A smooth wire is bent into a vertical circle of radius a. `A` bead `P` can slide smoothly on the wire. The circle is rotated about vertical diameter AB as axis with a speed omega as shown in figure. The bead `P` is ar rest w.r.t. the circular ring in the position shown. then `omega^(2)` is equal to:

Options

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

Answer

B

Explanation

As `, cos theta =(a)/(2a)`
`theta =60^(@)`
`:. N sin 60^(@) =mg`
`N cos 60^(@) = m ( omega^(2)a)/(2)`

`w.r.t.` wire
`:. tan 60^(@)=(2g)/(omega^(2)a), omega^(2)=(2g)/(asqrt(3))`