Two organ pipe , both closed at one end , have lengths `l and l + Delta l`. Neglect end corrections. If the velocity of sound in air is `V`, then the number of beats//s` is

Options

A. `(V)/( 4l)`
B. `(V)/( 2l)`
C. `(V)/( 4l^(2)) Delta l`
D. `(V)/( 2 l^(2)) Delta l`

Explanation

To solve the problem, we need to determine the number of beats per second produced by two organ pipes of lengths \( l \) and \( l + \Delta l \), closed at one end, when the velocity of sound in air is \( V \).

### Step-by-Step Solution:

1. **Understand the Fundamental Frequency of Closed Pipes**:
The fundamental frequency \( f \) of a closed organ pipe is given by the formula:
\[ f = \frac{V}{4L} \]
where \( L \) is the length of the pipe and \( V \) is the speed of sound in air.

2. **Calculate the Frequencies of Both Pipes**:
- For the first pipe of length \( l \):
\[ f_1 = \frac{V}{4l} \]
- For the second pipe of length \( l + \Delta l \):
\[ f_2 = \frac{V}{4(l + \Delta l)} \]

3. **Determine the Beat Frequency**:
The beat frequency \( B \) is the absolute difference between the two frequencies:
\[ B = |f_1 - f_2| = \left| \frac{V}{4l} - \frac{V}{4(l + \Delta l)} \right| \]

4. **Simplify the Expression**:
To simplify the expression, we can factor out \( \frac{V}{4} \):
\[ B = \frac{V}{4} \left| \frac{1}{l} - \frac{1}{l + \Delta l} \right| \]
Now, we can find a common denominator:
\[ B = \frac{V}{4} \left| \frac{(l + \Delta l) - l}{l(l + \Delta l)} \right| = \frac{V}{4} \left| \frac{\Delta l}{l(l + \Delta l)} \right| \]

5. **Assume \( \Delta l \) is Much Smaller than \( l \)**:
If \( \Delta l \) is much smaller than \( l \) (i.e., \( \Delta l \ll l \)), we can approximate \( l + \Delta l \) as \( l \):
\[ B \approx \frac{V}{4} \cdot \frac{\Delta l}{l^2} \]

6. **Final Expression for Beat Frequency**:
Thus, the number of beats per second is given by:
\[ B \approx \frac{V \Delta l}{4l^2} \]

### Conclusion:
The number of beats per second produced by the two organ pipes is:
\[ B = \frac{V \Delta l}{4l^2} \]