A bead of mass `m` is threaded on a smooth circular wire centre `O`, radius a, which is fixed in vertical plane. A light string of natural olength 'a', elastic constant `= (3mg)/(a)` and breaking strength `3mg` connects the bead to the lowest point `A` of the wire. The other end of the string is fixed to ring at point `B` near point `A`. The string is slaked initially. The bead is projected from `A` with speed `u`.

The smallest value `u_(0)` of `u` for which the bead will make complete revolutions of the wire will be

Options

A. `u_(0) = sqrt(5ga)`
B. `u_(0) = sqrt(6ga)`
C. `u_(0) = sqrt(7ga)`
D. `u_(0) = 2 sqrt(ga)`

Answer

C

Explanation

When particle is at highest position, the elastic force is downwards
`F_(l) = (2mg)/(a)(2a -a) =3mg`
it `v` is velocity at height point at `B`
`(m u_(0)^(2))/(a) = F_(l) + mg -N`
If `V =0`, then `KE` at lowest point `A` will be
`(1)/(2) m u_(0)^(2) = ["Elastic energy" + gPE] at B`
`= (1)/(2) ((3mg)/(a))a^(2) + mg2a`
`u_(0)^(2) = 7ga`.