A material has Poisson's ratio $0.5$. If a uniform rod of it suffers a longitudinal strain of $3 \times 10^{-3}$, what will be percentage increase in volume is .......... $\%$
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(d)
$\frac{\text { Lateral strain }}{\text { Longitudinal strain }}=\eta=0.5$
$\frac{-\Delta r / r}{\Delta l / l}=\frac{1}{2}$
$\frac{-2 \Delta r}{r}=\frac{\Delta l}{l}$
Magnitute wise both are equal but sign's would be different as both quantities cannot increase
Now volume $\propto$ area $\times$ length $v \propto r^2 \cdot L$
$\frac{\Delta V}{V}=\frac{2 \Delta r}{r}+\frac{\Delta L}{L}$
Substituting value of $\frac{\Delta L}{L}$
$\frac{\Delta V}{V}=0$
$\therefore$ No change in volume.
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