Q 11-06-149JEE MainJEE Main 2021 (24 Feb, Shift 1)Easy
Moment of inertia (M.I.) of four bodies, having same mass and radius, are reported as:
$I_1$ = M.I. of thin circular ring about its diameter,
$I_2$ = M.I. of circular disc about an axis perpendicular to disc and going through the centre,
$I_3$ = M.I. of solid cylinder about its axis and
$I_4$ = M.I. of solid sphere about its diameter.
Then:
Answer: (C) $I_1 = I_2 = I_3 > I_4$
$$I_1 = \tfrac12 MR^2,\quad I_2 = \tfrac12 MR^2,\quad I_3 = \tfrac12 MR^2,\quad I_4 = \tfrac25 MR^2$$
Since $\tfrac12 > \tfrac25$, $I_1 = I_2 = I_3 > I_4$.
Solution by Sreeraj P, M.Sc Physics