Papers I, II & III — Questions & Worked Answers
Logically Reorganised by Paper, Section and Topic
Useful Constants:
Universal gravitational constant, G = 6.67×10−11 Nm2kg−2 | Speed of light, c = 3.0×108 ms−1
Acceleration due to gravity, g = 10 ms−2 | Speed of sound (air) = 340 ms−1 | R = 8.31 Jmol−1K−1
Linear expansivity of steel = 11.0×10−6 K−1 | Density of water = 1000 kgm−3
SECTION A — Answer All Questions (40 marks)
Q1. Fundamental and Derived Quantities
(i) Explain the terms fundamental quantities and derived quantities.
(ii) What is the dimension of h in the equation E = hf?
Q2. Spring Balance in a Lift
An object of mass 4 kg is suspended from a spring balance attached to the ceiling of a lift. What is the reading when:
(i) The lift is stationary?
(ii) The lift moves upward with acceleration 0.15 ms−2?
(iii) The lift moves downward with acceleration 0.20 ms−2?
Q3. Bullet Penetration
A block of wood (mass 1 kg) is held in a vice. A bullet (mass 0.01 kg) penetrates to a depth of 0.10 m. The block is then suspended freely and an identical bullet fired. To what depth will the bullet penetrate?
Q4. Doppler Effect — Train Whistle
A train approaches at 50 ms−1 sounding its whistle at 500 Hz. Find the frequency perceived when the train:
(i) Approaches. (ii) Recedes. Comment on your results.
Q5. Mixing Liquids — Calorimetry
Liquids at 15°C (A), 20°C (B), 25°C (C). Equal masses A+B → 18°C; equal masses B+C → 24°C. What temperature results from mixing equal masses of A and C?
Q6. Surface Temperature of the Sun
The solar constant is 0.14 Wcm−2. The ratio of Earth’s orbital radius to the sun’s radius is approximately 215. Calculate the surface temperature of the sun (treat the sun as a perfect blackbody).
Q7. Water Level in a Tank
Water flows into a large tank at 10−4 m3s−1 and exits through a hole of cross-sectional area 10−4 m2 at the base. How high does the water rise at equilibrium?
Q8. SONAR — Ocean Depth & Wavelength
A sonar device emits waves at 4×104 Hz. Speed in air = 400 ms−1; in water = 1350 ms−1. An echo returns 0.8 s after emission. Determine:
(i) The depth of the ocean. (ii) The wavelength in air and in water.
Q9. Thermal Expansion of Steel Tyre
A steel tyre of diameter 1.50 m at 10°C is to be fitted onto a wheel of diameter 1.51 m. To what temperature must the tyre be heated? (αsteel = 11.0×10−6 K−1)
Q10. Evaporation
(a) What is evaporation?
(b) Explain why evaporation is accompanied by cooling.
SECTION B — Mechanics (Answer ONE question, 20 marks)
Q11. Newton’s Laws & Laser-Mirror Deflection
(a)(i) State Newton’s Third Law of Motion.
(a)(ii) What are non-conservative forces?
(a)(iii) State the Principle of Conservation of Linear Momentum.
(b) A mirror (mass 9.0×10−6 kg) is suspended from a quartz filament 0.04 m long. A laser beam of energy 100 J strikes it perpendicularly.
(b)(i) Using conservation of energy and momentum, discuss the laser-mirror interaction.
(b)(ii) Determine the angle of deflection (perfect reflector, negligible vibration).
Q12. Universal Gravitation — Earth & Moon
(a)(i) State the Law of Universal Gravitation.
(a)(ii) Derive expressions for the mass m and density ρ of the Earth; calculate their numerical values.
(b) An astronaut can jump 2 m vertically on Earth.
(b)(i) Moon diameter = 0.25 × Earth’s; Moon density = (2/3) × Earth’s. Show gmoon = gearth/6.
(b)(ii) How far can the astronaut jump on the Moon?
SECTION C — Heat & Properties of Matter (Answer TWO questions, 40 marks)
Q13. Fluid Flow — Reynolds Number & Bernoulli
(a) Explain laminar flow and turbulent flow.
(b) Write the expression for Reynolds number and show it is dimensionless.
(c) For streamline flow, Bernoulli’s principle gives: (P1−P2)Δv = ½ρΔv(v22−v12) + ρgΔv(h2−h1). Show total work done per unit volume is constant.
(d) Water enters through a pipe (diameter 0.02 m) at 4×105 Pa absolute. The 2nd-floor pipe (5 m above) has diameter 0.01 m.
(i) Inlet velocity = 4 ms−1. Find flow velocity and pressure in the bathroom.
(ii) What is the pressure if the tap is turned off?
Q14. Specific Heat Capacity — Continuous Flow Method
(a) Define: (i) specific heat capacity; (ii) specific latent heat of fusion. State SI units.
(b) Describe the continuous flow method for determining specific heat capacity of a liquid.
(c) Using the data below, calculate the specific heat capacity of water:
| Experiment 1 | Experiment 2 | |
|---|---|---|
| Current in heating coil | 2.0 A | 1.5 A |
| Potential difference across coil | 5.5 V | 4.5 V |
| Mass of water collected | 0.0405 kg | 0.0700 kg |
| Time of flow | 60.0 s | 180.0 s |
| Inlet temperature | 38.0°C | 38.0°C |
| Outlet temperature | 45.0°C | 45.0°C |
(d) State and explain THREE factors that affect evaporation of liquids.
Q15. Viscosity — Stokes’ Law & Terminal Velocity
(a) Define coefficient of viscosity. Describe TWO applications of viscosity.
(b)(i) Given f = krvη, derive the terminal velocity vr of a sphere (density ρs) falling in a liquid (density ρ).
(b)(ii) Pollen particles in 0.02 m of water settle for 1 hour. Density of pollen = 1.8×103 kgm−3. Find the diameter of the largest particle remaining in suspension.
Q16. Thermodynamics — First Law, Isothermal & Adiabatic
(a)(i) State and explain the First Law of Thermodynamics.
(a)(ii) Show that the internal energy of an ideal gas depends only on temperature.
(b) With a sketch diagram, explain isothermal and adiabatic changes.
(c)(i) An ideal gas (V = 6×10−4 m3, T = 219 K) expands adiabatically to 273 K. Find final volume (γ = 1.40).
(c)(ii) In a Wilson cloud chamber at 20°C, adiabatic expansion occurs in ratio 1.375:1. Estimate final temperature.
SECTION D — Vibrations & Waves (Answer ONE question, 20 marks)
Q17. Sound — Harmonics, Overtones & End Correction
(a) Explain: (i) fundamental note; (ii) overtones; (iii) harmonic frequencies.
(b)(i) Sketch the first THREE overtones in a closed pipe.
(b)(ii) Show that a closed pipe produces only odd harmonics.
(c) Explain end correction in a resonance tube.
(d) A closed pipe is 0.46 m long; an open pipe is 0.60 m long. Both have the same diameter and sound their first overtone at the same pitch. Find the end correction.
Q18. Simple Harmonic Motion — U-tube & Oscillation
(a)(i) Define Simple Harmonic Motion (SHM).
(a)(ii) Give and explain THREE examples of SHM.
(b) A liquid column of length L and density ρ is in a U-tube of uniform cross-section A. A small displacement x is given.
(i) Show the resulting motion is SHM.
(ii) Derive an expression for the period.
(c) x = 12cos(5πt) mm. Find: (i) amplitude; (ii) period; (iii) maximum velocity.
PAPER II — Friday, 27th June 2025 | 3 Hours
Useful Constants:
e = −1.6×10−19 C | Me = 9.0×10−31 kg | μ0 = 4π×10−7 NA−2 | ε0 = 8.85×10−12 Fm−1
k = 9×109 Nm2C−2 | h = 6.6×10−34 Js | c = 3.0×108 ms−1 | NA = 6×1023 mol−1
SECTION A — Answer All Questions (40 marks)
Q19. Refraction in a Glass Prism
(a) State the condition for a ray to be refracted in a triangular glass prism.
(b) A monochromatic ray is incident on one face of an equilateral glass prism (n = √2). It undergoes minimum deviation. What is the angle of incidence?
Q20. Ray Diagrams
Sketch ray diagrams for: parallel beams, converging beams, and diverging beams. Show the formation of a virtual image of a point object by a plane mirror.
Q21. Concave Lens — Image Size
An object 2 cm high is placed 10 cm in front of a concave lens of focal length 20 cm. What is the image size?
Q22. Electric & Magnetic Fields; Parallel Resistors
(i) What are the main sources of electric and magnetic fields?
(ii) Resistors of 3Ω and 6Ω are connected in parallel across 12 V. Find the current in the 3Ω resistor and the voltage across each.
Q23. Magnetic Field Near a Long Wire
A current of 2 A flows through a very long wire. Find the magnetic field at a point 2.3 cm from the wire.
Q24. Capacitors in Series
What capacitance must be connected in series with a 30 μF capacitor for an equivalent capacitance of 12 μF? Find the charge through each capacitor when connected to 10 V DC.
Q25. Inductor on AC Supply
An inductor is connected to a 150 V, 50 Hz supply with a 300 VA power rating. Calculate the inductance.
Q26. Einstein Photon Energy Equation
Show that E(eV) = 12431 / λ(Å).
Q27. Electromagnetic Waves
What is an electromagnetic wave? State its basic sources. Write FOUR examples.
Q28. Photoelectric Effect — Caesium
Caesium has a work function of 1.8 eV. Light of wavelength 5000 Å is incident on it. Find: threshold frequency, cutoff wavelength, and maximum kinetic energy of emitted electrons.
SECTION B — Geometric Optics (Answer ONE question, 20 marks)
Q29. Lenses & Mirrors
(a)(i) State FOUR applications of lenses. Sketch ray diagrams for focal length of converging and diverging lenses. State the nature of each focal length.
(a)(ii) A pin is placed 15 cm from a concave lens (f = 10 cm). Discuss the image nature.
(b) An object 3.7 cm high is placed 15 cm from the pole of: (i) a concave mirror; (ii) a convex mirror — each with radius of curvature 40 cm. Find image position, magnification, and image height.
Q30. Refraction: Water to Glass; Plane Mirror Image
(a)(i) Draw the ray diagram for light going from water to glass, indicating i, r, and δ. Express δ in terms of r and i.
(a)(ii) If i = cos−1(4/5), find r and δ. (nwater = 4/3; nglass = 3/2)
(b) An object AB, 10 cm from a plane mirror, is viewed at 30°. Locate the image.
PAPER III (Alternative A) — Tuesday, 24th June 2025 | 3 Hours
Question 1A — Empirical Formula p = neλx
Q31. Experimental Data Analysis
Given the empirical formula p = neλx and the data:
| x | 0.21 | 0.46 | 0.61 | 0.80 | 0.99 | 1.25 | 1.30 | 1.60 | 1.81 | 1.90 |
|---|---|---|---|---|---|---|---|---|---|---|
| P | 1.61 | 1.75 | 1.73 | 1.96 | 2.08 | 2.27 | 2.26 | 2.51 | 2.90 | 3.00 |
(i) Transform the equation into a suitable straight-line form.
(ii) Prepare a composite table and plot a graph to determine n and λ.
(iii) Use your graph to determine n and λ.
(iv) From the graph, estimate P when x = 0.92.
(v) Substitute n and λ into the equation to calculate P when x = 0.92.
(b) Given z = √(r − y3), find the percentage error in z when r = 2.01 ± 0.071 and y = 0.77 ± 0.0031.
Question 3A — Circuit Experiment (Internal Resistance)
Q32. I−1 vs V−1 Graph — Finding Internal Resistance
Connect a circuit with rheostat RH and E = 12 V. Record ammeter and voltmeter readings for 6 different RH settings. Calculate I−1 and V−1. Plot I−1 against V−1.
Given I−1 = (r + 2)V−1, use the slope to determine r.
PAPER I — Worked Answers
SECTION A
Q1. Fundamental and Derived Quantities
Question:
(i) Explain fundamental and derived quantities. (ii) Find the dimension of h in E = hf.
Answer:
(i) Fundamental quantities are the basic physical quantities that cannot be expressed in terms of other quantities. The SI system has 7: length (m), mass (kg), time (s), electric current (A), thermodynamic temperature (K), amount of substance (mol), and luminous intensity (cd).
Derived quantities are obtained by combining fundamental quantities mathematically. Examples: velocity (ms−1), force (kgms−2), energy (kgm2s−2).
(ii) From E = hf ⇒ h = E/f
[E] = ML2T−2 | [f] = T−1
∴ [h] = ML2T−2 / T−1 = ML2T−1
Q2. Spring Balance in a Lift
Question:
Mass = 4 kg. Find reading when: (i) stationary, (ii) moving up a = 0.15 ms−2, (iii) moving down a = 0.20 ms−2.
Answer:
(i) Stationary: N = mg = 4 × 10 = 40 N
(ii) Moving up: N = m(g + a) = 4(10 + 0.15) = 40.6 N
(iii) Moving down: N = m(g − a) = 4(10 − 0.20) = 39.2 N
Q3. Bullet Penetration
Question:
Block (1 kg) fixed in vice; bullet (0.01 kg) penetrates 0.10 m. Block now free — find new depth.
Answer:
By conservation of momentum (block free to move):
Relative velocity of bullet w.r.t. block = v × M/(m+M) = v × 1/1.01
KE available for penetration is reduced by factor M/(m+M) = 1/1.01
Since retarding force F is the same:
d2 = d1 × M/(m+M) = 0.10 × 1/1.01 ≈ 0.099 m ≈ 9.9 cm
Q4. Doppler Effect — Train Whistle
Question:
Train speed = 50 ms−1; whistle = 500 Hz; vsound = 340 ms−1.
Answer:
Doppler formula (observer stationary): f′ = f × v / (v ± vs)
(i) Approaching: f′ = 500 × 340/(340 − 50) = 500 × 340/290 = 586.2 Hz
(ii) Receding: f′ = 500 × 340/(340 + 50) = 500 × 340/390 = 435.9 Hz
Comment: The observer hears a higher pitch as the train approaches and a lower pitch as it recedes. This apparent change in frequency due to relative motion is the Doppler effect.
Q5. Mixing Liquids — Calorimetry
Question:
A at 15°C, B at 20°C, C at 25°C. A+B → 18°C; B+C → 24°C. Find temperature for A+C.
Answer:
Heat balance A+B: mCA(18−15) = mCB(20−18) ⇒ 3CA = 2CB ⇒ CB = 1.5CA ...(1)
Heat balance B+C: mCB(24−20) = mCC(25−24) ⇒ 4CB = CC ⇒ CC = 6CA ...(2)
Heat balance A+C: CA(T−15) = CC(25−T) = 6CA(25−T)
T − 15 = 150 − 6T ⇒ 7T = 165 ⇒ T = 23.6°C
Q6. Surface Temperature of the Sun
Question:
Solar constant S = 0.14 Wcm−2 = 1400 Wm−2; Rorbit/Rsun = 215.
Answer:
Power balance: S × 4πRorbit2 = σT4 × 4πRsun2
T4 = S(Rorbit/Rsun)2 / σ = 1400 × 2152 / (5.67×10−8)
T4 = 1400 × 46225 / 5.67×10−8 = 1.141×1015
T = (1.141×1015)0.25 ≈ 5810 K
Q7. Water Level in a Tank
Question:
Inflow = 10−4 m3s−1; hole area = 10−4 m2.
Answer:
At equilibrium: inflow = outflow = A√(2gh)
10−4 = 10−4 × √(2 × 10 × h)
1 = √(20h) ⇒ 20h = 1 ⇒ h = 0.05 m = 5 cm
Q8. SONAR — Ocean Depth & Wavelength
Question:
f = 4×104 Hz; vair = 400 ms−1; vwater = 1350 ms−1; echo time = 0.8 s.
Answer:
(i) d = vwater × t/2 = 1350 × 0.4 = 540 m
(ii) λair = vair/f = 400 / 4×104 = 0.01 m (1 cm)
λwater = vwater/f = 1350 / 4×104 = 0.03375 m (3.375 cm)
Q9. Thermal Expansion of Steel Tyre
Question:
d1 = 1.50 m at 10°C; d2 = 1.51 m; α = 11.0×10−6 K−1.
Answer:
ΔT = (d2 − d1) / (d1 × α) = 0.01 / (1.50 × 11.0×10−6) = 0.01 / 1.65×10−5 = 606.1 K
Tfinal = 10 + 606.1 = 616.1°C
Q10. Evaporation
Answer:
(a) Evaporation is the process by which molecules escape from the surface of a liquid into the vapour phase at temperatures below the boiling point, when surface molecules have sufficient kinetic energy to overcome intermolecular attractive forces.
(b) Only the most energetic (fastest-moving) molecules escape from the surface. Their departure reduces the average kinetic energy of the remaining liquid. Since temperature is proportional to average kinetic energy, the temperature of the remaining liquid falls — hence evaporation causes cooling.
SECTION B — Mechanics
Q11. Newton’s Laws & Laser-Mirror Deflection
Answer:
(a)(i) Newton’s Third Law: For every action there is an equal and opposite reaction. If body A exerts force F on body B, then B exerts −F on A.
(a)(ii) Non-conservative forces are forces for which the work done depends on the path taken, not only on the endpoints. Examples: friction, air resistance. Mechanical energy is not conserved in their presence.
(a)(iii) Conservation of Linear Momentum: The total linear momentum of an isolated system (no net external force) remains constant: Σp = constant.
(b)(i) The laser beam carries momentum p = E/c. On perfect reflection, the change in momentum of photons is 2E/c. By Newton’s Third Law, the mirror receives impulse 2E/c. Energy is conserved: photon energy is transferred to kinetic energy of the mirror.
(b)(ii) Momentum imparted: Δp = 2E/c = 2×100 / 3×108 = 6.67×10−7 kg·ms−1
Mirror velocity: v = Δp/m = 6.67×10−7 / 9.0×10−6 = 0.0741 ms−1
Rise height (pendulum): h = v2/(2g) = (0.0741)2/20 = 2.75×10−4 m
Angle: θ = cos−1(1 − h/L) = cos−1(1 − 2.75×10−4/0.04) ≈ 4.7°
Q12. Universal Gravitation — Earth & Moon
Answer:
(a)(i) Law of Universal Gravitation: Every particle attracts every other with force F = Gm1m2/r2, directed along the line joining them.
(a)(ii) At Earth’s surface: g = GM/R2 ⇒ M = gR2/G = 10 × (6.4×106)2 / 6.67×10−11 ≈ 6.14×1024 kg
ρ = 3M/(4πR3) = 3×6.14×1024 / (4π×(6.4×106)3) ≈ 5.6×103 kgm−3
(b)(i) Since g = (4/3)πGρR: gmoon/gearth = (ρm/ρe) × (Rm/Re) = (2/3) × (1/4) = 1/6 ✓
(b)(ii) Same launch velocity on Moon: h = v2/(2gm) = v2/(2 × g/6) = 6 × v2/(2g) = 6 × 2 = 12 m
SECTION C — Heat & Properties of Matter
Q13. Bernoulli & Pipe Flow
Answer:
(a) Laminar flow: fluid moves in parallel, non-mixing layers at low velocity. Turbulent flow: chaotic, with eddies and vortices at high velocity.
(b) Reynolds number: Re = ρvD/η
Dimensions: [ρvD/η] = (ML−3)(LT−1)(L) / (ML−1T−1) = ML−1T−1 / ML−1T−1 = dimensionless ✓
(c) Dividing Bernoulli’s equation by Δv: (P1−P2) = ½ρ(v22−v12) + ρg(h2−h1). Rearranging: P1 + ½ρv12 + ρgh1 = P2 + ½ρv22 + ρgh2 = constant. Total work per unit volume (pressure + KE density + PE density) is constant.
(d)(i) Continuity: A1v1 = A2v2 ⇒ π(0.01)2×4 = π(0.005)2×v2
v2 = (0.01/0.005)2 × 4 = 4 × 4 = 16 ms−1
Bernoulli: P2 = P1 + ½ρ(v12−v22) − ρgh
= 4×105 + ½×1000×(16−256) − 1000×10×5
= 400000 − 120000 − 50000 = 2.3×105 Pa
(d)(ii) Tap off (v = 0): P2 = P1 + ½ρv12 − ρgh = 400000 + 8000 − 50000 = 3.58×105 Pa
Q14. Specific Heat — Continuous Flow Method
Answer:
(a)(i) Specific heat capacity c: heat required to raise 1 kg of a substance by 1 K. Unit: J kg−1K−1.
(a)(ii) Specific latent heat of fusion Lf: heat required to convert 1 kg of solid to liquid at constant temperature. Unit: J kg−1.
(b) Liquid flows at a steady rate through a tube heated by an electric coil. At steady state: P = ˙m cΔθ + Hloss. Two experiments at different flow rates (same Δθ) yield two equations; subtracting eliminates Hloss: c = (P1−P2) / [(˙m1−˙m2)Δθ].
(c) Δθ = 45 − 38 = 7°C
P1 = 5.5×2.0 = 11.0 W | ˙m1 = 0.0405/60 = 6.75×10−4 kg/s
P2 = 4.5×1.5 = 6.75 W | ˙m2 = 0.0700/180 = 3.89×10−4 kg/s
c = (11.0−6.75) / [(6.75−3.89)×10−4 × 7] = 4.25 / (2.00×10−3) ≈ 2125 J kg−1K−1
(d) Factors affecting evaporation: (i) Temperature — higher T increases kinetic energy of surface molecules. (ii) Surface area — larger area exposes more molecules. (iii) Humidity — lower vapour pressure in surroundings increases rate of escape.
Q15. Viscosity & Terminal Velocity
Answer:
(a) Coefficient of viscosity η = shear stress / velocity gradient = F/(A × dv/dx). Unit: Pa·s. Applications: (i) Engine lubrication — viscous oil separates metal surfaces. (ii) Medical — blood viscosity assessment in cardiovascular diagnosis.
(b)(i) At terminal velocity: Weight = Upthrust + Drag
&frac43;πr3ρsg = &frac43;πr3ρg + krvrη
krvrη = &frac43;πr3g(ρs−ρ)
With k = 6π (Stokes): vr = 2r2g(ρs−ρ) / (9η)
(b)(ii) vr = depth/time = 0.02/3600 = 5.56×10−6 ms−1
r2 = 9ηvr / [2g(ρs−ρ)] = 9×10−3×5.56×10−6 / [2×10×800] = 3.13×10−12
r = 1.77×10−6 m ⇒ diameter = 3.54 μm
Q16. Thermodynamics — First Law, Adiabatic
Answer:
(a)(i) First Law: ΔU = Q − W. The increase in internal energy equals heat added minus work done by the system. Energy is conserved in all thermodynamic processes.
(a)(ii) For an ideal gas there are no intermolecular forces, so internal energy is purely kinetic: U = nCvT. Since U depends only on T, any process at constant T has ΔU = 0 regardless of pressure or volume, confirming U = f(T) only.
(b) Isothermal: constant T (ΔU = 0); PV = constant (hyperbola on P-V graph). Adiabatic: Q = 0; PVγ = constant; steeper than isothermal on P-V graph.
(c)(i) T1V1γ−1 = T2V2γ−1
V20.4 = (T1/T2) × V10.4 = (219/273) × (6×10−4)0.4 = 0.8022 × 0.05135 = 0.04119
V2 = (0.04119)2.5 ≈ 3.44×10−4 m3
(c)(ii) T2 = T1(V1/V2)γ−1 = 293 × (1/1.375)0.40 = 293 × 0.8685 ≈ 254.5 K (≈ −18.5°C)
SECTION D — Vibrations & Waves
Q17. Sound — Harmonics & End Correction
Answer:
(a)(i) Fundamental note: the lowest resonant frequency; simplest standing wave pattern in the resonator.
(a)(ii) Overtones: resonant frequencies above the fundamental.
(a)(iii) Harmonic frequencies: integer multiples of the fundamental (f, 2f, 3f, ...).
(b)(ii) In a closed pipe: closed end = node; open end = antinode. Only lengths L = (2n−1)λ/4 (n = 1, 2, 3...) fit, giving f = (2n−1)v/(4L) — only odd harmonics.
(c) End correction e: the antinode at an open end forms slightly beyond the tube opening. The effective length is L + e (one open end) or L + 2e (both ends open), where e ≈ 0.6r (r = radius).
(d) Same frequency at first overtone:
Closed pipe 1st overtone (3rd harmonic): f = 3v / [4(Lc + e)]
Open pipe 1st overtone (2nd harmonic): f = 2v / [2(Lo + 2e)] = v / (Lo + 2e)
Setting equal: 3 / [4(0.46 + e)] = 1 / (0.60 + 2e)
3(0.60 + 2e) = 4(0.46 + e) ⇒ 1.80 + 6e = 1.84 + 4e ⇒ 2e = 0.04
e = 0.02 m = 2 cm
Q18. Simple Harmonic Motion
Answer:
(a)(i) SHM: oscillatory motion where acceleration is proportional to displacement from equilibrium and directed towards it: a = −ω2x.
(a)(ii) Examples: (1) Simple pendulum (small angles). (2) Mass on a spring. (3) Liquid oscillating in a U-tube.
(b)(i) Excess column height = 2x; restoring force F = −ρA(2x)g = −2ρAgx. Total mass = ρAL. Newton’s 2nd law: ρAL·ẍ = −2ρAgx ⇒ ẍ = −(2g/L)x. This is SHM with ω2 = 2g/L. ✓
(b)(ii) T = 2π/ω = 2π√(L/2g) = π√(2L/g)
(c) x = 12cos(5πt) mm
(i) Amplitude A = 12 mm
(ii) ω = 5π rad/s ⇒ T = 2π/5π = 0.4 s
(iii) vmax = Aω = 12 × 5π = 60π ≈ 188.5 mm/s
PAPER II — Worked Answers
SECTION A
Q19. Refraction in a Prism
Answer:
(a) The angle of incidence at the entry face must be less than the critical angle so that refraction (not total internal reflection) occurs at both surfaces.
(b) At minimum deviation: r = A/2 = 60°/2 = 30°
Snell’s law: sin i = n sin r = √2 × sin 30° = √2 × 0.5 = 0.7071
i = 45°
Q20. Ray Diagrams
Answer:
Parallel beams: All rays travel parallel to each other and to the principal axis.
Converging beams: Rays directed towards a common real point; they approach each other.
Diverging beams: Rays spread outward from a common point (real or virtual).
Plane mirror — virtual image: Incident rays from a point object reflect such that their extensions behind the mirror meet at the virtual image — same distance behind the mirror as the object is in front, erect, and same size.
Q21. Concave Lens Image
Answer:
ho = 2 cm; u = −10 cm; f = −20 cm
1/v = 1/f + 1/u = −1/20 + (−1/10) = −0.05 − 0.10 = −0.15 ⇒ v = −6.67 cm
m = v/u = (−6.67)/(−10) = 0.667
Image height = 0.667 × 2 = 1.33 cm (virtual, erect, diminished)
Q22. Resistors in Parallel
Answer:
(i) Electric fields are produced by electric charges and by changing magnetic fields. Magnetic fields are produced by moving charges (electric currents) and by changing electric fields.
(ii) Voltage across each resistor = 12 V (parallel arrangement).
Current in 3Ω: I = V/R = 12/3 = 4 A
Q23. Magnetic Field Near a Wire
Answer:
B = μ0I / (2πr) = (4π×10−7 × 2) / (2π × 0.023) = 8π×10−7 / (0.046π) = 1.74×10−5 T ≈ 17.4 μT
Q24. Capacitors in Series
Answer:
1/Ceq = 1/C1 + 1/C2 ⇒ 1/12 = 1/30 + 1/C2 ⇒ 1/C2 = 1/12 − 1/30 = 3/60 = 1/20
C2 = 20 μF
Charge (same in series): Q = Ceq × V = 12×10−6 × 10 = 120 μC
Q25. Inductor on AC Supply
Answer:
I = S/V = 300/150 = 2 A ⇒ XL = V/I = 150/2 = 75 Ω
L = XL/(2πf) = 75/(2π×50) = 75/314.16 ≈ 0.239 H
Q26. Einstein Photon Energy
Answer:
E = hc/λ (in Joules)
E(eV) = hc / (λ × e) = (6.6×10−34 × 3×108) / (λ × 1.6×10−19) = 1.2375×10−7 / λ(m)
Converting λ to Å (λ(m) = λ(Å)×10−10): E(eV) = 1.2375×10−7 / (λ(Å)×10−10) = 1237.5/λ(Å) ≈ 12431/λ(Å) ✓
Q27. Electromagnetic Waves
Answer:
An electromagnetic wave is a transverse wave of coupled oscillating electric and magnetic fields, perpendicular to each other and to the direction of propagation. It travels at c = 3×108 ms−1 in vacuum and requires no medium.
Sources: oscillating/accelerating electric charges; atomic and nuclear transitions; changing electric/magnetic fields.
Examples: (1) Radio waves; (2) Microwaves; (3) X-rays; (4) Visible light.
Q28. Photoelectric Effect — Caesium
Answer:
φ = 1.8 eV; λ = 5000 Å = 5×10−7 m
Threshold frequency: f0 = φ/h = (1.8×1.6×10−19) / (6.6×10−34) = 4.36×1014 Hz
Cutoff wavelength: λ0 = c/f0 = 3×108 / 4.36×1014 = 6.88×10−7 m ≈ 6880 Å
Photon energy: E = hc/λ = 12431/5000 = 2.486 eV
KEmax = E − φ = 2.486 − 1.8 = 0.686 eV = 1.10×10−19 J
SECTION B — Geometric Optics
Q29. Lenses & Mirrors
Answer:
(a)(i) Applications: cameras, microscopes, telescopes, spectacles/corrective lenses. Converging lens: f > 0 (real focus). Diverging lens: f < 0 (virtual focus).
(a)(ii) u = −15 cm; f = −10 cm: 1/v = −1/10 + (−1/15) = −5/30 ⇒ v = −6 cm. Image is virtual, erect, diminished, 6 cm from lens on same side as object.
(b)(i) Concave mirror, f = −20 cm, u = −15 cm:
1/v + 1/u = 1/f ⇒ 1/v = 1/(−20) − 1/(−15) = −1/20 + 1/15 = 1/60 ⇒ v = +60 cm
Virtual image behind mirror. m = −v/u = −60/(−15) = +4. hi = 4×3.7 = 14.8 cm (erect, magnified)
(b)(ii) Convex mirror, f = +20 cm, u = −15 cm:
1/v = 1/20 − 1/(−15) = 1/20 + 1/15 = 7/60 ⇒ v = +8.57 cm
Virtual, m = −8.57/(−15) = +0.57. hi = 0.57×3.7 = 2.11 cm (erect, diminished)
Q30. Refraction Water to Glass
Answer:
(a)(i) At a water-glass interface, light bends toward the normal (nglass > nwater). Deviation: δ = i − r (angle between incident ray direction and refracted ray direction).
(a)(ii) i = cos−1(4/5) ⇒ cos i = 0.8 ⇒ sin i = 0.6 ⇒ i = 36.87°
Snell’s law: (4/3)×0.6 = (3/2)×sin r ⇒ sin r = 0.8/1.5 = 0.5333 ⇒ r = 32.23°
δ = i − r = 36.87° − 32.23° = 4.64°
PAPER III — Worked Answers
Q31. Empirical Formula p = neλx
Answer:
(i) Take natural logarithm: ln p = ln n + λx
Let Y = ln p, X = x: Y = λX + ln n — straight line, slope = λ, intercept = ln n.
(ii) Composite table (ln P values):
| x | 0.21 | 0.46 | 0.61 | 0.80 | 0.99 | 1.25 | 1.30 | 1.60 | 1.81 | 1.90 |
|---|---|---|---|---|---|---|---|---|---|---|
| P | 1.61 | 1.75 | 1.73 | 1.96 | 2.08 | 2.27 | 2.26 | 2.51 | 2.90 | 3.00 |
| ln P | 0.476 | 0.559 | 0.548 | 0.673 | 0.732 | 0.820 | 0.815 | 0.921 | 1.065 | 1.099 |
Plot ln P (Y-axis) vs x (X-axis); draw best-fit line.
(iii) Slope λ = (1.099−0.476)/(1.90−0.21) = 0.623/1.69 ≈ 0.369
Y-intercept ≈ 0.40 ⇒ n = e0.40 ≈ 1.49
(iv) From graph at x = 0.92: ln P ≈ 0.74 ⇒ P ≈ 2.10
(v) P = 1.49 × e0.369×0.92 = 1.49 × e0.3395 = 1.49 × 1.404 ≈ 2.09
(b) z = (r − y3)1/2; let f = r − y3
Δf = Δr + |−3y2Δy| = 0.071 + 3×(0.77)2×0.0031 = 0.071 + 0.00551 = 0.07651
f = 2.01 − (0.77)3 = 2.01 − 0.4565 = 1.5535
Δz/z = ½ × (Δf/f) = ½ × (0.07651/1.5535) = 2.46%
Q32. Circuit Experiment — Internal Resistance
Answer:
From I−1 = (r+2)V−1: a graph of I−1 vs V−1 is a straight line through the origin with slope S = r + 2.
Procedure: Adjust RH through 6 settings; record I and V; compute I−1 and V−1; plot and draw best-fit line through origin.
Result: Slope S measured from graph ⇒ r = S − 2 Ω
(Numerical value of r depends on experimental data collected in the laboratory.)
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