A disc with a flat small bottom beaker placed on it at a distance R from its center is revolving about an axis passing through the center and perpendicular to its plane with an angular velocity $ω$ . The coefficient of static friction between the bottom of the beaker and the surface of the disc is $μ$ . The beaker will revolve with the disc if :

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

$R \leq \frac{μg}{2 ω^{2}}$

$R \geq \frac{μg}{2 ω^{2}}$

$R \geq \frac{μg}{ω^{2}}$

$R \leq \frac{μg}{ω^{2}}$

answer is B.

(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

For beaker to move with disc

$f_{s} = {mω}^{2} R$

$We know that f_{s} \leq f_{smax}$

${mω}^{2} R \leq μmg$

$R \leq \frac{μg}{ω^{2}}$

Watch 3-min video & get full concept clarity

JEE

NEET

Foundation JEE

Foundation NEET

CBSE