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O Level Physics Practice Paper 2

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TuitionGoWhere Practice Paper - Physics O-Level

TuitionGoWhere Practice Paper (AI)

Subject: Physics (6091)
Level: O-Level
Topic: Electricity & Magnetism (Practice Paper - Version 2 of 5)
Duration: 1 hour
Total Marks: 50

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 this question paper.
The number of marks is given in brackets [ ] at the end of each question or part question.
You may use a calculator.
Take the acceleration of free fall, $g = 10 \text{ m/s}^2$ .
Assume standard mains voltage is $230 \text{ V}$ unless stated otherwise.

Section A: Multiple Choice & Short Structured Questions (20 Marks)

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

Answer: _________

2. A student rubs a polythene rod with a dry cloth. The rod becomes negatively charged. Which statement explains this observation?
[1]
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: _________

3. Fig. 3.1 shows two identical metal spheres, X and Y, on insulated stands. Sphere X is positively charged. Sphere Y is neutral. A grounded metal wire touches sphere Y while sphere X is brought close to it (without touching). The wire is then removed, followed by sphere X.
[2]

(a) State the final charge on sphere Y.

(b) Explain, in terms of electron movement, how sphere Y acquired this charge.

4. Fig. 4.1 shows a simple circuit containing a cell, a switch, a fixed resistor, and a thermistor.
[2]

(a) Define electromotive force (e.m.f.).

(b) The temperature of the thermistor increases. State and explain the effect on the current in the circuit.

5. A wire of length $L$ and cross-sectional area $A$ has a resistance of $8.0 \, \Omega$ .
[2]

(a) Calculate the resistance of a second wire made of the same material, with length $2L$ and cross-sectional area $A/2$ .

(b) State one factor, other than dimensions, that affects the resistance of a metallic conductor.

6. Fig. 6.1 shows the current-voltage (I-V) characteristic graph for a filament lamp.
[2]

(a) Explain why the graph is curved and not a straight line through the origin.

(b) Calculate the resistance of the lamp when the potential difference across it is $6.0 \text{ V}$ and the current is $0.5 \text{ A}$ .

7. Three resistors, each of resistance $6.0 \, \Omega$ , are connected in a circuit.
[2]

(a) Calculate the total resistance when they are connected in series.

(b) Calculate the total resistance when they are connected in parallel.

8. A battery of e.m.f. $12 \text{ V}$ and negligible internal resistance is connected to two resistors in series: $R_1 = 4.0 \, \Omega$ and $R_2 = 8.0 \, \Omega$ .
[2]

(a) Calculate the current in the circuit.

(b) Calculate the potential difference across $R_2$ .

9. Fig. 9.1 shows a potential divider circuit used to control the brightness of a lamp. The input voltage is $12 \text{ V}$ . The variable resistor is set such that the voltage across the lamp is $4.0 \text{ V}$ .
[2]

(a) If the resistance of the variable resistor is increased, what happens to the brightness of the lamp?

(b) Explain your answer in terms of voltage division.

10. An electric kettle is rated at $230 \text{ V}$ , $2.3 \text{ kW}$ .
[2]

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

(b) Suggest a suitable fuse rating for the plug: $3 \text{ A}$ , $5 \text{ A}$ , or $13 \text{ A}$ .

Section B: Structured Problems (20 Marks)

11. Fig. 11.1 shows a circuit containing a $12 \text{ V}$ battery, a switch S, a $4.0 \, \Omega$ resistor ( $R_1$ ), and a $6.0 \, \Omega$ resistor ( $R_2$ ) connected in parallel. An ammeter $A_1$ is in the main branch, and ammeter $A_2$ is in the branch with $R_2$ .
[5]

(a) Calculate the effective resistance of the parallel combination.
 
Answer: _______________ $\Omega$ [2]

(b) Calculate the reading on ammeter $A_1$ .
 
Answer: _______________ A [2]

12. A student investigates the relationship between the length of a resistance wire and its resistance. The wire has a uniform cross-sectional area. The student measures the resistance for different lengths $L$ . The results are plotted in Fig. 12.1.
[5]

(a) Describe the relationship between resistance $R$ and length $L$ shown in the graph.

_________________________________________________________________________ [1]

(b) The gradient of the graph is $0.05 \, \Omega/\text{cm}$ . Calculate the resistance of a $40 \text{ cm}$ length of this wire.
 
Answer: _______________ $\Omega$ [2]

(c) The student repeats the experiment with a wire of the same material but twice the cross-sectional area. On Fig. 12.1, sketch the new line of best fit. Label it 'Wire B'. [2]

13. Fig. 13.1 shows the wiring of a three-pin plug for a metal-cased electric heater.
[5]

(a) Identify the wires connected to terminals P, Q, and R.
P: ____________________
Q: ____________________
R: ____________________ [3]

(b) Explain the purpose of the earth wire (terminal P) in this appliance.

_________________________________________________________________________ [2]

14. An electric iron is rated at $230 \text{ V}$ , $1150 \text{ W}$ . It is used for $30$ minutes. The cost of electricity is $25$ cents per kWh.
[5]

(a) Calculate the energy consumed by the iron in kWh.
 
Answer: _______________ kWh [2]

(b) Calculate the cost of using the iron for this period.
 
Answer: _______________ cents [1]

(c) The iron has a double-insulated casing.
(i) What symbol is used to indicate double insulation?
_________________________________________________________________________ [1]
(ii) Why is an earth wire not required for this appliance?

_________________________________________________________________________ [1]

Section C: Electromagnetism & Induction (10 Marks)

15. Fig. 15.1 shows a straight wire carrying a current flowing vertically upwards. The wire passes through a horizontal card.
[3]

(a) Sketch the pattern of the magnetic field lines on the card. Include arrows to show the direction of the field. [2]

(b) State the rule used to determine the direction of the magnetic field.
_________________________________________________________________________ [1]

16. Fig. 16.1 shows a simple d.c. motor. It consists of a rectangular coil ABCD placed between the poles of a magnet.
[4]

(a) Explain why the coil rotates when a current flows through it.

_________________________________________________________________________ [2]

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

_________________________________________________________________________ [2]

17. Fig. 17.1 shows a bar magnet being pushed into a solenoid connected to a sensitive galvanometer.
[3]

(a) State what is observed on the galvanometer as the magnet moves into the solenoid.
_________________________________________________________________________ [1]

(b) State two changes that would increase the deflection on the galvanometer.

_________________________________________________________________________ [2]

Answers

TuitionGoWhere Practice Paper - Physics O-Level (Answer Key)

Topic: Electricity & Magnetism (Version 2)

Section A: Multiple Choice & Short Structured Questions

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

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

3.
(a) Negative [1]
(b) Explanation: The positive charge on X attracts electrons in Y. When Y is earthed, electrons flow from the ground to Y to neutralize the repelled positive charges (or simply: electrons are attracted from earth to Y due to the influence of X). When the earth is removed, these excess electrons are trapped on Y. [1]

4.
(a) Definition: The energy supplied by the source per unit charge passing through it. (Or: Work done by the source in driving a unit charge around a complete circuit). [1]
(b) Effect: Current increases. [1]
Explanation: As temperature increases, the resistance of the NTC thermistor decreases. Since $I = V/R$ , a lower total resistance results in a higher current. [1]

5.
(a) Calculation:
$R = \rho L / A$ .
New $R' = \rho (2L) / (A/2) = 4 (\rho L / A) = 4 \times 8.0 = 32.0 \, \Omega$ .
Answer: $32.0 \, \Omega$ [1]
(b) Factor: Temperature. [1]

6.
(a) Explanation: As the voltage/current increases, the temperature of the filament increases. This causes the metal ions to vibrate more vigorously, increasing the frequency of collisions with electrons, thus increasing resistance. [2]
(b) Calculation:
$R = V / I = 6.0 / 0.5 = 12 \, \Omega$ .
Answer: $12 \, \Omega$ [1]

7.
(a) Series: $R_{total} = R_1 + R_2 + R_3 = 6 + 6 + 6 = 18 \, \Omega$ .
Answer: $18 \, \Omega$ [1]
(b) Parallel:
$1/R_{total} = 1/6 + 1/6 + 1/6 = 3/6 = 1/2$ .
$R_{total} = 2.0 \, \Omega$ .
Answer: $2.0 \, \Omega$ [1]

8.
(a) Current:
$R_{total} = 4.0 + 8.0 = 12.0 \, \Omega$ .
$I = V / R = 12 / 12 = 1.0 \text{ A}$ .
Answer: $1.0 \text{ A}$ [1]
(b) Potential Difference:
$V_2 = I \times R_2 = 1.0 \times 8.0 = 8.0 \text{ V}$ .
Answer: $8.0 \text{ V}$ [1]

9.
(a) Brightness: Decreases (or lamp goes dimmer). [1]
(b) Explanation: Increasing the variable resistor increases its share of the voltage (voltage divider principle). Consequently, the voltage across the lamp decreases, reducing the current and power ( $P=VI$ ). [1]

10.
(a) Current:
$P = VI \Rightarrow I = P / V = 2300 / 230 = 10 \text{ A}$ .
Answer: $10 \text{ A}$ [1]
(b) Fuse: $13 \text{ A}$ [1] (Must be higher than operating current but closest standard value).

Section B: Structured Problems

11.
(a) Effective Resistance:
$\frac{1}{R_p} = \frac{1}{4} + \frac{1}{6} = \frac{3}{12} + \frac{2}{12} = \frac{5}{12}$ .
$R_p = \frac{12}{5} = 2.4 \, \Omega$ .
Answer: $2.4 \, \Omega$ [2]

(b) Ammeter $A_1$ (Total Current):
$I_{total} = V / R_p = 12 / 2.4 = 5.0 \text{ A}$ .
Answer: $5.0 \text{ A}$ [2]

(c) Power in $R_1$ :
Voltage across $R_1$ is $12 \text{ V}$ (parallel).
$P = V^2 / R = 12^2 / 4 = 144 / 4 = 36 \text{ W}$ .
Answer: $36 \text{ W}$ [1]

12.
(a) Relationship: Resistance is directly proportional to length ( $R \propto L$ ). [1]
(b) Calculation:
$R = \text{gradient} \times L = 0.05 \times 40 = 2.0 \, \Omega$ .
Answer: $2.0 \, \Omega$ [2]
(c) Sketch: A straight line through the origin with a gradient half that of the original line (since area doubled, resistance halved for same length). [2]

13.
(a) Identification:
P: Earth (Green/Yellow) [1]
Q: Neutral (Blue) [1]
R: Live (Brown) [1]
(Note: Pin positions may vary by diagram, but typically Top=Earth, Left=Neutral, Right=Live in standard diagrams. Accept based on standard color coding if diagram implies it).

(b) Purpose of Earth Wire:
It provides a low-resistance path to the ground. If the live wire touches the metal casing, a large current flows to earth, blowing the fuse and preventing the casing from becoming live/electrocuting the user. [2]

14.
(a) Energy in kWh:
Power $P = 1150 \text{ W} = 1.15 \text{ kW}$ .
Time $t = 30 \text{ min} = 0.5 \text{ h}$ .
$E = P \times t = 1.15 \times 0.5 = 0.575 \text{ kWh}$ .
Answer: $0.575 \text{ kWh}$ [2]

(b) Cost:
$\text{Cost} = 0.575 \times 25 = 14.375$ cents.
Answer: $14.4$ cents (or $14.38$ ) [1]

(c) Double Insulation:
(i) Symbol: A square within a square. [1]
(ii) Reason: The casing is made of insulating material (plastic), so it cannot become live even if internal wires loosen. Thus, no earth connection is needed for safety. [1]

Section C: Electromagnetism & Induction

15.
(a) Sketch: Concentric circles centered on the wire. Arrows pointing counter-clockwise (using Right-Hand Grip Rule for upward current). [2]
(b) Rule: Right-Hand Grip Rule. [1]

16.
(a) Rotation Explanation:
Current flowing through the coil in a magnetic field experiences a force (Motor Effect). Using Fleming's Left-Hand Rule, the forces on opposite sides of the coil are in opposite directions, creating a turning effect (torque). [2]

(b) Split-Ring Commutator Function:
It reverses the direction of the current in the coil every half rotation. This ensures that the forces on the coil sides always act in the same rotational direction, allowing continuous rotation. [2]

17.
(a) Observation: The galvanometer needle deflects (moves to one side). [1]

(b) Increases Deflection:

Move the magnet faster. [1]
Use a stronger magnet. [1]
(Alternative: Increase number of turns on the solenoid).