Combining uncertainties
Choose when to add absolute uncertainties, add percentages or include powers.
Young modulus practical questions
Calculate stress, strain and Young modulus from wire dimensions and force–extension data.
Practise
Extension–force gradientThis graph has extension vertically and force horizontally. Wire diameter 0.50 mm; original gauge length 1.20 m. Find Young modulus 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
| ΔF / N | Δx / mm |
|---|---|
| 3 | 0.15 |
| 6 | 0.3 |
| 9 | 0.45 |
| 12 | 0.6 |
| 15 | 0.75 |
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.
Larger graph
Graph values
| Point | Added force / N | Added extension / mm |
|---|---|---|
| 1 | 3 | 0.15 |
| 2 | 6 | 0.3 |
| 3 | 9 | 0.45 |
| 4 | 12 | 0.6 |
| 5 | 15 | 0.75 |
Hint
This gradient is 1/k in mm/N. Convert it to m/N; E = L/(A × SI gradient).
Worked solution
- Graph gradient = 1/20 mm/N = (1/20) × 10-3 m/N.
- E = 1.2/[A × (1/20) × 10-3] Pa, where A = πd²/4.
- Answer: 1.22 × 1011 Pa (3 significant figures).
Work through this dataset
- Read wire diameter
- Cross-sectional area
- Tensile stress
- Tensile strain
- Choose force–extension axes
- Force–extension gradient
- Young modulus from a gradient
- Extension–force gradient (current)
- Young modulus uncertainty
- Extension resolution
These tasks keep the same readings. Another question starts a new dataset. Your answers are checked separately.
Question type: Extension–force gradient
Report a problem
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