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Young modulus practical questions

Calculate stress, strain and Young modulus from wire dimensions and force–extension data.

Practise

Cross-sectional area

Original wire diameter is 0.50 mm. Find its cross-sectional area in m². Give 3 significant figures.

Practical setup and assumptions

A uniform circular metal wire has unloaded gauge length 1.00 m. A small preload removes slack; the table records additional force and additional extension from that reference.

The wire stays in its linear elastic region. Use the supplied ideal line and original cross-section. Treat the small preload’s change in length as negligible compared with the gauge length.

Readings for this dataset
Same-wire added loads and extensions
ΔF / NΔx / mm
20.1
40.2
60.3
80.4
100.5

The linked tasks use the same readings. These are generated practice data: treat them as exact unless an uncertainty is stated. Use the stated uncertainty rather than estimating it from scatter.

Use decimals or e notation, e.g. 4.2e-3. Enter the unit separately.

Hint

Use A = πd²/4 with diameter in metres.

Worked solution
  1. A = π(0.5 × 10-3)²/4 m².
  2. Answer: 1.96 × 10-7 m² (3 significant figures).
Next task in this dataset
Work through this dataset
  1. Read wire diameter
  2. Cross-sectional area (current)
  3. Tensile stress
  4. Tensile strain
  5. Choose force–extension axes
  6. Force–extension gradient
  7. Young modulus from a gradient
  8. Extension–force gradient
  9. Young modulus uncertainty
  10. Extension resolution

These tasks keep the same readings. Another question starts a new dataset. Your answers are checked separately.

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Question type: Cross-sectional area
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Stress, strain and Young modulus

Original area A = πd²/4. Tensile stress = force/A and tensile strain = extension/original length. In the linear elastic region, E = stress/strain = FL/(Ax).

Check the graph orientation

A force–extension gradient gives stiffness k, so E = kL/A. An extension–force gradient gives 1/k, so E = L/(A × gradient) after converting it to SI. Convert N/mm to N/m by multiplying by 1000; convert mm/N to m/N by multiplying by 0.001.

Preload and uncertainty

A small preload removes slack. These readings are added force and added extension from that reference, so the stress and strain tasks ask for increments. Use the unloaded gauge length; preload extension is assumed negligible compared with it. The wire remains in its linear elastic region.

For the stated first-order maximum method, add the k and L percentages and twice the diameter percentage. Extension resolution can be improved with an appropriate displacement gauge. This case is ideal data processing, not a complete loading or safety procedure.

A-level relevance

AQA 7408: 3.4.2.2; required practical 4 (selected); OCR A H556: 3.4.2(c–d) (selected). Tasks cover the selected skills listed above.

AQA mechanics and materials specification · OCR A specification · AQA practical handbook · OCR practical skills handbook

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