Second overtons frequency of a pipe and fourth harmonic frequency of an pipe are same. Then, choose the correct options.

Options

A. Fundamental frequency of closed pipe is more than the fundamental frequency of open pipe
B. First overtone frequency of closed pipe is more than the first overtone frequency of open pipe
C. Fifteenth harmonic frequency of closed pipe is equal to twelfth harmonic frequency of open pipe
D. Tenth haromic frequency of closed pipe is equal to eigth harmonic fequency of open pipe

Explanation

To solve the problem, we need to analyze the frequencies of the second overtone of a closed pipe and the fourth harmonic of an open pipe and determine the relationships between them.

### Step-by-Step Solution:

1. **Identify the Frequencies:**
- The second overtone frequency of a closed organ pipe is given by:
\[ f_{2} = \frac{5V}{4L_c} \]
- The fourth harmonic frequency of an open organ pipe is given by:
\[ f_{4} = \frac{4V}{2L_0} = \frac{2V}{L_0} \]

2. **Set the Frequencies Equal:**
Since it is given that the second overtone frequency of the closed pipe is equal to the fourth harmonic frequency of the open pipe, we can write:
\[ \frac{5V}{4L_c} = \frac{2V}{L_0} \]

3. **Cancel Common Terms:**
We can cancel \(V\) from both sides (assuming \(V \neq 0\)):
\[ \frac{5}{4L_c} = \frac{2}{L_0} \]

4. **Cross Multiply:**
Cross-multiplying gives:
\[ 5L_0 = 8L_c \]
Rearranging gives:
\[ \frac{L_0}{L_c} = \frac{8}{5} \]

5. **Analyze the Statements:**
Now, we will analyze the four statements provided in the question.

- **Statement 1:** The fundamental frequency of a closed pipe is more than that of an open pipe.
- Fundamental frequency of closed pipe:
\[ f_{1c} = \frac{V}{4L_c} \]
- Fundamental frequency of open pipe:
\[ f_{1o} = \frac{V}{2L_0} = \frac{V}{2 \cdot \frac{8}{5}L_c} = \frac{5V}{16L_c} \]
- Since \(f_{1o} > f_{1c}\), this statement is **incorrect**.

- **Statement 2:** The first overtone frequency of a closed pipe is more than that of an open pipe.
- First overtone of closed pipe:
\[ f_{1c} = \frac{3V}{4L_c} \]
- First overtone of open pipe:
\[ f_{1o} = \frac{2V}{L_0} = \frac{5V}{8L_c} \]
- Since \(f_{1c} > f_{1o}\), this statement is **correct**.

- **Statement 3:** The 15th harmonic frequency of a closed pipe is equal to the 12th harmonic frequency of an open pipe.
- 15th harmonic of closed pipe:
\[ f_{15c} = \frac{15V}{4L_c} \]
- 12th harmonic of open pipe:
\[ f_{12o} = \frac{12V}{2L_0} = \frac{12V}{2 \cdot \frac{8}{5}L_c} = \frac{15V}{4L_c} \]
- Thus, \(f_{15c} = f_{12o}\), this statement is **correct**.

- **Statement 4:** The 10th harmonic frequency of a closed pipe is equal to the 8th harmonic frequency of an open pipe.
- 10th harmonic of closed pipe:
\[ f_{10c} = \frac{10V}{4L_c} = \frac{5V}{2L_c} \]
- 8th harmonic of open pipe:
\[ f_{8o} = \frac{8V}{2L_0} = \frac{8V}{2 \cdot \frac{8}{5}L_c} = \frac{5V}{2L_c} \]
- Thus, \(f_{10c} = f_{8o}\), this statement is **correct**.

### Summary of Statements:
- Statement 1: Incorrect
- Statement 2: Correct
- Statement 3: Correct
- Statement 4: Correct

### Final Answer:
The correct options are Statements 2, 3, and 4.