There are two inclined planes A and B of same dimensions. A is sufficiently rough but B is smooth A round shaped object is released from rest at the top point of inclined plane A and acceleration the object is found to be a, and v1 is the velocity of centre of object when it reaches the bottom. When the same process is repeated with B then corresponding values are found to be `a_(2) and v_(2)`.

Options

A. `a_(1) gt a_(2)`
B. `a_(1) lt a_(2)`
C. `v_(1) gt v_(2)`
D. `v_(1)lt v_(2)`

Explanation

To solve the problem, we will analyze the motion of a round-shaped object on two inclined planes, A (rough) and B (smooth). We will determine the relationship between their accelerations and final velocities.

### Step-by-Step Solution:

1. **Understanding the System**:
- We have two inclined planes, A and B, with the same dimensions (height and angle).
- Plane A is rough, allowing the object to roll without slipping, while plane B is smooth, causing the object to slide without rolling.

2. **Acceleration on Plane A (Rough)**:
- For the rough inclined plane A, the object rolls down. The acceleration \( a_1 \) can be calculated using the formula:
\[ a_1 = \frac{g \sin \theta}{1 + \frac{I}{mR^2}} \]
where \( I \) is the moment of inertia of the object, \( m \) is its mass, \( R \) is its radius, and \( g \) is the acceleration due to gravity.

3. **Acceleration on Plane B (Smooth)**:
- For the smooth inclined plane B, the object slides down without rolling. The acceleration \( a_2 \) is given by:
\[ a_2 = g \sin \theta \]

4. **Comparing Accelerations**:
- Since \( a_1 \) includes a term in the denominator that is greater than 1 (due to the moment of inertia), we can conclude:
\[ a_1 < a_2 \]

5. **Velocity at the Bottom of Plane A**:
- The final velocity \( v_1 \) of the object when it reaches the bottom of plane A can be found using energy conservation:
\[ mgh = \frac{1}{2} mv_1^2 + \frac{1}{2} I \omega^2 \]
Since \( \omega = \frac{v_1}{R} \), we can substitute and rearrange to find \( v_1 \).

6. **Velocity at the Bottom of Plane B**:
- For plane B, the final velocity \( v_2 \) is calculated similarly:
\[ mgh = \frac{1}{2} mv_2^2 \]
This simplifies to:
\[ v_2 = \sqrt{2gh} \]

7. **Comparing Velocities**:
- Since part of the potential energy on plane A is converted into rotational kinetic energy, less energy is available for linear motion compared to plane B. Thus:
\[ v_1 < v_2 \]

### Conclusion:
- We find that:
- \( a_1 < a_2 \) (the acceleration on the rough plane is less than that on the smooth plane).
- \( v_1 < v_2 \) (the final velocity on the rough plane is less than that on the smooth plane).

### Final Answers:
- \( a_1 < a_2 \)
- \( v_1 < v_2 \)