$A$ 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

$A$ circular hole of radius $3 \text{ cm}$ is cut out from a uniform circular disc of radius $6 \text{ cm}$. The centre of the hole is at $3 \text{ cm}$ from the centre of the original disc. The distance of the centre of gravity of the resulting flat body from the centre of the original disc is: (in $\text{ cm}$)

The figure shows a uniform square plate from which four identical small squares have been removed from the corners. If square $1$ is removed,where will the center of mass $(C.M.)$ be located?

The position vector of the centre of mass $\vec{r}_{cm}$ of an asymmetric uniform bar of negligible area of cross-section as shown in the figure is:

From a circular disc of radius $R$,a square is cut out with a radius as its diagonal. The center of mass of the remaining part is at a distance (from the centre) of:

$A$ circular hole of radius $\left(\frac{a}{2}\right)$ is cut out of a circular disc of radius $'a'$ as shown in the figure. The centroid of the remaining circular portion with respect to point $'O'$ will be: