The moment of inertia of a uniform circular disc rotating about an axis perpendicular to the plane of the disc and passing through points O, A, B and C are $I_{o}$ , $I_{A}$ , $I_{B}$ and $I_{C}$ respectively, then

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By the parallel axis theorem, the moment of inertia of the disc about an axis perpendicular to its plane and passing through a point at a distance $d$ from the center is given by:

$I = I_{c m} + M d^{2}$

Here, the center of mass is at O, so $I_{c m} = I_{O}$ .

From the given figure, the distances of points O, A, B, and C from the center O follow the order:

$d_{O} < d_{A} < d_{B} < d_{C}$

Since the moment of inertia $I$ increases as the distance $d$ from the center of mass increases, the moments of inertia will follow the same order:

$I_{O} < I_{A} < I_{B} < I_{C}$

Answer: $I_{o} < I_{A} < I_{B} < I_{C}$

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