Interactive practical
Determining the Young modulus of a wire
Measure a wire with a micrometer, load and unload it, then determine Young modulus from a stress–strain graph.
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Determining Young modulus
CCEA 1210 — matched to your course.
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At a glance
Practical focus
- Measure extension, original dimensions and force to determine stress, strain and Young modulus.
Practical method
Aim
Determine the Young modulus of a metal wire by measuring extension for increasing loads.
Variables
Independent variable: load (tensile force)
Dependent variable: extension
Control variables: the same wire, original length and diameter; temperature; stay within the elastic region.
Apparatus
- Long thin copper or steel wire
- Clamp and pulley at the end of the bench
- Slotted masses
- Metre rule
- Micrometer
- Marker and vernier scale (or Searle's apparatus)
- Safety goggles
Method
- Measure the original length L from the clamp to the marker (about 2.5 m) with a metre rule.
- Measure the diameter with a micrometer at several points and in two perpendicular directions; take the mean and calculate A = \(\dfrac{\pi d^{2}}{4}\).
- Add a small load to straighten the wire and note the marker position.
- Add masses in steps, recording the marker position each time; remove them to check the wire returns to its original length.
- Plot force against extension. Young modulus E = \(\dfrac{\text{gradient} \times L}{A}\).
Sample results
L = 2.50 m, d = 0.30 mm, so A = 7.07 × 10⁻⁸ m².
| Force / N | Extension / mm |
|---|---|
| 10 | 1.77 |
| 20 | 3.54 |
| 30 | 5.31 |
| 40 | 7.07 |
Analysis
Force against extension is a straight line through the origin with gradient \(\dfrac{40}{0.00707}\) = 5.66 × 10³ N m⁻¹. E = \(\dfrac{\text{gradient} \times L}{A}\) = \(\dfrac{5.66 \times 10^{3} \times 2.50}{7.07 \times 10^{-8}}\) = 2.0 × 10¹¹ Pa, typical of steel.
Uncertainty
The diameter (0.30 ± 0.01 mm) has a 3.3% uncertainty, which doubles to 6.7% in A — the largest contribution. Length (±1 mm in 2.50 m) and extension (±0.05 mm in 7 mm, 0.7%) contribute much less, so E = (2.0 ± 0.2) × 10¹¹ Pa.
Errors and improvements
| Problem | Effect | Improvement |
|---|---|---|
| Wire not straight at the start | Early extensions too large | Add a small initial load and measure from there |
| Diameter measured once | Large error in A | Measure at several points and orientations |
| Loading past the elastic limit | Permanent extension | Check the wire returns on unloading |
Risk assessment
| Hazard | Risk | Control measure |
|---|---|---|
| Wire snapping | Eye injury | Wear goggles |
| Falling masses | Injury to feet | Use a box or cushion below |
Common mistakes
- Using the diameter instead of the radius in A = πr².
- Leaving extension in millimetres.
- Using the total length after loading instead of the original length.
Worked analysis
At 40 N the extension is 7.07 mm. Stress = \(\dfrac{40}{7.07 \times 10^{-8}}\) = 5.66 × 10⁸ Pa; strain = \(\dfrac{0.00707}{2.50}\) = 2.83 × 10⁻³; E = \(\dfrac{5.66 \times 10^{8}}{2.83 \times 10^{-3}}\) = 2.0 × 10¹¹ Pa.
Common misconception
Young modulus is a property of the material, not the wire: a thicker wire of the same metal extends less but has the same E.
Assessment coverage
Practice this practical
2 questions available, worth 4 marks in total.
Retrieval cards
What is the Young modulus?
Tensile stress ÷ tensile strain in the elastic region.
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How this practical fits your course
The label below uses the wording from your course specification.
- CCEA 1210 — Specification experiment: Learning outcome 4.1.5 – Young modulus experiment. The A2 deformation-of-solids learning outcomes require students to perform and describe an experiment to determine Young modulus. Checked 18 September 2026. View official course source