A circular hole of radius R/4 is made in a thin uniform disc having mass M and radius R, as shown in figure. The moment of inertia of the remaining portion of the disc about an axis passing through the point O and perpendicular to the plane of the disc is :

Let the mass per unit area of disc be σ
Now, moment of inertia of removed mass(M₂)=∬(σA')(R/4)²/2 + (σA')(3R/4)²
Using parallel axis theorem
M₂=(σA')(R²/32+9R²/16)
A'=π(R/4)²=πR²/16
M₂=(σπR²/16)(19R²/32)
Also, moment of inertia of complete disc is:
→M₁=((σA)×R²/2)
Effect moment of inertia =M₁−M₂
M₀=σ/2πR⁴−σπR⁴(19/32)
M₀=(σπR²).R²/2−(σπR⁴)(19/32)
M₀= (σπR²/2)(R²−(19R²/16))
M₀= (σπR²/2)(16R²-19R²)/16 = σπR⁴(237/512)
σπR²=M
M₀=237/512MR²