A uniform cube of mass `m` and slide`a` is resting in equilibrium on a rough `45^(@)` inclined surface. The distance point of application of normal reation measured from the lower edge of the cube is

Options

A. zero
B. `(a)/(3)`
C. `(a)/(sqrt2)`
D. `(a)/(4)`

Explanation

To solve the problem of finding the distance from the lower edge of the cube to the point of application of the normal reaction on a rough inclined surface, we will follow these steps:

### Step-by-Step Solution:

1. **Identify Forces Acting on the Cube**:
- The weight of the cube, \( mg \), acts vertically downwards.
- The normal force \( N \) acts perpendicular to the inclined surface.
- The frictional force \( f \) acts parallel to the inclined surface, opposing the tendency of the cube to slide down.

2. **Determine the Angles**:
- The inclined surface makes an angle \( \theta = 45^\circ \) with the horizontal.

3. **Break Down the Weight into Components**:
- The component of the weight acting down the incline is \( mg \sin \theta \).
- The component of the weight acting perpendicular to the incline is \( mg \cos \theta \).

4. **Equilibrium Conditions**:
- Since the cube is in equilibrium, the sum of forces in both the perpendicular and parallel directions must be zero.
- In the perpendicular direction:
\[ N = mg \cos \theta \]
- In the parallel direction:
\[ f = mg \sin \theta \]

5. **Static Friction**:
- The frictional force \( f \) is equal to \( mg \sin \theta \) because the cube is not moving.

6. **Calculate the Moments about the Center of the Cube**:
- Let \( x \) be the distance from the center of the cube to the point of application of the normal force.
- The moment due to friction about the center (C) is:
\[ \text{Moment due to friction} = f \cdot \frac{A}{2} = mg \sin \theta \cdot \frac{A}{2} \]
- The moment due to the normal force is:
\[ \text{Moment due to normal} = N \cdot x = mg \cos \theta \cdot x \]

7. **Set Up the Moment Equation**:
- Since the cube is in equilibrium, the sum of moments about point C must be zero:
\[ mg \sin \theta \cdot \frac{A}{2} - mg \cos \theta \cdot x = 0 \]

8. **Solve for \( x \)**:
- Rearranging the equation gives:
\[ mg \sin \theta \cdot \frac{A}{2} = mg \cos \theta \cdot x \]
- Dividing both sides by \( mg \) (which cancels out) and substituting \( \theta = 45^\circ \) (where \( \sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}} \)):
\[ \frac{A}{2} = x \]
- Thus, we find:
\[ x = \frac{A}{2} \]

9. **Determine the Distance from the Lower Edge**:
- The distance from the lower edge of the cube to the point of application of the normal reaction is:
\[ \text{Distance from lower edge} = A - x = A - \frac{A}{2} = \frac{A}{2} \]
- Since the normal force acts at the lower edge, the distance from the lower edge is effectively:
\[ 0 \]

### Final Answer:
The distance from the lower edge of the cube to the point of application of the normal reaction is \( 0 \).