$v \propto \frac{1}{{\sqrt \rho }}$
==>$\frac{{{v_1}}}{{{v_2}}} = \sqrt {\frac{{{\rho _2}}}{{{\rho _1}}}} = \sqrt {\frac{4}{1}} = 2:1$
(Useful information) : $\sqrt{167 R T}=640 j^{1 / 2} mole ^{-1 / 2} ; \sqrt{140 RT }=590 j ^{1 / 2} mole ^{-1 / 2}$. The molar masses $M$ in grams are given in the options. Take the value of $\sqrt{\frac{10}{ M }}$ for each gas as given there.)