Question

A material has Poisson's ratio of 0.5. lf a uniform rod suffers a longitudinal strain of $2\times {{10}^{-3}}$, the percentage increase in its volume is:
A. 2%
B. 0.5%
C. 4%
D. 0%

Answer 1

To find the percentage increase in volume, we use the relation between the volume change of a material due to strain and the Poisson’s ratio. The ratio of the change in length to the original length is termed as longitudinal strain.
Formula used:
The relationship between change in volume and Poisson’s ratio is given by,
$\dfrac{\Delta V}{V}=\left( 1-2\sigma \right)\left( \dfrac{\Delta L}{L} \right)$
where, $\Delta V$ is change in volume, V is original volume, $\sigma$ is Poisson’s ratio, $\Delta L$ is change in length and L is original length.

Complete answer:
Given,
$\sigma =0.5$
Longitudinal strain = $2\times {{10}^{-3}}$
Therefore,
$\dfrac{\Delta L}{L}=2\times {{10}^{-3}}$
Substituting these values in $\dfrac{\Delta V}{V}=\left( 1-2\sigma \right)\left( \dfrac{\Delta L}{L} \right)$, we get,
$\dfrac{\Delta V}{V}=\left( 1-2\sigma \right)\left( \dfrac{\Delta L}{L} \right)$
$\dfrac{\Delta V}{V}=\left( 1-2\left( 0.5 \right) \right)\left( 2\times {{10}^{-3}} \right)$
$\dfrac{\Delta V}{V}=\left( 1-1 \right)\left( 2\times {{10}^{-3}} \right)$
$\dfrac{\Delta V}{V}=0\times \left( 2\times {{10}^{-3}} \right)$
$\dfrac{\Delta V}{V}=0$
Since, the ratio between change in volume and original volume is zero, the percentage increase in volume will be 0%

The answer is option D.

Note:
The relationship between change in volume and Poisson’s ratio is derived as follows:
Volume of the rod is given by,
$V=\pi {{r}^{2}}l$
Where, r is radius and l is length.
Differentiating both sides, we get,
$dV=\pi {{r}^{2}}dl+\pi 2rldr$
Dividing both sides by volume, we get,
$\dfrac{dV}{V}=\dfrac{\pi {{r}^{2}}dl}{\pi {{r}^{2}}l}+\dfrac{\pi 2rldr}{\pi {{r}^{2}}l}$
Cancelling out common terms,
$\dfrac{dV}{V}=\dfrac{dl}{l}+\dfrac{2dr}{r}$
Now, Poisson’s ratio is given by,
$\sigma =\dfrac{-\left( \dfrac{dr}{r} \right)}{\dfrac{dl}{l}}$
Therefore,
$\dfrac{dr}{r}=-\sigma \dfrac{dl}{l}$
Substituting this value in $\dfrac{dV}{V}=\dfrac{dl}{l}+\dfrac{2dr}{r}$, we get,
$\dfrac{dV}{V}=\dfrac{dl}{l}+\dfrac{2dr}{r}$
$\dfrac{dV}{V}=\dfrac{dl}{l}-2\sigma \dfrac{dl}{l}$
$\dfrac{dV}{V}=\left( 1-2\sigma \right)\dfrac{dl}{l}$
The radius of the rod decreases as length increases, hence it is taken as negative.