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Two circular rings of equal mass (m) and radius (r) are placed side by side, touching each other. The moment of inertia of the system about tangential axis in the plane of system passing through point of contact of the rings is :

$\begin {array} {1 1} (1)\;\large\frac{3}{2}mr^2 & \quad (2)\;6\: mr^2 \\ (3)\;\large\frac{5}{2}mr^2 & \quad (4)\;3\:mr^2 \end {array}$

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(4) $3\: mr^2$
answered Nov 7, 2013 by pady_1

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