Two rings of the same mass and radius 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 the plane of one of the rings is (Take M =mass of each ring and R= radius of each ring)

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${MR}^{2}$

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

$2 {MR}^{2}$

$\frac{5}{2} {MR}^{2}$

answer is B.

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Detailed Solution

For one ring, the given axis is diameter. For the other ring, the given axis is ‘an axis passing through the centre and perpendicular to the plane’.

$∴ I = \frac{1}{2} {MR}^{2} + {MR}^{2} = \frac{3}{2} {MR}^{2}$

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