A hoop and a solid cylinder of same mass and radius are made of a permanent magnetic material with their magnetic moments along their respective axes. But the magnetic moment of the hoop is twice that of the 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:
Answer: (B) $T_h = T_c$
For small oscillations in a field, $T = 2\pi\sqrt{\dfrac{I}{mB}}$, where $I$ is about the axis of oscillation, a diameter (perpendicular to the magnetic moment).
Treating both as thin (the cylinder as a short disc): $I_h = \tfrac12MR^2$ and $I_c = \tfrac14MR^2$, so $I_h = 2I_c$. (About their own axes the ratio is also $MR^2 : \tfrac12MR^2 = 2$.) Also $m_h = 2m_c$:
$$\frac{T_h}{T_c} = \sqrt{\frac{I_h}{I_c}\cdot\frac{m_c}{m_h}} = \sqrt{2\times\frac12} = 1$$
So $T_h = T_c$.
Solution by Sreeraj P, M.Sc Physics