$A$ disc of mass $M$ with uniform surface mass density $\sigma$ is shown in the figure. The centre of mass of the quarter disc (the shaded area) is at the position $(\frac{x}{3} \frac{R}{\pi}, \frac{x}{3} \frac{R}{\pi})$,where $R$ is the radius of the disc and $x$ is ....... . (Round off to the nearest integer)

$A$ square plate of side $12 \ cm$ is shown in the figure. If a square of side $2 \ cm$ is cut from one of its corners,where will the center of mass of the remaining part be with respect to the center of the original square? The plate has uniform thickness and density.

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 uniform disk of radius $R$,a circular hole of radius $R/2$ is cut out. The centre of the hole is at $R/2$ from the centre of the original disc. Locate the centre of gravity of the resulting flat body.

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:

The centre of mass of a solid hemisphere of radius $8 \, cm$ is $X \, cm$ from the centre of the flat surface. Then the value of $X$ is $......$