Water is flowing at the rate of 3 km/hr through a circular pipe of 20 cm internal diameter into a circular cistern of diameter 10 m and depth 2 m. In how much time will the cistern be filled?

Let the cistern be filled in x hours.

Water column forms a cylinder of radius (r) = $\dfrac{\text{diameter}}{2} = \dfrac{20}{2} \times \dfrac{1}{100} = \dfrac{1}{10} \text{ m}.$

Given, water is flowing at the rate of 3 km/hr through a circular pipe.

= $\dfrac{3 \times 1000}{60 \times 60} = \dfrac{5}{6}$ m/s

Volume of water that flows in 1 second = Area of cross section of pipe × rate of flow of water

= πr² × rate of flow of water

$= \dfrac{22}{7} \times \Big(\dfrac{1}{10}\Big)^2 \times \dfrac{5}{6} \\[1em] = \dfrac{22}{7} \times \dfrac{1}{100} \times \dfrac{5}{6} \\[1em] = \dfrac{110}{4200} \\[1em] = \dfrac{11}{420}$

Given, diameter of cistern = 10 m

Radius (r) = $\dfrac{\text{diameter}}{2} = \dfrac{10}{2} = 5$ m

Depth of cistern = 2 m

Volume of cistern = πr²h

$= \dfrac{22}{7} \times 5^2 \times 2 \\[1em] = \dfrac{22}{7} \times 25 \times 2 \\[1em] = \dfrac{1100}{7}$

Required time = $\dfrac{\text{Volume of cistern}}{\text{volume of water flows in 1 second}}$

$= \dfrac{\dfrac{1100}{7}}{\dfrac{11}{420}} \\[1em] = \dfrac{1100}{7} \times \dfrac{420}{11} \\[1em] = \dfrac{462000}{77} \text{ seconds}\\[1em] = \dfrac{462000}{77} \times \dfrac{1}{60 \times 60}\text{ hours} \\[1em] = \dfrac{5}{3} \\[1em] = 1\dfrac{2}{3} \text{ hours.}$

= 1 hour 40 min.

Hence, cistern will be filled in 1 hour 40 min.