Q. If the moment of inertia of a disc about an axis tangential and parallel to its surface be $I$, then what will be the moment of inertia about the axis tangential but perpendicular to the surface?
A) $\frac{6}{5}I$
B) $\frac{3}{4}I$
C) $\frac{3}{2}I$
D) $\frac{5}{4}I$
Explanation
(a) MI of disc about tangent in a plane
$= \frac{5}{4}MR^{2} = I$
$\therefore\ mR^{2} = \frac{4}{5}I$
MI of disc about tangent $I$ to plane $I^{'} = \frac{3}{2}mR^{2}$
$\therefore\ I^{'} = \frac{3}{2}\left( \frac{4}{5}I \right)$
$= \frac{6}{5}I$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.