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To the free end of spring hanging from a rigid support, a block of mass m is hung and slowly allowed to come to its equilibrium position. Then stretching in the spring is d. If the same block is attached to the same spring and allowed to fall suddenly, the amount of stretching is : (force constant , k)

\[\begin {array} {1 1} (a)\;\frac{mg}{k} & \quad (b)\;2d \\ (c)\;\frac{mg}{3k} & \quad (d)\;4d \end {array}\]
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answered Nov 7, 2013 by pady_1

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