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:

$A$ 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?

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 $cm$)

$A$ thin uniform circular disc has mass $9M$ and radius $R$. $A$ small circular part of radius $R/3$ is cut from the disc as shown in the figure. Calculate the moment of inertia of the remaining part about an axis passing through the center $O$ and perpendicular to the plane of the disc.

From a uniform square plate,one-fourth part is removed as shown. The centre of mass of the remaining part will lie on:

$A$ 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: