Suppose we have a container that has a base of area $b$ and we fill it up with water.
Volume of water = $b \cdot h$, where $h$ is height.
Hence, $\mathrm{d}v/\mathrm{d}t = b \cdot \mathrm{d}h/\mathrm{d}t$.
The container has a small hole of area $a$ at the bottom corner and so water is constantly leaking.
From my research:
1) I found that $\mathrm{d}v/\mathrm{d}t = -a \cdot $ velocity of water.
2) Velocity = $\sqrt{2gh}$ where $g$ is the gravitational constant.
And so $\mathrm{d}v/\mathrm{d}t = -a \cdot \sqrt{2gh}$.
Initially I did this:
$\mathrm{d}v/\mathrm{d}t = -a \cdot \sqrt{2gh} = B * \mathrm{d}h/\mathrm{d}t$
$1/\sqrt{h}\mathrm{d}h = -a \sqrt{2g}/b \mathrm{d}t$
Integrate both sides
$2\sqrt{h} = -a \sqrt{2g}/b \cdot t$
$h = (a^2 \cdot g)/(2b^2) \cdot t^2$
But this doesn't make sense.
I realise though that the $h$ from $\mathrm{d}v/\mathrm{d}t = -a \cdot \sqrt{2gh}$ is also constantly changing but I'm still not sure what to do. I tried deriving the equation again but it did not get me anywhere.
I want to find the function how much height or volume has decreased after time t. Can someone help?
Best Answer
$$\huge {\rm Toricelli's\;Law}\\\large v=\sqrt{\frac{2{\rm gh}}{1-a^2/b^2}}\approx\sqrt{2{\rm gh}}\tag{$a\ll b\equiv (a/b)\to0$}$$ where $\rm a,b,v,g,h$ are cross-section area of hole ,base area of cylinder, velocity of water from hole, Acceleration due to gravity and Height of the liquid above the hole.
This can be derived using equation of continuity: $$av=bV$$ where $V$ is speed of horizontal surface of water. And Bernoulli's equation: $$\rm P+\frac12\rho v^2+\rho gh= constant$$ where $\rm P,\rho,v,g,h$ are Pressure,Density,Velocity,Acceleration due to gravity and Height of the point of the liquid where we are calculating the constant. I believe this can also be derived by Energy Conservation.