The area of a cross-section of steel wire is `0.5 cm^(-2)` and Young's modulus of steel is `2 xx 10^(11) N m^(-2)`. The force required to stretch by 0.5% of its length is

Options

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

Explanation

To find the force required to stretch a steel wire by 0.5% of its length, we can use the relationship between stress, strain, and Young's modulus.

### Step-by-step solution:

1. **Convert the area from cm² to m²:**
\[ \text{Area} = 0.5 \, \text{cm}^2 = 0.5 \times 10^{-4} \, \text{m}^2 \]

2. **Determine the strain:**
\[ \text{Strain} = \frac{\Delta L}{L} = 0.5\% = \frac{0.5}{100} = 0.005 \]

3. **Use Young's modulus to find stress:**
Young's modulus (Y) is defined as:
\[ Y = \frac{\text{Stress}}{\text{Strain}} \]
Rearranging this gives us:
\[ \text{Stress} = Y \times \text{Strain} \]
Substituting the values:
\[ Y = 2 \times 10^{11} \, \text{N/m}^2 \]
\[ \text{Stress} = 2 \times 10^{11} \, \text{N/m}^2 \times 0.005 = 1 \times 10^{9} \, \text{N/m}^2 \]

4. **Calculate the force using stress:**
Stress is also defined as:
\[ \text{Stress} = \frac{F}{A} \]
Rearranging gives:
\[ F = \text{Stress} \times A \]
Substituting the values:
\[ F = 1 \times 10^{9} \, \text{N/m}^2 \times 0.5 \times 10^{-4} \, \text{m}^2 \]
\[ F = 0.5 \times 10^{5} \, \text{N} = 5 \times 10^{4} \, \text{N} \]

### Final Answer:
The force required to stretch the steel wire by 0.5% of its length is \( 5 \times 10^{4} \, \text{N} \).