JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Laws of Motion

The bead feels only gravity and the wire's normal reaction, and on a circular wire that reaction points along the radius. Resolving, the vertical balance is $N\cos \theta =mg$ and the horizontal balance supplies the centripetal force $N\sin \theta =m{\omega }^2(r\sin \theta )$, where $\theta$ is the angle the radius makes with the vertical. The second line gives $N=m{\omega }^{2}r$, and feeding that into the first, $m{\omega }^{2}r\cos \theta =mg$, so ${\omega }^2=\frac{g}{r\cos \theta }$. At the position shown the horizontal offset is $\frac{r}{2}$, so $\sin \theta =\frac{1}{2}$ and $\cos \theta =\frac{\sqrt{3}}{2}$, giving ${\omega }^2=\frac{g}{r\cdot \frac{\sqrt{3}}{2}}=\frac{2g}{\sqrt{3}r}$.