Question

A circular conical reservoir has depth 20 feet and radius of the top 10 feet. water is leaking out so that the surface is falling at the rate of 1 2 ft hr .the rate, in cubic feet per hour, at which the water is leaving the reservoir when the water is 8 feet deep is

Answer 1

As per given condition we know that vertex angle of the cone is given as

$tan\theta = \frac{R}{H}$

so here we can say that vertex angle will remain constant

so here

$\frac{r}{y} = \frac{R}{H}$

$\frac{r}{8} = \frac{10}{20}$

$r = 4 feet$

now for the volume we can say

$V = \frac{1}{3}\pi r^2 y$

also we can say

$r = \frac{y}{2}$

so here we will have

$V = \frac{1}{3}\pi (\frac{y}{2})^2y = \frac{1}{12}\pi y^3$

now for volume flow rate

$Q = \frac{dV}{dt} = \frac{3}{12}\pi y^2\frac{dy}{dt}$

$Q = \frac{1}{4}\pi y^2 v_y$

now plug in all data

$Q = \frac{1}{4} \pi (8)^2 (12) ft^3/h$

$Q = 603.2 ft^3/h$