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Question

A smooth chain AB of mass m rests against a surface in the form of a quarter of a circle of radius R. If it is released from rest, the velocity of the chain after it comes over the horizontal part of the surface is

A
2gR
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B
gR
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C
2gR(12π)
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D
2gR(2π)
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Solution

The correct option is C 2gR(12π)


dm=(mπ/2).dθ=2mdθπ

h=R(1cosθ)

dUi=(dm)gh=2mgR(1cosθ)dθπ

Ui=900dUi=mgR(12π)
Uf=0
Decrease in PE = increase in KE.
UiUf=12mv2
mgR(12π)=12mv2
v=2gR(12π)

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