$A$ smaller cube with side $b$ (depicted by dashed lines) is excised from a bigger uniform cube with side $a$ as shown below,such that both cubes have a common vertex $P$. Let $X = a/b$. If the centre of mass of the remaining solid is at the vertex $O$ of the smaller cube,then $X$ satisfies:

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 $A$ of radius $1.5 r$ is removed from one edge of a uniform circular plate $B$ of radius $2 r$. The distance of the centre of mass of the remaining portion from the centre of the plate $B$ is

Consider a circular disc of radius $20 \ cm$ with centre located at the origin. $A$ circular hole of radius $5 \ cm$ is cut from this disc in such a way that the edge of the hole touches the edge of the disc. The distance of the centre of mass of the residual or remaining disc from the origin will be (in $cm$)

From a uniform disc of radius $R$,an equilateral triangle of side $\sqrt{3}R$ is cut as shown. The new position of the centre of mass is:

$A$ uniform square plate of side $L$ is placed in the $xy$-plane with its center at the origin $(0,0)$. Four identical small squares of side $a$ are removed from its four corners. If the small squares at $(L/2-a/2, L/2-a/2)$, $(-L/2+a/2, L/2-a/2)$, and $(-L/2+a/2, -L/2+a/2)$ are removed, where will the center of mass $(C.M.)$ of the remaining plate be located?