Answer:
Step-by-step explanation:
Given that a conical tank has height 3 m and radius 2 m at the top. Water flows in at a rate of 1.6m^3/min
We know in a cone the ratio of radius to height is constant equal to tangent of semivertical angle.
i.e. [tex]\frac{r}{h} =K[/tex]
Hence we get radius when height was 1.11 m is
[tex]\frac{2}{3} =\frac{r1}{1.1} \\r1 = 0.7333[/tex]
V' = 1.6
Volume of cone = [tex]\frac{1}{3} \pi r^2 h \\= \frac{1}{3} \pi (kh)^2 h\\=\frac{1}{3} \pi kh^3[/tex] where k = 2/3
[tex]V' = \pi k h^2 h'\\1.6 = \pi \frac{2}{3} (1.1)^2 h'\\1.6 = 0.8067 \pi h'\\h'=\frac{1.983}{\pi}[/tex] m/sec.
