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. 0.03
• B. 0.01
• C. 0.007
• D. 0.004

Answer 1

Answer: (D)

0.004

Answer 2

$\because$ Stress, $\frac{F}{\Delta A}=1 \%$ of $Y=\frac{Y}{100}$
But Young's modulus, $Y=\frac{\text { stress }}{\text { strain }}=\frac{\frac{F}{\Delta A}}{\frac{\Delta l}{l}}$
$\therefore Y=\frac{\frac{Y}{100}}{{\frac{\Delta l}{l}}} \,\,\left(\right.$ putting $\left.\frac{F}{\Delta A}=\frac{Y}{100}\right)$
$\therefore \frac{\Delta l}{l}=\frac{1}{100}$
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, $\frac{\Delta V}{V}=\frac{2 \Delta r}{r}+\frac{\Delta l}{l}$
$=\frac{2 \times(-0.3)}{100}+\frac{1}{100}$
$=\frac{1-0.6}{100}=\frac{0.4}{100}$
$\therefore \Delta V \% =0.4 \%$