For steel `Y=2xx10^(11) Nm^(-2)`. The force required to double the length of a steel wire of area 1 `cm^(2)` is

Options

A. `2 xx10^(7) N`
B. `2xx10^(6) N`
C. `2xx10^(8) N`
D. `2xx10^(5) N`

Explanation

To solve the problem, we need to determine the force required to double the length of a steel wire with a given Young's modulus and cross-sectional area.

### Step-by-Step Solution:

1. **Understand Young's Modulus**:
Young's modulus (Y) is defined as the ratio of stress (σ) to strain (ε). It can be expressed as:
\[ Y = \frac{\sigma}{\epsilon} \]
where:
- Stress (σ) = Force (F) / Area (A)
- Strain (ε) = Change in length (ΔL) / Original length (L)

2. **Determine the Strain for Doubling the Length**:
If the length of the wire is doubled, the strain (ε) can be calculated as:
\[ \epsilon = \frac{\Delta L}{L} = \frac{L}{L} = 1 \]
This means the wire is stretched to twice its original length.

3. **Calculate the Stress Required**:
Rearranging the formula for Young's modulus gives us:
\[ \sigma = Y \cdot \epsilon \]
Substituting the values:
\[ Y = 2 \times 10^{11} \, \text{N/m}^2 \quad \text{and} \quad \epsilon = 1 \]
Therefore:
\[ \sigma = 2 \times 10^{11} \, \text{N/m}^2 \cdot 1 = 2 \times 10^{11} \, \text{N/m}^2 \]

4. **Convert Area to SI Units**:
The cross-sectional area (A) is given as 1 cm². We need to convert this to square meters:
\[ A = 1 \, \text{cm}^2 = 1 \times 10^{-4} \, \text{m}^2 \]

5. **Calculate the Force**:
The force (F) can be calculated using the formula:
\[ F = \sigma \cdot A \]
Substituting the values:
\[ F = (2 \times 10^{11} \, \text{N/m}^2) \cdot (1 \times 10^{-4} \, \text{m}^2) \]
\[ F = 2 \times 10^{11} \times 10^{-4} = 2 \times 10^{7} \, \text{N} \]

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