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

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

The center of mass of three particles of masses $10 \ kg, 20 \ kg$,and $30 \ kg$ is at $(0, 0, 0)$. Where should a fourth particle of mass $40 \ kg$ be placed so that the center of mass of the system shifts to $(3, 3, 3)$?

Three identical spheres each of mass $M$ are placed at the corners of a right-angled triangle with mutually perpendicular sides equal to $3\,m$ each. Taking the point of intersection of mutually perpendicular sides as the origin,the magnitude of the position vector of the centre of mass of the system will be $\sqrt{x}\,m$. The value of $x$ is

$Assertion$ : The position of the centre of mass of a body depends upon the shape and size of the body.
$Reason$ : The centre of mass of a body always lies at the centre of the body.

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 $A$).