$A$ uniform square plate of side $a$ has a square hole of side $a/2$ cut out of it as shown in the figure. The hole is centered at $(a/4, a/4)$ from the center of the plate. The center of mass of the remaining part is:

In the figure shown,a hole of radius $2 \, cm$ is made in a semicircular disc of radius $6 \, cm$ at a distance $8 \, cm$ from the centre $C$ of the disc. The distance of the centre of mass of this system from point $C$ is ......... $cm$.

$A$ uniform square plate of side $a$ has a circular hole of diameter $a$ cut out from it as shown in the figure. The center of the hole is at $(a/4, a/4)$ from the center of the square. The distance of the center of mass of the resulting plate from the center of the square 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

As shown in the figure,when a spherical cavity (centred at $O$) of radius $1$ is cut out of a uniform sphere of radius $R$ (centred at $C$),the centre of mass of the remaining (shaded) part of the sphere is at $G$,i.e.,on the surface of the cavity. $R$ can be determined by the equation:

$A$ circular plate of radius $r$ is removed from a uniform circular plate $P$ of radius $4 r$ to form a hole. If the distance between the centre of the hole formed and the centre of the plate $P$ is $2 r$,then the distance of the centre of mass of the remaining portion from the centre of the plate $P$ is