The area of cross section of a steel wire `(Y=2.0 xx 10^(11) N//m^(2))` is `0.1 cm^(2)`. The force required to double is length will be

Options

A. `2xx10^(12)` N
B. `2xx10^(11)` N
C. `2xx10^(10)` N
D. `2xx10^(6)` N

Answer

To solve the problem of finding the force required to double the length of a steel wire, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Data**:
- Young's Modulus (Y) = \(2.0 \times 10^{11} \, \text{N/m}^2\)
- Area of Cross Section (A) = \(0.1 \, \text{cm}^2 = 0.1 \times 10^{-4} \, \text{m}^2\) (conversion from cm² to m²)
- Original Length (L) = L (let's denote it as L for simplicity)
- New Length = \(2L\) (since we want to double the length)

2. **Calculate Change in Length**:
- Change in Length (\(\Delta L\)) = New Length - Original Length = \(2L - L = L\)

3. **Use the Formula for Young's Modulus**:
- The formula for Young's Modulus is given by:
\[ Y = \frac{\text{Stress}}{\text{Strain}} \]
- Stress is defined as:
\[ \text{Stress} = \frac{F}{A} \]
- Strain is defined as:
\[ \text{Strain} = \frac{\Delta L}{L} \]

4. **Rearranging the Young's Modulus Formula**:
- We can express the force (F) in terms of Young's Modulus, area, change in length, and original length:
\[ Y = \frac{F/A}{\Delta L/L} \]
- Rearranging gives:
\[ F = Y \cdot A \cdot \frac{\Delta L}{L} \]

5. **Substituting Values**:
- Substitute the known values into the equation:
\[ F = (2.0 \times 10^{11} \, \text{N/m}^2) \cdot (0.1 \times 10^{-4} \, \text{m}^2) \cdot \frac{L}{L} \]
- Notice that \(L\) cancels out:
\[ F = (2.0 \times 10^{11}) \cdot (0.1 \times 10^{-4}) = 2.0 \times 10^{11} \cdot 10^{-5} = 2.0 \times 10^{6} \, \text{N} \]

6. **Final Answer**:
- The force required to double the length of the steel wire is:
\[ F = 2.0 \times 10^{6} \, \text{N} \]