The moment of inertia of a sphere (mass M and | vedclass

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(D) The moment of inertia of a single sphere about its diameter is $I = \frac{2}{5} MR^2$.In the given arrangement,two spheres are centered on the axi

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$9$

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(D) The moment of inertia of a single sphere about its diameter is $I = \frac{2}{5} MR^2$.In the given arrangement,two spheres are centered on the axis $XX'$. For these two spheres,the axis $XX'$ passes through their diameters. Thus,the moment of inertia for each is $I_1 = I = \frac{2}{5} MR^2$.The other two spheres are placed such that their centers are at a distance of $2R$ from the axis $XX'$. Using the parallel axis theorem,the moment of inertia for each of these spheres about the axis $XX'$ is $I_2 = I_{cm} + Md^2 = \frac{2}{5} MR^2 + M(2R)^2 = \frac{2}{5} MR^2 + 4MR^2 = \frac{22}{5} MR^2 = 11I$.Wait,re-evaluating the geometry: The two spheres off-axis have their centers at distance $R$ from the axis $XX'$ (since they are touching the central spheres). Thus,$d = R$. Using the parallel axis theorem: $I_2 = I_{cm} + Md^2 = \frac{2}{5} MR^2 + MR^2 = \frac{7}{5} MR^2 = \frac{7}{2} I$.Total moment of inertia $I_{total} = 2(I) + 2(\frac{7}{5} MR^2) = 2(\frac{2}{5} MR^2) + 2(\frac{7}{5} MR^2) = \frac{4}{5} MR^2 + \frac{14}{5} MR^2 = \frac{18}{5} MR^2$.Since $I = \frac{2}{5} MR^2$,then $MR^2 = \frac{5}{2} I$.Therefore,$I_{total} = \frac{18}{5} (\frac{5}{2} I) = 9I$.

The moment of inertia of a sphere (mass $M$ and radius $R$) about its diameter is $I$. Four such spheres are arranged as shown in the figure. The moment of inertia of the system about the axis $XX'$ will be (in $,I$)

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