Question

A material has Poisson’s ratio $0.2$. If a uniform rod of its suffers longitudinal strain $4.0\times10^{-3}$, calculate the percentage change in its volume.

• A. $0.15 \%$
• B. $0.02 \%$
• C. $0.24 \%$
• D. $0.48 \%$

Answer 1

Answer: (C)

As $\sigma$ $=-\frac{\Delta R/ R}{\Delta l /l}$
$\therefore $ $\quad$ $\frac{\Delta R}{R}$ $=-\sigma\frac{\Delta l}{l}$ $=-0.2\times4.0\times10^{-3}$ $=-0.8\times10^{-3}$
$V=\pi R^{2}l$
$\therefore $ $\quad$ $\frac{\Delta V}{V}\times100$ $=\left(2\frac{\Delta R}{R}+\frac{\Delta l}{l}\right)\times100$
$=\left[2\times\left(-0.8\times10^{-3}\right)+4.0\times10^{-3}\right]\times100$
$=2.4\times10^{-3}\times100=0.24 \%$.