Answer
The net work done on the water is zero in arrangement 1 and arrangement 4.
Work Step by Step
Let $m$ be the mass of the unit volume of water.
We can write an expression for the kinetic energy of the water:
$K = \frac{1}{2}mv^2$
Since the mass does not change as the unit volume of water flows through pipes of differing radii, the kinetic energy of the water depends on the speed.
We know that the water flows faster when the pipe radius is smaller, and the water flows slower when the pipe radius is larger.
In arrangement 1 and arrangement 4, the leftmost pipe radius is equal to the rightmost pipe radius.
Thus the speed of the water is equal in the leftmost pipe and the rightmost pipe.
Then the kinetic energy of the water in the leftmost pipe and the rightmost pipe is equal.
Therefore, the net work done on the water is zero in arrangement 1 and arrangement 4.
