Water is leaking out of an inverted conical tank at a rate of 9000 cubic centime

Question

Water is leaking out of an inverted conical tank at a rate of 9000 cubic centimeters per min at the same time that water is being pumped into the tank at a constant rate. The tank has height 9 meters and the diameter at the top is 3.5 meters. If the water level is rising at a rate of 20 centimeters per minute when the height of the water is 2.5 meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute

Explanation / Answer

Concept: (Rate at which water is being pumped in) - (rate at which water is being leaked) = (Rate of increase of the volume) Solution: (Rate at which water is being pumped in) = Ri = constant (rate at which water is being leaked) = Rl = 9000cc/min full height of the cone = 9 m full diameter of the cone = 3.5 m diameter when height is 2.5 -> (3.5/9)*2.5 m D=0.9722 m = 97.22 cm radius = 48.61 cm Rate of increase of the volume = (1/3)*Pi*radius*radius*(rate of increase of height) = (1/3)*3.14*48.61*48.61*20 cc/min = 49466.30 cc/min therefore : Ri - Rl = 49466.30 cc/min Ri = 49466.30 cc/min + Rl Ri = 49466.30 cc/min +9000cc/min = 58466.30 cc/min Rate at which water is being pumped in = 58466.30 cc/min (ANS)

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