Units and Measurements — NCERT Class 11 Physics

(a) Side = 1 cm = 10⁻² m Volume = (10⁻²)³ = 10⁻⁶ m³

(b) Radius r = 2.0 cm = 20 mm, Height h = 10.0 cm = 100 mm Surface area = 2πr(r + h) = 2 × 3.14 × 20 × (20 + 100) = 2 × 3.14 × 20 × 120 = 1.5 × 10⁴ mm²

(c) Speed = 18 km h⁻¹ = 18 × (1000/3600) = 5 m s⁻¹, so distance in 1 s = 5 m

(d) Relative density = 11.3, so density = 11.3 g cm⁻³ = 11.3 × 10³ = 1.13 × 10⁴ kg m⁻³

(a) 1 kg m² s⁻² = 10³ g × (10²)² cm² s⁻² = 10³ × 10⁴ = 10⁷ g cm² s⁻²

(b) 1 ly = 9.46 × 10¹⁵ m, so 1 m = 1/(9.46 × 10¹⁵) = 1.057 × 10⁻¹⁶ ly

(c) 3.0 m s⁻² = 3.0 × (10⁻³ km) × (3600)² h⁻² = 3.0 × 10⁻³ × 1.296 × 10⁷ = 3.888 × 10⁴ km h⁻²

(d) G = 6.67 × 10⁻¹¹ × (10²)³ cm³ × (10³)⁻¹ g⁻¹ s⁻² = 6.67 × 10⁻¹¹ × 10⁶ × 10⁻³ = 6.67 × 10⁻⁸ cm³ g⁻¹ s⁻²

1 calorie = 4.2 J = 4.2 kg m² s⁻²

In the new system: 1 kg = 1/α new mass unit, 1 m = 1/β new length unit, 1 s = 1/γ new time unit.

So, 1 kg m² s⁻² = (1/α)(1/β)² (1/γ)⁻² new units = α⁻¹ β⁻² γ² new units

Therefore, 1 calorie = 4.2 × α⁻¹ β⁻² γ² in the new system. Proved.

A dimensional quantity always has a number and a unit. The number alone tells us nothing without context. For example, calling the mass of an electron (9.1 × 10⁻³¹ kg) "small" is meaningless unless we compare it with something — it is small compared to a proton, but enormous compared to the Planck mass. Similarly, the speed of light (3 × 10⁸ m s⁻¹) is large compared to everyday speeds but small compared to the speed of gravitational waves. The magnitude of a dimensional quantity acquires meaning only in comparison with a reference standard.

Speed of light = 1 new length unit per second (c = 1) Time = 8 min 20 s = 480 + 20 = 500 s Distance = speed × time = 1 × 500 = 500 new length units

The least count determines precision: (a) Vernier callipers: LC = 1 mm/20 = 0.05 mm = 5 × 10⁻⁵ m (b) Screw gauge: LC = 1 mm/100 = 0.01 mm = 1 × 10⁻⁵ m (c) Optical instrument: LC ≈ wavelength of light ≈ 5 × 10⁻⁷ m

The optical instrument has the smallest least count, so (c) is the most precise.

Same as intext Q1 above.

Correct reading = Observed reading − Zero error = 3.28 − (−0.005) = 3.28 + 0.005 = 3.285 mm

Thickness = 2.01 cm = 0.0201 m

Area of two largest faces = 2(l × b) = 2(4.234 × 1.005) = 2 × 4.255 = 8.510 m² Area of side faces = 2(l × t) + 2(b × t) = 2(4.234 × 0.0201) + 2(1.005 × 0.0201) = 2(0.08510) + 2(0.02020) = 0.1702 + 0.04040 = 0.2106 m² Total surface area = 8.510 + 0.2106 ≈ 8.72 m² (3 significant figures, limited by thickness)

Volume = l × b × t = 4.234 × 1.005 × 0.0201 = 0.0855 m³ (3 significant figures)

(a) Mass of box = 2.300 kg = 2300.0 g Total mass = 2300.0 + 20.15 + 20.17 = 2340.32 g Least decimal places among the masses: box has 1 decimal place (2300.0) Total mass = 2340.3 g (or 2.340 kg)

(b) Difference = 20.17 − 20.15 = 0.02 g (2 significant figures)

Since P = a³b²/(c√d) = a³b²c⁻¹d⁻¹/²

% error in P = 3(% error in a) + 2(% error in b) + 1(% error in c) + (1/2)(% error in d) = 3(1) + 2(3) + 1(4) + (1/2)(2) = 3 + 6 + 4 + 1 = 14%

V = πr²L

% error in V = 2(% error in r) + (% error in L) = 2(1%) + 2% = 4%

Let F ∝ mᵃ vᵇ rᶜ

[MLT⁻²] = [M]ᵃ [LT⁻¹]ᵇ [L]ᶜ = [Mᵃ Lᵇ⁺ᶜ T⁻ᵇ]

Comparing dimensions:

• M: 1 = a → a = 1
• T: −2 = −b → b = 2
• L: 1 = b + c → 1 = 2 + c → c = −1

Therefore, F = k m¹ v² r⁻¹ = kmv²/r. Proved.

• Left side: s has dimensions [L]
• Right side: ut → [LT⁻¹][T] = [L] ✓
• (1/2)at² → [LT⁻²][T²] = [L] ✓

Both terms on the right have dimensions [L], same as the left side. The equation is dimensionally consistent.

[At²] must have dimensions of velocity [LT⁻¹] So [A][T²] = [LT⁻¹] → [A] = [LT⁻³]

[Bt] must have dimensions of velocity [LT⁻¹] So [B][T] = [LT⁻¹] → [B] = [LT⁻²]

Dimensions of A = [LT⁻³], Dimensions of B = [LT⁻²]

F = ma → m = F/a = F/(L/T²) = FT²/L

Dimensional formula for mass = [FT²L⁻¹]

Let K ∝ Iᵃ ωᵇ

K: [ML²T⁻²], I: [ML²], ω: [T⁻¹]

[ML²T⁻²] = [ML²]ᵃ [T⁻¹]ᵇ = [Mᵃ L²ᵃ T⁻ᵇ]

Comparing:

• M: 1 = a → a = 1
• L: 2 = 2a → a = 1 ✓
• T: −2 = −b → b = 2

Therefore K = kI¹ω² = kIω². Proved.

For a formula to be correct, the argument of sin/cos must be dimensionless.

(a) 2πt/T: [T]/[T] = dimensionless ✓ → Correct (b) vt: [LT⁻¹][T] = [L] — not dimensionless ✗ → Incorrect (c) t/a: [T]/[L] — not dimensionless ✗ → Incorrect (d) 2πt/T: dimensionless ✓, a/√2 has dimension [L], y has dimension [L] ✓ → Correct

Formulas (a) and (d) are correct.

The effective velocity of rain relative to the man has a horizontal component = wind velocity u and a vertical component = rain's terminal velocity v. For the man walking with velocity v_m:

tan φ = (horizontal component of relative velocity)/(vertical component)

If wind velocity is u and the rain falls vertically: tan φ = u/v (when man is stationary)

But if the man is walking with velocity v_m: tan φ = (u + v_m sin θ)/v (adjusting for the angle of wind)

The standard result: tan φ = u/v where u is the horizontal wind component and v is the vertical component of rain's velocity.

Measured value = (2.45 ± 0.02) cm

The true value lies between 2.45 − 0.02 and 2.45 + 0.02, i.e., between 2.43 cm and 2.47 cm.

μ is dimensionless (refractive index). So B/λ² must also be dimensionless.

[B]/[L²] = dimensionless → [B] = [L²]

Dimensions of B = [L²]