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 to it, we can follow these steps:

### Step 1: Identify the Given Values
- Length of the wire (L) = 2 m
- Stretch in the wire (ΔL) = 10 mm = 10 × 10^(-3) m = 0.01 m
- Mass attached (m) = 10 kg
- Acceleration due to gravity (g) = 10 m/s²

### Step 2: Calculate the Force (Weight) Acting on the Wire
The force (F) acting on the wire due to the attached mass can be calculated using the formula:
\[ F = m \cdot g \]
Substituting the values:
\[ F = 10 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 100 \, \text{N} \]

### Step 3: Use the Formula for Elastic Potential Energy
The elastic potential energy (U) stored in the wire when it is stretched can be calculated using the formula:
\[ U = \frac{1}{2} F \Delta L \]
Substituting the values we have:
\[ U = \frac{1}{2} \cdot 100 \, \text{N} \cdot 0.01 \, \text{m} \]

### Step 4: Calculate the Elastic Potential Energy
Now, we can calculate:
\[ U = \frac{1}{2} \cdot 100 \cdot 0.01 = \frac{100}{2} \cdot 0.01 = 50 \cdot 0.01 = 0.5 \, \text{J} \]

### Final Answer
The elastic potential energy gained by the wire is **0.5 Joules**.

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