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Materials: Stress, Strain and the Young Modulus Practice Problems with Solutions - A Level Year 12

A Level Year 12 | Physics | Materials: Stress, Strain and the Young Modulus
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Stress is force per unit area and strain is extension divided by original length, so this Year 12 sheet opens by separating the two deformations. The pascal is the stress unit, and Young modulus is recorded as stress ÷ strain.

Key Takeaways

Section A: Multiple Choice Questions (1 Mark Each)

Choose the correct option for each question.
1. Stress is defined as
(a) force per unit area(b) extension per unit length(c) force × length(d) energy per unit volume
2. Strain is defined as
(a) extension ÷ original length(b) force ÷ area(c) stress × area(d) length ÷ force
3. The unit of stress is the
(a) pascal (N/m²)(b) metre(c) newton(d) dimensionless
4. Strain is
(a) dimensionless (has no unit)(b) measured in metres(c) measured in pascals(d) measured in newtons
5. The Young modulus is given by
(a) stress ÷ strain(b) strain ÷ stress(c) stress × strain(d) stress + strain
6. Beyond the elastic limit, a material
(a) is permanently deformed(b) returns to its original shape(c) becomes stronger(d) stops stretching

Section B: Short Answer Type Questions (2 Marks Each)

Show all steps clearly.
7. Define stress.
8. Define strain.
9. Write the formula for the Young modulus.
10. What is the unit of the Young modulus?
11. What is meant by the elastic limit?
12. What is a brittle material?
13. What is a ductile material?
14. Write the formula for stress in terms of force and area.

Section C: Numericals & Word Problems (3 Marks Each)

Apply the concepts to solve the problems. Show all working.
15. A force of 200 N acts on a wire of cross-sectional area 2 × 10⁻⁶ m². Using stress = F ÷ A, find the stress.
16. A wire extends by 2 mm from an original length of 1000 mm. Using strain = extension ÷ original length, find the strain.
17. A material has a stress of 2 × 10⁸ Pa and a strain of 0.001. Using E = stress ÷ strain, find the Young modulus.
18. A wire of length 2 m extends by 4 mm. Using strain = extension ÷ original length, find the strain.
19. A force of 100 N is applied to a cross-sectional area of 1 × 10⁻⁶ m². Using stress = F ÷ A, find the stress.
20. A material has a Young modulus of 2 × 10¹¹ Pa and a stress of 2 × 10⁸ Pa. Using strain = stress ÷ E, find the strain.

Answer Key

1. a) force per unit area
2. a) extension ÷ original length
3. a) pascal (N/m²)
4. a) dimensionless (has no unit)
5. a) stress ÷ strain
6. a) is permanently deformed
7. Refer to solution guide.
8. Refer to solution guide.
9. Refer to solution guide.
10. Refer to solution guide.
11. Refer to solution guide.
12. Refer to solution guide.
13. Refer to solution guide.
14. Refer to solution guide.
15. Refer to solution guide.
16. Refer to solution guide.
17. Refer to solution guide.
18. Refer to solution guide.
19. Refer to solution guide.
20. Refer to solution guide.
End of Worksheet

People Also Ask

What is stress?

Force per unit area, option (a) in this worksheet (Q1).

What is strain?

Extension ÷ original length, option (a) in this worksheet (Q2).

What is the Young modulus?

Stress ÷ strain, option (a) in this worksheet (Q5).

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