A long spring when stretched by `x cm`, has a potential energy `U`. On increasing the stretching to `nx cm,` the potential energy stored in spring will be

Options

A. U/n
B. nU
C. `n^(2)U`
D. `U//n^(2)`

Answer

To solve the problem, we need to find the potential energy stored in a spring when it is stretched by `nx cm`, given that the potential energy stored when it is stretched by `x cm` is `U`.

### Step-by-Step Solution:

1. **Understand the Formula for Potential Energy in a Spring**:
The potential energy (U) stored in a spring when it is stretched or compressed by a distance `x` is given by the formula:
\[ U = \frac{1}{2} k x^2 \]
where `k` is the spring constant.

2. **Initial Potential Energy**:
When the spring is stretched by `x cm`, the potential energy is:
\[ U = \frac{1}{2} k x^2 \]
Let's denote this as Equation (1).

3. **Final Potential Energy**:
Now, if the spring is stretched to `nx cm`, the new potential energy (let's call it \( U_2 \)) will be:
\[ U_2 = \frac{1}{2} k (nx)^2 \]
This can be simplified to:
\[ U_2 = \frac{1}{2} k n^2 x^2 \]
Let's denote this as Equation (2).

4. **Relate the Two Energies**:
Now, we want to express \( U_2 \) in terms of \( U \). From Equation (1), we know that:
\[ U = \frac{1}{2} k x^2 \]
Therefore, we can substitute \( \frac{1}{2} k x^2 \) from Equation (1) into Equation (2):
\[ U_2 = n^2 \left( \frac{1}{2} k x^2 \right) = n^2 U \]

5. **Conclusion**:
Thus, the potential energy stored in the spring when stretched to `nx cm` is:
\[ U_2 = n^2 U \]

### Final Answer:
The potential energy stored in the spring when stretched to `nx cm` is \( n^2 U \).