Question

A material has Poisson's ratio $0.50$. If a uniform rod of it suffers a longitudinal strain of $2\times {10}^{-3}$, then the percentage change in volume is

• A. $0.6$
• B. $0.4$
• C. $0.2$
• D. zero

Answer 1

Answer: (D)

zero

Given: $\frac{dL}{L}=2\times {10}^{-3}$

Poisson's ratio, $\sigma =-\frac{\frac{dr}{r}}{\frac{dL}{L}}$

where, $r$ and $L$ are the radius and length of the rod respectively

$\Rightarrow$ $0.5=\frac{-\frac{dr}{r}}{2\times {10}^{-3}}$

$\Rightarrow \frac{dr}{r}=-{10}^{-3}$

Volume of the rod, $V=\pi r^{2}L$

Differentiating, we get, $dV=\pi (r^{2}dL+2Lrdr)$

$\Rightarrow \frac{dV}{V}\times 100=\frac{\pi (r^{2}dL+2rLdr)}{\pi r^{2}L}\times 100=(\frac{dL}{L}+2\frac{dr}{r})\times 100$

$\Rightarrow \frac{dV}{V}\times 100=[2\times {10}^{-3}+2(-{10}^{-3}\left)\right]\times 100=0$