I've been thinking about this for fun but sort of hit a brick wall.
Using a standard drinking straw, how many times does one have to dip the straw into a jar of, say, whisky (trapping the liquid in the straw) in order to fill a 4cl shot glass.
The volumes are trivial: $$V_J=\pi r_J^2\cdot H \quad V_S=\pi r_S^2\cdot H$$
with $V_J$ the volume of whisky in the jar, and $V_S$ the volume of whisky in the straw. $H$ is the hight of the whisky in both jar and straw.
I can see that the change in volume can be calculated from $$\Delta V=\pi(r_j^2-r_s^2)\cdot H$$
but the constantly changing volume and hight of the whisky is giving me problems. How can I account for this in order to find the number of dips n required with a jar of volume $V_j$ and whisky hight of $H$ to fill a glass of a given volume?
Hope my tag is ok. I had no idea where to put it!
It might be easier if you think of it in a different way. Instead of trying to sum up all the whisky you draw out of the jar, try to find how many dips it'll take for the volume of whisky in the jar to have decreased by the volume of the shot glass.
Notice that $V_{J}$ and $V_{S}$ are both proportional to $H$ by a constant factor, and hence they're proportional to each other. If the volume in the jar after $n$ dips is $V_{J,n}$, then you can find $H_{n}$ and hence $V_{S,n}$.
The change in volume in the jar is going to just be the volume held in the straw, $V_{S,n}$ and this will be proportional to the volume in the jar before the straw is dipped in, $V_{J,n}$. So the volume left in the jar after a dip will be a fixed fraction of the volume before the dip, hence you can express $V_{J,n+1}$ in terms of $V_{J,n}$:
You can use this to find a formula for $V_{J,n}$ in terms of $V_{J,0}$.
Then you just want the smallest integer $n$ where $V_{J,0}-V_{J,n} \ge 4\ \mathrm{cl}$.