A) \[y=\,\frac{mglL}{\pi {{r}^{2}}l}\]
B) \[y=\,\frac{mgL}{\pi rl}\]
C) \[y=\,\frac{mg{{\pi }^{2}}}{\pi L}\]
D) \[y=\,\frac{mgr}{\pi L}\]
Correct Answer: A
Solution :
\[Y=\frac{Stress}{Strain}=\frac{F/A}{l/L}=\frac{FL}{Al}\] \[=\frac{mgL}{\pi {{r}^{2}}l}\]
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