The fundamental frequency of a closed organ pipe of length `20 cm` is equal to the second overtone of an organ pipe open at both the ends. The length of organ pipe open at both the ends is

Options

A. `80 cm`
B. `100 cm`
C. `120 cm`
D. `140 cm`

Explanation

To solve the problem, we need to find the length of the organ pipe open at both ends (denoted as \( L_0 \)) given that the fundamental frequency of a closed organ pipe of length \( L_C = 20 \, \text{cm} \) is equal to the second overtone of an open organ pipe.

### Step-by-Step Solution:

1. **Understanding the Frequencies:**
- The fundamental frequency \( f_C \) of a closed organ pipe is given by the formula:
\[ f_C = \frac{V}{4L_C} \]
- The second overtone (which is the third harmonic) frequency \( f_O \) of an open organ pipe is given by:
\[ f_O = \frac{3V}{2L_0} \]

2. **Setting Up the Equation:**
- According to the problem, the fundamental frequency of the closed organ pipe is equal to the second overtone of the open organ pipe:
\[ f_C = f_O \]
- Substituting the formulas for \( f_C \) and \( f_O \):
\[ \frac{V}{4L_C} = \frac{3V}{2L_0} \]

3. **Canceling \( V \):**
- Since \( V \) (the speed of sound) is common on both sides, we can cancel it out:
\[ \frac{1}{4L_C} = \frac{3}{2L_0} \]

4. **Cross-Multiplying:**
- Cross-multiply to eliminate the fractions:
\[ 2L_0 = 12L_C \]

5. **Substituting the Length of Closed Pipe:**
- Substitute \( L_C = 20 \, \text{cm} \):
\[ 2L_0 = 12 \times 20 \]
\[ 2L_0 = 240 \]

6. **Solving for \( L_0 \):**
- Divide both sides by 2 to find \( L_0 \):
\[ L_0 = \frac{240}{2} = 120 \, \text{cm} \]

### Final Answer:
The length of the organ pipe open at both ends is \( L_0 = 120 \, \text{cm} \).

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