Question

A rod has poisson's ratio $0.2$. If a rod suffers a longitudinal strain of 2 x 10${}^{-3}$, then the percentage change in volume is

• A. $+0.12$
• B. $-0.12$
• C. $0.28$
• D. $-0.28$

Answer 1

The correct option is A $+0.12$

$\sigma =0.2$
$ϵx=2\times {10}^{-3}$
$0.2=-\frac{ϵy}{2\times {10}^{-3}}$
$-0.4\times {10}^{-3}=ϵy$
$Volume=\pi R^{2}L$

$lnV=ln\pi +2lnR+lnl$

Differentiating

$\frac{dV}{V}=0+2\frac{dr}{r}+\frac{dl}{l}$

$\frac{dV}{V}=changeinvolume=2\left(LateralStrain\right)+\left(LongitudinalStrain\right)$

$ChangeinVolume=2(-0.4\times {10}^{-3})+2\times {10}^{-3}$

$ChangeinVolume=(-0.8+2)\times {10}^{-3}$

$ChangeinVolume=0.12\times {10}^{-2}$

So option A is correct answer