Question

A farmer connects a pipe of internal diameter $20\mathrm{cm}$ form a canal into a cylindrical tank in her field, which is $10\mathrm{m}$ in diameter and $2\mathrm{m}$ deep.

If water flows through the pipe at the rate of $3\mathrm{km}/\mathrm{hr}$, in how much time will the tank be filled?

Answer 1

Step 1:Finding out the volume of water flowing and that of cylindrical tank.

Radius of Cylindrical tank $\left(\mathrm{r}\right)=5\mathrm{m}=500\mathrm{cm}[\mathrm{Since}\mathrm{radius}=\frac{\mathrm{diameter}}{2},1\mathrm{m}=100\mathrm{cm}]$

Depth of Cylindrical tank $\left(\mathrm{h}\right)=2\mathrm{m}=200\mathrm{cm}$

Speed of water flows in pipe $=3\mathrm{km}/\mathrm{hr}=(3,000\mathrm{m}/\mathrm{hr})[\because 1\mathrm{km}=1000\mathrm{m}]$

Volume of water flows in $1\mathrm{hr}={\mathrm{\pi r}}^2\mathrm{h}=\mathrm{\pi }\times 10\times 10\times 300000{\mathrm{cm}}^3$

Volume of cylindrical tank $={\mathrm{\pi r}}^2\mathrm{h}=\mathrm{\pi }\times 500\times 500\times 200{\mathrm{cm}}^3$

Step 2 : Finding the time taken to fill the tank.

$\mathrm{Time}\mathrm{taken}\mathrm{to}\mathrm{fill}\mathrm{the}\mathrm{tank}=\mathrm{Volume}\mathrm{of}\mathrm{cylindrical}\mathrm{tank}÷\mathrm{Volume}\mathrm{of}\mathrm{water}\mathrm{flow}\mathrm{in}1\mathrm{hr}$

$$\begin{gathered} \Rightarrow \mathrm{t}\times \left(30\mathrm{\pi }\right)={\mathrm{\pi r}}^2\mathrm{h} \\ \Rightarrow \mathrm{t}\times \left(30\mathrm{\pi }\right)=\mathrm{\pi }\times 5^2\times 2 \\ \Rightarrow \mathrm{t}=\frac{5\mathrm{x}5\mathrm{x}2}{30} \\ \Rightarrow \mathrm{t}=\frac{5}{3}\mathrm{hours} \\ \Rightarrow \mathrm{t}=\frac{5}{3}\times 60\mathrm{mins}[\mathrm{Since}1\mathrm{hr}=60\mathrm{mins}] \\ \Rightarrow \mathrm{t}=100\mathrm{minutes} \\ \Rightarrow \mathrm{t}=1\mathrm{hour}40\mathrm{minutes} \end{gathered}$$

Hence, The time required to fill the tank is $1\mathrm{hour}40\mathrm{minutes}$.