If the linear density of a rod of length $3 \ m$ varies as $\lambda = 2 + x$,then the position of the centre of mass of the rod is

The moment of inertia of an $HCl$ molecule about an axis passing through its centre of mass and perpendicular to the line joining the $H^+$ and $Cl^-$ ions is,if the interatomic distance is $1 \ \mathring A$:

An isolated particle of mass $m$ is moving in a horizontal plane $(x-y)$ along the $x$-axis,at a certain height above the ground. It suddenly explodes into two fragments of masses $m/4$ and $3m/4$. An instant later,the smaller fragment is at $y = +15 \ cm$. The larger fragment at this instant is at

$A$ $T$-shaped object with dimensions shown in the figure is lying on a smooth floor. $A$ force $\vec{F}$ is applied at point $P$ parallel to $AB$,such that the object has only translational motion without rotation. Find the location of $P$ with respect to $C$.

The variation of density of a cylindrical thick and long rod is $\rho = \rho_0 \frac{x^2}{L^2}$. The position of its centre of mass from the $x = 0$ end is:

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 $A$).