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Two identical solid spherical balls of mass M and radius R each are stuck on two ends of a rod of length 2R and mass M (see figure).The moment of inertia of the system about the axis passing perpendicularly through the centre of the rod is

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$10 M R^{2}$

$\frac{137}{3} M R^{2}$

$2 M R^{2}$

$\frac{137}{15} M R^{2}$

answer is C.

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Detailed Solution

$I_{ball} = \frac{2}{5} {MR}^{2} + M (2 R)^{2} = \frac{22}{5} {MR}^{2}$

$For two ball, I_{balls} = 2 \times \frac{22}{5} {MR}^{2} = \frac{44}{5} {MR}^{2}$

$I_{sys} = I_{balls} + I_{rod} = \frac{44}{5} {MR}^{2} + \frac{1}{12} M {(2 R)}^{2} = \frac{137}{15} {MR}^{2}$

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