PhysicsHelp.co.uk

Young modulus practical questions

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

Practise

Tensile stress

Wire diameter is 0.45 mm. For row 3, calculate the additional tensile stress in Pa. Give 3 significant figures.

Practical setup and assumptions

A uniform circular metal wire has unloaded gauge length 1.20 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

Additional stress = added force / original area. Use the circular area in m².

Worked solution
  1. A = π(0.45 × 10-3)²/4 m².
  2. Additional stress = 6/A Pa.
  3. Answer: 3.77 × 107 Pa (3 significant figures).
Next task in this dataset
Work through this dataset
  1. Read wire diameter
  2. Cross-sectional area
  3. Tensile stress (current)
  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.

Another question
Question type: Tensile stress
Report a problem

The question reference is attached automatically. Please leave out names and contact details.

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

Related practice