A block of mass m is pulled on an incline surface having coefficient of friction `mu=1 &` angle of inclination `theta=30^(@)`, with the horizontal, such that required external force is minimum. The angle made by this force with the incline is :

Options

A. `45^(@)`
B. `30^(@)`
C. `75^(@)`
D. `53^(@)`

Explanation

To solve the problem, we need to analyze the forces acting on the block on the incline and determine the angle made by the external force with the incline when the required external force is minimized.

### Step 1: Identify the forces acting on the block
- The weight of the block \( W = mg \) acts vertically downward.
- The normal force \( N \) acts perpendicular to the incline.
- The frictional force \( f \) acts parallel to the incline, opposing the motion.
- An external force \( F \) is applied at an angle \( \alpha \) with the incline.

### Step 2: Resolve the forces
- The weight can be resolved into two components:
- Parallel to the incline: \( W_{\parallel} = mg \sin \theta \)
- Perpendicular to the incline: \( W_{\perpendicular} = mg \cos \theta \)

- The external force \( F \) can also be resolved:
- Parallel to the incline: \( F_{\parallel} = F \cos \alpha \)
- Perpendicular to the incline: \( F_{\perpendicular} = F \sin \alpha \)

### Step 3: Write the equations of motion
Using the equilibrium conditions along the incline and perpendicular to the incline:

1. **Along the incline:**
\[ F \cos \alpha - f - mg \sin \theta = 0 \]
The frictional force \( f \) can be expressed as \( f = \mu N \).

2. **Perpendicular to the incline:**
\[ N + F \sin \alpha - mg \cos \theta = 0 \]

### Step 4: Express the normal force
From the second equation, we can express the normal force \( N \):
\[ N = mg \cos \theta - F \sin \alpha \]

### Step 5: Substitute \( N \) in the first equation
Substituting \( N \) into the first equation gives:
\[ F \cos \alpha - \mu (mg \cos \theta - F \sin \alpha) - mg \sin \theta = 0 \]

### Step 6: Rearranging the equation
Rearranging the equation leads to:
\[ F \cos \alpha + \mu F \sin \alpha = mg \sin \theta + \mu mg \cos \theta \]

### Step 7: Factor out \( F \)
Factoring out \( F \):
\[ F(\cos \alpha + \mu \sin \alpha) = mg (\sin \theta + \mu \cos \theta) \]

### Step 8: Find the minimum force condition
To minimize \( F \), we need to maximize the term \( \cos \alpha + \mu \sin \alpha \). This occurs when:
\[ \tan \alpha = \mu \]
Thus, the angle \( \alpha \) can be calculated as:
\[ \alpha = \tan^{-1}(\mu) \]

### Step 9: Substitute the value of \( \mu \)
Given \( \mu = 1 \):
\[ \alpha = \tan^{-1}(1) = 45^\circ \]

### Conclusion
The angle made by the external force with the incline when the required external force is minimized is:
\[ \alpha = 45^\circ \]