### JEE Main On 08 April 2018 Question 22

##### Question: From a uniform circular disc of radius R and mass 9 M, a small disc of radius $ \frac{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: [JEE Main Online 08-04-2018]

#### Options:

A) $ \text{10 M}{R^{2}} $

B) $ \frac{37}{9}\text{ M}{R^{2}} $

C) $ \text{4 M}{R^{2}} $

D) $ \frac{40}{9}\text{ M}{R^{2}} $

## Show Answer

#### Answer:

Correct Answer: C

#### Solution:

$ M=\frac{9M}{\pi R^{2}}\times \frac{\pi R^{2}}{9}=M $

$ I=\frac{(9M)R^{2}}{2}-[ \frac{M{{( \frac{R}{3} )}^{2}}}{2}+M{{( \frac{2R}{3} )}^{2}} ] $

$ I=4MR^{2} $