Four particles of masses $m_1 = 2m$,$m_2 = 4m$,$m_3 = m$,and $m_4$ are placed at the four corners of a square. What should be the value of $m_4$ so that the centre of mass of all the four particles is exactly at the centre of the square?

Two bodies $A$ and $B$ have masses $M$ and $m$ respectively,where $M > m$,and they are at a distance $d$ apart. Equal forces are applied to each of them so that they approach each other. The position where they hit each other is

Five masses are placed in a plane as shown in the figure. The coordinates of the centre of mass are nearest to:

$A$ uniform rod of mass $400 \ g$ and length $1 \ m$ is pivoted at one end. It is displaced by an angle of $60^o$. The increase in its potential energy is ....... $J$.

Three identical spheres each of mass $M$ are placed at the corners of a right-angled triangle with mutually perpendicular sides equal to $3\,m$ each. Taking the point of intersection of mutually perpendicular sides as the origin,the magnitude of the position vector of the centre of mass of the system will be $\sqrt{x}\,m$. The value of $x$ is

Obtain the general expression for the centre of mass of a system of $n$ distributed particles in three dimensions.