Start Free Mock LoginAlready registered? Login here

Physics - NLM Question with Solution | TestHub

PhysicsNLMMiscellaneousEasy2 minPYQ_2020

PhysicsEasysingle choice

Particles are released from rest at $A$ and slide down the smooth surface of height $h$ to a conveyor $B$ . The correct angular velocity $ω$ of the conveyor pulley of radius $r$ to prevent any sliding on the belt as the particles transfer to the conveyor is

Options:

Answer:

Solution:

First of all, we have to calculate the velocity of particles at the point B.

$∵$ Loss in P.E. = gain in K.E.

or $mgh = \frac{1}{2} {mv}^{2}$

$∴$ $v = \sqrt{2 gh}$

To prevent sliding, $v = r ω$

$∴ ω = \frac{v}{r} = \frac{\sqrt{2 gh}}{r}$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:NLMSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2020

Doubts & Discussion

Loading discussions...