A solid cylinder and a hollow cylinder, both of the same mass and same external diameter are released from the same height at the same time on an inclined plane. Both roll down without slipping. Which one will reach the bottom first ?

Options

A. Both together
B. Hollow cylinder
C. Solid cylinder
D. Both together only when angle of inclination of plane is `45^(@)`

Answer

To solve the problem of which cylinder reaches the bottom of the incline first, we need to analyze the motion of both the solid cylinder and the hollow cylinder as they roll down the inclined plane.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have two cylinders (one solid and one hollow) of the same mass and external diameter released from the same height on an inclined plane. Both are rolling without slipping.

2. **Acceleration of Rolling Objects**:
The acceleration \( a \) of a rolling object on an incline can be expressed as:
\[ a = \frac{g \sin \theta}{1 + \frac{I}{MR^2}} \]
where:
- \( g \) is the acceleration due to gravity,
- \( \theta \) is the angle of the incline,
- \( I \) is the moment of inertia of the object,
- \( M \) is the mass of the object,
- \( R \) is the radius of the object.

3. **Moments of Inertia**:
- For a **solid cylinder**, the moment of inertia \( I \) is given by:
\[ I_{solid} = \frac{1}{2} MR^2 \]
- For a **hollow cylinder** (or hollow ring), the moment of inertia \( I \) is:
\[ I_{hollow} = MR^2 \]

4. **Substituting Moments of Inertia**:
- For the solid cylinder:
\[ a_{solid} = \frac{g \sin \theta}{1 + \frac{\frac{1}{2} MR^2}{MR^2}} = \frac{g \sin \theta}{1 + \frac{1}{2}} = \frac{g \sin \theta}{\frac{3}{2}} = \frac{2g \sin \theta}{3} \]
- For the hollow cylinder:
\[ a_{hollow} = \frac{g \sin \theta}{1 + \frac{MR^2}{MR^2}} = \frac{g \sin \theta}{1 + 1} = \frac{g \sin \theta}{2} \]

5. **Comparing Accelerations**:
- We can see that:
\[ a_{solid} = \frac{2g \sin \theta}{3} \quad \text{and} \quad a_{hollow} = \frac{g \sin \theta}{2} \]
- To compare these, we can find a common denominator:
- \( a_{solid} = \frac{4g \sin \theta}{6} \)
- \( a_{hollow} = \frac{3g \sin \theta}{6} \)

6. **Conclusion**:
Since \( a_{solid} > a_{hollow} \), the solid cylinder has a greater acceleration than the hollow cylinder. Therefore, the solid cylinder will reach the bottom of the incline first.

### Final Answer:
The solid cylinder will reach the bottom first.