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 |

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 2 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 3 |

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 3 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 4 |

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 4 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 5 |

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 5 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 6 |

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 6 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 7 |

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 7 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 8 |

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 8 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 9 |

If the path of an irrigation canal is below the bed level of a natural stream, the type of cross-drainage structure provided is

Aqueduct | |

Level crossing | |

Sluice gate | |

Super passage |

Question 9 Explanation:

Irrigation canal below the bed level of a natural stream

\rightarrow Super passage

\rightarrow Super passage

Question 10 |

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 10 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}

There are 10 questions to complete.