O Level Physics Practice Paper 5

TuitionGoWhere Practice Paper - Physics O-Level

ANSWER KEY AND MARKING SCHEME

Paper: PRACTICE (Version 5 of 5)
Total Marks: 60

Section A: Short Answer and Structured Response (20 marks)

1. (a) Explain why the rod becomes negatively charged. [2]

Answer: When the polythene rod is rubbed with the woollen cloth, electrons are transferred from the cloth to the rod [1]. The rod gains electrons and therefore becomes negatively charged [1].

Marking notes:

• Award [1] for stating electrons transfer from cloth to rod.
• Award [1] for linking gain of electrons to negative charge.
• Accept: "Polythene has a greater affinity for electrons than wool."

1. (b) Explain why the uncharged foil is attracted to the charged rod. [2]

Answer: The negative charges on the rod repel the electrons in the aluminium foil, causing them to move to the far side of the foil [1]. The side of the foil nearest the rod becomes positively charged by induction. Since opposite charges attract, the foil is attracted to the rod [1].

Marking notes:

• Award [1] for describing charge separation/induction in the foil.
• Award [1] for stating attraction between opposite charges.
• Accept: "Charging by induction" with clear explanation.

2. (a) Sketch the magnetic field pattern around a bar magnet. [2]

Answer: Sketch should show:

• Field lines emerging from N pole and entering S pole [1].
• Lines curving around from N to S, showing correct direction (arrows from N to S) [1].
• Lines closer together near poles (stronger field) and not crossing.

Marking notes:

• Award [1] for correct shape and direction of field lines.
• Award [1] for correct labelling of N and S poles.
• Deduct [1] if lines cross or direction arrows are missing.

2. (b) State one difference between a permanent magnet and a temporary magnet. [1]

Answer: A permanent magnet retains its magnetism after the magnetising field is removed, whereas a temporary magnet loses its magnetism when the magnetising field is removed.

Marking notes:

• Award [1] for any valid difference.
• Accept: "Permanent magnets are made of hard magnetic materials (e.g., steel); temporary magnets are made of soft magnetic materials (e.g., soft iron)."
• Accept: "Permanent magnets are difficult to demagnetise; temporary magnets are easily demagnetised."

3. (a) Calculate the magnitude of the force acting on the wire. [2]

Answer: F = BIL [1] F = 0.40 × 3.0 × 0.15 F = 0.18 N [1]

Marking notes:

• Award [1] for correct formula F = BIL.
• Award [1] for correct answer with unit.
• Accept 0.18 N or 1.8 × 10⁻¹ N.

3. (b) State the rule used to determine the direction of this force. [1]

Answer: Fleming's left-hand rule.

Marking notes:

• Award [1] for correct name.
• Accept: "Fleming's left-hand motor rule."

4. (a) Explain why inserting the soft iron core increases the strength of the magnetic field. [2]

Answer: Soft iron is a ferromagnetic material that becomes strongly magnetised when placed in a magnetic field [1]. The magnetic domains in the soft iron align with the field produced by the solenoid, concentrating and strengthening the magnetic field lines [1].

Marking notes:

• Award [1] for stating soft iron becomes magnetised/concentrates field lines.
• Award [1] for linking to increased field strength.
• Accept reference to high magnetic permeability.

4. (b) State one practical application of an electromagnet. [1]

Answer: Any one of:

• Electric bell
• Relay switch / circuit breaker
• Magnetic crane for lifting scrap metal
• Electric motor
• Loudspeaker
• MRI scanner

Marking notes:

• Award [1] for any valid application.

5. (a) State what is observed on the galvanometer when the magnet is pushed quickly into the coil. [1]

Answer: The galvanometer needle deflects (in one direction), indicating an induced current/e.m.f.

Marking notes:

• Award [1] for stating deflection/reading on galvanometer.
• Accept: "A current is induced" or "The galvanometer shows a reading."

5. (b) State two ways in which the magnitude of the induced e.m.f. can be increased. [2]

Answer: Any two of:

• Move the magnet faster [1]
• Use a stronger magnet [1]
• Use a coil with more turns [1]
• Insert a soft iron core into the coil [1]

Marking notes:

• Award [1] each for any two valid methods, up to [2].

5. (c) State Lenz's law. [1]

Answer: The direction of the induced e.m.f. (or induced current) is such that it opposes the change in magnetic flux that produced it.

Marking notes:

• Award [1] for correct statement.
• Accept: "The induced current flows in a direction that opposes the change causing it."

Section B: Diagram and Data Interpretation (20 marks)

6. (a) Explain the purpose of the split-ring commutator in the d.c. motor. [2]

Answer: The split-ring commutator reverses the direction of the current in the coil every half-turn [1]. This ensures that the force on each side of the coil always acts in the same rotational direction, producing continuous rotation [1].

Marking notes:

• Award [1] for stating current reversal every half-turn.
• Award [1] for linking to continuous rotation in one direction.

6. (b) State two changes that could be made to increase the speed of rotation of the coil. [2]

Answer: Any two of:

• Increase the current in the coil [1]
• Use a stronger magnet / increase magnetic field strength [1]
• Increase the number of turns on the coil [1]
• Use a soft iron core in the coil [1]

Marking notes:

• Award [1] each for any two valid changes, up to [2].

6. (c) State the effect, if any, on the direction of rotation of the coil when the current direction is reversed. [1]

Answer: The direction of rotation reverses.

Marking notes:

• Award [1] for stating rotation reverses.
• Accept: "The coil rotates in the opposite direction."

7. (a) Calculate the output voltage of the transformer. [2]

Answer: Vₛ / Vₚ = Nₛ / Nₚ [1] Vₛ / 240 = 60 / 1200 Vₛ = 240 × (60 / 1200) = 240 × 0.05 = 12 V [1]

Marking notes:

• Award [1] for correct formula and substitution.
• Award [1] for correct answer with unit.
• Accept 12 V.

7. (b) Calculate the current in the primary coil when the lamp is operating at its rated power. [2]

Answer: For ideal transformer: VₚIₚ = VₛIₛ [1] Iₛ = P / Vₛ = 24 / 12 = 2.0 A Iₚ = (Vₛ × Iₛ) / Vₚ = (12 × 2.0) / 240 = 24 / 240 = 0.10 A [1]

Marking notes:

• Award [1] for correct method (using power conservation or turns ratio).
• Award [1] for correct answer with unit.
• Accept 0.10 A or 0.1 A.

7. (c) Explain why the core of the transformer is made of laminated soft iron. [2]

Answer: Soft iron is used because it is easily magnetised and demagnetised, reducing energy losses due to hysteresis [1]. The core is laminated (made of thin sheets insulated from each other) to reduce eddy currents, which would otherwise cause heating and energy loss [1].

Marking notes:

• Award [1] for explaining soft iron reduces hysteresis loss.
• Award [1] for explaining lamination reduces eddy current loss.
• Accept reference to "reducing energy losses" for either point.

8. (a) Calculate the voltmeter reading at 25 °C. [2]

Answer: Using potential divider: V_out = [R_fixed / (R_thermistor + R_fixed)] × V_supply [1] V_out = [5000 / (10000 + 5000)] × 6.0 V_out = (5000 / 15000) × 6.0 = (1/3) × 6.0 = 2.0 V [1]

Marking notes:

• Award [1] for correct potential divider formula.
• Award [1] for correct answer with unit.
• Accept 2.0 V.

8. (b) Calculate the new voltmeter reading at 10 °C. [2]

Answer: V_out = [5000 / (25000 + 5000)] × 6.0 [1] V_out = (5000 / 30000) × 6.0 = (1/6) × 6.0 = 1.0 V [1]

Marking notes:

• Award [1] for correct substitution.
• Award [1] for correct answer with unit.
• Accept 1.0 V.

8. (c) Explain, with reference to the potential divider principle, why the voltmeter reading changes as the temperature decreases. [2]

Answer: As temperature decreases, the resistance of the thermistor increases [1]. In a potential divider, the voltage across each resistor is proportional to its resistance. Since the thermistor's resistance increases, a larger fraction of the supply voltage is dropped across the thermistor, and a smaller fraction across the fixed resistor. Therefore, the voltmeter reading (across the fixed resistor) decreases [1].

Marking notes:

• Award [1] for stating thermistor resistance increases with decreasing temperature.
• Award [1] for explaining the effect on voltage distribution using potential divider principle.

9. (a) State how the graph shows that the filament lamp is a non-ohmic conductor. [1]

Answer: The I-V graph for the filament lamp is a curve (not a straight line through the origin), showing that current is not directly proportional to potential difference.

Marking notes:

• Award [1] for stating the graph is curved/non-linear.
• Accept: "The resistance changes with voltage/current."

9. (b) Explain why the resistance of the lamp increases as the current increases. [2]

Answer: As current increases, the filament gets hotter [1]. The increased temperature causes the metal ions in the filament to vibrate more vigorously, increasing the frequency of collisions between free electrons and the ions. This impedes the flow of electrons, so resistance increases [1].

Marking notes:

• Award [1] for linking increased current to increased temperature.
• Award [1] for explaining increased collisions/impeded electron flow.

9. (c) Calculate the resistance of the lamp at a potential difference of 4.0 V. [1]

Answer: R = V / I = 4.0 / 0.40 = 10 Ω [1]

Marking notes:

• Award [1] for correct answer with unit.
• Accept 10 Ω or 10.0 Ω.

Section C: Calculation and Problem Solving (20 marks)

10. (a) Calculate the current drawn by the kettle. [2]

Answer: P = VI [1] I = P / V = 1800 / 240 = 7.5 A [1]

Marking notes:

• Award [1] for correct formula.
• Award [1] for correct answer with unit.

10. (b) Calculate the energy required to heat the water. [2]

Answer: E = mcΔθ [1] E = 1.5 × 4200 × (100 - 25) E = 1.5 × 4200 × 75 E = 472 500 J (or 4.725 × 10⁵ J) [1]

Marking notes:

• Award [1] for correct formula and substitution.
• Award [1] for correct answer with unit.
• Accept 472.5 kJ or 473 kJ.

10. (c) Calculate the time taken to heat the water. [3]

Answer: Efficiency = Useful energy output / Total energy input × 100% 0.80 = 472 500 / (P × t) [1] P × t = 472 500 / 0.80 = 590 625 J [1] t = 590 625 / 1800 = 328.125 s ≈ 328 s (or 5 min 28 s) [1]

Marking notes:

• Award [1] for correct efficiency equation.
• Award [1] for calculating total energy input.
• Award [1] for correct time with unit.
• Accept 328 s, 5.5 min, or 5 min 28 s.

11. (a) Calculate the resistance of the wire. [1]

Answer: R = V / I = 4.0 / 0.50 = 8.0 Ω [1]

Marking notes:

• Award [1] for correct answer with unit.

11. (b) Calculate the new resistance of the replacement wire. [3]

Answer: R ∝ L / A (resistance is proportional to length and inversely proportional to cross-sectional area) [1] New length = 2 × original length → resistance doubles (×2) [1] New area = 0.5 × original area → resistance doubles (×2) Total factor = 2 × 2 = 4 New resistance = 4 × 8.0 = 32 Ω [1]

Marking notes:

• Award [1] for stating relationship R ∝ L/A.
• Award [1] for correct reasoning about length and area effects.
• Award [1] for correct answer with unit.
• Accept alternative method using R = ρL/A with consistent values.

11. (c) State one precaution the student should take to obtain reliable results. [1]

Answer: Any one of:

• Ensure all connections are tight/secure.
• Avoid heating the wire (switch off between readings).
• Take multiple readings and calculate average.
• Ensure the wire is straight and not coiled.
• Use appropriate range on meters.

Marking notes:

• Award [1] for any valid precaution.

12. (a) Calculate the total resistance of the two headlamps in parallel. [2]

Answer: 1/R_total = 1/R₁ + 1/R₂ [1] 1/R_total = 1/4.0 + 1/4.0 = 2/4.0 = 1/2.0 R_total = 2.0 Ω [1]

Marking notes:

• Award [1] for correct parallel resistance formula.
• Award [1] for correct answer with unit.
• Accept: "For two equal resistors in parallel, R_total = R/2 = 4.0/2 = 2.0 Ω."

12. (b) Calculate the total current drawn from the battery. [2]

Answer: I = V / R_total [1] I = 12 / 2.0 = 6.0 A [1]

Marking notes:

• Award [1] for correct formula.
• Award [1] for correct answer with unit.

12. (c) State and explain what happens to the brightness of the remaining headlamp when one fails. [2]

Answer: The brightness of the remaining headlamp remains unchanged [1]. In a parallel circuit, each branch receives the full supply voltage (12 V). When one lamp fails, the other lamp still has 12 V across it, so the current through it and its power dissipation remain the same [1].

Marking notes:

• Award [1] for stating brightness remains unchanged.
• Award [1] for correct explanation using parallel circuit principles.

13. (a) Calculate the power dissipated by the heater. [2]

Answer: P = V² / R [1] P = 240² / 30 = 57 600 / 30 = 1920 W (or 1.92 kW) [1]

Marking notes:

• Award [1] for correct formula.
• Award [1] for correct answer with unit.
• Accept 1920 W or 1.92 kW.

13. (b) Calculate the energy consumed in kilowatt-hours (kWh). [2]

Answer: E = P × t [1] P = 1.92 kW, t = 3.0 h E = 1.92 × 3.0 = 5.76 kWh [1]

Marking notes:

• Award [1] for correct formula and conversion to kW.
• Award [1] for correct answer with unit.

13. (c) Calculate the cost of using the heater for 3.0 hours. [1]

Answer: Cost = Energy × Rate = 5.76 × 0.25 = $1.44 [1]

Marking notes:

• Award [1] for correct answer with unit ($).
• Accept $1.44.

END OF ANSWER KEY

Mark Allocation Summary

Assessment Objectives Coverage

Markers should award partial credit for correct method even if final answer is incorrect. Consistent units and clear working should be rewarded.