Which of the points is the likely position of the centre of mass of the system shown in the figure?

$A$ thin bar of length $L$ has a mass per unit length $\lambda$ that increases linearly with distance $x$ from one end. If its total mass is $M$ and its mass per unit length at the lighter end $(x=0)$ is $\lambda_0$,then the distance of the centre of mass from the lighter end is:

Locate the centre of mass of the arrangement shown in the figure. The three rods are identical in mass and length.

Three identical uniform thin rods each of mass $m$ and length $L$ are arranged in the $XY$ plane as shown in the figure. $A$ fourth uniform thin rod of mass $3m$ is placed as shown in the figure in the $XY$ plane. The value of length of the fourth rod such that the centre of mass of all the four rods lies at the origin 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?

Two objects of masses $200 \text{ g}$ and $500 \text{ g}$ possess velocities $10 \hat{i} \text{ m/s}$ and $(3 \hat{i} + 5 \hat{j}) \text{ m/s}$ respectively. The velocity of their centre of mass in $\text{m/s}$ is: