The area of a cross-section of steel wire is `0.1 cm^(-2)` and Young's modulus of steel is `2 x 10^(11) N m^(-2)`. The force required to stretch by 0.1% of its length is

Options

A. 1000 N
B. 2000 N
C.  4000 N
D. 5000N

Explanation

To solve the problem, we need to calculate the force required to stretch a steel wire by 0.1% of its length using the given Young's modulus and the area of cross-section.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Area of cross-section, \( A = 0.1 \, \text{cm}^2 = 0.1 \times 10^{-4} \, \text{m}^2 = 1 \times 10^{-5} \, \text{m}^2 \)
- Young's modulus, \( Y = 2 \times 10^{11} \, \text{N/m}^2 \)
- Strain, \( \text{Strain} = \frac{\Delta L}{L} = 0.1\% = \frac{0.1}{100} = 0.001 \)

2. **Calculate the Stress:**
- According to the definition of Young's modulus:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L} \]
- Rearranging gives us:
\[ \text{Stress} = Y \times \text{Strain} \]
- Substitute the values:
\[ \text{Stress} = 2 \times 10^{11} \, \text{N/m}^2 \times 0.001 = 2 \times 10^{8} \, \text{N/m}^2 \]

3. **Calculate the Force:**
- Stress is also defined as:
\[ \text{Stress} = \frac{F}{A} \]
- Rearranging gives:
\[ F = \text{Stress} \times A \]
- Substitute the values:
\[ F = 2 \times 10^{8} \, \text{N/m}^2 \times 1 \times 10^{-5} \, \text{m}^2 = 2 \times 10^{3} \, \text{N} \]

4. **Final Result:**
- Therefore, the force required to stretch the wire by 0.1% of its length is:
\[ F = 2000 \, \text{N} \]