Q95 / 125 · BITSAT 2009 · Physics
A metallic rod breaks when strain produced is $0.2 \% .$ The Young's modulus of the material of the rod is $7 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2} .$ What should be its area of cross-section to support a load of $10^{4} \mathrm{~N}$ ?
- A$7.1 \times 10^{-8} \mathrm{~m}^{2}$
- B$7.1 \times 10^{-6} \mathrm{~m}^{2}$
- Correct answerC$7.1 \times 10^{-4} \mathrm{~m}^{2}$
- D$7.1 \times 10^{-2} \mathrm{~m}^{2}$
Solution
$\quad$ Maximum possible strain $=0.2 / 100$
$$
\begin{array}{l}
\therefore \mathrm{A}=\frac{\mathrm{F}}{\mathrm{Y} \times \text { strain }} \\
=\frac{10^{4} \times 100}{\left(7 \times 10^{9}\right) \times 0.2}=7.1 \times 10^{-4} \mathrm{~m}^{2}
\end{array}
$$
Hence, the correct answer is option: 'C'.