N molecules each of mass m of gas A and 2N molecules each of mass 2m of gas B are contained in the same vessel at temperature T. The mean square of the velocity of molecules of gas B is v² and the mean square of x component of the velocity of molecules of gas A is w² . The ratio w²/v² is

The correct option is 2/3.

Let's understand the step-by-step physical principles and calculations to arrive at this result.

1. Mean Square Velocity of Gas B
According to the kinetic theory of gases, the mean square velocity of molecules of a gas at absolute temperature T is given by the relation:
$v^2=\frac{3k_{B}T}{M}$
where $k_B$ is the Boltzmann constant and M is the mass of a single molecule of the gas.
For gas B, the mass of each molecule is $m_B=2m$. Therefore, the mean square velocity of molecules of gas B is:
$v^2=\frac{3k_{B}T}{2m}$     --- (Equation 1)

2. Mean Square of the x-component of Velocity of Gas A
For gas A, the mass of each molecule is m. The total mean square velocity of molecules of gas A is:
$v_A^2=\frac{3k_{B}T}{m}$
Since the gas molecules move randomly in all three dimensions, space is isotropic. This means the mean square velocities along the x, y, and z axes are equal:
$v_{Ax}^2=v_{Ay}^2=v_{Az}^2$
Therefore, the mean square of the x-component of the velocity is one-third of the total mean square velocity:
$v_{Ax}^2=\frac{1}{3}{v}_A^2=\frac{1}{3}\left(\frac{3k_{B}T}{m}\right)=\frac{k_{B}T}{m}$
Given that the mean square of the x-component of velocity of gas A is represented by $w^2$, we have:
$w^2=\frac{k_{B}T}{m}$     --- (Equation 2)

3. Calculating the Ratio
Dividing Equation 2 by Equation 1 to find the ratio $\frac{w^2}{v^2}$:
$\frac{w^2}{v^2}=\frac{\left(\frac{k_{B}T}{m}\right)}{\left(\frac{3k_{B}T}{2m}\right)}$
Simplifying the expression by cancelling the common terms $k_{B}T$ and $m$ from both the numerator and the denominator:
$\frac{w^2}{v^2}=\frac{1}{\left(\frac{3}{2}\right)}=\frac{2}{3}$