Water in a canal,$6 \, m$ wide and $1.5 \, m$ deep,is flowing with a speed of $10 \, km/h$. How much area will it irrigate in $30 \, minutes$,if $8 \, cm$ of standing water is needed?

(N/A) The cross-sectional area of the canal is $6 \, m \times 1.5 \, m = 9 \, m^2$.
The speed of water is $10 \, km/h = \frac{10000 \, m}{60 \, min} = \frac{500}{3} \, m/min$.
The volume of water that flows from the canal in $30 \, minutes$ is:
$V = \text{Area} \times \text{Speed} \times \text{Time} = 9 \, m^2 \times \frac{500}{3} \, m/min \times 30 \, min = 45000 \, m^3$.
Let the irrigated area be $A$. The volume of water required for irrigation is $A \times \text{depth}$.
Given depth $= 8 \, cm = 0.08 \, m$.
Equating the volumes:
$A \times 0.08 \, m = 45000 \, m^3$
$A = \frac{45000}{0.08} \, m^2 = 562500 \, m^2$.
Thus,the area irrigated in $30 \, minutes$ is $562500 \, m^2$.