O Level Physics Electricity Magnetism Quiz

O-Level Physics Quiz - Electricity Magnetism (Answer Key)

Section A: Multiple Choice

1. B
Reasoning: Soft iron is a magnetic material that can be easily magnetized and demagnetized. Steel retains magnetism (permanent magnet), which is undesirable for a crane that needs to release scrap. Copper and Aluminium are non-magnetic.

2. A
Reasoning: This is electrostatic induction. The negative rod repels electrons in the metal sphere to the far side, leaving the near side with a net positive charge. The attraction between the rod and the near positive side is stronger than the repulsion from the far negative side due to distance.

3. B
Reasoning: In a series circuit, the current is the same at all points. Potential difference and power depend on the individual resistance values, which may differ.

4. A
Reasoning: Using the transformer equation $\frac{V_p}{V_s} = \frac{N_p}{N_s}$:
$\frac{240}{V_s} = \frac{500}{100} = 5$
$V_s = \frac{240}{5} = 48 \, \text{V}$.

5. B
Reasoning: Using the Right-Hand Grip Rule: Thumb points in the direction of current (upwards), fingers curl in the direction of the magnetic field. Looking from above, the field is counter-clockwise. Option B shows counter-clockwise arrows.

6. D
Reasoning: Energy $E = I^2 R t$.
$E = (2)^2 \times 10 \times 30 = 4 \times 10 \times 30 = 1200 \, \text{J}$.
Alternatively, $P = I^2 R = 40 \, \text{W}$, $E = P t = 40 \times 30 = 1200 \, \text{J}$.

7. C
Reasoning: Power loss in cables is $P = I^2 R$. By increasing voltage ($V$), the current ($I = P/V$) is reduced for the same power transmitted. Lower current significantly reduces heat loss ($I^2 R$).

8. A
Reasoning: For an ohmic conductor at constant temperature, $V$ is directly proportional to $I$ ($V=IR$). This produces a straight line graph passing through the origin.

9. C
Reasoning: Calculate current $I = P/V$.
Kettle: $2000/230 \approx 8.7 \, \text{A}$
Hair dryer: $1200/230 \approx 5.2 \, \text{A}$
Lamp: $60/230 \approx 0.26 \, \text{A}$
Heater: $3000/230 \approx 13 \, \text{A}$
The fuse must be rated slightly higher than the operating current. A $3 \, \text{A}$ fuse is suitable for the lamp ($0.26 \, \text{A}$). The others would blow the fuse immediately.

10. C
Reasoning: As the magnet enters the solenoid, the changing magnetic flux induces a current in one direction. As it leaves, the flux changes in the opposite manner, inducing a current in the opposite direction.

Section B: Structured Questions

11.
(a) Total Resistance:
In series, $R_{total} = R_1 + R_2$
$R_{total} = 4 + 6 = 10 \, \Omega$
[1 mark for correct answer]

(b) Current:
Using Ohm's Law $V = IR \Rightarrow I = V/R$
$I = 12 / 10 = 1.2 \, \text{A}$
[1 mark for formula/substitution, 1 mark for answer with unit]

(c) Potential Difference across $R_2$:
$V_2 = I \times R_2$
$V_2 = 1.2 \times 6 = 7.2 \, \text{V}$
[1 mark for substitution, 1 mark for answer]

12.
(a) Rule:
Fleming’s Left-Hand Rule.
[1 mark]

(b) Continuous Rotation:

1. Current flows through the coil, creating forces on arms AB and CD in opposite directions (due to opposite current directions relative to the field), causing rotation.
2. When the coil passes the vertical position, the split-ring commutator reverses the direction of the current in the coil.
3. This reverses the direction of the forces, ensuring the torque continues to act in the same rotational direction, allowing continuous rotation.
   [1 mark for forces causing rotation, 1 mark for commutator reversing current, 1 mark for maintaining same direction of torque]

(c) Increase Speed:
Any two of:

1. Increase the current.
2. Increase the strength of the magnetic field (stronger magnets).
3. Increase the number of turns on the coil.
   [1 mark each, max 2]

13.
(a) Number of Turns:
$\frac{V_p}{V_s} = \frac{N_p}{N_s}$
$\frac{240}{12} = \frac{1000}{N_s}$
$20 = \frac{1000}{N_s} \Rightarrow N_s = \frac{1000}{20} = 50$ turns.
[1 mark for formula/substitution, 1 mark for answer]

(b) Primary Current:
Power in secondary $P_s = 24 \, \text{W}$.
Assuming 100% efficiency, $P_p = P_s = 24 \, \text{W}$.
$P_p = V_p I_p \Rightarrow 24 = 240 \times I_p$
$I_p = \frac{24}{240} = 0.1 \, \text{A}$.
[1 mark for efficiency concept, 1 mark for substitution, 1 mark for answer]

(c) Laminated Core:

1. To reduce eddy currents induced in the core.
2. Eddy currents cause energy loss through heating. Lamination increases resistance to these currents, reducing heat loss and improving efficiency.
   [1 mark for eddy currents, 1 mark for reducing heat/energy loss]

14.
(a) Resistance at $60^\circ\text{C}$:
From table, at $60^\circ\text{C}$, $V = 4.0 \, \text{V}$ and $I = 0.10 \, \text{A}$.
$R = V / I = 4.0 / 0.10 = 40 \, \Omega$.
[1 mark for substitution, 1 mark for answer]

(b) Relationship:
As temperature increases, resistance decreases. (Negative Temperature Coefficient).
[1 mark]

(c) Explanation:

1. As temperature increases, the number of charge carriers (free electrons) in the thermistor material increases significantly.
2. Although lattice vibrations increase (which tends to increase resistance), the increase in charge carriers dominates, leading to a decrease in overall resistance.
   [1 mark for increase in charge carriers, 1 mark for linking to decreased resistance]

15.
(a) Induction:

1. Sphere B is neutral but contains equal positive and negative charges.
2. When positive Sphere A approaches, it attracts electrons in Sphere B to the side nearest A, and repels positive charges to the far side.
3. The attractive force between A and the near negative side is stronger than the repulsive force from the far positive side (due to distance), resulting in net attraction.
   [1 mark for charge separation, 1 mark for correct distribution, 1 mark for net attraction explanation]

(b) Contact:

1. Upon touching, electrons flow from B to A (or charge shares).
2. Both spheres end up with the same type of charge (positive), as the total positive charge is distributed between them. They will now repel each other.
   [1 mark for same charge type, 1 mark for repulsion/sharing]

16.
(a) Color Code:
Brown.
[1 mark]

(b) Earth Wire Function:

1. The earth wire is connected to the metal casing of the appliance.
2. If a fault occurs (e.g., live wire touches the casing), the current flows through the earth wire (low resistance path) to the ground.
3. This large current blows the fuse/circuit breaker, disconnecting the live supply and preventing the user from receiving an electric shock.
   [1 mark for connection to casing, 1 mark for low resistance path/fuse blowing, 1 mark for safety/shock prevention]

(c) Switch in Live Wire:

1. Placing the switch in the live wire ensures that when the switch is off, the appliance is disconnected from the high voltage source.
2. If the switch were in the neutral wire, the appliance would still be at live potential even when off, posing a shock hazard if touched.
   [1 mark for disconnection from high voltage, 1 mark for safety/hazard explanation]

17.
(a) Total Resistance:
For parallel resistors: $\frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$
$\frac{1}{R_{total}} = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}$
$R_{total} = 2 \, \Omega$.
[1 mark for formula/setup, 1 mark for answer]

(b) Total Current:
$I = V / R_{total}$
$I = 12 / 2 = 6 \, \text{A}$.
[1 mark for substitution, 1 mark for answer]

(c) Brightness Change:

1. The brightness remains the same.
2. In a parallel circuit, each branch receives the full voltage of the source ($12 \, \text{V}$). Removing one branch does not change the voltage across or current through the other branches.
   [1 mark for 'same', 1 mark for explanation regarding voltage independence]

18.
(a) Principle:
An e.m.f. (or current) is induced in a conductor when it cuts magnetic field lines (or when there is a change in magnetic flux linkage).
[1 mark]

(b) Alternating Current:

1. As the coil rotates, side AB moves up then down relative to the magnetic field.
2. This reverses the direction of the induced current every half rotation, producing an alternating current.
   [1 mark for direction change, 1 mark for link to rotation]

(c) Increase Voltage:
Any two of:

1. Rotate the coil faster.
2. Use stronger magnets.
3. Increase the number of turns on the coil.
   [1 mark each, max 2]

19.
(a) Frequency:
The number of complete waves (or oscillations) passing a point per second.
[1 mark]

(b) Calculation:
Time period $T = 4 \, \text{cm} \times 2 \, \text{ms/cm} = 8 \, \text{ms} = 0.008 \, \text{s}$.
Frequency $f = 1 / T$
$f = 1 / 0.008 = 125 \, \text{Hz}$.
[1 mark for T calculation, 1 mark for formula, 1 mark for answer]

(c) Trace Change:

1. The amplitude (height) of the wave increases.
2. The wavelength (horizontal length) remains the same (since pitch/frequency is unchanged).
   [1 mark for amplitude increase, 1 mark for wavelength same]

20.
(a) Sequence:

1. Current flows in the low-voltage circuit, magnetizing the soft iron core.
2. The electromagnet attracts the soft iron armature.
3. The armature pivots, closing the contacts in the high-voltage circuit.
   [1 mark for each step]

(b) Advantage:
Allows a low-voltage/safe circuit to control a high-voltage/dangerous circuit. (Or: Allows switching from a distance).
[1 mark]

(c) Soft Iron:

1. Soft iron is easily magnetized and demagnetized.
2. This ensures the relay switches off quickly when the control current is removed (does not retain magnetism).
   [1 mark for easy magnetization/demagnetization, 1 mark for switching off capability]