Solid cylinders of radii `r_(1),r_(2)` and `r_(3)` roll down an inclined plane from the same place simultaneously. `r_(1)gtr_(2)gtr_(3)`, which one would reach the bottom first

Options

A. Cylinder of radius `r_(1)`
B. Cylinder of radius `r_(2)`
C. Cylinder of radius `r_(3)`
D. All the three cylinders simultaneously

Answer

To determine which solid cylinder reaches the bottom of the inclined plane first, we need to analyze the relationship between the radius of the cylinders and their moments of inertia, as well as how these factors affect their acceleration down the incline.

### Step-by-Step Solution:

1. **Understanding the Problem**: We have three solid cylinders with radii \( r_1, r_2, \) and \( r_3 \) such that \( r_1 > r_2 > r_3 \). We need to find out which cylinder reaches the bottom of the incline first.

2. **Moment of Inertia**: The moment of inertia \( I \) for a solid cylinder about its axis is given by the formula:
\[ I = \frac{1}{2} m r^2 \]
where \( m \) is the mass of the cylinder and \( r \) is its radius.

3. **Acceleration on the Incline**: When a solid cylinder rolls down an incline, its acceleration \( a \) can be expressed as:
\[ a = \frac{g \sin \theta}{1 + \frac{I}{m r^2}} \]
Substituting the moment of inertia for a solid cylinder:
\[ a = \frac{g \sin \theta}{1 + \frac{\frac{1}{2} m r^2}{m r^2}} = \frac{g \sin \theta}{1 + \frac{1}{2}} = \frac{g \sin \theta}{\frac{3}{2}} = \frac{2g \sin \theta}{3} \]

4. **Comparing Accelerations**: Since the acceleration is affected by the moment of inertia, we can see that:
\[ a \propto \frac{1}{1 + \frac{I}{m r^2}} \propto \frac{1}{1 + \frac{1}{2}} = \frac{2}{3} \]
This means that the larger the moment of inertia, the smaller the acceleration.

5. **Determining Moment of Inertia for Each Cylinder**:
- For cylinder 1 (radius \( r_1 \)): \( I_1 = \frac{1}{2} m r_1^2 \)
- For cylinder 2 (radius \( r_2 \)): \( I_2 = \frac{1}{2} m r_2^2 \)
- For cylinder 3 (radius \( r_3 \)): \( I_3 = \frac{1}{2} m r_3^2 \)

Since \( r_1 > r_2 > r_3 \), it follows that:
\[ I_1 > I_2 > I_3 \]

6. **Relating Acceleration to Moment of Inertia**: The acceleration is inversely proportional to the moment of inertia:
\[ a_1 < a_2 < a_3 \]
This indicates that the cylinder with the smallest moment of inertia (which corresponds to the smallest radius) will have the greatest acceleration.

7. **Conclusion**: Since \( r_3 \) is the smallest radius, it has the smallest moment of inertia and thus the largest acceleration. Therefore, the cylinder with radius \( r_3 \) will reach the bottom of the incline first.

### Final Answer:
The cylinder with radius \( r_3 \) will reach the bottom first.