# A wire is stretched by a certain amount under a load

A wire is stretched by a certain amount under a load. If the load and radius both are increased to four times, find the stretch caused in the wire.

∆l ∝ F/A
Given , ∆l’ ∝F’/A’
But F’ = 4F
A’ = $\pi$ 4r x 4r = 16 $\pi$ ${{r}^{2}}$
So , ∆l’ ∝ 4F / 16A

∆l’/ ∆l = 4F/F x A/16A = 1/4