A wire suspended vertically from one of itsends is strached by attached a weight of `200 N` to the lower end . The weight streches the wire by `1 mm` . Then the elastic energy stored in the wire is

Options

A. 0.1 J
B. 0.2 J
C. 10 J
D. 20 J

Explanation

To find the elastic energy stored in the wire when it is stretched by a weight, we can use the formula for elastic potential energy:

\[ U = \frac{1}{2} F x \]

where:
- \( U \) is the elastic potential energy,
- \( F \) is the force applied (in Newtons),
- \( x \) is the extension of the wire (in meters).

### Step-by-Step Solution:

1. **Identify the Force (F)**: The weight attached to the wire is given as \( 200 \, \text{N} \). Therefore, \( F = 200 \, \text{N} \).

2. **Identify the Extension (x)**: The wire is stretched by \( 1 \, \text{mm} \). We need to convert this to meters for consistency in units:
\[ x = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \]

3. **Substitute the Values into the Formula**: Now we can substitute the values of \( F \) and \( x \) into the formula for elastic potential energy:
\[ U = \frac{1}{2} \times 200 \, \text{N} \times 1 \times 10^{-3} \, \text{m} \]

4. **Calculate the Elastic Energy (U)**:
\[ U = \frac{1}{2} \times 200 \times 1 \times 10^{-3} \]
\[ U = 100 \times 10^{-3} = 0.1 \, \text{J} \]

Thus, the elastic energy stored in the wire is \( 0.1 \, \text{J} \).

### Final Answer:
The elastic energy stored in the wire is \( 0.1 \, \text{J} \).