Errors, uncertainties and graphs – A-level Physics exam questions with mark scheme

1Core4 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.

2Standard4 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.

3Hard5 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.

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4Hard5 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.

5Stretch5 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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