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In the given figure a ball of mass m is connected to an elastic spring of force constant K = mg/L through an inextensible string as shown. Now the ball is released from rest.

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The maximum potential energy stored in the spring during the fall is $mgL [2 - \sqrt{3}]$

The maximum velocity of mass m during fall is $\sqrt{3 gL}$

The maximum velocity of mass m during fall is $\sqrt{2 gL}$

The maximum potential energy stored in the spring during the fall is $mgL [2 + \sqrt{3}]$

answer is A, C.

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Detailed Solution

Using energy conservation between initial moment and equilibrium moment when

$\begin{array}{l} T = mg \\ \frac{1}{2} \frac{mg}{L} \times L^{2} + \frac{1}{2} {mv}^{2} = mg \times 2 L \\ v = \sqrt{3 gL} \end{array}$

If ball falls a maximum distance of L + x.

$\begin{array}{l} mg (L + x) = \frac{1}{2} \frac{mg}{L} x^{2} x = L (1 + \sqrt{3}) ∴ {PE}_{max} = mgL (2 + \sqrt{3}) \end{array}$

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