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 centre of disc is:

• $\frac{37}{9}M{R}^2$

• 4MR²

• $\frac{40}{9}M{R}^2$

• 10 MR²

B.

4MR²

let σ be the mass per unit area.

The total mass of the disc = σ x πR² = 9M
The mass of the circular disc out

$= \sigma x \pi {\left(\frac{R}{3}\right)}^2 = \sigma x \frac{\pi R^2}{9} = M$

Let us consider the above system as a complete disc of mass 9M and a negative mass M superimposed on it. Moment of Inertia (I₁₎ of the complete disc (9MR²)/2 about an axis passing through O and perpendicular to the plane of the disc.

M.I. of the cut-out portion about an axis passing through O and perpendicular to the plan of disc

$\frac{1}{2} x M x {\left(\frac{R}{3}\right)}^2$

Therefore, M.I (I2) of the cut out portion about an axis passing through O and perpendicular to the plane of disc

$= \left[\frac{1}{2} x M x {\left(\frac{R}{3}\right)}^2 + M x {\left(\frac{2R}{3}\right)}^2\right]$

Using perpendicular axis theorem

Therefore, the total M.I. of the system about an axis passing through O and perpendicular to the plane of the disc is
I = I₁ + I₂

$$\begin{gathered} = \frac{1}{2} 9 M{R}^2 - \left[\frac{1}{2} x M x {\left(\frac{R}{3}\right)}^2 + M x {\left(\frac{2R}{3}\right)}^2\right] \\ = \frac{9M{R}^2}{2}-\frac{9M{R}^2}{18} = \frac{(9-1) M{R}^2}{2} = 4M{R}^2 \end{gathered}$$