A long elastic spring is stretched by `2 cm` and its potential energy is `U`. If the spring is stretched by `10 cm`, the `PE`…… will be

Options

A. `5U`
B. `25U`
C. `U//5`
D. `U//20`

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 (PE) stored in a spring when it is stretched is given by the formula:

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

where:
- \( k \) is the spring constant,
- \( x \) is the amount of stretch.

### Step-by-Step Solution:

1. **Identify the initial conditions:**
- When the spring is stretched by \( x_1 = 2 \) cm, the potential energy is \( U \).

2. **Express the potential energy for the first case:**
- Using the formula for potential energy:
\[ U = \frac{1}{2} k (2 \text{ cm})^2 = \frac{1}{2} k (0.02 \text{ m})^2 \]

3. **Identify the new conditions:**
- Now, the spring is stretched by \( x_2 = 10 \) cm.

4. **Express the potential energy for the second case:**
- The potential energy when the spring is stretched by \( 10 \) cm is:
\[ PE_2 = \frac{1}{2} k (10 \text{ cm})^2 = \frac{1}{2} k (0.10 \text{ m})^2 \]

5. **Calculate the ratio of potential energies:**
- We can find the ratio of the potential energies:
\[ \frac{U}{PE_2} = \frac{\frac{1}{2} k (2 \text{ cm})^2}{\frac{1}{2} k (10 \text{ cm})^2} = \frac{(2 \text{ cm})^2}{(10 \text{ cm})^2} = \frac{4 \text{ cm}^2}{100 \text{ cm}^2} = \frac{4}{100} = \frac{1}{25} \]

6. **Express \( PE_2 \) in terms of \( U \):**
- Rearranging gives:
\[ PE_2 = 25U \]

### Final Result:
The potential energy when the spring is stretched by \( 10 \) cm is \( 25U \).