I was just wondering if anyone could help me figure out a question from my maths book that has been bugging me for days. I have a spherical tank with radius $R$, a hold in the bottom of the tank of radius $r$, and $h < R$ is the height of water above the bottom of the tank at time $t$.
I have worked out that the change dV in volume of water in the tank at a time interval $dt$ is given by $dV = \pi(2Rh - h^2)dh$
but i am now trying to prove, using Toricelli's law $v = 0.6\sqrt{2gh}$ that in a time interval $dt$, a volume of water equal to $ dV = \pi r^2vdt = 0.6\pi r^2 \sqrt{2gh} dt $ flows out of the tank.
I have been trying to figure this out for ages, but I honestly have no idea where to even start.
Thanks Corey
The volume of water that flows out satisfies ${dV \over dt} = \pi r^2 v$.
If you just substitute Toricelli's law you will get the result you are looking for. I have no idea what the other formula is (which is ${dV \over dh}$ by the way, not ${dV \over dt}$).