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A-Level · Physics

AP1 Measurements and their Errors

AQA 7408 · Calculator allowed · about 150 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

SI units, prefixes and estimation are the foundation of all physical measurements.

(a)State the SI base unit of: (i) mass, (ii) time, (iii) electric current.(3)

(b)A component in a circuit is 250 nm thick. Write this thickness in metres, using standard form.(1)

(c)A capacitor is labelled as having a capacitance of 4.7 nF. Express this value in picofarads (pF).(2)

(d)Estimate, to one significant figure, the diameter of a single human hair in metres. State the order of magnitude of your estimate.(2)

(Total for Question 1 is 8 marks)

Significant figures indicate the precision of a measured or calculated quantity.

(a)A student measures the diameter of a wire as 0.42 mm using a micrometer of resolution 0.01 mm. State the number of significant figures in this measurement.(1)

(b)Calculate the cross-sectional area of the wire in part (a), using A = π x d² / 4, giving your answer in m² to an appropriate number of significant figures.(3)

(c)A calculation gives a result of 12.3456 N. Round this value to 3 significant figures.(1)

(d)Explain why it is not appropriate to quote the final answer to a calculation to more significant figures than the least precise measurement used in that calculation.(2)

(Total for Question 2 is 7 marks)

Precision, accuracy and resolution describe different aspects of the quality of a measurement.

(a)Define the term 'resolution' of a measuring instrument.(1)

(b)Define the term 'accuracy' of a measurement.(1)

(c)Two students each measure the length of the same object several times. Student A's repeated readings are tightly clustered together but consistently 3 mm below the true length. Student B's readings are centred on the true length but are widely spread out. Describe what this tells you about the precision and accuracy of each student's set of results.(2)

(d)A metre ruler has a resolution of 1 mm. State the absolute uncertainty in a single length measurement made with this ruler.(1)

(Total for Question 3 is 5 marks)

A student measures the length of a metal rod using vernier calipers of resolution 0.1 mm, obtaining a reading of 45.0 mm.

(a)Calculate the percentage uncertainty in the single reading of 45.0 mm.(2)

(b)The student instead repeats the measurement 5 times, obtaining: 45.0 mm, 45.2 mm, 44.8 mm, 45.1 mm, 44.9 mm. Calculate the mean length and the absolute uncertainty using the range method (half the range of the readings).(3)

(c)Explain why the uncertainty found in part (b) is greater than the uncertainty found in part (a).(2)

(Total for Question 4 is 7 marks)

A rectangular metal plate has a measured length L = 12.0 ± 0.1 cm and width W = 8.0 ± 0.1 cm.

(a)Calculate the percentage uncertainty in L and the percentage uncertainty in W.(2)

(b)Calculate the area of the plate and express it with its absolute uncertainty.(4)

(c)The perimeter of the plate is found using P = 2(L + W). Calculate the absolute uncertainty in the perimeter.(2)

(Total for Question 5 is 8 marks)

A small metal sphere has a measured radius r = 2.50 ± 0.05 cm.

(a)Calculate the percentage uncertainty in r.(1)

(b)The volume of the sphere is given by V = (4/3) x π x r^3. Calculate the percentage uncertainty in V.(2)

(c)Calculate the volume of the sphere and express your answer with its absolute uncertainty, to an appropriate number of significant figures.(3)

(Total for Question 6 is 6 marks)

Errors in measurement can be systematic or random.

(a)State one difference between a systematic error and a random error.(2)

(b)A newton-meter (spring balance) reads 0.20 N when nothing is hung from it. Identify the type of error this illustrates and explain how the student should account for it.(2)

(c)A student times 20 complete oscillations of a pendulum, rather than timing a single oscillation, to determine its period. Explain how timing 20 oscillations reduces the percentage uncertainty in the period, compared with timing a single oscillation.(3)

(Total for Question 7 is 7 marks)

Required practical: A student determines the resistivity of a metal wire. The wire's diameter is measured at several points along its length using a micrometer of resolution 0.01 mm, giving a mean diameter d = 0.32 mm. The wire's length is l = 0.800 ± 0.001 m. A graph of potential difference V against current I for the wire gives a straight line through the origin of gradient R = 2.40 ± 0.05 ohm, where R is the resistance of the wire. Resistivity is given by ρ = R x A / l, where A is the cross-sectional area of the wire.

(a)Explain why the student measures the wire's diameter at several points along its length and uses the mean value in the calculation.(2)

(b)Calculate the cross-sectional area, A, of the wire, using A = π x d² / 4 and d = 0.32 mm. Give your answer in m^2.(3)

(c)The micrometer used has a resolution of 0.01 mm. Calculate the percentage uncertainty in the cross-sectional area, A.(3)

(d)Given R = 2.40 ± 0.05 ohm and l = 0.800 ± 0.001 m, calculate the percentage uncertainty in the resistivity, ρ = R x A / l.(3)

(e)Calculate the resistivity of the wire and express your answer with its absolute uncertainty, to an appropriate number of significant figures.(4)

(f)Evaluate how the student could improve the experimental procedure to reduce the overall percentage uncertainty in the calculated resistivity, referring to the measurements taken in this experiment.(6)

(Total for Question 8 is 21 marks)

Required practical: A student determines the density of a regular metal cylinder and of a small irregularly shaped stone.

(a)Describe a method for determining the volume of the small irregularly shaped stone using a displacement (eureka) can and a measuring cylinder.(3)

(b)A metal cylinder has a diameter of 1.50 cm (measured with vernier calipers) and a height of 4.00 cm (measured with a ruler). Calculate its volume, using V = π x r² x h.(3)

(c)The cylinder has a mass of 62.3 g. Calculate the density of the cylinder in kg/m^3.(3)

(d)The diameter is 1.50 ± 0.02 cm and the height is 4.00 ± 0.05 cm. Calculate the percentage uncertainty in the volume of the cylinder.(4)

(e)For the stone, the volume found by displacement is 15.0 ± 0.5 cm³ and the mass is 40.2 ± 0.1 g. Calculate the percentage uncertainty in the density of the stone, and state which measurement contributes the greater uncertainty.(4)

(Total for Question 9 is 17 marks)

A student investigates the extension, x, of a spring for a range of applied forces, F, and plots a graph of x against F. A best-fit straight line is drawn through the origin, along with the steepest and least steep straight lines that still pass through all the error bars on the data points.

(a)The best-fit line has a gradient of 24.5 (in units of N per m of extension per unit, i.e. N/m). The steepest acceptable line has a gradient of 26.0 N/m, and the least steep acceptable line has a gradient of 23.0 N/m. Calculate the percentage uncertainty in the gradient.(3)

(b)State the gradient with its absolute uncertainty, to an appropriate number of significant figures.(2)

(c)Explain why error bars are included on the graph, and how they are used to find the uncertainty in the gradient.(3)

(d)The spring constant, k, is defined by F = k x x. Suggest one reason why using the gradient of the graph gives a more reliable value of k than calculating k from a single pair of F and x readings.(2)

(Total for Question 10 is 10 marks)

A student uses a simple pendulum to determine the acceleration due to gravity, g, using T = 2 x π x √l/g, which rearranges to g = 4 x π² x l / T^2. The pendulum length is l = 0.850 ± 0.002 m. The time for 20 complete oscillations is measured as 37.0 ± 0.4 s.

(a)Show that the period, T, of the pendulum is 1.85 s.(2)

(b)Calculate the percentage uncertainty in the time for 20 oscillations, and hence state the percentage uncertainty in the period, T.(3)

(c)Calculate the percentage uncertainty in T^2.(1)

(d)The percentage uncertainty in l is 0.24%. Calculate the percentage uncertainty in g.(2)

(e)Calculate the value of g and its absolute uncertainty, to an appropriate number of significant figures.(4)

(f)The accepted value of g is 9.81 m/s^2. Comment on whether the student's result in part (e) is consistent with the accepted value.(2)

(Total for Question 11 is 14 marks)

Two students each determine the Young modulus of the same copper wire using an applied load that produces an extension of about 2 mm. Student A uses a single wire clamped horizontally, measuring the extension directly with a metre rule of resolution 1 mm. Student B uses Searle's apparatus, with two parallel wires (one as a reference) and a vernier/micrometer scale of resolution 0.01 mm to measure the extension.

(Total for Question 12 is 6 marks)

Mark scheme · AP1 Measurements and their Errors

Question 1

(a) B1 kilogram, kg
(a) B1 second, s
(a) B1 ampere, A
(a) Answer: (i) kilogram (kg); (ii) second (s); (iii) ampere (A).
(b) B1 2.5 x 10^-7 m (cao)
(b) Answer: 2.5 x 10^-7 m
(c) M1 recognises 1 nF = 10³ pF (i.e. multiplies by 1000)
(c) A1 4700 pF, or equivalently 4.7 x 10³ pF (cao)
(c) Answer: 4700 pF (4.7 x 10³ pF)
(d) B1 any sensible estimate in the range 5 x 10^-5 m to 2 x 10^-4 m
(d) B1 correctly states the order of magnitude of their estimate (e.g. 10^-4 m) oe
(d) Answer: Approximately 1 x 10^-4 m; order of magnitude 10^-4 m (any value from 5 x 10^-5 to 2 x 10^-4 m with a correctly stated matching order of magnitude is acceptable).

Question 2

(a) B1 2 (significant figures) cao
(a) Answer: 2 significant figures
(b) M1 converts diameter to metres: d = 4.2 x 10^-4 m
(b) M1 correct substitution and evaluation: A = π x (4.2 x 10^-4)² / 4 = 1.385 x 10^-7 m²
(b) A1 1.4 x 10^-7 m² (awrt, to 2 s.f., matching the 2 s.f. of the diameter)
(b) Answer: A = 1.4 x 10^-7 m²
(c) B1 12.3 N (cao)
(c) Answer: 12.3 N
(d) B1 the overall result cannot be known more precisely/certainly than its least precise input measurement oe
(d) B1 quoting extra significant figures would imply a false level of precision/accuracy that the data does not support
(d) Answer: A calculated result is only as precise as the least precise value used to find it; extra significant figures would falsely suggest greater precision than the measurements justify.

Question 3

(a) B1 the smallest change in the quantity being measured that causes a detectable/observable change in the instrument's reading oe
(a) Answer: The smallest change in a quantity that an instrument can detect.
(b) B1 how close a measured value is to the true (accepted) value of the quantity oe
(b) Answer: How close a measurement is to the true value of the quantity being measured.
(c) B1 Student A: precise (small spread/good repeatability) but not accurate, because of a systematic error offsetting all readings from the true value
(c) B1 Student B: accurate (mean value close to/matches the true value) but not precise, because of a large random error causing a wide spread of readings
(c) Answer: Student A is precise but not accurate (small spread, offset from true value by a systematic error). Student B is accurate but not precise (correct mean, but a large random spread).
(d) B1 ± 0.5 mm (half the smallest scale division) oe
(d) Answer: ± 0.5 mm

Question 4

(a) M1 identifies absolute uncertainty as half the resolution: ± 0.05 mm
(a) A1 0.11% (awrt)
(a) Answer: 0.11%
(b) M1 mean = (45.0 + 45.2 + 44.8 + 45.1 + 44.9) / 5 = 45.0 mm
(b) M1 uncertainty = half the range = (45.2 - 44.8) / 2 = 0.2 mm
(b) A1 45.0 ± 0.2 mm (cao)
(b) Answer: 45.0 ± 0.2 mm
(c) M1 the repeated readings show random variation (e.g. from positioning the calipers slightly differently each time) that a single reading cannot reveal
(c) A1 this random variation between repeats is larger than the instrument's resolution alone, so the range-based uncertainty is greater than the resolution-based uncertainty
(c) Answer: Repeating the measurement reveals random variation between readings that is bigger than the instrument's resolution, so the spread-based uncertainty in (b) exceeds the resolution-based uncertainty in (a).

Question 5

(a) B1 percentage uncertainty in L = (0.1/12.0) x 100 = 0.83%
(a) B1 percentage uncertainty in W = (0.1/8.0) x 100 = 1.25% (accept 1.3%)
(a) Answer: L: 0.83%; W: 1.25%
(b) M1 Area = 12.0 x 8.0 = 96.0 cm²
(b) M1 for multiplication, percentage uncertainties are added: %unc(Area) = 0.83% + 1.25% = 2.08% (ft from a)
(b) M1 absolute uncertainty = 96.0 x 0.0208 = 2.0 cm²
(b) A1 Area = 96 ± 2 cm² (cao)
(b) Answer: Area = (96 ± 2) cm²
(c) M1 for a sum, absolute uncertainties are added: uncertainty in (L+W) = 0.1 + 0.1 = 0.2 cm
(c) A1 multiplying by the constant 2 also multiplies the absolute uncertainty by 2: uncertainty in P = 0.4 cm (cao)
(c) Answer: ± 0.4 cm (P = 40.0 ± 0.4 cm)

Question 6

(a) B1 2.0% (cao)
(a) Answer: 2.0%
(b) M1 for a cubed quantity, the percentage uncertainty is multiplied by 3: %unc(V) = 3 x %unc(r)
(b) A1 6.0% (ft from a)
(b) Answer: 6.0%
(c) M1 V = (4/3) x π x (2.50)³ = 65.4 cm³
(c) M1 absolute uncertainty = 65.4 x 0.060 = 3.9 cm³ (ft from b)
(c) A1 V = (65 ± 4) cm³ (awrt, appropriate s.f.)
(c) Answer: V = (65 ± 4) cm³

Question 7

(a) B1 a systematic error affects all readings by the same amount and in the same direction (consistently too high or too low), and is not reduced by repeating/averaging
(a) B1 a random error causes readings to scatter unpredictably above and below the true value, and its effect can be reduced by repeating readings and averaging
(a) Answer: Systematic errors shift every reading consistently in the same direction and cannot be reduced by averaging; random errors scatter readings unpredictably and can be reduced by taking repeats and averaging.
(b) B1 this is a systematic (zero) error
(b) B1 the student should subtract 0.20 N from every subsequent reading (or re-zero/calibrate the instrument before use)
(b) Answer: It is a systematic (zero) error; the student should subtract 0.20 N from every reading taken (or zero the instrument before use).
(c) M1 the absolute uncertainty from human reaction time (e.g. approximately ± 0.2 s) when starting/stopping the stopwatch is essentially the same whichever quantity is timed
(c) M1 timing 20 oscillations gives a total time about 20 times larger than timing 1 oscillation, while the absolute uncertainty stays roughly fixed
(c) A1 so the percentage uncertainty (absolute uncertainty / measured value) is reduced by a factor of about 20 compared with timing a single oscillation
(c) Answer: The fixed reaction-time uncertainty becomes a much smaller fraction of the (about 20 times larger) total time, so the percentage uncertainty in the period is roughly 20 times smaller than if a single oscillation were timed.

Question 8

(a) B1 the wire's diameter is not perfectly uniform along its length, so a single reading is subject to random variation
(a) B1 taking a mean of several readings reduces the effect of this random variation, giving a more precise/representative value of the diameter
(a) Answer: A wire is not perfectly uniform in thickness, so measuring at several points and averaging reduces the effect of random variation and gives a more precise mean diameter.
(b) M1 converts diameter to metres: d = 3.2 x 10^-4 m
(b) M1 correct substitution and evaluation: A = π x (3.2 x 10^-4)² / 4 = 8.04 x 10^-8 m²
(b) A1 8.0 x 10^-8 m² (awrt, to 2 s.f., matching the precision of d)
(b) Answer: A = 8.0 x 10^-8 m²
(c) M1 absolute uncertainty in d = half the resolution = ± 0.005 mm
(c) M1 percentage uncertainty in d = (0.005/0.32) x 100 = 1.6%
(c) A1 since A depends on d², percentage uncertainty in A = 2 x 1.6% = 3.1% (ft)
(c) Answer: 3.1%
(d) M1 percentage uncertainty in R = (0.05/2.40) x 100 = 2.1%
(d) M1 percentage uncertainty in l = (0.001/0.800) x 100 = 0.13%
(d) A1 for a product/quotient, percentage uncertainties add: %unc(ρ) = 2.1% + 3.1% + 0.13% = 5.3% (ft from c)
(d) Answer: 5.3%
(e) M1 correct substitution: ρ = (2.40 x 8.04 x 10^-8) / 0.800
(e) M1 ρ = 2.41 x 10^-7 ohm m (or awrt, ft from b)
(e) M1 absolute uncertainty = 2.41 x 10^-7 x 0.053 = 1.3 x 10^-8 ohm m (ft from d)
(e) A1 ρ = (2.4 ± 0.1) x 10^-7 ohm m (awrt, appropriate s.f.)
(e) Answer: ρ = (2.4 ± 0.1) x 10^-7 ohm m
(f) L3 (5-6): A full, coherent evaluation that correctly identifies the diameter measurement as the dominant source of uncertainty (because it is doubled when calculating the area), proposes at least two valid, correctly justified improvements linked to specific measurements in the experiment, and reaches a clear, justified conclusion about which change would have the greatest effect.
(f) L2 (3-4): The answer identifies at least one valid improvement and links it correctly to a reduction in percentage uncertainty, but the evaluation is incomplete, lacks full justification, or does not identify the dominant source of uncertainty.
(f) L1 (1-2): A basic suggestion for improving the experiment (e.g. use better equipment) with little or no justification linking it to percentage uncertainty.
(f) L0 (0): No relevant content, or the answer does not relate to the question.
(f) Answer: Improving the diameter measurement (e.g. finer-resolution micrometer, more readings at different points/orientations, averaging) gives the greatest reduction in overall percentage uncertainty, since the diameter's percentage uncertainty is doubled in the area calculation and is the dominant contributor (3.1% of the total 5.3%).

Question 9

(a) B1 fill the displacement can with water until it overflows from the side spout, and allow the dripping to stop before starting
(a) B1 place an empty measuring cylinder under the spout, then lower the stone gently into the can on a thread until fully submerged, without touching the sides or base
(a) B1 the volume of water collected in the measuring cylinder equals the volume of the stone
(a) Answer: Fill the eureka can to the spout and let it stop dripping; place a measuring cylinder under the spout; lower the stone in on a thread until fully submerged; the volume of water collected equals the stone's volume.
(b) M1 radius = diameter/2 = 0.75 cm
(b) M1 correct substitution and evaluation: V = π x (0.75)² x 4.00
(b) A1 7.07 cm³ (awrt, cao)
(b) Answer: V = 7.07 cm³
(c) M1 density = mass/volume = 62.3/7.07 = 8.81 g/cm³ (ft from b)
(c) M1 converts g/cm³ to kg/m³ by multiplying by 1000
(c) A1 8810 kg/m³ (awrt, accept 8800-8820 kg/m³)
(c) Answer: 8810 kg/m³
(d) M1 percentage uncertainty in diameter = (0.02/1.50) x 100 = 1.33%
(d) M1 percentage uncertainty in height = (0.05/4.00) x 100 = 1.25%
(d) M1 since V depends on d² x h, percentage uncertainty is doubled for the diameter term: %unc(V) = (2 x 1.33%) + 1.25%
(d) A1 3.9% (awrt)
(d) Answer: 3.9%
(e) M1 percentage uncertainty in mass = (0.1/40.2) x 100 = 0.25%
(e) M1 percentage uncertainty in volume = (0.5/15.0) x 100 = 3.3%
(e) A1 percentage uncertainty in density = 0.25% + 3.3% = 3.6% (ft)
(e) B1 the volume measurement contributes the greater uncertainty, since 3.3% is much larger than 0.25%
(e) Answer: Percentage uncertainty in density = 3.6%; the volume (displacement) measurement contributes the greater uncertainty.

Question 10

(a) M1 uncertainty = half the difference between the steepest and least steep gradients = (26.0 - 23.0)/2 = 1.5 N/m
(a) M1 percentage uncertainty = (1.5/24.5) x 100
(a) A1 6.1% (awrt)
(a) Answer: 6.1%
(b) B1 central value stated as 24.5 N/m
(b) B1 uncertainty of 1.5 N/m quoted to the same precision (1 d.p.) as the central value: 24.5 ± 1.5 N/m
(b) Answer: 24.5 ± 1.5 N/m
(c) B1 each error bar shows the range of uncertainty in the measured value of that data point
(c) B1 the steepest and least steep straight lines that still pass through every error bar are drawn on the graph, and the gradient of each is calculated
(c) B1 the uncertainty in the gradient is half the difference between these two gradients
(c) Answer: Error bars show the uncertainty range on each point; drawing the steepest and shallowest lines that still pass through all the error bars, and taking half the difference between their gradients, gives the uncertainty in the gradient.
(d) B1 using many data points averages out the effect of random error/uncertainty in any one reading oe
(d) B1 any anomalous point is more easily identified/its effect is reduced, giving a more representative/reliable value of k than relying on a single pair of readings
(d) Answer: The gradient uses many data points, so random errors and any single anomalous reading have much less effect, giving a more reliable value of k than a single F-x pair.

Question 11

(a) M1 T = (time for 20 oscillations) / 20 = 37.0/20
(a) A1 1.85 s (cso, answer given)
(a) Answer: T = 1.85 s
(b) M1 percentage uncertainty in total time = (0.4/37.0) x 100 = 1.1%
(b) M1 dividing by the constant 20 to find T does not change the percentage uncertainty
(b) A1 percentage uncertainty in T = 1.1% (ft)
(b) Answer: 1.1%
(c) B1 2.2% (ft from b, doubled for the squared term)
(c) Answer: 2.2%
(d) M1 for a product/quotient, percentage uncertainties add: %unc(g) = %unc(l) + %unc(T²)
(d) A1 2.4% (awrt, ft from c)
(d) Answer: 2.4%
(e) M1 correct substitution: g = 4 x π² x 0.850 / (1.85)²
(e) M1 g = 9.80 m/s² (awrt)
(e) M1 absolute uncertainty = 9.80 x 0.024 = 0.24 m/s² (ft from d)
(e) A1 g = (9.80 ± 0.24) m/s² (awrt, appropriate s.f.; accept (9.8 ± 0.2) m/s²)
(e) Answer: g = (9.80 ± 0.24) m/s²
(f) B1 the range given by the result is 9.80 - 0.24 = 9.56 m/s² to 9.80 + 0.24 = 10.04 m/s² (ft from e)
(f) B1 since 9.81 m/s² lies within this range, the result is consistent with the accepted value
(f) Answer: Yes; 9.81 m/s² lies within the range 9.56 m/s² to 10.04 m/s², so the result is consistent with the accepted value.

Question 12

Level 3 (5-6): A full, coherent evaluation that calculates or clearly compares the percentage uncertainty in the extension for both methods, correctly identifies extension as the measurement that dominates the overall percentage uncertainty in the Young modulus for both students, and reaches a justified conclusion (with a valid supporting reason such as the use of a reference wire) that Student B's method is more precise.
Level 2 (3-4): The answer compares the resolutions or uncertainties of the two measuring instruments and reaches a conclusion about which method is more precise, but the reasoning is incomplete, lacks a numerical comparison of percentage uncertainty, or omits reference to the small size of the extension being the limiting factor.
Level 1 (1-2): A basic statement identifying one method as better (e.g. Searle's apparatus is more precise) with little or no supporting justification.
Level 0 (0): No relevant content, or the answer does not relate to the question.
Indicative content:
- The extension itself is small (about 2 mm) in both experiments, so the absolute uncertainty in measuring the extension is the critical factor in the overall percentage uncertainty of the Young modulus.
- Student A's metre rule has a resolution of 1 mm, giving an absolute uncertainty of about 0.5 mm on an extension of about 2 mm, a percentage uncertainty of roughly 25%.
- Student B's vernier/micrometer scale has a resolution of 0.01 mm, giving an absolute uncertainty of about 0.005-0.01 mm on a similar extension, a percentage uncertainty of only around 0.5%.
- Because the extension has by far the largest percentage uncertainty of any quantity in the Young modulus calculation, it dominates the overall percentage uncertainty in E for both students, but far more severely for Student A.
- Searle's apparatus also uses a reference wire under the same conditions (temperature, sag) as the test wire, reducing systematic errors that would otherwise arise from thermal expansion or the wire settling under load.
- Conclusion: Student B's method (Searle's apparatus) gives a more precise value of the Young modulus, because its much finer resolution greatly reduces the percentage uncertainty in the extension, the limiting measurement in this experiment.

Measurements and their Errors - Worksheets, Questions and Revision