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

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:

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.

$A$ 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 figure shows a composite system of two uniform rods of lengths as indicated. The coordinates of the centre of mass of the system of rods are ...........

From a circular disc of radius $R$,a triangular portion is cut as shown in the figure. The distance of the center of mass $(COM)$ of the remaining disc from the center of the disc $O$ is: