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

The correct option is 25V.

Step-by-Step Explanation:
The potential energy ($U$) stored in a stretched spring is given by the formula:
$U=\frac{1}{2}k{x}^2$
where:
- $k$ is the spring constant of the spring.
- $x$ is the displacement or stretch of the spring from its equilibrium position.

From this formula, we can see that for a given spring (where the spring constant $k$ remains constant), the potential energy is directly proportional to the square of the stretch:
$U\propto x^2$

Let $U_1$ be the initial potential energy when the spring is stretched by $x_1=2\text{cm}$. We are given:
$U_1=V$

Let $U_2$ be the potential energy when the spring is stretched by $x_2=10\text{cm}$.

We can set up the ratio of the potential energies in the two cases:
$\frac{U_2}{U_1}={\left(\frac{x_2}{x_1}\right)}^2$

Substitute the given values into the ratio:
$\frac{U_2}{V}={(\frac{10}{2}}^2$
$\frac{U_2}{V}={\left(5\right)}^2$
$\frac{U_2}{V}=25$

Solving for $U_2$:
$U_2=25V$

Thus, the potential energy of the spring when it is stretched by 10 cm will be 25V.