If the potential energy of a spring is `V` on stretching it by `2cm`, then its potential energy when it is stretched by `10cm` will be

Options

A. `V//25`
B. `5 V`
C. `V//5`
D. `25 V`

Explanation

To solve the problem, we need to understand the relationship between the potential energy stored in a spring and the amount it is stretched. The potential energy \( U \) stored in a spring is given by the formula:

\[ U = \frac{1}{2} k x^2 \]

where:
- \( U \) is the potential energy,
- \( k \) is the spring constant,
- \( x \) is the displacement from the equilibrium position.

### Step-by-Step Solution:

1. **Identify the given values:**
- The potential energy \( V \) when the spring is stretched by \( x_1 = 2 \) cm.
- We need to find the potential energy \( V_2 \) when the spring is stretched by \( x_2 = 10 \) cm.

2. **Write the potential energy equations:**
- For the first case (2 cm stretch):
\[ V = \frac{1}{2} k (2)^2 \]
- For the second case (10 cm stretch):
\[ V_2 = \frac{1}{2} k (10)^2 \]

3. **Express the potential energies in terms of the spring constant \( k \):**
- From the first equation:
\[ V = \frac{1}{2} k \cdot 4 \quad \Rightarrow \quad V = 2k \]
- From the second equation:
\[ V_2 = \frac{1}{2} k \cdot 100 \quad \Rightarrow \quad V_2 = 50k \]

4. **Relate the two potential energies:**
- We can express \( V_2 \) in terms of \( V \):
\[ V_2 = 50k = 25 \cdot 2k = 25V \]

5. **Conclusion:**
- Therefore, the potential energy when the spring is stretched by 10 cm is:
\[ V_2 = 25V \]

### Final Answer:
The potential energy when the spring is stretched by 10 cm will be \( 25V \).