A block of mass $$M$$ slides down on a rough inclined plane with constant velocity. The angle made by the incline plane with horizontal is $$\theta$$. The magnitude of the contact force will be :

A block of mass $$M$$ slides down a rough inclined plane with constant velocity, so its acceleration is zero and the net force on the block is zero. The forces acting on the block are the weight $$Mg$$ acting vertically downward, the normal reaction $$N$$ perpendicular to the inclined surface, and the friction force $$f$$ along the inclined surface (opposing motion and directed up the incline). The contact force between the block and the inclined surface is the resultant of the normal reaction and the friction force, and for equilibrium this contact force must balance the weight exactly, giving $$|\text{Contact force}| = Mg$$.

Verification: Along the incline one has $$f = Mg\sin\theta$$ and perpendicular to the incline $$N = Mg\cos\theta$$. Thus the magnitude of the contact force is $$\sqrt{N^2 + f^2} = \sqrt{M^2g^2\cos^2\theta + M^2g^2\sin^2\theta} = Mg\sqrt{\cos^2\theta + \sin^2\theta} = Mg\,. $$

Answer: Option A: $$Mg$$