A wire of length 10 m and cross-section are `10^(-6) m^(2)` is stretched with a force of 20 N. If the elongation is 1 mm, the Young's modulus of material of the wire will be

Options

A. `10^(10)N//m^(2)`
B. `10^(11)N//m^(2)`
C. `2xx10^(10)N//m^(2)`
D. `2xx10^(11)N//m^(2)`

Explanation

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

### Step 1: Identify the given values
- Length of the wire (L) = 10 m
- Cross-sectional area (A) = \(10^{-6} \, m^2\)
- Force applied (F) = 20 N
- Elongation (ΔL) = 1 mm = \(1 \times 10^{-3} \, m\)

### Step 2: Write the formula for Young's Modulus
Young's modulus (Y) is defined as the ratio of stress to strain:
\[ Y = \frac{\text{Stress}}{\text{Strain}} \]

### Step 3: Calculate Stress
Stress is defined as the force applied per unit area:
\[ \text{Stress} = \frac{F}{A} \]
Substituting the values:
\[ \text{Stress} = \frac{20 \, N}{10^{-6} \, m^2} = 20 \times 10^{6} \, N/m^2 = 2 \times 10^{7} \, N/m^2 \]

### Step 4: Calculate Strain
Strain is defined as the change in length divided by the original length:
\[ \text{Strain} = \frac{\Delta L}{L} \]
Substituting the values:
\[ \text{Strain} = \frac{1 \times 10^{-3} \, m}{10 \, m} = 0.1 \times 10^{-3} = 10^{-4} \]

### Step 5: Substitute Stress and Strain into Young's Modulus formula
Now we can substitute the values of stress and strain into the Young's modulus formula:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{2 \times 10^{7} \, N/m^2}{10^{-4}} = 2 \times 10^{7} \times 10^{4} \, N/m^2 \]
\[ Y = 2 \times 10^{11} \, N/m^2 \]

### Step 6: Conclusion
The Young's modulus of the material of the wire is:
\[ Y = 2 \times 10^{11} \, N/m^2 \]