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. `(2g)/(a)`
B. `(g)/(2a)`
C. `(2g)/(asqrt(3))`
D. `(gsqrt(3))/(2a)`

Answer

c

Explanation

`N = cos theta = mg `…(i)
`N sin theta = (m omega^(2)a)/(2)`…(ii)
from eqs(ii)(i),`tan theta = (omega^(2)a)/(2g) rArr omega^(2) = (2g tan theta)/(a)`
Now` sin theta = (a)/(2) xx(1)/(a) or theta = 30^(@)`
`or omega^(2) = (2g)/(sqrt(3a)`