The ratio of diameters of two wires of same material is $n : 1$. The length of wires are $4\, m$ each. On applying the same load, the increase in length of thin wire will be
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$1 \propto \frac{\mathrm{FL}}{\mathrm{r}^{2} \mathrm{Y}} \Rightarrow 1 \propto \frac{1}{\mathrm{r}^{2}}[\mathrm{F}, \mathrm{L} \text { and } \mathrm{Y} \text { are constant }]$
$\frac{1_{2}}{\mathrm{l}_{1}}=\left(\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}\right)^{2}=(\mathrm{n})^{2} \Rightarrow 1_{2}=\mathrm{n}^{2} \mathrm{l}_{1}$
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