A smooth wire of length $2 π r$ 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 $ω$ 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 $ω^{2}$ is equal to

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

$\frac{(g \sqrt{3})}{r}$

$\frac{2 g}{r}$

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

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

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Let $θ$ be the angle with the vertical, then $\sin θ = \frac{r / 2}{r} = \frac{1}{2}$

$\Rightarrow θ = \sin^{- 1} (\frac{1}{2}) = 30^{o} N \cos θ = m g N \sin θ = F_{c} = \frac{m ω^{2} r}{2} \Rightarrow \tan θ = \frac{ω^{2} r}{2 g} \Rightarrow ω^{2} = \frac{2 g \tan θ}{r} = \frac{2 g}{r \sqrt{3}}$

Hence, the correct answer is (C)

Watch 3-min video & get full concept clarity

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th