$A$ uniform square plate of side $a$ has a square hole of side $a/2$ cut out of it as shown in the figure. The hole is centered at $(a/4, a/4)$ from the center of the plate. The center of mass of the remaining part is:

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:

$A$ 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)

In the figure shown,a hole of radius $2 \, cm$ is made in a semicircular disc of radius $6 \, cm$ at a distance $8 \, cm$ from the centre $C$ of the disc. The distance of the centre of mass of this system from point $C$ is ......... $cm$.

$A$ uniform square plate is shown in the figure. Four identical small squares are removed from its corners. Where will the $C.M.$ (Center of Mass) be located after removing all four squares? Give the answer in terms of quadrants and axes.

$A$ circular plate of diameter '$a$' is kept in contact with a square plate of side '$a$' as shown. The density of the material and the thickness are same everywhere. The centre of mass of the composite system will be ...........