\includegraphics{figure_10}
A tank containing water is in the form of a cone with vertex \(C\). The axis is vertical and the semi-vertical angle is \(60°\), as shown in the diagram. At time \(t = 0\), the tank is full and the depth of water is \(H\). At this instant, a tap at \(C\) is opened and water begins to flow out. The volume of water in the tank decreases at a rate proportional to \(\sqrt{h}\), where \(h\) is the depth of water at time \(t\). The tank becomes empty when \(t = 60\).
- Show that \(h\) and \(t\) satisfy a differential equation of the form
$$\frac{dh}{dt} = -Ah^{-\frac{1}{2}},$$
where \(A\) is a positive constant. [4]
- Solve the differential equation given in part (i) and obtain an expression for \(t\) in terms of \(h\) and \(H\). [6]
- Find the time at which the depth reaches \(\frac{1}{2}H\). [1]
[The volume \(V\) of a cone of vertical height \(h\) and base radius \(r\) is given by \(V = \frac{1}{3}\pi r^2 h\).]