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, we need to analyze the forces acting on the cube resting on a 45-degree inclined surface and determine the point of application of the normal reaction force.

### Step-by-step Solution:

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

2. **Determine the Components of the Weight**:
- The component of the weight acting parallel to the incline is \( mg \sin \theta \).
- The component of the weight acting perpendicular to the incline is \( mg \cos \theta \).
- Since the incline is at \( 45^\circ \), we have \( \sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}} \).

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

4. **Moment About the Center of the Cube**:
- The normal force \( N \) acts at a distance \( x \) from the center of the cube.
- The moment due to the normal force about the center is \( N \cdot x \).
- The moment due to the frictional force about the center is \( f \cdot \frac{a}{2} \) (where \( a \) is the side length of the cube, and the distance from the center to the edge is \( \frac{a}{2} \)).
- Setting the moments equal for equilibrium:
\[ N \cdot x = f \cdot \frac{a}{2} \]

5. **Substituting the Forces**:
- Substitute \( N = mg \cos \theta \) and \( f = mg \sin \theta \):
\[ (mg \cos \theta) \cdot x = (mg \sin \theta) \cdot \frac{a}{2} \]
- Cancel \( mg \) from both sides:
\[ \cos \theta \cdot x = \sin \theta \cdot \frac{a}{2} \]

6. **Solving for \( x \)**:
- Rearranging gives:
\[ x = \frac{\sin \theta \cdot \frac{a}{2}}{\cos \theta} \]
- Since \( \tan \theta = \frac{\sin \theta}{\cos \theta} \):
\[ x = \frac{a}{2} \tan \theta \]
- For \( \theta = 45^\circ \), \( \tan 45^\circ = 1 \):
\[ x = \frac{a}{2} \]

7. **Finding the Distance from the Lower Edge**:
- The center of the cube is at a distance of \( \frac{a}{2} \) from the lower edge.
- Since \( x = \frac{a}{2} \), the point of application of the normal reaction force is at the lower edge of the cube:
\[ \text{Distance from the lower edge} = \frac{a}{2} - \frac{a}{2} = 0 \]

### Final Answer:
The distance of the point of application of the normal reaction measured from the lower edge of the cube is **0**.