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. `sqrt(2gR)`
B. `sqrt(gR)`
C. `sqrt(2gR (1-(2)/(pi)))`
D. `sqrt(2gR(2-pi))`

Answer

C

Explanation

Consider an elementary mass dm which makes an angle `d theta` at the centre of ring as in figure.


Mass of element is, `dm=(m)/(pi//2).d theta`
Gravitational potential energy of the element is,
`dU_(i) =(dm)gh =((2m)/(pi) d theta) gR (1-cos theta)`
So initial potential energy of the chain is
`U_(i) = int dU_(i)`
`U_(i) =(2mgR)/(pi) int_(0)^((pi)/(2)) (1-cos theta) d theta`
`U_(i) =(2mgR)/(pi) [theta-sin theta]_(0)^((pi)/(2))`
So, `U_(i) =(2mgR)/(pi) ((pi)/(2)-1)`
and, `U_(f)=0`
Let speed of chain on horizontal part is v.
By conservation of mechanical energy
`M. E_(i) = M. E_(f)`
`(2mgR)/(pi) ((pi)/(2)-1) =(1)/(2)mv^(2)`
`v= sqrt(2gR (1-(2)/(pi)))`