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 value of \omega ^{2} is equal to:

 

  • Option 1)

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

  • Option 2)

    2g/(r\sqrt{3})

  • Option 3)

    (g\sqrt{3})/r

  • Option 4)

    2g/r

Answers (1)

N \sin \Theta =m\frac{r}{2}\omega ^{2}

N\cos \Theta =mg

\tan \Theta =\frac{r\omega ^{2}}{2g}\cdots \cdots (1)

from geometry \tan \Theta =\frac{r}{2.\frac{\sqrt{3}}{2}r}\cdots \cdots (2)

equating (1) and (2)

\omega ^{2}=\frac{2g}{\sqrt{3}r}


Option 1)

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

Option 2)

2g/(r\sqrt{3})

Option 3)

(g\sqrt{3})/r

Option 4)

2g/r

Most Viewed Questions

Preparation Products

JEE Main Rank Booster 2022

Booster and Kadha Video Lectures, Unlimited Full Mock Test, Adaptive Time Table, 24x7 Doubt Chat Support,.

₹ 13999/- ₹ 9999/-
Buy Now
Knockout JEE Main 2022

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Live Classes(Classes will start from Last week of September), Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 34999/- ₹ 24999/-
Buy Now
Test Series JEE Main 2022

Chapter/Subject/Full Mock Tests for JEE Main, Personalized Performance Report, Weakness Sheet, Complete Answer Key,.

₹ 4999/-
Buy Now
Rank Booster JEE Main 2020

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 9999/- ₹ 2/-
Buy Now
Knockout JEE Main 2022

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Live Classes(Classes will start from Last week of September), Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 34999/- ₹ 24999/-
Buy Now
Boost your Preparation for JEE Main 2021 with Personlized Coaching
 
Exams
Articles
Questions