Q 11-06-215JEE MainJEE Main 2019 (8 Apr, Shift 1)Medium
A thin circular plate of mass $M$ and radius $R$ has its density varying as $\rho(r) = \rho_0 r$ with $\rho_0$ as constant and $r$ is the distance from its centre. The moment of inertia of the circular plate about an axis perpendicular to the plate and passing through its edge is $I = aMR^2$. The value of the coefficient $a$ is
Answer: (C) $\dfrac85$
Take rings of radius $r$ and width $dr$ (area $2\pi r\,dr$):
$$M = \int_0^R \rho_0 r\cdot2\pi r\,dr = \frac{2\pi\rho_0R^3}{3}$$
$$I_{centre} = \int_0^R \rho_0 r\cdot2\pi r\cdot r^2\,dr = \frac{2\pi\rho_0R^5}{5} = \frac35MR^2$$
By the parallel axis theorem, about a perpendicular axis through the edge:
$$I = \frac35MR^2 + MR^2 = \frac85MR^2 \Rightarrow a = \frac85$$
Solution by Sreeraj P, M.Sc Physics