Question

1. A force of 40N at the end of a wire of length 4m and diameter 2mm process and extension of 0.24mm. Calculate; (a) stress on the wire (b) strain in the wire (c) Young's Modulous for the material of the wire (π=3.14)

Answer 1

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Force = F = 40N

Length = L = 4m

Radius = Diameter/2 = 1mm = 10⁻³ m

Extension = ΔL = 0.24mm = 0.24×10⁻³ m

Area of C.S. = πr² = 3.14×10⁻⁶ m²

We know that,

Stress = σ = F/A

σ = 40/3.14×10⁻⁶

σ = 12.73×10⁶ N/m²

Strain = e = Change in dimension / Initial dimension

e = ΔL / L

e = 0.24×10⁻³ / 4

e = 0.6×10⁻³

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