Invisible waterfall

The free-falling water will not contribute to the weight. However, when the water arrives back to the container it will exert a downward force, that happens to be equal to the weight of the free falling water column. As a result, the weight is the same as before.

Details: Assuming the rate of flow (mass per unit time) is $f$ , the force of the impacting water is $F=fv$ , where $v=\sqrt{2gh}$ is the velocity at the impact. The mass of the free-falling water is $m=f t$ , where $t=\sqrt{2h/g}$ is the time-of-flight of a given droplet of water. The weight of the free-falling water is $F'=mg=gf\sqrt{2h/g}=f\sqrt{2gh}$ . It is clear that $F'=F$ . When the water is in steady flow the balance will show the same weight as it shows when there is no flow.

The center of mass of the system remains at rest, and is therefore not accelerating. Thus $F_{\text{net}} = 0$ .

Apart from gravity and the support of the scale, there are no external forces. Internal forces cancel by Newton's Third Law. (There is no need to figure out what all happens to the water inside the system!) Thus $F_{\text{scale}} = (-)F_g = mg$ .

In other words, the scale will read what it would ordinarily read.

Okay, finally, this has been put to rest. The answer is in fact 100 kg. Everything does balance out at the end.

(Erroneous argument follows here)

An hydraulics engineer is going to look at the bends in the pipe and see that there's going to be reactive forces in all of them. However, for a steady mass flow throughout, at each 90 degree bend there will be a reactive force that can be decomposed into equal forces in both the x and y right directions. Then all of the forces cancel out, except for the upward reactive force at the last bend. But this reactive force is based on some velocity of water leaving the nozzle, and when that water hits the tank again, it will be going at a higher velocity. Hence, there's going to be a net downward force. You could reverse everything and shoot a column of water right into the nozzle, and then drive a generator that maintains a steady flow through the pipes, with the same result. When liquid (or rocket gas) leaves a nozzle, the reaction force it generates is $Apv^2$ , where $A$ is the cross section area of the nozzle or pipe, $p$ is the mass density of the liquid (or gas), and $v$ is the velocity of the flow.

So, the correct answer is, "more than 100 kg"

Note: We are assuming that the pipe is completely filled with water everywhere, moving with a constant velocity.

Counterarguments are invited. Jury's still out on this one.

Except during start up the weight remains at 100 kg.

Below is a simple form of the steady-state control-volume equations for linear momentum. Here B stands for body-forces, gravity in our case, and S stands for surface forces such as pressure and stress. $\dot{m}$ is the mass flow rate (constant) and the V is velocity. In the accompanying figure I have drawn two control volumes. Picture these as having depth as well. The white CV encloses the entire apparatus. There is no in or out flow, so the forces (net) must be zero. In this case $F_{y,B}$ would be the weight of everything inside the control-volume and would point down. $F_{y,S}$ would be the force of the table acting on the bottom of the apparatus and would point up. Note that this is true regardless of the rate water is being pumped. The second control-volume, in yellow, cuts through the downward falling stream. Let the vertical cut have a height h. Now we do have fluid leaving the CV and entering the CV lower down. As was pointed out already, the reentering stream is going faster than the exiting stream due to the acceleration of gravity. But something else has changed as well: $F_{y,B}$ . There is less mass in the yellow CV than in the white CV. The shape of the falling stream has a cross section that decreases as the velocity increases. Since mass flow is constant and the velocity is going up, the area must decrease as $A = \frac{{{A_{out}}{V_{out}}}}{{\sqrt {V_{out}^2 - 2g\left( {{y_1} - {y_2}} \right)} }},\,\,\,\,\,\,\,\,\,\,0 \le {y_1} - {y_2} \le h\,\,$ This can be integrated to get a volume $\vartheta$ and a weight $\rho g\vartheta$ . The inlet velocity can be found using conservation of energy: ${V_{in}} = \sqrt {V_{out}^2 + 2gh}$ . Just remember that $\dot m = \rho {A_{out}}{V_{out}} = \rho {A_{in}}{V_{in}}$ . I am assuming density is constant. The integration is pretty basic. It turns out that the loss of weight is exactly equal to the increase in $\dot m{V_{in}}$ . I used two different CVs to try and demonstrate that one has to be careful when defining the CV. This was drilled into me as an engineering student.

it seems to be energy conservation problem since first P.E is converted into K.E but some energy must have gone through the water pump in form of heat or electrical energy

This solution is in my point of view. The water is being sucked into the pipe, where the water is flowing out from. Logically, the water is just going into up the 1 meter pipe, then coming back out again so it does not add or contribute to the weight, so it is 100 kg with or without the water flowing.

Put the complete apparatus inside a closed black box and imagine you don't know what's inside the box. Let's assume it's an electric pump with the wires coming out through a hole in the side of the box and connected to a power supply and switch. It would be a very interesting box if you could supply electrical power to it and have the box now weigh more or less than it did before. Maybe over time it may start to weigh less as some of the water evaporates out of the box.

Simply, there is nothing enters or leaves the pump loop

Running pump presumably powered by electricity, if the source is external it will add miniscule mass to the overall system, and the pump will weight a little more. If a source is internal - the pump will weight a little less - due to the fact that energy is leaving the system as heat.

It would have been awesome if the effect would be that of diminishing the weight. You could then setup the system in a way to almost diminish the whole weight and thus make something easier to lift etc.

When water is pulled up,the total amount o water in the system doesn't change. Just imagine it...when water goes into the pipe it is still being wieghed. Hence the total weight remains same , that is, 100 kg

The scale holds the foist too. So the water being in the foist does not change the weight.

The Free body diagram is unchanged hence no change in calculation and value of normal force (weight reading)