Q 11-06-143JEE MainJEE Main 2022 (28 Jul, Shift 1)Medium
Four identical discs each of mass $M$ and diameter $a$ are arranged in a small plane as shown in figure. If the moment of inertia of the system about $OO'$ is $\dfrac x4Ma^2$. Then, the value of $x$ will be ______ .
Numerical value type. Enter your answer.
Answer: 3
Each disc has radius $\dfrac a2$, and $OO'$ lies in the plane of the discs.
The top and bottom discs rotate about a diameter: $2\times\dfrac{M(a/2)^2}{4} = \dfrac{Ma^2}{8}$.
The two middle discs have centres at $\dfrac a2$ from the axis. By the parallel axis theorem each has $\dfrac{Ma^2}{16} + M\dfrac{a^2}{4} = \dfrac{5Ma^2}{16}$, so together $\dfrac{5Ma^2}{8}$.
$$I = \frac{Ma^2}{8} + \frac{5Ma^2}{8} = \frac34Ma^2 \Rightarrow x = 3$$
Solution by Sreeraj P, M.Sc Physics