$A$ uniform square plate of side $a$ has a square hole of side $a/2$ cut out of it as shown in the figure. The hole is centered at $(a/4, a/4)$ from the center of the plate. The center of mass of the remaining part 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 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

$A$ circular plate of uniform thickness has a diameter of $56 \ cm$. $A$ circular part of diameter $42 \ cm$ is removed from one edge. What is the position of the centre of mass of the remaining part in $cm$?

From a uniform disc of radius $R$,an equilateral triangle of side $\sqrt{3}R$ is cut as shown. The new position of the centre of mass is:

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 $^{\circ}$)