Question

A cylindrical water tank of diameter 1.4 m and height 2.1 m is being fed by a pipe of diameter 3.5 cm through which water flows at the rate of 2 metre per second. In how much time the tank will be filled?

Answer 1

$$\begin{gathered} \mathrm{Radius}\mathrm{of}\mathrm{the}\mathrm{cylindrical}\mathrm{tank}=0.7\mathrm{m} \\ \mathrm{Height}\mathrm{of}\mathrm{the}\mathrm{cylindrical}\mathrm{tank}=2.1\mathrm{m} \\ \mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{cylindrical}\mathrm{tank}=\mathrm{\pi }(0.7{)}^2(2.1){\mathrm{m}}^3 \\ \mathrm{Length}\mathrm{of}\mathrm{the}\mathrm{water}\mathrm{column}\mathrm{flown}\mathrm{from}\mathrm{the}\mathrm{pipe}\mathrm{in}1\mathrm{s}=2\mathrm{m} \\ \mathrm{Let}\mathrm{the}\mathrm{time}\mathrm{taken}\mathrm{to}\mathrm{completely}\mathrm{fill}\mathrm{the}\mathrm{water}\mathrm{tank}\mathrm{be}x\sec . \\ \mathrm{Length}\mathrm{of}\mathrm{the}\mathrm{water}\mathrm{column}\mathrm{flown}\mathrm{from}\mathrm{the}\mathrm{pipe}\mathrm{in}x\sec =2\mathit{}x\mathit{}\mathrm{m} \\ \mathrm{Radius}\mathrm{of}\mathrm{the}\mathrm{pipe}=1.75\mathrm{cm}=0.0175\mathrm{m} \\ \mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{water}\mathrm{column}\mathrm{flown}\mathrm{from}\mathrm{the}\mathrm{pipe}\mathrm{in}x\sec =\mathrm{\pi }(0.0175{)}^2(2x){\mathrm{m}}^3 \\ \mathrm{V}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}\mathrm{o}\mathrm{f}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{c}\mathrm{y}\mathrm{l}\mathrm{i}\mathrm{n}\mathrm{d}\mathrm{r}\mathrm{i}\mathrm{c}\mathrm{a}\mathrm{l}\tan \mathrm{k}=\mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{water}\mathrm{column}\mathrm{flown}\mathrm{from}\mathrm{the}\mathrm{pipe} \\ \mathrm{\pi }(0.7{)}^2(2.1)=\mathrm{\pi }(0.0175{)}^2(2x) \\ \mathrm{x}=\frac{0.7{)}^2(2.1)}{0.0175{)}^2(2)}=1680\sec =28\min \end{gathered}$$

Thus, the time required to fill the water tank is 28 min.