Mention the position of the centre of mass of particles of equal mass.

The distance between the carbon atom and the oxygen atom in a carbon monoxide molecule is $1.1 Å$. Given,mass of carbon atom is $12 amu$ and mass of oxygen atom is $16 amu$. Calculate the position of the centre of mass of the carbon monoxide molecule.

Four identical particles each of mass $m$ are kept at the four corners of a square of side $a$. If one of the particles is removed,the shift in the position of the centre of mass is

The distance of the centre of mass from end $A$ of a one-dimensional rod $(AB)$ having mass density $\rho = \rho_{0} \left(1 - \frac{x^{2}}{L^{2}}\right) \text{ kg/m}$ and length $L$ (in meters) is $\frac{3L}{\alpha} \text{ m}$. The value of $\alpha$ is $\ldots \ldots \ldots$ (where $x$ is the distance from end $A$).

Two point masses of $1.5 \ g$ and $2.5 \ g$ are placed at a distance of $16 \ cm$ from each other. If the center of mass is at a distance $x$ from the $1.5 \ g$ mass,then $x = \dots \ cm$.

Obtain an expression for the position vector of the centre of mass of a system of $n$ particles in one dimension.