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Question: A smooth chain PQ of mass M rests against a \(\frac { 1 } { 4 }\)th circular and smooth surface of r...

A smooth chain PQ of mass M rests against a 14\frac { 1 } { 4 }th circular and smooth surface of radius r. If released, its velocity to come over the horizontal part of the surface is

A

B

C

D

gr⁡(1−2π)\sqrt { \operatorname { gr } \left( 1 - \frac { 2 } { \pi } \right) }

Answer

Explanation

Solution

Work done = ∆KE

Mg (r−2rπ)=12Mv2\left( \mathrm { r } - \frac { 2 \mathrm { r } } { \pi } \right) = \frac { 1 } { 2 } \mathrm { Mv } ^ { 2 } ( of c.g. of the chain)

or u =