Holes in bucket Say a bucket of water has two holes equally close to the bottom, but one hole is larger.  As the water leaks out, the pressure at the holes decreases.  Should the amount of water leaking decrease proportionately at each hole--for example when the pressure is halved, would the amount of water leaking at EACH hole be half of what it previously was? 
 A: The amount of liquid leaking through the two holes stays constant throughout the process and is proportional to the area of the holes.  You can estimate the ratio of the flow through the two holes using one-dimensional theory. This means it is only valid for small holes where the change of parameters across the hole is negligible.

Assume an open container (hence everywhere the pressure $p_0$) with a constant cross-section $A_0$ and two holes with cross-sectional area $A_1$ and $A_2$ respectively.
With the incompressible one-dimensional continuity equation we find for the volume flux $Q_i = \frac{d V}{d t}$
$$ Q_0 = Q_1 + Q_2 $$
$$ A_0 u_0 = A_1 u_1 + A_2 u_2. $$
Apply Bernoulli's equation for a streamline from the water surface (or any other incompressible fluid) inside the container, that is sinking with velocity $u_0$, to the hole 1, where the water is exiting with velocity $u_1$,
$$ \frac{\rho u_0^2}{2} + \rho g h_0 + p_0 = \frac{\rho u_1^2}{2} + \rho g h_1 + p_0 $$
$$ \frac{u_0^2}{2} + g \underbrace{(h_0 - h_1)}_{h_{01}} = \frac{u_1^2}{2} \tag{1}\label{1}$$
and similarly for hole 2
$$ \frac{u_0^2}{2} + g \underbrace{(h_0 - h_2)}_{h_{02}} = \frac{u_2^2}{2}. \tag{2}\label{2} $$
If we subtract the two equations \eqref{2} - \eqref{1} we find
$$ g \underbrace{(h_1 - h_2)}_{h_{12}} = \frac{u_2^2}{2} - \frac{u_1^2}{2}. $$
As you can see if the centers of two holes are at the same height, meaning $h_{12} \approx 0$ the two velocities have to be equal. This result is independent of the liquid column and thus the flux of water through them $Q_i = A_i u_i$ at any given time only depends on the cross-sectional area.
