question_answer

A thin disc of mass M and radius R has mass per unit area \[\sigma (r)=k{{r}^{2}}\] where r is the distance from its centre. Its moment of inertia about an axis going through its centre of mass and perpendicular to its plane is : [JEE Main 10-4-2019 Morning]

A)  \[\frac{M{{R}^{2}}}{6}\]       

B)       \[\frac{M{{R}^{2}}}{3}\]

C)  \[\frac{2M{{R}^{2}}}{3}\]     

D)       \[\frac{M{{R}^{2}}}{2}\]

Correct Answer: C

Solution :

[c]

Mass of disc