Q. A hoop and a solid cylinder of same mass and radius are made of a permanent magnetic material with their 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 periods of hoop and cylinder are $T_h$ and $T_c$ respectively, then :

Solution:

$T = 2\pi \sqrt{\frac{I}{\mu B}}$
$ T_{h} = 2 \pi \sqrt{\frac{mR^{2}}{\left(2\mu\right)B}} $
$ T_{C} = 2\pi \sqrt{\frac{1/2mR^{2}}{\mu B}} $

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