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 ___________ $\mathrm{cm}$.
$\mathrm{F}=\mathrm{Y} . \mathrm{A} . \frac{\Delta \ell}{\ell}$
$\Delta \ell=\frac{\mathrm{F}}{\mathrm{Y} \cdot \mathrm{A} .} \ell$
$\Delta \ell=\frac{\mathrm{F} \ell}{\mathrm{Y} \cdot \pi \mathrm{r}^{2}}$
$\Delta \ell \propto \frac{\ell}{\mathrm{r}^{2}}$
$\frac{\Delta \ell_{2}}{\Delta \ell_{1}}=\left(\frac{\ell_{2}}{\ell_{1}}\right)\left(\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}\right)^{2}$
$=(2)\left(\frac{1}{2}\right)^{2}$
$\frac{\Delta \ell_{2}}{\Delta \ell_{1}}=\frac{1}{2}$
$\Delta \ell_{2}=\frac{\Delta \ell_{1}}{2}$
$=\frac{0.04}{2}$
$=0.02 \mathrm{~m}$
$\Delta \ell_{2}=2 \mathrm{~cm}$
Ans. $=2$
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