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Two rings of same radius (r) and mass (m) are placed such that their centres are at a common point and their planes are perpendicular to each other. The moment of inertia of the system about an axis passing through the centre and perpendicular to plane of one of the ring is

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$\frac{1}{2} {mr}^{2}$

mr²

$\frac{3}{2} {mr}^{2}$

2mr²

answer is C.

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Detailed Solution

$I_{1} = {mr}^{2} & I_{2} = \frac{{mr}^{2}}{2} I_{1} + I_{2} \Rightarrow {mr}^{2} + \frac{{mr}^{2}}{2} = \frac{3}{2} {mr}^{2}$

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