AQA AQA 2018-style practice: A wire of length 2.5 m and cross-sectional area 1.2 × 10^-7 m^2 extends by 1.8 mm under a load of 35 N. Calculate the Young modulus of the wire material.

Calculate the stress and strain, then divide. Stress = FA = 351.2 × 10^-7 = 2.92 × 10^8 Pa. Strain = LL = 1.8 × 10^-32.5 = 7.2 × 10^-4. Young modulus = stressstrain = 2.92 × 10^87.2 × 10^-4 = 4.1 × 10^11 Pa. Markers reward consistent SI units (extension in metres), correct calculation of stress and strain, and dividing in the right order.

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How do we measure the stiffness of a material independently of its shape?

The Young modulus as the ratio of tensile stress to tensile strain, the gradient of a stress-strain graph, the experiment to measure it for a wire, and interpreting stress-strain curves up to the breaking point.

A focused answer to AQA A-Level Physics 3.4.2.2, covering the Young modulus as the ratio of tensile stress to tensile strain, the gradient of a stress-strain graph, the experiment to measure it for a metal wire, and how to interpret stress-strain curves up to breaking.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Defining the Young modulus
From a stress-strain graph
Measuring the Young modulus of a wire
Interpreting the full curve
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What this dot point is asking

AQA specification point 3.4.2.2 wants you to define the Young modulus, find it from the gradient of a stress-strain graph, describe the experiment to measure it for a wire, and interpret stress-strain graphs including the elastic limit, yield and breaking points.

Defining the Young modulus

A large Young modulus means a stiff material that strains very little for a given stress; steel has a Young modulus of about $2 \times 10^{11} \text{ Pa}$ , far larger than rubber. Crucially, the Young modulus is a property of the material alone, independent of the shape or size of the sample, unlike the spring constant which also depends on the dimensions.

From a stress-strain graph

The area under the stress-strain graph (up to a given strain) equals the energy stored per unit volume of the material, with units of $\text{J m}^{-3}$ . This is the energy-density analogue of the area under a force-extension graph. Because stress is $\dfrac{F}{A}$ and strain is $\dfrac{\Delta L}{L}$ , multiplying them gives $\dfrac{F\Delta L}{AL} = \dfrac{\text{energy}}{\text{volume}}$ , confirming the area represents an energy density rather than a total energy. Comparing this with the elastic strain energy $\tfrac{1}{2}F\Delta L$ from the force-extension graph shows the two descriptions are consistent.

Measuring the Young modulus of a wire

Carrying out the wire experiment

Outline the standard required-practical method for finding the Young modulus of a metal wire.

step 1: Set up the wire

Clamp a long, thin wire of the test material horizontally over a pulley (or hang it vertically) with a marker and a fixed reference scale. Measure the original length $L$ with a ruler.

step 2: Measure the diameter

Use a micrometer to measure the wire's diameter at several points and at different orientations, then average and calculate the cross-sectional area $A = \pi r^2$ .

step 3: Load and record

Add known masses one at a time, recording the extension $\Delta L$ for each load $F$ . A long, thin wire gives larger extensions, reducing the percentage uncertainty.

step 4: Calculate and plot

For each load find the stress $\dfrac{F}{A}$ and strain $\dfrac{\Delta L}{L}$ , plot stress against strain, and take the gradient of the straight-line portion as the Young modulus.

Interpreting the full curve

Up the curve you meet the limit of proportionality (where Hooke's law ends), then the elastic limit (where permanent deformation begins), then (for ductile metals) the yield point where large plastic strain begins for little extra stress, and finally the breaking stress where the wire fractures.

Try this

Q1. State what a large Young modulus tells you about a material. [1 mark]

Cue. It is very stiff; it strains very little for a given stress.

Q2. Why is a long, thin wire chosen for the experiment? [2 marks]

Cue. Long and thin gives a larger extension, reducing the percentage uncertainty in the measured extension.

Q3. State what the gradient of the linear region of a stress-strain graph represents. [1 mark]

Cue. The Young modulus.

Exam-style practice questions

Practice questions written in the style of AQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AQA 20184 marksA wire of length

2.5 \text{ m}

and cross-sectional area

1.2 \times 10^{-7} \text{ m}^2

extends by

1.8 \text{ mm}

under a load of

35 \text{ N}

. Calculate the Young modulus of the wire material.

Show worked answer →

Calculate the stress and strain, then divide.

Stress $= \dfrac{F}{A} = \dfrac{35}{1.2 \times 10^{-7}} = 2.92 \times 10^8 \text{ Pa}$ .

Strain $= \dfrac{\Delta L}{L} = \dfrac{1.8 \times 10^{-3}}{2.5} = 7.2 \times 10^{-4}$ .

Young modulus $= \dfrac{\text{stress}}{\text{strain}} = \dfrac{2.92 \times 10^8}{7.2 \times 10^{-4}} = 4.1 \times 10^{11} \text{ Pa}$ .

Markers reward consistent SI units (extension in metres), correct calculation of stress and strain, and dividing in the right order.

AQA 20215 marksDescribe an experiment to determine the Young modulus of a metal in the form of a wire, stating the measurements taken, the graph plotted, and how the result is obtained.

Show worked answer →

Measure the original length $L$ of the wire with a ruler and its diameter at several points with a micrometer (average, then find $A = \pi r^2$ ). Clamp the wire and load it with known masses, recording the extension $\Delta L$ for each load $F$ using a marker against a fixed scale.

For each load, calculate stress $\dfrac{F}{A}$ and strain $\dfrac{\Delta L}{L}$ , and plot stress (y-axis) against strain (x-axis). The graph is a straight line through the origin in the elastic region.

The Young modulus is the gradient of this straight-line region. Using a long, thin wire makes the extension larger and so reduces the percentage uncertainty.

Markers reward measuring length and diameter, recording load and extension, plotting stress against strain, and taking the gradient of the linear region.

Related dot points

Sources & how we know this

AQA A-level Physics (7408) specification — AQA (2017)