The fundamental frequency of an organ pipe open at one end is 300 Hz. The frequency of `3^("rd")` overtone of this organ pipe is `100xx"n Hz"`. Find n.

Explanation

To solve the problem, we need to find the frequency of the third overtone of an organ pipe that is open at one end, given that the fundamental frequency is 300 Hz.

### Step-by-Step Solution:

1. **Understand the Organ Pipe Configuration**:
An organ pipe that is open at one end supports odd harmonics. The fundamental frequency (first harmonic) corresponds to the first mode of vibration.

2. **Identify the Fundamental Frequency**:
The fundamental frequency \( f_0 \) is given as 300 Hz.

3. **Determine the Wavelength for the Fundamental Frequency**:
For a pipe open at one end, the relationship between the length of the pipe \( L \), the wavelength \( \lambda \), and the fundamental frequency is given by:
\[ L = \frac{\lambda}{4} \]
Therefore, the wavelength for the fundamental frequency can be expressed as:
\[ \lambda_0 = \frac{v}{f_0} \]
where \( v \) is the speed of sound in air.

4. **Express the Fundamental Frequency in Terms of Length**:
Rearranging the equation gives:
\[ f_0 = \frac{v}{4L} \]
Since \( f_0 = 300 \, \text{Hz} \), we can write:
\[ 300 = \frac{v}{4L} \]

5. **Calculate the Wavelength for the Third Overtone**:
The third overtone corresponds to the 7th harmonic (since the harmonics for a pipe open at one end are given by \( f_n = (2n + 1)f_0 \)).
Thus, the frequency of the third overtone \( f_3 \) can be calculated as:
\[ f_3 = 7f_0 = 7 \times 300 \, \text{Hz} = 2100 \, \text{Hz} \]

6. **Express in the Required Form**:
The problem states that the frequency of the third overtone can be expressed as \( 100 \times n \, \text{Hz} \). Therefore:
\[ 2100 = 100 \times n \]
Solving for \( n \):
\[ n = \frac{2100}{100} = 21 \]

### Final Answer:
The value of \( n \) is **21**.