A

To find the angle \( \beta \) for minimum tension in the thread, we need to analyze the forces acting on the bar.

**Step 1: Identify forces acting on the bar**
- Weight of the bar acting downward: \( W = mg \)
- Normal force acting perpendicular to the inclined plane: \( N \)
- Frictional force acting parallel to the incline: \( F_f = kN \)
- Tension in the thread acting at angle \( \beta \) with respect to the incline.

**Step 2: Resolve the forces**
Resolving the weight \( mg \) into components:
- Perpendicular to the incline: \( mg \cos(\alpha) \)
- Parallel to the incline: \( mg \sin(\alpha)

The normal force is affected by the component of weight and the vertical component of tension:
\( N + T \sin(\beta) = mg \cos(\alpha) \)

**Step 3: Apply the equilibrium conditions**
In equilibrium (no acceleration), the net force along the incline must be zero:
\( T \cos(\beta) - mg \sin(\alpha) - F_f = 0 \)
Substituting \( F_f = kN \):
\( T \cos(\beta) - mg \sin(\alpha) - k(mg \cos(\alpha) - T \sin(\beta)) = 0 \)

Rearranging leads to an expression for \( T \).
For minimal tension, we can differentiate this equation with respect to \( \beta \) and set the derivative to zero:
\( \frac{dT}{d\beta} = 0 \).
This leads to the critical angle defined by the relationship between \( \alpha \) and coefficient of friction \( k \):
\( \tan(\beta) = \frac{k + \tan(\alpha)}{1} \), giving us the optimal angle.

Therefore, the angle \( \beta \) is found to be:
\( \tan(\beta) = k + \tan(\alpha) \)

This corresponds to the scenario where the tension is minimized. Hence, the value of \( \beta \) is defined concerning the angle of inclination \( \alpha \) and the coefficient of friction \( k \):
\( \beta = \tan^{-1}(k + \tan(\alpha)) \)

Therefore, the required answer is \( \beta \), depending on \( k \) and the inclination \( \alpha \).