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(D) Let $m$ be the mass of the bead and $N$ be the normal reaction force exerted by the wire on the bead.The radius of the circular path of the bead i

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$\frac{2g}{r\sqrt{3}}$

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(D) Let $m$ be the mass of the bead and $N$ be the normal reaction force exerted by the wire on the bead.The radius of the circular path of the bead is $R_{path} = r/2$.The forces acting on the bead are the gravitational force $mg$ (downwards) and the normal reaction $N$ (perpendicular to the wire).Resolving the forces into horizontal and vertical components:$N \sin 	heta = m R_{path} \omega^2 = m (r/2) \omega^2$ ... $(i)$$N \cos 	heta = mg$ ... (ii)Dividing $(i)$ by (ii),we get:$	an 	heta = \frac{(r/2) \omega^2}{g} = \frac{r \omega^2}{2g}$From the geometry of the circle,the distance from the center $O$ to the vertical axis is $r/2$. The angle $	heta$ that the radius vector $OP$ makes with the vertical is given by $\sin 	heta = \frac{r/2}{r} = 1/2$,so $	heta = 30^\circ$.Thus,$	an 30^\circ = \frac{1}{\sqrt{3}}$.Equating the two expressions for $	an 	heta$:$\frac{1}{\sqrt{3}} = \frac{r \omega^2}{2g}$$\omega^2 = \frac{2g}{r\sqrt{3}}$

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