The centre of mass of a system of particles does not depend on

$A$ rod of length $3 \; m$ has a linear mass density that is directly proportional to the distance $x$ from one of its ends. Find the position of its centre of gravity from that end. (in $; m$)

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

$A$ 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:

The position vectors of two particles of masses $1 \ kg$ and $3 \ kg$ are $\hat{i} + 2\hat{j} + \hat{k}$ and $-3\hat{i} - 2\hat{j} + \hat{k}$ respectively. The position vector of the center of mass of this system is:

In the $HCl$ molecule,the separation between the nuclei of the two atoms is about $1.27 \, \mathring{A}$ $(1 \, \mathring{A} = 10^{-10} \, m)$. The approximate location of the centre of mass of the molecule from the hydrogen atom,assuming the chlorine atom to be about $35.5$ times as massive as the hydrogen atom,is ....... $\mathring{A}$.