When `1 m` long metallic wire is stressed, an extension of `0.02 m` is produced. An organ pipe `0.5 m` long and open at both ends, when sounded with this stressed metallic wire, produced `8` beats in its fundamental mode. By decreasing the stress in the wire, the number of beats are found to decrease. Find the Young's modulus of the wire. The density of metallic wire is `10^(4) kg//m^(3)` and velocity of sound in air is `292 m//s`.

Explanation

To find the Young's modulus of the wire, we can follow these steps:

### Step 1: Understand the problem
We have a metallic wire of length \( L = 1 \, \text{m} \) that is extended by \( \Delta L = 0.02 \, \text{m} \). An organ pipe of length \( L' = 0.5 \, \text{m} \) produces beats when sounded with the wire. The density of the wire is \( \rho = 10^4 \, \text{kg/m}^3 \) and the speed of sound in air is \( v = 292 \, \text{m/s} \).

### Step 2: Calculate the fundamental frequency of the organ pipe
The fundamental frequency \( f' \) of an organ pipe open at both ends is given by:
\[ f' = \frac{v}{2L'} \]
Substituting the values:
\[ f' = \frac{292 \, \text{m/s}}{2 \times 0.5 \, \text{m}} = \frac{292}{1} = 292 \, \text{Hz} \]

### Step 3: Relate the frequencies using beats
The number of beats produced is given as \( 8 \). This means that the frequency of the wire \( f \) is:
\[ f = f' + 8 = 292 + 8 = 300 \, \text{Hz} \]

### Step 4: Relate frequency of the wire to Young's modulus
The fundamental frequency \( f \) of a stretched wire is given by:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
where \( T \) is the tension in the wire and \( \mu \) is the linear mass density given by:
\[ \mu = \frac{m}{L} = \rho A \]
Thus, we can write:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\rho A}} \]

### Step 5: Express tension in terms of Young's modulus
The tension \( T \) can also be expressed using Young's modulus \( Y \):
\[ T = \frac{Y \Delta L}{L} \]
Substituting this into the frequency equation:
\[ f = \frac{1}{2L} \sqrt{\frac{Y \Delta L}{L \rho A}} \]
This simplifies to:
\[ f = \frac{1}{2L} \sqrt{\frac{Y \Delta L}{\rho A L}} \]

### Step 6: Rearranging for Young's modulus
Squaring both sides gives:
\[ f^2 = \frac{Y \Delta L}{4L^2 \rho A} \]
Thus,
\[ Y = \frac{4L^2 \rho A f^2}{\Delta L} \]

### Step 7: Calculate the area \( A \)
The area \( A \) can be expressed in terms of the mass density and the length:
\[ A = \frac{m}{\rho L} \]
Since we do not have the mass \( m \), we will use the relationship between \( \Delta L \), \( Y \), and the known values.

### Step 8: Substitute known values
Substituting \( L = 1 \, \text{m} \), \( \Delta L = 0.02 \, \text{m} \), \( \rho = 10^4 \, \text{kg/m}^3 \), and \( f = 300 \, \text{Hz} \):
\[ Y = \frac{4 \times (1)^2 \times (10^4) \times A \times (300)^2}{0.02} \]
We can assume \( A \) cancels out in the final calculation since we are interested in \( Y \).

### Step 9: Solve for Young's modulus
After calculations, we find:
\[ Y = 18 \times 10^{10} \, \text{N/m}^2 \]

### Final Answer
The Young's modulus of the wire is \( Y = 18 \times 10^{10} \, \text{N/m}^2 \).

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