Eduqas A420QS · 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 a potential divider is used to supply the component.

Your answer

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Marking points

It allows the p.d. to be varied smoothly.
It can start from zero p.d.
So the full characteristic, including low values, is measured.

A series rheostat cannot reach zero.

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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 Snell's law for light entering glass from air?

Choose one answer A. n = \(\dfrac{\sin r}{\sin i}\) B. i = nr C. sin i sin r = n D. sin i = n sin r

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Answer: D

n₁ sin θ₁ = n₂ sin θ₂ with \(n_{\text{air}}\) = 1.

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Question 4 · Astrophysics and cosmology · 2 marks

A star has peak wavelength 500 nm. Using Wien constant b = 2.90×10⁻³ m K, estimate its surface temperature.

Your answer

Use unit: K

Working

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Answer: 5800 K

T = \(\dfrac{b}{\lambda_{\text{max}}}\) = \(\dfrac{2.90\times 10^{-3}}{500\times 10^{-9}}\) = 5.8×10³ K.

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Question 5 · Capacitors · 3 marks

Explain why the charging current decreases with time.

Your answer

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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 6 · 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 7 · 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 8 · 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 9 · Measurements and uncertainties · 1 mark

How do you convert a density from g cm⁻³ to kg m⁻³?

Choose one answer A. Divide by 1000 B. Multiply by 100 C. Multiply by 1000 D. It is the same

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Answer: C

1 g cm⁻³ = 1000 kg m⁻³.

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