A bead of mass `m` slides without friction on a vertical hoop of radius `R`. The bead moves under the combined influence of gravity and a spring of spring constant `k` attached to the bottom of the hoop. For simplicity assume, the equilibrium length of the spring to be zero. The bead is released at the top of the hoop with negligible speed as shown . The bead, on passing the bottom point will have a velocity of `:`

Options

A. `2sqrt(gR)`
B. `2sqrt(gR+(2kR^(2))/(m))`
C. `2sqrt(gR+(kR^(2))/(m))`
D. `sqrt(2gR+(kR^(2))/(m))`

Answer

C

Explanation

From mechanical energy conservation `:`
`mg(2R)+(1)/(2)k(2R)^(2)=(1)/(2)mv^(2)`
`v=2sqrt(gR+(kR^(2))/(m))`