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Secondary 4 Pure Physics Practice Paper 3

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TuitionGoWhere Practice Paper - Pure Physics Secondary 4

TuitionGoWhere Practice Paper (AI)
Subject: Pure Physics
Level: Secondary 4
Paper: Practice Paper 3 (Version 3 of 5)
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.
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 use a calculator.
Where appropriate, take the acceleration due to gravity $g = 10 \text{ m/s}^2$ .
The total mark 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.

1

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 120 V
C 480 V
D 960 V

Answer: □ [1]

2

An electric kettle rated 2.2 kW, 230 V is used for 15 minutes. What is the energy consumed in kWh?

A 0.55 kWh
B 1.10 kWh
C 2.20 kWh
D 33.0 kWh

Answer: □ [1]

3

A straight wire carrying a current of 5 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

Answer: □ [1]

4

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 An e.m.f. is induced when there is a change in magnetic flux linkage with a coil.
C An e.m.f. is induced only when a coil moves away from a magnet.
D An e.m.f. is induced when a magnet is stationary inside a coil.

Answer: □ [1]

5

A current-carrying coil experiences a turning effect in a magnetic field. Which factor does NOT affect the magnitude of this turning effect?

A The number of turns on the coil
B The current in the coil
C The resistance of the coil
D The magnetic flux density

Answer: □ [1]

6

The diagram shows a simple a.c. generator. At which position of the coil is the induced e.m.f. maximum?

<image_placeholder> id: Q6-fig1 type: diagram linked_question: Q6 description: Simple AC generator with rectangular coil rotating in uniform magnetic field. Show coil at four positions: (A) coil plane parallel to field lines, (B) coil plane at 45° to field lines, (C) coil plane perpendicular to field lines, (D) coil plane at 135° to field lines. labels: Magnetic field direction (N to S), coil sides labelled AB and CD, axis of rotation, slip rings and brushes values: Magnetic flux density B = 0.5 T, coil area = 0.02 m², angular velocity = 10 rad/s, number of turns = 50 must_show: Clear indication of coil orientation relative to magnetic field at each position </image_placeholder>

A Position A (coil plane parallel to field lines)
B Position B (coil plane at 45° to field lines)
C Position C (coil plane perpendicular to field lines)
D Position D (coil plane at 135° to field lines)

Answer: □ [1]

7

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 the transformer is 100% efficient, what is the secondary current?

A 0.125 A
B 0.5 A
C 2.0 A
D 8.0 A

Answer: □ [1]

8

Which of the following is the purpose of the earth wire in a household electrical appliance?

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

Answer: □ [1]

9

A magnetic field is directed into the page. A positively charged particle moves from left to right across the page. In which direction is the magnetic force on the particle?

A Into the page
B Out of the page
C Towards the top of the page
D Towards the bottom of the page

Answer: □ [1]

10

The power loss in transmission cables is given by $P = I^2 R$ . To reduce power loss, electricity is transmitted at high voltage. Which statement explains why this reduces power loss?

A High voltage increases the current for the same power
B High voltage decreases the current for the same power
C High voltage increases the resistance of the cables
D High voltage decreases the resistance of the cables

Answer: □ [1]

11

A solenoid carries a steady current. Which of the following will increase the magnetic field strength inside the solenoid?

A Decreasing the number of turns per unit length
B Decreasing the current
C Inserting a soft iron core
D Increasing the length of the solenoid while keeping total turns constant

Answer: □ [1]

12

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

<image_placeholder> id: Q12-fig1 type: diagram linked_question: Q12 description: CRO screen showing a sinusoidal waveform. The waveform spans 4 horizontal divisions for one complete cycle and has a peak-to-peak vertical height of 3 divisions. labels: Time-base: 5 ms/div, Y-gain: 2 V/div, horizontal divisions marked, vertical divisions marked values: One cycle = 4 horizontal divisions, peak-to-peak = 3 vertical divisions must_show: Clear sinusoidal waveform with grid lines, time-base and Y-gain settings visible </image_placeholder>

What is the frequency of the a.c. signal?

A 25 Hz
B 50 Hz
C 100 Hz
D 200 Hz

Answer: □ [1]

13

A wire of length 0.5 m carrying a current of 4 A is placed at an angle of 30° to a uniform magnetic field of flux density 0.6 T. What is the magnitude of the force on the wire?

A 0.3 N
B 0.6 N
C 1.0 N
D 1.2 N

Answer: □ [1]

14

In a household circuit, a 13 A fuse is used to protect a lighting circuit. The circuit has 8 lamps each rated 60 W, 230 V. Will the fuse blow when all lamps are switched on?

A Yes, because the total current is 2.1 A
B Yes, because the total current is 15.7 A
C No, because the total current is 2.1 A
D No, because the total current is 15.7 A

Answer: □ [1]

15

Lenz's law states that the direction of the induced current is such that it:

A Opposes the change producing it
B Assists the change producing it
C Is always clockwise
D Is always anticlockwise

Answer: □ [1]

16

A transformer is used to step down 240 V to 12 V for a low-voltage lighting system. The secondary coil has 60 turns. How many turns are on the primary coil?

A 5 turns
B 120 turns
C 1200 turns
D 2400 turns

Answer: □ [1]

17

The diagram shows a current-carrying conductor placed between the poles of a magnet.

<image_placeholder> id: Q17-fig1 type: diagram linked_question: Q17 description: Horseshoe magnet with N pole on left, S pole on right. A vertical wire passes through the uniform magnetic field between the poles. Current direction is into the page (shown as cross symbol). labels: N pole, S pole, magnetic field direction (left to right), current direction (into page), force direction arrow (to be determined) values: Magnetic flux density = 0.4 T, current = 3 A, length of wire in field = 0.08 m must_show: Clear 3D perspective showing field direction, current direction, and resulting force direction using Fleming's left-hand rule </image_placeholder>

Using Fleming's left-hand rule, what is the direction of the force on the conductor?

A Towards the N pole
B Towards the S pole
C Upwards
D Downwards

Answer: □ [1]

18

Which of the following appliances should be connected to the live wire through a switch?

A The earth wire
B The neutral wire
C The fuse
D The lamp

Answer: □ [1]

19

An a.c. generator produces a peak voltage of 20 V. What is the root-mean-square (r.m.s.) voltage?

A 10 V
B 14.1 V
C 20 V
D 28.3 V

Answer: □ [1]

20

A student investigates the force on a current-carrying wire in a magnetic field. She varies the current and measures the force. Which graph correctly shows the relationship between force and current?

<image_placeholder> id: Q20-fig1 type: graph linked_question: Q20 description: Four graph options showing Force (y-axis) vs Current (x-axis). Option A: straight line through origin with positive gradient. Option B: horizontal line. Option C: curve increasing exponentially. Option D: straight line with negative gradient. labels: Force / N (y-axis), Current / A (x-axis), origin marked values: Linear relationship F = BIL expected must_show: Four distinct graph shapes labelled A, B, C, D for multiple choice selection </image_placeholder>

A Graph A
B Graph B
C Graph C
D Graph D

Answer: □ [1]

Section B: Structured Questions [40 marks]

Answer all questions in the spaces provided.

21

A household has the following appliances connected to a 230 V mains supply:

Electric kettle: 2.2 kW
Microwave oven: 1.2 kW
Toaster: 0.9 kW
Refrigerator: 0.3 kW

The circuit is protected by a 30 A circuit breaker.

(a) Calculate the total power consumed when all four appliances operate simultaneously. [1]

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

(d) The refrigerator operates for 24 hours a day. Calculate the energy consumed by the refrigerator in one day, in kWh. [1]

(e) If electricity costs $0.28 per kWh, calculate the cost of running the refrigerator for 30 days. [2]

Total: [8]

22

The diagram shows a simple d.c. motor.

<image_placeholder> id: Q22-fig1 type: diagram linked_question: Q22 description: Simple DC motor with rectangular coil ABCD in uniform magnetic field between N and S poles. Commutator and brushes shown. Current direction indicated in coil sides. Axis of rotation through centre of coil. labels: N pole, S pole, magnetic field direction, coil sides AB and CD, commutator (split ring), carbon brushes, axis of rotation, current direction arrows in AB and CD values: Magnetic flux density B = 0.3 T, coil dimensions: 0.05 m × 0.08 m, number of turns = 20, current = 1.5 A must_show: Clear split-ring commutator, brushes, current direction in each coil side, magnetic field direction </image_placeholder>

(a) On the diagram, draw an arrow to show the direction of the force acting on side AB of the coil. Label this arrow F. [1]

(b) Explain why the coil experiences a turning effect (torque). [2]

(d) The coil has 20 turns, each of area $4.0 \times 10^{-3} \text{ m}^2$ . It carries a current of 1.5 A in a magnetic field of flux density 0.3 T. Calculate the maximum torque on the coil. [2]

(e) Suggest two modifications to increase the turning effect of the motor. [2]

Total: [8]

23

A step-down transformer is used to charge a mobile phone. The transformer has 1200 turns on the primary coil and 60 turns on the secondary coil. The primary coil is connected to a 240 V a.c. supply. The phone charger draws a current of 1.2 A from the secondary coil. The transformer is 85% efficient.

(a) Calculate the voltage across the secondary coil. [2]

(b) Calculate the current in the primary coil. [3]

(d) State one reason why transformers only work with a.c. and not d.c. [1]

Total: [8]

24

A student sets up an experiment to investigate electromagnetic induction. A bar magnet is dropped through a vertical copper tube. The time taken for the magnet to fall through the tube is measured. The experiment is repeated with a plastic tube of the same dimensions.

<image_placeholder> id: Q24-fig1 type: experimental_setup linked_question: Q24 description: Vertical copper tube and plastic tube side by side. Bar magnet dropped from same height through each tube. Magnet shown falling slower through copper tube. Induced eddy currents shown in copper tube. labels: Copper tube, plastic tube, bar magnet (N and S poles marked), direction of fall, induced eddy currents in copper tube (circular arrows), magnetic field lines values: Tube length = 0.5 m, tube inner diameter = 0.02 m, magnet mass = 0.05 kg, magnet magnetic moment = 0.8 A·m² must_show: Clear contrast between magnet falling through copper vs plastic tube, eddy currents visible in copper tube </image_placeholder>

(a) The magnet takes longer to fall through the copper tube than the plastic tube. Explain this observation using Faraday's law and Lenz's law. [4]

(b) The student observes that the magnet reaches a terminal velocity in the copper tube. Explain why terminal velocity is reached. [2]

(c) If the copper tube is replaced with a copper tube of the same length but with vertical slits cut along its length, how would the fall time change? Explain. [2]

Total: [8]

25

The diagram shows a wire frame PQRS placed in a uniform magnetic field directed into the page. Side PQ can slide along the frame. The frame is connected to a centre-zero galvanometer.

<image_placeholder> id: Q25-fig1 type: diagram linked_question: Q25 description: Rectangular wire frame PQRS with sliding side PQ. Magnetic field directed into page (cross symbols). Galvanometer connected in series with frame. PQ shown moving to the right with velocity v. labels: P, Q, R, S corners, magnetic field into page (×), velocity v arrow to the right on PQ, galvanometer G, current direction arrow values: Magnetic flux density B = 0.4 T, length PQ = 0.2 m, velocity v = 0.5 m/s, resistance of circuit = 2.0 Ω must_show: Clear sliding wire setup, field direction, motion direction, galvanometer </image_placeholder>

(a) Side PQ is moved to the right at a constant speed of 0.5 m/s. The length of PQ is 0.2 m and the magnetic flux density is 0.4 T.

(i) Calculate the magnitude of the induced e.m.f. [2]

(ii) Calculate the induced current in the circuit. The total resistance of the circuit is 2.0 Ω. [1]

(iii) State the direction of the induced current (P to Q or Q to P). Explain using Lenz's law. [2]

(b) If PQ is moved to the left at the same speed, how would the galvanometer deflection change? [1]

Total: [7]

Section C: Longer Structured Questions [20 marks]

Answer all questions in the spaces provided.

26

An overhead power transmission line carries electrical power from a power station to a substation. The power station generates 100 MW at 25 kV. A step-up transformer increases the voltage to 400 kV for transmission. The total resistance of the transmission cables is 10 Ω.

(a) Calculate the current in the transmission cables at 400 kV. [2]

(b) Calculate the power loss in the transmission cables. [2]

(d) If the voltage were not stepped up and power were transmitted at 25 kV, calculate the power loss. [2]

(e) Explain why high-voltage transmission is more efficient, with reference to your calculations. [2]

(f) Transformers are used to step up and step down voltages. Explain why transformers require alternating current to operate. [2]

(g) The step-up transformer has 500 turns on its primary coil. Calculate the number of turns on its secondary coil. [2]

(h) In practice, transformers are not 100% efficient. State two causes of energy loss in a transformer. [2]

Total: [16]

27

A cathode-ray oscilloscope (CRO) is used to analyse an a.c. signal from a signal generator. The CRO settings are: time-base = 2 ms/div, Y-gain = 5 V/div. The trace shows a sinusoidal waveform with a peak-to-peak height of 2.4 divisions and a period of 3.2 divisions.

<image_placeholder> id: Q27-fig1 type: diagram linked_question: Q27 description: CRO screen showing sinusoidal waveform. Grid with 10 horizontal divisions and 8 vertical divisions. Waveform spans 3.2 horizontal divisions per cycle and 2.4 vertical divisions peak-to-peak. labels: Time-base: 2 ms/div, Y-gain: 5 V/div, horizontal divisions, vertical divisions, peak-to-peak measurement, period measurement values: Period = 3.2 horizontal divisions, peak-to-peak = 2.4 vertical divisions must_show: Clear sinusoidal trace on calibrated grid, settings visible </image_placeholder>

(a) Determine the peak voltage of the signal. [2]

(b) Determine the r.m.s. voltage of the signal. [2]

(d) The signal generator is now connected to a resistor of 100 Ω. Calculate the mean power dissipated in the resistor. [2]

(e) Sketch on the diagram above how the trace would change if the time-base is changed to 5 ms/div and the Y-gain is changed to 2 V/div. Describe the changes. [3]

(f) Explain how a CRO can be used to measure the phase difference between two a.c. signals. [3]

Total: [14]

End of Paper

Total Marks: 80

Answers

TuitionGoWhere Practice Paper - Pure Physics Secondary 4 (Answer Key)

Subject: Pure Physics
Level: Secondary 4
Paper: Practice Paper 3 (Version 3 of 5)
Total Marks: 80

Section A: Multiple Choice Questions [20 marks]

1

Answer: A [1]

Explanation:
For a transformer, $\frac{V_s}{V_p} = \frac{N_s}{N_p}$ .
$V_s = V_p \times \frac{N_s}{N_p} = 240 \times \frac{100}{400} = 240 \times 0.25 = 60 \text{ V}$ .

Common mistake: Inverting the turns ratio gives 960 V (option D).

2

Answer: A [1]

Explanation:
Energy (kWh) = Power (kW) × Time (h)
Time = 15 minutes = 0.25 h
Energy = 2.2 × 0.25 = 0.55 kWh.

Common mistake: Using time in minutes (2.2 × 15 = 33) gives option D.

3

Answer: B [1]

Explanation:
Force on current-carrying wire: $F = BIL \sin\theta$
Wire is perpendicular to field, so $\theta = 90°$ , $\sin\theta = 1$ .
$F = 0.2 \times 5 \times 0.1 = 0.1 \text{ N}$ .

4

Answer: B [1]

Explanation:
Faraday's law: An e.m.f. is induced when there is a change in magnetic flux linkage with a coil. This can happen by moving a magnet, moving a coil, or changing the current in a nearby coil. Options A, C, and D describe specific cases but are not universally correct.

5

Answer: C [1]

Explanation:
Turning effect (torque) on a coil: $\tau = NBIA \sin\theta$
Depends on: Number of turns $N$ , Magnetic flux density $B$ , Current $I$ , Area $A$ .
Resistance of the coil does not directly affect the torque (though it affects current for a given voltage).

6

Answer: A [1]

Explanation:
Induced e.m.f. in a generator: $\mathcal{E} = N B A \omega \sin(\omega t)$
Maximum e.m.f. occurs when $\sin(\omega t) = 1$ , i.e., when the coil plane is parallel to the magnetic field lines (rate of change of flux is maximum). At this position, the coil sides are cutting field lines at the maximum rate.

Common mistake: Thinking maximum e.m.f. occurs when coil is perpendicular to field (maximum flux, but zero rate of change).

7

Answer: C [1]

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

8

Answer: B [1]

Explanation:
The earth wire provides a low-resistance path to ground if the live wire touches the metal casing. This causes a large current to flow, blowing the fuse or tripping the circuit breaker, disconnecting the appliance and preventing electric shock.

9

Answer: C [1]

Explanation:
Fleming's left-hand rule (for positive charge/current):

First finger (Field): Into the page
Second finger (Current): Left to right
Thumb (Force): Towards the top of the page

10

Answer: B [1]

Explanation:
Power transmitted $P = VI$ . For constant power, increasing voltage $V$ decreases current $I$ .
Power loss in cables $P_{\text{loss}} = I^2 R$ . Since $I$ decreases, $I^2 R$ decreases significantly.

11

Answer: C [1]

Explanation:
Magnetic field inside a solenoid: $B = \mu_0 n I$ where $n$ is turns per unit length.
Inserting a soft iron core increases the magnetic permeability, greatly increasing $B$ .
Options A, B, and D all decrease $B$ .

12

Answer: B [1]

Explanation:
Period $T = 4 \text{ div} \times 5 \text{ ms/div} = 20 \text{ ms} = 0.02 \text{ s}$
Frequency $f = \frac{1}{T} = \frac{1}{0.02} = 50 \text{ Hz}$ .

13

Answer: A [1]

Explanation:
$F = BIL \sin\theta = 0.6 \times 4 \times 0.5 \times \sin 30° = 0.6 \times 4 \times 0.5 \times 0.5 = 0.6 \text{ N}$
Wait: $0.6 \times 4 = 2.4$ , $2.4 \times 0.5 = 1.2$ , $1.2 \times 0.5 = 0.6 \text{ N}$ .
Correction: Answer is B (0.6 N).

Let me recalculate: $F = BIL \sin\theta = 0.6 \times 4 \times 0.5 \times 0.5 = 0.6 \text{ N}$ . Yes, B.

14

Answer: C [1]

Explanation:
Current per lamp: $I = \frac{P}{V} = \frac{60}{230} \approx 0.261 \text{ A}$
Total current for 8 lamps: $8 \times 0.261 = 2.09 \text{ A} \approx 2.1 \text{ A}$
Fuse rating is 13 A. Since 2.1 A < 13 A, the fuse will not blow.

15

Answer: A [1]

Explanation:
Lenz's law: The direction of the induced current is such that it opposes the change in magnetic flux that produced it. This is a consequence of conservation of energy.

16

Answer: C [1]

Explanation:
$\frac{V_p}{V_s} = \frac{N_p}{N_s}$
$N_p = N_s \times \frac{V_p}{V_s} = 60 \times \frac{240}{12} = 60 \times 20 = 1200 \text{ turns}$ .

17

Answer: C [1]

Explanation:
Fleming's left-hand rule:

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

18

Answer: D [1]

Explanation:
Switches and fuses must be connected in the live wire so that when opened/blown, the appliance is disconnected from the high voltage supply. The lamp (load) is connected to the live wire through the switch.

19

Answer: B [1]

Explanation:
$V_{\text{rms}} = \frac{V_{\text{peak}}}{\sqrt{2}} = \frac{20}{\sqrt{2}} = 20 \times 0.7071 \approx 14.1 \text{ V}$ .

20

Answer: A [1]

Explanation:
Force on current-carrying wire: $F = BIL \sin\theta$ . For constant $B$ , $L$ , $\theta$ , $F \propto I$ . This is a straight line through the origin with positive gradient.

Section B: Structured Questions [40 marks]

21

(a) Total power = 2.2 + 1.2 + 0.9 + 0.3 = 4.6 kW [1]

(b) $P = VI \Rightarrow I = \frac{P}{V} = \frac{4600}{230} = \mathbf{20 \text{ A}}$ [2]
(1 mark for correct substitution, 1 mark for correct answer with unit)

(c) Circuit breaker rating = 30 A. Total current = 20 A.
No, the circuit breaker will not trip because 20 A < 30 A. [2]
(1 mark for correct comparison, 1 mark for correct conclusion with reasoning)

(d) Energy = Power × Time = 0.3 kW × 24 h = 7.2 kWh [1]

(e) Daily cost = 7.2 × $0.28 =$ 2.016
30-day cost = $2.016 × 30 = **$ 60.48** [2]
(1 mark for daily energy cost, 1 mark for 30-day total)

22

(a) [1]
On the diagram: Arrow on side AB pointing upwards (using Fleming's left-hand rule: Field left-to-right, Current A-to-B, Force upwards).

(b) The coil experiences a turning effect because:

Current flows in opposite directions in sides AB and CD [1]
Forces on AB and CD are equal in magnitude but opposite in direction (Fleming's left-hand rule) [1]
These forces form a couple producing a torque about the axis of rotation.

(c) The split-ring commutator reverses the current direction in the coil every half-turn, ensuring the torque always acts in the same direction so the coil rotates continuously. [1]

(d) Maximum torque $\tau_{\text{max}} = N B I A$
$A = 0.05 \times 0.08 = 4.0 \times 10^{-3} \text{ m}^2$ (given)
$\tau_{\text{max}} = 20 \times 0.3 \times 1.5 \times 4.0 \times 10^{-3}$
$= 20 \times 0.3 \times 1.5 \times 0.004$
$= 20 \times 0.0018$
$= \mathbf{0.036 \text{ N·m}}$ [2]
(1 mark for correct formula and substitution, 1 mark for correct answer with unit)

(e) Any two of:

Increase the number of turns on the coil
Increase the current
Increase the magnetic flux density (stronger magnets or soft iron core)
Increase the area of the coil [2]
(1 mark each)

23

(a) $\frac{V_s}{V_p} = \frac{N_s}{N_p}$
$V_s = V_p \times \frac{N_s}{N_p} = 240 \times \frac{60}{1200} = 240 \times 0.05 = \mathbf{12 \text{ V}}$ [2]
(1 mark for correct ratio, 1 mark for correct answer with unit)

(b) For a transformer with efficiency $\eta$ :
$\eta = \frac{V_s I_s}{V_p I_p} \Rightarrow I_p = \frac{V_s I_s}{\eta V_p}$
$\eta = 85\% = 0.85$
$I_p = \frac{12 \times 1.2}{0.85 \times 240} = \frac{14.4}{204} = \mathbf{0.0706 \text{ A}}$ (or 70.6 mA) [3]
(1 mark for efficiency formula, 1 mark for correct substitution, 1 mark for correct answer with unit)

Common mistake: Forgetting to convert 85% to 0.85, giving $I_p = 0.06 \text{ A}$ .

(c) The core is laminated to reduce eddy currents.

Changing magnetic flux induces e.m.f. in the solid core, causing circulating eddy currents [1]
Laminations (thin insulated sheets) increase resistance to eddy current paths, reducing their magnitude and thus reducing energy loss as heat [1]

(d) Transformers require a changing magnetic flux to induce e.m.f. in the secondary coil (Faraday's law). D.C. produces a constant magnetic flux, so no e.m.f. is induced in the secondary (except momentarily at switch-on/off). [1]

24

(a) As the magnet falls through the copper tube:

The magnetic flux through the copper tube changes (increases as magnet approaches, decreases as it leaves) [1]
By Faraday's law, this changing flux induces an e.m.f. and eddy currents in the copper tube [1]
By Lenz's law, the direction of the induced eddy currents is such that they create a magnetic field opposing the change — i.e., opposing the motion of the magnet [1]
This results in an upward magnetic force on the magnet, opposing gravity, so the magnet falls slower than in the plastic tube (where no eddy currents can flow) [1]

(b) As the magnet falls:

Its speed increases, so the rate of change of flux increases [1]
This induces larger eddy currents and a larger opposing magnetic force [1]
The upward magnetic force increases until it balances the weight of the magnet
At this point, net force is zero, acceleration is zero, and the magnet reaches terminal velocity [1]
(Any two well-explained points for 2 marks)

(c) The fall time would decrease (magnet falls faster, closer to plastic tube time).
Explanation: Vertical slits break the circular paths for eddy currents, greatly increasing resistance to eddy current flow. This reduces the magnitude of eddy currents and thus the opposing magnetic force. The magnet experiences less magnetic braking. [2]
(1 mark for correct prediction, 1 mark for correct explanation)

25

(a)(i) Induced e.m.f. $\mathcal{E} = B L v$ (for wire moving perpendicular to field)
$\mathcal{E} = 0.4 \times 0.2 \times 0.5 = \mathbf{0.04 \text{ V}}$ [2]
(1 mark for correct formula, 1 mark for correct answer with unit)

(a)(ii) $I = \frac{\mathcal{E}}{R} = \frac{0.04}{2.0} = \mathbf{0.02 \text{ A}}$ (or 20 mA) [1]

(a)(iii) Direction: Q to P (or P to Q depending on exact setup — let's determine)
Using Fleming's right-hand rule (generator rule):

First finger (Field): Into page
Thumb (Motion): To the right
Second finger (Current): Q to P (upwards on the moving wire) [1]

Lenz's law explanation: The induced current creates a magnetic field that opposes the change. As PQ moves right, the area of the loop increases, increasing flux into the page. The induced current produces a field out of the page to oppose this increase. Using right-hand grip rule, current Q to P produces field out of page. [1]

(b) The galvanometer deflection would be in the opposite direction (same magnitude). [1]

(c) $\mathcal{E} = B L v$ . If $B$ is doubled and $v$ is halved: $\mathcal{E}_{\text{new}} = (2B) L (v/2) = B L v = \mathcal{E}_{\text{original}}$ .
The induced e.m.f. remains unchanged. [1]

Section C: Longer Structured Questions [20 marks]

26

(a) Power $P = V I \Rightarrow I = \frac{P}{V} = \frac{100 \times 10^6}{400 \times 10^3} = \frac{100 \times 10^3}{400} = \mathbf{250 \text{ A}}$ [2]
(1 mark for correct formula and substitution, 1 mark for correct answer with unit)

(b) Power loss $P_{\text{loss}} = I^2 R = (250)^2 \times 10 = 62,500 \times 10 = \mathbf{625,000 \text{ W}} = \mathbf{625 \text{ kW}}$ [2]
(1 mark for correct formula and substitution, 1 mark for correct answer with unit)

(c) Percentage loss = $\frac{P_{\text{loss}}}{P_{\text{generated}}} \times 100\% = \frac{625 \times 10^3}{100 \times 10^6} \times 100\% = \mathbf{0.625\%}$ [2]
(1 mark for correct formula, 1 mark for correct answer with unit)

(d) At 25 kV: $I = \frac{100 \times 10^6}{25 \times 10^3} = 4000 \text{ A}$
$P_{\text{loss}} = I^2 R = (4000)^2 \times 10 = 16 \times 10^6 \times 10 = \mathbf{160 \text{ MW}}$ [2]
(1 mark for correct current, 1 mark for correct power loss)

Note: Power loss (160 MW) exceeds generated power (100 MW) — this shows transmission

<stage5_exam_answers_md>

TuitionGoWhere Practice Paper - Pure Physics Secondary 4 (Answer Key)

Paper: Practice Paper 3 (Version 3 of 5)
Subject: Pure Physics
Level: Secondary 4
Total Marks: 80

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 = $2.2 \text{ kW} \times \frac{15}{60} \text{ h} = 0.55 \text{ kWh}$
3	B	$F = BIL = 0.2 \times 5 \times 0.1 = 0.1 \text{ N}$
4	B	Faraday's law: e.m.f. induced when magnetic flux linkage changes
5	C	Torque = $BINA$ ; resistance does not affect turning effect
6	A	Maximum e.m.f. when coil plane parallel to field (rate of flux change maximum)
7	C	$I_s = I_p \times \frac{N_p}{N_s} = 0.5 \times \frac{800}{200} = 2.0 \text{ A}$
8	B	Earth wire provides safety path if live touches metal casing
9	D	Fleming's left-hand rule: field into page, current left→right, force down
10	B	$P = VI$ ; high voltage → lower current for same power → lower $I^2R$ loss
11	C	Soft iron core concentrates magnetic flux, increasing field strength
12	B	Period = $4 \text{ div} \times 5 \text{ ms/div} = 20 \text{ ms}$ ; $f = \frac{1}{0.02} = 50 \text{ Hz}$
13	A	$F = BIL\sin\theta = 0.6 \times 4 \times 0.5 \times \sin30° = 0.6 \text{ N}$
14	C	Total current = $\frac{8 \times 60}{230} = 2.09 \text{ A} < 13 \text{ A}$ ; fuse will not blow
15	A	Lenz's law: induced current opposes the change producing it
16	C	$N_p = N_s \times \frac{V_p}{V_s} = 60 \times \frac{240}{12} = 1200 \text{ turns}$
17	D	Field left→right, current into page → force downwards (Fleming's left-hand rule)
18	D	Switch must be on live wire to control appliance; lamp is the appliance
19	B	$V_{\text{rms}} = \frac{V_{\text{peak}}}{\sqrt{2}} = \frac{20}{\sqrt{2}} \approx 14.1 \text{ V}$
20	A	$F = BIL$ ; force directly proportional to current → straight line through origin

Section B: Structured Questions [40 marks]

21

(a) Total power = $2.2 + 1.2 + 0.9 + 0.3 = \boxed{4.6 \text{ kW}}$ [1]

(b) $I = \frac{P}{V} = \frac{4600}{230} = \boxed{20 \text{ A}}$ [2]

(c) No, the circuit breaker will not trip. Total current (20 A) < circuit breaker rating (30 A). [2]

(d) Energy = $0.3 \text{ kW} \times 24 \text{ h} = \boxed{7.2 \text{ kWh}}$ [1]

(e) Cost = $7.2 \times 30 \times \$ 0.28 = \boxed{$60.48}$ [2]

Total: [8]

22

(a) Arrow on side AB pointing downwards (using Fleming's left-hand rule: field N→S, current A→B). [1]

(b) Current in AB and CD flows in opposite directions. Forces on AB and CD are equal, opposite, and not collinear → form a couple producing torque. [2]

(c) Reverses current direction in coil every half-turn to maintain continuous rotation in same direction. [1]

(d) $\tau_{\text{max}} = BINA = 0.3 \times 1.5 \times 20 \times (4.0 \times 10^{-3}) = \boxed{0.036 \text{ N·m}}$ [2]

(e) Any two: Increase number of turns; Increase current; Increase magnetic field strength (stronger magnets/soft iron core); Increase coil area. [2]

Total: [8]

23

(a) $\frac{V_s}{V_p} = \frac{N_s}{N_p}$ → $V_s = 240 \times \frac{60}{1200} = \boxed{12 \text{ V}}$ [2]

(b) $P_{\text{out}} = V_s I_s = 12 \times 1.2 = 14.4 \text{ W}$
$P_{\text{in}} = \frac{P_{\text{out}}}{0.85} = \frac{14.4}{0.85} = 16.94 \text{ W}$
$I_p = \frac{P_{\text{in}}}{V_p} = \frac{16.94}{240} = \boxed{0.0706 \text{ A}}$ [3]

(c) Laminations reduce eddy currents induced in the core, reducing heating/energy loss. [2]

(d) Transformers require changing magnetic flux to induce e.m.f. in secondary; d.c. produces constant flux. [1]

Total: [8]

24

(a) As magnet falls through copper tube, magnetic flux through tube changes → induces e.m.f. (Faraday's law) → eddy currents flow in tube. By Lenz's law, these currents create magnetic field opposing magnet's motion → upward magnetic force reduces net downward force → magnet accelerates slower than in plastic tube (no induced currents). [4]

(b) As speed increases, rate of flux change increases → larger eddy currents → larger opposing magnetic force. At terminal velocity, upward magnetic force = weight → net force zero → constant velocity. [2]

(c) Fall time decreases (magnet falls faster). Vertical slits break eddy current paths → reduced eddy currents → smaller opposing force → less retardation. [2]

Total: [8]

25

(a)(i) $\mathcal{E} = BLv = 0.4 \times 0.2 \times 0.5 = \boxed{0.04 \text{ V}}$ [2]

(a)(ii) $I = \frac{\mathcal{E}}{R} = \frac{0.04}{2.0} = \boxed{0.02 \text{ A}}$ [1]

(a)(iii) Direction: Q to P (anticlockwise in frame). As PQ moves right, flux into page increases. Induced current creates field out of page (opposing increase) → anticlockwise current → Q to P along PQ. [2]

(b) Galvanometer deflects in opposite direction (current reverses to P to Q). [1]

(c) Induced e.m.f. remains unchanged ( $\mathcal{E} \propto Bv$ ; $2B \times \frac{v}{2} = Bv$ ). [1]

Total: [7]

Section C: Longer Structured Questions [20 marks]

26

(a) $I = \frac{P}{V} = \frac{100 \times 10^6}{400 \times 10^3} = \boxed{250 \text{ A}}$ [2]

(b) $P_{\text{loss}} = I^2R = 250^2 \times 10 = \boxed{625 \text{ kW}}$ [2]

(c) $\% \text{ loss} = \frac{625 \times 10^3}{100 \times 10^6} \times 100\% = \boxed{0.625\%}$ [2]

(d) At 25 kV: $I = \frac{100 \times 10^6}{25 \times 10^3} = 4000 \text{ A}$
$P_{\text{loss}} = 4000^2 \times 10 = \boxed{160 \text{ MW}}$ [2]

(e) High voltage reduces current for same power ( $P=VI$ ). Power loss $= I^2R$ , so reducing current dramatically reduces losses. At 400 kV loss is 0.625%; at 25 kV loss would be 160% (impossible - more power lost than generated). [2]

(f) Transformers rely on changing magnetic flux to induce e.m.f. in secondary coil (Faraday's law). A.c. produces continuously changing flux; d.c. produces constant flux (no induction after initial switch-on). [2]

(g) $\frac{V_s}{V_p} = \frac{N_s}{N_p}$ → $N_s = 500 \times \frac{400}{25} = \boxed{8000 \text{ turns}}$ [2]

(h) Any two: Eddy current losses in core; Hysteresis losses in core; Copper losses ( $I^2R$ heating in windings); Flux leakage (not all flux links both coils). [2]

Total: [16]

27

(a) Time-base: 2 ms/div; Y-gain: 5 V/div
Period = $4 \text{ div} \times 2 \text{ ms/div} = 8 \text{ ms}$
Frequency $f = \frac{1}{T} = \frac{1}{8 \times 10^{-3}} = \boxed{125 \text{ Hz}}$ [2]

(b) Peak voltage = $3 \text{ div} \times 5 \text{ V/div} = 15 \text{ V}$
$V_{\text{rms}} = \frac{V_{\text{peak}}}{\sqrt{2}} = \frac{15}{\sqrt{2}} = \boxed{10.6 \text{ V}}$ [2]

(c) The trace shows a sinusoidal waveform → alternating current. [1]

(d) Peak-to-peak = $6 \text{ div} \times 5 \text{ V/div} = 30 \text{ V}$
Peak = 15 V (unchanged)
New Y-gain: peak = 2 div → $2 \times \text{new Y-gain} = 15$ → Y-gain = $\boxed{7.5 \text{ V/div}}$ [2]

(e) The CRO displays voltage vs time. It cannot directly measure current. To measure current, a known resistor is placed in circuit and voltage across it is measured ( $I = V/R$ ). [2]

Total: [9]

Grand Total: 80 marks

Marking Notes for Teachers

Section A: 1 mark per correct answer. No half marks.
Section B: Award marks for correct formulas, substitutions, and final answers with units. Allow ecf (error carried forward) where appropriate.
Section C: Longer questions require clear explanations with physics terminology. Award marks for key concepts mentioned.
Significant figures: Accept 2-3 significant figures unless exact value required.
Units: Deduct 1 mark per question for missing/incorrect units in final answer.