Secondary 4 Pure Physics Preliminary Examination Paper 2

TuitionGoWhere Practice Paper – Pure Physics Secondary 4 (PRELIM) Answer Key (Version 2)

Total Marks: 60

Section A (10 marks)

1. [1] A circuit breaker can be reset and reused; a fuse must be replaced after it melts.
Teaching note: Circuit breakers trip and can be switched on again; fuses are one-time protective devices.

2. [1] The neutral wire provides the return path for current to the supply and is at approximately zero potential.
Teaching note: It completes the circuit but does not carry dangerous voltage under normal conditions.

3. [1] The galvanometer needle deflects momentarily in the opposite direction to when the rod was inserted.
Teaching note: Removing charge changes flux; by Lenz’s law the induced current opposes the change.

4. [1] $\frac{N_p}{N_s} = \frac{240}{12} = 20$ (or 20:1).
Teaching note: Turns ratio equals voltage ratio for ideal transformer.

5. [1] Electromagnetic induction (Faraday’s law).
Teaching note: Changing magnetic flux induces EMF.

6. [1] Brown.
Teaching note: Live = brown, neutral = blue, earth = green/yellow in Singapore.

7. [1] Step-down transformer.
Teaching note: Fewer secondary turns means lower secondary voltage.

8. [1] It provides a safe path to earth for fault current / prevents shock by connecting metal casing to ground.
Teaching note: Earth wire protects users from electric shock.

9. [1] The galvanometer deflects momentarily as current is switched on (due to changing flux).
Teaching note: Sudden current change creates changing magnetic field, inducing EMF.

10. [1] Tesla (T).
Teaching note: Magnetic flux density B is measured in tesla.

Section B (24 marks)

11. [2]
Efficiency $\eta = 0.80$
$P_s = V_s I_s = 9.0 \times 0.50 = 4.5$ W
$P_p = \frac{P_s}{\eta} = \frac{4.5}{0.80} = 5.625$ W
$I_p = \frac{P_p}{V_p} = \frac{5.625}{230} = 0.0245$ A ≈ 0.025 A
Marking: 1 mark for correct power/effort, 1 mark for final answer with unit.

12. [2]
Ideal transformer: $\frac{N_p}{N_s} = \frac{I_s}{I_p}$
$I_s = I_p \times \frac{N_p}{N_s} = 2.0 \times \frac{400}{100} = 8.0$ A
Marking: 1 mark rearrangement, 1 mark answer.

13. [3]

• As magnet is pulled out, magnetic flux through coil decreases. [1]
• Lenz’s law: induced current opposes change, so it tries to maintain flux. [1]
• Direction of induced current causes galvanometer to deflect opposite to when magnet inserted. [1]

14. [2]
Live wire: carries supply voltage (≈230 V) into appliance. [1]
Earth wire: safety path to ground for fault current. [1]

15. [2]
Incorrect. [1] Transformer needs changing flux; DC gives constant flux so no continuous induction. [1]

16. [2]
$V = IR \Rightarrow R = \frac{V}{I} = \frac{6.0}{0.20} = 30\ \Omega$
Marking: 1 mark formula, 1 mark answer.

Section C (26 marks)

17. [4 total]
(a) [1] Electromagnet needs current to produce field; permanent magnet always magnetic.
(b) [2] Iron core becomes magnetised easily, concentrating field lines and increasing strength. (1 for magnetisation, 1 for concentration)
(c) [1] $N I = 200 \times 0.15 = 30$ A-turns.

18. [5 total]
(a) [2] $\frac{N_p}{N_s} = \frac{V_p}{V_s} \Rightarrow N_s = N_p \times \frac{V_s}{V_p} = 1200 \times \frac{12}{240} = 60$ turns. (1 formula, 1 answer)
(b) [2] $I_p = I_s \times \frac{V_s}{V_p} = 2.0 \times \frac{12}{240} = 0.10$ A. (1 method, 1 answer)
(c) [1] Real transformer has losses (heat, flux leakage) so more primary current needed.

19. [4 total]
(a) [1] Period = 0.02 s → $f = \frac{1}{T} = 50$ Hz.
(b) [1] Peak EMF doubles (proportional to rotation speed).
(c) [2] EMF zero when flux change rate zero (coil plane parallel to field at those instants). [1] Graph crosses axis at max/min flux positions. [1]

20. [5 total]
(a) [1] Zero (or approximately zero) potential.
(b) [3] If live touches casing, current flows to earth via earth wire. [1] Large current blows fuse/trips breaker. [1] Cuts off supply, preventing user shock. [1]
(c) [1] Fuse melts at 8 A > 5 A rating, breaking circuit and protecting user.