JEE Mains · Physics · STD 11 - Motion in plane
- JEE Mains 2019
- Hard
- MCQ
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
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(2g/\left( {r\sqrt 3 } \right)\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{l}
N\,\sin \,\theta = m\frac{r}{2}{\omega ^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)\\
N\,\cos \,\,\theta = mg\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)\\
\tan \,\,\theta \, - \frac{{r{\omega ^2}}}{{2g}}\\
\frac{r}{{2\frac{{\sqrt 3 \,r}}{2}}} = \frac{{r{\omega ^2}}}{{2g}}\,\,\,;\,\,{\omega ^2} = \frac{{2g}}{{\sqrt 3 \,r}}
\end{array}\)
N\,\sin \,\theta = m\frac{r}{2}{\omega ^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)\\
N\,\cos \,\,\theta = mg\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)\\
\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}}
\end{array}\)
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