For the deformation of a wire under tension, define (i) stress,
- (a)(i)For the deformation of a wire under tension, define (i) stress,[1]
- (a)(ii)(ii) strain.[1]
- (b)(i)A wire is fixed at one end so that it hangs vertically. The wire is given an extension x by suspending a load F from its free end. The variation of F with x is shown in [Figure 3.1]. The wire has cross-sectional area 9.4 × 10⁻⁸ m² and original length 2.5 m. Describe how measurements can be taken to determine accurately the cross-sectional area of the wire.[3]
- (b)(ii)Determine the Young modulus E of the material of the wire.[2]
- (b)(iii)Use [Figure 3.1] to calculate the increase in the energy stored in the wire when the load is increased from 2.0N to 4.0N.[2]
- (c)The wire in (b) is replaced by a new wire of the same material. The new wire has twice the length and twice the diameter of the old wire. The new wire also obeys Hooke's law. On [Figure 3.1], sketch the variation with extension x of the load F for the new wire from x = 0 to x = 0.80 mm.[2]
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Mark scheme
Marking points from the official mark scheme (Feb/Mar 2018). The mark or mark type is shown in brackets.
- [B1] force / (cross-sectional) area
- [B1] extension / original length
- [B1] measure / determine / find diameter
- [B1] using a micrometer / digital calipers
- [B1] several measurements in different places / along the wire / around the circumference (and average them)
- [C1] E = σ/ ε or E = FL / Ax or E = gradient × (L / A)
- E = (4 × 2.5) / (0.8 × 10⁻³) × (9.4 × 10⁻⁸)
- [A1] = 1.3 x 10^11 Pa
- [C1] E = 1/2Fx or E = 1/2kx^2 or E = area under graph
- E = 1/2 × (2+4) × 0.4 × 10^-3
- or E = (1/2 × 4 × 0.8×10⁻³) – (1/2 × 2 × 0.4×10⁻³)
- or E = [ 1/2 × 5000 × (0.8×10⁻³)² ] – [ 1/2 × 5000 × (0.4×10⁻³)² ]
- [A1] E = 1.2 × 10⁻³ J
- [M1] straight line from the origin and above the original line
- [A1] straight line passes through (0.80, 8.0)
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