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In the figure shown, pulley and spring are ideal. If k is spring constant of spring, the potential energy stored in it is $(m_{1} > m_{2})$

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$\frac{2 {m_{2}}^{2} g^{2}}{k}$

$\frac{1}{2} \frac{{(m_{1} - m_{2})}^{2} g^{2}}{k}$

$\frac{2 {m_{1}}^{2} g^{2}}{k}$

$\frac{{(m_{1} + m_{2})}^{2} g^{2}}{k}$

answer is A.

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Detailed Solution

$T = m_{1} g$

$kx = 2 T = 2 m_{1} g$

$x = \frac{2 m_{1} g}{k}$

$Energy stored = \frac{1}{2} {kx}^{2} = \frac{1}{2} k \frac{4 {m^{2}}_{1} g^{2}}{k^{2}} = \frac{2 {m^{2}}_{1} g^{2}}{k}$

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