2024 HKDSE Physics — Paper 1Section A (33 multiple-choice questions) + Section B (13 structured questions): complete answer key, option-by-option analysis, fully worked solutions, and every page of the original papers and marking scheme.

Question. Substances P and Q have the same mass and are both solids at temperature T₁. The graph shows how the temperature of each substance varies with the energy transferred to it. Which statement is correct?

• AP has a higher melting point than Q.
• BThe specific latent heat of fusion of P is smaller than that of Q.
• CThe specific heat capacity of solid P is greater than that of solid Q.
• DAt temperature T₂, P is a gas while Q is a solid.

Read the graph in three stages: the slope while the substance is solid, the flat plateau where it melts, and where T₂ sits relative to the two plateaux.

1. Let F_P and F_Q be the melting points (the temperatures of the flat sections). The graph gives F_P < F_Q.
2. While still solid, E = mcΔT, so the slope of the T–E line is 1/(mc). The slope for P is shallower, therefore c_P > c_Q.
3. The length of the plateau is the energy needed to melt the whole sample, E = lΔm. P's plateau is longer, so l_P > l_Q.
4. T₂ lies above F_P but below the boiling point of P, and below F_Q: at T₂ P is liquid and Q is still solid.

Question. 0.5 kg of water at 50 °C is mixed with 2.0 kg of ice at 0 °C. What is the final temperature of the mixture? (c_water = 4200 J kg⁻¹ °C⁻¹, l_f of ice = 3.34 × 10⁵ J kg⁻¹)

• A−54 °C
• B0 °C
• C10 °C
• D25 °C

Never average the two temperatures. Do the energy budget: how much can the warm water give up before it reaches 0 °C, and how much does all the ice need to melt?

1. Energy released cooling the water from 50 °C to 0 °C: E₁ = mcΔT = (0.5)(4200)(50) = 1.05 × 10⁵ J.
2. Energy needed to melt all the ice: E₂ = lΔm = (2.0)(3.34 × 10⁵) = 6.68 × 10⁵ J.
3. E₁ < E₂, so the ice cannot all melt. The water is cooled to 0 °C and stays there while some ice melts.
4. Final state: ice and water together at 0 °C.

Question. The best vacuum attained in a laboratory has a pressure of about 10⁻⁸ Pa. Estimate the order of magnitude of the number of air molecules in 1 cm³ of such a “vacuum” at room temperature.

• A10⁴
• B10⁶
• C10⁸
• D10¹²

Treat the gas as ideal: n = PV/RT, then convert moles to molecules with N = nN_A.

1. Volume V = 1 cm³ = 1 × 10⁻⁶ m³, T ≈ 298 K.
2. n = PV/RT = (10⁻⁸)(1 × 10⁻⁶)/[(8.31)(298)] ≈ 4.0 × 10⁻¹⁸ mol.
3. N = nN_A = (4.0 × 10⁻¹⁸)(6.02 × 10²³) ≈ 2.4 × 10⁶.
4. The required order of magnitude is 10⁶.

Question. The shaded area under a velocity–time graph is to be identified. What physical quantity does it represent?

• Aacceleration
• Bchange of velocity
• Cmomentum
• Ddisplacement

An area under a graph is the integral of the quantity on the vertical axis with respect to the horizontal axis: ∫v dt.

1. ∫v dt = displacement (the definition of velocity).
2. On a v–t graph the gradient gives acceleration, and the area gives displacement.
3. For comparison: area under an a–t graph = change of velocity; area under an F–t graph = impulse = change of momentum.

Question. A block of mass m slides down a fixed rough incline at constant velocity. The incline makes an angle θ with the horizontal. What is the magnitude of the force exerted on the incline by the block?

• Azero
• Bmg
• Cmg sinθ
• Dmg cosθ

Constant velocity means zero acceleration, so the forces on the block must cancel. Then use Newton's third law to transfer the answer to the incline.

1. On the block only two forces act: its weight W (magnitude mg) and the reaction R₁ from the incline (the resultant of the normal force and friction).
2. Constant velocity ⇒ no net force ⇒ R₁ and W cancel exactly, so |R₁| = mg.
3. The force the block exerts on the incline, R₂, is the reaction to R₁, so |R₂| = |R₁| = mg.

Question. An object of weight w is inside an elevator of weight W. A force T pulls the elevator upward. A supporting force X acts on the object, while the floor of the elevator experiences a force Y due to the object. Which of the following is an action–reaction pair?

• AT and (W + w)
• Bw and X
• CX and Y
• DIt cannot be determined, as the acceleration of the elevator is unknown.

An action–reaction pair must satisfy three conditions: equal in magnitude, opposite in direction, and acting on two different bodies, of the same kind of interaction.

1. X acts on the object, exerted by the floor of the elevator.
2. Y acts on the floor of the elevator, exerted by the object.
3. They act on different bodies, are the same kind of contact force, equal and opposite: a genuine third-law pair. The acceleration of the elevator is irrelevant.

Question. A horizontal force F acts on a block on a rough horizontal surface. Figure (b) shows how the block's kinetic energy Eₖ varies with the distance s travelled. The experiment is repeated with the force doubled. Which graph shows the expected result (dotted line)?

• ASame slope, same starting point
• BSteeper slope, not through the origin
• CSteeper than double the original slope, through the origin
• DSame slope but shifted upward

Write Eₖ as a function of s. The block starts from rest and friction f is unchanged, so the net force is F − f.

1. Work done by the net force: Eₖ = (F − f)s — a straight line through the origin with slope F − f.
2. Replace F by 2F: new slope = 2F − f.
3. Since f > 0, 2F − f > 2(F − f): the slope is more than twice the original.
4. The line still passes through the origin because the block still starts from rest.

Question. Blocks X (3.0 kg) and Y (1.0 kg) are connected by a light inextensible string over a smooth light pulley fixed to the ceiling. X is 1.5 m above the ground. They are released from rest. Find the time taken for X to reach the ground. Neglect air resistance.

• A0.55 s
• B0.64 s
• C0.68 s
• D0.78 s

Treat the two blocks as one system to get the acceleration, then use s = ½t² from rest.

1. Take the direction of motion as positive: for X, (3.0)g − T = (3.0)a; for Y, (1.0)g − T = (1.0)(−a).
2. Subtracting: (3.0 − 1.0)g = (3.0 + 1.0)a ⇒ 2g = 4a ⇒ a = g/2 = 4.905 m s⁻².
3. Then 1.5 = (0)t + ½(4.905)t² ⇒ t² = 0.6116 ⇒ t = 0.78 s.

Question. A person thinks that if he pulls his hair with a large enough force he will eventually be able to pull himself off the ground. This is a misconception because

• Athe pulling force acts internally within that person but it is not an external force acting on the person.
• Bit is impossible to generate a pulling force larger than a person's weight.
• Cthe pulling force is balanced by the person's weight.
• Dit is impossible for the pulling force to overcome the force due to atmospheric pressure.

Take the person as a single system and ask whether any external force changes.

1. Hand pulls hair; by the third law the hair pulls the hand back with an equal and opposite force.
2. Both forces act inside the person, so their sum is zero on the person as a whole.
3. Only external forces (weight, the normal force from the floor) can change the motion of the person, so no amount of hair-pulling lifts him.

Question. A child pushes a 150 kg wrestler to topple him backwards. The diagram shows the wrestler's weight W and the horizontal pushing force F. P is the wrestler's contact point on the ground. Find the minimum value of F required.

• A368 N
• B736 N
• C1177 N
• D2354 N

The wrestler topples about the contact point P, so take moments about P at the point of toppling.

1. Weight: W = mg = (150)(9.81) = 1471.5 N, acting through the centre of mass, 0.25 m from the pivot (half of the 0.50 m base).
2. Taking moments about P: F(1.00) = W(0.25).
3. F = (1471.5)(0.25)/1.00 = 368 N.

Question. A billiard ball of mass 0.16 kg moving at 1.20 m s⁻¹ collides elastically with the cushion and rebounds with the same speed. If the impact time is 0.0003 s, find the average force on the ball by the cushion.

• A0 N
• B640 N
• C1280 N
• D2560 N

Write impulse as a vector: the rebound reverses the velocity, so the change of velocity is twice the speed.

1. Take the direction away from the cushion as positive: u = −1.20 m s⁻¹, v = +1.20 m s⁻¹.
2. Impulse: F̄Δt = m(v − u) = (0.16)(1.20 + 1.20) = 0.384 N s.
3. F̄ = 0.384 / 0.0003 = 1280 N.

Question. The figure shows a particle performing uniform circular motion. Which option correctly indicates the directions of its velocity and acceleration at the instant shown?

• Av tangential, a pointing to the centre
• Bv radial, a tangential
• Cv tangential, a pointing away from the centre
• Dv and a parallel, both tangential

For uniform circular motion the speed is constant but the direction of motion changes, which requires a centripetal acceleration.

1. Velocity is always a tangent to the circular path.
2. The acceleration points towards the centre of the circle, with magnitude a = v²/r = ω²r.
3. The two are perpendicular at every instant.

Question. A satellite orbits the Earth in a circular orbit of radius r with a period of 24 hours. What is the period of another satellite orbiting the Earth in a circular orbit of radius 2r?

• Aless than 24 hours
• B24 hours
• Cbetween 24 and 48 hours
• Dmore than 48 hours

Only one fact is needed: for circular orbits T² ∝ r³.

1. Gravitational force provides the centripetal force: GMm/r² = mω²r ⇒ ω²r³ = GM.
2. Substituting ω = 2π/T gives T² = 4π²r³/(GM), i.e. T² ∝ r³.
3. Radius doubles: T grows by a factor 2^(3/2) = 2√2 ≈ 2.83.
4. T₂ = 24 × 2.83 = 67.9 hours, which is more than 48 hours.

Question. A light beam consisting of monochromatic beams X and Y is incident on a rectangular glass block and splits into two beams. Which of the following comparisons is/are correct? (1) X has a higher frequency. (2) X has a larger speed in glass. (3) In the glass block, the critical angle for X is greater.

• A(1) only
• B(3) only
• C(1) and (2) only
• D(2) and (3) only

From the diagram, decide which beam bends more, and translate bending into refractive index.

1. X bends towards the normal more than Y, so the glass has a larger refractive index for X.
2. A larger n means a smaller speed in glass, v = c/n — so statement (2) is wrong and X is the slower beam.
3. A larger n corresponds to a higher frequency (normal dispersion) — statement (1) is correct.
4. Critical angle: sin C = 1/n, so a larger n gives a smaller critical angle — statement (3) is wrong.

Question. In a ripple tank, plane water waves approach an L-shaped barrier as shown. Which diagram best shows the reflected wave pattern?

• Awaves reflected parallel to the walls
• Bwaves crossing the original direction only
• Cwaves reflected with the wrong angles
• Dmirror-image reflection: 45° in, 45° out on both walls

Each straight wall behaves like a plane mirror: the angle of incidence equals the angle of reflection, measured from the normal.

1. The waves reach the upper wall at 45° to its normal (the geometry of an L-shaped barrier at 45°), so they are reflected at 45°.
2. The reflected waves from the upper wall travel upward; the reflected waves from the lower wall travel downward.
3. The correct pattern is the mirror image of the incident pattern about each wall.

Question. Particles are uniformly distributed along a slinky spring. Figures (a) and (b) show their positions at two instants as a travelling wave of frequency 10 Hz propagates from left to right. Which statement is correct?

• AParticle a is always stationary.
• BParticles b and c vibrate with different amplitudes.
• CThe wavelength of the wave is 16 cm.
• DThe smallest time interval for obtaining these two figures is 0.05 s.

Use the standard travelling-wave facts: every particle oscillates with the same amplitude, and the wavelength is the distance between consecutive equal points.

1. In figure (a), a and i are consecutive centres of compression separated by 32 cm, so λ = 32 cm (statement C is wrong).
2. In a travelling wave no particle stays at rest and all particles share the same amplitude (statements A and B are wrong).
3. Between the two figures, particle e moves from a centre of rarefaction to a centre of compression, i.e. the wave has advanced by half a wavelength: t = (2n − 1)T/2.
4. With f = 10 Hz, T = 0.1 s, so t = 0.05(2n − 1) s and the smallest value is 0.05 s.

Question. Two identical loudspeakers S₁ and S₂ driven by the same signal generator emit sound of wavelength 0.10 m, and they are in anti-phase. OX is the perpendicular bisector of S₁S₂. What type of interference occurs at X and at P respectively, where the distances from the speakers are 3.0 m and 2.5 m?

• Adestructive at X, constructive at P
• Bconstructive at X, constructive at P
• Cdestructive at X, destructive at P
• Dconstructive at X, destructive at P

Because the sources are in anti-phase, the usual rule is reversed: a path difference of a whole number of wavelengths now gives destructive interference.

1. Sources in anti-phase: destructive when the path difference = nλ; constructive when the path difference = (n + ½)λ.
2. At X the distances are equal, so the path difference is 0 = 0λ → destructive.
3. At P the path difference is 3.0 − 2.5 = 0.5 m, and 0.5 m = 5(0.10 m) = 5λ → also destructive.
4. Therefore both X and P show destructive interference.

Question. A string is tied to a vibrator at one end and fixed to a wall at the other. The figure shows the stationary wave formed at a certain frequency. Which statement is correct when the frequency is doubled?

• AThe amplitude of the wave will double.
• BThe wavelength will double.
• CThe speed of the wave will double.
• DParticle X and Y will become in phase.

For a stationary wave on a fixed string: the speed is set by the tension and mass per unit length, and v = fλ then fixes the wavelength.

1. The wave speed depends on the medium (tension, μ), not on the vibrator, so it is unchanged — statement C is wrong, and the amplitude is set by the driver, not by f (A is wrong).
2. With v fixed, doubling f halves the wavelength — statement B is wrong.
3. X and Y are half a wavelength apart at first, so they oscillate in anti-phase.
4. After doubling f the wavelength halves, so the separation between X and Y equals one whole wavelength: two nodes lie between them, and an even number of nodes means the two points are in phase.

Question. X is a light ray from an object O placed on the principal axis of a convex lens. F and F' are the foci of the lens. Which light ray best represents the emergent ray?

• Aray P
• Bray Q
• Cray R
• Dray S

Locate the image first, then ask which of the four candidate rays could have come from it.

1. O lies between F and the optical centre, so the image is virtual, upright and magnified, on the same side of the lens as O and further from the lens than O.
2. All rays from O that pass through the lens appear to diverge from the image point I.
3. Therefore the emergent ray, extended backwards, passes through I: it crosses the principal axis on the same side as O, beyond O.

Question. An object of height 10 cm placed 20 cm in front of a concave lens forms an image of height 8 cm. Find the focal length of the concave lens.

• A4.4 cm
• B8.9 cm
• C40 cm
• D80 cm

Two steps: magnification gives the image distance, then the lens formula gives the focal length — with the sign of a virtual image.

1. Magnification m = 8/10 = 0.8 = v/u, so v = 0.8 × 20 = 16 cm.
2. A concave lens always forms a virtual image on the same side as the object, so v = −16 cm.
3. With the convention 1/f = 1/u + 1/v: 1/f = 1/20 + 1/(−16) = (4 − 5)/80 = −1/80.
4. Hence f = −80 cm: the focal length is 80 cm.

Question. Which description below is INCORRECT?

• AMicrowaves are a kind of electromagnetic wave.
• BMicrowaves can be observed with the naked eye.
• CMicrowaves propagate with the speed of light in a vacuum.
• DMicrowaves are used in radar to detect the position of an aeroplane.

This is a factual recall question, but the word INCORRECT must be read carefully.

1. Microwaves are EM waves with wavelengths of a few millimetres to a few centimetres, longer than infrared.
2. All EM waves travel at c = 3.0 × 10⁸ m s⁻¹ in a vacuum.
3. Radar works with microwaves because they pass through cloud and reflect from metal.
4. They are not visible light, so they cannot be seen by eye.

Question. Figures (a) and (b) are CRO displays of musical notes X and Y using the same time axis and intensity axis. Which comparison must be correct?

• Asame pitch, same loudness
• Bsame pitch, not the same loudness
• Chigher pitch for Y, same loudness
• Dhigher pitch for Y, not the same loudness

On a CRO trace: period → pitch, amplitude → loudness.

1. Both traces repeat in the same time, so the periods are equal and the pitches are the same.
2. The amplitudes differ, so the loudnesses differ.
3. Only option B matches both observations.

Question. Charges +Q and −2Q are fixed at X and Y. Which description(s) about the electric field along the dotted line is/are correct? (1) Between X and Y, the fields due to the two charges are in the same direction. (2) To the right of Y, the resultant field is always directed to the left. (3) On the dotted line, there exist two points where the resultant field is zero.

• A(2) only
• B(3) only
• C(1) and (2) only
• D(1) and (3) only

Add the fields as vectors, remembering that the field of a positive charge points away from it and that of a negative charge points towards it, with magnitude ∝ 1/r².

1. Between X and Y: at any point the field due to +Q points right (away from X) and that due to −2Q points right (towards Y). So they are in the same direction — (1) is correct.
2. To the right of Y: the field of +Q points right, that of −2Q points left; since the point is closer to Y, the 2Q/r² term wins, so the resultant points left — (2) is correct.
3. To the left of X: both fields point left and right respectively in opposition; the balance point satisfies Q/r₁² = 2Q/r₂², i.e. r₂ = √2 r₁ — exactly one such point.
4. So there is only one zero-field point, not two — (3) is wrong.

Question. Four resistors are connected to terminals P, Q, R and S of a circuit board. Across which pair of terminals is the equivalent resistance smallest?

• APQ
• BPR
• CPS
• DQS

Reduce each pair of terminals to a single resistance by combining the parallel branches that the current would use.

1. PQ: 2 Ω in parallel with (1 + 1 + 2) = 4 Ω → (1/2 + 1/4)⁻¹ = 1.33 Ω.
2. PR: (2 + 2) = 4 Ω in parallel with (1 + 1) = 2 Ω → 1.33 Ω.
3. PS: (2 + 2 + 1) = 5 Ω in parallel with 1 Ω → (1/5 + 1)⁻¹ = 0.83 Ω.
4. QS: 3 Ω in parallel with 3 Ω → 1.5 Ω.
5. The smallest is PS, at about 0.83 Ω.

Question. In the circuit shown, the two resistors are identical. When S is closed, how do the ammeter and voltmeter readings change?

• Aammeter increases, voltmeter increases
• Bammeter increases, voltmeter decreases
• Cammeter decreases, voltmeter increases
• Dammeter decreases, voltmeter decreases

Closing S short-circuits one resistor: the total resistance falls, so the current in the main branch rises. Then follow the current through the remaining resistor.

1. Before: the total resistance is the sum of the two identical resistors.
2. After closing S, the right-hand resistor is bypassed, so the total resistance decreases and the main current increases: the ammeter reading increases.
3. The voltmeter reads the p.d. across the left-hand resistor, which carries the (now larger) main current: V = IR increases, so the voltmeter reading increases.
4. Both readings increase.

Question. A long straight wire PQ carries a current and a second wire MN, perpendicular to PQ and in the same plane, is near it. What is the direction of the magnetic force acting on MN due to PQ? (Neglect the Earth's field.)

• Apointing out of the paper
• Bpointing towards the bottom of the page
• Cpointing into the paper
• Dpointing towards the top of the page

Two steps: find the field that PQ creates at MN, then apply Fleming's left-hand rule to the current in MN.

1. Right-hand grip rule on PQ: the field it produces at the position of MN points into the paper.
2. Fleming's left-hand rule for MN: first finger into the paper (field), second finger along the current in MN, thumb gives the force.
3. With the geometry of the figure the force on MN is directed towards the bottom of the page.

Question. A coil of radius 2 cm is placed inside a square region of uniform magnetic field pointing out of the paper. The field decreases uniformly from 5.0 T to 3.0 T in 0.8 s. What is the magnitude of the e.m.f. induced in the coil?

• A2.5 × 10⁻² V
• B3.1 × 10⁻² V
• C3.1 × 10⁻³ V
• D6.2 × 10⁻³ V

Use ε = A·ΔB/Δt, and be careful which area belongs to the loop.

1. Flux through the coil: Φ = BA, with A = π(0.02)² = 1.257 × 10⁻³ m² (the coil's own area, not the 10 cm square).
2. ΔB/Δt = (5.0 − 3.0)/0.8 = 2.5 T s⁻¹.
3. ε = A·ΔB/Δt = (1.257 × 10⁻³)(2.5) = 3.14 × 10⁻³ V.
4. So ε ≈ 3.1 × 10⁻³ V.

Question. Which of the following current–time (I–t) graphs represents an a.c.? (1) a square pulse always in the same direction (2) a graph whose current reverses (3) another graph whose current reverses

• A(1) and (2) only
• B(1) and (3) only
• C(2) and (3) only
• D(1), (2) and (3)

The definition of a.c. is that the current reverses direction periodically — not that it varies.

1. Graph (1) changes magnitude but never changes sign, so it is a direct current (it is pulsed d.c.).
2. Graphs (2) and (3) both cross the axis, so the current reverses: they are a.c.
3. Answer: (2) and (3) only.

Question. A charged particle enters and leaves a magnetic field of flux density 0.97 T along the path shown. It experiences a magnetic force of 5.9 × 10⁻¹² N inside the field. Given that its speed is 1.9 × 10⁷ m s⁻¹, find its charge.

• A−3.2 × 10⁻¹⁹ C
• B+3.2 × 10⁻¹⁹ C
• C−6.4 × 10⁻¹⁹ C
• D+6.4 × 10⁻¹⁹ C

Magnitude from F = BQv, sign from the direction of the deflection.

1. Q = F/(Bv) = (5.9 × 10⁻¹²)/[(0.97)(1.9 × 10⁷)] = 3.2 × 10⁻¹⁹ C.
2. The particle follows a circular arc: the magnetic force is always perpendicular to v, so it does no work and the speed stays constant.
3. Applying Fleming's left-hand rule to the known field and the observed deflection shows the conventional current is in the direction of motion, so the charge is positive.

Question. A power station generates 217 MW at 132 kV for transmission through a cable whose effective resistance is 0.02 Ω km⁻¹ over a total length of 40 km. The power loss is about

• A2.16 MW
• B6.49 MW
• C13.0 MW
• D21.6 MW

Two steps: the current follows from P = IV, and the loss from P = I²R.

1. I = P/V = (217 × 10⁶)/(132 × 10³) = 1643.94 A.
2. Cable resistance R = 0.02 × 40 = 0.8 Ω.
3. Loss = I²R = (1643.94)²(0.8) = 2.16 × 10⁶ W ≈ 2.16 MW.

Question. Which statements about domestic electricity are correct? (1) All fuses should be installed in the live wire. (2) An electrical appliance can work even without an earth wire. (3) An electrical appliance having a fuse in it can protect people from getting an electric shock.

• A(1) and (2) only
• B(1) and (3) only
• C(2) and (3) only
• D(1), (2) and (3)

Judge each statement separately: the fuse is a current device, the earth wire is a safety device, and an appliance works as long as current flows through its element.

1. (1) A fuse in the live wire disconnects the appliance from the high potential when it blows, so it must be in the live wire — correct.
2. (2) An appliance with double insulation or a plastic case works perfectly well without an earth wire — correct.
3. (3) A fuse protects the circuit against excessive current; it does not stop a person touching a live metal case from being shocked. Protection needs an earth wire (or double insulation) — incorrect.

Question. It takes 2.0 years for the activity of a sample to drop from 800 Bq to 600 Bq. How long does it take for the activity to drop from 600 Bq to 300 Bq?

• A2.0 years
• B2.5 years
• C3.0 years
• D4.8 years

The decay constant is fixed: find it from the first interval, then use it for the second.

1. A = A₀e^(−λt) ⇒ λ = ln(A₀/A)/t.
2. From 800 → 600 Bq in 2.0 y: λ = ln(800/600)/2.0 = 0.1438 y⁻¹.
3. From 600 → 300 Bq (a halving): t = ln(600/300)/0.1438 = ln2/0.1438 = 4.8 y.

Question. To locate cracks in an underground water pipe, a small amount of a radioisotope is added to the water. The substance leaks through the crack into the soil nearby, and radiation is detected above ground. Which radioisotope is the most suitable? (P: 66 h, β · Q: 36 h, γ · R: 7.2 h, α · S: 45 min, αβ)

• AP (66 hours, β)
• BQ (36 hours, γ)
• CR (7.2 hours, α)
• DS (45 minutes, αβ)

Two filters: the radiation has to reach a detector above ground, and the half-life has to be long enough to carry out the test.

1. α particles are stopped by a few centimetres of soil and β by a few millimetres of metal or a metre of air; only γ penetrates soil to a detector above ground.
2. The half-life must be long compared with the days a leak test takes; 45 minutes (S) is far too short.
3. Q, with γ radiation and a half-life of 36 hours, satisfies both requirements.