A light spring of length l and spring constant K is placed vertically on the table. A small ball of mass m is released on it. Height from the surface of the table at which the ball will have maximum velocity

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l−2mgK

Greater than l

l−mgK

answer is D.

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Detailed Solution

Acceleration a = mg - k∆xVelocity will be maximum when a = 0∴Δx=mgkSo height h = l−mgK

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