' $N$ ' molecules of gas $A$, each having mass ' $m$ ' and ' 2 N ' molecules of gas B , each of mass ' 2 m '…
' $N$ ' molecules of gas $A$, each having mass ' $m$ ' and ' 2 N ' molecules of gas B , each of mass ' 2 m ' are contained in the same vessel which is at constant temperature ' $T$ '. The mean square velocity of $B$ is $V^2$ and mean square of x -component of A is $\omega^2$. The value of $\frac{\omega^2}{V^2}$ is
$3: 2$
$2: 3$
$1: 2$
$2: 1$
Solution
Mean square velocity of molecule $=\frac{3 \mathrm{kT}}{\mathrm{m}}$
For gas A,
$x$ component of mean square velocity of molecule $=\omega^2$
$\therefore \quad$ Mean square velocity $=3 \omega^2=\frac{3 \mathrm{kT}}{\mathrm{m}}$...(i)
For gas B,
Mean square velocity $=\mathrm{V}^2=\frac{3 \mathrm{kT}}{2 \mathrm{~m}}$...(ii)
From (i) and (ii)
$\frac{3 \omega^2}{\mathrm{~V}^2}=\frac{3 \mathrm{kT}}{\mathrm{m}} \times \frac{2 \mathrm{~m}}{3 \mathrm{kT}}$
$\therefore \quad \frac{\omega^2}{\mathrm{~V}^2}=\frac{2}{3}$
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