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

Explanation

To solve the problem, we need to find the acceleration of a solid sphere sliding down an inclined plane without friction, given that its acceleration when rolling without slipping is `a`. Here’s a step-by-step solution:

### Step 1: Understand the scenario when the sphere rolls without slipping
When the solid sphere rolls down the inclined plane without slipping, the forces acting on it are:
- The gravitational force acting downwards, which can be resolved into two components:
- Perpendicular to the incline: \( mg \cos \theta \)
- Parallel to the incline: \( mg \sin \theta \)
- The frictional force acting up the incline.

### Step 2: Write the equation of motion for the rolling sphere
Using Newton's second law, the net force acting on the sphere along the incline can be expressed as:
\[ mg \sin \theta - f = ma \]
where \( f \) is the frictional force and \( a \) is the acceleration of the center of mass of the sphere.

### Step 3: Write the rotational motion equation
For rolling motion, the frictional force also causes a torque about the center of mass:
\[ f \cdot R = I \cdot \alpha \]
where \( R \) is the radius of the sphere, \( I \) is the moment of inertia of the sphere, and \( \alpha \) is the angular acceleration. For a solid sphere, \( I = \frac{2}{5} m R^2 \) and \( \alpha = \frac{a}{R} \) (since \( a = R \alpha \)).

Substituting for \( I \) and \( \alpha \):
\[ f \cdot R = \frac{2}{5} m R^2 \cdot \frac{a}{R} \]
This simplifies to:
\[ f = \frac{2}{5} m a \]

### Step 4: Substitute the expression for friction into the equation of motion
Now substitute \( f \) back into the equation of motion:
\[ mg \sin \theta - \frac{2}{5} m a = ma \]
Rearranging gives:
\[ mg \sin \theta = ma + \frac{2}{5} m a \]
\[ mg \sin \theta = \frac{7}{5} ma \]

### Step 5: Solve for the acceleration \( a \)
Canceling \( m \) from both sides:
\[ g \sin \theta = \frac{7}{5} a \]
Thus, the acceleration \( a \) when rolling without slipping is:
\[ a = \frac{5}{7} g \sin \theta \]

### Step 6: Analyze the case when the sphere slides without friction
When the sphere slides down the incline without friction, the only force acting along the incline is the gravitational component:
\[ mg \sin \theta = ma' \]
where \( a' \) is the acceleration of the sphere when sliding.

### Step 7: Solve for the acceleration \( a' \)
Canceling \( m \) gives:
\[ g \sin \theta = a' \]
Thus, the acceleration \( a' \) when sliding without friction is:
\[ a' = g \sin \theta \]

### Step 8: Relate \( a' \) to \( a \)
From our earlier result, we have:
\[ g \sin \theta = \frac{7}{5} a \]
Substituting this into the equation for \( a' \):
\[ a' = g \sin \theta = \frac{7}{5} a \]

### Conclusion
The acceleration of the sphere when it slides down the incline without friction is:
\[ a' = \frac{7}{5} a \]