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Geo

by beater » Thu Jan 01, 2009 7:00 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
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Re: Geo

by logitech » Thu Jan 01, 2009 7:07 pm
beater 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
So the tank volume: 72pi and height of the tank is 8 ft

72pi = Base x Height = Base x 8

So base is 72/8 = 9pi

Area of the base = pi r^2

9pi = pi r^2

r = 3ft

Since the tank is half full, the level will be up to its radius = 3ft

Hence, B
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by Sachindh » Thu Jan 01, 2009 9:26 pm
The volume of water = 36pi
36pi = pi*r^2*4
r=3

Now the volume of water is half of the volume of cylinder. Now when we tilt the cylinder to its side, the height will be equal to radius i.e. 3 ft.

Or in imagination - divide the cylinder in to four parts. shade the lower 2 parts as being fill with water. when you tilt with one side, the 1 shaded part will go to the third side or fill the lower 1/4 volume.

so answer choice is B
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