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:

$A$ 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

The $(x, y)$ coordinates (in $cm$) of the centre of mass of letter $E$ relative to the origin $O$,whose dimensions are shown in the figure are: (Take width of the letter $2 \ cm$ everywhere).

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

$A$ uniform circular disc has radius $r$. $A$ square portion of diagonal $r$ is cut from it. The centre of mass of the remaining disc from the centre of the disc is

$A$ sphere of diameter $r$ is cut from a sphere of radius $r$ such that the centre of mass of the remaining mass is at the maximum distance from the original centre. What is this distance?