Specification: AQA A-level Physics (7408) 3.1.2 | Paper: 1, 2 and 3 (tested throughout) | Data and formulae booklet: not given – learn these rules
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📌 KEY DEFINITIONS – learn them word for word
Precision and accuracy: a measurement is precise if repeat readings are close to each other, with only a small spread about the mean. A measurement is accurate if it is close to the true value. Precision says nothing about accuracy.
Repeatability and reproducibility: a measurement is repeatable if the same experimenter, using the same method and equipment, gets the same results. It is reproducible if the same results are obtained by a different experimenter or by using different equipment or techniques.
Random and systematic errors: random errors cause readings to spread unpredictably about the true value; repeating and averaging reduces their effect. Systematic errors make every reading differ from the true value by the same amount or in the same direction (for example a zero error); repeating does not remove them. The resolution of an instrument is the smallest change in the quantity that gives a change in the reading.
The equation
percentage uncertainty = (absolute uncertainty ÷ measured value) × 100% %x = (Δx ÷ x) × 100% and fractional uncertainty = absolute uncertainty ÷ measured value Δx ÷ x
| Quantity | Symbol | Unit |
|---|---|---|
| measured value (or mean of repeat readings) | x | the unit of the quantity (e.g. V, s, m) |
| absolute uncertainty | Δx | the same unit as x |
| fractional uncertainty | Δx ÷ x | no unit (a ratio) |
| percentage uncertainty | %x | per cent (%) |
Rearranged: Δx = (%x ÷ 100) × x | Resolution: analogue scale ± half the smallest division; digital display ± 1 in the last digit; or use the stated uncertainty | Repeat readings: uncertainty = ± half the range | Quote the uncertainty to 1 or 2 s.f. and the value to the same number of decimal places, e.g. 15.4 ± 0.2 s.
Use FIFA for every answer
- F – Formula: write the equation as it appears on the sheet.
- I – Insert: put in the numbers (convert to standard units first).
- F – Fix: rearrange to make the unknown the subject.
- A – Answer: calculate and always give the unit.
Core: Q1–4. Exam standard: Q5–8. Stretch: Q9–10.
Questions
Q1 (Core) A digital voltmeter reads 3.47 V. State the absolute uncertainty due to the resolution of the meter, and calculate the percentage uncertainty, to 2 significant figures.
Show solution
digital display, so the uncertainty is ± 1 in the last digit: ± 0.01 V
F: %V = (ΔV ÷ V) × 100%
I: %V = (0.01 ÷ 3.47) × 100%
F: %V is already the subject
A: 0.288…% = 0.29% (2 s.f.)
Q2 (Core) An analogue ammeter has scale divisions of 0.02 A. The pointer is at 0.86 A. State the absolute uncertainty in the reading, and calculate the fractional and percentage uncertainties, to 2 significant figures.
Show solution
analogue scale, so the uncertainty is ± half the smallest division = 0.02 ÷ 2 = ± 0.01 A
F: fractional uncertainty = ΔI ÷ I; %I = (ΔI ÷ I) × 100%
I: ΔI ÷ I = 0.01 ÷ 0.86 = 0.0116…
F: %I = 0.0116… × 100%
A: fractional uncertainty = 0.012; percentage uncertainty = 1.16…% = 1.2% (2 s.f.)
Q3 (Core) A student times a trolley rolling down a ramp four times and records 15.2 s, 15.6 s, 15.3 s and 15.5 s. Calculate the mean time, its absolute uncertainty from the spread of the readings, and the percentage uncertainty, to 2 significant figures.
Show solution
Step 1 – mean and range
mean = (15.2 + 15.6 + 15.3 + 15.5) ÷ 4 = 61.6 ÷ 4 = 15.4 s; range = 15.6 − 15.2 = 0.4 s
Step 2 – absolute uncertainty
uncertainty = ± half the range = 0.4 ÷ 2 = ± 0.2 s, so t = 15.4 ± 0.2 s
Step 3 – percentage uncertainty
F: %t = (Δt ÷ t) × 100%
I: %t = (0.2 ÷ 15.4) × 100%
A: 1.29…% = 1.3% (2 s.f.)
Q4 (Core) The mass of a block is recorded as 250 g with a percentage uncertainty of 0.40%. Calculate the absolute uncertainty in the mass, and write the mass with its uncertainty in kg.
Show solution
F: %m = (Δm ÷ m) × 100%
I: 0.40 = (Δm ÷ 250) × 100
F: Δm = (0.40 ÷ 100) × 250 = 1.0 g
Convert first: 250 g = 0.250 kg; 1.0 g = 0.0010 kg
A: Δm = 1.0 g, so m = 0.250 ± 0.001 kg
Watch out: converting the value but not the uncertainty. Both must be in the same unit, and the value and uncertainty should end at the same decimal place.
Q5 (Exam) A student measures g four times with a light gate and records 9.53, 9.55, 9.52 and 9.56 m s⁻². The accepted value is 9.81 m s⁻². (a) Calculate the mean, its absolute uncertainty and its percentage uncertainty, to 2 significant figures. (b) Calculate the percentage difference between the mean and the accepted value, to 2 significant figures. (c) Comment on the precision and accuracy of the results, and the type of error present.
Show solution
Step 1 – (a) mean and uncertainty
mean = (9.53 + 9.55 + 9.52 + 9.56) ÷ 4 = 38.16 ÷ 4 = 9.54 m s⁻²; range = 9.56 − 9.52 = 0.04 m s⁻², so Δg = ± 0.02 m s⁻²
F: %g = (Δg ÷ g) × 100%
I: %g = (0.02 ÷ 9.54) × 100%
A: g = 9.54 ± 0.02 m s⁻², percentage uncertainty 0.209…% = 0.21% (2 s.f.)
Step 2 – (b) percentage difference
percentage difference = (9.81 − 9.54) ÷ 9.81 × 100% = 0.27 ÷ 9.81 × 100% = 2.75…% = 2.8% (2 s.f.)
Step 3 – (c) comment
The readings are precise (very small spread) but not accurate: the accepted value lies well outside 9.52–9.56 m s⁻². All readings are low by a similar amount, which points to a systematic error, for example a delay in the electromagnet releasing the ball that adds the same extra time to every reading. Repeating the readings will not remove it.
Watch out: saying that small uncertainty means an accurate result. A small spread shows precision only; accuracy is judged by comparing with the true value.
Q6 (Exam) A student uses a micrometer (resolution 0.01 mm) to measure the diameter of a wire at five places: 0.45, 0.47, 0.44, 0.46 and 0.48 mm. Calculate the mean diameter and its uncertainty, and the percentage uncertainty to 2 significant figures. Give the diameter with its uncertainty in metres, in standard form.
Show solution
Step 1 – mean
mean = (0.45 + 0.47 + 0.44 + 0.46 + 0.48) ÷ 5 = 2.30 ÷ 5 = 0.46 mm
Step 2 – uncertainty
range = 0.48 − 0.44 = 0.04 mm, so half the range = ± 0.02 mm. This is larger than the resolution (± 0.01 mm), so use ± 0.02 mm
F: %d = (Δd ÷ d) × 100%
I: %d = (0.02 ÷ 0.46) × 100%
A: 4.34…% = 4.3% (2 s.f.)
Convert first: 1 mm = 10−3 m, so 0.46 mm = 4.6 × 10−4 m and 0.02 mm = 0.2 × 10−4 m
A: d = (4.6 ± 0.2) × 10−4 m
Watch out: always using the resolution as the uncertainty. When the spread of repeat readings is bigger than the resolution, the spread sets the uncertainty.
Q7 (Exam) When the jaws of a micrometer are closed it reads 0.03 mm instead of zero. A student then uses it to measure a sheet of metal and gets a reading of 1.27 mm. (a) Name this type of error and give the corrected thickness. (b) The resolution of the micrometer is 0.01 mm. Calculate the percentage uncertainty in the corrected thickness, to 2 significant figures. (c) Explain why taking more readings would not improve the accuracy.
Show solution
Step 1 – (a) correct the zero error
a zero error is a systematic error; corrected thickness = 1.27 − 0.03 = 1.24 mm
Step 2 – (b) percentage uncertainty
F: %x = (Δx ÷ x) × 100%
I: %x = (0.01 ÷ 1.24) × 100%
A: 0.806…% = 0.81% (2 s.f.)
Step 3 – (c)
every reading is 0.03 mm too large, so the mean of many readings is still 0.03 mm too large. Averaging reduces random error only; the zero error must be subtracted.
Watch out: adding the zero error instead of subtracting it. A positive zero reading means every measurement is too big.
Q8 (Exam) A student times 20 complete oscillations of a pendulum with a hand-held stopwatch and gets 35.9 s. Her reaction time gives an uncertainty of ± 0.2 s in each timing. (a) Calculate the period with its absolute uncertainty, and its percentage uncertainty to 2 significant figures. (b) Calculate the percentage uncertainty if she had timed only one oscillation (1.80 s), to 2 significant figures, and explain why timing many oscillations is better.
Show solution
Step 1 – (a) period and uncertainty
T = 35.9 ÷ 20 = 1.795 s; ΔT = 0.2 ÷ 20 = 0.01 s, so T = 1.80 ± 0.01 s (value rounded to 2 decimal places to match the uncertainty)
F: %T = (ΔT ÷ T) × 100%
I: %T = (0.2 ÷ 35.9) × 100% (the same as 0.01 ÷ 1.795 × 100%)
A: 0.557…% = 0.56% (2 s.f.)
Step 2 – (b) one oscillation
%T = (0.2 ÷ 1.80) × 100% = 11.1…% = 11% (2 s.f.)
the reaction-time uncertainty is the same however long the timing, so timing 20 oscillations makes it a 20 times smaller fraction of the time measured
Watch out: dividing 35.9 s by 20 but leaving the uncertainty at ± 0.2 s. The absolute uncertainty is divided by 20 too.
Q9 (Stretch) A student needs the thickness of one sheet of paper. Method A: measure a stack of 200 sheets with a millimetre ruler. The stack is 21 mm thick, and each of the two readings on the ruler (one at each end of the stack) has an uncertainty of ± 0.5 mm. Method B: measure one sheet with a micrometer, giving 0.11 ± 0.01 mm. Calculate the thickness of one sheet and its percentage uncertainty from each method, to 2 significant figures, and state which method is better.
Show solution
Step 1 – Method A – uncertainty in the stack
a length found from two readings has an uncertainty of 0.5 + 0.5 = ± 1 mm, so the stack is 21 ± 1 mm
Step 2 – Method A – one sheet
thickness = 21 ÷ 200 = 0.105 mm; uncertainty = 1 ÷ 200 = 0.005 mm, so t = 0.105 ± 0.005 mm
F: %t = (Δt ÷ t) × 100%
I: %t = (1 ÷ 21) × 100%
A: 4.76…% = 4.8% (2 s.f.)
Step 3 – Method B
%t = (0.01 ÷ 0.11) × 100% = 9.09…% = 9.1% (2 s.f.)
A: Method A is better: it has about half the percentage uncertainty. It assumes all the sheets have the same thickness and are pressed flat.
Watch out: using ± 0.5 mm for a ruler length. A length needs two readings (the zero end and the far end), so the uncertainty is ± 1 mm unless the zero is set exactly.
Q10 (Stretch) Two groups measure the speed of sound in air at 20 °C, where the accepted value is 343 m s⁻¹. Group A uses a resonance tube and gets 338, 341, 336 and 343 m s⁻¹. Group B uses two microphones and a data logger and gets 331, 332, 330 and 333 m s⁻¹. (a) For each group, calculate the mean, the absolute uncertainty and the percentage uncertainty to 2 significant figures, and write each result with its uncertainty to 1 significant figure. (b) Compare the results in terms of precision, accuracy, repeatability and reproducibility.
Show solution
Step 1 – (a) Group A
mean = (338 + 341 + 336 + 343) ÷ 4 = 1358 ÷ 4 = 339.5 m s⁻¹; range = 343 − 336 = 7 m s⁻¹, so uncertainty = ± 3.5 m s⁻¹
%v = (3.5 ÷ 339.5) × 100% = 1.03…% = 1.0% (2 s.f.); result 340 ± 4 m s⁻¹
Step 2 – (a) Group B
mean = (331 + 332 + 330 + 333) ÷ 4 = 1326 ÷ 4 = 331.5 m s⁻¹; range = 333 − 330 = 3 m s⁻¹, so uncertainty = ± 1.5 m s⁻¹
%v = (1.5 ÷ 331.5) × 100% = 0.452…% = 0.45% (2 s.f.); result 332 ± 2 m s⁻¹
Step 3 – (b) comparison
Precision: B is more precise (smaller spread). Accuracy: A is more accurate: 343 m s⁻¹ lies inside its range of 336–344 m s⁻¹, but outside B’s range of 330–334 m s⁻¹, so B has a systematic error.
Repeatability: each group’s readings are close, so each method is repeatable. Reproducibility: the two methods do not agree within their uncertainties, so the result has not been reproduced.
Watch out: calling B the better result because its uncertainty is smaller. A precise result can still be inaccurate.
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