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

From a homogeneous square plate,a triangle is cut out as shown in the figure. The side of the square is $a$,and the apex of the triangle is at the center of the square. The distance from the center of the square to the center of mass of the remainder of the plate 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:

$A$ 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

$A$ uniform rectangular thin sheet $ABCD$ of mass $M$ has length $a$ and breadth $b$,as shown in the figure. If the shaded portion $HBGO$ is cut off,the coordinates of the centre of mass of the remaining portion will be

$A$ disk of radius $R$ is removed from a larger disk of radius $2R$. The circumferences of both disks are tangent to each other. The center of mass of the new disk is at a distance $x$ from the center of the larger disk. Find the value of $x/R$.