2022 HKDSE Physics Practice Paper with Answers

(1) is incorrect: \(c_{\text{r.m.s.}} \propto \sqrt{T}\), so \(c_{\text{r.m.s.}}' = \sqrt{\frac{450}{300}} c_{\text{r.m.s.}} = \sqrt{1.5}\, c_{\text{r.m.s.}} \approx 1.22\, c_{\text{r.m.s.}}\), not \(1.5\).
(2) is correct: Average kinetic energy \(E_k \propto T\), so \(\frac{E_k'}{E_k} = \frac{450}{300} = 1.5\), which is an increase of \(50\%\).
(3) is correct: The collision frequency per unit area with the container walls is proportional to \(n \bar{v}\). Since number density \(n\) is constant in a fixed volume and average molecular speed \(\bar{v}\) increases, the frequency of collisions increases.

Award 1 mark for option D.

At the critical angle \(\theta_c = 60^\circ\), by Snell's law at the flat interface:
\[ n_{\text{glass}} \sin \theta_c = n_L \sin 90^\circ \]
\[ 1.50 \times \sin 60^\circ = n_L \times 1 \]
\[ n_L = 1.50 \times \frac{\sqrt{3}}{2} \approx 1.30 \]

Award 1 mark for option C.

According to Newton's third law of motion, the magnetic force exerted by wire \(Q\) on wire \(P\) and that exerted by wire \(P\) on wire \(Q\) form an action-and-reaction pair. Hence, they are equal in magnitude and opposite in direction. The force per unit length on wire \(Q\) must also be \(F\).

Award 1 mark for option A.

Terminal potential difference is given by \(V = E - Ir = E \frac{R}{R+r} = \frac{E}{1 + \frac{r}{R}}\). As \(R\) increases, \(\frac{r}{R}\) decreases, so \(V\) increases continuously towards \(E\).
Power dissipated in \(R\) is \(P = I^2 R = \frac{E^2 R}{(R+r)^2}\). This power reaches a maximum when \(R = r\) (maximum power transfer theorem). Therefore, as \(R\) increases from a value much less than \(r\) to much greater than \(r\), \(P\) increases to a maximum and then decreases.

Award 1 mark for option B.

Area of the coil \(A = \pi r^2 = \pi (0.04)^2 = 1.6\pi \times 10^{-3}\text{ m}^2\).
The rate of change of magnetic flux is:
\[ \frac{\Delta \Phi}{\Delta t} = A \frac{\Delta B}{\Delta t} = (1.6\pi \times 10^{-3}) \times \frac{0.50 - 0.10}{0.20} = (1.6\pi \times 10^{-3}) \times 2.0 = 3.2\pi \times 10^{-3}\text{ Wb s}^{-1} \]
Induced e.m.f. is:
\[ \varepsilon = N \frac{\Delta \Phi}{\Delta t} = 50 \times 3.2\pi \times 10^{-3} = 0.16\pi \approx 0.503\text{ V} \approx 0.50\text{ V} \]

Award 1 mark for option C.

The fringe separation formula is \(\Delta y = \frac{\lambda D}{a}\).
With the changes:
- \(\lambda' = 1.5\lambda\)
- \(D' = 2D\)
- \(a' = 0.5a\)

The new fringe separation is:
\[ \Delta y' = \frac{\lambda' D'}{a'} = \frac{(1.5\lambda)(2D)}{0.5a} = \frac{3.0}{0.5} \frac{\lambda D}{a} = 6\Delta y \]

Award 1 mark for option D.

Initial total momentum: \(p_i = m_1 u_1 + m_2 u_2 = 3 \times 4 + 5 \times 0 = 12\text{ kg m s}^{-1}\).
By conservation of momentum, common velocity \(v\) is:
\[ v = \frac{p_i}{m_1 + m_2} = \frac{12}{3 + 5} = 1.5\text{ m s}^{-1} \]
Initial kinetic energy \(E_{k,i} = \frac{1}{2} m_1 u_1^2 = \frac{1}{2} \times 3 \times 4^2 = 24\text{ J}\).
Final kinetic energy \(E_{k,f} = \frac{1}{2} (m_1 + m_2) v^2 = \frac{1}{2} \times 8 \times 1.5^2 = 9\text{ J}\).
Loss in kinetic energy \(\Delta E_k = 24 - 9 = 15\text{ J}\).

Award 1 mark for option C.

The count rate decreased from \(960\text{ cpm}\) to \(120\text{ cpm}\).
\[ \frac{120}{960} = \frac{1}{8} = \left(\frac{1}{2}\right)^3 \]
This corresponds to \(3\) half-lives.
Therefore, \(3 t_{1/2} = 18\text{ hours} \implies t_{1/2} = \frac{18}{3} = 6\text{ hours}\).

Award 1 mark for option B.

(1) is correct: \(c_{\text{rms}} = \sqrt{\frac{3RT}{M}} \propto \sqrt{T}\). When \(T\) doubles, \(c_{\text{rms}}\) increases by a factor of \(\sqrt{2}\).
(2) is correct: The average kinetic energy \(E_k = \frac{3}{2} k T \propto T\), so it doubles when \(T\) doubles.
(3) is incorrect: The collision rate per unit area is proportional to \(n \bar{v} \propto \frac{N}{V} \sqrt{T}\). Since volume is constant, collision frequency increases by a factor of \(\sqrt{2}\), not 2.

B (1 mark): Both statements (1) and (2) are correct, while statement (3) is incorrect.

For vertical equilibrium, the upward static friction \(f_s\) must balance the downward gravitational force: \(f_s = mg\).
The maximum static friction is \(f_{s,\max} = \mu_s N = \mu_s P\).
To prevent slipping:
\(mg \le \mu_s P \implies P \ge \frac{mg}{\mu_s} = \frac{4.0 \times 9.81}{0.50} = 78.48\text{ N} \approx 78.5\text{ N}\).

C (1 mark): Correct calculation of normal force balancing weight via static friction coefficient.

Taking the rebound direction as positive:
Initial velocity \(u = -15\text{ m s}^{-1}\), final velocity \(v = +15\text{ m s}^{-1}\).
Change in momentum \(\Delta p = m(v - u) = 0.20 \times (15 - (-15)) = 6.0\text{ N s}\).
Average force \(F = \frac{\Delta p}{\Delta t} = \frac{6.0}{0.040} = 150\text{ N}\).

C (1 mark): Correct application of impulse-momentum theorem \(F \Delta t = m \Delta v\).

The critical angle \(c\) for the glass-air interface is given by:
\(\sin c = \frac{1}{n} = \frac{1}{1.50} \approx 0.6667 \implies c \approx 41.8^\circ\).
Total internal reflection occurs when the angle of incidence \(i > c\).
Since \(45^\circ > 41.8^\circ\) and \(60^\circ > 41.8^\circ\), TIR occurs for (2) and (3), but not for (1) where \(38^\circ < 41.8^\circ\).

D (1 mark): Total internal reflection requires angle of incidence greater than the critical angle (\(41.8^\circ\)).

The fringe separation formula for double-slit interference is \(\Delta y = \frac{\lambda D}{a}\).
- Increasing \(\lambda\) increases \(\Delta y\) (statement (1) is correct).
- Increasing \(a\) decreases \(\Delta y\) (statement (2) is incorrect).
- Increasing \(D\) increases \(\Delta y\) (statement (3) is correct).

B (1 mark): Statements (1) and (3) both increase fringe width \(\Delta y\).

Total electrical power dissipated by a network connected across a fixed voltage \(V\) is \(P = \frac{V^2}{R_{\text{eq}}}\), which is maximized when the equivalent resistance \(R_{\text{eq}}\) is minimized.
- A: Three in series: \(R_{\text{eq}} = 3R \implies P = \frac{V^2}{3R}\)
- B: Three in parallel: \(R_{\text{eq}} = \frac{R}{3} \implies P = \frac{3V^2}{R}\)
- C: Two in series in parallel with one: \(R_{\text{eq}} = \frac{2R \times R}{2R + R} = \frac{2}{3}R \implies P = \frac{1.5V^2}{R}\)
- D: Two in parallel in series with one: \(R_{\text{eq}} = \frac{R}{2} + R = 1.5R \implies P = \frac{V^2}{1.5R} = \frac{2V^2}{3R}\)
Thus, the three-in-parallel combination has the smallest \(R_{\text{eq}}\) and highest power.

B (1 mark): Parallel network provides lowest equivalent resistance and therefore maximum total power.

For the electric field to be zero, the field vectors due to both charges must be equal in magnitude and opposite in direction.
Between \(x = 0\) and \(x = d\), both fields point in the \(+x\) direction, so they cannot cancel.
For \(x > d\), the field of \(-4q\) dominates because it is closer and has larger magnitude.
Thus, the zero point must lie on the side of the smaller magnitude charge: \(x < 0\).
Let \(x = -r\) (where \(r > 0\)):
\(E_1 = \frac{k q}{r^2}\) and \(E_2 = \frac{k(4q)}{(d+r)^2}\).
Setting \(E_1 = E_2\):
\(\frac{1}{r^2} = \frac{4}{(d+r)^2} \implies \frac{1}{r} = \frac{2}{d+r} \implies d + r = 2r \implies r = d\).
Therefore, \(x = -d\).

A (1 mark): Correctly locating and solving the neutral point along the axis outside the charges.

Area of the circular coil: \(A = \pi r^2 = \pi (0.10)^2 = 0.01\pi\text{ m}^2\).
Rate of change of magnetic field: \(\left|\frac{\Delta B}{\Delta t}\right| = \frac{0.80 - 0.20}{0.15} = \frac{0.60}{0.15} = 4.0\text{ T s}^{-1}\).
By Faraday's law of electromagnetic induction, the magnitude of induced e.m.f. is:
\(\varepsilon = N \frac{\Delta \Phi}{\Delta t} = N A \left|\frac{\Delta B}{\Delta t}\right| = 50 \times (0.01\pi) \times 4.0 = 2\pi \approx 6.28\text{ V}\).

D (1 mark): Correct calculation of induced e.m.f. \(\varepsilon = N A \frac{\Delta B}{\Delta t}\).

According to the ideal gas equation \( pV = nRT = \frac{m}{M}RT \), since the volume \( V \), temperature \( T \), and molar mass \( M \) remain constant, the mass of gas in the container is directly proportional to its pressure (\( m \propto p \)).

The remaining fraction of the gas is:
\[ \frac{m_2}{m_1} = \frac{p_2}{p_1} = \frac{0.90 \times 10^5}{1.20 \times 10^5} = 0.75 = 75\% \]

Therefore, the percentage of gas that has escaped is:
\[ 100\% - 75\% = 25\% \]

Award 1 mark for correct answer B.

By conservation of linear momentum:
\[ m_1 u_1 + m_2 u_2 = (m_1 + m_2) v \]
\[ (1.5)(4.0) + 0 = (1.5 + 2.5) v \implies v = 1.5 \text{ m s}^{-1} \]

Initial kinetic energy:
\[ E_{k1} = \frac{1}{2} m_1 u_1^2 = \frac{1}{2} (1.5)(4.0)^2 = 12.0 \text{ J} \]

Final kinetic energy:
\[ E_{k2} = \frac{1}{2} (m_1 + m_2) v^2 = \frac{1}{2} (4.0)(1.5)^2 = 4.5 \text{ J} \]

Loss in kinetic energy:
\[ \Delta E_k = 12.0 - 4.5 = 7.5 \text{ J} \]

Award 1 mark for correct answer B.

(1) Correct. The critical angle \( c \) is given by \( \sin c = \frac{n_2}{n_1} = \frac{1.20}{1.60} = 0.75 \implies c = \arcsin(0.75) \approx 48.6^\circ \).
(2) Correct. Speed of light in a medium is \( v = \frac{c}{n} \). Since \( n_2 < n_1 \), \( v_2 > v_1 \).
(3) Incorrect. Since the angle of incidence \( 60^\circ > c \), total internal reflection occurs, meaning all light is reflected (none is transmitted/refracted into medium 2).

Award 1 mark for correct answer B.

Current in the circuit is given by \( I = \frac{E}{R + r} \).
As \( R \) increases, the total resistance increases, so the circuit current \( I \) decreases.

1. The terminal potential difference across the battery is \( V = E - Ir \). Since \( I \) decreases, the lost volts \( Ir \) decrease, so \( V \) increases.
2. The rate of energy dissipation inside the battery is \( P_{\text{internal}} = I^2 r \). Since \( I \) decreases and \( r \) is constant, \( P_{\text{internal}} \) decreases.

Award 1 mark for correct answer A.

According to Faraday's law, an e.m.f. is induced whenever there is a change in the magnetic flux \( \Phi = B A \cos\theta \) linked with the loop.
(1) Correct. When \( B \) increases, magnetic flux \( \Phi \) increases, inducing an e.m.f.
(2) Correct. Rotating the loop about a diameter changes the angle \( \theta \) between the field and the area normal, thereby varying \( \Phi \) and inducing an e.m.f.
(3) Incorrect. Since the field is uniform and the loop remains entirely inside the field, the magnetic flux through the loop remains constant during planar translation; thus, no net e.m.f. is induced.

Award 1 mark for correct answer B.

The magnetic force provides the centripetal force:
\[ qvB = \frac{mv^2}{r} \implies r = \frac{mv}{qB} \]
For the second particle:
\[ r' = \frac{(4m)v}{(2q)B} = 2\left(\frac{mv}{qB}\right) = 2R \]

The period of circular motion is:
\[ T = \frac{2\pi r}{v} = \frac{2\pi m}{qB} \]
For the second particle:
\[ T' = \frac{2\pi (4m)}{(2q)B} = 2\left(\frac{2\pi m}{qB}\right) = 2T \]

Award 1 mark for correct answer A.

For a stationary wave on a string fixed at both ends with 3 antinodes, the length \( L \) equals 3 half-wavelengths:
\[ L = 3\left(\frac{\lambda}{2}\right) \implies \lambda = \frac{2L}{3} = \frac{2(1.20)}{3} = 0.80 \text{ m} \]

The wave speed \( v \) is:
\[ v = f \lambda = (150 \text{ Hz})(0.80 \text{ m}) = 120 \text{ m s}^{-1} \]

Award 1 mark for correct answer C.

Number of half-lives elapsed:
\[ n = \frac{24.0 \text{ days}}{8.0 \text{ days}} = 3 \]

Fraction of undecayed nuclei remaining:
\[ \frac{N}{N_0} = \left(\frac{1}{2}\right)^3 = \frac{1}{8} \]

Fraction of nuclei that have decayed:
\[ \frac{N_0 - N}{N_0} = 1 - \frac{1}{8} = \frac{7}{8} \]

Therefore, the ratio of decayed nuclei to remaining undecayed nuclei is:
\[ \frac{\text{decayed}}{\text{undecayed}} = \frac{7/8}{1/8} = 7 \]

Award 1 mark for correct answer C.

Using the ideal gas equation \(pV = NkT\):

For a container of fixed volume \(V\):
\[\frac{p_2}{p_1} = \frac{N_2 T_2}{N_1 T_1}\]

Given \(N_2 = 0.60 N_1\), \(T_1 = 300\text{ K}\), and \(T_2 = 450\text{ K}\):
\[p_2 = 1.20 \times 10^5 \times \left(0.60 \times \frac{450}{300}\right) = 1.20 \times 10^5 \times 0.90 = 1.08 \times 10^5\text{ Pa}\]

Award 1 mark for option B.

Taking the direction to the right as positive:
Initial velocity \(u = -6.0\text{ m s}^{-1}\)
Final velocity \(v = +4.0\text{ m s}^{-1}\)

Change in momentum:
\[\Delta p = m(v - u) = 0.20 \times (4.0 - (-6.0)) = 0.20 \times 10.0 = 2.0\text{ N s}\]

Average force:
\[F = \frac{\Delta p}{\Delta t} = \frac{2.0}{0.040} = 50\text{ N}\]

Award 1 mark for option C.

By Snell's Law across parallel boundaries:
\[n_1 \sin \theta_1 = n_2 \sin \theta_2 = n_3 \sin \theta_3\]

Given \(\theta_1 = 45^\circ\) and the ray is at grazing emergence into Medium 3 (\(\theta_3 = 90^\circ\)):
\[n_1 \sin 45^\circ = n_3 \sin 90^\circ\]
\[\frac{n_3}{n_1} = \sin 45^\circ = \frac{1}{\sqrt{2}} \approx 0.707\]

Award 1 mark for option A.

1. Terminal voltage across the battery is \(V = \mathcal{E} \frac{R}{R+r} = \frac{\mathcal{E}}{1 + r/R}\). As \(R\) increases, \(r/R\) decreases, so \(V\) strictly increases.

2. Power dissipated in \(R\) is \(P = I^2 R = \frac{\mathcal{E}^2 R}{(R+r)^2}\). According to the maximum power transfer theorem, \(P\) attains its maximum when \(R = r\). Therefore, as \(R\) increases from \(0.5r\) to \(2.0r\), the power \(P\) first increases (from \(0.5r\) to \(r\)) and then decreases (from \(r\) to \(2.0r\)).

Award 1 mark for option D.

(1) Correct: As the loop enters the field, inward magnetic flux increases. By Lenz's law, the induced current produces an outward magnetic field to oppose the change, which corresponds to an anticlockwise current.

(2) Correct: By Lenz's law / Fleming's left-hand rule, the magnetic force on the leading edge is directed opposite to the velocity vector, resisting entry.

(3) Correct: Induced emf \(\mathcal{E} = B L v\). Power dissipated is \(P = \frac{\mathcal{E}^2}{R} = \frac{B^2 L^2 v^2}{R} \propto v^2\). Doubling \(v\) makes \(P\) quadruple.

Award 1 mark for option D.

For constructive interference, path difference \(\Delta x = n\lambda\), where \(n\) is an integer.
The maximum possible path difference occurs when the observation point is along the line joining the sources, giving \(\Delta x_{\max} < d\) for points in front of the array (or \(\Delta x = d \sin\theta\) with \(-90^\circ < \theta < 90^\circ\)).
\[d = 1.6\text{ m}, \quad \lambda = 0.40\text{ m}\]
\[n_{\max} = \frac{d}{\lambda} = \frac{1.6}{0.40} = 4\]
Since \(\sin\theta = \pm 1\) (at \(\pm 90^\circ\)) corresponds to infinity along the baseline and is not in front of the speakers, the possible orders observable in front are \(n = 0, \pm 1, \pm 2, \pm 3\).
Thus, there are \(2 \times 3 + 1 = 7\) maxima in total.

Award 1 mark for option C.

Secondary voltage:
\[V_s = V_p \left(\frac{N_s}{N_p}\right) = 220 \times \left(\frac{120}{600}\right) = 44\text{ V}\]

Secondary current:
\[I_s = \frac{V_s}{R} = \frac{44\text{ V}}{22\text{ }\Omega} = 2.0\text{ A}\]

For an ideal transformer, input power equals output power (\(V_p I_p = V_s I_s\)):
\[I_p = I_s \left(\frac{N_s}{N_p}\right) = 2.0 \times \left(\frac{120}{600}\right) = 0.40\text{ A}\]

Award 1 mark for option B.

Net initial activity \(A_0 = 680 - 40 = 640\text{ counts min}^{-1}\).
Net activity after \(6.0\text{ h}\), \(A(t) = 200 - 40 = 160\text{ counts min}^{-1}\).

The fraction of remaining undecayed nuclei is:
\[\frac{A(t)}{A_0} = \frac{160}{640} = \frac{1}{4} = \left(\frac{1}{2}\right)^2\]

This corresponds to \(2\) half-lives in \(6.0\text{ hours}\).
Therefore, the half-life is:
\[t_{1/2} = \frac{6.0\text{ h}}{2} = 3.0\text{ hours}\]

Award 1 mark for option A.

For an ideal gas:
- The root-mean-square speed is given by \( c_{\text{rms}} = \sqrt{\frac{3RT}{M}} \propto \sqrt{T} \).
When the temperature changes from \( T_0 \) to \( 1.44 T_0 \), the new root-mean-square speed is \( c_{\text{rms}}' = \sqrt{1.44} c_{\text{rms}} = 1.20 c_{\text{rms}} \), which represents an increase of \( (1.20 - 1) \times 100\% = 20\% \). Thus, (1) is correct.

- From the ideal gas law \( pV = nRT \), since the volume \( V \) and number of moles \( n \) are constant, \( p \propto T \).
The new pressure is \( p' = 1.44 p_0 \), which represents an increase of \( (1.44 - 1) \times 100\% = 44\% \). Thus, (2) is correct.

- The average kinetic energy per molecule is given by \( E_k = \frac{3}{2} k_B T \propto T \).
The new average kinetic energy is \( E_k' = 1.44 E_k \), corresponding to an increase of \( 44\% \). Thus, (3) is correct.

Therefore, (1), (2), and (3) are all correct.

D (1 mark)
- (1) is correct: \( c_{\text{rms}} \propto \sqrt{T} \implies \sqrt{1.44} = 1.20 \implies +20\% \)
- (2) is correct: \( p \propto T \implies \frac{\Delta p}{p_0} = 0.44 = 44\% \)
- (3) is correct: \( \overline{E_k} \propto T \implies \frac{\Delta \overline{E_k}}{E_{k,0}} = 0.44 = 44\% \)

(a) Energy required \(Q = mc\Delta T = (0.75)(4200)(100 - 22) = 245700\text{ J}\).
Time \(t = \frac{Q}{P} = \frac{245700}{1800} = 136.5\text{ s}\).

(b) (i) Part of the electrical energy is used to raise the temperature of the kettle body / heating element, and some thermal energy is lost to the surrounding environment.
(ii) Energy input \(E_{in} = P t = 1800 \times 160 = 288000\text{ J}\).
Efficiency \(\eta = \frac{Q_{useful}}{E_{in}} \times 100\% = \frac{245700}{288000} \times 100\% = 85.3125\% \approx 85.3\%\).

(c) Total energy supplied in \(120\text{ s}\): \(E = P \times t = 1800 \times 120 = 216000\text{ J}\).
Mass of steam formed \(m_s = \frac{E}{l_v} = \frac{216000}{2.26 \times 10^6} \approx 0.095575\text{ kg} \approx 0.0956\text{ kg}\) (or \(95.6\text{ g}\)).

(a) \(Q = mc\Delta T = (0.75)(4200)(78) = 2.457 \times 10^5\text{ J}\) [1M]
\(t = \frac{245700}{1800} = 136.5\text{ s}\) (accept 137 s) [1A]

(b)(i) Heat absorbed by the kettle / heating element OR heat dissipated to the surroundings [1A]
(b)(ii) \(\eta = \frac{245700}{1800 \times 160}\) [1M]
\(= 85.3\%\) [1A]

(c) \(Q = ml_v \Rightarrow 1800 \times 120 = m (2.26 \times 10^6)\) [1M]
\(m = 0.0956\text{ kg}\) (accept \(9.56 \times 10^{-2}\text{ kg}\) or \(95.6\text{ g}\)) [1A]

(a) Convert temperature: \(T_1 = 27 + 273.15 = 300.15\text{ K}\) (or \(300\text{ K}\)).
Using \(pV = nRT\):
\(n = \frac{pV}{RT} = \frac{(2.5 \times 10^5)(8.0 \times 10^{-3})}{(8.31)(300)} = 0.80225\text{ mol} \approx 0.802\text{ mol}\).

(b) Internal energy of ideal monatomic gas is given by:
\(U = N E_k = \frac{3}{2} nRT = \frac{3}{2} pV = \frac{3}{2} (2.5 \times 10^5)(8.0 \times 10^{-3}) = 3000\text{ J}\).

(c)(i) For fixed volume, \(\frac{p_1}{T_1} = \frac{p_2}{T_2}\).
\(T_2 = 127 + 273 = 400\text{ K}\).
\(p_2 = p_1 \left(\frac{T_2}{T_1}\right) = 2.5 \times 10^5 \times \left(\frac{400}{300}\right) = 3.33 \times 10^5\text{ Pa}\).

(ii) Since average kinetic energy \(\frac{1}{2}m c_{\text{r.m.s.}}^2 = \frac{3}{2} k_B T\), we have \(c_{\text{r.m.s.}} \propto \sqrt{T}\).
Ratio \(\frac{c_{\text{r.m.s., new}}}{c_{\text{r.m.s., initial}}} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{400}{300}} = \sqrt{\frac{4}{3}} \approx 1.15\).
Thus, the r.m.s. speed increases by a factor of \(1.15\).

(a) \(T = 300\text{ K}\) [1M]
\(n = \frac{pV}{RT} = \frac{(2.5 \times 10^5)(8.0 \times 10^{-3})}{(8.31)(300)} = 0.802\text{ mol}\) [1A]

(b) \(U = \frac{3}{2} pV\) or \(\frac{3}{2} nRT\) [1M]
\(U = 3000\text{ J}\) (or \(3.0 \times 10^3\text{ J}\)) [1A]

(c)(i) \(\frac{p_1}{T_1} = \frac{p_2}{T_2} \Rightarrow \frac{2.5 \times 10^5}{300} = \frac{p_2}{400}\) [1M]
\(p_2 = 3.33 \times 10^5\text{ Pa}\) [1A]

(c)(ii) \(c_{\text{r.m.s.}} \propto \sqrt{T}\) [1M]
Ratio \(= \sqrt{\frac{400}{300}} = 1.15\) [1A]
Stating that r.m.s. speed increases [1A]

(a) Taking to the right as positive:
By conservation of linear momentum:
\(m_A u_A + m_B u_B = m_A v_A + m_B v_B\)
\((1.2)(2.5) + (0.8)(0) = (1.2) v_A + (0.8)(3.0)\)
\(3.0 = 1.2 v_A + 2.4\)
\(1.2 v_A = 0.60 \Rightarrow v_A = +0.50\text{ m s}^{-1}\) (to the right).

(b) Initial kinetic energy:
\(K_i = \frac{1}{2} m_A u_A^2 = \frac{1}{2}(1.2)(2.5)^2 = 3.75\text{ J}\).
Final kinetic energy:
\(K_f = \frac{1}{2} m_A v_A^2 + \frac{1}{2} m_B v_B^2 = \frac{1}{2}(1.2)(0.50)^2 + \frac{1}{2}(0.8)(3.0)^2 = 0.15 + 3.60 = 3.75\text{ J}\).
Since \(K_i = K_f\), total kinetic energy is conserved; hence the collision is elastic.

(c) (i) Change in momentum of \(B\): \(\Delta p_B = m_B (v_B - u_B) = 0.8(3.0 - 0) = 2.4\text{ kg m s}^{-1}\).
Average force \(F_{avg} = \frac{\Delta p}{\Delta t} = \frac{2.4}{0.060} = 40\text{ N}\) to the right.

(ii) Magnitude: \(40\text{ N}\); Direction: to the left.
Law: Newton's third law of motion.

(iii) Graph of elastic PE vs time: starting from zero at \(t=0\), rises smoothly to a peak value (at maximum compression, \(t \approx 0.03\text{ s}\)), then decreases symmetrically back to zero at \(t=0.060\text{ s}\).

(a) Conservation of momentum: \(1.2(2.5) = 1.2 v_A + 0.8(3.0)\) [1M]
\(v_A = 0.50\text{ m s}^{-1}\) (to the right) [1A]

(b) \(K_i = \frac{1}{2}(1.2)(2.5)^2 = 3.75\text{ J}\) AND \(K_f = \frac{1}{2}(1.2)(0.5)^2 + \frac{1}{2}(0.8)(3.0)^2 = 3.75\text{ J}\) [1M]
Statement: \(K_i = K_f\), thus elastic [1A]

(c)(i) \(F = \frac{\Delta p}{\Delta t} = \frac{0.8 \times 3.0}{0.060} = 40\text{ N}\) [1A]
(c)(ii) \(40\text{ N}\) to the left [1A], Newton's third law of motion [1A]
(c)(iii) Correct bell/inverted parabolic shape starting at 0, peaking, and returning to 0 at 0.06 s [1A]

(a) For vertical motion: taking downward as positive,
\(s_y = u_y t + \frac{1}{2} g t^2\)
\(45 = 0 + \frac{1}{2}(9.81) t^2\)
\(t^2 = \frac{90}{9.81} \approx 9.1743 \Rightarrow t \approx 3.0289\text{ s} \approx 3.03\text{ s}\).

(b) Horizontal distance \(s_x = u_x t = 12 \times 3.0289 \approx 36.35\text{ m} \approx 36.3\text{ m}\).

(c) Vertical velocity component upon impact:
\(v_y = u_y + g t = 0 + (9.81)(3.0289) \approx 29.71\text{ m s}^{-1}\).
Horizontal component \(v_x = 12\text{ m s}^{-1}\).
Resultant speed \(v = \sqrt{v_x^2 + v_y^2} = \sqrt{12^2 + 29.71^2} = \sqrt{144 + 882.9} \approx 32.05\text{ m s}^{-1} \approx 32.1\text{ m s}^{-1}\).
Angle with horizontal: \(\theta = \tan^{-1}\left(\frac{v_y}{v_x}\right) = \tan^{-1}\left(\frac{29.71}{12}\right) \approx 68.0^\circ\) below horizontal.

(d) The time of flight remains unchanged because the vertical motion is governed entirely by gravity and the initial vertical height and velocity (which is zero), which are independent of the horizontal launch speed.

(a) \(s_y = \frac{1}{2}gt^2 \Rightarrow 45 = \frac{1}{2}(9.81)t^2\) [1M]
\(t = 3.03\text{ s}\) (accept 3.0 s if using g=10) [1A]

(b) \(s_x = 12 \times 3.0289 = 36.3\text{ m}\) [1A]

(c) \(v_y = gt = 29.71\text{ m s}^{-1}\) [1M]
\(v = \sqrt{12^2 + 29.71^2} = 32.1\text{ m s}^{-1}\) [1A]
\(\theta = 68.0^\circ\) below horizontal [1A]

(d) Remains unchanged [1A]
Reason: Vertical acceleration and vertical displacement are unchanged / horizontal and vertical motions are independent [1A]

(a) Fringe separation \(\Delta y = \frac{\lambda D}{a}\).
\(\Delta y = \frac{(650 \times 10^{-9})(1.60)}{0.25 \times 10^{-3}} = 4.16 \times 10^{-3}\text{ m} = 4.16\text{ mm}\).

(b) (i) Since \(\Delta y \propto D\), as \(D\) increases, \(\Delta y\) increases (fringes become wider/more spaced out).
(ii) Since \(\Delta y \propto \lambda\) and \(\lambda_{green} < \lambda_{red}\), the fringe separation decreases.

(c) The two-source interference fringes will disappear. Instead, a single-slit diffraction pattern will be observed on the screen, characterized by a broad, intense central maximum flanked by much dimmer, narrower secondary maxima and minima on either side.

(d) Grating spacing \(d = \frac{1}{400 \times 10^3\text{ m}^{-1}} = 2.50 \times 10^{-6}\text{ m}\).
Using grating equation: \(d \sin \theta = n \lambda\).
For \(n = 2\):
\(\sin \theta_2 = \frac{2 \lambda}{d} = \frac{2 \times (650 \times 10^{-9})}{2.50 \times 10^{-6}} = \frac{1.30 \times 10^{-6}}{2.50 \times 10^{-6}} = 0.520\).
\(\theta_2 = \sin^{-1}(0.520) \approx 31.33^\circ \approx 31.3^\circ\).

(a) \(\Delta y = \frac{\lambda D}{a} = \frac{(650 \times 10^{-9})(1.60)}{0.25 \times 10^{-3}}\) [1M]
\(\Delta y = 4.16\text{ mm}\) (or \(4.16 \times 10^{-3}\text{ m}\)) [1A]

(b)(i) Increases / becomes wider [1A]
(b)(ii) Decreases / becomes narrower [1A]

(c) Double-slit interference fringes disappear [1A]; Single-slit diffraction pattern is formed with a bright central maximum and dimmer subsidiary maxima [1A]

(d) \(d = \frac{1\times 10^{-3}}{400} = 2.5 \times 10^{-6}\text{ m}\) [1M]
\(d \sin \theta = 2\lambda \Rightarrow \sin \theta_2 = 0.520\) [1M]
\(\theta_2 = 31.3^\circ\) [1A]

(a) (i) The fundamental mode consists of one single loop with Nodes (N) at the two fixed ends (\(x=0\) and \(x=1.20\text{ m}\)) and an Antinode (A) at the center (\(x=0.60\text{ m}\)).
(ii) For the fundamental mode, \(L = \frac{\lambda_1}{2} \Rightarrow \lambda_1 = 2L = 2(1.20) = 2.40\text{ m}\).
(iii) Wave speed \(v = f_1 \lambda_1 = (30)(2.40) = 72.0\text{ m s}^{-1}\).

(b) The third harmonic has frequency \(f_3 = 3 f_1 = 3 \times 30 = 90\text{ Hz}\).

(c) A stationary wave is the superposition of two identical progressive waves travelling in opposite directions. The energy carried by one wave to the right is equal to the energy carried by the other wave to the left, so energy is trapped between nodes and there is no net energy propagation along the string.

(a)(i) Correct sketch with 1 loop, nodes marked at ends and antinode at center [1A]
(a)(ii) \(\lambda_1 = 2L = 2.40\text{ m}\) [1A]
(a)(iii) \(v = f\lambda = 30 \times 2.40 = 72.0\text{ m s}^{-1}\) [1A]

(b) \(f_3 = 3 \times 30 = 90\text{ Hz}\) [1A]

(c) Formed by superposition of two identical waves travelling in opposite directions [1A]
Energy travels in opposite directions in equal amounts / energy is confined/stored between nodes [1A]
Hence no net transfer of energy along the medium [1A]

(a) (i) Parallel combination: \(R_p = \frac{R_2}{2} = \frac{6.0}{2} = 3.0\ \Omega\).
Total external resistance \(R_{ext} = R_1 + R_p = 4.5 + 3.0 = 7.5\ \Omega\).

(ii) Total circuit resistance \(R_{total} = R_{ext} + r = 7.5 + 1.5 = 9.0\ \Omega\).
Total current \(I = \frac{\mathcal{E}}{R_{total}} = \frac{12.0}{9.0} = 1.333\text{ A} \approx 1.33\text{ A}\).

(iii) Terminal voltage \(V = \mathcal{E} - I r = 12.0 - (1.333)(1.5) = 12.0 - 2.0 = 10.0\text{ V}\).
(Or \(V = I R_{ext} = 1.333 \times 7.5 = 10.0\text{ V}\)).

(b) Current through each \(R_2\) is \(I_2 = \frac{I}{2} = \frac{1.333}{2} = 0.6667\text{ A}\).
Power dissipated \(P_2 = I_2^2 R_2 = (0.6667)^2 \times 6.0 \approx 2.667 \times 10^{-1} \times 6.0 = 1.33\text{ W}\).
(Or \(V_p = I R_p = 1.333 \times 3.0 = 4.0\text{ V}\); \(P_2 = \frac{V_p^2}{R_2} = \frac{4.0^2}{6.0} = 2.67\text{ W}\) - wait: \(V_p = 4.0\text{ V}\), \(P = \frac{4.0^2}{6.0} = 2.67\text{ W}\). Note: \(I_2 = 4.0 / 6.0 = 0.667\text{ A}\), \(P_2 = (0.667)^2 \times 6.0 = 2.67\text{ W}\)).

(c) Removing one \(R_2\) resistor increases the parallel resistance from \(3.0\ \Omega\) to \(6.0\ \Omega\). Consequently, total external resistance \(R_{ext}\) increases from \(7.5\ \Omega\) to \(10.5\ \Omega\). This causes the total current \(I\) from the battery to decrease. Since terminal potential difference is \(V = \mathcal{E} - Ir\), a lower current \(I\) results in a smaller internal potential drop (\(Ir\)), so the terminal voltage \(V\) increases.

(a)(i) \(R_p = 3.0\ \Omega\), \(R_{ext} = 4.5 + 3.0 = 7.5\ \Omega\) [1A]
(a)(ii) \(I = \frac{12.0}{7.5 + 1.5} = 1.33\text{ A}\) (or \(\frac{4}{3}\text{ A}\)) [1M + 1A]
(a)(iii) \(V = 12.0 - (1.333)(1.5) = 10.0\text{ V}\) [1A]

(b) \(V_p = 1.333 \times 3.0 = 4.0\text{ V}\) or \(I_2 = 0.667\text{ A}\) [1M]
\(P = \frac{4.0^2}{6.0} = 2.67\text{ W}\) (accept 2.67 W) [1A]

(c) Increases [1A]
Reason: Total circuit resistance increases \(\rightarrow\) current \(I\) decreases [1A]
Smaller internal potential drop / loss (\(Ir\)) \(\rightarrow\) higher terminal voltage \(V = \mathcal{E} - Ir\) [1A]

(a) Using Fleming's right-hand rule (or Lenz's law): As the rod moves to the right, magnetic flux out of the page increases. To oppose this increase, the induced magnetic field must point into the page, which requires a clockwise current. Hence, current flows downwards through the rod, from \(P\) to \(Q\).

(b) Induced e.m.f. \(\mathcal{E} = B L v = (0.40\text{ T})(0.25\text{ m})(6.0\text{ m s}^{-1}) = 0.60\text{ V}\).

(c) (i) Total resistance of circuit \(R_{total} = R + r_{rod} = 2.50 + 0.50 = 3.00\ \Omega\).
Induced current \(I = \frac{\mathcal{E}}{R_{total}} = \frac{0.60}{3.00} = 0.20\text{ A}\).

(ii) Magnetic force on rod \(F_B = B I L = (0.40)(0.20)(0.25) = 0.020\text{ N}\) directed to the left (opposing motion).
To maintain constant velocity (zero net acceleration), the external force must balance the magnetic force:
\(F_{ext} = 0.020\text{ N}\) (to the right).

(d) Mechanical power input \(P_{mech} = F_{ext} v = (0.020\text{ N})(6.0\text{ m s}^{-1}) = 0.120\text{ W}\).
Total electrical power dissipated \(P_{elec} = I^2 R_{total} = (0.20)^2 (3.00) = (0.040)(3.00) = 0.120\text{ W}\).
Since \(P_{mech} = P_{elec} = 0.12\text{ W}\), mechanical work done per second is entirely converted into electrical energy dissipated as heat in the resistors, satisfying energy conservation.

(a) From \(P\) to \(Q\) [1A]

(b) \(\mathcal{E} = BLv = (0.40)(0.25)(6.0) = 0.60\text{ V}\) [1M + 1A]

(c)(i) \(I = \frac{\mathcal{E}}{R+r} = \frac{0.60}{3.00} = 0.20\text{ A}\) [1A]
(c)(ii) \(F = BIL = (0.40)(0.20)(0.25) = 0.020\text{ N}\) [1M + 1A]

(d) \(P_{mech} = F v = 0.020 \times 6.0 = 0.12\text{ W}\) [1M]
\(P_{elec} = I^2 R_{total} = (0.20)^2 \times 3.00 = 0.12\text{ W}\) [1M]
Conclusion: \(P_{mech} = P_{elec}\), energy is conserved [1A]

(a) Lenz's law states that the direction of the induced e.m.f. (or induced current) is always such as to oppose the magnetic flux change producing it.

(b) The magnetic flux is given by $\Phi = B A$. When the coil leaves the field at speed $v$, the rate of change of area is $\frac{\Delta A}{\Delta t} = L v$, where $L = 0.05\text{ m}$.
Induced e.m.f.:
$$\varepsilon = N B L v = 150 \times 0.40\text{ T} \times 0.05\text{ m} \times 1.2\text{ m s}^{-1} = 3.6\text{ V}$$

(c) As the coil moves out of the field, the magnetic flux pointing into the page decreases. By Lenz's law, the induced current produces a magnetic field pointing into the page to oppose this decrease. According to the right-hand grip rule, the direction of the induced current is clockwise.
Magnitude of the current:
$$I = \frac{\varepsilon}{R} = \frac{3.6\text{ V}}{2.4\ \Omega} = 1.5\text{ A}$$

(d) The magnetic force acting on the vertical side of the coil inside the field opposes the motion:
$$F_{\text{mag}} = N I L B = 150 \times 1.5\text{ A} \times 0.05\text{ m} \times 0.40\text{ T} = 4.5\text{ N}$$
Since the coil moves at a constant speed, the external pulling force must balance this magnetic force:
$$F_{\text{ext}} = F_{\text{mag}} = 4.5\text{ N}$$
(Alternatively, $P = I^2 R = (1.5)^2 \times 2.4 = 5.4\text{ W}$, $F_{\text{ext}} = \frac{P}{v} = \frac{5.4}{1.2} = 4.5\text{ N}$)

(e) The work done is converted into internal energy / thermal energy (Joule heating) in the coil.

(a) States that the induced current/e.m.f. opposes the change of magnetic flux / causes that produce it: [1A]

(b) Using $\varepsilon = N B L v$ or $\varepsilon = N \frac{\Delta \Phi}{\Delta t}$: [1M]
$\varepsilon = 150 \times 0.40 \times 0.05 \times 1.2 = 3.6\text{ V}$: [1A]

(c) Clockwise: [1A]
$I = \frac{\varepsilon}{R} = \frac{3.6}{2.4} = 1.5\text{ A}$: [1A]

(d) Using $F = N I L B$ or $F = \frac{I^2 R}{v}$: [1M]
$F = 4.5\text{ N}$: [1A]

(e) Thermal energy / internal energy / heat in the resistor/coil: [1A]

(a) Applying the ideal gas equation $p V = n R T$:
$$p = \frac{n R T}{V} = \frac{0.80\text{ mol} \times 8.31\text{ J mol}^{-1}\text{K}^{-1} \times 300\text{ K}}{0.025\text{ m}^3} = 79\,776\text{ Pa} \approx 7.98 \times 10^4\text{ Pa}$$

(b) (i) For an ideal monatomic gas, the average kinetic energy per molecule is $E_k = \frac{3}{2} k_B T$, so the total internal energy is:
$$U = \frac{3}{2} n R T = 1.5 \times 0.80\text{ mol} \times 8.31\text{ J mol}^{-1}\text{K}^{-1} \times 300\text{ K} = 2991.6\text{ J} \approx 2990\text{ J}$$

(ii) Since the root-mean-square speed satisfies $c_{\text{r.m.s.}} = \sqrt{\frac{3RT}{M}} \propto \sqrt{T}$:
$$\frac{c_{\text{new}}}{c_{\text{initial}}} = \sqrt{\frac{2T}{T}} = \sqrt{2} \approx 1.41$$

(c) As the temperature of the gas rises:
1. The average kinetic energy of the gas molecules increases, meaning molecules move at higher speeds on average.
2. Molecules collide with the walls of the cylinder more frequently.
3. The momentum change of each molecule during a collision with the wall is larger, exerting a greater average force per collision.
Consequently, the total average force exerted on the wall per unit area (pressure) increases.

(a) Using $p V = n R T$: [1M]
$p = 7.98 \times 10^4\text{ Pa}$ (accept $7.97 \times 10^4\text{ Pa}$ to $8.00 \times 10^4\text{ Pa}$): [1A]

(b) (i) Using $U = \frac{3}{2} n R T$ or $U = N E_k$: [1M]
$U = 2990\text{ J}$ (accept $2992\text{ J}$): [1A]
(ii) Ratio $= \sqrt{2} \approx 1.41$: [1A]

(c) Higher temperature leads to higher average molecular speed / kinetic energy: [1A]
Frequency of collisions between molecules and walls increases: [1A]
Average force per collision / rate of change of momentum per collision increases, hence pressure increases: [1A]

For a binary star system orbiting their common centre of mass:
(1) The centre of mass satisfies \(m_P r_P = m_Q r_Q\). Since \(m_P = 2 m_Q\), \(r_P = \frac{m_Q}{m_P} r_Q = \frac{1}{2} r_Q\). Statement (1) is correct.
(2) Both stars complete one revolution in the same amount of time so that they always remain on opposite sides of the centre of mass. Thus, \(T_P = T_Q\). Statement (2) is correct.
(3) The centripetal acceleration is \(a = \omega^2 r\). Since \(\omega\) is identical and \(r_P = \frac{1}{2} r_Q\), \(a_P = \frac{1}{2} a_Q\), so the centripetal acceleration of \(P\) is half that of \(Q\). Statement (3) is incorrect.

C (1 mark)

By Stefan's law, the luminosity of a spherical star is given by \(L = 4\pi R^2 \sigma T^4\).
Therefore:
\[\frac{L_A}{L_B} = \left(\frac{R_A}{R_B}\right)^2 \left(\frac{T_A}{T_B}\right)^4 = \left(\frac{R_A}{0.5 R_A}\right)^2 \left(\frac{6000}{3000}\right)^4 = (2)^2 \times (2)^4 = 4 \times 16 = 64\]
The apparent brightness is \(b = \frac{L}{4\pi d^2}\). Since \(b_A = b_B\):
\[\frac{L_A}{d_A^2} = \frac{L_B}{d_B^2} \implies \frac{d_A}{d_B} = \sqrt{\frac{L_A}{L_B}} = \sqrt{64} = 8\]
Thus, the ratio \(d_A : d_B = 8 : 1\).

D (1 mark)

From Einstein's photoelectric equation:
\[K_{\text{max}} = hf - \Phi \implies hf = K_{\text{max}} + \Phi\]
When the incident frequency is doubled to \(2f\):
\[K'_{\text{max}} = h(2f) - \Phi = 2(hf) - \Phi = 2(K_{\text{max}} + \Phi) - \Phi = 2K_{\text{max}} + \Phi\]

B (1 mark)

(1) In the Bohr model, the electron orbital speed \(v_n \propto 1/n\). When transitioning from \(n = 3\) to \(n = 1\), the speed and momentum increase, so the de Broglie wavelength \(\lambda = \frac{h}{p}\) decreases. (Or \(2\pi r_n = n\lambda_n \implies \lambda_n \propto n\), so \(\lambda\) decreases.) Thus (1) is correct.
(2) The electrical potential energy is \(U = -\frac{k e^2}{r}\). As \(n\) decreases from 3 to 1, the radius \(r\) decreases, making \(U\) more negative (i.e. decreasing). Thus (2) is correct.
(3) The orbital angular momentum is \(L_n = n \frac{h}{2\pi}\). As \(n\) decreases from 3 to 1, \(L\) decreases from \(\frac{3h}{2\pi}\) to \(\frac{h}{2\pi}\). Thus (3) is correct.

D (1 mark)

For a heat pump in heating mode:
\[\text{COP} = \frac{Q_H}{W} = 3.5\]
Given \(W = 2.0\text{ kW}\), the rate of heat delivered to the room is:
\[Q_H = 3.5 \times 2.0\text{ kW} = 7.0\text{ kW}\]
By the conservation of energy, \(Q_H = Q_C + W\), so the rate of heat extracted from the outdoor environment is:
\[Q_C = Q_H - W = 7.0\text{ kW} - 2.0\text{ kW} = 5.0\text{ kW}\]

B (1 mark)

The rate of heat conduction through a structure characterized by its \(U\)-value is:
\[\frac{Q}{t} = U A \Delta T\]
Substituting the given values:
\[\frac{Q}{t} = 0.60 \times 50 \times (22 - 7) = 30 \times 15 = 450\text{ W}\]

B (1 mark)

The intensity reflection coefficient \(\alpha\) at normal incidence is given by:
\[\alpha = \frac{I_r}{I_0} = \left(\frac{Z_2 - Z_1}{Z_2 + Z_1}\right)^2\]
Substituting the values:
\[\alpha = \left(\frac{7.80 \times 10^6 - 1.70 \times 10^6}{7.80 \times 10^6 + 1.70 \times 10^6}\right)^2 = \left(\frac{6.10}{9.50}\right)^2 \approx (0.6421)^2 \approx 0.412 = 41.2\%\]

C (1 mark)

The transmission of an X-ray beam through an attenuating medium follows the Beer-Lambert law:
\[I = I_0 e^{-\mu x}\]
The transmitted fraction is:
\[\frac{I}{I_0} = e^{-\mu x} = e^{-(0.23\text{ cm}^{-1})(6.0\text{ cm})} = e^{-1.38} \approx 0.2516 \approx 0.25\]

B (1 mark)

Luminosity $L = 4\pi R^2 \sigma T^4$. Since both stars have the same surface temperature $T$, the ratio of their luminosities is:
$$\frac{L_P}{L_Q} = \left(\frac{R_P}{R_Q}\right)^2 = 2^2 = 4$$

The apparent brightness $b$ of a star at distance $d$ is given by $b = \frac{L}{4\pi d^2}$. Thus, the ratio of apparent brightness is:
$$\frac{b_P}{b_Q} = \frac{L_P}{L_Q} \times \left(\frac{d_Q}{d_P}\right)^2 = 4 \times \left(\frac{10}{30}\right)^2 = 4 \times \frac{1}{9} = \frac{4}{9}$$

1A for selecting option A.

Using the Doppler shift formula for electromagnetic waves:
$$\frac{\Delta \lambda}{\lambda_0} = \frac{v}{c}$$
where $\Delta \lambda = 510\text{ nm} - 486\text{ nm} = 24\text{ nm}$, and $\lambda_0 = 486\text{ nm}$.

$$v = c \times \frac{\Delta \lambda}{\lambda_0} = (3.00 \times 10^8\text{ m s}^{-1}) \times \frac{24\text{ nm}}{486\text{ nm}} \approx 1.48 \times 10^7\text{ m s}^{-1} = 1.48 \times 10^4\text{ km s}^{-1}$$

1A for selecting option B.

According to Einstein's photoelectric equation:
$$e V_s = hf - \Phi \implies V_s = \frac{hf - \Phi}{e}$$

When the incident frequency is doubled to $2f$:
$$e V_s' = h(2f) - \Phi = 2hf - \Phi = 2(hf - \Phi) + \Phi = 2eV_s + \Phi$$

Dividing by $e$ gives:
$$V_s' = 2V_s + \frac{\Phi}{e}$$

Since the work function $\Phi > 0$, it follows that $V_s' > 2V_s$.

1A for selecting option C.

In the Bohr model, the quantization condition for orbital angular momentum is:
$$m v r = \frac{n h}{2\pi}$$

Since the orbital radius scales as $r_n \propto n^2$, the speed of the electron is:
$$v_n \propto \frac{n}{r_n} \propto \frac{n}{n^2} = \frac{1}{n}$$

The de Broglie wavelength is:
$$\lambda = \frac{h}{m v_n} \propto \frac{1}{v_n} \propto n$$

Therefore, the ratio of the de Broglie wavelengths is:
$$\frac{\lambda_3}{\lambda_1} = \frac{3}{1} = 3 : 1$$

1A for selecting option C.

The overall U-value of the composite wall is:
$$\frac{1}{U} = \frac{1}{U_1} + \frac{1}{U_2} = \frac{1}{2.0} + \frac{1}{0.50} = 0.50 + 2.0 = 2.5\text{ m}^2\text{ K W}^{-1}$$
$$U = \frac{1}{2.5} = 0.40\text{ W m}^{-2}\text{ K}^{-1}$$

The rate of heat transfer through the wall is:
$$\frac{Q}{t} = U A \Delta T = (0.40\text{ W m}^{-2}\text{ K}^{-1}) \times (12\text{ m}^2) \times (15\text{ K}) = 72\text{ W}$$

1A for selecting option A.

The swept area of the turbine blades is:
$$A = \pi r^2 = \pi (10)^2 = 100\pi\text{ m}^2 \approx 314.16\text{ m}^2$$

The maximum power available from the wind is:
$$P_{\text{wind}} = \frac{1}{2} \rho A v^3 = \frac{1}{2} \times 1.2 \times 314.16 \times (8.0)^3 = 0.6 \times 314.16 \times 512 = 96510\text{ W} = 96.51\text{ kW}$$

The overall efficiency is:
$$\eta = \frac{P_{\text{out}}}{P_{\text{wind}}} \times 100\% = \frac{36\text{ kW}}{96.51\text{ kW}} \times 100\% \approx 37.3\%$$

1A for selecting option B.

To correct the far point to infinity ($u = \infty$), the virtual image must be formed at the uncorrected far point $v = -2.0\text{ m}$:
$$P = \frac{1}{f} = \frac{1}{u} + \frac{1}{v} = \frac{1}{\infty} - \frac{1}{2.0} = -0.50\text{ D}$$

When viewing an object at the new near point $u_{\text{new}}$, the lens must form a virtual image at the uncorrected near point $v = -50\text{ cm} = -0.50\text{ m}$:
$$\frac{1}{u_{\text{new}}} + \frac{1}{-0.50} = -0.50$$
$$\frac{1}{u_{\text{new}}} = 2.0 - 0.50 = 1.50\text{ m}^{-1}$$
$$u_{\text{new}} = \frac{1}{1.50}\text{ m} \approx 0.67\text{ m} = 67\text{ cm}$$

1A for selecting option C.

The intensity reflection coefficient $\alpha$ is given by:
$$\alpha = \left( \frac{Z_2 - Z_1}{Z_2 + Z_1} \right)^2$$

Substituting the values:
$$\alpha = \left( \frac{1.70 \times 10^6 - 1.38 \times 10^6}{1.70 \times 10^6 + 1.38 \times 10^6} \right)^2 = \left( \frac{0.32}{3.08} \right)^2 = (0.1039)^2 \approx 0.0108 = 1.08\%$$

Assuming no absorption at the interface, the percentage of intensity transmitted is:
$$\text{Transmitted percentage} = 100\% - 1.08\% = 98.92\% \approx 98.9\%$$

1A for selecting option D.

(a) (i) Distance in parsecs:
\[ d = \frac{1}{p} = \frac{1}{0.040} = 25\text{ pc} \]
(ii) Distance in light-years:
\[ d = 25 \times 3.26 = 81.5\text{ ly} \]

(b) (i) Maximum wavelength shift:
\[ \Delta \lambda = 656.34\text{ nm} - 656.30\text{ nm} = 0.04\text{ nm} \]
By Doppler formula:
\[ \frac{\Delta \lambda}{\lambda_0} = \frac{v_r}{c} \]
\[ v_r = \frac{0.04 \times 10^{-9}}{656.30 \times 10^{-9}} \times (3.00 \times 10^8) \approx 1.8285 \times 10^4\text{ m s}^{-1} \approx 1.83 \times 10^4\text{ m s}^{-1} \text{ (or } 18.3\text{ km s}^{-1}\text{)} \]

(ii) Orbital period \(T = 4.0\text{ days} = 4.0 \times 24 \times 3600 = 3.456 \times 10^5\text{ s}\).
Since the orbit is edge-on, orbital speed \(v = v_r = 1.8285 \times 10^4\text{ m s}^{-1}\).
\[ v = \frac{2\pi r}{T} \implies r = \frac{v T}{2\pi} = \frac{(1.8285 \times 10^4)(3.456 \times 10^5)}{2\pi} \approx 1.006 \times 10^9\text{ m} \approx 1.01 \times 10^9\text{ m} \]

(c) (i) Relationship between absolute magnitude and luminosity:
\[ M_S - M_\odot = -2.5 \log_{10}\left(\frac{L_S}{L_\odot}\right) \]
\[ 1.8 - 4.8 = -3.0 = -2.5 \log_{10}\left(\frac{L_S}{L_\odot}\right) \]
\[ \log_{10}\left(\frac{L_S}{L_\odot}\right) = \frac{3.0}{2.5} = 1.2 \]
\[ \frac{L_S}{L_\odot} = 10^{1.2} \approx 15.85 \approx 15.8 \]

(ii) Using Stefan-Boltzmann Law \(L = 4\pi R^2 \sigma T^4\):
\[ \frac{L_S}{L_\odot} = \left(\frac{R_S}{R_\odot}\right)^2 \left(\frac{T_S}{T_\odot}\right)^4 \]
\[ 15.85 = \left(\frac{R_S}{R_\odot}\right)^2 \left(\frac{8500}{5800}\right)^4 \]
\[ 15.85 = \left(\frac{R_S}{R_\odot}\right)^2 \times (1.4655)^4 = \left(\frac{R_S}{R_\odot}\right)^2 \times 4.613 \]
\[ \left(\frac{R_S}{R_\odot}\right)^2 = \frac{15.85}{4.613} = 3.436 \implies \frac{R_S}{R_\odot} = \sqrt{3.436} \approx 1.85 \]
\[ R_S \approx 1.85 R_\odot \]

(a) (i) \(d = 25\text{ pc}\) [1A]
(ii) \(d = 81.5\text{ ly}\) (or \(81.5\text{--}82\text{ ly}\)) [1A]

(b) (i) Use of \(\frac{\Delta \lambda}{\lambda_0} = \frac{v}{c}\) [1M]
\(v_r = 1.83 \times 10^4\text{ m s}^{-1}\) (accept \(1.82 \times 10^4\) to \(1.83 \times 10^4\text{ m s}^{-1}\)) [1A]
(ii) Use of \(v = \frac{2\pi r}{T}\) with conversion of days to seconds [1M]
\(r = 1.01 \times 10^9\text{ m}\) (accept \(1.00 \times 10^9\) to \(1.01 \times 10^9\text{ m}\)) [1A]

(c) (i) Use of \(M_1 - M_2 = -2.5 \log_{10}(L_1/L_2)\) [1M]
\(\frac{L_S}{L_\odot} = 15.8\) (or \(15.9\)) [1A]
(ii) Use of \(L \propto R^2 T^4\) [1M]
\(R_S = 1.85 R_\odot\) (accept \(1.84 R_\odot\) to \(1.86 R_\odot\)) [1A]

(a) Acoustic impedance is the product of the density of the medium \(\rho\) and the speed of sound \(c\) in the medium (\(Z = \rho c\)).
SI unit: \(\text{kg m}^{-2}\text{s}^{-1}\) (or \(\text{Rayl}\)).

(b) (i) Acoustic impedance of lens:
\[ Z_{\text{lens}} = \rho_{\text{lens}} \times c_{\text{lens}} = 1050 \times 1640 = 1.722 \times 10^6\text{ kg m}^{-2}\text{s}^{-1} \approx 1.72 \times 10^6\text{ kg m}^{-2}\text{s}^{-1} \]

(ii) Acoustic impedance of aqueous humor:
\[ Z_{\text{aq}} = 1000 \times 1530 = 1.530 \times 10^6\text{ kg m}^{-2}\text{s}^{-1} \]
Intensity reflection coefficient:
\[ \alpha = \frac{(Z_{\text{lens}} - Z_{\text{aq}})^2}{(Z_{\text{lens}} + Z_{\text{aq}})^2} = \frac{(1.722 \times 10^6 - 1.530 \times 10^6)^2}{(1.722 \times 10^6 + 1.530 \times 10^6)^2} = \frac{(0.192)^2}{(3.252)^2} \approx 3.49 \times 10^{-3} \text{ (or } 0.349\%\text{)} \]

(c) (i) Distance from cornea to anterior lens surface:
\[ d_1 = \frac{c_{\text{aq}} \times t_1}{2} = \frac{1530 \times (4.0 \times 10^{-6})}{2} = 3.06 \times 10^{-3}\text{ m} = 3.06\text{ mm} \]

(ii) Time taken for ultrasound to travel through the lens and back:
\[ \Delta t = t_2 - t_1 = 9.0\ \mu\text{s} - 4.0\ \mu\text{s} = 5.0\ \mu\text{s} \]
Thickness of lens:
\[ d_{\text{lens}} = \frac{c_{\text{lens}} \times \Delta t}{2} = \frac{1640 \times (5.0 \times 10^{-6})}{2} = 4.10 \times 10^{-3}\text{ m} = 4.10\text{ mm} \]

(d) Higher frequency corresponds to a shorter wavelength, which reduces diffraction and provides a higher spatial / axial resolution to resolve fine structures of the eye. Since the eye is small, the greater attenuation associated with high frequency is not a significant limitation.

(a) \(Z = \rho c\) (product of density and speed of sound in medium) [1A]
Unit: \(\text{kg m}^{-2}\text{s}^{-1}\) or \(\text{Rayl}\) [1A]

(b) (i) \(Z_{\text{lens}} = 1.72 \times 10^6\text{ kg m}^{-2}\text{s}^{-1}\) [1A]
(ii) Use of \(\alpha = \frac{(Z_2 - Z_1)^2}{(Z_2 + Z_1)^2}\) [1M]
\(\alpha = 3.49 \times 10^{-3}\) (or \(0.35\%\)) (accept \(3.48 \times 10^{-3}\) to \(3.50 \times 10^{-3}\)) [1A]

(c) (i) Use of \(d = \frac{ct}{2}\) with \(c = 1530\text{ m s}^{-1}\) [1M]
\(d_1 = 3.06 \times 10^{-3}\text{ m}\) (or \(3.06\text{ mm}\)) [1A]
(ii) Use of \(\Delta t = 5.0\ \mu\text{s}\) and \(c = 1640\text{ m s}^{-1}\) [1M]
Thickness \(= 4.10 \times 10^{-3}\text{ m}\) (or \(4.10\text{ mm}\)) [1A]

(d) Higher frequency gives shorter wavelength, providing higher spatial resolution to distinguish fine details in the small eye [1A]