$N$molecules each of mass $m$of gas $A$and $2N$ molecules each of $2m$of gas $B$are contained in the vessel which is maintained at a temperature $T$. The mean square of velocity of the molecules of $B$type is denoted by ${\nu }^2$ and the mean square of the  x-component of the velocity of $A$ type is denoted by w². The ratio of $w^2$: $v^2$ is

The mean square velocity of gas molecules is given by 

$v^2=\frac{3kT}{m}$

For gas $A$, ${v^2}_A=\frac{3kT}{m}$  ……   $\left(i\right)$

For a gas molecule,

$v^2={v^2}_x+{v^2}_y+{v^2}_z=3{v^2}_x$                          This implies (${v^2}_x={v^2}_y={v^2}_z$)

or ${v^2}_x=\frac{v^2}{3}$

From eqn. (i), we get

$w^2={v^2}_x=\left[\frac{{\displaystyle \frac{3kT}{m}}}{3}\right]=\frac{kT}{m}$   ……..$\left(ii\right)$

For gas $B$, ${v^2}_B=v^2=\frac{3kT}{2m}$ ………$\left(iii\right)$

Dividing eqn. (ii) by eqn. (iii), we get

$\frac{w^2}{v^2}=\frac{{\displaystyle \frac{kT}{m}}}{{\displaystyle \frac{3kT}{2m}}}=\frac{2}{3}$