3. Liquid is pouring into a large vertical circular cylinder at a constant rate of $1600 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }$ and is leaking out of a hole in the base, at a rate proportional to the square root of the height of the liquid already in the cylinder. The area of the circular cross section of the cylinder is $4000 \mathrm {~cm} ^ { 2 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that at time $t$ seconds, the height $h \mathrm {~cm}$ of liquid in the cylinder satisfies the differential equation

$$\frac { \mathrm { d } h } { \mathrm {~d} t } = 0.4 - k \sqrt { } h , \text { where } k \text { is a positive constant. }$$

When $h = 25$, water is leaking out of the hole at $400 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }$.
\item Show that $k = 0.02$
\item Separate the variables of the differential equation

$$\frac { \mathrm { d } h } { \mathrm {~d} t } = 0.4 - 0.02 \sqrt { } h$$

to show that the time taken to fill the cylinder from empty to a height of 100 cm is given by

$$\int _ { 0 } ^ { 100 } \frac { 50 } { 20 - \sqrt { h } } \mathrm {~d} h$$

Using the substitution $h = ( 20 - x ) ^ { 2 }$, or otherwise,
\item find the exact value of $\int _ { 0 } ^ { 100 } \frac { 50 } { 20 - \sqrt { h } } \mathrm {~d} h$.
\item Hence find the time taken to fill the cylinder from empty to a height of 100 cm , giving your answer in minutes and seconds to the nearest second.\\[0pt]

\end{enumerate}

\hfill \mbox{\textit{SPS SPS FM Pure 2023 Q3 [13]}}