A load of 20 kg produces an extension of 1 mm in a wire 3 m in length and 1 mm in diameter.l Calculate Young's modulus of elesticity of wire.

Options

A. `3.25xx10^(10)Nm^(-2)`
B. `7.48xx10^(10)Nm^(2)`
C. `7.48xx10^(10)Nm^(-2)`
D. `7.48xx10^(-10)Nm^(-2)`

Answer

To calculate Young's modulus of elasticity for the given wire, we will follow these steps:

### Step 1: Identify the given values
- Load (mass) \( m = 20 \, \text{kg} \)
- Extension \( \Delta L = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \)
- Original length of the wire \( L = 3 \, \text{m} \)
- Diameter of the wire \( d = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \)

### Step 2: Calculate the force (F) applied on the wire
The force can be calculated using the formula:
\[ F = m \cdot g \]
where \( g \) (acceleration due to gravity) is approximately \( 9.81 \, \text{m/s}^2 \).

Substituting the values:
\[ F = 20 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 196.2 \, \text{N} \]

### Step 3: Calculate the cross-sectional area (A) of the wire
The area \( A \) can be calculated using the formula for the area of a circle:
\[ A = \pi r^2 \]
First, we need to find the radius \( r \):
\[ r = \frac{d}{2} = \frac{1 \times 10^{-3} \, \text{m}}{2} = 0.5 \times 10^{-3} \, \text{m} = 0.0005 \, \text{m} \]

Now substituting the radius into the area formula:
\[ A = \pi (0.0005 \, \text{m})^2 = \pi (0.00000025 \, \text{m}^2) \approx 7.85 \times 10^{-7} \, \text{m}^2 \]

### Step 4: Calculate the strain (ε)
Strain is defined as the change in length divided by the original length:
\[ \text{Strain} \, (\epsilon) = \frac{\Delta L}{L} = \frac{1 \times 10^{-3} \, \text{m}}{3 \, \text{m}} = \frac{1}{3000} \approx 3.33 \times 10^{-4} \]

### Step 5: Calculate the stress (σ)
Stress is defined as force per unit area:
\[ \text{Stress} \, (\sigma) = \frac{F}{A} = \frac{196.2 \, \text{N}}{7.85 \times 10^{-7} \, \text{m}^2} \approx 2.5 \times 10^{11} \, \text{N/m}^2 \]

### Step 6: Calculate Young's modulus (E)
Young's modulus is defined as the ratio of stress to strain:
\[ E = \frac{\sigma}{\epsilon} = \frac{2.5 \times 10^{11} \, \text{N/m}^2}{3.33 \times 10^{-4}} \approx 7.5 \times 10^{10} \, \text{N/m}^2 \]

### Final Answer
Thus, the Young's modulus of elasticity of the wire is approximately:
\[ E \approx 7.5 \times 10^{10} \, \text{N/m}^2 \]