$A$ circular plate of uniform thickness with a diameter of $56\, cm$ has its center at the origin. $A$ circular part with a diameter of $42\, cm$ is removed from one edge. What is the distance of the center of mass of the remaining part in $cm$?

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 $(x, y)$ 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).

$A$ uniform square plate is shown in the figure. Four identical small squares are removed from its corners. If squares $1, 2,$ and $3$ are removed,where will the center of mass $(C.M.)$ be located?

$A$ circular hole of radius $\left(\frac{a}{2}\right)$ is cut out of a circular disc of radius $'a'$ as shown in the figure. The centroid of the remaining circular portion with respect to point $'O'$ will be:

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