$A$ 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

$A$ circular disc of radius $R$ is cut from a thin metal plate. $A$ hole of radius $R/2$ is cut from this disc such that it touches the rim of the original disc. Find the distance of the center of mass of the remaining part from the center of the original disc.

$A$ 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

From a circle of radius $a,$ an isosceles right-angled triangle with the hypotenuse as the diameter of the circle is removed. The distance of the centre of gravity of the remaining portion from the centre of the circle is

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: