OCR H556 · course-matched practice

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Actual marks22

Questions10

Topics represented10

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

Question 1 · Electric circuits · 3 marks

Explain why the output across the fixed resistor rises with temperature.

Your answer

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Marking points

The thermistor's resistance falls as temperature rises.
In a potential divider, p.d. divides in proportion to resistance.
So a larger fraction of the supply p.d. is across the fixed resistor.

\(V_{\text{out}}\) = \(\dfrac{V_{\text{in}} \times R_{2}}{R_{1} + R_{2}}\).

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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 · 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.

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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 · 2 marks

A 0.20 m wire carries 3.0 A perpendicular to a 0.50 T field. Calculate the magnetic force.

Your answer

Use unit: N

Working

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Answer: 0.3 N

F = BIL = 0.50 × 3.0 × 0.20 = 0.30 N.

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Question 9 · Measurements and uncertainties · 1 mark

What is the main purpose of a pilot run?

Choose one answer A. To choose suitable ranges and check the method works before collecting data B. To get the final result C. To remove systematic error D. To replace repeats

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Answer: A

A pilot run refines the plan.

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Question 10 · Mechanics · 3 marks

Explain why a graph of h against t² gives a more reliable value of g than a single measurement.

Your answer

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Marking points

It uses many measurements, reducing the effect of random error.
A constant timing delay shifts the line but does not change the gradient.
Anomalous points can be identified and excluded.

Gradients remove systematic offsets.

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This paper-level result is for revision feedback and does not change your mastery score.