Let A and B be the two gases such that \(\frac{T_A}{M_A} = 4\frac{T_B}{M_B}\); where T is the temperature and M is the molecular mass. If v_A and v_B are the rms speeds, then the ratio \(\frac{v_A}{v_B}\) will be equal to : 

2

As \(v_{rms} = \sqrt{\frac{3RT}{M}}\)

\(\frac{v_A}{v_B} = \sqrt{\frac{T_A/T_B}{M_A/M_B}}\)

\(\frac{v_A}{v_B} = \sqrt{4}\)

\(\frac{v_A}{v_B} = 2 \) ... [As \(\frac{T_A}{T_B} = 4\frac{M_A}{M_B}\) is given ]