A 40 cm wire having a mass of 3.2 g is stretched between two fixed supports 40.05 cm apart. In its fundamental mode, the wire vibrates at 220 Hz. If the area of cross section of the wire is `1.0 mm^2`, find its Young modulus.

Options

A. `1.98xx10^(11) N//m^(2)`
B. `2.2xx10^(11) N//m^(2)`
C. `3.96xx10^(11) N//m^(2)`
D. `3.2xx10^(11) N//m^(2)`

Explanation

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

### Step 1: Calculate the change in length (ΔL)
The original length of the wire (L) is given as 40 cm, and the stretched length is 40.05 cm.

\[ \Delta L = \text{Stretched Length} - \text{Original Length} = 40.05 \, \text{cm} - 40 \, \text{cm} = 0.05 \, \text{cm} = 0.0005 \, \text{m} \]

### Step 2: Calculate the strain (ε)
Strain is defined as the change in length divided by the original length.

\[ \epsilon = \frac{\Delta L}{L} = \frac{0.0005 \, \text{m}}{0.4 \, \text{m}} = 0.00125 \]

### Step 3: Calculate the mass per unit length (μ)
The mass of the wire is given as 3.2 g, which we convert to kg:

\[ \text{Mass} = 3.2 \, \text{g} = 3.2 \times 10^{-3} \, \text{kg} \]

Now, we calculate the mass per unit length:

\[ \mu = \frac{\text{Mass}}{\text{Length}} = \frac{3.2 \times 10^{-3} \, \text{kg}}{0.4 \, \text{m}} = 8 \times 10^{-3} \, \text{kg/m} \]

### Step 4: Calculate the tension (T) using the frequency (f)
The frequency of the wire is given as 220 Hz. The formula for the fundamental frequency of a vibrating wire is:

\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]

Rearranging this to find T:

\[ T = (2Lf)^2 \mu \]

Substituting the values:

\[ T = (2 \times 0.4 \, \text{m} \times 220 \, \text{Hz})^2 \times (8 \times 10^{-3} \, \text{kg/m}) \]

Calculating \(2Lf\):

\[ 2Lf = 2 \times 0.4 \times 220 = 176 \, \text{m} \]

Now squaring it:

\[ (176)^2 = 30976 \, \text{m}^2 \]

Now substituting back to find T:

\[ T = 30976 \times 8 \times 10^{-3} = 247.808 \, \text{N} \]

### Step 5: Calculate the stress (σ)
Stress is defined as force per unit area. The area of cross-section is given as 1 mm², which we convert to m²:

\[ \text{Area} = 1 \, \text{mm}^2 = 1 \times 10^{-6} \, \text{m}^2 \]

Now we can calculate stress:

\[ \sigma = \frac{T}{A} = \frac{247.808 \, \text{N}}{1 \times 10^{-6} \, \text{m}^2} = 247808000 \, \text{N/m}^2 = 2.47808 \times 10^8 \, \text{N/m}^2 \]

### Step 6: Calculate Young's modulus (Y)
Young's modulus is defined as the ratio of stress to strain:

\[ Y = \frac{\sigma}{\epsilon} = \frac{2.47808 \times 10^8 \, \text{N/m}^2}{0.00125} \]

Calculating Y:

\[ Y = 1.982464 \times 10^{11} \, \text{N/m}^2 \approx 1.98 \times 10^{11} \, \text{N/m}^2 \]

### Final Answer:
The Young's modulus of the wire is approximately \(1.98 \times 10^{11} \, \text{N/m}^2\).

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