Physics - Heat & Thermo Question with Solution | TestHub

We know, ${\mathrm{V}}_{\mathrm{mms}}=\sqrt{\frac{3\mathrm{RT}}{\mathrm{M}}}$ $\Rightarrow \%$ increase in $$\begin{array}{l}{\mathrm{V}}_{\mathrm{rms}}=\frac{\sqrt{\frac{3\mathrm{R}{\mathrm{T}}_2}{\mathrm{M}}}-\sqrt{\frac{3{\mathrm{RT}}_1}{\mathrm{M}}}}{\sqrt{\frac{3{\mathrm{RT}}_1}{\mathrm{M}}}}\times 100\\ =\frac{\sqrt{{\mathrm{T}}_2}-\sqrt{{\mathrm{T}}_1}}{\sqrt{{\mathrm{T}}_1}}\times 100\\ =\frac{\sqrt{400}-\sqrt{300}}{\sqrt{300}}\times 100\\ =\frac{20-17.32}{17.32}\times 100=15.5\%\end{array}$$