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(D) Let the mass of each cylinder be $M' = M/4$.Consider the two cylinders perpendicular to the axis (labeled $I_1$ in the diagram). The axis passes t

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$ \frac{3}{8}MR^{2}+\frac{1}{6}ML^{2} $

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(D) Let the mass of each cylinder be $M' = M/4$.Consider the two cylinders perpendicular to the axis (labeled $I_1$ in the diagram). The axis passes through their centers perpendicular to their length. The moment of inertia of one such cylinder is $I_1 = \frac{M'R^2}{4} + \frac{M'L^2}{12}$.Consider the two cylinders parallel to the axis (labeled $I_2$ in the diagram). The axis is at a distance $L/2$ from their centers. The moment of inertia of one such cylinder about its own center (axis along length) is $\frac{M'R^2}{2}$. By the parallel axis theorem, $I_2 = \frac{M'R^2}{2} + M'(L/2)^2 = \frac{M'R^2}{2} + \frac{M'L^2}{4}$.The total moment of inertia is $I_{net} = 2I_1 + 2I_2 = 2(\frac{M'R^2}{4} + \frac{M'L^2}{12}) + 2(\frac{M'R^2}{2} + \frac{M'L^2}{4})$.$I_{net} = \frac{M'R^2}{2} + \frac{M'L^2}{6} + M'R^2 + \frac{M'L^2}{2} = \frac{3}{2}M'R^2 + \frac{2}{3}M'L^2$.Substituting $M' = M/4$: $I_{net} = \frac{3}{2}(M/4)R^2 + \frac{2}{3}(M/4)L^2 = \frac{3}{8}MR^2 + \frac{1}{6}ML^2$.

The moment of inertia of a square loop made of four uniform solid cylinders, each having radius $R$ and length $L$ $(R < L)$, about an axis passing through the midpoints of opposite sides, is (Take the mass of the entire loop as $M$):

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