A siphon is used to drain water from a tank with a steady frictionless flow.

See the diagram for the set up.

We can use Bernoulli equation to solve for the v. Here we can assume that the cross section area of the tube is much smaller than that of the beaker so we can treat the velocity that the water level drops in the beaker to be zero. Setting y=0 at the siphon's outlet, we have $$P_0+\rho gh=P_0+\frac 12\rho v^2$$

Plug in numbers and we can get v.

To find out the highest height the siphon can go, the question becomes equivalent to the following diagram. Basically, the liquid will evaporate into vapor gas and create a pressure — that is the lowest pressure it can get to. Lower than that, the liquid will boil to maintain such a pressure. For example, the liquid would instantaneously boil and evaporate in vacuum at room temperature. So in this question, the P at the top of the siphon or the test tube shown below would be equal to the vapor pressure.

With ||P_A=P_B=P_0||, we have ||P_A=P_0=\rho gy+P||.

Plug in the numbers and we can find y here.

See more at leiacademy.org.