Irrigation Engineering MCQ Quiz - Objective Question with Answer for Irrigation Engineering - Download Free PDF

Last updated on Jun 19, 2025

Latest Irrigation Engineering MCQ Objective Questions

Irrigation Engineering Question 1:

Direction: The item consists of two statements, one labelled as the 'Statement (I) and the other as 'Statement (II)'. You are to examine these two statements carefully and select the answers to these items using the codes given below:
Statement (1) : Cross drainage structure, aqueduct is constructed to negotiate an aligned channel over, below or at the same level of a stream.
Statement (II) : Aqueduct is constructed, when full supply level (FSL) of canal is much higher than high flood level (HFL) of a stream

  1. Both Statement (I) and Statement (II) are individually true and Statement (II) is the correct explanation of Statement (I)
  2. Both Statement (I) and Statement (II) are individually true but Statement (II) is not the correct explanation of Statement (I)
  3. Statement (I) is true but Statement (II) is false
  4. Statement (I) is false but Statement (II) is true

Answer (Detailed Solution Below)

Option 4 : Statement (I) is false but Statement (II) is true

Irrigation Engineering Question 1 Detailed Solution

Explanation:

  • Purpose of Cross Drainage Works (CDWs):
    Cross drainage works are constructed where an irrigation canal and a natural stream intersect. Their primary goal is to safely convey both canal water and stream flow without allowing one to interfere with the function of the other.

  • Function of an Aqueduct:
    An aqueduct carries the canal water over a drainage stream using a masonry or concrete trough supported by piers and abutments. It ensures uninterrupted canal supply even during high flood events in the underlying stream.

  • Aqueduct vs. Other CDWs:
    While an aqueduct carries the canal over the drainage, other CDWs like superpassages or syphons allow the stream to pass over or under the canal. Each is selected based on the relative elevations of the canal’s FSL and the stream’s HFL.

  • Design Basis of Aqueducts:
    Aqueducts are constructed when the canal’s Full Supply Level (FSL) is sufficiently higher than the stream’s High Flood Level (HFL), allowing gravity flow without pressure conditions in the canal.

  • Importance of FSL and HFL Relationship:
    The FSL of a canal is the designed maximum water level in the canal, while HFL of a stream is the highest water level expected during floods. Their relative positions dictate the type of CDW needed.

 Additional Information

  • False Understanding of Aqueducts Crossing at Same Level:
    An aqueduct never involves the canal and stream crossing at the same level; that condition is handled by a level crossing, which incorporates regulators to manage flow interaction.

  • Selection Criteria for CDW Type:
    Key factors in choosing the type of cross drainage structure include: relative FSL and HFL levels, topography, canal discharge, stream discharge, economic feasibility, and maintenance requirements.

  • Flow Regulation in Level Crossings (not Aqueducts):
    In level crossings, special regulators are provided to manage flow interference. Such systems are complex and not typically used unless FSL and HFL are close to the same level.

  • Importance of Freeboard in Aqueduct Design:
    Adequate freeboard (space between water surface and top of canal wall) must be provided in aqueducts to prevent overtopping during sudden rise in canal flow or wave action.

  • Canal Trough Design in Aqueducts:
    The canal trough in aqueducts must be designed to carry the full canal discharge and withstand static and dynamic forces. Materials used include reinforced concrete or stone masonry.

 F1 N.M M.P 31.07.19 D 5

F1 N.M M.P 31.07.19 D 4

Irrigation Engineering Question 2:

Khosla's theory of independent variables is used in the design of
a. Weirs and barrages
b. Cross regulators and head regulators
c. Modules

  1. a only
  2. c only
  3. Both a and b
  4. a, b and c

Answer (Detailed Solution Below)

Option 3 : Both a and b

Irrigation Engineering Question 2 Detailed Solution

Explanation:

  • Khosla’s Theory of Independent Variables is primarily applied in the design of subsurface flow structures to determine uplift pressures, exit gradients, and seepage paths below hydraulic structures.
  • Khosla's theory is extensively used in the design of weirs and barrages to evaluate seepage forces and ensure safety against piping and uplift.
  • Cross regulators and head regulators involve subsurface flow and potential uplift pressures, especially when located on permeable soils. Khosla’s theory is applied to design their foundations safely.
  • Modules are outlet structures used for water distribution in irrigation, and their design does not require Khosla’s theory. They are more concerned with flow control rather than seepage analysis.

 Additional Information

  • Khosla’s Theory of Independent Variables is an advancement over Bligh’s Creep Theory, developed for the design of subsurface flow under hydraulic structures.
  • It accounts for pressure distribution and exit gradients more accurately in pervious soils using mathematical solutions from potential theory. 

Weirs and Barrages:

  • These are major hydraulic structures built across rivers to divert or control flow.

  • When built on permeable foundations, seepage beneath the structure poses a risk of piping and uplift pressure.

  • Khosla’s theory helps determine:

    • Uplift pressures at various key points

    • Exit gradients at downstream end

    • Safe thickness of floor and cutoff depths

Head Regulators and Cross Regulators:

  • Found in canal systems to regulate flow between main canals and branches.

  • Subsurface flow analysis using Khosla’s theory ensures structural safety against uplift and seepage where these structures are located on sandy or alluvial foundations.

Modules:

  • These are flow measuring and regulating devices (like outlets).

  • Their function is hydraulic in nature, not structural against seepage.

  • Hence, they do not require analysis using Khosla’s theory.

Irrigation Engineering Question 3:

If duty is 1428 hectares/cumec and base period is 120 days for an irrigated crop, then the delta in meters is given by

  1. 1.381
  2. 0.017
  3. 0.726
  4. 102.805

Answer (Detailed Solution Below)

Option 3 : 0.726

Irrigation Engineering Question 3 Detailed Solution

Concept:

The relation between duty, base period and delta is given by:

\(\Delta=\frac{8.64 \times B}{D}\)

Calculation:

B = 120 days

D = 1428 hectares/cumec

\(\Delta=\frac{8.64 \times 120}{1428}=0.726m\)

Irrigation Engineering Question 4:

Match the list-1 with list-2 and select the correct answer using the codes given below,

List-1
(Name of the instrument)
List-2 (Purpose)
a. Current meter                 1. Control silt entry into the canal
b. Echo sounder                  2. Control water levels and flow rates in the canal
c. Head regulator               3. Depth of flow
d. Sluice gate                     4. Velocity of flow of water

  1. a-1, b-2, c-4, d-3
  2. a-1, b-3, c-2, d-4
  3. a-4, b-3, c-1, d-2
  4. a-2, b-4, c-1, d-3

Answer (Detailed Solution Below)

Option 3 : a-4, b-3, c-1, d-2

Irrigation Engineering Question 4 Detailed Solution

Explanation:

Current Meter

  • Used to measure the velocity of flowing water in open channels, rivers, streams, and pipes.

  • It typically consists of a rotating element (propeller or cups) that spins as water flows past; the speed of rotation correlates with the velocity.

  • Data collected with current meters helps in calculating discharge and understanding flow patterns, which are critical for hydraulic analysis and design.

Additional Information Echo Sounder

  • A depth-measuring instrument that operates on the principle of sonar.

  • It emits sound waves downward into the water; the time it takes for the sound to bounce back is measured and converted into depth.

  • Used extensively in hydrographic surveys, riverbed studies, and reservoir monitoring to determine water depth and underwater profiles.

Head Regulator

  • A hydraulic structure located at the head of a canal, where it controls the flow of water from the main water body (like a river or reservoir) into the canal system.

  • Designed to regulate water quantity entering the canal and also prevent excess silt from entering, thus maintaining water quality and reducing sedimentation issues.

  • Essential for effective management of irrigation systems.

Sluice Gate

  • A mechanical gate that controls water levels and flow rates in canals, dams, and waterways.

  • Operates by raising or lowering to adjust the opening through which water passes, thus managing flow and maintaining desired water levels.

  • Widely used for irrigation, flood control, and water regulation in civil engineering structures.

Irrigation Engineering Question 5:

Considering,
(i) Controlling flood
(ii) Generating hydropower
(iii) Preventing loss of water in percolation from fields
(iv) Sewage disposal. A reservoir is used for
Correct answer is

  1. (i) and (ii)
  2. (i) and (iii)
  3. Only (iii)
  4. (iii) and (iv)

Answer (Detailed Solution Below)

Option 1 : (i) and (ii)

Irrigation Engineering Question 5 Detailed Solution

Explanation:

A reservoir is an artificial or natural lake used to store water for various purposes. Two of the most important functions of reservoirs are:

  • Controlling floods — Reservoirs regulate river flow by storing excess water during heavy rainfall or snowmelt, releasing it gradually to prevent downstream flooding.

  • Generating hydropower — Reservoirs provide a constant supply of water to hydropower plants. The stored water is released through turbines to generate electricity, making use of the potential energy of the stored water.

 Additional Informatio

  • Multi-purpose reservoirs serve various functions such as irrigation, drinking water supply, recreation, and navigation, in addition to flood control and power generation.
  • Flood storage capacity of a reservoir is the volume available to store floodwaters temporarily.

  • Hydropower generation depends on both the head (height of water) and flow rate — both managed by the reservoir.

  • On the other hand, preventing percolation from fields and sewage disposal are not primary functions of a reservoir.

Top Irrigation Engineering MCQ Objective Questions

The minimum furrow grade to assure surface drainage is:

  1. 0.09%
  2. 0.02%
  3. 0.07%
  4. 0.05%

Answer (Detailed Solution Below)

Option 4 : 0.05%

Irrigation Engineering Question 6 Detailed Solution

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Furrow is a long, narrow irrigation trench made in the ground used for an optimal supply of water.

Furrows can be level and are very similar to long narrow basins. However, a minimum grade of 0.05% is recommended so that effective drainage can occur following irrigation or excessive rainfall.

Important points:

The maximum recommended furrow slope is 0.5% to avoid soil erosion. Furrows can be set when the mainland slope does not exceed 3%. Beyond this, there is a major risk of soil erosion following a breach in the furrow system. On steep land, terraces can also be constructed and furrows cultivated along the terraces.

What is the relation between Consumptive Irrigation Requirement (CIR), Net Irrigation Requirement (NIR), Field Irrigation Requirement (FIR) and Gross Irrigation Requirement (GIR)?

  1. CIR>FIR>GIR>NIR
  2. CIR>GIR>FIR>NIR
  3. GIR>FIR>CIR>NIR
  4. GIR>FIR>NIR>CIR

Answer (Detailed Solution Below)

Option 4 : GIR>FIR>NIR>CIR

Irrigation Engineering Question 7 Detailed Solution

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Explanation

We know,

a) Consumptive Irrigation Requirements (CIR) = Consumptive Use (Cu) – Effective Rainfall (Re)

b) Net Irrigation Requirement (NIR) = CIR + other requirements such as leaching

c) Field Irrigation Requirements (FIR) = NIR + Surface runoff losses + deep percolation losses

Or

\({\rm{FIR}} = \frac{{{\rm{NIR}}}}{{{\rm{Water\;application\;efficiency}}}}\)

d) Gross Irrigation Requirements (GIR) = FIR + Conveyance Losses (Seepage and Evaporation)

Or 

\({\rm{GIR}} = \frac{{{\rm{FIR}}}}{{{\rm{Water\;Conveyance\;efficiency}}}}\)

If you look through the above equations we can conclude that

 GIR > FIR, FIR > NIR, NIR > CIR 

∴ GIR > FIR > NIR > CIR

Which of the following is a plantation type of crop?

  1. Ground nut
  2. Tea
  3. Fodder
  4. Sugarcane

Answer (Detailed Solution Below)

Option 2 : Tea

Irrigation Engineering Question 8 Detailed Solution

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Concept:

The plantation crop refers to those crops which are cultivated on an extensive scale in an area. In this type of agriculture, single crop is raised on a large area. These crops include tea, coffee, rubber, cocoa, coconut, oil palm and cashew etc.

Other type of Crop categories in India:

1. Food Crops: Wheat, Maze, Rice, Millet, Pulses.

2. Cash Crops: Sugar cane, Tobacco, Jute, Oilseeds.

3. Horticulture Crops: Fruits and Vegetables.

Temporary spurs are also called:

  1. Weirs
  2. Canals
  3. Bunds
  4. Barrages

Answer (Detailed Solution Below)

Option 3 : Bunds

Irrigation Engineering Question 9 Detailed Solution

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Temporary spurs or bunds are temporary structures constructed every year after floods. A bund is a structure made to project flow from a riverbank into a stream or river with the aim of deflecting the flow away from the side of the river on which the groyne is built.

Important Points:

Bunds are temporary in nature whereas Weirs and barrages are permanent in nature.

In a weir, the water overflows the weir, but in a dam, the water overflows through a special place called a spillway. Weirs have traditionally been used to create mill ponds.

A barrage is a weir that has adjustable gates installed over top of it, to allow different water surface heights at different times.

Among the classification of canals based on alignment criteria, identify the canal in which the number of cross drainage works is maximum?

  1. Contour canal
  2. Side slope canal
  3. Detour canal
  4. Ridge canal

Answer (Detailed Solution Below)

Option 1 : Contour canal

Irrigation Engineering Question 10 Detailed Solution

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Based on Alignment Canals are classified into 3 categories. These are:

1. Ridge Canal, 2. Contour Canal and 3.  Side Slope Canal

Their characteristics are given below:

Ridge Canal ( Watershed canal)

Contour Canal ( Single Bank Canal)

Side Slope Canal

Aligned along the ridge or natural watershed Line

Aligned along the natural contour of the country

Aligned perpendicular to the contour of the country.

No Cross-Drainage work required

Maximum cross drainage work is required

No Cross Drainage work required.

Can irrigates on both sides of the ridge and hence, a large area can be cultivated

Can irrigate areas only on one side

Can irrigate areas only on one side

Calculate the water distribution efficiency, if the depths of penetration along the length of a border strip at an interval of 20 m are 1.5 m, 1.8 m and 2.1 m respectively.

  1. 0.6667
  2. 0.8333
  3. 0.8889
  4. 0.9767

Answer (Detailed Solution Below)

Option 3 : 0.8889

Irrigation Engineering Question 11 Detailed Solution

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Concept:

Water distribution efficiency \(= {{\rm{\eta }}_{\rm{d}}} = \left[ {1 - \frac{{\rm{y}}}{{{{\rm{d}}_{\rm{w}}}}}} \right] \times 100{\rm{\;}}\left( {{\rm{in\;\% }}} \right)\)

\({{\rm{d}}_{\rm{w}}} = \frac{{{{\rm{d}}_1} + {{\rm{d}}_2} + \ldots + {{\rm{d}}_{\rm{n}}}}}{{\rm{n}}},\) dw = average depth of water stored

\({\rm{y}} = \frac{{y_1 \ + \ y_2\ + \ ......\ +\ y_n}}{n}\), y = Average of absolute value of deviations

y1, y2,......., yn = absolute deviations from average depth of water stored

Calculation:

\({\rm{d_w}} = \frac{{1.5 + 1.8 + 2.1}}{3} = 1.8\)

y1 = 0.3 m, y2 = 1.8 – 1.8 = 0, y3 = 2.1 – 1.8 = 0.3 m

\({\rm{y}} = \frac{{0.3 + 0.3 + 0}}{3} = 0.2\)

\({{\rm{\eta }}_{\rm{d}}} = \left( {1 - \frac{{0.2}}{{1.8}}} \right) = \frac{8}{9} = 0.8889\)

What is the variation in duty of water from the head of a main canal (M) to that in the field (F)?

  1. ​Duty of water at M is always greater than duty of water at F
  2. ​Duty of water at M is always less than duty of water at F
  3. Duty of water at M is always equal to duty of water at F
  4. Duty of water at M can be greater or less than duty of water at F

Answer (Detailed Solution Below)

Option 2 : ​Duty of water at M is always less than duty of water at F

Irrigation Engineering Question 12 Detailed Solution

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Duty: It is the number of hectares of land irrigated for full growth of a given crop by a supply of 1 cumec of water continuously during the entire base period of that crop.

Duty of water changes from place to place, it will be maximum at the field and minimum at the head of the main canal.

Duty is the area that can be irrigated by the discharge of 1 cumec of water.

At the head of the canal, there are numerous losses to occur later which requires more amount of water to irrigate a particular field. However, if considered on the field, all losses have already occurred and a lesser amount of water is required to irrigate the same considered area.

F1 N.M M.P 22.07.19 D 1

The field capacity of a soil is 25%, its permanent wilting point is 15% and specific dry unit weight is 1.5 g/cc. If the depth of the root zone of a crop is 80 cm, the storage capacity of the soil is

  1. 8 cm
  2. 10 cm
  3. 12 cm
  4. 14 cm

Answer (Detailed Solution Below)

Option 3 : 12 cm

Irrigation Engineering Question 13 Detailed Solution

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Concept:

For an irrigation land:

Assignment 3 Subham SSC JE CE 24 JAN 18 Evening 9Q shubham D 1

SC = Saturation capacity, FC = Field capacity, OMC = Optimum moisture content, PWP = Permanent welting point and UWP = Ultimate welting point

1. Equivalent depth of water held at field capacity (x) = S × d × Fc

2. Equivalent depth of water held at PWP (x’) = S × d × (PWP)

3. Available moisture/storage capacity of soil (y) = S × d × (Fc - PWP)

4. Readily available moisture content (dw) = S × d × (Fc - OMC)

Given:

Field capacity (F. C) = 25 % = 0.25

Permanent willing point (PWP) = 15 % = 0.15

Depth of the root zone (d) = 80 cm

Dry Unit weight of soil (γd) = 1.5 g/cc

Calculations:

The storage capacity \(= \frac{{{\gamma _d} \times d \times \left( {F.C - PWP} \right)}}{{{\gamma _w}}}\)

The storage capacity \(= \frac{{1.5 \times 80 \times \left( {0.25 - 0.15} \right)}}{1} = 12\;cm\)

Hence, the storage capacity of the soil is 12 cm.

Major projects of irrigation are those which have irrigation potential of more than:

  1. 5,000 heactares
  2. 300 hectares
  3. 10,000 hectares
  4. 20,0000 hectares

Answer (Detailed Solution Below)

Option 3 : 10,000 hectares

Irrigation Engineering Question 14 Detailed Solution

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Concept:

Irrigation Projects in India are classified into three major aspects:

  • Major Irrigation Projects: A project having a cultivable command area(CCA) of more than 10,000 hectares.
  • Medium Irrigation Projects: A project having CCA between 2000 and 10,000 hectares.
  • Minor Irrigation: A project with is designed to irrigate an area of 2000 hectares or less is classified as a minor irrigation 


Note:

Whereas major and medium irrigation works are meant for tapping surface water (e.g., rivers), minor irrigation mainly involves groundwater development, e.g., tube-wells, boring works, etc.

The given figure represents ______.

F1 Abhishek M 12.2.21 Pallavi D4

  1. furrow method of irrigation
  2. classes of soil water
  3. base flow of water
  4. frequency of irrigation

Answer (Detailed Solution Below)

Option 4 : frequency of irrigation

Irrigation Engineering Question 15 Detailed Solution

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Explanation: 

Frequency of irrigation: It is defined as the ratio of available soil moisture depletion to rate of consumptive use. 

Screen Shot 2021-02-18 at 12.00.37 PM

Wilting coefficient: The moisture content at which plant can no longer extract the water from the soil for their growth and finally wilts up, It is known as the wilting coefficient. 

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