A bead of mass m is attached to one end of a spring of natural length R and spring constant `K=((sqrt(3)+1)mg)/(R )`. The other end of the spring is fixed at a point A on a smooth vertical ring of radius R as shown in fig. The normal reaction at B just after it is released to move is

Options

A. `(mg)/(2)`
B. `sqrt(3) mg`
C. `3sqrt(3) mg`
D. `(3sqrt(3)mg)/(2)`

Answer

C

Explanation

Let v ve the velocity of the particle when it makes `30^(@)` with the horizontal Then
` v cos 30^(@) = u cos 60^(@) implies v = 20 //sqrt(3)m//s. " Now "g cos 30^(@)=(v^(2))/(R) " So ,"R = (v^(2))/(g cos 30^(@))=15.4m`