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Exercise 1.1 – Fill in the blanks (Solved)

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(a)

Question:
The volume of a cube of side 1 cm is equal to _____ m³.

Solution:
Side of cube = 1 cm

$$\text{Volume}=(1\text{cm}{)}^3=1{\text{cm}}^3$$Since,

$$1\text{cm}={10}^{-2}\text{m}$$
$$1{\text{cm}}^3=({10}^{-2}{)}^3={10}^{-6}{\text{m}}^3$$

Answer:

$$\boxed{{\displaystyle 1\times {10}^{-6}{\text{m}}^3}}$$

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(b)

Question:
The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to _____ (mm)².

Solution:
Radius $r=2.0$
Height $h=10.0$

Surface area of a solid cylinder:

$$S=2\pi r(r+h)$$
$$S=2\pi \times 2\times (2+10)$$
$$S=48\pi {\text{cm}}^2$$

Convert cm² to mm²:

$$1{\text{cm}}^2=100{\text{mm}}^2$$
$$S=48\pi \times 100\approx 1.51\times {10}^4{\text{mm}}^2$$

Answer:

$$\boxed{{\displaystyle 1.51\times {10}^4(\text{mm}{)}^2}}$$

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(c)

Question:
A vehicle moving with a speed of 18 km h⁻¹ covers _____ m in 1 s.

Solution:

$$18{\text{km h}}^{-1}=\frac{18\times 1000}{3600}$$

$$=5{\text{m s}}^{-1}$$

Distance covered in 1 s:

$$=5\text{m}$$

Answer:

$$\boxed{{\displaystyle 5\text{m}}}$$

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(d)

Question:
The relative density of lead is 11.3. Its density is _____ g cm⁻³ or _____ kg m⁻³.

Solution:
Relative density =

$$\frac{\text{Density of substance}}{\text{Density of water}}$$

Density of water = 1 g cm⁻³

$$\text{Density of lead}=11.3\times 1=11.3{\text{g cm}}^{-3}$$

Convert to kg m⁻³:

$$1{\text{g cm}}^{-3}=1000{\text{kg m}}^{-3}$$$$11.3\times 1000=1.13\times {10}^4$$

Answer:

$$\boxed{{\displaystyle 11.3{\text{g cm}}^{-3}\text{ or }1.13\times {10}^4{\text{kg m}}^{-3}}}$$

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Exercise 1.2 – Fill in the blanks by suitable conversion of units (Solved)

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(a)

Question:
1 kg m² s⁻² = _____ g cm² s⁻².

Solution:$$1\text{kg}={10}^3\text{g}$$
$$1\text{m}={10}^2\text{cm}\Rightarrow 1{\text{m}}^2={10}^4{\text{cm}}^2$$
$$1{\text{kg m}}^2{\text{s}}^{-2}={10}^3\times {10}^4$$
$$={10}^7{\text{g cm}}^2{\text{s}}^{-2}$$

Answer:

$$\boxed{{\displaystyle {10}^7{\text{g cm}}^2{\text{s}}^{-2}}}$$

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Question 1.3

A calorie is a unit of heat (energy in transit) and it equals about 4.2 J where 1 J = 1 kg m² s⁻². Suppose we employ a system of units in which the unit of mass equals α kg, the unit of length equals β m, the unit of time is γ s. Show that a calorie has a magnitude $4.2{\alpha }^{-1}{\beta }^{-2}{\gamma }^2$ in terms of the new units.

Solution

We are given:

$$1\text{calorie}=4.2\text{J}$$

Also,

$$1\text{J}=1{\text{kg m}}^2{\text{s}}^{-2}$$

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Step 1: Write the new fundamental units

• New unit of mass = $\alpha \text{kg}$

• New unit of length = $\beta \text{m}$

• New unit of time = $\gamma \text{s}$

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Step 2: Express SI units in terms of new units

$$1\text{kg}={\alpha }^{-1}(\text{new mass unit})$$
$$1\text{m}={\beta }^{-1}(\text{new length unit})$$
$$1\text{s}={\gamma }^{-1}(\text{new time unit})$$

Step 3: Convert joule into the new system

$$1\text{J}=1{\text{kg m}}^2{\text{s}}^{-2}$$

Substitute the above relations:

$$1\text{J}=({\alpha }^{-1})({\beta }^{-1}{)}^2({\gamma }^{-1}{)}^{-2}$$
$$={\alpha }^{-1}{\beta }^{-2}{\gamma }^2$$

Step 4: Convert calorie

$$1\text{calorie}=4.2\text{J}$$
$$=4.2\times {\alpha }^{-1}{\beta }^{-2}{\gamma }^2$$

Final Answer

$$\boxed{{\displaystyle \text{1 calorie}=4.2{\alpha }^{-1}{\beta }^{-2}{\gamma }^2}}$$

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Question 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 :
(

Explanation

A dimensional quantity has physical dimensions such as length, mass, time, etc.
Calling such a quantity large or small has no meaning unless it is compared with a suitable standard or reference quantity of the same kind.

For example, saying that a mass is “large” is meaningless unless we specify large compared to what—a human body, a planet, or an atom.

Hence, every statement about size, mass, speed, or number must include a reference or comparison to be scientifically meaningful.

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Reframed Statements

(a) Atoms are very small objects.

Reframed:
Atoms are very small compared to ordinary macroscopic objects such as grains of sand or cells.

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(b) A jet plane moves with great speed.

Reframed:
A jet plane moves with great speed compared to road vehicles such as cars or trains.

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(c) The mass of Jupiter is very large.

Reframed:
The mass of Jupiter is very large compared to the mass of the Earth or other planets.

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(d) The air inside this room contains a large number of molecules.

Reframed:
The air inside this room contains a large number of molecules compared to the number of molecules in a small volume of air, such as a test tube.

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

Reframed:
A proton is much more massive than an electron (about 1836 times more massive).

✔️ (This statement is already meaningful because a clear comparison is given.)

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(f) The speed of sound is much smaller than the speed of light.

Reframed:
The speed of sound is much smaller than the speed of light in vacuum.

Question 1.5

A new unit of length is chosen such that the speed of light in vacuum is unity. What is the distance between the Sun and the Earth in terms of the new unit if light takes 8 min and 20 s to cover this distance?

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Solution

Step 1: Given data

Speed of light in vacuum:

$$c=1(\text{in the new system of units})$$

Time taken by light:

$$8\text{ min }20\text{ s}$$

Convert time into seconds:

$$8\text{ min }=8\times 60=480\text{ s}$$
$$\text{Total time}=480+20=500\text{ s}$$

Step 2: Use the formula for distance

$$\text{Distance}=\text{Speed}\times \text{Time}$$

Since $c=1$ in the new unit system:

$$\text{Distance}=1\times 500=500$$

Final Answer

$$\boxed{{\displaystyle \text{Distance between the Sun and the Earth}=500\text{ (new units of length)}}}$$

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Question 1.6

Which of the following is the most precise device for measuring length :
(a) a vernier callipers with 20 divisions on the sliding scale
(b) a screw gauge of pitch 1 mm and 100 divisions on the circular scale
(c) an optical instrument that can measure length to within a wavelength of light ?

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Solution

Precision of a measuring instrument depends on its least count.
The instrument with the smallest least count is the most precise.

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(a) Vernier callipers

For a standard vernier callipers:

• 1 main scale division (MSD) = 1 mm

• 20 vernier divisions = 19 MSD

Least count:

$$\text{LC}=1\text{MSD}-1\text{VSD}=1-\frac{19}{20}=0.05\text{mm}$$

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(b) Screw gauge

Given:

• Pitch = 1 mm

• Number of divisions = 100

Least count:

$$\text{LC}=\frac{\text{Pitch}}{\text{No. of divisions}}=\frac{1}{100}=0.01\text{mm}$$

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(c) Optical instrument

An optical instrument can measure length up to one wavelength of light.

Typical wavelength of visible light:

$$\lambda \approx 5\times {10}^{-7}\text{m}=0.0005\text{mm}$$

This is much smaller than the least count of both vernier callipers and screw gauge.

Answer

Since the optical instrument has the smallest least count, it is the most precise.

$$\boxed{{\displaystyle \text{(c) An optical instrument that can measure length to within a wavelength of light}}}$$

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Question 1.7

A student measures the thickness of a human hair by looking at it through a microscope of magnification 100. He makes 20 observations and finds that the average width of the hair in the field of view of the microscope is 3.5 mm. What is the estimate on the thickness of hair ?

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Solution

Step 1: Understand magnification

Magnification $M$ is given by:

$$M=\frac{\text{Apparent width}}{\text{Actual width}}$$

Given:

• Magnification, $M=100$

• Apparent (observed) width of hair = 3.5 mm

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Step 2: Find actual thickness of hair

$$\text{Actual thickness}=\frac{\text{Observed width}}{\text{Magnification}}$$

$$=\frac{3.5\text{mm}}{100}$$
$$=0.035\text{mm}$$

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Step 3: Convert into metre (optional)

$$0.035\text{mm}=3.5\times {10}^{-5}\text{m}$$

Answer

$$\boxed{{\displaystyle \text{Thickness of human hair}=0.035\text{mm}(=3.5\times {10}^{-5}\text{m})}}$$

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Question 1.8

Answer the following :

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(a)

You are given a thread and a metre scale. How will you estimate the diameter of the thread ?

Answer:

• Wind the thread closely and uniformly around a pencil or cylindrical rod for a large number of turns (say $n$turns), without leaving gaps.

• Measure the total length $L$ occupied by these $n$ turns using the metre scale.

• The diameter of the thread is then given by:

$$\text{Diameter of thread}=\frac{L}{n}$$

• Taking a large number of turns reduces error and gives a more accurate result.

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(b)

A screw gauge has a pitch of 1.0 mm and 200 divisions on the circular scale. Do you think it is possible to increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular scale ?

Answer:

No, it is not possible to increase the accuracy arbitrarily by increasing the number of divisions.

Explanation:

• Least count of screw gauge:

$$\text{LC}=\frac{\text{Pitch}}{\text{Number of divisions}}$$

• Increasing the number of divisions reduces the least count, but:

  • Mechanical limitations

  • Backlash error

  • Wear and tear

  • Manufacturing precision
    limit further improvement.

• Beyond a certain point, increasing divisions does not improve accuracy.

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(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 ?

Answer:

• Every measurement contains random errors.

• When a large number of measurements (like 100) are taken:

  • Random errors tend to cancel out.

  • The mean value becomes closer to the true value.

• With only 5 measurements:

  • Random errors have a larger effect.

  • The estimate is less reliable.

Hence, 100 measurements give a more accurate and reliable mean diameter than only 5 measurements.

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Question 1.9

The photograph of a house occupies an area of 1.75 cm² on a 35 mm slide. The slide is projected on to a screen, and the area of the house on the screen is 1.55 m². What is the linear magnification of the projector–screen arrangement?

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Solution

Step 1: Understand the relation between area and linear magnification

If linear magnification = $M$, then:

$$\text{Areal magnification}=M^2$$
$$M=\sqrt{\frac{\text{Area on screen}}{\text{Area on slide}}}$$

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Step 2: Convert all areas into the same unit

Given:

• Area on slide = $1.75{\text{cm}}^2$

• Area on screen = $1.55{\text{m}}^2$

Convert ${\text{m}}^2$ to ${\text{cm}}^2$:

$$1{\text{m}}^2={10}^4{\text{cm}}^2$$
$$1.55{\text{m}}^2=1.55\times {10}^4{\text{cm}}^2$$

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Step 3: Calculate linear magnification

$$M=\sqrt{\frac{1.55\times {10}^4}{1.75}}$$
$$=\sqrt{8857}$$
$$\approx 94$$

Answer

$$\boxed{{\displaystyle \text{Linear magnification}\approx 94}}$$

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Question 1.10

State the number of significant figures in the following :

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(a) 0.007 m²

Leading zeros are not significant.
Only 7 is significant.

Answer:

$$\boxed{{\displaystyle 1\text{ significant figure}}}$$

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(b) 2.64 × 10²⁴ kg

All digits in scientific notation are significant.
Digits: 2, 6, 4

Answer:

$$\boxed{{\displaystyle 3\text{ significant figures}}}$$

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(c) 0.2370 g cm⁻³

Leading zero is not significant.
Trailing zero after decimal is significant.

Digits: 2, 3, 7, 0

Answer:

$$\boxed{{\displaystyle 4\text{ significant figures}}}$$

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(d) 6.320 J

Trailing zero after decimal point is significant.

Digits: 6, 3, 2, 0

Answer:

$$\boxed{{\displaystyle 4\text{ significant figures}}}$$

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(e) 6.032 N m⁻²

Zero between non-zero digits is significant.

Digits: 6, 0, 3, 2

Answer:

$$\boxed{{\displaystyle 4\text{ significant figures}}}$$

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(f) 0.0006032 m²

Leading zeros are not significant.
Zero between 6 and 3 is significant.

Digits: 6, 0, 3, 2

Answer:

$$\boxed{{\displaystyle 4\text{ significant figures}}}$$

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Question 1.11

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

Solution

Given data

• Length, $l=4.234\text{m}$ → 4 significant figures

• Breadth, $b=1.005\text{m}$ → 4 significant figures

• Thickness, $t=2.01\text{cm}=0.0201\text{m}$ → 3 significant figures

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(a) Area of the sheet

$$\text{Area}=l\times b$$
$$=4.234\times 1.005$$
$$=4.25517{\text{m}}^2$$

Both length and breadth have 4 significant figures, so the result should be given to 4 significant figures.

Area (correct significant figures)

$$\boxed{{\displaystyle \text{Area}=4.255{\text{m}}^2}}$$

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(b) Volume of the sheet

$$\text{Volume}=l\times b\times t$$
$$=4.234\times 1.005\times 0.0201$$
$$=0.0855389{\text{m}}^3$$

The least number of significant figures among the quantities is 3 (thickness).

Volume (correct significant figures)

$$\boxed{{\displaystyle \text{Volume}=0.0855{\text{m}}^3}}$$

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Question 1.12

The mass of a box measured by a grocer’s balance is 2.30 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box. What is:
(a) the total mass of the box,
(b) the difference in the masses of the pieces, to correct significant figures?

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Solution

Given data

• Mass of box = 2.30 kg → 3 significant figures

• Mass of first gold piece = 20.15 g

• Mass of second gold piece = 20.17 g

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(a) Total mass of the box

Step 1: Convert gold masses into kg

$$20.15\text{g}=0.02015\text{kg}$$
$$20.17\text{g}=0.02017\text{kg}$$

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Step 2: Add all masses

$$\text{Total mass}=2.30+0.02015+0.02017$$
$$=2.34032\text{kg}$$

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Step 3: Apply significant-figure rule (addition)

In addition, the result should have the least number of decimal places among the quantities.

• Mass of box = 2.30 kg → 2 decimal places

So, round the result to 2 decimal places:

$$\boxed{{\displaystyle \text{Total mass}=2.34\text{kg}}}$$

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(b) Difference in the masses of the gold pieces

Step 1: Subtract the two masses

$$\text{Difference}=20.17\text{g}-20.15\text{g}$$
$$=0.02\text{g}$$

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Step 2: Apply significant-figure rule (subtraction)

Both values are correct to two decimal places, so the result should also be given to two decimal places.

$$\boxed{{\displaystyle \text{Difference in masses}=0.02\text{g}}}$$

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Question 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 :

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

Guess where to put the missing $c$.

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Solution

Step 1: Use dimensional analysis

• Speed $v$ has dimensions:

  $$[v]=L{T}^{-1}$$

• Therefore,

  $$[v^2]=L^2{T}^{-2}$$

The quantity inside the square root must be dimensionless.
But in the given expression:

$$(1-v^2)$$

the term $v^2$ has dimensions, so the expression is dimensionally incorrect.

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Step 2: Make the expression dimensionless

To cancel the dimensions of $v^2$, it must be divided by a quantity having the same dimensions, i.e. $c^2$, where $c$ is the speed of light.

Thus, the correct dimensionless term is:

$$\frac{v^2}{c^2}$$

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Step 3: Write the correct formula

Substituting this into the expression:

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

Answer

$$\boxed{{\displaystyle m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}}}$$

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Question 1.14

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

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Solution

Step 1: Given data

• Radius (size) of hydrogen atom:

$$r=0.5\mathring{\text{A}}=0.5\times {10}^{-10}\text{m}$$

• Number of atoms in one mole:

$$N=6.02\times {10}^{23}$$

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Step 2: Volume of one hydrogen atom

Assuming the atom to be spherical,

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

$$=\frac{4}{3}\pi (0.5\times {10}^{-10}{)}^3$$
$$=\frac{4}{3}\pi (0.125\times {10}^{-30})$$
$$=5.24\times {10}^{-31}{\text{m}}^3(\text{approximately})$$

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Step 3: Total atomic volume of one mole

$$\text{Total volume}=N\times \text{volume of one atom}$$
$$=6.02\times {10}^{23}\times 5.24\times {10}^{-31}$$
$$=3.15\times {10}^{-7}{\text{m}}^3$$

Answer

$$\boxed{{\displaystyle \text{Total atomic volume of one mole of hydrogen}\approx 3.1\times {10}^{-7}{\text{m}}^3}}$$

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Question. 1.15
One mole of an ideal gas at STP occupies 22.4 L. What is the ratio of molar volume to the atomic volume of a mole of hydrogen? (Size of hydrogen molecule ≈ 1 Å). Why is this ratio so large?

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Step 1: Molar volume of hydrogen gas at STP

$$V_{\text{molar}}=22.4\text{ L}=22.4\times {10}^{-3}{\text{ m}}^3$$

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Step 2: Volume of one hydrogen molecule

Size (diameter) of hydrogen molecule ≈ $1\mathring{\text{A}}={10}^{-10}\text{m}$

Assume the molecule to be spherical:

$$V_{\text{one molecule}}=\frac{4}{3}\pi r^3$$

$$r=\frac{1}{2}\times {10}^{-10}=0.5\times {10}^{-10}\text{m}$$
$$V_{\text{one molecule}}\approx \frac{4}{3}\pi (0.5\times {10}^{-10}{)}^3\approx 5.2\times {10}^{-31}{\text{m}}^3$$

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Step 3: Atomic (molecular) volume of one mole of hydrogen

Number of molecules in 1 mole:

$$N_A=6.02\times {10}^{23}$$
$$V_{\text{atomic (1 mole)}}=6.02\times {10}^{23}\times 5.2\times {10}^{-31}$$
$$V_{\text{atomic}}\approx 3.1\times {10}^{-7}{\text{m}}^3$$

Step 4: Required ratio

$$\text{Ratio}=\frac{V_{\text{molar}}}{V_{\text{atomic}}}=\frac{22.4\times {10}^{-3}}{3.1\times {10}^{-7}}$$
$$\boxed{{\displaystyle \text{Ratio}\approx 7.2\times {10}^4}}$$

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Question 1.16

Explanation of the Observation

When you look out of the window of a fast-moving train, the apparent motion of objects depends on their distance from you.

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1. Nearby objects (trees, houses, poles)

• These objects are close to the observer.

• As the train moves, your position with respect to these objects changes rapidly.

• This causes a large change in their angular position in a short time.

• Hence, nearby objects appear to move rapidly in the opposite direction to the train’s motion.

This effect is called relative motion.

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2. Distant objects (hills, mountains)

• These objects are very far away.

• Even though the train moves, the change in their angular position is very small.

• Therefore, they appear to move very slowly or seem almost stationary.

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3. Very distant objects (Moon, stars)

• These objects are at extremely large distances.

• The change in their position relative to the observer is negligible.

• As a result, they appear to be completely stationary or appear to move along with you.

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Question 1.17

The Sun is a hot plasma, so one might initially guess that its density should be like that of gases. Let us now check this guess by calculation using the given data.

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Given data

• Mass of the Sun,

  $$M=2.0\times {10}^{30}\text{ kg}$$

• Radius of the Sun,

  $$R=7.0\times {10}^8\text{ m}$$

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Step 1: Volume of the Sun

The Sun is approximately spherical.

$$V=\frac{4}{3}\pi R^3$$

$$V=\frac{4}{3}\pi (7.0\times {10}^8{)}^3$$
$$V\approx 1.44\times {10}^{27}{\text{m}}^3$$

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Step 2: Average density of the Sun

$$\rho =\frac{M}{V}$$
$$\rho =\frac{2.0\times {10}^{30}}{1.44\times {10}^{27}}$$

$$\rho \approx 1.4\times {10}^3{\text{kg m}}^{-3}$$

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Step 3: Interpretation

• Density of gases (at STP):

  $$\sim 1{\text{kg m}}^{-3}$$

• Density of liquids/solids (water):

  $$\sim {10}^3{\text{kg m}}^{-3}$$

The Sun’s average density ≈ 1400 kg m⁻³, which is:

much greater than gases
comparable to liquids and solids