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Comprehension

The moisture-holding capacity of the soil in a 100-hectare farm is 18 cm/m. The field to be irrigated when 50 percent of the available moisture in the root zone is depleted. The irrigation water is to be supplied by a pump working for 10 hours a day, and water application efficiency is 75%. Details of crops planned for cultivation are as follows

Crop

Root Zone Depth (m)

Peak rate of moisture use (mm/day)

X

1

5.0

Y

0.8

4.0


1. 40 hectares
2. 36 hectares
3. 30 hectares
4. 27 hectares

1 Answer

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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.

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