The linear mass density of a $3 \ m$ long rod varies as $\lambda = 2 + x$. The position of the center of mass of the rod is at:

$A$ rod of length $3\, m$ has a mass per unit length directly proportional to the distance $x$ from one of its ends. The center of gravity of the rod from that end will be at ........ $m$.

Four particles of masses $m, 2m, 3m$ and $4m$ are arranged at the corners of a parallelogram with each side equal to $a$ and one of the angles between two adjacent sides is $60^o$. The parallelogram lies in the $x-y$ plane with mass $m$ at the origin and $4m$ on the $x$-axis. The centre of mass of the arrangement will be located at

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:

What is the position of the centre of mass of symmetrical and homogeneous bodies?

The figure shows two cylindrical rods whose centers of mass are marked as $A$ and $B$. The line $AB$ divides the region into two parts: one containing point $O$ (region $1$) and the other containing point $O'$ (region $2$). Choose the correct option regarding the center of mass of the combined system.