A uniform metallic wire is elongated by 0.04 m

$\mathrm{y}=\frac{\mathrm{F} / \mathrm{A}}{\Delta \ell / \ell}$

$\Rightarrow \frac{F}{A}=y \frac{\Delta \ell}{\ell}$

$\Rightarrow \frac{F}{A}=y \times \frac{0.04}{\ell} \quad \ldots(1)$

When length $\backslash \&$ diameter is doubled.

$\Rightarrow \frac{F}{4 A}=y \times \frac{\Delta \ell}{2 \ell} \quad \cdots(2)$

$(1) \div(2)$

$\frac{\mathrm{F} / \mathrm{A}}{\mathrm{F} / 4 \mathrm{~A}}=\frac{\mathrm{y} \times \frac{0.04}{\ell}}{\mathrm{y} \times \frac{\Delta \ell}{2 \ell}}$

$4=\frac{0.04 \times 2}{\Delta \ell}$

$\Delta \ell=0.02$

$\Delta \ell=2 \times 10^{-2}$

$\therefore x=2$