Two identical 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 center 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

For $1 ball = \frac{2}{5} {MR}^{2} + M (2 R)^{2} = \frac{22}{5} {MR}^{2}$

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

For Rod, $I = M \frac{(2 R)^{2}}{12} = \frac{{MR}^{2}}{3}$

$I_{System} = I_{ball} + I_{rod} = (\frac{44}{5} + \frac{1}{3}) {MR}^{2} = \frac{137}{15} {MR}^{2}$

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