Doubt in derivation of Velocity of efflux (Torricelli's theorem) To find the velocity of efflux of fluid from a hole in a simple cylindrical  container we use the Bernoulli's equation.
$$P + ρv^2/2 + ρgh= constant$$ 
For our specific case we take the pressure at the hole to be equal to $P_{atm}$ and the pressure at the surface of the cylinder is also $P_{atm}$, but why? Shouldn't the pressure at the hole be equal to $P_{atm}$ + gauge pressure $ρgh$ where h is the depth of the hole?
 A: You see, the pressure in the water in the cylinder does not arise just because of the gravitational field. It must have a lower support, like the base of the cylinder, for the water "to feel that pressure". To see this, just consider a cylinder of water in free fall in the gravitational field. If we were living in a vacuum, the pressure in such a free falling cylinder of water would be $0$.
However, we're not. We have a cylinder of a fluid (water) in a fluid (air). This air is lower-supported, and hence has pressure which is $P_{atm}$ at the earth's surface. 
Once the water comes out of the hole, it has no pressure due to the gravitational field. The only pressure is $P_{atm}$ in this case, which is exerted by the ambient air.
A: Think about it this way – a block of mass m and a weighing machine under it are both in free fall. Sure, the block feels gravitational attraction, but the machine reads zero. Both have the same acceleration. Similarly, here both the top and hole are exposed to the atmosphere, so their pressure will be equal. 
Also, I think $P = dgh$ ($d$ is density) is valid only in fluid statics (fluid is stationary wrt container). 
^Note that I haven't read about his anywhere.. it's a wild guess – just like all points on a conductor need not be at the same potential when current is flowing. So, like I said, it's just an idea I got from electricity concepts.. I could be wrong.
A: Inside the container, in the very neighborhood of the exit hole (say a few hole diameters away), the fluid flow is accelerating toward the hole (say, like the hole is acting like a sink for the flow).  As the fluid velocity increases approaching the hole, the pressure decreases (as described by the Bernoulli equation).  Right at the hole, the fluid velocity is at the full flow velocity of the hole, and the pressure has decreased to atmospheric.  
So, to summarize:  Throughout most of the container, the pressure variation is hydrostatic.  But in very close proximity to the exit hole, there are rapid changes that occur in both the fluid flow velocity and the pressure.  Finally, at the exit hole, the full exit velocity has been achieved, and the pressure is atmospheric again.
A: 
As long as the fluid is flowing vertically, the velocity is quasi constant and the pressure increases as in static. But, towards the bottom of the tank, the speed must increase and the pressure decreases. The flow lines are also curved inward indicating that the pressure is higher on the edges of the container. At the exit of the hole, the flow is rarely already vertical. Its section contracts and the speed increases further. This is the famous "vena contracta" (I hope the English term is correct). From there, the flow is unidirectional. (It shrinks slightly as the speed increases).
There is a fluid mechanics theorem which states that for unidirectional flow, the pressure perpendicular to the direction of flow varies as in static. Since the pressure outside the vertical jet is atmospheric pressure, it is also equal to the atmospheric pressure inside the fluid beyond the "vena contracta".
In conclusion, the speed obtained by Toricelli's theorem is the speed after the contraction, at B on the diagram.  and not immediately at the exit of the hole. And the height to be taken into account is that from A to B.
Sorry for my poor english. It is not my native language.
