Question 1

(a)(i) What is meant by the term viscosity as applied in fluid dynamics?

Answer: Viscosity is the measure of a fluid's resistance to flow or deformation. It quantifies the internal friction between layers of fluid as they move relative to each other. A fluid with high viscosity (like honey) flows slowly, while one with low viscosity (like water) flows easily.

(a)(ii) Distinguish between streamline flow and turbulent flow of a liquid. Give two points.

Answer:

• Streamline flow: Fluid particles move in parallel layers, no mixing between layers; flow is smooth and orderly.
• Turbulent flow: Fluid particles move irregularly with mixing and swirling; flow is chaotic and contains eddies.

Extra points: Streamline flow occurs at low velocities (low Reynolds number), turbulent flow occurs at high velocities (high Reynolds number).

(b)(i) Identify the principle on which the continuity equation is based?

Answer: The continuity equation is based on the principle of conservation of mass — mass flowing into a section of pipe per unit time equals mass flowing out.

(b)(ii) Why does the velocity increase when water flowing in a broader pipe enters a narrow pipe?

Answer: According to the continuity equation (A1V1 = A2V2), if the cross-sectional area decreases, velocity must increase to maintain constant flow rate.

(c)(i) Briefly explain how the viscosities of two liquids can be compared.

Answer: By observing the rate of flow of each liquid through the same size capillary tube under identical conditions (e.g., using Poiseuille’s law). The liquid with the slower flow rate has a higher viscosity.

(c)(ii) Water flows through a horizontal tube of diameter 0.008 m and length 4 km at rate 20 liters/sec. Assuming only viscous resistance, estimate the pressure difference to maintain the flow.

Solution:

Use Poiseuille's law:
ΔP = (8ηLQ) / (πr⁴)

• η (viscosity) = 1×10⁻³ Ns/m² (given)
• L = 4000 m
• Q = 20 liters/s = 0.02 m³/s
• Diameter = 0.008 m → radius r = 0.004 m

Calculate:

Calculate denominator first: π × (0.004)⁴ = 3.14 × (2.56 × 10⁻¹⁰) ≈ 8.04 × 10⁻¹⁰

Numerator: 8 × 1×10⁻³ × 4000 × 0.02 = 0.64

So, ΔP = 0.64 / 8.04×10⁻¹⁰ ≈ 7.96 × 10⁸ Pa

Pressure difference required: Approximately 7.96 × 10⁸ Pascals (very high due to long tube and narrow radius).

(d) A horizontal pipe of diameter 20 cm has constriction of diameter 4 cm. Velocity and pressure before constriction are 2 m/s and 10 N/m² respectively. Find pressure at constriction.

Solution:

Use Bernoulli's equation and continuity:

• Area1 = π(0.1)² = 0.0314 m²
• Area2 = π(0.02)² = 0.00126 m²

Velocity at constriction V2 = (Area1/Area2) × V1 = (0.0314 / 0.00126) × 2 = 50 m/s

Bernoulli’s equation:

Assuming water density ρ = 1000 kg/m³:

Rearranged for P2:

Calculate:

• P1 = 10 N/m²
• 0.5 × 1000 × (2² - 50²) = 500 × (4 - 2500) = 500 × (-2496) = -1,248,000

P2 = 10 - 1,248,000 = -1,247,990 N/m² (negative pressure indicates vacuum, meaning the flow assumptions may be idealized)

Interpretation: Large increase in velocity at constriction causes a large pressure drop.

Question 2

(a)(i) State two distinctive properties between travelling and standing waves.

• Travelling waves move energy from one point to another; standing waves do not transfer energy along the medium but oscillate in fixed points (nodes and antinodes).
• Travelling waves have continuous wave fronts; standing waves have fixed nodes where displacement is zero.

(a)(ii) A sound wave travels in first medium with velocity v, in second medium with velocity 4v. Find wavelength in second medium.

Answer: Since frequency is constant, λ₂ = v₂ / f = 4v / f = 4 × λ₁.

(b)(i) State the principle of superposition as applied to wave motion.

Answer: When two or more waves meet, the resultant displacement at any point is the algebraic sum of the displacements of the individual waves.

(b)(ii) Analyse five differences between interference and diffraction of light waves.

(c) The transverse displacement of a string clamped at both ends is given by:
y(x, t) = 0.06 sin(273x) cos(120πt) (x and y in meters, t in seconds). Length = 1.5 m, mass = 3×10⁻² kg.

(c)(i) What type of wave does this equation represent?

Answer: It represents a standing wave formed by the superposition of two waves travelling in opposite directions on the string.

(c)(ii) Determine the tension in the string.

Solution:

Wave number k = 273 m⁻¹, angular frequency ω = 120π rad/s.

Wave speed v = ω / k = (120π) / 273 ≈ 1.38 m/s.

Mass per unit length μ = mass / length = 0.03 / 1.5 = 0.02 kg/m.

Tension T = μv² = 0.02 × (1.38)² = 0.02 × 1.9 = 0.038 N.

Tension in the string is approximately 0.038 Newtons.

Question 3

(a)(i) How do brittle materials differ from ductile materials?

• Brittle materials break suddenly without significant deformation (e.g., glass).
• Ductile materials deform plastically (stretch or bend) before breaking (e.g., copper).

(a)(ii) Figure 1 shows force F against extension e for two iron wires X and Y of same length. Which wire extends more under constant force? Why?

Answer: The wire with the steeper slope (larger gradient) is stiffer and extends less. The wire with gentler slope extends more. This depends on Young’s modulus and wire thickness.

(b)(i) A uniform iron bar diameter 8 mm, length 500 mm expands by 0.4 mm when heated uniformly. Later clamped and cooled. Calculate tension in bar.

Solution outline:

• Calculate strain due to thermal expansion: ε = ΔL / L = 0.4 / 500 = 8×10⁻⁴
• Stress = Young’s modulus × strain = E × ε = 2×10¹¹ × 8×10⁻⁴ = 1.6×10⁸ Pa
• Cross-sectional area A = πr² = π(0.004)² = 5.03×10⁻⁵ m²
• Tension force = stress × area = 1.6×10⁸ × 5.03×10⁻⁵ = 8048 N

Tension in bar is approximately 8.05 kN.

(b)(ii) Calculate increase in energy of brass bar length 0.2 m, area 1 cm² compressed with 49 N.

Solution:

• Strain energy U = (F² × L) / (2 × A × E)
• Convert area: 1 cm² = 1×10⁻⁴ m²
• Young's modulus for brass E = 1×10¹¹ Pa
• U = (49² × 0.2) / (2 × 1×10⁻⁴ × 1×10¹¹) = (2401 × 0.2) / (2 × 1×10⁷) = 480.2 / 2×10⁷ ≈ 2.4×10⁻⁵ J

Energy increase is about 2.4 × 10⁻⁵ Joules.

Question 4

(a)(i) Why are springs made of steel and not copper?

• Steel has a high Young's modulus (stiffness) which allows springs to store elastic energy efficiently.
• Copper is more ductile and less stiff, so it will deform permanently under stress.

(a)(ii) A copper rod of length 2 m and area 2.0 cm² is fastened end-to-end to a steel rod of area 1.0 cm². Equal and opposite pulls of 3×10⁴ N are applied. If elongations equal, find length L of steel rod. (E_steel = 1.2×10¹¹ N/m², E_copper = 2×10¹¹ N/m²)

Solution outline:

Elongation δ = FL / AE

For copper (Cu): δ_cu = F × 2 / (A_cu × E_cu)

For steel (St): δ_st = F × L / (A_st × E_st)

Set δ_cu = δ_st → solve for L:

Simplify (F cancels):

Calculate left side: 2 / (4×10⁷) = 5×10⁻⁸

Cross multiply:

L = 5×10⁻⁸ × 1.2×10⁷ = 0.6 m

Length of steel rod is 0.6 meters.

(b) Identify three important properties of equipotential surfaces.

• Equipotential surfaces are always perpendicular to electric field lines.
• No work is done moving a charge along an equipotential surface.
• Equipotential surfaces never intersect.

Question 5

(a)(i) Distinguish between magnetic flux density and magnetic field intensity.

• Magnetic flux density (B): Measure of magnetic field strength in a given area (Tesla). It relates to the force exerted on moving charges.
• Magnetic field intensity (H): Related to the magnetizing force produced by electric currents or magnets (Ampere per meter).

(a)(ii) Calculate maximum e.m.f induced in coil of 500 turns, area 4.0 cm², 50 revolutions per second in magnetic field 0.04 T.

Solution:

Maximum emf, E = N × A × B × ω, where ω = 2π × frequency

• N = 500 turns
• A = 4.0 cm² = 4.0 × 10⁻⁴ m²
• B = 0.04 T
• f = 50 Hz

Calculate ω:

ω = 2π × 50 = 314.16 rad/s

E = 500 × 4.0×10⁻⁴ × 0.04 × 314.16 = 2.5 Volts (approximately)

(b) A wire 2 m long carrying 10 A current is placed in magnetic field 0.15 T. Find force on wire when:

• (i) At right angle to field: F = BIL sinθ = 0.15 × 10 × 2 × sin90° = 3 N
• (ii) At 45°: F = 0.15 × 10 × 2 × sin45° = 3 × 0.707 = 2.12 N
• (iii) Along field (0°): F = 0 (sin0° = 0)

(c)(i) State Lenz's law of electromagnetic induction.

It states that the induced current in a conductor opposes the change in magnetic flux that produces it.

(c)(ii) Two identical wires R and S lie parallel 0.1 m apart. Current in R is 10 A opposite to 30 A in S. Find magnitude and direction of magnetic flux density at midpoint.

Solution outline:

• Magnetic field from wire: B = μ₀I / (2πr)
• μ₀ = 4π×10⁻⁷ T·m/A
• Distance to midpoint = 0.05 m

B_R = (4π×10⁻⁷)(10) / (2π×0.05) = 4×10⁻⁶ T

B_S = (4π×10⁻⁷)(30) / (2π×0.05) = 1.2×10⁻⁵ T

Since currents opposite, fields oppose. Net field = 1.2×10⁻⁵ - 4×10⁻⁶ = 8×10⁻⁶ T toward wire S.

Question 6

(a)(i) Analyse two drawbacks of Bohr's model of atom.

• Fails to explain spectra of atoms with more than one electron.
• Does not incorporate wave-particle duality and quantum mechanics principles.

(a)(ii) State two differences between Rutherford's model and Bohr's model.

• Rutherford model: electrons revolve in any orbit; Bohr model: electrons revolve in quantized energy levels.
• Bohr explains atomic spectra; Rutherford does not.

(b)(i) What would happen if electrons in atom were stationary?

Electrons would spiral into nucleus due to electrostatic attraction, making atoms unstable and matter as we know impossible.

(b)(ii) Electron requires 47.2 eV to excite from 2nd Bohr orbit; find Z (atomic number) in a nucleus with charge Ze.

Note: Energy to excite from 2nd orbit = 13.6 eV × Z² × (1/1² - 1/2²) = 13.6 Z² × (3/4)

Set equal: 47.2 = 13.6 × Z² × 3/4 = 10.2 Z² → Z² = 47.2 / 10.2 ≈ 4.63 → Z ≈ 2.15 (approx 2)

(c)(i) Which hydrogen spectrum series lies in visible region?

Balmer series (transitions to n=2) lies in visible spectrum.

(c)(ii) Hydrogen atoms emit spectral lines with frequency given by formula (Rydberg equation). Calculate highest frequency in Lyman series (transitions to n=1).

Highest frequency corresponds to transition from n=∞ to n=1:

Frequency = R_H × c = 3.29 × 10¹⁵ Hz (approximate value)