For increasing the length by 0.5mm in a steel wire of length 4m and area of cross section `5mm^(2)`, the force required is `(Y=11.90 xx 10^(12)N//m^(2))`

Options

A. `4 xx 10^(7)N`
B. `2 xx 10^(3)N`
C. `7.4 xx 10^(3)N`
D. `10^(3)N`

Explanation

To solve the problem, we will use the relationship between stress, strain, and Young's modulus. The formula we will use is:

\[ \text{Stress} = \text{Young's Modulus} \times \text{Strain} \]

Where:
- Stress (\( \sigma \)) is defined as \( \frac{F}{A} \) (Force per unit area)
- Strain (\( \epsilon \)) is defined as \( \frac{\Delta L}{L} \) (Change in length per original length)

### Step-by-Step Solution:

1. **Identify the given values:**
- Change in length (\( \Delta L \)) = 0.5 mm = \( 0.5 \times 10^{-3} \) m
- Original length (\( L \)) = 4 m
- Area of cross-section (\( A \)) = 5 mm² = \( 5 \times 10^{-6} \) m²
- Young's modulus (\( Y \)) = \( 11.90 \times 10^{12} \) N/m²

2. **Calculate the strain (\( \epsilon \)):**
\[ \epsilon = \frac{\Delta L}{L} = \frac{0.5 \times 10^{-3} \text{ m}}{4 \text{ m}} = \frac{0.5 \times 10^{-3}}{4} = 0.125 \times 10^{-3} \]

3. **Calculate the stress (\( \sigma \)) using Young's modulus:**
\[ \sigma = Y \times \epsilon = (11.90 \times 10^{12} \text{ N/m}^2) \times (0.125 \times 10^{-3}) \]
\[ \sigma = 11.90 \times 0.125 \times 10^{12 - 3} = 1.4875 \times 10^{10} \text{ N/m}^2 \]

4. **Relate stress to force and area:**
\[ \sigma = \frac{F}{A} \implies F = \sigma \times A \]
\[ F = (1.4875 \times 10^{10} \text{ N/m}^2) \times (5 \times 10^{-6} \text{ m}^2) \]
\[ F = 1.4875 \times 5 \times 10^{10 - 6} = 7.4375 \times 10^{4} \text{ N} \]

5. **Convert the force to a more convenient form:**
\[ F = 7.4375 \times 10^{4} \text{ N} = 7.44 \times 10^{3} \text{ N} \text{ (approximately)} \]

### Final Answer:
The force required to increase the length of the steel wire by 0.5 mm is approximately \( 7.44 \times 10^{3} \) N.