$A$ rod of length $5L$ is bent at a right angle,keeping one side length as $2L$. Find the position of the center of mass of the system. (Consider $L = 10 \ cm$)

$A$ 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:

$A$ 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 $; m$)

The centre of mass of three particles of masses $1 \ kg, 2 \ kg$ and $3 \ kg$ is at $(2, 2, 2)$. The position of the fourth mass of $4 \ kg$ to be placed in the system so that the new centre of mass is at $(0, 0, 0)$ is:

Two identical uniform rectangular blocks (with longest side $L$) and a solid sphere of radius $R$ are to be balanced at the edge of a heavy table such that the centre of the sphere remains at the maximum possible horizontal distance from the vertical edge of the table without toppling as indicated in the figure. If the mass of each block is $M$ and of the sphere is $M/2$,then the maximum distance $x$ that can be achieved is

For a system consisting of two objects,when can they be treated as point particles?