8.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{280ae2a5-7344-4ba3-907f-235fba3fd5b3-12_474_837_283_610}
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\caption{Figure 2}
\end{figure}
Figure 2 shows a cylindrical water tank. The diameter of a circular cross-section of the tank is 6 m . Water is flowing into the tank at a constant rate of \(0.48 \pi \mathrm {~m} ^ { 3 } \mathrm {~min} ^ { - 1 }\). At time \(t\) minutes, the depth of the water in the tank is \(h\) metres. There is a tap at a point \(T\) at the bottom of the tank. When the tap is open, water leaves the tank at a rate of \(0.6 \pi h \mathrm {~m} ^ { 3 } \mathrm {~min} ^ { - 1 }\).
- Show that \(t\) minutes after the tap has been opened
$$75 \frac { \mathrm {~d} h } { \mathrm {~d} t } = ( 4 - 5 h )$$
When \(t = 0 , h = 0.2\)
- Find the value of \(t\) when \(h = 0.5\)