$A$ slender uniform rod of length $L$ is balanced vertically at a point $P$ on a horizontal surface having some friction. If the top of the rod is displaced slightly to the right,the position of its centre of mass at the time when the rod becomes horizontal is:

Center of mass of a system of three particles of masses $1 \, kg, 2 \, kg, 3 \, kg$ is at the point $(1 \, m, 2 \, m, 3 \, m)$ and center of mass of another group of two particles of masses $2 \, kg$ and $3 \, kg$ is at point $(-1 \, m, 3 \, m, -2 \, m)$. Where should a $5 \, kg$ particle be placed so that the center of mass of the system of all these six particles shifts to the center of mass of the first system?

The position of the centre of mass of a cube of uniform density will be at

Three identical spheres,each of mass $1 \ kg$,are placed touching each other with their centres on a straight line. Their centres are marked $K, L$,and $M$ respectively. The distance of the centre of mass of the system from $K$ is

$A$ uniform rod of length $L$ and mass $M$ is placed along the $x$-axis with its one end at the origin. The centre of mass of the rod is located at:

An object comprises a uniform ring of radius $R$ and its uniform chord $AB$ (not necessarily made of the same material) as shown. Which of the following can not be the centre of mass of the object?