$A$ 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 hole of radius $3 \text{ cm}$ is cut out from a uniform circular disc of radius $6 \text{ cm}$. The centre of the hole is at $3 \text{ cm}$ from the centre of the original disc. The distance of the centre of gravity of the resulting flat body from the centre of the original disc is: (in $\text{ cm}$)

$A$ square plate of side $12 \ cm$ is shown in the figure. If a square of side $2 \ cm$ is cut from one of its corners,where will the center of mass of the remaining part be with respect to the center of the original square? The plate has uniform thickness and density.

$A$ 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

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 $cm$)

$A$ uniform square plate is shown in the figure. Four identical small squares are removed from its corners. Where will the $C.M.$ (Center of Mass) be located after removing all four squares? Give the answer in terms of quadrants and axes.