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. `5V`
C. `(V)/(5)`
D. `25V`

Explanation

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

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

where:
- \( V \) 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 information**:
- The potential energy \( V \) when the spring is stretched by \( 2 \, \text{cm} \).
- We need to find the potential energy when the spring is stretched by \( 10 \, \text{cm} \).

2. **Write the potential energy for the first case (2 cm)**:
\[ V = \frac{1}{2} k (2 \, \text{cm})^2 = \frac{1}{2} k (0.02 \, \text{m})^2 \]
This gives us the potential energy when the spring is stretched by \( 2 \, \text{cm} \).

3. **Write the potential energy for the second case (10 cm)**:
\[ V_1 = \frac{1}{2} k (10 \, \text{cm})^2 = \frac{1}{2} k (0.10 \, \text{m})^2 \]

4. **Express \( V_1 \) in terms of \( V \)**:
Since we know that potential energy is proportional to the square of the displacement, we can set up the ratio:
\[ \frac{V_1}{V} = \frac{(10 \, \text{cm})^2}{(2 \, \text{cm})^2} \]

5. **Calculate the ratio**:
\[ \frac{V_1}{V} = \frac{100 \, \text{cm}^2}{4 \, \text{cm}^2} = 25 \]

6. **Find \( V_1 \)**:
\[ V_1 = 25V \]

### Conclusion:
The potential energy when the spring is stretched by \( 10 \, \text{cm} \) will be \( 25V \).