Two molecules of a gas have speeds of $9\times {10}^6{\mathrm{ms}}^{-1}$ and $1\times {10}^6{\mathrm{ms}}^{-1}$respectively. What is the root mean square speed of these molecules?

For n-molecules, we know that

${\mathrm{v}}_{\mathrm{rms}}=\sqrt{\frac{{\mathrm{v}}_1^2+{\mathrm{v}}_2^2+{\mathrm{v}}_3^2+\cdots +{\mathrm{v}}_{\mathrm{n}}^2}{\mathrm{n}}}$

                   $\left[{\mathrm{v}}_{\mathrm{rms}}=\text{ root mean square velocity }\right]$

where ${\mathrm{v}}_1,{\mathrm{v}}_2,{\mathrm{v}}_3\dots ,{\mathrm{v}}_{\mathrm{n}},$are individual velocities of n-molecules of the gas.
For two molecules,

${\mathrm{v}}_{\mathrm{rms}}=\sqrt{\frac{{\mathrm{v}}_1^2+{\mathrm{v}}_2^2}{2}}\left[{\mathrm{v}}_{1,}{\mathrm{v}}_2\cdot {\mathrm{v}}_3\dots ,{\mathrm{v}}_{\mathrm{n}}\text{ are individual velocity }\right]$

${\text{ Given, v}}_1=9\times {10}^6\mathrm{m}/\mathrm{s} \mathrm{and} {\mathrm{v}}_2=1\times {10}^6\mathrm{m}/\mathrm{s}$

$\therefore {\mathrm{v}}_{\mathrm{rms}}=\sqrt{\frac{{\left(9\times {10}^6\right)}^2+{\left(1\times {10}^6\right)}^2}{2}}=\sqrt{41}\times {10}^6\mathrm{m}/\mathrm{s}$