An deal spring having force constant k is suspended from the rigid support and a block of mass Mis attached to its lower end. The mass is released from the natural length of the spring. Then the maximum extension in the spring is

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Solution

The correct option is B

$Potential energy = \frac{1}{2} k x^{2} = M g x \Rightarrow x = \frac{2 M g}{k}$

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