Answer:
1508527.582 cm³/min
Step-by-step explanation:
The net rate of flow dV/dt = flow rate in - flow rate out
Let flow rate in = k. Since flow rate out = 6800 cm³/min,
dV/dt = k - 6800
Now, the volume of a cone V = πr²h/3 where r = radius of cone and h = height of cone
dV/dt = d(πr²h/3)/dt = (πr²dh/dt)/3 + 2πrhdr/dt (since dr/dt is not given we assume it is zero)
So, dV/dt = (πr²dh/dt)/3
Let h = height of tank = 12 m, r = radius of tank = diameter/2 = 3/2 = 1.5 m, h' = height when water level is rising at a rate of 21 cm/min = 3.5 m and r' = radius when water level is rising at a rate of 21 cm/min
Now, by similar triangles, h/r = h'/r'
r' = h'r/h = 3.5 m × 1.5 m/12 m = 5.25 m²/12 m = 2.625 m = 262.5 cm
Since the rate at which the water level is rising is dh/dt = 21 cm/min, and the radius at that point is r' = 262.5 cm.
The net rate of increase of water is dV/dt = (πr'²dh/dt)/3
dV/dt = (π(262.5 cm)² × 21 cm/min)/3
dV/dt = (π(68906.25 cm²) × 21 cm/min)/3
dV/dt = 1447031.25π/3 cm³
dV/dt = 4545982.745/3 cm³
dV/dt = 1515327.582 cm³/min
Since dV/dt = k - 6800 cm³/min
k = dV/dt - 6800 cm³/min
k = 1515327.582 cm³/min - 6800 cm³/min
k = 1508527.582 cm³/min
So, the rate at which water is pumped in is 1508527.582 cm³/min
÷
. . .
÷
=11.5
