If two gases X and Y have their molecules travelling at the velocities in the ratio of `3 : 1`. The ration of their molecular mass `M_(x)//M_(y)` will be

Options

A. `1//9`
B. 9
C. 3
D. `1//3`

Explanation

To solve the problem, we need to find the ratio of the molecular masses \( \frac{M_x}{M_y} \) given the ratio of the velocities of the gases \( X \) and \( Y \) is \( 3:1 \).

### Step-by-Step Solution:

1. **Understanding the relationship between velocity and molecular mass**:
The root mean square velocity (\( v_{rms} \)) of a gas is given by the formula:
\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]
where \( R \) is the universal gas constant, \( T \) is the temperature, and \( M \) is the molecular mass of the gas.

2. **Setting up the equations for gases X and Y**:
Let \( v_x \) be the velocity of gas \( X \) and \( v_y \) be the velocity of gas \( Y \). According to the problem, we have:
\[ \frac{v_x}{v_y} = \frac{3}{1} \]
This implies:
\[ v_x = 3v_y \]

3. **Expressing velocities in terms of molecular masses**:
From the formula for \( v_{rms} \), we can write:
\[ v_x = \sqrt{\frac{3RT}{M_x}} \quad \text{and} \quad v_y = \sqrt{\frac{3RT}{M_y}} \]

4. **Setting up the ratio of velocities**:
Now, substituting the expressions for \( v_x \) and \( v_y \) into the ratio:
\[ \frac{v_x}{v_y} = \frac{\sqrt{\frac{3RT}{M_x}}}{\sqrt{\frac{3RT}{M_y}}} \]
This simplifies to:
\[ \frac{v_x}{v_y} = \sqrt{\frac{M_y}{M_x}} \]

5. **Equating the ratios**:
From the previous steps, we have:
\[ \frac{3}{1} = \sqrt{\frac{M_y}{M_x}} \]

6. **Squaring both sides**:
Squaring both sides to eliminate the square root gives:
\[ \left(\frac{3}{1}\right)^2 = \frac{M_y}{M_x} \]
Thus:
\[ \frac{9}{1} = \frac{M_y}{M_x} \]

7. **Finding the desired ratio**:
To find \( \frac{M_x}{M_y} \), we take the reciprocal of the above ratio:
\[ \frac{M_x}{M_y} = \frac{1}{9} \]

### Final Answer:
\[ \frac{M_x}{M_y} = \frac{1}{9} \]