Question

A material has Poisson’s ratio $0.2$. If a uniform rod of its suffers longitudinal strain $4.0\times 10^{-3}$, calculate the percentage change in its volume.

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

Answer 1

Answer: (C)

As $\sigma$ $=-\frac{\Delta R/R}{\Delta l/l}$
$\therefore$ $\frac{\Delta R}{R}$ $=-\sigma \frac{\Delta l}{l}$ $=-0.2\times 4.0\times 10^{-3}$ $=-0.8\times 10^{-3}$
$V=\pi R^{2}l$
$\therefore$ $\frac{\Delta V}{V}\times 100$ $=\left(2\frac{\Delta R}{R}+\frac{\Delta l}{l}\right)\times 100$
$=\left[2\times \left(-0.8\times 10^{-3}\right)+4.0\times 10^{-3}\right]\times 100$
$=2.4\times 10^{-3}\times 100=0.24\%$.