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Secondary 4 Pure Physics Preliminary Examination Paper 3

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TuitionGoWhere Exam Practice (AI) - Preliminary Examination

TuitionGoWhere Secondary School (AI)

Subject: Pure Physics
Level: Secondary 4
Paper: Preliminary Examination - Version 3
Duration: 1 hour 45 minutes
Total Marks: 60

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 in this booklet.
The number of marks is given in brackets [ ] at the end of each question or part question.
You may use an approved scientific calculator where appropriate.
Take the acceleration due to gravity, $g$ , to be $10 \text{ m/s}^2$ .
For calculations involving electricity, assume standard mains voltage is $230 \text{ V}$ unless stated otherwise.

Section A: Structured Questions (40 Marks)

Answer all questions in this section.

1. A student rubs a polythene rod with a woolen 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 resistor, and a lamp.

(a) Define electromotive force (e.m.f.). [1]

(b) The battery has an e.m.f. of $12 \text{ V}$ . The lamp has a resistance of $4.0 \, \Omega$ and the resistor has a resistance of $2.0 \, \Omega$ . Calculate the current flowing through the circuit when the switch is closed. [2]

3. A transformer is used to step down the voltage from $240 \text{ V}$ to $12 \text{ V}$ for a laptop charger. The primary coil has 2000 turns. (a) Calculate the number of turns on the secondary coil. [2]

(b) The laptop draws a current of $2.5 \text{ A}$ from the secondary coil. Assuming the transformer is $100\%$ efficient, calculate the current in the primary coil. [2]

4. Fig. 4.1 shows a current-carrying wire placed between the poles of a U-shaped magnet. The current flows into the page.

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

(b) Explain what happens to the magnitude of the force if: (i) The current in the wire is increased. [1] _________________________________________________________________________

   (ii) The magnet is replaced with a weaker magnet. [1]
   _________________________________________________________________________

5. A household electrical circuit includes a fuse and an earth wire for safety. (a) State the function of the live wire. [1]

(b) Explain how a fuse protects an appliance from excessive current. [2]

6. Two resistors, $R_1 = 6.0 \, \Omega$ and $R_2 = 3.0 \, \Omega$ , are connected in parallel across a $12 \text{ V}$ supply. (a) Calculate the combined resistance of the two resistors. [2]

(b) Calculate the total current supplied by the battery. [2]

7. Fig. 7.1 shows a coil of wire rotating in a magnetic field, connected to slip rings and brushes. This setup acts as an A.C. generator.

(a) State Faraday’s Law of Electromagnetic Induction. [2]

(b) Explain why an alternating current is produced in the external circuit. [2]

8. A student investigates the relationship between the length of a wire and its resistance. She uses wires of the same material and cross-sectional area but different lengths. (a) State the relationship between the length of the wire and its resistance. [1]

(b) The student measures a current of $0.5 \text{ A}$ and a potential difference of $3.0 \text{ V}$ across a $1.0 \text{ m}$ length of wire. Calculate the resistance of this wire. [1]

9. Fig. 9.1 shows the magnetic field pattern around a straight current-carrying wire. (a) Describe the shape of the magnetic field lines. [1]

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

10. A solar panel generates electricity for a home. (a) State the energy conversion that takes place in a solar panel. [1]

(b) The solar panel produces $200 \text{ W}$ of power. Calculate the energy generated in 5 hours. Give your answer in Joules. [2]

Section B: Free-Response Questions (20 Marks)

Answer all questions in this section.

11. A student sets up a circuit to measure the resistance of a filament lamp. The circuit includes a power supply, an ammeter, a voltmeter, a variable resistor, and the lamp. (a) Draw the circuit diagram for this setup. [3]

(b) The student obtains the following results:

Voltage (V)	Current (A)
2.0	0.5
4.0	0.8
6.0	1.0
8.0	1.1
10.0	1.2

(i) Plot a graph of Current (y-axis) against Voltage (x-axis). [3]

(ii) Explain why the graph is not a straight line. [2]

(iii) Calculate the resistance of the lamp when the voltage is $6.0 \text{ V}$ . [2]

12. Fig. 12.1 shows a simplified diagram of a circuit breaker used in a household distribution board.

(a) Explain the operation of a circuit breaker when a fault causes a large current to flow. [3]

(b) State one advantage of a circuit breaker over a fuse. [1]

(c) A kettle rated at $230 \text{ V}$ , $2000 \text{ W}$ is plugged into a socket. (i) Calculate the current flowing through the kettle. [2]

(ii) Suggest a suitable rating for the fuse or circuit breaker for this kettle. [1]

(iii) Explain why it is dangerous to use a fuse with a much higher rating than required. [2]

13. Electromagnetic induction is used in transformers to change voltage levels. (a) Explain why a transformer does not work with direct current (d.c.). [2]

(b) A power station generates electricity at $25,000 \text{ V}$ . This is stepped up to $400,000 \text{ V}$ for transmission. (i) Calculate the turns ratio ( $N_s : N_p$ ) of the transformer. [2]

(ii) Explain why high voltage is used for long-distance power transmission. [2]

(iii) If the current in the primary coil is $1000 \text{ A}$ , calculate the current in the secondary coil, assuming $100\%$ efficiency. [2]

Answers

TuitionGoWhere Exam Practice (AI) - Preliminary Examination Answer Key

Subject: Pure Physics
Level: Secondary 4
Paper: Preliminary Examination - Version 3

Section A: Structured Questions

1. (a) Electrons are transferred from the woolen 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, resulting in net attraction. [1]

2. (a) E.m.f. is the energy converted from non-electrical forms to electrical energy per unit charge passing through the source. [1] (Or: Work done per unit charge by the source.) (b) Total Resistance $R_T = 4.0 + 2.0 = 6.0 \, \Omega$ . [1] Current $I = V / R_T = 12 / 6.0 = 2.0 \text{ A}$ . [1] (c) $V_{lamp} = I \times R_{lamp} = 2.0 \times 4.0 = 8.0 \text{ V}$ . [1]

3. (a) $\frac{V_s}{V_p} = \frac{N_s}{N_p} \Rightarrow \frac{12}{240} = \frac{N_s}{2000}$ . [1] $N_s = \frac{12 \times 2000}{240} = 100$ turns. [1] (b) $V_p I_p = V_s I_s$ (100% efficiency). $240 \times I_p = 12 \times 2.5$ . [1] $I_p = \frac{30}{240} = 0.125 \text{ A}$ . [1] (c) Energy loss due to heating of coils (resistance) OR eddy currents in the core OR magnetic hysteresis. [1]

4. (a) Downwards. [1] (Using Fleming's Left-Hand Rule: Field N->S, Current Into Page, Force Down). (b) (i) Force increases. [1] ( $F \propto I$ ) (ii) Force decreases. [1] ( $F \propto B$ ) (c) D.C. Motor OR Loudspeaker OR Moving-coil ammeter. [1]

5. (a) Carries current from the supply to the appliance at high potential (230V). [1] (b) The fuse contains a thin wire with a low melting point. [1] If current exceeds the rating, the wire heats up and melts/blows, breaking the circuit. [1] (c) To provide a low-resistance path to earth if the live wire touches the metal casing. [1] This causes a large current to flow, blowing the fuse/tripping the breaker, preventing electric shock. [1]

6. (a) $\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{6} + \frac{1}{3} = \frac{1}{6} + \frac{2}{6} = \frac{3}{6} = \frac{1}{2}$ . [1] $R_T = 2.0 \, \Omega$ . [1] (b) $I = \frac{V}{R_T} = \frac{12}{2.0} = 6.0 \text{ A}$ . [2] (1 mark for formula/substitution, 1 mark for answer)

7. (a) The induced e.m.f. is directly proportional to the rate of change of magnetic flux linkage. [2] (Or: Magnitude of induced e.m.f. depends on rate of cutting of magnetic field lines). (b) As the coil rotates, the sides of the coil cut magnetic field lines in opposite directions during each half-turn. [1] This causes the direction of the induced current to reverse every half-rotation. [1] (c) 1. Increase speed of rotation. [1] 2. Increase strength of magnetic field OR Increase number of turns on coil. [1]

8. (a) Resistance is directly proportional to length. [1] (b) $R = \frac{V}{I} = \frac{3.0}{0.5} = 6.0 \, \Omega$ . [1] (c) $12.0 \, \Omega$ . [1] (Double length = double resistance)

9. (a) Concentric circles centered on the wire. [1] (b) Right-Hand Grip Rule. [1] (c) The direction of the magnetic field lines reverses (clockwise becomes anti-clockwise or vice versa). [1]

10. (a) Light (solar) energy to Electrical energy. [1] (b) $E = P \times t$ . $t = 5 \times 3600 = 18,000 \text{ s}$ . [1] $E = 200 \times 18,000 = 3,600,000 \text{ J}$ (or $3.6 \text{ MJ}$ ). [1] (c) Renewable / No greenhouse gas emissions during operation. [1]

Section B: Free-Response Questions

11. (a) Circuit Diagram:

Power supply symbol. [1]
Ammeter in series with lamp. [1]
Voltmeter in parallel with lamp. [1]
Variable resistor in series. [1] (Max 3 marks)

(b) (i) Graph:

Axes labeled correctly with units (V on x, A on y). [1]
Points plotted correctly. [1]
Smooth curve drawn through points (not straight line). [1]

(ii) As voltage/current increases, the temperature of the filament increases. [1] The resistance of the metal filament increases with temperature, so the ratio V/I increases (graph curves). [1]

(iii) At $V = 6.0 \text{ V}$ , $I = 1.0 \text{ A}$ . $R = \frac{V}{I} = \frac{6.0}{1.0} = 6.0 \, \Omega$ . [2]

12. (a) When a large current flows, the magnetic field generated by the electromagnet inside the breaker becomes strong enough. [1] It attracts the iron armature/switch mechanism. [1] This pulls the contacts apart, breaking the circuit and stopping the current flow. [1]

(b) Can be reset easily (no replacement needed) OR Faster response time. [1]

(ii) $10 \text{ A}$ or $13 \text{ A}$ fuse/breaker. [1] (Must be slightly higher than operating current).

(iii) If the fuse rating is too high, it will not blow when a moderate fault current flows. [1] This allows excessive current to continue flowing, which can overheat the appliance wiring and cause a fire. [1]

13. (a) Transformers rely on a changing magnetic field to induce an e.m.f. in the secondary coil. [1] Direct current produces a constant magnetic field, so there is no change in flux linkage and no induced e.m.f. [1]

(b) (i) $\frac{N_s}{N_p} = \frac{V_s}{V_p} = \frac{400,000}{25,000}$ . [1] Ratio = $16 : 1$ . [1]

(ii) High voltage reduces the current for the same power ( $P=IV$ ). [1] Lower current reduces energy loss due to heating in the transmission cables ( $P_{loss} = I^2 R$ ). [1]

(iii) $V_p I_p = V_s I_s$ . $25,000 \times 1000 = 400,000 \times I_s$ . [1] $I_s = \frac{25,000,000}{400,000} = 62.5 \text{ A}$ . [1]