Question

Calculate the percentage increases in length of a wire of diameter 2.5mm stretched by a force of 100kg weight . Young's modules of elasticity of wire is 12.5
.

10 to the power of 11

Answer 1

Hi,

Here is your answer,

Here,

            2r = 2.5 mm = 0.25
or         r = 0.125 cm

 Therefore,            a = πr² = 22/7 × (0.125)² sq.cm



                        F = 10 Kg = 100 × 1000 g 

                        F = 10 × 1000 × 980 dyne 

                        Y = 12.5 × 10¹¹ dyne/sq.cm

 As,                         →  Y= F × l/ a ×Δl

∴   ΔL/L = F/aY

Hence, Percentage increases in length


      = ΔL/L × 100 = F/aY × 100

      = (100×1000×980) × 7 × 100/ 22 × (0.125)² × 125 × 10¹¹  

      = 0.1812 Percent 



Hope it helps you !

Answer 2

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 %