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(D) The moment of inertia $(I)$ of the system about the axis $YY'$ is the sum of the moments of inertia of the three individual rings.$I = I_1 + I_2 +

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$\frac{7}{2}MR^2$

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(D) The moment of inertia $(I)$ of the system about the axis $YY'$ is the sum of the moments of inertia of the three individual rings.$I = I_1 + I_2 + I_3$For ring $1$,the axis $YY'$ passes through its diameter. The moment of inertia of a ring about its diameter is $I_1 = \frac{1}{2}MR^2$.For rings $2$ and $3$,the axis $YY'$ is tangent to the ring in its own plane. Using the parallel axis theorem,$I = I_{cm} + Md^2$,where $I_{cm} = \frac{1}{2}MR^2$ and $d = R$. Thus,$I_2 = I_3 = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$.Adding these together: $I = \frac{1}{2}MR^2 + \frac{3}{2}MR^2 + \frac{3}{2}MR^2 = \frac{7}{2}MR^2$.

Three rings each of mass $M$ and radius $R$ are arranged as shown in the figure. The moment of inertia of the system about $YY'$ will be

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