A block is at rest on an inclined plane making an angle `alpha` with the horizontal. As the angle of the incline is increased the block start slipping when the angle of inclination becomes `theta` then coefficient of friction is equal to

Options

A. `sintheta`
B. `costheta`
C. `tantheta`
D. Independent of `theta`

Answer

To solve the problem, we need to determine the coefficient of static friction (μ) when a block starts slipping on an inclined plane at an angle θ.

### Step-by-Step Solution:

1. **Identify Forces Acting on the Block:**
- The block experiences gravitational force (mg) acting downwards.
- This force can be resolved into two components:
- Perpendicular to the inclined plane: \( mg \cos \theta \)
- Parallel to the inclined plane: \( mg \sin \theta \)

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

3. **Static Friction Force:**
- The static friction force (F_s) acts up the incline to prevent the block from slipping. The maximum static friction force can be expressed as:
\[ F_s = \mu_s N \]
- Substituting the expression for the normal force, we get:
\[ F_s = \mu_s (mg \cos \theta) \]

4. **Setting Up the Equation:**
- At the point of slipping, the static friction force equals the component of gravitational force acting down the incline:
\[ \mu_s (mg \cos \theta) = mg \sin \theta \]

5. **Canceling Common Terms:**
- We can cancel \( mg \) from both sides (assuming \( m \neq 0 \)):
\[ \mu_s \cos \theta = \sin \theta \]

6. **Solving for Coefficient of Friction:**
- Rearranging the equation gives:
\[ \mu_s = \frac{\sin \theta}{\cos \theta} \]
- This simplifies to:
\[ \mu_s = \tan \theta \]

### Final Answer:
The coefficient of static friction (μ) when the block starts slipping is:
\[ \mu_s = \tan \theta \]