Problem Statement:
A circular disc of radius R and mass 9M is given. A small disc of radius R/3 is removed from the larger disc. The task is to find the moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through O.

Solution:

To solve this problem, we can consider the moment of inertia of the remaining disc as the difference between the moment of inertia of the original disc and the moment of inertia of the removed disc.

Moment of Inertia of the Original Disc:
The moment of inertia of a circular disc about an axis perpendicular to its plane and passing through its center is given by the formula:
I = (1/2) * M * R^2
Here, M is the mass of the disc and R is its radius.

In this case, the mass of the original disc is 9M and its radius is R. Therefore, the moment of inertia of the original disc is:
I_original = (1/2) * (9M) * R^2 = 4.5M * R^2

Moment of Inertia of the Removed Disc:
The moment of inertia of a circular disc about an axis perpendicular to its plane and passing through its center is given by the same formula as above:
I = (1/2) * m * r^2
Here, m is the mass of the disc and r is its radius.

In this case, the mass of the removed disc is M and its radius is R/3. Therefore, the moment of inertia of the removed disc is:
I_removed = (1/2) * (M) * (R/3)^2 = (1/18) * M * R^2

Moment of Inertia of the Remaining Disc:
The moment of inertia of the remaining disc can be calculated by subtracting the moment of inertia of the removed disc from the moment of inertia of the original disc:
I_remaining = I_original - I_removed
= 4.5M * R^2 - (1/18) * M * R^2
= (81/18) * M * R^2 - (1/18) * M * R^2
= (80/18) * M * R^2
= (40/9) * M * R^2

Therefore, the moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through O is (40/9) * M * R^2.