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

$A$ circular disc of radius $R$ is removed from one end of a bigger circular disc of radius $2R$. The centre of mass of the new disc is at a distance $\alpha R$ from the centre of the bigger disc. The value of $\alpha$ is

$A$ smaller cube with side $b$ (depicted by dashed lines) is excised from a bigger uniform cube with side $a$ as shown below,such that both cubes have a common vertex $P$. Let $X = a/b$. If the centre of mass of the remaining solid is at the vertex $O$ of the smaller cube,then $X$ satisfies:

$A$ uniform square plate of side $a$ has a circular hole of diameter $a/2$ cut out from it as shown in the figure. The center of the hole is at $a/4$ from the center of the square plate. The distance of the center of mass of the resulting plate from the center of the square is:

For the given uniform square lamina $ABCD$ whose centre is $O$,pick the incorrect statement.

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