Question

Water is flowing at a rate of $7\mathrm{m}/\mathrm{s}$ through a circular pipe whose internal diameter is $2\mathrm{cm}$ into a cylindrical tank the radius of whose base is $40\mathrm{cm}$. Determine the increase in water level in $\frac{1}{2}\mathrm{hours}$?

Answer 1

Step 1: Writing the given information

We have,

The internal diameter of the pipe$=2\mathrm{cm}$

Internal radius of the pipe$=1\mathrm{cm}$

The radius of the base of the tank$=40\mathrm{cm}$

Speed of water flowing out from the pipe$=7\mathrm{m}/\sec$

$$\begin{gathered} =7\times 100\times 60 \\ =42000\mathrm{cm}/\min [\mathrm{Since}1\mathrm{m}=100\mathrm{cm},1\mathrm{minute}=60\mathrm{seconds}] \end{gathered}$$

Step 2: Finding the volume of water flowing out in half an hour

$\begin{array}{rcl}\mathrm{Quantity}\mathrm{of}\mathrm{water}\mathrm{flows}\mathrm{out}\mathrm{from}\mathrm{pipe}\mathrm{in}1\min & =& {\mathrm{\pi r}}^2\mathrm{h}\\ & =& \mathrm{\pi }\times 1^2\times 42000{\mathrm{cm}}^3\end{array}$

$\begin{array}{rcl}\mathrm{Quantity}\mathrm{of}\mathrm{water}\mathrm{flows}\mathrm{out}\mathrm{from}\mathrm{pipe}\mathrm{in}30\min & =& {\mathrm{\pi r}}^2\mathrm{h}\times 30\\ & =& \mathrm{\pi }\times 1^2\times 42000\times 30\\ & =& 3960000{\mathrm{cm}}^3\end{array}$

Step 3: Finding the height of the water level in the tank

Let $\mathrm{H}$ be the height of the water level in the tank.

$$\begin{gathered} \mathrm{Quantity}\mathrm{of}\mathrm{water}\mathrm{filled}\mathrm{in}\mathrm{tank}\mathrm{in}30\min =3960000{\mathrm{cm}}^3 \\ \Rightarrow \mathrm{\pi }\times {40}^2\times \mathrm{H}=3960000 \\ \Rightarrow \mathrm{H}=3960000\times \frac{7}{22\times 1600} \\ \Rightarrow \mathrm{H}=787.5\mathrm{cm} \end{gathered}$$

Therefore, the water level increases to $787.5\mathrm{cm}$ in half-hour.