865Radius of rotation at 45° latitude is re√2. So net centrifugal force on the plank is mreω2√2. Now the Earth's surface at that point is tilted at an angle of 45° to the direction of the centrifugal force. So the component of the cemtrifuhal force along the Earth's surface is mreω22. This is equal to the friction force as the plank is at rest. So C is the answer.
