A rod of mass m is released on smooth horizontal surface making angle `theta` with horizontal. Then which of the following statement is correct ?

Options

A. Acceleration of rod along vertical is less than g
B. Acceleration of centre of mass along horizontal is zero.
C. Angular acceleration of rod is not constant
D. Momentum of the rod along vertical will remain constant

Answer

To solve the problem, we need to analyze the motion of the rod that is released on a smooth horizontal surface at an angle \(\theta\) with the horizontal.

### Step-by-Step Solution:

1. **Identify Forces Acting on the Rod**:
- The forces acting on the rod include the gravitational force acting downwards (weight \(mg\)) and the normal force acting upwards from the surface.
- Since the surface is smooth, there is no friction acting on the rod.

2. **Break Down the Gravitational Force**:
- The gravitational force can be resolved into two components:
- Vertical component: \(mg \cos \theta\)
- Horizontal component: \(mg \sin \theta\)

3. **Apply Newton's Second Law**:
- In the vertical direction:
- The normal force \(N\) balances the vertical component of the weight:
\[ N = mg \cos \theta \]
- Since there is no net force in the vertical direction, the vertical motion does not change, and thus the momentum in the vertical direction remains constant.

- In the horizontal direction:
- The horizontal component of the weight \(mg \sin \theta\) causes acceleration.
- The net force in the horizontal direction is given by:
\[ F_{\text{net}} = ma_x = mg \sin \theta \]
- Therefore, the horizontal acceleration \(a_x\) can be calculated as:
\[ a_x = g \sin \theta \]

4. **Determine the Angular Motion**:
- The rod is not rotating; it is simply translating along the surface.
- The angle \(\theta\) remains constant, which means there is no angular acceleration:
\[ \alpha = 0 \]

5. **Conclusion**:
- The correct statement regarding the motion of the rod is that the linear momentum in the vertical direction is constant, and the angular acceleration is zero.

### Final Answer:
The correct statement is that the linear momentum in the vertical direction is constant, and the angular acceleration of the rod is zero.