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?

The coordinates of the positions of particles of mass $7 \, g$,$4 \, g$,and $10 \, g$ are $(1, 5, -3) \, cm$,$(2, 5, 7) \, cm$,and $(3, 3, -1) \, cm$ respectively. The position of the centre of mass of the system would be:

Three identical spheres,each of mass $1 \ kg$,are kept as shown in the figure,touching each other,with their centres on a straight line. If their centres are marked $P, Q, R$ respectively,the distance of the centre of mass of the system from $P$ is

$A$ particle of mass $m$ is moving in the $xy$-plane at a certain height from the ground along the $x$-axis. It suddenly explodes into two fragments of masses $m/4$ and $3m/4$. At that instant,the smaller fragment is at $y = 15 \ cm$. At this instant,the larger fragment will be at $y = $ ........ $cm$.

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

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