$A$ 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 the 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

  • A
    $\frac{\sqrt{3}g}{2r}$
  • B
    $\frac{g\sqrt{3}}{r}$
  • C
    $\frac{2g}{r}$
  • D
    $\frac{2g}{r\sqrt{3}}$

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