中文（繁體）
中文（简体）
English
IGCSE Physics 0625 Paper 4 (10 Questions, 100 Marks, 75 Minutes)
IGCSE Physics 0625 Paper 4
IGCSE Physics 0625 Paper 4 practice paper (10 questions, 100 marks, 75 minutes): uniform acceleration and average speed, moments and pivot reaction, density and pressure, wav。
Download Question paper PDF | Answer key PDF
Paper details Level: IGCSE | Subject: Physics | Paper: 4 | Time: 75 minutes | Marks: 100 | Questions: 10
Section A Mechanics, Matter and Waves (50 marks)
Questions 1-2
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | 1. A car starts from rest and accelerates uniformly at 2.5 m/s² for 8 s, travels at constant speed for 12 s, then decelerates uniformly to rest in 5 s. (a) Find the maximum speed of the car. (3 marks) (b) Find the total distance travelled in 25 s. (4 marks) (c) Find the average speed for the whole journey. (3 marks) |
(a) v = 2.5 × 8 = 20 m/s. (3) (b) Stage 1 = ½ × 2.5 × 8² = 80 m; stage 2 = 20 × 12 = 240 m; stage 3 = ½ × 20 × 5 = 50 m; total = 370 m. (4) (c) Average speed = 370 ÷ 25 = 14.8 m/s. (3) Marking notes: one stage only scores 1 at most; omitting the deceleration stage loses 2. |
| 2 | 2. A uniform beam 2.4 m long weighs 60 N with its centre of gravity at the midpoint. It rests on a pivot 0.9 m from the left end, with a load W hung at the left end, and stays horizontal. (a) State the principle of moments. (2 marks) (b) Find W. (4 marks) (c) Find the upward reaction force of the pivot on the beam. (4 marks) |
(a) For a body in equilibrium, the sum of clockwise moments equals the sum of anticlockwise moments. (2) (b) Taking moments about the pivot: W × 0.9 = 60 × (1.2 − 0.9) = 18, so W = 20 N. (4) (c) Vertical forces balance: R = W + 60 = 80 N. (4) Marking notes: using 1.2 m as the moment arm scores 0; ignoring the beam weight loses 2. |
Questions 3-4
| # | Question | Answer and marking notes |
|---|---|---|
| 3 | 3. A rectangular metal block measures 0.20 m × 0.15 m × 0.10 m and has a mass of 4.8 kg. (a) Calculate the volume of the block. (3 marks) (b) Calculate the density of the metal. (3 marks) (c) The block is placed on a table on its smallest face. Calculate the pressure it exerts (g = 9.8 N/kg). (4 marks) |
(a) V = 0.20 × 0.15 × 0.10 = 0.003 m³. (3) (b) ρ = 4.8 ÷ 0.003 = 1600 kg/m³. (3) (c) Area = 0.15 × 0.10 = 0.015 m²; weight = 4.8 × 9.8 = 47.04 N; p = 47.04 ÷ 0.015 = 3136 Pa. (4) Marking notes: using mass instead of weight scores 0; using the largest face gives 313.6 Pa and loses 3. |
| 4 | 4. A source emits sound of frequency 170 Hz. The speed of sound in air is 340 m/s. (a) Calculate the wavelength. (3 marks) (b) Calculate the period to 3 significant figures. (3 marks) (c) A ship sends a pulse and receives the echo from the seabed after 0.60 s. Calculate the depth. (4 marks) |
(a) λ = v ÷ f = 340 ÷ 170 = 2 m. (3) (b) T = 1 ÷ 170 = 0.00588 s. (3) (c) Total path = 340 × 0.60 = 204 m; depth = 204 ÷ 2 = 102 m. (4) Marking notes: forgetting to halve gives 204 m and loses 2; writing 1/f without evaluating scores 1. |
Question 5
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | 5. A 500 W heater raises the temperature of 2.0 kg of water from 20 °C to 60 °C. The specific heat capacity of water is 4200 J/(kg °C). (a) Calculate the energy absorbed by the water. (3 marks) (b) Assuming no heat loss, calculate the time needed. (4 marks) (c) The actual efficiency is 80%. Calculate the actual time needed. (3 marks) |
(a) E = mcΔT = 2.0 × 4200 × 40 = 336000 J. (3) (b) t = E ÷ P = 336000 ÷ 500 = 672 s. (4) (c) t = 672 ÷ 0.80 = 840 s. (3) Marking notes: multiplying by 0.8 gives 538 s, a wrong direction and scores 0. |
Section B Electricity, Gases and Radioactivity (50 marks)
Questions 6-7
| # | Question | Answer and marking notes |
|---|---|---|
| 6 | 6. A 12 Ω and a 6 Ω resistor are connected in parallel, then in series with a 2 Ω resistor across a 12 V supply of negligible internal resistance. (a) Calculate the resistance of the parallel section. (3 marks) (b) Calculate the current drawn from the supply. (4 marks) (c) Calculate the current in the 12 Ω resistor and the total power of the circuit. (3 marks) |
(a) 1/R = 1/12 + 1/6 = 1/4, so R = 4 Ω. (3) (b) Total resistance = 4 + 2 = 6 Ω; I = 12 ÷ 6 = 2 A. (4) (c) P.d. across the parallel section = 2 × 4 = 8 V; I₁₂ = 8 ÷ 12 = 0.67 A; P = 12 × 2 = 24 W. (3) Marking notes: simply adding the resistors (18 Ω) scores 0. |
| 7 | 7. An ideal transformer has 1200 turns on the primary connected to 240 V a.c. and supplies a 24 W lamp at 12 V. (a) Calculate the number of secondary turns. (4 marks) (b) Calculate the secondary current. (3 marks) (c) Calculate the primary current assuming 100% efficiency. (3 marks) |
(a) Ns/Np = Vs/Vp, so Ns = 1200 × 12 ÷ 240 = 60 turns. (4) (b) Is = P ÷ Vs = 24 ÷ 12 = 2 A. (3) (c) Ip = P ÷ Vp = 24 ÷ 240 = 0.1 A. (3) Marking notes: inverting the ratio (24000 turns) scores 0; quoting the secondary current as primary scores 0. |
Questions 8-9
| # | Question | Answer and marking notes |
|---|---|---|
| 8 | 8. A fixed mass of gas at constant temperature has a volume of 300 cm³ at a pressure of 1.0 × 10⁵ Pa. (a) State the gas law that applies and calculate the volume when the pressure rises to 2.5 × 10⁵ Pa. (3 marks) (b) The volume is now kept constant while the temperature rises from 27 °C to 327 °C. Calculate the new pressure. (4 marks) (c) Explain why kelvin must be used in (b). (3 marks) |
(a) Boyle’s law, p₁V₁ = p₂V₂; V₂ = 1.0×10⁵ × 300 ÷ 2.5×10⁵ = 120 cm³. (3) (b) T₁ = 300 K, T₂ = 600 K; p₂ = p₁ × T₂/T₁ = 2.0e+05 Pa. (4) (c) Only kelvin is measured from absolute zero and is proportional to pressure; 0 °C is not zero pressure, so p/T = constant fails with celsius values. (3) Marking notes: substituting celsius directly gives 11.1 × 10⁵ Pa and loses 3. |
| 9 | 9. A radioactive source has a half-life of 8 days and an initial activity of 6400 Bq. (a) Calculate the activity after 24 days. (3 marks) (b) Calculate the time for the activity to fall to 400 Bq. (4 marks) (c) State how activity relates to the number of undecayed nuclei and explain why activity never reaches zero in a finite time. (3 marks) |
(a) 24 days = 3 half-lives; A = 6400 × (½)³ = 800 Bq. (3) (b) 400 = 6400 × (½)ⁿ gives n = 4, so t = 4 × 8 = 32 days. (4) (c) Activity is proportional to the number of undecayed nuclei; each half-life only halves it, so it approaches zero without ever reaching it in finite time. (3) Marking notes: the answer 32 days without working scores 2; saying the nuclei run out scores 0. |
Question 10
| # | Question | Answer and marking notes |
|---|---|---|
| 10 | 10. A crane lifts a 250 kg load through 12 m at constant speed in 20 s (g = 9.8 N/kg). (a) Calculate the work done on the load. (3 marks) (b) Calculate the useful output power. (3 marks) (c) The efficiency is 70%. Calculate the input power. (4 marks) |
(a) W = mgh = 250 × 9.8 × 12 = 29400 J. (3) (b) P = W ÷ t = 29400 ÷ 20 = 1470 W. (3) (c) Input power = 1470 ÷ 0.70 = 2100 W. (4) Marking notes: multiplying by 0.7 (1029 W) scores 0. |
Find a tutor — freeTutor Registration →
A TutorZone practice paper; original questions and passages, not publisher material.
