Correct Answer - Option 4 : 27 hectares
Concept:
The capacity of the pump is given by,
\(Capacity~of ~the~pump= \frac{{{\rm{Volume\;of\;water\;required}}}}{{Working\;time\;of\;pump}}\)
The frequency of irrigation is given by,
\(f_w =\frac{{dw}}{{{C_u}}} \)
dw - Depth of water to be supplied
cu - Daily consumptive use rate
The water application efficiency(η ) is given by
\({\rm{η }} = \frac{{Depth\;of\;water\;stored\;in\;root\;zone}}{{Depth\;of\;water\;applied~to ~soil}}\)
Calculation:
Given:
cu = 4 mm/day = 0.4 cm/day, ηa - 75%
Moisture holding capacity of soil = 18 cm/m × 0.8 = 14.4 cm
Allowable depletion of moisture = 50% of available moisture
Allowable depletion of moisture = 0.5 × 14.4 = 7.2m
Frequency of irrigation crop, \(\frac{{dw}}{{{C_u}}} = \frac{{7.2}}{0.4} = 18\;days\)
Hence water has to be supplied every 18 days
\({\rm{0.75}} = \frac{{7.2}}{{Depth\;of\;water\;applied~ to~ soil}}\)
Depth of water applied = 9.6 cm
Quantity of water required = 0.096 × A
\(\rm 40~litres/sec= \frac{{{\rm{0.096\times A}}}}{{18\times 10\times 60\times 60}}\)
A = 27 hectares.