AQA 7408 · course-matched practice

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Actual marks20

Questions9

Topics represented9

Your paper aims for about 20 marks, so the final total may be slightly higher.

Question 1 · Electric circuits · 3 marks

Explain why the terminal p.d. falls as the current increases.

Your answer

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Marking points

The cell has internal resistance r.
Lost volts Ir increase with current.
Terminal p.d. = ε − Ir decreases.

Energy is transferred inside the cell.

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Question 2 · Materials · 3 marks

Explain how the cross-sectional area of the wire is determined accurately.

Your answer

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Marking points

Measure the diameter with a micrometer.
Take readings at several points along the wire and in two perpendicular directions.
Use the mean diameter in A = \(\dfrac{\pi d^{2}}{4}\).

Area depends on d², so diameter errors double.

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Question 3 · Waves · 1 mark

What is the equation for double-slit fringe spacing?

Choose one answer A. w = \(\dfrac{\lambda s}{D}\) B. w = \(\dfrac{sD}{\lambda }\) C. w = \(\dfrac{\lambda D}{s}\) D. w = \(\dfrac{\lambda }{sD}\)

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Answer: C

Fringe spacing increases with λ and D and decreases with s.

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Question 4 · Capacitors · 3 marks

Explain why the charging current decreases with time.

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Marking points

As the capacitor charges, its p.d. rises.
The p.d. across the resistor (V₀ − \(V_{C}\)) falls.
So the current I = \(\dfrac{V_{0} - V_{C}}{R}\) decreases.

The capacitor opposes further charge.

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Question 5 · Electric fields · 2 marks

A 2.0 nC positive charge experiences a force of 6.0 × 10⁻⁴ N. Calculate the electric field strength.

Your answer

Use unit: N/C

Working

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Answer: 3.0e5 N/C

E = \(\dfrac{F}{Q}\) = \(\dfrac{6.0\times 10^{-4}}{2.0\times 10^{-9}}\) = 3.0×10⁵ N/C.

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Question 6 · Gravitational fields · 2 marks

Calculate the gravitational field strength 7.0 × 10⁶ m from the centre of a planet of mass 6.0 × 10²⁴ kg. Use G = 6.67 × 10⁻¹¹ N m² kg⁻².

Your answer

Use unit: N/kg

Working

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Answer: 8.17 N/kg

g = \(\dfrac{GM}{r^{2}}\) = \(\dfrac{6.67\times 10^{-11} \times 6.0\times 10^{24}}{(7.0\times 10^{6})^{2}}\) ≈ 8.17 N/kg.

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Question 7 · Magnetic fields and induction · 3 marks

State how the direction of the force on the wire is found and what reversing the current does.

Your answer

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Marking points

Fleming's left-hand rule: field, current, force/thrust.
Reversing the current reverses the force.
The balance reading then changes in the opposite direction.

Direction depends on current and field.

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Question 8 · Measurements and uncertainties · 2 marks

A length is measured as 5.0 ± 0.1 cm. Calculate the percentage uncertainty.

Your answer

Use unit: %

Working

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Answer: 2 %

Percentage uncertainty = \(\dfrac{0.1}{5.0}\) × 100 = 2%.

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Question 9 · Mechanics · 1 mark

What is the gradient of a graph of h (y-axis) against t² (x-axis) for an object falling from rest?

Choose one answer A. \(\dfrac{g}{2}\) B. g C. 2g D. \(\dfrac{1}{g}\)

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Answer: A

From h = ½gt², h is proportional to t² with gradient \(\dfrac{g}{2}\).

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