A smooth wire of length `2 pi r` is bent into a circle and kept in a vertical plane. A bead can slide smoothly on the wire. When the circle is rotating with angular speed `omega` about the vertical diameter `AB`, as shown in figure, the bead is at rest with respect to the circular ring at position `P` as shown. then the value of `omega^(2)` is equal to :
.

Options

A. `(sqrt(3)g)/(2r)`
B. `2g//r`
C. `2g//(rsqrt(3))`
D. `(gsqrt(3))//r`

Answer

C

Explanation

`Ncos tehta=m(r/2)omega^(2), N sin theta =mg`
`cot theta=(r omega^(2))/(2g), omega=(2g cot theta)/r=sqrt((2g)/(rsqrt(3)))`
`omega^(2)=(2g)/(sqrt(3)r)`