Exercises
Q1.1. Fill in the blanks
(a) The volume of a cube of side 1 cm is equal to ..... m³
(b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to ... (mm)²
(c) A vehicle moving with a speed of 18 km 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.
Ans.
(a) 1 cm = 10⁻² m → Volume = (10⁻²)³ = 10⁻⁶ m³
(b) Surface area of cylinder = 2πr(r + h) = 2π × 2.0(2.0 + 10.0) cm²
= 2π × 2.0 × 12.0 = 48π cm² ≈ 150.72 cm²
1 cm = 10 mm → 1 cm² = 100 mm²
= 150.72 × 100 = 1.5072 × 10⁴ mm²
(c) 18 km h⁻¹ = 18 × (1000/3600) m s⁻¹ = 5 m s⁻¹ → covers 5 m in 1 s
(d) Relative density = density of lead / density of water = 11.3
Density of water = 1 g cm⁻³ = 1000 kg m⁻³
Density of lead = 11.3 g cm⁻³ = 11.3 × 10³ kg m⁻³
Q1.2. Fill in the blanks by suitable conversion of units
(a) 1 kg m² s⁻² = ....g cm² s⁻²
(b) 1 m = ..... ly
(c) 3.0 m s⁻² = .... km h⁻²
(d) G = 6.67 × 10⁻¹¹ N m² (kg)⁻² = .... (cm)³ s⁻² g⁻¹
Ans.
(a) 1 kg = 10³ g; 1 m = 10² cm
1 kg m² s⁻² = 10³ g × (10² cm)² × s⁻² = 10³ × 10⁴ g cm² s⁻² = 10⁷ g cm² s⁻²
(b) Speed of light c = 3 × 10⁸ m/s; 1 light year = 9.46 × 10¹⁵ m
1 m = 1/(9.46 × 10¹⁵) ly ≈ 1.057 × 10⁻¹⁶ ly
(c) 1 m = 10⁻³ km; 1 s = 1/3600 h → 1 s⁻² = (3600)² h⁻²
0 m s⁻² = 3.0 × 10⁻³ km × (3600)² h⁻² = 3.0 × 10⁻³ × 1.296 × 10⁷ km h⁻² = 3.888 × 10⁴ km h⁻²
Answer
(d) G = 6.67 × 10⁻¹¹ N m² kg⁻²
1 N = 1 kg m s⁻² = 10³ g × 10² cm × s⁻² = 10⁵ g cm s⁻²
1 m² = 10⁴ cm²; 1 kg⁻¹ = 10⁻³ g⁻¹
G = 6.67 × 10⁻¹¹ × 10⁵ × 10⁴ × 10⁻³ cm³ s⁻² g⁻¹ = 6.67 × 10⁻⁸ cm³ s⁻² g⁻¹
3. Show that a calorie (= 4.2 J = 4.2 kg m² s⁻²) has a magnitude 4.2 α⁻¹ β⁻² γ² in a system where unit of mass = α kg, unit of length = β m, unit of time = γ s.
Answer
1 calorie = 4.2 J = 4.2 kg m² s⁻²
In new system: 1 kg = (1/α) new mass units; 1 m = (1/β) new length units; 1 s = (1/γ) new time units
⇒ 1 s⁻¹ = γ new time⁻¹
1 calorie = 4.2 × (1/α) × (1/β)² × (γ)² (in new units)
= 4.2 α⁻¹ β⁻² γ²
4. Explain: "To call a dimensional quantity 'large' or 'small' is meaningless without specifying a standard for comparison". Reframe the statements where necessary:
Answer
(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
Ans.
Any dimensional quantity is large or small only relative to some standard reference. Without specifying what it is being compared to, such a statement has no objective meaning.
(a) Atoms are very small objects compared to objects we observe in everyday life.
(b) A jet plane moves with a speed much greater than the fastest road vehicle.
(c) The mass of Jupiter is very large compared to the mass of Earth.
(d) The air in this room contains a large number of molecules compared to the number of people in the world.
(e) No reframing needed — already comparative (proton mass ≈ 1836 × electron mass).
(f) No reframing needed — already comparative.
5. A new unit of length is chosen such that the speed of light in vacuum is unity. What is the Sun-Earth distance in terms of the new unit if light takes 8 min 20 s to cover this distance?
Answer
Given: c = 1 new unit/s; time = 8 × 60 + 20 = 500 s
Distance = c × t = 1 × 500 = 500 new units of length
Q1.6. Which is the most precise device for measuring length
(a) vernier callipers with 20 divisions (b) screw gauge of pitch 1 mm and 100 divisions (c) optical instrument measuring to within wavelength of light?
Answer
(a) Least count = 1/20 mm = 0.05 mm
(b) Least count = 1/100 mm = 0.01 mm
(c) Least count ≈ λ_visible ≈ 4 × 10⁻⁷ m = 4 × 10⁻⁴ mm ≈ 0.0004 mm
The optical instrument (c) is most precise.
7. A student measures thickness of hair through a microscope of magnification 100. Average width observed = 3.5 mm. What is the estimate of thickness of hair?
Answer
Actual thickness = observed width / magnification = 3.5 / 100 = 0.035 mm = 3.5 × 10⁻⁵ m
Q1.8. Answer the following
(a) How to estimate diameter of a thread using a thread and metre scale?
(b) Can we increase screw gauge accuracy arbitrarily by increasing divisions?
(c) Why does 100 measurements give more reliable estimate than 5?
Ans.
(a) Wind the thread N times closely on a pencil. Measure total length L of the winding. Diameter = L/N.
(b) No. Mechanical imperfections, backlash, and human error in reading very small divisions impose a practical limit on accuracy regardless of the number of scale divisions.
(c) With more observations, random errors (equally likely positive or negative) tend to cancel out, reducing the net error in the mean. The mean of 100 measurements has much smaller random error than the mean of 5 measurements.
9. Area of house on 35 mm slide = 1.75 cm². Area on screen = 1.55 m². Find linear magnification of projector-screen arrangement.
Answer
Area on screen = 1.55 m² = 1.55 × 10⁴ cm²
Area magnification = (1.55 × 10⁴) / 1.75 = 8857
Linear magnification = √8857 ≈ 94.1
Q1.10. State the number of significant figures in
(a) 0.007 m² (b) 2.64 × 10²⁴ kg (c) 0.2370 g cm⁻³ (d) 6.320 J (e) 6.032 N m⁻² (f) 0.0006032 m²
Ans.
(a) 1 — only the digit 7 is significant
(b) 3 — digits 2, 6, 4
(c) 4 — digits 2, 3, 7, 0 (trailing zero after decimal is significant)
(d) 4 — digits 6, 3, 2, 0
(e) 4 — digits 6, 0, 3, 2 (zero between non-zero digits is significant)
(f) 4 — digits 6, 0, 3, 2
11. Length = 4.234 m, breadth = 1.005 m, thickness = 2.01 cm of a metal sheet. Give area and volume to correct significant figures.
Answer
Area = l × b = 4.234 × 1.005 = 4.25527 m²
⇒ 4.255 m² (4 sig figs)
t = 2.01 cm = 0.0201 m (3 sig figs)
Volume = 4.255 × 0.0201 = 0.085526 m³
⇒ 0.0855 m³ (3 sig figs, limited by thickness)
12. Mass of box = 2.30 kg. Two gold pieces: 20.15 g and 20.17 g are added. Find (a) total mass (b) difference in masses.
Answer
(a) Total = 2300 g + 20.15 g + 20.17 g = 2340.32 g
Box measured to nearest 10 g (2.30 kg accurate to 0.01 kg = 10 g), so total = 2.34 kg
(b) Difference = 20.17 − 20.15 = 0.02 g
13. A boy writes m = m₀ / (1 − v)^(1/2). Where does the missing c go?
Answer
The term inside the square root must be dimensionless. v has dimensions [L T⁻¹], so dividing by c (also [L T⁻¹]) makes it dimensionless. Also, it must be v² (to make the expression valid at v=0).
Correct formula: m = m₀ / √(1 − v²/c²)
14. Size of hydrogen atom ≈ 0.5 Å. Find total atomic volume in m³ of a mole of hydrogen atoms.
Answer
r = 0.5 Å = 0.5 × 10⁻¹⁰ m
Volume of 1 atom = (4/3)πr³ = (4/3) × 3.14159 × (0.5 × 10⁻¹⁰)³
= (4/3) × 3.14159 × 1.25 × 10⁻³¹ = 5.236 × 10⁻³¹ m³
Total volume = N_A × V_atom = 6.022 × 10²³ × 5.236 × 10⁻³¹ = 3.154 × 10⁻⁷ m³
15. Molar volume of ideal gas at STP = 22.4 L. Size of H₂ molecule ≈ 1 Å. Find ratio of molar volume to atomic volume of a mole of H. Why is it large?
Answer
r = 0.5 × 10⁻¹⁰ m; V_atom = 5.24 × 10⁻³¹ m³
Atomic volume of 1 mole = 6.022 × 10²³ × 5.24 × 10⁻³¹ = 3.15 × 10⁻⁷ m³
Molar volume = 22.4 L = 2.24 × 10⁻² m³
Ratio = (2.24 × 10⁻²) / (3.15 × 10⁻⁷) ≈ 7.1 × 10⁴
This ratio is large because in a gas at STP, molecules are far apart from each other — intermolecular distances are much larger than molecular dimensions. The actual volume of molecules is negligible compared to the volume occupied by the gas.
16. Why do nearby objects seem to move opposite to a moving train's direction, while distant objects seem stationary?
Answer
This is due to parallax. The angular velocity of an object at distance d as seen from the moving train is approximately v/d, where v is the train's speed. For nearby objects (small d), v/d is large, so they appear to sweep past quickly in the opposite direction. For very distant objects like mountains or the Moon (d
⇒ very large), v/d ≈ 0, so they appear stationary or seem to move with the observer.
17. In what range do you expect the mass density of the Sun? Check with mass = 2.0 × 10³⁰ kg, radius = 7.0 × 10⁸ m.
Answer
Although the Sun is in a plasma state at very high temperatures, its density can be calculated:
V = (4/3)πR³ = (4/3) × 3.14159 × (7.0 × 10⁸)³ = (4/3) × 3.14159 × 3.43 × 10²⁶ = 1.437 × 10²⁷ m³
ρ = M/V = (2.0 × 10³⁰) / (1.437 × 10²⁷) ≈ 1390 kg m⁻³ ≈ 1.4 × 10³ kg m⁻³
This is in the range of solids and liquids (water ≈ 10³ kg/m³), not gases (~1 kg/m³). Despite the high temperature, gravitational compression keeps the density high.
Important Extra Questions
Q. What are the seven fundamental quantities in SI? State their units.
Answer
The seven base quantities and their SI units are:
1. Length → metre (m)
2. Mass → kilogram (kg)
3. Time → second (s)
4. Electric current → ampere (A)
5. Thermodynamic temperature
⇒ kelvin (K)
6. Amount of substance → mole (mol)
7. Luminous intensity → candela (cd)
Q. What is dimensional analysis? State its applications and limitations.
Answer
Dimensional analysis uses the dimensions of physical quantities to establish relationships or check validity of equations.
Applications: (1) Check dimensional correctness of equations; (2) Derive relations between physical quantities; (3) Convert units between different systems.
Limitations: (1) Cannot determine dimensionless constants; (2) Cannot handle equations involving trigonometric or exponential functions; (3) Cannot confirm physical correctness, only dimensional consistency.
Q. Distinguish between systematic error and random error.
Ans.
Systematic error: Errors that are consistent and predictable, occurring in the same direction every time. Caused by faulty instruments, wrong technique, or incorrect calibration. Cannot be reduced by repeated measurements. Example: zero error in a vernier calliper.
Random error: Errors that vary unpredictably from measurement to measurement, equally likely to be positive or negative. Caused by fluctuations in experimental conditions. Can be reduced by taking more measurements and averaging.
Q. What is the principle of homogeneity of dimensions? Give an example.
Answer
The principle of homogeneity states that in a physically correct equation, each term must have the same dimensions. This is because you cannot add or equate quantities of different dimensions.
Example: Check v² = u² + 2as
[v²] = (m/s)² = m² s⁻²
[u²] = m² s⁻²
[2as] = (m s⁻²)(m) = m² s⁻²
All terms have same dimensions [M⁰ L² T⁻²], so the equation is dimensionally correct.
Q. What is the order of magnitude? Find the order of magnitude of: (a) 2.5 × 10¹⁰ (b) 8.6 × 10³
Answer
Order of magnitude is the power of 10 nearest to a number. For N = a × 10ⁿ: if a < 5, order = 10ⁿ; if a ≥ 5, order = 10ⁿ⁺¹.
(a) 2.5 × 10¹⁰: a = 2.5 < 5
⇒ order of magnitude = 10¹⁰
(b) 8.6 × 10³: a = 8.6 ≥ 5
⇒ order of magnitude = 10⁴