2025 HKDSE Physics Answers & Marking Scheme

First, find the wavelength: \(\lambda = \frac{v}{f} = \frac{12}{10} = 1.2\text{ m}\). The phase difference \(\Delta \phi\) is given by: \(\Delta \phi = \frac{\Delta x}{\lambda} \times 2\pi = \frac{0.6}{1.2} \times 2\pi = \pi\text{ rad}\).

Award 1 mark for the correct option (C). Correct calculation of wavelength (1.2 m) and phase difference (pi rad).

The wave speed on a string is \(v = \sqrt{T/\mu}\). When tension \(T\) is quadrupled, the speed becomes \(v' = 2v\). Initially, \(f_3 = 3 f_1 = 3 \frac{v}{2L}\), which means \(f_1 = \frac{1}{3}f_3\). With the new speed \(v'\), the new fundamental frequency is \(f_1' = \frac{v'}{2L} = \frac{2v}{2L} = 2 f_1 = \frac{2}{3} f_3\).

Award 1 mark for the correct option (C). Direct deduction using wave speed relation to tension and frequency formula for fixed strings.

When \(S\) is open, the total resistance is \(R_{\text{open}} = (R \parallel R) + R = 0.5R + R = 1.5R\). When \(S\) is closed, the third resistor is short-circuited, so \(R_{\text{closed}} = R \parallel R = 0.5R\). The ratio is \(\frac{R_{\text{open}}}{R_{\text{closed}}} = \frac{1.5R}{0.5R} = 3\).

Award 1 mark for the correct option (C). Correctly determines resistance when switch is open (1.5R) and closed (0.5R) and calculates ratio.

Resistance of bulb \(X\): \(R_X = \frac{V^2}{P_X} = \frac{12^2}{24} = 6\ \Omega\). Resistance of bulb \(Y\): \(R_Y = \frac{V^2}{P_Y} = \frac{12^2}{12} = 12\ \Omega\). The series equivalent resistance is \(R_{\text{total}} = R_X + R_Y = 18\ \Omega\). The total power consumed is \(P = \frac{V^2}{R_{\text{total}}} = \frac{12^2}{18} = 8\text{ W}\).

Award 1 mark for the correct option (B). Calculates resistance of each bulb and obtains the total power correctly.

Let the vertices of the prism be \(A\) (\(30^\circ\)), \(B\) (\(90^\circ\)), and \(C\) (\(60^\circ\)). The face opposite to the \(60^\circ\) angle is \(AB\). Since the light ray enters normally through \(AB\), it travels straight horizontally towards the hypotenuse \(AC\). The angle of incidence at the boundary \(AC\) is equal to \(30^\circ\). For total internal reflection, \(\theta_i \ge \theta_c \Rightarrow 30^\circ \ge \theta_c \Rightarrow \sin(30^\circ) \ge \frac{1}{n} \Rightarrow n \ge 2\). Therefore, the minimum refractive index is \(2.00\).

Award 1 mark for the correct option (D). Recognizes the angle of incidence is 30 degrees and applies the formula for critical angle correctly.

Let \(u\) and \(v\) be the object and image distances. Given \(u + v = 4.5f \Rightarrow v = 4.5f - u\). Using the lens formula: \(\frac{1}{u} + \frac{1}{4.5f - u} = \frac{1}{f}\), we get \(2u^2 - 9fu + 9f^2 = 0\), which factors to \((2u - 3f)(u - 3f) = 0\). So, \(u = 1.5f\) (which gives \(v = 3f\), \(m = v/u = 2\)) or \(u = 3f\) (which gives \(v = 1.5f\), \(m = v/u = 0.5\)).

Award 1 mark for the correct option (B). Formulates the quadratic equation and solves for object distance, then finds the corresponding magnifications.

The induced electromotive force (EMF) is \(\mathcal{E} = B L v = 0.8 \times 0.5 \times 4 = 1.6\text{ V}\). The induced current is \(I = \frac{\mathcal{E}}{R} = \frac{1.6}{2} = 0.8\text{ A}\). The magnetic force is \(F = B I L = 0.8 \times 0.8 \times 0.5 = 0.32\text{ N}\).

Award 1 mark for the correct option (B). Calculates induced EMF, current, and magnetic force correctly using the equations \(E = BLv\) and \(F = BIL\).

Since the magnetic force \(\vec{F} = q(\vec{v} \times \vec{B})\) is always perpendicular to the direction of motion (velocity \(\vec{v}\)), no work is done by the magnetic force on the proton. Therefore, statement (1) is correct. Since no work is done, the speed remains constant (statement 2 is incorrect) and the kinetic energy remains constant (statement 3 is correct).

Award 1 mark for the correct option (B). Analyzes the work done by magnetic forces and its consequences on speed and kinetic energy correctly.

According to Kepler's Third Law, \(T^2 \propto R^3\). Given \(\frac{T_A}{T_B} = 8\), we have \(\left(\frac{R_A}{R_B}\right)^3 = \left(\frac{T_A}{T_B}\right)^2 = 8^2 = 64 \Rightarrow \frac{R_A}{R_B} = 4\). The orbital speed is given by \(v = \sqrt{\frac{GM}{R}}\), so \(\frac{v_A}{v_B} = \sqrt{\frac{R_B}{R_A}} = \sqrt{\frac{1}{4}} = \frac{1}{2}\).

Award 1 mark for the correct option (B). Correct application of Kepler's Third Law to find radius ratio and orbital speed formula to find velocity ratio.

The energy of the emitted photon is given by \(\Delta E = \frac{hc}{\lambda}\). For hydrogen atom transitions: \(\Delta E_{3 \to 1} = E_0 \left(1 - \frac{1}{3^2}\right) = \frac{8}{9} E_0\), and \(\Delta E_{2 \to 1} = E_0 \left(1 - \frac{1}{2^2}\right) = \frac{3}{4} E_0\). Since \(\lambda \propto \frac{1}{\Delta E}\), the ratio is \(\frac{\lambda_1}{\lambda_2} = \frac{\Delta E_{2 \to 1}}{\Delta E_{3 \to 1}} = \frac{3/4}{8/9} = \frac{27}{32}\).

Award 1 mark for the correct option (B). Calculates energy differences and wavelengths correctly, then computes the accurate ratio.

The wave speed on a string is given by \(v = \sqrt{T/\mu}\). When tension \(T\) is quadrupled, the wave speed doubles (\(v_{new} = 2v\)). The initial frequency is the third harmonic, so \(f = 3 f_1\), which means the initial fundamental frequency is \(f_1 = f/3\). Since the fundamental frequency is \(f_1 = v / (2L)\), doubling the wave speed doubles the fundamental frequency. Therefore, the new fundamental frequency is \(2 f_1 = 2f/3\).

Award 1 mark for selecting correct option B. Method: Identify relations between wave speed, tension, and harmonics to deduce the ratio.

For Circuit 1, equivalent resistance \(R_{eq1} = R + R/2 = 1.5R\), so \(P_1 = V^2 / (1.5R) = 2V^2 / (3R)\). For Circuit 2, equivalent resistance \(R_{eq2} = (2R \times R) / (2R + R) = 2R/3\), so \(P_2 = V^2 / (2R/3) = 3V^2 / (2R)\). The ratio is \(P_2 / P_1 = (3/2) / (2/3) = 9/4\).

Award 1 mark for selecting correct option C. Method: Compute equivalent resistances for both configurations and find power ratio.

The critical angle \(\theta_c\) for total internal reflection at the glass-liquid interface is \(60^\circ\). Using Snell's law at critical condition: \(n_{glass} \sin \theta_c = n_{liquid}\), we get \(1.50 \times \sin 60^\circ = n_{liquid}\). Thus, \(n_{liquid} = 1.50 \times 0.866 = 1.30\).

Award 1 mark for selecting correct option C. Method: Apply critical angle equation for total internal reflection between two media.

The induced electromotive force in the loop as it is pulled out is \(\mathcal{E} = BLv\). The induced current is \(I = \mathcal{E} / R = BLv / R\). The magnetic force acting on the wire segment inside the magnetic field is \(F_B = I L B = B^2 L^2 v / R\). To maintain a constant speed, the external force must balance this magnetic force, so \(F_{ext} = B^2 L^2 v / R\).

Award 1 mark for selecting correct option C. Method: Relate induced emf, current, magnetic force, and force balance.

For statement (1): Orbital speed \(v = \sqrt{GM/r}\), so \(v_X / v_Y = \sqrt{r_Y / r_X} = 1 / \sqrt{2}\) (Correct). For statement (2): Kepler's third law states \(T^2 \propto r^3\), so \(T_X / T_Y = (r_X / r_Y)^{3/2} = 2^{3/2} = 2\sqrt{2}\) (Correct). For statement (3): Kinetic energy \(K = \frac{1}{2}mv^2 \propto v^2\). Since they have the same mass, \(K_X / K_Y = (v_X / v_Y)^2 = 1/2\) (Correct). Hence, all three statements are correct.

Award 1 mark for selecting correct option D. Method: Analyze orbital speed, Kepler's third law, and kinetic energy formulas.

The energy of the emitted photon is equal to the difference between the two energy levels: \(hf = E_i - E_f\). For transition \(3 \to 1\): \(hf_1 = -13.6 \times (1/9 - 1) = 13.6 \times 8/9\). For transition \(2 \to 1\): \(hf_2 = -13.6 \times (1/4 - 1) = 13.6 \times 3/4\). Thus, \(f_1 / f_2 = (8/9) / (3/4) = 32/27\).

Award 1 mark for selecting correct option C. Method: Apply energy level transition formula to calculate the photon frequency ratio.

Let \(T_f\) be the final equilibrium temperature. From conservation of energy: heat gained by water = heat lost by metal. \(m_w c_w (T_f - T_w) = m_m c_m (T_m - T_f)\). Substituting the values: \(0.20 \times 4200 \times (T_f - 20) = 0.50 \times 400 \times (80 - T_f)\). This simplifies to \(840 (T_f - 20) = 200 (80 - T_f) \implies 21(T_f - 20) = 5(80 - T_f)\). Thus, \(26 T_f = 420 + 400 = 820 \implies T_f \approx 31.5^\circ\text{C}\).

Award 1 mark for selecting correct option A. Method: Set up heat exchange equation and solve for final temperature.

Power is the rate of doing work. Since the power \(P\) is constant, the work done on the car in time \(t\) is \(W = Pt\). By the work-energy theorem, since the car starts from rest, this work equals its kinetic energy: \(Pt = \frac{1}{2}mv^2\). Solving for \(v\) yields \(v = \sqrt{\frac{2Pt}{m}}\).

Award 1 mark for selecting correct option C. Method: Apply work-energy theorem with constant power to relate speed and time.

Let East be the positive x-direction and North be the positive y-direction. The total initial momentum vector is \(\vec{p} = mu\hat{i} + 2mu\hat{j}\). The magnitude of the total momentum is \(p = \sqrt{(mu)^2 + (2mu)^2} = \sqrt{5}mu\). Since total mass after collision is \(M = 3m\), by conservation of momentum, the final speed \(v\) satisfies \(Mv = p \implies 3mv = \sqrt{5}mu \implies v = \frac{\sqrt{5}}{3}u\).

Award 1 mark for selecting correct option B. Method: Use 2D momentum conservation vector addition and divide by total mass.

First, convert temperatures to Kelvin: \(T_1 = 27 + 273 = 300\text{ K}\) and \(T_2 = 327 + 273 = 600\text{ K}\). The absolute temperature has doubled. For statement (1): The r.m.s. speed is proportional to \(\sqrt{T}\), so it increases by a factor of \(\sqrt{2}\), not 2 (Incorrect). For statement (2): For constant volume, pressure \(P \propto T\), so pressure doubles (Correct). For statement (3): The average kinetic energy is directly proportional to absolute temperature \(T\), so it doubles (Correct). Thus, (2) and (3) only are correct.

Award 1 mark for selecting correct option D. Method: Convert temperature to Kelvin and evaluate gas kinetic theory relations.

For statement (1): At maximum displacement, the velocity of the particle is zero. Hence, the particle at \(x = 2\text{ cm}\) is momentarily at rest. Statement (1) is correct.
For statement (2): The distance between the two particles is \(0.5\lambda\). Since a distance of \(\lambda\) corresponds to a phase difference of \(2\pi\text{ rad}\), a distance of \(0.5\lambda\) corresponds to \(\pi\text{ rad}\). Statement (2) is correct.
For statement (3): The wave travels to the right (positive \(x\)-direction). The particle at \(x = 2 + 0.25\lambda\text{ cm}\) is at a distance of a quarter-wavelength to the right of the wave crest. Since the wave crest propagates to the right, this particle is about to be reached by the crest and must move upwards (positive \(y\)-direction) to meet the upcoming peak. Statement (3) is correct.
Therefore, (1), (2) and (3) are all correct.

Award 1 mark for the correct option D. No marks are awarded for incorrect options or missing working.

Since \(n_X > n_Y\), total internal reflection occurs when the angle of incidence \(\theta\) is greater than or equal to the critical angle \(\theta_c\).
Just before total internal reflection occurs, the angle of incidence \(\theta\) is extremely close to \(\theta_c = \arcsin(n_Y / n_X)\), and the angle of refraction \(r\) approaches \(90^\circ\).
Therefore, the maximum possible angle of deviation is:
\(d_{\text{max}} = 90^\circ - \theta_c = 90^\circ - \arcsin(n_Y / n_X)\).

Award 1 mark for the correct option A.

1. In series connection:
Equivalent resistance of the external circuit: \(R_s = 3R\).
The current is \(I_s = \frac{\mathcal{E}}{R_s + r} = \frac{\mathcal{E}}{3R + R} = \frac{\mathcal{E}}{4R}\).
The power dissipated in the external circuit: \(P_s = I_s^2 R_s = \left(\frac{\mathcal{E}}{4R}\right)^2 (3R) = \frac{3\mathcal{E}^2}{16R}\).

2. In parallel connection:
Equivalent resistance of the external circuit: \(R_p = \frac{R}{3}\).
The current is \(I_p = \frac{\mathcal{E}}{R_p + r} = \frac{\mathcal{E}}{\frac{R}{3} + R} = \frac{\mathcal{E}}{\frac{4}{3}R} = \frac{3\mathcal{E}}{4R}\).
The power dissipated in the external circuit: \(P_p = I_p^2 R_p = \left(\frac{3\mathcal{E}}{4R}\right)^2 \left(\frac{R}{3}\right) = \frac{9\mathcal{E}^2}{16R^2} \cdot \frac{R}{3} = \frac{3\mathcal{E}^2}{16R}\).

Therefore, the ratio is \(P_s / P_p = 1\).

Award 1 mark for correct answer A.

When the bottom edge of the loop enters the magnetic field at a constant velocity \(v\), an induced e.m.f. \(\mathcal{E} = B w v\) is generated in the loop.
The induced current flowing through the loop is \(I = \frac{\mathcal{E}}{R} = \frac{B w v}{R}\).
The upward magnetic force acting on the bottom edge of the loop is \(F_B = B I w = B \left(\frac{B w v}{R}\right) w = \frac{B^2 w^2 v}{R}\).
Since the loop falls at terminal velocity, the magnetic force balances the gravitational force:
\(F_B = mg \implies \frac{B^2 w^2 v}{R} = mg \implies v = \frac{mgR}{B^2 w^2}\).

Award 1 mark for correct answer B.

1. The kinetic energy of the circular orbit is given by \(K = \frac{GMm}{2r}\).
If \(r\) decreases by \(\Delta r\), the change in kinetic energy is:
\(\Delta K \approx \frac{dK}{dr} (-\Delta r) = -\frac{GMm}{2r^2} (-\Delta r) = \frac{GMm\Delta r}{2r^2}\) (an increase).
2. The potential energy of the orbit is given by \(U = -\frac{GMm}{r}\).
If \(r\) decreases by \(\Delta r\), the change in potential energy is:
\(\Delta U \approx \frac{dU}{dr} (-\Delta r) = \frac{GMm}{r^2} (-\Delta r) = -\frac{GMm\Delta r}{r^2}\) (a decrease by \(\frac{GMm\Delta r}{r^2}\)).
Thus, kinetic energy increases by \(\frac{GMm\Delta r}{2r^2}\) and potential energy decreases by \(\frac{GMm\Delta r}{r^2}\).

Award 1 mark for correct answer A.

According to the Rydberg formula or energy levels of hydrogen atom, the transition energy is \(\Delta E = h c / \lambda \propto \left(\frac{1}{n_{\text{final}}^2} - \frac{1}{n_{\text{initial}}^2}\right)\).
For the \(3 \to 2\) transition:
\(\frac{1}{\lambda_0} = R \left(\frac{1}{2^2} - \frac{1}{3^2}\right) = R \left(\frac{1}{4} - \frac{1}{9}\right) = \frac{5}{36}R \implies \lambda_0 = \frac{36}{5R}\).

For the \(4 \to 3\) transition:
\(\frac{1}{\lambda_1} = R \left(\frac{1}{3^2} - \frac{1}{4^2}\right) = R \left(\frac{1}{9} - \frac{1}{16}\right) = \frac{7}{144}R \implies \lambda_1 = \frac{144}{7R}\).

Dividing \(\lambda_1\) by \(\lambda_0\):
\(\frac{\lambda_1}{\lambda_0} = \frac{144 / 7R}{36 / 5R} = \frac{144}{36} \cdot \frac{5}{7} = 4 \cdot \frac{5}{7} = \frac{20}{7}\).
Thus, \(\lambda_1 = \frac{20}{7}\lambda_0\).

Award 1 mark for correct answer B.

The formula for fringe separation is \(w = \frac{\lambda D}{a}\), where \(a\) is the slit separation.
Initially: \(y_1 = \frac{\lambda_1 D}{a}\).
Finally: \(y_2 = \frac{\lambda_2 (2D)}{a/2} = \frac{4 \lambda_2 D}{a}\).
Given \(y_2 = 3 y_1\):
\(\frac{4 \lambda_2 D}{a} = 3 \cdot \frac{\lambda_1 D}{a} \implies 4 \lambda_2 = 3 \lambda_1 \implies \frac{\lambda_2}{\lambda_1} = \frac{3}{4}\).

Award 1 mark for correct answer A.

The thin lens formula is:
\(\frac{1}{u} + \frac{1}{v} = \frac{1}{f}\).
Multiply both sides by \(v\):
\(\frac{v}{u} + 1 = \frac{v}{f}\).
For a real image formed by a convex lens, the linear magnification is \(m = \frac{v}{u}\).
Therefore:
\(m + 1 = \frac{v}{f} \implies m = \frac{1}{f}v - 1\).
This is in the form of a straight-line equation \(y = mx + c\) where the independent variable is \(v\).
Hence, the graph of \(m\) against \(v\) is a straight line with slope \(1/f\) and y-intercept \(-1\).

Award 1 mark for correct answer A.

For statement (1): The terminal voltage is given by \(V = \mathcal{E} \frac{R}{R+r} = \frac{\mathcal{E}}{1 + r/R}\). As \(R\) increases, \(r/R\) decreases, so \(V\) increases. Statement (1) is correct.
For statement (2): The power dissipated in the internal resistance is \(P_r = I^2 r = \left(\frac{\mathcal{E}}{R+r}\right)^2 r\). As \(R\) increases, the current \(I\) decreases, so \(P_r\) decreases. Statement (2) is incorrect.
For statement (3): The efficiency \(\eta\) of the circuit is given by \(\eta = \frac{I^2 R}{I^2(R+r)} = \frac{R}{R+r} = \frac{1}{1 + r/R}\). As \(R\) increases, \(1 + r/R\) decreases, so \(\eta\) increases. Statement (3) is correct.
Therefore, only (1) and (3) are correct.

Award 1 mark for correct answer B.

The r.m.s. voltage across the secondary coil is:
\(V_S = V \cdot \frac{N_S}{N_P}\).
The r.m.s. current in the secondary circuit is:
\(I_S = \frac{V_S}{R} = \frac{V}{R} \cdot \frac{N_S}{N_P}\).
For an ideal transformer, the input electrical power in the primary circuit equals the output power in the secondary circuit:
\(P_{\text{primary}} = P_{\text{secondary}} \implies V I_P = V_S I_S\).
Substituting \(V_S\) and \(I_S\):
\(I_P = \frac{V_S}{V} I_S = \left(\frac{N_S}{N_P}\right) \left(\frac{V}{R} \cdot \frac{N_S}{N_P}\right) = \frac{V}{R} \left(\frac{N_S}{N_P}\right)^2\).

Award 1 mark for correct answer B.

The relation between phase difference \(\Delta \phi\) and path difference \(\Delta x\) is given by \(\Delta \phi = \frac{2\pi \Delta x}{\lambda}\). Given \(\Delta \phi = \pi/3\text{ rad}\) and \(\Delta x = 0.15\text{ m}\), we have \(\frac{\pi}{3} = \frac{2\pi (0.15)}{\lambda}\). Solving for \(\lambda\) gives \(\lambda = 6 \times 0.15 = 0.9\text{ m}\). Using the wave equation \(v = f\lambda\), the wave speed is \(v = 10 \times 0.9 = 9.0\text{ m s}^{-1}\).

1 mark for the correct answer B. (Method: award 0.5 marks for finding the wavelength \(\lambda = 0.9\text{ m}\); award 0.5 marks for finding the wave speed \(v = 9.0\text{ m s}^{-1}\). No partial marks for MCQ).

When the switch \(S\) is open, \(R_3\) is disconnected. The circuit is a simple series connection of \(R_1\) and \(R_2\). The voltage across \(R_1\) is \(V_1 = 12 \times \frac{R_1}{R_1 + R_2} = 12 \times \frac{10}{10 + 20} = 4\text{ V}\), which is consistent with the given information. When the switch \(S\) is closed, \(R_2\) and \(R_3\) are in parallel, with an equivalent resistance \(R_p = \frac{R_2 R_3}{R_2 + R_3}\). The voltage across \(R_1\) becomes \(6\text{ V}\). Since the total voltage is \(12\text{ V}\), the voltage across the parallel combination is also \(12 - 6 = 6\text{ V}\). Since the voltages across \(R_1\) and \(R_p\) are equal, their resistances must be equal: \(R_p = R_1 = 10\ \Omega\). Thus, \rac{20 \times R_3}{20 + R_3} = 10 \Rightarrow 20 R_3 = 200 + 10 R_3 \Rightarrow 10 R_3 = 200 \Rightarrow R_3 = 20\ \Omega\.

1 mark for the correct answer C. (Method: award 0.5 marks for realizing that the parallel equivalent resistance \(R_p\) must equal \(R_1 = 10\ \Omega\); award 0.5 marks for calculating \(R_3 = 20\ \Omega\). No partial marks for MCQ).

For a satellite of mass \(m\) in a circular orbit of radius \(r\) around the Earth (mass \(M\)), the gravitational force provides the centripetal force: \(\frac{GMm}{r^2} = \frac{mv^2}{r}\), which gives \(mv^2 = \frac{GMm}{r}\). The kinetic energy is \(E_k = \frac{1}{2}mv^2 = \frac{GMm}{2r}\). Thus, kinetic energy is proportional to \(\frac{m}{r}\). For satellite \(X\), \(E = k \frac{m_X}{R}\) (where \(k = \frac{GM}{2}\)). For satellite \(Y\), \(E_Y = k \frac{m_Y}{r_Y} = k \frac{2m_X}{4R} = \frac{1}{2} \left(k \frac{m_X}{R}\right) = \frac{1}{2}E\).

1 mark for the correct answer C. (Method: award 0.5 marks for expressing kinetic energy in terms of \(G\), \(M\), \(m\), and \(r\); award 0.5 marks for setting up the ratio of \(E_Y / E_X = 1/2\). No partial marks for MCQ).

(a) For the third harmonic on a string fixed at both ends, the length of the string is \(L = 3\left(\frac{\lambda}{2}\right)\).\nTherefore, the wavelength is \(\lambda = \frac{2}{3}L = \frac{2}{3}(0.80) = 0.533\ \text{m}\).\nWave speed \(v = f\lambda = 150 \times 0.5333 = 80\ \text{m s}^{-1}\).\n\n(b) The wave speed is proportional to the square root of the tension (\(v \propto \sqrt{T}\)).\nWhen the tension is quadrupled, the wave speed doubles: \(v' = 2v = 160\ \text{m s}^{-1}\).\nThe fundamental wavelength is \(\lambda_1 = 2L = 2(0.80) = 1.60\ \text{m}\).\nThe new fundamental frequency is \(f_1' = \frac{v'}{\lambda_1} = \frac{160}{1.60} = 100\ \text{Hz}\).

(a) Use of \(L = 1.5\lambda\) to find wavelength (1M); Formula \(v = f\lambda\) and correct calculation of speed \(80\ \text{m s}^{-1}\) (2A).\n(b) Relate tension to speed change \(v' = 2v\) (1.67M); Calculate new fundamental frequency \(100\ \text{Hz}\) (1A).

(a) Wavelength \(\lambda = \frac{v}{f} = \frac{340}{680} = 0.50\ \text{m}\).\nPath difference at \(P\): \(\Delta r = 5.25 - 4.50 = 0.75\ \text{m}\).\nRatio \(\frac{\Delta r}{\lambda} = \frac{0.75}{0.50} = 1.5 = 1 + \frac{1}{2}\).\nSince the path difference is a half-integer multiple of the wavelength, destructive interference occurs.\n\n(b) As frequency increases, wavelength decreases. Currently at \(\Delta r = 1.5\lambda\), the next change in interference nature (from destructive to constructive) occurs when the path difference becomes the next integer multiple of the new wavelength \(\lambda'\):\n\(\Delta r = 2\lambda' \Rightarrow 0.75 = 2\lambda' \Rightarrow \lambda' = 0.375\ \text{m}\).\nNew frequency \(f' = \frac{v}{\lambda'} = \frac{340}{0.375} = 906.7\ \text{Hz}\) (or \(907\ \text{Hz}\)).

(a) Calculate wavelength \(\lambda = 0.50\ \text{m}\) (1M); Calculate path difference \(\Delta r = 0.75\ \text{m}\) (1M); Establish that \(\Delta r = 1.5\lambda\) (0.67M) and state it is destructive interference (1A).\n(b) Identify the condition for the next constructive interference \(\Delta r = 2\lambda'\) (1M); Correctly calculate the frequency as \(907\ \text{Hz}\) (1A).

(a) Using the terminal voltage formula: \(V = \frac{\varepsilon R}{R + r}\).\nFor \(R = 4.0\ \Omega\): \(6.0 = \frac{4.0\varepsilon}{4.0 + r} \Rightarrow 24.0 + 6.0r = 4.0\varepsilon \quad \text{--- (Eq 1)}\).\nFor \(R = 10.0\ \Omega\): \(7.5 = \frac{10.0\varepsilon}{10.0 + r} \Rightarrow 75.0 + 7.5r = 10.0\varepsilon \quad \text{--- (Eq 2)}\).\nFrom Eq 1, \(\varepsilon = 6.0 + 1.5r\).\nSubstitute into Eq 2: \(75.0 + 7.5r = 10.0(6.0 + 1.5r) = 60.0 + 15.0r\).\n\(15.0 = 7.5r \Rightarrow r = 2.0\ \Omega\).\nSubstituting back, \(\varepsilon = 6.0 + 1.5(2.0) = 9.0\ \text{V}\).\n\n(b) Maximum power transfer occurs when the external load resistance equals the internal resistance: \(R = r = 2.0\ \Omega\).\nMaximum power \(P_{\max} = \frac{\varepsilon^2}{4r} = \frac{9.0^2}{4(2.0)} = \frac{81.0}{8.0} = 10.125\ \text{W} \approx 10.1\ \text{W}\).

(a) Set up the simultaneous equations for both states (1M); Solve for internal resistance \(r = 2.0\ \Omega\) (1.67A); Solve for emf \(\varepsilon = 9.0\ \text{V}\) (1A).\n(b) State maximum power transfer condition \(R = r\) (1M); Calculate maximum power as \(10.1\ \text{W}\) (1A).

(a) Since the appliances are in parallel across \(220\ \text{V}\), they both operate at their rated powers.\nTotal power \(P_{\text{total}} = 1800 + 1200 = 3000\ \text{W}\).\nTotal current \(I = \frac{P_{\text{total}}}{V} = \frac{3000}{220} \approx 13.64\ \text{A}\).\nSince \(13.64\ \text{A} < 15\ \text{A}\), the current does not exceed the fuse rating. Thus, the fuse will not blow.\n\n(b) The power generated by the kettle is given by \(P = \frac{V^2}{R}\).\nSince the resistance \(R\) is assumed constant, \(P \propto V^2\).\nIf the voltage drops by \(10\%\), the new voltage is \(V' = 0.90V\).\nThe new power is \(P' = \frac{(0.90V)^2}{R} = 0.81 \left(\frac{V^2}{R}\right) = 0.81P\).\nPercentage decrease in power \(= \frac{P - P'}{P} \times 100\% = (1 - 0.81) \times 100\% = 19\%\).

(a) Compute total current \(13.6\ \text{A}\) (1.5M); Compare current with fuse rating and conclude 'will not blow' (1.5A).\n(b) Use the relation \(P \propto V^2\) (1M); Calculate the percentage decrease of \(19\%\) (1.67A).

(a) The critical angle \(\theta_c\) for glass-air interface is given by:\n\(\sin\theta_c = \frac{1}{n_{\text{glass}}} = \frac{1}{1.62}\).\n\(\theta_c = \sin^{-1}\left(\frac{1}{1.62}\right) \approx 38.12^\circ \approx 38.1^\circ\).\n\n(b) For the glass-liquid interface, the critical angle \(\theta_c'\) is given by:\n\(\sin\theta_c' = \frac{n_{\text{liquid}}}{n_{\text{glass}}} = \frac{1.35}{1.62} = 0.8333\).\n\(\theta_c' = \sin^{-1}(0.8333) \approx 56.44^\circ \approx 56.4^\circ\).\n\n(c) If the frequency of light is increased, the refractive index of glass \(n_{\text{glass}}\) increases due to normal dispersion (while the liquid is assumed unchanged or its change is negligible compared to the glass change, or simply based on standard dispersion of solid glass). From \(\sin\theta_c' = \frac{n_{\text{liquid}}}{n_{\text{glass}}}\), when \(n_{\text{glass}}\) increases, the ratio \(\frac{n_{\text{liquid}}}{n_{\text{glass}}}\) decreases, which means \(\sin\theta_c'\) decreases, and therefore the critical angle \(\theta_c'\) decreases.

(a) Formula \(\sin\theta_c = 1/n\) (1M); Correct angle \(38.1^\circ\) (1A).\n(b) Formula \(\sin\theta_c' = n_2/n_1\) (1M); Correct angle \(56.4^\circ\) (0.67A).\n(c) State that \(n_{\text{glass}}\) increases with frequency (1M); Conclude that critical angle decreases (1A).

(a) Magnification \(m = \frac{\text{height of image}}{\text{height of object}} = \frac{6.0}{3.0} = 2.0\).\nSince a real image is formed, \(m = \frac{v}{u} \Rightarrow 2.0 = \frac{v}{12.0} \Rightarrow v = 24.0\ \text{cm}\).\nUsing the lens formula: \\ \(\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\).\n\(\frac{1}{f} = \frac{1}{12.0} + \frac{1}{24.0} = \frac{3}{24.0} \Rightarrow f = 8.0\ \text{cm}\).\n\n(b) When the object is moved \(4.0\ \text{cm}\) closer, the new object distance is \(u' = 12.0 - 4.0 = 8.0\ \text{cm}\).\nSince \(u' = f = 8.0\ \text{cm}\), the object is located exactly at the principal focus of the lens.\nThe refracted rays from the lens will be parallel and will never meet on a screen. No real image is formed, and no screen position can capture a sharp image (or the image is at infinity, which means the screen would have to be moved infinitely far away).

(a) Find image distance \(v = 24.0\ \text{cm}\) (1M); Set up lens formula (1M); Correctly calculate \(f = 8.0\ \text{cm}\) (1A).\n(b) State that the new object distance is equal to the focal length (1M); Explain that refracted rays are parallel / image is at infinity, hence no real image can be formed on any screen (1.67A).

(a) Area of the square coil \(A = 0.10^2 = 0.01\ \text{m}^2\).\nRate of change of magnetic field \(\frac{\Delta B}{\Delta t} = \frac{0.20 - 0.80}{0.30} = -2.0\ \text{T s}^{-1}\).\nAccording to Faraday's Law, the induced emf is:\n\(\varepsilon = N \frac{\Delta \Phi}{\Delta t} = N A \frac{\Delta B}{\Delta t} = 50 \times 0.01 \times 2.0 = 1.0\ \text{V}\).\nInduced current \(I = \frac{\varepsilon}{R} = \frac{1.0}{2.0} = 0.50\ \text{A}\).\n\n(b) Direction: Clockwise.\nExplanation: The magnetic flux directed into the plane of the coil is decreasing. According to Lenz's Law, the induced current must create an induced magnetic field that opposes this decrease, meaning the induced field must point into the page. By the right-hand grip rule, the direction of the induced current is clockwise.

(a) Compute rate of change of flux or use formula \(\varepsilon = N A \Delta B/\Delta t\) (1M); Correctly calculate emf \(1.0\ \text{V}\) (1.67A); Correctly calculate current \(0.50\ \text{A}\) (1A).\n(b) State current is 'clockwise' (1M); Explain based on opposing the decrease of inward flux (1M).

(a) A geostationary satellite must remain stationary relative to the ground, so its orbital period must equal the Earth's rotation period (24 hours). The center of any circular orbit must coincide with the Earth's center of mass. To prevent the satellite from moving north and south relative to a ground observer, the orbital plane must lie in the equatorial plane.\n\n(b) The orbital period of the satellite is \(T = 24\ \text{hours} = 24 \times 3600\ \text{s} = 86400\ \text{s}\).\nThe centripetal force is provided by the gravitational force:\n\(\frac{G M m}{r^2} = m \left(\frac{2\pi}{T}\right)^2 r \Rightarrow M = \frac{4\pi^2 r^3}{G T^2}\).\nSubstituting the values:\n\(M = \frac{4\pi^2 \times (4.2 \times 10^7)^3}{6.67 \times 10^{-11} \times 86400^2}\).\n\(M = \frac{39.4784 \times 7.4088 \times 10^{22}}{6.67 \times 10^{-11} \times 7.465 \times 10^9} \approx 5.87 \times 10^{24}\ \text{kg}\) (accept \(5.9 \times 10^{24}\ \text{kg}\)).

(a) Explain that the center of the orbit is the Earth's center (1M); explain that matching the equatorial plane avoids north-south drift to keep it stationary relative to the ground (1M).\n(b) State period \(T = 86400\ \text{s}\) (1M); Equate gravitational force to centripetal force \(G M/r^2 = 4\pi^2 r/T^2\) (1.67M); Calculate mass \(M \approx 5.9 \times 10^{24}\ \text{kg}\) (1A).

(a) Energy difference of the transition:\n\(\Delta E = E_3 - E_1 = -1.51 - (-13.6) = 12.09\ \text{eV}\).\nIn Joules: \(\Delta E = 12.09 \times 1.60 \times 10^{-19}\ \text{J} = 1.9344 \times 10^{-18}\ \text{J}\).\nAccording to \(E = hf\), the frequency of the emitted photon is:\n\(f = \frac{\Delta E}{h} = \frac{1.9344 \times 10^{-18}}{6.63 \times 10^{-34}} \approx 2.92 \times 10^{15}\ \text{Hz}\).\n\n(b) For a photon to be absorbed by a hydrogen atom, its energy must exactly equal the difference between the initial ground state and an higher energy level (unless the energy is enough to ionize the atom, i.e., \(\ge 13.6\ \text{eV}\)).\nIf an \(11.5\ \text{eV}\) photon is absorbed, the final energy level of the electron would be \(-13.6 + 11.5 = -2.1\ \text{eV}\).\nLet us check if \(-2.1\ \text{eV}\) corresponds to any integer state \(n\):\nFor \(n=2\), \(E_2 = -3.4\ \text{eV}\).\nFor \(n=3\), \(E_3 = -1.51\ \text{eV}\).\nSince there is no discrete energy state at \(-2.1\ \text{eV}\), the photon cannot be absorbed.

(a) Calculate energy difference in Joules (1M); Apply formula \(E = hf\) (1M); Calculate frequency \(2.92 \times 10^{15}\ \text{Hz}\) (1A).\n(b) State the condition for photon absorption (must match discrete energy differences) (1M); Calculate target energy \(-2.1\ \text{eV}\) and point out that it does not correspond to any energy level of hydrogen, hence cannot be absorbed (1.67A).

(a) The gravitational force provides the centripetal force for the circular orbit:
\(F_g = F_c\)
\(G \frac{Mm}{r^2} = m \left(\frac{2\pi}{T}\right)^2 r\)
\(G \frac{M}{r^2} = \frac{4\pi^2 r}{T^2}\)
\(T^2 = \frac{4\pi^2 r^3}{GM}\)
Assumption: The exoplanet is spherical with a uniform mass distribution (or: the mass of the probe is negligible compared to the exoplanet).

(b)(i) Orbital radius \(r = R + h = 3.5 \times 10^6 + 1.5 \times 10^6 = 5.0 \times 10^6\text{ m}\).
Orbital period \(T = 4.0 \times 3600\text{ s} = 14400\text{ s}\).
Using \(M = \frac{4\pi^2 r^3}{G T^2}\):
\(M = \frac{4\pi^2 (5.0 \times 10^6)^3}{(6.67 \times 10^{-11})(14400)^2}\)
\(M = \frac{4.9348 \times 10^{21}}{1.3831 \times 10^{-2}} \approx 3.57 \times 10^{23}\text{ kg}\).

(b)(ii) Acceleration due to gravity on the surface:
\(g_s = \frac{GM}{R^2} = \frac{(6.67 \times 10^{-11})(3.568 \times 10^{23})}{(3.5 \times 10^6)^2} \approx 1.94\text{ m s}^{-2}\).

(c) When ejected backwards, the velocity of the landing module relative to the planet decreases. Since the orbital velocity decreases, the required centripetal force at that distance becomes less than the actual gravitational force acting on it. As a result, the gravitational pull exceeds the centripetal force, pulling the module into a lower, elliptical orbit that intersects the planet's atmosphere or surface, facilitating its descent.

(a)
- Equating gravitational force to centripetal force formula using \(T\). (1M)
- Correct derivation steps leading to \(T^2 = \frac{4\pi^2 r^3}{GM}\). (1A)
- State a valid assumption (spherical planet/uniform mass density/negligible satellite mass). (1A)

(b)(i)
- Correct calculation of orbital radius \(r = 5.0 \times 10^6\text{ m}\) and period \(T = 14400\text{ s}\). (1M)
- Substitution into derived formula. (1M)
- Correct answer: \(M = 3.57 \times 10^{23}\text{ kg}\) (accept \(3.56 \times 10^{23}\text{ kg}\) to \(3.58 \times 10^{23}\text{ kg}\)). (1A)

(b)(ii)
- Substitution into \(g = \frac{GM}{R^2}\) using exoplanet's radius \(R = 3.5 \times 10^6\text{ m}\). (1M)
- Correct answer: \(1.94\text{ m s}^{-2}\) (accept \(1.93\text{ m s}^{-2}\) to \(1.95\text{ m s}^{-2}\)). (1A)

(c)
- State that orbital velocity decreases. (1A)
- Explain that gravitational force is now larger than the required centripetal force. (1A)
- State that this causes the landing module to spiral downwards / move into a lower orbit. (1A)

(a)(i) According to Faraday's law of induction, the induced e.m.f. is given by:
\(\mathcal{E} = N \frac{\Delta \Phi}{\Delta t} = N B \frac{\Delta A}{\Delta t}\)
Since \(\Delta A = L (v \Delta t)\), we have:
\(\mathcal{E} = N B L v = 150 \times 0.40 \times 0.20 \times 3.0 = 36\text{ V}\). (Shown)

(a)(ii) Induced current \(I = \frac{\mathcal{E}}{R} = \frac{36}{2.5} = 14.4\text{ A}\).
As the coil enters the magnetic field, the upward magnetic flux through the coil increases. According to Lenz's law, the induced current must create a magnetic field pointing downwards (into the page) to oppose this increase. By the right-hand grip rule, the direction of the induced current must be clockwise as viewed from above.

(b) The magnetic force acting on the leading edge is:
\(F = N I L B = 150 \times 14.4 \times 0.20 \times 0.40 = 172.8\text{ N} \approx 173\text{ N}\).
Using Fleming's Left-Hand Rule (with current flowing to the right along the leading edge during the clockwise loop, and B pointing upwards), the magnetic force acts in the opposite direction to the velocity (resisting the motion of the coil).

(c) To keep the coil moving at a constant speed, the net force on it must be zero. Therefore, an external force equal in magnitude but opposite in direction to the magnetic force must be applied to balance it.
\(F_{\text{ext}} = 172.8\text{ N}\).
Mechanical power \(P_{\text{mech}} = F_{\text{ext}} v = 172.8 \times 3.0 = 518.4\text{ W} \approx 518\text{ W}\).
(Note: This equals the electrical power dissipated as heat in the resistor: \(P_{\text{elec}} = I^2 R = 14.4^2 \times 2.5 = 518.4\text{ W}\).)

(a)(i)
- Correct formula for induced e.m.f. \(\mathcal{E} = NBLv\) or equivalent Faraday's law expression. (1M)
- Correct substitution yielding exactly \(36\text{ V}\). (1A)

(a)(ii)
- Correct calculation of current \(I = 14.4\text{ A}\). (1A)
- State "clockwise". (1A)
- Explain using Lenz's law (opposing the increase in upward magnetic flux by producing a downward field). (1A)

(b)
- State the force formula \(F = NILB\) or equivalent. (1M)
- Calculate the magnitude: \(173\text{ N}\) (or \(172.8\text{ N}\)). (1A)
- State correct direction: opposite to the direction of motion (or to the left). (1A)

(c)
- Explain that net force must be zero (Newton's first law) to maintain constant speed, balancing the opposing magnetic force. (1M)
- Mechanical power formula: \(P = Fv\). (1M)
- Calculate mechanical power: \(518\text{ W}\) (or \(518.4\text{ W}\)). (1A)

(a)(i) Critical angle is the angle of incidence in an optically denser medium for which the angle of refraction in the optically less dense medium is \(90^\circ\).

(a)(ii) Using Snell's law:
\(n_1 \sin \theta_c = n_2 \sin 90^\circ\)
\(\sin \theta_c = \frac{n_2}{n_1} = \frac{1.42}{1.48} \approx 0.9595\)
\(\theta_c = \arcsin(0.9595) \approx 73.64^\circ \approx 73.6^\circ\).

(b)(i) Let \(\theta_r\) be the angle of refraction in the core. By Snell's law at the air-core interface:
\(n_0 \sin \theta_a = n_1 \sin \theta_r \implies \sin \theta_a = n_1 \sin \theta_r\) (since \(n_0 = 1.00\)).

At the core-cladding interface, the angle of incidence is \(\phi = 90^\circ - \theta_r\).
For total internal reflection to occur, we require:
\(\phi \ge \theta_c \implies \sin \phi \ge \sin \theta_c\)
Since \(\sin \phi = \sin(90^\circ - \theta_r) = \cos \theta_r\), we have:
\(\cos \theta_r \ge \frac{n_2}{n_1}\)
Using the identity \(\cos^2 \theta_r = 1 - \sin^2 \theta_r\):
\(1 - \sin^2 \theta_r \ge \frac{n_2^2}{n_1^2}\)
\(1 - \frac{\sin^2 \theta_a}{n_1^2} \ge \frac{n_2^2}{n_1^2}\)
\(\frac{\sin^2 \theta_a}{n_1^2} \le 1 - \frac{n_2^2}{n_1^2} = \frac{n_1^2 - n_2^2}{n_1^2}\)
\(\sin^2 \theta_a \le n_1^2 - n_2^2\)
For the maximum angle: \(\sin \theta_{a,\text{max}} = \sqrt{n_1^2 - n_2^2}\). (Shown)

(b)(ii)
\(\sin \theta_{a,\text{max}} = \sqrt{1.48^2 - 1.42^2} = \sqrt{2.1904 - 2.0164} = \sqrt{0.174} \approx 0.4171\)
\(\theta_{a,\text{max}} = \arcsin(0.4171) \approx 24.7^\circ\) (or \(24.65^\circ\)).

(c) Advantage: Higher bandwidth (can carry more data per second) / lower signal attenuation over long distances / immune to electromagnetic interference.
Disadvantage: More fragile and easily damaged when bent sharply / higher installation and splicing costs / requires specialized tools to join.

(a)(i)
- Define critical angle correctly (mentioning optically denser to less dense, and refraction angle is 90 degrees). (1A)

(a)(ii)
- Write \(\sin \theta_c = \frac{n_2}{n_1}\). (1M)
- Calculate \(73.6^\circ\) (accept \(73.6^\circ\) to \(73.7^\circ\)). (1A)

(b)(i)
- State \(\sin \theta_a = n_1 \sin \theta_r\) at the first interface. (1M)
- State the relation between the angles: \(\phi = 90^\circ - \theta_r\) and the condition for TIR: \(\sin \phi \ge \sin \theta_c = \frac{n_2}{n_1}\). (1M)
- Substitute \(\sin \phi = \cos \theta_r\) and relate to \(\sin \theta_r\) using \(\sin^2 \theta_r + \cos^2 \theta_r = 1\). (1M)
- Complete algebraic steps to show the required formula clearly. (1A)

(b)(ii)
- Correct substitution of values into formula. (1M)
- Correct answer: \(24.7^\circ\) (accept \(24.6^\circ\) to \(24.7^\circ\)). (1A)

(c)
- One valid advantage. (1A)
- One valid disadvantage. (1A)

By Kepler's Third Law, \(T^2 \propto R^3\), which means \(R \propto T^{2/3}\). Therefore, the ratio of the orbital radii is \(\frac{R_Q}{R_P} = \left(\frac{T_Q}{T_P}\right)^{2/3} = \left(\frac{8T}{T}\right)^{2/3} = 8^{2/3} = 4\). Thus, \(R_Q = 4R\).

B: 1 mark. Other options: 0 marks.

The initial mechanical energy of the satellite is \(E_i = K_i + U_i = K - 2K = -K\). In a stable circular orbit of radius \(2r\), the final kinetic energy is \(K_f = \frac{GMm}{4r} = 0.5K\), and the final potential energy is \(U_f = -\frac{GMm}{2r} = -K\). The final mechanical energy is \(E_f = K_f + U_f = 0.5K - K = -0.5K\). The work done by the thruster is \(\Delta E = E_f - E_i = -0.5K - (-K) = 0.5K\).

B: 1 mark. Other options: 0 marks.

According to the Stefan-Boltzmann law, \(L = 4\pi R^2 \sigma T^4\). Taking the ratio for Star X and Star Y: \(\frac{L_X}{L_Y} = \left(\frac{R_X}{R_Y}\right)^2 \left(\frac{T_X}{T_Y}\right)^4\). Substituting the given values: \(\frac{10^4}{10^{-4}} = \left(\frac{R_X}{R_Y}\right)^2 \left(\frac{3000}{12000}\right)^4 \implies 10^8 = \left(\frac{R_X}{R_Y}\right)^2 \left(\frac{1}{4}\right)^4 \implies 10^8 = \left(\frac{R_X}{R_Y}\right)^2 \frac{1}{256}\). Thus, \(\left(\frac{R_X}{R_Y}\right)^2 = 256 \times 10^8 = 2.56 \times 10^{10} \implies \frac{R_X}{R_Y} = 1.6 \times 10^5\).

A: 1 mark. Other options: 0 marks.

First, calculate the redshift \(z\): \(z = \frac{\Delta \lambda}{\lambda_0} = \frac{671.6 - 656.3}{656.3} = \frac{15.3}{656.3} \approx 0.023312\). The recessional velocity \(v\) is given by \(v = z c = 0.023312 \times 3.0 \times 10^5\text{ km s}^{-1} \approx 6994\text{ km s}^{-1}\). According to Hubble's law \(v = H_0 d\), the distance \(d\) is \(d = \frac{v}{H_0} = \frac{6994}{70} \approx 100\text{ Mpc}\).

C: 1 mark. Other options: 0 marks.

(1) is incorrect because stars with high initial masses (above \(8 M_{\odot}\)) will end as supernovae and form neutron stars or black holes, not white dwarfs. (2) is correct as the main sequence phase is defined by core hydrogen fusion. (3) is incorrect because although high-mass stars have more fuel, their nuclear fusion rate (and hence luminosity) is much higher, giving them a much shorter lifespan than low-mass stars. Therefore, only (2) is correct.

A: 1 mark. Other options: 0 marks.

In a binary star system, both stars orbit around their common center of mass with the same orbital period \(T\) and angular speed \(\omega\). (1) is correct: Since \(M_A r_A = M_B r_B\) and \(M_B = 2 M_A\), we get \(r_A = 2 r_B\). (2) is correct: Linear speed is \(v = r \omega\). Since \(r_A = 2 r_B\), we have \(v_A = 2 v_B\). (3) is incorrect: By Newton's third law, the gravitational forces they exert on each other must be equal in magnitude. Thus, (1) and (2) only are correct.

B: 1 mark. Other options: 0 marks.

The distance \(d\) in parsecs is given by \(d = \frac{1}{p}\), where \(p\) is the parallax angle in arcseconds. Thus, \(d = \frac{1}{0.04} = 25\text{ pc}\). Converting to light-years: \(25\text{ pc} = 25 \times 3.26\text{ light-years} = 81.5\text{ light-years}\).

C: 1 mark. Other options: 0 marks.

(1) is correct because the CMB is the remnant radiation from the early Hot Big Bang and its high isotropy confirms the cosmological principle. (2) is correct because the observed cosmic abundances of light isotopes (like helium-4 and deuterium) closely match Big Bang nucleosynthesis models. (3) is correct because the Hubble expansion of galaxies shows the universe was hotter and denser in the past, directly pointing to a primordial explosion. Thus, all three statements are valid evidence supporting the Big Bang theory.

D: 1 mark. Other options: 0 marks.

(a) First, convert the orbital period into seconds:\n\(T = 90 \text{ days} = 90 \times 24 \times 3600 \text{ s} = 7.776 \times 10^6 \text{ s}\)\nUsing Newton's law of gravitation for a circular orbit:\n\(\frac{G M m}{r^2} = m \left(\frac{2\pi}{T}\right)^2 r\)\n\(M = \frac{4\pi^2 r^3}{G T^2}\)\n\(M = \frac{4 \pi^2 (6.0 \times 10^{10})^3}{(6.67 \times 10^{-11}) (7.776 \times 10^6)^2} \approx 2.11 \times 10^{30} \text{ kg}\)\n\n(b) Using the mass-luminosity relation \(L \propto M^{3.5}\):\n\(\frac{L}{L_\odot} = \left(\frac{M}{M_\odot}\right)^{3.5}\)\n\(\frac{L}{3.8 \times 10^{26}} = \left(\frac{2.114 \times 10^{30}}{2.0 \times 10^{30}}\right)^{3.5}\)\n\(L = 3.8 \times 10^{26} \times (1.057)^{3.5} \approx 4.58 \times 10^{26} \text{ W}\) (or \(4.60 \times 10^{26} \text{ W}\) if using unrounded mass)\n\n(c) The radiant flux \(F\) received by the planet is:\n\(F = \frac{L}{4\pi r^2}\)\n\(F = \frac{4.58 \times 10^{26}}{4\pi (6.0 \times 10^{10})^2} \approx 1.01 \times 10^4 \text{ W m}^{-2}\)\n\n(d) The method is the radial velocity method (or Doppler spectroscopy).\nAs the planet and the star orbit their common center of mass, the star's periodic motion causes a Doppler shift in its light. When moving towards Earth, its spectral lines are blueshifted, and when moving away, they are redshifted. This periodic shift allows astronomers to detect the planet.

(a) \n- Convert period to seconds: \(T = 7.78 \times 10^6 \text{ s}\) (1M)\n- State Kepler's Third Law or express gravitational force as centripetal force: \(M = \frac{4\pi^2 r^3}{G T^2}\) (1M)\n- Correct answer: \(2.11 \times 10^{30} \text{ kg}\) (1A) [Accept: \(2.11 \times 10^{30}\) to \(2.12 \times 10^{30} \text{ kg}\)]\n\n(b)\n- State mass-luminosity relation: \(\frac{L}{L_\odot} = (\frac{M}{M_\odot})^{3.5}\) (1M)\n- Correct substitution of values (1M)\n- Correct answer: \(4.58 \times 10^{26} \text{ W}\) (1A) [Accept: \(4.58 \times 10^{26}\) to \(4.63 \times 10^{26} \text{ W}\)]\n\n(c)\n- State formula for radiant flux: \(F = \frac{L}{4\pi r^2}\) (1M)\n- Correct answer with unit: \(1.01 \times 10^4 \text{ W m}^{-2}\) (1A) [Accept: \(1.01 \times 10^4\) to \(1.02 \times 10^4 \text{ W m}^{-2}\)]\n\n(d)\n- Name of method: Radial velocity method / Doppler spectroscopy (1A)\n- Explain the process: Doppler blueshift/redshift of spectral lines due to stellar motion (1A)

From Einstein's photoelectric equation:
\(eV_s = hf - \Phi\)

For the first case:
\(eV = hf - \Phi\) --- (1)

For the second case:
\(e(2V) = 1.5hf - \Phi\) --- (2)

Multiplying equation (1) by 2:
\(2eV = 2hf - 2\Phi\) --- (3)

Equating (2) and (3):
\(1.5hf - \Phi = 2hf - 2\Phi\)
\(\Phi = 0.5hf\)

Therefore, the work function of the metal is \(0.50hf\).

Correct Option: B. Award 1 mark for selecting B.

In Bohr's model, the angular momentum of the electron is quantized:
\(m v r = \frac{n h}{2\pi}\)

This can be rewritten as:
\(2\pi r = \frac{n h}{m v} = n \lambda\)
where \(\lambda = \frac{h}{m v}\) is the de Broglie wavelength of the electron.

Since the radius of the \(n\)-th orbit is proportional to \(n^2\) (i.e., \(r_n \propto n^2\)), we have:
\(\lambda_n = \frac{2\pi r_n}{n} \propto \frac{n^2}{n} = n\)

Thus, \(\lambda \propto n\).

Correct Option: C. Award 1 mark for selecting C.

The kinetic energy \(K\) gained by a particle of charge \(q\) accelerated through potential difference \(V\) is:
\(K = qV\)

The momentum \(p\) of the particle is:
\(p = \sqrt{2mK} = \sqrt{2mqV}\)

The de Broglie wavelength is:
\(\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mqV}}\)

Since both the electron and the proton have the same magnitude of charge \(e\):
\(\lambda_e = \frac{h}{\sqrt{2 m_e eV}}\)
\(\lambda_p = \frac{h}{\sqrt{2 m_p eV}}\)

Taking the ratio:
\(\frac{\lambda_e}{\lambda_p} = \sqrt{\frac{m_p}{m_e}}\)

Correct Option: B. Award 1 mark for selecting B.

(1) is correct: The stopping potential \(V_s\) is determined by the frequency of the light \(f\) according to \(eV_s = hf - \Phi\). Since Beam P and Beam Q have the same stopping potential on the same metal plate, they have the same frequency. The saturation current is proportional to the intensity (photon rate) of the light; hence, Beam P has a higher intensity.
(2) is correct: Since the stopping potential of Beam R is more negative than that of Beam Q (i.e., \(|V_{s,R}| > |V_{s,Q}|\)), the photoelectrons emitted by Beam R have a higher maximum kinetic energy. This implies that Beam R must have a higher frequency.
(3) is correct: Since Beam P and Beam Q have the same stopping potential, and the stopping potential of Beam R is more negative than that of Beam Q, the stopping potential of Beam R is also more negative than that of Beam P. Thus, the maximum kinetic energy of photoelectrons emitted by Beam R is greater than that by Beam P.
Therefore, (1), (2), and (3) are all correct.

Correct Option: D. Award 1 mark for selecting D.

(1) is correct: The minimum wavelength of the continuous X-ray spectrum is given by \(\lambda_{\min} = \frac{hc}{eV}\). If the tube voltage is increased to \(1.5V\), the new minimum wavelength is \(\lambda_{\min}' = \frac{hc}{e(1.5V)} = \frac{2}{3}\lambda_{\min}\).
(2) is incorrect: The characteristic peaks correspond to transitions between core electronic levels of the target metal atoms, which depend only on the target material, not the accelerating voltage (as long as it exceeds the excitation threshold).
(3) is correct: When the tube voltage increases, the electrons bombarding the target have higher kinetic energy, leading to a higher rate of photon emission and thus higher overall X-ray intensity.
Therefore, (1) and (3) only are correct.

Correct Option: B. Award 1 mark for selecting B.

(1) is correct: Quantum tunneling is a wave-mechanical phenomenon where the wave function of an electron penetrates a thin potential barrier. Thus, it depends on the wave-particle duality of electrons.
(2) is incorrect: The tunneling current \(I\) decreases exponentially as the distance \(d\) between the tip and the sample surface increases (\(I \propto e^{-2\kappa d}\)).
(3) is correct: In addition to imaging, the probe tip of an STM can apply localized electrical pulses or forces to slide, pick up, or deposit individual atoms on a surface.
Therefore, (1) and (3) only are correct.

Correct Option: B. Award 1 mark for selecting B.

The momentum \(p\) of an electron with kinetic energy \(E\) is given by:
\(p = \sqrt{2mE}\)

The de Broglie wavelength of the electron is:
\(\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}}\)

If the kinetic energy is increased to \(4E\), the new wavelength \(\lambda'\) is:
\(\lambda' = \frac{h}{\sqrt{2m(4E)}} = \frac{1}{2}\lambda\)

For a small diffraction angle \(\theta\), we have \(\theta \approx \sin\theta = \frac{\lambda}{d}\).
The radius \(R\) of the diffraction ring on the screen is proportional to \(\theta\) (\(R \approx L \theta\)).
Therefore:
\(R \propto \lambda\)

Since the wavelength is halved, the radius of the rings decreases to \(\frac{1}{2}\) of the original radius.

Correct Option: C. Award 1 mark for selecting C.

The energy levels of a hydrogen atom are given by \(E_n = -\frac{13.6}{n^2}\text{ eV}\).
The energy of an emitted photon during transition is equal to the energy difference between the levels:
\(h f = E_{\text{initial}} - E_{\text{final}}\)

For the first transition from \(n=3\) to \(n=2\):
\(h f_1 = -13.6 \left( \frac{1}{3^2} - \frac{1}{2^2} \right) = 13.6 \left( \frac{1}{4} - \frac{1}{9} \right) = 13.6 \left( \frac{5}{36} \right)\)

For the second transition from \(n=2\) to \(n=1\):
\(h f_2 = -13.6 \left( \frac{1}{2^2} - \frac{1}{1^2} \right) = 13.6 \left( 1 - \frac{1}{4} \right) = 13.6 \left( \frac{3}{4} \right)\)

Taking the ratio:
\(\frac{f_1}{f_2} = \frac{13.6 \times (5/36)}{13.6 \times (3/4)} = \frac{5}{36} \times \frac{4}{3} = \frac{5}{27}\)

Correct Option: A. Award 1 mark for selecting A.

(a) Due to the Doppler effect, when one star moves towards Earth, its light is blueshifted (wavelength decreases). Simultaneously, the other star moves away from Earth, so its light is redshifted (wavelength increases). As they orbit, these spectral lines periodically shift in opposite directions, resulting in the observed spectral line splitting.

(b)(i) By definition of the centre of mass:
\(M_1 r_1 = M_2 r_2\)
Since \(d = r_1 + r_2\), we have \(r_2 = d - r_1\).
Substituting \(r_2\):
\(M_1 r_1 = M_2 (d - r_1)\)
\(M_1 r_1 = M_2 d - M_2 r_1\)
\((M_1 + M_2) r_1 = M_2 d\)
\(r_1 = \frac{M_2}{M_1 + M_2} d\)

(b)(ii) The gravitational force between the two stars provides the centripetal force for Star A:
\(F_G = F_c\)
\(\frac{G M_1 M_2}{d^2} = M_1 \omega^2 r_1\)
Since \(\omega = \frac{2\pi}{T}\):
\(\frac{G M_1 M_2}{d^2} = M_1 \left(\frac{2\pi}{T}\right)^2 r_1\)
Substitute \(r_1\) from (b)(i):
\(\frac{G M_1 M_2}{d^2} = M_1 \frac{4\pi^2}{T^2} \left(\frac{M_2}{M_1 + M_2} d\right)\)
Cancel out \(M_1 M_2\) from both sides:
\(\frac{G}{d^2} = \frac{4\pi^2 d}{T^2 (M_1 + M_2)}\)
Rearranging terms yields:
\(M_1 + M_2 = \frac{4\pi^2 d^3}{G T^2}

(b)(iii) Substitute the values into the formula:
\)M_1 + M_2 = \frac{4\pi^2 (8.0 \times 10^{11})^3}{(6.67 \times 10^{-11})(1.26 \times 10^8)^2}\)
\(M_1 + M_2 = \frac{4\pi^2 \times 5.12 \times 10^{35}}{(6.67 \times 10^{-11})(1.5876 \times 10^{16})}\)
\(M_1 + M_2 = \frac{2.021 \times 10^{37}}{1.059 \times 10^6}\)
\(M_1 + M_2 \approx 1.91 \times 10^{31} \text{ kg}\)

Part (a):
- Explaining blueshift for the star moving towards Earth (wavelength decreases) (1)
- Explaining redshift for the star moving away from Earth (wavelength increases) (1)
- Explaining that periodic motion causes the lines to split periodically in opposite directions (1)

Part (b)(i):
- Using the definition of centre of mass \(M_1 r_1 = M_2 r_2\) or equivalent (1)
- Substituting \(r_2 = d - r_1\) and correctly rearranging to show the expression (1)

Part (b)(ii):
- Stating that gravitational force provides centripetal force: \(F_G = F_c\) (1)
- Correct expression for centripetal force: \(F_c = M_1 \left(\frac{2\pi}{T}\right)^2 r_1\) (1)
- Correct substitution and algebraic manipulation to reach the final form (1)

Part (b)(iii):
- Correct substitution of values into the derived formula (1)
- Correct final answer with unit: \(1.91 \times 10^{31} \text{ kg}\) [accept range: \(1.90 \times 10^{31} \text{ kg}\) to \(1.92 \times 10^{31} \text{ kg}\)] (1)