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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

`sin^(-1)((R-h)/(R))`

B

`cos^(-1)((R-h)/(R))`

C

`tan^(-1)((R-h)/(R))`

D

`sec^(-1)((R-h)/(R))`

Text Solution

Verified by Experts

The correct Answer is:
B
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