AbeNeedsAnswers wrote: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 water in the tank is 4 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?

(A) 2
(B) 3
(C) 4
(D) 6
(E) 9

B

First we need to determine the radius of the circular base of the cylindrical tank. Recall that the volume of a cylinder is:

volume = π(radius)^2(height)

Since half of the capacity of the tank is 36Ï€ and the height of the water is 4 feet, the full capacity of the tank is 72Ï€ and the full height of the tank is 8. Thus:

72π = πr^2(8)

9 = r^2

3 = r

When the cylinder is on its side, the new height is represented by the diameter of the base. Since the cylinder is half full, the height of the water equals the radius of the base, which is 3.

Answer: B

Hi All,

We’re told that a cylindrical tank (re: a cylinder/tube) contains 36pi cubic feet of water and is filled to HALF of its capacity; when placed upright, the water reaches 4 feet high. We’re asked how high the water reaches when the tank is placed on its side.

One of the ‘keys’ to solving this question quickly is to realize that since the tank is half-full, regardless of which side the tank is laying on, the water will reach HALF of the height.

Volume of a cylinder is (pi)(Radius^2)(Height), so we can use that formula – along with what we know about the water – to figure out the radius of the tank…

V = (pi)(R^2)(H) =
36pi = (pi)(R^2)(4)
36 = (R^2)(4)
9 = R^2
R = 3

Thus, the radius of the tank is 3 feet and its diameter is 6 feet. When the cylinder is lying on its side, the water will go to the half-way point of the cylinder. Since the diameter is 6 feet, the half-way point would be the radius: 3 feet.

Final Answer: B

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