Secondary 4 Pure Physics Preliminary Examination Paper 5

TuitionGoWhere Practice Paper — Pure Physics Secondary 4

PRELIMINARY EXAMINATION — VERSION 5

Answer Key & Marking Scheme

Section A: Multiple Choice [10 marks]

1. B) 12 V [2 marks]

Working: $\frac{V_s}{V_p} = \frac{N_s}{N_p}$ $V_s = V_p \times \frac{N_s}{N_p} = 240 \times \frac{100}{2000} = 240 \times 0.05 = 12 \text{ V}$

2. C) 6 A [2 marks]

Working: $R_{total} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{6 \times 3}{6 + 3} = \frac{18}{9} = 2 \; \Omega$ $I = \frac{V}{R} = \frac{12}{2} = 6 \text{ A}$

3. C) To the left [2 marks]

Using the right-hand grip rule: grip the wire with the thumb pointing into the page (direction of current). The fingers curl anticlockwise when viewed from above. At a point directly above the wire, the field direction is to the left.

4. C) 2.0 A [2 marks]

Working: $\eta = \frac{V_s I_s}{V_p I_p}$ $0.80 = \frac{48 \times I_s}{240 \times 0.5}$ $0.80 = \frac{48 I_s}{120}$ $48 I_s = 0.80 \times 120 = 96$ $I_s = \frac{96}{48} = 2.0 \text{ A}$

5. C) Upward [2 marks]

Using Fleming's left-hand rule: First finger = field (into page), Second finger = current/velocity (right), Thumb = force (upward). For a positive charge, conventional current direction = direction of motion.

Section B: Structured Questions [30 marks]

6. [4 marks]

(a) $R_{total} = R_1 + R_2 = 4 + 8 = 12 \; \Omega$ [1 mark]

(b) $I = \frac{V}{R_{total}} = \frac{24}{12} = 2.0 \text{ A}$ [1 mark]

(c) $V_{R_2} = I \times R_2 = 2.0 \times 8 = 16 \text{ V}$ [2 marks: 1 for correct method, 1 for correct answer with unit]

7. [4 marks]

(a) $\frac{N_p}{N_s} = \frac{V_p}{V_s} = \frac{2400}{240} = \frac{10}{1}$ or 10:1 [1 mark]

(b) $P_{input} = V_p \times I_p = 2400 \times 0.8 = 1920 \text{ W}$ [1 mark]

(c) $P_{output} = \eta \times P_{input} = 0.90 \times 1920 = 1728 \text{ W}$ [1 mark] $I_s = \frac{P_{output}}{V_s} = \frac{1728}{240} = 7.2 \text{ A}$ [1 mark]

8. [3 marks]

(a) The force on side AB is directed downward (toward the bottom of the page). [1 mark]

(b) The current in side AB flows in the opposite direction to the current in side CD. [1 mark] By Fleming's left-hand rule, the forces on the two sides are in opposite directions, creating a couple/torque that causes the coil to rotate. [1 mark]

9. [4 marks]

(a) $I_{kettle} = \frac{P}{V} = \frac{2400}{240} = 10.0 \text{ A}$ [1 mark]

(b) $I_{iron} = \frac{1200}{240} = 5.0$ A; $I_{hairdryer} = \frac{1000}{240} = 4.17$ A; $I_{microwave} = \frac{800}{240} = 3.33$ A [1 mark for all correct] $I_{total} = 10.0 + 5.0 + 4.17 + 3.33 = 22.5$ A [1 mark]

(c) The fuse will not blow [1 mark] because the total current (22.5 A) is less than the fuse rating of 30 A. [Accept: justification based on comparison]

10. [3 marks]

(a) $F = BIL$ (when the wire is perpendicular to the field) [1 mark]

(b) $F = BIL = 0.08 \times 4.0 \times 0.50 = 0.16 \text{ N}$ [2 marks: 1 for substitution, 1 for correct answer with unit]

11. [4 marks]

(a) Lenz's law states that the direction of the induced current is such that it opposes the change producing it. [1 mark]

(b) As the North pole approaches the solenoid, the magnetic flux through the solenoid increases. [1 mark] By Lenz's law, the induced current must oppose this increase — so the end of the solenoid nearest the magnet must become a North pole to repel the approaching magnet. [1 mark] Using the right-hand grip rule, the current must flow from right to left through the galvanometer (i.e., the galvanometer needle deflects to the left). [1 mark]

12. [4 marks]

(a) [2 marks]

• Correctly labelled axes with units (V on y-axis, I on x-axis): 1 mark
• All points correctly plotted and a straight line of best fit drawn: 1 mark

(b) Gradient = $\frac{\Delta V}{\Delta I} = \frac{6.0 - 0.0}{1.20 - 0.00} = \frac{6.0}{1.20} = 5.0$ [1 mark] The gradient represents the resistance of the fixed resistor (5.0 Ω). [1 mark]

13. [4 marks]

(a) $F = Bqv = 0.02 \times 1.6 \times 10^{-19} \times 3.0 \times 10^6$ [1 mark] $F = 9.6 \times 10^{-15}$ N [1 mark]

(b) The electron follows a circular path in the plane perpendicular to the field. [1 mark] The magnetic force acts perpendicular to the velocity at all times, providing the centripetal force needed for circular motion. The force direction is always perpendicular to velocity, so speed is constant but direction changes continuously. [1 mark]

Section C: Free Response [20 marks]

14. [10 marks]

(a) $P = VI \implies I = \frac{P}{V} = \frac{200\,000}{5000} = 40$ A [2 marks: 1 for formula, 1 for correct answer]

(b) $P_{lost} = I^2 R = 40^2 \times 8 = 1600 \times 8 = 12\,800$ W = 12.8 kW [2 marks: 1 for formula/substitution, 1 for correct answer]

(c) $\% \text{ loss} = \frac{P_{lost}}{P_{total}} \times 100 = \frac{12\,800}{200\,000} \times 100 = 6.4\%$ [2 marks: 1 for method, 1 for correct answer]

(d) When the voltage is stepped up to 250,000 V, the current in the cables becomes: $I' = \frac{P}{V'} = \frac{200\,000}{250\,000} = 0.8$ A [1 mark] The power lost becomes: $P'_{lost} = I'^2 R = 0.8^2 \times 8 = 5.12$ W [1 mark] This is vastly less than the 12,800 W lost at low voltage. [1 mark] Explanation: Since $P_{lost} = I^2R$, reducing the current dramatically reduces the power lost. A step-up transformer increases voltage and reduces current for the same power, so transmitting at high voltage minimises $I^2R$ losses in the cables. [1 mark]

15. [10 marks]

(a) $\tau_{max} = BANI = 0.5 \times 0.02 \times 50 \times 2.0$ [1 mark for formula, 1 for substitution] $\tau_{max} = 1.0$ N m [1 mark]

(b) The split-ring commutator reverses the direction of current in the coil every half-turn. [1 mark] This ensures that the torque on the coil always acts in the same direction, allowing continuous rotation. [1 mark]

(c)(i) Net e.m.f. = Supply voltage − Back e.m.f. = $12.0 - 8.0 = 4.0$ V [1 mark] $I = \frac{V_{net}}{R} = \frac{4.0}{1.0} = 4.0$ A [1 mark]

(c)(ii) When the motor starts from rest, the back e.m.f. is zero because the coil is not yet moving through the magnetic field. [1 mark] The full supply voltage drives current through the coil, so the starting current is high ($I = V/R = 12/1 = 12$ A). [1 mark] As the coil speeds up, the rate of change of flux through the coil increases, inducing a larger back e.m.f. that opposes the supply voltage. The net voltage across the resistance decreases, so the current decreases. [1 mark]

END OF ANSWER KEY

Total: 60 marks