Four identical spheres each of mass $m$ are placed at the corners of a square of side $2 \, m$. Taking the point of intersection of the diagonals as the origin,the coordinates of the centre of mass are

In the figure shown,$ABC$ is a uniform wire. If the center of mass of the wire lies vertically below point $A$,then $\frac{BC}{AB}$ is close to:

Two objects $A$ and $B$ of masses $M$ and $m$ respectively,where $M > m$,are separated by a distance. They move towards each other under the influence of equal forces. At what point will they collide?

$A$ point mass $m_A$ is connected to a point mass $m_B$ by a massless rod of length $l$ as shown in the figure. It is observed that the ratio of the moment of inertia of the system about the two axes $AA$ and $BB$,which are parallel to each other and perpendicular to the rod,is $\frac{I_{BB}}{I_{AA}} = 3$. The distance of the centre of mass of the system from the mass $m_A$ is

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:

Two objects of masses $200 \text{ g}$ and $500 \text{ g}$ possess velocities $10 \hat{i} \text{ m/s}$ and $(3 \hat{i} + 5 \hat{j}) \text{ m/s}$ respectively. The velocity of their centre of mass in $\text{m/s}$ is: