- The reason behind buoyancy is the pressure difference in fluids. More specifically, the difference in hydrostatic pressure on different levels, since the hydrostatic pressure of water increases with depth ($p=\rho gh$ wher $h$ is the distance below the surface). The part of the body which is subject to higher hydrostatic pressure will be pushed more upwards than the part of the body subject to lower hydrostatic pressure will be pushed downwards. Also, in space there would be no buoyancy because there would be no gravity, hence no difference in hydrostatic pressure.
- The answer is actually A. Since, as you said, no water gets below the cube, there will be no force directed upwards due to pressure. The liquid will actually push the cube downwards, to the floor, so the force on the bottom of the vessel in contact with the cube will be the cube's weight + the weight of the liquid above.
- Yes, since the reason behind buoyancy is the pressure difference, if there's no water below the cube, there will actually be no buoyant force. The liquid will push the cube downwards.
- I would argue that the cube does actually displace water (if you removed it, the water level would decline), but in this case you're right when you treat it as a part of the vessel, since there's no water below it. Every body immersed in a fluid will displace some part of it (because the fluid was there in the first place, when you immersed the body it had to go somewhere), but the Archimedes principle saying the force on the body will be equal to the weight of the displaced fluid is only right when the body is completely immersed in it. Actually, the derivation of this law is fairly simple:
Let's consider a volume of water $V$, let it be in the shape of, for instance, a potato. The water is subject to two forces: its weight and the buoyant force. Since it doesn't move, we conclude that the buoyant force is equal the the weight of $V$ of water. Now, let's say we displace the water with a potato of the same shape and volume $V$ - it occupies exactly the same space as the water discussed. Since the buoyant force is the result of the difference in pressure and does not depend on any characteristics of the body, it will be exactly the same as it was with the water volume. That's because the pressure around the body doesn't change. That's why we conclude the bouyant force is equal to the weight of the displaced fluid.
In the elevator scenario, the elevator frame is getting accelerated; hence, the when you draw the free-body diagram, with respect to the elevator, the pseudo force acts downwards (opposite to the direction in which the frame is getting accelerated). Hence, the apparent weight increases as the pseudo force gets added up with the weight of the person.
Suppose the acceleration of the elevator is $a$ and the mass of the body is $m$, then the apparent weight of the body in the elevator frame is -
$$
N = m(a + g)
$$
In the second scenario, the buoyant force acts in the upward direction, because the buoyant force is always directed against the pressure gradient i.e, the direction in which pressure decreases. (Much like an electric field directed in the direction where the potential decreases)
Of course, the buoyant force exerted is equal to the weight of the fluid displaced by the body (Which is the Archimedes principle); but -
Drawing the FBD in the second case yields the weight of the body acting downwards, and the buoyant force acting upwards. This results in the weight decreasing (since the buoyant force is subtracted from the weight, not added up with it), and not increasing.
Say, the buoyant force acting on the body is $B$ and the actual weight is $W$, the net weight of the body (acting in the downward direction) then would be -
$$
W' = W - B
$$
Which is why the apparent weight of the body in the liquid decreases.
(This is considering that the density of the body is greater than the density of the liquid, in the case where it is opposite (the body doesn't sink; but floats partially), the signs of $W$ and $B$ are swapped and the net force is acting in the upward direction. In another scenario where the the weight of the the body is equal to the buoyant force, the net force on the body then is zero, hence it floats being completely submerged)
Keep in mind that a body loses weight in a liquid which is equal to the weight of the liquid displaced by it/equal to the buoyant force.
As for the bonus question, look into the answer to this question -
https://physics.stackexchange.com/a/296537/134658
Best Answer
"...because oil and water don't even mix so there's no displacement of water hence no buoyant force is exerted."
This is where you are misunderstanding. There is a displacement. Wood doesn't mix with water either, yet it displaces water and it floats. With oil, there is a slight depression of the lower surface, between the oil and water, where the displacement occurs.