IGCSE Physics 0625 Paper 6 Practice Paper (Alternative to Practical, 40 marks)

Level: Year 10–11 (IGCSE) | Subject: Physics 0625 | Time: 60 minutes | Total: 40 marks | Parts: 26

Paper at a glance

Paper	Time	Marks	Weighting
Paper 6 (Alternative to Practical)	1 h	40	20%

Questions

Question 1 (9 marks) — Simple pendulum: measuring g

(a) Describe how to use a pendulum and a stopwatch to measure the period accurately (including the number of oscillations timed and how the mean is treated). (2 marks)

(b) At which position of the bob should the stopwatch be started and stopped? Explain. (1 marks)

(c) The length is 0.800 m and 20 complete oscillations take 35.6 s. Find the period T (2 d.p.) and hence g from T = 2π√(L/g) (3 s.f.). (2 marks)

(d) If the experiment is repeated with five different lengths, which two quantities should be plotted? How is g found from the graph? (2 marks)

(e) If the reaction-time error is ±0.2 s, calculate the percentage error in timing 35.6 s (2 s.f.). (1 marks)

(f) What would be the effect on the percentage error of timing 10 oscillations instead of 20? (1 marks)

Question 2 (8 marks) — Current-voltage characteristic of a filament lamp

(a) How should the ammeter, voltmeter and variable resistor be connected? (2 marks)

(b) What should be done between readings, and why? (2 marks)

(c) At 3.0 V the current is 0.25 A; at 6.0 V it is 0.40 A. Calculate the resistance of the lamp at each p.d. (2 marks)

(d) Explain why the resistance of the lamp increases as the p.d. rises. (1 marks)

(e) Should the I–V graph for this lamp be a straight line or a curve? Describe its shape. (1 marks)

Question 3 (7 marks) — Density of an irregular solid

(a) Name the two pieces of apparatus needed to find the density of this solid. (1 marks)

(b) The solid has a mass of 84.0 g; the water level rises from 50.0 cm³ to 62.0 cm³. Find the volume of the solid and explain why it is lowered gently. (2 marks)

(c) Calculate the density of the solid in g/cm³ and in kg/m³. (2 marks)

(d) The smallest division of the cylinder is 0.5 cm³. Calculate the maximum percentage error in the volume (2 s.f.). (1 marks)

(e) Suggest one way to reduce this error. (1 marks)

Question 4 (8 marks) — Refraction through a glass block

(a) Describe how to trace a ray through a glass block and measure the angles of incidence and refraction. (2 marks)

(b) Readings: i = 30° gives r = 19°; i = 50° gives r = 31°. Calculate sin i / sin r for each (2 d.p.). (2 marks)

(c) Find the mean refractive index and hence the critical angle from sin c = 1/n (1 d.p.). (2 marks)

(d) Why use the narrow slit of a ray box and work in a darkened room? (1 marks)

(e) State one major source of error in this experiment. (1 marks)

Question 5 (8 marks) — Specific heat capacity of water

(a) Describe the electrical method for measuring the specific heat capacity of water (state the four quantities to measure). (2 marks)

(b) Why should the container be insulated (lagged or covered)? (1 marks)

(c) 0.500 kg of water is heated by a 12.0 V, 2.0 A heater for 300 s and its temperature rises by 3.4 °C. Calculate the energy supplied and the specific heat capacity (2 s.f.). (2 marks)

(d) Some heat is inevitably lost. Will the calculated c be too large or too small? Explain. (1 marks)

(e) Suggest two improvements to increase the accuracy. (2 marks)

Answers and marking notes

Question 1 — Simple pendulum: measuring g

(a) Measure the length twice and average it; let the bob swing through a small angle (about 5°); time 20 complete oscillations with a stopwatch and divide by 20 to get one period; repeat three times and average (1 mark each, max 2). [2 marks]

(b) At the equilibrium (lowest) position, because the bob moves fastest there, so the timing error is smallest. [1 marks]

(c) T = 35.6 / 20 = 1.78 s (1); g = 4π²L / T² = 4π² × 0.800 / 1.78² = 9.97 m/s² (1). [2 marks]

(d) Plot T² against L (1). The line is straight through the origin with gradient = 4π²/g, so g = 4π² / gradient (1). [2 marks]

(e) 0.2 / 35.6 × 100% = 0.56% (2 s.f.). [1 marks]

(f) The timing interval halves, so the same ±0.2 s error gives about double the percentage error, i.e. about 1.12%. [1 marks]

Question 2 — Current-voltage characteristic of a filament lamp

(a) Ammeter in series with the lamp; voltmeter in parallel across the lamp; variable resistor in series with the supply to vary the p.d. (1 mark each, max 2). [2 marks]

(b) Open the switch (or return the p.d. to zero) after each reading so the filament cools; otherwise the filament temperature keeps rising and the resistance changes, spoiling the curve (1 mark each). [2 marks]

(c) R = V/I: 3.0 / 0.25 = 12.00 Ω (1); 6.0 / 0.40 = 15.00 Ω (1). [2 marks]

(d) A larger current raises the filament temperature; the lattice ions vibrate more strongly, electrons collide more often, so the resistance increases. [1 marks]

(e) A curve: it starts at the origin and bends over as the p.d. increases (gradient decreasing). [1 marks]

Question 3 — Density of an irregular solid

(a) An electronic balance (for mass) and a measuring cylinder (or displacement can) for volume. [1 marks]

(b) Volume = 62.0 – 50.0 = 12.00 cm³ (1); lower it gently so no water splashes out (or air bubbles stick), which would make the volume reading too large (1). [2 marks]

(c) ρ = 84.0 / 12.00 = 7.00 g/cm³ (1); = 7.00 × 1000 = 7000 kg/m³ (1). [2 marks]

(d) 0.5 / 12.00 × 100% = 4.2% (2 s.f.: 4.2%). [1 marks]

(e) Use a narrower measuring cylinder (or the displacement method, or several identical solids to give a larger total volume) so the scale error is a smaller fraction. [1 marks]

Question 4 — Refraction through a glass block

(a) Draw round the block on paper; direct a narrow beam from a ray box at one face and mark two points on the incident and emergent rays, then join them; remove the block, draw the normal and measure both angles with a protractor (1 mark each, max 2). [2 marks]

(b) sin 30° / sin 19° = 1.54 (1); sin 50° / sin 31° = 1.49 (1). [2 marks]

(c) n = (1.54 + 1.49) / 2 = 1.51 (1); sin c = 1 / 1.51 = 0.662, c = 41.4° (1). [2 marks]

(d) The slit gives a thin, sharp ray and a darkened room makes it easier to see on the paper, both reducing errors in marking points and measuring angles. [1 marks]

(e) The protractor reads only to ±1° and the block edges do not coincide exactly with the traced outline (or the pencil line is thick), so the angles are uncertain. [1 marks]

Question 5 — Specific heat capacity of water

(a) Measure the mass of water m, the p.d. V and current I of the heater, the time t and the temperature rise Δθ, with E = VIt (0.5 mark each, max 2). [2 marks]

(b) To reduce heat loss to the surroundings so the measured temperature rise matches the energy actually supplied. [1 marks]

(c) E = VIt = 12.0 × 2.0 × 300 = 7200 J (1); c = E / (mΔθ) = 7200 / (0.500 × 3.4) = 4235 J/(kg °C) ≈ 4.2 × 10³ J/(kg °C) (1). [2 marks]

(d) Too large: heat loss makes Δθ smaller than it should be while E is taken from the meter readings, so c = E / (mΔθ) comes out too large. [1 marks]

(e) Use a lower power over a longer time to cut the proportion lost; plot θ against t and use the initial gradient; add lagging and a lid (any two, 1 mark each). [2 marks]

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