Question

A material has Poisson's ratio $0.5$. If a uniform rod made of the material suffers a 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 $0\%$

Given,
Longitudinal strain$\left(\frac{\Delta L}{L}\right)=2\times {10}^{-3}$
Poisson's ratio $\left(\sigma \right)=0.5$
Fractional change in volume of the cylindrical wire is given by,
$\frac{\Delta V}{V}=\frac{\Delta \left(\pi r^{2}L\right)}{\pi r^{2}L}=\frac{\Delta L}{L}+2\frac{\Delta r}{r}......\left(1\right)$
$\because$ $\sigma =-\frac{\left(\Delta r/r\right)}{\left(\Delta L/L\right)}$
or $\frac{\Delta r}{r}=-\sigma \frac{\Delta L}{L}.....\left(2\right)$
Using $\left(2\right)$ in $\left(1\right)$, we can write
$\frac{\Delta V}{V}=(1-2\sigma )\frac{\Delta L}{L}$
$\therefore \frac{\Delta V}{V}=2\times {10}^{-3}-2\times {10}^{-3}=0$
Hence, there is no fractional change in volume of the cylindrical wire.
Thus, option (c) is the correct answer.