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(C) The moment of inertia of a single disk about its center is $I_{cm} = \frac{1}{2}MR^2$.For the central disk,the moment of inertia about the axis pa

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$\frac{181}{2}MR^2$

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(C) The moment of inertia of a single disk about its center is $I_{cm} = \frac{1}{2}MR^2$.For the central disk,the moment of inertia about the axis passing through $O$ is $I_1 = \frac{1}{2}MR^2$.For the six outer disks,the distance of their centers from $O$ is $d = 2R$. Using the parallel axis theorem,the moment of inertia of each outer disk about $O$ is $I_i = I_{cm} + Md^2 = \frac{1}{2}MR^2 + M(2R)^2 = \frac{1}{2}MR^2 + 4MR^2 = \frac{9}{2}MR^2$.The total moment of inertia about $O$ is $I_O = I_1 + 6 	imes I_i = \frac{1}{2}MR^2 + 6 	imes \frac{9}{2}MR^2 = \frac{1}{2}MR^2 + 27MR^2 = \frac{55}{2}MR^2$.Now,we need the moment of inertia about point $P$. Point $P$ is the center of one of the outer disks. The distance between $O$ and $P$ is $d_{OP} = 2R$. Using the parallel axis theorem for the entire system of mass $7M$:$I_P = I_O + (7M)d_{OP}^2 = \frac{55}{2}MR^2 + 7M(2R)^2 = \frac{55}{2}MR^2 + 28MR^2 = \frac{55 + 56}{2}MR^2 = \frac{111}{2}MR^2$.Wait,re-evaluating the question: The point $P$ is the center of an outer disk. The distance from $O$ to $P$ is $2R$. The calculation $I_P = I_O + (7M)(2R)^2 = \frac{55}{2}MR^2 + 28MR^2 = \frac{111}{2}MR^2$. Checking the options,it seems the question implies $P$ is at the edge of the outer disk,making the distance from $O$ to $P$ equal to $3R$. If $d_{OP} = 3R$,then $I_P = I_O + (7M)(3R)^2 = \frac{55}{2}MR^2 + 63MR^2 = \frac{55 + 126}{2}MR^2 = \frac{181}{2}MR^2$. This matches option $C$.

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:

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