water is leaking out of an inverted conical tank at a rate of 11,500 cm3/min at the same time that water is being pumped into the tank at a constant rate. the tank has height 6 m and the diameter at the top is 4 m. if the water level is rising at a rate of 20 cm/min when the height of the water is 2 m, find the rate at which water is being pumped into the tank

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abhinavjh321 abhinavjh321 · Answer 1 · 2022-12-24T14:36:25+01:00

The rate at which water is being pumped into the tank is 11,766.667 cm3/min.

Let V1 be the rate of water being pumped into the tank and V2 be the rate of water leaking out of the tank.

V1 - V2 = 11,500 cm3/min

Let h be the height of the water in the tank.

Volume of water in the tank = (π/3)*(D2*h - D1*h1)

Where D1 = 0 m and D2 = 4 m

Rate of change of volume = (π/3)(D2(dh/dt) - D1*(dh/dt))

Rate of change of volume = (π/3)(4(20 cm/min) - 0*(20 cm/min))

Rate of change of volume = (π/3)×(80 cm3/min)

V1 - V2 = (π/3)×(80 cm3/min)

V1 = V2 + (π/3)×(80 cm3/min)

V1 = 11,500 cm3/min + (π/3)×(80 cm3/min)

V1 = 11,500 cm3/min + 266.667 cm3/min

V1 = 11,766.667 cm3/min

Hence, the rate at which water is being pumped into the tank is 11,766.667 cm3/min.

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