Four identical solid spheres each of mass 'm' and radius 'a' are placed with their centres on the four corners of a square of side 'b'. The moment of inertia of the system about one side of square where the axis of rotation is parallel to the plane of the square is :

A sphere of radius 'a' and mass 'm' rolls along a horizontal plane with constant speed v₀. It encounters an inclined plane at angle $$\theta$$ and climbs upward. Assuming that it rolls without slipping, how far up the sphere will travel?

Moment of inertia (M. I.) of four bodies, having same mass and radius, are reported as;

I₁ = M.I. of thin circular ring about its diameter,

I₂ = M.I. of circular disc about an axis perpendicular to disc and going through the centre,

I₃ = M.I. of solid cylinder about its axis and

I₄ = M.I. of solid sphere about its diameter.

Then :

The linear mass density of a thin rod AB of length L varies from A to B as
$$\lambda \left( x \right) = {\lambda _0}\left( {1 + {x \over L}} \right)$$, where x is the distance from A. If M is the mass of the rod then its moment of inertia about an axis passing through A and perpendicular to the rod is :