JEE Main Physics question
Moment of inertia of a ring of mass g and radius cm about an axis passing through its edge and parallel to its natural axis is
Answer: (C)
Solution
I_{edge}=I_{own}+mr^2=mr^2+mr^2=2mr^2=2(3)(1)^2=6 g-cm².
Moment of inertia of a ring of mass g and radius cm about an axis passing through its edge and parallel to its natural axis is
I_{edge}=I_{own}+mr^2=mr^2+mr^2=2mr^2=2(3)(1)^2=6 g-cm².