The ratio between the root mean square speed of $${H_2}$$ at $$50K$$  and that of $${O_2}$$ at $$800K$$  is,

Solution :
$$\eqalign{ & {\text{The expression of root mean square speed is}} \cr & {U_{rms}} = \sqrt {\frac{{3RT}}{M}} \cr & {\text{Hence,}} \cr & \frac{{{U_{rms}}\left( {{H_2}} \right)}}{{{U_{rms}}\left( {{O_2}} \right)}} = {\left[ {\frac{{\frac{{3R\left( {50K} \right)}}{{2g\,mo{l^{ - 1}}}}}}{{\frac{{3R\left( {800K} \right)}}{{32g\,mo{l^{ - 1}}}}}}} \right]^{\frac{1}{2}}} = 1 \cr} $$