O Level Physics Practice Paper 1

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]