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OG Closed cylindrical tank Q

by AbeNeedsAnswers » Tue Jul 25, 2017 8:09 pm
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
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by Jay@ManhattanReview » Tue Jul 25, 2017 11:29 pm
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
We know that the with half the volume, the height of the water level is 4 feet, thus, the height of the cylinder = 2*4 = 8 feet.

Since the volume of water = 36Ï€, which is half of the volume of the cylinder, the volume of the cylinder = 2*36Ï€ = 72Ï€ cubic feet

=> volume of the cylinder = πr^2h = 72π

8Ï€r^2 = 72Ï€; since h = 8

Thus, r = 3 feet

When the tank is placed on its side on level ground, the height of the surface of the water above the ground would be equal to the radius of the cylinder = 3 feet

The correct answer: B

Hope this helps!

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by Scott@TargetTestPrep » Mon Aug 14, 2017 1:17 pm
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

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Re: OG Closed cylindrical tank Q

by [email protected] » Sun Apr 18, 2021 3:21 pm
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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