The potential energy of a long spring when stretched by 2 cm is U. If the spring is stretched by 8 cm the potential energy stored in it is:-

Options

A. 4U
B. 8U
C. 16U
D. `U/4`

Explanation

To solve the problem, we will use the formula for the potential energy stored in a spring, which is given by:

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

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**:
- When the spring is stretched by \( x_1 = 2 \) cm, the potential energy is \( U \).
- We need to find the potential energy when the spring is stretched by \( x_2 = 8 \) cm.

2. **Write the expression for potential energy for both cases**:
- For the first stretch (2 cm):
\[ U_1 = \frac{1}{2} k (x_1)^2 = \frac{1}{2} k (2 \, \text{cm})^2 = \frac{1}{2} k (0.02 \, \text{m})^2 \]
- For the second stretch (8 cm):
\[ U_2 = \frac{1}{2} k (x_2)^2 = \frac{1}{2} k (8 \, \text{cm})^2 = \frac{1}{2} k (0.08 \, \text{m})^2 \]

3. **Relate the two potential energies**:
- We can express the ratio of the two potential energies:
\[ \frac{U_2}{U_1} = \frac{\frac{1}{2} k (x_2)^2}{\frac{1}{2} k (x_1)^2} = \frac{(x_2)^2}{(x_1)^2} \]
- Substituting the values:
\[ \frac{U_2}{U_1} = \frac{(8 \, \text{cm})^2}{(2 \, \text{cm})^2} = \frac{64 \, \text{cm}^2}{4 \, \text{cm}^2} = 16 \]

4. **Calculate \( U_2 \)**:
- Since \( U_1 = U \), we can write:
\[ U_2 = 16 U_1 = 16 U \]

5. **Final Answer**:
- The potential energy stored in the spring when stretched by 8 cm is:
\[ U_2 = 16U \]