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O Level Physics Practice Paper 3

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TuitionGoWhere Practice Paper - Physics O-Level

TuitionGoWhere Practice Paper (AI)

Field	Details
Subject:	Physics (6091)
Level:	O-Level
Paper:	Practice Paper — Version 3 of 5
Duration:	1 hour 45 minutes
Total Marks:	80

Name: _________________________ Class: _________________________ Date: _________________________

Instructions to Candidates

This paper consists of two sections: Section A and Section B.
Answer all questions in Section A.
In Section B, answer all questions.
Write your answers in the spaces provided.
Show all working clearly; marks are awarded for correct method.
State units in all final numerical answers.
Use g = 10 m/s² unless otherwise stated.
The number of marks is given in brackets [ ] at the end of each question or part question.
You are advised to spend about 55 minutes on Section A and 50 minutes on Section B.

Section A: Structured Questions (45 marks)

Answer all questions in this section.

Question 1: Static Electricity (6 marks)

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

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

(ii) State the charge acquired by the woollen cloth. Explain your answer. [1]

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

(i) Explain why the uncharged ball is attracted to the charged rod. [2]

(ii) After the ball touches the rod, it immediately moves away. Explain this observation. [1]

Question 2: Current of Electricity (8 marks)

(a) Define the term potential difference between two points in a circuit. [1]

(b) A charge of 240 C flows through a lamp in 4 minutes.

(i) Calculate the current flowing through the lamp. [2]

(ii) The potential difference across the lamp is 12 V. Calculate the resistance of the lamp. [2]

Potential Difference / V	0.0	2.0	4.0	6.0	8.0	10.0
Current / A	0.00	0.25	0.40	0.50	0.57	0.62

(i) Describe how the resistance of the filament lamp changes as the potential difference increases. [1]

(ii) Explain, in terms of the behaviour of particles in the filament, why the resistance changes in this way. [2]

Question 3: D.C. Circuits (9 marks)

A circuit consists of a 12 V battery connected to a 10 Ω fixed resistor in series with a parallel combination of a 15 Ω resistor and an unknown resistor R.

(a) Draw a clearly labelled circuit diagram of this arrangement. [2]

(b) The current drawn from the battery is 0.80 A.

(i) Calculate the total effective resistance of the circuit. [2]

(ii) Show that the effective resistance of the parallel combination is 5.0 Ω. [1]

(iii) Hence, calculate the value of the unknown resistor R. [2]

(c) The 15 Ω resistor is replaced with a thermistor. Describe and explain how the current drawn from the battery changes when the thermistor is heated. [2]

Question 4: Practical Electricity (7 marks)

(a) An electric kettle is rated at 240 V, 1800 W.

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

(ii) The kettle is used to heat water for 3 minutes. Calculate the energy consumed in kilowatt-hours (kWh). [2]

(b) The diagram below shows the three wires inside the cable of the kettle connected to a three-pin plug.

(i) State the colour of insulation on each of the three wires: Live, Neutral, and Earth. [1]

(ii) Explain the purpose of the earth wire in this appliance. [2]

Question 5: Magnetism and Electromagnetism (8 marks)

(a) A student places a bar magnet on a sheet of paper and sprinkles iron filings around it.

(i) Sketch the magnetic field pattern that the student would observe. Include arrows to show the direction of the field lines. [2]

(ii) State two properties of magnetic field lines. [2]

(b) A straight wire carrying a current of 3.0 A is placed perpendicular to a uniform magnetic field of flux density 0.40 T. The length of the wire within the field is 0.15 m.

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

(ii) State the rule used to determine the direction of the force on the wire. [1]

(iii) Describe one way to increase the magnitude of the force on the wire without changing the current or the magnetic field. [1]

Question 6: Electromagnetic Induction (7 marks)

(a) A bar magnet is pushed into a coil of wire connected to a sensitive centre-zero galvanometer. The galvanometer needle deflects to the right.

(i) Explain why an electromotive force (e.m.f.) is induced in the coil. [2]

(ii) State what happens to the galvanometer needle when the magnet is held stationary inside the coil. Explain your answer. [2]

(iii) The magnet is now pulled out of the coil rapidly. State and explain the direction of the galvanometer needle deflection compared to when the magnet was pushed in. [2]

(b) State one factor that affects the magnitude of the induced e.m.f. in the coil. [1]

Section B: Free Response Questions (35 marks)

Answer all questions in this section.

Question 7: Circuit Analysis and Design (12 marks)

A student designs a temperature-sensing circuit to switch on a cooling fan when the temperature exceeds 30 °C. The circuit uses a thermistor (resistance 500 Ω at 30 °C, decreasing with increasing temperature), a fixed resistor of 500 Ω, a 6.0 V battery, and a relay that activates when the voltage across it reaches 3.0 V or higher.

(a) Draw a circuit diagram showing how the thermistor, fixed resistor, battery, and relay coil should be connected so that the relay activates when the temperature rises above 30 °C. [3]

(b) (i) At exactly 30 °C, calculate the voltage across the thermistor. [2]

(ii) Explain why the relay does not activate at 30 °C. [1]

(iii) When the temperature rises to 35 °C, the thermistor resistance drops to 350 Ω. Show by calculation that the relay will now activate. [2]

(c) The student wishes to modify the circuit so that the fan switches on at a lower temperature of 25 °C. The thermistor resistance at 25 °C is 700 Ω. Suggest one change to the circuit that would achieve this, and explain your reasoning. [2]

(d) The relay coil has a resistance of 200 Ω. When the relay is activated, the current through the coil is 15 mA. Calculate the power dissipated in the relay coil. [2]

Question 8: Electromagnetic Devices (11 marks)

The diagram below represents a simple d.c. motor. A rectangular coil of wire is placed between the poles of a permanent magnet. The ends of the coil are connected to a battery via a split-ring commutator and carbon brushes.

(a) (i) Explain why a force acts on the sides of the coil that are perpendicular to the magnetic field. [2]

(ii) Using Fleming's left-hand rule, explain why the forces on the two perpendicular sides of the coil are in opposite directions. [2]

(iii) State the purpose of the split-ring commutator in the d.c. motor. [2]

(b) The coil consists of 50 turns of wire. Each side of the coil perpendicular to the field has a length of 4.0 cm. The magnetic flux density is 0.25 T, and the current in the coil is 2.0 A.

(i) Calculate the force acting on one of the perpendicular sides of the coil. [2]

(ii) The perpendicular distance from the axis of rotation to each side of the coil is 2.5 cm. Calculate the total turning moment (torque) acting on the coil when it is in the horizontal position. [2]

Question 9: Transformers and Power Transmission (12 marks)

A power station generates electricity at 25 kV. A step-up transformer increases this voltage to 400 kV for transmission over long distances. At a substation, a step-down transformer reduces the voltage to 240 V for domestic use.

(a) Explain why electrical energy is transmitted at very high voltages over long distances. [3]

(b) The step-up transformer has 800 turns on its primary coil.

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

(ii) The current in the primary coil is 1200 A. Assuming the transformer is 100% efficient, calculate the current in the secondary coil. [2]

(i) State two reasons why energy is lost in a practical transformer. [2]

(ii) Describe one design feature of a transformer that reduces energy losses. [1]

(d) The step-down transformer supplies a housing estate where the total power demand is 48 kW at 240 V.

(i) Calculate the total current drawn from the secondary coil of the step-down transformer. [2]

END OF PAPER

Answers

TuitionGoWhere Practice Paper - Physics O-Level

Answer Key and Marking Scheme — Version 3

Section A: Structured Questions (45 marks)

Question 1: Static Electricity (6 marks)

(a)(i) When the polythene rod is rubbed with the woollen cloth, electrons are transferred from the wool to the polythene rod [1]. The rod gains excess electrons, so it becomes negatively charged [1].

(a)(ii) The woollen cloth becomes positively charged [1]. It lost electrons to the rod, leaving it with a deficit of electrons / excess positive charge.

(b)(i) When the charged rod is brought near the uncharged ball, the charges in the ball separate (electrostatic induction) [1]. The negative charges (electrons) in the ball are repelled to the far side, leaving the near side positively charged. Since unlike charges attract, the ball is attracted to the negatively charged rod [1].

(b)(ii) When the ball touches the rod, electrons transfer from the rod to the ball, giving the ball a net negative charge [1]. The rod and ball are now both negatively charged, so they repel each other.

Question 2: Current of Electricity (8 marks)

(a) Potential difference between two points is the work done per unit charge to move charge between those two points [1].

(b)(i) t = 4 min = 240 s [1] I = Q / t = 240 / 240 = 1.0 A [1]

(b)(ii) R = V / I = 12 / 1.0 = 12 Ω [1] for formula, [1] for correct answer with unit.

(c)(i) The resistance of the filament lamp increases as the potential difference increases [1].

(c)(ii) As the potential difference increases, the current increases, causing the filament temperature to rise [1]. The metal atoms/ions in the filament vibrate more vigorously, which impedes the flow of electrons more, so resistance increases [1].

Question 3: D.C. Circuits (9 marks)

(a) Circuit diagram showing: 12 V battery, 10 Ω resistor in series, then a parallel branch with 15 Ω and R. All symbols correct and clearly labelled [2]. (Deduct 1 mark for missing labels or incorrect symbols.)

(b)(i) R_total = V / I = 12 / 0.80 = 15 Ω [1] for formula, [1] for correct answer.

(b)(ii) R_total = 10 + R_parallel = 15 Ω, so R_parallel = 15 - 10 = 5.0 Ω [1].

(b)(iii) 1/R_parallel = 1/15 + 1/R [1] 1/5 = 1/15 + 1/R 1/R = 1/5 - 1/15 = 3/15 - 1/15 = 2/15 R = 15/2 = 7.5 Ω [1]

(c) When the thermistor is heated, its resistance decreases [1]. This reduces the effective resistance of the parallel combination, which reduces the total circuit resistance. By Ohm's law (I = V/R), the current drawn from the battery increases [1].

Question 4: Practical Electricity (7 marks)

(a)(i) P = VI, so I = P / V = 1800 / 240 = 7.5 A [1] for formula, [1] for correct answer.

(a)(ii) Energy = Power × time = 1.8 kW × (3/60) h [1] = 1.8 × 0.05 = 0.09 kWh [1].

(b)(i) Live: Brown; Neutral: Blue; Earth: Green and Yellow stripes [1] for all three correct.

(b)(ii) The earth wire is connected to the metal casing of the kettle [1]. If a fault occurs and the live wire touches the metal casing, a large current flows through the earth wire to ground, which blows the fuse / trips the circuit breaker, disconnecting the appliance and preventing electric shock [1].

Question 5: Magnetism and Electromagnetism (8 marks)

(a)(i) Sketch showing field lines emerging from North pole, curving around, and entering South pole [1]. Arrows pointing from N to S [1].

(a)(ii) Any two from:

Field lines never cross each other [1]
Field lines are closer together where the field is stronger [1]
Field lines point from North to South outside the magnet [1]
Field lines form closed loops [1]

(b)(i) F = B I L = 0.40 × 3.0 × 0.15 [1] = 0.18 N [1].

(b)(ii) Fleming's left-hand rule [1].

(b)(iii) Increase the length of wire within the magnetic field [1] (or use a stronger magnet / increase number of turns of wire).

Question 6: Electromagnetic Induction (7 marks)

(a)(i) When the magnet moves into the coil, the magnetic flux / magnetic field lines passing through the coil change [1]. By Faraday's law, a changing magnetic flux induces an e.m.f. in the coil [1].

(a)(ii) The galvanometer needle returns to zero / shows no deflection [1]. When the magnet is stationary, there is no change in magnetic flux through the coil, so no e.m.f. is induced [1].

(a)(iii) The needle deflects to the left (opposite direction) [1]. By Lenz's law, the direction of the induced e.m.f. (and current) opposes the change causing it. Pulling the magnet out decreases the flux, so the induced current flows in the opposite direction to try to maintain the flux [1].

(b) Any one from:

Speed of relative motion between magnet and coil [1]
Number of turns in the coil [1]
Strength of the magnet [1]

Section B: Free Response Questions (35 marks)

Question 7: Circuit Analysis and Design (12 marks)

(a) Circuit diagram showing: 6 V battery, thermistor and 500 Ω fixed resistor connected in series forming a potential divider [1]. Relay coil connected across the fixed resistor [1]. Correct symbols and labels [1].

(b)(i) At 30 °C, R_thermistor = 500 Ω, R_fixed = 500 Ω. V across thermistor = (R_thermistor / (R_thermistor + R_fixed)) × V_supply [1] = (500 / 1000) × 6.0 = 3.0 V [1].

(b)(ii) The relay is connected across the fixed resistor, not the thermistor. The voltage across the fixed resistor is also 3.0 V, which is exactly the threshold. However, the relay needs more than 3.0 V to reliably activate, or the question context implies the relay activates when voltage across it reaches 3.0 V — at exactly 30 °C it is borderline and may not activate reliably [1]. (Accept: The relay activates when voltage across it is 3.0 V or higher; at 30 °C the voltage is exactly 3.0 V, so it is at threshold. The design intention is activation above 30 °C.)

(b)(iii) At 35 °C, R_thermistor = 350 Ω. V across fixed resistor = (500 / (350 + 500)) × 6.0 [1] = (500 / 850) × 6.0 = 0.588 × 6.0 = 3.53 V [1]. Since 3.53 V > 3.0 V, the relay activates.

(c) Replace the fixed 500 Ω resistor with a smaller resistance (e.g., 400 Ω) [1]. At 25 °C, R_thermistor = 700 Ω. With a 400 Ω fixed resistor, V across fixed resistor = (400 / (700 + 400)) × 6.0 = (400/1100) × 6.0 = 2.18 V (still below 3.0 V — so this doesn't work). Alternative correct answer: Connect the relay across the thermistor instead of the fixed resistor [1]. At 25 °C, V across thermistor = (700 / (700 + 500)) × 6.0 = 3.5 V, which exceeds 3.0 V, so the relay activates at 25 °C [1]. (Accept any valid modification with correct reasoning.)

(d) P = I²R [1] = (15 × 10⁻³)² × 200 = (2.25 × 10⁻⁴) × 200 = 0.045 W or 45 mW [1].

Question 8: Electromagnetic Devices (11 marks)

(a)(i) A current-carrying conductor placed in a magnetic field experiences a force [1]. This is due to the interaction between the magnetic field of the permanent magnet and the magnetic field created by the current in the wire [1].

(a)(ii) Using Fleming's left-hand rule: For one side, the current flows in one direction (e.g., into the page), and the magnetic field is from N to S. The force is directed upward [1]. For the opposite side, the current flows in the opposite direction (out of the page), while the magnetic field direction is the same. The force is therefore directed downward [1]. The opposite forces create a turning effect.

(a)(iii) The split-ring commutator reverses the direction of current in the coil every half rotation [1]. This ensures that the forces on the sides of the coil always act in the same rotational direction, maintaining continuous rotation [1].

(b)(i) For one turn: F = B I L = 0.25 × 2.0 × 0.040 = 0.020 N [1] For 50 turns: Total F = 50 × 0.020 = 1.0 N [1].

(b)(ii) Moment = Force × perpendicular distance from pivot [1] Total moment = 2 × F × d = 2 × 1.0 × 0.025 = 0.050 N m [1].

(c) Any one from:

Increase the number of turns on the coil [1]
Use a stronger magnet (increase magnetic flux density) [1]
Increase the area of the coil [1]
Wind the coil on a soft iron core to concentrate the magnetic field [1]

Question 9: Transformers and Power Transmission (12 marks)

(a) Electrical power transmitted is P = VI. For a given power, a higher voltage means a lower current [1]. Lower current reduces the heating effect in the transmission cables (power loss = I²R) [1]. This makes transmission more efficient and allows thinner, lighter, and cheaper cables to be used [1].

(b)(i) V_s / V_p = N_s / N_p [1] 400 000 / 25 000 = N_s / 800 16 = N_s / 800 N_s = 16 × 800 = 12 800 turns [1].

(b)(ii) For an ideal transformer: V_p I_p = V_s I_s [1] 25 000 × 1200 = 400 000 × I_s I_s = (25 000 × 1200) / 400 000 = 30 000 000 / 400 000 = 75 A [1].

(c)(i) Any two from:

Resistive heating in the coils (copper losses) [1]
Eddy currents induced in the iron core [1]
Hysteresis losses due to repeated magnetisation and demagnetisation of the core [1]
Flux leakage where not all magnetic flux links both coils [1]

(c)(ii) Any one from:

Use a laminated core to reduce eddy currents [1]
Use a soft iron core that is easily magnetised and demagnetised to reduce hysteresis losses [1]
Use thick copper wire for the coils to reduce resistance [1]

(d)(i) P = VI, so I = P / V [1] I = 48 000 / 240 = 200 A [1].

END OF ANSWER KEY