From a circular disc of radius $R$,a triangular portion is cut as shown in the figure. The distance of the center of mass $(COM)$ of the remaining disc from the center of the disc $O$ is:

The equilateral triangle $ABC$ is cut from a thin solid sheet of wood. $D, E$ and $F$ are the midpoints of its sides as shown and $G$ is the centre of the triangle. The moment of inertia of the triangle about an axis passing through $G$ and perpendicular to the plane of the triangle is $I_0$. If the smaller triangle $DEF$ is removed from $ABC$,the moment of inertia of the remaining figure about the same axis is $I$. Then

From a uniform disk of radius $R$,a circular hole of radius $R/2$ is cut out. The centre of the hole is at $R/2$ from the centre of the original disc. Locate the centre of gravity of the resulting flat body.

$A$ spherical cavity of radius $r$ is carved out of a uniform solid sphere of radius $R$ as shown in the figure below. The distance of the centre of mass of the resulting body from that of the solid sphere is given by

In the figure shown,find the distance of the centre of mass of a system of a uniform circular plate of radius $3R$ from $O$,in which a hole of radius $R$ is cut whose centre is at a distance of $2R$ from the centre of the large circular plate.

$A$ circular disc of radius $\alpha R$ has a smaller circular disc of radius $R$ removed from it such that their circumferences touch each other. The center of mass of the new disc is at a distance of $\alpha R$ from the center of the original large disc. What is the value of $\alpha$?