Water is flowing through a cylindrical pipe of internal diameter 2 cm, into a cylindrical tank of base radius 40 cm, at the rate of 0.4 m per second. Determine the rise in level of water in the tank in half an hour.
We have,
the internal radius of the cylindrical pipe, $r=\frac{2}{2}=1 \mathrm{~cm}$ and
the base radius of cylindrical tank, $R=40 \mathrm{~cm}$,
Also, the rate of water flow, $h=0.4 \mathrm{~m} / \mathrm{s}=40 \mathrm{~cm} / \mathrm{s}$
Let the rise in level of water be $H$.
Now,
The volume of water flowing out of the cylindrical pipe in $1 \mathrm{sec}=\pi r^{2} h=\pi \times 1 \times 1 \times 40=40 \pi \mathrm{cm}^{3}$
So, the volume of water flowing out of the cylindrical pipe in half an hour $(30 \mathrm{~min})=40 \pi \times 60 \times 30=72000 \pi \mathrm{cm}^{3}$
As,
Volume of water in the cylindrical tank = Volume of standing water in cylindrical pipe for half an hour
$\Rightarrow \pi R^{2} H=72000 \pi$
$\Rightarrow R^{2} H=72000$
$\Rightarrow 40 \times 40 \times H=72000$
$\Rightarrow H=\frac{72000}{40 \times 40}$
$\therefore H=45 \mathrm{~cm}$
So, the rise in level of water in the tank in half an hour is 45 cm.
Disclaimer: The answer given in the textbook is incorrect. The same has been corrected above.
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