Q 11-06-246JEE MainJEE Main 2018 (8 Apr)Medium
Seven identical circular planar disks, each of mass $M$ and radius $R$ are welded symmetrically as shown. The moment of inertia of the arrangement about the axis normal to the plane and passing through the point $P$ is:
Answer: (A) $\dfrac{181}{2}MR^2$
About the centre $O$: the middle disc gives $\dfrac{MR^2}{2}$; each of the six outer discs has its centre at $2R$ from $O$:
$$I_O = \frac{MR^2}{2} + 6\left(\frac{MR^2}{2} + M(2R)^2\right) = \frac{MR^2}{2} + 27MR^2 = \frac{55}{2}MR^2$$
$P$ is on the rim of an outer disc, at $3R$ from $O$. With total mass $7M$, by the parallel axis theorem:
$$I_P = \frac{55}{2}MR^2 + 7M(3R)^2 = \frac{55}{2}MR^2 + 63MR^2 = \frac{181}{2}MR^2$$
Solution by Sreeraj P, M.Sc Physics