N molecules each of mass m of gas A and 2 N molecules each of mass 2m of gas B are contained in the same vessel which is maintined at a temperature T. The mean square of the velocity of the molecules of B type is denoted by `v^(2)` and the mean square of the x-component of the velocity of a tye is denoted by `omega^(2)`. What is the ratio of `omega^(2)//v^(2) = ?`

Options

A. `3:2`
B. `1:3`
C. `2:3`
D. `1:1`

Explanation

To solve the problem, we need to find the ratio of the mean square of the x-component of the velocity of gas A (denoted as \( \omega^2 \)) to the mean square velocity of gas B (denoted as \( v^2 \)).

### Step-by-Step Solution:

1. **Mean Square Velocity of Gas B**:
The mean square velocity \( v^2 \) of gas B can be calculated using the formula:
\[ v^2 = \frac{3kT}{m_B} \]
where \( m_B \) is the mass of the molecules of gas B. Given that gas B has a mass of \( 2m \):
\[ v^2 = \frac{3kT}{2m} \]

2. **Mean Square of the x-component of Velocity of Gas A**:
For gas A, the mean square of the x-component of the velocity \( \omega^2 \) can be expressed as:
\[ \omega^2 = \frac{1}{3} v_A^2 \]
where \( v_A^2 \) is the mean square velocity of gas A. The mean square velocity \( v_A^2 \) for gas A (mass \( m \)) is given by:
\[ v_A^2 = \frac{3kT}{m} \]
Therefore, substituting this into the equation for \( \omega^2 \):
\[ \omega^2 = \frac{1}{3} \left(\frac{3kT}{m}\right) = \frac{kT}{m} \]

3. **Finding the Ratio \( \frac{\omega^2}{v^2} \)**:
Now, we can find the ratio of \( \omega^2 \) to \( v^2 \):
\[ \frac{\omega^2}{v^2} = \frac{\frac{kT}{m}}{\frac{3kT}{2m}} = \frac{kT}{m} \cdot \frac{2m}{3kT} \]
Simplifying this:
\[ \frac{\omega^2}{v^2} = \frac{2}{3} \]

### Final Answer:
The ratio \( \frac{\omega^2}{v^2} \) is \( \frac{2}{3} \).