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**Hint:**The restoring torque is the torque that occurs to return a twisted or rotating object to its original orientation.

Calculate the moment of inertia in both cases given in the question and consider which one is greater. Note that, the angular frequency is inversely proportional to the moment of inertia.

**Formula used:**

The parallel axis theorem states,$I = {I_{cm}} + M{l^2}$

${I_{cm}} = $ moment of inertia about the axis,

$M = $ mass of the object,

$l = $ the distance between the two axes.

For this case \[A\], $I = \dfrac{1}{3}m{L^2} + M{L^2}$

For this case \[B\], $I = \dfrac{1}{3}m{L^2} + \dfrac{1}{2}m{R^2} + M{(L + R)^2}$

$M = $ mass of the disc and

$R = $radius of the disc.

$m = $mass of the rod

$L = $ length of the rod.

**Complete step by step answer:**

There are two cases in this problem,

In this case \[A\], the disc does not rotate about its axis, hence according to the parallel axis theorem [$I = {I_{cm}} + M{l^2}$ ] the moment of inertia will be ${I_A} = \dfrac{1}{3}m{L^2} + M{L^2}$.

$M$ is the mass of the disc, $m$ Is the mass of the rod, and $L$ is the length of the rod.

In this case \[B\], the disc is rotated about its axis, hence the moment of inertia will be

$\Rightarrow {I_B} = \dfrac{1}{3}m{L^2} + \dfrac{1}{2}m{R^2} + M{(L + R)^2}$

$R$ is the radius of the disc.

So, we can see that the moment of inertia in the case \[A\] is greater than the case \[B\]. Since the angular frequency is inversely proportional to the moment of inertia.

Hence, the angular frequency in the case \[A\] is less than the case \[B\].

The restoring torque is the same for both cases.

**Hence, the correct answers in option $(A)$ and $(D)$.**

**Note:**The restoring torque is the torque that occurs to return a twisted or rotating object to its original orientation.

So it does not matter whether the disc rotates or not about its axis to calculate the restoring torque.

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