A pipe of length `l_(1)`, closed at one end is kept in a chamber of gas of density `rho_(1)`. A second pipe open at both ends is placed in a second chamber of gas of density `rho_(2)`. The compressibility of both the gases is equal. Calculate the length of the second pipe if frquency of first overtone in both the cases is equal.

Explanation

To solve the problem, we need to find the length of the second pipe (open at both ends) when the frequency of the first overtone in both pipes is equal. Let's break down the solution step by step.

### Step 1: Understand the Frequencies of the Pipes
For a pipe closed at one end (length \( l_1 \)), the frequency of the first overtone (which is the third harmonic) is given by:
\[ f_1 = \frac{3v}{4l_1} \]
where \( v \) is the speed of sound in the gas.

For a pipe open at both ends (length \( l_2 \)), the frequency of the first overtone (which is the second harmonic) is given by:
\[ f_2 = \frac{v}{l_2} \]

### Step 2: Set the Frequencies Equal
Since the frequencies of the first overtone in both cases are equal, we can set \( f_1 = f_2 \):
\[ \frac{3v}{4l_1} = \frac{v}{l_2} \]

### Step 3: Simplify the Equation
We can cancel \( v \) from both sides (assuming \( v \neq 0 \)):
\[ \frac{3}{4l_1} = \frac{1}{l_2} \]

### Step 4: Rearrange for \( l_2 \)
Now, we can rearrange the equation to find \( l_2 \):
\[ l_2 = \frac{4l_1}{3} \]

### Step 5: Consider the Effect of Density
The speed of sound in a gas is given by:
\[ v = \sqrt{\frac{E}{\rho}} \]
where \( E \) is the modulus of elasticity (related to compressibility) and \( \rho \) is the density of the gas. Since the compressibility of both gases is equal, we can say that:
\[ \frac{v_1}{v_2} = \sqrt{\frac{\rho_2}{\rho_1}} \]

### Step 6: Substitute into the Frequency Equation
Now substituting \( v_1 \) and \( v_2 \) into the frequency equations:
\[ f_1 = \frac{3\sqrt{\frac{E}{\rho_1}}}{4l_1} \]
\[ f_2 = \frac{\sqrt{\frac{E}{\rho_2}}}{l_2} \]

Setting these equal gives:
\[ \frac{3\sqrt{\frac{E}{\rho_1}}}{4l_1} = \frac{\sqrt{\frac{E}{\rho_2}}}{l_2} \]

### Step 7: Solve for \( l_2 \)
From this equation, we can isolate \( l_2 \):
\[ l_2 = \frac{4l_1}{3} \cdot \sqrt{\frac{\rho_2}{\rho_1}} \]

### Final Answer
Thus, the length of the second pipe is:
\[ l_2 = \frac{4l_1}{3} \sqrt{\frac{\rho_1}{\rho_2}} \]