2023 HKDSE Physics Practice Paper with Answers

(1) is correct: Shiny silver surfaces are poor emitters and good reflectors of radiation, which reduces heat transfer by radiation.
(2) is correct: Air is a poor conductor of heat (low thermal conductivity). When trapped to prevent convection currents, it acts as an effective thermal insulator against conduction.
(3) is correct: Black, matte surfaces have higher emissivity than shiny, polished surfaces at the same temperature and area, so they emit radiation at a higher rate.
Therefore, (1), (2), and (3) are all correct.

D (1 mark) for identifying that statements (1), (2), and (3) are all correct.

The equation of the straight line on the \(p-V\) plane is \(p(V) = p_0 - \frac{p_0}{4V_0}(V - V_0) = \frac{5}{4}p_0 - \frac{p_0}{4V_0}V\). According to the ideal gas law \(pV = nRT\), \(T \propto pV = \frac{5}{4}p_0 V - \frac{p_0}{4V_0}V^2\). This is a downward-opening quadratic function in \(V\). The maximum value of \(pV\) occurs at \(V = 2.5V_0\), where \(pV = 1.5625 p_0 V_0\). At the initial state \(P\), \(pV = p_0 V_0\). At the final state \(Q\), \(pV = 1.5 p_0 V_0\). Thus, the temperature increases first to a maximum and then decreases slightly to \(1.5\) times its initial value.

C (1 mark) for identifying that temperature increases to a maximum before decreasing.

Since the elevator is moving downwards and slowing down (decelerating), its acceleration is directed upwards. Taking the upward direction as positive, the equation of motion for the box is \(N - mg = ma\). Rearranging gives the normal reaction force \(N = m(g + a)\).

C (1 mark) for applying Newton's second law with upward acceleration.

When moving up the incline at constant speed, the driving force \(F\) required is \(F = f + mg\sin\theta = 600 + (1200)(9.81)(0.05) = 600 + 588.6 = 1188.6\text{ N}\). The power output is \(P = Fv = 1188.6 \times 20 = 23772\text{ W} \approx 23.8\text{ kW}\).

B (1 mark) for finding total force \(F = f + mg\sin\theta\) and calculating \(P = Fv\).

Let \(r\) be the angle of refraction at the top face. The angle of incidence at the vertical side face is \(90^\circ - r\). For total internal reflection to occur at the side face, \(\sin(90^\circ - r) \ge \frac{1}{n} = \frac{1}{1.30} \approx 0.7692\), which means \(\cos r \ge 0.7692\). Thus, \(\sin r = \sqrt{1 - \cos^2 r} \le \sqrt{1 - 0.7692^2} = \sqrt{1 - 0.5917} = 0.6390\). Using Snell's law at the top surface, \(1.00 \sin\theta = n \sin r = 1.30 \times 0.6390 = 0.8307\). Therefore, \(\theta \le \sin^{-1}(0.8307) \approx 56.2^\circ\).

D (1 mark) for calculating critical angle and corresponding maximum angle of incidence.

The distance between two adjacent nodes in a stationary wave equals half of the wavelength: \(\frac{\lambda}{2} = 15\text{ cm} = 0.15\text{ m} \implies \lambda = 0.30\text{ m}\). Using the wave equation \(v = f\lambda\), the frequency is \(f = \frac{v}{\lambda} = \frac{60}{0.30} = 200\text{ Hz}\).

B (1 mark) for using \(\lambda = 2 \times 0.15\text{ m}\) and \(f = v / \lambda\).

The equivalent resistance of the parallel pair is \(R_p = \frac{R \times R}{R + R} = \frac{R}{2}\). The total resistance of the circuit is \(R_{\text{total}} = R_p + R = \frac{R}{2} + R = \frac{3R}{2}\). By the potential divider principle, the potential difference across the parallel combination (and therefore the voltmeter reading) is \(V = \mathcal{E} \times \frac{R_p}{R_{\text{total}}} = \mathcal{E} \times \frac{R/2}{3R/2} = \frac{1}{3}\mathcal{E}\).

B (1 mark) for using potential divider rule to determine \(V = \mathcal{E}/3\).

The rod moves downward (velocity vector \(\vec{v}\) downwards). The magnetic field \(\vec{B}\) is directed northwards. By Fleming's right-hand rule (or by the magnetic force on positive mobile charges \(\vec{F} = q(\vec{v} \times \vec{B})\)), the force on positive charges is directed towards the east. Thus, positive charges accumulate at the east end of the rod, making the east end at a higher electrical potential. The magnitude of the induced e.m.f. is \(\mathcal{E} = B L v\).

B (1 mark) for determining the east end has higher potential and induced EMF is \(BLv\).

Total electrical energy supplied = \(P \times t = 60\text{ W} \times (4.0 \times 60\text{ s}) = 14400\text{ J}\).
Energy absorbed by the metal cylinder = \(14400 \times (1 - 0.15) = 12240\text{ J}\).
Using \(Q = mc\Delta T\):
\(12240 = 0.50 \times c \times (54 - 24)\)
\(12240 = 0.50 \times c \times 30 = 15 c\)
\(c = \frac{12240}{15} = 816\text{ J kg}^{-1}\text{ }^\circ\text{C}^{-1}\).

Correct option B: 1 mark.

Normal reaction force \(R = mg = 3.0 \times 9.81 = 29.43\text{ N}\).
Frictional force \(f_k = \mu_k R = 0.25 \times 29.43 = 7.35375\text{ N}\).
Net horizontal force \(F_{\text{net}} = F - f_k = 15.0 - 7.35375 = 7.64625\text{ N}\).
Acceleration \(a = \frac{F_{\text{net}}}{m} = \frac{7.64625}{3.0} = 2.54875\text{ m s}^{-2}\).
Using \(v^2 = u^2 + 2as\) with \(u = 0\):
\(v^2 = 2 \times 2.54875 \times 4.0 = 20.39\)
\(v = \sqrt{20.39} \approx 4.5155\text{ m s}^{-1} \approx 4.5\text{ m s}^{-1}\).

Correct option A: 1 mark.

Take right as positive.
Initial momentum \(p_i = (2.0)(3.0) + (4.0)(0) = +6.0\text{ kg m s}^{-1}\).
Final momentum \(p_f = (2.0)(-1.0) + (4.0)v_Q = -2.0 + 4.0 v_Q\).
By conservation of momentum: \(6.0 = -2.0 + 4.0 v_Q \implies v_Q = +2.0\text{ m s}^{-1}\). So (1) is correct.
Impulse on \(Q = \Delta p_Q = 4.0 \times (2.0 - 0) = 8.0\text{ N s}\). So (2) is correct.
Initial kinetic energy \(E_{k,i} = \frac{1}{2}(2.0)(3.0)^2 + 0 = 9.0\text{ J}\).
Final kinetic energy \(E_{k,f} = \frac{1}{2}(2.0)(-1.0)^2 + \frac{1}{2}(4.0)(2.0)^2 = 1.0 + 8.0 = 9.0\text{ J}\).
Since kinetic energy is conserved, the collision is elastic. So (3) is also correct.
Therefore, (1), (2) and (3) are all correct.

Correct option D: 1 mark.

Total internal reflection occurs when light travels from an optically denser medium to a less dense medium.
Using Snell's law at the critical angle \(c\):
\(n_{\text{glass}} \sin c = n_{\text{water}} \sin 90^\circ\)
\(1.50 \sin c = 1.33\)
\(\sin c = \frac{1.33}{1.50} \approx 0.8867\)
\(c = \arcsin(0.8867) \approx 62.5^\circ\).

Correct option C: 1 mark.

The fringe separation between adjacent bright fringes (or adjacent dark fringes) is \(\Delta y = \frac{\lambda D}{a}\).
\(\Delta y = \frac{(600 \times 10^{-9}\text{ m})(1.60\text{ m})}{0.24 \times 10^{-3}\text{ m}} = 4.0 \times 10^{-3}\text{ m} = 4.0\text{ mm}\).
The position of the \(m\)-th dark fringe is given by \(y_m = \left(m - \frac{1}{2}\right)\Delta y\).
For the 2nd dark fringe, \(m = 2\), so \(y_2 = 1.5 \Delta y = 1.5 \times 4.0\text{ mm} = 6.0\text{ mm}\) from the center.
The distance between the 2nd dark fringes on opposite sides of the central maximum is \(2 \times y_2 = 3 \Delta y = 3 \times 4.0\text{ mm} = 12.0\text{ mm}\).

Correct option C: 1 mark.

The current in the circuit is \(I = \frac{E}{R + r}\).
Terminal voltage \(V = E - Ir = E \frac{R}{R + r}\). As \(R\) increases, \(\frac{R}{R+r}\) increases, so the terminal voltage increases.
Power delivered to \(R\) is \(P = I^2 R = \frac{E^2 R}{(R + r)^2}\). Maximum power transfer occurs when \(R = r = 1.5\text{ }\Omega\).
For \(R > 1.5\text{ }\Omega\), increasing \(R\) causes the power dissipated in \(R\) to decrease.
Therefore, the terminal voltage increases while the power dissipated in \(R\) decreases.

Correct option B: 1 mark.

Area of the circular coil \(A = \pi r^2 = \pi (0.10)^2 = 0.01\pi\text{ m}^2\).
Initial magnetic flux linkage \(\Phi_1 = N B_1 A = 50 \times (+0.40) \times (0.01\pi) = 0.20\pi\text{ Wb}\).
Final magnetic flux linkage \(\Phi_2 = N B_2 A = 50 \times (-0.40) \times (0.01\pi) = -0.20\pi\text{ Wb}\).
Magnitude of change in flux linkage \(|\Delta (N\Phi)| = |\Phi_2 - \Phi_1| = 0.40\pi\text{ Wb}\).
Average induced e.m.f. \(\varepsilon = \frac{|\Delta (N\Phi)|}{\Delta t} = \frac{0.40\pi}{0.20} = 2\pi \approx 6.28\text{ V}\).

Correct option C: 1 mark.

Initial corrected count rate of isotope \(X = 1280 - 40 = 1240\text{ counts per minute}\).
Number of half-lives elapsed in \(24\text{ days} = \frac{24}{8.0} = 3\).
Corrected count rate after \(3\) half-lives = \(\frac{1240}{2^3} = \frac{1240}{8} = 155\text{ counts per minute}\).
Total measured count rate = corrected count rate + background count rate = \(155 + 40 = 195\text{ counts per minute}\).

Correct option B: 1 mark.

Statement (1) is correct: Dull, black surfaces have higher emissivity and absorptivity compared to shiny, reflective surfaces.
Statement (2) is correct: Convection currents rely on buoyancy driven by density differences, which requires the presence of a gravitational field.
Statement (3) is incorrect: An object in thermal equilibrium with its surroundings continues to emit radiation at the exact same rate as it absorbs radiation.

B (1 mark): Statements (1) and (2) only are correct.

Considering the whole connected system along the direction of motion:
Net accelerating force \(F_{\text{net}} = m_{\text{hanging}}g - F_{\text{applied}} = (2.0)(9.81) - 5.0 = 19.62 - 5.0 = 14.62\text{ N}\).
Total mass of system \(M = 3.0\text{ kg} + 2.0\text{ kg} = 5.0\text{ kg}\).
Acceleration \(a = \frac{F_{\text{net}}}{M} = \frac{14.62}{5.0} = 2.924\text{ m s}^{-2} \approx 2.92\text{ m s}^{-2}\).

B (1 mark): \(a = \frac{(2.0)(9.81) - 5.0}{3.0 + 2.0} = 2.92\text{ m s}^{-2}\).

Taking right as positive:
Initial momentum of \(X\): \(p_{X,i} = (0.20)(+6.0) = +1.20\text{ N s}\).
Final momentum of \(X\): \(p_{X,f} = (0.20)(-2.0) = -0.40\text{ N s}\).
Impulse on \(X = p_{X,f} - p_{X,i} = -0.40 - 1.20 = -1.60\text{ N s}\).
By Newton's third law, impulse acting on \(Y = +1.60\text{ N s}\).
Final velocity of \(Y\): \(v_Y = \frac{1.60}{0.40} = 4.0\text{ m s}^{-1}\).
Initial total kinetic energy \(E_{k,i} = \frac{1}{2}(0.20)(6.0)^2 = 3.60\text{ J}\).
Final total kinetic energy \(E_{k,f} = \frac{1}{2}(0.20)(2.0)^2 + \frac{1}{2}(0.40)(4.0)^2 = 0.40 + 3.20 = 3.60\text{ J}\).
Loss in kinetic energy \(= 3.60 - 3.60 = 0\text{ J}\).

B (1 mark): Magnitude of impulse \(= 1.60\text{ N s}\), loss of kinetic energy \(= 0\text{ J}\).

Using Snell's law at the interface:
\(n_{\text{liquid}} \sin(30.0^\circ) = n_{\text{glass}} \sin(20.0^\circ)\)
\(n_{\text{liquid}} (0.50) = 1.50 \sin(20.0^\circ)\)
\(n_{\text{liquid}} = 3.00 \sin(20.0^\circ) \approx 1.026\).
For light travelling from glass to liquid, the critical angle \(c\) is given by:
\(\sin c = \frac{n_{\text{liquid}}}{n_{\text{glass}}} = \frac{1.50 \sin(20.0^\circ) / \sin(30.0^\circ)}{1.50} = \frac{\sin(20.0^\circ)}{\sin(30.0^\circ)} = 2\sin(20.0^\circ) \approx 0.6840\)
\(c = \arcsin(0.6840) \approx 43.2^\circ\).

B (1 mark): Critical angle \(c = \arcsin\left(\frac{\sin 20.0^\circ}{\sin 30.0^\circ}\right) = 43.2^\circ\).

The fringe separation is given by \(\Delta y = \frac{\lambda D}{a}\).
(1) Halving \(a\) gives \(\Delta y' = \frac{\lambda D}{a/2} = 2\Delta y\). (Correct)
(2) Doubling \(D\) gives \(\Delta y' = \frac{\lambda (2D)}{a} = 2\Delta y\). (Correct)
(3) Doubling frequency \(f\) halves wavelength \(\lambda\) since \(v = f\lambda\), giving \(\Delta y' = \frac{(\lambda/2) D}{a} = \frac{1}{2}\Delta y\). (Incorrect)

B (1 mark): Statements (1) and (2) only are correct.

Current through the circuit is \(I = \frac{E}{R + r}\).
As \(R\) increases, current \(I\) decreases.
Terminal potential difference \(V = E - Ir\). Since \(I\) decreases, the internal potential drop \(Ir\) decreases, so terminal voltage \(V\) increases.
Power dissipated inside the battery \(P_{\text{internal}} = I^2 r\). Since \(I\) decreases and \(r\) is constant, \(P_{\text{internal}}\) decreases.

A (1 mark): Terminal voltage increases, power dissipated in the internal resistance decreases.

Statement (1) is correct: As the loop leaves the region, the magnetic flux into the page decreases. By Lenz's law, the induced current produces a magnetic field directed into the page to oppose the decrease, which corresponds to a clockwise current.
Statement (2) is correct: By Fleming's left-hand rule or Lenz's law, the magnetic force opposes the motion of the loop, thus pointing to the left.
Statement (3) is correct: Induced e.m.f. \(\mathcal{E} = BLv\), current \(I = \frac{BLv}{R}\), retarding force \(F = ILB = \frac{B^2 L^2 v}{R}\). The mechanical power is \(P = Fv = \frac{B^2 L^2 v^2}{R} \propto v^2\). When \(v\) doubles, \(P\) is multiplied by \(2^2 = 4\).

D (1 mark): All statements (1), (2), and (3) are correct.

Let the initial activity of \(Q\) be \(A_0\), then the initial activity of \(P\) is \(4A_0\).
Activity of \(P\) at time \(t\): \(A_P(t) = 4A_0 \left(\frac{1}{2}\right)^{t/4}\).
Activity of \(Q\) at time \(t\): \(A_Q(t) = A_0 \left(\frac{1}{2}\right)^{t/12}\).
Setting \(A_P(t) = A_Q(t)\):
\(4 \left(\frac{1}{2}\right)^{t/4} = \left(\frac{1}{2}\right)^{t/12}\)
\(2^2 \cdot 2^{-t/4} = 2^{-t/12}\)
\(2 - \frac{t}{4} = -\frac{t}{12}\)
\(2 = \frac{t}{4} - \frac{t}{12} = \frac{3t - t}{12} = \frac{2t}{12} = \frac{t}{6}\)
\(t = 12\text{ hours}\).

C (1 mark): \(t = 12\text{ hours}\).

Energy supplied by the heater is \(E = P t = 80\text{ W} \times (5.0 \times 60\text{ s}) = 24\,000\text{ J}\). Using \(E = m c \Delta T\), we have \(24\,000 = 0.50 \times c \times 12\), which yields \(c = \frac{24\,000}{6.0} = 4000\text{ J kg}^{-1}\,^\circ\text{C}^{-1}\).

C (1 mark)

Net horizontal force acting on the block is \(F_{\text{net}} = m a = 4.0 \times 2.5 = 10\text{ N}\). The frictional force is \(f = F - F_{\text{net}} = 18 - 10 = 8.0\text{ N}\). The normal reaction force is \(N = m g = 4.0 \times 9.81 = 39.24\text{ N}\). Thus, the coefficient of dynamic friction is \(\mu = \frac{f}{N} = \frac{8.0}{39.24} \approx 0.20\).

A (1 mark)

By conservation of linear momentum: \(m_P u_P + m_Q u_Q = m_P v_P + m_Q v_Q \implies (2.0)(3.0) + 0 = (2.0)(1.0) + (1.0) v_Q \implies v_Q = 4.0\text{ m s}^{-1}\) to the right. Statement (1) is correct.
Total initial KE = \(\frac{1}{2}(2.0)(3.0)^2 = 9.0\text{ J}\). Total final KE = \(\frac{1}{2}(2.0)(1.0)^2 + \frac{1}{2}(1.0)(4.0)^2 = 1.0 + 8.0 = 9.0\text{ J}\). KE is conserved, so statement (2) is correct.
Impulse on \(Q\) = \(\Delta p_Q = m_Q v_Q - 0 = 1.0 \times 4.0 = 4.0\text{ N s}\) to the right. Statement (3) is correct.

D (1 mark)

The wavelength is \(\lambda = \frac{v}{f} = \frac{12\text{ m s}^{-1}}{4.0\text{ Hz}} = 3.0\text{ m}\). The phase difference between two points separated by distance \(\Delta x = 1.5\text{ m}\) is \(\Delta \phi = \frac{\Delta x}{\lambda} \times 360^\circ = \frac{1.5}{3.0} \times 360^\circ = 180^\circ\).

D (1 mark)

Using the lens formula \(\frac{1}{u} + \frac{1}{v} = \frac{1}{f}\), with \(u = +30\text{ cm}\) and \(f = +20\text{ cm}\): \(\frac{1}{v} = \frac{1}{20} - \frac{1}{30} = \frac{1}{60} \implies v = +60\text{ cm}\). Since \(v > 0\), the image is real. The linear magnification is \(m = \frac{v}{u} = \frac{60}{30} = 2 > 1\), meaning the image is magnified and located at \(60\text{ cm}\) from the lens.

A (1 mark)

The equivalent resistance of the two parallel resistors is \(R_p = \frac{R}{2}\). The total resistance of the entire circuit is \(R_{\text{eq}} = R + R_p = R + \frac{R}{2} = \frac{3}{2}R\). The total power dissipated is \(P = \frac{E^2}{R_{\text{eq}}} = \frac{E^2}{\frac{3}{2}R} = \frac{2 E^2}{3 R}\).

B (1 mark)

The magnitude of the motional e.m.f. induced across a conductor moving perpendicular to a magnetic field is given by \(\varepsilon = B L v\). Substituting the given values: \(\varepsilon = 0.30\text{ T} \times 0.40\text{ m} \times 5.0\text{ m s}^{-1} = 0.60\text{ V}\).

C (1 mark)

Initial corrected count rate: \(R_0 = 960 - 40 = 920\text{ cpm}\). Corrected count rate at \(t = 18\text{ hours}\): \(R = 155 - 40 = 115\text{ cpm}\). The fraction remaining is \(\frac{R}{R_0} = \frac{115}{920} = \frac{1}{8} = \left(\frac{1}{2}\right)^3\). Thus, 3 half-lives have elapsed in \(18\text{ hours}\), so the half-life is \(T_{1/2} = \frac{18}{3} = 6.0\text{ hours}\).

C (1 mark)

Consider the entire system of total mass \(M_{\text{total}} = m_P + m_Q + m_R = 0.40 + 0.60 + 1.00 = 2.00\text{ kg}\).

The net accelerating force on the whole system is the weight of hanging block \(R\):
\[F_{\text{net}} = m_R g = 1.00 \times 9.81 = 9.81\text{ N}\]

The acceleration of the system is:
\[a = \frac{F_{\text{net}}}{M_{\text{total}}} = \frac{9.81\text{ N}}{2.00\text{ kg}} = 4.905\text{ m s}^{-2}\]

Since block \(P\) moves horizontally with acceleration \(a\) and does not slip on cart \(Q\), the only horizontal force acting on \(P\) is the static friction force \(f\) from the cart:
\[f = m_P a = 0.40\text{ kg} \times 4.905\text{ m s}^{-2} = 1.962\text{ N} \approx 1.96\text{ N}\]

A (1 mark): Correct application of Newton's second law to find system acceleration \(a = 4.905\text{ m s}^{-2}\) and net force on \(P\) as \(f = m_P a = 1.96\text{ N}\).

(a) Energy required to raise temperature:
\(E_1 = mc\Delta T = (0.80)(4200)(100 - 24) = 2.5536 \times 10^5\text{ J}\)
Time taken:
\(t_1 = \frac{E_1}{P} = \frac{2.5536 \times 10^5}{1800} \approx 141.87\text{ s} \approx 142\text{ s}\)

(b) Energy required to vaporize water:
\(E_2 = m_{vap} l_v = (0.050)(2.26 \times 10^6) = 1.13 \times 10^5\text{ J}\)
Additional time:
\(t_2 = \frac{E_2}{P} = \frac{1.13 \times 10^5}{1800} \approx 62.78\text{ s} \approx 62.8\text{ s}\)

(c) 1. Heat is lost to the cooler surroundings via conduction, convection, and radiation.
2. Some energy is absorbed to heat the container / heating element of the kettle itself.

(d) The steam leaving the spout directly is water in gaseous form, which is colorless and transparent. As the steam moves further away, it mixes with the cooler ambient air and cools down below \(100^\circ\text{C}\), condensing into tiny suspended water droplets which scatter light and appear as visible white mist.

(a) \(E_1 = mc\Delta T = 0.80 \times 4200 \times (100 - 24) = 2.5536 \times 10^5\text{ J}\) (1M)
\(t_1 = \frac{E_1}{P} = 142\text{ s}\) (1A)

(b) \(E_2 = ml_v = 0.050 \times 2.26 \times 10^6 = 1.13 \times 10^5\text{ J}\) (1M)
\(t_2 = \frac{E_2}{P} = 62.8\text{ s}\) (1A)

(c) Any TWO valid reasons (1A + 1A):
- Heat lost to the surrounding environment.
- Heat absorbed by the kettle / heating element.

(d) - Steam at the spout is invisible water vapour / gaseous state. (1A)
- Steam cools upon contact with cooler surrounding air. (1A)
- Steam condenses into tiny liquid water droplets forming mist. (1A)

(a) By conservation of linear momentum along the horizontal direction (taking right as positive):
\(m_A u_A + m_B u_B = m_A v_A + m_B v_B\)
\((0.60)(2.5) + 0 = (0.60)v_A + (0.40)(3.0)\)
\(1.50 = 0.60 v_A + 1.20\)
\(v_A = \frac{0.30}{0.60} = +0.50\text{ m s}^{-1}\)
Thus, velocity of cart \(A\) is \(0.50\text{ m s}^{-1}\) to the right.

(b) Initial kinetic energy:
\(E_{k,initial} = \frac{1}{2} m_A u_A^2 = \frac{1}{2}(0.60)(2.5)^2 = 1.875\text{ J}\)
Final kinetic energy:
\(E_{k,final} = \frac{1}{2} m_A v_A^2 + \frac{1}{2} m_B v_B^2 = \frac{1}{2}(0.60)(0.50)^2 + \frac{1}{2}(0.40)(3.0)^2 = 0.075 + 1.80 = 1.80\text{ J}\) (or \(1.875 - 1.800 = 0.075\text{ J}\) lost).
Since \(E_{k,final} < E_{k,initial}\), kinetic energy is not conserved, so the collision is inelastic.

(c) (i) \(F_{avg} = \frac{\Delta p_B}{\Delta t} = \frac{m_B v_B - 0}{\Delta t} = \frac{(0.40)(3.0)}{0.040} = 30\text{ N}\)
(ii) Magnitude is \(30\text{ N}\), directed to the left. Governed by Newton's third law of motion.

(d) Work-energy principle for cart \(B\):
\(f \cdot s = E_k \implies (1.2) s = \frac{1}{2}(0.40)(3.0)^2 = 1.80\text{ J} \implies s = \frac{1.80}{1.2} = 1.5\text{ m}\)

(a) \(m_A u_A + m_B u_B = m_A v_A + m_B v_B\) (1M)
\(v_A = 0.50\text{ m s}^{-1}\) to the right (1A)

(b) \(E_{k,i} = \frac{1}{2}(0.60)(2.5)^2 = 1.875\text{ J}\) (1M)
\(E_{k,f} = \frac{1}{2}(0.60)(0.50)^2 + \frac{1}{2}(0.40)(3.0)^2 = 1.80\text{ J}\) (1M)
State that kinetic energy is not conserved / \(E_{k,f} \neq E_{k,i}\), hence the collision is inelastic. (1A)

(c) (i) \(F = \frac{m_B v_B - m_B u_B}{\Delta t} = \frac{0.40 \times 3.0}{0.040} = 30\text{ N}\) (1M + 1A)
(ii) \(30\text{ N}\) to the left (1A); Newton's third law of motion (1A)

(d) \(f s = \frac{1}{2} m v^2 \implies s = 1.5\text{ m}\) (1A)

(a) The normal reaction force \(N\) between the car and track becomes zero (or gravity alone provides the necessary centripetal force).

(b) At the top:
\(N + mg = \frac{m v_{top}^2}{R}\)
For minimum speed, \(N = 0\):
\(mg = \frac{m v_{min}^2}{R} \implies v_{min} = \sqrt{gR} = \sqrt{(9.81)(0.40)} = \sqrt{3.924} \approx 1.9809\text{ m s}^{-1} \approx 1.98\text{ m s}^{-1}\).

(c) Taking the lowest point as reference level of gravitational potential energy:
\(E_{bottom} = E_{top}\)
\(\frac{1}{2} m v_{bottom}^2 = \frac{1}{2} m v_{top}^2 + mg(2R)\)
\(v_{bottom}^2 = v_{top}^2 + 4gR = gR + 4gR = 5gR\)
\(v_{bottom} = \sqrt{5gR} = \sqrt{5 \times 9.81 \times 0.40} = \sqrt{19.62} \approx 4.43\text{ m s}^{-1}\)

(d) (i) At the lowest point:
\(N - mg = \frac{m v^2}{R}\)
\(N = mg + \frac{m v^2}{R} = (0.25)(9.81) + \frac{(0.25)(5.0)^2}{0.40} = 2.4525 + 15.625 = 18.0775\text{ N} \approx 18.1\text{ N}\)
(ii) As the car rises, its speed decreases (due to conversion of KE to GPE) and the radial component of gravity increasingly contributes towards the center, so the normal force decreases continuously.

(a) Normal reaction \(N = 0\) / gravity provides all centripetal acceleration. (1A)

(b) \(mg = \frac{m v^2}{R}\) (1M)
\(v = \sqrt{9.81 \times 0.40} = 1.98\text{ m s}^{-1}\) (1A)

(c) Conservation of energy equation: \(\frac{1}{2}m v_b^2 = \frac{1}{2} m v_t^2 + mg(2R)\) (1M)
Correct substitution: \(v_b^2 = (1.98)^2 + 4(9.81)(0.40)\) (1M)
\(v_b = 4.43\text{ m s}^{-1}\) (1A)

(d) (i) \(N - mg = \frac{mv^2}{R}\) (1M)
\(N = 18.1\text{ N}\) (1A)
(ii) Decreases (1A)

(a) Using the wave equation:
\(v = f \lambda \implies \lambda = \frac{v}{f} = \frac{340}{680} = 0.50\text{ m}\)

(b) Using fringe separation formula \(\Delta y = \frac{\lambda D}{a}\):
\(\Delta y = \frac{(0.50)(6.00)}{1.50} = 2.00\text{ m}\)

(c) (i) At the perpendicular bisector (central position), the path difference is zero. With opposite phase, waves from \(S_1\) and \(S_2\) arrive completely out of phase (destructive interference), turning the central position into an intensity minimum.
(ii) The fringe separation remains unchanged because \(\lambda\), \(D\), and \(a\) are unchanged (the interference pattern simply shifts by \(\frac{\Delta y}{2}\)).

(d) 1. Sound waves reflect off classroom walls/ceiling/floor and arrive at the detector.
2. The amplitudes of the sound waves from \(S_1\) and \(S_2\) are slightly different at non-central minima due to different distances travelled, so they do not cancel out completely.

(a) \(v = f \lambda \implies \lambda = \frac{340}{680}\) (1M)
\(\lambda = 0.50\text{ m}\) (1A)

(b) \(\Delta y = \frac{\lambda D}{a} = \frac{0.50 \times 6.00}{1.50}\) (1M)
\(\Delta y = 2.00\text{ m}\) (1A)

(c) (i) The central position changes into a minimum / destructive interference occurs (1A) because zero path difference now corresponds to opposite phase arrival (1A).
(ii) Fringe separation remains unchanged / \(2.00\text{ m}\). (1A)

(d) Any TWO valid reasons (1A + 1A):
- Reflections from surrounding walls / environment (reverberation).
- Amplitudes of waves from the two sources are not identical at the detector position.

(a) Using Snell's law:
\(n_{air} \sin i = n_{glass} \sin r\)
\(1.00 \times \sin 40.0^\circ = 1.52 \times \sin r\)
\(\sin r = \frac{\sin 40.0^\circ}{1.52} = \frac{0.64279}{1.52} = 0.42289\)
\(r = \sin^{-1}(0.42289) \approx 25.018^\circ \approx 25.0^\circ\)

(b) For total internal reflection at glass-air interface:
\(\sin C = \frac{1}{n_{glass}} = \frac{1}{1.52} = 0.65789\)
\(C = \sin^{-1}(0.65789) \approx 41.14^\circ \approx 41.1^\circ\)

(c) (i) The ray originates from the center of the semicircle and travels along the radius. Therefore, it hits the curved surface at an angle of incidence of \(0^\circ\) (normal to the surface), so no refraction/bending occurs.
(ii) \(v = \frac{c}{n} = \frac{3.00 \times 10^8}{1.52} \approx 1.9737 \times 10^8\text{ m s}^{-1} \approx 1.97 \times 10^8\text{ m s}^{-1}\)

(d) No. Total internal reflection can only occur when light travels from an optically denser medium to an optically less dense medium (and angle of incidence exceeds critical angle). Here light travels from air to glass.

(a) \(1.00 \sin 40.0^\circ = 1.52 \sin r\) (1M)
\(r = 25.0^\circ\) (1A)

(b) \(\sin C = \frac{1}{1.52}\) (1M)
\(C = 41.1^\circ\) (1A)

(c) (i) Ray travels along normal / radius to the curved surface (1A), angle of incidence is \(0^\circ\) (1A).
(ii) \(v = \frac{c}{n} = \frac{3.00 \times 10^8}{1.52}\) (1M)
\(v = 1.97 \times 10^8\text{ m s}^{-1}\) (1A)

(d) No, light is travelling from less dense to denser medium. (1A)

(a) \(V = \mathcal{E} - Ir\)

(b) For \(R_1 = 5.0\text{ }\Omega\):
\(I_1 = \frac{V_1}{R_1} = \frac{4.0}{5.0} = 0.80\text{ A}\)
\(\mathcal{E} = V_1 + I_1 r \implies \mathcal{E} = 4.0 + 0.80 r\) ---(1)

For \(R_2 = 2.0\text{ }\Omega\):
\(I_2 = \frac{V_2}{R_2} = \frac{3.0}{2.0} = 1.50\text{ A}\)
\(\mathcal{E} = V_2 + I_2 r \implies \mathcal{E} = 3.0 + 1.50 r\) ---(2)

Equating (1) and (2):
\(4.0 + 0.80 r = 3.0 + 1.50 r \implies 0.70 r = 1.0 \implies r = \frac{1.0}{0.70} \approx 1.4286\text{ }\Omega\)
Wait, let's re-verify:
\(4.0 + 0.80 r = 3.0 + 1.50 r \implies 1.0 = 0.70 r \implies r = 1.43\text{ }\Omega\).
\(\mathcal{E} = 4.0 + 0.80(1.4286) = 5.14\text{ V}\).

(c) Maximum power transfer occurs when external resistance \(R = r = 1.43\text{ }\Omega\):
\(P_{max} = \frac{\mathcal{E}^2}{4r} = \frac{(5.143)^2}{4(1.4286)} = \frac{26.45}{5.714} \approx 4.63\text{ W}\)

(d) When short-circuited (\(R \approx 0\)), the total resistance is only \(r\), resulting in a very large current \(I = \frac{\mathcal{E}}{r}\). The rate of heat dissipation inside the battery (\(P = I^2 r\)) becomes extremely large, causing the battery to overheat, swell, catch fire, or explode.

(a) \(V = \mathcal{E} - Ir\) (1A)

(b) Finding currents: \(I_1 = 0.80\text{ A}\), \(I_2 = 1.50\text{ A}\) (1M)
Setting up simultaneous equations (1M)
\(r = 1.43\text{ }\Omega\) (1A)
\(\mathcal{E} = 5.14\text{ V}\) (1A)

(c) Using \(R = r\) for maximum power or \(P = I^2 R\) (1M)
\(P_{max} = 4.63\text{ W}\) (1A)

(d) - Extremely large current flows through the circuit. (1A)
- Rapid and excessive heat generation (\(I^2 r\)) leading to risk of burns, battery damage, or fire. (1A)

(a) Faraday's law states that the magnitude of the induced e.m.f. in a circuit is directly proportional to the time rate of change of magnetic flux linkage through the circuit.

(b) (i) In time \(\Delta t\), area change \(\Delta A = w v \Delta t\).
Rate of change of flux:
\(\mathcal{E} = \frac{\Delta \Phi}{\Delta t} = \frac{B \Delta A}{\Delta t} = B w v = (0.35)(0.20)(4.0) = 0.28\text{ V}\).

(ii) Induced current:
\(I = \frac{\mathcal{E}}{R} = \frac{0.28}{0.14} = 2.0\text{ A}\).
By Lenz's law, as the inward magnetic flux decreases, the induced current produces an inward magnetic field to oppose the decrease. By the right-hand grip rule, the induced current flows in a clockwise direction.

(iii) By Fleming's left-hand rule, the vertical wire inside the field carrying upward current experiences a magnetic force directed to the left (opposing the motion). To maintain a constant velocity, the net horizontal force must be zero, hence an equal and opposite external pulling force directed to the right is needed.

(c) \(F_{ext} = F_B = B I w = (0.35)(2.0)(0.20) = 0.14\text{ N}\)

(d) Mechanical power: \(P_{mech} = F_{ext} v = (0.14)(4.0) = 0.56\text{ W}\).
Electrical power: \(P_{elec} = I^2 R = (2.0)^2(0.14) = 0.56\text{ W}\).
Since \(P_{mech} = P_{elec} = 0.56\text{ W}\), energy is conserved.

(a) Induced e.m.f. is proportional to the rate of change of magnetic flux (linkage). (1A)

(b) (i) \(\mathcal{E} = Bwv\) (1M)
\(\mathcal{E} = 0.35 \times 0.20 \times 4.0 = 0.28\text{ V}\) (1A)
(ii) \(I = \frac{\mathcal{E}}{R} = \frac{0.28}{0.14} = 2.0\text{ A}\) (1A); Clockwise (1A)
(iii) Induced current experiences a magnetic force to the left opposing the motion (Lenz's law) (1A); external force to the right balances this magnetic force to maintain constant speed (1A).

(c) \(F = B I w = 0.35 \times 2.0 \times 0.20\) (1M)
\(F = 0.14\text{ N}\) (1A)

(d) Show \(P_{mech} = 0.14 \times 4.0 = 0.56\text{ W}\) and \(P_{elec} = 2.0^2 \times 0.14 = 0.56\text{ W}\), which are equal. (1A)

(a) \(^{222}_{86}\text{Rn} \rightarrow \text{ }^{218}_{84}\text{Po} + \text{ }^4_2\text{He}\) (or \(^4_2\alpha\))

(b) Half-life in seconds:
\(t_{1/2} = 3.82 \times 24 \times 3600\text{ s} = 3.30048 \times 10^5\text{ s}\)
\(\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.69315}{3.30048 \times 10^5} \approx 2.1001 \times 10^{-6}\text{ s}^{-1} \approx 2.10 \times 10^{-6}\text{ s}^{-1}\)

(c) (i) \(A_0 = \lambda N_0 \implies N_0 = \frac{A_0}{\lambda} = \frac{6.4 \times 10^5}{2.1001 \times 10^{-6}} \approx 3.047 \times 10^{11} \approx 3.05 \times 10^{11}\)

(ii) Number of half-lives:
\(n = \frac{t}{t_{1/2}} = \frac{19.1\text{ days}}{3.82\text{ days}} = 5.0\)
\(A = A_0 \left(\frac{1}{2}\right)^n = 6.4 \times 10^5 \times \left(\frac{1}{2}\right)^5 = \frac{6.4 \times 10^5}{32} = 2.0 \times 10^4\text{ Bq}\)

(d) Being a gas, radon can easily be inhaled into the respiratory tract and lungs. Since alpha particles have high ionizing power, they cause severe localized ionization damage to internal lung cells / DNA, significantly increasing the risk of lung cancer.

(a) \(^{222}_{86}\text{Rn} \rightarrow \text{ }^{218}_{84}\text{Po} + \text{ }^4_2\text{He}\) (1A)

(b) \(t_{1/2} = 3.82 \times 86400 = 3.30 \times 10^5\text{ s}\) (1M)
\(\lambda = \frac{\ln 2}{t_{1/2}} = 2.10 \times 10^{-6}\text{ s}^{-1}\) (1A)

(c) (i) \(N_0 = \frac{A_0}{\lambda} = \frac{6.4 \times 10^5}{2.10 \times 10^{-6}}\) (1M)
\(N_0 = 3.05 \times 10^{11}\) (1A)
(ii) \(A = 6.4 \times 10^5 \times (0.5)^5\) (1M)
\(A = 2.0 \times 10^4\text{ Bq}\) (1A)

(d) - Radon is an inhaled gas that enters the lungs directly. (1A)
- Alpha particles have high ionizing power / direct exposure inside lung tissue damages DNA / cells. (1A)

(a) Faraday's law states that the induced e.m.f. in a circuit is directly proportional to the time rate of change of magnetic flux linkage (i.e. \(\varepsilon = -\frac{\Delta \Phi}{\Delta t}\)).

(b) (i) Polarity: N-pole (or North pole).
By Lenz's law, the induced current flows in such a direction as to oppose the change causing it (the approaching N-pole of the magnet). Thus, a North pole is induced at the top of the solenoid to repel the falling magnet and oppose its motion.

(ii) Under gravity, the magnet accelerates as it falls, so its speed when exiting the bottom of the solenoid is greater than when entering the top. A higher speed means the magnetic flux through the coil changes over a shorter time interval (i.e. greater rate of change of magnetic flux linkage \(\frac{\Delta \Phi}{\Delta t}\)), producing a larger peak e.m.f.

(c) (i) When the circuit is closed, the induced e.m.f. drives an induced current. By Lenz's law, the magnetic field produced by this current exerts an upward magnetic force on the falling magnet, opposing its downward motion. Therefore, the net downward force is \(F_{\text{net}} = mg - F_{\text{magnetic}} < mg\), which gives an acceleration \(a = \frac{F_{\text{net}}}{m} < g\).

(ii) Total vertical distance fallen from release to exiting the bottom of the solenoid:
\[ s = 0.45\text{ m} + 0.10\text{ m} = 0.55\text{ m} \]
Loss in gravitational potential energy:
\[ \Delta E_p = m g s = (0.050)(9.81)(0.55) = 0.269775\text{ J} \]
By conservation of energy:
\[ \Delta E_p = E_k + E_{\text{thermal}} \]
\[ 0.269775 = \frac{1}{2} m v^2 + 0.035 \]
\[ \frac{1}{2}(0.050)v^2 = 0.269775 - 0.035 = 0.234775\text{ J} \]
\[ 0.025 v^2 = 0.234775 \implies v^2 \approx 9.391 \]
\[ v = \sqrt{9.391} \approx 3.06\text{ m s}^{-1} \]

(a) State that the induced e.m.f. is proportional to the rate of change of magnetic flux (linkage). [1A]

(b)(i) State N-pole / north polarity. [1A]
Explain that the induced current opposes the approach of the magnet by creating a repulsive force / opposing magnetic field (Lenz's law). [1A]

(b)(ii) State that the magnet travels faster when leaving the coil than when entering. [1A]
Explain that a higher speed leads to a higher rate of change of magnetic flux linkage (thus larger e.m.f.). [1A]

(c)(i) State that an induced current creates an upward opposing magnetic force on the magnet. [1A]
State that the net downward force is less than the weight \(mg\), so acceleration \(a < g\). [1A]

(c)(ii) Correct application of conservation of energy: \(mgs = \frac{1}{2}mv^2 + E_{\text{thermal}}\) with \(s = 0.55\text{ m}\). [1M]
Correct calculated value of speed \(v = 3.06\text{ m s}^{-1}\) (accept 3.06 to 3.07 m s⁻¹). [1A]

Using the Doppler formula for electromagnetic waves: \(\frac{\Delta \lambda}{\lambda_0} = \frac{v}{c}\). \(\Delta \lambda = 515\text{ nm} - 500\text{ nm} = 15\text{ nm}\). Thus, \(v = \frac{15}{500} \times 3.00 \times 10^8 = 9.00 \times 10^6\text{ m s}^{-1}\). Since the observed wavelength is longer (redshifted), the galaxy is moving away from Earth.

Award 1 mark for the correct option (D).

By Stefan-Boltzmann law, luminosity \(L = 4\pi R^2 \sigma T^4\). Therefore, \(\frac{L_P}{L_Q} = \left(\frac{R_P}{R_Q}\right)^2 \left(\frac{T_P}{T_Q}\right)^4 = \left(\frac{2}{6}\right)^2 \left(\frac{8000}{4000}\right)^4 = \left(\frac{1}{3}\right)^2 (2)^4 = \frac{16}{9} \approx 1.78\).

Award 1 mark for the correct option (B).

Ground state energy is \(E_1 = -13.6\text{ eV}\). The energy of the excited state is \(E_n = -13.6\text{ eV} + 12.75\text{ eV} = -0.85\text{ eV}\). Solving \(-\frac{13.6}{n^2} = -0.85\) gives \(n^2 = 16 \implies n = 4\). The number of possible emission transitions among the 4 levels is \(\frac{n(n-1)}{2} = \frac{4 \times 3}{2} = 6\).

Award 1 mark for the correct option (C).

From Einstein's photoelectric equation: \(E_1 = hf - \Phi \implies hf = E_1 + \Phi\). For frequency \(2f\): \(E_2 = h(2f) - \Phi = 2(hf) - \Phi = 2(E_1 + \Phi) - \Phi = 2E_1 + \Phi\). Since the work function \(\Phi > 0\), it follows that \(E_2 > 2E_1\).

Award 1 mark for the correct option (B).

Total thermal resistance \(R = \frac{d_1}{\kappa_1} + \frac{d_2}{\kappa_2} = \frac{0.15}{1.2} + \frac{0.05}{0.04} = 0.125 + 1.25 = 1.375\text{ m}^2\text{ K W}^{-1}\). The overall \(U\)-value is \(U = \frac{1}{R} = \frac{1}{1.375} \approx 0.73\text{ W m}^{-2}\text{ K}^{-1}\).

Award 1 mark for the correct option (A).

The power generated by a wind turbine is proportional to the swept area \(A = \pi r^2\) and the cube of the wind speed \(v^3\), so \(P \propto r^2 v^3\). Therefore, \(\frac{P_{\text{new}}}{P} = \left(\frac{2r}{r}\right)^2 \left(\frac{1.5v}{v}\right)^3 = (4)(3.375) = 13.5\). The new power output is \(13.5 P\).

Award 1 mark for the correct option (C).

The intensity reflection coefficient is \(\alpha = \left(\frac{Z_2 - Z_1}{Z_2 + Z_1}\right)^2 = \left(\frac{1.70 - 1.38}{1.70 + 1.38}\right)^2 = \left(\frac{0.32}{3.08}\right)^2 \approx 0.0108 = 1.08\%\). Therefore, the percentage transmitted is \(100\% - 1.08\% = 98.92\% \approx 98.9\%\).

Award 1 mark for the correct option (D).

The lens needs to form a virtual image of an object at \(u = +25\text{ cm} = +0.25\text{ m}\) at the person's near point \(v = -80\text{ cm} = -0.80\text{ m}\). Lens power \(P = \frac{1}{f} = \frac{1}{u} + \frac{1}{v} = \frac{1}{0.25} - \frac{1}{0.80} = 4.00 - 1.25 = +2.75\text{ D}\). A positive power corresponds to a converging lens.

Award 1 mark for the correct option (B).

By Stefan-Boltzmann law, luminosity $L = 4\pi R^2 \sigma T^4$. Therefore, $\frac{L_P}{L_\odot} = \left(\frac{R_P}{R_\odot}\right)^2 \left(\frac{T_P}{T_\odot}\right)^4 = (3)^2 \times \left(\frac{12000}{6000}\right)^4 = 9 \times 2^4 = 9 \times 16 = 144$.

Correct answer D (1 mark).

Using the Doppler formula for electromagnetic waves $\frac{\Delta \lambda}{\lambda_0} \approx \frac{v}{c}$, we have $\Delta \lambda = 669.4 - 656.3 = 13.1\text{ nm}$. Then $v = c \times \frac{\Delta \lambda}{\lambda_0} = 3.00 \times 10^8 \times \frac{13.1}{656.3} \approx 5.99 \times 10^6\text{ m s}^{-1} = 5990\text{ km s}^{-1}$ moving away from Earth.

Correct answer B (1 mark).

From Einstein's photoelectric equation, $E_k = hf - \Phi = e V_s$. Since the photon energy $hf$ and work function $\Phi$ do not depend on intensity, both $E_k$ and $V_s$ remain unchanged, making (1) incorrect and (2) correct. Doubling the light intensity doubles the number of incident photons per second, which doubles the number of emitted photoelectrons per second, thus doubling the saturation photocurrent, so (3) is correct.

Correct answer C (1 mark).

The energy difference is $\Delta E = E_4 - E_2 = -\frac{13.6}{16} - \left(-\frac{13.6}{4}\right) = -0.85 - (-3.40) = 2.55\text{ eV}$. In joules, $\Delta E = 2.55 \times 1.60 \times 10^{-19}\text{ J} = 4.08 \times 10^{-19}\text{ J}$. The emitted wavelength is $\lambda = \frac{hc}{\Delta E} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{4.08 \times 10^{-19}} \approx 4.88 \times 10^{-7}\text{ m} = 488\text{ nm}$.

Correct answer A (1 mark).

The thermal resistance per unit area (R-value) of the composite wall is $R_{\text{tot}} = \frac{d_1}{k_1} + \frac{d_2}{k_2} = \frac{0.10}{0.60} + \frac{0.050}{0.040} = 0.1667 + 1.250 = 1.4167\text{ m}^2\text{ K W}^{-1}$. The rate of heat flow per unit area is $\frac{Q}{A \cdot t} = \frac{\Delta T}{R_{\text{tot}}} = \frac{32 - 22}{1.4167} \approx 7.06\text{ W m}^{-2} \approx 7.1\text{ W m}^{-2}$.

Correct answer B (1 mark).

The sweeping area of the turbine blades is $A = \pi r^2 = \pi (12)^2 \approx 452.39\text{ m}^2$. The total kinetic power available in the wind is $P_{\text{wind}} = \frac{1}{2} \rho A v^3 = \frac{1}{2} \times 1.2 \times 452.39 \times (8.0)^3 \approx 138974\text{ W} \approx 139\text{ kW}$. With an efficiency of $35\%$, the electrical power produced is $P_{\text{elec}} = 0.35 \times 139\text{ kW} \approx 48.6\text{ kW} \approx 49\text{ kW}$.

Correct answer C (1 mark).

The intensity reflection coefficient $\alpha$ is given by $\alpha = \left(\frac{Z_2 - Z_1}{Z_2 + Z_1}\right)^2 = \left(\frac{6.00 \times 10^6 - 1.63 \times 10^6}{6.00 \times 10^6 + 1.63 \times 10^6}\right)^2 = \left(\frac{4.37}{7.63}\right)^2 \approx (0.5727)^2 \approx 0.328 = 32.8\% \approx 33\%$.

Correct answer D (1 mark).

From the attenuation formula $I = I_0 e^{-\mu x}$, we have $\frac{I}{I_0} = 0.25$. Therefore, $e^{-0.25 x} = 0.25 \implies -0.25 x = \ln(0.25) \approx -1.3863 \implies x = \frac{1.3863}{0.25} \approx 5.55\text{ cm}$ (or using half-value thickness: two half-value thicknesses are needed to reduce intensity to $25\%$, where $\text{HVT} = \frac{\ln 2}{0.25} \approx 2.77\text{ cm}$, so $x = 2 \times 2.77\text{ cm} = 5.54\text{ cm}$).

Correct answer B (1 mark).

(a) (i) Energy of an incident photon:
\[ E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{380 \times 10^{-9}} = 5.234 \times 10^{-19}\text{ J} \]
Converting to \(\text{eV}\):
\[ E = \frac{5.234 \times 10^{-19}}{1.60 \times 10^{-19}} = 3.271\text{ eV} \]
Using Einstein's photoelectric equation:
\[ E_{k,\text{max}} = hf - \Phi = 3.271\text{ eV} - 2.30\text{ eV} = 0.971\text{ eV} \approx 0.97\text{ eV} \]

(ii) Stopping potential \(V_s\):
\[ eV_s = E_{k,\text{max}} \implies V_s = 0.971\text{ V} \approx 0.97\text{ V} \]

(b) (i) The stopping potential remains unchanged.
Explanation: The stopping potential depends only on the maximum kinetic energy \(E_{k,\text{max}}\) of the emitted electrons, which is determined solely by the frequency (or wavelength) of the light and the work function of the metal, neither of which changes with intensity.

(ii) The saturation photocurrent is doubled.
Explanation: Doubling the light intensity doubles the number of photons striking the cathode per unit time. Since each photoelectron is ejected by a single photon, the rate of emission of photoelectrons doubles, leading to twice the saturation current.

(c) (i) Kinetic energy in joules:
\[ E_k = 54.0 \times 1.60 \times 10^{-19} = 8.64 \times 10^{-18}\text{ J} \]
Momentum \(p = \sqrt{2m_e E_k}\):
\[ p = \sqrt{2 \times (9.11 \times 10^{-31}) \times (8.64 \times 10^{-18})} = 3.968 \times 10^{-24}\text{ kg m s}^{-1} \]
de Broglie wavelength:
\[ \lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{3.968 \times 10^{-24}} = 1.671 \times 10^{-10}\text{ m} \approx 0.167\text{ nm} \]

(ii) The formation of rings is due to diffraction/interference, which confirms that electrons possess wave properties (wave-particle duality).

(a)(i)
- Calculation of photon energy \(E = 3.27\text{ eV}\) (or \(5.23 \times 10^{-19}\text{ J}\)) [1M]
- \(E_{k,\text{max}} = 0.97\text{ eV}\) (accept 0.96 to 0.98 eV) [1A]

(a)(ii)
- \(V_s = 0.97\text{ V}\) (accept 0.971 V, or e.c.f. from (a)(i)) [1A]

(b)(i)
- Stating that \(V_s\) is unchanged [1A]
- Explaining that \(E_{k,\text{max}}\) depends on the frequency/wavelength of light (or work function), not the intensity [1A]

(b)(ii)
- Stating that saturation current doubles [1A]
- Explaining that higher intensity means more photons per second, ejecting more photoelectrons per second [1A]

(c)(i)
- Correct formula for momentum \(p = \sqrt{2mE_k}\) or de Broglie relation \(\lambda = \frac{h}{p}\) [1M]
- \(\lambda = 1.67 \times 10^{-10}\text{ m}\) (or \(0.167\text{ nm}\), accept 1.66 to 1.68 × 10⁻¹⁰ m) [1A]

(c)(ii)
- Mentioning diffraction / interference demonstrating wave properties / wave nature of matter [1A]

(a) (i) Under steady-state conditions, the rate of heat transfer per unit area through the brick equals that through the polystyrene:
\[ \frac{k_1 (T_1 - T_i)}{d_1} = \frac{k_2 (T_i - T_2)}{d_2} \]
Substitute the values:
\[ \frac{0.60 (36.0 - T_i)}{0.150} = \frac{0.030 (T_i - 24.0)}{0.050} \]
\[ 4.0 (36.0 - T_i) = 0.60 (T_i - 24.0) \]
\[ 144.0 - 4.0 T_i = 0.60 T_i - 14.4 \]
\[ 4.60 T_i = 158.4 \implies T_i = 34.43^\circ\text{C} \approx 34.4^\circ\text{C} \]

(ii) Rate of heat conduction \(\frac{Q}{t}\):
\[ \frac{Q}{t} = \frac{k_1 A (T_1 - T_i)}{d_1} = \frac{0.60 \times 45.0 \times (36.0 - 34.435)}{0.150} \approx 282\text{ W} \]
(Alternatively, using total thermal resistance \(R = \frac{d_1}{k_1} + \frac{d_2}{k_2} = \frac{0.150}{0.60} + \frac{0.050}{0.030} = 0.25 + 1.667 = 1.917\text{ m}^2\text{ K W}^{-1}\):
\[ \frac{Q}{t} = \frac{A (T_1 - T_2)}{R} = \frac{45.0 \times 12.0}{1.9167} = 281.7\text{ W} \approx 282\text{ W} \])

(b) (i) Luminous flux of one fluorescent tube \(= 36\text{ W} \times 70\text{ lm W}^{-1} = 2520\text{ lm}\).
Total luminous flux \(\Phi_{\text{total}} = 4 \times 2520\text{ lm} = 10080\text{ lm}\) (or \(1.01 \times 10^4\text{ lm}\)).

(ii) Total power consumed by the 4 fluorescent tubes \(= 4 \times 36\text{ W} = 144\text{ W}\).
Total electrical power required by LED lamps:
\[ P_{\text{LED}} = \frac{\Phi_{\text{total}}}{\text{efficacy}} = \frac{10080\text{ lm}}{105\text{ lm W}^{-1}} = 96\text{ W} \]
Total electrical power saved:
\[ \Delta P = 144\text{ W} - 96\text{ W} = 48\text{ W} \]

(c) Installing external shading devices (e.g. overhangs, louvres) / using low-emissivity (Low-E) double glazing / painting external walls with light-coloured or reflective coatings.

(a)(i)
- Equating heat flux across the two layers: \(\frac{k_1 (T_1 - T_i)}{d_1} = \frac{k_2 (T_i - T_2)}{d_2}\) [1M]
- Correct substitution of values [1M]
- \(T_i = 34.4^\circ\text{C}\) (accept 34.4 to 34.5 °C) [1A]

(a)(ii)
- Correct formula for rate of heat transfer [1M]
- \(\frac{Q}{t} = 282\text{ W}\) (accept 281 W to 283 W, or e.c.f. from (a)(i)) [1A]

(b)(i)
- \(\text{Total luminous flux} = 4 \times 36 \times 70\) [1M]
- \(\Phi = 10080\text{ lm}\) (or \(1.01 \times 10^4\text{ lm}\)) [1A]

(b)(ii)
- Power of LED \(= \frac{10080}{105} = 96\text{ W}\) [1M]
- Power saved \(= 144 - 96 = 48\text{ W}\) [1A]

(c)
- Any valid passive design feature: external shading devices (fins/overhangs) / reflective exterior paint / low-e coating on windows / green roof [1A]