Two molecules of a gas have speeds of $9 \mathrm{x} {10}^6 {\mathrm{ms}}^{-\mathrm{l}}$ 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

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

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

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

For two molecules, ${\mathrm{v}}_{\mathrm{rms}} = \sqrt{\frac{{{\mathrm{v}}^2}_1+{{\mathrm{v}}^2}_2}{2}}$

[${\mathrm{v}}_1, {\mathrm{v}}_2, {\mathrm{v}}_3,...{\mathrm{v}}_{\mathrm{n}} \mathrm{are} \mathrm{individual} \mathrm{velocity}]$

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

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