Correct Answer - Option 2 : 67 litres/sec
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:
Area of the irrigation for crop 'X' = 36 hectare
cu = 0.5 cm/day, ηa - 75%
Moisture holding capacity of soil = 18 cm/m × 1 = 18 cm
Allowable depletion of moisture = 50% of available moisture
Allowable depletion of moisture = 0.5 × 18 = 9 cm
Frequency of irrigation crop, \(\frac{{dw}}{{{C_u}}} = \frac{{9}}{0.5} = 18\;days\)
Hence water has to be supplied every 18 days
\({\rm{0.75}} = \frac{{9}}{{Depth\;of\;water\;applied~ to~ soil}}\)
Depth of water applied = 12 cm
Quantity of water required = 0.12 × 36 × 104 = 43200 m3
\(Capacity ~of~ pump= \frac{{{\rm{43200\times 10^3}}}}{{18\times 10\times 60\times 60}} \)
⇒ 66.67 litres/sec