Measurements & errors (uncertainties) – A-level Physics exam questions with mark scheme

1Core3 marks

The Young modulus of a material has the unit Pa.

(a) Show that the pascal can be expressed in SI base units as kg m$^{-1}$ s$^{-2}$. [2]

(b) The Young modulus of a polymer is 3.2 GPa. Write this value in Pa, using standard form. [1]

(a) Pa = N m$^{-2}$ (from pressure = force / area)(1)
(a) N = kg m s$^{-2}$ (from $F = ma$) so Pa = kg m s$^{-2}$ $\times$ m$^{-2}$ = kg m$^{-1}$ s$^{-2}$ (must see both steps; answer given)(1)
(b) $3.2 \times 10^{9}$ Pa(1)
Answer: (a) Pa = N m$^{-2}$ = kg m s$^{-2}$ m$^{-2}$ = kg m$^{-1}$ s$^{-2}$ (b) $3.2\times10^{9}$ Pa
Examiner tip: In 'show that' unit questions every substitution must be written out — just stating the final unit scores nothing.

2Core4 marks

A student uses a micrometer screw gauge to measure the diameter of a wire. When the jaws are closed, the micrometer reads 0.02 mm.

(a) State the type of error this causes and explain how the student should correct her readings. [2]

(b) State how the student should reduce the effect of random errors on her value for the diameter. [1]

(a) systematic error / zero error(1)
(a) subtract 0.02 mm from every reading allow 'correct each reading for the zero error' or 'reset the micrometer to read zero before measuring'; reject 'add 0.02 mm'(1)
(b) repeat the measurement at several points along the wire (and in different orientations) and calculate the mean allow 'take repeat readings and find the mean'; reject 'use a more precise instrument' alone(1)
(c) the repeated measurements are close together / have very little spread about the mean allow 'consistent' / 'fluctuate only slightly about a mean value'; reject 'close to the true value' (that is accuracy)(1)
Answer: (a) systematic (zero) error; subtract 0.02 mm from each reading (b) repeat at different places/orientations and average (c) little spread in repeated readings
Examiner tip: Precise means little spread in repeats; accurate means close to the true value — swapping these is the most common lost mark.

3Standard4 marks

A student times 10 oscillations of a pendulum five times. Her results are:

14.2 s 14.6 s 14.3 s 14.5 s 14.4 s

(a) Calculate the time period $T$ of the pendulum. [1]

(b) Calculate the percentage uncertainty in $T$. Use half the range of the readings as the absolute uncertainty. [3]

(a) mean of 10 oscillations = 14.4 s so $T = 1.44$ s(1)
(b) absolute uncertainty = half range = (14.6 − 14.2)/2 = 0.2 s (for 10 oscillations) or 0.02 s in T(1)
(b) percentage uncertainty = $\dfrac{0.2}{14.4} \times 100$ or $\dfrac{0.02}{1.44} \times 100$(1)
(b) 1.4% allow 1.39%(1)
Answer: (a) $T = 1.44$ s (b) 1.4%
Examiner tip: Dividing by the number of oscillations reduces the absolute uncertainty but not the percentage uncertainty — both shrink in the same ratio.

4Hard5 marks

A student determines the density of a metal cylinder. Her measurements are:

mass $m = 48.2 \pm 0.1$ g
diameter $d = 2.50 \pm 0.01$ cm
length $L = 3.60 \pm 0.05$ cm

Calculate the density of the metal in kg m$^{-3}$ and its absolute uncertainty. Give your answer to an appropriate number of significant figures. [5]

volume $V = \pi \left(\dfrac{2.50}{2}\right)^2 \times 3.60 = 17.7$ cm$^3$ ($1.77 \times 10^{-5}$ m$^3$)(1)
$\rho = \dfrac{48.2}{17.67} = 2.73$ g cm$^{-3}$ = 2730 kg m$^{-3}$ allow 2728(1)
percentage uncertainty in $d^2$ = $2 \times 0.4\%$ = 0.8%(1)
total percentage uncertainty = 0.21% + 0.8% + 1.39% = 2.4%(1)
absolute uncertainty = $0.024 \times 2728 = 65$ kg m$^{-3}$, quoted to 1 s.f. as 70, so $\rho = 2730 \pm 70$ kg m$^{-3}$ allow 2728 ± 65; ecf from their percentage(1)
Answer: $\rho = (2.73 \pm 0.07) \times 10^{3}$ kg m$^{-3}$
Examiner tip: The diameter is squared in the volume, so its percentage uncertainty must be doubled before adding.

5Core3 marks

Estimate the kinetic energy of an adult sprinter running at top speed. Show any values you assume. [3]

sensible mass stated: 50–100 kg(1)
sensible top speed stated: 8–12 m s$^{-1}$(1)
$E_k = \frac{1}{2}mv^2$ used to give a value in the range 1600 J – 7200 J, e.g. $\frac{1}{2}\times 70 \times 10^2 = 3500$ J (to 1 or 2 s.f.)(1)
Answer: about $4 \times 10^{3}$ J (e.g. 3500 J for 70 kg at 10 m s$^{-1}$)
Examiner tip: In estimate questions write down every value you assume — the marks are for the assumptions as well as the answer.

6Hard5 marks

A student investigates how the period $T$ of a simple pendulum depends on its length $l$. Theory predicts $T^2 = \dfrac{4\pi^2}{g}\,l$.

She plots $T^2$ against $l$ with error bars. The gradient of her line of best fit is 4.02 s$^2$ m$^{-1}$. The steepest line that still passes through all the error bars has a gradient of 4.18 s$^2$ m$^{-1}$.

(a) Calculate a value for $g$. [2]

(b) Determine the absolute uncertainty in $g$. [3]

(a) $g = \dfrac{4\pi^2}{\text{gradient}} = \dfrac{4\pi^2}{4.02}$(1)
(a) 9.82 m s$^{-2}$ allow 9.8(1)
(b) percentage uncertainty in gradient = $\dfrac{4.18 - 4.02}{4.02} \times 100 = 4.0\%$(1)
(b) percentage uncertainty in $g$ is the same, 4.0%(1)
(b) absolute uncertainty = $0.040 \times 9.82 = 0.39$ m s$^{-2}$ ($\approx \pm 0.4$ m s$^{-2}$) allow 0.4; allow 0.38 from $9.82 - \dfrac{4\pi^2}{4.18} = 9.82 - 9.44$; ecf from (a)(1)
Answer: (a) $g = 9.82$ m s$^{-2}$ (b) $\pm 0.4$ m s$^{-2}$
Examiner tip: Uncertainty in a gradient comes from the difference between best and worst acceptable lines, divided by the best gradient.

7Stretch5 marks

A student measures the length of a pendulum as $l = 0.800 \pm 0.002$ m and its period as $T = 1.80 \pm 0.02$ s.

She uses $T = 2\pi\sqrt{\dfrac{l}{g}}$ to calculate a value for $g$.

Deduce whether her result is consistent with the accepted value of $g = 9.81$ m s$^{-2}$. [5]

$g = \dfrac{4\pi^2 l}{T^2}$(1)
$g = \dfrac{4\pi^2 \times 0.800}{1.80^2} = 9.75$ m s$^{-2}$(1)
percentage uncertainties: $l$: $\dfrac{0.002}{0.800}=0.25\%$ and $T^2$: $2 \times \dfrac{0.02}{1.80} = 2.2\%$ allow max/min method: one of $g_{max} = \dfrac{4\pi^2\times0.802}{1.78^2} = 9.99$ or $g_{min} = \dfrac{4\pi^2\times0.798}{1.82^2} = 9.51$ m s$^{-2}$ correctly calculated(1)
total = 0.25% + 2.2% = 2.5%, so absolute uncertainty = $\pm 0.24$ m s$^{-2}$, range 9.51 to 9.99 m s$^{-2}$ allow 2.47%; max/min method: both 9.51 and 9.99 (range) for this mark(1)
9.81 lies within this range so the result is consistent (conclusion must follow from their range); allow comparison of percentage difference $\frac{9.81-9.75}{9.81} \approx 0.6\%$ with the 2.5% uncertainty(1)
Answer: $g = 9.75 \pm 0.24$ m s$^{-2}$ (9.51–9.99), which includes 9.81, so consistent
Examiner tip: 'Consistent' means the accepted value lies inside your uncertainty range — you must quote the range to earn the conclusion mark.

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