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Three identical spherical shells, each of mass M and radius r are placed as shown in figure. Consider an axis XX^’ 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 XX^’ axis is

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

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

$3 M r^{2}$

$4 M r^{2}$

answer is C.

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Detailed Solution

Moment of inertia of shell 1 along diameter $I_{d i a m e t e r} = \frac{2}{3} M R^{2}$

Moment of inertia of shell 2 = m. i. of shell 3 = $I_{\tan g e n t i a l} = \frac{2}{3} M R^{2} + M R^{2} = \frac{5}{3} M R^{2}$

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So, I of the system along $x x^{1}$ $= I_{d i a m e t e r} + (I_{t a n g e n t i a l}) \times 2$

Or, $I_{t o t a l} = \frac{2}{3} M R^{2} + (\frac{5}{3} M R^{2}) \times 2$ $= \frac{12}{3} M R^{2} = 4 M R^{2}$

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