What force is required to stretch a steel wire to double its length when its area of cross section 1 cm2 and Young.s modulus `2xx10^(11)N//m^(2)`.

Answer

To find the force required to stretch a steel wire to double its length, we can use the formula related to Young's modulus (Y). The relationship is given by:

\[ Y = \frac{\text{Stress}}{\text{Strain}} \]

Where:
- Stress = \(\frac{F}{A}\) (Force per unit area)
- Strain = \(\frac{\Delta L}{L}\) (Change in length per original length)

Given:
- Young's Modulus, \(Y = 2 \times 10^{11} \, \text{N/m}^2\)
- Area of cross-section, \(A = 1 \, \text{cm}^2 = 1 \times 10^{-4} \, \text{m}^2\)
- The wire is stretched to double its length, so if the original length is \(L\), then \(\Delta L = L\).

### Step 1: Calculate Strain
Since the wire is stretched to double its length:
\[ \Delta L = L \]
Thus, the strain is:
\[ \text{Strain} = \frac{\Delta L}{L} = \frac{L}{L} = 1 \]

### Step 2: Calculate Stress
Using the relationship of Young's modulus:
\[ Y = \frac{\text{Stress}}{\text{Strain}} \]
Rearranging gives us:
\[ \text{Stress} = Y \times \text{Strain} \]
Substituting the values:
\[ \text{Stress} = (2 \times 10^{11} \, \text{N/m}^2) \times 1 = 2 \times 10^{11} \, \text{N/m}^2 \]

### Step 3: Calculate Force
Now, we can use the formula for stress to find the force:
\[ \text{Stress} = \frac{F}{A} \]
Rearranging gives:
\[ F = \text{Stress} \times A \]
Substituting the values:
\[ F = (2 \times 10^{11} \, \text{N/m}^2) \times (1 \times 10^{-4} \, \text{m}^2) \]
Calculating this gives:
\[ F = 2 \times 10^{11} \times 10^{-4} = 2 \times 10^{7} \, \text{N} \]

### Final Answer
The force required to stretch the steel wire to double its length is:
\[ F = 2 \times 10^{7} \, \text{N} \]