When a solid sphere rolls without slipping down an inclined plane making an angle `theta` with the horizontal, the acceleration of its centre of mass is `a`. If the same sphere slides without friction, its.

Options

A. `(7//2)` a
B. `(5//7)` a
C. `(7//5)` a
D. `(5//2)` a

Answer

To solve the problem, we need to find the relationship between the acceleration of a solid sphere rolling down an inclined plane without slipping and the acceleration of the same sphere sliding down the plane without friction.

### Step-by-Step Solution:

1. **Understand the Motion**:
- When a solid sphere rolls down an inclined plane without slipping, it exhibits both translational and rotational motion.
- The acceleration of the center of mass of the rolling sphere is denoted as \( a \).

2. **Acceleration of a Rolling Sphere**:
- The formula for the acceleration of a rolling body is given by:
\[ a = \frac{g \sin \theta}{1 + \frac{k^2}{r^2}} \]
- Here, \( g \) is the acceleration due to gravity, \( \theta \) is the angle of the incline, \( k \) is the radius of gyration, and \( r \) is the radius of the sphere.

3. **Moment of Inertia of a Solid Sphere**:
- The moment of inertia \( I \) of a solid sphere about its center is:
\[ I = \frac{2}{5} m r^2 \]
- The radius of gyration \( k \) is calculated as:
\[ k = \sqrt{\frac{I}{m}} = \sqrt{\frac{\frac{2}{5} m r^2}{m}} = \sqrt{\frac{2}{5}} r \]

4. **Substituting \( k \) into the Acceleration Formula**:
- Substitute \( k \) into the acceleration formula:
\[ a = \frac{g \sin \theta}{1 + \frac{\left(\sqrt{\frac{2}{5}} r\right)^2}{r^2}} = \frac{g \sin \theta}{1 + \frac{2}{5}} = \frac{g \sin \theta}{\frac{7}{5}} = \frac{5 g \sin \theta}{7} \]

5. **Acceleration of a Sliding Sphere**:
- If the sphere slides down the incline without friction, its acceleration is simply the component of gravitational acceleration along the incline:
\[ a_s = g \sin \theta \]

6. **Finding the Ratio of Accelerations**:
- The ratio of the acceleration of the sliding sphere to the rolling sphere is:
\[ \frac{a_s}{a} = \frac{g \sin \theta}{\frac{5 g \sin \theta}{7}} = \frac{7}{5} \]
- Therefore, we can express the acceleration of the sliding sphere in terms of the acceleration of the rolling sphere:
\[ a_s = \frac{7}{5} a \]

### Final Result:
If the acceleration of the rolling solid sphere is \( a \), then the acceleration of the sliding solid sphere is:
\[ a_s = \frac{7}{5} a \]