From a uniform circular disc of radius R and mass 9M, a small disc of radius R/3 is removed as shown in the figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through the centre of the disc is?

Given: Mass of disc 9M
Mass of removed part m = 9M(πR²)/(π(R/3)²) = M
Moment of inertia of Disc without hole about axis perpendicular to the plane passing through center is
I₁ = 9MR²/2
Moment if inertia of hole about axis passing through center of disc
I₂ = M(R/3)²/2 + M(2R/3)²
The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through centre of disc
I = I₁ - I₂
I = 9MR²/2 - [M(R/3)²/2 + M(2R/3)²]
= 4MR²