Question

A material has Poisson's ratio $0.5$. If a uniform rod made of it suffers a longitudinal strain $2\times {10}^{-3}$. What is the percentage increase in volume?

Answer 1

We know that, Volume $V=\pi r^{2}L$

Therefore, $\frac{\Delta V}{V}=\frac{\Delta L}{L}+2\frac{\Delta r}{r}=\frac{\Delta L}{L}-2\sigma \frac{\Delta L}{L}$
$\because \sigma =\frac{\text{Lateral strain}}{\text{Longitudional strain}}=\frac{\frac{-dr}{r}}{\frac{dL}{L}}$

where, r and L are the radius and length of the rod respectively
$\frac{\Delta V}{V}=2\times {10}^{-3}-2\times {10}^{-3}=0$

$\therefore$ % increase in volume $=0\%$.