JEE PYQ: Rotational Motion Question 51
Question 51 - 2019 (10 Apr Shift 1)
A thin disc of mass $M$ and radius $R$ has mass per unit area $\sigma(r) = kr^2$ 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:
(1) $\frac{MR^2}{3}$
(2) $\frac{2MR^2}{3}$
(3) $\frac{MR^2}{6}$
(4) $\frac{MR^2}{2}$