7.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{0e7cab3d-c1e6-4420-93b4-eca5af704432-07_591_730_294_735}
\captionsetup{labelformat=empty}
\caption{Diagram not drawn to scale}
\end{center}
\end{figure}

Figure 4\\[0pt]
[ The volume of a cone of base radius $r$ and height $h$ is $\frac { 1 } { 3 } \pi r ^ { 2 } h$ ]\\
Figure 4 shows a container in the shape of an inverted right circular cone which contains some water.

The cone has an internal base radius of 2.5 m and a vertical height of 4 m .\\
At time $t$ seconds

\begin{itemize}
  \item the height of the water is $h \mathrm {~m}$
  \item the volume of the water is $V \mathrm {~m} ^ { 3 }$
  \item the water is modelled as leaking from a hole at the bottom of the container at a rate of
\end{itemize}

$$\left( \frac { \pi } { 512 } \sqrt { h } \right) m ^ { 3 } s ^ { - 1 }$$
\begin{enumerate}[label=(\alph*)]
\item Show that, while the water is leaking

$$h ^ { \frac { 3 } { 2 } } \frac { \mathrm {~d} h } { \mathrm {~d} t } = - \frac { 1 } { 200 }$$

Given that the container was initially full of water
\item find an equation, in terms of $h$ and $t$, to model this situation.
\end{enumerate}

\hfill \mbox{\textit{SPS SPS FM 2020 Q7 [8]}}