A body of mass M kg rests on a plane inclined at an angle $\theta$ to the horizontal. The component of the weight of the body parallel to the plane is

Analyze the question: A body of mass M sits on an incline of angle θ. We need the part of its weight that acts along (parallel to) the slope.

Concept: The weight W = Mg acts vertically downward. On an incline, we resolve it into two perpendicular components:

• Parallel to the plane: this pulls the body down the slope.
• Perpendicular to the plane: this presses the body into the surface and is balanced by the normal reaction.

Solution: The angle between the weight (vertical) and the perpendicular to the plane is equal to the angle of the incline, θ. So:

• Component perpendicular to the plane = Mg cos θ
• Component parallel to the plane = Mg sin θ

Check with limiting cases:

• If θ = 0° (flat surface), there is no tendency to slide, so the parallel component should be 0. sin 0° = 0 ✓.
• If θ = 90° (vertical), the whole weight acts along the "plane", so the component should be Mg. sin 90° = 1 ✓.

So the parallel component is Mg sin θ.

Answer: A. Mg sin θ.

Tip: For an incline, remember "Slope Sine": the component along the slope uses sin θ, and the one perpendicular uses cos θ. Quick check: at θ = 0, nothing slides, so the parallel part must be zero, which only sin gives.