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 maintained at a temperature T. The mean square velocity of molecules of B type is denoted by ` V_(2)` and the mean square velocity of A type is denoted by `V_(1)` then `(V_(1))/(V_(2))` is

Options

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

Explanation

To solve the problem, we need to find the ratio of the mean square velocities of two gases, A and B, contained in the same vessel at temperature T.

### Step-by-Step Solution:

1. **Understand the Mean Square Velocity Formula**:
The mean square velocity \( V^2 \) of a gas is given by the formula:
\[ V^2 = \frac{3kT}{m} \]
where \( k \) is the Boltzmann constant, \( T \) is the temperature, and \( m \) is the mass of the gas molecules.

2. **Calculate Mean Square Velocity for Gas A**:
For gas A, the mass of each molecule is \( m \). Therefore, the mean square velocity \( V_1^2 \) for gas A is:
\[ V_1^2 = \frac{3kT}{m} \]

3. **Calculate Mean Square Velocity for Gas B**:
For gas B, the mass of each molecule is \( 2m \). Therefore, the mean square velocity \( V_2^2 \) for gas B is:
\[ V_2^2 = \frac{3kT}{2m} \]

4. **Find the Ratio of Mean Square Velocities**:
We need to find the ratio \( \frac{V_1^2}{V_2^2} \):
\[ \frac{V_1^2}{V_2^2} = \frac{\frac{3kT}{m}}{\frac{3kT}{2m}} = \frac{3kT}{m} \cdot \frac{2m}{3kT} \]
Simplifying this expression:
\[ \frac{V_1^2}{V_2^2} = \frac{2}{1} = 2 \]

5. **Conclusion**:
The ratio of the mean square velocities \( \frac{V_1}{V_2} \) is the square root of the ratio of the squares:
\[ \frac{V_1}{V_2} = \sqrt{2} \]

However, since the question asks for \( \frac{V_1}{V_2} \) in terms of the squares, we can state that:
\[ \frac{V_1}{V_2} = \sqrt{2} \text{ (not directly asked, but useful to note)} \]

The final answer for \( \frac{V_1^2}{V_2^2} \) is:
\[ \frac{V_1^2}{V_2^2} = 2 \]