Two organ (open) pipes of lengths `50cm` and `51cm` produce `6` beats/s. Then the speed of sound is nearly

Options

A. `300m//s`
B. `306m//s`
C. `303m//s`
D. `350m//s`

Explanation

To solve the problem, we need to find the speed of sound using the information provided about the two organ pipes. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the relationship between frequency and length of the pipe
The frequency of sound produced by an open pipe is given by the formula:
\[ f = \frac{v}{2L} \]
where \( f \) is the frequency, \( v \) is the speed of sound, and \( L \) is the length of the pipe.

### Step 2: Define the frequencies of the two pipes
Let:
- \( L_1 = 50 \, \text{cm} = 0.5 \, \text{m} \)
- \( L_2 = 51 \, \text{cm} = 0.51 \, \text{m} \)

The frequencies of the two pipes can be expressed as:
- \( f_1 = \frac{v}{2L_1} = \frac{v}{2 \times 0.5} = \frac{v}{1} = v \)
- \( f_2 = \frac{v}{2L_2} = \frac{v}{2 \times 0.51} = \frac{v}{1.02} \)

### Step 3: Calculate the difference in frequencies
The difference in frequencies (which corresponds to the beat frequency) is given as:
\[ \Delta f = f_1 - f_2 = v - \frac{v}{1.02} \]
This difference is equal to the beat frequency of 6 beats/s:
\[ \Delta f = 6 \, \text{Hz} \]

### Step 4: Set up the equation
Substituting the expression for \(\Delta f\):
\[ 6 = v - \frac{v}{1.02} \]

### Step 5: Simplify the equation
Factor out \( v \):
\[ 6 = v \left(1 - \frac{1}{1.02}\right) \]
Calculating \( 1 - \frac{1}{1.02} \):
\[ 1 - \frac{1}{1.02} = \frac{1.02 - 1}{1.02} = \frac{0.02}{1.02} \]
So we have:
\[ 6 = v \cdot \frac{0.02}{1.02} \]

### Step 6: Solve for \( v \)
Rearranging gives:
\[ v = 6 \cdot \frac{1.02}{0.02} \]
Calculating the right side:
\[ v = 6 \cdot 51 = 306 \, \text{m/s} \]

### Conclusion
The speed of sound is nearly:
\[ \boxed{306 \, \text{m/s}} \]