Q. The moment of inertia of thin circular disc about an axis passing through its centre and perpendicular to its plane is $I$. Then, the moment of inertia of the disc about an axis parallel to its diameter and touching the edge of the rim is
A) $I$
B) $2I$
C) $\frac{3}{2}I$
D) $\frac{5}{2}I$
Explanation
(d) Moment of inertia of a circular disc about an axis passing through centre of gravity and perpendicular to its plane
$I = \frac{1}{2}\ MR^{2}$ …(i)
From Eq.(i)
$MR^{2} = 2I$
Then, moment of inertia of disc about tangent in a plane
$= \frac{5}{4}MR^{2} = \frac{5}{4}(2I) = \frac{5}{2}I$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.