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Secondary 4 Pure Physics Preliminary Examination Paper 3

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TuitionGoWhere Practice Paper — Pure Physics Secondary 4

TuitionGoWhere Secondary School (AI)

Subject: Pure Physics (6091)
Level: Secondary 4
Paper: Preliminary Examination — Electricity & Magnetism
Duration: 1 hour 15 minutes
Total Marks: 60
Version: 3 of 5

Name: _________________________
Class: _________________________
Date: _________________________

Instructions to Candidates

This paper consists of three sections: Section A, Section B, and Section C.
Answer all questions.
Write your answers in the spaces provided.
Show all working for calculation questions. Marks are awarded for correct method and final answer.
The number of marks is given in brackets [ ] at the end of each question or part question.
You may use a scientific calculator.
Take g = 10 m/s² where required.

Section A: Structured Questions (20 marks)

Answer all questions in this section.

1. A student rubs a polythene rod with a woollen cloth. The rod becomes negatively charged.

(a) Explain, in terms of electron transfer, how the polythene rod becomes negatively charged. [2]

(b) The negatively charged rod is brought near a small, uncharged aluminium foil ball suspended by an insulating thread. The ball is attracted to the rod.

Explain why the uncharged ball is attracted to the negatively charged rod. [2]

2. A circuit consists of a 12 V battery connected to a resistor of resistance 6.0 Ω.

(a) Calculate the current flowing through the resistor. [2]

(b) Calculate the charge that passes through the resistor in 5.0 minutes. [2]

3. A student investigates the I-V characteristic of a filament lamp. The results are shown in the table below.

Voltage / V	0	1.0	2.0	3.0	4.0	5.0	6.0
Current / A	0	0.12	0.18	0.22	0.25	0.27	0.29

(a) On the grid below, plot a graph of current against voltage. [3]

(Grid space provided — draw a smooth curve through the points)

(b) Using your graph, explain how the resistance of the filament lamp changes as the voltage increases. [2]

(c) Explain why the resistance of the filament lamp changes in this way. [2]

4. An electric kettle is rated at 2200 W, 240 V.

(a) Calculate the current drawn by the kettle when operating at its rated voltage. [2]

(b) The kettle is used to heat 1.5 kg of water from 25 °C to 100 °C. The specific heat capacity of water is 4200 J/(kg·°C).

Calculate the minimum time required to heat the water, assuming no energy losses. [3]

Section B: Circuit Analysis (20 marks)

Answer all questions in this section.

5. Figure 5.1 shows a circuit with two resistors connected in parallel to a 6.0 V battery.

(Circuit diagram: 6.0 V battery connected to a 4.0 Ω resistor and a 12.0 Ω resistor in parallel.)

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

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

(c) Calculate the current flowing through the 4.0 Ω resistor. [2]

(d) Calculate the power dissipated in the 12.0 Ω resistor. [2]

6. A potential divider circuit consists of a 10.0 Ω fixed resistor and a variable resistor connected in series across a 9.0 V supply. The output voltage is taken across the fixed resistor.

(a) Draw a labelled circuit diagram of this potential divider. [2]

(Space for diagram)

(b) The variable resistor is set to 20.0 Ω. Calculate the output voltage across the fixed resistor. [3]

(c) State what happens to the output voltage when the resistance of the variable resistor is decreased. Explain your answer. [2]

7. A household electrical circuit is protected by a 13 A fuse. The mains supply is 240 V.

(a) State the function of the fuse in the circuit. [1]

(b) Calculate the maximum power that can be safely drawn from this circuit. [2]

(c) An electric heater rated at 3000 W, 240 V is connected to this circuit. Determine whether the 13 A fuse is suitable for this heater. Show your working. [2]

Section C: Electromagnetism & Induction (20 marks)

Answer all questions in this section.

8. A straight wire carrying a current is placed between the poles of a permanent magnet, as shown in Figure 8.1.

(Diagram: Wire perpendicular to magnetic field between N and S poles.)

(a) State the direction of the force acting on the wire. You may use Fleming's left-hand rule to help you. [1]

(b) State two ways in which the magnitude of the force on the wire can be increased. [2]

(c) The wire has a length of 0.15 m within the magnetic field. The magnetic flux density is 0.80 T and the current in the wire is 2.5 A.

Calculate the force acting on the wire when the wire is perpendicular to the magnetic field. [2]

9. A student investigates electromagnetic induction using a bar magnet and a coil of wire connected to a sensitive galvanometer.

(a) The student pushes the north pole of the magnet into the coil. State what is observed on the galvanometer. [1]

(b) The magnet is now held stationary inside the coil. State what is observed on the galvanometer. Explain your answer. [2]

(c) The student then pulls the magnet out of the coil quickly. State and explain how the galvanometer reading compares with that in part (a). [3]

10. A step-down transformer is used to operate a 12 V, 24 W lamp from a 240 V mains supply. The transformer is 80% efficient.

(a) State what is meant by a "step-down" transformer. [1]

(b) Calculate the turns ratio (Nₚ : Nₛ) for an ideal transformer operating between 240 V and 12 V. [2]

(c) Calculate the current in the secondary coil when the lamp is operating at its rated power. [2]

(d) Calculate the current in the primary coil, taking into account the 80% efficiency. [3]

(e) Explain why the transformer is not 100% efficient. State one cause of energy loss in a practical transformer. [2]

— END OF PAPER —

Answers

TuitionGoWhere Practice Paper — Pure Physics Secondary 4

Answer Key and Marking Scheme

Paper: Preliminary Examination — Electricity & Magnetism
Version: 3 of 5
Total Marks: 60

Section A: Structured Questions (20 marks)

1. A student rubs a polythene rod with a woollen cloth. The rod becomes negatively charged.

(a) Explain, in terms of electron transfer, how the polythene rod becomes negatively charged. [2]

Answer:

When the polythene rod is rubbed with the woollen cloth, electrons are transferred from the wool to the polythene rod. [1]
The polythene rod gains electrons and therefore has an excess of negative charge, making it negatively charged. [1]

Marking notes:

Award [1] for stating electrons transfer from wool to polythene.
Award [1] for linking excess electrons to negative charge.
Accept: "Polythene gains electrons from the wool" for both marks if clearly stated.

(b) The negatively charged rod is brought near a small, uncharged aluminium foil ball suspended by an insulating thread. The ball is attracted to the rod.

Explain why the uncharged ball is attracted to the negatively charged rod. [2]

Answer:

The negatively charged rod repels electrons in the aluminium ball to the far side of the ball (electrostatic induction). [1]
The side of the ball nearer to the rod becomes positively charged. Since unlike charges attract, the ball is attracted to the negatively charged rod. [1]

Marking notes:

Award [1] for describing induction (electrons move away from rod / positive charge induced on near side).
Award [1] for linking induced positive charge to attraction.
Accept: "The rod induces opposite charges on the ball; the positive side is closer so attraction occurs."

2. A circuit consists of a 12 V battery connected to a resistor of resistance 6.0 Ω.

(a) Calculate the current flowing through the resistor. [2]

Answer:

Using Ohm's Law: V = IR
I = V / R = 12 / 6.0 = 2.0 A [1 for correct substitution, 1 for correct answer with units]

Marking notes:

Award [1] for correct formula/substitution.
Award [1] for correct answer: 2.0 A.
Accept 2 A.

(b) Calculate the charge that passes through the resistor in 5.0 minutes. [2]

Answer:

Using Q = I × t
t = 5.0 min = 5.0 × 60 = 300 s [1 for time conversion]
Q = 2.0 × 300 = 600 C [1 for correct answer with units]

Marking notes:

Award [1] for converting minutes to seconds (300 s).
Award [1] for correct answer: 600 C.
Allow ecf from part (a) if current is incorrect but method is correct.

3. A student investigates the I-V characteristic of a filament lamp.

(a) On the grid below, plot a graph of current against voltage. [3]

Answer:

Correct plotting of all 7 points (±½ small square) [2]
Smooth curve drawn through points (not dot-to-dot) [1]

Marking notes:

Award [2] for all points correctly plotted; [1] if 5-6 points correct.
Award [1] for smooth curve (should show decreasing gradient as V increases).
Curve must pass through origin.

(b) Using your graph, explain how the resistance of the filament lamp changes as the voltage increases. [2]

Answer:

The gradient of the I-V graph decreases as voltage increases. [1]
Since resistance R = V/I = 1/gradient, the resistance of the filament lamp increases as the voltage increases. [1]

Marking notes:

Award [1] for noting gradient decreases / curve flattens.
Award [1] for linking decreasing gradient to increasing resistance.
Accept: "Resistance increases because the current does not increase proportionally with voltage."

(c) Explain why the resistance of the filament lamp changes in this way. [2]

Answer:

As current increases, the filament gets hotter. [1]
The increased temperature causes increased vibration of the metal ions in the filament, which increases the frequency of collisions between free electrons and ions, increasing resistance. [1]

Marking notes:

Award [1] for linking to temperature increase.
Award [1] for explaining increased ion vibration → more electron-ion collisions → higher resistance.
Accept reference to "positive ions vibrate more vigorously, impeding electron flow."

4. An electric kettle is rated at 2200 W, 240 V.

(a) Calculate the current drawn by the kettle when operating at its rated voltage. [2]

Answer:

Using P = IV
I = P / V = 2200 / 240 = 9.17 A (or 9.2 A) [1 for substitution, 1 for correct answer with units]

Marking notes:

Award [1] for correct formula/substitution.
Award [1] for correct answer: 9.17 A or 9.2 A (accept 9.2 A to 2 sf).

(b) The kettle is used to heat 1.5 kg of water from 25 °C to 100 °C. The specific heat capacity of water is 4200 J/(kg·°C).

Calculate the minimum time required to heat the water, assuming no energy losses. [3]

Answer:

Energy required: Q = mcΔθ
Q = 1.5 × 4200 × (100 - 25) = 1.5 × 4200 × 75 = 472,500 J [1]
Power = Energy / time, so t = Q / P
t = 472,500 / 2200 = 214.8 s ≈ 215 s (or 3 min 35 s) [1 for method, 1 for correct answer]

Marking notes:

Award [1] for correct calculation of energy (472,500 J).
Award [1] for using t = E/P.
Award [1] for correct answer: 215 s (accept 214.8 s or 3.58 min).
Allow ecf if energy calculation is incorrect but method is correct.

Section B: Circuit Analysis (20 marks)

5. Figure 5.1 shows a circuit with two resistors connected in parallel to a 6.0 V battery.

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

Answer:

For parallel resistors: 1/R_eff = 1/R₁ + 1/R₂
1/R_eff = 1/4.0 + 1/12.0 = 3/12.0 + 1/12.0 = 4/12.0 = 1/3.0 [1]
R_eff = 3.0 Ω [1]

Marking notes:

Award [1] for correct formula and substitution.
Award [1] for correct answer: 3.0 Ω.

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

Answer:

Using Ohm's Law: I = V / R_eff
I = 6.0 / 3.0 = 2.0 A [1 for substitution, 1 for correct answer with units]

Marking notes:

Award [1] for correct formula/substitution.
Award [1] for correct answer: 2.0 A.
Allow ecf from part (a).

(c) Calculate the current flowing through the 4.0 Ω resistor. [2]

Answer:

In parallel, voltage across each resistor = 6.0 V
I₄ = V / R = 6.0 / 4.0 = 1.5 A [1 for method, 1 for correct answer with units]

Marking notes:

Award [1] for recognising voltage across resistor = 6.0 V.
Award [1] for correct answer: 1.5 A.

(d) Calculate the power dissipated in the 12.0 Ω resistor. [2]

Answer:

Current through 12.0 Ω resistor: I₁₂ = V / R = 6.0 / 12.0 = 0.5 A [1]
Power: P = I²R = (0.5)² × 12.0 = 0.25 × 12.0 = 3.0 W [1]
Alternative: P = V²/R = (6.0)² / 12.0 = 36 / 12.0 = 3.0 W

Marking notes:

Award [1] for correct current through 12 Ω resistor (0.5 A) or correct formula.
Award [1] for correct answer: 3.0 W.
Accept any valid method (P = IV, P = I²R, P = V²/R).

6. A potential divider circuit consists of a 10.0 Ω fixed resistor and a variable resistor connected in series across a 9.0 V supply.

(a) Draw a labelled circuit diagram of this potential divider. [2]

Answer:

Correct series arrangement of fixed resistor and variable resistor across 9.0 V supply [1]
Output voltage labelled correctly across the fixed resistor [1]

Marking notes:

Award [1] for correct circuit layout (series connection, battery symbol correct).
Award [1] for clear labelling of output voltage (V_out) across the 10.0 Ω fixed resistor.
Variable resistor symbol must be shown (arrow through resistor or box with arrow).

(b) The variable resistor is set to 20.0 Ω. Calculate the output voltage across the fixed resistor. [3]

Answer:

Total resistance: R_total = 10.0 + 20.0 = 30.0 Ω [1]
Current in circuit: I = V / R_total = 9.0 / 30.0 = 0.30 A [1]
Output voltage: V_out = I × R_fixed = 0.30 × 10.0 = 3.0 V [1]
Alternative: V_out = (R_fixed / R_total) × V_supply = (10.0 / 30.0) × 9.0 = 3.0 V

Marking notes:

Award [1] for total resistance (30.0 Ω).
Award [1] for current (0.30 A) or correct ratio method.
Award [1] for correct answer: 3.0 V.

(c) State what happens to the output voltage when the resistance of the variable resistor is decreased. Explain your answer. [2]

Answer:

The output voltage increases. [1]
When the variable resistor decreases, the total resistance decreases, so the current increases. Since V_out = I × R_fixed and R_fixed is constant, the output voltage increases. OR: The fixed resistor now has a larger proportion of the total resistance, so a larger share of the supply voltage appears across it. [1]

Marking notes:

Award [1] for stating output voltage increases.
Award [1] for valid explanation (current increases OR proportion argument).

7. A household electrical circuit is protected by a 13 A fuse. The mains supply is 240 V.

(a) State the function of the fuse in the circuit. [1]

Answer:

The fuse melts and breaks the circuit if the current exceeds its rated value, protecting the appliance and wiring from overheating / preventing electrical fires.

Marking notes:

Award [1] for stating that the fuse breaks the circuit when current is too high / prevents excessive current.
Accept: "Protects against overcurrent" or "Melts to break circuit when current exceeds 13 A."

(b) Calculate the maximum power that can be safely drawn from this circuit. [2]

Answer:

Using P = IV
P_max = 13 × 240 = 3120 W [1 for substitution, 1 for correct answer with units]

Marking notes:

Award [1] for correct formula/substitution.
Award [1] for correct answer: 3120 W (or 3.12 kW).

(c) An electric heater rated at 3000 W, 240 V is connected to this circuit. Determine whether the 13 A fuse is suitable for this heater. Show your working. [2]

Answer:

Current drawn by heater: I = P / V = 3000 / 240 = 12.5 A [1]
Since 12.5 A < 13 A, the fuse is suitable (the current is below the fuse rating). [1]

Marking notes:

Award [1] for calculating current (12.5 A).
Award [1] for correct conclusion with reasoning (12.5 A < 13 A, so fuse is suitable).
Accept: "Yes, because the heater draws 12.5 A which is less than the 13 A fuse rating."

Section C: Electromagnetism & Induction (20 marks)

8. A straight wire carrying a current is placed between the poles of a permanent magnet.

(a) State the direction of the force acting on the wire. You may use Fleming's left-hand rule to help you. [1]

Answer:

The force acts perpendicular to both the magnetic field direction and the current direction. (Direction depends on specific diagram orientation — accept: upward / downward / into page / out of page as appropriate to the diagram shown.)

Marking notes:

Award [1] for correct direction relative to the diagram.
If diagram shows conventional current direction and field direction, answer must be consistent with Fleming's left-hand rule.

(b) State two ways in which the magnitude of the force on the wire can be increased. [2]

Answer:

Any two from:
- Increase the current in the wire. [1]
- Use a stronger magnet (increase magnetic flux density). [1]
- Increase the length of wire within the magnetic field. [1]

Marking notes:

Award [1] for each correct method (max 2).
Do not accept "increase voltage" unless linked to increased current.

(c) The wire has a length of 0.15 m within the magnetic field. The magnetic flux density is 0.80 T and the current in the wire is 2.5 A.

Calculate the force acting on the wire when the wire is perpendicular to the magnetic field. [2]

Answer:

Using F = BIL (for wire perpendicular to field)
F = 0.80 × 2.5 × 0.15 = 0.30 N [1 for substitution, 1 for correct answer with units]

Marking notes:

Award [1] for correct formula and substitution.
Award [1] for correct answer: 0.30 N (or 0.3 N).

9. A student investigates electromagnetic induction using a bar magnet and a coil of wire connected to a sensitive galvanometer.

(a) The student pushes the north pole of the magnet into the coil. State what is observed on the galvanometer. [1]

Answer:

The galvanometer needle deflects momentarily (in one direction).

Marking notes:

Award [1] for "deflects momentarily" or "shows a momentary reading."
Must indicate deflection is temporary/transient.

(b) The magnet is now held stationary inside the coil. State what is observed on the galvanometer. Explain your answer. [2]

Answer:

The galvanometer shows no deflection (needle remains at zero). [1]
When the magnet is stationary, there is no change in magnetic flux linking the coil. According to Faraday's law, an EMF is induced only when there is a change in magnetic flux. [1]

Marking notes:

Award [1] for stating no deflection / zero reading.
Award [1] for explaining no change in magnetic flux → no induced EMF.

(c) The student then pulls the magnet out of the coil quickly. State and explain how the galvanometer reading compares with that in part (a). [3]

Answer:

The galvanometer needle deflects in the opposite direction. [1]
The deflection is larger (greater magnitude). [1]
Explanation: Pulling the magnet out causes a change in magnetic flux in the opposite direction, inducing an EMF in the opposite direction (Lenz's law). The faster motion causes a greater rate of change of magnetic flux, inducing a larger EMF (Faraday's law). [1]

Marking notes:

Award [1] for stating opposite direction.
Award [1] for stating larger deflection.
Award [1] for explaining opposite direction (flux decreasing vs increasing) and larger magnitude (faster rate of change of flux).
Accept reference to Lenz's law and Faraday's law.

10. A step-down transformer is used to operate a 12 V, 24 W lamp from a 240 V mains supply. The transformer is 80% efficient.

(a) State what is meant by a "step-down" transformer. [1]

Answer:

A step-down transformer has fewer turns on the secondary coil than on the primary coil, so the output (secondary) voltage is lower than the input (primary) voltage.

Marking notes:

Award [1] for stating secondary voltage is less than primary voltage OR Nₛ < Nₚ.

(b) Calculate the turns ratio (Nₚ : Nₛ) for an ideal transformer operating between 240 V and 12 V. [2]

Answer:

For an ideal transformer: Vₚ / Vₛ = Nₚ / Nₛ
Nₚ / Nₛ = 240 / 12 = 20 : 1 [1 for method, 1 for correct ratio]

Marking notes:

Award [1] for correct formula/substitution.
Award [1] for correct ratio: 20 : 1 (Nₚ : Nₛ).

(c) Calculate the current in the secondary coil when the lamp is operating at its rated power. [2]

Answer:

Using P = IV
Iₛ = P / Vₛ = 24 / 12 = 2.0 A [1 for substitution, 1 for correct answer with units]

Marking notes:

Award [1] for correct formula/substitution.
Award [1] for correct answer: 2.0 A.

(d) Calculate the current in the primary coil, taking into account the 80% efficiency. [3]

Answer:

Efficiency η = (Output power / Input power) × 100%
Output power = Vₛ × Iₛ = 12 × 2.0 = 24 W [1]
Input power = Output power / η = 24 / 0.80 = 30 W [1]
Iₚ = P_input / Vₚ = 30 / 240 = 0.125 A [1]

Marking notes:

Award [1] for output power (24 W) or recognising P_out = VₛIₛ.
Award [1] for input power (30 W) using efficiency.
Award [1] for correct primary current: 0.125 A (or 0.13 A).
Accept alternative method using η = (VₛIₛ) / (VₚIₚ) rearranged to Iₚ = (VₛIₛ) / (η × Vₚ).

(e) Explain why the transformer is not 100% efficient. State one cause of energy loss in a practical transformer. [2]

Answer:

Energy is lost due to: [1 for any one cause]
- Resistance in the coils (copper losses) — heat generated by current flowing through the wire.
- Eddy currents induced in the iron core — these circulating currents produce heat.
- Hysteresis losses — energy required to repeatedly magnetise and demagnetise the core.
- Magnetic flux leakage — not all flux from the primary links with the secondary coil.
Because of these losses, the output power is always less than the input power. [1]

Marking notes:

Award [1] for stating one valid cause of energy loss.
Award [1] for linking loss to reduced output power / efficiency < 100%.
Accept any valid cause with brief explanation.

— END OF ANSWER KEY —