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

Option C is the correct answer.

Step-by-step explanation:

We equation for elongation

`\Delta L=(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=(PL)/(5* 10^(-5)* 2* 10^(11))\\\\P=10^7N`

Option C is the correct answer.