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

Hint: 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.