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A metal wire of length $L$ is suspended vertically from a rigid support. When a body of mass $M$ is attached to the lower end of wire and released suddenly with the wire in its natural length. The cross sectional area of the wire is A and Young's modulus of its material is Y . Then

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maximum elongation in the wire is $\frac{M g L}{2 A Y}$

maximum elongation in the wire is $\frac{2 M g L}{A Y}$

the maximum elastic potential energy stored in the wire is $\frac{2 M^{2} g^{2} L^{2}}{A^{2} Y}$

the maximum elastic potential energy stored in the wire is $\frac{4 M^{2} g^{2} L^{2}}{A^{2} Y}$

answer is B, C.

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Detailed Solution

Mg $x_{m a x}$ = $\frac{1}{2} \frac{A Y}{L} {x_{m a x}}^{2} \Rightarrow x_{m a x} = \frac{2 M g L}{A Y}$

$S t r a i n \in = \frac{2 M g}{A Y} U_{m a x} = \frac{1}{2} Y \in^{2} = \frac{2 M^{2} g^{2} L^{2}}{A^{2} Y}$

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