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

Options

A. (a) `sqrt2gR`
B. (b) `sqrt(gR)`
C. (c) `sqrt(2gR(1-(2)/(pi)))`
D. (d) `sqrt(2gR(2-pi))`

Answer

C

Explanation

`dm((m)/(pi//(2)) dtheta = (((2m)/(pi))) d theta`
`h = R(1 - costheta)`

`dU_(i) =(dm) gh=(2mgR)/(pi)(1-cos theta) d theta`
`:. U_(i) =int_(0)^(pi//2)dU_(i) =(2mgR)/(pi)((pi)/(2) -1)`
`:. =mgR(1-(2)/(pi))`
Now, `U_(i) + K_(i) =U_(f) K_(f)`
`:. mgR(1-(2)/(pi)) =0 +1/2mv^(2)`
or `v=sqrt(2gR(1-(2)/(pi)))`