The area of a cross section of steel wire is `0.1cm^2` and Young's modulus of steel is `2xx10^(11)Nm^-2`. The force required to strech by `0.1%` of its length is

Options

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

Explanation

To solve the problem, we will use the formula for Young's modulus and the relationship between stress, strain, and force.

### Step-by-Step Solution:

1. **Convert the Area**:
The area of the cross-section is given as \(0.1 \, \text{cm}^2\). We need to convert this to square meters for consistency in SI units.
\[ 0.1 \, \text{cm}^2 = 0.1 \times 10^{-4} \, \text{m}^2 = 1 \times 10^{-5} \, \text{m}^2 \]

2. **Identify Young's Modulus**:
The Young's modulus of steel is given as \(2 \times 10^{11} \, \text{N/m}^2\).

3. **Calculate Strain**:
The strain is defined as the change in length divided by the original length. The problem states that the wire is stretched by \(0.1\%\) of its length. This can be expressed as:
\[ \text{Strain} = \frac{\Delta L}{L} = 0.1\% = \frac{0.1}{100} = 0.001 \]

4. **Use Young's Modulus Formula**:
The formula for Young's modulus \(Y\) is given by:
\[ Y = \frac{\text{Stress}}{\text{Strain}} \]
Stress is defined as:
\[ \text{Stress} = \frac{F}{A} \]
Therefore, we can rewrite the Young's modulus formula as:
\[ Y = \frac{F/A}{\Delta L/L} \]

5. **Rearranging for Force**:
Rearranging the formula to find the force \(F\):
\[ F = Y \cdot A \cdot \frac{\Delta L}{L} \]

6. **Substituting Values**:
Now, substituting the known values into the equation:
\[ F = (2 \times 10^{11} \, \text{N/m}^2) \cdot (1 \times 10^{-5} \, \text{m}^2) \cdot (0.001) \]

7. **Calculating the Force**:
\[ F = 2 \times 10^{11} \cdot 1 \times 10^{-5} \cdot 0.001 \]
\[ F = 2 \times 10^{11} \cdot 10^{-8} \]
\[ F = 2 \times 10^{3} \, \text{N} = 2000 \, \text{N} \]

### Final Answer:
The force required to stretch the steel wire by \(0.1\%\) of its length is \(2000 \, \text{N}\).