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 circular disc of radius $a$ is taken. $A$ circular portion of radius $b$ has been removed from it as shown in the figure. If the centre of the hole is at a distance $c$ from the centre of the disc,the distance $x_2$ of the centre of mass of the remaining part from the initial centre of mass $O$ is given by

$A$ circular disc of radius $R$ is removed from one end of a bigger circular disc of radius $2R$. The centre of mass of the new disc is at a distance $\alpha R$ from the centre of the bigger disc. The value of $\alpha$ is

$A$ circular portion of radius $R_2$ has been removed from one edge of a circular disc of radius $R_1$. The correct expression for the centre of mass for the remaining portion of the disc is

$A$ 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:

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).