A hoop and a solid cylinder of same mass and radius are made of a permanent magnetic material with 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:

We first recall the relation between the restoring torque acting on a magnetic dipole and the angular acceleration it produces. A magnetic dipole of moment $$\mu$$ kept in a uniform magnetic field $$B$$ experiences a torque

$$\tau = \mu B\sin\theta,$$

where $$\theta$$ is the angle between $$\vec\mu$$ and $$\vec B$$. For very small oscillations, $$\sin\theta \approx \theta$$, so the torque becomes

$$\tau \;=\; -\,\mu B\,\theta.$$

(The minus sign only indicates that the torque is restoring.)

In rotational dynamics the torque is also related to the angular acceleration $$\alpha$$ by

$$\tau = I\alpha = I\frac{d^{2}\theta}{dt^{2}},$$

where $$I$$ is the moment of inertia about the axis of rotation. Combining the two expressions, we obtain

$$I\frac{d^{2}\theta}{dt^{2}} = -\,\mu B\,\theta.$$

This is the differential equation of simple harmonic motion, with the square of the angular frequency given by

$$\omega^{2} = \frac{\mu B}{I}.$$

Hence the time-period of small oscillations is

$$T = 2\pi\sqrt{\frac{I}{\mu B}}.$$

Now we analyse the two given bodies one by one.

Hoop (ring)
We are told that the mass of the hoop is $$M$$ and the radius is $$R$$. About its own axis the moment of inertia is well known:

$$I_h = MR^{2}.$$

The magnetic moment of the hoop is twice that of the solid cylinder, so if we denote the cylinder’s magnetic moment by $$\mu_c$$, then

$$\mu_h = 2\mu_c.$$

Substituting these values in the period formula, we find

$$T_h = 2\pi\sqrt{\frac{I_h}{\mu_h B}} = 2\pi\sqrt{\frac{MR^{2}}{(2\mu_c)B}} = 2\pi\sqrt{\frac{MR^{2}}{2\mu_c B}}.$$

Solid cylinder
For a solid cylinder of the same mass $$M$$ and radius $$R$$, the moment of inertia about its own axis is

$$I_c = \frac{1}{2}MR^{2}.$$

Its magnetic moment is simply $$\mu_c.$$
Putting these quantities into the period formula gives

$$T_c = 2\pi\sqrt{\frac{I_c}{\mu_c B}} = 2\pi\sqrt{\frac{\dfrac{1}{2}MR^{2}}{\mu_c B}} = 2\pi\sqrt{\frac{MR^{2}}{2\mu_c B}}.$$

We see that the square-root expressions for $$T_h$$ and $$T_c$$ are identical:

$$T_h = 2\pi\sqrt{\frac{MR^{2}}{2\mu_c B}} \quad \text{and} \quad T_c = 2\pi\sqrt{\frac{MR^{2}}{2\mu_c B}}.$$

Therefore

$$T_h = T_c.$$

Hence, the correct answer is Option B.