c) A digital watch

(d) An atomic clock

Given reason for your answer.

## Show Answer

**Answer**A wall clock can measure time correctly upto one second. A stop watch can measure time correctly upto a fraction of a second. A digital watch can measure time up to a fraction of second. An atomic clock can measure time most precisely as its precision is $1 s$ in $10^{13} s$.

**27.** The distance of a galaxy is of the order of $10^{25} m$. Calculate the order of magnitude of time taken by light to reach us from the galaxy.

## Show Answer

**Answer**
Given, distance of the galaxy $=10^{25} m$

$ \text { Speed of light }=3 \times 10^{8} m / s $

Hence, time taken by light to reach us from galaxy is,

$ \begin{aligned} t & =\frac{\text { Distance }}{\text { Speed }}=\frac{10^{25}}{3 \times 10^{8}} \approx \frac{1}{3} \times 10^{17} \\ & =\frac{10}{3} \times 10^{16}=3.33 \times 10^{16} s \end{aligned} $

**28.** The vernier scale of a travelling microscope has 50 divisions which coincide with 49 main scale divisions. If each main scale division is $0.5 mm$, calculate the minimum inaccuracy in the measurement of distance.

## Show Answer

**Thinking Process**

Inaccuracy will be measured by difference of MSD and 1VSD, where MSD = main scale division and VSD = verneir scale division.

**Answer**
By question, it is given that

$ \begin{aligned} 50 VSD & =49 MSD \\ 1 MSD & =\frac{50}{49} VSD \\ 1 VSD & =\frac{49}{50} MSD \\ \text { ccuracy } & =1 MSD-1 VSD \\ & =1 MSD-\frac{49}{50} MSD=\frac{1}{50} MSD \end{aligned} $

$ \text { Minimum inaccuracy }=1 MSD-1 VSD $

Given,

Hence,

$ \begin{aligned} 1 MSD & =0.5 mm \\ \text { minimum inaccuracy } & =\frac{1}{50} \times 0.5 mm=\frac{1}{100}=0.01 mm \end{aligned} $

**29.** During a total solar eclipse the moon almost entirely covers the sphere of the sun. Write the relation between the distances and sizes of the sun and moon.

## Show Answer

**Answer**

Consider the diagram given below

$R_{m e}=$ Distance of moon from earth

$R_{\text {se }}=$ Distance of sun from earth

Let angle made by sun and moon is $\theta$, we can write

$ \theta=\frac{A_{\text {sun }}}{R_se^{2}}=\frac{A_{\text {moon }}}{R_{\text {me }}^{2}} $

Here,

$A_{\text {sun }}=$ Area of the sun

$A_{\text {moon }}=$ Area of the moon

$ \begin{matrix} \Rightarrow & & \theta & =\frac{\pi R_s^{2}}{R_{s e}^{2}}=\frac{\pi R_m^{2}}{R_{m e}^{2}} \\ \Rightarrow & & (\frac{R_{s}}{R_{s e}})^{2} & =(\frac{R_{m}}{R_{m e}})^{2} \\ \Rightarrow & \frac{R_{s}}{R_{s e}} & =\frac{R_{m}}{R_{m e}} \\ \Rightarrow & \frac{R_{s}}{R_{m}} & =\frac{R_{s e}}{R_{m e}} \end{matrix} $

(Here, radius of sun and moon represents their sizes respectively)

**30.** If the unit of force is $100 N$, unit of length is $10 m$ and unit of time is $100 s$, what is the unit of mass in this system of units?

## Show Answer

**Thinking Process**

First write dimension of each quantity and then relate them.

**Answer**
Dimension of force $F=[MLT^{-2}]=100 N$

$\begin{aligned} \text { Length }(L) & =[L]=10 m \\ Time(t) & =[T]=100 s\end{aligned}$

Substituting values of $L$ and $T$ from Eqs. (ii) and (iii) in Eq. (i), we get

$ M \times 10 \times(100)^{-2}=100 $

$\Rightarrow \quad \frac{M \times 10}{100 \times 100}=100$

$\Rightarrow \quad M=100 \times 1000 kg$

$M=10^{5} kg$

**31.** Give an example of

(a) a physical quantity which has a unit but no dimensions

(b) a physical quantity which has neither unit nor dimensions

(c) a constant which has a unit

(d) a constant which has no unit

## Show Answer

**Answer**
(a) Plane angle $\theta=\frac{L}{r}$ radian

its unit is radian but has no dimensions

(b) Strain $=\frac{\Delta L}{L}=\frac{\text { Change in length }}{\text { length }}$

It has neither unit nor dimensions

(c) Gravitational constant $(G)=6.67 \times 10^{-11} N-m^{2} / kg^{2}$

(d) Reynold’s number is a constant which has no unit.

**32.** Calculate the length of the arc of a circle of radius $31.0 cm$ which subtends an angle of $\frac{\pi}{6}$ at the centre.

## Show Answer

**Answer**
We know that angle

$ \begin{aligned} & \theta=\frac{l}{r} \text { radian } \\ & \theta=\frac{\pi}{6}=\frac{l}{31} cm \end{aligned} $

Given,

Hence,

$ \text { length }=l=31 \times \frac{\pi}{6} cm=\frac{31 \times 3.14}{6} cm=16.22 cm $

Rounding off to three significant figures it would be $16.2 cm$.

**33.** Calculate the solid angle subtended by the periphery of an area of $1 cm^{2}$ at a point situated symmetrically at a distance of $5 cm$ from the area.

## Show Answer

**Answer**
We know that solid angle $\Omega=\frac{\text { Area }}{(\text { Distance }^{2}.}$

$ =\frac{1 cm^{2}}{(5 cm)^{2}}=\frac{1}{25}=4 \times 10^{-2} \text { steradian } $

$(\because.$ Area $=1 cm^{2}$, distance $.=5 cm)$

Note We should not confuse, solid angle with plane angle $\theta=\frac{l}{r}$ radian.

**34.** The displacement of a progressive wave is represented by $y=A \sin (\omega t-k x)$, where $x$ is distance and $t$ is time. Write the dimensional formula of (i) $\omega$ and (ii) $k$.

## Show Answer

**Thinking Process**

In solving these type of questions, we should apply principle of homogeneity of dimensions.

**Answer**
Now, by the principle of homogeneity, i.e., dimensions of LHS and RHS should be equal, hence

$[LHS]=[RHS]$

$\Rightarrow$

$[L]=[A]=L$

As $\omega t-k x$ should be dimensionless, $[\omega t]=[k x]=1$

$\begin{matrix} \Rightarrow & {[\omega] T=[k] L=1}\end{matrix} $

$\Rightarrow \quad[\omega]=T^{-1}$ and $[k]=L^{-1}$

**35.** Time for 20 oscillations of a pendulum is measured as $t_1=39.6 s$; $t_2=39.9 s$ and $t_3=39.5 s$. What is the precision in the measurements? What is the accuracy of the measurement?

## Show Answer

**Thinking Process**

We will apply formula for mean value, absolute error as well as mean absolute error.

**Answer**
Given, $\quad t_1=39.6 s, t_2=39.9 s$ and $t_3=39.5 s$

Least count of measuring instrument $=0.1 s$

(As measurements have only one decimal place)

Precision in the measurement $=$ Least count of the measuring instrument $=0.1 s$

Mean value of time for 20 oscillations is given by

$ \begin{aligned} t & =\frac{t_1+t_2+t_3}{3} \\ & =\frac{39.6+39.9+39.5}{3}=39.7 s \end{aligned} $

Absolute errors in the measurements

$ \begin{aligned} & \Delta t_1=t-t_1=39.7-39.6=0.1 s \\ & \Delta t_2=t-t_2=39.7-39.9=-0.2 s \\ & \Delta t_3=t-t_3=39.7-39.5=0.2 s \\ & \text { Mean absolute error }= \frac{|\Delta t_1|+|\Delta t_2|+|\Delta t_3|}{3} \\ &= \frac{0.1+0.2^{2}+0.2}{3} \\ &= \frac{0.5}{3}=0.17 \approx 0.2 \quad \text { (rounding off upto one decimal place) } \end{aligned} $

$\therefore \quad$ Accuracy of measurement $= \pm 0.2 s$

### Long Answer Type Questions

**36.** A new system of units is proposed in which unit of mass is $\alpha kg$, unit of length $\beta m$ and unit of time $\gamma s$. How much will $5 J$ measure in this new system?

## Show Answer

**Thinking Process**

For solving this question, we will apply the formula for a system of unit $u, n u=$ constant.

**Answer**
We know that dimension of energy $=[ML^{2} T^{-2}]$

Let $M_1, L_1, T_1$ and $M_2, L_2, T_2$ are units of mass, length and time in given two systems.

$ \begin{matrix} \therefore \quad & M_1=1 kg, L_1=1 m, T_1=1 s \\ M_2 & =\alpha kg, L_2=\beta m, T_2=\gamma s \end{matrix} $

The magnitude of a physical quantity remains the same, whatever be the system of units of its measurement i.e. $n_1 u_1=n_2 u_2$

$ \begin{aligned} \Rightarrow \quad n_2 & =n_1 \frac{u_1}{u_2}=n_1 \frac{[M_1 L_1^{2} T_1^{-2}]}{[M_2 L_2^{2} T_2^{-2}]}=5[\frac{M_1}{M_2}] \times[\frac{L_1}{L_2}]^{2} \times[\frac{T_1}{T_2}]^{-2} \\ & =5[\frac{1}{\alpha} kg] \times[\frac{1}{\beta} m]^{2} \times[\frac{1}{\gamma} s]^{-2} \\ & =5 \times \frac{1}{\alpha} \times \frac{1}{\beta^{2}} \times \frac{1}{\gamma^{-2}} \\ n_2 & =\frac{5 \gamma^{2}}{\alpha \beta^{2}} \end{aligned} $

Thus, new unit of energy will be $\frac{\gamma^{2}}{\alpha \beta^{2}}$.

**37.** The volume of a liquid flowing out per second of a pipe of length $l$ and radius $r$ is written by a student as $V=\frac{\pi}{8} \frac{p r^{4}}{\eta l}$ where $p$ is the pressure difference between the two ends of the pipe and $\eta$ is coefficent of viscosity of the liquid having dimensional formula $[ML^{-1} T^{-1}]$. Check whether the equation is dimensionally correct.

## Show Answer

**Thinking Process**

If dimensions of LHS of an equation is equal to dimensions of RHS, then equation is said to be dimensionally correct.

**Answer**
The volume of a liquid flowing out per second of a pipe is given by $V=\frac{\pi}{8} \frac{p r^{4}}{\eta l}$

Dimension of V= $\frac{\text { Dimension of volume }}{\text { Dimension of time }}=\frac{[L^{3}]}{[T]}=[L^{3} T^{-1}]$

$(\because V$ is the volume of liquid flowing out per second)

Dimension of p =$[ML^{-1} T^{-2}] $

Dimension of $\eta =[ML^{-1} T^{-1}] $

Dimension of l =[L]

Dimension of r =[L]

Dimensions of LHS, $[V]=\frac{[L^{3}]}{[T]}=[L^{3} T^{-1}]$

Dimensions of RHS, $\frac{[ML^{-1} T^{-2}] \times[L^{4}]}{[ML^{-1} T^{-1}] \times[L]}=[L^{3} T^{-1}]$

As dimensions of LHS is equal to the dimensions of RHS.

Therefore, equation is correct dimensionally.

**38.** A physical quantity $X$ is related to four measurable quantities $a, b, c$ and $d$ as follows $X=a^{2} b^{3} c^{5 / 2} d^{-2}$. The percentage error in the measurement of $a, b, c$ and $d$ are $1 %, 2 %, 3 %$ and $4 %$, respectively. What is the percentage error in quantity $X$ ? If the value of $X$ calculated on the basis of the above relation is 2.763 , to what value should you round off the result.

## Show Answer

**Thinking Process**

We will apply the formula for percentage error in quantity $x$, as $\frac{\Delta x}{x} \times 100$.

**Answer**
Given, physical quantity is $X=a^{2} b^{3} c^{5 / 2} d^{-2}$

Maximum percentage error in $X$ is

$ \begin{aligned} \frac{\Delta X}{X} \times 100 & = \pm[2(\frac{\Delta a}{a} \times 100)+3(\frac{\Delta b}{b} \times 100)+\frac{5}{2}(\frac{\Delta c}{c} \times 100)+2(\frac{\Delta d}{d} \times 100)] \\ & = \pm[2(1)+3(2)+\frac{5}{2}(3)+2(4)] % \\ & = \pm[2+6+\frac{15}{2}+8]= \pm 23.5 % \end{aligned} $

$\therefore$ Percentage error in quantity $X= \pm 23.5 %$

Mean absolute error in $X= \pm 0.235= \pm 0.24 \quad$ (rounding-off upto two significant digits)

The calculated value of $x$ should be round-off upto two significant digits.

$\therefore \quad X=2.8$

**39.** In the expression $P=E l^{2} m^{-5} G^{-2}, E, m, l$ and $G$ denote energy, mass, angular momentum and gravitational constant, respectively. Show that $P$ is a dimensionless quantity.

## Show Answer

**Thinking Process**

A dimensionless quantity will have dimensional formula as $[M^{0} L^{0} T^{0}]$.

**Answer**
Given, expression is

$ \begin{aligned} P & =E L^{2} m^{-5} G^{-2} \\ {[E] } & =[M L^{2} T^{-2}] \\ {[m] } & =[M] \\ {[L] } & =[M L^{2} T^{-1}] \\ {[G] } & =[M^{-1} L^{3} T^{-2}] \end{aligned} $

where $E$ is energy

$m$ is mass

$L$ is angular momentum

$G$ is gravitational constant

Substituting dimensions of each term in the given expression,

$ \begin{aligned} {[P] } & =[ML^{2} T^{-2}] \times[ML^{2} T^{-1}]^{2} \times[M]^{-5} \times[M^{-1} L^{3} T^{-2}]^{-2} \\ & =[M^{1+2-5+2} L^{2+4-6} T^{-2-2+4}]=[M^{0} L^{0} T^{0}] \end{aligned} $

Therefore, $P$ is a dimensionless quantity.

**40.** If velocity of light $c$, Planck’s constant $h$ and gravitational constant $G$ are taken as fundamental quantities, then express mass, length and time in terms of dimensions of these quantities.

## Show Answer

**Thinking Process**

In this problem, we have to apply principle of homogeneity of dimensions that is LHS and RHS of an equation will have same dimensions.

**Answer**
We know that, dimensions of

$ (h)=[ML^{2} T^{-1}] $

Dimensions of

$(c)=[L T^{-1}]$

Dimensions of gravitational constant

$ (G)=[M^{-1} L^{3} T^{-2}] $

(i) Let

$m \propto c^{x} h^{y} G^{z}$

$\Rightarrow \quad m=k c^{x} h^{y} G^{z}$

where, $k$ is a dimensionless constant of proportionality.

Substituting dimensions of each term in Eq. (i), we get

$ \begin{aligned} {[ML^{0} T^{0}] } & =[LT^{-1}]^{x} \times[ML^{2} T^{-1}]^{y}[M^{-1} L^{3} T^{-2}]^{z} \\ & =[M^{y-z} L^{x+2 y+3 z} T^{-x-y-2 z}] \end{aligned} $

Comparing powers of same terms on both sides, we get

$ \begin{matrix} y-z=1 \\ x+2 y+3 z=0 \\ -x-y-2 z=0 \end{matrix} $

Adding Eqs. (ii), (iii) and (iv), we get

$ 2 y=1 \Rightarrow y=\frac{1}{2} $

Substituting value of $y$ in Eq. (ii), we get

$ z=-\frac{1}{2} $

From Eq. (iv)

$ x=-y-2 z $

Substituting values of $y$ and $z$, we get

$ x=-\frac{1}{2}-2(-\frac{1}{2})=\frac{1}{2} $

Putting values of $x, y$ and $z$ in Eq. (i), we get

$ m=k c^{1 / 2} h^{1 / 2} G^{-1 / 2} $

$ \Rightarrow \quad m=k \sqrt{\frac{c h}{G}} $

(ii) Let

$ L \propto c^{x} h^{y} G^{z} $

$ \Rightarrow \quad L=k c^{x} h^{y} G^{z} $

where, $k$ is a dimensionless constant.

Substituting dimensions of each term in Eq. (v), we get

$ \begin{aligned} {[M^{0} LT^{0}] } & =[LT^{-1}]^{x} \times[ML^{2} T^{-1}]^{y} \times[M^{-1} L^{3} T^{-2}]^{z} \\ & =[M^{y-z} L^{x+2 y+3 z} T^{-x-y-2 z}] \end{aligned} $

On comparing powers of same terms, we get

$ \begin{matrix} y-z=0 \\ x+2 y+3 z=1 \\ -x-y-2 z=0 \end{matrix} $

Adding Eqs. (vi), (vii) and (viii), we get

$\begin{matrix} & & 2 y=1 \\ y & y & =\frac{1}{2}\end{matrix} $

Substituting value of $y$ in Eq. (vi), we get

$ z=\frac{1}{2} $

From Eq. (viii),

$ x=-y-2 z $

Substituting values of $y$ and $z$, we get

$ x=-\frac{1}{2}-2(\frac{1}{2})=-\frac{3}{2} $

Putting values of $x, y$ and $z$ in Eq. (v), we get

$ \begin{aligned} & L=k c^{-3 / 2} h^{1 / 2} G^{1 / 2} \\ & L=k \sqrt{\frac{h G}{c^{3}}} \end{aligned} $

(iii) Let $T \propto c^{x} h^{y} G^{z}$

$ \Rightarrow \quad T=k c^{x} h^{y} G^{z} $

where, $k$ is a dimensionless constant.

Substituting dimensions of each term in Eq. (ix), we get

$ \begin{aligned} {[M^{0} L^{0} T.} & =[LT^{-1}]^{x} \times[ML^{2} T^{-1}]^{y} \times[M^{-1} L^{3} T^{-2}]^{z} \\ & =[M^{y-z} L^{x+2 y+3 z} T^{-x-y-2 z}] \end{aligned} $

On comparing powers of same terms, we get

$ \begin{matrix} y-z=0 \\ x+2 y+3 z=0 \\ -x-y-2 z=1 \end{matrix} $

Adding Eqs. (x), (xi) and (xii), we get

$ \begin{aligned} & 2 y & =1 \\ \Rightarrow & y & =\frac{1}{2} \end{aligned} $

Substituting value of $y$ in Eq. ( $x$ ), we get

$ z=y=\frac{1}{2} $

From Eq. (xii),

$ x=-y-2 z-1 $

Substituting values of $y$ and $z$, we get

$ x=-\frac{1}{2}-2(\frac{1}{2})-1=-\frac{5}{2} $

Putting values of $x, y$ and $z$ in Eq. (ix), we get

$ \begin{aligned} T & =k c^{-5 / 2} h^{1 / 2} G^{1 / 2} \\ T & =k \sqrt{\frac{h G}{c^{5}}} \end{aligned} $

**41.** An artificial satellite is revolving around a planet of mass $M$ and radius $R$, in a circular orbit of radius $r$. From Kepler’s third law about the period of a satellite around a common central body, square of the period of revolution $T$ is proportional to the cube of the radius of the orbit $r$. Show using dimensional analysis, that $T=\frac{k}{R} \sqrt{\frac{r^{3}}{g}}$, where $k$ is a dimensionless constant and $g$ is acceleration due to gravity.

## Show Answer

**Thinking Process**

In this problem, we have to apply Kepler’s third law, $T^{2} \propto a^{3}$ i.e., square of time period $(T^{2})$ of a satellite revolving around a planet, is proportional to cube of the radius of the orbit $(a^{3})$.

**Answer**
By Kepler’s third law, $\quad T^{2} \propto r^{3} \Rightarrow T \propto r^{3 / 2}$

We know that $T$ is a function of $R$ and $g$.

Let

$ \begin{aligned} & T \propto r^{3 / 2} R^{a} g^{b} \\ & T=k r^{3 / 2} R^{a} g^{b} \end{aligned} $

where, $k$ is a dimensionless constant of proportionality.

Substituting the dimensions of each term in Eq. (i), we get

$ \begin{aligned} {[M^{0} L^{0} T] } & =k[L]^{3 / 2}[L]^{a}[LT^{-2}]^{b} \\ & =k[L^{a+b+3 / 2} T^{-2 b}] \end{aligned} $

On comparing the powers of same terms, we get

$ \begin{aligned} a+b+3 / 2 & =0 \\ -2 b & =1 \Rightarrow b=-1 / 2 \end{aligned} $

From Eq. (ii), we get

$ a-1 / 2+3 / 2=0 \Rightarrow a=-1 $

Substituting the values of $a$ and $b$ in Eq. (i), we get

$ \begin{matrix} \Rightarrow & T & =k r^{3 / 2} R^{-1} g^{-1 / 2} \\ T & =\frac{k}{R} \sqrt{\frac{r^{3}}{g}} \end{matrix} $

Note When we are applying formulae, we should be careful about $r$ (radius of orbit) and $R$ (radius of planet).

**42.** In an experiment to estimate the size of a molecule of oleic acid $1 mL$ of oleic acid is dissolved in $19 mL$ of alcohol. Then $1 mL$ of this solution is diluted to $20 mL$ by adding alcohol. Now, 1 drop of this diluted solution is placed on water in a shallow trough. The solution spreads over the surface of water forming one molecule thick layer. Now, lycopodium powder is sprinkled evenly over the film and its diameter is measured. Knowing the volume of the drop and area of the film we can calculate the thickness of the film which will give us the size of oleic acid molecule.

Read the passage carefully and answer the following questions

(a) Why do we dissolve oleic acid in alcohol?

(b) What is the role of lycopodium powder?

(c) What would be the volume of oleic acid in each $mL$ of solution prepared?

(d) How will you calculate the volume of $n$ drops of this solution of oleic acid?

(e) What will be the volume of oleic acid in one drop of this solution?

## Show Answer

**Answer**
(a) Oleic acid does not dissolve in water hence, it is dissolved in alcohol.

(b) Lycopodium powder spreads over the entire surface of water when it is sprinkled evenly. When a drop of prepared solution is dropped on water, oleic acid does not dissolve in water. Instead it spreads on the water surface pushing the lycopodium powder away to clear a circular area where the drop falls. We can therefore, measure the area over which oleic acid spreads.

(c) In each $mL$ of solution prepared volume of oleic acid $=\frac{1}{20} mL \times \frac{1}{20}=\frac{1}{400} mL$

(d) Volume of $n$ drops of this solution of oleic acid can be calculated by means of a burette and measuring cylinder and measuring the number of drops.

(e) If $n$ drops of the solution make $1 mL$, the volume of oleic acid in one drop will be $\frac{1}{(400) n} mL$.

**43.** (a) How many astronomical units (AU) make 1 parsec?

(b) Consider a sunlike star at a distance of 2 parsecs. When it is seen through a telescope with 100 magnification, what should be the angular size of the star? Sun appears to be $(1 / 2)^{\circ}$ from the earth. Due to atmospheric fluctuations, eye cannot resolve objects smaller than 1 arc minute.

(c) Mars has approximately half of the earth’s diameter. When it is closest to the earth it is at about $1 / 2 AU$ from the earth. Calculate what size it will appear when seen through the same telescope.

## Show Answer

**Answer**
(a) By definition,

1 parsec $=$ Distance at which $1 AU$ long arc subtends an angle of $1 s$.

$ \begin{aligned} \therefore & 1 \text { parsec } & =(\frac{1 AU}{1 arcsec}) \\ \therefore & 1 deg & =3600 arcsec \\ \therefore & 1 \text { parsec } & =\frac{\pi}{3600 \times 180} rad \\ & 1 \text { parsec } & =\frac{3600 \times 180}{\pi} AU=206265 AU \approx 2 \times 10^{5} AU \end{aligned} $

(b) Sun’s diameter is $(\frac{1}{2})^{\circ}$ at $1 AU$.

Therefore, at 1 parsec, star is $\frac{1 / 2}{2 \times 10^{5}}$ degree in diameter $=15 \times 10^{-5}$ arc min.

With 100 magnification, it should look $15 \times 10^{-3}$ arcmin. However, due to atmospheric fluctuations, it will still look of about 1 arcmin. It cannot be magnified using telescope.

(c) Given that

$ \frac{D_{\text {mars }}}{D_{\text {earth }}}=\frac{1}{2} $

where $D$ represents diameter.

From answer 25(e)

$\begin{matrix} \text { we know that, } & \frac{D_{\text {earth }}}{D_{\text {sun }}}=\frac{1}{100} \\ \therefore & \frac{D_{\text {mars }}}{D_{\text {sun }}}=\frac{1}{2} \times \frac{1}{100}\end{matrix} $

[from Eq. (i)]

$ \begin{aligned} & \text { At } 1 AU \text { sun’s diameter }=(\frac{1}{2})^{\circ} \\ & \therefore \quad \text { mar’s diameter }=\frac{1}{2} \times \frac{1}{200}=\frac{1}{400} \\ & \text { At } \frac{1}{2} AU, \text { mar’s diameter }=\frac{1}{400} \times 2^{\circ}=(\frac{1}{200})^{\circ} \end{aligned} $

With 100 magnification, Mar’s diameter $=\frac{1}{200} \times 100^{\circ}=(\frac{1}{2})^{\circ}=30^{\prime}$

This is larger than resolution limit due to atmospheric fluctuations. Hence, it looks magnified.

**44.** Einstein’s mass-energy relation emerging out of his famous theory of relativity relates mass $(m)$ to energy $(E)$ as $E=m c^{2}$, where $c$ is speed of light in vacuum. At the nuclear level, the magnitudes of energy are very small. The energy at nuclear level is usually measured in MeV, where $1 MeV=1.6 \times 10^{-13} J$; the masses are measured in unified atomic mass unit (u) where, $1 u=1.67 \times 10^{-27} kg$.

(a) Show that the energy equivalent of $1 u$ is $931.5 MeV$.

(b) A student writes the relation as $1 u=931.5 MeV$. The teacher points out that the relation is dimensionally incorrect. Write the correct relation.

## Show Answer

**Thinking Process**

In this problem, we have to apply Einstein’s mass-energy relation. $E=m c^{2}$, to calculate the energy equivalent of the given mass.

**Answer**
(a) We know that

$ \text { Applying } \begin{aligned} 1 amu & =1 u=1.67 \times 10^{-27} kg \\ E & =m c^{2} \\ \text { Energy } & =E=(1.67 \times 10^{-27})(3 \times 10^{8})^{2} J \\ & =1.67 \times 9 \times 10^{-11} J \\ E & =\frac{1.67 \times 9 \times 10^{-11}}{1.6 \times 10^{-13}} MeV=939.4 MeV \approx 931.5 MeV \end{aligned} $

(b) The dimensionally correct relation is

$1 amu \times c^{2}=1 u \times c^{2}=931.5 MeV$