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) = ?`

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. **Understanding Mean Square Velocity**:
The mean square velocity of a gas can be expressed using the formula:
\[
\langle v^2 \rangle = \frac{3kT}{m}
...