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O Level Physics Practice Paper 3

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Questions

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TuitionGoWhere Practice Paper - Physics O-Level

TuitionGoWhere Secondary School (AI)

Subject: Physics
Level: O-Level
Paper: PRACTICE - Electricity & Magnetism
Duration: 1 hour 15 minutes
Total Marks: 60
Version: 3 of 5

Name: _________________________
Class: _________________________
Date: _________________________

Instructions to Candidates

  1. This paper consists of three sections: Section A, Section B, and Section C.
  2. Answer all questions.
  3. Write your answers in the spaces provided.
  4. Show all working clearly; marks are awarded for method.
  5. Include units in all final numerical answers.
  6. Use appropriate significant figures (2–3 s.f.) unless otherwise stated.
  7. The number of marks is given in brackets [ ] at the end of each question or part question.
  8. You may use a scientific calculator.

Section A: Short Answer and Data Interpretation

[20 marks]

Answer all questions in this section.

Question 1

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

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

(b) The charged rod is brought near a small piece of uncharged aluminium foil. The foil is attracted to the rod. Explain why this happens. [2]

[Total: 4 marks]

Question 2

Fig. 2.1 shows two bar magnets placed on a flat surface with their north poles facing each other.

    N ──────── S          N ──────── S

Fig. 2.1

(a) On Fig. 2.1, sketch the magnetic field pattern between the two north poles. Use arrows to indicate the direction of the field lines. [2]

(b) State what is meant by a neutral point in a magnetic field. [1]

(c) A student places a plotting compass at the midpoint between the two north poles. Describe and explain the behaviour of the compass needle at this location. [2]

[Total: 5 marks]

Question 3

A student investigates the I–V characteristic of a filament lamp. The circuit used is shown in Fig. 3.1.

    + ───[A]───[Lamp]───[V]─── -

Fig. 3.1

The student records the following data:

Voltage / VCurrent / A
0.00.00
1.00.12
2.00.18
3.00.22
4.00.25
5.00.27
6.00.29

(a) Calculate the resistance of the filament lamp when the voltage across it is 2.0 V. [2]

(b) Calculate the resistance of the filament lamp when the voltage across it is 6.0 V. [2]

(c) Using your answers to (a) and (b), describe how the resistance of the filament lamp changes as the voltage increases. Explain this change in terms of the behaviour of electrons in the metal filament. [3]

(d) On the axes below, sketch the I–V characteristic graph for the filament lamp. Label both axes clearly. [2]

I/A
^
|
|
|
|
+------------------> V/V

[Total: 9 marks]

Question 4

Fig. 4.1 shows a simple d.c. motor.

(a) State the direction of the force acting on side AB of the coil when the current flows as shown. Use Fleming's left-hand rule to explain your answer. [2]

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

[Total: 4 marks]

Question 5

A transformer has 500 turns on its primary coil and 50 turns on its secondary coil. The primary coil is connected to a 240 V a.c. mains supply.

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

(b) The transformer is used to operate a device that requires a current of 2.0 A at the secondary voltage. Assuming the transformer is 100% efficient, calculate the current drawn from the mains supply. [2]

(c) Explain why the core of the transformer is made of laminated soft iron. [2]

[Total: 6 marks]

Section B: Structured Questions

[24 marks]

Answer all questions in this section.

Question 6

Fig. 6.1 shows a circuit containing a battery of e.m.f. 12 V, a fixed resistor R of resistance 8.0 Ω, and a thermistor T. The thermistor is placed in a water bath whose temperature can be varied.

    + ───[12 V]───[R = 8.0 Ω]───[T]─── -

Fig. 6.1

The resistance of the thermistor at various temperatures is given in the table below.

Temperature / °CResistance of thermistor / Ω
10240
20150
3090
4055
5035

(a) (i) Calculate the total resistance of the circuit when the temperature of the water bath is 20 °C. [1]

(ii) Calculate the current in the circuit at this temperature. [2]

(iii) Calculate the potential difference across the fixed resistor R at 20 °C. [2]

(b) The temperature of the water bath is increased from 20 °C to 50 °C.

(i) State and explain what happens to the current in the circuit. [2]

(ii) State and explain what happens to the potential difference across the thermistor. [2]

(c) A voltmeter is connected across the fixed resistor R. Describe how this circuit could be used as a temperature-sensing device. [2]

[Total: 11 marks]

Question 7

Fig. 7.1 shows an a.c. generator (alternator).

(a) Explain how an e.m.f. is induced in the coil as it rotates in the magnetic field. [3]

(b) On the axes below, sketch the voltage–time graph for the output of the a.c. generator. Label the axes and indicate one complete cycle. [2]

Voltage/V
^
|
|
|
|
+------------------> time/s
|
|
|

(c) State two changes that could be made to the generator to increase the maximum induced e.m.f. [2]

[Total: 7 marks]

Question 8

Fig. 8.1 shows a circuit containing a 6.0 V battery, a lamp L, and two resistors connected in parallel. The resistors have values 12 Ω and 4.0 Ω.

    + ───[6.0 V]───[L]───┬───[12 Ω]───┬─── -
                         └───[4.0 Ω]──┘

Fig. 8.1

(a) Calculate the combined resistance of the two resistors in parallel. [2]

(b) The current through the lamp is 0.50 A. Calculate the resistance of the lamp. [2]

(c) Calculate the power dissipated by the 4.0 Ω resistor. [2]

[Total: 6 marks]

Section C: Free Response Questions

[16 marks]

Answer all questions in this section.

Question 9

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

(a) Describe what is observed on the galvanometer when:

(i) the north pole of the magnet is pushed quickly into the coil; [1]

(ii) the magnet is held stationary inside the coil; [1]

(iii) the magnet is pulled quickly out of the coil. [1]

(b) State Lenz's law of electromagnetic induction. [2]

(c) The student repeats the experiment but moves the magnet into the coil more slowly. State and explain how the galvanometer reading differs from that in (a)(i). [2]

(d) The student replaces the single coil with a coil having twice the number of turns. The magnet is moved into the coil at the same speed as in (a)(i). State and explain how the galvanometer reading differs. [2]

[Total: 9 marks]

Question 10

Fig. 10.1 shows a circuit that can be used to switch on a lamp automatically when it gets dark. The circuit contains a light-dependent resistor (LDR), a fixed resistor R, a relay, and a 6.0 V battery.

    + ───[6.0 V]───┬───[LDR]───┬─── -
                   └───[R]────┘
                         │
                    [Relay Coil]
                         │
                    ┌────┴────┐
                    │  Switch  │
                    └────┬────┘
                         │
                    ┌────┴────┐
                    │   Lamp   │
                    └────┬────┘
                         │
                    [240 V a.c.]

Fig. 10.1

The relay switch closes when the potential difference across the relay coil reaches 3.0 V. The resistance of the LDR in bright light is 200 Ω and in darkness is 2000 Ω.

(a) Explain why the potential difference across the relay coil changes as the light level changes. [2]

(b) The fixed resistor R has a value of 600 Ω. Determine, by calculation, whether the lamp will be switched on when it is dark. [3]

(c) Suggest and explain one change that could be made to the circuit so that the lamp switches on at a higher light level (i.e., earlier in the evening). [2]

[Total: 7 marks]

END OF PAPER

Answers

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TuitionGoWhere Practice Paper - Physics O-Level

Answer Key and Marking Scheme

Paper: PRACTICE - Electricity & Magnetism
Version: 3 of 5
Total Marks: 60

Section A: Short Answer and Data Interpretation

Question 1

(a) Answer: When the polythene rod is rubbed with the woollen cloth, electrons are transferred from the cloth to the rod. The rod gains electrons and therefore becomes negatively charged. [2]

Marking:

  • 1 mark: electrons transferred from cloth to rod / rod gains electrons
  • 1 mark: negative charge due to excess electrons

(b) Answer: When the negatively charged rod is brought near the uncharged aluminium foil, electrons in the foil are repelled to the far side of the foil. This leaves the near side of the foil with a net positive charge (induced charge). The positive induced charge is attracted to the negative rod, so the foil moves towards the rod. [2]

Marking:

  • 1 mark: electrons repelled / charge separation / induced charge described
  • 1 mark: attraction between unlike charges / positive side attracted to negative rod

Question 2

(a) Answer: Field lines should curve away from each other between the two north poles, showing repulsion. Arrows should point away from each north pole. A neutral point should be shown at the midpoint where the fields cancel. [2]

Marking:

  • 1 mark: correct field line pattern showing repulsion between like poles
  • 1 mark: arrows correctly indicating direction away from north poles

(b) Answer: A neutral point is a point in a magnetic field where the resultant magnetic field strength is zero / where the magnetic fields from two or more sources cancel each other out. [1]

Marking:

  • 1 mark: correct definition (resultant field = zero / fields cancel)

(c) Answer: The compass needle will not point in any particular direction / will point randomly / will not show a definite direction. This is because at the neutral point, the resultant magnetic field is zero, so there is no net magnetic force to align the compass needle. [2]

Marking:

  • 1 mark: compass shows no definite direction / points randomly
  • 1 mark: because resultant magnetic field is zero at the neutral point

Question 3

(a) Answer: R = V / I = 2.0 / 0.18 = 11.1 Ω ≈ 11 Ω [2]

Marking:

  • 1 mark: correct formula R = V/I
  • 1 mark: correct answer with unit (11.1 Ω or 11 Ω)

(b) Answer: R = V / I = 6.0 / 0.29 = 20.7 Ω ≈ 21 Ω [2]

Marking:

  • 1 mark: correct formula and substitution
  • 1 mark: correct answer with unit (20.7 Ω or 21 Ω)

(c) Answer: The resistance of the filament lamp increases as the voltage increases. This is because as the current increases, the filament gets hotter. The increased temperature causes the metal ions in the filament to vibrate more vigorously, which increases the frequency of collisions between the flowing electrons and the ions. This impedes the flow of electrons, resulting in higher resistance. [3]

Marking:

  • 1 mark: resistance increases with voltage
  • 1 mark: filament temperature increases / filament gets hotter
  • 1 mark: increased ion vibrations cause more electron collisions / greater opposition to electron flow

(d) Answer: Graph should show a curve starting from the origin, initially steep (low resistance), then becoming less steep (higher resistance) as voltage increases. Axes labelled: y-axis "Current / A", x-axis "Voltage / V". [2]

Marking:

  • 1 mark: correct curved shape (gradient decreasing with increasing V)
  • 1 mark: axes correctly labelled with quantities and units

Question 4

(a) Answer: The force on side AB is upwards (or downwards, depending on diagram orientation). Using Fleming's left-hand rule: the First finger points in the direction of the magnetic Field (N to S), the seCond finger points in the direction of the Current, and the thuMb points in the direction of the force (Motion). [2]

Marking:

  • 1 mark: correct direction stated
  • 1 mark: correct application of Fleming's left-hand rule explained

(b) Answer: The split-ring commutator reverses the direction of the current in the coil every half rotation. This ensures that the force on each side of the coil always acts in the same rotational direction, allowing the coil to rotate continuously in one direction rather than oscillating back and forth. [2]

Marking:

  • 1 mark: reverses current every half rotation
  • 1 mark: ensures continuous rotation in one direction / prevents oscillation

Question 5

(a) Answer: V_s / V_p = N_s / N_p → V_s = V_p × (N_s / N_p) = 240 × (50 / 500) = 240 × 0.1 = 24 V [2]

Marking:

  • 1 mark: correct formula V_s/V_p = N_s/N_p
  • 1 mark: correct answer 24 V

(b) Answer: For an ideal transformer: V_p × I_p = V_s × I_s → I_p = (V_s × I_s) / V_p = (24 × 2.0) / 240 = 48 / 240 = 0.20 A [2]

Marking:

  • 1 mark: correct formula V_p × I_p = V_s × I_s (or power input = power output)
  • 1 mark: correct answer 0.20 A

(c) Answer: The core is made of soft iron because it is easily magnetised and demagnetised, which improves the efficiency of magnetic flux linkage between the coils. The core is laminated (made of thin sheets insulated from each other) to reduce eddy currents induced in the core. Eddy currents would cause energy loss as heat, reducing the transformer's efficiency. [2]

Marking:

  • 1 mark: soft iron for easy magnetisation and demagnetisation / good magnetic flux linkage
  • 1 mark: lamination reduces eddy currents / reduces energy loss as heat

Section B: Structured Questions

Question 6

(a)(i) Answer: At 20 °C, R_thermistor = 150 Ω. Total resistance R_total = R + R_thermistor = 8.0 + 150 = 158 Ω [1]

Marking:

  • 1 mark: correct total resistance 158 Ω

(a)(ii) Answer: I = V / R_total = 12 / 158 = 0.0759 A ≈ 0.076 A [2]

Marking:

  • 1 mark: correct formula I = V/R
  • 1 mark: correct answer 0.076 A (accept 0.0759 A)

(a)(iii) Answer: V_R = I × R = 0.0759 × 8.0 = 0.607 V ≈ 0.61 V [2]

Marking:

  • 1 mark: correct formula V = IR
  • 1 mark: correct answer 0.61 V (accept 0.607 V)

(b)(i) Answer: The current in the circuit increases. As the temperature increases from 20 °C to 50 °C, the resistance of the thermistor decreases (from 150 Ω to 35 Ω). This reduces the total resistance of the circuit. Since the e.m.f. of the battery is constant, by Ohm's law (I = V/R), a decrease in total resistance causes an increase in current. [2]

Marking:

  • 1 mark: current increases
  • 1 mark: explanation linking decreased thermistor resistance → decreased total resistance → increased current

(b)(ii) Answer: The potential difference across the thermistor decreases. As the thermistor's resistance decreases, a smaller proportion of the total voltage is dropped across it (voltage divider principle). The p.d. across the fixed resistor increases, so the p.d. across the thermistor must decrease since the sum equals the battery e.m.f. [2]

Marking:

  • 1 mark: p.d. across thermistor decreases
  • 1 mark: explanation using voltage divider / proportion of total resistance

(c) Answer: The voltmeter reading across R changes with temperature. As temperature increases, the thermistor resistance decreases, the current increases, and the p.d. across R increases. The voltmeter can be calibrated to read temperature directly. For example, a higher voltmeter reading corresponds to a higher temperature. [2]

Marking:

  • 1 mark: voltmeter reading changes with temperature / described relationship
  • 1 mark: voltmeter can be calibrated to display temperature

Question 7

(a) Answer: As the coil rotates, the sides of the coil cut through the magnetic field lines. This changes the magnetic flux linking the coil. According to Faraday's law, a changing magnetic flux induces an e.m.f. in the coil. The induced e.m.f. drives a current if the circuit is complete. The direction of the induced e.m.f. reverses every half rotation as each side of the coil moves alternately up and down through the field. [3]

Marking:

  • 1 mark: coil cuts magnetic field lines / changing magnetic flux
  • 1 mark: changing flux induces e.m.f. (Faraday's law)
  • 1 mark: e.m.f. direction reverses every half rotation / explanation of alternating output

(b) Answer: Graph should show a sinusoidal waveform (sine curve) with positive and negative halves. Axes labelled: y-axis "Voltage / V" or "e.m.f. / V", x-axis "Time / s". One complete cycle should be indicated (e.g., from one peak to the next corresponding peak). [2]

Marking:

  • 1 mark: correct sinusoidal shape with positive and negative values
  • 1 mark: axes correctly labelled and one complete cycle indicated

(c) Answer: Any two from:

  1. Increase the number of turns on the coil
  2. Use a stronger magnet / increase the magnetic field strength
  3. Rotate the coil at a higher speed / frequency
  4. Insert a soft iron core into the coil [2]

Marking:

  • 1 mark each for any two correct changes (max 2 marks)

Question 8

(a) Answer: 1/R_parallel = 1/12 + 1/4.0 = 1/12 + 3/12 = 4/12 → R_parallel = 12/4 = 3.0 Ω [2]

Marking:

  • 1 mark: correct formula 1/R = 1/R₁ + 1/R₂
  • 1 mark: correct answer 3.0 Ω

(b) Answer: Total circuit resistance R_total = V / I = 6.0 / 0.50 = 12 Ω. The parallel combination has resistance 3.0 Ω. Since the lamp and parallel resistors are in series: R_lamp = R_total - R_parallel = 12 - 3.0 = 9.0 Ω [2]

Marking:

  • 1 mark: correct total resistance 12 Ω
  • 1 mark: correct lamp resistance 9.0 Ω

(c) Answer: Voltage across parallel combination = I × R_parallel = 0.50 × 3.0 = 1.5 V. Current through 4.0 Ω resistor: I = V / R = 1.5 / 4.0 = 0.375 A. Power = I² × R = (0.375)² × 4.0 = 0.1406 × 4.0 = 0.56 W. Alternative: P = V² / R = (1.5)² / 4.0 = 2.25 / 4.0 = 0.56 W. [2]

Marking:

  • 1 mark: correct voltage across parallel combination (1.5 V) or correct current through 4.0 Ω resistor (0.375 A)
  • 1 mark: correct power 0.56 W (accept 0.5625 W)

Section C: Free Response Questions

Question 9

(a)(i) Answer: The galvanometer needle deflects (e.g., to the right), indicating an induced current. [1]

Marking:

  • 1 mark: galvanometer deflects / shows a reading

(a)(ii) Answer: The galvanometer shows zero deflection / the needle returns to the centre. [1]

Marking:

  • 1 mark: zero deflection / no reading

(a)(iii) Answer: The galvanometer needle deflects in the opposite direction (e.g., to the left) compared to (a)(i). [1]

Marking:

  • 1 mark: deflection in opposite direction

(b) Answer: Lenz's law states that the direction of the induced e.m.f. (and hence the induced current) is such that it opposes the change in magnetic flux that produced it. [2]

Marking:

  • 1 mark: induced e.m.f./current opposes the change
  • 1 mark: reference to change in magnetic flux

(c) Answer: The galvanometer deflection is smaller. When the magnet moves more slowly, the rate of change of magnetic flux through the coil is smaller. By Faraday's law, the induced e.m.f. is proportional to the rate of change of flux, so a slower movement produces a smaller induced e.m.f. and hence a smaller current. [2]

Marking:

  • 1 mark: smaller deflection / smaller reading
  • 1 mark: slower movement → smaller rate of change of flux → smaller induced e.m.f.

(d) Answer: The galvanometer deflection is larger. A coil with twice the number of turns experiences twice the total change in magnetic flux linkage for the same magnet movement. By Faraday's law, the induced e.m.f. is proportional to the number of turns (and the rate of change of flux linkage), so doubling the turns doubles the induced e.m.f., resulting in a larger current and greater deflection. [2]

Marking:

  • 1 mark: larger deflection / larger reading
  • 1 mark: more turns → greater flux linkage change → larger induced e.m.f.

Question 10

(a) Answer: The LDR and fixed resistor R form a potential divider. The p.d. across the relay coil is the same as the p.d. across R. As the light level changes, the resistance of the LDR changes. In bright light, the LDR has low resistance, so most of the battery voltage is dropped across R, giving a high p.d. across the relay coil. In darkness, the LDR has high resistance, so a smaller proportion of the voltage is dropped across R, giving a lower p.d. across the relay coil. [2]

Marking:

  • 1 mark: LDR and R form a potential divider
  • 1 mark: changing LDR resistance changes voltage distribution / p.d. across R changes

(b) Answer: In darkness, R_LDR = 2000 Ω. Total resistance = 2000 + 600 = 2600 Ω. Circuit current I = V / R_total = 6.0 / 2600 = 0.00231 A. P.d. across R (and relay coil) = I × R = 0.00231 × 600 = 1.38 V. Since 1.38 V < 3.0 V, the relay switch does not close. The lamp will NOT be switched on. [3]

Marking:

  • 1 mark: correct calculation of current or voltage ratio
  • 1 mark: correct p.d. across relay coil (1.38 V or 1.4 V)
  • 1 mark: correct conclusion that lamp is NOT switched on (with comparison to 3.0 V threshold)

(c) Answer: Increase the value of the fixed resistor R. With a larger R, the p.d. across R (and the relay coil) will be larger for the same LDR resistance. This means the relay coil will reach the 3.0 V threshold at a higher LDR resistance, which corresponds to a higher light level (brighter conditions). The lamp will therefore switch on earlier in the evening. [2]

Marking:

  • 1 mark: increase the value of R (or decrease the LDR resistance / use a different LDR)
  • 1 mark: explanation linking change to earlier switching at higher light level

END OF ANSWER KEY