The rate at which water flows out of a drain in the bottom of a certain tank is proportional to the height of water in the tank. The tank is a vertical cylinder with cross-sectional area of 1.0 m², so that every 1 cm in height represents 10 L. If the flow is 10 L/min (i.e. 1 cm/min) when the water level is 200 cm, how long will it take for the level to go from 200 cm to 20 cm? A) 1060 minutes B) 180 minutes C) 7 minutes D) 461 minutes

Question

Quizplus · Accepted Answer

Let's call the current height of the water in the tank H. We know that when H = 200 cm, the flow rate is 10 L/min.
Let k be the constant of proportionality, so we can write:
rate = kH
When H = 200 cm, we have:
10 L/min = k(200 cm)
k = (10 L/min)/(200 cm) = 0.05 L/cm/min

Now, we want to find how long it takes for the water level to go from H = 200 cm to H = 20 cm. We can set up a differential equation:
dH/dt = -kH
This is a separable differential equation, so we can integrate both sides:
∫(1/H) dH = -∫k dt
ln(H) = -kt + C
Using the initial condition that H = 200 cm when t = 0, we can solve for C:
ln(200) = 0 + C
C = ln(200)
So the equation for H as a function of t is:
H = e^(-kt + ln(200)) = 200e^(-kt)

Now we can find how long it takes for H to decrease from 200 to 20 cm:
20 = 200e^(-kt)
e^(-kt) = 1/10
-kt = ln(1/10)
t = (1/k)ln(10) = (1/0.05)ln(10) ≈ 461 minutes

Therefore, the answer is D) 461 minutes.

The Rate at Which Water Flows Out of a Drain