Two solid cylinders P and Q of same mass and same radius start rolling down a fixed inclined plane from the same height at the same time. Cylinder P has most of its mass concentrated near its surface, while Q has most its mass concentrated near the axis. Which statement(s) is (are) correct?

Options

A. Both cylinders `P` and `Q` reach the ground at the same time
B. Cylinder `P` has larger linear acceleration than cylinder `Q`
C. Both cylinders reach the ground with same translational kinetic energy
D. Cylinder `Q` reaches the ground with larger angular speed.

Explanation

To solve the problem, we need to analyze the motion of the two cylinders, P and Q, as they roll down the inclined plane. We will consider their moments of inertia and how that affects their linear and angular accelerations.

### Step-by-Step Solution:

1. **Understanding the Moment of Inertia**:
- Cylinder P has most of its mass concentrated near its surface, which means it has a larger moment of inertia (I_P).
- Cylinder Q has most of its mass concentrated near its axis, resulting in a smaller moment of inertia (I_Q).
- Since both cylinders have the same mass (m) and radius (r), we can conclude that I_P > I_Q.

**Hint**: Recall that the moment of inertia depends on the distribution of mass relative to the axis of rotation. More mass further from the axis increases the moment of inertia.

2. **Applying Newton's Second Law**:
- For both cylinders rolling down the incline, the forces acting on them include gravitational force (mg sin θ) and frictional force (F).
- The translational motion can be described by the equation:
\[ mg \sin \theta - F = ma \]
- For rotational motion, we have:
\[ F \cdot r = I \cdot \alpha \]
- Since the cylinders are rolling without slipping, we have the relationship:
\[ a = \alpha r \]

**Hint**: Remember that the frictional force is what allows the cylinders to roll without slipping, and it is crucial for the relationship between linear and angular acceleration.

3. **Relating Linear and Angular Acceleration**:
- Substituting \( \alpha = \frac{a}{r} \) into the rotational equation gives:
\[ F \cdot r = I \cdot \frac{a}{r} \]
- Rearranging this gives:
\[ F = \frac{I \cdot a}{r^2} \]

**Hint**: This step is important to connect the forces acting on the cylinders with their moments of inertia.

4. **Combining the Equations**:
- Substituting \( F \) back into the translational motion equation:
\[ mg \sin \theta - \frac{I \cdot a}{r^2} = ma \]
- Rearranging gives:
\[ mg \sin \theta = ma + \frac{I \cdot a}{r^2} \]
- Factoring out \( a \):
\[ mg \sin \theta = a \left( m + \frac{I}{r^2} \right) \]
- Thus, the linear acceleration \( a \) is given by:
\[ a = \frac{mg \sin \theta}{m + \frac{I}{r^2}} \]

**Hint**: This equation shows how the moment of inertia affects the acceleration of the cylinders. A larger moment of inertia results in a smaller acceleration.

5. **Comparing Accelerations**:
- Since \( I_P > I_Q \), it follows that:
\[ a_P < a_Q \]
- Therefore, cylinder P will have a smaller linear acceleration than cylinder Q.

**Hint**: Remember that the acceleration is inversely related to the moment of inertia.

6. **Conclusion on Time and Velocity**:
- Since both cylinders start from the same height and roll down the same incline, the time taken to reach the bottom will be longer for cylinder P than for cylinder Q.
- Consequently, the final velocities will also differ, with cylinder Q reaching a higher velocity than cylinder P.

**Hint**: The relationship between acceleration and velocity is direct; higher acceleration leads to higher final velocity over the same distance.

### Final Statements:
- Cylinder P will take more time to reach the bottom than cylinder Q.
- Cylinder Q will have a larger linear velocity and angular speed when they reach the bottom.

### Summary of Correct Statements:
- Cylinder P has less linear acceleration than cylinder Q.
- Cylinder Q reaches the ground with a larger angular speed than cylinder P.