The problem asks for the command area of a tubewell, given its discharge rate, daily operating hours, irrigation interval, and irrigation depth.
Total Hours = Daily Operating Hours $\times$ Number of Days
Total Hours = $20$ h/day $\times 20$ days = $400$ h
Total Volume ($V$) = Discharge ($Q$) $\times$ Total Hours
$V = 40$ m$^3$ h$^{-1} \times 400$ h = $16000$ m$^3$
$d = 8$ cm = $0.08$ m
Area ($A$) = Total Volume ($V$) / Depth ($d$)
$A = 16000$ m$^3$ / $0.08$ m = $200000$ m$^2$
Area in ha = Area in m$^2$ / $10000$
Area in ha = $200000$ m$^2$ / $10000$ m$^2$/ha = $20$ ha
The command area of the tubewell is calculated to be 20 ha.
Irrigation water having SAR value of 20, is called
The ratio between the areas of a crop irrigated and the quantity of water required during its entire period of growth is known as
Alkaline soils are reclaimed by
For crops like potato and sugar-beets, the commonly adopted method of surface irrigation is