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Q.

A mass M is gently hung from one end of a light wire suspended from ceiling through other end. The elastic energy stored in the wire would be (Y : Young’s modulus, A: area of cross-section, l: length of wire)

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a

$\frac{M^{2} g^{2} l}{3 AY}$

b

$\frac{Mgl}{2}$

c

$\frac{M^{2} g^{2} l}{2 AY}$

d

$\frac{Mgl}{4}$

answer is C.

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Detailed Solution

$\begin{array}{l} Y = \frac{(F / A)}{(Δ / ll)} \\ Energy = \frac{1}{2} stress \times strain \times volume \\ Y = \frac{stress}{strain} \Rightarrow strain = \frac{stress}{Y} = \frac{Mg}{AY} \\ \Rightarrow Energy = \frac{1}{2} \times \frac{Mg}{A} \times \frac{Mg}{AY} \times Al \\ = \frac{M^{2} g^{2} l}{2 AY} \end{array}$

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