Question

What is the percentage increase in length of a wire of diameter 2.5mm?

Answer 1

Given a wire of length L,  

  Diameter = 2.5 mm.   =>  Radius R = 1.25 mm = 1.25 * 10⁻³ m

  Cross section area = A = π R² = π * 1.25² * 10⁻⁶ m²

  Young's Modulus  Y = 12.5 * 10¹¹ dynes/cm² = 1.25 * 10¹¹  N/m²

Tensile force applied on the wire along the length 

  F = 100 kg weight = 100 * g Newtons 

      = 1, 000 Newtons    (g = 10 m/s²).

  Stress = σ = F / A = 1, 000 /[1.25² π * 10⁻⁶]  Pa

  Strain = ε = ΔL / L 

Young's law:   Young's modulus Y = stress / strain

       Y = σ / ε

       ε = σ / Y

          = 1, 000 / [1.25² π * 10⁻⁶ * 1.25 * 10¹¹]

          = 1.63 * 10⁻³

% increase in the length of wire = ΔL / L * 100 = 100 * ε

    = 0.163 %