A material has poissons ratio 0.5. If a uniform rod of it suffers a longitudinal strain of 2 × 10–3, then the percentage increase in its volume is

# A material has poissons ratio 0.5. If a uniform rod of it suffers a longitudinal strain of 2 × 10–3, then the percentage increase in its volume is

1. A

0%

2. B

10%

3. C

20%

4. D

5%

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### Solution:

Given,

$\frac{\mathrm{dL}}{\mathrm{L}}=2×{10}^{-3}$

The relation between a material's volume change caused by strain and the poison's ratio is as follows:

$\mathrm{\sigma }=\frac{\frac{\mathrm{dr}}{\mathrm{\sigma r}}}{\frac{\mathrm{dL}}{\mathrm{L}}}$

$0.5=\frac{-\frac{\mathrm{dr}}{\mathrm{r}}}{2×{10}^{-3}}$

$\frac{\mathrm{dr}}{\mathrm{r}}=-{10}^{-3}$

We know that volume of the rod is,

$\mathrm{V}={\mathrm{\pi r}}^{2}\mathrm{L}$

Differentiating,

$\mathrm{dV}=\mathrm{\pi }\left({\mathrm{r}}^{2}\mathrm{dL}+2\mathrm{Lrdr}\right)$

$\frac{\mathrm{dV}}{\mathrm{V}}×100=\frac{\mathrm{\pi }\left({\mathrm{r}}^{2}\mathrm{dL}+2\mathrm{rLdr}\right)}{{\mathrm{\pi r}}^{2}\mathrm{L}}×100=\left(\frac{\mathrm{dL}}{\mathrm{L}}+2\frac{\mathrm{dr}}{\mathrm{r}}\right)×100$

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

Hence the correct answer is 0%.

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