Centre of mass $(C.M.)$ of three particles of masses $1 \,kg, 2 \,kg$ and $3 \,kg$ lies at the point $(1, 2, 3)$ and $C.M.$ of another system of particles of $3 \,kg$ and $2 \,kg$ lies at the point $(-1, 3, -2)$. Where should we put a particle of mass $5 \,kg$ so that the $C.M.$ of the entire system lies at the $C.M.$ of the first system?

Two particles of masses $1\,kg$ and $3\,kg$ have position vectors $2\hat{i} + 3\hat{j} + 4\hat{k}$ and $-2\hat{i} + 3\hat{j} - 4\hat{k}$ respectively. The position vector of the centre of mass 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

In an $HCl$ molecule,the distance between the nuclei of the two atoms is $1.27 \ \mathring A$. The $Cl$ atom is approximately $35.5$ times heavier than the $H$ atom. The center of mass of this molecule will be at a distance of approximately ....... $\mathring A$ from the center of the $H$ atom.

The distance of the centre of mass of a solid uniform cone from its vertex is $z_0$. If the radius of its base is $R$ and its height is $h$,then $z_0$ is equal to:

The velocities of three particles of masses $20 \ g$,$30 \ g$ and $50 \ g$ are $10 \hat{i}$,$10 \hat{j}$ and $10 \hat{k}$,respectively. The velocity of the centre of mass of the three particles is