Question 1 |

A reservoir with a live storage of 300 million cubic metre irrigates 40000 hectares ( 1 hectare= 10^{4} \mathrm{~m}^{2}) of a crop with two fillings of the reservoir. If the base period of the crop is 120 days, the duty for this crop (in hectares per cumec, round off to integer) will then be ________

691 | |

127 | |

985 | |

456 |

Question 1 Explanation:

\begin{aligned} \text { Live storage } &=300 \mathrm{Mm}^{3} \\ \text { Area } &=40000 \text { hectare }\\ \text {Since 2 filling,} & \text{ so volume of water needed,} \\ &=600 \mathrm{Mm}^{3}\\ B&=120 days \\\text { Duty } &=\frac{8.64 B}{\Delta} \\ \Delta &=\frac{600 \times 10^{6}}{40000 \times 10^{4}}=1.5 \mathrm{~m} \\ \text { Duty } &=\frac{8.64 \times 120}{1.5}=691.2 \text { ha/cumec } \end{aligned}

Question 2 |

Crops are grown in a field having soil, which has field capacity of 30% and permanent
wilting point of 13%. The effective depth of root zone is 80 cm. Irrigation water is supplied
when the average soil moisture drops to 20%. Consider density of the soil as 1500 kg/m^3
and density of water as 1000 kg/m^3. If the daily consumptive use of water for the crops
is 2 mm, the frequency of irrigating the crops (in days), is

7 | |

10 | |

11 | |

13 |

Question 2 Explanation:

As per GATE Official Answer Key MTA (Marks to All)

FC=30%

PWP=13%

\begin{aligned} d_w&=\frac{\gamma _d}{\gamma _w}d \times (FC-OMC) \\ &= \frac{1500}{1000}\times 80 (0.3-0.2)\\ &= 12cm=120mm \end{aligned}

Consumptive use = 2 mm/day

So, frequency of irrigation =120/2 = 60day

FC=30%

PWP=13%

\begin{aligned} d_w&=\frac{\gamma _d}{\gamma _w}d \times (FC-OMC) \\ &= \frac{1500}{1000}\times 80 (0.3-0.2)\\ &= 12cm=120mm \end{aligned}

Consumptive use = 2 mm/day

So, frequency of irrigation =120/2 = 60day

Question 3 |

The data for an agricultural field for a specific month are given below:

Pan Evaporation = 100 mm

Effective Rainfall = 20 mm (after deducting losses due to runoff and deep percolation)

Crop Coefficient = 0.4

Irrigation Efficiency = 0.5

The amount of irrigation water (in mm) to be applied to the field in that month, is

Pan Evaporation = 100 mm

Effective Rainfall = 20 mm (after deducting losses due to runoff and deep percolation)

Crop Coefficient = 0.4

Irrigation Efficiency = 0.5

The amount of irrigation water (in mm) to be applied to the field in that month, is

0 | |

20 | |

40 | |

80 |

Question 3 Explanation:

Water required by the crop = 100 x 0.4 = 40mm

Effective rainfall =20mm

Additional water required =20mm

Amount of water required after accounting irrigation efficiency =\frac{20}{0.5}=40mm

Effective rainfall =20mm

Additional water required =20mm

Amount of water required after accounting irrigation efficiency =\frac{20}{0.5}=40mm

Question 4 |

The command area of a canal grows only one crop, i.e., wheat. The base period of wheat is 120 days and its total water requirement, \Delta, is 40 cm. If the canal discharge is 2 m^3/s, the area, in hectares, rounded off to the nearest integer, which could be irrigated (neglectingall losses) is _______

5184 | |

4282 | |

2364 | |

9852 |

Question 4 Explanation:

Given data:

Base period, B = 120 days

Delta of crop, \Delta = 40 cm

Discharge, Q = 2 m^3/s

Area to be irrigated, A = ?

Duty of water, \Delta =\frac{864 \times B}{\Delta } ha/cumec

and Area to be irrigated; A = Q x D

\Rightarrow A = 2 m^3/s \times \frac{864 \times 120}{40}\; ha/cumec

A = 5184\; ha

Base period, B = 120 days

Delta of crop, \Delta = 40 cm

Discharge, Q = 2 m^3/s

Area to be irrigated, A = ?

Duty of water, \Delta =\frac{864 \times B}{\Delta } ha/cumec

and Area to be irrigated; A = Q x D

\Rightarrow A = 2 m^3/s \times \frac{864 \times 120}{40}\; ha/cumec

A = 5184\; ha

Question 5 |

The intensity of irrigation for the Kharif season is 50% for an irrigation project with culturable command area of 50,000 hectares. The duty for the Kharif season is 1000 hectare/cumec. Assuming transmission loss of 10%, the required discharge (in cumec, up to two decimal places) at the head of the canal is ______

27.78 | |

14.56 | |

56.47 | |

280.36 |

Question 5 Explanation:

Culturable command area =50000ha

Intensity of irrigation for kharif season =50 %

\therefore Area under kharif =25000ha

Duty for kharif season =1000 ha/cumec

Duty =\frac{\text { Area }}{\text { Discharge }}

\therefore Discharge at the head of field

\begin{aligned} Q &=\frac{25000 \mathrm{ha}}{1000 \mathrm{ha} / \mathrm{cumec}} \\ &=25 \mathrm{cumec} \end{aligned}

Transmission/conveyance loss =10 %

\therefore \quad \eta_{\text {conveyance }}=90 %

Discharge at the head of canal

\begin{array}{l} =\frac{25}{0.9} \text { cumec } \\ =27.78 \text { cumec } \end{array}

Intensity of irrigation for kharif season =50 %

\therefore Area under kharif =25000ha

Duty for kharif season =1000 ha/cumec

Duty =\frac{\text { Area }}{\text { Discharge }}

\therefore Discharge at the head of field

\begin{aligned} Q &=\frac{25000 \mathrm{ha}}{1000 \mathrm{ha} / \mathrm{cumec}} \\ &=25 \mathrm{cumec} \end{aligned}

Transmission/conveyance loss =10 %

\therefore \quad \eta_{\text {conveyance }}=90 %

Discharge at the head of canal

\begin{array}{l} =\frac{25}{0.9} \text { cumec } \\ =27.78 \text { cumec } \end{array}

Question 6 |

The culturable command area of a canal is 10,000 ha. The area grows only two crops-rice in the Kharif season and wheat in the Rabi season. The design discharge of the canal is based on the rice requirements, which has an irrigated area of 2500 ha, base period of 150 days and delta of 130 cm. The maximum permissible irrigated area (in ha) for wheat, with a base period of 120 days and delta of 50 cm, is

2600 | |

5200 | |

2500 | |

5000 |

Question 6 Explanation:

\begin{aligned} \mathrm{CCA} &=10000 \mathrm{ha} \\ \text { For rice, } \Delta_{r}&=130 \mathrm{cm}=1.3 \mathrm{m} \\ A_{r} & =2500 \mathrm{ha} \\ B_{r} & =150 \text { days } \\ \text{Duty,}\quad D_{r} & =\frac{8.64 B_{r}}{\Delta_{r}}=\frac{8.64 \times 150}{1.30} \\ D_{r} & =996.923 \mathrm{ha} / \mathrm{cumecs} \\ D_{r} & =\frac{A}{Q} \\ Q & =\frac{2500}{996.923} \mathrm{m}^{3} / \mathrm{s}=2.5077 \mathrm{m}^{3} / \mathrm{s}\\ \text{For wheat also,} &\text{discharge is }2.5077 \mathrm{m}^{3} / \mathrm{s}\\ B_{w} &=120 \text { days } \\ \Delta &=50 \mathrm{cm} \\ A_{w} &=Q \times D_{w} \\ D_{w} &=\frac{8.64 \times B_{w}}{\Delta_{w}} \\ &=\frac{8.64 \times 120}{0.5}=2073.6 \\ \therefore \quad A_{w} &=2.5077 \times 2073.6 \\ &=5199.966 \mathrm{ha} \simeq 5200 \mathrm{ha} \end{aligned}

Question 7 |

A field channel has cultivable commanded area of 2000 hectares. The intensities of irrigation for gram and wheat are 30% and 50% respectively. Gram has a kor period of 18 days, kor depth of 12 cm, while wheat has a kor period of 18 days and a kor depth of 15 cm. The discharge (in m^{3}/s) required in the field channel to supply water to the commanded area during the kor period is_____.

0.8 | |

1.2 | |

1.4 | |

1.8 |

Question 7 Explanation:

Gram and wheat are Rabi crops.

So both are grown in the same period.

Discharge required for gram,

Q_{g}=\frac{A_{g}}{D_{g}}=\frac{2000 \times 0.3}{8.64 \times \frac{18}{0.12}}=0.4630 \mathrm{m}^{3} / \mathrm{s}

Discharge required for wheat,

Q_{w}=\frac{A_{w}}{D_{w}}=\frac{2000 \times 0.5}{8.64 \times \frac{18}{0.15}}=0.9645 \mathrm{m}^{3} / \mathrm{s}

Total discharge =Q_{g}+Q_{w}=1.4275 \mathrm{m}^{3} / \mathrm{s}

So both are grown in the same period.

Discharge required for gram,

Q_{g}=\frac{A_{g}}{D_{g}}=\frac{2000 \times 0.3}{8.64 \times \frac{18}{0.12}}=0.4630 \mathrm{m}^{3} / \mathrm{s}

Discharge required for wheat,

Q_{w}=\frac{A_{w}}{D_{w}}=\frac{2000 \times 0.5}{8.64 \times \frac{18}{0.15}}=0.9645 \mathrm{m}^{3} / \mathrm{s}

Total discharge =Q_{g}+Q_{w}=1.4275 \mathrm{m}^{3} / \mathrm{s}

Question 8 |

The two columns below show some parameters and their possible values.

Which of the following options matches the parameters and the values correctly?

Which of the following options matches the parameters and the values correctly?

P-I, Q-II, R-III, S-IV | |

P-III, Q-VI, R-I, S-V | |

P-I, Q-V, R-VI, S-II | |

P-III, Q-II, R-V, S-IV |

Question 8 Explanation:

Gross Command Area is measured in hectares.

(P-III)

Duty has a unit of ha/cumec. (R-I)

Delta of wheat can be practically 40 cm

(S- V)

(P-III)

Duty has a unit of ha/cumec. (R-I)

Delta of wheat can be practically 40 cm

(S- V)

Question 9 |

Irrigation water is to be provided to a crop in a field to bring the moisture content of the soil from the existing 18% to the field capacity of the soil at 28%. The effective root zone of the crop is 70 cm. If the densities of the soil and water are 1.3 g/cm^2 and 1.0 g/cm^3respectively, the depth of irrigation water (in mm) required for irrigating the crop is ________

91 | |

81 | |

71 | |

61 |

Question 9 Explanation:

Given,

Root zone depth,

d=70 \mathrm{cm}

Field capacity,

F_{C}=28 \%

Existing moisture content,

w=18 \%

Density of soil,

\gamma=1.3 \mathrm{gm} / \mathrm{cm}^{3}

Density of water,

\gamma_{w}=1.0 \mathrm{gm} / \mathrm{cm}^{3}

Depth of irrigation water required,

\begin{aligned} d_{w} &=\frac{\gamma}{\gamma_{w}} d\left(F_{c}-w\right) \\ &=\frac{1.3}{1.0} \times(70 \times 10)(28 \%-18 \%) \\ &=1.3 \times(70 \times 10) \times \frac{10}{100} \\ &=91 \mathrm{mm} \end{aligned}

Root zone depth,

d=70 \mathrm{cm}

Field capacity,

F_{C}=28 \%

Existing moisture content,

w=18 \%

Density of soil,

\gamma=1.3 \mathrm{gm} / \mathrm{cm}^{3}

Density of water,

\gamma_{w}=1.0 \mathrm{gm} / \mathrm{cm}^{3}

Depth of irrigation water required,

\begin{aligned} d_{w} &=\frac{\gamma}{\gamma_{w}} d\left(F_{c}-w\right) \\ &=\frac{1.3}{1.0} \times(70 \times 10)(28 \%-18 \%) \\ &=1.3 \times(70 \times 10) \times \frac{10}{100} \\ &=91 \mathrm{mm} \end{aligned}

Question 10 |

The transplantation of rice requires 10 days and total depth of water required during transplantation
is 48 cm. During transplantation, there is an effective rainfall (useful for irrigation) of 8 cm. The
duty of irrigation water (in hectares/cumec) is:

612 | |

216 | |

300 | |

108 |

Question 10 Explanation:

\begin{aligned} D &=864 \mathrm{B} / \Delta \mathrm{ha} / \mathrm{cumecs} \\ &=\frac{864 \times 10}{(48-8) \times 10}=216 \mathrm{ha} / \mathrm{cumecs} \end{aligned}

There are 10 questions to complete.