Q 11-06-208JEE MainJEE Main 2019 (10 Apr, Shift 1)Medium
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:
Answer: (D) $\dfrac{2MR^2}{3}$
Use rings of radius $r$ and width $dr$, mass $dm = kr^2\cdot2\pi r\,dr$:
$$M = 2\pi k\int_0^R r^3\,dr = \frac{\pi kR^4}{2},\qquad I = 2\pi k\int_0^R r^5\,dr = \frac{\pi kR^6}{3}$$
$$\frac IM = \frac{2R^2}{3} \;\Rightarrow\; I = \frac{2MR^2}{3}$$
Solution by Sreeraj P, M.Sc Physics