Q 11-06-128JEE MainJEE Main 2023 (1 Feb, Shift 2)Easy
Moment of inertia of a disc of mass $M$ and radius $R$ about any of its diameters is $\dfrac{MR^2}{4}$. The moment of inertia of this disc about an axis normal to the disc and passing through a point on its edge will be $\dfrac x2MR^2$. The value of $x$ is ______.
Numerical value type. Enter your answer.
Answer: 3
Perpendicular axis theorem: about the central normal axis $I=\dfrac{MR^2}{2}$. Parallel axis to the edge: $\dfrac{MR^2}{2}+MR^2=\dfrac32MR^2$, so $x=3$.
Solution by Sreeraj P, M.Sc Physics