The potential energy of a particle of mass 1 kg moving in X-Y plane is given by `U=(12x+5y)` joules, where x an y are in meters. If the particle is initially at rest at origin, then select incorrect alternative :-

Options

A. Magnitude of its acceleration is `13 m//s^(2)`
B. It never crosses y-axis
C. It never crosses x-axis
D. It moves in a circle.

Answer

To solve the problem, we need to analyze the potential energy function given and derive the force acting on the particle, then determine the motion of the particle based on the force.

### Step-by-Step Solution:

1. **Identify the Potential Energy Function**:
The potential energy \( U \) is given as:
\[ U = 12x + 5y \text{ joules} \]
where \( x \) and \( y \) are in meters.

2. **Calculate the Force**:
The force \( \mathbf{F} \) acting on the particle can be derived from the potential energy using the relation:
\[ \mathbf{F} = -\nabla U \]
where \( \nabla U \) is the gradient of the potential energy.

We can express the gradient in terms of its components:
\[ \nabla U = \left( \frac{\partial U}{\partial x} \hat{i} + \frac{\partial U}{\partial y} \hat{j} \right) \]

- Calculate \( \frac{\partial U}{\partial x} \):
\[ \frac{\partial U}{\partial x} = 12 \]

- Calculate \( \frac{\partial U}{\partial y} \):
\[ \frac{\partial U}{\partial y} = 5 \]

Thus,
\[ \nabla U = 12 \hat{i} + 5 \hat{j} \]

Therefore, the force is:
\[ \mathbf{F} = -\nabla U = -12 \hat{i} - 5 \hat{j} \text{ Newtons} \]

3. **Determine the Acceleration**:
Using Newton's second law, \( \mathbf{F} = m \mathbf{a} \), where \( m = 1 \text{ kg} \):
\[ \mathbf{a} = \frac{\mathbf{F}}{m} = -12 \hat{i} - 5 \hat{j} \text{ m/s}^2 \]

4. **Calculate the Magnitude of Acceleration**:
The magnitude of the acceleration \( |\mathbf{a}| \) can be calculated as:
\[ |\mathbf{a}| = \sqrt{(-12)^2 + (-5)^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \text{ m/s}^2 \]

5. **Analyze the Motion**:
Since the particle starts from rest at the origin (0,0) and has a constant negative acceleration, it will move in the direction of the force, which is in the negative x and y directions. This means the particle will move straight towards the third quadrant of the coordinate system.

6. **Select the Incorrect Alternative**:
The options likely include statements about the particle's motion. The correct statements would be:
- The particle will not cross the y-axis (since it is moving left).
- The particle will not cross the x-axis (since it is moving downward).
- The particle does not move in a circle (it moves in a straight line).

Therefore, the incorrect alternative is the one that states that the particle moves in a circle.