MHTCET 11 May 2024 Shift 2 Paper

Concept:At the same temperature, the mean square velocity of gas molecules depends on the molecular mass and is shared equally along the three coordinate axes.Explanation:For gas A, the mean square of the x-component of velocity is $\omega^2$.The total mean square velocity is the sum of the squares of the three components, so $\overline{v_A^2} = 3\omega^2$.From kinetic theory, $\overline{v_A^2} = \frac{3kT}{m}$.Thus, $3\omega^2 = \frac{3kT}{m}$, which gives $\omega^2 = \frac{kT}{m}$.For gas B, the mean square velocity is given as $V^2$.Since the mass of each molecule of gas B is $2m$, we have $V^2 = \frac{3kT}{2m}$.Dividing the two results, $\frac{\omega^2}{V^2} = \frac{kT/m}{3kT/(2m)} = \frac{2}{3}$.Answer:The required value of $\frac{\omega^2}{V^2}$ is $\frac{2}{3}$, which is the ratio $2 : 3$.Therefore, the correct option is B.