The string holding the lift given below is cut. What will be the motion of the ring (as seen from inside the lift) initially at rest on top of the incline? The incline surface is rough.

Stay at its initial position

When the string is cut, the entire system will be in free fall. The only force acting on the ring is in the vertically downward direction. When this is observed from inside the lift, even the observer is in free fall, so according to him the net force on the ring is zero and hence, it will stay at rest at its initial position. Note that for an observer on the ground the ring will be falling vertically downwards.