A smooth chain of mass m rests against a surface in the form of a quarter of circle of radius ‘r’. If it is released from, the velocity of chain after it cmes over the horizontal part of the surface is

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2gr

2gr(2−π)

2gr1−2π

answer is D.

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Detailed Solution

W=12mv2−u2mghi−hf=12mv2−02gr−2rπ=12v2∵dm=mdθπ/2, h=R1-cosθ;U= ∫0π/2du= ∫0π/2dmghv=2gr1−2π