Question

If a uniform rod made of material having poisson's ratio $0.5$, suffers longitudinal strain of $2\times {10}^{-3}$, what is the percentage increase in volume?

• A. $2\%$
• B. $4\%$
• C. $0\%$
• D. $5\%$

Answer 1

The correct option is C $0\%$

We know that,
$\frac{\Delta V}{V}=\frac{\Delta \left(\pi r^{2}L\right)}{\pi r^{2}L}$
$=\frac{\pi r^2\Delta L+\pi L2r.\Delta r}{\pi r^{2}L}$
$\Rightarrow \frac{\Delta V}{V}=\frac{\Delta L}{L}+2\frac{\Delta r}{r}.....\left(1\right)$
Now, Poisson's ratio is,
$\mu =-\frac{\left(\Delta r/r\right)}{\left(\Delta L/L\right)}$
or $\frac{\Delta r}{r}=-\mu \frac{\Delta L}{L}$
Given,
Longitudinal strain, $\frac{\Delta L}{L}=2\times {10}^{-3}$
$\therefore \frac{\Delta r}{r}=-0.5\times 2\times {10}^{-3}=-1\times {10}^{-3}$
Hence, from eq$\left(1\right)$, we get
$\frac{\Delta V}{V}=2\times {10}^{-3}-2\times {10}^{-3}=0$
$\therefore \%$ increase in volume $=0$