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Almost every meaningful physical result is calculated from several measured quantities rather than read directly off an instrument. The resistivity of a wire is found from R, L and A; the acceleration due to gravity is found from a measured period and length; the wavelength of light is found from a diffraction-grating spacing and an angle. Each input has its own uncertainty, and those uncertainties combine to give the uncertainty in the final result. The rules for that combination — sometimes called error propagation — are the technical heart of every required-practical write-up at A-Level.
This lesson sets out the three core rules (addition/subtraction, multiplication/division, powers), works through the dominant-contribution analysis that examiners regularly ask about, and shows the full pipeline from raw measurements to a properly-quoted final result with absolute and percentage uncertainty.
Spec mapping: This lesson addresses content from AQA 7408 §3.1 on the combination of uncertainties through addition and subtraction (sum of absolutes), multiplication and division (sum of percentages), and powers (percentage uncertainty multiplied by the power), together with the use of the resulting absolute or percentage uncertainty in a final reported result (refer to the official AQA specification document for exact wording).
Synoptic links:
- Required practical 6 (resistivity): the canonical example — combine uncertainties in R, L and d to get the uncertainty in ρ = RA/L; the diameter contribution dominates because of the d² term.
- Required practical 1 (free fall): combine uncertainties in distance and time to get uncertainty in g; the time contribution dominates because of the t² term.
- Mathematical skills: the rules here are the high-school version of the partial-derivative formulae used at university; understanding the algebra makes the undergraduate transition smoother.
For a result y calculated from independent measurements a, b, c, ... with absolute uncertainties Δa, Δb, Δc, ..., the rules for finding the uncertainty Δy depend on the form of the calculation.
For y = a + b or y = a − b:
Δy = Δa + Δb
Add the absolute uncertainties, regardless of whether you are adding or subtracting the values themselves.
Why? A measurement a could be at its maximum (a + Δa) while b is at its minimum (b − Δb); the resulting y = (a + Δa) + (b − Δb) lies within Δa + Δb of (a + b) in the worst case. The same worst-case logic applies to subtraction: if y = a − b and a is high while b is low, then y is too high by Δa + Δb.
For y = a × b, y = a / b, y = ab/c, or any product/quotient combination:
%Δy = %Δa + %Δb + %Δc + ...
Add the percentage uncertainties.
Why? Take y = ab. The fractional change in y from a small change in a is Δa/a, and similarly Δb/b. To first order, the fractional changes add: Δy/y ≈ Δa/a + Δb/b. Multiplying both sides by 100% gives the percentage rule.
For y = aⁿ (where n can be a positive or negative integer, or a fraction):
%Δy = |n| × %Δa
The percentage uncertainty is multiplied by the absolute value of the power.
Why? y = aⁿ implies dy/da = naⁿ⁻¹, so Δy = n aⁿ⁻¹ Δa, and Δy/y = n Δa/a. Taking the absolute value handles negative powers (e.g. y = 1/a² has n = −2, and the percentage uncertainty rule gives |−2| × %Δa = 2 × %Δa).
When a formula involves all three operations, apply the rules in the natural order — typically convert each multiplicative/power term to its percentage uncertainty, sum them, then convert back to absolute uncertainty for the final answer.
flowchart TD
M[Measured quantities<br/>a, b, c with Δa, Δb, Δc] --> CHOOSE{Operation in formula?}
CHOOSE -->|Addition or subtraction| ABS[Add absolute uncertainties<br/>Δy = Δa + Δb]
CHOOSE -->|Multiplication or division| PCT[Add percentage uncertainties<br/>%Δy = %Δa + %Δb]
CHOOSE -->|Power y = aⁿ| POW[Multiply percentage uncertainty by n<br/>%Δy = n × %Δa]
ABS --> FIN[Combine into final Δy]
PCT --> FIN
POW --> FIN
FIN --> REP[Report y ± Δy with matched sig figs]
style FIN fill:#27ae60,color:#fff
style REP fill:#1f4e79,color:#fff
Two lengths are measured: L₁ = 25.0 ± 0.1 cm and L₂ = 15.0 ± 0.1 cm. Find L = L₁ + L₂.
L = 25.0 + 15.0 = 40.0 cm.
ΔL = ΔL₁ + ΔL₂ = 0.1 + 0.1 = 0.2 cm.
L = (40.0 ± 0.2) cm.
The difference between two nearly-equal temperatures is sometimes needed: T = T_hot − T_cold. If T_hot = 25.4 ± 0.2 °C and T_cold = 24.6 ± 0.2 °C, then T = 0.8 ± 0.4 °C. The percentage uncertainty in T is (0.4/0.8) × 100 = 50% — enormous, even though each input had only 0.8% uncertainty.
This is the cancellation trap: subtracting nearly-equal quantities amplifies the relative uncertainty dramatically. Whenever possible, redesign the experiment to measure the difference directly (a differential thermometer) rather than subtracting two large numbers.
Find the area of a rectangle: A = L × W, with L = 12.4 ± 0.2 cm and W = 8.5 ± 0.1 cm.
%ΔL = (0.2 / 12.4) × 100 = 1.6%
%ΔW = (0.1 / 8.5) × 100 = 1.2%
A = 12.4 × 8.5 = 105.4 cm²
%ΔA = 1.6 + 1.2 = 2.8%
ΔA = 2.8% × 105.4 = 2.95 cm² ≈ 3 cm²
A = (105 ± 3) cm² (rounded to match the precision of the uncertainty).
The volume of a sphere V = (4/3)πr³, with r = 2.50 ± 0.05 cm.
%Δr = (0.05 / 2.50) × 100 = 2.0%
%ΔV = 3 × %Δr = 6.0% (the 4/3 and π are exact and contribute no uncertainty).
V = (4/3)π(2.50)³ = 65.45 cm³
ΔV = 6.0% × 65.45 = 3.9 cm³ ≈ 4 cm³
V = (65 ± 4) cm³.
A student measures:
Calculate ρ = RA/L = R π r² / L with its percentage and absolute uncertainty.
Step 1: percentage uncertainties on each input.
%ΔR = (0.05 / 2.15) × 100 = 2.3%
%ΔL = (0.001 / 0.750) × 100 = 0.13%
%Δr = (5 × 10⁻⁶ / 1.40 × 10⁻⁴) × 100 = 3.6%
Step 2: handle the r² term using the power rule.
%Δ(r²) = 2 × %Δr = 2 × 3.6 = 7.1%
Step 3: sum the percentage uncertainties (R, r², L all in product/quotient).
%Δρ = %ΔR + %Δ(r²) + %ΔL = 2.3 + 7.1 + 0.13 = 9.5%
Step 4: calculate ρ.
A = π r² = π × (1.40 × 10⁻⁴)² = 6.16 × 10⁻⁸ m²
ρ = RA/L = 2.15 × 6.16 × 10⁻⁸ / 0.750 = 1.77 × 10⁻⁷ Ω m
Step 5: convert percentage uncertainty to absolute.
Δρ = 9.5% × 1.77 × 10⁻⁷ = 0.17 × 10⁻⁷ Ω m ≈ 0.2 × 10⁻⁷ Ω m
Final answer: ρ = (1.8 ± 0.2) × 10⁻⁷ Ω m.
(Note that the value is rounded to 2 sig figs to match the precision of the uncertainty.)
A student drops a ball from h = 1.50 ± 0.01 m and times the fall as t = 0.55 ± 0.05 s. Using h = ½gt², find g.
%Δh = (0.01 / 1.50) × 100 = 0.67%
%Δt = (0.05 / 0.55) × 100 = 9.1%
Rearranging: g = 2h / t² → %Δg = %Δh + 2 × %Δt = 0.67 + 18.2 = 18.9%
g = 2 × 1.50 / 0.55² = 9.92 m s⁻²
Δg = 18.9% × 9.92 = 1.87 m s⁻²
g = (10 ± 2) m s⁻². Roughly consistent with the accepted value 9.81 m s⁻², but with such large uncertainty that any value between 8 and 12 m s⁻² is consistent. The time measurement dominates the uncertainty — replacing the stopwatch with light gates would tighten the result.
A student determines Young's modulus E of a copper wire using the standard A-Level apparatus. The relationship is E = (F L₀) / (A × e), where F is the tension in the wire, L₀ is the unstretched length, A is the cross-sectional area = π(d/2)², and e is the extension. The measurements are:
The aim is to find E with its absolute uncertainty and identify which input dominates.
Step 1 — percentage uncertainty in each input.
%ΔF = (0.01 / 19.62) × 100 = 0.051%
%ΔL₀ = (0.001 / 1.250) × 100 = 0.080%
%Δd = (0.01 / 0.32) × 100 = 3.13%
%Δe = (0.05 / 0.85) × 100 = 5.88%
Step 2 — handle the d² term using the power rule (since A = π(d/2)² ∝ d²).
%ΔA = 2 × %Δd = 2 × 3.13 = 6.25%
Step 3 — combine via the product/quotient rule for E = FL₀ / (A × e):
%ΔE = %ΔF + %ΔL₀ + %ΔA + %Δe = 0.051 + 0.080 + 6.25 + 5.88 = 12.26%
Step 4 — calculate E.
A = π(d/2)² = π(1.6 × 10⁻⁴)² = 8.04 × 10⁻⁸ m²
E = (F L₀) / (A × e) = (19.62 × 1.250) / (8.04 × 10⁻⁸ × 8.5 × 10⁻⁴) = 24.525 / (6.834 × 10⁻¹¹) = 3.59 × 10¹¹ Pa
Step 5 — convert percentage uncertainty to absolute.
ΔE = 12.26% × 3.59 × 10¹¹ = 0.44 × 10¹¹ Pa ≈ 0.4 × 10¹¹ Pa
Final answer: E = (3.6 ± 0.4) × 10¹¹ Pa.
The accepted Young's modulus of copper is around 1.3 × 10¹¹ Pa, so the experimental result is markedly higher than expected — almost certainly because of an unidentified systematic error (perhaps the wire is brass rather than copper, or the measured extension is contaminated by yielding of the support clamp). This kind of comparison-with-literature step is what AQA practical write-ups call the conclusion-evaluation move, and it earns AO3 marks distinct from the uncertainty-calculation marks.
Step 6 — identify the dominant contribution.
The percentage contributions in step 3 are: F (0.051%), L₀ (0.080%), A (6.25%), e (5.88%). The two largest are A (which is 2 × %Δd) and the extension e itself. Together they account for 12.13 percentage points of the 12.26% total — over 99% of the uncertainty in E. The diameter and the extension are essentially co-dominant.
Implications for experimental redesign. A modest improvement to either d or e propagates roughly linearly into ΔE; halving %Δd (e.g. by using a 0.001 mm micrometer instead of 0.01 mm) would shave 3.1% off the combined total. Halving %Δe (e.g. by using a longer wire so the same load produces a larger extension, scaling e in proportion to L₀) would shave 2.9%. Doing both would bring %ΔE from 12.3% down to roughly 6%, halving the uncertainty in the final value. This kind of "where is my next mark worth investing?" reasoning is exactly the design-evaluation skill that an A* candidate is expected to deploy on practical-endorsement work.
A standard AQA exam question asks "which measurement contributes the largest percentage uncertainty?" or "suggest one modification to reduce the percentage uncertainty in the result". The answer is always to look at the biggest term in the sum of percentage uncertainties.
In the resistivity example: %Δρ = 2.3% + 7.1% + 0.13%. The r² term (7.1%) dominates, and the diameter has the largest single contribution (3.6% × 2 = 7.1%). The advice to the student is to measure the diameter more accurately — e.g. use a screw-gauge micrometer with finer resolution, or take more diameter readings at different points along the wire and average.
In the free-fall example: %Δg = 0.67% + 18.2%. The time-squared term (18.2%) overwhelms the height term (0.67%). Replace the stopwatch with light gates.
The general rule. The variable raised to the highest power is usually the dominant source of uncertainty. Diameter (squared in A = πr²) is the canonical culprit in many practicals; time (squared in h = ½gt², or squared via T² in pendulum work) is the canonical culprit in another large class.
These two terms are easily confused.
Percentage uncertainty is calculated from the spread of repeat readings (or instrument resolution) and tells you how precise your measurement is.
Percentage error is calculated by comparing your experimental value to an accepted theoretical value: %error = |experimental − accepted| / accepted × 100%.
If %error ≤ %uncertainty, the experimental result is consistent with the accepted value — your data does not disagree with theory.
If %error > %uncertainty, the experimental result disagrees with the accepted value at the quoted uncertainty level, and either the theory is wrong or there is an unidentified systematic error in the experiment.
Worked example. The free-fall experiment above gives g = (9.9 ± 1.9) m s⁻². Accepted value g = 9.81 m s⁻². %error = |9.9 − 9.81| / 9.81 × 100 = 0.9%. Since 0.9% < 18.9%, the result is consistent with the accepted value despite its huge uncertainty.
Specimen question modelled on the AQA paper format. Total: 9 marks.
"A student determines the density of a metal sphere by measuring its mass m, and its diameter d. The measurements are:
m = 32.4 ± 0.1 g
d = 1.95 ± 0.02 cm
The density is given by ρ = m / V where V = (4/3)π(d/2)³.
(a) Calculate the percentage uncertainty in m. (1 mark)
(b) Calculate the percentage uncertainty in V. (3 marks)
(c) Calculate the density and quote the result with its absolute uncertainty. (3 marks)
(d) Suggest, with reasoning, one modification that would have the largest effect on reducing the uncertainty in ρ. (2 marks)"
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