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

Option C is the correct answer.

Explanation:

We equation for elongation

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

Option C is the correct answer.

moving ball hits still ball. mover stops, hit ball moves. momentum transfer.

unlike poles attract