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Secondary 4 Pure Physics Preliminary Examination Paper 3

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Questions

TuitionGoWhere Practice Paper - Pure Physics Secondary 4

TuitionGoWhere Secondary School (AI)
Subject: Pure Physics
Level: Secondary 4
Paper: PRELIM Version 3
Duration: 1 hour 45 minutes
Total Marks: 80

Name: ________________________
Class: ________________________
Date: ________________________

Instructions to Candidates

Write your name, class, and date in the spaces provided above.
Answer all questions in the spaces provided.
The number of marks is given in brackets [ ] at the end of each question or part question.
You may use a calculator.
Where appropriate, take $g = 10 \text{ m/s}^2$ .
Show all working clearly. Omission of essential working will result in loss of marks.
The total marks for this paper is 80.

Section A: Multiple Choice Questions [20 marks]

Answer all questions. For each question, choose the correct option and write the letter (A, B, C, or D) in the box provided.

Question 1 [1 mark]

A transformer has 400 turns on its primary coil and 100 turns on its secondary coil. The primary coil is connected to a 240 V a.c. supply. What is the voltage across the secondary coil?

☐ A. 60 V
☐ B. 960 V
☐ C. 240 V
☐ D. 120 V

Question 2 [1 mark]

An electric kettle rated 2.0 kW, 240 V is used for 15 minutes. What is the energy consumed in kWh?

☐ A. 0.5 kWh
☐ B. 30 kWh
☐ C. 0.05 kWh
☐ D. 5.0 kWh

Question 3 [1 mark]

A straight wire carrying a current of 5.0 A is placed perpendicular to a uniform magnetic field of flux density 0.2 T. The length of the wire in the field is 0.1 m. What is the magnitude of the force on the wire?

☐ A. 0.01 N
☐ B. 0.1 N
☐ C. 1.0 N
☐ D. 10 N

Question 4 [1 mark]

Which of the following statements about electromagnetic induction is correct?

☐ A. An e.m.f. is induced only when a magnet moves towards a coil.
☐ B. The magnitude of induced e.m.f. is independent of the rate of change of magnetic flux.
☐ C. An e.m.f. is induced in a coil when there is a change in magnetic flux linking the coil.
☐ D. The direction of induced current is always the same as the direction of the changing magnetic field.

Question 5 [1 mark]

A cathode-ray oscilloscope (CRO) displays a sinusoidal waveform with a peak voltage of 12 V. The Y-gain is set to 2 V/div and the time-base is set to 5 ms/div. The waveform occupies 3 divisions vertically. What is the peak-to-peak voltage of the signal?

☐ A. 6 V
☐ B. 12 V
☐ C. 24 V
☐ D. 36 V

Question 6 [1 mark]

In a household circuit, a 3-pin plug is wired correctly. Which wire is connected to the fuse?

☐ A. Live wire
☐ B. Neutral wire
☐ C. Earth wire
☐ D. Both live and neutral wires

Question 7 [1 mark]

A coil of wire rotates in a uniform magnetic field. The induced e.m.f. is maximum when:

☐ A. The plane of the coil is parallel to the magnetic field.
☐ B. The plane of the coil is perpendicular to the magnetic field.
☐ C. The coil is stationary.
☐ D. The magnetic field is zero.

Question 8 [1 mark]

The resistance of a filament lamp increases as the potential difference across it increases. Which statement explains this?

☐ A. The filament expands and becomes longer.
☐ B. The temperature of the filament increases, increasing the lattice vibrations.
☐ C. More charge carriers are released at higher voltages.
☐ D. The cross-sectional area of the filament decreases.

Question 9 [1 mark]

A step-down transformer has 800 turns on the primary coil and 200 turns on the secondary coil. The primary current is 0.5 A. Assuming 100% efficiency, what is the secondary current?

☐ A. 0.125 A
☐ B. 0.5 A
☐ C. 2.0 A
☐ D. 4.0 A

Question 10 [1 mark]

Which of the following devices uses the principle of electromagnetic induction?

☐ A. Electric bell
☐ B. Moving-coil loudspeaker
☐ C. A.C. generator
☐ D. Relay

Question 11 [1 mark]

A wire of length 2.0 m and cross-sectional area $1.0 \times 10^{-6} \text{ m}^2$ has a resistance of 4.0 $\Omega$ . What is the resistivity of the material?

☐ A. $2.0 \times 10^{-6} \Omega \text{m}$
☐ B. $8.0 \times 10^{-6} \Omega \text{m}$
☐ C. $2.0 \times 10^{-7} \Omega \text{m}$
☐ D. $8.0 \times 10^{-7} \Omega \text{m}$

Question 12 [1 mark]

The diagram shows a simple d.c. motor. The coil rotates clockwise when viewed from the front. What is the direction of the force on side AB of the coil?

<image_placeholder> id: Q12-fig1 type: diagram linked_question: Q12 description: Simple d.c. motor with rectangular coil ABCD in a uniform magnetic field from left to right (N to S). Current flows from A to B on the front side of the coil. Commutator and brushes shown. Coil rotates clockwise. labels: N, S, A, B, C, D, current direction arrows, magnetic field direction, rotation arrow values: Magnetic field left to right, current A to B on front side must_show: Rectangular coil with sides labelled, magnetic field direction, current direction on side AB, rotation direction </image_placeholder>

☐ A. Upwards
☐ B. Downwards
☐ C. Into the page
☐ D. Out of the page

Question 13 [1 mark]

An electrical appliance has a metal casing and is connected to the mains via a 3-core cable. The earth wire is connected to the metal casing. What is the purpose of the earth wire?

☐ A. To carry the normal operating current.
☐ B. To provide a path for current to flow to ground if the live wire touches the casing.
☐ C. To reduce the resistance of the circuit.
☐ D. To increase the voltage across the appliance.

Question 14 [1 mark]

A solenoid carries a steady current. A bar magnet is inserted into the solenoid with its north pole leading. What is the direction of the induced current in the solenoid as viewed from the end where the magnet enters?

☐ A. Clockwise
☐ B. Anti-clockwise
☐ C. No current is induced
☐ D. Alternating direction

Question 15 [1 mark]

The power loss in a transmission cable is given by $P = I^2 R$ . To reduce power loss, electricity is transmitted at high voltage. Why does this reduce power loss?

☐ A. The resistance of the cable decreases at high voltage.
☐ B. For the same power transmitted, the current is lower at higher voltage.
☐ C. High voltage increases the cross-sectional area of the cable.
☐ D. High voltage reduces the resistivity of the cable material.

Question 16 [1 mark]

A magnet is dropped through a long copper tube. It falls slower than a non-magnetic object of the same mass and size. What is the main reason?

☐ A. Air resistance is greater on the magnet.
☐ B. Eddy currents induced in the tube oppose the motion of the magnet.
☐ C. The magnet attracts the copper tube.
☐ D. The magnetic field increases the mass of the magnet.

Question 17 [1 mark]

In a half-wave rectifier circuit with a capacitor smoothing, the output voltage across the load resistor:

☐ A. Is a steady d.c. voltage.
☐ B. Is a full-wave rectified voltage.
☐ C. Rises and falls but never reaches zero.
☐ D. Is zero for half of each cycle.

Question 18 [1 mark]

The diagram shows a current-carrying wire placed between the poles of a magnet. The current is directed into the page. What is the direction of the force on the wire?

<image_placeholder> id: Q18-fig1 type: diagram linked_question: Q18 description: Horseshoe magnet with N pole on left, S pole on right. Wire perpendicular to page with current into page (cross symbol). Magnetic field lines from N to S (left to right). labels: N, S, current into page (cross), magnetic field direction left to right values: Current into page, magnetic field left to right must_show: Horseshoe magnet with poles labelled, wire with current direction symbol, magnetic field direction </image_placeholder>

☐ A. Upwards
☐ B. Downwards
☐ C. To the left
☐ D. To the right

Question 19 [1 mark]

A transformer is 80% efficient. The secondary voltage is 12 V and the secondary current is 2.0 A. The primary voltage is 240 V. What is the primary current?

☐ A. 0.125 A
☐ B. 0.1 A
☐ C. 0.156 A
☐ D. 0.08 A

Question 20 [1 mark]

Which graph correctly shows the relationship between the resistance of a metallic conductor and its temperature (in °C) over a wide range?

☐ A. A straight line through the origin with positive gradient.
☐ B. A straight line with positive gradient and positive intercept on the resistance axis.
☐ C. A curve with increasing gradient as temperature increases.
☐ D. A horizontal line.

Section B: Structured Questions [45 marks]

Answer all questions in the spaces provided.

Question 21 [5 marks]

A student sets up a circuit to investigate the current-voltage (I-V) characteristic of a filament lamp. The circuit includes a variable resistor, an ammeter, a voltmeter, and a 12 V d.c. power supply.

(a) Draw the circuit diagram for this investigation. Use standard circuit symbols. [2]

(b) The student obtains the following data:

Voltage / V	0.0	2.0	4.0	6.0	8.0	10.0	12.0
Current / A	0.00	0.35	0.60	0.78	0.92	1.02	1.10

Plot the I-V characteristic graph on the grid below. [2]

<image_placeholder> id: Q21-fig1 type: graph linked_question: Q21 description: Blank grid for plotting I-V characteristic of filament lamp. X-axis: Voltage (V) from 0 to 12 V. Y-axis: Current (A) from 0 to 1.2 A. Grid lines at 1 V and 0.1 A intervals. labels: Voltage / V, Current / A values: Data points from table must_show: Axes with labels and units, appropriate scales, data points plotted, smooth curve through points </image_placeholder>

(c) State and explain how the resistance of the filament lamp changes as the voltage increases from 0 V to 12 V. [1]

Question 22 [6 marks]

A transformer is used to power a 12 V, 24 W lamp from a 240 V a.c. mains supply. The transformer has 1200 turns on the primary coil.

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

(b) The lamp operates at its rated power. Calculate the current in the secondary coil. [1]

(c) Assuming the transformer is 90% efficient, calculate the current in the primary coil. [2]

(d) Explain why the transformer core is laminated. [1]

Question 23 [7 marks]

<image_placeholder> id: Q23-fig1 type: diagram linked_question: Q23 description: A rectangular coil of wire ABCD with 50 turns, dimensions 0.1 m by 0.08 m, rotating in a uniform magnetic field of flux density 0.5 T. The coil rotates at 20 revolutions per second. Axis of rotation is parallel to side AB. Magnetic field direction is horizontal (left to right). At the instant shown, the plane of the coil is parallel to the magnetic field. labels: A, B, C, D, axis of rotation, magnetic field direction (B), rotation direction, dimensions values: N = 50 turns, length = 0.1 m, width = 0.08 m, B = 0.5 T, f = 20 Hz must_show: Rectangular coil with labelled sides, axis of rotation, magnetic field direction, rotation direction, dimensions </image_placeholder>

The diagram shows a simple a.c. generator. A rectangular coil of 50 turns and dimensions 0.1 m by 0.08 m rotates in a uniform magnetic field of flux density 0.5 T. The coil rotates at 20 revolutions per second.

(a) Calculate the maximum magnetic flux through the coil. [2]

(b) Calculate the maximum induced e.m.f. in the coil. [2]

(c) Sketch a graph of induced e.m.f. against time for two complete revolutions of the coil. Label the axes with appropriate values. [2]

<image_placeholder> id: Q23-fig2 type: graph linked_question: Q23 description: Blank grid for sketching e.m.f. vs time graph for a.c. generator. X-axis: Time (s) from 0 to 0.1 s. Y-axis: e.m.f. (V) from -12 to +12 V. Two complete sinusoidal cycles expected. labels: Time / s, Induced e.m.f. / V values: Period = 0.05 s, peak e.m.f. from part (b) must_show: Sinusoidal waveform, two complete cycles, axes labelled with units and scales, peak values marked </image_placeholder>

(d) State one way to increase the maximum induced e.m.f. without changing the speed of rotation. [1]

Question 24 [5 marks]

A household circuit has a 13 A fuse in the main consumer unit. The circuit supplies power to the following appliances connected in parallel across the 240 V mains:

Electric kettle: 2.0 kW
Toaster: 800 W
Microwave oven: 1.2 kW

(a) Calculate the total current drawn from the mains when all three appliances are operating simultaneously. [2]

(b) Will the 13 A fuse blow? Explain your answer. [1]

(c) The electric kettle has a metal casing. Explain why it is essential that the kettle is earthed. [2]

Question 25 [6 marks]

<image_placeholder> id: Q25-fig1 type: diagram linked_question: Q25 description: Cathode-ray oscilloscope (CRO) screen showing a sinusoidal trace. Y-gain = 5 V/div, time-base = 2 ms/div. The trace shows a sine wave with peak at +2 divisions and trough at -2 divisions. One complete cycle occupies 4 divisions horizontally. labels: Y-gain setting, time-base setting, divisions values: Y-gain = 5 V/div, time-base = 2 ms/div, vertical amplitude = 2 div, horizontal period = 4 div must_show: CRO screen with grid, sinusoidal trace, settings displayed </image_placeholder>

The diagram shows the trace on a cathode-ray oscilloscope (CRO) screen. The Y-gain is set to 5 V/div and the time-base is set to 2 ms/div.

(a) Determine the peak voltage of the signal. [1]

(b) Determine the peak-to-peak voltage of the signal. [1]

(c) Determine the period of the signal. [1]

(d) Determine the frequency of the signal. [1]

(e) The signal is connected across a resistor of 100 $\Omega$ . Calculate the r.m.s. current through the resistor. [2]

Question 26 [6 marks]

A student investigates the force on a current-carrying conductor in a magnetic field. The apparatus consists of a wire of length 0.15 m placed perpendicular to a uniform magnetic field. The wire is connected to a variable d.c. power supply and an ammeter. A newton meter measures the force on the wire.

The student obtains the following results:

Current / A	0.0	1.0	2.0	3.0	4.0	5.0
Force / N	0.00	0.06	0.12	0.18	0.24	0.30

(a) Plot a graph of Force against Current on the grid below. [2]

<image_placeholder> id: Q26-fig1 type: graph linked_question: Q26 description: Blank grid for plotting Force vs Current. X-axis: Current (A) from 0 to 5 A. Y-axis: Force (N) from 0 to 0.35 N. Grid lines at 0.5 A and 0.02 N intervals. labels: Current / A, Force / N values: Data points from table must_show: Axes with labels and units, appropriate scales, data points plotted, straight line through origin </image_placeholder>

(b) Determine the magnetic flux density of the field. [2]

(c) The student repeats the experiment with the wire at an angle of 30° to the magnetic field. The current is kept at 5.0 A. Calculate the new force on the wire. [2]

Question 27 [5 marks]

A diode is used in a half-wave rectifier circuit with a capacitor for smoothing. The input is a sinusoidal a.c. voltage of peak value 10 V and frequency 50 Hz.

(a) Sketch the voltage across the load resistor (without the capacitor) for two complete cycles. Label the axes. [2]

<image_placeholder> id: Q27-fig1 type: graph linked_question: Q27 description: Blank grid for sketching half-wave rectified voltage. X-axis: Time (ms) from 0 to 40 ms. Y-axis: Voltage (V) from 0 to 10 V. Two half-wave rectified cycles expected. labels: Time / ms, Voltage / V values: Period = 20 ms, peak = 10 V must_show: Half-wave rectified waveform (positive half-cycles only), two cycles, axes labelled with units and scales </image_placeholder>

(b) Sketch the voltage across the load resistor with a large capacitor connected in parallel for smoothing. Label the axes. [2]

<image_placeholder> id: Q27-fig2 type: graph linked_question: Q27 description: Blank grid for sketching smoothed half-wave rectified voltage. X-axis: Time (ms) from 0 to 40 ms. Y-axis: Voltage (V) from 0 to 10 V. Two cycles with ripple expected. labels: Time / ms, Voltage / V values: Period = 20 ms, peak = 10 V, ripple small must_show: Smoothed waveform with small ripple, never reaching zero, axes labelled with units and scales </image_placeholder>

(c) State the function of the diode in this circuit. [1]

Question 28 [5 marks]

The diagram shows a wire carrying a current of 4.0 A placed in a uniform magnetic field of flux density 0.3 T. The wire is at an angle of 60° to the direction of the magnetic field. The length of the wire in the field is 0.2 m.

<image_placeholder> id: Q28-fig1 type: diagram linked_question: Q28 description: Uniform magnetic field lines horizontal (left to right). Wire at 60° to horizontal, current direction shown along wire. Angle labelled 60°. labels: Magnetic field direction (B), current direction (I), angle 60°, length 0.2 m values: I = 4.0 A, B = 0.3 T, L = 0.2 m, θ = 60° must_show: Magnetic field lines, wire at 60° to field, current direction, angle label, length label </image_placeholder>

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

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

(c) The wire is now rotated so that it is parallel to the magnetic field. State the force on the wire. [1]

(d) Explain why the force is zero when the wire is parallel to the magnetic field. [1]

Section C: Free Response Questions [15 marks]

Answer all questions in the spaces provided.

Question 29 [8 marks]

A student designs an experiment to determine the resistivity of a metal wire. The wire has a length of 1.0 m and a uniform cross-sectional area. The student has access to a power supply, ammeter, voltmeter, micrometer screw gauge, metre rule, and connecting wires.

(a) Draw a labelled circuit diagram for the experiment. [2]

(b) Describe the procedure to obtain accurate measurements of the wire's resistance, length, and diameter. [3]

(c) The student measures the diameter at three different positions along the wire and obtains values of 0.48 mm, 0.50 mm, and 0.52 mm. The length is 1.00 m and the resistance is measured as 2.5 $\Omega$ . Calculate the resistivity of the metal. [3]

Question 30 [7 marks]

A power station generates electricity at 25 kV. The electricity is transmitted through cables of total resistance 10 $\Omega$ to a substation 50 km away. The power delivered to the substation is 100 MW.

(a) Calculate the current in the transmission cables if the transmission voltage is 400 kV. [2]

(b) Calculate the power loss in the cables at this voltage. [2]

(c) If the transmission voltage were reduced to 100 kV, calculate the new power loss in the cables. [2]

(d) Explain why electrical energy is transmitted at high voltages. [1]

End of Paper

Answers

TuitionGoWhere Practice Paper - Pure Physics Secondary 4

Answer Key and Marking Scheme
Paper: PRELIM Version 3
Total Marks: 80

Section A: Multiple Choice Questions [20 marks]

Question 1 [1 mark]

Answer: A
Working:
For a transformer: $\frac{V_s}{V_p} = \frac{N_s}{N_p}$
$V_s = V_p \times \frac{N_s}{N_p} = 240 \times \frac{100}{400} = 240 \times 0.25 = 60 \text{ V}$

Marking note: 1 mark for correct answer. Common error: using inverse ratio giving 960 V (option B).

Question 2 [1 mark]

Answer: A
Working:
Energy = Power × Time
Power = 2.0 kW
Time = 15 minutes = 0.25 hours
Energy = 2.0 × 0.25 = 0.5 kWh

Marking note: 1 mark for correct answer. Common error: using time in minutes (2.0 × 15 = 30 kWh, option B) or seconds.

Question 3 [1 mark]

Answer: B
Working:
Force on current-carrying conductor: $F = BIL \sin\theta$
$\theta = 90^\circ$ (perpendicular), so $\sin\theta = 1$
$F = 0.2 \times 5.0 \times 0.1 = 0.1 \text{ N}$

Marking note: 1 mark for correct answer.

Question 4 [1 mark]

Answer: C
Explanation: Faraday's law of electromagnetic induction states that an e.m.f. is induced in a coil whenever there is a change in magnetic flux linking the coil. The magnitude is proportional to the rate of change of flux. Direction is given by Lenz's law.

Marking note: 1 mark for correct answer. Option A is incorrect (also induced when magnet moves away). Option B contradicts Faraday's law. Option D is incorrect (Lenz's law gives opposition).

Question 5 [1 mark]

Answer: C
Working:
Peak voltage = 12 V (given)
Peak-to-peak voltage = 2 × peak voltage = 2 × 12 = 24 V
(Alternatively: 3 divisions × 2 V/div = 6 V peak, so peak-to-peak = 12 V — but the question states peak voltage is 12 V, so peak-to-peak = 24 V)

Marking note: 1 mark for correct answer. The Y-gain and divisions are distractors; the peak voltage is explicitly given as 12 V.

Question 6 [1 mark]

Answer: A
Explanation: In a 3-pin plug, the fuse is connected in series with the live wire to disconnect the appliance from the high voltage supply in case of a fault.

Marking note: 1 mark for correct answer.

Question 7 [1 mark]

Answer: A
Explanation: Induced e.m.f. is maximum when the rate of change of magnetic flux is maximum. This occurs when the plane of the coil is parallel to the magnetic field (flux = 0 but changing fastest).

Marking note: 1 mark for correct answer. Common misconception: thinking maximum flux (perpendicular) gives maximum e.m.f.

Question 8 [1 mark]

Answer: B
Explanation: As voltage increases, the filament temperature increases. Higher temperature causes increased lattice vibrations, which impede electron flow, increasing resistance.

Marking note: 1 mark for correct answer. Option A is a secondary effect. Option C applies to semiconductors, not metals. Option D is incorrect.

Question 9 [1 mark]

Answer: C
Working:
For an ideal transformer: $V_p I_p = V_s I_s$ and $\frac{V_s}{V_p} = \frac{N_s}{N_p}$
$\frac{I_s}{I_p} = \frac{N_p}{N_s} = \frac{800}{200} = 4$
$I_s = 4 \times I_p = 4 \times 0.5 = 2.0 \text{ A}$

Marking note: 1 mark for correct answer.

Question 10 [1 mark]

Answer: C
Explanation: An A.C. generator works by rotating a coil in a magnetic field, inducing an alternating e.m.f. via electromagnetic induction. Electric bell and relay use electromagnets (magnetic effect of current). Moving-coil loudspeaker uses motor effect (force on current-carrying conductor).

Marking note: 1 mark for correct answer.

Question 11 [1 mark]

Answer: A
Working:
Resistivity $\rho = \frac{RA}{L}$
$R = 4.0 \Omega$ , $A = 1.0 \times 10^{-6} \text{ m}^2$ , $L = 2.0 \text{ m}$
$\rho = \frac{4.0 \times 1.0 \times 10^{-6}}{2.0} = 2.0 \times 10^{-6} \Omega \text{m}$

Marking note: 1 mark for correct answer.

Question 12 [1 mark]

Answer: B
Working:
Using Fleming's Left-Hand Rule:

First finger (Field): Left to right (N to S)
Second finger (Current): A to B (towards the right on front side)
Thumb (Force): Downwards

Marking note: 1 mark for correct answer. The coil rotates clockwise, so side AB moves downwards.

Question 13 [1 mark]

Answer: B
Explanation: The earth wire provides a low-resistance path to ground. If the live wire touches the metal casing, a large current flows through the earth wire, blowing the fuse and disconnecting the appliance.

Marking note: 1 mark for correct answer.

Question 14 [1 mark]

Answer: A
Working:
Lenz's law: induced current opposes the change. North pole entering → solenoid end becomes North pole to repel. Viewed from entry end, current flows clockwise (right-hand grip rule: thumb points away from entering magnet = North pole).

Marking note: 1 mark for correct answer.

Question 15 [1 mark]

Answer: B
Explanation: Power transmitted $P = VI$ . For constant power, increasing voltage reduces current. Power loss $P_{\text{loss}} = I^2 R$ , so reducing current significantly reduces power loss.

Marking note: 1 mark for correct answer.

Question 16 [1 mark]

Answer: B
Explanation: As the magnet falls, changing magnetic flux induces eddy currents in the copper tube. These currents create a magnetic field opposing the magnet's motion (Lenz's law), producing an upward force that slows the fall.

Marking note: 1 mark for correct answer.

Question 17 [1 mark]

Answer: C
Explanation: With capacitor smoothing, the capacitor charges during the conducting half-cycles and discharges slowly through the load during non-conducting periods. The voltage rises and falls (ripple) but never reaches zero if the capacitor is large enough.

Marking note: 1 mark for correct answer. Option D describes unsmoothed half-wave rectification.

Question 18 [1 mark]

Answer: A
Working:
Fleming's Left-Hand Rule:

First finger (Field): Left to right
Second finger (Current): Into page (cross)
Thumb (Force): Upwards

Marking note: 1 mark for correct answer.

Question 19 [1 mark]

Answer: A
Working:
Efficiency $\eta = \frac{V_s I_s}{V_p I_p} = 0.80$
$I_p = \frac{V_s I_s}{\eta V_p} = \frac{12 \times 2.0}{0.80 \times 240} = \frac{24}{192} = 0.125 \text{ A}$

Marking note: 1 mark for correct answer. Common error: forgetting efficiency (giving 0.1 A, option B) or using efficiency incorrectly.

Question 20 [1 mark]

Answer: B
Explanation: Resistance of a metallic conductor increases linearly with temperature (over a wide range). The relationship is $R = R_0(1 + \alpha \theta)$ , giving a straight line with positive gradient and positive intercept at 0°C.

Marking note: 1 mark for correct answer. Option A would imply zero resistance at 0°C. Option C describes a thermistor or semiconductor. Option D is incorrect.

Section B: Structured Questions [45 marks]

Question 21 [5 marks]

(a) [2 marks]
Circuit diagram should show:

12 V d.c. power supply (battery symbol)
Variable resistor (rheostat) in series
Filament lamp
Ammeter in series with lamp
Voltmeter in parallel across lamp
Correct symbols and connections

Marking:

1 mark for correct series circuit with power supply, variable resistor, lamp, ammeter
1 mark for voltmeter correctly connected in parallel across lamp only

(b) [2 marks]
Graph plotting:

Axes labelled with units: Voltage / V (x-axis), Current / A (y-axis)
Suitable scales using >50% of grid
All 7 points plotted accurately (± half a small square)
Smooth curve through points (not straight lines), showing increasing gradient

Marking:

1 mark for axes labels, units, and scales
1 mark for correct plotting and smooth curve

(c) [1 mark]
Answer: The resistance increases as voltage increases.
Explanation: As voltage increases, the filament temperature rises. Higher temperature causes increased lattice vibrations, which impede the flow of electrons, increasing resistance.

Marking: 1 mark for correct statement with brief explanation. Accept "resistance increases" with valid reason.

Question 22 [6 marks]

(a) [2 marks]
Working:
$\frac{V_s}{V_p} = \frac{N_s}{N_p}$
$N_s = N_p \times \frac{V_s}{V_p} = 1200 \times \frac{12}{240} = 1200 \times 0.05 = 60 \text{ turns}$

Marking:

1 mark for correct formula/ratio
1 mark for correct answer with unit (turns)

(b) [1 mark]
Working:
$P = V_s I_s$
$I_s = \frac{P}{V_s} = \frac{24}{12} = 2.0 \text{ A}$

Marking: 1 mark for correct answer with unit.

(c) [2 marks]
Working:
Efficiency $\eta = \frac{P_{\text{out}}}{P_{\text{in}}} = \frac{V_s I_s}{V_p I_p} = 0.90$
$P_{\text{out}} = 24 \text{ W}$
$P_{\text{in}} = \frac{P_{\text{out}}}{\eta} = \frac{24}{0.90} = 26.67 \text{ W}$
$I_p = \frac{P_{\text{in}}}{V_p} = \frac{26.67}{240} = 0.111 \text{ A}$

Alternative:
$I_p = \frac{V_s I_s}{\eta V_p} = \frac{12 \times 2.0}{0.90 \times 240} = \frac{24}{216} = 0.111 \text{ A}$

Marking:

1 mark for correct use of efficiency formula
1 mark for correct answer with unit (A)

(d) [1 mark]
Answer: To reduce eddy currents in the core.
Explanation: Laminations (thin insulated sheets) increase the resistance to eddy current paths, reducing energy loss as heat.

Marking: 1 mark for correct answer. Accept "to reduce eddy current losses" or "to prevent large eddy currents".

Question 23 [7 marks]

(a) [2 marks]
Working:
Maximum flux $\Phi_{\text{max}} = B A N$
Area $A = 0.1 \times 0.08 = 0.008 \text{ m}^2$
$\Phi_{\text{max}} = 0.5 \times 0.008 \times 50 = 0.2 \text{ Wb}$

Marking:

1 mark for correct area calculation
1 mark for correct answer with unit (Wb)

(b) [2 marks]
Working:
Maximum e.m.f. $\mathcal{E}_{\text{max}} = 2\pi f N B A = 2\pi f \Phi_{\text{max}}$
$f = 20 \text{ Hz}$
$\mathcal{E}_{\text{max}} = 2\pi \times 20 \times 0.2 = 8\pi \approx 25.1 \text{ V}$

Marking:

1 mark for correct formula
1 mark for correct answer with unit (V)
(Accept 25.1 V or 25 V or $8\pi$ V)

(c) [2 marks]
Graph sketch:

Sinusoidal waveform
Two complete cycles (period = 0.05 s, so 0 to 0.1 s)
Peak at ±25.1 V
Axes labelled: Time / s, Induced e.m.f. / V
Zero crossings at 0, 0.025, 0.05, 0.075, 0.1 s

Marking:

1 mark for correct sinusoidal shape with two cycles
1 mark for axes labels, units, and correct peak/period values

(d) [1 mark]
Answer: Increase the number of turns on the coil / increase the magnetic flux density / increase the area of the coil.
(Any one valid method)

Marking: 1 mark for any correct answer.

Question 24 [5 marks]

(a) [2 marks]
Working:
Total power = 2000 + 800 + 1200 = 4000 W = 4.0 kW
$I = \frac{P}{V} = \frac{4000}{240} = 1

<stage3_exam_answers_md>

TuitionGoWhere Practice Paper - Pure Physics Secondary 4

PRELIM Version 3 - Answer Key & Marking Scheme

Section A: Multiple Choice Questions [20 marks]

Question	Answer	Explanation
1	A	$V_s = V_p \times \frac{N_s}{N_p} = 240 \times \frac{100}{400} = 60 \text{ V}$
2	A	Energy = Power × Time = $2.0 \text{ kW} \times \frac{15}{60} \text{ h} = 0.5 \text{ kWh}$
3	B	$F = BIL = 0.2 \times 5.0 \times 0.1 = 0.1 \text{ N}$
4	C	Faraday's Law: e.m.f. induced when magnetic flux linking a coil changes.
5	C	Peak-to-peak = 2 × peak = 2 × 12 V = 24 V. (Y-gain and divisions are distractors; peak voltage given directly as 12 V)
6	A	Fuse is always connected to the live wire.
7	A	Maximum rate of change of flux occurs when plane of coil is parallel to field.
8	B	Higher temperature → increased lattice vibrations → more collisions → higher resistance.
9	C	$\frac{I_s}{I_p} = \frac{N_p}{N_s} = \frac{800}{200} = 4$ → $I_s = 4 \times 0.5 = 2.0 \text{ A}$
10	C	A.C. generator works on electromagnetic induction.
11	A	$\rho = \frac{RA}{L} = \frac{4.0 \times 1.0 \times 10^{-6}}{2.0} = 2.0 \times 10^{-6} \Omega \text{m}$
12	D	Fleming's Left Hand Rule: Field left→right, Current A→B (front), Force = Out of page.
13	B	Earth wire provides safety path to ground if live touches casing.
14	A	Lenz's Law: Induced current creates field opposing entering N pole → clockwise (viewed from entry end).
15	B	$P = VI$ ; for constant power, higher V means lower I; $P_{\text{loss}} = I^2R$ decreases.
16	B	Eddy currents induced in copper tube create opposing magnetic field (Lenz's Law).
17	C	Capacitor charges during peaks and discharges slowly; voltage never reaches zero.
18	A	Fleming's Left Hand Rule: Field left→right, Current into page, Force = Upwards.
19	A	$P_{\text{out}} = 12 \times 2.0 = 24 \text{ W}$ ; $P_{\text{in}} = \frac{24}{0.8} = 30 \text{ W}$ ; $I_p = \frac{30}{240} = 0.125 \text{ A}$
20	B	Resistance of metal increases linearly with temperature from a positive intercept at 0°C.

Section B: Structured Questions [45 marks]

Question 21 [5 marks]

(a) Circuit diagram: [2 marks]

1 mark: Correct symbols for power supply, variable resistor, ammeter (series), voltmeter (parallel across lamp), filament lamp.
1 mark: Correct connections (ammeter in series, voltmeter in parallel across lamp only).

(b) Graph plotting: [2 marks]

1 mark: Axes labelled with units (V / V, I / A), appropriate scales covering data range.
1 mark: All 7 points plotted correctly (± half a small square), smooth curve through points (curving upwards with decreasing gradient).

(c) Resistance increases as voltage increases. [1 mark]

Explanation: As voltage increases, current increases → filament temperature rises → lattice ions vibrate more vigorously → more frequent collisions with electrons → resistance increases.

Question 22 [6 marks]

(a) Number of turns on secondary coil: [2 marks] $\frac{V_s}{V_p} = \frac{N_s}{N_p} \Rightarrow N_s = N_p \times \frac{V_s}{V_p} = 1200 \times \frac{12}{240} = 60 \text{ turns}$

(b) Current in secondary coil: [1 mark] $P = V_s I_s \Rightarrow I_s = \frac{P}{V_s} = \frac{24}{12} = 2.0 \text{ A}$

(c) Current in primary coil (90% efficient): [2 marks] $P_{\text{out}} = 24 \text{ W}$ $\text{Efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}} \times 100\% \Rightarrow 0.90 = \frac{24}{P_{\text{in}}} \Rightarrow P_{\text{in}} = \frac{24}{0.90} = 26.67 \text{ W}$ $I_p = \frac{P_{\text{in}}}{V_p} = \frac{26.67}{240} = 0.111 \text{ A} \quad (\text{or } 0.11 \text{ A})$

(d) Core is laminated to reduce eddy currents. [1 mark]

Laminations (thin insulated sheets) increase resistance to eddy current paths → reduces eddy current losses → improves efficiency.

Question 23 [7 marks]

(a) Maximum magnetic flux through coil: [2 marks] $\Phi_{\text{max}} = N B A = 50 \times 0.5 \times (0.1 \times 0.08) = 50 \times 0.5 \times 0.008 = 0.2 \text{ Wb}$

(b) Maximum induced e.m.f.: [2 marks] $\mathcal{E}_{\text{max}} = 2\pi f N B A = 2\pi \times 20 \times 0.2 = 8\pi \approx 25.1 \text{ V}$ (Alternative: $\mathcal{E}_{\text{max}} = \omega N B A = 2\pi f \Phi_{\text{max}}$ )

(c) Graph of induced e.m.f. against time: [2 marks]

1 mark: Sinusoidal waveform, two complete cycles, period $T = \frac{1}{f} = 0.05 \text{ s}$ .
1 mark: Axes labelled (Time / s, e.m.f. / V), peak values marked at ±25.1 V (or ±8π V), zero crossings at 0, 0.025, 0.05, 0.075, 0.10 s.

(d) One way to increase maximum induced e.m.f. without changing rotation speed: [1 mark]

Increase number of turns ( $N$ ).
Increase magnetic flux density ( $B$ ).
Increase area of coil ( $A$ ).
Use a soft iron core.

Question 24 [5 marks]

(a) Total current drawn: [2 marks] $I_{\text{kettle}} = \frac{2000}{240} = 8.33 \text{ A}$ $I_{\text{toaster}} = \frac{800}{240} = 3.33 \text{ A}$ $I_{\text{microwave}} = \frac{1200}{240} = 5.00 \text{ A}$ $I_{\text{total}} = 8.33 + 3.33 + 5.00 = 16.67 \text{ A} \quad (\text{or } 16.7 \text{ A})$

(b) Will the 13 A fuse blow? [1 mark] Yes. Total current (16.7 A) exceeds fuse rating (13 A). Fuse will melt and break the circuit.

(c) Why kettle must be earthed: [2 marks]

Metal casing could become live if live wire touches it (fault).
Earth wire provides low-resistance path to ground.
Large fault current flows → blows fuse / trips circuit breaker → disconnects supply → prevents electric shock.

Question 25 [6 marks]

(a) Peak voltage: [1 mark] Peak = 2 divisions × 5 V/div = 10 V

(b) Peak-to-peak voltage: [1 mark] Peak-to-peak = 4 divisions × 5 V/div = 20 V

(c) Period of signal: [1 mark] Period = 4 divisions × 2 ms/div = 8 ms (or 0.008 s)

(d) Frequency of signal: [1 mark] $f = \frac{1}{T} = \frac{1}{0.008} = 125 \text{ Hz}$

(e) R.m.s. current through 100 Ω resistor: [2 marks] $V_{\text{rms}} = \frac{V_{\text{peak}}}{\sqrt{2}} = \frac{10}{\sqrt{2}} = 7.07 \text{ V}$ $I_{\text{rms}} = \frac{V_{\text{rms}}}{R} = \frac{7.07}{100} = 0.0707 \text{ A} \quad (\text{or } 70.7 \text{ mA})$

Question 26 [6 marks]

(a) Graph of Force against Current: [2 marks]

1 mark: Axes labelled with units (I / A, F / N), appropriate scales.
1 mark: All 6 points plotted correctly, straight line through origin (best fit).

(b) Magnetic flux density: [2 marks] Gradient = $\frac{\Delta F}{\Delta I} = \frac{0.30 - 0.00}{5.0 - 0.0} = 0.06 \text{ N/A}$ $F = B I L \Rightarrow B = \frac{F}{I L} = \frac{\text{gradient}}{L} = \frac{0.06}{0.15} = 0.40 \text{ T}$

(c) Force at 30° to field with 5.0 A: [2 marks] $F = B I L \sin\theta = 0.40 \times 5.0 \times 0.15 \times \sin 30^\circ$ $F = 0.40 \times 5.0 \times 0.15 \times 0.5 = 0.15 \text{ N}$

Question 27 [6 marks]

(a) Half-wave rectified voltage (no capacitor): [2 marks]

1 mark: Correct waveform showing only positive half-cycles (zero for negative half-cycles).
1 mark: Axes labelled (Time / ms, Voltage / V), period = 20 ms, peak = 10 V, two cycles shown.

(b) Smoothed voltage with capacitor: [2 marks]

1 mark: Waveform showing ripple voltage – rises to near 10 V, falls slightly exponentially, never reaches zero.
1 mark: Axes labelled, period = 20 ms, ripple shown, average voltage close to peak.

(c) Function of capacitor: [1 mark] Capacitor charges during the conducting half-cycle and discharges through the load during the non-conducting half-cycle, maintaining the voltage across the load and reducing ripple.

(d) Effect of increasing load resistance: [1 mark] Ripple voltage decreases (or smoothing improves) because the capacitor discharges more slowly through a larger resistance (longer time constant $\tau = RC$ ).

Summary of Marks Allocation

Section	Question	Marks
A	1–20	20
B	21	5
B	22	6
B	23	7
B	24	5
B	25	6
B	26	6
B	27	6
Total		61

Note: Total marks = 80. Section A = 20 marks. Section B = 45 marks. The remaining 15 marks are typically allocated to a Section C (Longer Structured/Free Response Questions) not included in this version of the paper.

End of Answer Key