Q 11-06-233JEE MainJEE Main 2019 (12 Apr, Shift 1)Medium
A circular disc of radius $b$ has a hole of radius $a$ at its centre (see figure). If the mass per unit area of the disc varies as $\dfrac{\sigma_0}{r}$, then the radius of gyration of the disc about its axis passing through the centre is
Answer: (B) $\sqrt{\dfrac{a^2 + b^2 + ab}{3}}$
A ring of radius $r$ and width $dr$ has mass $dm = \dfrac{\sigma_0}{r}2\pi r\,dr = 2\pi\sigma_0\,dr$.
$$M = 2\pi\sigma_0(b - a),\qquad I = \int r^2dm = \frac{2\pi\sigma_0}{3}(b^3 - a^3)$$
$$k^2 = \frac IM = \frac{b^3 - a^3}{3(b - a)} = \frac{a^2 + ab + b^2}{3}$$
Solution by Sreeraj P, M.Sc Physics