A wire of `10m` long and `1mm^(2)` area of cross section is strechted by a force of `20N`. If the elongation is `2mm` the young's modulus of the material of the wire (in `Pa`) is

Options

A. `1xx10^(9)`
B. `2xx10^(-9)`
C. `1xx10^(11)`
D. `1xx10^(12)`

Explanation

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

### Step 1: Identify the given values
- Length of the wire (L) = 10 m
- Cross-sectional area (A) = 1 mm² = \(1 \times 10^{-6}\) m² (conversion from mm² to m²)
- Force applied (F) = 20 N
- Elongation (ΔL) = 2 mm = \(2 \times 10^{-3}\) m (conversion from mm to m)

### Step 2: Calculate Stress
Stress is defined as the force applied per unit area. The formula for stress (σ) is:
\[ \sigma = \frac{F}{A} \]
Substituting the values:
\[ \sigma = \frac{20 \, \text{N}}{1 \times 10^{-6} \, \text{m}^2} = 20 \times 10^{6} \, \text{Pa} = 2 \times 10^{7} \, \text{Pa} \]

### Step 3: Calculate Strain
Strain is defined as the change in length per original length. The formula for strain (ε) is:
\[ \epsilon = \frac{\Delta L}{L} \]
Substituting the values:
\[ \epsilon = \frac{2 \times 10^{-3} \, \text{m}}{10 \, \text{m}} = 0.0002 \]

### Step 4: Calculate Young's Modulus
Young's modulus (Y) is defined as the ratio of stress to strain. The formula for Young's modulus is:
\[ Y = \frac{\sigma}{\epsilon} \]
Substituting the values:
\[ Y = \frac{2 \times 10^{7} \, \text{Pa}}{0.0002} = 1 \times 10^{11} \, \text{Pa} \]

### Final Answer
The Young's modulus of the material of the wire is:
\[ Y = 1 \times 10^{11} \, \text{Pa} \]

---