HKDSE Physics Paper 1A Multiple-Choice Practice Paper (30 Questions, 30 Marks, 45 Minutes)
- 2026-11-02
HKDSE Physics Paper 1A MC
HKDSE Physics Paper 1A multiple-choice practice paper: 30 original questions in 45 minutes on kinematics, forces and moments, momentum, waves, circuits, power.
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Paper details Level: S6 (DSE) | Subject: Physics | Time: 45 minutes | Marks: 30 | Questions: 30
Section A Mechanics and motion (10 marks)
Questions 1-10 (1 mark each)
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | A car starts from rest and accelerates uniformly at 4 m s⁻² for 6 s. Find its final speed. (1 mark) A. 24 m s⁻¹ | B. 20 m s⁻¹ | C. 12 m s⁻¹ | D. 30 m s⁻¹ |
Answer: A. 24 m s⁻¹ Why: v = u + at = 0 + 4 × 6 = 24 m s⁻¹. |
| 2 | A car travelling at 15 m s⁻¹ brakes with a deceleration of 3 m s⁻² for 4 s. Find its speed afterwards. (1 mark) A. 0 m s⁻¹ | B. 3 m s⁻¹ | C. 12 m s⁻¹ | D. 27 m s⁻¹ |
Answer: B. 3 m s⁻¹ Why: v = u + at = 15 − 3 × 4 = 3 m s⁻¹. |
| 3 | An object starts from rest and accelerates uniformly at 2 m s⁻². Find its displacement in 5 s. (1 mark) A. 10 m | B. 50 m | C. 20 m | D. 25 m |
Answer: D. 25 m Why: s = ½at² = ½ × 2 × 5² = 25 m. |
| 4 | A car travelling at 20 m s⁻¹ decelerates uniformly at 5 m s⁻² to rest. Find the braking distance. (1 mark) A. 80 m | B. 4 m | C. 40 m | D. 20 m |
Answer: C. 40 m Why: s = u² ÷ (2a) = 400 ÷ 10 = 40 m. |
| 5 | A 2 kg object experiences a net force of 12 N. Find its acceleration. (1 mark) A. 10 m s⁻² | B. 6 m s⁻² | C. 24 m s⁻² | D. 3 m s⁻² |
Answer: B. 6 m s⁻² Why: a = F ÷ m = 12 ÷ 2 = 6 m s⁻². |
| 6 | A 5 kg instrument rests on the Moon where g = 1.6 N kg⁻¹. Find its weight. (1 mark) A. 50 N | B. 5 N | C. 16 N | D. 8 N |
Answer: D. 8 N Why: W = mg = 5 × 1.6 = 8 N; lunar gravity is weaker, so weight is not equal to mass. |
| 7 | A 0.5 kg ball moves at 8 m s⁻¹. Find its momentum. (1 mark) A. 4 kg m s⁻¹ | B. 8 kg m s⁻¹ | C. 16 kg m s⁻¹ | D. 2 kg m s⁻¹ |
Answer: A. 4 kg m s⁻¹ Why: p = mv = 0.5 × 8 = 4 kg m s⁻¹. |
| 8 | A net force of 20 N acts on a body for 0.3 s. Find the impulse. (1 mark) A. 0.6 N s | B. 1.5 N s | C. 6 N s | D. 60 N s |
Answer: C. 6 N s Why: Impulse = F × t = 20 × 0.3 = 6 N s. |
| 9 | A 15 N force acts 0.4 m from a pivot. Find the minimum force at 0.3 m on the other side to balance it. (1 mark) A. 15 N | B. 11.3 N | C. 6 N | D. 20 N |
Answer: D. 20 N Why: Moments: 15 × 0.4 = F × 0.3, so F = 6 ÷ 0.3 = 20 N. |
| 10 | A force of 600 N acts uniformly on an area of 0.04 m². Find the pressure. (1 mark) A. 24000 Pa | B. 15000 Pa | C. 24 Pa | D. 1500 Pa |
Answer: B. 15000 Pa Why: P = F ÷ A = 600 ÷ 0.04 = 15000 Pa. |
Section B Waves and electricity (10 marks)
Questions 11-20 (1 mark each)
| # | Question | Answer and marking notes |
|---|---|---|
| 11 | A wave of frequency 250 Hz has a wavelength of 1.2 m. Find its speed. (1 mark) A. 300 m s⁻¹ | B. 30 m s⁻¹ | C. 208.3 m s⁻¹ | D. 240 m s⁻¹ |
Answer: A. 300 m s⁻¹ Why: v = fλ = 250 × 1.2 = 300 m s⁻¹. |
| 12 | An alternating current has a frequency of 50 Hz. Find its period. (1 mark) A. 0.2 s | B. 0.5 s | C. 0.02 s | D. 50 s |
Answer: C. 0.02 s Why: T = 1 ÷ f = 1 ÷ 50 = 0.02 s. |
| 13 | Find the speed of light in glass of refractive index 1.5 (c = 3.0 × 10⁸ m s⁻¹). (1 mark) A. 4.5 × 10⁸ m s⁻¹ | B. 1.5 × 10⁸ m s⁻¹ | C. 3.0 × 10⁸ m s⁻¹ | D. 2.0 × 10⁸ m s⁻¹ |
Answer: D. 2.0 × 10⁸ m s⁻¹ Why: v = c ÷ n = 3.0 × 10⁸ ÷ 1.5 = 2.0 × 10⁸ m s⁻¹. |
| 14 | Three resistors of 4 Ω, 6 Ω and 8 Ω are connected in series. Find the total resistance. (1 mark) A. 2.2 Ω | B. 12 Ω | C. 18 Ω | D. 6 Ω |
Answer: C. 18 Ω Why: In series resistances add: 4 + 6 + 8 = 18 Ω. |
| 15 | Two 6 Ω resistors are connected in parallel. Find the equivalent resistance. (1 mark) A. 3 Ω | B. 12 Ω | C. 6 Ω | D. 1.5 Ω |
Answer: A. 3 Ω Why: 1/R = 1/6 + 1/6 = 1/3, so R = 3 Ω. |
| 16 | A current of 0.5 A flows through a 24 Ω resistor. Find the potential difference across it. (1 mark) A. 24 V | B. 12 V | C. 48 V | D. 6 V |
Answer: B. 12 V Why: V = IR = 0.5 × 24 = 12 V. |
| 17 | An appliance runs on 12 V with a current of 2 A. Find its power. (1 mark) A. 6 W | B. 14 W | C. 48 W | D. 24 W |
Answer: D. 24 W Why: P = VI = 12 × 2 = 24 W. |
| 18 | A 1.5 kW kettle is used for 4 hours. How much energy does it consume in kWh? (1 mark) A. 24 kWh | B. 0.4 kWh | C. 6 kWh | D. 1.5 kWh |
Answer: C. 6 kWh Why: E = Pt = 1.5 kW × 4 h = 6 kWh. |
| 19 | A current of 0.4 A flows for 300 s. Find the charge that passes. (1 mark) A. 120 C | B. 12 C | C. 750 C | D. 300 C |
Answer: A. 120 C Why: Q = It = 0.4 × 300 = 120 C. |
| 20 | A transformer has 200 turns on the primary and 50 on the secondary. The primary voltage is 240 V. Find the secondary voltage. (1 mark) A. 48 V | B. 60 V | C. 240 V | D. 960 V |
Answer: B. 60 V Why: V₂ = 240 × 50 ÷ 200 = 60 V, a step-down transformer. |
Section C Heat, atoms and astronomy (10 marks)
Questions 21-30 (1 mark each)
| # | Question | Answer and marking notes |
|---|---|---|
| 21 | 0.5 kg of water (specific heat capacity 4200 J kg⁻¹ K⁻¹) rises by 20 K. Find the energy absorbed. (1 mark) A. 21000 J | B. 84000 J | C. 42000 J | D. 8400 J |
Answer: C. 42000 J Why: E = mcΔT = 4200 × 0.5 × 20 = 42000 J. |
| 22 | 0.2 kg of ice at 0 °C melts completely (specific latent heat 3.34 × 10⁵ J kg⁻¹). Find the energy needed. (1 mark) A. 334000 J | B. 16700 J | C. 33400 J | D. 66800 J |
Answer: D. 66800 J Why: E = mL = 0.2 × 3.34 × 10⁵ = 66800 J. |
| 23 | A heater supplies 42000 J in 60 s. Find its power. (1 mark) A. 700 W | B. 70 W | C. 4200 W | D. 7000 W |
Answer: A. 700 W Why: P = E ÷ t = 42000 ÷ 60 = 700 W. |
| 24 | A machine takes 500 J of input and gives 150 J of useful output. Find its efficiency. (1 mark) A. 50% | B. 30% | C. 70% | D. 15% |
Answer: B. 30% Why: Efficiency = 150 ÷ 500 × 100% = 30%. |
| 25 | A radionuclide has a half-life of 4 days. Starting with 80 g, how much remains after 12 days? (1 mark) A. 40 g | B. 5 g | C. 10 g | D. 20 g |
Answer: C. 10 g Why: 12 ÷ 4 = 3 half-lives, so 80 ÷ 2³ = 10 g. |
| 26 | A source has an initial count rate of 640 s⁻¹. What is the count rate after 2 half-lives? (1 mark) A. 320 s⁻¹ | B. 80 s⁻¹ | C. 480 s⁻¹ | D. 160 s⁻¹ |
Answer: D. 160 s⁻¹ Why: Halving twice: 640 ÷ 2 ÷ 2 = 160 s⁻¹. |
| 27 | A 4 kg body moves at 6 m s⁻¹. Find its kinetic energy. (1 mark) A. 24 J | B. 72 J | C. 144 J | D. 36 J |
Answer: B. 72 J Why: KE = ½mv² = ½ × 4 × 6² = 72 J. |
| 28 | A 3 kg object is raised 5 m (g = 10 N kg⁻¹). Find the gain in gravitational potential energy. (1 mark) A. 150 J | B. 15 J | C. 50 J | D. 1500 J |
Answer: A. 150 J Why: PE = mgh = 3 × 10 × 5 = 150 J. |
| 29 | A radio signal takes 500 s to travel from the Earth to the Moon (c = 3.0 × 10⁸ m s⁻¹). Find the distance. (1 mark) A. 6.0 × 10¹¹ m | B. 1.5 × 10¹⁴ m | C. 1.5 × 10¹¹ m | D. 1.5 × 10⁸ m |
Answer: C. 1.5 × 10¹¹ m Why: d = ct = 3.0 × 10⁸ × 500 = 1.5 × 10¹¹ m. |
| 30 | A metal block has a mass of 240 g and a volume of 30 cm³. Find its density. (1 mark) A. 2 g cm⁻³ | B. 8 g cm⁻³ | C. 4 g cm⁻³ | D. 12 g cm⁻³ |
Answer: B. 8 g cm⁻³ Why: ρ = m ÷ V = 240 ÷ 30 = 8 g cm⁻³. |
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