8. 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 \(4000 \mathrm {~cm} ^ { 2 }\).

1. 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 \(400 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\).
2. Show that \(k = 0.02\)
3. 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,
4. find the exact value of \(\int _ { 0 } ^ { 100 } \frac { 50 } { 20 - \sqrt { h } } \mathrm {~d} h\).
5. 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.