Two circular rings each of mass M and radius R are attached to each other at their rims and their planes are perpendicular to each other as shown in the figure. The moment of inertia of system about a tangent to one of the rings and perpendicular to its plane as shown in the figure is

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$\frac{11}{4} M R^{2}$

$\frac{3}{4} M R^{2}$

$\frac{13 M R^{2}}{4}$

$\frac{13}{2} M R^{2}$

answer is C.

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Detailed Solution

$I_{1} = M {(2 R)}^{2} + \frac{M R^{2}}{2} f o r t h e v e r t i c a l r i n g I_{2} = M {(R)}^{2} + M {(R)}^{2} = 2 M R^{2} f o r t h e h o r i z o n t a l r i n g T o t a l m o m e n t o f i n e r t i a l I_{1} + I_{2} = \frac{13 M R^{2}}{2}$

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