INSTRUCTIONS
- This paper consists of six (6) 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 booklet(s).
- The following information may be useful:
(a) Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)(b) Pi, \( \pi = 3.14 \, \text{s} \)(c) Avogadro's Number, \( N_A = 6.0 \times 10^{23} \, \text{mol}^{-1} \)(d) Density of water = \( 10^3 \, \text{kg/m}^3 \)(e) Charge of electron = \( 1.6 \times 10^{-19} \, \text{C} \)(f) 1 Year = \( 3.15 \times 10^7 \, \text{s} \)(g) 1 MeV = \( 1.6 \times 10^{-13} \, \text{J} \)(h) Permeability of free space, \( \mu_0 = 4\pi \times 10^{-7} \, \text{Hm}^{-1} \)(i) Mass of electron \( m_e = 9.1 \times 10^{-31} \, \text{kg} \)(j) Permittivity of free space, \( \epsilon_0 = 8.854 \times 10^{-12} \, \text{Fm}^{-1} \)(k) Relative permittivity of air \( \epsilon_r = 1 \)(l) Surface tension of water, \( T = 0.072 \, \text{Nm}^{-1} \)(m) Mass of \( \frac{7}{H} = 3.345 \times 10^{-27} \, \text{kg}, \, \frac{3}{H} = 5.008 \times 10^{-27} \, \text{kg}, \, \frac{4}{He} = 6.647 \times 10^{-27} \, \text{kg} \) and \( \frac{1}{n} = 1.675 \times 10^{-27} \, \text{kg} \)
Solution 1(a):
Two factors determining the magnitude of viscous force are:
- Coefficient of viscosity (η): The internal friction property of the fluid itself.
- Velocity gradient (dv/dx): The rate of change of velocity with distance perpendicular to flow direction.
According to Newton's law of viscosity: \( F = \eta A \frac{dv}{dx} \)
Where η = coefficient of viscosity, A = area, dv/dx = velocity gradient.
Solution 1(b):
Limitations of Stokes' Law:
- Only valid for small spherical particles moving at low velocities (creeping flow, Re < 0.1).
- Assumes fluid is infinite in extent (no wall effects).
- Assumes particle is rigid and smooth.
- Not applicable for turbulent flow or high Reynolds numbers.
Importances of Stokes' Law:
- Used to determine viscosity of fluids (falling sphere viscometer).
- Used to determine terminal velocity of falling particles.
- Applied in sedimentation analysis (Stokes' law settling).
- Used in designing separation processes (cyclones, centrifuges).
- Important in atmospheric science (fall of raindrops, dust particles).
Area 1 (Wide tube A) → Narrow tube C → Area 1 (Wide tube B)
Points 1 and 2 indicated for Bernoulli's theorem application
Solution 1(c):
Applying Bernoulli's theorem between points 1 and 2:
\( P_1 + \frac{1}{2}\rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho gh_2 \)
For horizontal venturi: h₁ = h₂, so:
\( P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2 \)
\( P_1 - P_2 = \frac{1}{2}\rho (v_2^2 - v_1^2) \) ...(1)
From manometer reading: \( P_1 - P_2 = \rho g h \) ...(2)
Equating (1) and (2): \( \rho g h = \frac{1}{2}\rho (v_2^2 - v_1^2) \)
\( g h = \frac{1}{2}(v_2^2 - v_1^2) \) ...(3)
From continuity equation: \( A_1 v_1 = A_2 v_2 = Q \)
So: \( v_1 = \frac{Q}{A_1} \) and \( v_2 = \frac{Q}{A_2} \)
Substituting in (3): \( g h = \frac{1}{2}\left(\frac{Q^2}{A_2^2} - \frac{Q^2}{A_1^2}\right) \)
\( 2g h = Q^2\left(\frac{1}{A_2^2} - \frac{1}{A_1^2}\right) \)
\( 2g h = Q^2\left(\frac{A_1^2 - A_2^2}{A_1^2 A_2^2}\right) \)
\( Q^2 = \frac{2g h A_1^2 A_2^2}{A_1^2 - A_2^2} \)
\( Q = A_1 A_2 \sqrt{\frac{2g h}{A_1^2 - A_2^2}} \)
Hence proved.
Solution 1(d)(i):
Using Torricelli's theorem: \( v = \sqrt{2gh} \)
Where h = height of water above orifice = 5 m
g = 9.8 m/s²
\( v = \sqrt{2 \times 9.8 \times 5} = \sqrt{98} = 9.899 \, \text{m/s} \)
Therefore, initial speed from orifice is approximately 9.9 m/s.
Solution 1(d)(ii):
The water emerges horizontally from orifice at height H = 5 m (platform height).
Horizontal velocity v_x = 9.9 m/s (from part i)
Vertical motion: Initial vertical velocity v_y = 0, falling from height 5 m
Time to fall: \( t = \sqrt{\frac{2H}{g}} = \sqrt{\frac{2 \times 5}{9.8}} = \sqrt{1.0204} = 1.01 \, \text{s} \)
Vertical velocity at impact: \( v_y = gt = 9.8 \times 1.01 = 9.898 \, \text{m/s} \)
Resultant velocity: \( v = \sqrt{v_x^2 + v_y^2} = \sqrt{9.9^2 + 9.898^2} = \sqrt{98.01 + 97.96} = \sqrt{195.97} = 14.0 \, \text{m/s} \)
Therefore, speed when striking ground is approximately 14.0 m/s.
Solution 2(a)(i):
A dextro-rotatory substance rotates the plane of polarization of plane polarized light to the right (clockwise) when viewed looking toward the light source.
Solution 2(a)(ii):
A laevo-rotatory substance rotates the plane of polarization of plane polarized light to the left (counter-clockwise) when viewed looking toward the light source.
Solution 2(a)(iii):
An optically active substance has the ability to rotate the plane of polarization of plane polarized light. This property is important for analyzing molecular structure and concentration in polarimetry.
Solution 2(a)(iv):
Double refraction (birefringence) is the splitting of a light ray into two rays (ordinary and extraordinary) with different polarizations and velocities. This property is used in Nicol prisms and other polarizing devices to produce plane polarized light.
Solution 2(b)(i):
Polaroid: A synthetic material that produces plane polarized light by selective absorption (dichroism). It's used as a polarizer or analyzer.
Polarimeter: An instrument used to measure the angle of rotation of plane polarized light by optically active substances.
Solution 2(b)(ii):
Plane of vibration: The plane containing the electric field vector and direction of propagation.
Plane of polarization: The plane perpendicular to the plane of vibration (contains magnetic field vector and direction of propagation).
Solution 2(b)(iii):
Ordinary light (unpolarized): Light in which electric field vectors vibrate in all possible directions perpendicular to direction of propagation.
Plane polarized light: Light in which electric field vectors vibrate in only one plane perpendicular to direction of propagation.
Solution 2(c):
A Nicol prism is constructed from a calcite crystal (calcium carbonate) which exhibits double refraction:
- A calcite crystal is cut diagonally into two parts along a plane perpendicular to the principal section.
- The two parts are cemented together with Canada balsam, which has a refractive index (1.55) between the ordinary (1.658) and extraordinary (1.486) refractive indices of calcite.
- The prism is cut such that the optic axis is parallel to the end faces and at an angle to the sides.
- When unpolarized light enters, it splits into ordinary (o-ray) and extraordinary (e-ray).
- The o-ray undergoes total internal reflection at the Canada balsam interface because its refractive index (1.658) is greater than Canada balsam (1.55).
- The e-ray passes through as it has lower refractive index (1.486) than Canada balsam.
- Thus, the emerging light is plane polarized.
Solution 2(d)(i):
When the glass plate is silvered on its front surface:
- The fringe pattern becomes sharper and more distinct.
- The contrast between bright and dark fringes increases.
- This happens because silvering increases reflection, making the interference more pronounced.
- Multiple reflections are enhanced, improving fringe visibility.
Solution 2(d)(ii):
When sodium lamp (monochromatic light) is replaced by white light:
- Colored fringes are observed instead of dark and bright fringes.
- Central spot may appear dark or colored depending on phase conditions.
- Different colors interfere constructively at different thicknesses, creating rainbow-like patterns.
- Only a few fringes are visible due to overlapping of different colored patterns.
Solution 2(d)(iii):
When a transparent liquid is introduced between lens and plate:
- The fringe pattern contracts (rings become smaller).
- The radius of the rings decreases.
- This happens because the optical path difference changes: \( \mu t \) instead of \( t \), where μ is refractive index of liquid.
- For same optical path difference, actual thickness t is smaller, so rings are closer to center.
Solution 3(a)(i):
When the top of a capillary tube is closed:
- Initially, liquid rises due to capillary action as in open tube.
- As liquid rises, air above it gets compressed, increasing pressure.
- This increased air pressure opposes further rise of liquid.
- Equilibrium is reached when: Capillary force = Weight of liquid column + Force due to compressed air.
- Final height is less than in open capillary tube.
Solution 3(a)(ii):
Rain drops are spherical due to:
- Surface tension: Liquid tends to minimize its surface area for given volume.
- For given volume, sphere has minimum surface area.
- Surface tension acts like an elastic skin, pulling liquid into spherical shape.
- In free fall, air resistance can flatten larger drops, but small drops remain nearly spherical.
Solution 3(b)(i):
Brick walls are plastered with cement because:
- Smooth surface: Provides smooth finish for painting or decoration.
- Weather protection: Protects bricks from rain, moisture, and weathering.
- Strength: Increases structural strength and durability.
- Thermal insulation: Provides additional insulation.
- Aesthetic appeal: Improves appearance of walls.
Solution 3(b)(ii):
Capillary rise/depression: \( h = \frac{2\gamma \cos\theta}{\rho g r} \)
For water: γ = 0.072 N/m, θ ≈ 0°, cosθ ≈ 1, ρ = 1000 kg/m³, g = 9.8 m/s²
For narrow tube (r₁ = 6.5×10⁻⁴ m): \( h_1 = \frac{2 \times 0.072 \times 1}{1000 \times 9.8 \times 6.5 \times 10^{-4}} = 0.0226 \, \text{m} \)
For wide tube (r₂ = 1.24×10⁻³ m): \( h_2 = \frac{2 \times 0.072 \times 1}{1000 \times 9.8 \times 1.24 \times 10^{-3}} = 0.01185 \, \text{m} \)
Given: h₁' - h₂' = 0.2 m (observed difference)
True pressure difference: \( P = \rho g (h_1' - h_2' + h_2 - h_1) \)
\( P = 1000 \times 9.8 \times (0.2 + 0.01185 - 0.0226) \)
\( P = 9800 \times 0.18925 = 1854.65 \, \text{Pa} \)
Solution 3(c)(i):
Surface tension: The property of liquid surface to behave like a stretched elastic membrane, minimizing its surface area. It is defined as the force per unit length acting perpendicular to an imaginary line drawn on the liquid surface.
S.I. unit: Newton per meter (N/m) or Joule per square meter (J/m²).
Solution 3(c)(ii):
Let radius of each small drop = r
Volume of 64 drops = 64 × (4/3)πr³
Let radius of combined drop = R
Volume conserved: (4/3)πR³ = 64 × (4/3)πr³
R³ = 64r³ ⇒ R = 4r
Surface energy = Surface tension × Surface area
For 64 drops: E₁ = 64 × γ × 4πr² = 256πγr²
For single drop: E₂ = γ × 4πR² = γ × 4π(4r)² = γ × 4π × 16r² = 64πγr²
Ratio E₁/E₂ = (256πγr²)/(64πγr²) = 4
Therefore, ratio is 4:1
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