The variation of density of a cylindrical thick and long rod, is $\rho  = {\rho _0}\frac{{{x^2}}}{{{L^2}}}$ , then position of its centre of mass from $x = 0$ end is

A circular disc of radius $R$ is removed from a bigger circular disc of radius $2R$ such that the circumferences of the discs coincide. The centre of mass of the new disc is $\frac{\alpha}{R}$ form the centre of the bigger disc. The value of a is $\alpha $ is

Where will be the centre of mass on combining two masses $m$ and $M$ $(M>m)$

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

In the figure shown $ABC$ is a uniform wire . If centre of mass of wire lies vertically below point $A$ , then $\frac{{BC}}{{AB}}$ is close to 

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