JEE MAIN - Physics (2019 - 10th January Evening Slot - No. 13)
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 Th and Tc respectively, then -
Th = 1.5 Tc
Th = Tc
Th = 2Tc
Th = 0.5 Tc
Explanation
T = $$2\pi \sqrt {{1 \over {\mu B}}} $$
Th = $$2\pi \sqrt {{{m{R^2}} \over {\left( {2\mu } \right)B}}} $$
TC = $$2\pi \sqrt {{{1/2m{R^2}} \over {\mu B}}} $$
Th = $$2\pi \sqrt {{{m{R^2}} \over {\left( {2\mu } \right)B}}} $$
TC = $$2\pi \sqrt {{{1/2m{R^2}} \over {\mu B}}} $$
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