MetadataPastPaper.sampleTitle

Using the formula for resistors in parallel: \(\frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2}\). Substituting the given values: \(\frac{1}{R_{\text{total}}} = \frac{1}{12} + \frac{1}{6} = \frac{1}{12} + \frac{2}{12} = \frac{3}{12} = \frac{1}{4}\). Therefore, \(R_{\text{total}} = 4\ \Omega\).

1 mark for the correct answer (A).

Work done is calculated using the formula: \(\text{Work done} = \text{Force} \times \text{distance}\). Here, the force is equal to the weight of the load: \(\text{Work done} = 20\text{ N} \times 3.0\text{ m} = 60\text{ J}\). Note that the time of 5.0 seconds is not needed for calculating work done (it would be used to calculate power).

1 mark for the correct answer (B).

The work done in stretching a spring (elastic potential energy stored) is calculated using the formula: \(E_e = \frac{1}{2} k e^2\). Substituting the given values: \(E_e = 0.5 \times 25\text{ N/m} \times (0.20\text{ m})^2 = 12.5 \times 0.04 = 0.50\text{ J}\).

1 mark for the correct answer (A).

As temperature increases, the average kinetic energy of the gas particles increases, so they move faster. This causes them to collide with the walls of the container more frequently and with greater force, resulting in a higher pressure. The volume is fixed, so density remains constant, and individual gas particles do not expand.

1 mark for the correct answer (B).

First, calculate the area of the base in contact with the ground: \(\text{Area} = 0.30\text{ m} \times 0.40\text{ m} = 0.12\text{ m}^2\). Then, use the pressure formula: \(\text{Pressure} = \frac{\text{Force}}{\text{Area}} = \frac{48\text{ N}}{0.12\text{ m}^2} = 400\text{ Pa}\).

1 mark for the correct answer (D).

Use the equation of motion: \(v^2 - u^2 = 2as\). Here, \(u = 4.0\text{ m/s}\), \(v = 12.0\text{ m/s}\), and \(s = 32\text{ m}\). Rearranging for \(a\): \(a = \frac{v^2 - u^2}{2s} = \frac{12.0^2 - 4.0^2}{2 \times 32} = \frac{144 - 16}{64} = \frac{128}{64} = 2.0\text{ m/s}^2\).

1 mark for the correct answer (B).

Initial momentum of the trolley is: \(p_i = m \times u = 5.0\text{ kg} \times 4.0\text{ m/s} = 20\text{ kg m/s}\). The impulse applied is: \(\text{Impulse} = F \times t = 15\text{ N} \times 3.0\text{ s} = 45\text{ N s}\). The change in momentum is equal to the impulse, so: \(p_f = p_i + \text{Impulse} = 20 + 45 = 65\text{ kg m/s}\).

1 mark for the correct answer (C).

Using the transformer equation: \(\frac{V_p}{V_s} = \frac{N_p}{N_s}\). Rearranging for \(V_s\): \(V_s = V_p \times \frac{N_s}{N_p} = 60\text{ V} \times \frac{800}{200} = 60 \times 4 = 240\text{ V}\).

1 mark for the correct answer (C).

As the temperature of a thermistor increases, its resistance decreases. Because the thermistor and the fixed resistor are in series, the decreased resistance of the thermistor means it takes a smaller share of the cell's potential difference. Thus, the potential difference across the fixed resistor must increase.

1 mark for selecting the correct option B.

To find \(R\):
1. Calculate the total resistance of the circuit: \(R_{\text{total}} = \frac{V}{I} = \frac{12.0\text{ V}}{1.5\text{ A}} = 8.0\ \Omega\).
2. Calculate the equivalent resistance of the parallel branch: \(R_{\text{parallel}} = R_{\text{total}} - R_{\text{series}} = 8.0\ \Omega - 5.0\ \Omega = 3.0\ \Omega\).
3. Use the formula for two identical parallel resistors: \(\frac{1}{R_{\text{parallel}}} = \frac{2}{R}\), which gives \(R = 2 \times R_{\text{parallel}} = 2 \times 3.0\ \Omega = 6.0\ \Omega\).

- Recall and use of \(R = \frac{V}{I}\) to find the total resistance of the circuit (1 mark).
- Correct value for total resistance: \(8.0\ \Omega\) (1 mark).
- Subtraction of the series resistor from the total resistance to find parallel branch resistance: \(8.0 - 5.0 = 3.0\ \Omega\) (1 mark).
- Realisation that two identical resistors in parallel have a combined resistance of half of individual resistance: \(R = 2 \times R_{\text{parallel}}\) (1 mark).
- Correct final answer for \(R = 6.0\ \Omega\) (1 mark).

1. Set the initial direction (to the right) as positive. Initial momentum: \(p_i = m \times u = 0.80\text{ kg} \times 3.0\text{ m/s} = +2.4\text{ kg m/s}\).
2. Rebound velocity is in the opposite direction (to the left), so \(v = -2.0\text{ m/s}\). Final momentum: \(p_f = m \times v = 0.80\text{ kg} \times (-2.0\text{ m/s}) = -1.6\text{ kg m/s}\).
3. Calculate the change in momentum: \(\Delta p = p_f - p_i = -1.6 - 2.4 = -4.0\text{ kg m/s}\) (magnitude is \(4.0\text{ kg m/s}\)).
4. Use the force formula: \(F = \frac{\Delta p}{\Delta t} = \frac{-4.0\text{ kg m/s}}{0.16\text{ s}} = -25\text{ N}\).
5. The negative sign indicates that the force acts in the direction opposite to the initial motion, which is to the left.

- Calculate initial and final momentum values (with opposite signs or correct handling of direction): \(2.4\text{ kg m/s}\) and \(-1.6\text{ kg m/s}\) (1 mark).
- Calculate change in momentum magnitude: \(4.0\text{ kg m/s}\) (1 mark; award only 1 mark total for momentum change if direction was ignored resulting in \(0.8\text{ kg m/s}\)).
- Recall and substitute into \(F = \frac{\Delta p}{t}\) (1 mark).
- Correct magnitude of the force: \(25\text{ N}\) (1 mark).
- Correct direction stated: to the left / opposite to initial direction (1 mark).

(a) Calculate specific latent heat of fusion (\(L_f\)):
- Energy supplied to melt: \(E_1 = \text{Power} \times t_1 = 60\text{ W} \times 120\text{ s} = 7200\text{ J}\).
- Use \(E = m \times L_f \implies L_f = \frac{E_1}{m} = \frac{7200\text{ J}}{0.20\text{ kg}} = 36000\text{ J/kg}\) (or \(36\text{ kJ/kg}\)).

(b) Calculate specific heat capacity (\(c\)):
- Energy supplied to heat the liquid: \(E_2 = \text{Power} \times t_2 = 60\text{ W} \times 80\text{ s} = 4800\text{ J}\).
- Use \(E = m \times c \times \Delta\theta \implies c = \frac{E_2}{m \times \Delta\theta} = \frac{4800\text{ J}}{0.20\text{ kg} \times 15\text{ }^\circ\text{C}} = \frac{4800}{3} = 1600\text{ J/kg}^\circ\text{C}\).

- Calculate electrical energy transferred during melting: \(7200\text{ J}\) (1 mark).
- Use \(E = mL\) to find specific latent heat of fusion: \(36000\text{ J/kg}\) (or \(36\text{ kJ/kg}\)) (1 mark).
- Calculate electrical energy transferred during heating: \(4800\text{ J}\) (1 mark).
- Recall and rearrange \(E = mc\Delta\theta\) to make \(c\) the subject (1 mark).
- Correct final value for specific heat capacity: \(1600\text{ J/kg}^\circ\text{C}\) (1 mark).

1. Find the extension under \(6.0\text{ N}\): \(e_1 = 17.0\text{ cm} - 12.0\text{ cm} = 5.0\text{ cm} = 0.05\text{ m}\).
2. Calculate the spring constant (\(k\)): \(k = \frac{F}{e_1} = \frac{6.0\text{ N}}{0.05\text{ m}} = 120\text{ N/m}\).
3. Find the new extension when total length is \(21.0\text{ cm}\): \(e_2 = 21.0\text{ cm} - 12.0\text{ cm} = 9.0\text{ cm} = 0.09\text{ m}\).
4. Calculate the stored elastic potential energy (\(E_e\)): \(E_e = \frac{1}{2} k e_2^2 = 0.5 \times 120\text{ N/m} \times (0.09\text{ m})^2 = 60 \times 0.0081 = 0.486\text{ J}\).

- Calculate first extension in metres: \(0.05\text{ m}\) (1 mark).
- Recall and use Hooke's law \(F = ke\) to find \(k = 120\text{ N/m}\) (1 mark).
- Calculate second extension in metres: \(0.09\text{ m}\) (1 mark).
- Recall \(E_e = \frac{1}{2} k e^2\) and substitute correct values (1 mark).
- Correct calculation of energy: \(0.486\text{ J}\) (accept \(0.49\text{ J}\)) (1 mark).

(a) Calculate the final velocity (\(v\)):
- Use the equation of motion: \(v^2 - u^2 = 2as \implies v^2 = u^2 + 2as\).
- Substitute the values: \(v^2 = (5.0\text{ m/s})^2 + 2 \times 2.0\text{ m/s}^2 \times 36\text{ m} = 25 + 144 = 169\).
- Thus, \(v = \sqrt{169} = 13\text{ m/s}\).

(b) Calculate the time taken (\(t\)):
- Use the definition of acceleration: \(a = \frac{v - u}{t} \implies t = \frac{v - u}{a}\).
- Substitute the values: \(t = \frac{13\text{ m/s} - 5.0\text{ m/s}}{2.0\text{ m/s}^2} = \frac{8.0}{2.0} = 4.0\text{ s}\).

- Recall equation of motion \(v^2 - u^2 = 2as\) (1 mark).
- Substitute values correctly to get \(v^2 = 169\) (1 mark).
- Calculate final velocity: \(13\text{ m/s}\) (1 mark).
- Recall or rearrange a formula connecting time to other quantities, e.g. \(a = \frac{v-u}{t}\) (1 mark).
- Calculate time: \(4.0\text{ s}\) (1 mark).

(a) Calculate output voltage (\(V_s\)):
- Use the transformer equation: \(\frac{V_p}{V_s} = \frac{N_p}{N_s}\).
- Rearrange and substitute: \(V_s = V_p \times \frac{N_s}{N_p} = 230\text{ V} \times \frac{64}{800} = 18.4\text{ V}\).

(b) Calculate primary current (\(I_p\)):
- For a 100% efficient transformer, power input = power output: \(V_p \times I_p = V_s \times I_s\).
- Rearrange and substitute: \(I_p = \frac{V_s \times I_s}{V_p} = \frac{18.4\text{ V} \times 3.0\text{ A}}{230\text{ V}} = \frac{55.2\text{ W}}{230\text{ V}} = 0.24\text{ A}\).
- Alternatively, use current turn ratio: \(I_p = I_s \times \frac{N_s}{N_p} = 3.0\text{ A} \times \frac{64}{800} = 0.24\text{ A}\).

- Recall and use turn ratio equation \(\frac{V_p}{V_s} = \frac{N_p}{N_s}\) (1 mark).
- Rearrange to solve for \(V_s\) (1 mark).
- Correct value of secondary voltage: \(18.4\text{ V}\) (1 mark).
- Recall and use power conservation relationship \(V_p I_p = V_s I_s\) (or equivalent ratio) (1 mark).
- Correct value of primary current: \(0.24\text{ A}\) (1 mark).

(a) Pressure due to water alone:
- Use \(p = h \rho g\).
- \(p = 1.5\text{ m} \times 1000\text{ kg/m}^3 \times 10\text{ N/kg} = 15000\text{ Pa}\) (or \(1.5 \times 10^4\text{ Pa}\)).

(b) Total force on the base:
- Total pressure at base: \(p_{\text{total}} = p_{\text{atmospheric}} + p_{\text{water}} = 100000\text{ Pa} + 15000\text{ Pa} = 115000\text{ Pa}\).
- Area of base: \(A = \text{length} \times \text{width} = 0.80\text{ m} \times 0.50\text{ m} = 0.40\text{ m}^2\).
- Use \(F = p \times A\):
- \(F = 115000\text{ Pa} \times 0.40\text{ m}^2 = 46000\text{ N}\).

- Recall and use \(p = h \rho g\) (1 mark).
- Correct pressure due to water: \(15000\text{ Pa}\) (1 mark).
- Calculate area of the base: \(0.40\text{ m}^2\) (1 mark).
- Calculate total pressure at the bottom of the container: \(115000\text{ Pa}\) (1 mark).
- Correct final calculation of force: \(46000\text{ N}\) (1 mark).

1. Calculate weight of crate: \(F = m \times g = 250\text{ kg} \times 10\text{ N/kg} = 2500\text{ N}\).
2. Calculate work done: \(W = F \times d = 2500\text{ N} \times 12\text{ m} = 30000\text{ J}\).
3. Calculate useful power output of the motor: \(P_{\text{out}} = \frac{W}{t} = \frac{30000\text{ J}}{10\text{ s}} = 3000\text{ W}\).
4. Convert efficiency to a decimal: \(75\% = 0.75\).
5. Use efficiency formula to find total power input: \(\text{Efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}} \implies P_{\text{in}} = \frac{3000\text{ W}}{0.75} = 4000\text{ W}\).

- Calculate the weight of the crate: \(2500\text{ N}\) (1 mark).
- Calculate work done in raising the crate: \(30000\text{ J}\) (1 mark).
- Recall and use equation for useful power: \(P = \frac{\text{Work}}{\text{time}}\) (1 mark).
- Calculate useful power output: \(3000\text{ W}\) (1 mark).
- Recall efficiency formula and calculate electrical power input: \(4000\text{ W}\) (1 mark).

**(a)**
- Set up a circuit consisting of a DC power supply, an ammeter connected in series with the thermistor, and a voltmeter connected in parallel across the thermistor.
- Place the thermistor inside a beaker of cold water.
- Use a thermometer to measure the temperature of the water.
- Place the beaker on a tripod and gauze and heat it gently using a Bunsen burner (or use an electric water bath) to vary the temperature from \(20^\circ\text{C}\) up to around \(80^\circ\text{C}\).
- Take readings of the current (\(I\)) and potential difference (\(V\)) at regular temperature intervals (e.g., every \(10^\circ\text{C}\)).
- At each temperature, calculate resistance using \(R = \frac{V}{I}\).
- Safety precaution: Use insulated tongs or heatproof mats when handling hot beakers; avoid splashing water onto electrical connections.

**(b)**
Using the formula:
\(R = \frac{V}{I}\)
\(R = \frac{4.5}{0.015} = 300\ \Omega\)

**(c)**
- As temperature increases, the resistance of a thermistor decreases.
- Since \(I = \frac{V}{R}\) and the potential difference remains constant, the current in the circuit will increase.

**(a) Level of Response (6 marks)**
- **Level 3 (5-6 marks)**: A detailed, coherent description of a logical experiment. Includes a complete circuit description (ammeter in series, voltmeter in parallel), a clear method for heating water and measuring temperature, and explains how resistance is calculated with appropriate safety measures.
- **Level 2 (3-4 marks)**: A mostly complete description. Mentions measuring current and voltage, heating the thermistor, and calculating resistance, but lacks some detail or safety considerations.
- **Level 1 (1-2 marks)**: Fragmented description. Identifies some components (e.g., thermometer, thermistor) or says resistance changes, but does not provide a working experimental method.

**(b) [2 marks total]**
- 1 mark for correct substitution: \(\frac{4.5}{0.015}\)
- 1 mark for correct evaluation with units: \(300\ \Omega\) (or ohms)

**(c) [2 marks total]**
- 1 mark for stating that resistance decreases as temperature increases.
- 1 mark for stating that current increases as a result.

**(a)**
- Newton's third law states that when one object exerts a force on another, the second object exerts an equal and opposite force on the first.
- During the collision, the force exerted by car A on car B is equal in magnitude and opposite in direction to the force exerted by car B on car A.

**(b)**
Let the direction to the right be positive (+), and left be negative (-).
- Initial velocity of car A, \(u_A = +4.0\text{ m/s}\)
- Initial velocity of car B, \(u_B = -2.0\text{ m/s}\)

Total initial momentum:
\(p_{\text{initial}} = (m_A \times u_A) + (m_B \times u_B)\)
\(p_{\text{initial}} = (0.50 \times 4.0) + (0.30 \times (-2.0))\)
\(p_{\text{initial}} = 2.0 - 0.60 = 1.4\text{ kg m/s}\)

Since momentum is conserved:
\(p_{\text{final}} = p_{\text{initial}} = 1.4\text{ kg m/s}\)

The cars stick together, so their combined mass is:
\(m_{\text{total}} = 0.50 + 0.30 = 0.80\text{ kg}\)

Combined final velocity \(v\):
\(v = \frac{p_{\text{final}}}{m_{\text{total}}} = \frac{1.4}{0.80} = 1.75\text{ m/s}\)

Since the value is positive, the direction of movement is to the right.

**(c)**
Using force = change in momentum / time:
For car B:
Initial momentum of car B = \(0.30 \times (-2.0) = -0.60\text{ kg m/s}\)
Final momentum of car B = \(0.30 \times 1.75 = 0.525\text{ kg m/s}\)
Change in momentum of car B = \(0.525 - (-0.60) = 1.125\text{ kg m/s}\)
Force \(F = \frac{\Delta p}{t} = \frac{1.125}{0.15} = 7.5\text{ N}\)

(Alternatively, using acceleration of car B:
\(a = \frac{v - u}{t} = \frac{1.75 - (-2.0)}{0.15} = \frac{3.75}{0.15} = 25\text{ m/s}^2\)
\(F = m \times a = 0.30 \times 25 = 7.5\text{ N}\))

**(a) [3 marks total]**
- 1 mark for stating Newton's third law (forces are equal in size and opposite in direction when two bodies interact).
- 1 mark for applying it to the cars: Car A exerts a force on Car B, and Car B exerts an equal force on Car A.
- 1 mark for stating that these forces act in opposite directions.

**(b) [5 marks total]**
- 1 mark for calculating the initial momentum of Car A: \(0.5 \times 4.0 = 2.0\text{ kg m/s}\).
- 1 mark for calculating the initial momentum of Car B with correct negative sign (due to opposite direction): \(0.3 \times (-2.0) = -0.6\text{ kg m/s}\) (total initial momentum = \(1.4\text{ kg m/s}\)).
- 1 mark for calculating the total mass of combined cars: \(0.80\text{ kg}\).
- 1 mark for calculating the magnitude of velocity: \(1.75\text{ m/s}\) (or \(1.8\text{ m/s}\)).
- 1 mark for specifying the correct direction: to the right.

**(c) [2 marks total]**
- 1 mark for using the correct relationship: \(F = \frac{\Delta p}{t}\) or \(F = ma\) with appropriate substitutions (even if using wrong velocity from part b).
- 1 mark for correct final force: \(7.5\text{ N}\) (allow ecf from part b).

**(a)**
Structure of a transformer:
- It consists of a primary coil and a secondary coil wound around a laminated soft iron core.
- In a step-up transformer, there are more turns on the secondary coil than on the primary coil.

How it operates:
- An alternating potential difference is applied across the primary coil.
- This produces an alternating current in the primary coil.
- The alternating current creates a continuously changing (alternating) magnetic field in the iron core.
- The iron core guides this changing magnetic field to the secondary coil.
- The changing magnetic field passes through the secondary coil, inducing an alternating potential difference across its ends (electromagnetic induction).
- Because there are more turns on the secondary coil, the induced potential difference is greater than the input potential difference.

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

Rearranging to find \(V_s\):
\(V_s = V_p \times \frac{N_s}{N_p}\)
\(V_s = 230 \times \frac{2400}{120}\)
\(V_s = 230 \times 20 = 4600\text{ V}\)

**(c)**
- A higher potential difference means that for the same electrical power transmitted, the current flowing through the transmission cables is much lower (\(P = IV\)).
- A lower current reduces the heating effect in the transmission cables (\(P = I^2 R\)), meaning less thermal energy is wasted to the surroundings, which increases efficiency.

**(a) Level of Response (6 marks)**
- **Level 3 (5-6 marks)**: Clear and accurate description of the structure (coils and core, step-up having more turns on secondary) and detailed explanation of how it operates. Explains the role of AC, the changing magnetic field in the core, and electromagnetic induction in the secondary coil.
- **Level 2 (3-4 marks)**: Describes the structure and mentions the changing magnetic field, but may omit the role of the iron core or have minor gaps in the explanation of induction.
- **Level 1 (1-2 marks)**: Simple points, such as mentioning primary and secondary coils, or stating that a transformer changes voltage.

**(b) [2 marks total]**
- 1 mark for correct substitution: \(V_s = 230 \times \frac{2400}{120}\) (or equivalent ratio).
- 1 mark for correct calculation: \(4600\text{ V}\) (allow \(4.6\text{ kV}\)).

**(c) [2 marks total]**
- 1 mark for stating that increasing voltage decreases the current in the transmission wires (for the same power).
- 1 mark for explaining that a lower current reduces energy loss as heat/thermal energy in the cables (improving efficiency).

**(a)**
- Set up a clamp stand with a clamp holding the top of the spring.
- Secure a metre ruler vertically next to the spring using another clamp.
- Take an initial reading of the position of the bottom of the spring on the ruler (unstretched length).
- Add a known weight (e.g., \(1\text{ N}\)) to the hanger and record the new position of the bottom of the spring.
- Calculate the extension by subtracting the original length from the new length (\(\text{extension} = \text{new length} - \text{original length}\)).
- To ensure accuracy/precision: Read the ruler at eye level to avoid parallax error; use a fiducial marker (pointer) attached to the bottom of the spring to make reading easier.
- Repeat the experiment and calculate average extensions for each load.

**(b)**
First, calculate the extension in metres:
\(\text{Extension } x = 22.5\text{ cm} - 15.0\text{ cm} = 7.5\text{ cm} = 0.075\text{ m}\)

Using Hooke's Law:
\(F = kx\)
\(k = \frac{F}{x} = \frac{6.0\text{ N}}{0.075\text{ m}} = 80\text{ N/m}\)
(Alternatively: in \(\text{N/cm}\): \(k = \frac{6.0}{7.5} = 0.8\text{ N/cm}\)).

**(c)**
Using the equation for elastic potential energy:
\(E_e = \frac{1}{2} k x^2\)
\(E_e = 0.5 \times 80\text{ N/m} \times (0.075\text{ m})^2\)
\(E_e = 40 \times 0.005625 = 0.225\text{ J}\)

**(a) [4 marks total]**
- 1 mark for describing how the vertical ruler is set up next to the spring.
- 1 mark for explaining that extension is calculated as the loaded length minus the unstretched length.
- 1 mark for detail on ensuring precision/accuracy (e.g., reading at eye level to avoid parallax error, using a pointer/fiducial marker).
- 1 mark for repeating the experiment to find a mean / check for anomalies.

**(b) [3 marks total]**
- 1 mark for calculating the correct extension: \(7.5\text{ cm}\) or \(0.075\text{ m}\).
- 1 mark for correct calculation of spring constant: \(80\) (if using metres) or \(0.8\) (if using cm).
- 1 mark for correct unit corresponding to value: \(\text{N/m}\) or \(\text{N/cm}\).

**(c) [3 marks total]**
- 1 mark for using the correct equation: \(E_e = \frac{1}{2} k x^2\).
- 1 mark for correct substitution: \(E_e = 0.5 \times 80 \times (0.075)^2\) (allow ecf from part b).
- 1 mark for correct evaluation: \(0.225\text{ J}\) (or matching ecf value).

**(a)**
- Specific heat capacity is the energy required to raise the temperature of \(1\text{ kg}\) of a substance by \(1^\circ\text{C}\) without changing its state.
- Specific latent heat is the energy required to change the state of \(1\text{ kg}\) of a substance with no change in temperature.

**(b)**
(i) As the ice warms up from \(-10^\circ\text{C}\) to \(0^\circ\text{C}\), the thermal energy increases the kinetic energy of the water molecules, causing them to vibrate faster about their fixed positions.
(ii) While melting at \(0^\circ\text{C}\), the energy is used to break intermolecular bonds between the molecules. The molecules move out of their fixed lattice positions to become free to move past each other, while their kinetic energy (and temperature) remains constant.

**(c)**
First, calculate the energy supplied by the heater during the melting process:
\(\text{Time } t = 22\text{ minutes} = 22 \times 60 = 1320\text{ s}\)
\(\text{Energy } E = \text{Power } P \times t\)
\(E = 50\text{ W} \times 1320\text{ s} = 66,000\text{ J}\)

Next, use the specific latent heat formula:
\(E = m \times L\)
\(L = \frac{E}{m}\)
\(L = \frac{66,000\text{ J}}{0.20\text{ kg}}\)
\(L = 330,000\text{ J/kg}\) (or \(330\text{ kJ/kg}\) or \(3.3 \times 10^5\text{ J/kg}\))

**(a) [2 marks total]**
- 1 mark for defining specific heat capacity as involving a change in temperature (per kg).
- 1 mark for defining specific latent heat as involving a change in state at a constant temperature (per kg).

**(b) [3 marks total]**
- 1 mark for (i): stating that molecules gain kinetic energy / vibrate faster.
- 1 mark for (ii): stating that bonds between molecules are broken.
- 1 mark for (ii): stating that molecules are freed from their fixed positions to flow/slide over each other (or kinetic energy/temperature remains constant).

**(c) [5 marks total]**
- 1 mark for converting time to seconds: \(22 \times 60 = 1320\text{ s}\).
- 1 mark for using \(E = P \times t\): \(50 \times 1320 = 66,000\text{ J}\).
- 1 mark for using the correct formula: \(L = \frac{E}{m}\).
- 1 mark for substituting values: \(L = \frac{66,000}{0.20}\).
- 1 mark for correct calculation with unit: \(330,000\text{ J/kg}\) (or \(3.3 \times 10^5\text{ J/kg}\) or \(330\text{ kJ/kg}\)).

To investigate the resistance of a filament lamp: 1. Circuit Connection: Connect a power supply, an ammeter, a filament lamp, and a variable resistor in a series loop. Connect a voltmeter in parallel across the filament lamp. 2. Obtaining Measurements: Switch on the circuit and record the initial current (I) from the ammeter and potential difference (V) from the voltmeter. Adjust the variable resistor to a new position to change the potential difference and current, then record the new readings. Repeat this process for at least 5 different settings to obtain a wide range of values. 3. Calculating Resistance: Use the equation \(R = \frac{V}{I}\) to calculate the resistance of the lamp at each potential difference. A graph of resistance against potential difference or current can be plotted to show how the resistance of the filament lamp increases as the current (and temperature) increases.

Level 3 (5-6 marks): Clear, detailed description of a complete workable circuit (ammeter in series, voltmeter in parallel across the lamp, variable resistor or variable power supply). Clear method explaining how to vary the current/PD to collect multiple pairs of readings. Correct formula \(R = \frac{V}{I}\) given for finding resistance. Level 2 (3-4 marks): Most of the circuit is correct (e.g., minor error in meter placement). Describes how to vary the PD/current to get some readings and mentions calculating resistance. Level 1 (1-2 marks): Mentions some correct circuit components (e.g., lamp and power supply). Identifies that current and potential difference are measured, or mentions the resistance formula. Level 0 (0 marks): No response or no relevant physics.