A uniform circular disc has a sector of angle `90^(@)` removed from it. Mass of the remaining disc is M. Write the moment of inertia of the remaining disc about the axis `x x` shown in figure (Radius is R)

From a uniform circular disc of radius R and mass 9M, a small disc of radius (R )/(3) is removed as shown in the figure. The moment of intertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through centre of the disc is

From a circular disc of radius R and mass 9 M , a small disc of radius R/3 is removed from the disc. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through O is

A thin disc of mass 9M and radius R from which a disc of radius R/3 is cut shown in figure. Then moment of inertia of the remaining disc about O, perpendicular to the plane of disc is -

Consider moment of inertia of a uniform thin circular disc about a diametrical axis of the disc. There is a theorem which helps to find the moment of inertia of the disc about another axis parallel to this axis. Give the statement of this theorem.

One quarter of the disc of mass m is removed. If r be the radius of the disc, the new moment of inertia is

From a circular disc of radius R and mass 9M, a small disc of radius R/3 is removed as shown in figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through O is

From a circular disc of radius R and mass 9 M , a small disc of mass M and radius ( R)/(3) is removed concentrically. What will be the moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through the centre?