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A hoop and a disc of same mass and radius are made of a permanent magnetic material with their magnetic moments 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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$T_{h} = 0 .5 T_{c}$

$T_{h} = T_{c}$

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

$T_{h} = 1 .5 T_{c}$

answer is B.

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Detailed Solution

$T = 2 π \sqrt{\frac{I}{mB}} ∴ \frac{T_{h}}{T_{c}} = \sqrt{\frac{I_{h} m_{c}}{I_{c} m_{n}}} = \sqrt{\frac{\frac{1}{2} {MR}^{2} m_{c}}{\frac{1}{4} {MR}^{2} 2 m_{c}}} = 1 \Rightarrow T_{h} = T_{c}$

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