A mass placed on an inclined place is just in equilibrium. If `mu` is coefficient of friction of the surface, then maximum inclination of the plane with the horizontal is

Options

A. `tan^(-1) mu`
B. `tan^(-1)(mu//2)`
C. `sin^(-1) mu`
D. `cos^(-1)mu`

Explanation

To solve the problem of finding the maximum inclination of an inclined plane with the horizontal when a mass is just in equilibrium, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Forces Acting on the Mass**:
- The weight of the mass (W = mg) acts vertically downward.
- This weight can be resolved into two components:
- Perpendicular to the inclined plane: \( W_{\perp} = mg \cos \theta \)
- Parallel to the inclined plane: \( W_{\parallel} = mg \sin \theta \)

2. **Determine the Normal Force**:
- The normal force (N) acting on the mass is equal to the perpendicular component of the weight:
\[ N = mg \cos \theta \]

3. **Calculate the Frictional Force**:
- The frictional force (f) acting on the mass, which opposes the motion down the incline, is given by:
\[ f = \mu N = \mu (mg \cos \theta) \]

4. **Set Up the Equilibrium Condition**:
- For the mass to be in equilibrium, the net force acting parallel to the inclined plane must be zero. Therefore, we can set up the equation:
\[ W_{\parallel} - f = 0 \]
- Substituting the expressions for \( W_{\parallel} \) and \( f \):
\[ mg \sin \theta - \mu (mg \cos \theta) = 0 \]

5. **Simplify the Equation**:
- Cancel \( mg \) from both sides (assuming \( m \neq 0 \)):
\[ \sin \theta = \mu \cos \theta \]

6. **Rearrange to Find the Relationship**:
- Dividing both sides by \( \cos \theta \):
\[ \tan \theta = \mu \]

7. **Find the Maximum Inclination Angle**:
- To find the angle \( \theta \), take the inverse tangent:
\[ \theta = \tan^{-1}(\mu) \]

### Final Result:
Thus, the maximum inclination of the plane with the horizontal is:
\[ \theta = \tan^{-1}(\mu) \]