From a uniform square plate,one-fourth part is removed as shown. The centre of mass of the remaining part will lie on:

$A$ circular hole of radius $\frac{R}{4}$ is made in a thin uniform disc having mass $M$ and radius $R$,as shown in the figure. The moment of inertia of the remaining portion of the disc about an axis passing through the point $O$ and perpendicular to the plane of the disc 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 $^{\circ}$)

$A$ 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 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

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: