Q 11-06-183JEE MainJEE Main 2020 (7 Jan, Shift 2)Medium
The mass per unit area of a circular disc of radius $a$ depends on the distance $r$ from its centre as $\sigma(r) = A + Br$. The moment of inertia of the disc about the axis perpendicular to the plane and passing through its centre is:
Answer: (A) $2\pi a^4\left(\dfrac A4 + \dfrac{aB}{5}\right)$
Take a thin ring of radius $r$ and width $dr$: $dm = (A + Br)2\pi r\,dr$.
$$I = \int_0^a r^2(A + Br)2\pi r\,dr = 2\pi\left(\frac{Aa^4}{4} + \frac{Ba^5}{5}\right) = 2\pi a^4\left(\frac A4 + \frac{aB}{5}\right)$$
Solution by Sreeraj P, M.Sc Physics