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Three identical spherical shells, each of mass m and radius r are placed as shown in figure. Consider an axis X X' which is touching to two shells and passing through diameter of third shell. Moment of inertia of the system consisting of these three spherical shells about X X' axis is

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$\frac{16}{5} m r^{2}$

$4 m r^{2}$

$\frac{11}{5} m r^{2}$

$3 m r^{2}$

answer is B.

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Detailed Solution

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Total moment of inertia of the system comprises one shell about its diameter and two shells about the tangent.

So, $\frac{2}{3} {MR}^{2} + (\frac{2}{3} {MR}^{2} + {MR}^{2}) 2$

$\frac{2}{3} {MR}^{2} + \frac{4}{3} {MR}^{2} + 2 {MR}^{2} = 4 {MR}^{2}$

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