A cubical block of mass M and edge a slides down a rougg inclined plane of inclination `theta` with a uniform velocity. The torque of the normal force on the block about its centre has magnitude.

Explanation

To solve the problem of finding the torque of the normal force on a cubical block sliding down a rough inclined plane, we can follow these steps:

### Step-by-Step Solution

1. **Identify the Forces Acting on the Block**:
- The forces acting on the block are its weight (mg), the normal force (N), and the frictional force (F). The weight can be resolved into two components: one parallel to the incline (mg sin θ) and one perpendicular to the incline (mg cos θ).

2. **Understand the Condition of Uniform Velocity**:
- Since the block is sliding down with uniform velocity, the net force acting along the incline is zero. This means that the frictional force (F) must balance the component of the gravitational force acting down the incline:
\[ F = mg \sin \theta \]

3. **Determine the Torque Due to the Frictional Force**:
- The torque (τ) due to the frictional force about the center of mass of the block can be calculated. The frictional force acts at the bottom edge of the block, which is at a distance of \( \frac{a}{2} \) from the center of mass. Therefore, the torque due to friction is:
\[ \tau_{\text{friction}} = F \cdot \frac{a}{2} \]
- Substituting the expression for frictional force:
\[ \tau_{\text{friction}} = (mg \sin \theta) \cdot \frac{a}{2} \]

4. **Analyze the Torque Due to the Normal Force**:
- The normal force acts perpendicular to the surface of the incline. The distance from the center of mass to the line of action of the normal force is also \( x \), which we need to find. However, since the block is in equilibrium (uniform velocity), the net torque about the center of mass must be zero:
\[ \tau_{\text{normal}} + \tau_{\text{friction}} = 0 \]
- Therefore, we can express the torque due to the normal force as:
\[ \tau_{\text{normal}} = -\tau_{\text{friction}} \]

5. **Calculate the Magnitude of the Torque of the Normal Force**:
- Since we have established that the torque due to the normal force balances the torque due to friction, we can write:
\[ \tau_{\text{normal}} = -\left(mg \sin \theta \cdot \frac{a}{2}\right) \]
- The magnitude of the torque of the normal force about the center of mass is:
\[ |\tau_{\text{normal}}| = mg \sin \theta \cdot \frac{a}{2} \]

### Final Answer
The magnitude of the torque of the normal force on the block about its center is:
\[ |\tau_{\text{normal}}| = \frac{mg \sin \theta \cdot a}{2} \]