2022 HKDSE Physics Answers & Marking Scheme

(1) is incorrect because when the magnet is far away or exactly at the center of the ring, the rate of change of magnetic flux is zero, so the induced current is zero, and its acceleration is exactly equal to \(g\). (2) is correct because as the magnet enters, there is repulsion (downward force on the ring), and as it leaves, there is attraction (downward force on the ring). (3) is correct because a cut prevents a complete loop of induced current from flowing, so there is no magnetic force and the acceleration is always \(g\).

1 mark for the correct answer C.

Initially, clockwise current \(I\) in side \(RS\) flows to the left. Since \(\vec{B}\) is into the page, the magnetic force on \(RS\) is directed downwards: \(F_1 = IaB\) (downwards). Forces on vertical sides cancel, and the top side is outside the field. The scale reading is \(T_1 = mg + IaB\). When current is reversed (counter-clockwise), current in \(RS\) flows to the right. The magnetic force on \(RS\) is directed upwards: \(F_2 = IaB\) (upwards). The scale reading becomes \(T_2 = mg - IaB\). The change in the scale reading is \(T_1 - T_2 = 2IaB\).

1 mark for the correct answer B.

For an ideal transformer, \(V_{out, max} = \frac{N_2}{N_1} V_0\). Doubling \(N_1\) halves \(V_{out, max}\) (as \(V_0\) and \(N_2\) are unchanged). From Faraday's law, \(V_1(t) = N_1 \frac{d\Phi}{dt} \implies V_0 \sin(\omega t) = N_1 \frac{d\Phi}{dt}\), which gives \(\Phi(t) = -\frac{V_0}{N_1 \omega} \cos(\omega t)\). Thus, the maximum magnetic flux \(\Phi_{max} = \frac{V_0}{N_1 \omega}\). Doubling both \(N_1\) and \(\omega\) makes \(\Phi_{max}' = \frac{V_0}{(2N_1)(2\omega)} = \frac{1}{4} \Phi_{max}\).

1 mark for the correct answer B.

Let each bulb have resistance \(R\). With switch \(S\) closed, the parallel combination of \(Y\) and \(Z\) has resistance \(0.5R\). The total resistance is \(1.5R\). The voltage across \(X\) is \(\frac{2}{3}V\), and across \(Z\) is \(\frac{1}{3}V\). When \(S\) is opened, bulb \(Y\) is disconnected. The circuit becomes a simple series combination of \(X\) and \(Z\), so the total resistance is \(2R\). The voltage across both \(X\) and \(Z\) becomes \(0.5V\). Thus, the voltage across \(X\) decreases from \(0.67V\) to \(0.5V\) (brightness decreases), and the voltage across \(Z\) increases from \(0.33V\) to \(0.5V\) (brightness increases).

1 mark for the correct answer B.

The terminal voltage \(V = \mathcal{E} \frac{R}{R+r}\). From the first case: \(8 = \mathcal{E} \frac{4}{4+r} \implies \mathcal{E} = 8 + 2r\) (Eq. 1). From the second case: \(10 = \mathcal{E} \frac{10}{10+r} \implies \mathcal{E} = 10 + r\) (Eq. 2). Equating (1) and (2): \(8 + 2r = 10 + r \implies r = 2\ \Omega\). Substituting \(r = 2\ \Omega\) back gives \(\mathcal{E} = 12\text{ V}\).

1 mark for the correct answer A.

Since the container is insulated and rigid, \(Q=0\) and \(W=0\) so the total internal energy \(U\) is conserved: \(U_i = U_f\). For a monoatomic ideal gas, \(U = \frac{3}{2} n R T\). Initial internal energy \(U_i = \frac{3}{2} (1) R T_0 + \frac{3}{2} (2) R (2T_0) = \frac{15}{2} R T_0\). Final internal energy \(U_f = \frac{3}{2} (1+2) R T_f = \frac{9}{2} R T_f\). Setting \(U_i = \frac{15}{2} R T_0 = \frac{9}{2} R T_f\) yields \(T_f = \frac{15}{9} T_0 = \frac{5}{3} T_0 \approx 1.67 T_0\).

1 mark for the correct answer C.

The absolute temperatures are \(T_1 = 27 + 273 = 300\text{ K}\) and \(T_2 = 327 + 273 = 600\text{ K}\). Thus, \(T_2 / T_1 = 2\). Since the volume is fixed, pressure \(P \propto T\), so \(P\) is doubled. The root-mean-square speed \(v_{rms} \propto \sqrt{T}\), so \(v_{rms}\) increases by a factor of \(\sqrt{2}\).

1 mark for the correct answer B.

The slope of \(\sin\theta_2\) against \(\sin\theta_1\) is \(\frac{n_X}{n_Y} = 0.75 < 1\), meaning \(n_X < n_Y\). Since \(n_X < n_Y\), the speed of light \(v = c/n\) is greater in X than in Y, so (1) is correct. Total internal reflection occurs when light travels from a denser medium to a rarer medium, so Y to X is correct, thus (2) is correct. The critical angle \(\theta_c\) is given by \(\sin\theta_c = \frac{n_X}{n_Y} = 0.75 \implies \theta_c \approx 48.6^\circ\), so (3) is correct.

1 mark for the correct answer D.

The fringe width is given by \(w = \frac{\lambda D}{a}\). We have \(w_1 = \frac{\lambda_1 D}{a}\) and \(w_2 = \frac{\lambda_2 D}{a/2} = \frac{2 \lambda_2 D}{a}\). Since \(w_2 = 3 w_1\), we get \(\frac{2 \lambda_2 D}{a} = 3 \frac{\lambda_1 D}{a} \implies \frac{\lambda_2}{\lambda_1} = 1.5\).

1 mark for the correct answer B.

The lens formula is \(\frac{1}{u} + \frac{1}{v} = \frac{1}{f}\). Since \(m = v/u \implies v = m u\). Substituting into the lens formula gives \(\frac{1}{u} + \frac{1}{mu} = \frac{1}{f}\). Multiplying by \(u\) yields \(1 + \frac{1}{m} = \frac{u}{f} \implies \frac{1}{m} = \frac{1}{f}u - 1\). This is a straight line equation where the slope is \(1/f\) and the horizontal intercept (where \(1/m = 0\)) is \(u = f\).

1 mark for the correct answer A.

When the loop enters the magnetic field, an electromotive force (e.m.f.) is induced in the leading edge of the loop:
\(\mathcal{E} = B a v = 0.5 \times 0.1 \times 2 = 0.1\text{ V}\).
The induced current in the loop is:
\(I = \frac{\mathcal{E}}{R} = \frac{0.1}{0.2} = 0.5\text{ A}\).
The magnetic force acting on the leading edge (opposing the motion) is:
\(F_B = B I a = 0.5 \times 0.5 \times 0.1 = 0.025\text{ N}\).
To maintain a constant velocity, the external force must balance this magnetic force, so \(F_{\text{ext}} = 0.025\text{ N}\).

Award 1 mark for the correct answer B. No marks are awarded for incorrect options.

The induced e.m.f. in a rotating conductor can be found by using its average speed. The velocity of the pivoted end is 0, and the velocity of the free end is \(v_{\text{max}} = \omega L\). The average speed of the rod is:
\(v_{\text{avg}} = \frac{0 + \omega L}{2} = \frac{1}{2} \omega L\).
Using the formula for induced e.m.f.:
\(\mathcal{E} = B L v_{\text{avg}} = \frac{1}{2} B \omega L^2 = \frac{1}{2} \times 0.2 \times 10 \times (0.5)^2 = 0.25\text{ V}\).

Award 1 mark for the correct answer B. No marks are awarded for incorrect options.

According to Lenz's law, the induced current in the copper ring always opposes the change in magnetic flux that produces it.
1. When the magnet falls towards the ring, the magnetic flux increases. The induced current in the ring creates a magnetic field that repels the magnet, resulting in an upward force. Thus, the downward acceleration \(a_1 < g\).
2. After the magnet passes through the ring and falls away, the magnetic flux decreases. The induced current creates a magnetic field that attracts the magnet, resulting in an upward force to oppose its departure. Thus, the downward acceleration \(a_2 < g\).
Therefore, both accelerations are less than \(g\).

Award 1 mark for the correct answer A. No marks are awarded for incorrect options.

1. Terminal voltage: \(V = E - I r\). Since \(I = \frac{E}{R+r}\), as \(R\) increases, the current \(I\) decreases, which makes \(V\) increase.
2. Power dissipated in the variable resistor: \(P = I^2 R = \frac{E^2 R}{(R+r)^2}\). According to the maximum power transfer theorem, \(P\) reaches a maximum when \(R = r\). Therefore, as \(R\) increases from a very small value (starting from \(R < r\) to \(R > r\)), \(P\) first increases and then decreases.

Award 1 mark for the correct answer B. No marks are awarded for incorrect options.

First, calculate the equivalent resistance of \(R_2\) and \(R_3\) in parallel:
\(R_{23} = \frac{R_2 R_3}{R_2 + R_3} = \frac{6 \times 12}{6 + 12} = 4\ \Omega\).
The total external resistance is:
\(R_{\text{ext}} = R_1 + R_{23} = 4 + 4 = 8\ \Omega\).
Since \(R_1\) is connected in series with the parallel combination, the current flowing through \(R_1\) is the total circuit current \(I\).
From \(P_1 = I^2 R_1\):
\(16 = I^2 \times 4 \implies I = 2\text{ A}\).
The total resistance of the circuit (including internal resistance) is:
\(R_{\text{total}} = R_{\text{ext}} + r = 8 + 2 = 10\ \Omega\).
Therefore, the e.m.f. of the battery is:
\(E = I R_{\text{total}} = 2 \times 10 = 20\text{ V}\).

Award 1 mark for the correct answer C. No marks are awarded for incorrect options.

According to the ideal gas law \(pV = nRT\):
For the initial state: \(p_1 = 1.0 \times 10^5\text{ Pa}\), \(T_1 = 27 + 273 = 300\text{ K}\), number of moles is \(n_1\).
For the final state: \(p_2\) is the final pressure, \(T_2 = 327 + 273 = 600\text{ K}\), and \(n_2 = 0.5 n_1\).
Since the volume \(V\) is constant:
\(\frac{p_2 V}{p_1 V} = \frac{n_2 R T_2}{n_1 R T_1} \implies \frac{p_2}{p_1} = \frac{n_2}{n_1} \times \frac{T_2}{T_1}\).
Substitute the values:
\(\frac{p_2}{1.0 \times 10^5} = 0.5 \times \frac{600}{300} = 1\).
Therefore, \(p_2 = 1.0 \times 10^5\text{ Pa}\).

Award 1 mark for the correct answer B. No marks are awarded for incorrect options.

According to the kinetic theory of gases, the root-mean-square (r.m.s.) speed of gas molecules is given by:
\(v_{\text{rms}} = \sqrt{\frac{3RT}{M}}\).
At the same temperature \(T\), the r.m.s. speed is inversely proportional to the square root of the molar mass \(M\):
\(v_{\text{rms}} \propto \frac{1}{\sqrt{M}}\).
Therefore, the ratio is:
\(\frac{v_{\text{rms, Y}}}{v_{\text{rms, X}}} = \sqrt{\frac{M_X}{M_Y}}\).
Given that \(M_X = 4 M_Y\):
\(\frac{v_{\text{rms, Y}}}{v_{\text{rms, X}}} = \sqrt{\frac{4}{1}} = 2\).
Thus, the ratio is \(2 : 1\).

Award 1 mark for the correct answer C. No marks are awarded for incorrect options.

For light to be prevented from escaping, any light ray reaching the surface outside the disc must undergo total internal reflection. The critical angle \(\theta_c\) is given by:
\(\sin\theta_c = \frac{1}{n} = \frac{1}{1.25} = 0.8\).
This represents a 3-4-5 right-angled triangle, where:
\(\cos\theta_c = \sqrt{1 - 0.8^2} = 0.6\),
\(\tan\theta_c = \frac{\sin\theta_c}{\cos\theta_c} = \frac{0.8}{0.6} = \frac{4}{3}\).
From the geometry, the radius of the disc \(r\) and the depth of liquid \(H\) are related by:
\(r = H \tan\theta_c = 1.2 \times \frac{4}{3} = 1.6\text{ m}\).

Award 1 mark for the correct answer C. No marks are awarded for incorrect options.

Let \(u\) and \(v\) be the object distance and image distance respectively for the first position of the lens.
The sum of these distances is the distance between the object and the screen:
\(u + v = L = 90\text{ cm}\).
Due to the symmetry of the lens equation, the second position of the lens interchanges the object and image distances. Thus, the distance between the two lens positions is:
\(|v - u| = d = 30\text{ cm}\).
Assume \(v > u\), we can solve the system of equations:
\(v + u = 90\)
\(v - u = 30\)
Adding them gives \(2v = 120 \implies v = 60\text{ cm}\).
Subtracting them gives \(2u = 60 \implies u = 30\text{ cm}\).
Using the thin lens formula:
\(\frac{1}{f} = \frac{1}{u} + \frac{1}{v} = \frac{1}{30} + \frac{1}{60} = \frac{3}{60} = \frac{1}{20}\).
Therefore, \(f = 20\text{ cm}\).

Award 1 mark for the correct answer B. No marks are awarded for incorrect options.

The fringe width \(w\) in double-slit interference is given by:
\(w = \frac{\lambda D}{d}\).
Let the new wavelength be \(\lambda'\), the new slit separation be \(d' = \frac{d}{2}\), and the new screen distance be \(D' = 2D\).
The new fringe width is:
\(w' = \frac{\lambda' D'}{d'} = \frac{\lambda' (2D)}{\frac{d}{2}} = 4 \frac{\lambda' D}{d}\).
For the fringe width to remain unchanged (\(w' = w\)):
\(4 \frac{\lambda' D}{d} = \frac{\lambda D}{d} \implies \lambda' = \frac{\lambda}{4}\).

Award 1 mark for the correct answer A. No marks are awarded for incorrect options.

Since the process is carried out slowly and the cylinder is immersed in an ice-water bath, the temperature of the gas remains constant at 0 degrees Celsius (isothermal process). For an ideal gas, internal energy depends only on its temperature. Since the temperature is constant, the average kinetic energy of the gas molecules and the internal energy of the gas remain unchanged. Thus, (1) is incorrect and (3) is correct. According to Boyle's law (\(pV = \text{constant}\)), as the volume is halved, the pressure of the gas doubles. Since the average speed of the molecules is constant (due to constant temperature), the increase in pressure is caused by the increase in the number density of the molecules, which increases the frequency of collisions on the cylinder walls. Thus, (2) is correct. Therefore, (2) and (3) only are correct.

Correct Answer: C (1 mark). No partial marks.

Let the resistance of each bulb be \(R\) and the source voltage be \(V\). When \(S\) is closed, the parallel combination of \(Y\) and \(Z\) has an equivalent resistance of \(R/2\). The total resistance is \(R + R/2 = 1.5R\). The current through bulb X is \(I_X = \frac{V}{1.5R} = \frac{2V}{3R}\), so the potential difference across bulb X is \(V_X = I_X \times R = \frac{2V}{3}\). The potential difference across bulb Y is \(V_Y = V - V_X = \frac{V}{3}\). When \(S\) is opened, bulb Z is disconnected. The circuit is now bulb X and bulb Y in series with a total resistance of \(2R\). The current becomes \(I' = \frac{V}{2R}\). The potential difference across bulb X becomes \(V_X' = \frac{V}{2}\). Since \(\frac{V}{2} < \frac{2V}{3}\), bulb X becomes dimmer. The potential difference across bulb Y becomes \(V_Y' = \frac{V}{2}\). Since \(\frac{V}{2} > \frac{V}{3}\), bulb Y becomes brighter. Thus, X becomes dimmer and Y becomes brighter.

Correct Answer: A (1 mark). No partial marks.

(1) Entering: As the loop enters the magnetic field, the magnetic flux into the page increases. According to Lenz's law, the induced current must create an outward magnetic field to oppose this increase. By the right-hand grip rule, the induced current is anticlockwise. (2) Fully inside: When the loop is entirely inside the uniform magnetic field, the magnetic flux through the loop remains constant. No electromotive force is induced, so the current is zero. (3) Leaving: As the loop leaves the magnetic field, the magnetic flux into the page decreases. By Lenz's law, the induced current must create an inward magnetic field to oppose this decrease, resulting in a clockwise current. Thus, the sequence is: Anticlockwise, Zero, Clockwise.

Correct Answer: A (1 mark). No partial marks.

The critical angle \(\theta_c\) of the glass prism is given by \(\sin\theta_c = \frac{1}{n} = \frac{1}{1.5} \approx 0.667\), which gives \(\theta_c \approx 41.8^\circ\). When the light ray enters normally through one of the shorter faces, it does not bend. At the hypotenuse, the normal to the surface makes an angle of \(45^\circ\) with the incident ray. Thus, the angle of incidence on the hypotenuse is \(45^\circ\). Since \(45^\circ > 41.8^\circ\), total internal reflection occurs at the hypotenuse. The angle of reflection is also \(45^\circ\). The reflected ray travels downwards, perpendicular to the other shorter face, and emerges from it normally without deviation. The final emerging ray is perpendicular to the original incident ray, meaning it has been deflected by \(90^\circ\). Thus, the total angle of deviation is \(90^\circ\).

Correct Answer: C (1 mark). No partial marks.

The net force acting on the charge is given by the Lorentz force formula: \(\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})\). The electric force is \(\vec{F}_E = q\vec{E} = qE \hat{i}\) (along the +x direction). The magnetic force is \(\vec{F}_B = q(\vec{v} \times \vec{B}) = q(v \hat{k} \times B \hat{j}) = -qvB \hat{i}\) (along the -x direction). Thus, the net force is \(\vec{F} = (qE - qvB)\hat{i}\). Depending on the magnitudes of E, v, and B, this force can point along the +x or -x direction, or can even be zero. In all cases, the net force must lie along the x-axis. Therefore, option C is correct.

Correct Answer: C (1 mark). No partial marks.

The root-mean-square (r.m.s.) speed of gas molecules is given by \(v_{\text{rms}} = \sqrt{\frac{3RT}{M}}\), where \(T\) is the absolute temperature and \(M\) is the molar mass of the gas. For Helium in container A, \(v_{\text{rms}, A} = \sqrt{\frac{3R \times 300}{4}}\). For Oxygen in container B, \(v_{\text{rms}, B} = \sqrt{\frac{3R \times 600}{32}}\). The ratio is: \(\frac{v_{\text{rms}, A}}{v_{\text{rms}, B}} = \sqrt{\frac{300 / 4}{600 / 32}} = \sqrt{\frac{75}{18.75}} = \sqrt{4} = 2\). Therefore, the ratio is 2 : 1.

Correct Answer: C (1 mark). No partial marks.

The maximum power that the circuit can safely support is: \(P_{\text{max}} = V \times I_{\text{fuse}} = 220\text{ V} \times 13\text{ A} = 2860\text{ W}\). Since the air-conditioner is already consuming 2200 W, the remaining allowable power is: \(P_{\text{remaining}} = 2860\text{ W} - 2200\text{ W} = 660\text{ W}\). Let's evaluate each option: (1) Total power = \(350\text{ W} + 100\text{ W} = 450\text{ W} < 660\text{ W}\) (Can be switched on). (2) Total power = \(800\text{ W} > 660\text{ W}\) (Cannot be switched on). (3) Total power = \(150\text{ W} + 500\text{ W} = 650\text{ W} < 660\text{ W}\) (Can be switched on). Therefore, only (1) and (3) can be switched on.

Correct Answer: C (1 mark). No partial marks.

The fringe width \(w\) is given by \(w = \frac{\lambda D}{d}\). For the red laser: \(w = \frac{650 \times 10^{-9} \times D}{d}\). For the green laser: \(w' = \frac{520 \times 10^{-9} \times 1.5D}{d}\). Taking the ratio of \(w'\) to \(w\): \(\frac{w'}{w} = \frac{520 \times 1.5}{650} = \frac{780}{650} = 1.2\). Thus, the new fringe width is \(1.2w\).

Correct Answer: C (1 mark). No partial marks.

(1) As the copper ring falls, the magnetic flux through the area enclosed by the ring changes. According to Faraday's law of electromagnetic induction, an electromotive force (emf) is induced in the ring, which appears across the gap. Hence, (1) is correct. (2) Since the ring has a gap, the electrical circuit is open, so no induced current can flow through the ring. Hence, (2) is incorrect. (3) Because there is no induced current, there is no induced magnetic field produced by the ring, and thus no magnetic force acts on the ring to oppose its motion. The only force acting on the ring is gravity, so its acceleration is always equal to the acceleration due to gravity g. Hence, (3) is correct. Therefore, (1) and (3) only are correct.

Correct Answer: C (1 mark). No partial marks.

According to the potential divider equation, when the LDR is in the dark: \(V_{\text{out}} = V_{\text{supply}} \times \frac{R_{\text{LDR}}}{R + R_{\text{LDR}}}\) which gives \(8 = 12 \times \frac{80}{R + 80}\). Solving this: \(8(R + 80) = 960 \implies 8R + 640 = 960 \implies 8R = 320 \implies R = 40\text{ k}\Omega\). Under bright light, the resistance of the LDR becomes 10 k\(\Omega\). The new output voltage is: \(V_{\text{out}}' = 12 \times \frac{10}{R + 10} = 12 \times \frac{10}{40 + 10} = 12 \times \frac{10}{50} = 2.4\text{ V}\). Thus, the output voltage under bright light is 2.4 V.

Correct Answer: D (1 mark). No partial marks.

The average translational kinetic energy of gas molecules is directly proportional to the absolute temperature, i.e., \(E_k = \frac{3}{2} k_B T \propto T\). Therefore, the ratio of the final average kinetic energy to the initial average kinetic energy is \(\frac{1.5T}{T} = 1.5\).

The root-mean-square speed of gas molecules is proportional to the square root of the absolute temperature, i.e., \(v_{\text{rms}} = \sqrt{\frac{3RT}{M}} \propto \sqrt{T}\). Therefore, the ratio of the final root-mean-square speed to the initial root-mean-square speed is \(\sqrt{1.5} \approx 1.22\).

Thus, the correct answer is B.

Award 1 mark for the correct answer (B). No mark is awarded for other options.

Let \(R\) be the resistance of each bulb and \(V\) be the e.m.f. of the source.

When switch \(S\) is closed, the equivalent resistance of the parallel branch (bulbs \(Y\) and \(Z\)) is \(R_{YZ} = R/2\).
The total resistance of the circuit is \(R_{\text{total}} = R_X + R_{YZ} = R + R/2 = 1.5R\).
The total current is \(I = \frac{V}{1.5R} = \frac{2V}{3R}\).
The current through bulb \(X\) is \(I_X = \frac{2V}{3R}\).
The current through bulb \(Y\) is \(I_Y = \frac{1}{2} I = \frac{V}{3R}\).

When switch \(S\) is opened, bulb \(Z\) is disconnected.
The total resistance of the circuit becomes \(R_{\text{total}}' = R_X + R_Y = 2R\).
The new total current is \(I' = \frac{V}{2R}\).
The current through bulb \(X\) becomes \(I_X' = \frac{V}{2R}\). Since \(I_X' < I_X\), bulb \(X\) becomes dimmer.
The current through bulb \(Y\) becomes \(I_Y' = \frac{V}{2R}\). Since \(I_Y' > I_Y\), bulb \(Y\) becomes brighter.

Therefore, the correct answer is C.

Award 1 mark for the correct answer (C). No mark is awarded for other options.

According to the right-hand grip rule, a clockwise current in the outer loop produces a magnetic field pointing into the paper at its center.

As the current \(I\) in the outer loop decreases, the magnetic flux pointing into the paper through the inner loop decreases.

By Lenz's law, the induced current in the inner loop must oppose this change by generating a magnetic field pointing into the paper.

According to the right-hand grip rule, this requires a clockwise induced current in the inner loop.

Hence, the magnetic field at the center points into the paper, and the induced current is clockwise. Option A is correct.

Award 1 mark for the correct answer (A). No mark is awarded for other options.

(a) By Charles' Law, at constant pressure: \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\). Given \(V_2 = 2 V_1\), we have \(T_2 = T_1 \times \frac{V_2}{V_1} = 300 \times 2 = 600 \text{ K}\).

(b) By the Pressure Law, at constant volume: \(\frac{P_2}{T_2} = \frac{P_3}{T_3}\). Therefore, \(T_3 = T_2 \times \frac{P_3}{P_2} = 600 \times \frac{0.8 \times 10^5}{1.0 \times 10^5} = 480 \text{ K}\).

(c) Since the internal energy of an ideal gas depends only on its temperature, and the final temperature (\(480 \text{ K}\)) is greater than the initial temperature (\(300 \text{ K}\)), the internal energy of the gas overall increases.

(a)
- Applying Charles' Law: 1M
- Correct substitution: 1M
- Final temperature \(T_2 = 600 \text{ K}\): 1A

(b)
- Applying Pressure Law: 1M
- Correct substitution of pressures: 1M
- Final temperature \(T_3 = 480 \text{ K}\): 1A

(c)
- Stating that internal energy increases: 1.2A
- Reasoning based on temperature increase (\(480 \text{ K} > 300 \text{ K}\)): 1.2M

(a) The critical angle \(c\) is given by \(\sin c = \frac{1}{n} = \frac{1}{1.50}\). Therefore, \(c = \sin^{-1}\left(\frac{1}{1.50}\right) \approx 41.8^\circ\).

(b) The light ray enters the short face normally, so it passes undeflected. It hits the hypotenuse interface at an angle of incidence of \(60^\circ\). Since the angle of incidence (\(60^\circ\)) is greater than the critical angle (\(41.8^\circ\)), total internal reflection (TIR) occurs. The reflected ray hits the other short face at an angle of incidence of \(30^\circ\). Since \(30^\circ < 41.8^\circ\), the ray refracts into the air. Using Snell's law: \(n \sin 30^\circ = 1.0 \sin \theta \Rightarrow 1.50 \times 0.5 = \sin \theta \Rightarrow \sin \theta = 0.75 \Rightarrow \theta = 48.6^\circ\). The path undergoes TIR at the hypotenuse (deflected by \(180^\circ - 2 \times 60^\circ = 60^\circ\)) and refracts at the exit face (deflected by \(48.6^\circ - 30^\circ = 18.6^\circ\)). The total angle of deviation is \(60^\circ + 18.6^\circ = 78.6^\circ\).

(a)
- Applying formula for critical angle: 1.2M
- Correct value of \(c = 41.8^\circ\): 1.2A

(b)
- Identification of TIR at hypotenuse (with reasoning that \(60^\circ > 41.8^\circ\)): 1.5M
- Correct application of Snell's Law at the exiting boundary: 1.5M
- Finding exit angle \(\theta = 48.6^\circ\): 1.5A
- Calculating the total deviation angle \(78.6^\circ\): 1.5A

(a) The induced electromotive force \(\mathcal{E}\) is given by: \(\mathcal{E} = B L v = 0.80 \times 0.20 \times 4.0 = 0.64 \text{ V}\).

(b) First, find the induced current: \(I = \frac{\mathcal{E}}{R} = \frac{0.64}{0.50} = 1.28 \text{ A}\). The magnetic force on the leading wire is: \(F_B = B I L = 0.80 \times 1.28 \times 0.20 = 0.2048 \text{ N}\). Since the speed is constant, the external force must balance this magnetic force: \(F_{ext} = F_B \approx 0.205 \text{ N}\).

(c) As the frame is pulled out, the inward magnetic flux through the loop decreases. According to Lenz's law, the induced current must generate an inward magnetic field to oppose this decrease. By the right-hand grip rule, this corresponds to a clockwise current.

(a)
- Applying \(\mathcal{E} = B L v\): 1.5M
- Correct value with units: 1.5A

(b)
- Calculating current \(I = 1.28 \text{ A}\): 1M
- Calculating magnetic force \(F_B = B I L\): 1M
- Stating \(F_{ext} = F_B = 0.205 \text{ N}\): 1A

(c)
- Stating 'clockwise': 1A
- Explanation mentioning decreasing inward flux: 0.7M
- Explanation mentioning opposition to change (Lenz's Law): 0.7M

(a) Distance from each wire to point \(P\) is \(r = \frac{0.15}{2} = 0.075 \text{ m}\). By the right-hand grip rule, wire \(X\) (upwards current) produces a magnetic field directed into the page at \(P\). Wire \(Y\) (downwards current) also produces a magnetic field directed into the page at \(P\).
Using \(B = \frac{\mu_0 I}{2 \pi r}\):
\(B_X = \frac{4\pi \times 10^{-7} \times 6.0}{2 \pi \times 0.075} = 1.60 \times 10^{-5} \text{ T}\) (into the page).
\(B_Y = \frac{4\pi \times 10^{-7} \times 4.0}{2 \pi \times 0.075} = 1.07 \times 10^{-5} \text{ T}\) (into the page).
Since both fields are in the same direction, the total magnetic field \(B_{total} = B_X + B_Y = 2.67 \times 10^{-5} \text{ T}\) directed into the page.

(b) The force per unit length is given by: \(\frac{F}{L} = \frac{\mu_0 I_X I_Y}{2 \pi d} = \frac{4\pi \times 10^{-7} \times 6.0 \times 4.0}{2 \pi \times 0.15} = 3.20 \times 10^{-5} \text{ N m}^{-1}\). Since the currents flow in opposite directions, the force is repulsive.

(a)
- Determining the directions of both magnetic fields (both into the page): 1M
- Applying \(B = \frac{\mu_0 I}{2 \pi r}\) to calculate individual fields: 1.5M
- Summing fields to obtain total \(B_{total} = 2.67 \times 10^{-5} \text{ T}\): 1A
- Specifying correct direction (into the page): 0.5A

(b)
- Applying the formula \(\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}\): 2M
- Calculating magnitude \(3.20 \times 10^{-5} \text{ N m}^{-1}\): 1.4A
- Correctly identifying the force as 'repulsive': 1A

(a) The equivalent resistance of the parallel branch is: \(R_p = \frac{R_2 \times R_3}{R_2 + R_3} = \frac{6.00 \times 3.00}{6.00 + 3.00} = 2.00 \Omega\).
The total external resistance of the circuit is: \(R_{ext} = R_1 + R_p = 4.50 + 2.00 = 6.50 \Omega\).

(b) The total resistance of the whole circuit is: \(R_{total} = R_{ext} + r = 6.50 + 1.50 = 8.00 \Omega\).
The total current from the battery is: \(I = \frac{\mathcal{E}}{R_{total}} = \frac{12.0}{8.00} = 1.50 \text{ A}\).
The terminal voltage is: \(V_{terminal} = \mathcal{E} - I r = 12.0 - (1.50 \times 1.50) = 9.75 \text{ V}\).

(c) The voltage across the parallel branch is: \(V_p = I R_p = 1.50 \times 2.00 = 3.00 \text{ V}\).
The power dissipated in \(R_3\) is: \(P_3 = \frac{V_p^2}{R_3} = \frac{3.00^2}{3.00} = 3.00 \text{ W}\).

(a)
- Correct parallel combination formula: 1M
- Value \(R_{ext} = 6.50 \Omega\): 1A

(b)
- Determining the total resistance \(8.00 \Omega\) and current \(1.50 \text{ A}\): 1.4M
- Formula for terminal voltage \(V = \mathcal{E} - I r\): 1M
- Calculating \(9.75 \text{ V\): 1A

(c)
- Finding the voltage across \(R_3\) (\(3.00 \text{ V}\)): 1M
- Applying \(P = \frac{V^2}{R}\) or equivalent power formulas: 1M
- Finding the final power \(3.00 \text{ W}\): 1A

(a) Taking the direction to the right as positive:
Using conservation of momentum: \(m_A u_A + m_B u_B = m_A v_A + m_B v_B\).
\(0.40 \times 3.0 + 0 = 0.40 v_A + 0.20 \times 4.0\)
\(1.2 = 0.40 v_A + 0.80 \Rightarrow 0.40 v_A = 0.40 \Rightarrow v_A = 1.0 \text{ m s}^{-1}\) (to the right).

(b) Calculate total kinetic energy before and after collision:
Initial kinetic energy \(K_i = \frac{1}{2} m_A u_A^2 = \frac{1}{2} \times 0.40 \times 3.0^2 = 1.80 \text{ J}\).
Final kinetic energy \(K_f = \frac{1}{2} m_A v_A^2 + \frac{1}{2} m_B v_B^2 = \frac{1}{2} \times 0.40 \times 1.0^2 + \frac{1}{2} \times 0.20 \times 4.0^2 = 0.20 + 1.60 = 1.80 \text{ J}\).
Since \(K_i = K_f\), kinetic energy is conserved, and therefore the collision is elastic.

(c) Since the air track is frictionless, there are no external horizontal forces acting on the two-glider system. By Newton's second law, since \(F_{ext} = 0\), the rate of change of momentum is zero, so the total momentum remains constant.

(a)
- Applying conservation of momentum equation: 1M
- Correct substitutions: 1M
- Value \(1.0 \text{ m s}^{-1}\) and direction 'right': 1A

(b)
- Calculating initial kinetic energy: 1M
- Calculating final kinetic energy and showing \(K_i = K_f\): 1M
- Concluding the collision is elastic: 1A

(c)
- Stating that net external force is zero: 1.2M
- Relating zero net external force to conservation of momentum: 1.2M

(a) The work done by the pulling force is: \(W_{pull} = F \times s = 120 \times 8.00 = 960 \text{ J}\).

(b) The normal force perpendicular to the incline is: \(N = m g \cos 30^\circ = 15.0 \times 9.81 \times \cos 30^\circ \approx 127.44 \text{ N}\).
The kinetic friction is: \(f_k = \mu_k N = 0.20 \times 127.44 \approx 25.49 \text{ N}\).
The work done against friction is: \(W_{f} = f_k \times s = 25.49 \times 8.00 \approx 203.9 \text{ J} \approx 204 \text{ J}\).

(c) The work done by gravity (potential energy change) is: \(\Delta U_g = m g h = m g s \sin 30^\circ = 15.0 \times 9.81 \times 8.00 \times \sin 30^\circ = 588.6 \text{ J}\).
By the Work-Energy Theorem:
\(W_{net} = \Delta K \Rightarrow W_{pull} - W_{f} - \Delta U_g = \frac{1}{2} m v^2\)
\(960 - 203.9 - 588.6 = \frac{1}{2} \times 15.0 \times v^2\)
\(167.5 = 7.50 v^2 \Rightarrow v^2 \approx 22.33 \Rightarrow v \approx 4.73 \text{ m s}^{-1}\).

(a)
- Calculating work done by pull force (\(W = F s\)): 1M
- Finding \(960 \text{ J}\): 1A

(b)
- Determining normal force \(N = m g \cos 30^\circ\): 1M
- Calculating friction \(f_k = \mu_k N\): 1M
- Calculating work done against friction \(204 \text{ J}\): 1.4A

(c)
- Finding potential energy gain \(588.6 \text{ J}\): 1M
- Applying conservation of energy or Work-Energy Theorem: 1M
- Finding the final velocity \(4.73 \text{ m s}^{-1}\): 1A

(a) The orbital radius is: \(r = R + h = 3.40 \times 10^6 + 4.00 \times 10^6 = 7.40 \times 10^6 \text{ m}\).
The gravitational force provides the centripetal force:
\(\frac{G M m}{r^2} = \frac{m v^2}{r} \Rightarrow v = \sqrt{\frac{G M}{r}}\)
\(v = \sqrt{\frac{6.67 \times 10^{-11} \times 6.42 \times 10^{23}}{7.40 \times 10^6}} = \sqrt{\frac{4.282 \times 10^{13}}{7.40 \times 10^6}} \approx 2405.5 \text{ m s}^{-1} \approx 2.41 \times 10^3 \text{ m s}^{-1}\).

(b) The period \(T\) is given by: \(T = \frac{2 \pi r}{v}\).
\(T = \frac{2 \pi \times 7.40 \times 10^6}{2405.5} \approx 1.933 \times 10^4 \text{ s}\).
Converting to hours:
\(T_{hours} = \frac{1.933 \times 10^4}{3600} \approx 5.37 \text{ hours}\).

(a)
- Finding total orbital radius \(r = 7.40 \times 10^6 \text{ m}\): 1M
- Equating gravitational force to centripetal force: 1M
- Correct formula for \(v\): 1M
- Finding speed \(2.41 \times 10^3 \text{ m s}^{-1}\): 1A

(b)
- Using relationship \(T = \frac{2\pi r}{v}\): 2M
- Finding period in seconds: 1M
- Correct conversion to hours (\(5.37 \text{ hours}\)): 1.4A

(a) The total energy supplied by the heater is: \(E = P \times t = 150 \times (4.00 \times 60) = 36000 \text{ J}\).
The temperature rise is: \(\Delta T = 58.0 - 22.0 = 36.0 \text{ K}\).
Assuming no heat loss: \(E = m_l c_l \Delta T \Rightarrow 36000 = 0.45 \times c_l \times 36.0\).
\(c_l = \frac{36000}{16.2} \approx 2220 \text{ J kg}^{-1} \text{ K}^{-1}\).

(b) The heat absorbed by the aluminum container is: \(Q_{container} = m_{al} c_{al} \Delta T = 0.12 \times 900 \times 36.0 = 3888 \text{ J}\).
The energy absorbed by the liquid is: \(Q_{liquid} = E - Q_{container} = 36000 - 3888 = 32112 \text{ J}\).
Therefore, the corrected specific heat capacity is:
\(Q_{liquid} = m_l c_{corr} \Delta T \Rightarrow 32112 = 0.45 \times c_{corr} \times 36.0\).
\(c_{corr} = \frac{32112}{16.2} \approx 1980 \text{ J kg}^{-1} \text{ K}^{-1}\).

(a)
- Finding total thermal energy supplied (\(36000 \text{ J}\)): 1.5M
- Applying \(Q = m c \Delta T\): 1.5M
- Calculating specific heat capacity \(2220 \text{ J kg}^{-1} \text{ K}^{-1}\): 1A

(b)
- Finding heat absorbed by aluminum container (\(3888 \text{ J}\)): 1.5M
- Finding corrected heat for the liquid (\(32112 \text{ J}\)): 1.5M
- Calculating corrected specific heat capacity \(1980 \text{ J kg}^{-1} \text{ K}^{-1}\): 1.4A

(a) The heat required to melt the ice is:
\(Q_{melt} = m_{ice} L_f = 0.08 \times 3.34 \times 10^5 = 2.672 \times 10^4 \text{ J} \approx 2.67 \times 10^4 \text{ J}\).

(b) The maximum heat that can be released by water when it cools down to \(0.0^\circ\text{C}\) is:
\(Q_{max} = m_w c_w \Delta T = 0.30 \times 4200 \times (25.0 - 0) = 31500 \text{ J}\).
Since \(Q_{max} > Q_{melt}\) (\(31500 \text{ J} > 26720 \text{ J}\)), the heat from the water is more than enough to melt all the ice. Therefore, all the ice will melt, and the final temperature \(T_f\) will be above \(0.0^\circ\text{C}\).
By conservation of energy:
Heat gained by ice (melting + heating) = Heat lost by water
\(Q_{melt} + m_{ice} c_w (T_f - 0) = m_w c_w (25.0 - T_f)\)
\(26720 + 0.08 \times 4200 \times T_f = 0.30 \times 4200 \times (25.0 - T_f)\)
\(26720 + 336 T_f = 1260 \times (25.0 - T_f) = 31500 - 1260 T_f\)
\((336 + 1260) T_f = 31500 - 26720\)
\(1596 T_f = 4780 \Rightarrow T_f \approx 2.99^\circ\text{C}\).

(a)
- Using \(Q = m L\): 1.2M
- Correct answer with units (\(2.67 \times 10^4 \text{ J}\)): 1.2A

(b)
- Determining the maximum energy from cooling water (\(31500 \text{ J}\)) and proving all ice melts: 1.5M
- Writing energy conservation equation (\(Q_{melt} + m_i c_w T_f = m_w c_w (25 - T_f)\)): 1.5M
- Correct substitution: 1.5M
- Finding the final temperature \(2.99^\circ\text{C}\) (accept \(3.0^\circ\text{C}\)): 1.5A

The lifetime of a star \(\tau\) is proportional to its mass \(M\) (available fuel) divided by its luminosity \(L\) (rate of fuel consumption): \(\tau \propto \frac{M}{L}\). Given \(L \propto M^{3.5}\), we have \(\tau \propto \frac{M}{M^{3.5}} = M^{-2.5}\). Therefore, \(\frac{\tau_1}{\tau_2} = \left(\frac{M_2}{M_1}\right)^{2.5} = 3^{2.5} \approx 15.6\).

1 mark for selecting C. Correctly relating lifetime to \(\tau \propto M / L\) is required.

First, find the redshift \(z\): \(z = \frac{\Delta \lambda}{\lambda_0} = \frac{512.5 - 500.0}{500.0} = \frac{12.5}{500.0} = 0.025\). The recession speed \(v = z c = 0.025 \times 3.0 \times 10^5\text{ km s}^{-1} = 7500\text{ km s}^{-1}\). According to Hubble's Law, \(v = H_0 d\), so the distance \(d = \frac{v}{H_0} = \frac{7500}{75} = 100\text{ Mpc}\).

1 mark for B. Method involves calculating redshift, speed, and using Hubble's law.

From Einstein's photoelectric equation, \(e V = \frac{hc}{\lambda} - \phi\). Thus, \(e V_0 = \frac{hc}{\lambda} - \phi\) (Equation 1), and \(2.5 e V_0 = \frac{hc}{\lambda/2} - \phi = 2 \frac{hc}{\lambda} - \phi\) (Equation 2). From Equation 1, we get \(\frac{hc}{\lambda} = e V_0 + \phi\). Substitute this into Equation 2: \(2.5 e V_0 = 2(e V_0 + \phi) - \phi = 2 e V_0 + \phi \implies \phi = 0.5 e V_0\).

1 mark for B.

The energy level is given by \(E_n = -\frac{E_0}{n^2}\). For \(n=4 \to n=2\), the energy emitted is \(E = E_4 - E_2 = -E_0 \left(\frac{1}{16} - \frac{1}{4}\right) = \frac{3}{16} E_0\), so \(E_0 = \frac{16}{3} E\). For \(n=3 \to n=1\), the energy emitted is \(E' = E_3 - E_1 = -E_0 \left(\frac{1}{9} - 1\right) = \frac{8}{9} E_0\). Thus, \(E' = \frac{8}{9} \times \frac{16}{3} E = \frac{128}{27} E \approx 4.74 E\).

1 mark for D.

(1) is effective because a lower shading coefficient reduces solar heat gain through glass windows, reducing the OTTV. (2) is effective because a lighter color reduces solar absorption on opaque walls, lowering the equivalent temperature difference (\(TD_{eq}\)), thus lowering the OTTV. (3) is ineffective because windows generally have higher heat transfer and solar heat gain coefficients than opaque walls, so increasing window area increases the overall OTTV.

1 mark for B.

The power available from the wind is \(P_{wind} = \frac{1}{2} \rho A v^3\), where \(A = \pi L^2\) is the swept area. The electrical power output is \(P = \eta P_{wind} \propto L^2 v^3\). Therefore, the new power output is \(P' \propto (2L)^2 \times (0.5v)^3 = 4 L^2 \times 0.125 v^3 = 0.5 P\).

1 mark for B.

To view an object at \(u = 25\text{ cm} = 0.25\text{ m}\), the corrective lens must form a virtual image at the person's near point, so \(v = -80\text{ cm} = -0.80\text{ m}\). Using the lens formula: \(P = \frac{1}{f} = \frac{1}{u} + \frac{1}{v} = \frac{1}{0.25} + \frac{1}{-0.80} = 4 - 1.25 = +2.75\text{ D}\).

1 mark for B. Note that positive power represents a converging lens.

The reflection coefficient \(\alpha\) (fraction of intensity reflected) is given by \(\alpha = \frac{(Z_2 - Z_1)^2}{(Z_2 + Z_1)^2}\). Substitute \(Z_1 = 1.5 \times 10^6\text{ kg m}^{-2}\text{ s}^{-1}\) and \(Z_2 = 6.0 \times 10^6\text{ kg m}^{-2}\text{ s}^{-1}\): \(\alpha = \frac{(6.0 - 1.5)^2}{(6.0 + 1.5)^2} = \left(\frac{4.5}{7.5}\right)^2 = (0.6)^2 = 0.36 = 36\%\).

1 mark for A.

The escape velocity from the surface of the planet is \(v_{esc} = \sqrt{\frac{2GM}{R}}\). The probe orbits at a distance \(r = R + h = 2R\) from the center of the planet. The orbital speed is \(v_{orb} = \sqrt{\frac{GM}{r}} = \sqrt{\frac{GM}{2R}}\). Thus, the ratio of the escape velocity to the orbital speed is \(\frac{v_{esc}}{v_{orb}} = \frac{\sqrt{2GM/R}}{\sqrt{GM/2R}} = \sqrt{4} = 2.00\).

1 mark for B.

The kinetic energy gained is \(K = qV\). The momentum is \(p = \sqrt{2mK} = \sqrt{2mqV}\). The de Broglie wavelength is \(\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mqV}}\). The ratio of the wavelengths is \(\frac{\lambda_e}{\lambda_{\alpha}} = \sqrt{\frac{m_{\alpha} q_{\alpha}}{m_e q_e}} = \sqrt{\frac{(4m)(2e)}{m \cdot e}} = \sqrt{8} \approx 2.83\).

1 mark for C.

According to Stefan-Boltzmann law, the luminosity of a star is given by \(L = 4\pi R^2 \sigma T^4\). Comparing the star with the Sun: \(\frac{L}{L_\odot} = \left(\frac{R}{R_\odot}\right)^2 \left(\frac{T}{T_\odot}\right)^4\). Given that \(L/L_\odot = 81\) and \(T/T_\odot = 2\), we can substitute these values: \(81 = \left(\frac{R}{R_\odot}\right)^2 \times 2^4\), which simplifies to \(81 = 16 \left(\frac{R}{R_\odot}\right)^2\). Thus, \(\left(\frac{R}{R_\odot}\right)^2 = \frac{81}{16} = 5.0625\), which gives \(\frac{R}{R_\odot} = \sqrt{5.0625} = 2.25\). Therefore, the correct option is B.

Award 1 mark for the correct answer B. No marks for any incorrect options or unanswered questions.

Using the relationship between redshift and recession velocity for non-relativistic speeds: \(v = zc = 0.04 \times (3.0 \times 10^5\text{ km s}^{-1}) = 12000\text{ km s}^{-1}\). According to Hubble's Law, \(v = H_0 d\), where \(d\) is the distance: \(12000\text{ km s}^{-1} = (70\text{ km s}^{-1}\text{ Mpc}^{-1}) \times d\). This gives \(d = \frac{12000}{70}\text{ Mpc} \approx 171.4\text{ Mpc}\). Rounding to the nearest integer gives 171 Mpc, which corresponds to option B.

Award 1 mark for the correct answer B. No marks for any incorrect options or unanswered questions.

The maximum wind power available to a wind turbine is given by \(P_{\text{wind}} = \frac{1}{2} \rho A v^3\), where \(A = \pi L^2\) is the swept area of the blades. This gives: \(P = \eta P_{\text{wind}} = \frac{1}{2} \eta \rho \pi L^2 v^3\). So, the electrical power is proportional to \(L^2 v^3\). When \(L\) increases to \(1.2 L\) and \(v\) increases to \(1.5 v\): \(P' \propto (1.2 L)^2 \times (1.5 v)^3 = 1.44 L^2 \times 3.375 v^3 = 4.86 \times (L^2 v^3)\). Therefore, the new electrical power is \(4.86 P\), which corresponds to option D.

Award 1 mark for the correct answer D. No marks for any incorrect options or unanswered questions.

The rate of heat conduction is given by \(\frac{Q}{t} = \frac{k A \Delta T}{d}\). Here, \(k = 0.80\text{ W m}^{-1}\text{ K}^{-1}\), \(A = 4.0\text{ m}^2\), \(\Delta T = 35 - 23 = 12^\circ\text{C} = 12\text{ K}\), and \(d = 3\text{ mm} = 3 \times 10^{-3}\text{ m}\). Substituting these values: \(\frac{Q}{t} = \frac{0.80 \times 4.0 \times 12}{3 \times 10^{-3}} = \frac{38.4}{3 \times 10^{-3}} = 12800\text{ W} = 12.8\text{ kW}\). Therefore, the correct option is B.

Award 1 mark for the correct answer B. No marks for any incorrect options or unanswered questions.

The intensity after passing through multiple layers is given by: \(I = I_0 e^{-\mu_1 x_1 - \mu_2 x_2}\). Here, for soft tissue: \(\mu_1 = 0.30\text{ cm}^{-1}\), \(x_1 = 3.0\text{ cm}\), so \(\mu_1 x_1 = 0.30 \times 3.0 = 0.90\). For bone: \(\mu_2 = 1.20\text{ cm}^{-1}\), \(x_2 = 1.5\text{ cm}\), so \(\mu_2 x_2 = 1.20 \times 1.5 = 1.80\). Therefore, the total exponent is: \(-(\mu_1 x_1 + \mu_2 x_2) = -(0.90 + 1.80) = -2.70\). The fraction transmitted is: \(\frac{I}{I_0} = e^{-2.70} \approx 0.0672\). Expressing this as a percentage: \(0.0672 \times 100\% \approx 6.7\%\). Therefore, the correct option is B.

Award 1 mark for the correct answer B. No marks for any incorrect options or unanswered questions.

According to Einstein's photoelectric equation: \(hf = \Phi + K\) (Equation 1) where \(\Phi\) is the work function. This gives \(K = hf - \Phi\). For the second case: \(h(1.5f) = \Phi + 2K\) (Equation 2). Substitute \(K\) from Equation 1 into Equation 2: \(1.5hf = \Phi + 2(hf - \Phi)\) which simplifies to \(1.5hf = \Phi + 2hf - 2\Phi\), and thus \(1.5hf = 2hf - \Phi\). Solving for \(\Phi\) gives \(\Phi = 2hf - 1.5hf = 0.5hf\). Therefore, the correct option is B.

Award 1 mark for the correct answer B. No marks for any incorrect options or unanswered questions.

(a) (i) Rate of solar energy input:
\(P_{\text{in}} = I \times A = 800 \times 3.0 = 2400\text{ W}\) (or \(\text{J s}^{-1}\))

(ii) Rate of heat energy absorbed by water:
\(P_{\text{out}} = \frac{\Delta Q}{\Delta t} = \frac{m}{\Delta t} c \Delta T\)
\(P_{\text{out}} = 0.040 \times 4200 \times (31 - 20) = 1848\text{ W}\)

(iii) Efficiency of the collector:
\(\eta = \frac{P_{\text{out}}}{P_{\text{in}}} \times 100\% = \frac{1848}{2400} \times 100\% = 77\%\) (or \(0.77\))

(b)
1. **Greenhouse Effect (Radiation loss reduction)**: The transparent glass cover allows short-wavelength solar radiation to pass through to heat the absorber plate. The hot plate emits long-wavelength infrared radiation, to which glass is opaque. Thus, the heat is trapped inside, reducing radiation loss to the surroundings.
2. **Convection & Conduction reduction**: The layer of air trapped between the glass cover and the absorber plate acts as a poor conductor (insulator). This stagnant air layer also prevents free convection currents from carrying heat rapidly to the outside air.

(a) (i)
- Correct formula or substitution: \(800 \times 3.0\) (1M)
- Correct answer with unit: \(2400\text{ W}\) (or \(2.4\text{ kW}\)) (1A)

(ii)
- Correct formula/substitution: \(0.040 \times 4200 \times (31-20)\) (1M)
- Correct answer with unit: \(1848\text{ W}\) (or \(1.85\text{ kW}\)) (1A)

(iii)
- Calculation of efficiency ratio: \(1848 / 2400\) (1M)
- Correct answer: \(77\%\) (or \(0.77\)) (1A)

(b)
- Glass traps long-wavelength infrared radiation / Greenhouse effect explained (2M)
- Trapped air reduces heat loss through convection or conduction (2M)

(a) Acoustic impedance is defined as \(Z = \rho v\), where \(rho\) is the density of the medium and \(v\) is the speed of sound in the medium.
SI Unit: \(\text{kg m}^{-2}\text{ s}^{-1}\).

(b) Reflection coefficient (fraction of intensity reflected):
\(\alpha = \left( \frac{Z_2 - Z_1}{Z_2 + Z_1} \right)^2\)
\(\alpha = \left( \frac{7.80 \times 10^6 - 1.70 \times 10^6}{7.80 \times 10^6 + 1.70 \times 10^6} \right)^2 = \left( \frac{6.10}{9.50} \right)^2 \approx 0.412\) (or \(41.2\%\))

(c) The time interval represents the round-trip travel of the ultrasound pulse in the muscle layer.
\(2d = v \times t\)
\(2d = 1540 \times (24.0 \times 10^{-6}) = 0.03696\text{ m}\)
\(d = 0.01848\text{ m} \approx 0.0185\text{ m}\) (or \(1.85\text{ cm}\) / \(18.5\text{ mm}\))

(d) Air has a very low acoustic impedance compared to human tissue, resulting in a very large impedance mismatch at the transducer-air-skin boundary. This causes almost all (nearly \(100\%\)) of the ultrasound waves to be reflected before entering the body. The coupling gel has an acoustic impedance matching that of tissue, eliminating air gaps and allowing ultrasound to transmit efficiently into the patient's body.

(a)
- Correct definition of acoustic impedance: product of density and speed of sound in medium (1M)
- Correct unit: \(\text{kg m}^{-2}\text{ s}^{-1}\) (1A)

(b)
- Correct formula for reflection coefficient (1M)
- Substitution of values: \(\left( \frac{7.80 - 1.70}{7.80 + 1.70} \right)^2\) (1M)
- Correct answer: \(0.412\) (or \(41.2\%\)) (1A)

(c)
- Recognizing round-trip distance \(2d\) (1M)
- Correct substitution of speed and time: \(1540 \times 24.0 \times 10^{-6}\) (1M)
- Correct thickness: \(0.0185\text{ m}\) (accept \(1.85\text{ cm}\) or \(18.5\text{ mm}\)) (1A)

(d)
- Explain large impedance mismatch with air causing high reflection (1M)
- Explain gel matches impedance of tissue to allow transmission (1M)