Q. A $1m$ long rod has a mass of $0.12\ kg$. What is the moment of inertia about an axis passing through the centre and perpendicular to the length of rod
A) $0.01kg$-$m^{2}$
B) $0.001kg$-$m^{2}$
C) $1kg$-$m^{2}$
D) $10kg$-$m^{2}$
Explanation
(a) $I = \frac{ML^{2}}{12} = \frac{0.12 \times 1^{2}}{12} = 0.01kg - m^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.