Question 1 |

In the context of cross-drainage structures, the correct statement(s) regarding
the relative positions of a natural drain (stream/river) and an irrigation canal,
is/are

In an aqueduct, natural drain water goes under the irrigation canal, whereas in a super-passage, natural drain water goes over the irrigation canal. | |

In a level crossing, natural drain water goes through the irrigation canal. | |

In an aqueduct, natural drain water goes over the irrigation canal, whereas in a super-passage, natural drain water goes under the irrigation canal. | |

In a canal syphon, natural drain water goes through the irrigation canal. |

Question 1 Explanation:

In Aqueduct and Syphon aqueduct under irrigation
canal or canal passes over natural drain.

In Super passage and Canal syphon or simply syphon, natural drain passes over irrigation canal or canal goes under natural drain.

In level crossing, natural drain water goes through the irrigation canal.

In Super passage and Canal syphon or simply syphon, natural drain passes over irrigation canal or canal goes under natural drain.

In level crossing, natural drain water goes through the irrigation canal.

Question 2 |

During a particular stage of the growth of a crop, the consumptive use of water is
2.8 mm/day. The amount of water available in the soil is 50 % of the maximum
depth of available water in the root zone. Consider the maximum root zone depth
of the crop as 80 mm and the irrigation efficiency as 70%.

The interval between irrigation (in days) will be _________. (round off to the nearest integer)

The interval between irrigation (in days) will be _________. (round off to the nearest integer)

14 | |

45 | |

58 | |

78 |

Question 2 Explanation:

MTA-Marks to ALL

Frequency = \frac{\text{Depth of water needed}}{\text{Consumptive use}}=\frac{40}{2.8}=14.28 days

Frequency = \frac{\text{Depth of water needed}}{\text{Consumptive use}}=\frac{40}{2.8}=14.28 days

Question 3 |

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 3 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 4 |

An unlined canal under regime conditions along with a silt factor of 1 has a width of flow 71.25m. Assuming the unlined canal as a wide channel, the corresponding average depth of flow (in m, round off to two decimal places) in the canal will be ______________

1.25 | |

4.58 | |

3.82 | |

2.94 |

Question 4 Explanation:

R = D (for wide rectangular channel)

\begin{aligned} &\begin{aligned} A f^{2} &=140\left(\frac{2}{5} f R\right)^{5 / 2} \\ (B D) f^{2} &=140\left(\frac{2}{5} f \times D\right)^{5 / 2} \\ (71.25 \times D) \times 1 &=140\left(\frac{2}{5} \times 1 \times D\right)^{5 / 2} \\ D \times 0.5089 &=\left(\frac{2}{5}\right)^{5 / 2} \times(D)^{5 / 2} \\ D^{3 / 2} &=5.029 \\ \text { or } \qquad \qquad \qquad \qquad D &=2.94 \mathrm{~m} \end{aligned} \end{aligned}

Question 5 |

A concrete dam holds 10 m of static water as shown in the figure (not drawn to the
scale). The uplift assumed to vary linearly from full hydrostatic head at the heel, to zero
at the toe of dam. The coefficient of friction between the dam and foundation soil is 0.45.
Specific weights of concrete and water are 24 kN/m^3 and 9.81 kN/m^3, respectively.

For NO sliding condition, the required minimum base width B (in m, round off to two decimal places) is __________.

For NO sliding condition, the required minimum base width B (in m, round off to two decimal places) is __________.

12.45 | |

18.26 | |

14.12 | |

15.87 |

Question 5 Explanation:

\begin{aligned} \mu &=0.45\\ \gamma_{conc} &=24 kN/m^3\\ B_{min. sliding}&=\frac{10}{0.45(2.4-1)}\\ &=15.873m \end{aligned}

Question 6 |

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 6 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 7 |

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 7 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 8 |

At the foot of a spillway, water flows at a depth of 23 cm with a velocity of 8.1 m/s, as shown in the figure.

The flow enters as an M-3 profile in the long wide rectangular channel with bed slope = \frac{1}{1800} and Manning's n= 0.015. A hydraulic jump is formed at a certain distance from the foot of the spillway. Assume the acceleration due to gravity, g = 9.81 m/s^2. Just before the hydraulic jump, the depth of flow y_1 (in m, round off to 2 decimal places) is _______

The flow enters as an M-3 profile in the long wide rectangular channel with bed slope = \frac{1}{1800} and Manning's n= 0.015. A hydraulic jump is formed at a certain distance from the foot of the spillway. Assume the acceleration due to gravity, g = 9.81 m/s^2. Just before the hydraulic jump, the depth of flow y_1 (in m, round off to 2 decimal places) is _______

0.24 | |

0.42 | |

0.82 | |

0.96 |

Question 8 Explanation:

\begin{aligned} y &=0.23m \\ V&=8.1m/s \\ q&=Vy=0.23 \times 8.1 \\ &= 1.863 m^3/s-m\\ S_0&=\frac{1}{800} \\ D&=0.015 \end{aligned}

y_n= Normal depth of flow

R = y for wide rectangular channel

By Manning's equation

\begin{aligned} q&=\frac{y_n}{n}R^{2/3}S_0^{1/2} \\ 1.863&=\frac{y_n^{5/3}}{0.015} \times \left ( \frac{1}{1800} \right )^{1/2} \\ y_n&= 1.108m \end{aligned}

y_1 is conjugate depth of y_0,

\begin{aligned} \frac{y_1}{y_n}&=\frac{-1+\sqrt{1+8Fr_n^2}}{2} \\ \because &\;\;\left ( Fr_n^2=\frac{q^2}{gy_n^3} \right ) \\ y_1&=\frac{-1+\sqrt{1+8 \times \frac{1.863^2}{9.81 \times 1.108^3}}}{2} \times 1.108\\ &=0.418m\simeq 0.42m \end{aligned}

Question 9 |

A confined aquifer of 15 m constant thickness is sandwiched between two aquicludes as shown in the figure.

The heads indicated by two piezometers P and Q are 55.2 m and 34.1 m, respectively. The aquifer has a hydraulic conductivity of 80 m/day and its effective porosity is 0.25. If the distance between the piezometers is 2500 m, the time taken by the water to travel through the aquifer from piezometer location P to Q (in days,round off to 1 decimal place) is______

The heads indicated by two piezometers P and Q are 55.2 m and 34.1 m, respectively. The aquifer has a hydraulic conductivity of 80 m/day and its effective porosity is 0.25. If the distance between the piezometers is 2500 m, the time taken by the water to travel through the aquifer from piezometer location P to Q (in days,round off to 1 decimal place) is______

925.6 | |

654.4 | |

369.6 | |

254.2 |

Question 9 Explanation:

Discharge velocity,

\begin{aligned} V&=ki\\ V&=k\frac{h_1-h_2}{L}\\ &=80 \left ( \frac{55.2-34.1}{2500} \right )\\ &=0.6752 m/day\\ &\text{Porosity}, n =0.25\\ &\therefore \;\;\text{Seepage velocity,}\\ V_s&=\frac{V}{n}=\frac{0.6752}{0.25}\\ &=2.7008 m/day\\ \text{Time taken}&=\frac{L}{V_s}\\ &=\frac{2500}{2.7008}=925.65 days \end{aligned}

\begin{aligned} V&=ki\\ V&=k\frac{h_1-h_2}{L}\\ &=80 \left ( \frac{55.2-34.1}{2500} \right )\\ &=0.6752 m/day\\ &\text{Porosity}, n =0.25\\ &\therefore \;\;\text{Seepage velocity,}\\ V_s&=\frac{V}{n}=\frac{0.6752}{0.25}\\ &=2.7008 m/day\\ \text{Time taken}&=\frac{L}{V_s}\\ &=\frac{2500}{2.7008}=925.65 days \end{aligned}

Question 10 |

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 10 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

There are 10 questions to complete.