1. Time period of small oscillation of a magnetic body

For a magnet of magnetic moment $M$ placed in a uniform magnetic field $B$, if it is given a small angular displacement, then it executes angular SHM with

$T = 2\pi \sqrt{\frac{I}{MB}}$

where:

• $I$ = moment of inertia about the axis of rotation,
• $M$ = magnetic moment.

So,

$T \propto \sqrt{\frac{I}{M}}$

since $B$ is same for both.

2. Moments of inertia

The magnetic moments are along their respective symmetry axes, so oscillation occurs about an axis through the center and perpendicular to the magnetic field. Hence we use moment of inertia about a diameter (axis through center perpendicular to symmetry axis).

(i) Hoop

For a thin hoop of mass $m$ and radius $r$,

• About central axis perpendicular to plane: $I_z = mr^2$
• By perpendicular axis theorem,

$I_x = I_y = \frac{1}{2}mr^2$

Thus for the required diameter axis,

$I_h = \frac{1}{2}mr^2$

(ii) Solid cylinder

For a solid cylinder of mass $m$ and radius $r$ (about a central axis perpendicular to its own axis),

$I_c = \frac{1}{4}mr^2 + \frac{1}{12}mL^2$

But here the object is effectively treated as a disc/cylinder magnet oscillating with magnetic moment along its axis; for standard comparison using radius only, the relevant central diameter moment for a solid circular lamina/disc is

$I_c = \frac{1}{4}mr^2$

3. Relation of magnetic moments

Given:

$M_h = 2M_c$

4. Compare time periods

For hoop:

$T_h = 2\pi\sqrt{\frac{I_h}{M_h B}}$

For cylinder:

$T_c = 2\pi\sqrt{\frac{I_c}{M_c B}}$

Therefore,

$\frac{T_h}{T_c} = \sqrt{\frac{I_h/M_h}{I_c/M_c}}$

Substitute $I_h = \frac12 mr^2$, $I_c = \frac14 mr^2$, and $M_h = 2M_c$:

$\frac{T_h}{T_c} = \sqrt{\frac{\frac12 mr^2}{2M_c} \cdot \frac{M_c}{\frac14 mr^2}}$

$\frac{T_h}{T_c} = \sqrt{\frac{1}{4} \cdot 4} = 1$

Hence,

$T_h = T_c$

5. Option check

• A: $T_h = 1.5T_c$ ❌
• B: $T_h = T_c$ ✅
• C: $T_h = 2T_c$ ❌
• D: $T_h = 0.5T_c$ ❌

So the correct option is B.