Q. The moment of inertia of a circular ring of mass $1\ kg$ about an axis passing through centre and perpendicular to its plane is $4kg$-$m^{2}$. The diameter of the ring is
A) $2\ m$
B) $4\ m$
C) $5\ m$
D) $6\ m$
Explanation
(b) Moment of inertia of a ring about an axis passing through the centre $= MR^{2} = 1 \times R^{2} = 4$, as $M = 1kg$
$\therefore R = 2m,$ Diameter $D = 2R = 4m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.