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 $=\left(\frac{V_2-V_1}{V_1}\right)\times 100$
$=\left(\frac{\pi r_2^2{l}_2-\pi r_1^2{l}_1}{\pi r_1^2{l}_1}\right)\times 100=\left(\frac{r_2^2{l}_2}{r_1^2{l}_1}-1\right)\times 100$
$=\left[(0.9999{)}^2\times 1.001-1\right]\times 100=0.08\%$