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O Level Physics Practice Paper 4

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TuitionGoWhere Practice Paper - Physics O-Level

TuitionGoWhere Practice Paper (AI)

Subject: Physics (6091)
Level: O-Level
Paper: Practice Paper (Version 4 of 5)
Topic Focus: Electricity & Magnetism
Duration: 1 hour 15 minutes
Total Marks: 50

Name: ________________________
Class: ________________________
Date: ________________________

Instructions to Candidates

Write your name, class, and date in the spaces above.
Answer all questions.
Write your answers in the spaces provided on the question paper.
You may use a scientific calculator.
Take the acceleration of free fall, $g = 10 \text{ m/s}^2$ .
The total mark for this paper is 50.

Section A: Multiple Choice & Short Structured Questions [20 marks]

1. Which of the following correctly describes the direction of conventional current and electron flow in a metallic wire connected to a battery?

Option	Conventional Current	Electron Flow
A	Positive to Negative	Negative to Positive
B	Negative to Positive	Positive to Negative
C	Positive to Negative	Positive to Negative
D	Negative to Positive	Negative to Positive

Answer: ______ [1]

2. A student rubs a polythene rod with a dry 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.

Answer: ______ [1]

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

(Imagine diagram: Field lines originate from X and terminate at Y. Lines are denser near X than near Y.)

Which statement is correct?

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

Answer: ______ [1]

4. A wire of length $L$ and cross-sectional area $A$ has a resistance of $8.0 \, \Omega$ . A second wire is made of the same material but has length $2L$ and cross-sectional area $0.5A$ . What is the resistance of the second wire?

A. $2.0 \, \Omega$
B. $8.0 \, \Omega$
C. $16.0 \, \Omega$
D. $32.0 \, \Omega$

Answer: ______ [1]

5. In the circuit shown, the battery has an electromotive force (e.m.f.) of $12 \text{ V}$ and negligible internal resistance. The resistors are $R_1 = 4.0 \, \Omega$ and $R_2 = 8.0 \, \Omega$ connected in series.

What is the potential difference across $R_2$ ?

A. $4.0 \text{ V}$
B. $6.0 \text{ V}$
C. $8.0 \text{ V}$
D. $12 \text{ V}$

Answer: ______ [1]

6. Which component has an I-V characteristic curve that starts with a steep gradient and then becomes less steep as voltage increases?

A. Fixed resistor
B. Filament lamp
C. Diode
D. Thermistor (NTC)

Answer: ______ [1]

7. A household appliance is rated at $2.4 \text{ kW}$ , $240 \text{ V}$ . Which fuse rating is most suitable for the plug of this appliance?

A. $3 \text{ A}$
B. $5 \text{ A}$
C. $10 \text{ A}$
D. $13 \text{ A}$

Answer: ______ [1]

8. Why is the core of a transformer made of soft iron?

A. It is a good electrical conductor.
B. It is easily magnetized and demagnetized.
C. It retains magnetism permanently.
D. It has high electrical resistance.

Answer: ______ [1]

9. A transformer has 500 turns on the primary coil and 100 turns on the secondary coil. The primary voltage is $240 \text{ V}$ . Assuming the transformer is 100% efficient, what is the secondary voltage?

A. $12 \text{ V}$
B. $48 \text{ V}$
C. $1200 \text{ V}$
D. $2400 \text{ V}$

Answer: ______ [1]

10. High voltage is used for the transmission of electrical energy over long distances primarily to:

A. Increase the current in the cables.
B. Reduce the resistance of the cables.
C. Reduce the energy loss due to heating in the cables.
D. Increase the frequency of the alternating current.

Answer: ______ [1]

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

(Imagine diagram: Rectangular coil between N and S poles, connected to a split-ring commutator and brushes.)

(a) State the function of the split-ring commutator.
...........................................................................................................................
........................................................................................................................... [1]

(b) State two ways to increase the speed of rotation of the motor.

.......................................................................................................................
....................................................................................................................... [2]

12. A negatively charged rod is brought near a neutral metal sphere resting on an insulating stand. The sphere is then earthed momentarily with a finger, and the earth connection is removed before the rod is taken away.

(a) Name this method of charging.
........................................................................................................................... [1]

(b) State the final charge on the sphere.
........................................................................................................................... [1]

13. Define the term electromotive force (e.m.f.) of a battery.
...........................................................................................................................
...........................................................................................................................
........................................................................................................................... [2]

14. In the circuit below, three identical lamps are connected to a battery. Switch S is initially open.

(Imagine diagram: Lamp A in series with a parallel combination of Lamp B and Lamp C. Switch S is in series with Lamp C only.)

(a) When switch S is closed, what happens to the brightness of Lamp A?
........................................................................................................................... [1]

(b) Explain your answer.
...........................................................................................................................
........................................................................................................................... [2]

15. A thermistor is connected in series with a fixed resistor and a battery. A voltmeter is connected across the fixed resistor.

(a) State what happens to the resistance of the NTC thermistor as the temperature increases.
........................................................................................................................... [1]

(b) Explain what happens to the reading on the voltmeter as the temperature increases.
...........................................................................................................................
...........................................................................................................................
........................................................................................................................... [2]

Section B: Structured Problems [30 marks]

16. 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 table shows her results for a wire of length $0.50 \text{ m}$ .

Potential Difference / V	Current / A
0.0	0.00
2.0	0.40
4.0	0.78
6.0	1.15
8.0	1.48

(a) Calculate the resistance of the wire when the potential difference is $2.0 \text{ V}$ .
 Resistance = ______________________ $\Omega$ [2]

(b) Calculate the resistance of the wire when the potential difference is $8.0 \text{ V}$ .
 Resistance = ______________________ $\Omega$ [2]

(c) Explain why the resistance changes as the potential difference increases.
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
........................................................................................................................... [3]

(d) On the axes below, sketch the I-V graph for this wire. Label the axes clearly.

(Space for sketch)
 [2]

17. The diagram shows a circuit used to control a heater. The circuit includes a thermistor, a variable resistor, a relay switch, and a heater connected to a separate high-voltage supply.

(Imagine diagram: Control circuit with battery, thermistor, variable resistor in series. Relay coil is in parallel with the thermistor. The relay switch controls the heater circuit.)

(a) Explain how the relay switch allows a low-voltage control circuit to switch on a high-voltage heater.
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
........................................................................................................................... [3]

(b) The thermistor is an NTC thermistor. Explain what happens to the current in the relay coil as the temperature of the surroundings decreases.
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
........................................................................................................................... [3]

(c) Suggest one practical application for this type of circuit.
........................................................................................................................... [1]

18. A transformer is used to step down the voltage from $240 \text{ V}$ to $12 \text{ V}$ to power a lamp. The lamp is rated at $12 \text{ V}$ , $24 \text{ W}$ .

(a) Calculate the current flowing through the lamp when it is operating at normal brightness.
 Current = ______________________ A [2]

(b) The primary coil has 4000 turns. Calculate the number of turns on the secondary coil.
 Number of turns = ______________________ [2]

(c) Assuming the transformer is 100% efficient, calculate the current in the primary coil.
 Current = ______________________ A [2]

(d) In reality, transformers are not 100% efficient. State two reasons for energy loss in a transformer.

.......................................................................................................................
....................................................................................................................... [2]

19. A straight wire carrying a current is placed between the poles of a U-shaped magnet. The wire experiences a force.

(a) State the rule used to determine the direction of this force.
........................................................................................................................... [1]

(b) The current in the wire is $5.0 \text{ A}$ . The length of the wire in the magnetic field is $0.10 \text{ m}$ . The magnetic flux density is $0.20 \text{ T}$ . The wire is perpendicular to the magnetic field.
Calculate the magnitude of the force on the wire.
(Note: Formula $F = BIL$ may be used if known, otherwise describe the proportionality. For O-Level, usually qualitative or given formula. Here, calculate using standard formula.)
 Force = ______________________ N [2]

.......................................................................................................................
....................................................................................................................... [2]

20. A student wants to determine the electrical energy consumed by an electric kettle. The kettle is rated at $2.0 \text{ kW}$ .

(a) Calculate the energy consumed in kilowatt-hours (kWh) if the kettle is used for 15 minutes.
 Energy = ______________________ kWh [2]

(b) If the cost of electricity is $25$ cents per kWh, calculate the cost of using the kettle for this 15-minute period.
 Cost = ______________________ cents [2]

(c) Explain why the actual energy supplied by the mains might be slightly higher than the energy calculated in (a) if the voltage drops slightly below $240 \text{ V}$ , assuming the kettle is used to boil a fixed amount of water from room temperature.
(Hint: Consider power and time.)
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
........................................................................................................................... [3]

End of Paper

Answers

TuitionGoWhere Practice Paper - Physics O-Level

Answer Key & Marking Scheme
Version: 4 of 5
Topic: Electricity & Magnetism

Section A: Multiple Choice & Short Structured Questions [20 marks]

1. A
Conventional current flows from positive to negative. Electrons (negative charge) flow from negative to positive. [1]

2. C
Polythene gains electrons from the cloth. Electrons are the mobile charge carriers in solids. [1]

3. A
Field lines go from Positive to Negative. Density of lines indicates field strength/charge magnitude. Denser at X means X has larger charge. [1]

4. D
$R = \rho L / A$ . New $R' = \rho (2L) / (0.5A) = 4 (\rho L / A) = 4 \times 8.0 = 32.0 \, \Omega$ . [1]

5. C
Total $R = 4 + 8 = 12 \, \Omega$ . Current $I = V/R = 12/12 = 1 \text{ A}$ . $V_{R2} = I \times R_2 = 1 \times 8 = 8 \text{ V}$ . Alternatively, voltage divider: $V_{R2} = \frac{8}{4+8} \times 12 = 8 \text{ V}$ . [1]

6. B
Filament lamp: As V increases, I increases, temperature increases, resistance increases. Gradient of I-V graph (which represents 1/R) decreases. [1]

7. D
$P = VI \Rightarrow I = P/V = 2400/240 = 10 \text{ A}$ . Fuse must be rated slightly higher than operating current. 13 A is the next standard size above 10 A. [1]

8. B
Soft iron is easily magnetized and demagnetized, allowing the magnetic field in the core to change rapidly with the AC current. [1]

9. B
$V_s / V_p = N_s / N_p \Rightarrow V_s / 240 = 100 / 500 \Rightarrow V_s = 240 \times 0.2 = 48 \text{ V}$ . [1]

10. C
$P_{loss} = I^2 R$ . High voltage allows lower current for the same power transmitted ( $P=VI$ ), thus reducing $I^2 R$ losses. [1]

11.
(a) To reverse the direction of current in the coil every half rotation [1], ensuring the coil continues to rotate in the same direction (maintains torque direction). [1]
(b) Any two:

Increase the current.
Use stronger magnets (increase magnetic field strength).
Increase the number of turns on the coil. [2]

12.
(a) Charging by induction. [1]
(b) Positive. [1]
(Electrons are repelled by the negative rod to earth. When earth is removed, sphere is left with net positive charge.)

13.
The work done by the source (battery) in driving a unit charge around the complete circuit. [1]
OR
The energy converted from chemical (or other) form to electrical energy per unit charge. [1]
(Must mention "per unit charge" or "work done/energy per coulomb".) [2]

14.
(a) Increases (gets brighter). [1]
(b) Closing S adds Lamp C in parallel with B. This decreases the total resistance of the parallel section. [1]
Therefore, the total resistance of the whole circuit decreases. [1]
Total current from the battery increases. Since Lamp A is in the main branch, the current through A increases, so it gets brighter. [1]
(Award marks for logical chain: R total down -> I total up -> I through A up.) [2]

15.
(a) Resistance decreases. [1]
(b) As temperature increases, resistance of thermistor decreases. [1]
This causes the total resistance of the circuit to decrease, so the current in the circuit increases. [1]
Since $V = IR$ for the fixed resistor (and R is constant), the potential difference across it increases. [1]
(Alternatively: Voltage divider principle. Thermistor takes less share of voltage, so fixed resistor takes more.) [2]

Section B: Structured Problems [30 marks]

16.
(a) $R = V / I = 2.0 / 0.40 = 5.0 \, \Omega$ . [2]
(1 mark for formula/substitution, 1 mark for answer.)

(b) $R = V / I = 8.0 / 1.48 \approx 5.41 \, \Omega$ . [2]
(Accept 5.4 or 5.40. 1 mark for formula/substitution, 1 mark for answer.)

(c) As the potential difference (and current) increases, the electrical energy dissipated heats up the wire/filament. [1]
The temperature of the wire increases. [1]
As temperature increases, the metal ions vibrate more vigorously, causing more frequent collisions with the flowing electrons, which increases the resistance. [1] [3]

(d) Graph:

Axes labeled: Y-axis "Current / A", X-axis "Potential Difference / V". [1]
Curve starts steep at origin and curves towards the V-axis (gradient decreases). [1] [2]

17.
(a) A small current in the control circuit flows through the relay coil. [1]
This magnetizes the soft iron core of the relay, attracting the iron armature/switch. [1]
This closes the contacts in the high-voltage heater circuit, allowing a large current to flow through the heater. [1] [3]

(b) As temperature decreases, the resistance of the NTC thermistor increases. [1]
Since the thermistor and variable resistor are in series (or depending on specific diagram, usually thermistor is part of a potential divider), if the relay is in parallel with the thermistor:
Correction based on standard potential divider logic for "heater on when cold":
Usually, for a heater to turn on when cold, the relay should activate when the thermistor resistance is high.
If the relay coil is in parallel with the thermistor: As T decreases, $R_{therm}$ increases. In a series circuit with a fixed resistor, the voltage across the thermistor increases ( $V = V_{source} \times \frac{R_{therm}}{R_{total}}$ ). [1]
Therefore, the potential difference across the relay coil increases. [1]
This causes the current in the relay coil to increase, activating the switch. [1]
(Note: If the diagram implied the relay was in series with the thermistor, the current would decrease. However, standard control circuits use a potential divider. The question asks for the effect on current in the coil. If the coil is across the thermistor, V increases, so I in coil increases.) [3]

18.
(a) $P = VI \Rightarrow I = P / V = 24 / 12 = 2.0 \text{ A}$ . [2]

(b) $\frac{V_s}{V_p} = \frac{N_s}{N_p} \Rightarrow \frac{12}{240} = \frac{N_s}{4000}$ .
$N_s = \frac{12}{240} \times 4000 = 0.05 \times 4000 = 200$ turns. [2]

(c) $P_{in} = P_{out}$ (100% efficient). $P_{in} = 24 \text{ W}$ .
$I_p = P_{in} / V_p = 24 / 240 = 0.10 \text{ A}$ . [2]
(Alternatively: $I_p / I_s = N_s / N_p \Rightarrow I_p / 2 = 200 / 4000 \Rightarrow I_p = 2 \times 0.05 = 0.1 \text{ A}$ .)

(d) Any two:

Heating of coils (due to resistance of copper wire).
Eddy currents in the core (causing heating).
Magnetization and demagnetization of the core (hysteresis loss).
Leakage of magnetic flux (not all flux links both coils). [2]

19.
(a) Fleming's Left-Hand Rule. [1]

(b) $F = B I L$
$F = 0.20 \times 5.0 \times 0.10$
$F = 0.10 \text{ N}$ . [2]
(1 mark for formula/substitution, 1 mark for answer.)

Reverse the direction of the current.
Reverse the direction of the magnetic field (swap N and S poles). [2]

20.
(a) Time in hours = $15 / 60 = 0.25 \text{ h}$ .
Power in kW = $2.0 \text{ kW}$ .
Energy = $P \times t = 2.0 \times 0.25 = 0.50 \text{ kWh}$ . [2]

(b) Cost = $0.50 \times 25 = 12.5$ cents. [2]

(c) If voltage drops, the power output of the kettle decreases ( $P = V^2/R$ ). [1]
To boil the same amount of water, the same amount of thermal energy is required. [1]
Since Power is lower, the time taken to boil the water will increase ( $t = E/P$ ). [1]
The question asks about energy supplied. If we assume the kettle switches off automatically when boiling, the energy supplied is roughly the same (ignoring heat loss over time). However, if the question implies "energy drawn from mains to complete the task", and considering heat losses over a longer time:
Alternative interpretation for higher marks:
Actually, if $V$ drops, $P$ drops. The kettle takes longer. Heat loss to surroundings occurs over a longer period. Therefore, more total energy must be supplied to compensate for the extra heat loss to the surroundings during the longer boiling time. [3]
(Accept: Lower V -> Lower P -> Longer time -> More heat loss to surroundings -> More energy needed.)