## Units And Measurement Exercises

### 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 $=1 ~cm^{3}$

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

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

Hence, the volume of a cube of side $1 ~cm$ 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,

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

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

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

**(c)** Using the conversion,

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

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

Therefore, distance can be obtained using the relation:

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

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

**(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 $=1 ~g / ~cm^{3}$

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

$ =11.3 \times 1=11.3 g / ~cm^{3} $

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

$1 ~cm^{3}=10^{-6} ~m^{3}$

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

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

**1.2**Fill in the blanks by suitable conversion of units

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

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

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

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

## Show Answer

**Answer**

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

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

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

$=10^{3} ~g \times 10^{4} cm^{2} \times 1 ~s^{-2}=10^{7} ~g cm^{2} s^{-2}$

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

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

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

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

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

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

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

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

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

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

**(d)** $1 ~N=1 ~kg m s^{-2}$

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

$1 ~m^{3}=10^{6} ~cm^{3}$

$\therefore 6.67 \times 10^{-11} ~N m^{2} kg^{-2}=6.67 \times 10^{-11} \times(1 ~kg m s^{-2})(1 m^{2})(1 s^{-2})$

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

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

New unit of mass $=\alpha kg$

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

In terms of the new unit of length,

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

And, in terms of the new unit of time,

$ \begin{aligned} & 1 s=\frac{1}{\gamma}=\gamma^{-1} \\ & 1 s^{2}=\gamma^{-2} \\ & 1 s^{-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{aligned} $

**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 $8 \mathrm{~min}$ 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, $t=8 \min 20 s=500 s$

$\therefore$ 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)

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

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

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

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

$=0.00001 cm$

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 $=3.5 ~mm$

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

**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 $=1.75 cm^{2}$

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

$=1.55 \times 10^{4} cm^{2}$

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 $0.007 ~m^{2}$.

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 $0.2370 ~g cm^{-3}$.

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 $6.320 ~J$.

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 $6.032 ~Nm^{-2}$.

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

**(f)**

The given quantity is $0.0006032 ~m^{2}$.

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, $l=4.234 ~m$

Breadth of sheet, $b=1.005 ~m$

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

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 $=2(l \times b+b \times h+h \times l)$

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

$=2(4.25517+0.02620+0.08510)$

$=2 \times 4.360$

$=8.72 ~m^{2}$

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

$=4.234 \times 1.005 \times 0.0201$

$=0.0855 ~m^{3}$

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 $=2.300 ~kg$

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

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

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

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 $2.3 ~kg$.

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

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 :

$m=\frac{m_{O}}{\left(1-v^{2}\right)^{1 / 2}}$

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 $\mathring{A}$ : $1 \mathring{A}=10^{-10} \mathrm{~m}$. The size of a hydrogen atom is about $0.5 \mathring{A}$. What is the total atomic volume in $\mathrm{m}^{3}$ of a mole of hydrogen atoms?

## Show Answer

**Answer**

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

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 $1 \mathring{A}$ ). Why is this ratio so large?

## Show Answer

**Answer**

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

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,

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

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

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.