A farmer connects a pipe of internal diameter 25 cm from a canal into a cylindrical tank in his field, which is 12 m in diameter and 2.5 m deep.

Solution:

We have,

the radius of the cylindrical tank, $R=\frac{12}{2}=6 \mathrm{~m}=600 \mathrm{~cm}$,

the depth of the tank, $H=2.5 \mathrm{~m}=250 \mathrm{~cm}$,

the radius of the cylindrical pipe, $r=\frac{25}{2}=12.5 \mathrm{~cm}$,

speed of the flowing water, $h=3.6 \mathrm{~km} / \mathrm{h}=\frac{360000 \mathrm{~m}}{3600 \mathrm{~s}}=100 \mathrm{~cm} / \mathrm{s}$

Now,

Volume of water flowing out from the pipe in a hour $=\pi r^{2} h$

$=\frac{22}{7} \times 12.5 \times 12.5 \times 100 \mathrm{~cm}^{3}$

Also,

Volume of the $\operatorname{tank}=\pi R^{2} H$

$=\frac{22}{7} \times 600 \times 600 \times 250 \mathrm{~cm}^{3}$

So, the time taken to fill the $\operatorname{tank}=\frac{\text { Volume of the tank }}{\text { Volume of water flowing out from the pipe in a hour }}$

$=\frac{\frac{22}{7} \times 600 \times 600 \times 250}{\frac{22}{7} \times 12.5 \times 12.5 \times 100}$

$=5760 \mathrm{~s}$

$=\frac{5760}{3600}$

$=1.6 \mathrm{~h}$

Also, the cost of water $=0.07 \times \frac{22}{7} \times 6 \times 6 \times 2.5=₹ 19.80$