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Secondary 4 Pure Physics Preliminary Examination Paper 5

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TuitionGoWhere Practice Paper - Pure Physics Secondary 4

TuitionGoWhere Secondary School (AI)

PRELIMINARY EXAMINATION — VERSION 5


Subject:	Pure Physics
Level:	Secondary 4
Paper:	Paper 2 (Structured & Free Response)
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.
Show all working clearly. Marks are awarded for correct working even if the final answer is wrong.
The number of marks for each question is shown in brackets [ ].
You may use a calculator.
Essential working must be shown for full credit on calculation questions.

Section A: Multiple Choice [10 marks]

Questions 1–5

Each question is worth 2 marks. Choose the most accurate answer and write the letter in the space provided.

1. A step-down transformer has 2000 turns on the primary coil and 100 turns on the secondary coil. The primary voltage is 240 V. What is the secondary voltage?

A) 4.8 V B) 12 V C) 48 V D) 4800 V

Answer: ______________

2. A 6 Ω resistor and a 3 Ω resistor are connected in parallel across a 12 V battery. What is the total current drawn from the battery?

A) 2 A B) 4 A C) 6 A D) 8 A

Answer: ______________

3. A straight wire carries a current into the page (×). What is the direction of the magnetic field at a point directly above the wire?

A) Upward B) Downward C) To the left D) To the right

Answer: ______________

4. A transformer has an efficiency of 80%. The primary voltage is 240 V and the primary current is 0.5 A. If the secondary voltage is 48 V, what is the secondary current?

A) 1.0 A B) 1.5 A C) 2.0 A D) 2.5 A

Answer: ______________

5. A positively charged particle enters a uniform magnetic field directed into the page. The particle's initial velocity is to the right. In which direction is the initial magnetic force on the particle?

A) Into the page B) Out of the page C) Upward D) Downward

Answer: ______________

Section B: Structured Questions [30 marks]

Questions 6–13

6. [4 marks]

The diagram below shows a simple circuit with a 24 V battery (negligible internal resistance), a 4 Ω resistor (R₁), and an 8 Ω resistor (R₂) connected in series.

+24V ---[4Ω]---[8Ω]--- (back to battery)

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

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

7. [4 marks]

A transformer is used to step down a voltage from 2400 V to 240 V for household use. The primary coil carries a current of 0.8 A. The transformer is 90% efficient.

(a) State the ratio of the number of turns on the primary coil to the number of turns on the secondary coil. [1 mark]

(b) Calculate the power input to the primary coil. [1 mark]

8. [3 marks]

The figure shows a rectangular coil placed between the poles of a magnet, with the current flowing as indicated.

    N                    S
    |                    |
    |   +-------+        |
    |   |  coil |        |
    |   +-------+        |
    |                    |
    ↑ B-field direction →

(a) Using Fleming's left-hand rule, state the direction of the force on side AB of the coil (the side nearest the North pole). [1 mark]

(b) Explain why the coil rotates when current flows through it. [2 marks]

9. [4 marks]

A household circuit is protected by a 30 A fuse. The following appliances are connected in parallel to the 240 V mains supply:

Appliance	Power Rating
Kettle	2400 W
Iron	1200 W
Hair Dryer	1000 W
Microwave	800 W

(a) Calculate the current drawn by the kettle. [1 mark]

(b) Calculate the total current drawn when all four appliances are switched on simultaneously. [2 marks]

10. [3 marks]

A wire of length 0.50 m carries a current of 4.0 A and is placed perpendicular to a uniform magnetic field of flux density 0.08 T.

(a) State the equation for the force on a current-carrying conductor in a magnetic field. [1 mark]

(b) Calculate the magnitude of the force on the wire. [2 marks]

11. [4 marks]

The diagram shows a solenoid connected to a sensitive centre-zero galvanometer. A bar magnet is moved towards the solenoid as shown.

  N → → →  [solenoid]  ← galvanometer

(a) State Lenz's law. [1 mark]

(b) Explain, using Lenz's law, the direction of the induced current through the galvanometer as the magnet approaches. [3 marks]

12. [4 mains]

A student sets up a circuit to investigate the relationship between the potential difference (V) across a fixed resistor and the current (I) through it. The results are shown in the table below:

V / V	0.0	1.0	2.0	3.0	4.0	5.0	6.0
I / A	0.00	0.20	0.40	0.60	0.80	1.00	1.20

(a) Plot a graph of V (y-axis) against I (x-axis) on the grid provided. [2 marks]

(b) Determine the gradient of the line. State what the gradient represents. [2 marks]

13. [4 marks]

An electron travelling horizontally at 3.0 × 10⁶ m/s enters a uniform magnetic field of flux density 0.02 T directed vertically upward. The charge of an electron is −1.6 × 10⁻¹⁹ C and its mass is 9.1 × 10⁻³¹ kg.

(a) Calculate the magnetic force on the electron. **[2 marks]]

(b) Describe the path of the electron in the magnetic field. Explain your answer. [2 marks]

Section C: Free Response [20 marks]

Questions 14–15

14. [10 marks]

A power station generates electricity at 5000 V. The electricity is transmitted through cables to a town 20 km away. The total resistance of the transmission cables is 8 Ω. The power station delivers 200 kW of power.

(a) Calculate the current in the transmission cables. [2 marks]

(b) Calculate the power lost as heat in the transmission cables. [2 marks]

(d) A step-up transformer at the power station increases the voltage to 250,000 V before transmission. Explain, with reference to your calculations in (a)–(c), why transmitting electricity at high voltages reduces energy loss. [4 marks]

15. [10 marks]

The figure shows a simple d.c. motor consisting of a single rectangular coil ABCD placed between the poles of a permanent magnet. The coil has 50 turns, each of area 0.02 m². The magnetic flux density is 0.5 T. A current of 2.0 A flows through the coil.

        N                 S
        |                 |
   A ---|--- B            |
   |    |    |            |
   D ---|--- C            |
        |                 |
   (commutator at left)

(a) Calculate the maximum torque on the coil. [3 marks]

(b) Explain the function of the split-ring commutator in a d.c. motor. [2 marks]

(c) The motor is used to lift a load. When the motor is running at full speed, the back e.m.f. induced in the coil is 8.0 V. The supply voltage is 12.0 V and the resistance of the coil is 1.0 Ω.

(i) Calculate the current through the coil when the motor is running at full speed. [2 marks]

(ii) Explain why the current decreases when the motor speeds up from rest. [3 marks]

END OF PAPER

Answers

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