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 value of \( n \) given that the fundamental frequency of an organ pipe open at one end is 300 Hz and the frequency of the 3rd overtone is \( 100n \) Hz.

### Step-by-Step Solution:

1. **Identify the Fundamental Frequency**:
The fundamental frequency \( f_1 \) of the organ pipe is given as:
\[ f_1 = 300 \, \text{Hz} \]

2. **Understand the Frequencies of Overtones**:
For an organ pipe that is open at one end, the frequencies of the overtones can be calculated using the formula:
\[ f_n = (2n + 1) \cdot f_1 \]
where \( n \) is the overtone number (starting from 0 for the fundamental frequency).

3. **Determine the Frequency of the 3rd Overtone**:
The 3rd overtone corresponds to \( n = 3 \):
\[ f_3 = (2 \cdot 3 + 1) \cdot f_1 \]
Simplifying this gives:
\[ f_3 = (6 + 1) \cdot 300 = 7 \cdot 300 = 2100 \, \text{Hz} \]

4. **Relate the 3rd Overtone Frequency to \( 100n \)**:
According to the problem, the frequency of the 3rd overtone is also given as:
\[ f_3 = 100n \]
Setting the two expressions for \( f_3 \) equal to each other:
\[ 2100 = 100n \]

5. **Solve for \( n \)**:
To find \( n \), divide both sides by 100:
\[ n = \frac{2100}{100} = 21 \]

### Final Answer:
The value of \( n \) is:
\[ \boxed{21} \]