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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 2026
Version 5 of 5

Subject: Pure Physics
Level: Secondary 4
Paper: 2 (Structured Questions)
Duration: 1 hour 15 minutes
Total Marks: 60

Name: __________________________
Class: __________________________
Date: __________________________

INSTRUCTIONS TO CANDIDATES

Write your name, class, and date in the spaces provided.
Answer all questions.
Write your answers in the spaces provided on the question paper.
The number of marks is given in brackets [ ] at the end of each question or part question.
You may lose marks if you do not show your working or if you do not use appropriate units.
Take the acceleration of free fall, $g = 10 \text{ m/s}^2$ .

Section A

Answer all questions in this section.

1. A student investigates the properties of electric charges using two suspended polystyrene balls coated with conducting paint. (a) Ball A is positively charged. Ball B is negatively charged. State what happens to the balls when they are brought close to each other but do not touch.

   _________________________________________________________________________ [1]

(b) The student touches Ball A with a neutral metal rod connected to the earth. Explain, in terms of electron movement, what happens to the charge on Ball A.

   _________________________________________________________________________
   _________________________________________________________________________ [2]

2. Fig. 2.1 shows a simple circuit containing a battery, a switch, a resistor, and a lamp.

(a) Define the term electromotive force (e.m.f.).

   _________________________________________________________________________
   _________________________________________________________________________ [2]

(b) The battery has an e.m.f. of 12 V. When the switch is closed, the potential difference across the lamp is 8.0 V. Calculate the potential difference across the resistor.

   Potential difference = __________________________ V [1]

3. A transformer is used to step down the voltage from 240 V to 12 V for a low-voltage lamp. The primary coil has 1000 turns. (a) Calculate the number of turns on the secondary coil.

   Number of turns = __________________________ [2]

(b) The lamp operates at a power of 24 W. Assuming the transformer is 100% efficient, calculate the current in the primary coil.

   Current = __________________________ A [2]

4. Fig. 4.1 shows a current-carrying wire placed between the poles of a U-shaped magnet. The current flows into the page.

(a) State the direction of the force acting on the wire.

   Direction: __________________________ [1]

(b) State two changes that would increase the magnitude of this force. 1. _____________________________________________________________________ 2. _____________________________________________________________________ [2]

5. A household electrical circuit includes a fuse and an earth wire for safety. (a) State the function of the live wire.

   _________________________________________________________________________ [1]

(b) Explain why the fuse is connected to the live wire and not the neutral wire.

   _________________________________________________________________________
   _________________________________________________________________________
   _________________________________________________________________________ [2]

6. A coil of wire is connected to a sensitive centre-zero galvanometer. A bar magnet is pushed into the coil. (a) State what is observed on the galvanometer.

   _________________________________________________________________________ [1]

(b) The magnet is then held stationary inside the coil. State and explain what is observed on the galvanometer.

   Observation: ___________________________________________________________
   Explanation: ___________________________________________________________
   _________________________________________________________________________ [2]

7. Two resistors, $R_1 = 4.0 \, \Omega$ and $R_2 = 6.0 \, \Omega$ , are connected in series to a 10 V battery. (a) Calculate the total resistance of the circuit.

   Total resistance = __________________________ $\Omega$ [1]

(b) Calculate the current flowing through the circuit.

   Current = __________________________ A [2]

8. Fig. 8.1 shows the input and output voltage waveforms of an alternating current (a.c.) generator. (a) State one advantage of using a.c. over direct current (d.c.) for long-distance power transmission.

   _________________________________________________________________________ [1]

(b) The frequency of the a.c. supply is 50 Hz. Calculate the time period of one complete cycle.

   Time period = __________________________ s [1]

9. A student plots a graph of current ( $I$ ) against potential difference ( $V$ ) for a filament lamp. (a) Sketch the shape of the graph on the axes below.

   ```
   I / A
   |
   |
   |
   |
   |
   |________________________ V / V
   ```
   [2]

(b) Explain why the graph has this shape.

   _________________________________________________________________________
   _________________________________________________________________________
   _________________________________________________________________________ [2]

10. A charged sphere is placed in a uniform electric field. (a) Define electric field strength.

    _________________________________________________________________________
    _________________________________________________________________________ [2]

(b) Draw the electric field lines around an isolated positive point charge.
    
    ```
          .
       .     .
     .    +    .
       .     .
          .
    ```
    [1]

Section B

Answer all questions in this section.

11. Fig. 11.1 shows a simple d.c. motor.

(a) Explain why the coil rotates when a current flows through it.

   _________________________________________________________________________
   _________________________________________________________________________
   _________________________________________________________________________
   _________________________________________________________________________ [3]

(b) State the function of the split-ring commutator in the motor.

   _________________________________________________________________________
   _________________________________________________________________________ [2]

(c) Suggest two ways to increase the speed of rotation of the motor. 1. _____________________________________________________________________ 2. _____________________________________________________________________ [2]

12. A transformer is used in a laptop charger. The input voltage is 230 V a.c. and the output voltage is 19 V d.c. (after rectification). The transformer itself steps down the a.c. voltage. (a) The primary coil has 5000 turns. Calculate the number of turns on the secondary coil.

   Number of turns = __________________________ [2]

(b) The laptop draws a current of 3.0 A at 19 V. The transformer is 90% efficient. Calculate the current in the primary coil.

   Current = __________________________ A [3]

   Soft iron: ______________________________________________________________
   _________________________________________________________________________
   Laminated: _____________________________________________________________
   _________________________________________________________________________ [2]

13. Fig. 13.1 shows a circuit with three resistors. $R_1 = 2.0 \, \Omega$ , $R_2 = 3.0 \, \Omega$ , and $R_3 = 6.0 \, \Omega$ . $R_2$ and $R_3$ are connected in parallel, and this combination is connected in series with $R_1$ . The battery e.m.f. is 12 V and has negligible internal resistance.

(a) Calculate the combined resistance of $R_2$ and $R_3$ .

   Combined resistance = __________________________ $\Omega$ [2]

(b) Calculate the total resistance of the circuit.

   Total resistance = __________________________ $\Omega$ [1]

   Total current = __________________________ A [2]

(d) Calculate the potential difference across $R_1$ .

   Potential difference = __________________________ V [1]

14. A student investigates the relationship between the length of a wire and its resistance. (a) State the relationship between the length of a wire and its resistance, assuming constant cross-sectional area and temperature.

   _________________________________________________________________________ [1]

(b) The student measures the resistance of a 1.0 m length of wire as 5.0 $\Omega$ . Calculate the resistance of a 0.4 m length of the same wire.

   Resistance = __________________________ $\Omega$ [1]

   _________________________________________________________________________
   _________________________________________________________________________
   _________________________________________________________________________ [2]

15. Fig. 15.1 shows a cathode ray oscilloscope (CRO) screen displaying a waveform. The time-base is set to 2 ms/cm and the Y-gain is set to 5 V/cm.

(a) The peak of the waveform is 3 cm from the center line. Calculate the peak voltage.

   Peak voltage = __________________________ V [1]

(b) One complete cycle occupies 4 cm horizontally. Calculate the frequency of the signal.

   Frequency = __________________________ Hz [2]

Section C

Answer all questions in this section.

16. A high-voltage transmission line carries a current of 500 A at a voltage of 400,000 V. (a) Calculate the power transmitted.

   Power = __________________________ W [2]

(b) The total resistance of the transmission lines is 10 $\Omega$ . Calculate the power loss in the transmission lines due to heating.

   Power loss = __________________________ W [2]

   _________________________________________________________________________
   _________________________________________________________________________
   _________________________________________________________________________
   _________________________________________________________________________ [2]

17. Fig. 17.1 shows a circuit breaker connected in a household circuit.

(a) Describe how a circuit breaker operates when the current exceeds the rated value.

   _________________________________________________________________________
   _________________________________________________________________________
   _________________________________________________________________________
   _________________________________________________________________________ [3]

(b) State one advantage of a circuit breaker over a fuse.

   _________________________________________________________________________ [1]

18. A coil of wire is rotated in a magnetic field to generate electricity. (a) State Faraday’s Law of Electromagnetic Induction.

   _________________________________________________________________________
   _________________________________________________________________________ [2]

(b) Explain why the induced e.m.f. is zero when the plane of the coil is perpendicular to the magnetic field lines.

   _________________________________________________________________________
   _________________________________________________________________________
   _________________________________________________________________________ [2]

19. Fig. 19.1 shows a potential divider circuit. The input voltage is 12 V. $R_1$ is a fixed resistor of 100 $\Omega$ . $R_2$ is a variable resistor.

(a) Calculate the output voltage when $R_2$ is set to 200 $\Omega$ .

   Output voltage = __________________________ V [2]

(b) Explain what happens to the output voltage if the resistance of $R_2$ is increased.

   _________________________________________________________________________
   _________________________________________________________________________ [2]

20. A student wants to determine the resistance of an unknown resistor $R_x$ . (a) Draw a circuit diagram showing how you would connect a battery, an ammeter, a voltmeter, and the resistor $R_x$ to measure its resistance.

[3]

(b) State the formula used to calculate the resistance from the readings.

   _________________________________________________________________________ [1]

   _________________________________________________________________________
   _________________________________________________________________________ [1]

END OF PAPER

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Answers

TuitionGoWhere Practice Paper - Pure Physics Secondary 4

ANSWER KEY & MARKING SCHEME
PRELIMINARY EXAMINATION 2026
Version 5 of 5

Subject: Pure Physics
Level: Secondary 4
Paper: 2 (Structured Questions)

Section A

1. (a) They attract each other. [1] (b) Electrons flow from the earth to Ball A [1]. This neutralizes the positive charge (or reduces the positive charge) [1].

2. (a) E.m.f. is the energy supplied by the source per unit charge passing through it [1]. It converts chemical energy (or other forms) into electrical energy [1]. (b) $12 - 8 = 4.0$ V [1].

3. (a) $\frac{N_s}{N_p} = \frac{V_s}{V_p}$ [1] $N_s = 1000 \times \frac{12}{240} = 50$ turns [1]. (b) $P_{in} = P_{out} = 24$ W [1] $I_p = \frac{P}{V_p} = \frac{24}{240} = 0.1$ A [1].

4. (a) Upwards (or towards the top of the page) [1]. (b) 1. Increase the current [1]. 2. Increase the magnetic field strength (or use a stronger magnet) [1]. (Accept: Increase length of wire in field)

5. (a) Carries current from the supply to the appliance [1]. (b) If a fault occurs, the live wire carries the high potential [1]. The fuse melts/breaks the circuit, disconnecting the appliance from the high voltage [1]. If the fuse were on the neutral, the appliance would still be live even if the fuse blew, posing a shock hazard [1]. (Max 2 marks)

6. (a) The needle deflects (momentarily) [1]. (b) Observation: The needle returns to zero / no deflection [1]. Explanation: There is no change in magnetic flux linkage when the magnet is stationary [1]. Hence, no e.m.f. is induced [1].

7. (a) $R_T = R_1 + R_2 = 4.0 + 6.0 = 10.0 \, \Omega$ [1]. (b) $I = \frac{V}{R} = \frac{10}{10} = 1.0$ A [2]. (1 mark for formula/substitution, 1 mark for answer)

8. (a) A.C. voltage can be easily stepped up or down using transformers [1], allowing for high voltage transmission which reduces energy loss. (b) $T = \frac{1}{f} = \frac{1}{50} = 0.02$ s [1].

9. (a) Curve starting from origin with decreasing gradient (concave down) [2]. (1 mark for correct shape, 1 mark for passing through origin) (b) As voltage/current increases, the temperature of the filament increases [1]. The resistance of the metal increases with temperature [1]. Therefore, the ratio $V/I$ increases, causing the gradient ( $I/V$ ) to decrease [1]. (Max 2 marks)

10. (a) Electric field strength is the force experienced per unit positive charge placed at that point [2]. (b) Radial lines pointing outwards from the charge [1].

Section B

11. (a) When current flows through the coil, it creates a magnetic field [1]. This interacts with the external magnetic field from the magnets [1]. Forces act on opposite sides of the coil in opposite directions (Fleming's Left Hand Rule), creating a turning effect/moment [1]. (b) It reverses the direction of the current in the coil every half rotation [1]. This ensures that the forces on the coil always act in the same rotational direction, allowing continuous rotation [1]. (c) 1. Increase the current [1]. 2. Increase the strength of the magnetic field [1]. (Accept: Increase number of turns on coil)

12. (a) $\frac{N_s}{N_p} = \frac{V_s}{V_p}$ [1] $N_s = 5000 \times \frac{19}{230} \approx 413$ turns [1]. (Accept 413.04) (b) Output Power $P_{out} = V_s I_s = 19 \times 3.0 = 57$ W [1]. Efficiency $\eta = \frac{P_{out}}{P_{in}} \Rightarrow P_{in} = \frac{P_{out}}{\eta}$ [1] $P_{in} = \frac{57}{0.90} = 63.33$ W [1]. $I_p = \frac{P_{in}}{V_p} = \frac{63.33}{230} \approx 0.275$ A [1]. (Accept 0.27 - 0.28 A) (Note: If student uses $V_p I_p = V_s I_s$ directly without efficiency, max 1 mark for method) (c) Soft iron: It is easily magnetized and demagnetized, reducing energy loss due to hysteresis [1]. Laminated: To reduce eddy currents in the core, which reduces heating/energy loss [1].

13. (a) $\frac{1}{R_{23}} = \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{3} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}$ [1] $R_{23} = 2.0 \, \Omega$ [1]. (b) $R_T = R_1 + R_{23} = 2.0 + 2.0 = 4.0 \, \Omega$ [1]. (c) $I_T = \frac{V}{R_T} = \frac{12}{4.0} = 3.0$ A [2]. (1 mark for formula/sub, 1 mark for ans) (d) $V_1 = I_T \times R_1 = 3.0 \times 2.0 = 6.0$ V [1].

14. (a) Resistance is directly proportional to length [1]. (b) $R = 5.0 \times \frac{0.4}{1.0} = 2.0 \, \Omega$ [1]. (c) As temperature increases, the metal ions/atoms vibrate with greater amplitude [1]. This increases the frequency of collisions between free electrons and the ions [1], impeding the flow of electrons (increasing resistance) [1]. (Max 2 marks)

15. (a) Peak Voltage = $3 \text{ cm} \times 5 \text{ V/cm} = 15$ V [1]. (b) Time for one cycle = $4 \text{ cm} \times 2 \text{ ms/cm} = 8 \text{ ms} = 0.008$ s [1]. Frequency $f = \frac{1}{T} = \frac{1}{0.008} = 125$ Hz [1].

Section C

16. (a) $P = VI = 400,000 \times 500 = 200,000,000$ W or $2 \times 10^8$ W [2]. (b) $P_{loss} = I^2 R = (500)^2 \times 10 = 250,000 \times 10 = 2,500,000$ W or $2.5 \times 10^6$ W [2]. (c) High voltage allows for lower current for the same power transmitted ( $P=VI$ ) [1]. Power loss is proportional to the square of the current ( $P=I^2R$ ) [1]. Therefore, lower current significantly reduces energy loss as heat in the cables [1]. (Max 2 marks)

17. (a) When current exceeds the rated value, the magnetic field in the electromagnet becomes strong enough [1] to attract the iron armature [1]. This pulls the contacts apart, breaking the circuit [1]. (b) It can be reset immediately after the fault is cleared (no need to replace a fuse) [1]. OR It responds faster to overcurrent.

18. (a) The induced e.m.f. is directly proportional to the rate of change of magnetic flux linkage [2]. (b) When the plane of the coil is perpendicular to the field, the sides of the coil are moving parallel to the magnetic field lines [1]. Therefore, they do not cut the magnetic field lines [1]. Hence, the rate of change of flux linkage is zero, and induced e.m.f. is zero [1]. (Max 2 marks)

19. (a) $V_{out} = V_{in} \times \frac{R_2}{R_1 + R_2}$ [1] $V_{out} = 12 \times \frac{200}{100 + 200} = 12 \times \frac{200}{300} = 12 \times \frac{2}{3} = 8.0$ V [1]. (b) If $R_2$ increases, the fraction $\frac{R_2}{R_1 + R_2}$ increases [1]. Therefore, the output voltage increases [1].

20. (a) Diagram must show: - Battery/Power supply [1] - Ammeter in series with resistor $R_x$ [1] - Voltmeter in parallel with resistor $R_x$ [1] - Variable resistor in series (optional for diagram marks but required for c) (b) $R = \frac{V}{I}$ [1]. (c) To vary the current and voltage to take multiple readings for an average / to prevent overheating of the resistor [1].

END OF MARKING SCHEME