Solution :
The diameter of the tank = 3 m = 300 cm
Radius of the tank = 1.5 m = 150 cm
Height of the tank = 8 m = 800 cm
The rate of increase of height of water level = 26 cm/min
∴ [tex]$\frac{dh}{dt} = 20 \ cm/min$[/tex]
The water is leaking at a rate of = 10700 cubic centimeters per min
∴ rate out = 10,700 [tex]$cm^3 / cm $[/tex]
At time 't', the height of water level is 'h' with radius 'r' and volume 'V'.
By the property of similarity of triangles we can write
[tex]$\frac{h}{r} = \frac{800}{150} $[/tex]
[tex]$\frac{h}{r} = \frac{16}{3} $[/tex]
[tex]$r = \frac{3}{16}h$[/tex]
Volume of water at time 't' is given as
[tex]$V= \frac{1}{3} \pi r^2 h $[/tex]
[tex]$V= \frac{1}{3} \pi \left(\frac{3}{16}h\right)^2 h $[/tex]
[tex]$V =\frac{3}{256} \pi h^3$[/tex]
Now differentiating w.r.t 't'
[tex]$\frac{dv}{dt} = \frac{9}{256} \pi h^2 \frac{dh}{dt}$[/tex]
The rate of increase in tank = rate of inward flow - rate of outward flow
[tex]$\frac{dv}{dt} $[/tex] = rate of inward flow - 10,700
rate of inflow = 10,700 + [tex]$ \frac{9}{256} \pi h^2 \frac{dh}{dt}$[/tex]
Substituting the values we get
rate of inflow = 10,700 + [tex]$ \frac{9}{256} \times 3.14 \times (250)^2 \times 26$[/tex]
∴ rate of inward flow = [tex]$1.9 \times 10^5 \ cm^3/min$[/tex]
