O Level Physics Practice Paper 1

TuitionGoWhere Practice Paper – Physics O-Level

ANSWER KEY AND MARKING SCHEME

Version 1 of 5 | Electricity & Magnetism | Total: 60 marks

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

1. State the principle of conservation of charge. [1 mark]

Answer: Charge cannot be created or destroyed; it can only be transferred from one body to another. / The total electric charge in an isolated system remains constant.

Award [1] for a correct statement conveying that total charge is constant / charge is transferred, not created or destroyed.

2. (a) Explain, in terms of electron transfer, how the rod becomes negatively charged. [2 marks]

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

Award [1] for stating electrons are transferred from cloth to rod; [1] for stating rod gains electrons → negatively charged.

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

Answer: The negative charges on the rod repel electrons in the aluminium foil to the far side of the foil (charging by induction). The side of the foil nearest the rod becomes positively charged. Since unlike charges attract, the foil is attracted to the rod.

Award [1] for describing induction / separation of charges in the foil; [1] for stating attraction between unlike charges (positive foil side and negative rod).

3. Define the term electromotive force (e.m.f.). [1 mark]

Answer: The electromotive force (e.m.f.) is the work done per unit charge by a source (such as a battery) in driving charge around a complete circuit. / The energy converted from other forms to electrical energy per unit charge.

Award [1] for a correct definition referencing work done / energy converted per unit charge.

4. (a) Calculate the total resistance of the circuit. [1 mark]

Answer: R_total = R₁ + R₂ = 4.0 + 8.0 = 12.0 Ω

Award [1] for correct answer with unit.

(b) Calculate the current flowing in the circuit. [2 marks]

Answer: I = V / R_total = 12 / 12.0 = 1.0 A

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

(c) Calculate the potential difference across the 8.0 Ω resistor. [2 marks]

Answer: V = I × R = 1.0 × 8.0 = 8.0 V

Award [1] for correct formula/substitution; [1] for correct answer with unit.

5. (a) Plot graph of current against voltage. [3 marks]

Answer: Graph should show:

• Correctly labelled axes: Current / A (y-axis), Voltage / V (x-axis)
• Appropriate scales
• All six points plotted accurately
• Smooth curve drawn through points (curve should show decreasing gradient as voltage increases, indicating increasing resistance)

Award [1] for correct axes and scales; [1] for all points plotted correctly; [1] for smooth curve (not straight lines connecting points).

(b) Explain whether the filament lamp is an ohmic conductor. [2 marks]

Answer: The filament lamp is NOT an ohmic conductor. An ohmic conductor has a constant resistance, which would produce a straight-line graph through the origin (I ∝ V). The graph for the filament lamp is curved, showing that the resistance changes (increases) as the voltage/current increases.

Award [1] for stating it is not ohmic; [1] for explanation referencing the curved graph / non-constant resistance.

(c) Calculate the resistance of the filament lamp when the voltage is 6.0 V. [2 marks]

Answer: At V = 6.0 V, I = 0.50 A (from table) R = V / I = 6.0 / 0.50 = 12 Ω

Award [1] for correct reading from table; [1] for correct calculation with unit.

(d) Explain why the resistance of the filament lamp changes as voltage increases. [2 marks]

Answer: As the voltage/current increases, the filament gets hotter. The increased temperature causes the metal atoms in the filament to vibrate more vigorously, which increases the resistance to electron flow (more collisions between electrons and vibrating atoms).

Award [1] for linking to increased temperature; [1] for explaining increased atomic vibrations → more collisions → higher resistance.

Section B: Calculation and Circuit Analysis (20 marks)

6. (a) Calculate the maximum power that can be supplied without blowing the fuse. [2 marks]

Answer: P_max = V × I_max = 240 × 13 = 3120 W

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

(b) Determine whether the kettle can be safely used. [2 marks]

Answer: Current drawn by kettle: I = P / V = 2200 / 240 = 9.17 A Since 9.17 A < 13 A (the fuse rating), the kettle can be safely used on this circuit.

Award [1] for calculating current; [1] for correct conclusion with comparison to fuse rating.

7. (a) Calculate the resistance of the wire. [2 marks]

Answer: R = ρL / A = (1.1 × 10⁻⁶ × 1.50) / (3.0 × 10⁻⁸) = (1.65 × 10⁻⁶) / (3.0 × 10⁻⁸) = 55 Ω

Award [1] for correct formula and substitution; [1] for correct answer with unit.

(b) Calculate the current flowing through the wire. [2 marks]

Answer: I = V / R = 6.0 / 55 = 0.109 A ≈ 0.11 A

Award [1] for correct formula; [1] for correct answer with unit (accept 0.11 A or 0.109 A).

(c) Calculate the new total current when both halves are connected in parallel. [3 marks]

Answer: Each half: length = 0.75 m, so R_half = ρL / A = (1.1 × 10⁻⁶ × 0.75) / (3.0 × 10⁻⁸) = 27.5 Ω For two identical resistors in parallel: 1/R_total = 1/27.5 + 1/27.5 = 2/27.5 R_total = 27.5 / 2 = 13.75 Ω I_total = V / R_total = 6.0 / 13.75 = 0.436 A ≈ 0.44 A

Alternative: I through each half = 6.0 / 27.5 = 0.218 A; I_total = 2 × 0.218 = 0.436 A

Award [1] for calculating resistance of one half; [1] for calculating total parallel resistance or current per branch; [1] for correct total current with unit.

8. (a) Calculate output voltage across fixed resistor at 20 °C. [2 marks]

Answer: At 20 °C: R_thermistor = 30 kΩ, R_fixed = 10 kΩ V_out = [R_fixed / (R_fixed + R_thermistor)] × V_supply = [10 / (10 + 30)] × 9.0 = (10/40) × 9.0 = 2.25 V

Award [1] for correct potential divider formula; [1] for correct answer with unit.

(b) Calculate output voltage across fixed resistor at 60 °C. [2 marks]

Answer: At 60 °C: R_thermistor = 5.0 kΩ, R_fixed = 10 kΩ V_out = [10 / (10 + 5.0)] × 9.0 = (10/15) × 9.0 = 6.0 V

Award [1] for correct substitution; [1] for correct answer with unit.

(c) Explain whether the fan will be on at 20 °C or 60 °C. [3 marks]

Answer: The fan activates when V_out > 6.0 V. At 20 °C: V_out = 2.25 V (< 6.0 V) → fan is OFF. At 60 °C: V_out = 6.0 V (= 6.0 V) → fan is at threshold / just ON (or OFF if strictly > 6.0 V required).

As temperature increases, the thermistor resistance decreases (NTC thermistor), so V_out across the fixed resistor increases. The fan will be ON at higher temperatures (60 °C and above) and OFF at lower temperatures (20 °C).

Award [1] for comparing each voltage to 6.0 V threshold; [1] for correct conclusion about 20 °C (OFF) and 60 °C (ON/at threshold); [1] for explaining the relationship between temperature, thermistor resistance, and output voltage.

9. (a) Calculate the current drawn by the heater. [2 marks]

Answer: I = P / V = 1500 / 240 = 6.25 A

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

(b) Calculate the cost of using the heater for 30 days. [3 marks]

Answer: Daily energy consumption: E = P × t = 1.5 kW × 3.0 h = 4.5 kWh Energy for 30 days: E_total = 4.5 × 30 = 135 kWh Cost = 135 × $0.28 =$37.80

Award [1] for correct daily energy in kWh; [1] for total energy over 30 days; [1] for correct cost with unit ($).

Section C: Electromagnetism and Applications (20 marks)

10. (a) Calculate the magnitude of the force on the wire. [2 marks]

Answer: F = BIL = 0.40 × 3.0 × 0.15 = 0.18 N

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

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

Answer: Fleming's left-hand rule.

Award [1] for correct rule name.

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

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

Award [1] for stating it reverses current every half-turn; [1] for explaining this maintains continuous rotation in one direction.

(b) State two ways to increase the speed of rotation of the motor. [2 marks]

Answer: Any two from:

• Increase the current in the coil
• Increase the strength of the magnetic field (use stronger magnets)
• Increase the number of turns on the coil
• Increase the area of the coil

Award [1] for each valid method (max 2).

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

Answer: V_s / V_p = N_s / N_p V_s = V_p × (N_s / N_p) = 240 × (50 / 500) = 240 × 0.1 = 24 V

Award [1] for correct transformer equation; [1] for correct answer with unit.

(b) State one reason why the core of a transformer is laminated. [1 mark]

Answer: To reduce eddy currents in the core, which would cause energy loss as heat. / To improve efficiency by minimising eddy current losses.

Award [1] for any valid reason referencing eddy currents / energy losses.

(c) Calculate the current drawn from the mains supply. [2 marks]

Answer: For 100% efficiency: V_p × I_p = V_s × I_s I_p = (V_s × I_s) / V_p = (24 × 4.0) / 240 = 96 / 240 = 0.40 A

Award [1] for correct power equation; [1] for correct answer with unit.

13. (a) Explain why the galvanometer deflects when the magnet moves. [2 marks]

Answer: When the magnet moves into the solenoid, there is a changing magnetic flux (field) through the solenoid. This changing flux induces an e.m.f. across the solenoid (Faraday's law of electromagnetic induction), which drives a current through the circuit, causing the galvanometer to deflect.

Award [1] for referencing changing magnetic flux/field; [1] for linking to induced e.m.f. and current.

(b) State what happens when the magnet is held stationary inside the solenoid. Explain. [2 marks]

Answer: The galvanometer needle returns to zero / shows no deflection. When the magnet is stationary, there is no change in magnetic flux through the solenoid. Without a changing flux, no e.m.f. is induced and no current flows.

Award [1] for stating no deflection / returns to zero; [1] for explaining no change in flux → no induced e.m.f.

(c) State and explain two differences observed when the magnet is pulled out faster. [4 marks]

Answer: Difference 1: The galvanometer deflects in the opposite direction (to the left). Explanation: The magnet is moving out of the solenoid, so the magnetic flux is decreasing (rather than increasing). By Lenz's law, the induced e.m.f. and current act in the opposite direction to oppose this decrease.

Difference 2: The deflection is larger (greater magnitude). Explanation: The faster speed means a greater rate of change of magnetic flux. According to Faraday's law, a greater rate of change of flux induces a larger e.m.f., resulting in a larger current and greater galvanometer deflection.

Award [1] for each difference stated; [1] for each correct explanation (max 4). Accept other valid differences such as needle returns to zero more quickly.

END OF ANSWER KEY

Total: 60 marks