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Answer :
A water tank in the form of an inverted right circular cone is 20 ft across the top and 15 ft deep. If the surface of the water is 5 ft below the top of the tank, the work done in pumping the water to the top of the tank is (C). 217,817 ft-lbn is correct option.
To find the work done in pumping the water to the top of the tank, we can use the concept of work done against gravity, which is given by the formula:
Work = Force × Distance
The tank is in the form of an inverted right circular cone, so we can use the formula for the volume of a cone:
Volume = (1/3)πr²h
where r is the radius of the base and h is the height of the cone.
Given:
Diameter of the top of the tank = 20 ft, so radius (r) = 20/2 = 10 ft
Depth of the tank = 15 ft
Surface of the water is 5 ft below the top, so the height of the water (h) = 15 ft - 5 ft = 10 ft
Now, let's calculate the volume of water in the tank:
Volume = (1/3)π(10²)(10) = (1/3)π(100)(10) = (1/3)π(1000) = 1000π cubic feet
Weight = Volume × Density × Gravity
The density of water is approximately 62.4 lb/ft³, and the acceleration due to gravity is approximately 32.2 ft/s².
Weight = 1000π × 62.4 lb/ft³ × 32.2 ft/s²
Now, let's calculate the work done in pumping this water to the top of the tank, which is a distance of 5 ft:
Work = Weight × Distance
Work = [1000π × 62.4 lb/ft³ × 32.2 ft/s²] × 5 ft
Work = 1000π × 62.4 lb/ft³ × 32.2 ft/s² × 5 ft
Now, calculate this value:
Work ≈217,817.44 lb·ft
So, the work done in pumping the water to the top of the tank is approximately 217,817 .44 lb·ft.
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