Originally posted by gentlerock:
Sorry ah, sometimes I just like to ask stupid questions for fun.
There is a plain empty room, except for a 4 foot straight metal pipe in the middle of the room. This metal pipe is 4 feet long, with a 44 mm internal diameter. The pipe was installed vertically, with half its length embedded in the floor and the other half sticking out. Inside this pipe there is a ping pong ball with 40 mm diameter.
You're given the following tools:
- A 5-foot length of sewing thread.
- A 2 litre pail.
- A straight piece of wood, 2 feet long with cross section 15mm x 10mm.
- A mechanical pencil with one piece of lead.
How would you extract the ping pong ball, by just using the above given tools?
A 2-litre pail would be useless if it's empty...
Internal diameter of pipe is 44mm which translates into 2*pi*radius => radius of pipe is 2.2/pi cm. Hence, the cross-sectional area of pipe is pi*radius^2 = (2.2^2)/pi cm^2. Therefore, the internal volume of the pipe is 4 feet (or 121.92 cm) * (2.2^2)/pi cm^2 which is approximately 189 cm^3.
Since 1 litre = 1000 cm^3, we would require at least 189 ml of water to bring the ping pong ball out.
Since the question didn't state that the 2-litre pail contains water, I assume it didn't. As for the rest of the tools, I simply have no idea.
Therefore, I give up.