The diagram represents a vertical cylindrical water cooler of constant cross-sectional area A - HSC - SSCE Mathematics Extension 2 - Question 7 - 2002 - Paper 1

Starting from the derived equation $\frac{dy}{dt} = -\frac{k}{A} \sqrt{y},$ we can separate variables and rewrite it as:

$\frac{dy}{\sqrt{y}} = -\frac{k}{A} dt.$ Integrating both sides, we have:

$\int \frac{1}{\sqrt{y}} dy = -\frac{k}{A} \int dt.$ This yields:

$2\sqrt{y} = -\frac{k}{A} t + C.$ Using the initial condition when t = 0, y = y0, we find C. Substituting this back into the equation gives us:

$y = y0 \left(1 - \frac{kt}{2Ay0}\right).$ If we denote the total time to drain the cooler by T (for half-drain, T = 10 seconds here), we find that over time 0 ≤ t ≤ T, it simplifies to:

$y = y0\left(1 - \frac{t}{T}\right).$

Given that it takes 10 seconds for half the water to drain, we note that this corresponds to when:

$y = \frac{y0}{2}.$ Using the equation derived previously:

$y = y0\left(1 - \frac{t}{T}\right)$ where t = 10 seconds, we can solve for T:

$\frac{y0}{2} = y0\left(1 - \frac{10}{T}\right) \Rightarrow \frac{1}{2} = 1 - \frac{10}{T} \Rightarrow \frac{10}{T} = \frac{1}{2} \Rightarrow T = 20 \text{ seconds}.$ Thus, the time to empty the entire cooler is 20 seconds.