IGCSE Physics 0625 Paper 6 Practice Paper (Practical Skills, 10 Questions, 100 Marks)

2026-11-02

IGCSE Physics 0625

IGCSE Physics 0625 Paper 6 practical skills: micrometer, density, Hooke law, pendulum, resistance, heat, refraction, lenses, echoes, gas law. Answers with notes.

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Paper details　Level: IGCSE　|　Subject: Physics (0625)　|　Time: 60 minutes　|　Marks: 100　|　Questions: 10

Section A: Measurement and density (40 marks)

Answer questions 1 to 4. Each question carries 10 marks.

#	Question	Answer and marking notes
1	Q1 A micrometer is used to measure the diameter of a wire three times: 0.48, 0.49 and 0.48 mm. The micrometer has a zero error of +0.02 mm. (a) Calculate the mean reading. (b) Find the diameter after correcting for the zero error. (c) Calculate the cross-sectional area of the wire (m², 3 s.f.). (d) If each reading has an instrument uncertainty of ±0.01 mm, find the percentage uncertainty in the diameter and in the area.	(a) (0.48 + 0.49 + 0.48) / 3 = 0.483 mm (2 marks). (b) A positive zero error is subtracted: 0.483 − 0.02 = 0.463 mm (2 marks). (c) r = 0.463 / 2 = 0.2317 mm = 2.317e-04 m, so A = πr² = 1.686e-07 m² (3 marks). (d) Diameter: 0.01 / 0.463 × 100 = 2.2%; the area depends on the square of the diameter, so its percentage uncertainty doubles to 4.3% (3 marks).
2	Q2 An irregular metal block has a mass of 62.4 g. When it is lowered into a measuring cylinder containing 50.0 cm³ of water, the level rises to 74.0 cm³. (a) Calculate the volume of the block. (b) Calculate its density (g cm⁻³). (c) Express the density in kg m⁻³. (d) The mass is ±0.1 g and each volume reading is ±0.5 cm³; find the percentage uncertainty in the density.	(a) 74.0 − 50.0 = 24.0 cm³ (2 marks). (b) ρ = 62.4 / 24.0 = 2.60 g cm⁻³ (3 marks). (c) 2.60 × 1000 = 2600 kg m⁻³ (2 marks). (d) Mass 0.16% + volume (±1.0 cm³ over two readings) 4.2% = 4.3%; the volume term dominates (3 marks).
3	Q3 A 0.200 kg mass is hung on a spring and its length changes from 12.0 cm to 15.9 cm (g = 9.81 N kg⁻¹). (a) Calculate the weight of the mass. (b) Calculate the extension in metres. (c) Find the spring constant (N m⁻¹). (d) Find the elastic potential energy stored and calculate the period of oscillation of this system.	(a) W = 0.200 × 9.81 = 1.96 N (2 marks). (b) 15.9 − 12.0 = 3.9 cm = 0.039 m (2 marks). (c) k = 1.96 / 0.039 = 50.3 N m⁻¹ (3 marks). (d) E = ½Fx = ½ × 1.96 × 0.039 = 0.038 J; T = 2π√(m/k) = 0.396 s (3 marks).
4	Q4 A simple pendulum of length 0.800 m is used to measure g; 20 complete oscillations take 35.8 s. (a) Calculate the period. (b) Use T = 2π√(L/g) to calculate g. (c) Calculate the percentage difference from the accepted value 9.81 m s⁻². (d) Explain why 20 oscillations are timed instead of one, and suggest one way to reduce the error.	(a) T = 35.8 / 20 = 1.79 s (2 marks). (b) g = 4π²L / T² = 9.86 m s⁻² (4 marks). (c) \|9.86 − 9.81\| / 9.81 × 100 = 0.48% (2 marks). (d) The reaction-time error at start and stop is fixed, so timing many oscillations divides it by 20; improvements: measure length to the centre of the bob, use a light gate, or keep the amplitude below 10° (2 marks).

Section B: Electricity and heat (30 marks)

Answer questions 5 to 7. Each question carries 10 marks.

#	Question	Answer and marking notes
5	Q5 A 1.20 m length of wire of diameter 0.38 mm carries a current of 0.30 A with a potential difference of 2.40 V across it. (a) Calculate the resistance. (b) Calculate the cross-sectional area of the wire (m²). (c) Calculate the resistivity of the metal (Ω m, 3 s.f.). (d) Explain why a micrometer is used for the diameter instead of a ruler.	(a) R = V / I = 2.40 / 0.30 = 8.0 Ω (2 marks). (b) r = 0.38 / 2 = 0.19 mm = 1.9 × 10⁻⁴ m, so A = πr² = 1.134e-07 m² (3 marks). (c) ρ = RA / L = 8.0 × 1.134e-07 / 1.20 = 7.56e-07 Ω m (3 marks). (d) The diameter is a fraction of a millimetre and a ruler scale of 1 mm cannot be read to 0.01 mm; a micrometer reads to 0.01 mm, greatly reducing the percentage uncertainty in the area (2 marks).
6	Q6 A 24 W heater heats 0.100 kg of water for 300 s; the temperature rises from 21.0 °C to 36.8 °C. (a) Calculate the energy supplied by the heater. (b) State the energy gained by the water, assuming no losses. (c) Hence calculate the specific heat capacity of water (J kg⁻¹ K⁻¹). (d) The accepted value is 4200 J kg⁻¹ K⁻¹ — calculate the percentage error and give one reason why the result may be too large.	(a) E = Pt = 24 × 300 = 7,200 J (3 marks). (b) With no losses the water gains 7,200 J (1 mark). (c) c = E / (mΔT) = 7,200 / (0.100 × 15.8) = 4,557 J kg⁻¹ K⁻¹ (3 marks). (d) Error = (4,557 − 4200) / 4200 × 100 = 8.5%; causes: heat lost to the air and container, or a ΔT that is too small because the thermometer was read before the temperature settled — both make c too large (3 marks).
7	Q7 A ray of light passes from air into a transparent plastic at an angle of incidence of 40°, and the angle of refraction is 25°. (a) State Snell law. (b) Calculate the refractive index of the plastic (3 s.f.). (c) Calculate the critical angle for light travelling from the plastic to air. (d) Explain why arcs of radius greater than 5 cm are drawn when measuring the angles.	(a) n = sin i / sin r (2 marks). (b) n = sin 40° / sin 25° = 1.52 (3 marks). (c) sin c = 1 / 1.52, so c = 41.1° (3 marks). (d) A larger radius makes the arc longer for the same angular error, so the relative error in measuring the chord is smaller; it also reduces errors from the pin holes and the thickness of the pencil line (2 marks).

Section C: Optics, waves and gases (30 marks)

Answer questions 8 to 10. Each question carries 10 marks.

#	Question	Answer and marking notes
8	Q8 An object is placed 30.0 cm in front of a converging lens and a sharp image forms on a screen 15.0 cm behind the lens. (a) Use 1/f = 1/u + 1/v to calculate the focal length. (b) Calculate the magnification. (c) Describe the nature of the image. (d) Explain how the experiment ensures the image is as sharp as possible.	(a) 1/f = 1/30.0 + 1/15.0 = 0.1000 cm⁻¹, so f = 10.0 cm (4 marks). (b) m = v / u = 15.0 / 30.0 = 0.50 (3 marks). (c) Because the image is caught on a screen it is real; it is inverted and diminished (0.50×) (2 marks). (d) Keep the object and lens fixed and move the screen until the edges of the image are sharpest, re-checking after each small adjustment and working in dim light to reduce stray light (1 mark).
9	Q9 An ultrasound detector sends a pulse towards a wall and receives the echo after 0.0240 s; take the speed of sound as 1540 m s⁻¹. (a) Calculate the total distance travelled by the pulse. (b) Calculate the distance from the detector to the wall. (c) If the timing uncertainty is ±0.0002 s, calculate the percentage uncertainty in the distance. (d) Explain why ultrasound rather than audible sound is used for this measurement.	(a) s = vt = 1540.0 × 0.0240 = 36.96 m (3 marks). (b) The pulse travels there and back, so d = 36.96 / 2 = 18.48 m (3 marks). (c) 0.83%, i.e. ±0.15 m (2 marks). (d) Ultrasound has a short wavelength and is highly directional, so the beam stays narrow and the echo is strong; it is also inaudible, so it causes no disturbance (2 marks).
10	Q10 A fixed mass of gas at room temperature has a volume of 60 cm³ at a pressure of 100 kPa. (a) Calculate the product pV. (b) If the volume is reduced to 40 cm³ at constant temperature, calculate the new pressure. (c) Describe how a syringe and a pressure sensor could be used to test this relationship. (d) Explain why the piston must be pushed slowly.	(a) pV = 100 × 60 = 6,000 kPa cm³ (3 marks). (b) p₂ = 6,000 / 40 = 150 kPa (3 marks). (c) Seal a fixed mass of gas in a syringe connected to a pressure sensor; reduce the volume in steps, record the pressure each time and calculate pV for every reading (or plot p against 1/V). If the product stays roughly constant, the relationship is verified (2 marks). (d) Rapid compression does work on the gas and raises its temperature, so the “constant temperature” condition fails; pushing slowly lets the heat escape and keeps the change isothermal (2 marks).

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