A particle moving on x - axis has potential energy `U = 2 - 20x + 5x^(2)` joule along x - axis. The particle is released at `x = -3`. The maximum value of `x` will be (`x` is in metre)

Options

A. `5m`
B. `3 m`
C. `7 m`
D. `8 m`

Explanation

To solve the problem, we need to find the maximum value of \( x \) for a particle moving along the x-axis with the given potential energy function \( U = 2 - 20x + 5x^2 \) joules. The particle is released from \( x = -3 \).

### Step-by-Step Solution:

1. **Identify the Potential Energy Function**:
The potential energy \( U \) is given as:
\[ U(x) = 2 - 20x + 5x^2 \]

2. **Calculate the Force**:
The force \( F \) acting on the particle can be found using the relation:
\[ F_x = -\frac{dU}{dx} \]
We need to differentiate \( U \) with respect to \( x \):
\[ \frac{dU}{dx} = 0 - 20 + 10x = 10x - 20 \]
Therefore, the force is:
\[ F_x = - (10x - 20) = 20 - 10x \]

3. **Find the Equilibrium Position**:
The equilibrium position occurs when the force is zero:
\[ 20 - 10x = 0 \]
Solving for \( x \):
\[ 10x = 20 \implies x = 2 \]

4. **Determine the Maximum Displacement**:
The particle is released from \( x = -3 \). The equilibrium position is at \( x = 2 \). The distance from the equilibrium position to the release point is:
\[ \text{Distance} = 2 - (-3) = 5 \text{ meters} \]
Since the particle will oscillate around the equilibrium position, the maximum displacement from the equilibrium position will also be 5 meters in the positive direction.

5. **Calculate the Maximum Value of \( x \)**:
The maximum value of \( x \) will be:
\[ x_{\text{max}} = 2 + 5 = 7 \text{ meters} \]

### Final Answer:
The maximum value of \( x \) is \( \boxed{7} \) meters.