A-LevelPhysicsDeformation of solids9702/21 · Oct/Nov 2017 · Q45 Marks
A wire is designed to ensure that its strain does not exceed 4.0 × 10⁻⁴ when a force of 8.0kN is applied. The Young modulus of the metal of the wire is 2.1 × 10¹¹ Pa. It may be assumed that the wire obeys Hooke's law. For a force of 8.0 kN, calculate, for the wire,
- (a)Define strain.[1]
- (b)(i)the maximum stress,[2]
- (b)(ii)the minimum cross-sectional area.[2]
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2017). The mark or mark type is shown in brackets.
- [B1] (strain =) extension / original length
- [C1] E = σ/ε
- [A1] maximum stress = 2.1 × 10¹¹ x 4.0 × 10⁻⁴ = 8.4 x 10⁷ Pa
- [C1] σ = F/A
- [A1] minimum area = 8.0 × 10³/8.4 × 10⁷ = 9.5 x 10⁻⁵m²
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