$A$ circular plate $A$ of radius $1.5 r$ is removed from one edge of a uniform circular plate $B$ of radius $2 r$. The distance of the centre of mass of the remaining portion from the centre of the plate $B$ is

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

The equilateral triangle $ABC$ is cut from a thin solid sheet of wood. $D, E$ and $F$ are the midpoints of its sides as shown and $G$ is the centre of the triangle. The moment of inertia of the triangle about an axis passing through $G$ and perpendicular to the plane of the triangle is $I_0$. If the smaller triangle $DEF$ is removed from $ABC$,the moment of inertia of the remaining figure about the same axis is $I$. Then

$A$ 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:

$A$ regular square plate is shown in the figure. Four identical squares are removed from its corners. Where will the $C.M.$ (Centre of Mass) be located if squares $1$ and $2$ are removed?

$A$ ring is formed by joining two uniform semi-circular rings $ABC$ and $ADC$. The mass of $ABC$ is thrice that of $ADC$. If the ring is hinged to a fixed support at $A$,it can rotate freely in a vertical plane. Find the value of $\tan \theta$,where $\theta$ is the angle made by the line $AC$ with the vertical in equilibrium.