A uniform metallic wire is elongated by $0.04 \mathrm{~m}$ when subjected to a linear force $\mathrm{F}$.
The elongation, if its length and diameter is doubled andsubjected to the same force will be $\mathrm{cm}$.
$\mathrm{y}=\frac{\mathrm{F} / \mathrm{A}}{\Delta \ell / \ell}$
$\Rightarrow \frac{F}{A}=y \frac{\Delta \ell}{\ell}$
$\Rightarrow \frac{F}{A}=y \times \frac{0.04}{\ell} \quad \ldots(1)$
When length $\backslash \&$ diameter is doubled.
$\Rightarrow \frac{F}{4 A}=y \times \frac{\Delta \ell}{2 \ell} \quad \cdots(2)$
$(1) \div(2)$
$\frac{\mathrm{F} / \mathrm{A}}{\mathrm{F} / 4 \mathrm{~A}}=\frac{\mathrm{y} \times \frac{0.04}{\ell}}{\mathrm{y} \times \frac{\Delta \ell}{2 \ell}}$
$4=\frac{0.04 \times 2}{\Delta \ell}$
$\Delta \ell=0.02$
$\Delta \ell=2 \times 10^{-2}$
$\therefore x=2$