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(1−2π)
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D
√2gR(2−π)
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Solution
The correct option is C√2gR(1−2π)
dm=(mπ/2).dθ=2mdθπ
h=R(1−cosθ)
dUi=(dm)gh=2mgR(1−cosθ)dθπ
∴Ui=∫90∘0dUi=mgR(1−2π) Uf=0
Decrease in PE = increase in KE. Ui−Uf=12mv2 mgR(1−2π)=12mv2 v=√2gR(1−2π)