For a certain pipe, three successive resonant frequencies are observed at 300 Hz, 420 Hz and 540 Hz.
The speed of sound in air is 340 m/s. The pipe is a

Options

A. closed pipe of length `11/12` m
B. open pipe of length `11/12` m
C. closed pipe of length ` 17/12` m
D. open pipe of length `17/12` m

Explanation

To determine the type of pipe and its length based on the given resonant frequencies, we can follow these steps:

### Step 1: Identify the Frequencies
The three successive resonant frequencies are given as:
- \( f_1 = 300 \, \text{Hz} \)
- \( f_2 = 420 \, \text{Hz} \)
- \( f_3 = 540 \, \text{Hz} \)

### Step 2: Calculate the Differences Between Frequencies
Calculate the differences between the successive frequencies:
- \( f_2 - f_1 = 420 \, \text{Hz} - 300 \, \text{Hz} = 120 \, \text{Hz} \)
- \( f_3 - f_2 = 540 \, \text{Hz} - 420 \, \text{Hz} = 120 \, \text{Hz} \)

### Step 3: Determine the Type of Pipe
Since the differences between successive frequencies are equal (120 Hz), this indicates that the pipe is likely a closed pipe (one end closed). In a closed pipe, only odd harmonics are present, which explains why we observe these frequencies without even harmonics.

### Step 4: Use the Formula for Resonant Frequencies
For a closed pipe, the resonant frequencies can be expressed as:
\[ f_n = \frac{(2n - 1)V}{4L} \]
where:
- \( n \) is the harmonic number (1, 2, 3, ...)
- \( V \) is the speed of sound in air (340 m/s)
- \( L \) is the length of the pipe.

### Step 5: Relate the Frequencies to the Length of the Pipe
From the difference of the first two frequencies:
\[ f_2 - f_1 = \frac{V}{4L} \]
Substituting the values:
\[ 120 = \frac{340}{4L} \]

### Step 6: Solve for Length \( L \)
Rearranging the equation to find \( L \):
\[ 4L \cdot 120 = 340 \]
\[ L = \frac{340}{480} = \frac{17}{24} \, \text{m} \]

### Conclusion
The type of pipe is a closed pipe, and its length is:
\[ L = \frac{17}{24} \, \text{m} \]