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IAL Physics (Edexcel) · Unit 6 · Topic 6.2A2≈ 26 min read

6.2 Uncertainties and logarithmic graphs (A2)

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A2 practical skills (Unit 6, WPH16): the data-analysis half of the unit. Single-reading uncertainties and linearising with y = mx + c are AS skills (3.1); planning, precision and accuracy are on 6.1; the core practicals themselves are on 6.3.

What this page covers

8 points
  • Combining uncertainties: absolute uncertainties for sums and differences
  • Percentage uncertainties for products and quotients, and multiplying by the power
  • Percentage difference from an accepted value, compared with the percentage uncertainty, to judge agreement
  • Error bars, steepest and shallowest gradient lines, and the uncertainty in a gradient and an intercept
  • Natural and base-10 logarithms, the rules you need, and significant figures in logs
  • Plotting ln y against x to test y = Aekx (capacitor discharge, radioactive decay, absorption)
  • Plotting log y against log x to test y = kxn and find n and k
  • Units and significant figures for log columns, axes, gradients and intercepts

In one minute

6 points
  • At A2 you carry uncertainties all the way into the final answer: add absolute uncertainties for sums and differences, add percentage uncertainties for products and quotients, and use the result to judge whether the answer agrees with an accepted value.
  • A quantity raised to a power n contributes n times its percentage uncertainty (a square root is the power ½). A small difference between two large readings can have a large percentage uncertainty.
  • Percentage difference = |experimental − accepted| ÷ accepted × 100%. If it is smaller than the percentage uncertainty, the result agrees with the accepted value within its uncertainty.
  • Draw error bars, then the steepest and shallowest lines that still pass through them. Uncertainty in the gradient ≈ (steepest − shallowest) ÷ 2.
  • y = Aekx → plot ln y against x: gradient k, intercept ln A. y = kxn → plot log y against log x: gradient n, intercept log k.
  • A log has no unit: label axes ln (V / V). Give a log to as many decimal places as the number has significant figures.

Contents

13 sections
  1. How do you combine uncertainties when quantities are added or subtracted?
  2. How do you combine uncertainties for products, quotients and powers?
  3. How do you decide whether a result agrees with an accepted value?
  4. How do you find the uncertainty in a gradient and an intercept?
  5. What do you need to know about logarithms?
  6. How do you test an exponential relationship with a log graph?
  7. How do you test a power law with a log–log graph?
  8. Exam tips for 6.2
  9. Common misconceptions
  10. How this topic connects
  11. Key equations
  12. Check your understanding
  13. Specification checklist

How do you combine uncertainties when quantities are added or subtracted?

At AS you found the uncertainty of one reading at a time. At A2 most results are calculated from several readings, so you need rules for combining uncertainties. These rules give a cautious, worst-case estimate. Start with a result that is a sum or a difference of two measured values. Each reading could be out by its full uncertainty, and the two errors could happen to push the result the same way. That is true even for a subtraction: if the first reading is slightly too high and the second slightly too low, the difference is too large by both amounts together. So the absolute uncertainties add.

Sums and differences. If \(y = a + b\) or \(y = a - b\), then \(\Delta y = \Delta a + \Delta b\), where \(\Delta a\) and \(\Delta b\) are absolute uncertainties in the same unit as a and b. This is a practical-skills rule, not a formula on the data sheet.

Typical cases: a temperature rise, an extension from two scale readings, and a count rate corrected for background. Work out the absolute uncertainty first; turn it into a percentage only if the result goes on into a product or quotient.

The danger is a small difference between two large readings: the absolute uncertainty stays the same size but becomes a much larger fraction of the result. Design the experiment so the difference is as large as practical.

10152025303540455055temperature / °Cstart 18.5 ± 0.5end 46.0 ± 0.5smallest possible rise 26.5 °Clargest possible rise 28.5 °C
The rise can be anywhere from 26.5 °C to 28.5 °C, so it is 27.5 °C ± 1.0 °C: absolute uncertainties add even though the readings are subtracted.
Worked example 1 Easy · Calculate

A block is heated electrically. A thermometer with 1 °C divisions reads 18.5 °C at the start and 46.0 °C at the end; each reading is ± 0.5 °C. Calculate the temperature rise, its absolute uncertainty and its percentage uncertainty, and compare the last with the percentage uncertainty of the final reading alone.

Step 1 The rise is a difference: Δθ = 46.0 − 18.5 = 27.5 °C.

Step 2 Add the absolute uncertainties: 0.5 + 0.5 = ± 1.0 °C.

Step 3 Percentage uncertainty: 1.0 ÷ 27.5 × 100 = 3.6%. The 46.0 °C reading alone is only 0.5 ÷ 46.0 × 100 = 1.1%.

Δθ = 27.5 °C ± 1.0 °C (3.6%), over three times the 1.1% of one reading. Heating for longer would cut the percentage, because the ± 1.0 °C stays the same.

How do you combine uncertainties for products, quotients and powers?

For products and quotients the percentage uncertainties combine. If a is 2% too big and b is 3% too big, ab is about 5% too big. If b is 3% too small and on the bottom of a fraction, a/b is again about 5% too big: uncertainties never cancel.

Products, quotients and powers. If \(y = \dfrac{a^p b^q}{c^r}\), then \[\%y = p(\%a) + q(\%b) + r(\%c)\] with each power used as a positive number. Then turn the total back into an absolute uncertainty: \(\Delta y = y \times \%y \div 100\). This is a practical-skills rule to learn.
  • Powers. A square doubles the percentage uncertainty, a cube triples it and a square root halves it. For small uncertainties \((1+p)^2 \approx 1 + 2p\), which is where the factor comes from.
  • Exact numbers (2, π, ½, a count of 20 oscillations) have no uncertainty and contribute nothing.
  • A constant multiplier changes the value but not the percentage uncertainty: T = (time for 20 swings) ÷ 20 has the same percentage uncertainty as the time.
  • Mixed calculations such as \(c = E \div m\Delta\theta\): absolute uncertainty of the difference first, then its percentage, then add percentages.
  • Rounding. Give the final absolute uncertainty to 1 (or 2) significant figures and the value to the same decimal place.

The quantity with the highest power usually dominates the total, so it deserves the most precise instrument: for a wire, measure the diameter with a micrometer at several places and average.

Worked example 2 Medium · Calculate

For a wire, R = 6.4 Ω ± 0.1 Ω, diameter d = 0.28 mm ± 0.01 mm and length l = 0.850 m ± 0.002 m. Calculate the resistivity ρ = RA/l, where A = πd²/4, and its absolute uncertainty.

Step 1 The value: A = π(0.28 × 10⁻³)² ÷ 4 = 6.16 × 10⁻⁸ m², so ρ = 6.4 × 6.16 × 10⁻⁸ ÷ 0.850 = 4.64 × 10⁻⁷ Ω m.

Step 2 Percentage uncertainty of each measurement: %R = 0.1 ÷ 6.4 × 100 = 1.6%; %d = 0.01 ÷ 0.28 × 100 = 3.6%; %l = 0.002 ÷ 0.850 × 100 = 0.24%.

Step 3 d is squared, so its percentage counts twice; add everything: %ρ = 1.6 + 2 × 3.6 + 0.24 = 8.9%.

Step 4 Back to an absolute uncertainty: Δρ = 8.9% × 4.64 × 10⁻⁷ = 0.41 × 10⁻⁷ Ω m.

ρ = (4.6 ± 0.4) × 10⁻⁷ Ω m. The uncertainty is quoted to 1 s.f., so the value is rounded to the same decimal place (4.6, not 4.64).

012345678contribution to %ρR: 1.6%d (×2): 7.1%l: 0.24%
Worked example 2: the diameter is the least precise measurement and is squared, so it supplies 7.1 of the 8.9 percentage points. Improving the length measurement would make almost no difference.
Common error

Subtracting the percentage uncertainty of a quantity on the bottom of a fraction. Multiplying or dividing, its percentage (times its power) is added.

How do you decide whether a result agrees with an accepted value?

Once a result has an uncertainty, you can ask whether it is any good. Two different percentages answer two different questions.

Percentage difference \(= \dfrac{|\text{experimental value} - \text{accepted value}|}{\text{accepted value}} \times 100\%\). It measures how accurate the result is (how close to the accepted value). The percentage uncertainty measures how precise it is (how wide its range). Learn this; it is not on the data sheet.
  • Percentage difference < percentage uncertainty: the accepted value lies inside the range of your result, so the result agrees with (is consistent with) it within the experimental uncertainty.
  • Percentage difference > percentage uncertainty: the accepted value lies outside your range. The uncertainties you estimated cannot explain the gap, so there is probably a systematic error, or an uncertainty you have not allowed for. Say what it might be and which way it pushes the result.
  • Equivalent check: does the accepted value fall inside value ± uncertainty?
  • With no accepted value (two methods, or a tolerance such as "470 μF ± 10%"), compare the two ranges: if they overlap, the values are consistent.
Worked example 3 Medium · Calculate, deduce

A student finds the p.d. at which a red LED just starts to conduct, V = 1.74 V ± 0.05 V. The LED's wavelength is given as 630 nm ± 10 nm. She uses h = eVλ/c. (a) Calculate h and its percentage uncertainty. (b) Decide whether her value agrees with the accepted value, 6.63 × 10⁻³⁴ J s, and comment.

(a) e and c are constants with no uncertainty: h = 1.60 × 10⁻¹⁹ × 1.74 × 630 × 10⁻⁹ ÷ 3.00 × 10⁸ = 5.85 × 10⁻³⁴ J s. Add the percentage uncertainties: 0.05 ÷ 1.74 × 100 + 10 ÷ 630 × 100 = 2.9% + 1.6% = 4.5%.

(b) Percentage difference: (6.63 − 5.85) ÷ 6.63 × 100 = 12%, more than twice the 4.5% uncertainty. The value does not agree: (5.85 ± 0.26) × 10⁻³⁴ J s reaches only 6.11. A systematic error is likely. Judging the "switch-on" p.d. by eye gives a value that is too low, which makes h too low.

5.45.65.866.26.46.66.87h / 10⁻³⁴ J saccepted 6.63LED result 5.85 ± 0.26a result that agrees: 6.50 ± 0.30
Worked example 3 as a number line. The LED range stops short of 6.63; a range that includes the accepted value (teal) agrees with it.

How do you find the uncertainty in a gradient and an intercept?

Most Unit 6 results come from the gradient or intercept of a straight-line graph, so the uncertainty is found from the graph itself.

  • Error bars. A vertical bar through each point, ± the uncertainty in y (horizontal too if x is uncertain), from the resolution or half-range of repeats.
  • Best-fit line. As at AS, with points scattered evenly about it; it should pass through (almost) all the error bars.
  • Steepest and shallowest lines. The steepest and least steep straight lines that still pass through all the error bars. In practice they usually join the bottom of the first bar to the top of the last bar, and the top of the first to the bottom of the last.
  • Uncertainty in the gradient ≈ (steepest − shallowest) ÷ 2. Another common method uses |best − worst|, with whichever extreme line differs more from the best fit; either way, show the gradients you used.
  • Uncertainty in the intercept: half the difference between the intercepts of the steepest and shallowest lines. A long extrapolation makes it far less certain than the gradient.
  • Percentage uncertainty in the gradient = uncertainty ÷ best gradient × 100%; it feeds into anything calculated from the gradient.
Worked example 4 Hard · Determine

Air is sealed in a flask of fixed volume and heated in a water bath. Pressure p is read from a gauge with an uncertainty of ± 1.0 kPa.

θ / °C2030405060708090
p / kPa107.4110.8114.7118.1122.0125.4129.1133.0

The best-fit line passes through (20 °C, 107.3 kPa) and (90 °C, 132.8 kPa). (a) Calculate its gradient and the temperature at which it predicts p = 0. (b) Use the steepest and shallowest lines to find the uncertainties. (c) Compare the result with −273 °C.

102030405060708090100100105110115120125130135140θ / °Cp / kPasteepestshallowestbest fit
The steepest line joins (20 °C, 106.4 kPa) to (90 °C, 134.0 kPa); the shallowest joins (20 °C, 108.4 kPa) to (90 °C, 132.0 kPa). Both still pass through every ± 1.0 kPa error bar.

(a) gradient = (132.8 − 107.3) ÷ 70 = 0.364 kPa °C⁻¹. p = 0 when θ = 20 − 107.3 ÷ 0.364 = −275 °C.

(b) Steepest: (134.0 − 106.4) ÷ 70 = 0.394; shallowest: (132.0 − 108.4) ÷ 70 = 0.337. Uncertainty = (0.394 − 0.337) ÷ 2 = ± 0.029 kPa °C⁻¹ (7.8%). Extrapolate each line to p = 0: steepest 20 − 106.4 ÷ 0.394 = −250 °C; shallowest 20 − 108.4 ÷ 0.337 = −302 °C, so the intercept is uncertain by ± 26 °C.

(c) Result (−275 ± 26) °C. Percentage difference from −273 °C = 0.7%, far inside the ± 9% range: the data agree with −273 °C.

−200−100100−2020406080100120140Oθ / °Cp / kPa−302−250−275data
The same three lines extended to p = 0. The gradients differ by only 0.057 kPa °C⁻¹, but more than 290 °C beyond the data their intercepts are 52 °C apart: long extrapolation magnifies small gradient differences.

What do you need to know about logarithms?

Exponential and power-law relationships give curves, which are hard to judge by eye. Taking logs turns both into straight lines whose gradient and intercept give the constants. First, the log skills themselves.

  • Two bases. ln uses base e (≈ 2.718): \(\ln e^x = x\). log (or lg) uses base 10: \(\log 10^x = x\). Use the one the question uses, and stay with it.
  • The three rules (true for either base): \(\log(ab) = \log a + \log b\), \(\log(a/b) = \log a - \log b\), \(\log(a^n) = n\log a\).
  • Units. Only a pure number has a log, so divide by the unit first: ln (V / V), log (T / s). Logs have no unit.
  • Significant figures. Give a log to the same number of decimal places as the number has significant figures. 5.83 V (3 s.f.) → ln 5.83 = 1.763; 0.64 V (2 s.f.) → ln 0.64 = −0.45. The digits before the decimal point only say where the decimal point was.
  • Error bars on a log graph. \(\Delta(\ln y) \approx \Delta y \div y\), the fractional uncertainty. For 2.38 V ± 0.05 V: ln 2.43 = 0.888 and ln 2.33 = 0.846 either side of ln 2.38 = 0.867, about ± 0.021. The bars grow as y gets smaller.
  • Changing base. \(\log y = 0.434\ln y\). So if you plot log y for an exponential, the gradient is 0.434k, not k: another reason to use ln for exponentials.
RelationshipPlot yagainst xGradientIntercept
y = Aekx (exponential)ln yxk (unit: 1 / unit of x)ln A
y = kxn (power law)log y (or ln y)log x (or ln x)n (no unit)log k (or ln k)
V = V0e−t/RCln Vt−1/RCln V0
N = N0e−λt (or activity, count rate)ln Nt−λln N0
T = 2π√(l/g)log Tlog l0.5log (2π/√g)
Exam tip

Before plotting, take logs of both sides of the equation in writing and line the result up under y = mx + c. That one line tells you what to plot, what the gradient means and what the intercept means, and it is often asked for on its own.

How do you test an exponential relationship with a log graph?

An exponential change has a constant ratio in equal intervals of x: the same fraction is lost every 15 s, every half-life, every millimetre of absorber. A falling curve could be many things, so the curve alone is never convincing.

0153045607590105120012345678910t / sV / Vhalves in 31.5 s31.5 s31.5 s
The constant-ratio property: the p.d. halves every 31.5 s (= ln 2 × RC) whatever the starting value. Equal times, equal fractions lost.

Take natural logs of \(y = Ae^{kx}\):

\[\ln y = \ln A + kx\]

This matches y = mx + c with ln y against x. A straight line means the relationship is exponential; the gradient is k and the intercept ln A, so \(A = e^{\text{intercept}}\), with the unit of y.

  • For a decay k is negative: the gradient gives the decay constant (−λ) or −1/RC.
  • Half-life (time to halve) = ln 2 ÷ |gradient|.
  • Correct the data first if the physics needs it: subtract the background count from a count rate, or the room temperature from a cooling temperature, before taking logs. Uncorrected data give a curve on the log graph.
05101520253035400123456t / minln (count rate / min⁻¹)background not subtractedbackground (20 min⁻¹) subtracted
A source with a background of 20 counts per minute. Uncorrected, the ln graph curves towards ln 20; corrected, it is straight with gradient −0.1 min⁻¹ = −λ.
Worked example 5 Hard · Show that, determine

A capacitor discharges through a 100 kΩ resistor; a data logger records the p.d. V across it. The capacitor is labelled 470 μF ± 10%.

t / sV / Vln (V / V)
08.952.192
156.431.861
304.611.528
453.321.200
602.380.867
751.710.536
901.230.207
1050.88−0.13
1200.64−0.45

(a) Show that a graph of ln V against t should be a straight line. (b) Use the graph to determine the capacitance. (c) Decide whether the result is consistent with the label.

0153045607590105120012345678910t / sV / VV₀/e = 3.29 V at t = RC ≈ 45 s
The discharge data on ordinary axes: exponential, or some other falling curve? The log graph settles it. After one time constant (about 45 s) V has fallen to 3.29 V, 1/e of 8.94 V.
0153045607590105120−0.50.511.522.5t / sln (V / V)Δt = 120 sΔ ln V = −2.64
The same data as ln (V / V) against t: a straight line, so the discharge is exponential. The triangle spans the whole line, from 2.19 at 0 s to −0.45 at 120 s. Error bars of ± 0.05 V become ± 0.05/V in ln V, so they grow as V falls.

(a) Take natural logs of \(V = V_0 e^{-t/RC}\): \(\ln V = \ln V_0 - \dfrac{1}{RC}t\), which matches y = c + mx. A straight line of gradient −1/RC and intercept ln V₀.

(b) Gradient from the triangle, values read from the line: (−0.45 − 2.19) ÷ (120 − 0) = −0.0220 s⁻¹. RC = 1 ÷ 0.0220 = 45.5 s, so C = 45.5 ÷ (100 × 10³) = 455 μF (4.55 × 10⁻⁴ F).

(c) Percentage difference from the label: (470 − 455) ÷ 470 × 100 = 3.2% < 10%: consistent with the label.

The intercept, 2.19, gives V₀ = e2.19 = 8.94 V, matching the first reading. The time for the p.d. to halve is ln 2 × RC = 31.5 s.

How do you test a power law with a log–log graph?

For \(y = kx^n\) with n unknown, y against x is a curve you cannot read n from. Take logs of both sides:

\[\log y = n\log x + \log k\]

Plot log y against log x; a straight line confirms a power law.

  • The gradient is the power n and has no unit. Either base works and gives the same n; just use the same base on both axes.
  • The intercept is log k, but only where log x = 0. The log x axis rarely starts at 0, so calculate it: log k = log y − n log x using a point on the line, then \(k = 10^{\log k}\).
  • If theory predicts the power (0.5 for a pendulum, −2 for an inverse-square law), compare your n with it using its uncertainty from the extreme lines.
  • If you already know n, a direct linear graph (T² against l) is simpler; use log–log when n itself is the question.
Worked example 6 Hard · Determine

Astronomers measure the orbital radius r and period T of five moons of a distant planet. They expect T = krn.

r / 10⁸ mT / 10⁴ slog (r / m)log (T / s)
1.203.538.0794.548
1.957.328.2904.865
3.1014.78.4915.167
4.8528.78.6865.458
7.4054.18.8695.733

(a) Use the graph to determine n. (b) For circular orbits, T = (2π/√(GM)) r1.5. Determine the mass M of the planet.

0123456780102030405060r / 10⁸ mT / 10⁴ sT ∝ rⁿ ?
On ordinary axes the moon data curve upwards. From this graph alone you cannot tell whether the power is 1.5, 1.4 or 2.
88.18.28.38.48.58.68.78.88.994.44.64.855.25.45.65.8log (r / m)log (T / s)Δ log r = 0.75Δ log T = 1.126
log (T / s) against log (r / m) is a straight line of gradient 1.50, so T ∝ r1.5. The triangle uses points read from the line, not data points.

(a) Gradient from the triangle: n = (5.705 − 4.579) ÷ (8.85 − 8.10) = 1.126 ÷ 0.75 = 1.50.

(b) log k from a point on the line, with n = 1.50: log k = 4.579 − 1.50 × 8.10 = −7.571, so k = 10−7.571 = 2.69 × 10⁻⁸ s m−1.5. Match k to 2π/√(GM) and rearrange: M = 4π² ÷ (Gk²) = 4π² ÷ (6.67 × 10⁻¹¹ × (2.69 × 10⁻⁸)²) = 8.2 × 10²⁶ kg.

M ∝ 1/k² = 10−2 log k, so an error of 0.01 in log k changes M by about 5%: keep three decimal places in logs.

0123456789−8−6−4−2246log (r / m)log (T / s)true intercept log k = −7.571axis drawn at 8.04.43 is NOT log k
The data occupy only the solid part, log r ≈ 8 to 9. The intercept you need is at log r = 0, far off the usual graph, so calculate it from c = y − mx.
Common error

Reading the "intercept" of a log–log graph where the line meets the vertical axis drawn at log x = 8.0. That is not log k; log k is the value at log x = 0. Calculate it from c = y − mx.

Exam tips for 6.2

Exam tip 1 — say which rule you are using

Write "percentage uncertainties add for a product" or "absolute uncertainties add for a difference", then show the numbers. Remember to multiply by the power: a squared diameter counts twice.

Exam tip 2 — finish the comparison

When comparing with an accepted value, calculate both percentages, compare them in one sentence ("4.5% is less than 12%"), and give a conclusion: agrees within uncertainty, or does not, with a likely systematic error and its direction.

Exam tip 3 — steepest and shallowest lines

Make both lines pass through every error bar, label them, and show the two gradients you read. Half their difference is the uncertainty; quote it to 1 or 2 significant figures.

Exam tip 4 — log columns

Head the column ln (V / V), give each value to as many decimal places as the reading has significant figures, and keep the minus sign for values below 1.

Exam tip 5 — units of gradients on log graphs

For ln y against t, the gradient has the unit s⁻¹ (1 / unit of x). For log y against log x, the gradient has no unit. The intercept ln A has no unit, but A does.

Common misconceptions

MisconceptionWhy it is wrongCorrect idea
For a difference, the uncertainties subtract.Each reading could be out in the opposite direction, which widens the range of the difference.Absolute uncertainties always add for sums and differences.
A quantity on the bottom of a fraction reduces the overall uncertainty.Dividing by an uncertain number makes the result more uncertain, not less.Add the percentage uncertainties for products and quotients alike.
Squaring a quantity squares its percentage uncertainty.For small uncertainties, (1 + p)² ≈ 1 + 2p.Multiply the percentage uncertainty by the power.
A small percentage difference means the result is precise.Percentage difference measures closeness to the accepted value (accuracy), not spread.Compare percentage difference (accuracy) with percentage uncertainty (precision).
A log graph only "makes the line straight".Its gradient and intercept are the constants of the relationship.Match the log equation to y = mx + c to read k, n or A.
Logs should be given to the same number of significant figures as the data.The figures before the decimal point of a log only fix the power of ten.Give the log the same number of decimal places as the data has significant figures.

How this topic connects

3.1 Measurements, uncertainties and graphs (AS, Unit 3)

Half the resolution, half the range, single percentage uncertainties and linearising to y = mx + c are the starting point for everything on this page.

Capacitors (Unit 4) and nuclear decay, oscillations and gravitation (Unit 5)

Capacitor discharge, Q = Q₀e−t/RC, is the classic ln graph. ln (count rate) against t gives the decay constant; log–log graphs test T ∝ l0.5 and T ∝ r1.5; error bars and extreme lines judge absolute zero.

6.3 The A2 core practicals

Capacitor discharge, gas pressure, gamma absorption by lead and resonance all finish with the analysis practised here. Planning, precision and accuracy are on 6.1.

Key equations

EquationUseIn the formulae list?
Δy = Δa + Δb for y = a ± buncertainty of a sum or difference (absolute uncertainties, same unit as y)no — learn it
%y = p(%a) + q(%b) for y = apbq (or ap/bq)uncertainty of a product, quotient or powerno — learn it
percentage difference = |exp − accepted| ÷ accepted × 100%judging accuracy and agreementno — learn it
Δ(gradient) ≈ (steepest − shallowest) ÷ 2uncertainty in a gradient (or intercept)no — learn it
ln y = ln A + kxtesting y = Aekx; gradient k in 1 / unit of xno — derive it by taking logs
log y = n log x + log ktesting y = kxn; gradient n has no unitno — derive it by taking logs
Q = Q0e−t/RC; N = N0e−λtthe exponential relationships most often linearised (RC in s, λ in s⁻¹)yes

Check your understanding

1. a = 2.0 ± 0.1 and b = 5.0 ± 0.2. What is the percentage uncertainty in ab, and in a/b?

%a = 0.1 ÷ 2.0 × 100 = 5% and %b = 0.2 ÷ 5.0 × 100 = 4%. So 5% + 4% = 9% for both: percentage uncertainties add for products and for quotients.

2. A length has a percentage uncertainty of 2%. What is the percentage uncertainty in its cube, and in its square root?

3 × 2% = 6% for the cube; ½ × 2% = 1% for the square root.

3. Two lengths are 25.3 cm ± 0.1 cm and 24.1 cm ± 0.1 cm. Find their difference and its percentage uncertainty. What does this suggest?

1.2 cm ± 0.2 cm, which is 0.2 ÷ 1.2 × 100 = 17%. Small differences between large readings are very uncertain; measure the difference directly or make it larger.

4. You suspect y = kxn with unknown n. What do you plot, and how do you find n and k?

log y against log x. A straight line confirms the power law; its gradient is n; log k = log y − n log x for any point on the line, so k = 10log k.

5. A reading is 3.40 V. Write ln (V / V) to the correct precision.

3.40 has 3 significant figures, so the log gets 3 decimal places: ln 3.40 = 1.224.

6. A result is (−275 ± 26) °C and the accepted value is −273 °C. Does it agree? Would it still agree if the uncertainty were ± 1 °C?

Percentage difference = 2 ÷ 273 × 100 = 0.7% < 26 ÷ 275 × 100 = 9.5%, so it agrees. With ± 1 °C the range −276 °C to −274 °C excludes −273 °C, so it would not agree.

Specification checklist

  • (analysis) I can find the uncertainty of a sum or difference by adding absolute uncertainties.
  • (analysis) I can find the percentage uncertainty of a product, quotient or power (adding percentages times powers) and convert it back to an absolute uncertainty.
  • (analysis) I can give a final value and its uncertainty to consistent, sensible significant figures, with the correct unit.
  • (evaluation) I can compare a percentage difference with the percentage uncertainty to decide whether a result agrees with an accepted value.
  • (analysis) I can draw error bars and steepest and shallowest lines and find the uncertainty in a gradient and an intercept.
  • (analysis) I can use the rules of logarithms with base e and base 10, and give logs to the correct number of decimal places.
  • (analysis) I can plot ln y against x to test an exponential relationship and find its constants.
  • (analysis) I can plot log y against log x to test a power law and find n and k.
  • (evaluation) I can identify the largest contribution to the uncertainty and suggest how to reduce it.