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A large, cylindrical water tank with diameter 3.60 m is on a platform 2.00 m above the ground. The vertical tank is open to the air and the depth of the water in the tank is 2.00 m. There is a hole with diameter 0.520 cm in the side of the tank just above the bottom of the tank. The hole is plugged with a cork. You remove the cork and collect in a bucket the water that flows out the hole. Part A When 1.00 gal of water flows out of the tank, what is the change in the height of the water in the tank?

Answer :

cjmejiab

To solve this problem it is necessary to apply the concepts related to the geometry of a cylindrical tank and its respective definition.

The volume of a tank is given by

[tex]V = \frac{\pi d^2}{4}h[/tex]

Where

d = Diameter

h = Height

Considering that there are two stages, let's define the initial and final volume as,

[tex]V_0 = \frac{\pi d^2}{4}H[/tex]

[tex]V_f = \frac{\pi d^2}{4}h[/tex]

We know as well by definition that

[tex]1gal = 3.785*10^{-3}m^3[/tex]

Then we have for the statement that

[tex]V_f = V_0 -1gal[/tex]

[tex]V_f = V_0 - 3.785*10^{-3}[/tex]

Replacing the previous data

[tex]\frac{\pi d^2}{4}h = \frac{\pi d^2}{4}H- 3.785*10^{-3}[/tex]

[tex]\frac{\pi (3.6)^2}{4}h = \frac{\pi (3.6)^2}{4}(2)- 3.785*10^{-3}[/tex]

Solving to get h,

[tex]h = 1.99963m[/tex]

Therefore the change is

[tex]\Delta h = H-h[/tex]

[tex]\Delta h = 2- 1.99963[/tex]

[tex]\Delta h = 3.7*10^{-4}m=0.37mm[/tex]

Therefore te change in the height of the water in the tank is 0.37mm

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