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: A. `(2g)/(a)`B. `(2g)/(asqrt3)`C. `(gsqrt3)/(a)`D. `(2a)/(gsqrt3)`

A smooth wire is bent into a vertical circle of radius a. `A` bead `P`

1.	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: A. `(2g)/(a)`B. `(2g)/(asqrt3)`C. `(gsqrt3)/(a)`D. `(2a)/(gsqrt3)`
Answer» Correct Answer - B As, `costheta=(a)/(2a)` `theta=60^(@)` :. `Nsin60^(@)=mg` `Nsin60^(@)=m(omega^(2)a)/(2)` :. `tan60^(@)=(2g)/(omega^(2)a)/(2)` implies `omega^(2)=(2g)/(asqrt(3))` .

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