Five beats per second are produced on vibrating two closed organ pipes simultaneously. If the ratio of their lengths is `21 : 20 `, then their frequencies will be

Options

A. `105Hz` and `100Hz`
B. `105Hz` and `110Hz`
C. `100Hz` and `105Hz`
D. `110Hz` and `105Hz`

Explanation

To solve the problem, we need to find the frequencies of two closed organ pipes given that they produce 5 beats per second and their lengths are in the ratio of 21:20.

### Step-by-Step Solution:

1. **Understanding the relationship between frequency and length:**
The frequency of a closed organ pipe is given by the formula:
\[ f = \frac{v}{4L} \]
where \( f \) is the frequency, \( v \) is the speed of sound in air, and \( L \) is the length of the pipe.

2. **Setting up the lengths and frequencies:**
Let the lengths of the two pipes be \( L_1 \) and \( L_2 \). Given the ratio of their lengths:
\[ \frac{L_1}{L_2} = \frac{21}{20} \]
We can express \( L_1 \) and \( L_2 \) in terms of a common variable \( k \):
\[ L_1 = 21k \quad \text{and} \quad L_2 = 20k \]

3. **Finding the frequencies:**
Using the formula for frequency:
\[ f_1 = \frac{v}{4L_1} = \frac{v}{4 \times 21k} = \frac{v}{84k} \]
\[ f_2 = \frac{v}{4L_2} = \frac{v}{4 \times 20k} = \frac{v}{80k} \]

4. **Calculating the beat frequency:**
The beat frequency is given by the absolute difference between the two frequencies:
\[ |f_1 - f_2| = 5 \text{ Hz} \]
Substituting the expressions for \( f_1 \) and \( f_2 \):
\[ \left| \frac{v}{84k} - \frac{v}{80k} \right| = 5 \]
Simplifying this gives:
\[ \left| \frac{v(80 - 84)}{84 \times 80k} \right| = 5 \]
\[ \left| \frac{-4v}{6720k} \right| = 5 \]
\[ \frac{4v}{6720k} = 5 \]
\[ 4v = 5 \times 6720k \]
\[ v = \frac{5 \times 6720k}{4} \]

5. **Finding the individual frequencies:**
Now, substituting \( v \) back into the frequency equations:
\[ f_1 = \frac{5 \times 6720k}{4 \times 84k} = \frac{5 \times 6720}{336} = 100 \text{ Hz} \]
\[ f_2 = \frac{5 \times 6720k}{4 \times 80k} = \frac{5 \times 6720}{320} = 105 \text{ Hz} \]

6. **Final frequencies:**
Thus, the frequencies of the two pipes are:
\[ f_1 = 100 \text{ Hz} \quad \text{and} \quad f_2 = 105 \text{ Hz} \]

### Conclusion:
The frequencies of the two closed organ pipes are 100 Hz and 105 Hz.