Statement-I : When a sphere and a solid cylinder are allowed to roll down an inclined plane, the sphere will reach the ground first even if the mass and radius of the two bodies are different.
Statement-II : The acceleration of the body rolling down the inclined plane is directly proportional to the radius of the rolling body.

Options

A. If both Statement-I and Statement-II are true, and Statement - II is the correct explanation of Statement– I.
B. If both Statement-I and Statement-II are true but Statement - II is not the correct explanation of Statement – I.
C. If Statement-I is true but Statement-II is false.
D. If Statement-I is false but Statement-II is true.

Answer

To solve the problem, we need to analyze both statements provided in the question.

### Step 1: Understand the Motion of Rolling Bodies
When a solid sphere and a solid cylinder roll down an inclined plane, they experience both translational and rotational motion. The acceleration of each body depends on its moment of inertia.

### Step 2: Calculate the Moment of Inertia
The moment of inertia (I) for the two objects is given by:
- For a solid sphere: \( I = \frac{2}{5} m r^2 \)
- For a solid cylinder: \( I = \frac{1}{2} m r^2 \)

### Step 3: Derive the Acceleration of the Center of Mass
The acceleration of the center of mass (a) for a rolling object down an incline can be expressed as:
\[ a = \frac{g \sin \theta}{1 + \frac{I}{m r^2}} \]
Substituting the moments of inertia:
- For the sphere:
\[ a_{\text{sphere}} = \frac{g \sin \theta}{1 + \frac{\frac{2}{5} m r^2}{m r^2}} = \frac{g \sin \theta}{1 + \frac{2}{5}} = \frac{g \sin \theta}{\frac{7}{5}} = \frac{5g \sin \theta}{7} \]
- For the cylinder:
\[ a_{\text{cylinder}} = \frac{g \sin \theta}{1 + \frac{\frac{1}{2} m r^2}{m r^2}} = \frac{g \sin \theta}{1 + \frac{1}{2}} = \frac{g \sin \theta}{\frac{3}{2}} = \frac{2g \sin \theta}{3} \]

### Step 4: Compare the Accelerations
Now we can compare the two accelerations:
- \( a_{\text{sphere}} = \frac{5g \sin \theta}{7} \)
- \( a_{\text{cylinder}} = \frac{2g \sin \theta}{3} \)

To compare these, we can find a common denominator:
- \( a_{\text{sphere}} = \frac{15g \sin \theta}{21} \)
- \( a_{\text{cylinder}} = \frac{14g \sin \theta}{21} \)

Since \( \frac{15g \sin \theta}{21} > \frac{14g \sin \theta}{21} \), we conclude that:
\[ a_{\text{sphere}} > a_{\text{cylinder}} \]
This means the sphere accelerates faster than the cylinder and will reach the ground first.

### Step 5: Analyze Statement II
Statement II claims that the acceleration of the body rolling down the inclined plane is directly proportional to the radius of the rolling body. This is incorrect because the acceleration depends on the moment of inertia (which involves the radius) but is not directly proportional to the radius itself. The acceleration is influenced by the shape of the object (through the moment of inertia) rather than being directly proportional to the radius.

### Conclusion
- **Statement I** is true: The sphere reaches the ground first.
- **Statement II** is false: The acceleration is not directly proportional to the radius.

### Final Answer
- Statement I is true.
- Statement II is false.