The question statement appears truncated after

"A thin disk of mass $M$ and radius $R$ ( ..."

but from the options, this is clearly a comparison of two physical pendulum arrangements (Case A and Case B) involving the same rod and disk, with different orientations of the disk.

The standard interpretation is:

• A rod of length $L$ and mass $m$ is pivoted at one end.
• A thin disk of mass $M$ and radius $R$ is attached at the free end in two different ways:
  • Case A: disk lies in the plane of oscillation.
  • Case B: disk is oriented perpendicular to that plane.

In both cases, the center of mass position remains the same, so the restoring torque is expected to be the same, while the moment of inertia changes, so angular frequency changes.

1. Restoring torque for small angular displacement

For a physical pendulum,

$au = - \left( \text{total mass} \right) g \,(\text{distance of COM from pivot})\,\theta$

for small angular displacement $\theta$.

The restoring torque depends only on:

1. total weight distribution through the center of mass, and
2. distance of the combined center of mass from the pivot.

Changing the orientation of the disk does not change the location of its center of mass. Therefore, the combined center of mass of rod + disk is the same in both cases.

Hence,

$\tau_A = \tau_B$

So:

• Option A is correct
• Option B is incorrect

2. Angular frequency of small oscillation

For a physical pendulum,

$\omega = \sqrt{\frac{K}{I_p}}$

where

$K = (\text{total mass})g(\text{distance of COM from pivot})$

Since restoring torque factor $K$ is same in both cases, comparison of angular frequency reduces to comparison of the moment of inertia about the pivot.

Thus,

$\omega \propto \frac{1}{\sqrt{I_p}}$

So the case with larger moment of inertia has smaller angular frequency.

3. Compare moment of inertia of the disk in the two cases

The rod contributes the same moment of inertia in both cases, so only the disk part matters.

Let the disk center be at distance $L$ from the pivot. By parallel axis theorem,

$I_{p,\text{disk}} = I_{\text{center}} + ML^2$

Now compare $I_{\text{center}}$ for the two orientations.

Case A

If the disk oscillates about a diameter in its plane,

$I_{\text{center},A} = \frac{1}{4}MR^2$

Case B

If the disk oscillates about an axis perpendicular to its plane through its center,

$I_{\text{center},B} = \frac{1}{2}MR^2$

Therefore,

$I_{p,A} = I_{\text{rod}} + ML^2 + \frac{1}{4}MR^2$

$I_{p,B} = I_{\text{rod}} + ML^2 + \frac{1}{2}MR^2$

So,

$I_{p,A} < I_{p,B}$

Hence,

$\omega_A > \omega_B$

Therefore:

• Option C is correct
• Option D is incorrect

4. Final derived answer

The correct options are:

$\boxed{A,\ C}$

5. Comparison with stored correct answer

Stored correct answer: $A, D$

My derived answer is $A, C$.

I disagree with the stored answer.

Reason: if restoring torque is same in both cases, then angular frequency depends inversely on $\sqrt{I}$. Since the disk has larger moment of inertia in Case B, Case B must have smaller angular frequency. Thus,

$\omega_A > \omega_B$

not the reverse.

So the stored answer likely has an error in the angular-frequency comparison.