A wire of length 5 m is stretched by 1 mm by a force of 1 N. The value of energy (in mJ) stored in the wire is

Options

A. 0.5
B. 0.05
C. 5
D. 50

Explanation

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

\[ \text{Energy} = \frac{1}{2} \times \text{Force} \times \text{Extension} \]

### Step-by-step solution:

1. **Identify the given values:**
- Length of the wire, \( L = 5 \, \text{m} \)
- Extension of the wire, \( \Delta L = 1 \, \text{mm} = 0.001 \, \text{m} \)
- Force applied, \( F = 1 \, \text{N} \)

2. **Substitute the values into the energy formula:**
\[ \text{Energy} = \frac{1}{2} \times F \times \Delta L \]
\[ \text{Energy} = \frac{1}{2} \times 1 \, \text{N} \times 0.001 \, \text{m} \]

3. **Calculate the energy:**
\[ \text{Energy} = \frac{1}{2} \times 1 \times 0.001 = \frac{0.001}{2} = 0.0005 \, \text{J} \]

4. **Convert joules to millijoules:**
Since \( 1 \, \text{J} = 1000 \, \text{mJ} \),
\[ 0.0005 \, \text{J} = 0.0005 \times 1000 \, \text{mJ} = 0.5 \, \text{mJ} \]

5. **Final answer:**
The energy stored in the wire is \( 0.5 \, \text{mJ} \).