A cylinder will slip on an inclined plane of inclination angle $$\theta $$ if the coefficient of static friction between plane and cylinder is A. less than $$\frac{1}{3}\tan \theta $$ B. less than $$\frac{2}{3}\tan \theta $$ C. less than $$\frac{1}{3}\sin \theta $$ D. less than $$\frac{2}{3}\sin \theta $$

The correct answer is: A. less than $\frac{1}{3}\tan \theta$

The coefficient of static friction is the ratio of the maximum static friction force to the normal force. The normal force is the force perpendicular to the surface of contact. The maximum static friction force is the maximum force that can be applied to an object without causing it to move.

When an object is placed on an inclined plane, the force of gravity pulls it down the plane. The normal force is the force that the plane exerts on the object to prevent it from falling through the plane. The static friction force is the force that opposes the motion of the object down the plane.

If the coefficient of static friction is less than $\frac{1}{3}\tan \theta$, then the static friction force will be less than the force of gravity. In this case, the object will slide down the plane.

If the coefficient of static friction is greater than $\frac{1}{3}\tan \theta$, then the static friction force will be greater than the force of gravity. In this case, the object will not slide down the plane.

The other options are incorrect because they do not take into account the coefficient of static friction.