`v_(1) and v_(2)` are the velocities of sound at the same temperature in two monoatomic gases of densities `rho_(1) and rho_(2)` respectively. If `(rho_(1))/(rho_(2))=(1)/(4)` then the ratio of velocities `v_(1) and v_(2)` is

Options

A. `1:2`
B. `4:1`
C. `2:1`
D. `1:4`

Answer

To solve the problem, we need to find the ratio of the velocities of sound in two monoatomic gases given their densities. Let's break it down step by step:

### Step 1: Understand the relationship between velocity of sound and density
The velocity of sound \( v \) in a medium is given by the formula:
\[ v = \sqrt{\frac{P \cdot \gamma}{\rho}} \]
where:
- \( P \) is the pressure,
- \( \gamma \) is the adiabatic index (ratio of specific heats, \( C_p/C_v \)),
- \( \rho \) is the density of the gas.

### Step 2: Write the expressions for the velocities of the two gases
For the first gas:
\[ v_1 = \sqrt{\frac{P \cdot \gamma}{\rho_1}} \]
For the second gas:
\[ v_2 = \sqrt{\frac{P \cdot \gamma}{\rho_2}} \]

### Step 3: Find the ratio of the velocities
To find the ratio \( \frac{v_1}{v_2} \), we divide the two expressions:
\[ \frac{v_1}{v_2} = \frac{\sqrt{\frac{P \cdot \gamma}{\rho_1}}}{\sqrt{\frac{P \cdot \gamma}{\rho_2}}} \]
This simplifies to:
\[ \frac{v_1}{v_2} = \sqrt{\frac{\rho_2}{\rho_1}} \]

### Step 4: Substitute the given density ratio
We are given that:
\[ \frac{\rho_1}{\rho_2} = \frac{1}{4} \]
From this, we can find \( \frac{\rho_2}{\rho_1} \):
\[ \frac{\rho_2}{\rho_1} = 4 \]

### Step 5: Calculate the ratio of velocities
Now substituting this back into our equation for the ratio of velocities:
\[ \frac{v_1}{v_2} = \sqrt{4} = 2 \]

### Step 6: Final ratio
Thus, the ratio of the velocities \( v_1 : v_2 \) is:
\[ v_1 : v_2 = 2 : 1 \]

### Conclusion
The final answer is that the ratio of the velocities \( v_1 \) and \( v_2 \) is \( 2 : 1 \).

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