Q. Five particles of mass $2\ kg$ are attached to the rim of a circular disc of radius $0.1\ m$ & negligible mass. Moment of inertia of the system about the axis passing through the centre of the disc & perpendicular to its plane is
A) $1\ kg$-$m^{2}$
B) $0.1\ kg$-$m^{2}$
C) $2\ kg$-$m^{2}$
D) $0.2\ kg$-$m^{2}$
Explanation
(b) As the mass of disc is negligible therefore only moment of inertia of five particles will be considered
$I = \sum_{}^{}{mr^{2} = 5mr^{2} = 5 \times 2 \times (0.1)^{2} = 0.1kg - m^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.