question_answer

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 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 : [JEE Main 12-4-2019 Afternoon]

A) \[\left( g\sqrt{3} \right)/r\]                    

B) \[\frac{\sqrt{3}g}{2r}\]

C) \[2g/r\] 

D)   \[2g/\left( r\sqrt{3} \right)\]

Correct Answer: D

Solution :

\[N\sin \theta =m\frac{r}{2}{{\omega }^{2}}\]                                          ?.(1) \[N\cos \theta =mg\]                                        ?.(2) \[\tan \theta =\frac{r{{\omega }^{2}}}{2g}\] \[\frac{r}{2\frac{\sqrt{3}r}{2}}=\frac{r{{\omega }^{2}}}{2g}\] \[{{\omega }^{2}}=\frac{2g}{\sqrt{3}r}\]