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(C) The moment of inertia of a solid sphere of mass $m$ and radius $r$ about its diameter is $I_{cm} = \frac{2}{5} mr^2$.For sphere $1$,the axis $YY^1

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$\frac{16}{5} mr^2$

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(C) The moment of inertia of a solid sphere of mass $m$ and radius $r$ about its diameter is $I_{cm} = \frac{2}{5} mr^2$.For sphere $1$,the axis $YY^1$ passes through its center (diameter). Thus,$I_1 = \frac{2}{5} mr^2$.For spheres $2$ and $3$,the axis $YY^1$ is tangent to them. The distance of the center of these spheres from the axis $YY^1$ is $r$. Using the parallel axis theorem,$I = I_{cm} + md^2$,where $d = r$.So,$I_2 = I_3 = \frac{2}{5} mr^2 + mr^2 = \frac{7}{5} mr^2$.The total moment of inertia of the system is $I_{total} = I_1 + I_2 + I_3 = \frac{2}{5} mr^2 + \frac{7}{5} mr^2 + \frac{7}{5} mr^2 = \frac{16}{5} mr^2$.

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