What is the force requiredto stretch a steel wire of `1 cm^(2)` cross-section to 1.1 times its length ? `(Y = 2 xx 10^(11) N//m^(2))`

Options

A. `2 xx 10^(3) N`
B. `2 xx 10^(-6) N`
C. `2 xx 10^(-7) N`
D. `2 xx 10^(6) N`

Explanation

To solve the problem of finding the force required to stretch a steel wire of 1 cm² cross-section to 1.1 times its original length, we can use the formula relating Young's modulus (Y), stress, and strain.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Cross-sectional area (A) = 1 cm² = \(1 \times 10^{-4} \, m²\)
- Young's modulus (Y) = \(2 \times 10^{11} \, N/m²\)
- The wire is stretched to 1.1 times its original length, which means:
- If the original length is \(L\), the new length \(L' = 1.1L\).
- The change in length (\(\Delta L\)) is given by:
\[ \Delta L = L' - L = 1.1L - L = 0.1L \]

2. **Calculate Strain:**
- Strain (\(e\)) is defined as the change in length per unit original length:
\[ e = \frac{\Delta L}{L} = \frac{0.1L}{L} = 0.1 \]

3. **Calculate Stress:**
- Stress (\(\sigma\)) is defined as the force (F) applied per unit area (A):
\[ \sigma = \frac{F}{A} \]

4. **Relate Young's Modulus to Stress and Strain:**
- Young's modulus (Y) is defined as the ratio of stress to strain:
\[ Y = \frac{\sigma}{e} \]
- Rearranging this gives:
\[ \sigma = Y \cdot e \]

5. **Substitute Values:**
- Substitute the values of \(Y\) and \(e\):
\[ \sigma = (2 \times 10^{11} \, N/m²) \cdot (0.1) = 2 \times 10^{10} \, N/m² \]

6. **Calculate Force (F):**
- Now substitute the stress back into the stress formula to find the force:
\[ \sigma = \frac{F}{A} \implies F = \sigma \cdot A \]
- Substitute the values of \(\sigma\) and \(A\):
\[ F = (2 \times 10^{10} \, N/m²) \cdot (1 \times 10^{-4} \, m²) = 2 \times 10^{6} \, N \]

### Final Answer:
The force required to stretch the steel wire is \(2 \times 10^{6} \, N\).