Q 11-06-239JEE MainJEE Main 2018 (16 Apr, Shift 1)Easy
A thin circular disk is in the $xy$ plane as shown in the figure. The ratio of its moment of inertia about $z$ and $z'$ axes will be:
Answer: (C) $1 : 3$
The $z$-axis passes through the centre perpendicular to the disc: $I_z = \dfrac12MR^2$.
The $z'$-axis is parallel to it through a point on the rim, a distance $R$ away. By the parallel axis theorem,
$$I_{z'} = \frac12MR^2 + MR^2 = \frac32MR^2$$
$$I_z : I_{z'} = 1 : 3$$
Solution by Sreeraj P, M.Sc Physics