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

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$T_{h} = 0.5 T_{c}$

$T_{h} = 2 T_{c}$

$T_{h} = 1.5 T_{c}$

$T_{h} = T_{c}$

answer is D.

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Detailed Solution

Time period of a oscillating dipole,

$T = 2 π \sqrt{\frac{I}{μ B}} T_{h} = 2 π \sqrt{\frac{M R^{2}}{(2 μ) B}}, T_{c} = 2 π \sqrt{\frac{(\frac{1}{2}) M R^{2}}{μ B}}$

Hence, $T_{h} = T_{c}$

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