O Level Physics Practice Paper 3

TuitionGoWhere Practice Paper - Physics O-Level

Answer Key and Marking Scheme — Version 3

Section A: Structured Questions (45 marks)

Question 1: Static Electricity (6 marks)

(a)(i) When the polythene rod is rubbed with the woollen cloth, electrons are transferred from the wool to the polythene rod [1]. The rod gains excess electrons, so it becomes negatively charged [1].

(a)(ii) The woollen cloth becomes positively charged [1]. It lost electrons to the rod, leaving it with a deficit of electrons / excess positive charge.

(b)(i) When the charged rod is brought near the uncharged ball, the charges in the ball separate (electrostatic induction) [1]. The negative charges (electrons) in the ball are repelled to the far side, leaving the near side positively charged. Since unlike charges attract, the ball is attracted to the negatively charged rod [1].

(b)(ii) When the ball touches the rod, electrons transfer from the rod to the ball, giving the ball a net negative charge [1]. The rod and ball are now both negatively charged, so they repel each other.

Question 2: Current of Electricity (8 marks)

(a) Potential difference between two points is the work done per unit charge to move charge between those two points [1].

(b)(i) t = 4 min = 240 s [1] I = Q / t = 240 / 240 = 1.0 A [1]

(b)(ii) R = V / I = 12 / 1.0 = 12 Ω [1] for formula, [1] for correct answer with unit.

(c)(i) The resistance of the filament lamp increases as the potential difference increases [1].

(c)(ii) As the potential difference increases, the current increases, causing the filament temperature to rise [1]. The metal atoms/ions in the filament vibrate more vigorously, which impedes the flow of electrons more, so resistance increases [1].

Question 3: D.C. Circuits (9 marks)

(a) Circuit diagram showing: 12 V battery, 10 Ω resistor in series, then a parallel branch with 15 Ω and R. All symbols correct and clearly labelled [2]. (Deduct 1 mark for missing labels or incorrect symbols.)

(b)(i) R_total = V / I = 12 / 0.80 = 15 Ω [1] for formula, [1] for correct answer.

(b)(ii) R_total = 10 + R_parallel = 15 Ω, so R_parallel = 15 - 10 = 5.0 Ω [1].

(b)(iii) 1/R_parallel = 1/15 + 1/R [1] 1/5 = 1/15 + 1/R 1/R = 1/5 - 1/15 = 3/15 - 1/15 = 2/15 R = 15/2 = 7.5 Ω [1]

(c) When the thermistor is heated, its resistance decreases [1]. This reduces the effective resistance of the parallel combination, which reduces the total circuit resistance. By Ohm's law (I = V/R), the current drawn from the battery increases [1].

Question 4: Practical Electricity (7 marks)

(a)(i) P = VI, so I = P / V = 1800 / 240 = 7.5 A [1] for formula, [1] for correct answer.

(a)(ii) Energy = Power × time = 1.8 kW × (3/60) h [1] = 1.8 × 0.05 = 0.09 kWh [1].

(b)(i) Live: Brown; Neutral: Blue; Earth: Green and Yellow stripes [1] for all three correct.

(b)(ii) The earth wire is connected to the metal casing of the kettle [1]. If a fault occurs and the live wire touches the metal casing, a large current flows through the earth wire to ground, which blows the fuse / trips the circuit breaker, disconnecting the appliance and preventing electric shock [1].

Question 5: Magnetism and Electromagnetism (8 marks)

(a)(i) Sketch showing field lines emerging from North pole, curving around, and entering South pole [1]. Arrows pointing from N to S [1].

(a)(ii) Any two from:

• Field lines never cross each other [1]
• Field lines are closer together where the field is stronger [1]
• Field lines point from North to South outside the magnet [1]
• Field lines form closed loops [1]

(b)(i) F = B I L = 0.40 × 3.0 × 0.15 [1] = 0.18 N [1].

(b)(ii) Fleming's left-hand rule [1].

(b)(iii) Increase the length of wire within the magnetic field [1] (or use a stronger magnet / increase number of turns of wire).

Question 6: Electromagnetic Induction (7 marks)

(a)(i) When the magnet moves into the coil, the magnetic flux / magnetic field lines passing through the coil change [1]. By Faraday's law, a changing magnetic flux induces an e.m.f. in the coil [1].

(a)(ii) The galvanometer needle returns to zero / shows no deflection [1]. When the magnet is stationary, there is no change in magnetic flux through the coil, so no e.m.f. is induced [1].

(a)(iii) The needle deflects to the left (opposite direction) [1]. By Lenz's law, the direction of the induced e.m.f. (and current) opposes the change causing it. Pulling the magnet out decreases the flux, so the induced current flows in the opposite direction to try to maintain the flux [1].

(b) Any one from:

• Speed of relative motion between magnet and coil [1]
• Number of turns in the coil [1]
• Strength of the magnet [1]

Section B: Free Response Questions (35 marks)

Question 7: Circuit Analysis and Design (12 marks)

(a) Circuit diagram showing: 6 V battery, thermistor and 500 Ω fixed resistor connected in series forming a potential divider [1]. Relay coil connected across the fixed resistor [1]. Correct symbols and labels [1].

(b)(i) At 30 °C, R_thermistor = 500 Ω, R_fixed = 500 Ω. V across thermistor = (R_thermistor / (R_thermistor + R_fixed)) × V_supply [1] = (500 / 1000) × 6.0 = 3.0 V [1].

(b)(ii) The relay is connected across the fixed resistor, not the thermistor. The voltage across the fixed resistor is also 3.0 V, which is exactly the threshold. However, the relay needs more than 3.0 V to reliably activate, or the question context implies the relay activates when voltage across it reaches 3.0 V — at exactly 30 °C it is borderline and may not activate reliably [1]. (Accept: The relay activates when voltage across it is 3.0 V or higher; at 30 °C the voltage is exactly 3.0 V, so it is at threshold. The design intention is activation above 30 °C.)

(b)(iii) At 35 °C, R_thermistor = 350 Ω. V across fixed resistor = (500 / (350 + 500)) × 6.0 [1] = (500 / 850) × 6.0 = 0.588 × 6.0 = 3.53 V [1]. Since 3.53 V > 3.0 V, the relay activates.

(c) Replace the fixed 500 Ω resistor with a smaller resistance (e.g., 400 Ω) [1]. At 25 °C, R_thermistor = 700 Ω. With a 400 Ω fixed resistor, V across fixed resistor = (400 / (700 + 400)) × 6.0 = (400/1100) × 6.0 = 2.18 V (still below 3.0 V — so this doesn't work). Alternative correct answer: Connect the relay across the thermistor instead of the fixed resistor [1]. At 25 °C, V across thermistor = (700 / (700 + 500)) × 6.0 = 3.5 V, which exceeds 3.0 V, so the relay activates at 25 °C [1]. (Accept any valid modification with correct reasoning.)

(d) P = I²R [1] = (15 × 10⁻³)² × 200 = (2.25 × 10⁻⁴) × 200 = 0.045 W or 45 mW [1].

Question 8: Electromagnetic Devices (11 marks)

(a)(i) A current-carrying conductor placed in a magnetic field experiences a force [1]. This is due to the interaction between the magnetic field of the permanent magnet and the magnetic field created by the current in the wire [1].

(a)(ii) Using Fleming's left-hand rule: For one side, the current flows in one direction (e.g., into the page), and the magnetic field is from N to S. The force is directed upward [1]. For the opposite side, the current flows in the opposite direction (out of the page), while the magnetic field direction is the same. The force is therefore directed downward [1]. The opposite forces create a turning effect.

(a)(iii) The split-ring commutator reverses the direction of current in the coil every half rotation [1]. This ensures that the forces on the sides of the coil always act in the same rotational direction, maintaining continuous rotation [1].

(b)(i) For one turn: F = B I L = 0.25 × 2.0 × 0.040 = 0.020 N [1] For 50 turns: Total F = 50 × 0.020 = 1.0 N [1].

(b)(ii) Moment = Force × perpendicular distance from pivot [1] Total moment = 2 × F × d = 2 × 1.0 × 0.025 = 0.050 N m [1].

(c) Any one from:

• Increase the number of turns on the coil [1]
• Use a stronger magnet (increase magnetic flux density) [1]
• Increase the area of the coil [1]
• Wind the coil on a soft iron core to concentrate the magnetic field [1]

Question 9: Transformers and Power Transmission (12 marks)

(a) Electrical power transmitted is P = VI. For a given power, a higher voltage means a lower current [1]. Lower current reduces the heating effect in the transmission cables (power loss = I²R) [1]. This makes transmission more efficient and allows thinner, lighter, and cheaper cables to be used [1].

(b)(i) V_s / V_p = N_s / N_p [1] 400 000 / 25 000 = N_s / 800 16 = N_s / 800 N_s = 16 × 800 = 12 800 turns [1].

(b)(ii) For an ideal transformer: V_p I_p = V_s I_s [1] 25 000 × 1200 = 400 000 × I_s I_s = (25 000 × 1200) / 400 000 = 30 000 000 / 400 000 = 75 A [1].

(c)(i) Any two from:

• Resistive heating in the coils (copper losses) [1]
• Eddy currents induced in the iron core [1]
• Hysteresis losses due to repeated magnetisation and demagnetisation of the core [1]
• Flux leakage where not all magnetic flux links both coils [1]

(c)(ii) Any one from:

• Use a laminated core to reduce eddy currents [1]
• Use a soft iron core that is easily magnetised and demagnetised to reduce hysteresis losses [1]
• Use thick copper wire for the coils to reduce resistance [1]

(d)(i) P = VI, so I = P / V [1] I = 48 000 / 240 = 200 A [1].

END OF ANSWER KEY