A conical funnel stands with its vertex pointing down, directly above an empty cylindrical jar, as…

Set-up. At depth $h$ the water in the funnel is a small cone similar to the funnel, so $\dfrac{r}{h} = \dfrac{12}{24}$ and $r = \dfrac h2$. Its volume is $V = \dfrac13 \pi r^2 h = \dfrac{\pi h^3}{12}$, and $\dfrac{dV}{dt} = \dfrac{\pi h^2}{4}\,\dfrac{dh}{dt}$.

(a) With $h = 12\ \text{cm}$ and $\dfrac{dh}{dt} = -0.5\ \text{cm/s}$: $\dfrac{dV}{dt} = \dfrac{\pi \times 144}{4} \times (-0.5) = -18 \pi\ \text{cm}^3/\text{s}$. Water leaves at $18 \pi \approx 56.55\ \text{cm}^3/\text{s}$.

(b) The jar's water has volume $64\pi y$ at depth $y$, and gains water at the rate the funnel loses it: $64\pi\,\dfrac{dy}{dt} = 18 \pi$, so $\dfrac{dy}{dt} = \frac{9}{32} = 0.28125 \approx 0.281\ \text{cm/s}$.

(c) $r = \dfrac h2$, so $\dfrac{dr}{dt} = \dfrac12\,\dfrac{dh}{dt} = -0.25\ \text{cm/s}$: the radius shrinks at $0.25\ \text{cm/s}$.

(d) The surface has area $A = \pi r^2$, so $\dfrac{dA}{dt} = 2\pi r\,\dfrac{dr}{dt} = 2\pi \times 6 \times (-0.25) = -3 \pi$. It shrinks at $3 \pi \approx 9.42\ \text{cm}^2/\text{s}$.

(e) Now $\dfrac{\pi h^2}{4} \times 2 = 18\pi$, so $h^2 = 36$ and $h = 6\ \text{cm}$. The level falls faster as the funnel empties, because the same volume per second comes from a smaller surface.