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Particles are released from rest A and s...

Particles are released from rest A and slide down the smooth surface of hight h to a conveyor B. The correct angular veleocity `omega` of the coneyor pulley of radius r to prevent any sliding on the belt as the particles transfer to the conveyor is

A

`sqrt(2gh)`

B

`(2gh)/(r)`

C

`sqrt(2gh)/(r)`

D

`(2gh^(2))/(r^(2))`

Text Solution

Verified by Experts

(c) First of all, we have to calculate the velocity of particles at the point B.
`:.` Loss in PE =gain in KE or mgh `=(1)/(2)mmv^(2)`
`:.` To Prevent sliding , `v=r_(omega)`
`:.` `omega=(v)/(r)=sqrt(2gh)/(r)`
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