A wheel `A` starts rolling up a rough inclined plane and another identical wheel `B` starts rolling up a smooth plane having same inclination with the horizontal. If initial velocity of both the wheels is same then:

Options

A. `A` stops ascending earlier than `B`
B. kinetic energy of `B` never becomes zero
C. maximum height ascended by `A` is less than that, by `B`
D. friction acting on `A` remains constant during the round trip

Explanation

To solve the problem, we will analyze the motion of both wheels A and B as they roll up their respective inclined planes.

### Step 1: Understanding the System
We have two identical wheels:
- Wheel A is rolling up a rough inclined plane.
- Wheel B is rolling up a smooth inclined plane.

Both wheels start with the same initial linear velocity \( v_0 \) and angular velocity \( \omega_0 \).

### Step 2: Forces Acting on the Wheels
For Wheel A (on the rough plane):
- The forces acting on it include:
- Gravitational force component along the incline: \( mg \sin \theta \)
- Normal force: \( N = mg \cos \theta \)
- Frictional force: \( f \) (acting up the incline since it opposes the motion)

For Wheel B (on the smooth plane):
- The forces acting on it include:
- Gravitational force component along the incline: \( mg \sin \theta \)
- Normal force: \( N = mg \cos \theta \)
- There is no frictional force acting on Wheel B because the surface is smooth.

### Step 3: Equations of Motion
For Wheel A, the net force acting on it as it rolls up the incline can be expressed as:
\[ m a_A = -mg \sin \theta - f \]
Where \( a_A \) is the linear acceleration of Wheel A.

For Wheel B, since there is no friction, the equation simplifies to:
\[ m a_B = -mg \sin \theta \]
Where \( a_B \) is the linear acceleration of Wheel B.

### Step 4: Rolling Condition
For Wheel A, it will eventually achieve pure rolling motion, which means:
\[ a_A = \alpha R \]
Where \( \alpha \) is the angular acceleration and \( R \) is the radius of the wheel.

For Wheel B, since there is no friction, it will not experience any torque, and thus its angular velocity will remain constant while its linear velocity decreases.

### Step 5: Energy Considerations
Using conservation of energy:
- For Wheel A:
\[ \frac{1}{2} I \omega^2 + \frac{1}{2} mv^2 = mgh_A \]
Where \( h_A \) is the height reached by Wheel A.

- For Wheel B:
\[ \frac{1}{2} mv^2 = mgh_B \]
Where \( h_B \) is the height reached by Wheel B.

### Step 6: Comparing Heights
Since Wheel A loses both translational and rotational kinetic energy while Wheel B only loses translational kinetic energy, we can conclude that:
\[ h_A > h_B \]
This means Wheel A will ascend to a greater height than Wheel B.

### Conclusion
- Wheel A will stop ascending later than Wheel B because it can convert both its translational and rotational kinetic energy into potential energy.
- The maximum height ascended by Wheel A is greater than that of Wheel B.