Answer
$\frac{9}{20 \pi} $ $ft/min$
Work Step by Step
Ref to the fig
$\frac{BD}{AB}=\frac{24}{10}=\frac{CD}{r}$
Let $CD=h$ $\Longrightarrow\frac{h}{r}=\frac{12}{5}$
$ 12r=5h$
$r=\frac{5}{12} h$ ................... eq (1)
Let $h$ be the height of the water tank and r be the radius of the same.
Let $V$ be the volume of the conical water tank.
$V=\frac{1}{3} \pi r^2 h$ .................. eq (2)
Putting equation (1) in equation (2)
$V=\frac{1}{3} \pi (\frac{5h}{12})^2 \times h=\frac{25 \pi h^3}{432}$
Taking derivative with respect to $t$
$\frac{dV}{dt}=\frac{d ( \frac{25 \pi h^3}{432} ) }{dt}= \frac{25 \pi}{432}\frac{dh^3}{dt} = \frac{25 \pi}{432}\frac{dh^3}{dh} \frac{dh}{dt}=\frac{25 \pi \times 3 h^2}{432} \frac{dh}{dt} $
$\frac{dV}{dt}=\frac{25 \pi \times 3 h^2}{432} \frac{dh}{dt} $ ..................... eq (3)
Putting $ \frac{dV}{dt}=20 $ $ ft^3/min$, $h=16$ ft in equation (3)
$ 20 =\frac{25 \pi \times 3 \times 16^2}{432} \frac{dh}{dt} = \frac{ 19200\pi}{432} \frac{dh}{dt}= \frac{400\pi}{9}\frac{dh}{dt}$
$\frac{dh}{dt}= \frac{9}{400\pi}\times 20=\frac{9}{20 \pi}$ $ ft/min$
