Q 11-06-247JEE MainJEE Main 2018 (8 Apr)Medium
From a uniform circular disc of radius $R$ and mass $9M$, a small disc of radius $\dfrac R3$ 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:
Answer: (B) $4MR^2$
The removed disc has $\dfrac19$ of the area, so its mass is $M$; its centre is $\dfrac{2R}{3}$ from the centre.
$$I_{\text{removed}} = \frac12M\left(\frac R3\right)^2 + M\left(\frac{2R}{3}\right)^2 = \frac{MR^2}{18} + \frac{4MR^2}{9} = \frac{MR^2}{2}$$
$$I = \frac12(9M)R^2 - \frac{MR^2}{2} = 4MR^2$$
Solution by Sreeraj P, M.Sc Physics