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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A-Level Physics

Determining Young modulus

Pearson Edexcel 9PH0 — matched to your course.

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Practical focus

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

Method

  1. Measure the original length L from the clamp to the marker (about 2.5 m) with a metre rule.
  2. Measure the diameter with a micrometer at several points and in two perpendicular directions; take the mean and calculate A = \(\dfrac{\pi d^{2}}{4}\).
  3. Add a small load to straighten the wire and note the marker position.
  4. Add masses in steps, recording the marker position each time; remove them to check the wire returns to its original length.
  5. 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 / NExtension / mm
101.77
203.54
305.31
407.07

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

ProblemEffectImprovement
Wire not straight at the startEarly extensions too largeAdd a small initial load and measure from there
Diameter measured onceLarge error in AMeasure at several points and orientations
Loading past the elastic limitPermanent extensionCheck the wire returns on unloading

Risk assessment

HazardRiskControl measure
Wire snappingEye injuryWear goggles
Falling massesInjury to feetUse a box or cushion below

Common mistakes

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

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What is the Young modulus?

Tensile stress ÷ tensile strain in the elastic region.

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