Uncertainties: Absolute, Fractional and Percentage Revision Notes | A Level AQA Physics

1. Why Uncertainty Matters in Physics

Every measurement you make in physics carries a degree of doubt. A ruler marked in millimetres cannot tell you whether a length is exactly 23.0 mm or somewhere between 22.5 mm and 23.5 mm. That doubt has a name: uncertainty.

Uncertainty is not a mistake or a sign of poor technique. It is an honest statement about the limits of your measuring instrument and your method. AQA examiners expect you to handle it with precision — both in practicals and in calculation questions.

There are three forms you need to master: absolute, fractional, and percentage uncertainty. Each one tells you something slightly different, and knowing when to use each is the key skill this lesson builds.

2. Absolute, Fractional, and Percentage Uncertainty

Picture a digital voltmeter displaying 4.52 V. The last digit represents 0.01 V, so the absolute uncertainty is ±0.01 V. The reading is reported as $4.52 \pm 0.01$ V. For an analogue ammeter with 0.2 A divisions, the absolute uncertainty would be $\pm 0.1$ A — half the smallest division.

When you take repeated readings, the absolute uncertainty can also be estimated as half the range: $\Delta x = \frac{x_{\text{max}} - x_{\text{min}}}{2}$. This approach captures the spread of your data rather than just the instrument resolution.

Fractional uncertainty is the bridge between absolute uncertainty and percentage uncertainty. A fractional uncertainty of 0.02 means the measurement could be off by 2% of its value — regardless of what units are involved. This makes it ideal for combining uncertainties across different quantities.

Notice that percentage uncertainty is the fractional uncertainty multiplied by 100. The two forms carry identical information — percentage uncertainty is often more intuitive when comparing sources of error in an experiment.

3. Representing Uncertainty in a Final Answer

Once you have calculated an uncertainty, you need to present your final answer correctly. The absolute uncertainty must be rounded to one significant figure (occasionally two if the leading digit is 1), and the measured value must be rounded to match the same decimal place as the uncertainty.

For example, if a calculation gives a resistance of 47.38 Ω with an absolute uncertainty of 2.4 Ω, you round the uncertainty to 2 Ω and then round the value to match: $47 \pm 2$ Ω. Writing $47.38 \pm 2.4$ Ω implies a false precision — the uncertainty already tells you the last two digits of 47.38 are meaningless.

4. Combining Uncertainties

Most physics quantities are not measured directly — they are calculated from two or more measured values. Each measurement carries its own uncertainty, and those uncertainties must be combined to find the overall uncertainty in the derived quantity. The rules depend on the mathematical operation involved.

The addition rule for subtraction is worth pausing on. When you subtract two similar-sized quantities, the absolute uncertainties still add — but the result can be small, making the percentage uncertainty very large. This is why measuring a small difference between two large quantities is experimentally risky.

For powers, the rule follows directly from the multiplication rule. If $Q = A^3$, that is the same as $A \times A \times A$, so the fractional uncertainty adds three times: $\frac{\Delta Q}{Q} = 3 \times \frac{\Delta A}{A}$. A 2% uncertainty in a radius becomes a 6% uncertainty in a volume calculated from $r^3$.

Worked example: Combining Uncertainties in a Derived Quantity

A student measures the diameter of a wire as $d = 0.40 \pm 0.02$ mm. Calculate the cross-sectional area $A = \pi (d/2)^2$ and find the absolute uncertainty in $A$. Give your answer in mm².

1. $r = d/2 = 0.40/2 = 0.20$ mm $\Delta r = \Delta d / 2 = 0.02/2 = 0.01$ mm — Dividing by a constant (2) divides the absolute uncertainty by the same constant. No fractional uncertainty rule needed here — this is a scale factor, not a separate measurement.
2. $\frac{\Delta r}{r} = \frac{0.01}{0.20} = 0.05$ — This is the fractional uncertainty in the radius. We need it because the next step involves a power.
3. $\frac{\Delta(r^2)}{r^2} = 2 \times \frac{\Delta r}{r} = 2 \times 0.05 = 0.10$ — The power rule: multiply the fractional uncertainty by the index. Squaring doubles the fractional uncertainty. The constant $\pi$ has no uncertainty, so it does not contribute.
4. $A = \pi r^2 = \pi \times (0.20)^2 = \pi \times 0.20 \times 0.20 = \pi \times 0.04 \approx 0.1257$ mm² — 0.20 × 0.20 = 0.04 (expanded longhand to avoid errors). Multiplying by π gives approximately 0.1257 mm².
5. $\Delta A = 0.10 \times 0.1257 \approx 0.013$ mm² — Absolute uncertainty = fractional uncertainty × calculated value. Round to 1 significant figure: 0.013 mm².
6. $A = 0.13 \pm 0.01$ mm² — The uncertainty rounds to 0.01 mm² (1 s.f.), so the area is rounded to the same decimal place: 0.13 mm². This is the correctly represented final answer. [M1] power rule applied; [M1] correct fractional uncertainty; [A1] final answer with matched decimal places.

Answer: $A = 0.13 \pm 0.01$ mm²

5. A Multi-Step Combination: Worked Exam Example

Now that the individual rules are clear, exam questions typically chain them together. You might measure several quantities, each with its own uncertainty, and need to propagate all of them through a formula. The strategy is to work through the formula step by step, applying the correct rule at each stage.

Worked example: Multi-Step Uncertainty Combination

A student measures the following quantities to calculate the resistance $R = V/I$: $V = 6.0 \pm 0.2$ V $I = 0.30 \pm 0.01$ A Calculate $R$ and express it with its absolute uncertainty.

1. $R = \frac{V}{I} = \frac{6.0}{0.30} = 20$ Ω — Straightforward division. [M1] for correct formula and substitution.
2. $\frac{\Delta V}{V} \times 100\% = \frac{0.2}{6.0} \times 100\% = 3.33\%$ — Percentage uncertainty in the voltage reading.
3. $\frac{\Delta I}{I} \times 100\% = \frac{0.01}{0.30} \times 100\% = 3.33\%$ — Percentage uncertainty in the current reading. Notice both sources contribute equally here.
4. $\% \text{ uncertainty in } R = 3.33\% + 3.33\% = 6.67\%$ — For $R = V/I$, add the percentage uncertainties of numerator and denominator. [M1] for correct combination rule.
5. $\Delta R = \frac{6.67}{100} \times 20 = 1.33$ Ω $\approx 1$ Ω — Absolute uncertainty = (fractional uncertainty) × value. Round to 1 significant figure: 1 Ω.
6. $R = 20 \pm 1$ Ω — The uncertainty is 1 Ω (1 s.f.), and the value 20 Ω is already at the units place — decimal places match. [A1] for correctly represented final answer.

Answer: $R = 20 \pm 1$ Ω

Notice how the two sources of uncertainty contributed equally in that example. In many experiments, one source dominates — identifying the dominant uncertainty is a higher-order skill that AQA rewards in evaluation questions.

7. Exam Technique and Mark-Winning Strategies

AQA uncertainty questions follow predictable patterns. Recognising those patterns before you sit the exam means you can allocate your working efficiently and avoid the traps that cost marks even when students understand the physics.

When a question says 'calculate the percentage uncertainty', show the fractional uncertainty calculation explicitly before multiplying by 100. Examiners award method marks for the correct ratio, even if arithmetic slips occur. When a question says 'express your answer with its uncertainty', the mark is specifically for the ± notation with matched decimal places — do not omit it.

For questions that ask you to 'evaluate' which measurement contributes most to the overall uncertainty, calculate each source's percentage uncertainty separately, apply any power multipliers, then compare them explicitly. State a conclusion: 'The uncertainty in [quantity] dominates because its percentage uncertainty of X% is larger than that of [other quantity] at Y%.' That reasoning chain is what earns the evaluation mark.

Worked example: Exam-Style: Uncertainty in a Power Calculation

A student measures the period of a pendulum as $T = 1.80 \pm 0.04$ s. The length is calculated using $L = g T^2 / (4\pi^2)$, where $g = 9.81$ m/s² (assumed exact). Calculate $L$ and express it with its absolute uncertainty. [4 marks]

1. $L = \frac{9.81 \times (1.80)^2}{4\pi^2}$ $(1.80)^2 = 1.80 \times 1.80 = 3.24$ $4\pi^2 \approx 4 \times 9.8696 = 39.478$ $L = \frac{9.81 \times 3.24}{39.478} = \frac{31.784}{39.478} \approx 0.805$ m — Expanding 1.80² longhand: 1.80 × 1.80 = 3.24. [M1] for correct substitution.
2. $\frac{\Delta T}{T} = \frac{0.04}{1.80} = 0.0222$ — This is the fractional uncertainty in the period measurement.
3. $\frac{\Delta(T^2)}{T^2} = 2 \times 0.0222 = 0.0444$ — The period is squared in the formula, so multiply the fractional uncertainty by 2. Constants $g$, 4, and $\pi^2$ are assumed exact and contribute no uncertainty. [M1] for correct power rule.
4. $\Delta L = 0.0444 \times 0.805 = 0.0357$ m $\approx 0.04$ m — Absolute uncertainty = fractional uncertainty × calculated value. Round to 1 significant figure: 0.04 m. [M1] for correct conversion to absolute uncertainty.
5. $L = 0.80 \pm 0.04$ m — The uncertainty is 0.04 m (2 decimal places), so the value is rounded to 2 decimal places: 0.80 m. [A1] for correctly represented final answer with matched decimal places.

Answer: $L = 0.80 \pm 0.04$ m

Every spec requirement in this lesson — defining the three forms of uncertainty, representing uncertainty in a final answer, and combining absolute and percentage uncertainties — maps directly onto the calculation and evaluation skills tested in AQA Paper 1 and Paper 2. Practise identifying the operation type first, then apply the matching rule, and always finish by checking your decimal places match between value and uncertainty.