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If there is no torsion in the suspension...

If there is no torsion in the suspension thread, then the time period of a magnet executing SHM is

A

`T = 2pi sqrt((MB)/(I))`

B

`T= 2pi sqrt((I)/(MB))`

C

`T= (1)/(2pi) sqrt((I)/(MB))`

D

`T= (1)/(2pi) sqrt((M)/(I))`

Text Solution

Verified by Experts

The correct Answer is:
B

`T= 2pi sqrt((I)/(MB))`
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