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

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

The $\text{'}y\text{'}$ co-ordinate of the centre of mass of the system of three rods of length $\text{'}2a\text{'}$ and two rods of length $\text{'}a\text{'}$ as shown in the figure is (Assume all rods to be of uniform and equal density)

A segment of angle $\mathrm{\theta }$ is cut from a half disc of symmetrically as shown. If the centre of mass of the remaining part is at a distance $a$ from $O$ and the centre of mass of the original disc was at distance $d$, then it can be definitely said that,

A solid ball of mass $m$ and radius $R$ is released from the position shown in a large hollow fixed shell of same mass $m$ and radius $3R$ as shown in the figure. The displacement of the centre of mass of the system from its initial position when the solid ball touches the lower surface of the hollow shell is (centres of both the spheres coincide initially):

In the figure shown a semicircular area is removed from a uniform square plate of side $l$ and mass (before removing) $m.$ The $x-$coordinate of centre of mass of remaining portion is (The origin is at the centre of square)