The Poisson's ratio of a material is $0.5$. If a force is applied to a wire of this material, there is a decrease in the cross-sectional area by $4 \%$. The percentage increase in the length is ........ $\%$
Medium
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(d)
$\frac{\text { Lateral strain }}{\text { Longitudinal strain }}=\eta$
$\frac{\Delta r / r}{\Delta l / I}=0.5$
Substitute $\Delta r / r=2 / 100$
$\frac{\Delta l}{l}=\frac{4}{100}$
$\therefore\% \text { increase }=\frac{\Delta l}{l} \times 100=4 \%$
$\because A \propto r^2$
So $\frac{\Delta A}{A}=\frac{2 \Delta r}{r}$
$\frac{4}{100}=2 \times \frac{\Delta r}{r}$
$\frac{2}{100}=\frac{\Delta r}{r}$
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