Q133 / 155 · WBJEE 2018 · Physics
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$0.7 \%$
- Correct answerD$0.4 \%$
Solution
$\because$ Stress, $\frac{F}{\Delta A}=1 \%$ of $Y=\frac{Y}{100}$
But Young's modulus, $Y=\frac{\text { stress }}{\text { strain }}=\frac{\frac{F}{\Delta A}}{\frac{\Delta l}{l}}$
$\therefore \quad Y=\frac{\frac{Y}{100}}{\frac{\Delta}{l}} \quad\left(\right.$ putting $\left.\frac{F}{\Delta A}=\frac{Y}{100}\right)$
$\therefore \quad \frac{\Delta l}{l}=\frac{1}{100}$
Poisson's ratio, $\sigma=\frac{-\frac{\Delta r}{r}}{\frac{\Delta l}{l}}$
$\therefore \quad \frac{\Delta r}{r}=-\sigma \cdot \frac{\Delta l}{l}=\frac{-0.3}{100}$
$$
\frac{\Delta r}{r}=\frac{-0.3}{100}
$$
$\therefore$ Change in volume, $\frac{\Delta V}{V}=\frac{2 \Delta r}{r}+\frac{\Delta l}{l}$
$$
\begin{array}{l}
=\frac{2 \times(-0.3)}{100}+\frac{1}{100} \\
=\frac{1-0.6}{100}=\frac{0.4}{100}
\end{array}
$$
$\Delta V \%=0.4 \%$
Hence, the correct answer is option: 'D'.