A block is kept on a smooth inclined plane of angle of inclination `theta` that moves wuth a contact acceleration so that the block does not slide relative to the inclined plane. If the inclined plane stops the normal contact force offered by the plane on the block changes by a factor

Options

A. `tan theta`
B. `tan ^(2) theta`
C. `cos ^(2) theta`
D. `cot theta`

Explanation

To solve the problem, we will analyze the forces acting on the block on the inclined plane and how they change when the inclined plane stops.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Block**:
- The block experiences gravitational force \( mg \) acting downwards.
- The normal force \( N \) acts perpendicular to the surface of the inclined plane.

2. **Consider the Motion of the Inclined Plane**:
- The inclined plane is moving with a constant acceleration \( a \).
- Since the block does not slide relative to the inclined plane, it is in a non-inertial frame of reference.

3. **Analyze the Forces in the Non-Inertial Frame**:
- When the inclined plane accelerates, the effective gravitational force acting on the block can be modified due to the acceleration of the inclined plane.
- The effective force acting on the block in the direction perpendicular to the inclined plane can be expressed as:
\[ N \cos(\theta) = mg - ma \sin(\theta) \]
- Here, \( ma \sin(\theta) \) is the pseudo force acting on the block due to the acceleration of the inclined plane.

4. **Determine the Normal Force \( N_1 \)**:
- Rearranging the equation gives:
\[ N = \frac{mg - ma \sin(\theta)}{\cos(\theta)} \]

5. **Consider the Case When the Inclined Plane Stops**:
- When the inclined plane stops, the acceleration \( a \) becomes zero.
- Thus, the normal force \( N_1 \) when the inclined plane is moving with acceleration becomes:
\[ N_1 = \frac{mg}{\cos(\theta)} \]

6. **Normal Force Without Acceleration \( N_2 \)**:
- When the inclined plane is not accelerating (stopped), the normal force \( N_2 \) acting on the block is given by:
\[ N_2 = mg \cos(\theta) \]

7. **Finding the Ratio of Normal Forces**:
- Now, we can find the change in the normal force when the inclined plane stops:
\[ \frac{N_2}{N_1} = \frac{mg \cos(\theta)}{\frac{mg}{\cos(\theta)}} = \cos^2(\theta) \]

8. **Conclusion**:
- Therefore, the normal contact force offered by the plane on the block changes by a factor of \( \cos^2(\theta) \).

### Final Answer:
The normal contact force changes by a factor of \( \cos^2(\theta) \) (Option 3).