Question

Calculate the force required to double the length of a steel wire of area of cross section 5 x 10-5 m2? (Y=2x 1011N m2) (a) 10-5 N (b) 10-7 N n's D (c) 107 N (d) 105 N

Answer 1

Answer:

Option C is the correct answer.

Explanation:

We equation for elongation

      $\Delta L=\frac{PL}{AE}$

Here we need to find load required,

We need to double the wire, that is ΔL = 2L - L = L

A = 5 x 10⁻⁵ m²

E = 2 x 10¹¹ N/m²

Substituting

     $L=\frac{PL}{5\times 10^{-5}\times 2\times 10^{11}}\\\\P=10^7N$

Option C is the correct answer.