2024 Cambridge IGCSE Physics (0625) 模擬試題連答案詳解

1. **Identify the phases of motion:**
- **Phase 1: Constant acceleration**
Initial speed u = 0 m/s, acceleration a₁ = 2.0 m/s², time t₁ = 6.0 s.
Final speed v = u + a₁ t₁ = 0 + 2.0 × 6.0 = 12 m/s.
Distance travelled d₁ = 0.5 × (u + v) × t₁ = 0.5 × 12 × 6.0 = 36 m.

- **Phase 2: Constant speed**
Speed v = 12 m/s, time t₂ = 12 s.
Distance travelled d₂ = v × t₂ = 12 × 12 = 144 m.

- **Phase 3: Constant deceleration**
Initial speed v = 12 m/s, final speed = 0 m/s, deceleration a₃ = 3.0 m/s².
Time to stop t₃ = v / a₃ = 12 / 3.0 = 4.0 s.
Distance travelled d₃ = 0.5 × v × t₃ = 0.5 × 12 × 4.0 = 24 m.

2. **Calculate total distance:**
Total distance = d₁ + d₂ + d₃ = 36 + 144 + 24 = 204 m.

1 mark for correct answer B.
- Award 1 mark for correct calculation of total distance of 204 m.
- Reject incorrect options resulting from failing to divide by 2 for accelerating/decelerating phases.

1. Let k be the spring constant of a single spring. Under load W, the extension is x = W/k.
2. **Parallel section:** Two springs in parallel have an effective spring constant of 2k. The load shared by this combination is 2W. The extension of this parallel section is: x_parallel = 2W / (2k) = W/k = x.
3. **Series section:** The single spring in series supports the full load of 2W. Its extension is: x_series = 2W / k = 2x.
4. **Total extension:**
x_total = x_parallel + x_series = x + 2x = 3x.

1 mark for correct answer C.
- Award 1 mark for calculating the total extension as 3x.
- Reject other options where parallel spring combinations or series extensions are miscalculated.

1. **Calculate total electrical energy input (E_in):**
E_in = Power × time = 1200 W × (5.0 × 60 s) = 1200 × 300 = 360,000 J.

2. **Calculate useful potential energy output (E_out):**
E_out = mgh = 1800 kg × 9.8 m/s² × 8.0 m = 141,120 J.

3. **Calculate efficiency:**
Efficiency = (E_out / E_in) × 100% = (141,120 / 360,000) × 100% ≈ 39.2%.

This rounds to 39%.

1 mark for correct answer C.
- Award 1 mark for showing that the efficiency is approximately 39%.
- Reject options where minutes are not converted to seconds, or kilowatt is not converted to watt.

1. **Total energy supplied by the heater (E):**
E = Power × time = 48 W × (5.0 × 60 s) = 14,400 J.

2. **Thermal energy absorbed by the block (Q):**
Since 15% is lost, 85% is absorbed.
Q = 0.85 × 14,400 = 12,240 J.

3. **Specific heat capacity (c):**
Q = m c ΔT
12,240 = 1.5 × c × (36 - 20)
12,240 = 1.5 × c × 16 = 24 c
c = 12,240 / 24 = 510 J / (kg °C).

1 mark for correct answer B.
- Award 1 mark for the correct calculation of 510 J / (kg °C) taking the 15% heat loss into account.
- Reject 600 J / (kg °C) which assumes no thermal loss.

1. **Calculate the critical angle (c) of the glass-air boundary:**
sin(c) = 1 / n = 1 / 1.50 ≈ 0.667
c = sin⁻¹(0.667) ≈ 41.8°.

2. **Compare the angle of incidence (i = 45°) with the critical angle:**
Since i > c (45° > 41.8°), total internal reflection occurs.

3. **Determine the angle of reflection:**
By the law of reflection, the angle of reflection equals the angle of incidence, which is 45°.

1 mark for correct answer D.
- Award 1 mark for determining that total internal reflection occurs because the angle of incidence exceeds the critical angle, resulting in a reflection angle of 45°.

1. **Find the potential difference across the fixed resistor (V_resistor):**
V_resistor = V_supply - V_thermistor = 9.0 V - 3.0 V = 6.0 V.

2. **Find the circuit current (I):**
I = V_resistor / R_resistor = 6.0 V / 120 Ω = 0.050 A.

3. **Calculate the resistance of the thermistor (R_thermistor):**
R_thermistor = V_thermistor / I = 3.0 V / 0.050 A = 60 Ω.

Alternatively, using the potential divider ratio:
R_thermistor / R_resistor = V_thermistor / V_resistor => R_thermistor / 120 = 3.0 / 6.0 = 0.5 => R_thermistor = 60 Ω.

1 mark for correct answer B.
- Award 1 mark for correctly determining the thermistor resistance as 60 Ω.
- Reject 40 Ω, which is a common mistake resulting from dividing by total supply voltage instead of fixed resistor voltage.

1. **Calculate the secondary voltage (V_s):**
V_s / V_p = N_s / N_p => V_s = 240 V × (150 / 600) = 60 V.

2. **Calculate the secondary current (I_s):**
I_s = V_s / R = 60 V / 15 Ω = 4.0 A.

3. **Calculate the primary current (I_p) using power conservation for an ideal transformer (P_p = P_s):**
V_p I_p = V_s I_s => I_p = I_s × (V_s / V_p) = 4.0 A × (60 V / 240 V) = 1.0 A.

1 mark for correct answer B.
- Award 1 mark for calculating the primary current as 1.0 A.
- Reject 4.0 A (secondary current) or 16 A (incorrect step-up current scaling).

1. **Redshift definition:** Redshift is the observed increase in the wavelength of electromagnetic radiation emitted by stars or galaxies that are moving away from the observer.
2. Since the galaxy is receding (moving away from Earth), the light waves are stretched out, meaning the detected wavelength is longer than the emitted wavelength.
3. Therefore, the detected wavelength is longer and the galaxy is moving away from Earth.

1 mark for correct answer D.
- Award 1 mark for selecting the option stating that wavelength increases (longer) and motion is away from Earth.

The total distance travelled is equal to the area under the velocity-time graph. The area can be split into three sections: 1) A trapezium from \(t = 0\) to \(t = 4.0\text{ s}\): \(\text{Area}_1 = \frac{2.0 + 10.0}{2} \times 4.0 = 24.0\text{ m}\). 2) A rectangle from \(t = 4.0\text{ s}\) to \(t = 7.0\text{ s}\): \(\text{Area}_2 = 10.0 \times (7.0 - 4.0) = 30.0\text{ m}\). 3) A triangle from \(t = 7.0\text{ s}\) to \(t = 9.0\text{ s}\): \(\text{Area}_3 = \frac{1}{2} \times 10.0 \times (9.0 - 7.0) = 10.0\text{ m}\). The total distance is the sum of these areas: \(24.0 + 30.0 + 10.0 = 64.0\text{ m}\).

1 mark for the correct calculation of the total distance (64 m).

First, find the extension of the spring under the \(4.0\text{ N}\) load: \(15.0\text{ cm} - 12.0\text{ cm} = 3.0\text{ cm}\). Since Hooke's law is obeyed, extension is directly proportional to load. The extension \(x\) for a \(10.0\text{ N}\) load is: \(x = 3.0\text{ cm} \times \frac{10.0\text{ N}}{4.0\text{ N}} = 7.5\text{ cm}\). The new length is the original unstretched length plus the extension: \(12.0\text{ cm} + 7.5\text{ cm} = 19.5\text{ cm}\).

1 mark for the correct length calculation (19.5 cm).

Useful work done in lifting the load: \(W = F \times d = 120\text{ N} \times 5.0\text{ m} = 600\text{ J}\). Useful power output of the motor: \(P_{\text{out}} = \frac{W}{t} = \frac{600\text{ J}}{4.0\text{ s}} = 150\text{ W}\). Efficiency of the motor: \(\text{Efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}} \times 100\% = \frac{150\text{ W}}{200\text{ W}} \times 100\% = 75\%\).

1 mark for the correct calculation of efficiency (75%).

The electrical energy supplied to the block is: \(E = P \times t = 40\text{ W} \times (3.0 \times 60\text{ s}) = 7200\text{ J}\). The temperature rise of the metal is: \(\Delta \theta = 38^\circ\text{C} - 20^\circ\text{C} = 18^\circ\text{C}\). Using the formula \(E = m c \Delta \theta\), the specific heat capacity \(c\) is: \(c = \frac{E}{m \Delta \theta} = \frac{7200\text{ J}}{0.50\text{ kg} \times 18^\circ\text{C}} = 800\text{ J / (kg }^\circ\text{C)}\).

1 mark for the correct specific heat capacity (800 J / (kg °C)).

The relationship between the refractive index \(n\) and the critical angle \(c\) is given by: \(\sin(c) = \frac{1}{n}\). Substituting \(n = 1.50\): \(\sin(c) = \frac{1}{1.50} \approx 0.667\). Therefore, \(c = \sin^{-1}(0.667) \approx 41.8^\circ\), which rounds to \(42^\circ\).

1 mark for the correct critical angle (42°).

First, calculate the combined resistance of the parallel combination: \(R_p = \frac{R_1 R_2}{R_1 + R_2} = \frac{3.0 \times 6.0}{3.0 + 6.0} = \frac{18.0}{9.0} = 2.0\ \Omega\). Next, calculate the total resistance of the series circuit: \(R_{\text{total}} = R_p + R_3 = 2.0\ \Omega + 4.0\ \Omega = 6.0\ \Omega\). The total current flowing from the power supply is: \(I = \frac{V}{R_{\text{total}}} = \frac{12\text{ V}}{6.0\ \Omega} = 2.0\text{ A}\). Since the \(4.0\ \Omega\) resistor is in series with the parallel pair, the entire total current of \(2.0\text{ A}\) flows through it.

1 mark for the correct current value (2.0 A).

The activity decreases from \(800\text{ counts/s}\) to \(100\text{ counts/s}\). The fraction of remaining activity is: \(\frac{100}{800} = \frac{1}{8}\). Since \(\frac{1}{8} = \left(\frac{1}{2}\right)^3\), exactly three half-lives (\(3 \times T_{1/2}\)) have passed. Therefore: \(3 \times T_{1/2} = 12\text{ days}\), which gives \(T_{1/2} = 4\text{ days}\).

1 mark for the correct calculation of half-life (4 days).

Redshift shows that the observed wavelength of light has shifted to longer wavelengths, meaning the galaxies are moving away from Earth. According to Hubble's law, the recession speed of a galaxy is directly proportional to its distance from Earth.

1 mark for identifying the direction of motion and the direct proportionality to distance.

The total distance is the area under the speed-time graph. Area 1 (acceleration phase): \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5.0\text{ s} \times 20\text{ m/s} = 50\text{ m}\). Area 2 (constant speed phase): \(\text{base} \times \text{height} = 10\text{ s} \times 20\text{ m/s} = 200\text{ m}\). Area 3 (deceleration phase): \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4.0\text{ s} \times 20\text{ m/s} = 40\text{ m}\). Total distance = \(50 + 200 + 40 = 290\text{ m}\).

1 mark for calculating the total area under the graph correctly to obtain 290 m.

For a single spring, a \(4.0\text{ N}\) load causes an extension of \(15.0\text{ cm} - 12.0\text{ cm} = 3.0\text{ cm}\). This gives a spring constant of \(k = \frac{4.0\text{ N}}{3.0\text{ cm}} = \frac{4}{3}\text{ N/cm}\). When two identical springs are connected in parallel, the effective spring constant is doubled: \(k_{\text{eff}} = 2k = \frac{8}{3}\text{ N/cm}\). For a load of \(12.0\text{ N}\), the extension is \(e = \frac{F}{k_{\text{eff}}} = \frac{12.0}{\frac{8}{3}} = 4.5\text{ cm}\). The new length is \(12.0\text{ cm} + 4.5\text{ cm} = 16.5\text{ cm}\).

1 mark for the correct calculation of parallel extension and new length of 16.5 cm.

The useful work done is equal to the increase in gravitational potential energy: \(E_p = mgh = 80 \times 9.8 \times 15 = 11760\text{ J}\). The useful power output is \(P_{\text{out}} = \frac{11760\text{ J}}{6.0\text{ s}} = 1960\text{ W}\). The power input is \(2.8\text{ kW} = 2800\text{ W}\). The efficiency is \(\frac{P_{\text{out}}}{P_{\text{in}}} \times 100\% = \frac{1960}{2800} \times 100\% = 70\%\).

1 mark for calculating useful power output and dividing by input power to get 70%.

Thermal energy gained by the water = \(m_w \times c_w \times \Delta T_w = 0.80\text{ kg} \times 4200\text{ J/(kg}^\circ\text{C)} \times (25 - 20)^\circ\text{C} = 16800\text{ J}\). Assuming no heat loss, thermal energy lost by the metal block = \(16800\text{ J}\). Therefore, \(m_m \times c_m \times \Delta T_m = 16800\text{ J}\), which gives \(0.60\text{ kg} \times c_m \times (95 - 25)^\circ\text{C} = 16800\text{ J}\). This simplifies to \(42 \times c_m = 16800\), so \(c_m = 400\text{ J/(kg}^\circ\text{C)}\).

1 mark for equating heat lost to heat gained and calculating specific heat capacity of metal as 400 J/(kg°C).

The relationship between refractive index \(n\) and critical angle \(c\) is given by \(\sin(c) = \frac{1}{n}\). Here, \(\sin(c) = \frac{1}{1.50} = 0.667\), which gives \(c \approx 41.8^\circ\). Total internal reflection occurs when light travels from a more dense to a less dense medium and the angle of incidence is greater than the critical angle. Since the angle of incidence is \(45^\circ\), which is greater than \(41.8^\circ\), total internal reflection occurs.

1 mark for correctly calculating the critical angle as 41.8° and identifying that TIR occurs.

First, calculate the combined resistance of the parallel pair: \(R_p = \frac{R_1 R_2}{R_1 + R_2} = \frac{6.0 \times 12}{6.0 + 12} = \frac{72}{18} = 4.0\ \Omega\). The total resistance of the series circuit is \(R_{\text{total}} = R_{\text{thermistor}} + R_p = 8.0 + 4.0 = 12.0\ \Omega\). The current in the circuit is \(I = \frac{V}{R_{\text{total}}} = \frac{12\text{ V}}{12.0\ \Omega} = 1.0\text{ A}\). The potential difference across the parallel combination is \(V_p = I \times R_p = 1.0\text{ A} \times 4.0\ \Omega = 4.0\text{ V}\).

1 mark for calculating the parallel resistance and using the potential divider method to find 4.0 V.

Subtract the background count rate to find the source's actual count rate. Initial corrected count rate = \(344 - 24 = 320\text{ counts/minute}\). Corrected count rate after \(6.0\text{ hours}\) = \(64 - 24 = 40\text{ counts/minute}\). The fraction of activity remaining is \(\frac{40}{320} = \frac{1}{8}\). Since \(\frac{1}{8} = \left(\frac{1}{2}\right)^3\), exactly 3 half-lives have elapsed in \(6.0\text{ hours}\). Therefore, the half-life is \(\frac{6.0\text{ hours}}{3} = 2.0\text{ hours}\).

1 mark for subtracting background count rates to find corrected counts, determining that 3 half-lives passed, and finding half-life is 2.0 hours.

Hubble's Law states that the recessional velocity \(v\) of a galaxy is directly proportional to its distance \(d\) from Earth, described by the equation \(v = H_0 d\), where \(H_0\) is the Hubble constant.

1 mark for identifying the directly proportional relationship between recession speed and distance.

Step 1: Find the maximum speed reached. Since the trolley starts from rest and accelerates at \(2.0\text{ m/s}^2\) for \(3.0\text{ s}\), the maximum speed is \(v = a \times t = 2.0 \times 3.0 = 6.0\text{ m/s}\). Step 2: Calculate the distance traveled in each stage. First stage (accelerating): \(\text{distance}_1 = \text{average speed} \times \text{time} = 3.0\text{ m/s} \times 3.0\text{ s} = 9.0\text{ m}\). Second stage (constant speed): \(\text{distance}_2 = 6.0\text{ m/s} \times 4.0\text{ s} = 24.0\text{ m}\). Third stage (decelerating): \(\text{distance}_3 = \text{average speed} \times \text{time} = 3.0\text{ m/s} \times 2.0\text{ s} = 6.0\text{ m}\). Step 3: Calculate total distance and total time. \(\text{Total distance} = 9.0 + 24.0 + 6.0 = 39.0\text{ m}\). \(\text{Total time} = 3.0 + 4.0 + 2.0 = 9.0\text{ s}\). Step 4: Calculate the average speed. \(\text{Average speed} = \frac{39.0\text{ m}}{9.0\text{ s}} \approx 4.3\text{ m/s}\).

Award 1 mark for the correct calculation of total distance (39.0 m) and total time (9.0 s) leading to the correct average speed of 4.3 m/s.

Since the plank is pivoted at its center (which is \(1.0\text{ m}\) from each end), the \(80\text{ N}\) weight acts at a distance of \(1.0\text{ m}\) to the left of the pivot, creating an anticlockwise moment of \(80\text{ N} \times 1.0\text{ m} = 80\text{ N m}\). The \(30\text{ N}\) weight acts at a distance of \(1.0\text{ m}\) to the right of the pivot, creating a clockwise moment of \(30\text{ N} \times 1.0\text{ m} = 30\text{ N m}\). For rotational equilibrium, the sum of clockwise moments must equal the sum of anticlockwise moments. Because the anticlockwise moment is greater, the plank's own weight (\(100\text{ N}\)) must act to the right of the pivot at a distance \(d\) to create an additional clockwise moment. Setting up the moment equation: \(80\text{ N m} = 30\text{ N m} + (100\text{ N} \times d)\). Solving for \(d\): \(50 = 100d \implies d = 0.50\text{ m}\).

Award 1 mark for calculating the correct distance of 0.50 m using the principle of moments.

Step 1: Calculate the useful work output (increase in gravitational potential energy). \(\text{GPE} = mgh = 25\text{ kg} \times 9.8\text{ N/kg} \times 12\text{ m} = 2940\text{ J}\). Step 2: Calculate the total electrical energy input. \(\text{Energy} = V \times I \times t = 24\text{ V} \times 25\text{ A} \times 6.0\text{ s} = 3600\text{ J}\). Step 3: Calculate the efficiency. \(\text{Efficiency} = \frac{\text{Useful energy output}}{\text{Total energy input}} \times 100\% = \frac{2940}{3600} \times 100\% \approx 81.7\%\), which rounds to \(82\%\).

Award 1 mark for the correct calculation of output work (2940 J) and input energy (3600 J) leading to 82% efficiency.

Step 1: Calculate total thermal energy supplied by the heater: \(Q_{\text{supplied}} = P \times t = 40\text{ W} \times (5.0 \times 60\text{ s}) = 12\,000\text{ J}\). Step 2: Calculate temperature rise: \(\Delta T = 68^\circ\text{C} - 20^\circ\text{C} = 48^\circ\text{C}\). Step 3: Use the formula \(Q = mc\Delta T\) to calculate \(c\): \(c = \frac{12\,000}{0.50 \times 48} = 500\text{ J/(kg }^\circ\text{C)}\thinspace\). Step 4: Because thermal energy was lost to the surroundings, the actual energy absorbed by the block to raise its temperature was less than \(12\,000\text{ J}\). Therefore, the true specific heat capacity is lower than \(500\text{ J/(kg }^\circ\text{C)}\), meaning the calculated value is higher than the true value.

Award 1 mark for calculating the specific heat capacity as 500 J/(kg °C) and correctly identifying that thermal energy loss causes the calculated value to be higher than the true value.

Diffraction increases when the wavelength of the wave is comparable to or larger than the gap width. Statement 1: Decreasing the frequency increases the wavelength (\(\lambda = v/f\)), which increases the wavelength relative to the gap width, thus increasing diffraction. Statement 2: Decreasing the gap width directly increases diffraction because the gap becomes closer in size to the wavelength. Statement 3: Changing the amplitude of the wave does not affect the wavelength or gap width, and therefore has no effect on diffraction. Hence, only 1 and 2 are correct.

Award 1 mark for identifying that both decreasing frequency (increasing wavelength) and decreasing the gap width increase diffraction, while amplitude has no effect.

Step 1: Calculate the equivalent resistance of the parallel combination (\(R_p\)): \(\frac{1}{R_p} = \frac{1}{6.0} + \frac{1}{12} = \frac{3}{12} \implies R_p = 4.0\ \Omega\). Step 2: Calculate the total resistance of the circuit: \(R_{\text{total}} = R_1 + R_p = 4.0\ \Omega + 4.0\ \Omega = 8.0\ \Omega\). Step 3: Calculate the total circuit current: \(I = \frac{V}{R_{\text{total}}} = \frac{12\text{ V}}{8.0\ \Omega} = 1.5\text{ A}\). Step 4: Calculate the potential difference across the parallel combination: \(V_p = I \times R_p = 1.5\text{ A} \times 4.0\ \Omega = 6.0\text{ V}\).

Award 1 mark for correctly calculating the parallel resistance as 4.0 ohms and using the potential divider ratio or current to find the potential difference of 6.0 V.

Step 1: Calculate the secondary voltage (\(V_s\)) using the transformer equation: \(V_s = V_p \times \frac{N_s}{N_p} = 230\text{ V} \times \frac{100}{400} = 57.5\text{ V}\). Step 2: Use Ohm's law to find the current in the secondary coil (\(I_s\)): \(I_s = \frac{V_s}{R} = \frac{57.5\text{ V}}{5.75\ \Omega} = 10\text{ A}\). Step 3: For an ideal transformer, input power equals output power (\(V_p \times I_p = V_s \times I_s\)), or use the current relationship: \(I_p = I_s \times \frac{N_s}{N_p} = 10\text{ A} \times \frac{100}{400} = 2.5\text{ A}\).

Award 1 mark for the correct primary current of 2.5 A by calculating secondary voltage and secondary current first.

Let the constant background radiation count rate be \(B\). The initial corrected count rate of the source is \(800 - B\). A time of \(3.0\text{ hours}\) corresponds to exactly \(3\) half-lives (since \(1.0\text{ half-life} = 1.0\text{ hour}\)). After 3 half-lives, the corrected count rate of the source is reduced by a factor of \(2^3 = 8\), so it becomes \(\frac{800 - B}{8}\). The measured count rate after \(3.0\text{ hours}\) is \(170\text{ counts/minute}\), which includes the background, so: \(\text{measured count rate} = \text{corrected count rate} + B \implies 170 = \frac{800 - B}{8} + B\). Multiply the entire equation by 8: \(1360 = 800 - B + 8B \implies 1360 = 800 + 7B\). Subtract 800 from both sides: \(560 = 7B \implies B = 80\text{ counts/minute}\).

Award 1 mark for setting up the equation incorporating 3 half-lives and solving for a background count rate of 80 counts/minute.

The motion can be represented on a speed-time graph. The area under a speed-time graph represents the total distance travelled. The shape of the graph is a trapezium with parallel sides of length \(8.0\text{ s}\) (the time spent at constant speed) and \(12.0\text{ s}\) (the total time of travel, which is \(2.0\text{ s} + 8.0\text{ s} + 2.0\text{ s} = 12.0\text{ s}\)).\
\
The area \(A\) of a trapezium is given by:\
\(A = \frac{1}{2} \times (a + b) \times h\)\
\
Substituting the given values:\
\(50 = \frac{1}{2} \times (8.0 + 12.0) \times v\)\
\(50 = 10 \times v\)\
\(v = 5.0\text{ m/s}\)\
\
Therefore, the maximum speed \(v\) is \(5.0\text{ m/s}\).

1 mark for the correct calculation of maximum speed, matching option B.

First, calculate the extension produced by the \(6.0\text{ N}\) load:\
\(\text{Extension } x_1 = 18.0\text{ cm} - 15.0\text{ cm} = 3.0\text{ cm}\)\
\
According to Hooke's Law, extension is directly proportional to the applied load:\
\(\frac{F_1}{x_1} = \frac{F_2}{x_2}\)\
\
Substitute the values to find the new extension \(x_2\) under a \(10.0\text{ N}\) load:\
\(\frac{6.0\text{ N}}{3.0\text{ cm}} = \frac{10.0\text{ N}}{x_2}\)\
\(2.0\text{ N/cm} = \frac{10.0\text{ N}}{x_2}\)\
\(x_2 = 5.0\text{ cm}\)\
\
To find the new total length, add this extension to the original unstretched length:\
\(\text{Total length} = 15.0\text{ cm} + 5.0\text{ cm} = 20.0\text{ cm}\).

1 mark for the correct calculation of the total length, matching option A.

First, calculate the electrical energy supplied by the heater:\
\(E = P \times t = 40\text{ W} \times (5.0 \times 60)\text{ s} = 40 \times 300 = 12\,000\text{ J}\)\
\
Next, find the change in temperature:\
\(\Delta T = 68^\circ\text{C} - 20^\circ\text{C} = 48^\circ\text{C}\)\
\
Using the formula for specific heat capacity \(c\):\
\(E = m c \Delta T\)\
\(12\,000 = 0.50 \times c \times 48\)\
\(12\,000 = 24 c\)\
\(c = \frac{12\,000}{24} = 500\text{ J / (kg }^\circ\text{C)}\)\
\
This matches option B.

1 mark for the correct calculation of specific heat capacity, matching option B.

To find the angle of refraction \(r\), use Snell's Law:\
\(n = \frac{\sin i}{\sin r}\)\
\(1.5 = \frac{\sin 40^\circ}{\sin r}\)\
\(\sin r = \frac{\sin 40^\circ}{1.5} \approx \frac{0.643}{1.5} \approx 0.429\)\
\(r \approx \sin^{-1}(0.429) \approx 25^\circ\)\
\
To find the speed of light in glass \(v\):\
\(n = \frac{c}{v}\)\
\(1.5 = \frac{3.0 \times 10^8\text{ m/s}}{v}\)\
\(v = \frac{3.0 \times 10^8}{1.5} = 2.0 \times 10^8\text{ m/s}\)\
\
Thus, the angle of refraction is \(25^\circ\) and the speed of light in the glass is \(2.0 \times 10^8\text{ m/s}\), which matches option A.

1 mark for the correct calculation of angle of refraction and speed of light in glass, matching option A.

First, calculate the equivalent resistance of the parallel pair (\(R_1 = 3.0\,\Omega\) and \(R_2 = 6.0\,\Omega\)):\
\(R_p = \frac{R_1 R_2}{R_1 + R_2} = \frac{3.0 \times 6.0}{3.0 + 6.0} = \frac{18.0}{9.0} = 2.0\,\Omega\)\
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Next, calculate the total resistance of the circuit:\
\(R_{total} = R_p + R_3 = 2.0\,\Omega + 4.0\,\Omega = 6.0\,\Omega\)\
\
Now, find the total current \(I_{total}\) drawn from the power supply:\
\(I_{total} = \frac{V}{R_{total}} = \frac{18\text{ V}}{6.0\,\Omega} = 3.0\text{ A}\)\
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Calculate the potential difference \(V_p\) across the parallel combination:\
\(V_p = I_{total} \times R_p = 3.0\text{ A} \times 2.0\,\Omega = 6.0\text{ V}\)\
\
Finally, find the current \(I_2\) in the \(6.0\,\Omega\) resistor:\
\(I_2 = \frac{V_p}{R_2} = \frac{6.0\text{ V}}{6.0\,\Omega} = 1.0\text{ A}\).

1 mark for calculating the correct current through the resistor, matching option A.

Using the transformer turns ratio equation:\
\(\frac{V_p}{V_s} = \frac{N_p}{N_s}\)\
\(\frac{240}{12} = \frac{800}{N_s}\)\
\(20 = \frac{800}{N_s} \Rightarrow N_s = 40\text{ turns}\)\
\
For an ideal transformer (100% efficiency), the electrical power input equals power output:\
\(V_p \times I_p = V_s \times I_s\)\
\(240 \times I_p = 12 \times 4.0\)\
\(240 \times I_p = 48\)\
\(I_p = \frac{48}{240} = 0.20\text{ A}\)\
\
This corresponds to option A.

1 mark for the correct number of turns and primary current, matching option A.

First, convert the total elapsed time to minutes:\
\(1.0\text{ hour} = 60\text{ minutes}\)\
\
Next, calculate the number of half-lives that have passed:\
\(\text{Number of half-lives} = \frac{60\text{ minutes}}{15\text{ minutes}} = 4\)\
\
Calculate the count rate after each half-life:\
- Start: \(960\text{ counts per minute}\)\
- After 1 half-life: \(480\text{ counts per minute}\)\
- After 2 half-lives: \(240\text{ counts per minute}\)\
- After 3 half-lives: \(120\text{ counts per minute}\)\
- After 4 half-lives: \(60\text{ counts per minute}\)\
\
This matches option A.

1 mark for calculating the correct final count rate, matching option A.

Redshift is the increase in the observed wavelength of electromagnetic radiation from distant galaxies. The further away a galaxy is from Earth, the greater the redshift of its light, which shows that more distant galaxies are moving away at higher speeds. This is the basis of Hubble's Law (\(v = H_0 d\)) and supports the theory of the expanding Universe. Thus, option B is correct.

1 mark for identifying the correct relationship between redshift, distance, and recessional velocity, matching option B.

(a) Acceleration is given by the gradient of the velocity-time graph: \( a = \frac{v - u}{t} = \frac{80\text{ m/s} - 0}{8.0\text{ s}} = 10\text{ m/s}^2 \). (b) Total distance fallen is the area under the velocity-time graph: Area of triangle (0 to 8.0 s) = \( \frac{1}{2} \times \text{base} \times \text{height} = 0.5 \times 8.0\text{ s} \times 80\text{ m/s} = 320\text{ m} \). Area of rectangle (8.0 to 10.0 s) = \( \text{width} \times \text{height} = (10.0 - 8.0)\text{ s} \times 80\text{ m/s} = 160\text{ m} \). Total distance = \( 320\text{ m} + 160\text{ m} = 480\text{ m} \). (c) Weight of the capsule: \( W = mg = 120\text{ kg} \times 9.8\text{ m/s}^2 = 1176\text{ N} \). Net force on the capsule: \( F_{\text{net}} = W - R = 1176\text{ N} - 360\text{ N} = 816\text{ N} \). Acceleration: \( a = \frac{F_{\text{net}}}{m} = \frac{816\text{ N}}{120\text{ kg}} = 6.8\text{ m/s}^2 \).

(a) [2 marks] 1 mark for correct formula \( a = \Delta v / t \) or substitution. 1 mark for correct final answer with unit \( 10\text{ m/s}^2 \). (b) [3 marks] 1 mark for calculating the triangular area (320 m). 1 mark for calculating the rectangular area (160 m). 1 mark for adding them to show 480 m. (c) [3.8 marks] 1 mark for calculating weight \( W = 1176\text{ N} \) (or 1200 N if using 10). 1 mark for calculating net force \( F_{\text{net}} = 816\text{ N} \) (or 840 N). 1 mark for using Newton's second law \( a = F/m \). 0.8 marks for the correct final acceleration \( 6.8\text{ m/s}^2 \) (or \( 7.0\text{ m/s}^2 \) if \( g = 10\text{ m/s}^2 \) was used) with units.

(a) For a body in equilibrium, the sum of the clockwise moments about a pivot is equal to the sum of the anticlockwise moments about the same pivot. (b) Weight of the rule: \( W = mg = 0.15\text{ kg} \times 9.8\text{ m/s}^2 = 1.47\text{ N} \). (c) (i) Tension force: \( T = kx = 25\text{ N/m} \times 0.12\text{ m} = 3.0\text{ N} \). (ii) Taking moments about the pivot at 30 cm: The weight of the rule acts downwards at its centre of gravity (50 cm mark). Distance from pivot = \( 50 - 30 = 20\text{ cm} = 0.20\text{ m} \). This creates a clockwise moment: \( M_W = 1.47\text{ N} \times 0.20\text{ m} = 0.294\text{ N m} \). The spring force acts downwards at 90 cm. Distance from pivot = \( 90 - 30 = 60\text{ cm} = 0.60\text{ m} \). This creates a clockwise moment: \( M_T = 3.0\text{ N} \times 0.60\text{ m} = 1.80\text{ N m} \). Force \( F \) acts downwards at 10 cm. Distance from pivot = \( 30 - 10 = 20\text{ cm} = 0.20\text{ m} \). This creates an anticlockwise moment: \( M_F = F \times 0.20\text{ m} \). Equating clockwise and anticlockwise moments: \( F \times 0.20 = 0.294 + 1.80 = 2.094\text{ N m} \implies F = \frac{2.094}{0.20} = 10.47\text{ N} \approx 10.5\text{ N} \).

(a) [2 marks] 1 mark for 'sum of clockwise moments equals sum of anticlockwise moments', 1 mark for 'about a pivot / for a system in equilibrium'. (b) [2 marks] 1 mark for formula \( W = mg \), 1 mark for correct calculation of \( 1.47\text{ N} \). (c) (i) [2.3 marks] 1 mark for Hooke's Law formula \( F = kx \), 1 mark for substitution, 0.3 marks for correct answer \( 3.0\text{ N} \). (ii) [2.5 marks] 1 mark for identifying the correct distances of all forces from the pivot (20 cm, 20 cm, and 60 cm). 1 mark for correct moment balance equation: \( F \times 0.20 = (1.47 \times 0.20) + (3.0 \times 0.60) \). 0.5 marks for correct final force \( 10.5\text{ N} \) (or \( 10.47\text{ N} \)).

(a) Gravitational potential energy lost per second: \( \frac{\Delta E_p}{\Delta t} = \frac{mgh}{t} = 150\text{ kg/s} \times 9.8\text{ m/s}^2 \times 45\text{ m} = 66,150\text{ W} = 6.6\text{ kW} \times 10 = 66\text{ kW} \). (b) Kinetic energy entering the turbine per second: \( \frac{E_k}{t} = \frac{1}{2} \left(\frac{m}{t}\right) v^2 = 0.5 \times 150\text{ kg/s} \times (28\text{ m/s})^2 = 75 \times 784 = 58,800\text{ W} = 58.8\text{ kW} \). (c) Some mechanical energy is converted to thermal energy due to friction and turbulence as water flows through the pipe. (d) Total input power to the system is the GPE lost per second = 66,150 W. Electrical output power = 42 kW = 42,000 W. Efficiency = \( \frac{\text{Power Output}}{\text{Power Input}} \times 100\% = \frac{42,000}{66,150} \times 100\% = 63.49\% \approx 63.5\% \).

(a) [3 marks] 1 mark for formula \( \Delta E_p = mgh \). 1 mark for substituting values. 1 mark for correct answer of \( 66\text{ kW} \) (or \( 66,150\text{ W} \) or \( 6.6 \times 10^4\text{ W} \)). (b) [2.8 marks] 1 mark for kinetic energy formula \( E_k = 0.5 m v^2 \). 1 mark for substitution. 0.8 marks for correct answer of \( 58.8\text{ kW} \) (or \( 5.9 \times 10^4\text{ W} \)). (c) [1 mark] 1 mark for explaining that work is done against resistive forces (friction/drag), converting mechanical energy into thermal energy. (d) [2 marks] 1 mark for formula \( \text{efficiency} = \text{useful output} / \text{total input} \). 1 mark for correct calculation of \( 63.5\% \) (accept \( 63\% \) to \( 64\% \)).

(a) Energy \( E = P \times t = 120\text{ W} \times (5.0 \times 60\text{ s}) = 120 \times 300\text{ s} = 36,000\text{ J} = 3.6 \times 10^4\text{ J} \). (b) Specific heat capacity is given by: \( c = \frac{E}{m \Delta \theta} \). Here, \( E = 36,000\text{ J} \), \( m = 0.40\text{ kg} \), and \( \Delta \theta = 80^\circ\text{C} - 20^\circ\text{C} = 60^\circ\text{C} \). \( c = \frac{36,000}{0.40 \times 60} = \frac{36,000}{24} = 1500\text{ J/(kg }^\circ\text{C)} \). (c) Energy supplied during melting: \( E_{\text{melt}} = P \times t_{\text{melt}} = 120\text{ W} \times (12.0 \times 60\text{ s}) = 120 \times 720\text{ s} = 86,400\text{ J} \). Specific latent heat of fusion: \( L_f = \frac{E_{\text{melt}}}{m} = \frac{86,400\text{ J}}{0.40\text{ kg}} = 216,000\text{ J/kg} = 2.16 \times 10^5\text{ J/kg} \approx 2.2 \times 10^5\text{ J/kg} \).

(a) [2 marks] 1 mark for time conversion \( 5.0\text{ mins} = 300\text{ s} \), 1 mark for calculation showing \( 3.6 \times 10^4\text{ J} \). (b) [3.8 marks] 1 mark for formula \( c = E / (m \Delta \theta) \). 1 mark for \( \Delta \theta = 60^\circ\text{C} \). 1 mark for value \( 1500 \). 0.8 marks for correct unit \( \text{J/(kg }^\circ\text{C)} \) or \( \text{J kg}^{-1}\text{ }^\circ\text{C}^{-1} \). (c) [3 marks] 1 mark for calculating energy during melting \( E_{\text{melt}} = 86,400\text{ J} \). 1 mark for formula \( L_f = E / m \). 1 mark for correct final answer \( 2.2 \times 10^5\text{ J/kg} \) or \( 216,000\text{ J/kg} \).

(a) Using Snell's law: \( n = \frac{\sin i}{\sin r} \implies 1.52 = \frac{\sin 42.0^\circ}{\sin r} \implies \sin r = \frac{\sin 42.0^\circ}{1.52} = \frac{0.6691}{1.52} = 0.4402 \implies r = \sin^{-1}(0.4402) = 26.1^\circ \). (b) Speed of light in glass: \( v = \frac{c}{n} = \frac{3.00 \times 10^8\text{ m/s}}{1.52} = 1.9737 \times 10^8\text{ m/s} \approx 1.97 \times 10^8\text{ m/s} \). (c) Critical angle equation: \( \sin c = \frac{1}{n} = \frac{1}{1.52} = 0.6579 \implies c = \sin^{-1}(0.6579) = 41.1^\circ \). Since the angle of incidence inside the glass is \( 45^\circ \), which is greater than the critical angle of \( 41.1^\circ \), and the light is traveling towards a less optically dense medium, total internal reflection will occur.

(a) [3 marks] 1 mark for Snell's law formula \( n = \sin i / \sin r \). 1 mark for rearranging and substituting. 1 mark for correct angle \( 26.1^\circ \) (allow \( 26.0^\circ \) to \( 26.2^\circ \)). (b) [2.8 marks] 1 mark for formula \( v = c / n \). 1 mark for substitution. 0.8 marks for correct speed \( 1.97 \times 10^8\text{ m/s} \) with unit. (c) [3 marks] 1 mark for critical angle equation \( \sin c = 1/n \). 1 mark for finding critical angle \( 41.1^\circ \) (allow \( 41^\circ \)). 1 mark for comparing \( 45^\circ > 41.1^\circ \) and concluding total internal reflection will occur.

(a) Combined resistance of the parallel branch: \( R_p = \frac{R_2 \times R_3}{R_2 + R_3} = \frac{12.0 \times 6.0}{12.0 + 6.0} = \frac{72.0}{18.0} = 4.0\ \Omega \). (b) Total equivalent resistance of the circuit: \( R_t = R_1 + R_p = 8.0\ \Omega + 4.0\ \Omega = 12.0\ \Omega \). Total current: \( I = \frac{V}{R_t} = \frac{12.0\text{ V}}{12.0\ \Omega} = 1.00\text{ A} \). (c) Potential difference across \( R_1 \): \( V_1 = I \times R_1 = 1.00\text{ A} \times 8.0\ \Omega = 8.0\text{ V} \). (d) Potential difference across the parallel branch: \( V_p = V - V_1 = 12.0\text{ V} - 8.0\text{ V} = 4.0\text{ V} \). Power dissipated in \( R_3 \): \( P_3 = \frac{V_p^2}{R_3} = \frac{4.0^2}{6.0} = \frac{16.0}{6.0} = 2.67\text{ W} \).

(a) [2 marks] 1 mark for correct parallel resistance formula or substitution. 1 mark for correct answer \( 4.0\ \Omega \). (b) [2.8 marks] 1 mark for finding total resistance \( 12.0\ \Omega \). 1 mark for Ohm's law substitution \( I = V/R_t \). 0.8 marks for correct total current \( 1.00\text{ A} \). (c) [2 marks] 1 mark for formula \( V = IR \). 1 mark for correct potential difference \( 8.0\text{ V} \). (d) [2 marks] 1 mark for identifying potential difference across parallel branch is 4.0 V. 1 mark for correct power calculation \( 2.67\text{ W} \) (or \( 2.7\text{ W} \)) using \( P = V^2 / R \) or \( P = I^2 R \).

(a) An alternating current (a.c.) flows through the primary coil, which creates a continuously changing magnetic field in the soft iron core. This changing magnetic field passes through the secondary coil, cutting across its turns. This changing magnetic flux induces an alternating electromotive force (e.m.f. / voltage) in the secondary coil by electromagnetic induction. Because the secondary coil has fewer turns than the primary, the induced secondary voltage is less than the primary voltage (step-down). (b) Using the transformer equation: \( \frac{V_s}{V_p} = \frac{N_s}{N_p} \implies V_s = 230\text{ V} \times \frac{40}{800} = 11.5\text{ V} \). (c) Current in the secondary coil: \( I_s = \frac{V_s}{R} = \frac{11.5\text{ V}}{4.6\ \Omega} = 2.5\text{ A} \). For a 100% efficient transformer: \( P_p = P_s \implies V_p I_p = V_s I_s \implies I_p = \frac{V_s I_s}{V_p} = \frac{11.5\text{ V} \times 2.5\text{ A}}{230\text{ V}} = 0.125\text{ A} \).

(a) [3.8 marks] 1 mark for a.c. creating a changing magnetic field. 1 mark for iron core linking magnetic field to secondary coil. 1 mark for changing magnetic field inducing an e.m.f./voltage in secondary coil. 0.8 marks for mentioning less turns results in step-down of voltage. (b) [2 marks] 1 mark for transformer equation substitution. 1 mark for correct secondary voltage \( 11.5\text{ V} \). (c) [3 marks] 1 mark for calculating secondary current \( I_s = 2.5\text{ A} \). 1 mark for power equality relation \( V_p I_p = V_s I_s \) or \( I_p = I_s (N_s / N_p) \). 1 mark for correct primary current \( 0.125\text{ A} \).

(a) Redshift is the increase in the observed wavelength (or decrease in frequency) of light received from an astronomical object that is moving away from the observer. (b) It indicates that distant galaxies are moving away from the Earth (receding), which supports the expansion of the universe. (c) Wavelength change: \( \Delta \lambda = 672.4\text{ nm} - 656.3\text{ nm} = 16.1\text{ nm} \). Laboratory wavelength: \( \lambda_0 = 656.3\text{ nm} \). Recession speed: \( v = c \times \frac{\Delta \lambda}{\lambda_0} = 3.00 \times 10^8\text{ m/s} \times \frac{16.1\text{ nm}}{656.3\text{ nm}} = 3.00 \times 10^8 \times 0.02453 = 7.359 \times 10^6\text{ m/s} \approx 7.36 \times 10^6\text{ m/s} \). (d) From Hubble's Law: \( v = H_0 d \implies d = \frac{v}{H_0} = \frac{7.36 \times 10^6\text{ m/s}}{2.2 \times 10^{-18}\text{ s}^{-1}} = 3.345 \times 10^{24}\text{ m} \approx 3.3 \times 10^{24}\text{ m} \).

(a) [2 marks] 1 mark for increase in observed wavelength of light. 1 mark for stating this happens when the source is moving away from the observer. (b) [1 mark] 1 mark for stating that galaxies are moving away / receding. (c) [3.8 marks] 1 mark for calculating \( \Delta \lambda = 16.1\text{ nm} \). 1 mark for using the formula and substituting values. 1 mark for the numerical calculation. 0.8 marks for the correct answer \( 7.36 \times 10^6\text{ m/s} \) with unit. (d) [2 marks] 1 mark for rearranging Hubble's Law formula \( d = v / H_0 \) and substituting values. 1 mark for correct final answer \( 3.3 \times 10^{24}\text{ m} \) (accept \( 3.35 \times 10^{24}\text{ m} \)).

**Part (a)**
Faraday's law of electromagnetic induction states that the induced electromotive force (e.m.f.) in a conductor is directly proportional to the rate of change of magnetic flux linkage (or the rate of cutting of magnetic flux lines).

**Part (b)**
(i) The slip rings rotate with the coil and provide continuous electrical connection to the external circuit without tangling the connecting wires. The stationary carbon brushes rub against the rotating slip rings to complete the circuit and transfer the induced current to the external load.
(ii) Any two of the following modifications will increase the maximum output e.m.f.:
- Increase the speed of rotation of the coil.
- Use a stronger magnet (increase the magnetic field strength).
- Increase the number of turns on the coil.
- Increase the cross-sectional area of the coil.

**Part (c)**
(i) Using the transformer equation:
\(\frac{V_s}{V_p} = \frac{N_s}{N_p}\)

Rearranging for the secondary voltage \(V_s\):
\(V_s = V_p \times \frac{N_s}{N_p} = 12\text{ V} \times \frac{3000}{150} = 240\text{ V}\)

(ii) A steady direct current (d.c.) in the primary coil creates a constant magnetic field (constant magnetic flux). A transformer relies on a changing magnetic field to induce an e.m.f. in the secondary coil. Since there is no change in magnetic flux linkage, no e.m.f. is induced.

**(a)**
- Induced e.m.f. / current is proportional to the rate of change of magnetic flux linkage (or rate of cutting of magnetic flux lines) [1]
- Explicit mention of 'rate of' change or 'rate of' cutting [1]

**(b)(i)**
- Slip rings rotate with the coil to prevent the external circuit wires from tangling [1]
- Brushes provide continuous sliding electrical contact with the rotating slip rings [1]

**(b)(ii)**
- Any two of: higher speed of rotation, stronger magnets / stronger magnetic field, more turns in the coil, larger coil area [2]
*(1 mark for each correct modification, max 2)*

**(c)(i)**
- Correct calculation: \(240\text{ V}\) [1]

**(c)(ii)**
- d.c. produces a constant/steady magnetic field [1]
- There is no change in magnetic flux linkage in the secondary coil, so no e.m.f. is induced [1]

1. The temperature is halfway between 84 and 85, which is \(84.5^\circ\text{C}\).
2. Parallax error is avoided by looking perpendicularly (at eye level) to the graduated scale of the thermometer.
3. \(\Delta\theta_A = 84.5 - 66.0 = 18.5^\circ\text{C}\) and \(\Delta\theta_B = 84.5 - 55.5 = 29.0^\circ\text{C}\).
4. Beaker A is more effective because its overall temperature drop is smaller (\(18.5^\circ\text{C} < 29.0^\circ\text{C}\)).
5. Any two from: initial volume of water, initial temperature of water, material/surface area of beakers, ambient room temperature.
6. Side heat loss is reduced by wrapping the beakers with lagging (e.g. bubble wrap); bottom heat loss is reduced by placing the beakers on a wooden or cork insulating mat.

1. Correctly reading the thermometer scale: 84.5 °C [1 mark]
2. Precaution: View scale perpendicularly / at eye level [1 mark]
3. Correct calculations: \(\Delta\theta_A = 18.5^\circ\text{C}\) [1 mark] and \(\Delta\theta_B = 29.0^\circ\text{C}\) [1 mark]
4. Correct identification of Beaker A [1 mark] and correct justification referring to smaller temperature drop [1 mark]
5. Two correct control variables: [1 mark] each
6. Two valid improvements: [1 mark] each (e.g., lagging/insulating the sides, placing on an insulating stand/mat)

1. Measuring from the normal line NN' to the rays: \(i = 45^\circ\) and \(r = 28^\circ\).
2. \(\sin(60.0^\circ) = 0.866 \approx 0.87\) and \(\sin(35.0^\circ) = 0.5736 \approx 0.57\).
3. Graph details: The points plotted are (0.33, 0.50), (0.47, 0.71), and (0.57, 0.87). A straight line of best fit is drawn passing through the origin.
4. Gradient calculation: \(\text{Gradient} = \frac{\Delta \sin i}{\Delta \sin r} = \frac{0.87}{0.57} = 1.53\). Any value between 1.45 and 1.55 is acceptable.
5. Precaution: Space pins as far apart as possible (minimum 5 cm) to minimize alignment error, or ensure pins are pushed in vertically and aligned by looking at their bases.

1. Correctly measured angles: \(i = 45^\circ\) [1 mark] and \(r = 28^\circ\) [1 mark] (allow \(\pm 2^\circ\))
2. Correct values of \(\sin i = 0.87\) [1 mark] and \(\sin r = 0.57\) [1 mark]
3. Graph evaluation: Axes labeled with quantities and sensible linear scale [1 mark], accurate plotting of the three data points [1 mark], straight line of best fit drawn using a ruler [1 mark]
4. Correct gradient working shown with coordinates from the line [1 mark] and correct calculation of refractive index in the range 1.45 to 1.55 [1 mark]
5. Valid precaution: spacing pins far apart (> 5 cm) or viewing pin bases [1 mark]

1. Reading ammeter: 0.4 A + 4 divisions \(\times\) 0.01 A = 0.44 A. Reading voltmeter: 2.20 V.
2. \(R = \frac{V}{I} = \frac{2.20}{0.44} = 5.0\ \Omega\).
3. Plotting the points: (20.0, 1.1), (40.0, 2.2), (60.0, 3.3), (80.0, 4.4), (100.0, 5.5). All points lie perfectly on a straight line passing through (0,0).
4. Direct proportion is verified because the graph is a straight line and it passes through the origin (0,0).
5. Passing current through the wire causes it to heat up. An increase in temperature would increase the wire's resistance, rendering the test unfair.

1. Correct ammeter reading of 0.44 A [1 mark] and voltmeter reading of 2.2 V / 2.20 V [1 mark]
2. Correct calculation of resistance with 2 significant figures and unit (5.0 \(\Omega\)) [1 mark]
3. Graph scale sensible and covering >50% of grid [1 mark], correct labeling with units [1 mark], accurate plotting of all five points [1 mark], straight line drawn with a ruler [1 mark]
4. Correctly stating direct proportion [1 mark] and justifying because the line is straight and passes through the origin [1 mark]
5. Correct reasoning about avoiding temperature/resistance increase due to heating [1 mark]

1. The arrow must align with the top and bottom of the main coiled part of the spring (excluding the hooks/loops at the end).
2. Extensions: \(38 - 25 = 13\text{ mm}\), \(50 - 25 = 25\text{ mm}\), \(63 - 25 = 38\text{ mm}\), \(75 - 25 = 50\text{ mm}\), \(88 - 25 = 63\text{ mm}\).
3. Graph plotting: \(F\) on the vertical axis (0 to 5), \(e\) on the horizontal axis (0 to 70). The points to plot are (0,0), (13, 1.0), (25, 2.0), (38, 3.0), (50, 4.0), (63, 5.0). A straight line of best fit is drawn.
4. Gradient calculation: Using points (0,0) and (63, 5.0) on the line, \(\text{gradient} = \frac{5.0 - 0}{63 - 0} = 0.0794 \approx 0.079\text{ N/mm}\). Accept values in the range 0.077 to 0.082 N/mm.
5. To avoid parallax error: Read the ruler scale with the eye at the same horizontal level as the top/bottom of the coil; keep the rule as close to the spring as possible; or use a set square.

1. Arrow drawn correctly spanning the main body / coiled part of the spring only [1 mark]
2. All five extension values calculated correctly [2 marks] (deduct 1 mark if one calculation is wrong)
3. Graph plotting: Correct labels on axes with units [1 mark], linear scale with points taking up at least half the page [1 mark], accurate plotting [1 mark], straight best-fit line through origin [1 mark]
4. Gradient: Triangle method shown on graph covering at least half of the line [1 mark] and correct calculation with unit N/mm or N/m [1 mark]
5. Valid precaution against parallax error: line of sight perpendicular, or rule close to spring, or use of a set square [1 mark]