Edexcel International A Level Physics · Unit 6 · Practical Skills in Physics II
Uncertainties and logarithmic graphs (A2)
28 free, high-quality past-paper-style questions on Uncertainties and logarithmic graphs (A2) for Edexcel IAL Physics, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Physics notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.28 new this month · 171 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked
1IALP-U6-2-B-01Unit 6 · A2Easy4 marks
A pointer is fixed to the bottom of a spring hanging beside a vertical metre rule. Before a load is added the pointer is at 236 mm; with the load it is at 318 mm. Each reading has an uncertainty of ± 1 mm.
Figure: pointer positions against the vertical metre rule (readings 236 mm and 318 mm)
(a)
Calculate the extension of the spring.
(1)
(b)
State the absolute uncertainty in the extension.
(1)
(c)
Calculate the percentage uncertainty in the extension.
(1)
(d)
Explain why this is much larger than the percentage uncertainty in the reading of 318 mm.
(1)
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Answer
Mark
Notes
Part (a)
318 − 236 = 82 mm
✓ 1
Part (b)
the absolute uncertainties add for a difference: 1 + 1 = ± 2 mm
✓ 1
Part (c)
2 ÷ 82 × 100 = 2.4%
✓ 1
Part (d)
the extension is a small difference between two larger readings: the uncertainty (doubled) is compared with 82 mm rather than 318 mm, which alone gives only 0.31%
✓ 1
Common error: Subtracting the uncertainties for a difference (1 − 1 = 0). For sums and differences the absolute uncertainties always add.
A student measures a solid brass cylinder with vernier calipers and a balance: diameter d = 12.0 mm ± 0.1 mm; length h = 25.4 mm ± 0.1 mm; mass m = 24.3 g ± 0.1 g. The density is ρ = 4m / πd2h.
(a)
Calculate the density of the brass in kg m−3.
(2)
(b)
Calculate the percentage uncertainty in the density.
(2)
(c)
A data book gives 8.60 × 103 kg m−3 for this brass. Decide whether the student's value agrees with it.
(2)
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Answer
Mark
Notes
Part (a)
V = π × 1.202 × 2.54 ÷ 4 = 2.87 cm3
✓ 1
ρ = 24.3 ÷ 2.873 = 8.46 g cm−3 = 8.46 × 103 kg m−3
A student finds the density of a small stone. She reads the water level in a measuring cylinder (Figure 1(a)), lowers the stone in so that it is completely covered (Figure 1(b)) and reads the level again. Each reading has an uncertainty of ± 0.5 cm3. The mass of the stone is 68.4 g ± 0.1 g.
(a)
Use the figures to find the volume of the stone.
(1)
(b)
State the absolute uncertainty in the volume, explaining how you obtained it.
(1)
(c)
Calculate the percentage uncertainty in the volume.
(1)
(d)
Calculate the density of the stone and give it with its absolute uncertainty.
(2)
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Answer
Mark
Notes
Part (a)
67 − 42 = 25 cm3
✓ 1
Part (b)
the volume is a difference of two readings, so their absolute uncertainties add: 0.5 + 0.5 = ± 1 cm3
✓ 1
Part (c)
1 ÷ 25 × 100 = 4.0%
✓ 1
Part (d)
ρ = 68.4 ÷ 25 = 2.7 g cm−3 (2.736)
✓ 1
percentage uncertainty = 4.0 + 0.15 = 4.1%, which is 0.11 g cm−3, so (2.7 ± 0.1) g cm−3
✓ 1
Exam tip: Find the absolute uncertainty of a difference first, then turn it into a percentage before combining it with the uncertainty of a quotient.
Four students each measure the speed of sound in air using a different method. The figure shows each result with its uncertainty, together with the accepted value of 340 m s−1 for the room temperature.
(a)
State which students' results agree with the accepted value.
(1)
(b)
For student B, calculate the percentage uncertainty and the percentage difference from the accepted value, and use them to explain whether the result agrees.
(2)
(c)
Student D has the most precise result, but it does not agree with the accepted value. Suggest what this indicates about D's method.
(1)
(d)
Explain why student C's result, although it agrees with the accepted value, is the least useful.
(1)
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Answer
Mark
Notes
Part (a)
A and C: their ranges (327–343 and 312–344 m s−1) include 340 m s−1
3.5% is greater than 1.1%, so B does not agree with the accepted value
✓ 1
Part (c)
a systematic error (e.g. a zero error, or a distance consistently measured to the wrong point) has shifted all of D's readings in one direction; repeating does not remove it
✓ 1
Part (d)
its uncertainty is ± 16 m s−1 (about 5%), the largest, so it is imprecise: almost any reasonable value would “agree” with it
✓ 1
Exam tip: A result agrees with an accepted value when the accepted value lies inside its uncertainty range; equivalently, when the percentage difference is smaller than the percentage uncertainty.
A student finds the Young modulus of a metal wire using E = 4FL / πd2x. Her readings are: load F = 40.0 N, original length L = 2.000 m, diameter d = 0.50 mm and extension x = 2.00 mm. The bar chart shows the percentage uncertainty in each of these measurements.
(a)
Explain why the diameter contributes 2.0% to the percentage uncertainty in E, not 1.0%.
(1)
(b)
Calculate the percentage uncertainty in E.
(1)
(c)
Calculate E and give it with its absolute uncertainty.
(2)
(d)
State which measurement the student should improve first, and suggest how.
(1)
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Answer
Mark
Notes
Part (a)
d is squared in the equation, so its percentage uncertainty is multiplied by 2: 2 × 1.0% = 2.0%
5.1% of 2.04 × 1011 = 0.10 × 1011 Pa, so (2.0 ± 0.1) × 1011 Pa
✓ 1
Part (d)
the extension (largest contribution, 2.5%): use a longer wire or a larger load so the extension is bigger, or measure it with a vernier scale
✓ 1
Common error: Adding the bar heights as they stand (0.1 + 1.0 + 0.5 + 2.5 = 4.1%). Any quantity raised to a power must be multiplied by that power first.
Light from a laser passes through a double slit of separation s = 0.30 mm ± 0.01 mm. Fringes are seen on a screen at a distance D = 2.40 m ± 0.01 m from the slits. The figure shows the fringes against a millimetre scale; the centre of each fringe can be judged to ± 0.5 mm. The wavelength is given by λ = ws/D, where w is the fringe spacing.
(a)
Use the figure to find the distance from the centre of fringe 1 to the centre of fringe 11, with its absolute uncertainty.
(2)
(b)
Calculate the fringe spacing w and its percentage uncertainty.
(1)
(c)
Calculate the wavelength of the light.
(1)
(d)
Calculate the percentage uncertainty in the wavelength.
(1)
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Answer
Mark
Notes
Part (a)
62 − 14 = 48 mm
✓ 1
difference of two readings: 0.5 + 0.5 = ± 1 mm
✓ 1
Part (b)
10 spacings: w = 48 ÷ 10 = 4.8 mm ± 0.1 mm; 1 ÷ 48 × 100 = 2.1%
percentage difference = (1124 − 900) ÷ 900 × 100 = 25%, much greater than 6.3%, so it does not agree
✓ 1
systematic error: energy is lost to the surroundings, so the temperature rise is smaller than it should be and c comes out too large — insulate the block
✓ 1
Exam tip: Deal with the difference first (add absolute uncertainties), then convert it to a percentage and add it to the other percentages.
A capacitor is charged to different potential differences V and the charge Q on it is measured each time with a coulombmeter. Each value of Q has an uncertainty of ± 2 μC.
V / V
Q / μC
1.0
22
2.0
45
3.0
66
4.0
89
5.0
110
6.0
133
A graph of Q against V is plotted with error bars. The best-fit line passes through (1.0 V, 22.3 μC) and (6.0 V, 132.7 μC). The steepest line that passes through all the error bars joins (1.0 V, 20 μC) to (6.0 V, 135 μC); the shallowest joins (1.0 V, 24 μC) to (6.0 V, 131 μC).Figure: Q against V with error bars of ± 2 μC, best-fit, steepest and shallowest lines
(a)
State what the gradient of the graph represents.
(1)
(b)
Calculate the gradient of the best-fit line.
(1)
(c)
Use the steepest and shallowest lines to find the uncertainty in the capacitance.
(2)
(d)
The capacitor is labelled 22 μF ± 10%. Comment on whether the result is consistent with the label.
A Geiger–Müller tube is placed near a sample of a short-lived radioactive isotope. Before the sample is brought in, the background count rate is measured as 0.40 s−1. The total count rate C is then measured at intervals.
t / s
C / s−1
0
12.4
50
7.85
100
4.95
150
3.25
200
2.13
250
1.49
300
1.06
Figure: total count rate against time, with the background level
(a)
Explain why the background count rate must be subtracted before the data are analysed.
(1)
(b)
Calculate ln (A / s−1) for t = 150 s, where A is the corrected count rate. Give your answer to an appropriate number of decimal places.
(2)
(c)
A graph of ln (A / s−1) against t is a straight line through (0 s, 2.487) and (300 s, −0.409). Determine the decay constant.
(2)
(d)
Calculate the half-life of the isotope.
(1)
(e)
State how the graph shows that the decay is exponential.
(1)
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Answer
Mark
Notes
Part (a)
only the count rate from the sample decays exponentially; the background is a constant extra amount, and leaving it in would make the ln graph curve
✓ 1
Part (b)
A = 3.25 − 0.40 = 2.85 s−1
✓ 1
ln 2.85 = 1.047 (3 d.p., as 2.85 has 3 s.f.)
✓ 1
Part (c)
gradient = (−0.409 − 2.487) ÷ 300 = −0.00965 s−1
✓ 1
λ = −gradient = 9.65 × 10−3 s−1
✓ 1
Part (d)
ln 2 ÷ 9.65 × 10−3 = 72 s (71.8 s)
✓ 1
Part (e)
ln A = ln A0 − λt is linear in t, so a straight line (with negative gradient) shows A = A0e−λt
✓ 1
Common error: Taking logs of the total count rate. Correct for background first; otherwise the points curve away from a straight line at long times.
In a darkened room, a light meter measures the intensity I at different distances d from a small lamp.
d / m
I / W m−2
0.100
86.6
0.150
39.1
0.200
22.3
0.300
10.1
0.400
5.7
0.500
3.7
The student suspects that I = kdn and plots log (I / W m−2) against log (d / m).
(a)
Calculate log (d / m) and log (I / W m−2) for d = 0.300 m.
(1)
(b)
The best-fit line passes through (−1.000, 1.938) and (−0.300, 0.566). Calculate n.
(2)
(c)
The steepest and shallowest lines through the error bars have gradients −2.06 and −1.87. Decide whether the data support an inverse-square law.
(2)
(d)
Suggest one reason why n might not be exactly −2 in this experiment.
(1)
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Answer
Mark
Notes
Part (a)
log 0.300 = −0.523; log 10.1 = 1.004
✓ 1
Part (b)
gradient = (0.566 − 1.938) ÷ (−0.300 − (−1.000))
✓ 1
= −1.372 ÷ 0.700 = −1.96
✓ 1
Part (c)
uncertainty = (2.06 − 1.87) ÷ 2 = ± 0.10 (0.095), so n = −1.96 ± 0.10
✓ 1
the range −2.06 to −1.86 includes −2, so the data are consistent with an inverse-square law
✓ 1
Part (d)
the lamp is not a point source at short distances / distance measured to the wrong point of the lamp or sensor (systematic) / reflections from walls or background light add to I
✓ 1
Model sketch
Figure: log I against log d: straight line, gradient n ≈ −2
Exam tip: On a log–log graph the gradient is the power and has no unit. Both axes must use the same base.
A straight horizontal wire carrying a current I lies at right angles to the uniform field between two magnets resting on a top-pan balance. The force F on the wire is found from the change in the balance reading. The length of wire in the field is L = 5.0 cm ± 0.1 cm. The graph shows F against I with error bars, a best-fit line and the steepest and shallowest lines that pass through all the error bars.
(a)
Explain why the gradient of the graph is equal to BL, where B is the magnetic flux density.
(1)
(b)
Determine the gradient of the best-fit line.
(2)
(c)
Use the steepest and shallowest lines to find the percentage uncertainty in the gradient.
(2)
(d)
Calculate B and its absolute uncertainty.
(2)
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Answer
Mark
Notes
Part (a)
F = BIL, so F is proportional to I with gradient F/I = BL
uncertainty = (3.17 − 2.83) ÷ 2 = 0.17 mN A−1, which is 0.17 ÷ 3.00 × 100 = 5.6%
✓ 1
Part (d)
B = 3.00 × 10−3 ÷ 0.050 = 0.060 T
✓ 1
percentage uncertainty = 5.6 + 2.0 = 7.6%, so B = (0.060 ± 0.005) T
✓ 1
Exam tip: The uncertainty in a gradient is a percentage like any other: add it to the percentage uncertainties of the other quantities you combine it with.
A card is fixed to the bob of a pendulum so that air resistance makes the oscillations die away. The amplitude A is measured every 10 s with a scale behind the pendulum, to the nearest 0.1 cm. It is suggested that A = A0e−kt. The graph shows ln (A / cm) against time t, with a best-fit line.
(a)
At t = 30 s the amplitude was 6.5 cm. Calculate the value of ln (A / cm) that was plotted, to an appropriate number of decimal places.
(1)
(b)
Explain how the graph supports the suggested relationship.
(1)
(c)
Determine k, giving its unit.
(2)
(d)
Determine the initial amplitude A0.
(1)
(e)
Calculate the time taken for the amplitude to halve.
(1)
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Answer
Mark
Notes
Part (a)
ln 6.5 = 1.87 (2 d.p., as 6.5 cm has 2 s.f.)
✓ 1
Part (b)
ln A = ln A0 − kt, so the relationship predicts a straight line of gradient −k; the points lie close to a straight line
✓ 1
Part (c)
gradient = (1.28 − 2.48) ÷ 60 = −0.0200 s−1
✓ 1
k = 0.0200 s−1 (allow 0.0195–0.0205)
✓ 1
Part (d)
intercept = ln A0 = 2.48, so A0 = e2.48 = 11.9 cm (allow 11.7–12.2)
✓ 1
Part (e)
ln 2 ÷ 0.0200 = 35 s (34.7 s)
✓ 1
Common error: Quoting the intercept 2.48 as the starting amplitude. The intercept is ln A0; undo the log with ex.
Water drains from a burette through a long, narrow glass tube attached to its tap. The graph shows the height h of the water surface above the outlet against time t.
(a)
Use readings from the graph at equal time intervals to show that h decreases exponentially with t.
(2)
(b)
Calculate the decay constant λ in h = h0e−λt.
(1)
(c)
A graph of ln (h / cm) against t is to be plotted from the same data. State the values expected for its gradient and its intercept on the ln (h / cm) axis.
(2)
(d)
Calculate the time for h to fall to 2.5 cm.
(1)
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Answer
Mark
Notes
Part (a)
t = 0, 60, 120, 180 s give h = 40, 20, 10, 5 cm
✓ 1
in each equal interval of 60 s, h falls by the same factor (½): a constant ratio in equal times is the property of an exponential decay
✓ 1
Part (b)
λ = ln 2 ÷ 60 = 0.0116 s−1
✓ 1
Part (c)
gradient = −λ = −0.0116 s−1
✓ 1
intercept = ln 40 = 3.69 (3.689)
✓ 1
Part (d)
t = ln (40 ÷ 2.5) ÷ 0.01155 = ln 16 ÷ 0.01155 = 240 s (four halvings of 60 s)
✓ 1
Exam tip: Without plotting logs you can test for an exponential by checking that equal time intervals give the same ratio of values, not the same difference.
A steel wire is stretched between two bridges with a tension of 40.0 N. The frequency f of its fundamental vibration is measured for different lengths L between the bridges. It is suggested that f = kLn. The graph shows log (f / Hz) against log (L / m), with a best-fit line.
(a)
Explain how n and k can be found from this graph.
(2)
(b)
Determine n.
(2)
(c)
Determine k, giving its unit.
(1)
(d)
For the fundamental, f = v/2L, where v is the speed of transverse waves on the wire. Use your value of k to find the mass per unit length μ of the wire.
(2)
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Answer
Mark
Notes
Part (a)
log f = n log L + log k, so the gradient is n
✓ 1
the value of log f where log L = 0 is log k, so k = 10intercept
✓ 1
Part (b)
gradient = (1.68 − 2.28) ÷ (0.00 − (−0.60))
✓ 1
= −1.00 (allow −0.96 to −1.04)
✓ 1
Part (c)
log k = 1.68, so k = 101.68 = 47.9 Hz m (allow 46.8–49.0)
A string is set vibrating by a vibration generator at 50.0 Hz ± 0.5 Hz. It passes over a pulley and supports a hanging mass of 0.200 kg ± 0.001 kg. The figure shows the stationary wave on the string and a metre rule fixed above it. The position of each node can be judged to ± 0.5 cm. g = 9.81 N kg−1.
(a)
Use the figure to find the wavelength of the wave on the string and its percentage uncertainty.
(2)
(b)
Calculate the speed of the waves on the string and its percentage uncertainty.
(2)
(c)
The speed is given by v = √(T/μ), where T is the tension and μ the mass per unit length. Calculate μ and its percentage uncertainty.
(2)
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Answer
Mark
Notes
Part (a)
distance between the end nodes = 105.0 − 15.0 = 90.0 cm ± 1.0 cm; three loops are 1.5 wavelengths, so λ = 90.0 ÷ 1.5 = 60.0 cm
Exam tip: Measure across as many loops as possible: the ± 1.0 cm uncertainty is then shared over a longer distance, and dividing by the exact number of loops does not change the percentage.
A sealed plastic container is pumped up and then leaks slowly through a pinhole. The excess pressure Δp above atmospheric pressure is expected to follow Δp = Δp0e−kt. A student plots the graph shown, using logarithms to base 10.
(a)
Show that the graph should be a straight line and express its gradient in terms of k.
(2)
(b)
Determine k.
(2)
(c)
Determine the initial excess pressure Δp0.
(1)
(d)
Calculate the time taken for the excess pressure to halve.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
ln Δp = ln Δp0 − kt; dividing by ln 10 = 2.303 gives log Δp = log Δp0 − (k/2.303)t, a straight line against t
✓ 1
gradient = −k/2.303 (= −0.434k)
✓ 1
Part (b)
gradient = (1.04 − 1.90) ÷ 40 = −0.0215 min−1
✓ 1
k = 2.303 × 0.0215 = 0.0495 min−1 (allow 0.048–0.051)
A load of mass m oscillates on a spring. Because the spring itself moves, the period is expected to follow T2 = (4π2/k)(m + me), where k is the stiffness and me is a constant “effective mass” of the spring. Each T2 value has an uncertainty of ± 0.008 s2.
m / kg
T2 / s2
0.050
0.112
0.100
0.188
0.150
0.270
0.200
0.345
0.250
0.427
0.300
0.504
On a graph of T2 against m: the best-fit line passes through (0.050 kg, 0.111 s2) and (0.300 kg, 0.504 s2); the steepest line through the error bars passes through (0.050 kg, 0.104 s2) and (0.300 kg, 0.512 s2); the shallowest through (0.050 kg, 0.120 s2) and (0.300 kg, 0.496 s2).Figure: T² against m with error bars of ± 0.008 s², best-fit, steepest and shallowest lines
(a)
Explain how k and me can be found from the graph.
(2)
(b)
Determine k from the best-fit line.
(2)
(c)
Use the steepest and shallowest lines to find the uncertainty in k.
(1)
(d)
Determine me from the best-fit line and comment on its reliability.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
gradient = 4π2/k, so k = 4π2 ÷ gradient
✓ 1
T2 = 0 when m = −me: the line meets the m-axis at −me
gradients 1.632 and 1.504 give k = 24.2 and 26.2 N m−1, so ± 1.0 N m−1
✓ 1
Part (d)
intercept on the m-axis = 0.050 − 0.111 ÷ 1.572 = −0.021 kg, so me = 0.021 kg
✓ 1
the steepest and shallowest lines give 0.014 kg and 0.030 kg, about ± 0.008 kg (nearly 40%): the intercept is far less certain than the gradient because it is an extrapolation
✓ 1
Exam tip: To find an intercept on the x-axis, set y = 0 in y = mx + c: x = −c/m. Do it for the best, steepest and shallowest lines.
Hot water in a beaker cools in a room at 22.0 °C. A student records its temperature θ every 3 minutes.
t / min
θ / °C
0
80.0
3
71.8
6
65.1
9
58.9
12
53.9
15
49.3
18
45.7
It is suggested that the temperature excess above the room, θ − 22.0 °C, decreases exponentially with time.Figure: the temperature readings, with the room temperature shown dashed
(a)
Explain why ln (θ − 22.0) rather than ln θ should be plotted against t.
(2)
(b)
Calculate ln (θ − 22.0) for t = 9 min, where θ is in °C.
(1)
(c)
The graph is a straight line through (0 min, 4.059) and (18 min, 3.162). Determine the constant k.
(2)
(d)
Calculate the time taken for the temperature excess to halve.
(1)
(e)
Use the graph to predict the time at which the water reaches 30.0 °C, and comment on the reliability of this prediction.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
the exponential relationship applies to the excess temperature: (θ − 22.0) = Δθ0e−kt, so ln (θ − 22.0) = ln Δθ0 − kt, a straight line
✓ 1
θ itself tends to 22.0 °C, not zero, so ln θ against t would not be a straight line
✓ 1
Part (b)
ln (58.9 − 22.0) = ln 36.9 = 3.608
✓ 1
Part (c)
gradient = (3.162 − 4.059) ÷ 18 = −0.0498 min−1
✓ 1
k = 0.0498 min−1
✓ 1
Part (d)
ln 2 ÷ 0.0498 = 13.9 min
✓ 1
Part (e)
ln 8.0 = 2.079 = 4.059 − 0.0498t, so t = 40 min; this is an extrapolation well beyond the 18 min of data, so it is less reliable
✓ 1
Common error: Taking logs of the temperature in °C. Subtract the room temperature first: it is the excess that decays towards zero.
The power P supplied to a small lamp is measured at several filament temperatures T, the temperatures being found from the filament's resistance.
T / K
P / W
1200
2.70
1500
6.20
1800
12.2
2100
21.6
2400
35.4
If all the energy left the filament as radiation from a constant area, P = kT4 would be expected.
(a)
Explain why a graph of log P against log T is a better choice than a graph of P against T4 for testing whether P = kT4.
(2)
(b)
Calculate log (P / W) for T = 1800 K.
(1)
(c)
The best-fit line passes through (3.080, 0.435) and (3.380, 1.549), using log (T / K) and log (P / W). The steepest and shallowest acceptable lines have gradients 3.85 and 3.57. Determine n with its uncertainty.
(2)
(d)
Discuss whether the data support P = kT4, and suggest a reason for any difference.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
if P = kTn, log P = n log T + log k, so the gradient gives the power n directly
✓ 1
a P–T4 graph assumes the power is 4; the log–log graph tests it (and gives n with an uncertainty)
✓ 1
Part (b)
log 12.2 = 1.086
✓ 1
Part (c)
n = (1.549 − 0.435) ÷ (3.380 − 3.080) = 3.71
✓ 1
uncertainty = (3.85 − 3.57) ÷ 2 = ± 0.14, so n = 3.71 ± 0.14
✓ 1
Part (d)
the range 3.57 to 3.85 does not include 4 (percentage difference 7%, uncertainty 3.8%), so the data do not support P ∝ T4
✓ 1
some energy also leaves by conduction through the supports and leads (more important at low temperature) / the temperatures found from resistance have a systematic error
✓ 1
Model sketch
Figure: log P against log T: straight line, gradient n
Exam tip: When theory predicts a power, quote your n with its uncertainty and check whether the predicted value lies inside that range.
A plastic ruler is clamped to a bench with a length L overhanging. The free end is flicked and the frequency f of its vibration is measured with a microphone and data logger for eight values of L. It is suggested that f = kLn, where k and n are constants.
(a)
Explain how the student should process and plot the data to test this relationship and to find n and k, including how to estimate the uncertainty in n.
(6)
Show mark scheme
Answer
Mark
Notes
Indicative content:
—
take logs: log f = n log L + log k
✓ 1
calculate log (f / Hz) and log (L / m) for each reading, to as many decimal places as the data have significant figures
✓ 1
plot log f on the y-axis against log L on the x-axis; if the points lie on a straight line, the relationship is supported
✓ 1
n = gradient, found from a large triangle on the best-fit line
✓ 1
log k = log f − n log L for a point on the line (the intercept at log L = 0), and k = 10log k
✓ 1
add error bars, draw the steepest and shallowest lines through them, and take half the difference of their gradients as the uncertainty in n
✓ 1
Exam tip: In a 6-mark explanation, each step of the analysis is a separate point: the log equation, the processed columns, the axes, the gradient, the constant and the uncertainty.
In a ripple tank, the speed v of long-wavelength surface waves is measured for different water depths d. For shallow water the theory predicts v = √(gd). The graph shows ln (v / m s−1) against ln (d / m), with a best-fit line. Error bars (not shown) give steepest and shallowest acceptable lines with gradients 0.53 and 0.47.
(a)
Show that the theory predicts a straight line of gradient 0.5 and state what the intercept on the ln (v / m s−1) axis would represent.
(2)
(b)
Determine the gradient of the best-fit line.
(1)
(c)
Use the steepest and shallowest gradients to decide whether the data support the predicted power of ½.
(1)
(d)
The intercept cannot be read directly from the graph. Use the best-fit line to determine a value for g.
(2)
(e)
Calculate the percentage difference between this value and 9.81 m s−2 and suggest one reason why the experimental value is lower.
(1)
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Answer
Mark
Notes
Part (a)
v2 = gd, so 2 ln v = ln g + ln d and ln v = ½ ln d + ½ ln g: straight line, gradient 0.5
gradient = 0.50 ± 0.03, range 0.47 to 0.53, which includes 0.5, so the data support v ∝ d½
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Part (d)
c = y − mx = −0.28 − 0.500 × (−2.80) = 1.12
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½ ln g = 1.12, so g = e2.24 = 9.4 m s−2 (9.39; allow 9.0–9.8, or ECF from a gradient within 0.49–0.51)
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Part (e)
(9.81 − 9.39) ÷ 9.81 × 100 = 4.3%; e.g. every depth was measured slightly too large (a systematic error, such as a zero error on the depth gauge): since g = v2/d, too large a d gives too small a g / friction with the floor of the tank slows the waves slightly in such shallow water, so v is a little less than √(gd)
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Common error: Reading the value where the line meets the left-hand edge of the grid as the intercept. That edge is at ln d = −5.6, not 0; use c = y − mx with a point on the line.
A Hall probe measures the magnetic flux density B on the axis of a short, strong bar magnet at distances d from its centre. The student does not know whether B varies exponentially with d or as a power of d, so the same data are plotted in two ways (Figure 1(a) and Figure 1(b)). Error bars (not shown) give steepest and shallowest acceptable lines on Figure 1(b) with gradients −3.12 and −2.86.
(a)
Explain which relationship the graphs support.
(2)
(b)
Use Figure 1(b) to determine n.
(2)
(c)
State n with its uncertainty and say whether it is consistent with an inverse-cube law.
(1)
(d)
Use the line in Figure 1(b) to estimate B at d = 12.0 cm, and comment on how reliable this estimate is.
(2)
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Answer
Mark
Notes
Part (a)
if B = B0e−λd, ln B against d would be straight; Figure 1(a) is curved, so B does not vary exponentially
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if B = kdn, ln B against ln d is straight; Figure 1(b) is straight, so B follows a power law
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Part (b)
gradient = (0.40 − 3.70) ÷ (2.40 − 1.30)
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= −3.30 ÷ 1.10 = −3.00 (allow −2.95 to −3.05)
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Part (c)
uncertainty = (3.12 − 2.86) ÷ 2 = 0.13, so n = −3.00 ± 0.13; the range −3.13 to −2.87 includes −3, so it is consistent
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Part (d)
ln 12.0 = 2.485; ln B = 3.70 − 3.00 × (2.485 − 1.30) = 0.145, so B = e0.145 = 1.16 mT
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12.0 cm lies outside the measured range (4–10 cm), so this is an extrapolation and less reliable; the Earth's field (about 0.05 mT) is also a larger fraction of such a small reading
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Exam tip: To choose between ekx and xn, plot ln y against x and ln y against ln x: the one that gives a straight line identifies the relationship.
The resistance R of a thermistor is measured at eight temperatures T between 283 K and 353 K. The resistance is expected to follow R = R∞eβ/T, where R∞ and β are constants. The ohmmeter readings fluctuate, so each value of R is uncertain by ± 5%. The graph shows ln (R / Ω) against 1/T with error bars, a best-fit line, and steepest and shallowest lines whose gradients are 3.63 × 103 K and 3.37 × 103 K.
(a)
Explain why ln R is plotted against 1/T, and state what the gradient represents.
(1)
(b)
Show that a ± 5% uncertainty in R gives error bars of about ± 0.05 on ln (R / Ω).
(1)
(c)
Determine β from the best-fit line.
(2)
(d)
The supplier gives β = 3.43 × 103 K. Decide whether the student's value agrees with this.
(2)
(e)
Use the best-fit line to find the resistance of the thermistor when 1/T = 3.00 × 10−3 K−1.
(1)
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Answer
Mark
Notes
Part (a)
ln R = ln R∞ + β(1/T), which is a straight line against 1/T with gradient β
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Part (b)
ln (1.05R) − ln R = ln 1.05 = 0.049 (and ln 0.95 = −0.051), so about ± 0.05 whatever the value of R
A new cell labelled 1.5 V is connected to a variable resistor and an ammeter, with a voltmeter across the cell. The graph shows the terminal potential difference V against the current I, with error bars, a best-fit line and the steepest and shallowest lines that pass through all the error bars.
(a)
Explain how the e.m.f. ε and the internal resistance r are found from the graph.
(1)
(b)
Determine ε and r from the best-fit line.
(2)
(c)
Use the steepest and shallowest lines to find the absolute uncertainty in r and in ε.
(2)
(d)
Decide whether the value of ε agrees with the 1.5 V on the label, and suggest a reason for your conclusion.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
V = ε − Ir: the intercept on the V axis (at I = 0) is ε and the gradient is −r
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Part (b)
ε = 1.56 V (allow 1.55–1.57 V)
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gradient = (0.44 − 1.56) ÷ 1.40 = −0.80 Ω, so r = 0.80 Ω (allow 0.78–0.82 Ω)
intercepts 1.60 V and 1.52 V, so uncertainty in ε = ± 0.04 V
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Part (d)
percentage difference = (1.56 − 1.50) ÷ 1.50 × 100 = 4.0%, greater than the 2.6% uncertainty (0.04 ÷ 1.56), so it does not agree: 1.50 V lies outside 1.52–1.60 V
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1.5 V is only the nominal value printed on the label; the e.m.f. of a fresh cell can be a little above its nominal value, so a measured value slightly above 1.5 V is not unexpected
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Exam tip: The steepest and shallowest lines give the uncertainty in the intercept as well as in the gradient: read where each meets the axis and halve the difference.
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