Two identical rings each of mass m with their planes mutually perpendicular, radius R are welded at their point of contact O. If the system is free to rotate about an axis passing through the point P perpendicular to the plane of the paper the moment of inertia of the system about this axis is equal to :
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a
6.5 mR2
b
12 mR2
c
6 mR2
d
11.5 mR2
answer is D.
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Detailed Solution
Required moment of inertia :I=mR2+mR2+mR22+m(3R)2=11.5mR2