Water in a canal, 5.4 m wide and 1.8 m deep, is flowing with a speed of 25 km/hr.

Solution:

Width of the canal $=5.4 \mathrm{~m}$

Depth of the canal $=1.8 \mathrm{~m}$

Height of the standing water needed for irrigation $=10 \mathrm{~cm}=0.1 \mathrm{~m}$

Speed of the flowing water $=25 \mathrm{~km} / \mathrm{h}=\frac{25000}{60}=\frac{1250}{3} \mathrm{~m} / \mathrm{min}$

Volume of water flowing out of the canal in $1 \mathrm{~min}$

$=$ Area of opening of canal $\times \frac{1250}{3}$

$=5.4 \times 1.8 \times \frac{1250}{3}$

$=4050 \mathrm{~m}^{3}$

$\therefore$ Volume of water flowing out of the canal in $40 \mathrm{~min}=40 \times 4050 \mathrm{~m}^{3}=162000 \mathrm{~m}^{3}$

Now,

Area of irrigation

$=\frac{\text { Volume of water flowing out from canal in } 40 \mathrm{~min}}{\text { Height of the standing water needed for irrigation }}$

$=\frac{162000}{0.1}$

$=1620000 \mathrm{~m}^{2}$

$=162$ hectare $\quad\left(\because 1\right.$ hectare $\left.=10000 \mathrm{~m}^{2}\right)$

Thus, the area irrigated in 40 minutes is 162 hectare.