A ring 'A' of mass 'm' is attached to a stretched spring of force constant `K`, which is fixed at `C` on a smooth vertical circular track of radius `R`. Points `A` and `C` are diametrically opposite. When the ring slips form rest in the track to point `B`, making an angle of `30^@` with `AC`. `(/_ ACB = 30^@)` spring becomes unstretched. Find the velocity of the ring at `B`.
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Options

A. `[(KR^(2))/(2m)(2- sqrt(3))^(2) + gR sqrt(3)]^((1)/(2))`
B. `[(KR^(2))/(m)(2- sqrt(3))^(2) +gR]^((1)/(2))`
C. `[(2KR^(2))/(m) (2- sqrt(3))^(2) + gR sqrt(3)]^((1)/(2))`
D. `[(KR^(2))/(2m)(sqrt(2)-1)^(2) + gR]^((1)/(2))`.

Answer

B

Explanation

Decrease in elastic `PE` + Decrease in `PE` Increase in `KE`
`(1)/(2)Kx^(2)+mg (AD)=(1)/(2)mv^(2)`
`x = AC -CB = 2R -2R cos 30^@`
=`R(2-sqrt(3))(As angleCBA=90^@)`
`AD=AB cos 60^@`
=`(AC sin 30^@) cos 60^@ = (R)/(2)`
So, `(1)/(2) KR^(2)(1-sqrt(3))^(2)+mg"R/(2) = (1)/(2) mv^(2)`
`v = [(KR^(2))/(m)(2-sqrt(3))^(2)+ gR]^((1)/(2))`.