Question

A farmer connects a pipe of internal diameter 25 cm from a canal into a cylindrical tank in his field, which is 12 m in diameter and 2.5 m deep. If water flows through the pipe at the rate of 3.6 km/hr, then in how much time will the tank be filled? Also, find the cost of water if the canal department charges at the rate of ₹0.07 per m³. [CBSE 2009C]

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{cylindrical}\mathrm{tank},R=\frac{12}{2}=6\mathrm{m}=600\mathrm{cm}, \\ \mathrm{the}\mathrm{depth}\mathrm{of}\mathrm{the}\mathrm{tank},H=2.5\mathrm{m}=250\mathrm{cm}, \\ \mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{cylindrical}\mathrm{pipe},r=\frac{25}{2}=12.5\mathrm{cm}, \\ \mathrm{speed}\mathrm{of}\mathrm{the}\mathrm{flowing}\mathrm{water},h=3.6\mathrm{km}/\mathrm{h}=\frac{360000\mathrm{m}}{3600\mathrm{s}}=100\mathrm{cm}/\mathrm{s} \\ \mathrm{Now}, \\ \mathrm{Volume}\mathrm{of}\mathrm{water}\mathrm{flowing}\mathrm{out}\mathrm{from}\mathrm{the}\mathrm{pipe}\mathrm{in}\mathrm{a}\mathrm{hour}=\mathrm{\pi }{r}^{2}h \\ =\frac{22}{7}\times 12.5\times 12.5\times 100{\mathrm{cm}}^3 \\ \mathrm{Also}, \\ \mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{tank}=\mathrm{\pi }{R}^{2}H \\ =\frac{22}{7}\times 600\times 600\times 250{\mathrm{cm}}^3 \\ \mathrm{So},\mathrm{the}\mathrm{time}\mathrm{taken}\mathrm{to}\mathrm{fill}\mathrm{the}\mathrm{tank}=\frac{\mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{tank}}{\mathrm{Volume}\mathrm{of}\mathrm{water}\mathrm{flowing}\mathrm{out}\mathrm{from}\mathrm{the}\mathrm{pipe}\mathrm{in}\mathrm{a}\mathrm{hour}} \\ =\frac{{\displaystyle \frac{22}{7}}\times 600\times 600\times 250}{{\displaystyle \frac{22}{7}}\times 12.5\times 12.5\times 100} \\ =5760\mathrm{s} \\ =\frac{5760}{3600} \\ =1.6\mathrm{h} \\ \mathrm{Also},\mathrm{the}\mathrm{cost}\mathrm{of}\mathrm{water}=0.07\times \frac{22}{7}\times 6\times 6\times 2.5=₹19.80 \end{gathered}$$