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

Options

A. `T_(h)=T_(c)`
B. `T_(h)=1.5T_(c)`
C. `T_(h)=2T_(c)`
D. `T_(h)=0.5T_(c)`

Answer

A

Explanation

`T= 2pi sqrt((I)/(MB)) , (T_(h))/(T_(c))= sqrt((I_(h))/(M_(h))xx(M_(c))/(I_(c)))`
`I_(h)=MR^(2), I_(c)=(MR^(2))/(2), M_(h)=2M_(c)`
Hence `(T_(h))/(T_(c))= sqrt(2xx(1)/(2))=1`
`T_(h)=T_(c)`