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A hoop and a solid cylinder of same mass...

A hoop and a solid cylinder of same mass and radius are made of a permanent magnetic parallel to their respective magnetic moment parallel to their respective axes. But the magnetic moment of hoop is twice of solid cylinder. They are placed in a uniform magnetic field in such a manner that their magnetic moments make a small angle with the field. If the oscillation period of hoop and cylinder are `T_(h) and T_(c)` respectively, then

A

`T_(h) = T_(c )`

B

`T_(h) = 0.5 T_(c )`

C

`T_(h) =2T_(c )`

D

`T_(h) =1.5 T_(c )`

Text Solution

Verified by Experts

The correct Answer is:
A
`T = 2pi sqrt(I/(MB))`
`(T_(H))/(T_(C))=sqrt((I_(H))/(I_(C)))sqrt((M_(C))/(M_(H)))=sqrt(((MR^(2))/(MR^(2)))/2) sqrt((M/2)/M)=1`
`rArr T_(H) = T_(C )`.
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