Question

The stress along the length of a rod (with rectangular cross section) is $1\%$ of the Young's modulus of its material. What is the approximate percentage of change of its volume ? (Poisson's ratio of the material of the rod is $0.3$)

• A. 3%
• B. 1%
• C. 0.7%
• D. 0.4%

Answer 1

Answer: (D)

$\because$ Stress, $\frac{F}{\Delta A}=1\%$ of $Y=\frac{Y}{100}$
But Young's modulus, $Y=\frac{\text{ stress }}{\text{ strain }}=\frac{\frac{F}{\Delta A}}{\frac{\Delta l}{l}}$
$\therefore Y=\frac{\frac{Y}{100}}{\frac{\Delta l}{l}}($ putting $\frac{F}{\Delta A}=\frac{Y}{100})$
$\therefore \frac{\Delta l}{l}=\frac{1}{100}$
Poisson's ratio, $\sigma =\frac{-\frac{\Delta r}{r}}{\frac{\Delta l}{l}}$
$\therefore \frac{\Delta r}{r}=-\sigma \cdot \frac{\Delta l}{l}=\frac{-0.3}{100}$
$\frac{\Delta r}{r}=\frac{-0.3}{100}$
$\therefore$ Change in volume, $\frac{\Delta V}{V}=\frac{2\Delta r}{r}+\frac{\Delta l}{l}$
$=\frac{2\times (-0.3)}{100}+\frac{1}{100}$
$=\frac{1-0.6}{100}=\frac{0.4}{100}$
$\therefore \Delta V\%=0.4\%$