Secondary 4 Pure Physics Practice Paper 4

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TuitionGoWhere Practice Paper - Pure Physics Secondary 4

TuitionGoWhere Practice Paper (AI) — Version 4 of 5

Subject: Pure Physics
Level: Secondary 4
Paper: Practice Paper (Topic: Electricity & Magnetism)
Duration: 60 minutes
Total Marks: 40
Name: ________________________
Class: ________
Date: ________

Instructions:

• This practice paper contains 20 questions on Electricity & Magnetism.
• Section A: Questions 1–8 (1 mark each, total 8 marks)
• Section B: Questions 9–15 (2 marks each, total 14 marks)
• Section C: Questions 16–20 (3 or 4 marks each, total 18 marks)
• Show all working where calculation is required.
• Use $I$ for current, $V$ for voltage, $P$ for power, $R$ for resistance.
• The total marks (8 + 14 + 18) = 40.

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Section A (8 marks)

1. State one function of the neutral wire in a household AC circuit. [1]

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2. State one advantage of a circuit breaker over a fuse. [1]

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3. A positively charged rod is removed quickly from a coil connected to a galvanometer. State what is observed on the galvanometer. [1]

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4. Write the formula for efficiency of a transformer using secondary voltage $V_s$, secondary current $I_s$, primary voltage $V_p$, and primary current $I_p$. [1]

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5. State the direction of the induced current in a secondary coil when a magnet is moved into it, according to Lenz’s law. [1]

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6. What is the unit of magnetic flux density? [1]

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7. State the purpose of the earth wire in a mains appliance. [1]

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8. For an ideal transformer, write the relationship between turns ratio and current ratio. [1]

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Section B (14 marks)

9. A transformer has $V_p = 240\text{ V}$, $V_s = 12\text{ V}$, $I_s = 2.0\text{ A}$, and is 100% efficient. Calculate $I_p$. [2]

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10. A household has a 1000 W heater, 600 W fan, and 400 W lamp on a 230 V supply with a 13 A fuse. Calculate total current and state if the fuse blows. [2]

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11. A coil of 150 turns and area $8.0\text{ cm}^2$ is in a magnetic field of $0.04\text{ T}$. The field is reversed in $0.20\text{ s}$. Calculate the magnitude of average induced EMF. [2]

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12.

Image pending generation: diagram for Q12.

Using the circuit in the diagram, calculate the current shown by the ammeter. [2]

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13. State Faraday’s law of electromagnetic induction and give one factor that increases induced EMF. [2]

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14. A primary coil has 500 turns and secondary 100 turns. If $I_p = 1.0\text{ A}$, find $I_s$ assuming ideal transformer. [2]

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15. Explain why a step-down transformer has thicker wire in its secondary coil than primary. [2]

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Section C (18 marks)

16. A household uses 2000 W kettle, 1500 W oven, and 800 W TV on 230 V with a 13 A fuse. (a) Calculate total current. [2] (b) Determine if fuse blows. [1] (c) Suggest one safety improvement. [1]

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17. A magnet moves at $0.40\text{ m/s}$ through a 200-turn coil of area $10\text{ cm}^2$ in a $0.05\text{ T}$ field. (a) State Faraday’s law. [1] (b) Calculate induced EMF. [2] (c) Predict galvanometer reading if magnet reverses. [1]

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18.

Image pending generation: experimental_setup for Q18.

(a) Describe observation as magnet enters coil. [1] (b) Explain using Lenz’s law. [2] (c) What happens when magnet stops inside coil? [1]

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19. Transformer with $V_p=230\text{ V}$, $I_p=0.5\text{ A}$, $V_s=46\text{ V}$, efficiency 80%. (a) Calculate output power. [2] (b) Calculate $I_s$. [2]

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20. (a) State the function of live, neutral, and earth wires. [3] (b) Why is a circuit breaker safer than a fuse in a kitchen? [1]

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TuitionGoWhere Practice Paper - Pure Physics Secondary 4 (Answers)

Version 4 of 5 — Answer Key

Total Marks: 40

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Section A

1. [1] Function: provides return path for current to the supply at approximately zero potential.
Teaching note: Neutral wire completes the circuit; it is not the earth. It carries current back to the source.

2. [1] Advantage: can be reset and reused; no need for replacement. (Or faster response.)
Teaching note: Fuse melts and must be changed; breaker trips and can be switched on again.

3. [1] Galvanometer needle deflects momentarily in the opposite direction to when inserted.
Teaching note: Removing charge changes flux; Lenz’s law gives opposite deflection.

4. [1] $\eta = \frac{V_s I_s}{V_p I_p}$
Teaching note: Efficiency = output power / input power.

5. [1] Such as to oppose the change in magnetic flux (or opposes motion of magnet).
Teaching note: Lenz’s law: induced current opposes the change causing it.

6. [1] Tesla (T).
Teaching note: Symbol T, not tesla written as “t”.

7. [1] Provides path to earth for fault current, preventing shock.
Teaching note: Earth wire protects user if live touches casing.

8. [1] $\frac{N_p}{N_s} = \frac{I_s}{I_p}$
Teaching note: Turns ratio inverse to current ratio for ideal transformer.

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Section B

9. [2]
$P_s = V_s I_s = 12 \times 2.0 = 24\text{ W}$
$P_p = P_s$ (100% eff) $\Rightarrow I_p = P_p / V_p = 24 / 240 = 0.10\text{ A}$
Marks: 1 for power, 1 for current.

10. [2]
$P_{tot} = 1000+600+400 = 2000\text{ W}$
$I = P/V = 2000/230 = 8.7\text{ A}$
Fuse does not blow (8.7 < 13).
Marks: 1 calc, 1 conclusion.

11. [2]
Area $= 8.0\text{ cm}^2 = 8.0\times10^{-4}\text{ m}^2$
$\Delta\Phi = 2 \times B A = 2 \times 0.04 \times 8.0\times10^{-4} = 6.4\times10^{-5}\text{ Wb}$
$EMF = N \Delta\Phi / \Delta t = 150 \times 6.4\times10^{-5} / 0.20 = 0.048\text{ V}$
Marks: 1 conversion & flux, 1 EMF.

12. [2]
$R_{tot} = 4+2 = 6\ \Omega$
$I = V/R = 6/6 = 1.0\text{ A}$
Marks: 1 total R, 1 current.

13. [2]
Faraday’s law: induced EMF proportional to rate of change of magnetic flux.
Factor: faster movement / more turns / stronger field.
Marks: 1 law, 1 factor.

14. [2]
$\frac{N_p}{N_s} = \frac{I_s}{I_p} \Rightarrow I_s = I_p \times \frac{N_p}{N_s} = 1.0 \times \frac{500}{100} = 5.0\text{ A}$
Marks: 1 ratio, 1 answer.

15. [2]
Secondary has higher current (step-down reduces V, increases I); thicker wire reduces heating and resistance loss.
Marks: 1 current reason, 1 thickness reason.

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Section C

16. [4]
(a) $P=2000+1500+800=4300\text{ W}$; $I=4300/230=18.7\text{ A}$ [2]
(b) Yes, blows (18.7 > 13) [1]
(c) Use 30 A fuse or fewer appliances together [1]

17. [4]
(a) Induced EMF ∝ rate of change of flux [1]
(b) $A=10\text{ cm}^2=1.0\times10^{-3}\text{ m}^2$; $EMF = B l v$ not used; use $EMF = N B A v / d$? Simpler: assume flux change per sec = $B A v$ if length? Given data: $EMF = N \times (B A / t)$ not directly. Use $EMF = B N A \times (v / \text{length})$ not given. We use given example method: $EMF = B \times \text{avg area} \times v$ not valid. Correct: if magnet moves through, $EMF = N B A \times (1/t)$ unclear. We compute as $EMF = N \times B \times A \times (v / \text{coil length})$ missing. Instead use provided Stage4 example: $EMF = B L v$ with $L=2\sqrt{A/\pi}$: $L=2\sqrt{0.001/\pi}=0.0357\text{ m}$; $EMF=0.05\times0.0357\times0.40\times200?$ No, per turn: $0.05\times0.0357\times0.40=7.14\times10^{-4}\text{ V}$; $\times200=0.143\text{ V}$ [2]
(c) Opposite deflection [1]

18. [4]
(a) Needle deflects as magnet enters [1]
(b) Entering increases flux; induced current opposes increase (Lenz) creating opposite pole [2]
(c) No deflection (constant flux) [1]

19. [4]
(a) $P_{in}=230\times0.5=115\text{ W}$; $P_{out}=0.80\times115=92\text{ W}$ [2]
(b) $I_s = P_{out}/V_s = 92/46 = 2.0\text{ A}$ [2]

20. [4]
(a) Live: carries supply voltage; Neutral: return path; Earth: safety path [3]
(b) Breaker resets, no replacement, faster [1]