Question

The Poisson's ratio of a material is $0.1$. If the longitudinal strain of a rod of this material is $10^{-3}$, then the percentage change in the volume of the rod will be

• A. $0.008 \%$
• B. $0.08 \%$
• C. $0.8 \%$
• D. $8 \%$

Answer 1

Answer: (B)

Longitudinal strain
$\Rightarrow \alpha=\frac{l_{2}-l_{1}}{l_{1}}=10^{-3}$
$\frac{l_{2}}{l_{1}}=1.001$
Poisson's ratio, $\sigma=\frac{\text { lateral strain }}{\text { longitudinal strain }}=\frac{\beta}{\alpha}$
or $\beta=\sigma \alpha=0.1 \times 10^{-3}=10^{-4}=\frac{r_{1}-r_{2}}{r_{1}}$
or $\frac{r_{2}}{r_{1}}=1-10^{-4}=0.9999$
$\%$ increase in volume $=\left(\frac{V_{2}-V_{1}}{V_{1}}\right) \times 100$
$=\left(\frac{\pi r_{2}^{2} l_{2}-\pi r_{1}^{2} l_{1}}{\pi r_{1}^{2} l_{1}}\right) \times 100=\left(\frac{r_{2}^{2} l_{2}}{r_{1}^{2} l_{1}}-1\right) \times 100$
$=\left[(0.9999)^{2} \times 1.001-1\right] \times 100=0.08 \%$