5 beat/s are produced on blowing together two closed organ pipes of the same diameter but of different lengths. If shorter pipe of length 10cm and speed of sound in air 300m/s , then the length of longer pipe , will be

Options

A. 10.06cm
B. 11.22cm
C. 15cm
D. 30cm

Explanation

To solve the problem step-by-step, we will use the information given about the two closed organ pipes and the concept of beats produced due to the interference of sound waves from the pipes.

### Step 1: Understand the Beat Frequency
When two sound waves of slightly different frequencies interfere, they produce a phenomenon known as beats. The beat frequency (Δn) is given by the absolute difference between the frequencies of the two waves. Here, Δn = 5 beats/s.

### Step 2: Frequency of Closed Organ Pipes
The frequency (n) of a closed organ pipe is given by the formula:
\[ n = \frac{V}{4L} \]
where:
- \( V \) is the speed of sound in air (300 m/s),
- \( L \) is the length of the pipe.

### Step 3: Set Up the Equation for Beat Frequency
For two pipes with lengths \( L_1 \) and \( L_2 \), the frequencies can be expressed as:
\[ n_1 = \frac{V}{4L_1} \]
\[ n_2 = \frac{V}{4L_2} \]

The beat frequency is given by:
\[ \Delta n = |n_1 - n_2| = 5 \]

Substituting the expressions for \( n_1 \) and \( n_2 \):
\[ \Delta n = \left| \frac{V}{4L_1} - \frac{V}{4L_2} \right| = 5 \]

### Step 4: Substitute Known Values
We know:
- \( L_1 = 10 \text{ cm} = 0.1 \text{ m} \)
- \( V = 300 \text{ m/s} \)

Substituting these values into the equation:
\[ 5 = \left| \frac{300}{4 \times 0.1} - \frac{300}{4L_2} \right| \]

### Step 5: Simplify the Equation
Calculating \( n_1 \):
\[ n_1 = \frac{300}{4 \times 0.1} = \frac{300}{0.4} = 750 \text{ Hz} \]

Now substituting \( n_1 \) into the beat frequency equation:
\[ 5 = \left| 750 - \frac{300}{4L_2} \right| \]

### Step 6: Solve for \( L_2 \)
This gives us two cases to consider:
1. \( 750 - \frac{300}{4L_2} = 5 \)
2. \( 750 - \frac{300}{4L_2} = -5 \)

**Case 1:**
\[ 750 - \frac{300}{4L_2} = 5 \]
\[ \frac{300}{4L_2} = 745 \]
\[ 4L_2 = \frac{300}{745} \]
\[ L_2 = \frac{300}{4 \times 745} = \frac{300}{2980} \approx 0.10067 \text{ m} \]

**Case 2:**
\[ 750 - \frac{300}{4L_2} = -5 \]
\[ \frac{300}{4L_2} = 755 \]
\[ 4L_2 = \frac{300}{755} \]
\[ L_2 = \frac{300}{4 \times 755} = \frac{300}{3020} \approx 0.09934 \text{ m} \]

### Step 7: Convert to Centimeters
Now converting \( L_2 \) to centimeters:
- For Case 1: \( L_2 \approx 10.06 \text{ cm} \)
- For Case 2: \( L_2 \approx 9.93 \text{ cm} \)

Since we are looking for the longer pipe, we take:
\[ L_2 \approx 10.06 \text{ cm} \]

### Final Answer
The length of the longer pipe is approximately **10.06 cm**.