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Secondary 4 Pure Physics Preliminary Examination Paper 1
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Questions
TuitionGoWhere Practice Paper - Pure Physics Secondary 4
PRELIMINARY EXAMINATION (Version 1)
TuitionGoWhere Secondary School (AI)
| Subject: | Pure Physics (6091) |
| Level: | Secondary 4 |
| Paper: | PRELIM Paper 2 – Structured Questions |
| Duration: | 1 hour 30 minutes |
| Total Marks: | 60 |
Name: _______________________________
Class: _______________________________
Date: _______________________________
Instructions to Candidates
- This paper consists of Section A (20 marks) and Section B (40 marks).
- 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.
- Take g = 10 m/s² unless otherwise stated.
- You may use a scientific calculator.
Section A: Short Structured Questions
[20 marks]
Answer all questions in this section.
Question 1: Static Electricity
(a) State the SI unit of electric charge.
__________________________________________________________________________________ [1]
(b) A student rubs a polythene rod with a woollen cloth. The rod becomes negatively charged.
(i) Explain, in terms of particle movement, how the rod becomes negatively charged.
__________________________________________________________________________________ [2]
(ii) The negatively charged rod is brought near a small, uncharged piece of aluminium foil lying on a table. The foil is attracted to the rod. Explain why this happens.
__________________________________________________________________________________ [2]
Question 2: Current of Electricity
A charge of 48 C flows through a lamp in 2 minutes.
(a) Define electric current.
__________________________________________________________________________________ [1]
(b) Calculate the current flowing through the lamp.
__________________________________________________________________________________ [2]
(c) The lamp has a resistance of 15 Ω. Calculate the potential difference across the lamp.
__________________________________________________________________________________ [2]
Question 3: D.C. Circuits
Two resistors, 4.0 Ω and 12.0 Ω, are connected in parallel across a 6.0 V battery.
(a) Calculate the combined resistance of the two resistors.
__________________________________________________________________________________ [2]
(b) Calculate the total current drawn from the battery.
__________________________________________________________________________________ [1]
(c) State the potential difference across the 12.0 Ω resistor. Explain your answer.
__________________________________________________________________________________ [2]
Question 4: Practical Electricity
A household electric kettle is rated at 240 V, 1800 W.
(a) Calculate the current drawn by the kettle when operating at its rated voltage.
__________________________________________________________________________________ [1]
(b) The kettle is used to heat water for 5 minutes. Calculate the electrical energy consumed in joules.
__________________________________________________________________________________ [2]
(c) State one advantage of using a circuit breaker rather than a fuse to protect the kettle circuit.
__________________________________________________________________________________ [1]
(d) State the function of the earth wire connected to the metal body of the kettle.
__________________________________________________________________________________ [1]
Section B: Extended Structured Questions
[40 marks]
Answer all questions in this section.
Question 5: Magnetism and Electromagnetism
Figure 5.1 shows a bar magnet placed near a plotting compass.
N S
[=======|===================] ○
bar magnet compass
Figure 5.1
(a) On Figure 5.1, draw the magnetic field pattern around the bar magnet. Show the direction of the field lines. [2]
(b) The compass needle points in a particular direction. Explain why.
__________________________________________________________________________________ [2]
(c) A student winds a coil of insulated copper wire around a soft iron core and connects it to a d.c. supply to make an electromagnet.
(i) State one way the student could increase the strength of the electromagnet.
__________________________________________________________________________________ [1]
(ii) Explain why soft iron is used as the core material rather than steel.
__________________________________________________________________________________ [2]
(iii) State one practical application of an electromagnet.
__________________________________________________________________________________ [1]
Question 6: Force on a Current-Carrying Conductor
Figure 6.1 shows a straight wire placed between the poles of a permanent magnet. The wire is connected to a d.c. supply.
N
[ ]
==================== wire
[ ]
S
Figure 6.1
(a) When the switch is closed, a current flows through the wire. State what happens to the wire and explain why.
__________________________________________________________________________________ [3]
(b) State two ways the direction of the force on the wire could be reversed.
-
-
________________________________________________________________________________ [2]
(c) This principle is used in a d.c. motor. State the function of the split-ring commutator in a d.c. motor.
__________________________________________________________________________________ [2]
Question 7: Electromagnetic Induction
A student investigates electromagnetic induction using a coil of wire connected to a sensitive centre-zero galvanometer and a bar magnet.
(a) The student pushes the north pole of the magnet quickly into the coil. State what is observed on the galvanometer.
__________________________________________________________________________________ [1]
(b) The magnet is held stationary inside the coil. State what is observed on the galvanometer. Explain your answer.
__________________________________________________________________________________ [2]
(c) The student then pulls the magnet quickly out of the coil. State and explain what is observed on the galvanometer.
__________________________________________________________________________________ [2]
(d) State two ways the student could increase the magnitude of the induced e.m.f. without changing the magnet.
-
-
________________________________________________________________________________ [2]
Question 8: Transformers
A step-down transformer is used to operate a 12 V, 24 W lamp from a 240 V a.c. mains supply. The transformer has 1200 turns on the primary coil.
(a) Explain why the transformer will not work if connected to a d.c. supply.
__________________________________________________________________________________ [2]
(b) Calculate the number of turns on the secondary coil.
__________________________________________________________________________________ [2]
(c) Calculate the current in the secondary coil when the lamp is operating at its rated power.
__________________________________________________________________________________ [2]
(d) The transformer has an efficiency of 80%. Calculate the current in the primary coil.
__________________________________________________________________________________ [3]
(e) State one reason why a transformer is not 100% efficient.
__________________________________________________________________________________ [1]
Question 9: Electricity and Magnetism – Integrated Application
A small a.c. generator is used to light a lamp. The generator consists of a rectangular coil that rotates between the poles of a permanent magnet. The coil is connected to an external circuit via slip rings and carbon brushes.
(a) Explain how an e.m.f. is induced in the coil as it rotates.
__________________________________________________________________________________ [3]
(b) Sketch a graph of the output voltage against time for one complete rotation of the coil. Label the axes clearly. [2]
(c) The generator is connected to a lamp with a resistance of 8.0 Ω. At a particular instant, the output voltage is 6.0 V. Calculate the power dissipated in the lamp at that instant.
__________________________________________________________________________________ [2]
(d) State one difference between the slip rings used in an a.c. generator and the split-ring commutator used in a d.c. motor.
__________________________________________________________________________________ [1]
END OF PAPER
This paper was generated by TuitionGoWhere (AI) for practice purposes. It is not an official examination paper.
Answers
TuitionGoWhere Practice Paper - Pure Physics Secondary 4
PRELIMINARY EXAMINATION (Version 1) – ANSWER KEY
TuitionGoWhere Secondary School (AI)
Section A: Short Structured Questions
Question 1: Static Electricity
(a) Coulomb (C) ✓ [1]
(b)(i) Electrons are transferred from the woollen cloth to the polythene rod [1]. The rod gains excess electrons, giving it a net negative charge [1]. ✓✓ [2]
(b)(ii) The negatively charged rod repels electrons in the aluminium foil to the far side of the foil [1]. The side of the foil nearest the rod becomes positively charged (by induction). Since unlike charges attract, the foil is attracted to the rod [1]. ✓✓ [2]
Question 2: Current of Electricity
(a) Electric current is the rate of flow of electric charge. ✓ [1]
(b) I = Q / t I = 48 / (2 × 60) [1] I = 48 / 120 = 0.40 A [1] ✓✓ [2]
(c) V = IR V = 0.40 × 15 [1] V = 6.0 V [1] ✓✓ [2]
Question 3: D.C. Circuits
(a) 1/R = 1/R₁ + 1/R₂ 1/R = 1/4.0 + 1/12.0 = 3/12.0 + 1/12.0 = 4/12.0 [1] R = 12.0/4 = 3.0 Ω [1] ✓✓ [2]
(b) I = V/R = 6.0/3.0 = 2.0 A ✓ [1]
(c) The potential difference across the 12.0 Ω resistor is 6.0 V [1]. In a parallel circuit, the potential difference across each branch is equal to the supply voltage [1]. ✓✓ [2]
Question 4: Practical Electricity
(a) P = IV → I = P/V I = 1800/240 = 7.5 A ✓ [1]
(b) E = Pt E = 1800 × (5 × 60) [1] E = 1800 × 300 = 540,000 J (or 540 kJ) [1] ✓✓ [2]
(c) A circuit breaker can be reset and reused, whereas a fuse must be replaced after it blows. (Accept: Circuit breaker responds faster to overcurrent / more convenient) ✓ [1]
(d) The earth wire provides a low-resistance path for current to flow to ground if a fault occurs (e.g., live wire touches the metal body), preventing electric shock to the user. ✓ [1]
Section B: Extended Structured Questions
Question 5: Magnetism and Electromagnetism
(a) Field lines drawn from N to S outside the magnet, curving around from N to S [1]. Arrows on lines pointing away from N and toward S. Lines closer together near the poles (stronger field) [1]. ✓✓ [2]
(b) The compass needle is a small magnet. Its north pole is attracted to the south pole of the bar magnet and repelled by the north pole [1]. The needle aligns with the direction of the magnetic field at that point [1]. ✓✓ [2]
(c)(i) Increase the current in the coil / increase the number of turns in the coil / insert a soft iron core (if not already present). (Any one) ✓ [1]
(c)(ii) Soft iron is easily magnetised and demagnetised (it is a soft magnetic material) [1]. This means the electromagnet can be switched on and off quickly. Steel is a hard magnetic material that retains magnetism, so it would not work well for an electromagnet that needs to be controlled [1]. ✓✓ [2]
(c)(iii) Electric bell / relay / magnetic crane / loudspeaker / door lock. (Any one reasonable application) ✓ [1]
Question 6: Force on a Current-Carrying Conductor
(a) The wire experiences a force and moves (upwards or downwards depending on current and field directions) [1]. When a current-carrying conductor is placed in a magnetic field, the magnetic field of the current interacts with the permanent magnetic field [1]. This produces a force on the conductor (Fleming's left-hand rule / motor effect) [1]. ✓✓✓ [3]
(b)
- Reverse the direction of the current (by reversing the battery connections) [1].
- Reverse the direction of the magnetic field (by swapping the magnet poles) [1]. ✓✓ [2]
(c) The split-ring commutator reverses the direction of the current in the coil every half rotation [1]. This ensures that the force on each side of the coil always acts in the same rotational direction, keeping the coil rotating continuously [1]. ✓✓ [2]
Question 7: Electromagnetic Induction
(a) The galvanometer needle deflects momentarily (in one direction). ✓ [1]
(b) The galvanometer shows no deflection (needle remains at zero) [1]. When the magnet is stationary, there is no change in magnetic flux (magnetic field lines) through the coil. By Faraday's law, an e.m.f. is induced only when there is a change in magnetic flux [1]. ✓✓ [2]
(c) The galvanometer needle deflects momentarily in the opposite direction [1]. Removing the magnet causes a change in magnetic flux through the coil in the opposite sense. By Lenz's law, the induced e.m.f. (and current) opposes the change causing it, so the deflection is opposite to that in part (a) [1]. ✓✓ [2]
(d)
- Increase the number of turns in the coil [1].
- Move the magnet faster (increase the speed of insertion/removal) [1]. (Accept: Use a stronger magnet / insert a soft iron core into the coil) ✓✓ [2]
Question 8: Transformers
(a) A transformer works on the principle of electromagnetic induction, which requires a changing magnetic flux [1]. A d.c. supply produces a constant current and therefore a constant magnetic field in the primary coil. With no changing flux, no e.m.f. is induced in the secondary coil [1]. ✓✓ [2]
(b) Vₛ/Vₚ = Nₛ/Nₚ Nₛ = Nₚ × Vₛ/Vₚ Nₛ = 1200 × 12/240 [1] Nₛ = 1200 × 0.05 = 60 turns [1] ✓✓ [2]
(c) P = IV → Iₛ = P/Vₛ Iₛ = 24/12 [1] Iₛ = 2.0 A [1] ✓✓ [2]
(d) Efficiency η = (VₛIₛ) / (VₚIₚ) 0.80 = (12 × 2.0) / (240 × Iₚ) [1] 0.80 = 24 / (240 × Iₚ) 240 × Iₚ = 24/0.80 = 30 [1] Iₚ = 30/240 = 0.125 A (or 0.13 A) [1] ✓✓✓ [3]
(e) Energy is lost as heat due to resistance in the coils (copper losses) / energy lost due to eddy currents induced in the iron core / energy lost due to magnetisation and demagnetisation of the core (hysteresis loss) / flux leakage. (Any one) ✓ [1]
Question 9: Electricity and Magnetism – Integrated Application
(a) As the coil rotates, the angle between the plane of the coil and the magnetic field changes continuously [1]. This causes the magnetic flux (field lines) passing through the coil to change continuously [1]. By Faraday's law, a changing magnetic flux induces an e.m.f. across the ends of the coil [1]. ✓✓✓ [3]
(b) Graph should show:
- A sinusoidal (sine wave) curve [1]
- Voltage axis (y-axis) labelled "Voltage" or "e.m.f." with positive and negative values
- Time axis (x-axis) labelled "Time"
- One complete cycle shown, crossing zero twice [1] ✓✓ [2]
(c) P = V²/R P = (6.0)²/8.0 [1] P = 36/8.0 = 4.5 W [1] ✓✓ [2]
(d) Slip rings allow the coil to maintain continuous electrical connection while producing alternating current (the connections do not reverse). A split-ring commutator reverses the connections every half rotation to produce direct current in the external circuit. (Accept any valid comparison) ✓ [1]
END OF ANSWER KEY
Marking Notes:
- Marks are awarded for correct method and final answer in calculation questions.
- Accept alternative correct phrasings for explanation questions.
- For graph sketching, axes must be labelled and shape must be approximately correct.
- Numerical answers should be given to 2 or 3 significant figures unless exact.
- Units must be included in final answers for calculation questions.
This answer key was generated by TuitionGoWhere (AI) for practice purposes.