Two point masses $M$ each are placed at $(L, 0)$ and $(-L, 0)$. $A$ third point mass $M$ is uniformly rotating on the circle $x^2 + y^2 = L^2$. The equation of the path traced by the $COM$ of the $3$ point masses is:

Two particles of masses $m_1$ and $m_2$ are separated by a distance $d$. The shift in the centre of mass when the two particles are interchanged is:

Three masses of $2\,kg$,$4\,kg$,and $4\,kg$ are placed at the three points $(1, 0, 0)$,$(1, 1, 0)$,and $(0, 1, 0)$ respectively. The position vector of its center of mass is:

The balls $A, B$ and $C$ of masses $50 \ g, 100 \ g$ and $150 \ g$,respectively,are placed at the vertices of an equilateral triangle. The length of each side is $1 \ m$. If $A$ is placed at $(0,0)$ and $B$ is placed at $(1,0) \ m$,find the coordinates $(x, y)$ for the centre of mass of this system of the balls.

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

Seven identical homogeneous bricks,each of length $L$,are arranged as shown in the figure. Each brick is displaced with respect to the one in contact by $\frac{L}{10}$. Calculate the $x$-coordinate of the centre of mass of this system relative to the origin $O$ as shown.