A wire 2 m in length suspended vertically stretches by. 10 mm when mass of 10 kg is attached to the lower end. The elastic potential energy gain by the wire is (take g = `10 m//s^(2)`)

Options

A. 0.5 J
B. 5 J
C. 50 J
D. 500 J

Answer

To find the elastic potential energy gained by the wire when a mass is attached, we can follow these steps:

### Step 1: Identify the given values
- Length of the wire, \( L = 2 \, \text{m} \)
- Stretch of the wire, \( \Delta L = 10 \, \text{mm} = 10 \times 10^{-3} \, \text{m} \)
- Mass attached, \( m = 10 \, \text{kg} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

### Step 2: Calculate the force applied to the wire
The force \( F \) due to the weight of the mass can be calculated using the formula:
\[ F = m \cdot g \]
Substituting the values:
\[ F = 10 \, \text{kg} \times 10 \, \text{m/s}^2 = 100 \, \text{N} \]

### Step 3: Use the formula for elastic potential energy
The elastic potential energy \( U \) stored in the wire when it stretches is given by the formula:
\[ U = \frac{1}{2} F \Delta L \]
Substituting the values we found:
\[ U = \frac{1}{2} \times 100 \, \text{N} \times 10 \times 10^{-3} \, \text{m} \]

### Step 4: Calculate the elastic potential energy
Now, calculate \( U \):
\[ U = \frac{1}{2} \times 100 \times 0.01 = \frac{1}{2} \times 1 = 0.5 \, \text{J} \]

### Final Answer
The elastic potential energy gained by the wire is \( 0.5 \, \text{J} \).

---