A sphere of mass $M$ is welded at center of a massless rod and suspended with the help of two wires of same length $L$ and cross-sectional area $A$ with Young’s modulus $Y_{1} a n d Y_{2}$ respectively. Assuming the rod stays horizontal, find the elastic potential energy of the system will be:

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$\frac{M^{2} g^{2} L}{2 A (Y_{1} + Y_{2})}$

$\frac{M^{2} g^{2} L (Y_{1} + Y_{2})}{2 A Y_{1} Y_{2}}$

$\frac{M^{2} g^{2} L}{A (Y_{1} + Y_{2})}$

$\frac{2 M^{2} g^{2} L}{A (Y_{1} + Y_{2})}$

answer is C.

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Detailed Solution

Equivalent spring constant of a wire is given by
$K = \frac{Y A}{L}$
$K_{e q} = K_{1} + K_{2}$
$\frac{Y (2 A)}{L} = \frac{Y_{1} A}{L} + \frac{Y_{2} A}{L}$
Or $Y = \frac{Y_{1} + Y_{2}}{2}$
The system can be replaced as if suspended from a single wire of Young’s modulus Y.
$U = \frac{1}{2} \times s t r e s s \times s t r a i n \times v o l u m e$
$= \frac{1}{2} \times \frac{{(s t r e s s)}^{2}}{Y} A L$
$= \frac{1}{2} {[\frac{M g}{A}]}^{2} \times \frac{1}{Y} \times A L = \frac{M^{2} g^{2} L}{2 A (Y_{1} + Y_{2})}$