Q. A circular thin disc of mass $2\ kg$ has a diameter $0.2\ m$. Calculate its moment of inertia about an axis passing through the edge and perpendicular to the plane of the disc (in $kg - m^{2}$)
A) 0.01
B) 0.03
C) 0.02
D) 3
Explanation
(b) $\frac{3}{2}MR^{2} = \frac{3}{2} \times 2 \times (0.1)^{2} = 0.03kg - m^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.