When the temperature of a gas is raised from 30° C to 90° C, the percentage increase in the r.m.s. velocity  of the molecules will be :

$\begin{array}{l}{V}_{rms}\propto \sqrt{T}{T}_1=30+273=303K\\ T_2=90+273=363K\\ \frac{V_1}{V_2}=\sqrt{\frac{T_1}{T_2}}=\sqrt{\frac{303}{363}}\\ \frac{V_2-{\text{\hspace{0.17em}V}}_1}{V_1}\times 100=\left(\sqrt{\frac{363}{303}}-1\right)\times 100=9.5\%\end{array}$