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# A material has poisson's ratio $0.50$. If a uniform rod of it suffers a longitudinal strain of $2×{10}^{-3}$, then the percentage change in volume is

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Solution

## Step 1. Given data: Poisson's ratio $\left(\alpha \right)$= $0.50$ Longitudinal strain $\left(\frac{dl}{l}\right)$ = $2×{10}^{-3}$Step 2. Finding percentage change in volumeWe know, Poisson's ratio = $-\frac{dr}{r}}{dl}{l}}$where $r$ and $l$ are the radius and length of the given rod respectively.$-\frac{dr}{r}}{dl}{l}}=.50\phantom{\rule{0ex}{0ex}}dr}{r}=-.50×dl}{l}\phantom{\rule{0ex}{0ex}}=-.50×2×{10}^{-3}\phantom{\rule{0ex}{0ex}}=-{10}^{-3}$Hence, $dr}{r}=-{10}^{-3}$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 $\frac{dV}{V}×100=\frac{\mathrm{\pi }\left({\mathrm{r}}^{2}\mathrm{dl}+2\mathrm{l}.\mathrm{rdr}\right)}{{\mathrm{\pi r}}^{2}\mathrm{l}}×100=\left(\frac{dl}{l}+2\frac{dr}{r}\right)×100$ $\frac{dV}{V}×100=\left(\left(2×{10}^{-3}\right)+2\left(-{10}^{-3}\right)\right)×100=0$Thus, the percentage change in volume is zero.

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