A wire suspended vertically is stretched by a 20 kgf. applied to its free end. The increase in length of the wire is 2 mm. The energy stored in the wire is `(g=10ms^(-2))`:

Options

A. 0.1 J
B. 0.2 J
C. 0.4 J
D. 5 J

Explanation

To solve the problem step by step, we need to calculate the energy stored in the wire when a force is applied to it. Here’s how we can do that:

### Step 1: Determine the Force Applied
The force applied to the wire is due to the weight of the mass hanging from it. The weight (force) can be calculated using the formula:
\[ F = m \cdot g \]
where:
- \( m = 20 \, \text{kg} \) (mass)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)

Substituting the values:
\[ F = 20 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 200 \, \text{N} \]

### Step 2: Convert the Increase in Length to Meters
The increase in length of the wire is given as 2 mm. We need to convert this to meters for our calculations:
\[ \Delta L = 2 \, \text{mm} = \frac{2}{1000} \, \text{m} = 0.002 \, \text{m} \]

### Step 3: Calculate the Elastic Potential Energy
The elastic potential energy (E) stored in the wire can be calculated using the formula:
\[ E = \frac{1}{2} \cdot F \cdot \Delta L \]
Substituting the values we found:
\[ E = \frac{1}{2} \cdot 200 \, \text{N} \cdot 0.002 \, \text{m} \]

### Step 4: Simplify the Calculation
Now, we perform the calculation:
\[ E = \frac{1}{2} \cdot 200 \cdot 0.002 = 100 \cdot 0.002 = 0.2 \, \text{J} \]

### Final Answer
The energy stored in the wire is:
\[ E = 0.2 \, \text{J} \]