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).

From a circular disc of radius $R$,a square is cut out with a radius as its diagonal. The center of mass of the remaining part is at a distance (from the centre) of:

$A$ circular disc of radius $\alpha R$ has a smaller circular disc of radius $R$ removed from it such that their circumferences touch each other. The center of mass of the new disc is at a distance of $\alpha R$ from the center of the original large disc. What is the value of $\alpha$?

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:

$A$ circular hole of radius $\left(\frac{a}{2}\right)$ is cut out of a circular disc of radius $'a'$ as shown in the figure. The centroid of the remaining circular portion with respect to point $'O'$ will be:

From a uniform circular disc of radius $2 \,cm$ (its centre of mass is at $O$), a circular portion of radius $1 \,cm$ is removed such that the shift in centre of mass is maximum. The disc is now rotated by an angle $\theta$ about an axis perpendicular to its plane and passing through $O$. If the magnitude of displacement of the new centre of mass is $\frac{1}{\sqrt{3}} \,cm$, then the $\theta$ is (in $^{\circ}$)