O Level Physics Practice Paper 2

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

Topic: Electricity and Magnetism
Version: 2 of 5

Section A: Multiple Choice & Short Structured Questions

1. B
Conventional current is defined as flow of positive charge (positive to negative). Electrons are negative and flow opposite to conventional current (negative to positive). [1]

2. C
Polythene gains electrons from wool. Electrons are negatively charged. Transfer of electrons causes the charge. [1]

3. A
Field lines go from Positive to Negative. Density of lines indicates strength. More lines near P implies larger magnitude. [1]

4. D
$R = \rho L / A$. New $R' = \rho (2L) / (A/2) = 4 (\rho L / A) = 4R$. [1]

5. B
As voltage increases, current increases, heating the filament. Higher temperature increases lattice vibrations, increasing resistance. Gradient of I-V is $1/R$, so decreasing gradient means increasing R. [1]

6. 8.0 V
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: $V_{R2} = \frac{8}{4+8} \times 12 = 8 \text{ V}$. [2]

7. The work done by the source in driving a unit charge around a complete circuit. (Or: Energy converted from non-electrical to electrical form per unit charge). [1]

8. 2.5 W
$P = I^2 R = (0.5)^2 \times 10 = 0.25 \times 10 = 2.5 \text{ W}$. [2]

9. Feature: Earth wire (or Fuse).
Function: Earth wire provides a low-resistance path to ground if live wire touches casing, preventing electric shock. Fuse melts if current exceeds rating, breaking circuit to prevent fire.
(Accept either valid pair) [2]

10. To reverse the direction of current in the coil every half rotation, ensuring the torque acts in the same direction and the coil rotates continuously. [1]

Section B: Structured Questions

11. (a)

• Voltage across fixed resistor $V_R = 2.0 \text{ V}$.
• Current in circuit $I = V_R / R = 2.0 / 500 = 0.004 \text{ A}$.
• Voltage across thermistor $V_T = 6.0 - 2.0 = 4.0 \text{ V}$.
• Resistance of thermistor $R_T = V_T / I = 4.0 / 0.004 = 1000 \, \Omega$.
  • [1] for current calculation.
  • [1] for voltage across thermistor.
  • [1] for final resistance.
  • Answer: $1000 \, \Omega$ [3]

(b)

• Statement: The voltmeter reading increases. [1]
• Explanation: As temperature increases, the resistance of the NTC thermistor decreases. This decreases the total resistance of the circuit, increasing the current. Since $V = IR$ for the fixed resistor, and $I$ increases, $V$ across the fixed resistor increases. (Alternatively: Voltage divider principle; as $R_T$ decreases, it takes a smaller share of the voltage, so the fixed resistor takes a larger share). [1] [2]

12. (a) (i) L1 and L2 light up; L3 is off. L1 and L2 are in series. [1] (ii) L1, L2, and L3 light up. L2 and L3 are in parallel, and this combination is in series with L1. [1]

(b)

• When S2 closes, L3 is added in parallel to L2.
• The effective resistance of the parallel section decreases.
• Therefore, the total resistance of the whole circuit decreases.
• Since $I = V/R_{total}$, the total current from the battery increases.
• Since L1 is in the main branch, the current through L1 increases, so it becomes brighter.
  • [1] for identifying resistance decrease.
  • [1] for identifying current increase.
  • [1] for linking current to brightness. [3]

13. (a)

• Formula: $\frac{V_p}{V_s} = \frac{N_p}{N_s}$
• Substitution: $\frac{240}{12} = \frac{2000}{N_s}$
• Calculation: $20 = \frac{2000}{N_s} \Rightarrow N_s = \frac{2000}{20} = 100$.
  • Answer: 100 turns [2]

(b)

• Formula: $V_p I_p = V_s I_s$ (100% efficiency)
• Substitution: $240 \times I_p = 12 \times 2.0$
• Calculation: $240 I_p = 24 \Rightarrow I_p = \frac{24}{240} = 0.1 \text{ A}$.
  • Answer: 0.1 A [2]

(c)

• Transformers rely on electromagnetic induction.
• Induction requires a changing magnetic field (or changing magnetic flux linkage).
• D.C. produces a constant magnetic field, so there is no change in flux linkage, and no e.m.f. is induced in the secondary coil. [2]

14. (a) Fleming’s Left-Hand Rule. [1]

(b) Any two of:

1. Increase the current.
2. Increase the strength of the magnetic field (use stronger magnets).
3. Increase the length of the wire in the field. [2]

(c) The direction of the force is reversed. [1]

Section C: Free Response Questions

15. (a)

• $P = VI \Rightarrow I = P/V$
• $I = 2400 \text{ W} / 240 \text{ V} = 10 \text{ A}$.
  • Answer: 10 A [2]

(b)

• $R = V/I$ or $P = V^2/R$
• $R = 240 / 10 = 24 \, \Omega$.
  • Answer: 24 $\Omega$ [2]

(c)

• Time $t = 5 \text{ min} = 5/60 \text{ h} = 1/12 \text{ h}$.
• Power $P = 2.4 \text{ kW}$.
• Energy $E = P \times t = 2.4 \times (1/12) = 0.2 \text{ kWh}$.
  • Answer: 0.2 kWh [2]

(d)

• The normal operating current is 10 A.
• A 3 A or 5 A fuse would melt immediately during normal operation.
• A 13 A fuse allows the 10 A current to flow but will melt if the current becomes dangerously high (e.g., due to a fault), providing safety. [2]

16. (a)

• As the coil rotates, the sides of the coil cut through the magnetic field lines.
• This causes a change in magnetic flux linkage through the coil.
• According to Faraday’s Law, this changing flux induces an e.m.f. (and current if the circuit is closed) in the coil. [3]

(b) (i) 4 V [1] (ii)

• Period $T = 0.02 \text{ s}$.
• Frequency $f = 1/T = 1/0.02 = 50 \text{ Hz}$.
  • Answer: 50 Hz [2]

(c) (i) Peak voltage doubles (increases). [1] (ii) Frequency doubles (increases). [1]

17. (a)

• Diagram must show:
  • Battery symbol.
  • Three resistors in parallel branches.
  • Ammeter in series with the battery (main line) to measure total current.
  • Voltmeter in parallel across the battery (or across the parallel combination).
  • Correct symbols.
  • [1] for parallel arrangement.
  • [1] for correct ammeter position.
  • [1] for correct voltmeter position. [3]

(b)

• Formula: $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$
• $\frac{1}{R_{eq}} = \frac{1}{10} + \frac{1}{10} + \frac{1}{10} = \frac{3}{10}$
• $R_{eq} = \frac{10}{3} = 3.33 \, \Omega$.
  • Answer: $3.33 \, \Omega$ (or $3.3 \, \Omega$) [2]

(c)

• Greater than. [1]
• Explanation: In series, resistances add up ($R_{total} = R_1 + R_2 + R_3 = 30 \, \Omega$). In parallel, the total resistance is less than the smallest individual resistor. Adding resistors in series increases the total length of the resistive path, increasing resistance. [1] [2]

18. (a)

• Steel is a hard magnetic material (retains magnetism).
• Correction/Refinement: Actually, for an electromagnet that needs to switch on/off, soft iron is preferred. Steel is a permanent magnet material.
• Re-reading the question: "Explain why steel is used... rather than iron." This is a trick question or implies a specific context where retention is needed? No, standard scrap yard electromagnets use soft iron cores so they lose magnetism when current is off to drop the load.
• Wait, looking at standard O-Level questions: Usually, they ask why soft iron is used. If the question asks why steel is used, it might be a "false premise" check or referring to the scrap being steel?
• Let's assume the question meant: "Why is soft iron used for the core?" OR "Why is the core not made of steel?"
• Standard Answer for Electromagnet Core: Soft iron is used because it is easily magnetized and demagnetized. Steel retains magnetism (becomes a permanent magnet), which would prevent the scrap from dropping when the current is switched off.
• If the prompt strictly asks "Why is steel used... rather than iron", it is likely a typo in the generated question for "Why is soft iron used... rather than steel". I will answer based on the correct physics principle for electromagnets.
• Answer: Actually, steel is not typically used for the core of a lifting electromagnet because it retains magnetism. Soft iron is used. If the question implies the scrap is steel, that's different.
• Let's correct the interpretation: The question likely intends to test the difference between soft magnetic materials (iron) and hard magnetic materials (steel).
• Revised Answer for (a): The question contains a common misconception check. Steel is not suitable for the core of an electromagnet intended for lifting and dropping because it is a hard magnetic material and retains magnetism (becomes a permanent magnet) after the current is switched off. This would make it difficult to release the scrap. Soft iron is preferred because it loses its magnetism quickly.
  • Note to marker: If student says "Steel is used because it is strong mechanically," that is weak. The magnetic property is key. If the question implies the core is steel, the student should identify this is poor design or explain why iron is better.
  • Let's provide the standard "Why Iron is better" answer as the intended learning point.
  • Answer: Steel is a hard magnetic material and retains magnetism. This is undesirable for a lifting electromagnet as the load would not drop when switched off. Soft iron is preferred as it loses magnetism easily. [2]

(b)

1. Increase the number of turns on the coil.
2. Increase the current flowing through the coil. [2]

(c)

• The magnetic field can be switched on and off by switching the current.
• This allows the scrap to be picked up (current on) and dropped (current off) easily.
• A permanent magnet cannot be switched off. [2]

19. (a)

• Power loss in cables is due to heating: $P_{loss} = I^2 R$.
• To transmit a fixed power $P = VI$, increasing voltage $V$ reduces the current $I$.
• Lower current significantly reduces power loss ($I^2$ dependence) and allows for thinner/cheaper cables. [2]

(b)

• $P = V_p I_p$
• $500 \times 10^6 = 25 \times 10^3 \times I_p$
• $I_p = \frac{500 \times 10^6}{25 \times 10^3} = \frac{500,000}{25} = 20,000 \text{ A}$.
  • Answer: 20,000 A (or 20 kA) [2]

(c)

• $V_p I_p = V_s I_s$
• $500 \text{ MW} = 400 \text{ kV} \times I_s$
• $500 \times 10^6 = 400 \times 10^3 \times I_s$
• $I_s = \frac{500,000}{400} = 1,250 \text{ A}$.
  • Answer: 1,250 A [2]

20. (a) Circular path (or arc of a circle). [1]

(b)

• The magnetic force acts perpendicular to the direction of motion (velocity).
• Work done = Force $\times$ distance $\times \cos(90^\circ) = 0$.
• Since no work is done, the kinetic energy (and thus speed) remains constant. The force only changes the direction, not the magnitude of velocity. [2]

(c) Left (using Fleming's Left Hand Rule: Field into page, Current/Motion upwards $\rightarrow$ Force to the Left). [1]