Q 11-06-111JEE MainJEE Main 2023 (25 Jan, Shift 1)Easy
$I_{CM}$ is moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc. $I_{AB}$ is its moment of inertia about an axis AB perpendicular to plane and parallel to axis CM at a distance $\dfrac23R$ from center, where $R$ is the radius of the disc. The ratio of $I_{AB}$ and $I_{CM}$ is $x:9$. The value of $x$ is ______.
Numerical value type. Enter your answer.
Answer: 17
Parallel axis theorem:
$$I_{AB}=\frac12MR^2+M\left(\frac23R\right)^2=MR^2\left(\frac12+\frac49\right)=\frac{17}{18}MR^2$$
$\dfrac{I_{AB}}{I_{CM}}=\dfrac{17/18}{1/2}=\dfrac{17}{9}$, so $x=17$.
Solution by Sreeraj P, M.Sc Physics