2024 HKDSE Physics Practice Paper with Answers

Let \(P\) be the power of the heater, \(c\) be the specific heat capacity, and \(L_v\) be the specific latent heat of vaporization.

During heating:
\(Q_1 = P \times (5.0 \times 60) = m c \Delta T = (0.50) c (140 - 80) = 30 c\)
Thus, \(P \times 300 = 30 c \implies c = 10 P\).

During vaporization of half the mass:
\(Q_2 = P \times (15.0 \times 60) = \left(\frac{1}{2} m\right) L_v = (0.25) L_v\)
Thus, \(P \times 900 = 0.25 L_v \implies L_v = 3600 P\).

Therefore, the ratio is:
\(\frac{L_v}{c} = \frac{3600 P}{10 P} = 360\text{ K}\) (or \(360^\circ\text{C}\)).

1 mark for correct option B. Award 0 marks for incorrect options.

For the first stage (\(t = 0\) to \(4.0\text{ s}\)):
\(v_1 = u + a_1 t_1 = 0 + (2.0)(4.0) = 8.0\text{ m s}^{-1}\)
\(s_1 = \frac{1}{2} a_1 t_1^2 = \frac{1}{2}(2.0)(4.0)^2 = 16.0\text{ m}\)

For the second stage (\(t = 4.0\text{ s}\) to \(8.0\text{ s}\), \(\Delta t = 4.0\text{ s}\)):
\(s_2 = v_1 \Delta t + \frac{1}{2} a_2 (\Delta t)^2 = (8.0)(4.0) + \frac{1}{2}(-1.0)(4.0)^2 = 32.0 - 8.0 = 24.0\text{ m}\)

Total displacement at \(t = 8.0\text{ s}\):
\(s = s_1 + s_2 = 16.0 + 24.0 = 40.0\text{ m}\).

1 mark for correct option C. Award 0 marks for incorrect options.

In both cases, the total mass is \(m = 2.0 + 3.0 = 5.0\text{ kg}\), so the common acceleration is:
\(a = \frac{F}{m} = \frac{15}{5.0} = 3.0\text{ m s}^{-2}\).

In Case 1, the contact force \(N_1\) acts on \(Y\) to accelerate it:
\(N_1 = m_Y a = 3.0 \times 3.0 = 9.0\text{ N}\).

In Case 2, the contact force \(N_2\) acts on \(X\) to accelerate it:
\(N_2 = m_X a = 2.0 \times 3.0 = 6.0\text{ N}\).

Therefore, \(N_1 = 9.0\text{ N}\) and \(N_2 = 6.0\text{ N}\).

1 mark for correct option B. Award 0 marks for incorrect options.

Along the line joining two coherent in-phase sources, a stationary wave pattern is established. The separation between adjacent destructive interference points (nodes/minima) is \(\frac{\lambda}{2}\).

\(\frac{\lambda}{2} = 0.20\text{ m} \implies \lambda = 0.40\text{ m}\).

Using the wave equation \(v = f \lambda\):
\(f = \frac{v}{\lambda} = \frac{340}{0.40} = 850\text{ Hz}\).

1 mark for correct option B. Award 0 marks for incorrect options.

Let \(D = 60\text{ cm}\) be the distance between the object and the screen, and \(d = 20\text{ cm}\) be the distance between the two lens positions.

For the first position:
\(u = \frac{D - d}{2} = \frac{60 - 20}{2} = 20\text{ cm}\)
\(v = D - u = 60 - 20 = 40\text{ cm}\)

Using the lens formula \(\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\):
\(\frac{1}{f} = \frac{1}{20} + \frac{1}{40} = \frac{3}{40}\text{ cm}^{-1}\)
\(f = \frac{40}{3}\text{ cm} \approx 13.3\text{ cm}\).

1 mark for correct option B. Award 0 marks for incorrect options.

The terminal voltage is given by \(V = \mathcal{E} \frac{R}{R + r}\). As \(R\) increases, \(\frac{R}{R + r}\) increases monotonically towards 1, so \(V\) increases continuously.

The power delivered to the load is \(P = I^2 R = \frac{\mathcal{E}^2 R}{(R + r)^2}\). According to the maximum power transfer theorem, \(P\) is zero at \(R=0\), reaches a maximum when \(R = r\), and approaches zero as \(R \to \infty\). Thus, \(P\) first increases to a maximum and then decreases.

1 mark for correct option B. Award 0 marks for incorrect options.

According to Faraday's law of electromagnetic induction, the magnitude of the induced e.m.f. \(\mathcal{E}\) is:
\(\mathcal{E} = N \frac{\Delta \Phi}{\Delta t} = N A \left|\frac{\Delta B}{\Delta t}\right|\)
\(\mathcal{E} = 50 \times 0.040 \times 0.20 = 0.40\text{ V}\)

The rate of heat dissipation (power) in the coil is:
\(P = \frac{\mathcal{E}^2}{R} = \frac{(0.40)^2}{2.0} = \frac{0.16}{2.0} = 0.080\text{ W}\).

1 mark for correct option B. Award 0 marks for incorrect options.

The activity decreases from \(1600\text{ Bq}\) to \(200\text{ Bq}\), which is a factor of:
\(\frac{200}{1600} = \frac{1}{8} = \left(\frac{1}{2}\right)^3\).
This corresponds to 3 half-lives.
Therefore, \(3 \times t_{1/2} = 18\text{ days} \implies t_{1/2} = 6\text{ days}\).

A further \(12\text{ days}\) corresponds to \(\frac{12}{6} = 2\) more half-lives.
The activity will become:
\(A = 200 \times \left(\frac{1}{2}\right)^2 = 200 \times \frac{1}{4} = 50\text{ Bq}\).

1 mark for correct option B. Award 0 marks for incorrect options.

The energy supplied by the heater during the melting process is given by:
\[ E = P \Delta t = 120\text{ W} \times 180\text{ s} = 21\,600\text{ J} \]
Using the relation \(E = m l_f\):
\[ l_f = \frac{E}{m} = \frac{21\,600\text{ J}}{0.40\text{ kg}} = 5.4 \times 10^4\text{ J kg}^{-1} \]

B (1 mark)
Award 1 mark for the correct option B. No partial marks.

From the ideal gas equation \(PV = nRT\), we have:
\[ P = \frac{nRT}{V} \]
Let the initial pressure, volume, and temperature be \(P_1\), \(V_1\), and \(T_1\) respectively.
The new temperature is \(T_2 = 1.20 T_1\) and the new volume is \(V_2 = 0.90 V_1\).
The new pressure \(P_2\) is:
\[ P_2 = \frac{n R (1.20 T_1)}{0.90 V_1} = \frac{1.20}{0.90} P_1 = \frac{4}{3} P_1 \approx 1.333 P_1 \]
Percentage change in pressure:
\[ \frac{P_2 - P_1}{P_1} \times 100\% = \left(\frac{4}{3} - 1\right) \times 100\% = +33.3\% \]
Thus, the pressure increases by \(33.3\%\).

C (1 mark)
Award 1 mark for the correct option C. No partial marks.

Resolving forces vertically for the stationary block:
\[ N + F\sin\theta = W \implies N = W - F\sin\theta \]
As \(\theta\) increases from \(0^\circ\) to \(60^\circ\), \(\sin\theta\) increases, so \(N\) decreases.

Resolving forces horizontally for equilibrium:
\[ f = F\cos\theta \]
As \(\theta\) increases from \(0^\circ\) to \(60^\circ\), \(\cos\theta\) decreases, so \(f\) decreases.

Therefore, both the normal reaction force and the friction force decrease.

D (1 mark)
Award 1 mark for the correct option D. No partial marks.

For circular orbital motion around a central mass \(M\):
\[ \frac{G M m}{r^2} = \frac{m v^2}{r} \implies v = \sqrt{\frac{GM}{r}} \]
When the radius becomes \(4R\):
\[ v' = \sqrt{\frac{GM}{4R}} = \frac{1}{2} \sqrt{\frac{GM}{R}} = 0.5 v \]
By Kepler's Third Law, \(T^2 \propto r^3\):
\[ T' = T \left(\frac{4R}{R}\right)^{3/2} = T (4)^{3/2} = 8T \]
Thus, the speed is \(0.5 v\) and the period is \(8T\).

A (1 mark)
Award 1 mark for the correct option A. No partial marks.

The distance between two adjacent nodes in a stationary wave is equal to half a wavelength:
\[ \frac{\lambda}{2} = 30\text{ cm} = 0.30\text{ m} \implies \lambda = 0.60\text{ m} \]
The wave speed \(v\) is given by the wave equation:
\[ v = f \lambda = 50\text{ Hz} \times 0.60\text{ m} = 30\text{ m s}^{-1} \]

C (1 mark)
Award 1 mark for the correct option C. No partial marks.

Let the angle of incidence be \(i = 60^\circ\), the angle of reflection be \(r' = i = 60^\circ\), and the angle of refraction be \(r\).
Since the reflected ray and refracted ray are mutually perpendicular:
\[ r' + 90^\circ + r = 180^\circ \implies r = 90^\circ - 60^\circ = 30^\circ \]
Applying Snell's law:
\[ n = \frac{\sin i}{\sin r} = \frac{\sin 60^\circ}{\sin 30^\circ} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3} \approx 1.73 \]

B (1 mark)
Award 1 mark for the correct option B. No partial marks.

Analyzing the circuit:
1. The current is \(I = \frac{\mathcal{E}}{R + r}\). As \(R\) increases, \(R + r\) increases, so \(I\) decreases. (Statement (1) is correct.)
2. The terminal potential difference across the battery is \(V = \mathcal{E} - Ir\). Since \(I\) decreases, the internal drop \(Ir\) decreases, so \(V\) increases. (Statement (2) is correct.)
3. The power dissipated inside the battery is \(P_{\text{int}} = I^2 r\). Since \(I\) decreases and \(r\) is constant, \(P_{\text{int}}\) decreases. (Statement (3) is incorrect.)

Therefore, only statements (1) and (2) are correct.

A (1 mark)
Award 1 mark for the correct option A. No partial marks.

First, calculate the corrected count rates by subtracting the background count rate:
- Initial corrected count rate \(A_0 = 420 - 20 = 400\text{ cpm}\).
- Corrected count rate at \(t = 6.0\text{ hours}\): \(A(t) = 120 - 20 = 100\text{ cpm}\).

The ratio of corrected count rate is:
\[ \frac{A(t)}{A_0} = \frac{100}{400} = \frac{1}{4} = \left(\frac{1}{2}\right)^2 \]
This means that exactly \(2\) half-lives have elapsed in \(6.0\text{ hours}\).
\[ 2 \times T_{1/2} = 6.0\text{ hours} \implies T_{1/2} = 3.0\text{ hours} \]

B (1 mark)
Award 1 mark for the correct option B. No partial marks.

Calculate the energy released when water cools from \(30\ ^\circ\text{C}\) to \(0\ ^\circ\text{C}\):
\(E_1 = m_w c_w \Delta T = (0.50)(4200)(30 - 0) = 63000\text{ J}\).

Calculate the energy required to warm the ice from \(-10\ ^\circ\text{C}\) to \(0\ ^\circ\text{C}\):
\(E_2 = m_{\text{ice}} c_{\text{ice}} \Delta T = (0.20)(2100)(0 - (-10)) = 4200\text{ J}\).

The energy available to melt the ice is:
\(E_{\text{available}} = 63000 - 4200 = 58800\text{ J}\).

The energy required to melt all \(0.20\text{ kg}\) of ice is:
\(E_{\text{melt}} = m_{\text{ice}} l_f = (0.20)(3.34 \times 10^5) = 66800\text{ J}\).

Since \(E_{\text{available}} < E_{\text{melt}}\), not all the ice melts. The ice-water mixture remains in thermal equilibrium at \(0\ ^\circ\text{C}\).

Correct Answer: B (1 mark)

For the initial setup:
\(mg = (M + m)a \implies a = \frac{m}{M + m}g\)
\(T = Ma = \frac{Mm}{M + m}g\).

When the hanging mass is \(2m\):
\(a' = \frac{2m}{M + 2m}g = \frac{2(M+m)}{M+2m} a = \left(1 + \frac{M}{M+2m}\right) a\).
Since \(M > 0\), \(1 < \frac{2(M+m)}{M+2m} < 2\), so \(a < a' < 2a\).

Similarly, the tension is \(T' = M a'\), which means \(T < T' < 2T\).

Therefore, the acceleration increases but is less than \(2a\), and the tension increases but is less than \(2T\).

Correct Answer: B (1 mark)

The orbital radius from the center of the planet is \(r = R + 2R = 3R\).

At the surface of the planet, \(g = \frac{GM}{R^2} \implies GM = gR^2\).

For circular orbit, the centripetal force is provided by gravitation:
\(\frac{m v^2}{r} = \frac{GMm}{r^2} \implies v = \sqrt{\frac{GM}{r}} = \sqrt{\frac{gR^2}{3R}} = \sqrt{\frac{gR}{3}}\).

Correct Answer: B (1 mark)

Using the lens formula \(\frac{1}{u} + \frac{1}{v} = \frac{1}{f}\):
\(\frac{1}{15} + \frac{1}{v} = \frac{1}{10} \implies \frac{1}{v} = \frac{1}{10} - \frac{1}{15} = \frac{1}{30} \implies v = +30\text{ cm}\).

The refracted rays converge towards a point \(30\text{ cm}\) behind the lens.
Since the plane mirror is located \(20\text{ cm}\) behind the lens, it intercepts the rays at a distance of \(30 - 20 = 10\text{ cm}\) before they converge.
Thus, the virtual object for the plane mirror is \(10\text{ cm}\) behind the mirror.
The plane mirror reflects the rays to form a real image at \(10\text{ cm}\) in front of the mirror.
The position of the final image is \(20\text{ cm} - 10\text{ cm} = 10\text{ cm}\) behind the lens.

Correct Answer: A (1 mark)

For the third harmonic of a string fixed at both ends, there are 3 antinodes, so:
\(L = 3\left(\frac{\lambda}{2}\right) \implies 1.2 = \frac{3\lambda}{2} \implies \lambda = 0.80\text{ m}\).

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

Correct Answer: C (1 mark)

(1) \(I = \frac{\mathcal{E}}{R + r}\). Since \(r\) is fixed and \(R \ge 0\), \(I\) is maximum when \(R = 0\). Statement (1) is correct.
(2) \(V_R = I R = \mathcal{E}\frac{R}{R + r} = \frac{\mathcal{E}}{1 + r/R}\). As \(R\) increases, \(r/R\) decreases, so \(V_R\) increases. Statement (2) is correct.
(3) \(P_R = I^2 R = \frac{\mathcal{E}^2 R}{(R + r)^2}\). According to the maximum power transfer theorem, \(P_R\) is maximum when \(R = r\). Statement (3) is correct.

Thus, (1), (2) and (3) are correct.

Correct Answer: D (1 mark)

(1) As the loop enters the field, inward magnetic flux increases. By Lenz's law, the induced current produces an opposing outward magnetic field, which corresponds to an anticlockwise current. Statement (1) is correct.
(2) Induced e.m.f. is \(\mathcal{E} = Bwv\), giving an induced current \(I = \frac{\mathcal{E}}{R} = \frac{Bwv}{R}\). The magnetic force opposing the entry is \(F = B I w = \frac{B^2 w^2 v}{R}\). To maintain constant velocity, the external pulling force must balance this magnetic force, so \(F_{\text{ext}} = \frac{B^2 w^2 v}{R}\). Statement (2) is correct.
(3) \(\frac{\Delta \Phi}{\Delta t} = B \frac{\Delta A}{\Delta t} = B w v\), which is constant. Statement (3) is correct.

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

Correct Answer: D (1 mark)

Let \(N_0\) be the initial number of nuclei of \(X\).
At time \(t\), \(N_X + N_Y = N_0\).
Given \(\frac{N_Y}{N_X} = 7 \implies N_Y = 7 N_X\).

Therefore,
\(N_X + 7 N_X = 8 N_X = N_0 \implies N_X = \frac{1}{8} N_0 = \left(\frac{1}{2}\right)^3 N_0\).

This shows that \(3\) half-lives have elapsed.
\(t = 3 \times 6.0\text{ hours} = 18.0\text{ hours}\).

Correct Answer: B (1 mark)

The rate of heat supplied is constant: \(\frac{\Delta Q}{\Delta t} = P\). The mass of liquid vaporized in time \(\Delta t\) is \(\Delta m_{\text{vap}} = \frac{P \Delta t}{l_v}\). Therefore, the remaining mass is \(m = m_0 - \left(\frac{P}{l_v}\right)t\), which is a straight line with a constant negative slope.

Award 1 mark for the correct option B. Award 0 marks for incorrect options.

The total mass of the system is \(M + m\), so the common acceleration is \(a = \frac{F}{M + m}\). The only horizontal force acting on the block \(m\) is the static friction \(f\), which provides its acceleration: \(f = m a = \frac{m}{M + m}F\).

Award 1 mark for the correct option A. Award 0 marks for incorrect options.

For a satellite in circular orbit of radius \(R\) around a mass \(M_p\), the gravitational force provides the centripetal force: \(\frac{G M_p m}{R^2} = \frac{m v^2}{R} \implies v = \sqrt{\frac{G M_p}{R}}\). Thus, \(\frac{v_A}{v_B} = \sqrt{\frac{r_B}{r_A}} = \sqrt{\frac{4r}{r}} = 2\).

Award 1 mark for the correct option C. Award 0 marks for incorrect options.

Linear magnification \(m = \frac{v}{u} = 2.0 \implies v = 2u\). Using the lens formula \(\frac{1}{u} + \frac{1}{v} = \frac{1}{f}\): \(\frac{1}{u} + \frac{1}{2u} = \frac{1}{15} \implies \frac{3}{2u} = \frac{1}{15} \implies u = 22.5\text{ cm}\). Then \(v = 2(22.5) = 45\text{ cm}\). The distance between the object and the screen is \(D = u + v = 22.5 + 45 = 67.5\text{ cm}\).

Award 1 mark for the correct option D. Award 0 marks for incorrect options.

At maximum displacement (crest), the particle is momentarily at rest, so its velocity is zero. The restoring force is at its maximum directed downwards toward the equilibrium position, so the acceleration of \(P\) is at its maximum and directed downwards.

Award 1 mark for the correct option B. Award 0 marks for incorrect options.

Initially, total external resistance \(R_1 = 4\,\Omega\), total resistance \(R_{\text{total}} = 4 + 2 = 6\,\Omega\). Current \(I_1 = \frac{12}{6} = 2\text{ A}\). Terminal voltage \(V_1 = 12 - 2(2) = 8\text{ V}\). Total power \(P_1 = E I_1 = 12 \times 2 = 24\text{ W}\). When another \(4\,\Omega\) resistor is connected in parallel, equivalent external resistance becomes \(R_2 = 2\,\Omega\). Total resistance \(R_{\text{total}}' = 2 + 2 = 4\,\Omega\). New current \(I_2 = \frac{12}{4} = 3\text{ A}\). New terminal voltage \(V_2 = 12 - 3(2) = 6\text{ V}\) (decreases). New total power \(P_2 = E I_2 = 12 \times 3 = 36\text{ W}\) (increases).

Award 1 mark for the correct option A. Award 0 marks for incorrect options.

The change in magnetic flux density is \(\Delta B = 0.80 - 0.20 = 0.60\text{ T}\). The induced e.m.f. is \(\varepsilon = \frac{\Delta \Phi}{\Delta t} = \frac{A \Delta B}{\Delta t} = \frac{0.05 \times 0.60}{0.30} = 0.10\text{ V}\). The induced current is \(I = \frac{\varepsilon}{R} = \frac{0.10\text{ V}}{2.0\,\Omega} = 0.050\text{ A} = 50\text{ mA}\).

Award 1 mark for the correct option C. Award 0 marks for incorrect options.

Let \(N_0\) be the initial number of nuclei for both. At time \(t\), \(N_X(t) = N_0 \left(\frac{1}{2}\right)^{t/4}\) and \(N_Y(t) = N_0 \left(\frac{1}{2}\right)^{t/12}\). We are given \(\frac{N_Y(t)}{N_X(t)} = 4 = 2^2\). Thus, \(\frac{2^{-t/12}}{2^{-t/4}} = 2^{t/4 - t/12} = 2^{t/6} = 2^2\). Therefore, \(\frac{t}{6} = 2 \implies t = 12\text{ days}\).

Award 1 mark for the correct option B. Award 0 marks for incorrect options.

Let \(I = \frac{\mathcal{E}}{R+r}\) be the current in the circuit.

- Statement (1): The terminal voltage is given by \(V = \mathcal{E} - Ir\). As \(R\) increases, the total resistance \(R+r\) increases, so the current \(I\) decreases. Consequently, the potential drop across the internal resistance \(Ir\) decreases, and the terminal voltage \(V\) increases. Thus, (1) is correct.

- Statement (2): The power dissipated in the external resistor is \(P = I^2 R = \frac{\mathcal{E}^2 R}{(R+r)^2}\). According to the maximum power transfer theorem, \(P\) attains its maximum value when \(R = r\). As \(R\) increases beyond \(r\), the power \(P\) decreases. Thus, (2) is correct.

- Statement (3): The total power supplied by the cell is \(P_{\text{total}} = \mathcal{E} I\). Since the e.m.f. \(\mathcal{E}\) is constant and the current \(I\) decreases as \(R\) increases, \(P_{\text{total}}\) decreases. Thus, (3) is correct.

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

D (1 mark)
- Award 1 mark for the correct option D.
- Deduct 0 marks for incorrect attempts.