Problem Statement
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 for irrigation?
Verified Solution & Marking Scheme
Calculate Length of Water Column in 30 Minutes
Speed of water $= 10$ km/h $= 10,000$ m/h.
Time $= 30$ minutes $= \frac{30}{60} = 0.5$ hour.
Length of water flowing in 30 minutes ($L$):
$L = 10,000 \times 0.5 = 5,000 \text{ m}$
Compute Total Volume of Water Discharged
The cross-section of the canal is rectangular with:
- Width ($b$) $= 6$ m
- Depth ($h$) $= 1.5$ m
Volume of water flowing in 30 minutes ($V$):
$V = L \times b \times h = 5000 \times 6 \times 1.5 = 5000 \times 9 = 45,000 \text{ m}^3$
Relate Volume to Irrigated Area and Standing Height
Let the irrigated area be $A$ m$^2$.
Standing water height needed $= 8$ cm $= \frac{8}{100}$ m $= 0.08$ m.
Volume of standing water required $= A \times 0.08$ m$^3$.
Equating volumes:
$A \times 0.08 = 45,000$
$A = \frac{45,000}{0.08} = \frac{4,500,000}{8} = 562,500 \text{ m}^2$
Converting to hectares ($1 \text{ hectare} = 10,000 \text{ m}^2$):
$A = \frac{562,500}{10,000} = 56.25 \text{ hectares}$.