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Secondary 4 Pure Physics Preliminary Examination Paper 3

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TuitionGoWhere Preliminary Examination Practice – Pure Physics (Secondary 4)

TuitionGoWhere Secondary School (AI)


Subject:	Pure Physics
Level:	Secondary 4
Paper:	Paper 2 – Structured & Free Response
Version:	3 of 5
Duration:	1 hour 45 minutes (105 minutes)
Total Marks:	60

Name: ___________________________ Class: ________ Date: _______________

Instructions to Candidates

Write your name, class, and date in the spaces provided above.
Answer ALL questions in the spaces provided.
Write in dark blue or black pen. You may use a pencil for diagrams.
Do not use correction fluid.
The number of marks is shown in brackets [ ] at the end of each question or part-question.
You may use a calculator.
Essential working must be shown for calculation questions. Answers without working may not receive full credit.
The total mark for this paper is 60.

Section A – Short Structured Questions [20 marks]

Answer all questions 1–10. Each question carries 2 marks unless otherwise stated.

1. A step-down transformer has a primary voltage of 240 V and a secondary voltage of 12 V. The secondary coil supplies a current of 4.0 A to a lamp. The transformer is 80% efficient.

(a) Calculate the power output of the transformer. [1]

(b) Hence, or otherwise, calculate the current in the primary coil. [1]

2. State two differences between the magnetic field pattern around a straight current-carrying wire and that around a solenoid. [2]

3. A rectangular coil of 50 turns, each of area 0.020 m², is placed in a uniform magnetic field of flux density 0.40 T. The plane of the coil is perpendicular to the field.

(a) Calculate the magnetic flux through one turn of the coil. [1]

(b) The coil is rotated so that its plane becomes parallel to the field. State the new magnetic flux through the coil. [1]

4. A straight wire carrying a current of 5.0 A is placed perpendicular to a uniform magnetic field. The length of wire in the field is 0.08 m and the magnetic flux density is 0.25 T.

(a) Calculate the force on the wire. [1]

(b) State the direction of the force relative to the current and the magnetic field. [1]

5. A student connects a circuit with a 6.0 V battery of negligible internal resistance and two resistors in series: R₁ = 4.0 Ω and R₂ = 8.0 Ω.

(a) Calculate the total resistance of the circuit. [1]

(b) Calculate the current flowing through R₂. [1]

6. State Lenz's law. [2]

7. A lamp is labelled "12 V, 24 W". Calculate:

(a) the resistance of the lamp when operating normally. [1]

(b) the energy consumed by the lamp in 30 minutes, in kilowatt-hours. [1]

8. The figure shows a wire XY placed between the poles of a magnet. When a current flows through the wire from X to Y, the wire experiences a force directed into the page.

State the direction of the magnetic field between the poles. Explain your reasoning. [2]

9. A household circuit is protected by a 30 A fuse. At 240 V, calculate the maximum total power of appliances that can be safely connected to this circuit. [2]

10. A d.c. motor consists of a coil placed in a magnetic field. State two ways to increase the turning effect (torque) on the coil. [2]

Section B – Structured & Application Questions [25 marks]

Answer all questions 11–15.

11. A transformer is used to step down the mains voltage of 240 V to 12 V for a doorbell.

The primary coil has 2000 turns.

(a) Calculate the number of turns on the secondary coil. [2]

(b) The doorbell draws a current of 0.50 A from the secondary coil. Assuming the transformer is 100% efficient, calculate the current drawn from the mains supply. [2]

(c) In practice, the transformer is not 100% efficient. Explain one source of energy loss in a transformer and how it can be reduced. [1]

12. The figure below shows a simple d.c. motor. The coil ABCD is placed between two magnetic poles. Side AB is on the left and side CD is on the right. Current flows from terminal P through the coil to terminal Q.

(a) State the direction of the force on side AB. Explain your answer using Fleming's left-hand rule. [2]

(b) State the direction of the force on side CD. [1]

13. A student investigates electromagnetic induction using a bar magnet and a solenoid connected to a sensitive centre-zero galvanometer.

(a) Describe what the student should do to induce a current in the solenoid. [1]

(b) The student pushes the north pole of the magnet into the solenoid. The galvanometer needle deflects to the right. When the magnet is pulled out, the needle deflects to the left. Explain these observations using Lenz's law. [3]

14. The figure shows a circuit containing a 12.0 V battery (negligible internal resistance), a 3.0 Ω resistor, and a parallel combination of a 6.0 Ω resistor and a 12.0 Ω resistor.

(a) Calculate the equivalent resistance of the parallel combination. [2]

(b) Calculate the total current drawn from the battery. [2]

15. A power station generates electrical power at 25 000 W at a voltage of 5000 V. The power is transmitted through cables of total resistance 10 Ω to a factory.

(a) Calculate the current in the transmission cables. [1]

(b) Calculate the power lost in the cables. [1]

(d) Explain why electrical power is transmitted at high voltages rather than low voltages. [2]

Section C – Free Response & Data Interpretation [15 marks]

Answer all questions 16–20.

16. A student sets up a circuit to investigate the relationship between the current through a resistor and the potential difference across it. The student varies the variable power supply and records the readings from an ammeter and a voltmeter.

The results are shown in the table below:

Potential Difference / V	0.0	1.0	2.0	3.0	4.0	5.0	6.0
Current / A	0.00	0.10	0.20	0.30	0.40	0.50	0.60

(a) On the grid provided, plot a graph of potential difference (y-axis) against current (x-axis). [3]

(b) Determine the gradient of your graph. State what the gradient represents. [2]

(d) The student replaces the resistor with a filament lamp. Without recalculating, describe how the graph would differ from the one you have drawn. Explain why. [3]

17. Read the following passage and answer the questions that follow.

Electrical energy is generated at power stations and transmitted over long distances to homes and industries. At the power station, a step-up transformer increases the voltage from 11 000 V to 275 000 V for transmission. At the consumer end, step-down transformers reduce the voltage to 240 V for household use. The transmission cables have a total resistance of 50 Ω.

(a) Explain why a step-up transformer is used at the power station. Your answer should include a relevant calculation to support your explanation. [4]

(b) Calculate the current in the transmission cables when the power station delivers 55 MW of power. [2]

(d) State one advantage of using a.c. rather than d.c. for long-distance power transmission. [1]

18. The figure shows a current-carrying wire placed between two magnetic poles. The wire is free to move.

(a) On the diagram, draw an arrow to show the direction of the force on the wire. Label the arrow F. [1]

(b) The wire has a length of 0.10 m within the magnetic field and carries a current of 2.0 A. The magnetic flux density is 0.50 T. Show that the force on the wire is 0.10 N. [2]

(d) Suggest one practical application of the force on a current-carrying conductor in a magnetic field. [1]

19. A student has a 240 V mains supply and wants to connect three identical lamps, each rated "12 V, 6 W", to the supply.

(a) Calculate the resistance of one lamp. [1]

(b) Draw a circuit diagram showing how the three lamps should be connected to a 12 V battery so that each lamp operates at its rated power. [1]

(i) the total resistance of the three lamps in series. [1]

(ii) the current through each lamp. [1]

(iii) the power dissipated by each lamp. Comment on whether the lamps will operate normally. [2]

20. A simple a.c. generator consists of a rectangular coil rotating in a uniform magnetic field.

(a) Explain how an e.m.f. is generated in the coil as it rotates. Your answer should refer to the change in magnetic flux linkage. [3]

(b) The coil has 80 turns, each of area 0.025 m², and rotates at a frequency of 50 Hz in a magnetic field of flux density 0.10 T. Calculate the maximum e.m.f. generated. [3]

(c) On the axes below, sketch a graph of e.m.f. against time for one complete rotation of the coil. Label the axes and indicate the maximum e.m.f. value. [2]

(d) State two ways to increase the frequency of the output e.m.f. [2]

END OF PAPER

Total: 60 marks

Answers

TuitionGoWhere Preliminary Examination Practice – Pure Physics (Secondary 4)

Answer Key – Version 3 of 5

Section A – Short Structured Questions [20 marks]

(a) Power output = V_s × I_s = 12 × 4.0 = 48 W [1]

(b) Efficiency: η = P_out / P_in 0.80 = 48 / P_in P_in = 48 / 0.80 = 60 W

I_p = P_in / V_p = 60 / 240 = 0.25 A [1]

Marking note: Award 1 mark for correct P_out. Award 1 mark for correct I_p with working. Common error: using 80 instead of 0.80 for efficiency — penalise once only.

The field around a straight wire consists of concentric circles; the field inside a solenoid is uniform/straight and parallel. [1]
The field around a straight wire is non-uniform (stronger closer to wire); the field inside a solenoid is uniform (constant strength). [1]

Marking note: Award 1 mark for each valid difference. Accept any two valid differences.

(a) Φ = B × A = 0.40 × 0.020 = 8.0 × 10⁻³ Wb (or 0.0080 Wb) [1]

(b) When the plane is parallel to the field, the normal to the coil is perpendicular to B, so the flux is 0 Wb [1]

Marking note: Award 1 mark for correct calculation with unit. Award 1 mark for zero flux with correct reasoning.

(a) F = BIL = 0.25 × 5.0 × 0.08 = 0.10 N [1]

(b) The force is perpendicular to both the current direction and the magnetic field direction. [1]

Marking note: Award 1 mark for correct magnitude with unit. Award 1 mark for stating perpendicular to both.

(a) R_total = R₁ + R₂ = 4.0 + 8.0 = 12.0 Ω [1]

(b) I = V / R_total = 6.0 / 12.0 = 0.50 A [1]

Marking note: In a series circuit, the current is the same through all components. Award 1 mark each for correct answers.

Lenz's law states that the direction of the induced current is such that it opposes the change in magnetic flux (or the motion) that produced it. [2]

Marking note: Award 2 marks for a complete statement. Award 1 mark for a partial statement (e.g., "opposes the change" without reference to flux/motion).

(a) P = V² / R, so R = V² / P = 12² / 24 = 144 / 24 = 6.0 Ω [1]

(b) E = P × t = 24 × (30 × 60) = 24 × 1800 = 43 200 J

In kWh: E = 0.024 kW × 0.5 h = 0.012 kWh [1]

Marking note: Award 1 mark for correct resistance. Award 1 mark for correct energy in kWh. Accept 43 200 J as alternative for part (b) if unit is correct.

The magnetic field is directed from top to bottom (or from N to S, vertically downward). [1]

By Fleming's left-hand rule: the force is into the page, the current is from X to Y (left to right), so the field must be vertically downward to produce a force into the page. [1]

Marking note: Award 1 mark for correct direction. Award 1 mark for correct reasoning using Fleming's left-hand rule.

P = V × I = 240 × 30 = 7200 W (or 7.2 kW) [2]

Marking note: Award 2 marks for correct answer with unit. Award 1 mark for correct method with arithmetic error.

10.

Any two of the following:

Increase the current in the coil
Increase the number of turns in the coil
Increase the magnetic field strength (use stronger magnets)
Increase the area of the coil

[1 mark each, total 2 marks]

Marking note: Award 1 mark for each valid method, maximum 2 marks.

Section B – Structured & Application Questions [25 marks]

11.

(a) V_p / V_s = N_p / N_s 240 / 12 = 2000 / N_s N_s = 2000 × 12 / 240 = 100 turns [2]

(b) For 100% efficiency: V_p × I_p = V_s × I_s 240 × I_p = 12 × 0.50 I_p = 6.0 / 240 = 0.025 A [2]

(c) Energy loss due to resistance (heating) in the copper windings. This can be reduced by using thicker wire (lower resistance) or by using wire with lower resistivity. [1]

Alternative acceptable answer: Eddy current losses in the iron core — reduced by using a laminated core.

Marking note: Award 2 marks for correct N_s with working. Award 2 marks for correct I_p with working. Award 1 mark for identifying a valid loss and a valid reduction method.

12.

(a) Using Fleming's left-hand rule: the magnetic field is from N to S (left to right), current in AB flows from A to B (downwards), so the force on AB is into the page (or towards the back). [2]

(b) The force on side CD is out of the page (or towards the front). [1]

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

Marking note: Award 2 marks for correct direction with correct application of Fleming's left-hand rule. Award 1 mark for correct direction on CD. Award 2 marks for explaining the function of the commutator.

13.

(a) The student should move the magnet relative to the solenoid (e.g., push the magnet into the solenoid or pull it out). [1]

(b) When the north pole is pushed in, the magnetic flux through the solenoid increases. By Lenz's law, the induced current flows in a direction to oppose this increase — it creates a north pole at the end nearest the approaching magnet to repel it. This causes the galvanometer to deflect to the right. [1.5]

When the magnet is pulled out, the flux decreases. The induced current flows to oppose this decrease — it creates a south pole at the end nearest the receding magnet to attract it. The current is in the opposite direction, so the galvanometer deflects to the left. [1.5]

Move the magnet faster
Use a stronger magnet
Increase the number of turns on the solenoid

[0.5 mark each, total 1 mark]

Marking note: Award 1 mark for correct action. Award 3 marks for explanation using Lenz's law (1.5 for each case). Award 0.5 for each valid method to increase e.m.f.

14.

(a) 1/R_parallel = 1/6.0 + 1/12.0 = 2/12 + 1/12 = 3/12 = 1/4 R_parallel = 4.0 Ω [2]

(b) R_total = 3.0 + 4.0 = 7.0 Ω I_total = V / R_total = 12.0 / 7.0 = 1.71 A (or 12/7 A) [2]

I_6Ω = V_parallel / 6.0 = (48/7) / 6.0 = 48/42 = 1.14 A (or 8/7 A) [1]

Marking note: Award 2 marks for correct R_parallel. Award 2 marks for correct I_total. Award 1 mark for correct current through 6Ω resistor.

15.

(a) I = P / V = 25000 / 5000 = 5.0 A [1]

(b) P_loss = I²R = 5.0² × 10 = 25 × 10 = 250 W [1]

(d) At high voltage, the current in the cables is lower for the same power transmitted (since P = VI). [1]

Since power loss = I²R, a lower current means much less power is wasted as heat in the cables. [1]

Marking note: Award 1 mark each for parts (a)–(c). Award 2 marks for explanation in part (d): 1 mark for lower current, 1 mark for reduced power loss.

Section C – Free Response & Data Interpretation [15 marks]

16.

(a) Graph: Straight line through the origin passing through points (0, 0), (0.10, 1.0), (0.20, 2.0), (0.30, 3.0), (0.40, 4.0), (0.50, 5.0), (0.60, 6.0). [3]

Marking scheme:

[1] Correctly labelled axes with units (V on y-axis, I on x-axis)
[1] Correct scale and all points plotted correctly
[1] Straight line of best fit drawn through the origin

(b) Gradient = rise / run = 6.0 / 0.60 = 10 [1]

The gradient represents the resistance of the resistor (R = V/I). [1]

This is Ohm's law. [1]

(d) The graph for a filament lamp would be a curve (not a straight line), starting steep and becoming less steep as current increases. [1]

This is because as the current increases, the filament heats up, causing its resistance to increase. [1]

Since R = V/I and R increases, the ratio V/I increases, so the graph curves upward (gradient increases). [1]

Marking note: Award 3 marks for graph. Award 2 for gradient and meaning. Award 2 for relationship and law name. Award 3 for description and explanation of lamp graph.

17.

(a) A step-up transformer increases the voltage, which reduces the current in the transmission cables for the same power. [1]

Since P = VI, for a given power P, if V increases, I decreases. [1]

Power loss in cables = I²R, so a lower current means significantly less power is lost as heat. [1]

Example calculation: If P = 275 MW at 275 000 V: I = P/V = 275 × 10⁶ / 275 000 = 1000 A

If transmitted at 11 000 V instead: I = 275 × 10⁶ / 11 000 = 25 000 A

Power loss at 275 kV: I²R = 1000² × 50 = 50 MW Power loss at 11 kV: I²R = 25000² × 50 = 31 250 MW (far exceeds generated power) [1]

(b) I = P / V = 55 × 10⁶ / 275 000 = 200 A [2]

Percentage loss = (P_loss / P_total) × 100 = (2.0 / 55) × 100 = 3.64% [2]

Marking note for (c): Award 1 mark for correct P_loss, 2 marks for correct percentage.

(d) The voltage of a.c. can be easily stepped up or stepped down using transformers, which is not possible (or much more difficult) with d.c. [1]

Marking note: Award 4 marks for part (a): 1 mark for reduced current, 1 mark for I²R relationship, 1 mark for explanation, 1 mark for supporting calculation. Award 2 marks for part (b). Award 3 marks for part (c). Award 1 mark for part (d).

18.

(a) The force arrow should be drawn perpendicular to both the current and the magnetic field, following Fleming's left-hand rule. Direction depends on the diagram, but the arrow should be correctly oriented. [1]

(b) F = BIL = 0.50 × 2.0 × 0.10 = 0.10 N ✓ [2]

(d) Any one of: electric motor, loudspeaker, galvanometer, electromagnetic relay [1]

Marking note: Award 1 mark for correct arrow direction. Award 2 marks for correct working and answer. Award 2 marks for correct new force. Award 1 mark for valid application.

19.

(a) R = V² / P = 12² / 6 = 144 / 6 = 24 Ω [1]

(b) The three lamps should be connected in parallel across the 12 V battery. [1]

(Circuit diagram: battery connected to three parallel branches, each containing one lamp.)

(c)

(i) R_total = 24 + 24 + 24 = 72 Ω [1]

(ii) I = V / R_total = 240 / 72 = 3.33 A (or 10/3 A) [1]

(iii) P_each = I² × R = (10/3)² × 24 = (100/9) × 24 = 2400/9 = 266.7 W [1]

Comment: Each lamp dissipates 266.7 W, which is far greater than its rated power of 6 W. The lamps will burn out immediately (or be destroyed). [1]

Marking note: Award 1 mark for correct resistance. Award 1 mark for correct parallel circuit diagram. Award 1 mark each for (i) and (ii). Award 2 marks for (iii): 1 for calculation, 1 for comment.

20.

(a) As the coil rotates, the magnetic flux linkage through the coil changes continuously. [1]

When the coil is perpendicular to the field, flux linkage is maximum; when parallel, it is zero. [1]

By Faraday's law, the changing flux linkage induces an e.m.f. in the coil. [1]

(b) Maximum e.m.f.: ε₀ = B × A × N × ω

ω = 2πf = 2π × 50 = 100π rad/s

ε₀ = 0.10 × 0.025 × 80 × 100π = 0.10 × 0.025 × 80 × 314.16 = 0.10 × 0.025 × 25132.8 = 0.10 × 628.32 = 62.8 V (or 20π V) [3]

Marking note: Award 1 mark for correct formula, 1 mark for correct ω, 1 mark for correct final answer.

(c) Graph: Sinusoidal wave starting at zero, peaking at 62.8 V at T/4, returning to zero at T/2, going to −62.8 V at 3T/4, and returning to zero at T. [2]

Marking scheme:

[1] Correct sinusoidal shape
[1] Correct amplitude (62.8 V) and period labelled

(d) Any two of:

Increase the speed of rotation (frequency)
Increase the number of rotations per second

[1 mark each, total 2 marks]

Marking note: Award 3 marks for part (a). Award 3 marks for part (b). Award 2 marks for part (c). Award 2 marks for part (d).

END OF ANSWER KEY

Total: 60 marks