Let the length of wire = L
Radius = r
When a force F is applied, then increase in length = $\left(l\right)$
Since, Young’s modulus
$Y=\frac{\frac{F}{\pi r^2}}{\frac{l}{L}}\Rightarrow Y=\frac{FL}{\pi r^{2}l}$ …(i)
For another wire,
Length of wire, L’ = 2L
Radius of wire, r’ = 2r
Applied force, F’ = 2F
$\therefore Y=\frac{F^{\text{'}}{L}^{\text{'}}}{\pi r^2{l}^{\text{'}}}$ where, $l$ ’ = elongation in length
$=\frac{2F.2L}{\pi {\left(2r\right)}^2.c^{\text{'}}}=\frac{4FL}{4\pi r^2{l}^{\text{'}}}$
$Y=\frac{FL}{\pi r^2{l}^{\text{'}}}$ ….(ii)
From Eqs. (i) and (ii), we have $l$ ’ = $l$