OCR A Level Physics A: Practical Endorsement skills, apparatus and techniques (H556) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover OCR A Level Physics A (H556), section 1.2 (1.2.1 practical skills and 1.2.2 apparatus and techniques), from the OCR specification, Version 3.0 (March 2026). Read the study guide first and keep the revision notes open. Hub: OCR A Level Physics. Printable checklist.

Questions

1. A wire has a diameter of about 0.60 mm. (a) Calculate the percentage uncertainty if the diameter is measured with a rule to ±0.5 mm. [1] (b) Name a more suitable instrument, give its resolution, and calculate the percentage uncertainty for an uncertainty of ±0.01 mm. [2]

2. A pendulum completes 30 oscillations in 52.2 s. The uncertainty in the total time is ±0.2 s. (a) Calculate the period T. [1] (b) Calculate the absolute and percentage uncertainty in T. [2] (c) Explain how a fiducial marker helps the timing. [1]

3. (a) A micrometer shows 4.5 mm on the sleeve and 37 on the thimble. With the jaws closed it reads −0.02 mm. Find the corrected diameter. [2] (b) A vernier caliper (resolution 0.02 mm) shows 23 mm on the main scale with the 13th vernier mark aligned. Find the reading. [1] (c) Give two techniques that improve a wire diameter measurement. [2]

4. A student finds the resistivity of a wire. Six micrometer readings of the diameter (zero error zero) are 0.34, 0.35, 0.34, 0.36, 0.35 and 0.34 mm. The length is 0.800 ± 0.002 m. A p.d. of 1.50 V gives a current of 0.300 A. (a) Calculate the mean diameter. [1] (b) Calculate the cross-sectional area. [2] (c) Calculate the resistance and the resistivity. [3] (d) The diameter has an uncertainty of ±0.01 mm and the resistance ±2 %. Estimate the percentage and absolute uncertainty in the resistivity. [2] (e) State where the voltmeter and ammeter are connected. [1]

5. An oscilloscope is set to 2.0 V/div and 0.5 ms/div. The trace is 3.2 divisions from peak to trough, and one cycle covers 4.0 divisions horizontally. (a) Find the peak p.d. [2] (b) Find the frequency. [2] (c) Which setting change would show more cycles? [1]

6. A card of width 0.020 m passes light gate 1 in 0.016 s and light gate 2 in 0.008 s. The time between the gates is 0.40 s. (a) Calculate both speeds. [2] (b) Calculate the acceleration. [2] (c) Give one advantage of light gates over a stopwatch. [1]

7. (a) An LED needs 2.4 V at 18 mA from a 9.0 V supply. Calculate the series resistance, then the current when the nearest larger standard value of 390 Ω is used. [2] (b) A 5.0 V supply drives a divider with 3.0 kΩ above and 2.0 kΩ below the output. A 2.0 kΩ load is connected across the lower resistor. Calculate the new output p.d. [2] (c) State two checks that show the LED circuit is correct. [2]

8. Stationary microwaves form between a transmitter and a metal plate. A probe finds seven successive minima over 8.4 cm. (a) Calculate the wavelength. [2] (b) Calculate the frequency (c = 3.00 × 10⁸ m/s). [2] (c) Why is the minima spacing half a wavelength? [1] (d) Why measure across seven minima, not two? [1]

9. A laser shines through a diffraction grating of 500 lines per mm. The first-order maximum is 0.498 m from the central maximum on a screen 1.50 m from the grating. (a) Find the grating spacing d. [1] (b) Find the angle of the first-order maximum. [1] (c) Calculate the wavelength. [2] (d) State two safety precautions. [2]

10. A Geiger-Muller tube records 150 counts in 300 s with no source. With a source it records 2610 counts in 60 s, and 1305 counts in 60 s with an absorber added. (a) Explain why the first measurement is taken. [1] (b) Calculate the corrected count rate with and without the absorber. [2] (c) Find the fraction transmitted by the absorber. [1] (d) Estimate the percentage uncertainty in the 2610 count. [2] (e) Give two safe handling rules for the source. [2]

11. (a) A 50 Hz signal is recorded with a data logger. Suggest a sampling rate and justify it. [2] (b) Write a citation for a web page, listing the details you would record. [2] (c) Give two checks that a website is reliable. [2]

Answers

1. (a) 0.5 / 0.60 × 100 = 83 % [1] (b) A micrometer [1] with resolution 0.01 mm: 0.01 / 0.60 × 100 = 1.7 % [1] Examiner insight: Answers gain credit when the instrument is paired with its resolution and a calculation, not just “more accurate”.

2. (a) T = 52.2 / 30 = 1.74 s [1] (b) Absolute: 0.2 / 30 = 0.0067 s [1]; percentage: 0.0067 / 1.74 = 0.38 % [1] (c) It marks a fixed position, such as the equilibrium position, so every count starts and stops at the same point of the swing. [1] Examiner insight: Dividing the total time uncertainty by the number of oscillations is the step to show clearly.

3. (a) 4.5 + 0.37 = 4.87 mm [1]; subtract the zero error: 4.87 − (−0.02) = 4.89 mm [1] (b) 23 + 13 × 0.02 = 23.26 mm [1] (c) Measure at several positions and orientations, then average [1]; use the ratchet so the wire is not compressed [1] Examiner insight: A negative zero error is subtracted, so the correction increases the reading; sign errors lose credit.

4. (a) Sum = 2.08, mean = 0.3467 mm (0.347 mm) [1] (b) A = πd²/4 [1] = π × (0.3467 × 10⁻³)² / 4 = 9.44 × 10⁻⁸ m² [1] (c) R = 1.50 / 0.300 = 5.00 Ω [1]; ρ = RA/L = 5.00 × 9.44 × 10⁻⁸ / 0.800 [1] = 5.9 × 10⁻⁷ Ω m [1] (d) d is squared: 2 × (0.01 / 0.3467) = 5.8 %; length 0.25 %; R 2 %; total 8.0 % [1]; absolute = 0.08 × 5.9 × 10⁻⁷ = ±0.5 × 10⁻⁷ Ω m [1] (e) Ammeter in series with the wire; voltmeter in parallel across the wire. [1] Examiner insight: Keep the unrounded mean until the end; squaring the diameter means its percentage uncertainty is counted twice.

5. (a) Peak-to-trough = 3.2 × 2.0 = 6.4 V [1]; peak = 3.2 V [1] (b) T = 4.0 × 0.5 = 2.0 ms [1]; f = 1 / 0.0020 = 500 Hz [1] (c) Increase the time-base setting (for example 1 ms/div). [1] Examiner insight: Answers lose credit when the peak-to-trough reading is quoted as the amplitude or the time-base is not converted from ms.

6. (a) v₁ = 0.020 / 0.016 = 1.25 m/s [1]; v₂ = 0.020 / 0.008 = 2.5 m/s [1] (b) a = (2.5 − 1.25) / 0.40 [1] = 3.1 m/s² [1] (c) There is no human reaction-time error. [1] Examiner insight: Quote the gate interval and the card width from the question; the acceleration uses the time between gates, not either gate time.

7. (a) R = (9.0 − 2.4) / 0.018 = 367 Ω [1]; with 390 Ω, I = 6.6 / 390 = 16.9 mA [1] (b) Parallel combination: 2.0 kΩ and 2.0 kΩ is 1.0 kΩ [1]; V = 5.0 × 1.0 / (3.0 + 1.0) = 1.25 V [1] (c) Check the LED polarity before switching on [1]; measure the p.d. across the LED with a voltmeter, expecting about 2.4 V [1] Examiner insight: Rounding to a larger standard value keeps the current safe; say so, and show the new current.

8. (a) Spacing = 8.4 / 6 = 1.4 cm [1]; λ = 2 × 1.4 = 2.8 cm [1] (b) f = c / λ = 3.00 × 10⁸ / 0.028 [1] = 1.07 × 10¹⁰ Hz [1] (c) Moving the probe by λ/2 changes the round-trip path of the reflected wave by λ, so the interference pattern repeats. [1] (d) The fixed uncertainty in locating a minimum is shared over six spacings, so the percentage uncertainty is smaller. [1] Examiner insight: Count spacings, not minima; seven minima give six gaps.

9. (a) d = 1 × 10⁻³ / 500 = 2.00 × 10⁻⁶ m [1] (b) θ = tan⁻¹(0.498 / 1.50) = 18.4° [1] (c) λ = d sin θ [1] = 2.00 × 10⁻⁶ × sin 18.4° = 630 nm [1] (d) Any two: never look into the beam or its reflection; remove reflective objects; stop the beam with a screen; use the lowest suitable laser power. [1] [1] Examiner insight: Use the angle from the tangent, not the screen distance, when the screen is flat; safety answers must be specific to lasers.

10. (a) To find the background count rate, which must be subtracted from the source readings. [1] (b) Background = 150 / 300 = 0.50 s⁻¹; no absorber 2610 / 60 = 43.5, so 43.0 s⁻¹ [1]; with absorber 1305 / 60 = 21.75, so 21.3 s⁻¹ [1] (c) Fraction = 21.25 / 43.0 = 0.49 [1] (d) Uncertainty is about √2610 = 51 [1]; 51 / 2610 = 2.0 % [1] (e) Any two: keep the source in its container when not in use; handle with tongs; point it away from people; keep exposure time short and distance large. [1] [1] Examiner insight: Subtract the background before comparing the two readings.

11. (a) A rate of 1000 samples per second [1] gives 20 samples per 20 ms cycle, enough to show the waveform shape [1] (b) Author or organisation, year, page title, web address and date accessed [1], placed beside the information it supports [1] (c) Any two: it comes from a recognised body or textbook; it names its author or date; it agrees with a second source. [1] [1] Examiner insight: A citation needs enough detail for someone else to find the page.

Where marks are usually lost

• Giving a percentage uncertainty without dividing by the measured value.
• Forgetting the zero error, or applying it with the wrong sign.
• Using the wrong number of intervals (minima, fringes, oscillations).
• Reading an oscilloscope without converting divisions into volts or seconds.
• Not stating that background was subtracted.
• Writing “be careful” instead of a specific hazard and control.
• Connecting a voltmeter in the current path, or an ammeter across a component.
• Quoting too many significant figures for the data.

Next steps

• Review with the revision notes and the study guide.
• Practise planning and analysis in Development of practical skills practice.
• Course hub: OCR A Level Physics; checklist.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OCR, A Level Specification Physics A (H556), Version 3.0 (March 2026), for first assessment in 2017: sections 1.2.1, 1.2.2 and 5g.