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A smooth hemispherical bowl of radius R=...

A smooth hemispherical bowl of radius R=0.1 m is rotating about its own axis (which is vertical) with an angular velocity 0. A particle of mass `10^(-2)` kg on the inner surface of the bowl at an angled position `theta` (with vertical) is also rotating with same '`omega`'. The particle is about a height 'h' from the bottom of thebowl.
The value of '`theta`' in temis of 'R' and h' is

A

`h= R- ((g)/(omega^(2)))`

B

`h= R- gomega^(2)`

C

`R= h-((g)/(omega^(2)))`

D

`R= h- g omega^(2)`

Text Solution

Verified by Experts

The correct Answer is:
A
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