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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 and subjected to the same force will be $\qquad$ $\mathrm{cm}$.
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$2 \mathrm{~cm}$
$\begin{aligned} & \mathrm{F}=\mathrm{Y} . \mathrm{A} \cdot \frac{\Delta \ell}{\ell} \\ & \Delta \ell=\frac{\mathrm{F}}{\text { Y.A. }} . \ell \\ & \Delta \ell=\frac{\mathrm{F} \cdot \ell}{\mathrm{Y} . \pi \mathrm{r}^2} \\ & \Delta \ell \propto \frac{\ell}{\mathrm{r}^2} \\ & \frac{\Delta \ell_2}{\Delta \ell_1}=\left(\frac{\ell_2}{\ell_1}\right)\left(\frac{\mathrm{r}_1}{\mathrm{r}_2}\right)^2=\text { (2) }\left(\frac{1}{2}\right)^2 \\ & \frac{\Delta \ell_2}{\Delta \ell_1}=\frac{1}{2} \\ & \Delta \ell_2=\frac{\Delta \ell_1}{2} \\ & =\frac{0.04}{2}=0.02 \mathrm{~m} \\ & \end{aligned}$
$\Delta \ell_2=2 \mathrm{~cm}$
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