Secondary 4 Pure Physics Preliminary Examination Paper 4

TuitionGoWhere Practice Paper - Pure Physics Secondary 4

Answer Key — PRELIM Paper 2 (Version 4 of 5)

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Section A: Multiple Choice [10 marks]

1. B [1]

Working: Using the transformer equation: V_s / V_p = N_s / N_p → 12 / 240 = N_s / 800 → N_s = (12 × 800) / 240 = 40 turns.

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2. B [1]

Explanation: The magnetic field around a straight current-carrying wire forms concentric circles around the wire. The direction is determined by the right-hand grip rule (thumb in direction of current, fingers curl in direction of field).

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3. A [1]

Working: F = BIL = 0.5 × 3 × 0.2 = 0.3 N.

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4. D [1]

Explanation: Moving the magnet away from the coil at constant speed will produce a deflection, but the question asks what will not increase the deflection. Moving the magnet away produces a deflection in the opposite direction, not an increase in the original deflection. However, the key point is that options A, B, and C all increase the rate of change of magnetic flux, increasing deflection. Option D changes the direction but does not increase the magnitude of deflection compared to moving it in. Note: This question tests understanding that deflection magnitude depends on rate of flux change; moving away at the same speed gives the same magnitude of deflection but opposite direction. The answer is D because it does not increase the deflection — it reverses it.

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5. B [1]

Working: P = VI → I = P / V = 2000 / 240 = 8.33 A.

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6. C [1]

Explanation: Using Fleming's left-hand rule: for force into the page, with magnetic field from N to S (top to bottom), the current must flow left to right. Therefore, S pole is above and N pole is below, with current flowing left to right. Answer: C.

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7. C [1]

Working: Efficiency η = (V_s × I_s) / (V_p × I_p) → 0.80 = (48 × I_s) / (240 × 2) → 0.80 = 48 I_s / 480 → I_s = (0.80 × 480) / 48 = 8 A.

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8. B [1]

Explanation: The SI unit for magnetic flux density (magnetic field strength) is the Tesla (T). Weber (Wb) is the unit for magnetic flux. Henry (H) is the unit for inductance.

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9. A [1]

Explanation: Using Fleming's left-hand rule: the particle is positively charged, so conventional current direction is east (thumb). Magnetic field is upward (second finger). The force (first finger) points north. Answer: A.

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10. C [1]

Explanation: The electromagnetic spectrum in order of increasing wavelength: gamma rays → X-rays → ultraviolet → visible light → infrared → microwaves → radio waves. Infrared has wavelength longer than visible light but shorter than microwaves.

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Section B: Structured Questions [30 marks]

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11.

(a) Faraday's law of electromagnetic induction: The magnitude of the induced e.m.f. in a circuit is directly proportional to the rate of change of magnetic flux linkage through the circuit. [2]

Marking notes:

• 1 mark for "induced e.m.f. is proportional to rate of change of magnetic flux (linkage)"
• 1 mark for stating it is the magnitude / direct proportionality
• Accept: "The induced e.m.f. is proportional to the rate at which the magnetic flux through the coil changes"

(b)(i) Explanation of deflection: As the north pole moves into the coil, the magnetic flux through the coil changes (increases). By Faraday's law, this changing flux induces an e.m.f. in the coil, which drives a current through the galvanometer, causing the needle to deflect. [2]

Marking notes:

• 1 mark for identifying that the magnetic flux through the coil is changing
• 1 mark for linking this to induced e.m.f. / current via Faraday's law

(b)(ii) When magnet is stationary: The galvanometer shows zero deflection. When the magnet is stationary, there is no change in magnetic flux through the coil. By Faraday's law, no e.m.f. is induced, so no current flows and the galvanometer does not deflect. [2]

Marking notes:

• 1 mark for stating zero / no deflection
• 1 mark for explaining that flux is not changing, so no e.m.f. is induced

(b)(iii) Direction when magnet is removed: The galvanometer deflects to the left (opposite direction). As the magnet is pulled out, the magnetic flux through the coil decreases. By Lenz's law, the induced current flows in a direction to oppose this change (i.e., to try to maintain the flux), which means the induced current is in the opposite direction to when the magnet was pushed in. [2]

Marking notes:

• 1 mark for stating the deflection is to the left / opposite direction
• 1 mark for explaining using Lenz's law or the reversal of flux change direction

[Total: 8 marks]

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12.

(a) The current in side AB flows from A to B (or right to left, depending on diagram orientation — accept either with correct reasoning). [1]

Note: For a clockwise rotation, side AB must experience a downward force. Using Fleming's left-hand rule with field from N to S (left to right), current must flow from A to B (into the page direction for side AB).

(b) Explanation using Fleming's left-hand rule:

• The magnetic field (second finger) points from the N pole to the S pole (e.g., left to right).
• The current in side AB (thumb) flows in the appropriate direction.
• The force (first finger) points downward.
• Therefore, side AB experiences a downward force. [3]

Marking notes:

• 1 mark for correctly identifying the magnetic field direction
• 1 mark for correctly identifying the current direction in AB
• 1 mark for correctly applying Fleming's left-hand rule to show the force is downward

(c) Function of the split-ring commutator: The split-ring commutator reverses the direction of the current in the coil every half-turn. This ensures that the force on each side of the coil always acts in the same direction of rotation, allowing the coil to rotate continuously in one direction. [2]

Marking notes:

• 1 mark for stating it reverses the current direction every half-turn
• 1 mark for explaining that this maintains continuous rotation in one direction

(d) Two ways to increase speed of rotation:

1. Increase the current in the coil (by increasing the supply voltage)
2. Use a stronger magnet (increase magnetic field strength)
3. Increase the number of turns on the coil
4. Increase the area of the coil

(Accept any two of the above) [2]

Marking notes: 1 mark for each valid suggestion.

[Total: 8 marks]

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13.

(a) Turns ratio:

N_s / N_p = V_s / V_p = 250 000 / 2500 = 100 [2]

Marking notes:

• 1 mark for correct substitution
• 1 mark for correct answer (100:1 or simply 100)

(b) Why step-up transformers are used: Step-up transformers increase the voltage for transmission. Since power loss in transmission lines is given by P_loss = I²R, increasing the voltage reduces the current for the same power transmitted (P = VI), thereby reducing energy lost as heat in the cables. This makes power transmission more efficient. [2]

Marking notes:

• 1 mark for stating that it reduces current for the same power
• 1 mark for linking this to reduced power loss (I²R losses)

(c) Secondary current calculation:

Efficiency η = (V_s × I_s) / (V_p × I_p)

0.95 = (250 000 × I_s) / (2500 × 40)

0.95 = (250 000 × I_s) / 100 000

I_s = (0.95 × 100 000) / 250 000

I_s = 95 000 / 250 000

I_s = 0.38 A [3]

Marking notes:

• 1 mark for writing the efficiency equation
• 1 mark for correct substitution
• 1 mark for correct answer (0.38 A)

Common mistake: Forgetting to convert efficiency percentage to decimal (using 95 instead of 0.95).

[Total: 7 marks]

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14.

(a) Force when perpendicular:

F = BIL = 0.08 × 4 × 0.5 = 0.16 N [2]

Marking notes:

• 1 mark for correct formula and substitution
• 1 mark for correct answer with unit (N)

(b) Force at 30°:

F = BIL sin θ = 0.08 × 4 × 0.5 × sin 30° = 0.08 × 4 × 0.5 × 0.5 = 0.08 N [2]

Marking notes:

• 1 mark for using F = BIL sin θ
• 1 mark for correct answer (0.08 N)

(c) Force when parallel: The force is zero. When the wire is parallel to the magnetic field, the angle between the current and the field is 0°, and sin 0° = 0. Therefore, F = BIL sin 0° = 0. Alternatively, no magnetic flux is cut when the wire moves parallel to the field. [2]

Marking notes:

• 1 mark for stating zero force
• 1 mark for correct reason (sin 0° = 0, or no flux is cut)

[Total: 6 marks]

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Section C: Application and Data-Based Questions [20 marks]

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15.

(a) Current drawn by each appliance (using I = P / V):

Appliance	Power Rating	Current Drawn
Refrigerator	150 W	150/240 = 0.625 A
Electric Oven	3000 W	3000/240 = 12.5 A
Television	120 W	120/240 = 0.5 A
Washing Machine	2000 W	2000/240 = 8.33 A

Marking notes: 1 mark for each correct current value. Accept 0.63 A, 12.5 A, 0.5 A, 8.3 A. [4]

(b) Will the fuse blow?

Total current = 0.625 + 12.5 + 0.5 + 8.33 = 21.955 A

Since 21.955 A < 30 A, the fuse will not blow. [3]

Marking notes:

• 1 mark for calculating total current correctly
• 1 mark for comparing with fuse rating
• 1 mark for correct conclusion

(c) Why a fuse is necessary: A fuse protects the circuit and appliances from excessive current. If the current exceeds a safe value (due to a short circuit or overload), the fuse wire melts and breaks the circuit, preventing damage to appliances and reducing the risk of fire. [2]

Marking notes:

• 1 mark for stating it protects against excessive current / overload / short circuit
• 1 mark for explaining that it melts and breaks the circuit

(d) Difference between fuse and circuit breaker:

• A fuse must be replaced after it blows, whereas a circuit breaker can be reset.
• A fuse contains a wire that melts; a circuit breaker uses an electromagnetic mechanism to trip the switch.

(Accept any valid difference) [1]

[Total: 10 marks]

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16.

(a) Kinetic energy gained:

KE = eV = 1.6 × 10⁻¹⁹ × 5000 = 8.0 × 10⁻¹⁶ J [2]

Marking notes:

• 1 mark for using KE = eV
• 1 mark for correct answer (8.0 × 10⁻¹⁶ J)

(b) Speed of electron:

KE = ½mv²

8.0 × 10⁻¹⁶ = ½ × 9.11 × 10⁻³¹ × v²

v² = (2 × 8.0 × 10⁻¹⁶) / (9.11 × 10⁻³¹)

v² = 1.6 × 10⁻¹⁵ / 9.11 × 10⁻³¹

v² = 1.756 × 10¹⁵

v = √(1.756 × 10¹⁵)

v ≈ 4.19 × 10⁷ m/s [3]

Marking notes:

• 1 mark for using KE = ½mv²
• 1 mark for correct substitution and rearrangement
• 1 mark for correct answer (accept 4.2 × 10⁷ m/s)

(c) Magnetic force on electron:

F = Bqv = 0.02 × 1.6 × 10⁻¹⁹ × 4.19 × 10⁷

F = 0.02 × 1.6 × 10⁻¹⁹ × 4.19 × 10⁷

F ≈ 1.34 × 10⁻¹³ N [2]

Marking notes:

• 1 mark for using F = Bqv
• 1 mark for correct answer (accept 1.3 × 10⁻¹³ N to 1.34 × 10⁻¹³ N)

(d) Path of the electron: The electron follows a circular path (or curved path). Since the electron enters the magnetic field perpendicular to the field direction, the magnetic force acts perpendicular to the velocity at all times, providing a centripetal force. This causes the electron to move in a circular path. [2]

Marking notes:

• 1 mark for stating circular / curved path
• 1 mark for explaining that the force is always perpendicular to velocity (centripetal force)

(e) Direction of magnetic force: Using Fleming's left-hand rule (remembering the electron is negatively charged, so conventional current is opposite to electron motion): If the electron moves horizontally and the field is into the page, the force is directed vertically downward (or upward, depending on the specific direction of motion — the key is applying the left-hand rule correctly with reversed current direction for negative charge). [1]

Note: The exact direction depends on the specific orientation given in the figure. The student should apply Fleming's left-hand rule with the conventional current direction opposite to the electron's velocity.

[Total: 10 marks]

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Summary of Marks

Section	Marks
A: Multiple Choice	10
B: Structured Questions	30
C: Application & Data-Based	20
Total	60

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Common Mistakes to Watch For

1. Transformer efficiency: Forgetting to convert percentage to decimal (e.g., using 80 instead of 0.80).
2. Fleming's left-hand rule: Confusing it with the right-hand rule. Remember: Left hand = Force (motor effect), Right hand = Induction.
3. Electron charge: Forgetting that electrons are negatively charged when applying Fleming's left-hand rule — conventional current direction is opposite to electron flow.
4. Units: Omitting units in final answers (A, N, V, T, etc.).
5. Magnetic force formula: Using F = BIL instead of F = BIL sin θ when the wire is not perpendicular to the field.
6. Faraday's vs. Lenz's law: Confusing the two — Faraday's law gives the magnitude of induced e.m.f., Lenz's law gives the direction.