A closed cylindrical tank contains 36π cubic feet of water and is filled to…
2025
A closed cylindrical tank contains 36π 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.
2.5
- C.
3
- D.
4.5
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Correct answer: C
Concept: For an upright cylinder, volume = π × r2 × height, and because the cross-sectional area is constant along the axis, the fraction of the volume filled equals the fraction of the height filled. For a cylinder lying on its side, that simple proportion no longer holds in general — the volume-height relationship becomes a circular-segment calculation. There is one important exception: when the cylinder is exactly HALF full by volume, the water surface must sit exactly at the horizontal diameter, because any diameter always splits a circle into two equal-area halves. So in this half-full case only, the water's height above the ground equals the radius.
Let the tank's radius be r and its total height be H. Standing upright and half full by volume, the water height is 4 ft; since (for the upright case) volume-fraction equals height-fraction, this 4 ft is half of H: H ÷ 2 = 4, so H = 8 ft.
The volume of water when upright is area × height = π × r2 × 4, and this equals the given volume: π × r2 × 4 = 36π.
Dividing both sides by 4π: r2 = 9, so r = 3 ft.
On its side, the tank is still half full by volume, so — by the concept above — the water surface sits exactly at the horizontal diameter, meaning its height above the ground equals the radius: 3 ft.
Cross-check: the tank's total volume is π × r2 × H = π × 9 × 8 = 72π cubic feet; half of that is 36π cubic feet, which matches the given water volume — confirming r = 3 is consistent.