Q. Two circular iron discs are of the same thickness. The diameter of $A$ is twice that of $B$. The moment of inertia of $A$ as compared to that of $B$ is
A) Twice as large
B) Four times as large
C) 8 times as large
D) 16 times as large
Explanation
(d) M.I. of disc $= \frac{1}{2}mR^{2} = \frac{1}{2}\left( \pi R^{2}t \right)\rho R^{2} = \frac{1}{2}\pi R^{4}t\rho$ [$\rho =$ density, $t =$ thickness]
If discs are made of same material and same thickness then $I \propto R^{4} \propto (Diameter)^{4}$
$\therefore\frac{I_{A}}{I_{B}} = \left( \frac{D_{A}}{D_{B}} \right)^{4} = \left( \frac{2}{1} \right)^{4} = \frac{16}{1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.