You have a cylinderical beaker completely filled with water, and having no markings on it. You have to somehow empty two thirds of water, without using anything else.
Some info about beaker:
It has a small hole at the centre of its base
It has a transparent wall
Assume perfectionism exists (but not enough to judge when exactly one thirds of water will remain if you let the water leave through the hole or pour it out.)
Follow these steps:
First:
Spin the cylinder rapidly with the hole at its base facing downward. Let the water drain until the vortex looks like this:
A cone of water height H and radius R has now been removed from the cylinder of water height H radius R. Since the volume ratio of a cone to a cylinder with the same height and radius is one-third, you've drained 1/3 of the water.
Second:
Mark the current height of the water with your finger, then tilt the cylinder, with your finger on the bottom holding the spot, until the water level is flat between your finger and the top of the base, like this:
You have now drained half the remaining water, meaning:
2/3 total has been drained!
An answer has already been accepted and the $1/3$ has been commented to be $2/3$ but here is my answer anyway.
From trial and error I think this may have something to do with:
The refractive index of water which is almost exactly $4/3$.
Look at the reflection of the hole in the opposite wall of the glass, aligned with the rim.
When the glass is more than $1/3$ full, the reflection is below the water line.
When the glass is less than $1/3$ full, the reflection is above the water line.
When the glass is exactly $1/3$ full, the reflection is exactly on the water line.
I haven't worked out the physics - it's an experimental guess.
