Secondary 4 Pure Physics Preliminary Examination Paper 2

TuitionGoWhere Practice Paper — Pure Physics Secondary 4

PRELIMINARY EXAMINATION — VERSION 2

Answer Key & Marking Scheme

Section A: Multiple Choice [10 marks]

Question 1 [2]

Answer: B) 12 V

Working: Using the transformer equation: V_s / V_p = N_s / N_p

V_s = V_p × (N_s / N_p) = 240 × (100 / 2000) = 240 × 0.05 = 12 V

Marking notes:

• Award 2 marks for correct answer B.
• Award 0 marks for any other option.
• No partial marks for this multiple-choice question.

Question 2 [2]

Answer: C) 2.0 A

Working: Using efficiency: η = (V_s × I_s) / (V_p × I_p)

0.80 = (48 × I_s) / (240 × 0.50)

0.80 = (48 × I_s) / 120

48 × I_s = 0.80 × 120 = 96

I_s = 96 / 48 = 2.0 A

Marking notes:

• Award 2 marks for correct answer C.
• Award 0 marks for any other option.

Question 3 [2]

Answer: B) Concentric circles around the wire, with direction given by the right-hand grip rule

Marking notes:

• Award 2 marks for correct answer B.
• Award 0 marks for any other option.

Question 4 [2]

Answer: C) Increasing the area of the coil

Marking notes:

• Award 2 marks for correct answer C.
• The induced e.m.f. depends on the rate of change of magnetic flux. Increasing the area increases the flux linkage, hence increasing the induced e.m.f.
• Award 0 marks for any other option.

Question 5 [2]

Answer: C) live wire

Marking notes:

• Award 2 marks for correct answer C.
• The fuse must be in the live wire so that when it blows, the circuit is disconnected from the high-voltage supply, preventing electric shock.
• Award 0 marks for any other option.

Section B: Structured Questions [30 marks]

Question 6 [3]

(a) [1] In a step-up transformer, the secondary voltage is greater than the primary voltage, whereas in a step-down transformer, the secondary voltage is less than the primary voltage.

(b) [1] In a step-up transformer, the number of turns on the secondary coil is greater than the number of turns on the primary coil, whereas in a step-down transformer, the secondary has fewer turns than the primary.

(c) [1] A transformer operates on the principle of electromagnetic induction (or mutual induction).

Marking notes:

• Award 1 mark each for (a), (b), and (c).
• For (a) and (b), accept any valid, clearly stated difference.
• For (c), accept "mutual induction" or "electromagnetic induction".

Question 7 [4]

(a) [2] Turns ratio:

V_s / V_p = N_s / N_p

N_s / N_p = 12 / 240 = 1 / 20 (or 0.05)

Marking:

• 1 mark for correct formula or substitution.
• 1 mark for correct answer (1:20 or 0.05).

(b) [2] Primary current (100% efficient):

V_p × I_p = V_s × I_s

240 × I_p = 12 × 2.0

I_p = 24.0 / 240 = 0.10 A

Marking:

• 1 mark for correct formula or substitution.
• 1 mark for correct answer with unit (0.10 A).

Common mistakes:

• Forgetting to include the unit (A) — deduct 1 mark.
• Confusing primary and secondary values.

Question 8 [3]

(a) [1] The current flows from B to A (upward in the wire).

Reasoning: Using Fleming's left-hand rule: the force is into the page, the magnetic field goes from N to S (left to right), so the current must be upward (from B to A).

(b) [1] Any one of:

• Increase the current in the wire
• Use a stronger magnet (increase magnetic field strength)
• Increase the length of the wire in the magnetic field

(c) [1] Electric motor (or d.c. motor)

Marking notes:

• Award 1 mark each for (a), (b), and (c).
• For (b), accept any valid method.

Question 9 [4]

(a) [2] Initial magnetic flux:

Φ = B × A × cos θ

When the coil is perpendicular to the field, the normal to the coil is parallel to B, so θ = 0°.

Φ_initial = B × A = 0.50 × 0.020 = 0.010 Wb

Marking:

• 1 mark for correct formula.
• 1 mark for correct answer with unit (0.010 Wb).

(b) [2] Average e.m.f.:

When the coil is parallel to the field, the normal is perpendicular to B, so Φ_final = 0.

Change in flux linkage = N × ΔΦ = 50 × (0.010 − 0) = 0.50 Wb

e.m.f. = Δ(NΦ) / Δt = 0.50 / 0.10 = 5.0 V

Marking:

• 1 mark for calculating change in flux linkage correctly.
• 1 mark for correct final answer with unit (5.0 V).

Common mistakes:

• Forgetting to multiply by number of turns (N = 50) — this is a critical error; deduct 1 mark.
• Using wrong angle for initial flux.

Question 10 [3]

Expected answer (3 marks):

• Power transmitted: P = V × I
• For a given power, increasing the voltage reduces the current (I = P / V)
• Power lost in cables: P_loss = I² × R
• Since P_loss depends on I², reducing the current greatly reduces the energy lost as heat
• Therefore, transmitting at high voltage minimises energy loss in the cables

Marking scheme:

• 1 mark for stating that high voltage means low current for the same power.
• 1 mark for stating that power loss = I²R (or equivalent reasoning).
• 1 mark for concluding that lower current means less power lost as heat.

Common mistakes:

• Simply stating "less energy is lost" without explaining why — award only 1 mark.
• Confusing voltage and current — no marks if the reasoning is fundamentally wrong.

Question 11 [4]

(a) [2] For the coil to rotate clockwise:

• Side AB (on the left) must experience a force downward.
• Using Fleming's left-hand rule: field (N→S, left to right), force (downward) → current must be from A to B (into the page at side AB, but since AB is the left side, current flows from A at the top to B at the bottom — i.e., downward).

Marking:

• 1 mark for correct direction (A to B, or downward in side AB).
• 1 mark for correct explanation referencing Fleming's left-hand rule.

(b) [2] Any two of:

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

Marking: 1 mark each, maximum 2 marks.

Question 12 [3]

(a) [2] Current:

V = I × R

I = V / R = 12 / 4.0 = 3.0 A

Marking:

• 1 mark for correct formula or substitution.
• 1 mark for correct answer with unit (3.0 A).

(b) [1] Charge:

Q = I × t = 3.0 × 30 = 90 C

Marking:

• 1 mark for correct answer with unit (90 C).

Question 13 [3]

Lenz's Law [1]: The direction of the induced current is such that it opposes the change producing it (or opposes the change in magnetic flux).

Explanation [2]:

• As the magnet falls through the copper tube, the changing magnetic flux induces eddy currents in the tube.
• By Lenz's Law, the eddy currents create a magnetic field that opposes the motion of the magnet (i.e., exerts an upward force on the falling magnet).
• As the magnet speeds up, the opposing force increases until it equals the weight of the magnet.
• At this point, the net force is zero and the magnet falls at a constant (terminal) velocity.

Marking scheme:

• 1 mark for correct statement of Lenz's Law.
• 1 mark for explaining that eddy currents are induced and create an opposing force.
• 1 mark for explaining that terminal velocity is reached when the opposing force equals the weight.

Question 14 [3]

(a) [2] Total power:

P_total = 2400 + 1200 + (3 × 60) = 2400 + 1200 + 180 = 3780 W

Total current:

I_total = P_total / V = 3780 / 240 = 15.75 A

Marking:

• 1 mark for correct total power.
• 1 mark for correct total current with unit (15.75 A or 15.8 A).

(b) [1] Yes, the fuse will blow because the total current drawn (15.75 A) exceeds the fuse rating of 15 A.

Marking:

• 1 mark for correct conclusion with valid reasoning.

Section C: Free Response / Application [20 marks]

Question 15 [5]

(a) [2] Current at 250 V:

P = V × I

I = P / V = 10 000 / 250 = 40 A

Marking:

• 1 mark for correct formula.
• 1 mark for correct answer with unit (40 A).

(b) [1] Power lost at 250 V:

P_loss = I² × R = 40² × 4.0 = 1600 × 4.0 = 6400 W

Marking:

• 1 mark for correct answer with unit (6400 W).

(c) [2] Power lost at 2500 V:

New current: I = P / V = 10 000 / 2500 = 4.0 A

P_loss = I² × R = 4.0² × 4.0 = 16 × 4.0 = 64 W

Marking:

• 1 mark for calculating the new current correctly.
• 1 mark for correct final answer with unit (64 W).

Common mistakes:

• Using the same current (40 A) for part (c) — this is a critical error.
• Forgetting to square the current in P = I²R.

Question 16 [5]

(a) [2] Peak voltage:

From the trace: peak is 2 divisions above the centre line.

V_peak = 2 × 5.0 = 10.0 V

Marking:

• 1 mark for reading 2 divisions correctly from the diagram.
• 1 mark for correct answer with unit (10.0 V).

(b) [2] Frequency:

From the trace: one complete cycle spans 30 ms (3 divisions on time-base at 10 ms/div).

Period T = 30 ms = 30 × 10⁻³ s = 0.030 s

f = 1 / T = 1 / 0.030 = 33.3 Hz (or 33 Hz)

Marking:

• 1 mark for correct period reading.
• 1 mark for correct frequency with unit (33.3 Hz).

(c) [1] No, this a.c. supply is not suitable for charging a 12 V d.c. battery because the peak voltage (10.0 V) is less than the 12 V required, and the supply is alternating current, not direct current. A battery charger requires d.c. and a voltage higher than the battery voltage.

Marking:

• 1 mark for correct conclusion with valid reasoning.

Question 17 [5]

(a) [2] To induce a current:

• The student should move the bar magnet into or out of the solenoid (relative motion between the magnet and solenoid).
• Alternatively, the student could move the solenoid towards or away from the stationary magnet.

Marking:

• 1 mark for describing relative motion.
• 1 mark for specifying that the magnet must be moved into or out of the solenoid (not held stationary).

(b) [2] Any two of:

• Move the magnet faster (increase speed of movement)
• Use a stronger magnet
• Increase the number of turns on the solenoid

Marking: 1 mark each, maximum 2 marks.

(c) [1] The end of the solenoid nearest the magnet becomes a north pole.

Explanation: By Lenz's Law, the induced current opposes the change. Since the north pole is approaching, the solenoid creates a north pole at the nearest end to repel the approaching magnet and oppose the increase in flux.

Marking:

• 1 mark for correct polarity (north) with correct explanation referencing Lenz's Law.

Question 18 [5]

(a) [1] Input power:

P_in = V_in × I_in = 240 × 0.050 = 12.0 W

Marking:

• 1 mark for correct answer with unit (12.0 W).

(b) [2] Output power:

η = P_out / P_in

0.90 = P_out / 12.0

P_out = 0.90 × 12.0 = 10.8 W

Marking:

• 1 mark for correct formula or substitution.
• 1 mark for correct answer with unit (10.8 W).

(c) [2] Output current:

P_out = V_out × I_out

I_out = P_out / V_out = 10.8 / 5.0 = 2.16 A

Marking:

• 1 mark for correct formula or substitution.
• 1 mark for correct answer with unit (2.16 A).

Common mistakes:

• Using input power instead of output power in part (c) — deduct 1 mark.
• Forgetting to convert efficiency percentage to decimal — this would give P_out = 1080 W, which is clearly wrong.

Summary of Marks

© TuitionGoWhere Secondary School (AI) — Preliminary Examination Practice, Version 2 of 5 — Answer Key