The Young's modulus of a wire of length 2m and area of cross section `1 mm^(2)` is `2 xx 10^(11) N//m^(2)`. The work done in increasing its length by 2mm is

Options

A. 0.1 J
B. 0.2 J
C. 0.02 J
D. 0.4 J

Explanation

To solve the problem, we need to calculate the work done in increasing the length of a wire using the given Young's modulus, initial length, area of cross-section, and the increase in length.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Length of the wire, \( L = 2 \, \text{m} \)
- Area of cross-section, \( A = 1 \, \text{mm}^2 = 1 \times 10^{-6} \, \text{m}^2 \)
- Young's modulus, \( Y = 2 \times 10^{11} \, \text{N/m}^2 \)
- Increase in length, \( \Delta L = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m} \)

2. **Use the Formula for Work Done:**
The work done \( W \) in stretching a wire is given by the formula:
\[ W = \frac{1}{2} \cdot \frac{Y \cdot A \cdot (\Delta L)^2}{L} \]

3. **Substitute the Values into the Formula:**
\[ W = \frac{1}{2} \cdot \frac{(2 \times 10^{11}) \cdot (1 \times 10^{-6}) \cdot (2 \times 10^{-3})^2}{2} \]

4. **Calculate \( (\Delta L)^2 \):**
\[ (\Delta L)^2 = (2 \times 10^{-3})^2 = 4 \times 10^{-6} \, \text{m}^2 \]

5. **Substitute \( (\Delta L)^2 \) into the Work Done Formula:**
\[ W = \frac{1}{2} \cdot \frac{(2 \times 10^{11}) \cdot (1 \times 10^{-6}) \cdot (4 \times 10^{-6})}{2} \]

6. **Simplify the Expression:**
\[ W = \frac{1}{2} \cdot \frac{(2 \times 4) \times 10^{11} \times 10^{-12}}{2} \]
\[ W = \frac{1}{2} \cdot \frac{8 \times 10^{-1}}{2} \]
\[ W = \frac{8 \times 10^{-1}}{4} = 2 \times 10^{-1} = 0.2 \, \text{J} \]

7. **Final Result:**
The work done in increasing the length of the wire by 2 mm is \( W = 0.2 \, \text{J} \).