A closed cylindrical tank contains 36pi cubic feet of water and is filled to half its capacity. When the tank is placed upright on its circular base on level ground, the height of the waer in the tank is 2 feet. When the tank is placed on its side on level ground, what is the height, in feet, of the surface of the water above the ground?

Water volume:
pi * r^2 * h = 36 * pi
so r^2 * h = 36
Substitute water height h = 2 to get
r^2 = 18
so r = √18 = 3√2

[On it's side the water runs to full height which is double before, so H = 2h = 4 (but this is not needed in the question)].

As it is filled to half capacity, this means the water reaches the centre line (diameter) of the tank, which is at a height of the radius, r = 3√2 = 4.242640687...

So the closest answer is (C) 4

The answer is B.

Sak32 wrote:The answer is B.

OK, let's try radius = 3:

Volume of water/pi = r^2 * 2h / 2 (half because it's half full)
Substitute r = 3 and h = 2

V/pi = 3^2 * 2 = 18

This is not 36.

Therefore. may I propose that the wording of the question is wrong.
Instead of "A closed cylindrical tank contains 36pi cubic feet of water and is filled to half its capacity", it should read "A closed cylindrical tank can contain 36pi cubic feet of water and is filled to half its capacity". Now the answer 3.

Otherwise, the question states that it actually contains 36pi cubic feet off water, and is half full - implying that the tank could contain 72pi cubic feet.

With the question re-worded to mean that the tank has 36pi capacity, which is only half full of water:

Water volume:
pi * r^2 * h = 18 * pi
so r^2 * h = 18
Substitute water height h = 2 to get
r^2 = 9
so r = √9 = 3

Using h=4

pi * r^2 * h = 36 * pi
so r^2 * h = 36
Substitute water height h =4 to get
r^2 = 9
so r = √9 = 3