A ring, cylinder and solid sphere are placed on the top of a rough incline on which the sphere can just roll without slipping. When all of them are released at the same instant from the same position, then

Options

A. all of them reach the ground at the same instant
B. the sphere reaches first and the ring at the last
C. the sphere reaches first and the cylinder and ring reach together
D. none of the above

Explanation

To solve the problem of determining which object (ring, cylinder, or solid sphere) reaches the bottom of a rough incline first when released from the same height, we need to analyze the motion of each object considering both translational and rotational dynamics.

### Step-by-Step Solution:

1. **Identify Forces Acting on the Objects**:
Each object experiences gravitational force acting down the incline (mg sin θ) and frictional force acting up the incline. The frictional force is necessary for rolling without slipping.

2. **Set Up the Equations of Motion**:
For each object, we can establish the following equations:
- Translational motion: \( mg \sin \theta - F = ma \)
- Rotational motion: \( F \cdot r = I \cdot \alpha \)

Here, \( F \) is the frictional force, \( I \) is the moment of inertia, \( r \) is the radius, \( a \) is the linear acceleration, and \( \alpha \) is the angular acceleration.

3. **Relate Linear and Angular Acceleration**:
For rolling without slipping, the relationship between linear acceleration \( a \) and angular acceleration \( \alpha \) is given by:
\[ a = \alpha r \]

4. **Substitute Angular Acceleration**:
From the rotational equation, we can express \( F \) in terms of \( a \):
\[ F = \frac{I \cdot a}{r^2} \]
Substituting this into the translational motion equation gives:
\[ mg \sin \theta - \frac{I \cdot a}{r^2} = ma \]

5. **Solve for Acceleration \( a \)**:
Rearranging the equation leads to:
\[ mg \sin \theta = ma + \frac{I \cdot a}{r^2} \]
\[ mg \sin \theta = a \left( m + \frac{I}{r^2} \right) \]
Thus, the acceleration \( a \) can be expressed as:
\[ a = \frac{mg \sin \theta}{m + \frac{I}{r^2}} \]

6. **Calculate Moment of Inertia for Each Object**:
- Ring: \( I = mr^2 \)
- Cylinder: \( I = \frac{1}{2} mr^2 \)
- Solid Sphere: \( I = \frac{2}{5} mr^2 \)

7. **Substitute Moments of Inertia**:
- For the ring:
\[ a_{\text{ring}} = \frac{mg \sin \theta}{m + \frac{mr^2}{r^2}} = \frac{g \sin \theta}{1 + 1} = \frac{g \sin \theta}{2} \]
- For the cylinder:
\[ a_{\text{cylinder}} = \frac{mg \sin \theta}{m + \frac{1}{2}mr^2/r^2} = \frac{g \sin \theta}{1 + \frac{1}{2}} = \frac{g \sin \theta}{\frac{3}{2}} = \frac{2g \sin \theta}{3} \]
- For the solid sphere:
\[ a_{\text{sphere}} = \frac{mg \sin \theta}{m + \frac{2}{5}mr^2/r^2} = \frac{g \sin \theta}{1 + \frac{2}{5}} = \frac{g \sin \theta}{\frac{7}{5}} = \frac{5g \sin \theta}{7} \]

8. **Compare Accelerations**:
- \( a_{\text{ring}} = \frac{g \sin \theta}{2} \)
- \( a_{\text{cylinder}} = \frac{2g \sin \theta}{3} \)
- \( a_{\text{sphere}} = \frac{5g \sin \theta}{7} \)

The order of acceleration from highest to lowest is:
\[ a_{\text{sphere}} > a_{\text{cylinder}} > a_{\text{ring}} \]

9. **Conclusion**:
Since the object with the highest acceleration reaches the bottom first, the solid sphere will reach the bottom first, followed by the cylinder, and lastly the ring.

### Final Answer:
The order in which they reach the bottom is: **Sphere > Cylinder > Ring**.