Question

Water flows out through a circular pipe whose internal diameter is $2\mathrm{cm}$ at the rate of $6\mathrm{metre}\mathrm{per}\mathrm{second}$ into a cylindrical tank the radius of whose base is $60\mathrm{cm}$. How much will the level of water rise in $30\mathrm{minutes}$?

Answer 1

Step 1: Finding out the volume of water that flows out through pipe

We have,

Internal diameter of the pipe$=2\mathrm{cm}$
So, its radius$=1\mathrm{cm}=\frac{1}{100}\mathrm{m}$
It is given that, water that flows out through the pipe in $6\mathrm{m}/\mathrm{s}$
So,

$\mathrm{Volume}\mathrm{of}\mathrm{water}\mathrm{that}\mathrm{flows}\mathrm{out}\mathrm{through}\mathrm{the}\mathrm{pipe}\mathrm{in}1\sec ={\mathrm{\pi r}}^2\mathrm{h}$

$=\mathrm{\pi }\times {\left(\frac{1}{100}\right)}^2\times 6{\mathrm{m}}^3$
$\therefore \mathrm{In}30\mathrm{minutes},$ $\mathrm{Volume}\mathrm{of}\mathrm{water}\mathrm{flow}=\mathrm{\pi }\times \frac{1}{100\times 100}\times 6\times 30\times 60{\mathrm{m}}^3$
This must be equal to the volume of water that rises in the cylindrical tank after $30\mathrm{minutes}$ and height up to which it rises say $\mathrm{h}$.
Step 2: Finding out the volume of tank

Radius of tank $=60\mathrm{cm}=\frac{60}{100}\mathrm{m}$
$\begin{array}{rcl}\mathrm{Volume}\mathrm{of}\mathrm{tank}& =& {\mathrm{\pi r}}^2\mathrm{h}\\ & =& \mathrm{\pi }{\left(\frac{60}{100}\right)}^2\mathrm{h}\end{array}$

Step 3: Finding the rise in level of water

$$\begin{gathered} \mathrm{Volume}\mathrm{of}\mathrm{water}\mathrm{that}\mathrm{flows}\mathrm{out}=\mathrm{Volume}\mathrm{of}\mathrm{tank} \\ \Rightarrow \mathrm{\pi }\times \frac{1}{100\times 100}\times 6\times 30\times 60=\mathrm{\pi }\times {\left(\frac{60}{100}\right)}^2\times \mathrm{h} \\ \Rightarrow \frac{6\times 30\times 30}{100\times 100}=\frac{60\times 60}{100\times 100}\mathrm{h} \\ \Rightarrow \mathrm{h}=\frac{3\times 36}{36} \\ \Rightarrow \mathrm{h}=3\mathrm{m} \end{gathered}$$
Hence, required height will be $3\mathrm{m}.$