A 40 cm long wire having a mass 3.1 gm and area of cross section `1mm^(2)` is stretched between the support 40.05 cm apart. In its fundamental mode , it vibrates with a frequency `1000//64 Hz.` Find the young's modulus of the wire in the form `X xx 108 N-m^(2)` and fill value of X.

Answer

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

### Step 1: Convert the given values to standard units
- Length of the wire, \( L = 40.05 \, \text{cm} = 0.4005 \, \text{m} \)
- Mass of the wire, \( m = 3.1 \, \text{g} = 3.1 \times 10^{-3} \, \text{kg} \)
- Area of cross-section, \( A = 1 \, \text{mm}^2 = 1 \times 10^{-6} \, \text{m}^2 \)
- Frequency, \( f = \frac{1000}{64} \, \text{Hz} \)

### Step 2: Calculate the mass per unit length (\( \mu \))
\[ \mu = \frac{m}{L} = \frac{3.1 \times 10^{-3} \, \text{kg}}{0.4005 \, \text{m}} \approx 7.74 \times 10^{-3} \, \text{kg/m} \]

### Step 3: Use the formula for fundamental frequency
The formula for the fundamental frequency is given by:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
Rearranging this to find the tension \( T \):
\[ T = (2Lf)^2 \mu \]

### Step 4: Substitute the values into the tension formula
First, calculate \( f \):
\[ f = \frac{1000}{64} \approx 15.625 \, \text{Hz} \]

Now substitute \( L \), \( f \), and \( \mu \) into the tension formula:
\[ T = (2 \times 0.4005 \times 15.625)^2 \times (7.74 \times 10^{-3}) \]

Calculating \( T \):
\[ T \approx (12.500)^2 \times (7.74 \times 10^{-3}) \approx 156.25 \times 7.74 \times 10^{-3} \approx 1.213 \, \text{N} \]

### Step 5: Calculate Young's modulus (\( Y \))
Young's modulus is given by:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{\frac{T}{A}}{\frac{\Delta L}{L}} \]

Where:
- Stress \( = \frac{T}{A} \)
- Strain \( = \frac{\Delta L}{L} \)

Assuming the extension \( \Delta L = 0.05 \, \text{cm} = 0.0005 \, \text{m} \):
\[ \text{Strain} = \frac{0.0005}{0.4005} \approx 0.001247 \]

Calculating stress:
\[ \text{Stress} = \frac{1.213}{1 \times 10^{-6}} = 1.213 \times 10^6 \, \text{N/m}^2 \]

### Step 6: Substitute into Young's modulus formula
\[ Y = \frac{1.213 \times 10^6}{0.001247} \approx 9.73 \times 10^8 \, \text{N/m}^2 \]

### Final Answer
In the form \( X \times 10^8 \, \text{N/m}^2 \), we find:
\[ X \approx 9.73 \]