Relation Between Volumetric Strain, Lateral Strain And Poisson’s Ratio MCQ - Practice Questions with Answers

Let us long rod have a length L and radius ' r ', then the volume of this rod $=\pi r^{2}L$.
Now, Differentiating both the sides of (1), we get

$$dV=\pi r^{2}dL+\pi 2rLdr$$

Now, dividing both the sides by volume of the rod, i.e., $\pi r^{2}L$, we get,

$$\frac{dV}{V}=\frac{\pi r^{2}dL}{\pi r^{2}L}+\frac{\pi 2rLdr}{\pi r^{2}L}=\frac{dL}{L}+2\frac{dr}{r}\dots ($$

So we can say that,
Volumetric strain $=$ Longitudinal strain +2 (Lateral strain)
Also, equation(2) can be written as,

$$\begin{array}{r}\Rightarrow \frac{dV}{V}=\frac{dL}{L}-2\sigma \frac{dL}{L}=(1-2\sigma )\frac{dL}{L}\\ \text{ This is because, }[\sigma =\frac{-dr/r}{dL/L}\Rightarrow \frac{dr}{r}=-\sigma \frac{dL}{L}]\end{array}$$
Special case -
- When $\sigma =0.5$, then $dV=0$. It means that the substance is incompressible, so there is no change in volume.
- If a material having $\sigma =0$, it means the lateral strain is zero. So, when a substance is stretched its length increases without any decrease in diameter. For example - cork has $\sigma =0$. Also, in this case, change in volume is maximum.