A block of mass `m` rests on a fixed incline plane of inclination `theta` with horizontal Assume friction is large enough to make the block stationary. Then choose correct alternative (s)

Options

A. Maximum value of horizontal component of friction is mg
B. angle `theta` at which horizontal component of friction is maximum is `(pi)/(4)`
C. Maximum value of horizontal component of friction is `(mg)/(2)`
D. Angle `theta` at which horizontal component of friction is maximum is `(pi)/(3)`

Explanation

To solve the problem, we need to analyze the forces acting on the block resting on the inclined plane. Let's go through the steps systematically:

### Step 1: Identify the Forces Acting on the Block
The forces acting on the block are:
- The weight of the block, \( mg \), acting vertically downward.
- The normal force \( N \) acting perpendicular to the surface of the incline.
- The frictional force \( F \) acting parallel to the incline, opposing the motion of the block.

### Step 2: Resolve the Weight into Components
The weight \( mg \) can be resolved into two components:
- The component parallel to the incline: \( mg \sin \theta \)
- The component perpendicular to the incline: \( mg \cos \theta \)

### Step 3: Set Up the Equilibrium Condition
Since the block is stationary, the net force acting on it must be zero. Therefore, the frictional force must balance the component of the weight acting down the incline:
\[ F = mg \sin \theta \tag{1} \]

### Step 4: Determine the Horizontal Component of the Frictional Force
The frictional force \( F \) can also be resolved into horizontal and vertical components. The horizontal component \( F_H \) can be expressed as:
\[ F_H = F \cos \theta \]
Substituting equation (1) into this, we get:
\[ F_H = (mg \sin \theta) \cos \theta \]
Using the trigonometric identity \( \sin 2\theta = 2 \sin \theta \cos \theta \), we can rewrite this as:
\[ F_H = \frac{mg}{2} \sin 2\theta \tag{2} \]

### Step 5: Find the Maximum Horizontal Component of Friction
To find the maximum horizontal component of friction, we need to maximize \( \sin 2\theta \). The maximum value of \( \sin 2\theta \) is 1, which occurs when:
\[ 2\theta = 90^\circ \quad \Rightarrow \quad \theta = 45^\circ \]
Thus, substituting this into equation (2):
\[ F_H^{max} = \frac{mg}{2} \cdot 1 = \frac{mg}{2} \]

### Step 6: Conclusion
The maximum horizontal component of friction is \( \frac{mg}{2} \), and this occurs at an angle \( \theta = 45^\circ \).

### Final Answers
- The maximum horizontal component of friction is \( \frac{mg}{2} \).
- The angle at which this maximum occurs is \( 45^\circ \).

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