Q. The moment of inertia of a circular ring about an axis passing through its centre and normal to its plane is $200\ g \times cm^{2}$. Then its moment of inertia about a diameter is
A) $400\ g \times cm^{2}$
B) $300\ g \times cm^{2}$
C) $200\ g \times cm^{2}$
D) $100\ g \times cm^{2}$
Explanation
(d) $I_{z} = I_{x} + I_{y}$
$200 = I_{D} + I_{D} = 2I_{D}$
$\therefore I_{D} = 100g \times cm^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.