Water is flowing through a cylindrical pipe of internal diameter 2 cm,

Solution:

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.