2024 Edexcel AS Level Physics (8PH0) 模擬試題連答案詳解

First, we rearrange the equation for resistivity: \(\rho = \frac{R A}{l}\). The units of resistance \(R\) can be derived from potential difference divided by current: \(R = \frac{V}{I}\). Since potential difference is work done per unit charge, \(V = \frac{W}{Q} = \frac{W}{I t}\), the SI base units for potential difference are: \([V] = \frac{\text{kg m}^2 \text{s}^{-2}}{\text{A s}} = \text{kg m}^2 \text{s}^{-3} \text{A}^{-1}\). Consequently, the SI base units for resistance are: \([R] = \frac{\text{kg m}^2 \text{s}^{-3} \text{A}^{-1}}{\text{A}} = \text{kg m}^2 \text{s}^{-3} \text{A}^{-2}\). Substituting this back into the resistivity equation gives: \([\rho] = \frac{\text{kg m}^2 \text{s}^{-3} \text{A}^{-2} \times \text{m}^2}{\text{m}} = \text{kg m}^3 \text{s}^{-3} \text{A}^{-2}\).

1 mark: Correctly identifies the SI base units of resistivity as \(\text{kg m}^3 \text{s}^{-3} \text{A}^{-2}\). Reject options with incorrect powers of metres or seconds.

The initial kinetic energy is given by \(E_k = \frac{1}{2} m u^2\). At its maximum height, the vertical component of the velocity is zero, and the projectile travels only with its horizontal component of velocity, which remains constant: \(v_x = u \cos(60^\circ)\). Therefore, the kinetic energy at maximum height is: \(E_{\text{max}} = \frac{1}{2} m v_x^2 = \frac{1}{2} m (u \cos(60^\circ))^2 = \frac{1}{2} m u^2 \cos^2(60^\circ)\). The ratio of the kinetic energy at maximum height to the initial kinetic energy is: \(\frac{E_{\text{max}}}{E_k} = \cos^2(60^\circ) = (0.5)^2 = 0.25\).

1 mark: Correctly identifies the ratio of the kinetic energies as 0.25. Reject other numerical values.

At terminal velocity, the net force acting on the bead is zero. The forces acting upwards are the upthrust \(U = \rho_l V g\) and the viscous drag force \(F = 6 \pi \eta r v\). The force acting downwards is the weight of the bead \(W = \rho_s V g\), where \(V = \frac{4}{3} \pi r^3\) is the volume of the sphere. Setting the upward and downward forces equal: \(W = U + F \implies \rho_s V g = \rho_l V g + 6 \pi \eta r v\). Rearranging gives: \(6 \pi \eta r v = (\rho_s - \rho_l) V g\). Substituting \(V = \frac{4}{3} \pi r^3\): \(6 \pi \eta r v = (\rho_s - \rho_l) \frac{4}{3} \pi r^3 g\). Solving for \(v\): \(v = \frac{2 r^2 g (\rho_s - \rho_l)}{9 \eta}\).

1 mark: Correctly identifies the derivation of terminal velocity with the density difference term as \(\rho_s - \rho_l\).

The terminal potential difference is given by \(V = E - I r\), which can also be written in terms of the external resistance as \(E = V + I r = V + \frac{V r}{R}\). For the first case: \(E = 1.6 + \frac{1.6 r}{4.0} = 1.6 + 0.4 r\). For the second case: \(E = 1.8 + \frac{1.8 r}{9.0} = 1.8 + 0.2 r\). Equating the two expressions for \(E\): \(1.6 + 0.4 r = 1.8 + 0.2 r \implies 0.2 r = 0.2 \implies r = 1.0\ \Omega\).

1 mark: Correctly calculates the value of internal resistance as \(1.0\ \Omega\) by equating terms of the e.m.f. loop equation.

Einstein's photoelectric equation is \(h f = \phi + E_k\), which can be written as \(\frac{h c}{\lambda} = \phi + E_k\). When the wavelength is halved, the photon energy is doubled: \(E_{\text{photon}}' = \frac{h c}{\lambda / 2} = 2 \frac{h c}{\lambda}\). Substituting the initial relation into this gives: \(E_{\text{photon}}' = 2(\phi + E_k) = 2\phi + 2E_k\). The new maximum kinetic energy is: \(E_k' = E_{\text{photon}}' - \phi = (2\phi + 2E_k) - \phi = 2E_k + \phi\). Since the work function \(\phi\) is a positive physical constant, the new maximum kinetic energy \(E_k'\) must be greater than \(2 E_k\).

1 mark: Correctly identifies that the new maximum kinetic energy must be greater than \(2 E_k\) because of the constant positive value of the work function \(\phi\).

The elastic strain energy stored in a stretched wire is given by \(W = \frac{1}{2} F \Delta x\), where \(F\) is the force causing the extension. The Young modulus \(E\) is defined as \(E = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{\Delta x/L} = \frac{F L}{A \Delta x}\). Rearranging this for force gives \(F = \frac{E A \Delta x}{L}\). Substituting this expression for \(F\) into the work equation yields: \(W = \frac{1}{2} \left(\frac{E A \Delta x}{L}\right) \Delta x = \frac{E A (\Delta x)^2}{2L}\).

1 mark: Correctly identifies the formula for the stored elastic strain energy in terms of the given parameters.

According to Newton's second law of motion, force is equal to the rate of change of momentum: \(F = \frac{\Delta p}{\Delta t}\). The initial momentum is \(p_i = m v\) and the final momentum is \(p_f = -m u\). The change in momentum is \(\Delta p = p_f - p_i = -m u - m v = -m(u + v)\). The magnitude of this change in momentum is \(|\Delta p| = m(u + v)\). Thus, the magnitude of the average force is \(F = \frac{m(u + v)}{\Delta t}\).

1 mark: Correctly identifies the sum of the magnitudes of initial and final momentum to find the overall change in momentum, yielding option C.

The diffraction grating equation is \(d \sin \theta = n \lambda\). We are given: wavelength \(\lambda = 500\text{ nm} = 5.0 \times 10^{-7}\text{ m}\), order of maximum \(n = 2\), and diffraction angle \(\theta = 30^\circ\). Substituting these values into the equation: \(d \sin(30^\circ) = 2 \times (5.0 \times 10^{-7}\text{ m}) \implies 0.5 d = 1.0 \times 10^{-6}\text{ m} \implies d = 2.0 \times 10^{-6}\text{ m}\). The number of lines per millimetre \(N_{\text{mm}}\) is: \(N_{\text{mm}} = \frac{10^{-3}\text{ m}}{d} = \frac{10^{-3}\text{ m}}{2.0 \times 10^{-6}\text{ m}} = 500\text{ lines/mm}\).

1 mark: Correctly calculates \(d = 2.0 \times 10^{-6}\text{ m}\) and converts this to 500 lines per millimetre.

For part (a):
1. Identify the horizontal velocity component: \(v_x = 15\text{ m s}^{-1}\) (constant if we neglect air resistance).
2. Calculate the vertical velocity component \(v_y\) using the equations of motion: \(v_y^2 = u_y^2 + 2as\). Since \(u_y = 0\text{ m s}^{-1}\), \(v_y^2 = 2 \times 9.81\text{ m s}^{-2} \times 35\text{ m} = 686.7\text{ m}^2\text{ s}^{-2}\). Thus, \(v_y = \sqrt{686.7} \approx 26.2\text{ m s}^{-1}\).
3. Use Pythagoras' theorem to find the magnitude of the final velocity: \(v = \sqrt{v_x^2 + v_y^2} = \sqrt{15^2 + 26.2^2} = \sqrt{225 + 686.7} = \sqrt{911.7} \approx 30.2\text{ m s}^{-1}\).

For part (b):
1. Air resistance acts in the direction opposite to the stone's motion, opposing both vertical and horizontal components.
2. This resistive force does work against the motion of the stone, transferring kinetic energy to thermal energy of the surroundings.
3. Consequently, both the horizontal and vertical velocity components are reduced, resulting in a lower final velocity magnitude compared to the no-air-resistance case.

Part (a) [4 marks]:
- [1M] Selection and use of \(v^2 = u^2 + 2as\) with vertical values (\(a = 9.81\text{ m s}^{-2}\)).
- [1M] Correct calculation of vertical velocity component (\(v_y = 26.2\text{ m s}^{-1}\)).
- [1M] Application of Pythagoras' theorem to find resultant velocity.
- [1M] Correct final answer: \(30.2\text{ m s}^{-1}\) (accept \(30\text{ m s}^{-1}\) to 2 s.f.).

Part (b) [3.4 marks]:
- [1M] Identification that air resistance opposes the direction of motion / acts as a drag force.
- [1M] Explanation that work is done against air resistance, transferring kinetic energy to thermal energy.
- [1.4M] Concluding statement that both velocity components are reduced, leading to a smaller final velocity.

For part (a):
1. The potential difference across the thermistor is \(V_{th} = V_{total} - V_R = 9.0\text{ V} - 3.6\text{ V} = 5.4\text{ V}\).
2. The current \(I\) in the circuit can be calculated from the fixed resistor: \(I = \frac{V_R}{R} = \frac{3.6\text{ V}}{1500\text{ }\Omega} = 2.40 \times 10^{-3}\text{ A}\).
3. Since the components are in series, the same current flows through the thermistor. The resistance of the thermistor is \(R_{th} = \frac{V_{th}}{I} = \frac{5.4\text{ V}}{2.40 \times 10^{-3}\text{ A}} = 2250\text{ }\Omega = 2.25\text{ k}\Omega\).

For part (b):
1. As temperature increases, the charge carrier density in the NTC thermistor increases, causing its resistance to decrease.
2. The total resistance of the series circuit decreases, so the current \(I\) in the circuit increases (\(I = V / R_{total}\)).
3. Since the resistance \(R\) of the fixed resistor remains constant, the potential difference across it (\(V_R = I \times R\)) must increase.

Part (a) [4 marks]:
- [1M] Calculation of potential difference across the thermistor: \(5.4\text{ V}\).
- [1M] Calculation of circuit current: \(2.4 \times 10^{-3}\text{ A}\).
- [1M] Correct use of Ohm's law to determine thermistor resistance: \(R_{th} = V_{th} / I\).
- [1M] Correct final resistance: \(2250\text{ }\Omega\) (or \(2.25\text{ k}\Omega\) / \(2.3\text{ k}\Omega\)).

Part (b) [3.4 marks]:
- [1M] Stating that the resistance of an NTC thermistor decreases as temperature increases.
- [1M] Stating that total circuit resistance decreases and therefore circuit current increases.
- [1.4M] Concluding that because current increases and \(R\) of the fixed resistor is constant, \(V = IR\) increases (or explaining via potential divider ratio).

For part (a):
1. Calculate the energy of an incident photon using \(E = hf\):
\(E = (6.63 \times 10^{-34}\text{ J s}) \times (7.2 \times 10^{14}\text{ Hz}) = 4.77 \times 10^{-19}\text{ J}\).
2. Convert the work function of zinc from electronvolts to Joules:
\(\Phi = 4.3\text{ eV} \times (1.60 \times 10^{-19}\text{ J eV}^{-1}) = 6.88 \times 10^{-19}\text{ J}\).
3. Compare the photon energy \(E\) with the work function \(\Phi\):
Since \(E < \Phi\) (\(4.77 \times 10^{-19}\text{ J} < 6.88 \times 10^{-19}\text{ J}\)), the individual incident photons do not have enough energy to liberate an electron from the zinc surface. Therefore, photoelectric emission does not occur.

For part (b):
1. The threshold frequency \(f_0\) corresponds to the minimum frequency required for emission: \(h f_0 = \Phi\).
2. The maximum wavelength \(\lambda_0\) is related to this threshold frequency: \(\lambda_0 = \frac{c}{f_0} = \frac{hc}{\Phi}\).
3. Calculate \(\lambda_0\):
\(\lambda_0 = \frac{(6.63 \times 10^{-34}\text{ J s}) \times (3.00 \times 10^8\text{ m s}^{-1})}{6.88 \times 10^{-19}\text{ J}} = 2.89 \times 10^{-7}\text{ m}\) (or \(289\text{ nm}\)).

Part (a) [4 marks]:
- [1M] Use of \(E = hf\) to calculate photon energy (\(4.77 \times 10^{-19}\text{ J}\)).
- [1M] Conversion of work function to Joules (\(6.88 \times 10^{-19}\text{ J}\)).
- [1M] Comparison of photon energy and work function (\(E < \Phi\)).
- [1M] Clear concluding statement that emission does not occur because the energy of one photon is less than the work function.

Part (b) [3.4 marks]:
- [1M] Use of \(\Phi = \frac{hc}{\lambda_0}\) or finding threshold frequency first.
- [1M] Rearranging formula correctly to make \(\lambda_0\) the subject.
- [1.4M] Correct calculation of \(\lambda_0 = 2.89 \times 10^{-7}\text{ m}\) (accept \(2.9 \times 10^{-7}\text{ m}\)).

For part (a):
1. Calculate the tensile force (tension) \(F = mg = 4.5\text{ kg} \times 9.81\text{ m s}^{-2} = 44.15\text{ N}\).
2. Calculate the stress \(\sigma = \frac{F}{A} = \frac{44.15\text{ N}}{1.5 \times 10^{-7}\text{ m}^2} = 2.94 \times 10^8\text{ Pa}\).
3. Calculate the strain \(\epsilon = \frac{\Delta L}{L} = \frac{0.58 \times 10^{-3}\text{ m}}{2.4\text{ m}} = 2.42 \times 10^{-4}\).
4. Calculate the Young modulus \(E = \frac{\sigma}{\epsilon} = \frac{2.94 \times 10^8\text{ Pa}}{2.42 \times 10^{-4}} = 1.21 \times 10^{11}\text{ Pa}\) (or \(1.2 \times 10^{11}\text{ Pa}\)).

For part (b):
1. Elastic behaviour means the material returns to its original shape and length when the load is removed.
2. On an atomic scale, the tensile force pulls the atoms slightly further apart, increasing their separation against interatomic attractive forces.
3. This does work against the interatomic forces, which stores potential energy in the bonds between the atoms. When the force is removed, this stored elastic potential energy is released, returning the atoms to their original equilibrium positions.

Part (a) [4 marks]:
- [1M] Correct calculation of force \(F = 44.15\text{ N}\) (allow use of \(g = 9.8\text{ m s}^{-2}\)).
- [1M] Calculation of stress \(\sigma = 2.94 \times 10^8\text{ Pa}\) or strain \(\epsilon = 2.42 \times 10^{-4}\).
- [1M] Correct formula for Young modulus \(E = \frac{FL}{A\Delta L}\) and substitution of values.
- [1M] Correct final answer: \(1.2 \times 10^{11}\text{ Pa}\) (accept \(1.21 \times 10^{11}\text{ Pa}\)).

Part (b) [3.4 marks]:
- [1M] Definition of elastic behaviour (returns to original dimensions when force is removed).
- [1M] Explanation that work is done to change the separation of atoms against interatomic forces.
- [1.4M] Explanation that this work is stored as potential energy in the interatomic bonds.

For part (a):
1. At terminal velocity, the forces are balanced: Weight = Upthrust + Drag force.
\(W = m_s g = \rho_s V g\)
\(U = \rho_f V g\)
\(D = 6 \pi \eta r v\) (Stokes' Law)
2. Formulate the equation:
\((\rho_s - \rho_f) V g = 6 \pi \eta r v\)
Since the sphere has volume \(V = \frac{4}{3}\pi r^3\):
\((\rho_s - \rho_f) \frac{4}{3}\pi r^3 g = 6 \pi \eta r v \implies v = \frac{2 r^2 g (\rho_s - \rho_f)}{9 \eta}
3. Substitute the values:
\)v = \frac{2 \times (1.2 \times 10^{-3}\text{ m})^2 \times 9.81\text{ m s}^{-2} \times (7800\text{ kg m}^{-3} - 1260\text{ kg m}^{-3})}{9 \times 1.41\text{ Pa s}}\)
\(v = \frac{2 \times 1.44 \times 10^{-6} \times 9.81 \times 6540}{12.69} = \frac{0.1848}{12.69} \approx 0.0146\text{ m s}^{-1}\).

For part (b):
1. At a higher temperature, the viscosity \(\eta\) of the liquid (glycerol) decreases because the molecules have more kinetic energy and can slide past each other more easily, reducing cohesive forces.
2. Since terminal velocity \(v\) is inversely proportional to viscosity (\(v \propto \frac{1}{\eta}\)), a lower viscosity results in a smaller drag force at any given speed.
3. Therefore, the sphere must accelerate to a higher speed before the drag force plus upthrust equals the weight, leading to a higher terminal velocity.

Part (a) [4 marks]:
- [1M] Identification of force balance: Weight = Upthrust + Drag.
- [1M] Correct substitution of formulas leading to \(v = \frac{2 r^2 g (\rho_s - \rho_f)}{9 \eta}\).
- [1M] Correct substitution of values with consistent units (e.g. converting \(1.2\text{ mm}\) to meters).
- [1M] Correct final answer: \(0.0146\text{ m s}^{-1}\) (accept \(0.015\text{ m s}^{-1}\)).

Part (b) [3.4 marks]:
- [1M] Stating that higher temperature decreases the viscosity of a liquid.
- [1M] Linking the decrease in viscosity to a decrease in the drag force at a given velocity.
- [1.4M] Concluding that a higher terminal velocity is reached because the sphere must travel faster for the drag force to balance the net downward force.

For part (a):
1. Find the cross-sectional area \(A\) from the diameter \(d = 0.38 \times 10^{-3}\text{ m}\):
\(A = \frac{\pi d^2}{4} = \frac{\pi \times (0.38 \times 10^{-3}\text{ m})^2}{4} \approx 1.134 \times 10^{-7}\text{ m}^2\).
2. Calculate the resistivity \(\rho\) using \(\rho = \frac{RA}{L}\) with \(R = 4.20\text{ }\Omega\) and \(L = 0.850\text{ m}\):
\(\rho = \frac{4.20\text{ }\Omega \times 1.134 \times 10^{-7}\text{ m}^2}{0.850\text{ m}} \approx 5.60 \times 10^{-7}\text{ }\Omega\text{ m}\).

For part (b):
1. Percentage uncertainty in length \(L\):
\(\%\Delta L = \frac{0.1}{85.0} \times 100 \approx 0.118\%\).
2. Percentage uncertainty in resistance \(R\):
\(\%\Delta R = \frac{0.05}{4.20} \times 100 \approx 1.190\%\).
3. Percentage uncertainty in diameter \(d\):
\(\%\Delta d = \frac{0.01}{0.38} \times 100 \approx 2.632\%\).
4. Since \(A \propto d^2\), the percentage uncertainty in area \(A\) is twice the percentage uncertainty in \(d\):
\(\%\Delta A = 2 \times \%\Delta d = 2 \times 2.632\% \approx 5.264\%\).
5. Total percentage uncertainty in resistivity \(\rho\):
\(\%\Delta \rho = \%\Delta R + \%\Delta A + \%\Delta L = 1.190\% + 5.264\% + 0.118\% = 6.572\% \approx 6.6\%\).

Part (a) [4 marks]:
- [1M] Calculation of cross-sectional area \(A = 1.13 \times 10^{-7}\text{ m}^2\).
- [1M] Correct recall and rearrangement of \(\rho = RA/L\).
- [1M] Correct substitution of converted quantities (\(L = 0.850\text{ m}\), \(d = 3.8 \times 10^{-4}\text{ m}\)).
- [1M] Correct value for resistivity: \(5.6 \times 10^{-7}\text{ }\Omega\text{ m}\) (accept \(5.60 \times 10^{-7}\text{ }\Omega\text{ m}\)).

Part (b) [3.4 marks]:
- [1M] Calculation of percentage uncertainty in \(R\) (\(1.2\%\)) and \(L\) (\(0.12\%\)).
- [1M] Calculation of percentage uncertainty in \(d\) (\(2.6\%\)) and multiplying by 2 to get percentage uncertainty in area \(A\) (\(5.3\%\)).
- [1.4M] Addition of percentage uncertainties to give final value of \(6.6\%\) (accept \(6.5\%\) to \(6.7\%\)).

For part (a):
1. Calculate the mass per unit length \(\mu\) of the string:
\(\mu = \frac{m}{L} = \frac{1.3 \times 10^{-3}\text{ kg}}{0.65\text{ m}} = 2.0 \times 10^{-3}\text{ kg m}^{-1}\).
2. Calculate the wave speed \(v\) on the string using \(v = \sqrt{\frac{T}{\mu}}\):
\(v = \sqrt{\frac{85\text{ N}}{2.0 \times 10^{-3}\text{ kg m}^{-1}}} = \sqrt{42500} \approx 206.2\text{ m s}^{-1}\).
3. For the fundamental frequency, the wavelength \(\lambda\) is twice the length of the string: \(\lambda = 2L = 2 \times 0.65\text{ m} = 1.30\text{ m}\).
4. Calculate the fundamental frequency \(f = \frac{v}{\lambda}\):
\(f = \frac{206.2\text{ m s}^{-1}}{1.30\text{ m}} \approx 158.6\text{ Hz} \approx 159\text{ Hz}\).

For part (b):
1. The wave speed \(v\) is proportional to the square root of the tension \(T\) (\(v \propto \sqrt{T}\)).
2. Since the string length is constant, the wavelength of the fundamental mode remains unchanged (\(\lambda = 2L\)).
3. Because \(f = \frac{v}{\lambda}\), the fundamental frequency is also proportional to the square root of the tension (\(f \propto \sqrt{T}\)).
4. Therefore, doubling the tension increases the fundamental frequency by a factor of \(\sqrt{2} \approx 1.41\), resulting in a new frequency of \(159 \times 1.41 \approx 224\text{ Hz}\).

Part (a) [4 marks]:
- [1M] Calculation of mass per unit length \(\mu = 2.0 \times 10^{-3}\text{ kg m}^{-1}\).
- [1M] Use of \(v = \sqrt{T/\mu}\) to calculate wave speed \(v = 206\text{ m s}^{-1}\).
- [1M] Linking fundamental mode wavelength to string length: \(\lambda = 1.30\text{ m}\).
- [1M] Correct final frequency: \(159\text{ Hz}\) (accept \(158.6\text{ Hz}\) to \(160\text{ Hz}\)).

Part (b) [3.4 marks]:
- [1M] Stating that wave speed \(v\) is proportional to \(\sqrt{T}\).
- [1M] Explaining that the wavelength remains constant as the string length is unchanged.
- [1.4M] Concluding that fundamental frequency increases by a factor of \(\sqrt{2}\) (or about \(1.41\)) to \(224\text{ Hz}\).

Part (a): The student should use a micrometer screw gauge to measure the diameter of the wire. They should take measurements at several different points along the length of the wire and at different orientations at each point to account for any non-circularity. A mean diameter \( d \) is calculated, and the cross-sectional area \( A \) is calculated using the equation \( A = \frac{\pi d^2}{4} \).

Part (b): First, find the cross-sectional area: \( A = \frac{\pi (0.28 \times 10^{-3} \text{ m})^2}{4} = 6.16 \times 10^{-8} \text{ m}^2 \). The Young modulus is defined as \( E = \frac{\text{Stress}}{\text{Strain}} = \frac{F / A}{\Delta x / L} = \left(\frac{F}{\Delta x}\right) \frac{L}{A} \). The term \( \frac{F}{\Delta x} \) is the gradient of the graph. Substituting the values gives: \( E = (3.42 \times 10^3 \text{ N m}^{-1}) \times \frac{2.03 \text{ m}}{6.16 \times 10^{-8} \text{ m}^2} = 1.13 \times 10^{11} \text{ Pa} \), which is approximately \( 1.1 \times 10^{11} \text{ Pa} \).

Part (c): Find the individual percentage uncertainties:
- Percentage uncertainty in \( L = \frac{0.01}{2.03} \times 100\% = 0.49\% \).
- Percentage uncertainty in \( d = \frac{0.02}{0.28} \times 100\% = 7.14\% \).
- Since \( A \propto d^2 \), the percentage uncertainty in area \( A \) is \( 2 \times 7.14\% = 14.28\% \).
- Overall percentage uncertainty in \( E \) is the sum of the percentage uncertainties of the quantities: \( \%\Delta E = \%\Delta \text{gradient} + \%\Delta L + \%\Delta A = 3.0\% + 0.49\% + 14.28\% = 17.77\% \), which rounds to \( 18\% \).

Part (a) [3 Marks]:
- Correctly identifies the use of a micrometer screw gauge and measures at different positions and orientations along the wire (1)
- Explains how to find the mean diameter (1)
- Correctly identifies the formula \( A = \frac{\pi d^2}{4} \) to calculate cross-sectional area (1)

Part (b) [3 Marks]:
- Calculates correct cross-sectional area \( A = 6.16 \times 10^{-8} \text{ m}^2 \) (allow \( 6.2 \times 10^{-8} \text{ m}^2 \)) (1)
- Recalls and rearranges formula for Young modulus in terms of gradient: \( E = \text{gradient} \times \frac{L}{A} \) (1)
- Correctly substitutes values to yield \( 1.13 \times 10^{11} \text{ Pa} \) (or equivalent) and shows it is approximately \( 1.1 \times 10^{11} \text{ Pa} \) (1)

Part (c) [4 Marks]:
- Calculates percentage uncertainty in length \( L \) as \( 0.49\% \) (1)
- Calculates percentage uncertainty in diameter \( d \) as \( 7.14\% \) (1)
- Doubles percentage uncertainty of diameter to find uncertainty in area \( A \) as \( 14.28\% \) (1)
- Sums all percentage uncertainties to obtain \( 17.8\% \) or \( 18\% \) (1)

Part (a): Final velocity \( v = \frac{\text{length of card}}{\text{time}} = \frac{0.050 \text{ m}}{0.022 \text{ s}} = 2.27 \text{ m s}^{-1} \).
Percentage uncertainty in length of card \( = \frac{0.1}{5.0} \times 100\% = 2.0\% \).
Percentage uncertainty in time \( = \frac{0.001}{0.022} \times 100\% = 4.55\% \).
Total percentage uncertainty in \( v = 2.0\% + 4.55\% = 6.55\% \).
Absolute uncertainty in \( v = 2.27 \text{ m s}^{-1} \times 0.0655 = 0.15 \text{ m s}^{-1} \).
Thus, \( v = 2.27 \pm 0.15 \text{ m s}^{-1} \).

Part (b): Applying the equations of motion for constant acceleration, \( v^2 = u^2 + 2as \). Since the trolley is released from rest, the initial velocity \( u = 0 \), which simplifies the equation to \( v^2 = 2as \). Comparing this to the equation of a straight line through the origin, \( y = mx \), where \( y = v^2 \) and \( x = s \), the gradient of the graph, \( m \), must represent \( 2a \).

Part (c): From the graph gradient, \( 2a = 3.48 \text{ m s}^{-2} \implies a = 1.74 \text{ m s}^{-2} \).
The driving force down the slope is the component of weight parallel to the slope: \( F_{\text{drive}} = mg \sin\theta \).
Applying Newton's second law along the slope: \( mg \sin\theta - f = ma \), where \( f \) is the resistive force.
Rearranging for \( f \): \( f = m(g \sin\theta - a) \).
Substitute the values: \( f = 0.450 \text{ kg} \times (9.81 \text{ m s}^{-2} \times \sin(12^\circ) - 1.74 \text{ m s}^{-2}) \).
\( g \sin(12^\circ) = 9.81 \times 0.2079 = 2.04 \text{ m s}^{-2} \).
\( f = 0.450 \times (2.04 - 1.74) = 0.450 \times 0.30 = 0.135 \text{ N} \) (accept values in the range of \( 0.13 \text{ N} \) to \( 0.14 \text{ N} \)).

Part (a) [3 Marks]:
- Correct calculation of velocity \( v = 2.27 \text{ m s}^{-1} \) (1)
- Correct addition of percentage uncertainties to give \( 6.5\% \) or \( 6.6\% \) (1)
- Correct absolute uncertainty calculated as \( \pm 0.15 \text{ m s}^{-1} \) (1)

Part (b) [3 Marks]:
- References the equation of motion \( v^2 = u^2 + 2as \) (1)
- Explains that \( u = 0 \) simplifies the equation to \( v^2 = 2as \) (1)
- Compares with \( y = mx + c \) to demonstrate that plotting \( v^2 \) on the y-axis against \( s \) on the x-axis gives a straight line through the origin with a gradient equal to \( 2a \) (1)

Part (c) [4 Marks]:
- Correctly determines acceleration \( a = 1.74 \text{ m s}^{-2} \) from the gradient (1)
- Correctly identifies the component of weight parallel to the slope as \( mg \sin\theta \) (1)
- Sets up Newton's second law equation \( mg \sin\theta - f = ma \) or equivalent (1)
- Calculates the final resistive force as \( 0.135 \text{ N} \) (accept values from \( 0.13 \text{ N} \) to \( 0.14 \text{ N} \)) (1)

The formula for the critical angle \(\theta_c\) at a boundary between two media is given by \(\sin\theta_c = \frac{n_2}{n_1}\), where \(n_1\) is the refractive index of the denser medium (glass, \(1.62\)) and \(n_2\) is the refractive index of the rarer medium (liquid). Rearranging the formula for \(n_2\) gives: \(n_2 = n_1 \sin\theta_c = 1.62 \times \sin(58^\circ) \approx 1.62 \times 0.848 = 1.37\).

1 mark for the correct option (A).

The elastic strain energy stored in a stretched wire is given by \(\text{Energy} = \frac{1}{2} F \Delta x\). From the definition of Young modulus, \(E = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{\Delta x/L} = \frac{F L}{A \Delta x}\). Rearranging for \(F\) gives \(F = \frac{E A \Delta x}{L}\). Substituting this expression for \(F\) into the energy equation gives: \(\text{Energy} = \frac{1}{2} \left(\frac{E A \Delta x}{L}\right) \Delta x = \frac{E A (\Delta x)^2}{2L}\).

1 mark for the correct option (B).

According to Einstein's photoelectric equation, \(E_k = \frac{hc}{\lambda} - \Phi\), where \(\Phi\) is the work function of the metal. If the wavelength is halved, the energy of the incident photons doubles. The new maximum kinetic energy \(E_k'\) is: \(E_k' = \frac{hc}{\lambda/2} - \Phi = 2\left(\frac{hc}{\lambda}\right) - \Phi\). Since \(\frac{hc}{\lambda} = E_k + \Phi\), substituting this in gives: \(E_k' = 2(E_k + \Phi) - \Phi = 2E_k + \Phi\). Because the work function \(\Phi\) is a positive value, \(2E_k + \Phi > 2E_k\).

1 mark for the correct option (C).

The forces acting on a bead at terminal velocity are weight, upthrust, and viscous drag. Weight \(W = \frac{4}{3}\pi r^3 \rho_s g\) and upthrust \(U = \frac{4}{3}\pi r^3 \rho_f g\). Viscous drag is given by Stokes' Law: \(F_D = 6\pi \eta r v\). Equating forces: \(W - U = F_D \Rightarrow \frac{4}{3}\pi r^3 g (\rho_s - \rho_f) = 6\pi \eta r v\). This simplifies to show that the terminal velocity \(v\) is proportional to \(r^2\). Therefore, doubling the radius increases the terminal velocity by a factor of \(2^2 = 4\).

1 mark for the correct option (B).

For a string fixed at both ends, the frequency of the \(n\)-th harmonic is given by \(f_n = n f_1\), where \(f_1\) is the fundamental frequency (first harmonic). We are given that the frequency of the second harmonic (\(n=2\)) is \(120\text{ Hz}\), so \(f_2 = 2 f_1 = 120\text{ Hz}\), which gives \(f_1 = 60\text{ Hz}\). The fifth harmonic (\(n=5\)) frequency is \(f_5 = 5 f_1 = 5 \times 60\text{ Hz} = 300\text{ Hz}\).

1 mark for the correct option (B).

When combining uncertainties, fractional or percentage uncertainties add for multiplication and division. If a value is raised to a power, its percentage uncertainty is multiplied by that power. For resistivity \(\rho = \frac{\pi R d^2}{4 L}\), the percentage uncertainty is: \(\frac{\Delta \rho}{\rho} = \frac{\Delta R}{R} + 2\left(\frac{\Delta d}{d}\right) + \frac{\Delta L}{L}\). Substituting the values: \(\frac{\Delta \rho}{\rho} = 2\% + 2(1.5\%) + 1\% = 2\% + 3\% + 1\% = 6.0\%\).

1 mark for the correct option (C).

Tensile stress is defined as \(\sigma = \frac{F}{A}\), where \(F\) is the applied force (load) and \(A\) is the cross-sectional area. The load \(F\) is the same for both wires. The cross-sectional area of a wire is given by \(A = \frac{\pi d^2}{4}\). Since wire X has half the diameter of wire Y (\(d_X = 0.5 d_Y\)), its area is \(A_X = \frac{\pi (0.5 d_Y)^2}{4} = 0.25 A_Y\). The stress in wire X is therefore \(\sigma_X = \frac{F}{0.25 A_Y} = 4 \frac{F}{A_Y} = 4 \sigma_Y\). The ratio \(\frac{\sigma_X}{\sigma_Y} = 4\).

1 mark for the correct option (D).

The diffraction grating equation for the first-order maximum is \(d \sin\theta = \lambda\), which gives \(\sin\theta = \frac{\lambda}{d}\). The grating spacing \(d\) is inversely proportional to the number of lines per millimetre \(N\) (i.e., \(d \propto \frac{1}{N}\)). Therefore, \(\sin\theta\) is proportional to both \(\lambda\) and \(N\), meaning \(\sin\theta \propto \lambda N\). When the wavelength becomes \(1.2\lambda\) and the lines per millimetre becomes \(0.8N\), the new sine of the angle \(\sin\theta'\) is proportional to \((1.2\lambda)(0.8N) = 0.96 \lambda N\). Hence, \(\sin\theta' = 0.96 \sin\theta\).

1 mark for the correct option (B).

Part (a): The limit of proportionality is the maximum stress or force up to which Hooke's law is obeyed (stress is directly proportional to strain). The elastic limit is the maximum stress or force a material can experience and still return to its original length when the load is removed. Beyond the limit of proportionality, the relationship becomes non-linear but deformation can still be elastic. Beyond the elastic limit, plastic (permanent) deformation occurs. Part (b): Stress \( \sigma = \frac{F}{A} = \frac{45.0\text{ N}}{1.80 \times 10^{-7}\text{ m}^2} = 2.50 \times 10^8\text{ Pa} \). Strain \( \varepsilon = \frac{\Delta L}{L_0} = \frac{0.0150\text{ m}}{2.50\text{ m}} = 0.00600 \). Young modulus \( E = \frac{\sigma}{\varepsilon} = \frac{2.50 \times 10^8}{0.00600} = 4.17 \times 10^{10}\text{ Pa} \). Part (c): Breaking force \( F_{\text{break}} = \text{Breaking stress} \times A = 3.20 \times 10^8\text{ Pa} \times 1.80 \times 10^{-7}\text{ m}^2 = 57.6\text{ N} \). During plastic deformation, the polymer chains slide past one another, breaking intermolecular bonds. These chains do not return to their original configurations when the load is removed, resulting in permanent elongation.

Part (a) [3.0 marks total]: 1 mark for defining limit of proportionality as limit of Hooke's law. 1 mark for defining elastic limit as the point beyond which permanent deformation occurs. 1 mark for stating that beyond proportionality it may still be elastic but non-linear. Part (b) [3.0 marks total]: 1 mark for calculating stress. 1 mark for calculating strain. 1 mark for final Young modulus value \(4.17 \times 10^{10}\text{ Pa}\) (accept \(4.2 \times 10^{10}\text{ Pa}\)). Part (c) [3.6 marks total]: 1 mark for calculation of breaking force \(57.6\text{ N}\). 1.6 marks for description of polymer chains sliding past each other and permanent rearrangement/breaking of weak intermolecular bonds.

Part (a): The ray enters along the normal to the curved surface (angle of incidence is zero degrees at the curved surface). Therefore, no refraction (bending) occurs at this first boundary. Part (b): Refractive index \( n = \frac{\sin(i)}{\sin(r)} = \frac{\sin(48.0^\circ)}{\sin(29.5^\circ)} = \frac{0.7431}{0.4924} = 1.51 \). Speed of light \( v = \frac{c}{n} = \frac{3.00 \times 10^8\text{ m s}^{-1}}{1.509} = 1.99 \times 10^8\text{ m s}^{-1} \). Part (c): Critical angle \( \theta_c = \sin^{-1}\left(\frac{1}{n}\right) = \sin^{-1}\left(\frac{1}{1.51}\right) = 41.5^\circ \). Since the new angle of incidence within the glass (\(45.0^\circ\)) is greater than the critical angle (\(41.5^\circ\)), total internal reflection occurs. No refracted ray exists in the air, and all light is reflected back into the glass obeying the law of reflection (at \(45.0^\circ\)).

Part (a) [2.0 marks total]: 1 mark for stating the ray is parallel to the normal / perpendicular to the boundary. 1 mark for concluding the angle of refraction is zero. Part (b) [4.0 marks total]: 1 mark for Snell's law formula. 1 mark for refractive index value \(1.51\). 1 mark for wave speed formula. 1 mark for speed value \(1.99 \times 10^8\text{ m s}^{-1}\). Part (c) [3.6 marks total]: 1 mark for critical angle equation. 1 mark for calculating \(41.5^\circ\). 1.6 marks for identifying total internal reflection, stating no light refracts, and all light reflects at \(45.0^\circ\).

Part (a): Stokes' law is valid for: 1. Laminar (non-turbulent) flow of fluid around the sphere. 2. A perfectly spherical object. 3. A fluid of infinite extent (or negligible wall effects). Part (b): At terminal velocity, three forces act on the sphere: weight (\(W\)) acting downwards, upthrust (\(U\)) acting upwards, and viscous drag (\(F_d\)) acting upwards. The balanced force equation is: \( W = U + F_d \). Part (c): Volume of sphere \( V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (2.00 \times 10^{-3}\text{ m})^3 = 3.351 \times 10^{-8}\text{ m}^3 \). Weight \( W = \rho_s V g = 7800 \times 3.351 \times 10^{-8} \times 9.81 = 2.564 \times 10^{-3}\text{ N} \). Upthrust \( U = \rho_f V g = 920 \times 3.351 \times 10^{-8} \times 9.81 = 3.024 \times 10^{-4}\text{ N} \). Viscous drag \( F_d = W - U = 2.262 \times 10^{-3}\text{ N} \). From Stokes' law: \( F_d = 6\pi\eta r v \), so \( \eta = \frac{F_d}{6\pi r v} = \frac{2.262 \times 10^{-3}}{6\pi \times (2.00 \times 10^{-3}) \times 0.150} = 0.400\text{ Pa s} \) (or \(0.400\text{ N s m}^{-2}\)).

Part (a) [2.0 marks total]: 1 mark for mentioning laminar flow. 1 mark for mentioning spherical body / small speed / negligible wall effects. Part (b) [3.6 marks total]: 1.6 marks for identifying all three forces (weight downwards, upthrust upwards, drag upwards). 2 marks for stating the equilibrium condition: \(W = U + F_d\). Part (c) [4.0 marks total]: 1 mark for volume calculation. 1 mark for expressing viscous drag force \(F_d = V g (\rho_s - \rho_f)\). 1 mark for substituting values into Stokes' law. 1 mark for final viscosity \(0.400\text{ Pa s}\) (accept range \(0.395\) to \(0.405\)).

Part (a): A wave is produced by the generator and travels along the wire. It reflects at the fixed boundaries. The incident and reflected waves travel in opposite directions with the same frequency and amplitude, superposing. Nodes are formed by destructive interference (out of phase) and antinodes are formed by constructive interference (in phase). Part (b): Wave speed \( v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{18.0\text{ N}}{1.50 \times 10^{-3}\text{ kg m}^{-1}}} = \sqrt{12000} = 109.54\text{ m s}^{-1} \). For the third harmonic, the length of the wire contains \(1.5\) wavelengths: \( L = \frac{3\lambda}{2} \Rightarrow \lambda = \frac{2L}{3} = \frac{2 \times 1.20\text{ m}}{3} = 0.800\text{ m} \). Frequency \( f = \frac{v}{\lambda} = \frac{109.54}{0.800} = 136.9\text{ Hz} \approx 137\text{ Hz} \). Part (c): Since wave speed \( v \propto \sqrt{T} \) and \( f = \frac{v}{\lambda} \), where wavelength \( \lambda \) remains constant (determined by the wire length), the frequency is proportional to \( \sqrt{T} \). If tension is doubled, the speed increases by a factor of \( \sqrt{2} \), so the frequency increases by a factor of \( \sqrt{2} \) (new frequency is \(136.9 \times \sqrt{2} = 194\text{ Hz}\)).

Part (a) [3.6 marks total]: 1 mark for mentioning reflection at the boundary. 1 mark for stating that incident and reflected waves superpose/interfere. 1.6 marks for explaining nodes (destructive, out of phase) and antinodes (constructive, in phase). Part (b) [3.0 marks total]: 1 mark for wave speed calculation \(110\text{ m s}^{-1}\). 1 mark for relating wavelength to length for 3rd harmonic (\(\lambda = 0.80\text{ m}\)). 1 mark for frequency \(137\text{ Hz}\) (accept \(136.9\text{ Hz}\)). Part (c) [3.0 marks total]: 1 mark for relating speed to tension (\(v \propto \sqrt{T}\)). 1 mark for stating wavelength is constant. 1 mark for concluding frequency increases by factor of \(\sqrt{2}\) / to \(194\text{ Hz}\).

Part (a): According to wave theory, wave energy is delivered continuously. A wave of any frequency would eventually deliver enough energy to cause electron emission if left long enough. In the photon model, light consists of packets of energy (photons). The interaction between a photon and an electron is a one-to-one event. If the energy of a single photon \( E = hf \) is less than the work function \( \Phi \), no emission occurs, regardless of intensity, defining a threshold frequency \( f_0 = \frac{\Phi}{h} \). Part (b): Energy of incident photon \( E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34}\text{ J s} \times 3.00 \times 10^8\text{ m s}^{-1}}{220 \times 10^{-9}\text{ m}} = 9.04 \times 10^{-19}\text{ J} \). Work function in Joules \( \Phi = 4.30 \times 1.60 \times 10^{-19}\text{ J} = 6.88 \times 10^{-19}\text{ J} \). Maximum kinetic energy \( E_{\text{k,max}} = hf - \Phi = 9.04 \times 10^{-19} - 6.88 \times 10^{-19} = 2.16 \times 10^{-19}\text{ J} \). Part (c): Stopping potential \( V_s \) is given by \( e V_s = E_{\text{k,max}} \). Therefore, \( V_s = \frac{2.16 \times 10^{-19}\text{ J}}{1.60 \times 10^{-19}\text{ C}} = 1.35\text{ V} \).

Part (a) [3.6 marks total]: 1.6 marks for explaining why wave theory predicts continuous buildup of energy (and hence no threshold). 2 marks for explaining the one-to-one photon-electron interaction and relating photon energy to work function. Part (b) [3.0 marks total]: 1 mark for calculating photon energy in Joules. 1 mark for converting work function to Joules. 1 mark for final kinetic energy \(2.16 \times 10^{-19}\text{ J}\) (accept \(2.2 \times 10^{-19}\text{ J}\)). Part (c) [3.0 marks total]: 1.5 marks for equating maximum kinetic energy to \(e V_s\). 1.5 marks for final stopping potential \(1.35\text{ V}\).

Part (a)
- According to the photon model of light, electromagnetic radiation consists of packets of energy called photons, where energy \(E = hf\).
- The interaction between photons and surface electrons is a one-to-one process.
- If the frequency of the radiation is below the threshold frequency (which is the case for visible light), the energy of a single photon is less than the work function of the material, so no electrons can be emitted.
- Increasing the intensity of visible light only increases the number of photons per second, not the energy of individual photons, so still no emission occurs. UV photons have higher frequency, hence enough energy to overcome the work function.

Part (b)(i)
- Convert work function to Joules: \(\Phi = 5.0\text{ eV} = 5.0 \times 1.60 \times 10^{-19}\text{ J} = 8.0 \times 10^{-19}\text{ J}\).
- Use \(\Phi = h f_0\):
\(f_0 = \frac{8.0 \times 10^{-19}}{6.63 \times 10^{-34}} = 1.21 \times 10^{15}\text{ Hz}\) (accept \(1.2 \times 10^{15}\text{ Hz}\)).

Part (b)(ii)
- Calculate energy of the incident UV photon:
\(E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{1.80 \times 10^{-7}} = 1.105 \times 10^{-18}\text{ J}\).
- Use Einstein's photoelectric equation:
\(h f = \Phi + E_{k(\text{max})}\)
\(E_{k(\text{max})} = E - \Phi = 1.105 \times 10^{-18}\text{ J} - 8.0 \times 10^{-19}\text{ J} = 3.05 \times 10^{-19}\text{ J}\) (accept \(3.1 \times 10^{-19}\text{ J}\)).

Part (c)
- As photoelectrons are emitted, the dust particles lose negative charge and acquire a net positive charge.
- The lunar surface itself also becomes positively charged by the same mechanism, leading to electrostatic repulsion that overcomes gravity and causes levitation.

Part (a) (Max 4 marks):
- (MP1) reference to photon model / energy of photon depends on frequency or \(E = hf\). [1]
- (MP2) emission is a 1-to-1 interaction between a photon and an electron. [1]
- (MP3) threshold frequency defined, or state that visible light photons have energy less than the work function \(\Phi\). [1]
- (MP4) statement that increasing intensity of visible light only increases number of photons, not energy of each photon / UV photons have higher frequency so energy exceeds \(\Phi\). [1]

Part (b)(i) (Max 2 marks):
- (MP1) correct conversion of \(5.0\text{ eV}\) to Joules (\(8.0 \times 10^{-19}\text{ J}\)). [1]
- (MP2) correct calculation of frequency \(1.2 \times 10^{15}\text{ Hz}\). [1]

Part (b)(ii) (Max 4 marks):
- (MP1) Use of \(E = hc/\lambda\) with correct substitution. [1]
- (MP2) Obtains photon energy \(1.1 \times 10^{-18}\text{ J}\). [1]
- (MP3) Correct subtraction of work function in Joules. [1]
- (MP4) Correct value for kinetic energy of \(3.1 \times 10^{-19}\text{ J}\) or \(3.05 \times 10^{-19}\text{ J}\). [1]

Part (c) (Max 2 marks):
- (MP1) Dust particles lose electrons and become positively charged. [1]
- (MP2) Repelled by other positively charged dust particles or by the positively charged surface of the Moon. [1]

Part (a)
- The critical angle \(\theta_c\) is given by \(\sin \theta_c = \frac{n_2}{n_1}\), where \(n_1\) is the refractive index of glass and \(n_2\) is the refractive index of the outer medium.
- Since the refractive index of water (\(n \approx 1.33\)) is higher than that of air (\(n \approx 1.00\)), the critical angle increases when water is present on the outer surface of the glass.
- The angle of incidence of the infrared beam is designed to be greater than the glass-air critical angle but less than the glass-water critical angle.
- Therefore, when dry, total internal reflection occurs and all light reaches the LDR. When wet, the angle of incidence is less than the new critical angle, so refraction occurs, light escapes into the water droplets, and less light is reflected to the LDR.

Part (b)(i)
- For the dry windshield:
\(R_{\text{LDR}} = 400\ \Omega\)
\(V_{\text{out}} = V_{\text{in}} \times \frac{R_{\text{LDR}}}{R + R_{\text{LDR}}} = 9.0\text{ V} \times \frac{400\ \Omega}{1200\ \Omega + 400\ \Omega} = 9.0 \times 0.25 = 2.25\text{ V}\).
This is approximately \(2.3\text{ V}\).
- For the wet windshield:
\(R_{\text{LDR}} = 2.8\text{ k}\Omega = 2800\ \Omega\)
\(V_{\text{out}} = 9.0\text{ V} \times \frac{2800\ \Omega}{1200\ \Omega + 2800\ \Omega} = 9.0 \times 0.70 = 6.3\text{ V}\).
This is exactly \(6.3\text{ V}\).

Part (b)(ii)
- Set \(V_{\text{out}} = 5.0\text{ V}\):
\(5.0\text{ V} = 9.0\text{ V} \times \frac{R_{\text{LDR}}}{1200\ \Omega + R_{\text{LDR}}}\)
- Multiply through by \((1200 + R_{\text{LDR}})\):
\(5.0 \times 1200 + 5.0 R_{\text{LDR}} = 9.0 R_{\text{LDR}}\)
\(6000 = 4.0 R_{\text{LDR}}\)
\(R_{\text{LDR}} = 1500\ \Omega = 1.5\text{ k}\Omega\).

Part (a) (Max 4 marks):
- (MP1) use of \(\sin \theta_c = n_2 / n_1\) or statement that critical angle depends on outer medium's refractive index. [1]
- (MP2) state that refractive index of water is greater than air, so critical angle increases when wet. [1]
- (MP3) state that angle of incidence was larger than the dry critical angle (causing TIR), but becomes smaller than the wet critical angle. [1]
- (MP4) refraction occurs / light escapes into water, resulting in lower intensity reflected to LDR. [1]

Part (b)(i) (Max 4 marks):
- (MP1) Uses potential divider formula with dry resistance. [1]
- (MP2) Obtains \(2.25\text{ V}\) (shown as \(\approx 2.3\text{ V}\)). [1]
- (MP3) Uses potential divider formula with wet resistance. [1]
- (MP4) Obtains \(6.3\text{ V}\). [1]

Part (b)(ii) (Max 4 marks):
- (MP1) Set up equation \(5.0 = 9.0 \times \frac{R_{\text{LDR}}}{1200 + R_{\text{LDR}}}\) (or alternative via current: \(I = V / R = (9.0 - 5.0)/1200 = 3.33 \times 10^{-3}\text{ A}\)). [1]
- (MP2) Correct algebraic rearrangement step (e.g., \(6000 + 5 R = 9 R\) or \(R_{\text{LDR}} = 5.0 / I\)). [1]
- (MP3) Correct calculation showing \(4.0 R_{\text{LDR}} = 6000\). [1]
- (MP4) Final answer of \(1.5\text{ k}\Omega\) or \(1500\ \Omega\) (with units). [1]