Q153 / 180 · KCET 2022 · Physics
A metallic rod breaks when strain produced is $0.2 \%$. The Young's modulus of the material of the $\operatorname{rod} 7 \times 10^9 \mathrm{~N} / \mathrm{m}^2$. The area of crosssection to support a load of $10^4 \mathrm{~N}$ is
- A$7.1 \times 10^{-6} \mathrm{~m}^2$
- Correct answerB$7.1 \times 10^{-4} \mathrm{~m}^2$
- C$7.1 \times 10^{-2} \mathrm{~m}^2$
- D$7.1 \times 10^{-8} \mathrm{~m}^2$
Solution
Given, Youngs' modulus, $Y=7 \times 10^9 \mathrm{~N} / \mathrm{m}^2$ Load, $F=10^4 \mathrm{~N}$
We know that.
$$
\begin{aligned}
Y & =\frac{F l}{A \Delta l} \\
\Rightarrow \quad A & =\frac{F l}{Y \Delta l}=\frac{F}{Y} \times \frac{1}{\left(\frac{\Delta l}{l}\right)} \\
& =\frac{10^4}{7 \times 10^9} \times \frac{1}{0.2}=7.1 \times 10^{-4} \mathrm{~m}^2
\end{aligned}
$$
Hence, the correct answer is option: 'B'.