Category: Irrigation Engineering Objective Questions

250+ TOP MCQs on Factor Controlling Occurrence of Ground Water and Answers

Irrigation Engineering Questions and Answers for Entrance exams on “Factor Controlling Occurrence of Ground Water”.

1. An irrigation project is classified as a major project when the CCA involved in the project is more than ___________
a) 2000 hectares
b) 5000 hectares
c) 10000 hectares
d) 2500 hectares
Answer: c
Clarification: Culturable Command Area is the basis for the design of watercourse and the basis for the design of an irrigation project. The irrigation schemes in India are classified into three parts viz. Minor, Medium and Major Irrigation schemes depending upon the areas involved. Major irrigation scheme is the one where CCA involved in the project is greater than 10,000 hectares.

2. A minor irrigation scheme serves up to ________________
a) 100 hectares
b) 500 hectares
c) 1000 hectares
d) 2000 hectares
Answer: d
Clarification: The irrigation schemes in India are classified into three parts depending upon the areas involving culturable command area.

Type of scheme Areas involving CCA
1. Minor irrigation scheme < 2000 hectares
2. Medium irrigation scheme 2000 to 10000 hectares
3. Major irrigation scheme > 10000 hectares

3. Energy is required in the utilisation of _____________
a) groundwater
b) surface water
c) both groundwater and surface water
d) capillary water
Answer: a
Clarification: For irrigation purposes, groundwater is largely tapped in India through wells and tube wells. Manual, wind, diesel or electric power can be used for lifting water from open wells. The subsequent development in the technique of tapping groundwater is the use of tube wells which requires diesel or electric power.

4. Which of the following property of geological formation represents its water storage capacity?
a) Permeability
b) Porosity
c) Both permeability and porosity
d) Transmissibility
Answer: b
Clarification: Porosity is the quantitative measurement of the interstices of voids present in the rock. In other words, it is the maximum amount of water that can be stored in the rock.

5. The zone of aeration in a groundwater profile does not include ___________
a) capillary zone
b) soil water zone
c) intermediate zone
d) saturation zone
Answer: d
Clarification: Depending upon the number of interstices present, the aeration zone is divided into three classes. The capillary fringe is a continuation of the zone of saturation and does contain some interstitial water. Soil zone is the depth from the surface penetrated by the roots of vegetation and the remaining intermediate part is intermediate zone.

6. In which of the following zone the stresses are beyond the elastic limits?
a) Zone of rock fracture
b) Zone of rock flowage
c) Zone of saturation
d) Zone of aeration
Answer: b
Clarification: In the zone of rock flowage, interstices are absent because the stresses are beyond the elastic limits. The rocks remain in a state of plastic flow and water present in this zone is known as internal water.

7. The rate of flow of water through ground strata can be estimated by _____________
a) Manning’s formula
b) Strickler’s formula
c) Dupuit’s equation
d) Darcy’s formula
Answer: d
Clarification: A French Scientist Mr. H Darcy on the basis of experimental evidence gave a law governing the discharge through soils. According to the law, the discharge is directly proportional to the head loss and the area of cross-section of the soil and inversely proportional to the length of the soil sample.

8. The relation between Transmissibility (T) and Permeability (K) for an aquifer of depth d is _______
a) K = T.d
b) T = K.d
c) T = K.log d
d) T = ln (Kd)
Answer: b
Clarification: Transmissibility is measured by the coefficient of transmissibility (T) and was introduced by Theiss. It can be defined as the rate of flow of water through an aquifer of unit width and full-depth under a hydraulic gradient and at a temperature of 20°C. The relation between K and T is given as T=K.d.

9. Darcy’s law is valid when the flow is ___________
a) laminar and steady
b) non-uniform
c) turbulent
d) both laminar and turbulent
Answer: a
Clarification: The Darcy’s law has been demonstrated to be valid only for laminar flow conditions because the flow in sands, silts, and clays is invariably laminar.
Mathematically, v = K.i where v = discharge velocity, K = coefficient of permeability and I = hydraulic gradient.

10. The coefficient of permeability indicates the ease with which water can flow through a soil mass. The soil type which has less permeability is __________
a) gravelly soil
b) clayey soil
c) sandy soil
d) both sandy and gravel soil
Answer: b
Clarification: The approximate average value of the coefficient of permeability in clayey soil is 0.04 x 10-5 cm/sec. The permeability coefficient of gravel soil is of the order of 4cm/sec and for sandy soil is 0.04 cm/sec.

11. Which of the following is the most important zone for a groundwater hydraulic engineer?
a) Zone of rock fracture
b) Zone of rock flowage
c) Zone of saturation
d) Zone of aeration
Answer: c
Clarification: This is the most important zone for a hydraulic engineer as this water has to be tapped out and the water in this zone is under hydrostatic pressure. In this zone, water exists within the interstice which is nothing but groundwater.

12. Which zone contains water that is under molecular attraction?
a) Zone of rock fracture
b) Zone of rock flowage
c) Zone of saturation
d) Zone of aeration
Answer: d
Clarification: Zone of aeration is the space above the water table and below the surface and the water exists in this zone by molecular attraction. The water in this zone is not at hydrostatic pressure and the gravity water moves through this zone.

Irrigation Engineering for Entrance exams,

250+ TOP MCQs on Canal Falls – Location and Answers

Irrigation Engineering Multiple Choice Questions on “Canal Falls – Location”.

1. Canal drops are provided to _______________
a) dissipate the excess energy
b) increase the driving head of flow of water
c) dissipate excess land slope
d) dissipate inadequate land slope
Answer: c
Clarification: when a canal fall or drop is provided at a suitable section, it brings down the canal bed line. In this process, water comes down with a great force and all the potential energy is converted into kinetic energy. The excess energy of the flow is destroyed with some suitable energy dissipation method.

2. For a canal that irrigates the area directly the fall should be provided at a location FSL outstrips the G.L before the bed of the canal comes into the filling.
a) True
b) False
Answer: a
Clarification: This is done in case of branch canals and distributary channels which irrigate the area directly, the FSL is fixed and marked at the head of the off-taking channel and outlets. In the case of the main canal which does not irrigate the area directly, the site of the fall is determined by considerations of economy.

3. In case of branch canals and distributary channels, the falls are located with consideration to _________________
a) command area
b) topography
c) cost economy
d) availability of earth material
Answer: a
Clarification: In the case of branch canals and distributary canals, the procedure is to mark the FSL required on the L-section of the canal. It is done so as to cover all the commanded points that decide the suitable locations for fall in canal FSL and in canal beds.

4. By providing a natural drop in one drop _______________________
a) the quantity of unbalanced earthwork decreases
b) the quantity of unbalanced earthwork increases
c) the quantity of balanced earthwork increases
d) number of fall increases
Answer: b
Clarification: The number of fall decreases and the quantity of unbalanced earthwork increases by providing a larger drop in one step. All these factors have to be worked out before deciding the locations and extent of falls.

5. A canal fall is provided when the available natural slope is flatter than the designed bed slope of the channel.
a) True
b) False
Answer: b
Clarification: When the ground has a steep slope heavy earth filling is required to construct the canal with a flatter bed slope. The difference between the natural slope and the designed bed slope is adjusted by constructing vertical falls or drops in the canal at a suitable interval.

6. Consider the following statements.
I. Possibility of combining some other structure with a fall eg. regulator, road bridge etc.
II. Cutting and filling are required below and above the fall should not be necessarily equal.
III. The command should not be reduced due to the lowering of F.S.L and the fall may be located below the outlets.
Which of the following statement is correct for site selection of a fall?
a) I only
b) III only
c) I and II
d) I and III
Answer: d
Clarification: The unbalanced earthwork is quite expensive so the excavation and filling on two sides of the fall should be balanced. The considerations of the economy in ‘cost of cutting and filling’ versus ‘cost of fall’ form a basis in deciding the site of a fall in case of the main canal.

250+ TOP MCQs on Classification of Rivers and Answers

Irrigation Engineering Multiple Choice Questions on “Classification of Rivers”.

1. Depending on the topography of the river basin, into how many classifications the river reaches are divided?
a) 3
b) 4
c) 5
d) 1
Answer: a
Clarification: The classifications given on the basis of topography of river reaches are river in hills (upper reaches), rivers in plains (lower reaches), and tidal rivers.

2. The rivers in hills are further classified into how many groups?
a) 4
b) 2
c) 3
d) 6
Answer: b
Clarification: Before entering the plains the upper reaches of the rivers flow through the hills, hence the name river in hills. These are further subdivided into a rocky river stage, and boulder river stage.

3. Which stage of the river is formed by erosion?
a) Aggrading
b) Tidal Rivers
c) Boulder River Stage
d) Rocky River Stage
Answer: d
Clarification: In this type of stage of the river the flow is formed by rapid erosion. These river reaches have high steep with swift flow, and form rapids along their courses. The bed load carried by these reaches cannot be determined by usual bed load transportation formulas.

4. What type of river stage widens the bed?
a) Degrading
b) Incised River Stage
c) Boulder River Stage
d) Stable Type
Answer: c
Clarification: The river bed in these reaches is created by itself, consists a mixture of boulders, gravels, shingles, and alluvial sand deposits. In latter stage, the river flows through deep well defined beds and wider floodplains. Here the river flows in a straight course. Due to the floods, the boulders and material travels downward and gets deposited there after floods. Because of the opposition to the flow due to these obstructions the river takes new course and thus widens the river bed.

5. In what reaches the river flows in a zigzag manner?
a) Boulder River Stage
b) Lower Reaches
c) Degrading
d) Tidal Rivers
Answer: b
Clarification: The chief character of the river in these reaches is that it flows in a zigzag manner called meandering. This meandering is due to the difference in carrying the sediment which is similar to the bed load from one bank to the other bank of the river.

6. Into how many groups the rivers in floodplains can further be divided?
a) 5
b) 6
c) 4
d) 3
Answer: a
Clarification: River in the floodplains is further divided into aggrading, degrading, stable, braided, and deltaic types.

7. In the meandering process on which side of the bank erosion takes place?
a) Convex Side
b) Bottom of the River
c) Concave Side
d) Due to Sediment Load
Answer: c
Clarification: In these reaches there is a constant erosion of the river bed on the concave side and the deposition of the eroded soil on the convex side of the successive bends or between two successive bends.

8. In which type of river building up of slope is present?
a) Deltaic
b) Stable type
c) Degrading Type
d) Aggrading Type
Answer: d
Clarification: This type of river is formed due to the silting action which increases the bed slope. This silting action is due to various factors like heavy sediment load, construction of obstructions like dam or a weir, and sudden intrusion from a tributary.

9. In which type of river there is a reduction of the available slope?
a) Braided
b) Stable Type
c) Aggrading Type
d) Degrading Type
Answer: d
Clarification: This type of river is formed due to the constant erosion to reduce and finally dissipate available excess land scape. It is found below a dam or weir or barrage or above a cut off.

10. What type of river does the diagram represent?
irrigation-engineering-questions-answers-classification-rivers-q10
a) Braided River
b) Aggrading Type River
c) Degrading type River
d) Stable Type River
Answer: a
Clarification: This is braided river as the flow is around alluvial islands. These islands are developed due deposition of coarser material which cannot be transported under the existing conditions of the flow, consisting of coarse and fine material.

11. A river before joining the sea or ocean gets divided into branches.
a) False
b) True
Answer: b
Clarification: As a river approaches the sea or ocean, the velocity of the flow gets reduced, and in turn silting action takes place resulting in rising of water levels and the formation of new channels. These channels or branches multiply in numbers as the river approaches the sea or ocean.

12. A river which does not change its alignment, slope is called a stable river.
a) True
b) False
Answer: a
Clarification: In this type of river changes like silting action, scouring etc in alignment or slope occurs, but these changes are negligible may fail to produce any changes in the regime of the channel. In this type most of the sediment is brought to the sea or ocean.

250+ TOP MCQs on Stable Channels Design in India and Answers

Irrigation Engineering Multiple Choice Questions on “Stable Channels Design in India”.

1. Design of alluvial channels in India is based on Kennedy and Lacey theories.
a) True
b) False
Answer: a
Clarification: The direct accurate mathematical solution based on the resistance equations given by Chezy’s formula and Manning’s formula is very complicated in Indian based alluvial channels. So, therefore hypothetical theories given by Kennedy and Lacey are used, as these theories are based on experiments and experience from existing channels over the years.

2. On what condition does the resistance equations of Chezy’s formula and manning’s formula are applicable?
a) τo
b) τc
c) τo < τc
d) τo > τc
Answer: c
Clarification: The average shear stress (τo) which is acting on the boundary of an alluvial channel should be less than the critical shear stress (τc), then the channel shape remains unchanged, and hence the channel is considered having rigid boundary. So based on this condition Chezy’s formula and Manning’s formula can be applied.

3. What is the problem in India for artificial channels?
a) Formation of Depressions
b) Formation of Alluvial Soil
c) Untimely Rains
d) Improper Usage of Channels
Answer: b
Clarification: In prehistoric periods, the area starting from the Himalayas to Vindhya mountains is used to in the form of depressions with water flowing over it. But over the ages these depressions got filled with fine silt particles, thereby forming into alluvial soil. So the rivers present in this area of north India flow in alluvial soil carrying silt. So, therefore the artificial channels carrying water from such rivers thus have to carry silt sediment.

4. The velocity of flow in a channel should not more or less, but instead it should be adequate.
a) True
b) False
Answer: a
Clarification: The given velocity of flow of a channel and certain depth can carry in suspension, some amount of silt. But if the velocity is more and depth is not fully charged with silt it will erode the bed and sides of the channel. And if the velocity is less, the silt cannot be carried in suspension by the flow, hence it is dropped.

5. What is the effect of scouring in channels?
a) The Channel Section gets reduced
b) Reduces Discharge Capacity
c) Improper Working of Channel
d) Breaching of Canal Banks
Answer: d
Clarification: Scouring is not a rare phenomenon in channels. It is very understandable and can be controlled or avoided by proper designing of channels. Scouring causes loss of command in channels and lowers full supply level of channel. This causes breaching of canal banks and in turn causes failure of foundations of irrigation structures.

6. What is the effect of silting in channels?
a) Reduced Discharge Capacity of Channel
b) Causes Loss of Command
c) Breaching of Canal Banks
d) Failure of Irrigation Structures
Answer: a
Clarification: The silting problem is very common in any kind of artificial channels, or we can say there is no natural or artificial channel with the silting problem. Silting interferes with the proper working of a channel, as it causes reduction in the channel section due to siltation, which thereby reduces the discharge capacity of the channel.

7. Which type of state is achievable in artificial channels but not in rivers?
a) Turbulent State
b) Laminar State
c) Regime State
d) Uniform State
Answer: c
Clarification: The channel is said to be in regime state, when the flow of the channel is such that silting and scouring effects need no special attention. This state is not easy to achieve in rivers but can be achieved in artificial channels, by proper designing of the channels.

8. On what basis for designing the regime state is obtained?
a) Silt and Velocity of Flow
b) Silt
c) Velocity
d) Depth of Channel
Answer: a
Clarification: The basis for designing a channel to be in regime state, whatever the amount of silt has entered the channel has to be kept in suspension, so that it does not settle down and get deposited in the channel anywhere. And the velocity of the flow in the channel should be such that it does not produce silt by eroding the banks and beds of the channel.

9. What vital factor is not considered in regime theories during the design of regime channels?
a) Velocity of Flow in the Channel
b) Type of Flow in the Channel
c) Quantity of Sediment Entering a channel
d) Area in which the Channel flows
Answer: c
Clarification: The quantity of sediment load entering the channel from the head works plays an important role in designing the regime channels, as it controls the cross-section and shape of the regime channel. Moreover the design of regime channel is not complete until provisions are made for the effects produced by the actual quantity of sediment load present in the channel.

10. According to Kennedy the silting action in channels is due to?
a) Generation of Eddies
b) Slope of the Bed
c) Area of Channel
d) Type of Flow in the Channel
Answer: a
Clarification: From the observations of his research, he concluded that the silting action in channels is due to the generation of eddies, rising to the surface. These eddies are in turn generated by the friction of flowing water with the channel surface.

11. Based on his research what factor is given by Kennedy for free silting and scouring actions in a channel?
a) Critical Velocity (Vo)
b) Bed Slope of Channel
c) Hydraulic Mean Depth
d) Rugosity Coefficient
Answer: a
Clarification: Eddies affect the formation of silting action in a channel brief investigation shows that the vertical component of these eddies try to move sediment up and the weight of sediment tries to bring it down, thus keeping the sediment in suspension. So, silting is avoided if sufficient velocity is generated to form eddies and keep the sediment in suspension. This velocity is called critical velocity.

12. Given to design the irrigation channel to carry 80 cumecs of discharge. The channel slope is 1 in 3000. The critical velocity ratio is 1.3. Take rugosity coefficient as 0.021.
a) Depth = 3 m, Base width = 15 m
b) Depth = 3.5 m, Base width = 15.5 m
c) Depth = 3.3 m, Base width = 15.2 m
d) Depth = 3.2 m, Base width = 15.1 m
Answer: d
Clarification: Given Q = 80 cumecs, S = 1/3000, m = 1.3, n = 0.021
Critical velocity (Vo) = 0.55my0.64 (y = depth of the channel)
= 0.55 x 1.3 x (2)0.64
= 1.114 m/sec
Area (A) = Q/Vo = 80 / 1.114 = 71.8 m2
Now assume side slope as (1/2): 1(1/2H: V)
Now, A = y (b + y x 1/2)
71.8 = 2(b + 2 x 1/2)
71.8 = 2(b + 1)
b = 34.9 m
Perimeter (P) = b + 2 x √5/2y
= 34.9 + 4.47
= 39.37 m
Now R = A/P = 71.8 / 39.37 = 1.82 m
V = ((1/n + (23 + 0.00155/s)) / (1 + (23 + 0.00155/s)n/(sqrt{RR})))(sqrt{RSRS})
V = 0.78 m/sec
Now calculating for critical velocity,
Vo = 0.55 x 1.3 x y0.64
= 0.715 x y0.64
Now assume y = 2.5 m
Vo = 0.715 x (2.5)0.64 = 1.28 m/sec
A = 80 / 1.28 = 62.5 m2
62.5 = 2.5 (b + 1/2(2.5))
b = 23.75 m
P = 23.75 + 2 x √5/2(2.5) = 29.35 m
R = A/P = 2.13
√R = 1.46
V = 75.27/1.4 x (1.46/54.8) = 1.432 m/sec
V > V0
Now assume y = 3.2 m
Vo = 0.715 x (3.2)0.64 = 1.51 m/sec
A = 80 / 1.5 = 53.33 m2
53.33 = 3.2 (b + 1.6)
b= 15.1 m
P = 15.1 + √5(3.2)
P = 22.2 m
R = A/P = 53.33/22.2 = 2.4 m
√R = 1.56
V = (75.27/1.4) x (1.56/54.8)
= 1.53 m/sec
Therefore V = Vo
Hence use the depth equal to 3.2 m and base width 15.1 m with slope 1/2:1.

13. Design a regime channel for a discharge of 90 cumecs and silt factor is 1.3 use Lacey’s Theory.
a) Depth = 2.2 m, Base width = 40.5 m, Slope = 1/4670
b) Depth = 2.12 m, Base width = 40.42 m, Slope = 1/4675
c) Depth = 2.135 m, Base width = 40.3 m, Slope = 1/4676
d) Depth = 2.123 m, Base width = 40.1 m, Slope = 1/4672
Answer: c
Clarification: Given Q = 90 cumecs, f = 1.3
V = [Qf2/140]1/6 = 1.014 m/sec
A = Q/V = 90/1.014 = 88.76 m2
R = 5/2 x V2/f = 2 m
P = 4.75√Q = 45.06 m
For a trapezoidal channel with 1/2H: 1V slopes
P = b +√5y and A = (b + y/2) y
45.06 = b + √5y, 88.76 = (b + y/2) y
b = 45.06 – 2.24y, substitute this value in area equation
88.76 = (45.06 – 2.24y + y/2) y
Y = 2.135 m
b = 45.06 – 2.24 x 2.135 = 40.3 m
S = (f5/3 / 3340Q1/6) = 1/4676.

14. According to Lacey what factor is needed to have a true regime in an artificial channel?
a) Flow is Uniform
b) Critical Velocity = Actual Velocity
c) No Silting or Scouring Action
d) Lining of the Channel Bed
Answer: a
Clarification: For a channel to be in regime condition it should not have silting and scouring action. To obtain this the silt entering the channel must be carried away through the channel, by the channel section. The artificial channel to be in a regime it must have a fixed slope and fixed section. So, therefore to achieve this flow in the channel should uniform flow.

250+ TOP MCQs on Water Requirements of Crops – Effective Rainfall and Answers

Irrigation Engineering Multiple Choice Questions on “Water Requirements of Crops – Effective Rainfall”.

1. Which of the following statement is not correct about Effective rainfall?
a) It doesn’t take into consideration precipitation lost through deep percolation
b) It satisfies evapotranspiration needs of the crop
c) It includes surface runoff loss
d) It doesn’t include surface runoff water loss
Answer: c
Clarification: Effective rainfall is that portion of precipitation that remains in the soil and is available for consumptive use. It is available to meet the evapotranspiration needs of the crop and does not include percolation losses and surface runoff losses.

2. What is the field irrigation requirement (FIR)?
a) The water required to meet the evaporation needs of a crop
b) Amount of water required to delivered at the field to meet evapotranspiration and leaching needs
c) Amount of water required to meet the net irrigation requirements plus water lost due to surface runoff and percolation
d) Amount of water required to meet the field irrigation requirements plus water lost in conveyance through the canal system
Answer: c
Clarification: FIR can be defined as the water required meeting the net irrigation requirements plus the amount of water lost as surface runoff and through deep percolation.
Mathematically, FIR = NIR / Na where Na = application efficiency and NIR = Net irrigation requirement.

3. What is the correct representation for the calculation of NIR?
a) Cu – Re
b) CIR + leaching losses
c) FIR + conveyance losses
d) FIR + leaching losses + conveyance losses
Answer: b
Clarification: Net irrigation requirement (NIR) can be defined as the amount of irrigation water required to be delivered at the field to meet the evapotranspiration and leaching losses.
NIR = (Cu – Re) + amount of water required for leaching = CIR + leaching losses requirement.

4. Determine the Gross irrigation requirement for an irrigation system if consumptive use of crop is 200 cm in terms of depth and the rainfall excess is 30 cm. assume water application efficiency = 50% and conveyance efficiency = 90%.
a) 370 cm
b) 380.79 cm
c) 355.56 cm
d) 377.77 cm
Answer: d
Clarification: Given, Cu = 200 cm, Re = 30 cm, Na = 50%, Nc = 90%
NIR = CIR + losses = (200-300) + 0 = 170 cm
FIR = NIR / Na = 170/0.5 = 340 cm
GIR = FIR / Nc = 340/0.9 = 377.77 cm.

5. An outlet irrigates an area of 40 ha. The discharge required at this outlet to meet the evapotranspiration requirement of 40 mm occurring uniformly in 40 days neglecting other field losses is _________
a) 4.52 litres / sec
b) 2.31 litres / sec
c) 4.01 litres / sec
d) 1.52 litres / sec
Answer: a
Clarification: Volume of water required for evapo-transpiration = 40 x 104 x 40 x 10-3 x 103 = 16 x 106 litres
Hence, the discharge required at outlet
= (16 x 106) / (40 x 86400)
= 4.52 litres / sec.

6. The moisture holding capacity of the soil in a 100-hectare farm is 20 cm/m. The root zone depth of the crop is 1.0 m and peak rate to moisture use is 5 mm/day. The field is to be irrigated when 40% of the available moisture in the root zone is depleted. Calculate the frequency of irrigation.
a) 16 days
b) 18 days
c) 20 days
d) 25 days
Answer: a
Clarification: The moisture holding capacity is 20 cm/m and the root zone depth = 1 m
Moisture holding capacity of soil for crop = 20 x 1 = 20 cm
Allowable depletion of moisture = 40% of 20 cm = 8 cm
Consumptive use of water= 5 mm/day = 0.5 cm/day
The frequency of irrigation = 8 / 0.5 = 16 days.

7. What is the correct formula for Gross irrigation requirement?
a) Cu – Re
b) CIR + leaching losses
c) NIR/Na
d) FIR/Nc
Answer: d
Clarification: It is defined as the amount of water required to meet the field irrigation requirements plus the amount of irrigation water lost in conveyance through the canal system by evaporation and seepage.
Mathematically, GIR = FIR/Nc.

8. The following data gives the value of consumptive uses and effective rainfalls for the periods shown against them for a Jowar crop. Determine NIR of this crop neglecting other losses.

Dates Cu (in mm) Re (in mm)
16-31 October 37 30.8
1-30 November 84.2 20.4
1-31 December 154.9 6.7
1-31 January 188.1 2.4
1-10 February 13.3 1.0

a) 41.60 cm
b) 41.60 mm
c) 44 cm
d) 44 mm
Answer: a
Clarification: NIR = consumptive use – effective rainfall.

Dates Cu (in mm) Re (in mm) NIR = Cu – Re
16-31 October 37 30.8 6.2
1-30 November 84.2 20.4 63.8
1-31 December 154.9 6.7 148.2
1-31 January 188.1 2.4 185.7
1-10 February 13.3 1.0 12.3

Summation of all values of NIR = 416.2 mm = 41.62 cm.

250+ TOP MCQs on Arch Dams – Design and Answers

Irrigation Engineering Multiple Choice Questions on “Arch Dams – Design”.

1. Which of the following forces is the least important in the design of arch dams?
a) Reservoir water force
b) Uplift pressure
c) Temperature stresses
d) Ice load
Answer: b
Clarification: The uplift pressure in an arch dam is generally neglected because it is small due to the narrow base width of its body. Ice load, Temperature stresses and yield stresses are quite important in such dams and are thoroughly examined.

2. Which of the following theory is the most accurate method to design arch?
a) The thin cylinder theory
b) The theory of elastic arches
c) The trial load method
d) Creep theory
Answer: c
Clarification: Thin cylinder theory can only be relied upon for rough estimation of dimensions of the arches. The Elastic theory does not take into account the thickness of arch rings and the restrictions at the abutments. The Trial load method removes the above objections and leads to an accurate estimation of the stresses in arches.

3. Which among the following is not an assumption of the ‘Trial Load Analysis’ method of design of arch dams?
a) Plane sections normal to the axis remain plane after flexure
b) Any horizontal arch ring is independent of the arch rings below and above
c) Modulus of elasticity of concrete and rock foundation is equal for compression and tension
d) Stresses are proportional to strains
Answer: b
Clarification: In Trial load method, the dam is divided into two systems of elements i.e. the horizontal arches and the vertical cantilevers. Each element is supposed to take a percentage of the load due to silt and water pressure. The actual distribution of the load between these elements is then arrived at by several trials.

4. The thin cylinder theory for designing arch dams is based only on ___________________
a) temperature stresses
b) ice pressures
c) yield stresses
d) hydrostatic water pressure
Answer: d
Clarification: In thin cylinder theory analysis, the design is based only on hydrostatic water pressure. The theory does not take into account the shear and bending stresses in the arch, temperature stresses, and ice pressures which are quite important in arch dams.

5. The temperature stresses producing worst effects in the design of arch dams are caused by ______________
a) rise in temperature
b) fall in temperature
c) both rise and fall
d) no effect in the design
Answer: b
Clarification: The dam moves upstream during the summer and downstream during the winter due to the internal stresses caused by the temperature changes. In stress analysis, the low temperature becomes quite important since these stresses act additive to the reservoir water pressure.

6. The stresses due to rib-shortening become quite important in the case of _____________
a) long thin arch dams
b) thick arch dams of small angle
c) both thin and thick arch dams
d) all types of arch dams
Answer: b
Clarification: In the case of an arch dam, the ends being fixed the distance between them does not change and this restriction of the change in the span causes additional stresses known as rib-shortening stresses. These stresses are small in a long thin arch but in the case of a thick arch of small-angle, they are appreciable.

7. Which of the following is not the basis for the design of an arch dam?
a) Thin cylinder theory
b) Trial load theory
c) Unit column theory
d) Elastic theory
Answer: c
Clarification: Arch dams are designed on the basis of any one of the following three methods–
• Thin cylinder theory
• Theory of Elastic arches
• Trial load method
Out of these methods, the trial load method is generally adopted nowadays and it gives precise results.

8. According to thin cylinder theory, the volume of concrete required for an arch dam would be minimum if the central angle is ____________
a) 130° 34’
b) 133° 34’
c) 143° 34’
d) 153° 34’
Answer: b
Clarification: The most economical central angle is equal to 133°-34’ and such an angle can be adopted only at one place in constant radius arch dam as there is considerable variation with height due to narrow V-shape of the valley. It is therefore considered to keep the economical angle of 133°-34’ at about mid-height for a minimum value of concrete.

9. The best design of an arch dam is when __________________
a) all horizontal water loads are transferred horizontally to the abutments
b) the dam is safe against sliding at various levels
c) the load is divided between the arches and cantilevers and the deflections at the conjugal points being equal
d) the deflections of the cantilevers are equal at different points
Answer: c
Clarification: In the trial load method, deflections for the arches are calculated at the points where the cantilevers cross it in each trial. If the deflection calculated at these points from cantilever elements is within reasonable limits and same, then the assumed distribution of load is correct. If not, then the calculations are repeated using a different set of load distribution until the condition is satisfied.

10. At a given elevation, the thickness of the arch ring is proportional to the radius of the arch.
a) True
b) False
Answer: a
Clarification: The thickness of the arch ring (t) under limits is given by –
t = ϒw . h. r / fc where r is the radius of the arch, h is depth, fc is the allowable compressive stress for the arch material and ϒw is the unit weight of water.
It is clear from the above equation that the thickness of the arch is proportional to the radius of the arch at a given elevation.