In a wire, when elongation is 2cm, energy stored is E. If it is stretched by 10 cm, then the energy stored will be

Options

A. E
B. 2 E
C. 4 E
D. 5 E

Explanation

To solve the problem, we need to understand the relationship between the elongation of a wire and the energy stored in it. The energy stored in a wire when it is stretched can be calculated using the formula:

\[ U = \frac{1}{2} F \Delta L \]

where:
- \( U \) is the energy stored,
- \( F \) is the force applied,
- \( \Delta L \) is the change in length (elongation).

### Step-by-Step Solution:

1. **Identify the initial conditions**:
- Given that when the elongation \( \Delta L_1 = 2 \) cm, the energy stored is \( E \).

2. **Relate energy and elongation**:
- Using the energy formula:
\[ E = \frac{1}{2} F \Delta L_1 \]
- Substitute \( \Delta L_1 = 2 \) cm:
\[ E = \frac{1}{2} F \times 2 \]
- Simplifying gives:
\[ E = F \]

3. **Determine the new elongation**:
- Now, we are stretching the wire to \( \Delta L_2 = 10 \) cm.

4. **Calculate the new energy stored**:
- Using the same energy formula for the new elongation:
\[ U = \frac{1}{2} F \Delta L_2 \]
- Substitute \( \Delta L_2 = 10 \) cm:
\[ U = \frac{1}{2} F \times 10 \]

5. **Substituting the value of \( F \)**:
- From step 2, we know \( F = E \):
\[ U = \frac{1}{2} E \times 10 \]
- Simplifying gives:
\[ U = 5E \]

6. **Conclusion**:
- Therefore, the energy stored when the wire is stretched by 10 cm is \( 5E \).

### Final Answer:
The energy stored when the wire is stretched by 10 cm is \( 5E \).