Answer
$\frac{20}{9 \pi} $ cm/s
Work Step by Step
Ref to the figure
$AB=4 $cm= Upper radius of the cup
$FD= 2$ cm =Lower radius of the cup
$BD=6$ cm = Height of the cup
$CD=3 $ cm= Coffee level of the cup
$DE=(h-6)$ cm = Height of the cone that had been cut to form existing shape of the cup.
In the right-angled triangle $ABE$
$\frac{BE}{AB}=\frac{h}{4}=\frac{DE}{FD}=\frac{h-6}{2}$Or
$\frac{h}{4}=\frac{h-6}{2}$ Or
$2h=4h-24$Or
$2h-4h=-24$ Or
$-2h=-24$Or
$h=\frac{-24}{-2}=12$cm at $r=4$cm
$\Longrightarrow$
$\frac{BE}{AB}=\frac{h}{r}=\frac{12}{4}=3$Or
$\frac{h}{r}=3$
$\Longrightarrow$
$h=3r$ Or
$r=\frac{h}{3}$ ..................... eq (1)
Let $V$ be the volume of the cone.
Then
$V=\frac{1}{3} \pi r^2 h $ ............... eq (2)
Putting equation (1) in equation (2)
$V=\frac{1}{3} \pi (\frac{h}{3})^2 \times h= \frac{h^3 \pi}{27} $
Taking derivative with respect to time t
$\frac{dV}{dt}=\frac{ d ( \frac{h^3 \pi}{27} )}{dt}= \frac{ \pi}{27} \frac{d(h^3)}{dt}=\frac{\pi}{27}\frac{d(h^3)}{dh} \frac{dh}{dt}= \frac{3h^2\pi}{27}\frac{dh}{dt}= \frac{h^2\pi}{9} \frac{dh}{dt}$ .............. eq(3)
Also from fig
$CE=CD+DE=3+h-6=h-3$
Putting $h=12 \Longrightarrow$
$CE= 12-3=9$ =level of rising coffee level=h
Putting $ \frac{dV}{dt}=20 cm^3/s$ and $h=9$ in equation (3)
$ 20= \frac{9^2 \pi}{9}\frac{dh}{dt}= 9\pi\frac{dh}{dt}$
$ \Longrightarrow\frac{dh}{dt}=\frac{20}{9\pi}$ cm/s
