Problem Statement:
We have a disc of mass M and radius R, and a rod of mass M and length L = 2R. We need to find the moment of inertia of this system about an axis passing through the center of the disc and perpendicular to its plane.

Solution:
To find the moment of inertia of the system, we need to consider the individual moments of inertia of the disc and the rod, and then add them up.

Moment of Inertia of the Disc:
The moment of inertia of a disc about an axis perpendicular to its plane passing through its center is given by the formula:

I_disc = (1/2) * M * R^2

Since the axis we are considering passes through the center of the disc, the moment of inertia of the disc is (1/2) * M * R^2.

Moment of Inertia of the Rod:
The moment of inertia of a rod about an axis passing through its center and perpendicular to its length is given by the formula:

I_rod = (1/12) * M * L^2

In this case, the length of the rod is L = 2R. Substituting this value, we get:

I_rod = (1/12) * M * (2R)^2
= (1/12) * M * 4R^2
= (1/3) * M * R^2

Since the axis we are considering passes through the center of the rod, the moment of inertia of the rod is (1/3) * M * R^2.

Total Moment of Inertia:
To find the total moment of inertia of the system, we add the individual moments of inertia of the disc and the rod:

I_total = I_disc + I_rod
= (1/2) * M * R^2 + (1/3) * M * R^2
= (3/6) * M * R^2 + (2/6) * M * R^2
= (5/6) * M * R^2

Therefore, the moment of inertia of the system about the axis passing through the center of the disc and perpendicular to its plane is (5/6) * M * R^2.

Any one can do this ...plz help me