Question 1

1 Fill in the blanks

(a) The volume of a cube of side $1\mathrm{cm}$ is equal to $\dots \mathrm{m}^3$

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

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

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

Accepted Answer

(a) Volume of cube = side^3 = (1 cm)^3 = (1 × 10^{-2} m)^3 = 1 × 10^{-6} m^3

(b) Surface area of cylinder = 2\pi r (r + h)
= 2 × 3.1416 × 2 cm × (2 cm + 10 cm) = 2 × 3.1416 × 2 × 12 = 150.8 cm^2
Convert to mm^2: 1 cm^2 = 100 mm^2
So, 150.8 cm^2 = 150.8 × 100 = 15080 mm^2

(c) Speed = 18 km/h = 18 × (1000 m / 3600 s) = 5 m/s
Distance covered in 1 s = 5 m

(d) Relative density (RD) = density of lead / density of water
Density of water = 1 g/cm^3
So, density of lead = RD × density of water = 11.3 × 1 = 11.3 g/cm^3
In kg/m^3: 1 g/cm^3 = 1000 kg/m^3
So, density = 11.3 × 1000 = 11300 kg/m^3 — Step-by-step conversions and formula applications:
(a) Convert cm to m and cube the length.
(b) Use formula for surface area of cylinder, then convert cm^2 to mm^2.
(c) Convert km/h to m/s and calculate distance.
(d) Use relative density definition and convert units.

Question 2

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) $1\mathrm{m} = \dots \mathrm{ly}$

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

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

Accepted Answer

(a) 1 kg m^2 s^{-2} = 1 × (10^3 g) × (10^2 cm)^2 × s^{-2} = 1 × 10^7 g cm^2 s^{-2}

(b) 1 m = 1 / (9.46 × 10^{15}) ly ≈ 1.057 × 10^{-16} ly

(c) 3.0 m s^{-2} = 3.0 × (3600)^2 km h^{-2} = 3.0 × 12960000 = 3.888 × 10^{7} km h^{-2}

(d) G = 6.67 × 10^{-11} N m^2 kg^{-2}
1 N = 1 kg m s^{-2}
Convert units:
1 m = 100 cm, 1 kg = 1000 g
So,
G = 6.67 × 10^{-11} × (100)^3 cm^3 s^{-2} (1000)^{-1} g^{-1} = 6.67 × 10^{-8} cm^3 s^{-2} g^{-1} — Conversions:
(a) kg to g (1 kg = 1000 g), m to cm (1 m = 100 cm), square the length conversion.
(b) Light year (ly) in meters is approx 9.46 × 10^{15} m.
(c) Convert m/s^2 to km/h^2 using (1 m/s^2 = 12960 km/h^2).
(d) Convert G from SI units to cgs units using unit conversions.

Question 3

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

Accepted Answer

Given: 1 calorie = 4.2 J
1 J = 1 kg m^2 s^{-2}
In new units:
Mass unit = α kg
Length unit = β m
Time unit = γ s

Express 1 J in new units:
1 J = 1 kg m^2 s^{-2}
= (1/α) mass units × (1/β)^2 length units^2 × (1/γ)^(-2) time units^{-2}
= (1/α) × (1/β^2) × (γ^2) in new units

Therefore, 1 calorie = 4.2 J = 4.2 × α^{-1} × β^{-2} × γ^{2} in new units. — Energy dimensionally is M L^2 T^{-2}.
Replacing M by α mass units, L by β length units, T by γ time units,
Energy unit = α × β^2 × γ^{-2} in SI units.
So, 1 J = 1/(α β^2 γ^{-2}) new units = α^{-1} β^{-2} γ^{2} new units.
Multiply by 4.2 for calorie.

Question 4

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.

Accepted Answer

The statement means that 'large' or 'small' are relative terms and only meaningful when compared to a reference or standard.

Reframed statements:
(a) Atoms are very small compared to everyday objects.
(b) A jet plane moves with great speed compared to a car.
(c) The mass of Jupiter is very large compared to Earth.
(d) The air inside this room contains a large number of molecules compared to a vacuum.
(e) A proton is much more massive than an electron (mass of proton is about 1836 times that of electron).
(f) The speed of sound is much smaller than the speed of light (speed of sound ~340 m/s, speed of light ~3×10^8 m/s). — The terms 'large' and 'small' are relative and require a standard for meaningful comparison.
Each statement is reframed to specify the comparison standard.

Question 5

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?

Accepted Answer

Given:
Speed of light c = 1 new unit
Time taken by light to travel from Sun to Earth = 8 min 20 s = 8 × 60 + 20 = 500 s

Distance = speed × time = 1 × 500 = 500 new units

Therefore, the distance between Sun and Earth is 500 units in the new length unit system. — Since speed of light is defined as 1 unit, distance = time × speed = time × 1 = time in seconds.
Convert time to seconds and multiply.

Question 6

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 $1 \, 	ext{mm}$ and 100 divisions on the circular scale
(c) an optical instrument that can measure length to within a wavelength of light?

Accepted Answer

Option (c) is the most precise device.

Explanation:
- Vernier callipers typically have precision up to 0.01 cm or 0.1 mm.
- Screw gauge with 1 mm pitch and 100 divisions can measure up to 0.01 mm.
- Optical instruments can measure lengths to within the wavelength of light (~500 nm or 0.0005 mm), which is more precise than mechanical devices. — Precision depends on the least count.
Optical instruments can measure smaller lengths than mechanical devices due to smaller wavelength scale.

Question 7

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

Accepted Answer

Given:
Magnification, M = 100
Observed width = 3.5 mm

Actual thickness = Observed width / Magnification = 3.5 mm / 100 = 0.035 mm = 35 micrometers

Therefore, the estimated thickness of the hair is 0.035 mm or 35 μm. — Magnification relates observed size to actual size by:
Actual size = Observed size / Magnification.
Apply this formula to find thickness.

Question 8

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 $1.0 \, 	ext{mm}$ 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?

Accepted Answer

(a) Wind the thread around a pencil or a thin rod multiple times (say N turns). Measure the total length L of the thread on the pencil using the metre scale. Then, diameter of the thread = L / N.

(b) Increasing the number of divisions on the circular scale reduces the least count and improves accuracy to some extent. However, beyond a limit, mechanical imperfections, backlash, and human error limit the accuracy. So, accuracy cannot be increased arbitrarily.

(c) Taking 100 measurements reduces random errors and gives a better average value, improving reliability and precision compared to only 5 measurements. — (a) Using multiple turns averages out errors and allows measurement with a less precise scale.
(b) Mechanical and practical limitations restrict accuracy.
(c) Larger sample size reduces statistical errors.

Question 9

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

Accepted Answer

Given:
Area on slide, A1 = 1.75 cm^2 = 1.75 × 10^{-4} m^2
Area on screen, A2 = 1.55 m^2

Linear magnification, M = \sqrt{\frac{A2}{A1}} = \sqrt{\frac{1.55}{1.75 	imes 10^{-4}}} = \sqrt{8857.14} ≈ 94.1

Therefore, the linear magnification is approximately 94. — Linear magnification is the ratio of linear dimensions, so it is the square root of area magnification.
Calculate area magnification = A2 / A1, then take square root.

Question 10

10 State the number of significant figures in the following :

(a) $0.007 \, 	ext{m}^2$
(b) $2.64 	imes 10^{24} \, 	ext{kg}$
(c) $0.2370 \, 	ext{g} \, 	ext{cm}^{-3}$
(d) $6.320 \, 	ext{J}$
(e) $6.032 \, 	ext{N} \, 	ext{m}^{-2}$
(f) $0.0006032 \, 	ext{m}^2$

Accepted Answer

(a) 2 significant figures (7 and 0 are zeros before first non-zero digit, so not significant)
(b) 3 significant figures (2, 6, 4)
(c) 4 significant figures (2, 3, 7, 0; trailing zero after decimal is significant)
(d) 4 significant figures (6, 3, 2, 0)
(e) 4 significant figures (6, 0, 3, 2)
(f) 4 significant figures (6, 0, 3, 2) — Rules:
- Leading zeros are not significant.
- Trailing zeros after decimal point are significant.
- All non-zero digits are significant.

Question 11

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

Accepted Answer

Given:
Length, L = 4.234 m (4 significant figures)
Breadth, B = 1.005 m (4 significant figures)
Thickness, T = 2.01 cm = 0.0201 m (3 significant figures)

Area = L × B = 4.234 × 1.005 = 4.25517 m^2
Significant figures limited by least precise measurement (4 sig figs)
So, Area = 4.255 m^2

Volume = L × B × T = 4.234 × 1.005 × 0.0201 = 0.0856 m^3
Least number of significant figures is 3 (from thickness)
So, Volume = 0.0856 m^3 — Multiplication results are rounded to the least number of significant figures among the factors.
Convert thickness to meters before calculation.

Question 12

12 The mass of a box measured by a grocer's balance is $2.30 \, 	ext{kg}$. Two gold pieces of masses $20.15 \, 	ext{g}$ and $20.17 \, 	ext{g}$ 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?

Accepted Answer

(a) Convert masses to same units:
Mass of box = 2.30 kg = 2300 g
Mass of gold pieces = 20.15 g and 20.17 g

Total mass = 2300 + 20.15 + 20.17 = 2340.32 g
Expressed with correct significant figures (box mass has 3 sig figs, gold pieces 4 sig figs)
Total mass = 2340 g = 2.34 kg (3 significant figures)

(b) Difference = 20.17 g - 20.15 g = 0.02 g
Difference has 1 significant figure (since last digit uncertain), so difference = 0.02 g — Add masses after converting to same units.
Round off final answer according to least number of significant figures.
Difference is limited by least precise measurement.

Question 13

13 A famous relation in physics relates 'moving mass' $m$ to the 'rest mass' $m_{\mathrm{e}}$ 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:

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

Guess where to put the missing $c$.

Accepted Answer

The correct formula is:

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

The missing $c$ should be inside the square root denominator as $v^2 / c^2$ to make the expression dimensionally consistent. — In special relativity, the Lorentz factor is $ \gamma = \frac{1}{\sqrt{1 - v^2/c^2}} $.
The boy's formula misses the division by $c^2$ inside the square root.
Without $c$, the expression is dimensionally incorrect.

Question 14

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

Accepted Answer

Given:
Size of hydrogen atom = 0.5 Å = 0.5 × 10^{-10} m = 5 × 10^{-11} m

Atomic volume (volume of one atom) = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (5 	imes 10^{-11})^3
= \frac{4}{3} \pi (1.25 	imes 10^{-31}) = 5.24 	imes 10^{-31} m^3

Number of atoms in 1 mole = Avogadro's number = 6.022 × 10^{23}

Total atomic volume = atomic volume × number of atoms
= 5.24 × 10^{-31} × 6.022 × 10^{23} = 3.16 × 10^{-7} m^3 — Calculate volume of one atom assuming spherical shape.
Multiply by Avogadro's number to get total volume of one mole.

Question 15

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

Accepted Answer

Given:
Molar volume = 22.4 L = 22.4 × 10^{-3} m^3
Size of hydrogen molecule = 1 Å = 1 × 10^{-10} m

Atomic volume of one molecule = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (1 	imes 10^{-10})^3 = 4.19 	imes 10^{-30} m^3

Total atomic volume for one mole = 4.19 × 10^{-30} × 6.022 × 10^{23} = 2.52 × 10^{-6} m^3

Ratio = molar volume / atomic volume = \frac{22.4 	imes 10^{-3}}{2.52 	imes 10^{-6}} \approx 8889

This ratio is very large because gases have large intermolecular spaces; atoms/molecules occupy very small volume compared to the volume occupied by the gas. — Calculate atomic volume for one mole and compare with molar volume.
Large ratio indicates gas molecules are far apart in space.

Units And Measurement

Units And Measurement — Study Notes

1.1 Introduction

1.2 The international system of units

1.3 Significant figures

Practice Questions — Units And Measurement

All 7 Chapters in Physics Part-I