As per given condition we know that vertex angle of the cone is given as
[tex]tan\theta = \frac{R}{H}[/tex]
so here we can say that vertex angle will remain constant
so here
[tex]\frac{r}{y} = \frac{R}{H}[/tex]
[tex]\frac{r}{8} = \frac{10}{20}[/tex]
[tex]r = 4 feet[/tex]
now for the volume we can say
[tex]V = \frac{1}{3}\pi r^2 y[/tex]
also we can say
[tex]r = \frac{y}{2}[/tex]
so here we will have
[tex]V = \frac{1}{3}\pi (\frac{y}{2})^2y = \frac{1}{12}\pi y^3[/tex]
now for volume flow rate
[tex]Q = \frac{dV}{dt} = \frac{3}{12}\pi y^2\frac{dy}{dt}[/tex]
[tex]Q = \frac{1}{4}\pi y^2 v_y[/tex]
now plug in all data
[tex]Q = \frac{1}{4} \pi (8)^2 (12) ft^3/h [/tex]
[tex]Q = 603.2 ft^3/h[/tex]
