Water is flowing at the rate of 8 m per second through a circular pipe whose internal diameter is 2 cm, into a cylindrical tank, the radius of whose base is 40 cm. Determine the increase in the water level in 30 minutes.

Internal radius of circular pipe = $\dfrac{\text{Diameter}}{2} = \dfrac{2}{2} = 1 \text{ cm.}$

Volume of water flowing through pipe per second = Area of cross section of pipe × rate of flow of water

= πr² × 8 m/s

= πr² × 8 × 100 cm/s

$= \dfrac{22}{7} \times (1)^2 \times 800 \\[1em] = \dfrac{22}{7} \times 800 \\[1em] = \dfrac{17600}{7} \text{ cm}^3/s$

Volume of cylindrical tank = πR²h

$= \dfrac{22}{7} \times 40^2 \times \text{h} \\[1em] = \dfrac{22}{7} \times 1600 \times \text{h} \\[1em] = \dfrac{35200}{7} \times \text{h}$

Volume of water flowing through pipe in 30 minutes = Volume of cylindrical tank

(1 minute = 60 second so, 30 minutes = 30 × 60 = 1800 s)

$\Rightarrow 1800 \times \dfrac{17600}{7} = \dfrac{35200}{7} \times \text{h} \\[1em] \Rightarrow \dfrac{31680000}{7} = \dfrac{35200}{7} \times \text{h} \\[1em] \Rightarrow \text{h} = \dfrac{31680000}{7} \times \dfrac{7}{35200} \\[1em] \Rightarrow \text{h} = 900 \text{ cm.} \\[1em] \Rightarrow \text{h} = 9 \text{ m.}$

Hence, the increase in the water level in 30 minutes is 9 m.