Two particles of masses $m_1$ and $m_2$ are separated by a distance $d$. The shift in the centre of mass when the two particles are interchanged is:

$A$ rigid body consists of a $3 \ kg$ mass and a $2 \ kg$ mass connected by a massless rod. The $3 \ kg$ mass is at $\vec{r}_1 = (2 \hat{i} + 5 \hat{j}) \ m$ and the $2 \ kg$ mass is at $\vec{r}_2 = (4 \hat{i} + 2 \hat{j}) \ m$. Find the length of the rod and the coordinates of the center of mass.

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

Four particles of masses $m_1 = 2m$,$m_2 = 4m$,$m_3 = m$,and $m_4$ are placed at the four corners of a square. What should be the value of $m_4$ so that the centre of mass of all the four particles is exactly at the centre of the square?

The distance of the centre of mass from end $A$ of a one-dimensional rod $(AB)$ having mass density $\rho = \rho_{0} \left(1 - \frac{x^{2}}{L^{2}}\right) \text{ kg/m}$ and length $L$ (in meters) is $\frac{3L}{\alpha} \text{ m}$. The value of $\alpha$ is $\ldots \ldots \ldots$ (where $x$ is the distance from end $A$).

Give the location of the centre of mass of a $(i)$ sphere,$(ii)$ cylinder,$(iii)$ ring,and $(iv)$ cube,each of uniform mass density. Does the centre of mass of a body necessarily lie inside the body?