INSTRUCTIONS
- This paper consists of six (06) questions.
- Answer five (5) questions.
- Each question carries twenty (20) marks.
- Mathematical tables and non-programmable calculators may be used.
- Cellular phones and any unauthorized materials are not allowed in the examination room.
- Write your examination number on every page of your answer sheet/booklet(s).
- The following information may be useful:
a) Acceleration due to gravity \( g = 9.8 \, \text{m/s}^2 \)b) Density of water, \( \rho_w = 10^3 \, \text{kg} \, \text{m}^{-3} \)c) Density of air, \( \rho_{\text{air}} = 1.29 \, \text{kg/m}^3 \)d) Density of mercury \( = 13.6 \times 10^3 \, \text{kg/m}^3 \)e) Coefficient of viscosity of water \( = 1.0 \times 10^{-3} \, \text{Ns} \, \text{m}^{-2} \)f) Velocity of sound in air, \( V = 330 \, \text{m/s} \)g) Coefficient of linear expansion of copper \( = 8 \times 10^{-6} \, \text{°C} \)h) Bulk modulus of elasticity of copper \( = 3.6 \times 10^{11} \, \text{N/m}^2 \)i) Surface tension of water \( = 7.2 \times 10^{-2} \, \text{N/m} \)j) Permeability of free space, \( \mu_0 = 4 \pi \times 10^{-7} \, \text{Hm}^{-1} \)k) Permittivity of free space, \( \epsilon_0 = 8.854 \times 10^{-12} \, \text{Fm}^{-1} \)l) Mass of \( \frac{3}{2} \text{Ca} = 39.962589u \)m) Mass of neutron, \( M_n = 1.008665u \)n) Mass of proton, \( M_p = 1.007825u \)o) Universal gas constant, \( R = 8.31 \, \text{J/molK} \)p) Young modulus of aluminium, \( Y_{al} = 7 \times 10^{10} \, \text{Nm}^{-2} \)q) Boltzman constant \( (K) = 1.38 \times 10^{-23} \, \text{J/K} \)r) Electronic charge, \( e = 1.6 \times 10^{-19} \, \text{C} \)s) Avogadro's number, \( N_A = 6.0 \times 10^{23} \, \text{mole} \)t) Mass of electron, \( M_e = 9.1 \times 10^{-31} \, \text{kg} \)u) Planck's constant, \( h = 6.62 \times 10^{-34} \, \text{Js} \)v) Pie, \( \pi = 3.14 \)
Solution 1(a)(i):
Raindrops do not acquire very high velocities due to:
- Air resistance (drag force): As a raindrop falls, it experiences upward drag force that increases with velocity.
- Terminal velocity: When drag force equals weight, net force becomes zero and velocity becomes constant.
- Small size: Raindrops are small, so their weight is relatively small compared to drag force.
- Shape: The spherical shape creates significant drag.
Typically, raindrops reach terminal velocities of about 9 m/s for small drops and up to 12 m/s for larger drops.
Solution 1(a)(ii):
Let: m = mass of sphere, ρₛ = density of sphere, ρₗ = density of fluid, r = radius, η = viscosity, vₜ = terminal velocity
Weight = buoyant force + viscous drag at terminal velocity:
\( mg = \rho_l g V + 6\pi\eta r v_t \)
\( \rho_s g V = \rho_l g V + 6\pi\eta r v_t \)
\( v_t = \frac{2r^2 g (\rho_s - \rho_l)}{9\eta} \)
Initial acceleration: aᵢ = g - (ρₗ/ρₛ)g = g(1 - ρₗ/ρₛ)
Average acceleration: aₐᵥ = aᵢ/2 = g(1 - ρₗ/ρₛ)/2
Time to reach vₜ: t = vₜ/aₐᵥ
\( t = \frac{2r^2 g (\rho_s - \rho_l)}{9\eta} \times \frac{2}{g(1 - \rho_l/\rho_s)} \)
\( t = \frac{4r^2 (\rho_s - \rho_l)}{9\eta (1 - \rho_l/\rho_s)} \)
Since (1 - ρₗ/ρₛ) = (ρₛ - ρₗ)/ρₛ
\( t = \frac{4r^2 (\rho_s - \rho_l)}{9\eta} \times \frac{\rho_s}{(\rho_s - \rho_l)} = \frac{4r^2 \rho_s}{9\eta} \)
Thus, time t is independent of fluid density ρₗ.
Solution 1(b)(i):
Deep water runs slowly because:
- Continuity equation: For steady flow, A₁v₁ = A₂v₂ (constant).
- In deeper sections, the cross-sectional area (A) is larger.
- To maintain constant flow rate (discharge), velocity (v) must decrease when area increases.
- Bernoulli's principle: In deeper sections, pressure is higher and velocity is lower.
- Friction with river bed also affects flow velocity distribution.
Solution 1(b)(ii):
Velocity head hᵥ = v²/2g
Given: hᵥ = 10 cm = 0.1 m, g = 9.8 m/s²
\( v^2 = 2gh_v = 2 \times 9.8 \times 0.1 = 1.96 \)
\( v = \sqrt{1.96} = 1.4 \, \text{m/s} \)
Therefore, the speed of flow is 1.4 m/s.
Solution 1(c)(i):
Flow rate Q = 8 × 10³ m³/s
Cross-sectional area A = 40 × 10⁻⁴ m² = 4 × 10⁻³ m²
Flow velocity v = Q/A = (8 × 10³)/(4 × 10⁻³) = 2 × 10⁶ m/s
This seems unusually high - likely the flow rate should be 8 × 10⁻³ m³/s
If Q = 8 × 10⁻³ m³/s, then v = (8 × 10⁻³)/(4 × 10⁻³) = 2 m/s (more realistic)
Solution 1(c)(ii):
Part 1: Total pressure = static pressure + dynamic pressure
Static pressure Pₛ = 3 × 10⁴ Pa
From part (i), v = 2 m/s, ρ = 1000 kg/m³
Dynamic pressure = ½ρv² = ½ × 1000 × 2² = 2000 Pa
Total pressure Pₜ = Pₛ + ½ρv² = 3 × 10⁴ + 2000 = 3.2 × 10⁴ Pa
Part 2: New total pressure Pₜ' = 3.6 × 10⁴ Pa
Static pressure remains Pₛ = 3 × 10⁴ Pa (horizontal pipe)
Dynamic pressure = Pₜ' - Pₛ = (3.6 - 3) × 10⁴ = 6000 Pa
½ρv² = 6000
v² = (2 × 6000)/1000 = 12
v = √12 = 3.46 m/s
Therefore, new velocity is 3.46 m/s.
Solution 1(d):
This involves Poiseuille's equation for viscous flow through pipes.
Given: L₁ = 21 cm, L₂ = 7 cm, d₂ = 0.07 cm, Total ΔP = 14 cm water = ρgh = 1000 × 9.8 × 0.14 = 1372 Pa
Let d₁ = 0.225 cm
For series flow: Q = constant, ΔP = ΔP₁ + ΔP₂
Poiseuille's law: \( Q = \frac{\pi \Delta P r^4}{8\eta L} \)
For same Q: \( \frac{\Delta P_1 r_1^4}{L_1} = \frac{\Delta P_2 r_2^4}{L_2} \)
Also: ΔP = ΔP₁ + ΔP₂ = 1372 Pa
r₁ = 0.225/2 = 0.1125 cm = 1.125 × 10⁻³ m
r₂ = 0.07/2 = 0.035 cm = 3.5 × 10⁻⁴ m
\( \frac{\Delta P_1}{\Delta P_2} = \frac{L_1 r_2^4}{L_2 r_1^4} = \frac{21 \times (3.5 \times 10^{-4})^4}{7 \times (1.125 \times 10^{-3})^4} \)
Calculate ratio: ≈ 0.085
So ΔP₁ = 0.085 ΔP₂
ΔP₁ + ΔP₂ = 0.085ΔP₂ + ΔP₂ = 1.085ΔP₂ = 1372
ΔP₂ = 1372/1.085 = 1265 Pa
ΔP₁ = 1372 - 1265 = 107 Pa
In cm water: ΔP₂ = 1265/(1000×9.8) = 0.129 m = 12.9 cm water
ΔP₁ = 107/(1000×9.8) = 0.0109 m = 1.09 cm water
Solution 2(a)(i):
The Doppler principle for sound waves states that the apparent frequency of a sound wave changes when there is relative motion between the source of sound and the observer.
When source and observer approach each other, the apparent frequency increases (higher pitch).
When source and observer recede from each other, the apparent frequency decreases (lower pitch).
Formula: \( f' = f \left( \frac{v \pm v_o}{v \mp v_s} \right) \) where v is speed of sound, vₒ is observer speed, vₛ is source speed.
Solution 2(a)(ii):
We cannot hear echo in a typical room because:
- Distance requirement: For distinct echo, the reflecting surface must be at least 17 meters away (for 0.1 second delay between direct sound and echo).
- Room size: Most rooms are smaller than this, so reflected sound reaches ear within 0.1 second, merging with direct sound as reverberation.
- Absorption: Room surfaces (furniture, carpets, curtains) absorb sound, reducing reflection intensity.
- Multiple reflections: In small rooms, many reflections occur quickly, creating reverberation rather than distinct echo.
Solution 2(a)(iii):
No, the man at the center will not observe Doppler effect because:
- There is no radial component of relative motion between source and observer.
- The source is moving in a circle around the observer, but at constant distance.
- The tangential motion does not affect frequency as there's no component along the line joining source and observer.
- Only radial motion (changing distance) causes Doppler effect for sound waves.
Solution 2(b)(i):
Transverse waves:
- Light waves (electromagnetic waves)
- Waves on a string
- Water waves (surface waves)
- S-waves (secondary seismic waves)
Longitudinal waves:
- Sound waves in air
- Spring waves (compression waves)
- P-waves (primary seismic waves)
- Ultrasound waves
Solution 2(b)(ii):
For closed pipe: Fundamental f₁ = v/4L꜀
Overtones: f₃ (third overtone) = 7f₁ = 7v/4L꜀
For open pipe: Fundamental f₁' = v/2Lₒ
First overtone f₂' = 2f₁' = v/Lₒ
Given: 7v/4L꜀ = v/Lₒ
7/4L꜀ = 1/Lₒ
Lₒ/L꜀ = 4/7
Ratio Lₒ:L꜀ = 4:7
Solution 2(c)(i):
Conditions for sustainable interference:
- Sources must be coherent (constant phase relationship).
- Sources must have same frequency (monochromatic).
- Superposing waves must have same polarization.
- Amplitudes should be comparable for observable interference.
- Path difference should be within coherence length.
- Medium should be isotropic and homogeneous.
Solution 2(c)(ii):
Diffraction grating equation: d sinθ = nλ
d = 200 μm = 200 × 10⁻⁶ m = 2 × 10⁻⁴ m
λ = 600 nm = 600 × 10⁻⁹ m = 6 × 10⁻⁷ m
Maximum n occurs when sinθ = 1: n_max = d/λ = (2 × 10⁻⁴)/(6 × 10⁻⁷) = 333.33
So maximum order n = 333 (integer less than 333.33)
Number of diffracted beams = 2n_max + 1 (including zero order) = 2×333 + 1 = 667 beams
But visible beams limited by θ = 90°, so beams with θ ≤ 90° only.
Solution 2(d)(i):
For Young's double slit: Fringe separation β = λD/d
Where: λ = wavelength, D = distance to screen, d = slit separation
If d is doubled (d' = 2d), then β' = λD/2d = β/2
Therefore, fringe separation is halved when slit separation is doubled.
The fringes become closer together, more densely packed.
Solution 2(d)(ii):
Brewster's law: tan θ_B = n₂/n₁
For air to glass: n₁ = 1 (air), n₂ ≈ 1.5 (typical glass)
tan θ_B = 1.5/1 = 1.5
θ_B = tan⁻¹(1.5) ≈ 56.3°
At Brewster angle, reflected light is completely polarized perpendicular to plane of incidence.
Solution 3(a)(i):
In a dropper, water doesn't come out unless the bulb is pressed because:
- Atmospheric pressure: Atmospheric pressure (≈ 10⁵ Pa) pushes down on water surface in dropper.
- Surface tension: Creates additional pressure difference across curved meniscus.
- Capillary action: In narrow tube, adhesive forces between water and glass create upward force.
- When bulb is pressed, air pressure inside increases, overcoming atmospheric pressure + surface tension forces.
- When bulb is released, pressure decreases, atmospheric pressure pushes water back up.
Solution 3(a)(ii):
Initial: r₁ = 2.4 × 10⁻⁴ m, P₁ = 10⁵ Pa, γ = 0.08 N/m
For soap bubble: Excess pressure ΔP = 4γ/r
Inside pressure P_in = P_out + 4γ/r
Initially: P_in₁ = 10⁵ + 4×0.08/(2.4×10⁻⁴) = 10⁵ + 1333.33 = 101333.33 Pa
Final: r₂ = r₁/2 = 1.2 × 10⁻⁴ m
Let final cylinder pressure = P₂
P_in₂ = P₂ + 4×0.08/(1.2×10⁻⁴) = P₂ + 2666.67 Pa
For isothermal compression of air inside bubble: P₁V₁ = P₂V₂
V ∝ r³, so V₂/V₁ = (r₂/r₁)³ = (1/2)³ = 1/8
P_in₁ × (4/3)πr₁³ = P_in₂ × (4/3)πr₂³
101333.33 × r₁³ = P_in₂ × (r₁/2)³
101333.33 × 8 = P_in₂
P_in₂ = 810666.64 Pa
P₂ = P_in₂ - 2666.67 = 810666.64 - 2666.67 = 807999.97 Pa ≈ 8.08 × 10⁵ Pa
Solution 3(b)(i):
If the top of a capillary tube is closed:
- Liquid will initially rise to the same height as open tube (due to capillary action).
- As liquid rises, air above it is compressed, increasing pressure.
- This increased pressure opposes further rise.
- Final height will be less than in open capillary.
- Equilibrium: Capillary force = Weight of liquid column + Force due to compressed air pressure.
Solution 3(b)(ii):
Capillary rise: Upward force = 2πrγ cosθ (for water θ ≈ 0°, cosθ ≈ 1)
Given: Upward force = 75 × 10⁻⁴ N = 0.0075 N
Surface tension of water γ = 7.2 × 10⁻² N/m
2πrγ = 0.0075
r = 0.0075/(2π × 7.2 × 10⁻²) = 0.0075/(0.4524) = 0.0166 m = 1.66 cm
Therefore, inner radius is approximately 1.66 cm.
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