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Questions on uncertainty and error
Absolute uncertainty is the ± amount, in the measurement's own units, within which the true value is expected to lie. Percentage uncertainty is that amount divided by the value, times 100. Random and systematic errors produce uncertainty differently. Repeats reduce random scatter; a systematic error shifts every reading the same way and repeating readings does not reduce it. Error bars give the uncertainty in a gradient.
20 original questions · 65 marks · the uncertainty and error notes · Measurements
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Distinguish between random errors and systematic errors.
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Random errors scatter readings unpredictably above and below the true value, and are reduced by repeating and averaging (1). Systematic errors shift all readings the same way, a zero error for example, and are not reduced by repeating (1).A length is measured as 2.50 ± 0.05 m. Calculate the percentage uncertainty.
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Percentage uncertainty = (0.05/2.50) × 100 (1)
= 2.0% (1)Distinguish between the precision and the accuracy of a set of measurements.
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Precision is how close repeated readings are to each other, that is how small the scatter is (1). Accuracy is how close readings are to the true value (1). Measurements can be precise but inaccurate if a systematic error is present.State what is meant by the resolution of an instrument, and give the resolution of a digital stopwatch that displays times to 0.01 s.
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Resolution is the smallest change in the quantity that the instrument can display (1). For this stopwatch the resolution is 0.01 s (1).A student's repeated measurements of a resistance agree closely with each other. A second student, using a different meter, obtains values that agree closely with the first student's. State the term that describes each of these two observations.
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The first student's results are repeatable: the same experimenter with the same equipment gets consistent results (1). The agreement between the two students, using different equipment, shows the results are reproducible (1).Two lengths, 3.0 ± 0.1 m and 5.0 ± 0.2 m, are added. State the total length and its absolute uncertainty.
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When adding, absolute uncertainties add (1)
Total = 8.0 ± 0.3 m (1)A rectangle has length 10.0 ± 0.2 cm and width 5.0 ± 0.1 cm. Calculate its area and the absolute uncertainty in the area.
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Percentage uncertainties: length 2%, width 2% (1)
For a product they add, so the area uncertainty is 4% (1)
Area = 10.0 × 5.0 = 50 cm2 (1)
Uncertainty = 4% of 50 = ±2 cm2 (1)A length of 45 mm is measured with a ruler of 1 mm resolution, giving a reading uncertainty of ±0.5 mm. Calculate the percentage uncertainty.
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Uncertainty is ±0.5 mm on a reading of 45 mm (1)
Percentage uncertainty = (0.5/45) × 100 (1)
= 1.1% (1)Four repeated readings of a length are 20.1, 20.3, 20.2 and 20.4 mm. Calculate the mean and estimate its uncertainty from the range.
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Mean = (20.1 + 20.3 + 20.2 + 20.4)/4 (1)
Mean = 20.25 mm (1)
Uncertainty ≈ range/2 = (20.4 − 20.1)/2 = ±0.15 mm (1)With its jaws fully closed, a micrometer reads −0.03 mm. With a wire between the jaws it reads 0.36 mm. (a) Name the type of error the closed-jaw reading reveals. (b) Determine the diameter of the wire. (c) Explain why taking repeat readings would not remove this error.
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(a) A systematic (zero) error (1). (b) Corrected diameter = 0.36 − (−0.03) = 0.39 mm (1). (c) The offset shifts every reading the same way, so averaging repeats leaves the shift unchanged; the correction must be applied instead (1).To find the extension of a rubber cord, a student measures its stretched length as 482 ± 1 mm and its unstretched length as 476 ± 1 mm. (a) State the extension and its absolute uncertainty. (b) Calculate the percentage uncertainty in the extension. (c) Suggest one change to the experiment that would reduce this percentage uncertainty.
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(a) When subtracting, absolute uncertainties add: extension = 6 ± 2 mm (1). (b) Evidence of use of (uncertainty/value) × 100, i.e. (2/6) × 100 (1); percentage uncertainty = 33% (1). (c) Use a larger load so the extension is bigger (or measure the extension directly between fixed marks): the same ±2 mm is then a smaller fraction of the value (1).A student determines g from g = 4π2l/T2 using a simple pendulum, with l = 0.650 ± 0.002 m and T = 1.62 ± 0.02 s. Calculate the percentage uncertainty in the value of g.
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Percentage uncertainty in l = (0.002/0.650) × 100 = 0.3% (1). T is squared, so its percentage uncertainty counts twice: 2 × (0.02/1.62) × 100 = 2.5% (1). Total = 0.3 + 2.5 = 2.8% (1).The radius of a circle is 4.0 ± 0.1 cm. Calculate the area and its absolute uncertainty (area = πr2).
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Percentage uncertainty in r = (0.1/4.0) × 100 = 2.5% (1)
Area ∝ r2, so its percentage uncertainty = 2 × 2.5% = 5% (1)
Area = π × 4.02 = 50.3 cm2 (1)
Uncertainty = ±2.5 cm2 (1)A density is found from ρ = m/V, with m = 0.500 ± 0.005 kg and V = 2.0 × 10−4 ± 0.1 × 10−4 m3. Calculate the density and its absolute uncertainty.
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Percentage uncertainties: mass 1%, volume 5% (1)
For a quotient they add, giving 6% (1)
ρ = 0.500/(2.0 × 10−4) = 2500 kg m−3 (1)
Uncertainty = 6% = ±150 kg m−3 (1)State how you would reduce (a) a random error and (b) a systematic error in an experiment.
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(a) Reduce random error by repeating readings and taking a mean (1), and by using instruments of finer resolution (1). (b) Reduce systematic error by checking for and correcting zero errors, calibrating the instrument, and improving the technique (1).A student investigating a falling ball plots a graph with error bars on every point. The best-fit line has gradient 0.204 s2 m−1. The steepest and shallowest lines that still pass through all the error bars have gradients 0.212 s2 m−1 and 0.196 s2 m−1. Determine the absolute and percentage uncertainty in the gradient, and quote the gradient with its uncertainty.
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Uncertainty = half the difference between the extreme gradients (1): (0.212 − 0.196)/2 = ±0.008 s2 m−1 (1). Percentage uncertainty = (0.008/0.204) × 100 = 3.9% (1). Gradient = 0.204 ± 0.008 s2 m−1 (1).A supplier claims that a batch of ball bearings has diameter 5.00 mm with a tolerance of ±1%. A student measures one bearing four times with a micrometer, reading 5.11, 5.09, 5.07 and 5.09 mm. Deduce whether the bearing meets the supplier's claim.
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Mean = (5.11 + 5.09 + 5.07 + 5.09)/4 = 5.09 mm (1), with uncertainty = half the range = ±0.02 mm (1). The claim allows 5.00 ± 0.05 mm, i.e. 4.95 to 5.05 mm (1). Even the smallest reading (5.07 mm) lies above 5.05 mm, so the measured range is wholly outside the tolerance and the bearing does not meet the claim (1).A student measures g by timing a ball falling a height h and using g = 2h/t2, with h = 1.250 ± 0.005 m and t = 0.505 ± 0.005 s. Determine the percentage uncertainty each measurement contributes to g, and state which single measurement most needs improving.
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h contributes (0.005/1.250) × 100 = 0.4% (1). t is squared, so it contributes 2 × (0.005/0.505) × 100 = 2.0% (1). The timing contributes about five times the percentage uncertainty of the height, so the measurement of t most needs improving (1).A student varies the length l of a simple pendulum, measures the period T for each length, and plots T2 on the y-axis against l on the x-axis, with error bars on the T2 values. The period is given by T2 = 4π2l/g. Describe fully how the student should use the graph to determine a value for g and its percentage uncertainty.
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Draw the best-fit straight line and measure its gradient using a large triangle (1). Comparing with T2 = (4π2/g)l, the gradient equals 4π2/g (1), so g = 4π2/gradient (1). Draw the steepest and shallowest lines that still pass through all the error bars and measure their gradients (1). The uncertainty in the gradient is half the difference between these two gradients (1). The percentage uncertainty in g equals the percentage uncertainty in the gradient, (Δgradient/gradient) × 100, since g is inversely proportional to the gradient (1).A student measures the diameter of a wire at five points along its length, using a micrometer of resolution 0.01 mm. The wire is to be used in an experiment that needs its cross-sectional area, A = πd2/4. Determine the mean diameter the student should quote, with its absolute and percentage uncertainty, and hence the percentage uncertainty in the cross-sectional area.
reading diameter d / mm 1 0.41 2 0.43 3 0.52 4 0.42 5 0.44 Mark scheme
Reading 3, 0.52 mm, is anomalous, lying far outside the 0.41 mm to 0.44 mm spread of the other four, and is excluded before any mean is taken (1)
Mean of the four remaining readings = (0.41 + 0.43 + 0.42 + 0.44)/4 = 0.425 mm (1)
Uncertainty = half the range of those four = (0.44 − 0.41)/2 = ±0.015 mm, which is larger than the ±0.005 mm that the micrometer's resolution alone would give, so the spread is what sets the uncertainty (1)
Percentage uncertainty in d = (0.015/0.425) × 100 = 3.5% (1)
A goes as d2, so the power rule doubles the percentage: uncertainty in A = 2 × 3.53 = 7.1% (accept 7.0%) (1)
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