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A spherical bowl of radius R rotates abo...

A spherical bowl of radius R rotates about the verical diameter with angular velocity `omega`.The bowl contains a small object inside and in absence of friction, this object takes up a position inside the bowl such that its radius vector makes an angle `theta` with the vertical (see figure). Then

A

`omega=sqrt(g//r cos theta)`

B

`omega=2 pi g//r`

C

`omega = sqrt((g cos theta)/r)`

D

`omega=sqrt(r cos theta xx g )`

Text Solution

Verified by Experts

The correct Answer is:
A
Reaction N has component in vertical direction equal to the weight of the object . The horizontal component of N provides the centripetal force. Now { `omega` can be calculated }
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