Test SeriesFree PYQsExamsPricing

JEE Main Physics question

PhysicsSystems of Particles and Rotational MotionJEE Main

The moment of inertia of a solid sphere about an axis passing through its centre of gravity is . Then its radius of gyration about a parallel axis at a distance from the first axis is

Answer: (B)

Solution

I = I_{cm}+M(2R)^2 = \dfrac25MR^2+4MR^2=\dfrac{22}{5}MR^2 = MK^2 \Rightarrow K=\sqrt{\dfrac{22}{5}}\,R.

Question ID VK-158501