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

Answer 1

Step 1. Given data:

Poisson's ratio $\left(\alpha \right)$= $0.50$

Longitudinal strain $\left(\frac{dl}{l}\right)$ = $2\times {10}^{-3}$

Step 2. Finding percentage change in volume

We know, Poisson's ratio = $-\frac{{\displaystyle \raisebox{1ex}{$dr$}\!\left/ \!\raisebox{-1ex}{$r$}\right.}}{{\displaystyle \raisebox{1ex}{$dl$}\!\left/ \!\raisebox{-1ex}{$l$}\right.}}$

where $r$ and $l$ are the radius and length of the given rod respectively.

$$\begin{gathered} -\frac{{\displaystyle \raisebox{1ex}{$dr$}\!\left/ \!\raisebox{-1ex}{$r$}\right.}}{{\displaystyle \raisebox{1ex}{$dl$}\!\left/ \!\raisebox{-1ex}{$l$}\right.}}=.50 \\ \raisebox{1ex}{$dr$}\!\left/ \!\raisebox{-1ex}{$r$}\right.=-.50\times \raisebox{1ex}{$dl$}\!\left/ \!\raisebox{-1ex}{$l$}\right. \\ =-.50\times 2\times {10}^{-3} \\ =-{10}^{-3} \end{gathered}$$

Hence, $\raisebox{1ex}{$dr$}\!\left/ \!\raisebox{-1ex}{$r$}\right.=-{10}^{-3}$

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

Differentiating partially, we get

$dV=\mathrm{\pi }\left({\mathrm{r}}^2\mathrm{dl}+2\mathrm{l}.\mathrm{rdr}\right)$

Or $\frac{dV}{V}\times 100=\frac{\mathrm{\pi }\left({\mathrm{r}}^2\mathrm{dl}+2\mathrm{l}.\mathrm{rdr}\right)}{{\mathrm{\pi r}}^2\mathrm{l}}\times 100=\left(\frac{dl}{l}+2\frac{dr}{r}\right)\times 100$

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

Thus, the percentage change in volume is zero.