Question

A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank which is 10 m in diameter and 2 m deep. If the water flows through the pipe at the rate of 4 km/hr, then in how much time will the tank be filled completely?

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \mathrm{Internal}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{pipe},r=\frac{20}{2}=10\mathrm{cm}=0.1\mathrm{m}, \\ \mathrm{Radius}\mathrm{of}\mathrm{the}\mathrm{cylindrical}\mathrm{tank},R=\frac{10}{2}=5\mathrm{m}\mathrm{and} \\ \mathrm{Height}\mathrm{of}\mathrm{the}\mathrm{cylindrical}\mathrm{tank},H=2\mathrm{m} \\ \mathrm{Also},\mathrm{the}\mathrm{speed}\mathrm{of}\mathrm{the}\mathrm{water}\mathrm{flow}\mathrm{in}\mathrm{the}\mathrm{pipe},h=4\mathrm{km}/\mathrm{hr}=\frac{4\times 1000\mathrm{m}}{1\mathrm{hr}}=4000\mathrm{m}/\mathrm{hr} \\ \mathrm{Now}, \\ \mathrm{The}\mathrm{volume}\mathrm{of}\mathrm{the}\mathrm{water}\mathrm{flowing}\mathrm{out}\mathrm{of}\mathrm{the}\mathrm{pipe}\mathrm{in}\mathrm{a}\mathrm{hour}=\mathrm{\pi }{r}^{2}h \\ =\frac{22}{7}\times 0.1\times 0.1\times 4000 \\ =\frac{880}{7}{\mathrm{m}}^3 \\ \mathrm{And}, \\ \mathrm{The}\mathrm{volume}\mathrm{of}\mathrm{the}\mathrm{cylindrical}\mathrm{tank}=\mathrm{\pi }{R}^{2}H \\ =\frac{22}{7}\times 5\times 5\times 2 \\ =\frac{1100}{7}{\mathrm{m}}^3 \\ \mathrm{So}, \\ \mathrm{The}\mathrm{time}\mathrm{taken}\mathrm{to}\mathrm{fill}\mathrm{the}\mathrm{tank}=\frac{\mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{cylindrical}\mathrm{tank}}{\mathrm{Volume}\mathrm{of}\mathrm{water}\mathrm{flowing}\mathrm{out}\mathrm{of}\mathrm{the}\mathrm{pipe}\mathrm{in}\mathrm{a}\mathrm{hour}} \\ =\frac{\left({\displaystyle \frac{1100}{7}}\right)}{\left({\displaystyle \frac{880}{7}}\right)} \\ =\frac{1100}{880} \\ =\frac{5}{4}\mathrm{hr} \\ =1\frac{1}{4}\mathrm{hr} \\ =1\mathrm{hr}\mathrm{and}\frac{1}{4}\times 60\min \\ =1\mathrm{hr}15\min \end{gathered}$$

So, the tank will be completely filled in 1 hour 15 minutes.