A hollow sphere of mass m and radius R is rolling downdard on a rough inclined plane of inclination `theta`. If the coefficient of friction between the hollow sphere and incline is `mu`, then

Options

A. Friction opposes its translation
B. Friction supports rotation motion
C. On decreasing `theta`, frictional force decreases
D. All of these

Answer

To solve the problem of a hollow sphere rolling down a rough inclined plane, we will analyze the forces acting on the sphere and how they relate to the motion.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Sphere**:
- The gravitational force acting downward: \( F_g = mg \)
- The normal force acting perpendicular to the inclined plane: \( N = mg \cos(\theta) \)
- The frictional force acting up the incline: \( F_f = \mu N = \mu mg \cos(\theta) \)

2. **Resolve the Gravitational Force**:
- The component of gravitational force acting down the incline:
\[ F_{\text{down}} = mg \sin(\theta) \]

3. **Apply Newton's Second Law**:
- For translational motion down the incline:
\[ mg \sin(\theta) - F_f = ma \]
- Substituting \( F_f \):
\[ mg \sin(\theta) - \mu mg \cos(\theta) = ma \]

4. **Consider Rotational Motion**:
- The frictional force also provides torque for the rotational motion of the sphere. The torque (\( \tau \)) about the center of mass is given by:
\[ \tau = F_f \cdot R = \mu mg \cos(\theta) \cdot R \]
- The moment of inertia (\( I \)) of a hollow sphere is:
\[ I = \frac{2}{3} m R^2 \]
- The angular acceleration (\( \alpha \)) is related to the linear acceleration (\( a \)) by:
\[ a = R \alpha \]
- Therefore, we can express the torque in terms of angular acceleration:
\[ \tau = I \alpha = \frac{2}{3} m R^2 \cdot \frac{a}{R} = \frac{2}{3} m R a \]

5. **Set Up the Equations**:
- From the torque equation:
\[ \mu mg \cos(\theta) R = \frac{2}{3} m R a \]
- Cancel \( mR \) (assuming \( R \neq 0 \)):
\[ \mu g \cos(\theta) = \frac{2}{3} a \]

6. **Solve for Acceleration**:
- Rearranging gives:
\[ a = \frac{3}{2} \mu g \cos(\theta) \]

### Conclusion:
The acceleration of the hollow sphere rolling down the inclined plane is given by:
\[ a = \frac{3}{2} \mu g \cos(\theta) \]