Answer:
The rate at which water is being pumped into the tank = 289252.68 cm³/min or 2.893 x 10^(5) cm³/min
Step-by-step explanation:
In the question, we are given the following;
diameter = 4 m = 400cm
radius = 200 cm
height (h) = 6 m = 600 cm
dh/dt = 20 cm/min
rate out = 10000 cm³/min
Now, by property of similar triangles, we have;
h/r = 6/2
r = h/3
Now, volume at time is given as volume of cone which is;
v = (1/3)πr²h
Substitute h/3 for r;
= (1/3)π(h/3)²h
= (1/27)(πh³)
dv/dt = (3/27)(π)h² dh/dt = (1/9)(π)h² dh/dt
(rate of water into tank) - 10000 = (1/9)(πh²)dh/dt
rate into tank = 10000 + (1/9)(πh²)dh/dt
Substituting the relevant values we get;
rate into tank = 10000 + (1/9)(π) (200²)(20)
= 10000 + 800000(π)/9 = 10000 + 279252.68 = 289252.68 cm³/min or 2.893 x 10^(5) cm³/min
