$A$ circular disc of radius $R$ has a uniform thickness. $A$ circular hole of diameter equal to the radius of the disc has been cut out as shown. The distance of the centre of mass of the remaining disc from point $O$ is

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:

$A$ uniform thin metal plate of mass $10 \ kg$ with the dimensions shown in the figure is given. The ratio of the $x$ and $y$ coordinates of the center of mass of the plate is $\frac{n}{9}$. The value of $n$ is:

The coordinates of the centre of mass of a uniform $L$-shaped plate of mass $3 \ kg$ shown in the figure is:

$A$ 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 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