JEE MAIN - Physics (2023 - 6th April Morning Shift - No. 22)
A steel rod has a radius of $$20 \mathrm{~mm}$$ and a length of $$2.0 \mathrm{~m}$$. A force of $$62.8 ~\mathrm{kN}$$ stretches it along its length. Young's modulus of steel is $$2.0 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2}$$. The longitudinal strain produced in the wire is _____________ $$\times 10^{-5}$$
Answer
25
Explanation
$$
\begin{aligned}
& \text { Strain }=\frac{\text { stress }}{Y}=\frac{\frac{62.8 \times 10^3}{\pi \times(0.02)^2}}{2 \times 10^{11}} \\\\
& =\frac{62.8 \times 10^3}{3.14 \times 4 \times 10^{-4} \times 2 \times 10^{11}} \\\\
& =2.5 \times 10^{-4} \\\\
& =25 \times 10^{-5}
\end{aligned}
$$
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