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Secondary 4 Pure Physics Practice Paper 3

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TuitionGoWhere Practice Paper - Pure Physics Secondary 4

TuitionGoWhere Practice Paper (AI)
Version: 3 of 5
Subject: Pure Physics
Level: Secondary 4
Paper: Practice Paper (Electricity & Magnetism Focus)
Duration: 1 hour 15 minutes
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.
All working must be clearly shown.
You may use a scientific calculator.
Take the acceleration of free fall, $g = 10 \text{ m/s}^2$ .
Assume the density of water is $1000 \text{ kg/m}^3$ and specific heat capacity of water is $4200 \text{ J/(kg}^\circ\text{C)}$ unless stated otherwise.

Section A: Structured Questions (30 Marks)

Answer all questions in this section.

1. A student rubs a polythene rod with a wool 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, uncharged piece of paper. The paper is attracted to the rod. Explain why this attraction occurs. [2]

2. Figure 2.1 shows a simple circuit containing a battery, a switch, and two identical lamps, L1 and L2, connected in series.

(a) State the relationship between the current flowing through L1 and the current flowing through L2. [1]

(b) The battery provides an electromotive force (e.m.f.) of 12 V. The potential difference across L1 is measured to be 5 V. Calculate the potential difference across L2. [1]

Answer: _______________ V

(c) Explain the difference between electromotive force (e.m.f.) and potential difference (p.d.) in terms of energy conversion. [2]

3. A heating element in an electric kettle has a resistance of $40 \, \Omega$ . It is connected to a $240 \text{ V}$ mains supply. (a) Calculate the current flowing through the heating element. [2]

Answer: _______________ A

(b) Calculate the power dissipated by the heating element. [2]

Answer: _______________ W

4. Figure 4.1 shows a transformer used to step down the voltage for a laptop charger.

Primary coil turns ( $N_p$ ) = 2000
Secondary coil turns ( $N_s$ ) = 100
Primary voltage ( $V_p$ ) = $240 \text{ V}$

(a) Calculate the secondary voltage ( $V_s$ ), assuming the transformer is ideal. [2]

Answer: _______________ V

(b) The laptop draws a current of $2.0 \text{ A}$ from the secondary coil. Assuming the transformer is 100% efficient, calculate the current in the primary coil. [2]

Answer: _______________ A

5. A student investigates the magnetic field pattern around a straight current-carrying wire. (a) Describe the shape of the magnetic field lines around the wire. [1]

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

6. Figure 6.1 shows a simple d.c. motor.

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

(b) Explain why the coil continues to rotate in the same direction. [2]

7. A solenoid is connected to a sensitive centre-zero galvanometer. A bar magnet is pushed into the solenoid. (a) State what is observed on the galvanometer as the magnet moves into the solenoid. [1]

(b) State what is observed on the galvanometer when the magnet is held stationary inside the solenoid. [1]

8. In a household wiring system, appliances are connected in parallel rather than in series. (a) State two advantages of connecting appliances in parallel. [2]

(b) Explain the purpose of the earth wire in a three-pin plug for a metal-cased appliance. [2]

9. A fuse is rated at $5 \text{ A}$ . (a) Explain what happens to the fuse if a current of $8 \text{ A}$ flows through it. [2]

(b) Why is it dangerous to replace a $5 \text{ A}$ fuse with a $13 \text{ A}$ fuse in a circuit designed for $5 \text{ A}$ ? [2]

10. An electric heater is rated at $2 \text{ kW}$ . It is used for 3 hours. (a) Calculate the electrical energy consumed in kilowatt-hours (kWh). [2]

Answer: _______________ kWh

(b) If the cost of electricity is $\$ 0.25$ per kWh, calculate the cost of using the heater for this duration. [1]

Answer: $_______________

Section B: Free-Response Questions (20 Marks)

Answer all questions in this section.

11. A student sets up a circuit to investigate the relationship between the current ( $I$ ) and potential difference ( $V$ ) for a filament lamp and a fixed resistor. The results are plotted on the graph in Figure 11.1.

(a) Identify which line (A or B) represents the fixed resistor and which represents the filament lamp. [1] Line A: ________________________ Line B: ________________________

(b) Explain the shape of the graph for the filament lamp in terms of temperature and resistance. [3]

Answer: _______________ $\Omega$

12. Figure 12.1 shows a circuit containing a $12 \text{ V}$ battery, a switch S, a $4 \, \Omega$ resistor ( $R_1$ ), and a $6 \, \Omega$ resistor ( $R_2$ ) connected in series.

(a) Calculate the total resistance of the circuit when switch S is closed. [1]

Answer: _______________ $\Omega$

(b) Calculate the current flowing through the circuit. [2]

Answer: _______________ A

Answer: _______________ V

(d) A third resistor ( $R_3$ ) of $12 \, \Omega$ is now connected in parallel with $R_2$ . (i) State and explain the effect on the total resistance of the circuit. [2] _________________________________________________________________________ _________________________________________________________________________

   (ii) State and explain the effect on the current flowing through $R_1$. [2]
   _________________________________________________________________________
   _________________________________________________________________________

13. Electromagnetic induction is used in electricity generation. (a) Describe the structure and operation of an a.c. generator. Include in your answer: * The main components. * How an alternating e.m.f. is produced. * The energy conversion involved. [6]

(b) State two factors that affect the magnitude of the induced e.m.f. in the generator. [2]

14. High-voltage transmission lines are used to transmit electrical power over long distances. (a) Explain why electrical power is transmitted at high voltage. [3]

(b) A power station generates $500 \text{ MW}$ of power at $25 \text{ kV}$ . This voltage is stepped up to $400 \text{ kV}$ for transmission. (i) Calculate the current in the transmission lines at $400 \text{ kV}$ . [2]

   Answer: _______________ A

   (ii) If the total resistance of the transmission lines is $10 \, \Omega$, calculate the power loss in the lines. [2]
   
   
   
   Answer: _______________ W

End of Paper

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Answers

TuitionGoWhere Practice Paper - Pure Physics Secondary 4 (Answer Key)

Version: 3 of 5
Subject: Pure Physics
Level: Secondary 4

Section A: Structured Questions

1. (a) Electrons are transferred from the wool cloth to the polythene rod. [1]
The rod gains excess electrons, giving it a net negative charge. [1]

(b) The negative 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 current is the same in both lamps. [1]

(b) $V_{total} = V_1 + V_2$
$12 = 5 + V_2$
$V_2 = 7 \text{ V}$ [1]

(c) e.m.f. is the energy converted from non-electrical forms (e.g., chemical) to electrical energy per unit charge passing through the source. [1]
p.d. is the energy converted from electrical energy to other forms (e.g., heat, light) per unit charge passing through a component. [1]

3. (a) $I = V / R$
$I = 240 / 40$
$I = 6.0 \text{ A}$ [2] (1 mark for formula/substitution, 1 mark for answer)

(b) $P = V \times I$ (or $P = V^2 / R$ )
$P = 240 \times 6.0$
$P = 1440 \text{ W}$ [2] (1 mark for formula/substitution, 1 mark for answer)

4. (a) $V_s / V_p = N_s / N_p$
$V_s / 240 = 100 / 2000$
$V_s = 240 \times (1/20)$
$V_s = 12 \text{ V}$ [2]

(b) For ideal transformer: $P_{in} = P_{out}$ or $V_p I_p = V_s I_s$
$240 \times I_p = 12 \times 2.0$
$240 I_p = 24$
$I_p = 24 / 240$
$I_p = 0.1 \text{ A}$ [2]

5. (a) Concentric circles centered on the wire. [1]

(b) Right-hand grip rule. [1]

(c) 1. Increase the current flowing through the wire. [1]
2. Decrease the distance from the wire (or use a soft iron core if coiled, but for straight wire, just current/distance). [1]

6. (a) To reverse the direction of current in the coil every half rotation. [1]
This ensures the torque acts in the same direction, allowing continuous rotation. [1]

(b) The split-ring commutator swaps the contacts with the brushes every half turn. [1]
This reverses the current direction in the coil, so the force on each arm of the coil remains in the same rotational direction relative to the pivot. [1]

7. (a) The needle deflects (momentarily) in one direction. [1]

(b) The needle returns to zero (no deflection). [1]

(c) As the magnet moves, the magnetic flux linking the solenoid changes (increases). [1]
According to Faraday’s Law, a changing magnetic flux induces an e.m.f. (and current) in the coil. [1]

8. (a) 1. Each appliance receives the full mains voltage (230/240 V). [1]
2. Appliances can be switched on/off independently; if one fails, others continue to work. [1]

(b) The earth wire provides a low-resistance path to the ground. [1]
If the live wire touches the metal casing, a large current flows to earth, blowing the fuse/tripping the breaker, preventing electric shock to the user. [1]

9. (a) The fuse wire heats up due to the excessive current ( $I^2 R$ heating). [1]
The wire melts/breaks, breaking the circuit and stopping the current flow. [1]

(b) The circuit wiring is designed to safely carry up to 5 A. [1]
A 13 A fuse will not blow until the current exceeds 13 A, which may cause the wiring to overheat and start a fire before the fuse blows. [1]

10. (a) Energy ( $E$ ) = Power ( $P$ ) $\times$ Time ( $t$ )
$E = 2 \text{ kW} \times 3 \text{ h}$
$E = 6 \text{ kWh}$ [2]

(b) Cost = $6 \text{ kWh} \times \$ 0.25/\text{kWh} $Cost =$ $1.50$ [1]

Section B: Free-Response Questions

11. (a) Line A: Fixed Resistor [0.5]
Line B: Filament Lamp [0.5]
(Note: Line A is straight/linear; Line B curves) [1]

(b) As the potential difference increases, the current increases, causing the filament temperature to rise. [1]
The increased temperature causes the metal ions in the filament to vibrate more vigorously. [1]
This increases the frequency of collisions between free electrons and ions, increasing the resistance. Hence, the graph curves (gradient decreases). [1]

(c) From Line A (Resistor): At $V = 6.0 \text{ V}$ , read current $I$ .
(Assuming standard graph where $R$ is constant, e.g., if $I=0.6\text{A}$ at $6\text{V}$ )
$R = V / I$
$R = 6.0 / I$
(Example calculation: If $I=0.6\text{A}$ , $R = 10 \, \Omega$ ).
[1 mark for formula, 1 mark for correct value based on graph reading].

12. (a) $R_{total} = R_1 + R_2 = 4 + 6 = 10 \, \Omega$ [1]

(b) $I = V / R_{total} = 12 / 10 = 1.2 \text{ A}$ [2]

(d) (i) The total resistance decreases. [1]
Adding a resistor in parallel provides an additional path for current, reducing the overall opposition to flow. The combined resistance of parallel resistors is less than the smallest individual resistor. [1]

(ii) The current through $R_1$ increases. [1]
Since the total resistance of the circuit decreases, the total current drawn from the battery increases ( $I = V/R_{total}$ ). Since $R_1$ is in series with the battery, the total current flows through it. [1]

13. (a) Structure: Consists of a coil (armature) rotating in a magnetic field (between permanent magnets or electromagnets). Slip rings and carbon brushes connect the coil to the external circuit. [2]
Operation: As the coil rotates, it cuts the magnetic field lines. The magnetic flux linkage through the coil changes continuously. [1]
This induces an e.m.f. in the coil (Faraday’s Law). [1]
As the coil rotates past the vertical position, the side of the coil moving up moves down, reversing the direction of the induced current every half rotation. This produces an alternating current (a.c.). [1]
Energy Conversion: Mechanical energy (kinetic energy of rotation) is converted into electrical energy. [1]

(b) Any two of: [2]

Speed of rotation (frequency).
Strength of the magnetic field.
Number of turns on the coil.
Area of the coil.

14. (a) Power loss in transmission lines is given by $P_{loss} = I^2 R$ . [1]
To transmit a fixed power $P = VI$ , increasing the voltage $V$ reduces the current $I$ . [1]
Since power loss is proportional to the square of the current ( $I^2$ ), reducing the current significantly reduces the energy lost as heat in the cables. [1]

(b) (i) $P = 500 \text{ MW} = 500 \times 10^6 \text{ W}$
$V = 400 \text{ kV} = 400 \times 10^3 \text{ V}$
$I = P / V = (500 \times 10^6) / (400 \times 10^3)$
$I = 1250 \text{ A}$ [2]

(ii) $P_{loss} = I^2 R$
$P_{loss} = (1250)^2 \times 10$
$P_{loss} = 1,562,500 \times 10$
$P_{loss} = 15,625,000 \text{ W}$ or $15.625 \text{ MW}$ [2]