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TuitionGoWhere Exam Practice (AI)

O-Level Physics Practice Paper - Electricity & Magnetism (Version 5)

Subject: Physics (6091)
Level: O-Level
Paper: Practice Paper 5 (Topic: Electricity & Magnetism)
Duration: 1 hour 15 minutes
Total Marks: 60

Name: ________________________
Class: ________________________
Date: ________________________

Instructions to Candidates

Write your name, class, and date in the spaces provided.
Answer all questions.
Write your answers in the spaces provided on the question paper.
You may use an approved scientific calculator.
Assume the acceleration of free fall, $g = 10 \text{ m/s}^2$ , where necessary.
A copy of the Periodic Table is printed on page 12 (not included in this extract).
At the end of the examination, fasten all your work securely together.
The number of marks is given in brackets [ ] at the end of each question or part question.

Section A: Multiple Choice & Short Structured Questions

Answer all questions in this section.

1. Which of the following statements correctly describes the direction of conventional current and electron flow in a metallic conductor?

A. Conventional current flows from negative to positive; electrons flow from positive to negative.
B. Conventional current flows from positive to negative; electrons flow from negative to positive.
C. Both conventional current and electrons flow from positive to negative.
D. Both conventional current and electrons flow from negative to positive.

[1]

2. A student rubs a plastic rod with a cloth. The rod becomes negatively charged. Which statement explains this observation?

A. Positive charges move from the cloth to the rod.
B. Positive charges move from the rod to the cloth.
C. Electrons move from the cloth to the rod.
D. Electrons move from the rod to the cloth.

[1]

3. The diagram below shows the electric field lines around two point charges, X and Y.

(Diagram Description: Field lines originate from X and terminate at Y. The lines are denser near X than near Y.)

Which of the following correctly identifies the signs and relative magnitudes of the charges?

A. X is positive, Y is negative; magnitude of X > magnitude of Y.
B. X is positive, Y is negative; magnitude of X < magnitude of Y.
C. X is negative, Y is positive; magnitude of X > magnitude of Y.
D. X is negative, Y is positive; magnitude of X < magnitude of Y.

[1]

4. A wire of length $L$ and cross-sectional area $A$ has a resistance of $R$ . A second wire is made of the same material but has length $2L$ and cross-sectional area $A/2$ .

What is the resistance of the second wire?

A. $R/4$
B. $R/2$
C. $2R$
D. $4R$

[1]

5. The I-V characteristic graph for a component is shown below.

(Diagram Description: A curve starting at the origin, with gradient decreasing as V increases.)

Identify the component and explain the shape of the graph.

Component: ________________________
Explanation: _________________________________________________________________________

[2]

6. In the circuit shown, the battery has an e.m.f. of 12 V and negligible internal resistance. Resistors $R_1 = 4 \, \Omega$ and $R_2 = 6 \, \Omega$ are connected in series.

Calculate the potential difference across $R_2$ .

Answer: ________________________ V [2]

7. A thermistor and a fixed resistor are connected in series to a 10 V supply. The thermistor is an NTC (Negative Temperature Coefficient) type.

Describe and explain what happens to the voltmeter reading across the fixed resistor when the temperature of the thermistor increases.

[3]

8. The diagram shows a simple d.c. motor.

(Diagram Description: A rectangular coil between magnetic poles, connected to a split-ring commutator and brushes.)

(a) State the function of the split-ring commutator.

[1]

(b) Suggest two changes that would increase the speed of rotation of the coil.

[2]

9. A transformer has 500 turns on the primary coil and 100 turns on the secondary coil. The primary coil is connected to a 240 V a.c. supply.

(a) Calculate the secondary voltage.

Answer: ________________________ V [2]

(b) Explain why this transformer cannot be used with a d.c. supply.

[2]

10. High-voltage transmission lines are used to transmit electrical power over long distances.

Explain why high voltage is used, referring to energy loss in the cables.

[3]

Section B: Structured Problems

Answer all questions in this section.

11. A student investigates the resistance of a wire. She measures the current through the wire and the potential difference across it for different lengths of the wire. The results are shown in the table.

Length / m	Potential Difference / V	Current / A
0.20	1.2	0.60
0.40	2.4	0.60
0.60	3.6	0.60
0.80	4.8	0.60

(a) Calculate the resistance of the 0.40 m length of wire.

Answer: ________________________ $\Omega$ [2]

(b) Plot a graph of Resistance (y-axis) against Length (x-axis) on the grid provided below.
(Note: In a real exam, a grid would be provided. Here, describe the expected trend.)
Describe the relationship between resistance and length based on the data.

[2]

(c) The student repeats the experiment with a wire of the same material but twice the cross-sectional area.
Predict the resistance of the 0.40 m length of this new wire.

Answer: ________________________ $\Omega$ [1]

12. The circuit diagram below shows a lighting system for a garden. It includes an LDR (Light Dependent Resistor), a variable resistor, a transistor switch, and a relay that controls a lamp.

(Diagram Description: Voltage divider with LDR and variable resistor feeding the base of an NPN transistor. The collector drives a relay coil. The relay switch controls a separate lamp circuit.)

(a) Explain how the resistance of the LDR changes as it gets darker.

[1]

(b) Explain why the lamp turns on when it gets dark. Refer to the potential difference at the base of the transistor.

[4]

[1]

13. A bar magnet is dropped through a long vertical copper tube.

(a) Explain why the magnet falls slower than it would in free fall. Refer to electromagnetic induction and Lenz’s Law.

[4]

(b) If the copper tube is replaced with a plastic tube of identical dimensions, describe the motion of the magnet.

[1]

14. An electric kettle is rated at 240 V, 2.4 kW.

(a) Calculate the current flowing through the kettle when it is operating normally.

Answer: ________________________ A [2]

(b) Calculate the resistance of the heating element.

Answer: ________________________ $\Omega$ [2]

Answer: ________________________ kWh [2]

15. The diagram shows a simple a.c. generator.

(Diagram Description: A coil rotating in a magnetic field, connected to slip rings and brushes.)

(a) On the axes below, sketch the graph of the induced e.m.f. against time for one complete rotation of the coil, starting from the position where the coil is horizontal (parallel to field lines).
(Note: Describe the shape: Sinusoidal wave starting at zero, reaching peak, returning to zero, negative peak, back to zero.)

[2]

(b) State two factors that would increase the maximum induced e.m.f.

[2]

Section C: Free Response & Application

Answer all questions in this section.

16. A student builds a circuit to measure the resistance of a filament lamp. The circuit includes a battery, an ammeter, a voltmeter, a variable resistor, and the lamp.

(a) Draw the circuit diagram for this experiment.

[3]

(b) The student obtains the following results:

V / V	I / A
2.0	0.40
4.0	0.70
6.0	0.90
8.0	1.05
10.0	1.15

Calculate the resistance of the lamp at 2.0 V and at 10.0 V.

Resistance at 2.0 V: ________________________ $\Omega$
Resistance at 10.0 V: ________________________ $\Omega$ [2]

[3]

17. A household wiring system uses parallel circuits for appliances.

(a) State two advantages of connecting appliances in parallel rather than in series.

[2]

(b) Explain the function of the earth wire in a three-pin plug for a metal-cased appliance.

[3]

[2]

18. The diagram shows a current-carrying wire placed between the poles of a U-shaped magnet. The current flows into the page.

(Diagram Description: N pole on left, S pole on right. Wire in center. Current direction 'X' indicates into page.)

(a) Use Fleming’s Left-Hand Rule to determine the direction of the force on the wire.
Direction: ________________________ [1]

(b) The wire experiences a force of 0.05 N. The magnetic flux density is 0.2 T, and the length of the wire in the field is 0.1 m.
Calculate the current flowing through the wire.

Answer: ________________________ A [2]

(ii) The poles of the magnet are swapped.

[2]

19. A transformer is used to step down the voltage from 11,000 V to 240 V for domestic use. The primary coil has 5,000 turns.

(a) Calculate the number of turns on the secondary coil.

Answer: ________________________ turns [2]

(b) The transformer is assumed to be 100% efficient. If the current in the secondary coil is 50 A, calculate the current in the primary coil.

Answer: ________________________ A [2]

[1]

20. An electrostatic precipitator is used in a factory chimney to remove ash particles from smoke.

(a) Explain how the ash particles become charged.

[2]

(b) Explain how the charged ash particles are removed from the smoke.

[3]

[1]

End of Paper

Answers

TuitionGoWhere Exam Practice (AI) - Answer Key

O-Level Physics Practice Paper - Electricity & Magnetism (Version 5)

Total Marks: 60

Section A: Multiple Choice & Short Structured Questions

1. B
Conventional current is defined as flowing from positive to negative. Electrons, being negatively charged, flow from negative to positive. [1]

2. C
Plastic is an insulator. Rubbing causes electron transfer. Since the rod becomes negative, it has gained electrons from the cloth. [1]

3. A
Field lines go from Positive to Negative. Therefore X is (+) and Y is (-). The density of lines indicates field strength; denser lines near X imply a larger magnitude of charge. [1]

4. D
$R = \rho \frac{L}{A}$ .
New $R' = \rho \frac{2L}{A/2} = \rho \frac{4L}{A} = 4R$ . [1]

5.
Component: Filament Lamp (or Lamp) [1]
Explanation: As voltage/current increases, the temperature of the filament increases. This causes the metal ions to vibrate more, increasing collisions with electrons, thus increasing resistance. The graph curves because R is not constant. [1]

6.
Total Resistance $R_T = 4 + 6 = 10 \, \Omega$ .
Current $I = V/R_T = 12/10 = 1.2 \text{ A}$ .
$V_{R2} = I \times R_2 = 1.2 \times 6 = 7.2 \text{ V}$ .
Alternatively, using voltage divider: $V_{R2} = \frac{6}{4+6} \times 12 = 7.2 \text{ V}$ .
Answer: 7.2 V [2]

As temperature increases, resistance of NTC thermistor decreases. [1]
This causes the total resistance of the series circuit to decrease, so the current in the circuit increases. [1]
Since $V = IR$ for the fixed resistor (R is constant), the potential difference across the fixed resistor increases. [1]

8.
(a) To reverse the direction of current in the coil every half rotation, ensuring the torque acts in the same direction for continuous rotation. [1]
(b) Any two of:

Increase the current.
Increase the strength of the magnetic field.
Increase the number of turns on the coil.
Increase the area of the coil. [2]

9.
(a) $\frac{V_s}{V_p} = \frac{N_s}{N_p}$
$V_s = 240 \times \frac{100}{500} = 240 \times 0.2 = 48 \text{ V}$ .
Answer: 48 V [2]
(b) A transformer works on the principle of electromagnetic induction, which requires a changing magnetic field. D.C. produces a constant magnetic field, so no e.m.f. is induced in the secondary coil. [2]

10.

Power loss in cables is given by $P_{loss} = I^2 R$ . [1]
For a fixed power transmission ( $P=VI$ ), increasing Voltage ( $V$ ) decreases Current ( $I$ ). [1]
Since loss is proportional to $I^2$ , a smaller current significantly reduces energy loss as heat in the transmission lines. [1]

Section B: Structured Problems

11.
(a) $R = V/I = 2.4 / 0.60 = 4.0 \, \Omega$ .
Answer: 4.0 $\Omega$ [2]
(b) Graph: Straight line through the origin.
Relationship: Resistance is directly proportional to length. [2]
(c) $R \propto 1/A$ . If area doubles, resistance halves.
$R_{new} = 4.0 / 2 = 2.0 \, \Omega$ .
Answer: 2.0 $\Omega$ [1]

12.
(a) Resistance of LDR increases as light intensity decreases (gets darker). [1]
(b)

As it gets dark, LDR resistance increases. [1]
This increases the potential difference across the LDR in the voltage divider. [1]
Consequently, the potential difference across the variable resistor (connected to the base) decreases? Correction: Standard circuit usually has LDR at the top or bottom.
Assumption based on standard "dark sensor" circuit: LDR is usually the bottom resistor in the divider feeding the base, or top. Let's assume standard configuration: LDR is $R_1$ (top) and Variable Resistor is $R_2$ (bottom). If LDR is top, $V_{base}$ drops. If LDR is bottom, $V_{base}$ rises.
Standard Answer Logic: Usually, the LDR is placed such that when dark (high R), the voltage at the base rises above 0.7V to switch on the transistor.
Let's assume LDR is in the lower part of the divider (between Base and Ground).

Dark $\rightarrow$ $R_{LDR}$ increases.
Voltage across LDR ( $V_{base}$ ) increases (Voltage divider rule: larger share of V). [1]
When $V_{base} > 0.7 \text{ V}$ , the transistor switches on, allowing current to flow through the relay coil. [1]
The relay magnetizes, closing the switch for the lamp circuit. [1]
(Note: If the diagram showed LDR at the top, the explanation would involve the variable resistor voltage dropping, which wouldn't turn it on. So LDR must be at the bottom or part of a bridge. Accept logical consistency.)

(c) To protect the transistor from the high back-e.m.f. (induced voltage) generated when the relay coil is switched off. [1]

13.
(a)

As the magnet falls, the magnetic flux through the copper tube changes. [1]
This induces eddy currents in the copper tube (Faraday's Law). [1]
According to Lenz's Law, the direction of these induced currents creates a magnetic field that opposes the change causing it (the motion of the magnet). [1]
This creates an upward magnetic force on the magnet, opposing gravity, thus slowing its fall. [1]
(b) The magnet falls with acceleration $g$ (free fall) because plastic is an insulator, so no eddy currents are induced, and there is no opposing magnetic force. [1]

14.
(a) $P = VI \Rightarrow I = P/V = 2400 / 240 = 10 \text{ A}$ .
Answer: 10 A [2]
(b) $R = V/I = 240 / 10 = 24 \, \Omega$ . (Or $R = V^2/P$ ).
Answer: 24 $\Omega$ [2]
(c) Time $t = 5 \text{ min} = 5/60 \text{ h} = 1/12 \text{ h}$ .
Power $P = 2.4 \text{ kW}$ .
$E = P \times t = 2.4 \times (1/12) = 0.2 \text{ kWh}$ .
Answer: 0.2 kWh [2]

15.
(a) Graph: Sinusoidal wave. Starts at 0. Goes to positive peak at $T/4$ . Zero at $T/2$ . Negative peak at $3T/4$ . Zero at $T$ . [2]
(b) Any two of:

Increase speed of rotation (frequency).
Increase magnetic field strength.
Increase number of turns on the coil.
Increase area of the coil. [2]

Section C: Free Response & Application

16.
(a) Diagram must show:

Battery/Power supply.
Ammeter in series with the lamp.
Voltmeter in parallel with the lamp.
Variable resistor in series (to change V).
Correct symbols. [3]
(b)
At 2.0 V: $R = 2.0 / 0.40 = 5.0 \, \Omega$ .
At 10.0 V: $R = 10.0 / 1.15 \approx 8.7 \, \Omega$ .
Answers: 5.0 $\Omega$ , 8.7 $\Omega$ [2]
(c)

As voltage increases, current increases, causing the filament to get hotter. [1]
The metal ions in the filament lattice vibrate with greater amplitude. [1]
This increases the frequency of collisions between free electrons and ions, increasing resistance. [1]

17.
(a) Any two of:

Each appliance receives the full mains voltage (240V).
Appliances can be switched on/off independently.
If one appliance fails (blows), others continue to work.
Total resistance decreases, allowing more current for more devices. [2]
(b)
The earth wire connects the metal casing to the ground. [1]
If the live wire touches the casing, a large current flows to earth through the low-resistance earth wire. [1]
This large current blows the fuse/trips the breaker, disconnecting the supply and preventing electric shock to the user. [1]
(c)
The fuse contains a thin wire that melts when the current exceeds its rating. [1]
At 15 A, the fuse wire heats up due to $I^2R$ heating and melts, breaking the circuit and stopping the current flow. [1]

18.
(a) Downwards (or towards the bottom of the page).
Check: Field N(left) to S(right). Current Into page. Left Hand Rule: First finger Right, Second finger In, Thumb points Down. [1]
(b) $F = BIL \Rightarrow I = F / (BL)$ .
$I = 0.05 / (0.2 \times 0.1) = 0.05 / 0.02 = 2.5 \text{ A}$ .
Answer: 2.5 A [2]
(c)
(i) Force direction reverses (Upwards). [1]
(ii) Force direction reverses (Upwards). [1]

19.
(a) $\frac{N_s}{N_p} = \frac{V_s}{V_p} \Rightarrow N_s = 5000 \times \frac{240}{11000}$ .
$N_s = 5000 \times 0.021818... \approx 109.09$ .
Since turns must be integer, typically round to nearest whole number: 109 turns. (Accept 109 or 110 depending on sig fig rules, but 109 is precise calculation). [2]
(b) $V_p I_p = V_s I_s$ (100% efficient).
$11000 \times I_p = 240 \times 50$ .
$I_p = 12000 / 11000 = 1.09 \text{ A}$ .
Answer: 1.09 A [2]
(c) Any one of:

Heating of coils (resistance).
Eddy currents in the core.
Magnetization/demagnetization of the core (hysteresis).
Flux leakage. [1]

20.
(a)

Smoke particles pass through a grid/corona discharge wire. [1]
They gain electrons (or ions) and become negatively charged. [1]
(b)
The charged particles are attracted to positively charged collector plates. [1]
They stick to the plates due to electrostatic attraction. [1]
The plates are periodically vibrated/shaken to dislodge the ash into a collection hopper. [1]
(c) Any one of:
Photocopier / Laser Printer.
Electrostatic spray painting.
Insecticide spraying. [1]