The coordinates of the center of mass of a uniform flag-shaped lamina (thin flat plate) of mass $4 \; kg$ (the coordinates of the same are shown in the figure) 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:

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 $A$).

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.

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

$A$ straight rod of length $L$ has one of its ends at the origin and the other at $x = L$. If the mass per unit length of the rod is given by $\lambda = Ax$ (where $A$ is a constant),then where is its center of mass from the origin?