A sphere can roll on a surface inclined at an angle `theta` if the friction coefficient is more than `(2)/(7) g sin theta`. Suppose the friction coefficient is `(1)/(7) g sin theta`, and a sphere is released from rest on the incline,

Options

A. it will stay at rest
B. it wil make pure translation motion
C. it will translate and rotate about the centre
D. the angular momentum of the sphere about its centre will remain constant

Answer

To solve the problem, we need to analyze the forces acting on the sphere when it is placed on an inclined plane with an angle \( \theta \) and a friction coefficient of \( \frac{1}{7} g \sin \theta \).

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Sphere**:
- The gravitational force acting down the incline: \( F_{\text{gravity}} = mg \sin \theta \)
- The frictional force acting up the incline: \( F_{\text{friction}} \)

2. **Determine the Condition for Rolling**:
- For the sphere to roll without slipping (pure rolling), the frictional force must provide enough torque to satisfy the rolling condition. The condition for pure rolling is given by:
\[ a = \alpha r \]
where \( a \) is the linear acceleration of the center of mass, \( \alpha \) is the angular acceleration, and \( r \) is the radius of the sphere.

3. **Calculate the Maximum Frictional Force**:
- The maximum static frictional force can be expressed as:
\[ F_{\text{friction}} \leq \mu_s N \]
where \( \mu_s \) is the coefficient of static friction and \( N \) is the normal force. The normal force on the incline is:
\[ N = mg \cos \theta \]
Therefore, the maximum frictional force is:
\[ F_{\text{friction}} \leq \frac{1}{7} g \sin \theta \cdot mg \cos \theta = \frac{1}{7} mg \sin \theta \cos \theta \]

4. **Evaluate the Forces**:
- The net force acting on the sphere along the incline when it is released is:
\[ F_{\text{net}} = mg \sin \theta - F_{\text{friction}} \]
- The net force also relates to the linear acceleration \( a \) of the sphere:
\[ F_{\text{net}} = ma \]
- Thus, we can write:
\[ mg \sin \theta - F_{\text{friction}} = ma \]

5. **Determine the Angular Acceleration**:
- The torque due to friction about the center of mass is:
\[ \tau = F_{\text{friction}} \cdot r \]
- This torque causes an angular acceleration \( \alpha \):
\[ \tau = I \alpha \]
where \( I \) is the moment of inertia of the sphere, given by \( I = \frac{2}{5} m r^2 \).

6. **Conclusion**:
- Since the friction coefficient is less than \( \frac{2}{7} g \sin \theta \), the sphere will not roll without slipping. Instead, it will translate and rotate, but not perform pure rolling.
- Thus, the correct option is that the sphere will translate and rotate.

### Final Answer:
The sphere will translate and rotate, but it will not perform pure rolling.