The distance between the two atoms in an $HCl$ molecule is approximately $1.27 \mathring{A}$ $(1 \mathring{A} = 10^{-10} \text{ m})$. The chlorine atom is $35.5$ times heavier than the hydrogen atom. The position of the center of mass from the hydrogen atom is approximately ..... $\mathring{A}$.

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?

Two uniform plates of the same thickness and area but of different materials,one shaped like an isosceles triangle and the other shaped like a rectangle,are joined together to form a composite body as shown in the figure. If the centre of mass of the composite body is located at the mid-point of their common side,then the ratio between the masses of the triangle to that of the rectangle 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 $A$).

Two particles of mass $5\, kg$ and $10\, kg$ respectively are attached to the two ends of a rigid rod of length $1\, m$ with negligible mass. The centre of mass of the system from the $5\, kg$ particle is nearly at a distance of $..........\, cm$.

To find the centre of mass of a rigid body,why is it not possible to know $\sum m_i \vec{r}_i$ for all the particles?