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O Level Physics Practice Paper 1

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Questions

TuitionGoWhere Practice Paper - Physics O-Level

TuitionGoWhere Practice Paper (AI)

Subject: Physics
Level: O-Level (6091)
Paper: Practice Paper - Version 1 of 5
Topic: Electricity and Magnetism (Topics 13–19)
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.
The number of marks is given in brackets [ ] at the end of each question or part question.
Show all working in calculations. Use $g = 10 \text{ m/s}^2$ where applicable, though not required for this specific topic.
You may use a scientific calculator.

Section A: Structured Questions (40 Marks)

Answer all questions in this section.

1. A student rubs a polythene rod with a dry cloth. The rod becomes negatively charged. (a) Explain, in terms of electron transfer, how the rod becomes negatively charged. [2]

(b) The charged rod is brought near a small piece of neutral paper. The paper is attracted to the rod. Explain why this attraction occurs. [2]

2. Fig. 2.1 shows a simple circuit containing a battery, a switch, a fixed resistor $R$ , and a thermistor $T$ .

(Imagine Fig 2.1: Series circuit with Battery, Switch, Resistor R, Thermistor T)

(a) State how the resistance of the thermistor changes as the temperature increases. [1]

(b) The temperature of the thermistor increases. State and explain the effect on the reading of the ammeter in the circuit. [2]

3. A filament lamp is rated at $6.0 \text{ V}$ , $0.5 \text{ A}$ . (a) Calculate the resistance of the lamp when it is operating at its normal brightness. [2]

(b) The current-voltage (I-V) characteristic of the filament lamp is a curve, not a straight line. Explain why the resistance of the lamp changes as the voltage increases. [2]

4. Fig. 4.1 shows two resistors, $R_1 = 4.0 \, \Omega$ and $R_2 = 6.0 \, \Omega$ , connected in parallel across a $12 \text{ V}$ supply.

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

(b) Calculate the current flowing through resistor $R_1$ . [2]

5. A potential divider circuit is used to control the voltage across a component. It consists of a $12 \text{ V}$ battery connected in series with a fixed resistor of $100 \, \Omega$ and a variable resistor set to $200 \, \Omega$ . The output voltage is taken across the variable resistor.

(a) Calculate the output voltage. [2]

(b) State one practical application of a potential divider circuit using a Light Dependent Resistor (LDR). [1]

6. An electric kettle is rated at $240 \text{ V}$ , $2.4 \text{ kW}$ . (a) Calculate the current flowing through the kettle when it is switched on. [2]

(b) Calculate the electrical energy converted to heat energy if the kettle is used for 3 minutes. [2]

(c) Suggest a suitable fuse rating for the plug of this kettle from the following options: $3 \text{ A}$ , $5 \text{ A}$ , $13 \text{ A}$ . Explain your choice. [2]

7. Fig. 7.1 shows a bar magnet with its North pole on the left and South pole on the right.

(a) Sketch the magnetic field lines around the bar magnet. Include at least four lines and indicate the direction with arrows. [2] (Space for sketch)

(b) Describe how you would use a plotting compass to determine the direction of the magnetic field at a point near the North pole. [2]

8. A straight wire carries a current flowing vertically upwards. The wire is placed in a uniform magnetic field directed horizontally from West to East.

(a) State the direction of the force acting on the wire. [1]

(b) State two ways to increase the magnitude of the force on the wire. [2]

9. Fig. 9.1 shows a simple d.c. motor.

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

(b) State two changes that would cause the coil to rotate in the opposite direction. [2]

10. A transformer has 500 turns on the primary coil and 100 turns on the secondary coil. The primary coil is connected to a $240 \text{ V}$ a.c. supply.

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

(b) The transformer is assumed to be 100% efficient. If the current in the primary coil is $0.5 \text{ A}$ , calculate the current in the secondary coil. [2]

Section B: Free Response Questions (20 Marks)

Answer all questions in this section.

11. A student investigates the relationship between the length of a resistance wire and its resistance. The student uses a wire of uniform cross-sectional area.

(a) State the hypothesis for this investigation. [1]

(b) Describe the experimental procedure the student should follow to obtain reliable results. Your answer should include:

The circuit diagram description (or sketch description).
The measurements to be taken.
How the resistance is calculated. [4]

(c) The student plots a graph of Resistance ( $R$ ) against Length ( $L$ ). Describe the expected shape of this graph. [1]

(d) Another student repeats the experiment using a wire with twice the cross-sectional area but the same material and length. Compare the resistance of this new wire to the original wire. [2]

12. Electromagnetic induction is the principle behind electrical generators.

(a) State Faraday’s Law of electromagnetic induction. [2]

(b) Fig. 12.1 shows a coil of wire connected to a sensitive galvanometer. A bar magnet is pushed into the coil.

(i) Explain why the galvanometer needle deflects. [2]

(ii) State two ways to increase the magnitude of the deflection. [2]

13. High voltage transmission is used to distribute electrical energy over long distances.

(a) Explain why electrical energy is transmitted at high voltages. [2]

(b) A power station generates electricity at $25 \text{ kV}$ . A step-up transformer increases this to $400 \text{ kV}$ for transmission. (i) Calculate the ratio of the number of turns on the secondary coil to the number of turns on the primary coil. [2]

(ii) If the current in the primary coil is $4000 \text{ A}$ , calculate the current in the secondary coil, assuming the transformer is ideal. [2]

Answers

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

Topic: Electricity and Magnetism
Version: 1 of 5

Section A: Structured Questions

1. (a) Electrons are transferred from the dry cloth to the polythene rod. [1] The rod gains excess electrons, giving it a net negative charge. [1] (b) The negative charge on the rod repels electrons in the paper to the far side, leaving the near side positively charged (induction). [1] The attractive force between the rod and the near positive side is stronger than the repulsive force from the far negative side (due to distance), resulting in net attraction. [1]

2. (a) The resistance of the thermistor decreases. [1] (b) The total resistance of the circuit decreases. [1] According to Ohm’s Law ( $I = V/R$ ), since voltage is constant and resistance decreases, the current increases. [1]

3. (a) $R = V / I$ [1] $R = 6.0 / 0.5 = 12 \, \Omega$ [1] (b) As voltage increases, current increases, causing the filament to heat up. [1] The increased temperature causes the metal ions in the lattice to vibrate more vigorously, increasing the frequency of collisions with electrons, thus increasing resistance. [1]

4. (a) $\frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2}$ [1] $\frac{1}{R_{total}} = \frac{1}{4} + \frac{1}{6} = \frac{3}{12} + \frac{2}{12} = \frac{5}{12}$ $R_{total} = \frac{12}{5} = 2.4 \, \Omega$ [1] (b) $I_1 = V / R_1$ [1] $I_1 = 12 / 4.0 = 3.0 \text{ A}$ [1]

5. (a) $V_{out} = V_{in} \times \frac{R_{variable}}{R_{fixed} + R_{variable}}$ [1] $V_{out} = 12 \times \frac{200}{100 + 200} = 12 \times \frac{200}{300} = 12 \times \frac{2}{3} = 8 \text{ V}$ [1] (b) Automatic street lighting / Night light / Camera light meter. [1] (Any valid LDR application)

6. (a) $P = VI \Rightarrow I = P / V$ [1] $I = 2400 \text{ W} / 240 \text{ V} = 10 \text{ A}$ [1] (b) $E = P \times t$ [1] $t = 3 \times 60 = 180 \text{ s}$ $E = 2400 \times 180 = 432,000 \text{ J}$ (or $432 \text{ kJ}$ ) [1] (c) $13 \text{ A}$ fuse. [1] The operating current is $10 \text{ A}$ . A $3 \text{ A}$ or $5 \text{ A}$ fuse would blow immediately. A $13 \text{ A}$ fuse allows the normal current to flow but protects against excessive current. [1]

7. (a) Lines emerge from North pole and enter South pole. [1] Lines do not cross. Arrows point from N to S. At least 4 lines drawn correctly. [1] (b) Place the plotting compass near the North pole. [1] Mark the position of the North pole of the compass needle. Move the compass so its South pole is at the previous mark. Repeat to trace a line. The direction is indicated by the North pole of the compass. [1]

8. (a) Vertically Out of the page / Upwards (depending on orientation definition, but strictly: Current Up, Field East -> Force is North if using standard cardinal directions on a horizontal plane, or Out of page if Field is Left-Right and Current is Up-Down on paper). Correction for standard 2D diagram interpretation: If Current is Up (paper plane) and Field is West-to-East (paper plane, Left-to-Right), Force is Into the page (using Fleming's Left Hand Rule: First finger East, Second finger Up, Thumb points Into Page). Acceptable Answer: Into the page (or perpendicular to both current and field). [1] (b) 1. Increase the current. [1] 2. Increase the magnetic field strength (or use stronger magnets). [1] (Also accept: Increase length of wire in field)

9. (a) It reverses the direction of the current in the coil every half rotation. [1] This ensures that the force on the arms of the coil always acts in the same rotational direction, allowing continuous rotation. [1] (b) 1. Reverse the polarity of the magnetic field (swap N and S poles). [1] 2. Reverse the direction of the current supply. [1]

10. (a) $\frac{V_s}{V_p} = \frac{N_s}{N_p}$ [1] $V_s = 240 \times \frac{100}{500} = 240 \times 0.2 = 48 \text{ V}$ [1] (b) $V_p I_p = V_s I_s$ (for 100% efficiency) [1] $240 \times 0.5 = 48 \times I_s$ $120 = 48 I_s$ $I_s = 120 / 48 = 2.5 \text{ A}$ [1] (c) Transformers rely on a changing magnetic field to induce an EMF in the secondary coil. [1] D.C. produces a constant magnetic field, so there is no change in flux linkage and no induced EMF. [1]

Section B: Free Response Questions

11. (a) Resistance is directly proportional to the length of the wire. [1] (b) Circuit: Connect the resistance wire in series with an ammeter and a power supply. Connect a voltmeter in parallel across the specific length of the wire being tested. Use a jockey or crocodile clips to vary the length. [1] Measurements: Measure the length $L$ of the wire between the contacts. Record the current $I$ from the ammeter and voltage $V$ from the voltmeter. [1] Calculation: Calculate resistance using $R = V/I$ for each length. [1] Reliability: Repeat readings for each length and take an average. Ensure the wire does not heat up significantly (use low current). [1] (c) A straight line passing through the origin. [1] (d) Resistance is inversely proportional to cross-sectional area ( $R \propto 1/A$ ). [1] Since the area is doubled, the resistance will be half that of the original wire. [1]

12. (a) The magnitude of the induced EMF is directly proportional to the rate of change of magnetic flux linkage. [2] (b) (i) As the magnet moves, the magnetic field lines cut the coil (or magnetic flux through the coil changes). [1] This induces an EMF, which drives a current through the galvanometer, causing deflection. [1] (ii) 1. Move the magnet faster. [1] 2. Use a stronger magnet (or increase the number of turns on the coil). [1] (c) The reading is zero. [1] There is no change in magnetic flux linkage because the magnet is stationary relative to the coil. [1]

13. (a) To reduce energy loss due to heating in the transmission cables. [1] Since $P_{loss} = I^2 R$ , transmitting at high voltage allows for lower current for the same power, significantly reducing heat loss. [1] (b) (i) $\frac{N_s}{N_p} = \frac{V_s}{V_p}$ [1] Ratio $= \frac{400 \text{ kV}}{25 \text{ kV}} = 16$ [1] (ii) $V_p I_p = V_s I_s$ [1] $25 \times 4000 = 400 \times I_s$ $100,000 = 400 I_s$ $I_s = 100,000 / 400 = 250 \text{ A}$ [1]