1. The value of consumptive use may be different for different crops and different for the same crop at different times and places.
a) True
b) False
Answer: a
Explanation: Consumptive use is defined as the total amount of water used by the plants in transpiration and evaporation from soils, in specific time. So, therefore the values of consumptive use vary for different crops, and vary for same crops at different places or times.

2. What is the process of a plant called, through which it leaves water?
a) Photosynthesis
b) Transpiration
c) Evapotranspiration
d) Chlorosis
Answer: b
Explanation: Transpiration is the process of the plant through which water leaves the plant, through its leaves as water vapor and enters the atmosphere.

3. Transpiration process is an integral part of the main process called photosynthesis.
a) True
b) False
Answer: a
Explanation: Yes, photosynthesis is an important process of the plant through which the plant produces carbohydrates for its growth, and it is during this process transpiration occurs as an integral process in the whole photosynthesis process.

4. Which protein in the leaf of the plant utilize carbon dioxide and produce carbohydrates?
a) Chloroplasts
b) Stomata
c) Xylene
d) Leaf
Answer: a
Explanation: Water enters the leaves of the plant through the roots, where photosynthesis happens. During this, air enters the stomata of the leaves. The protein chloroplasts present here takes the carbon dioxide in air and uses it to produce carbohydrates.

5. Total transpiration about 95% occurs during the day alone.
a) True
b) False
Answer: a
Explanation: Preparation of food, i.e carbohydrates is done during daylight alone, as photosynthesis process occurs only during sunlight. Therefore most of the transpiration occurs during the day alone.

6. On which factor does the transpiration loss also depends on?
a) Available Moisture
b) Type of Soil
c) Type of Irrigation
d) Method of Irrigation
Answer: a
Explanation: Availability of moisture also affects the transpiration losses because plants transpire moisture according to its availability. That is plants transpire less moisture when moisture is scarce, and transpire more moisture when it is more available.

7. On what factor does transpiration ratio depends on?
a) Water
b) Soil
c) Air
d) Moisture
Answer: a
Explanation: Transpiration ratio is directly proportional to water requirement of the plant. So, therefore if amount of water increases the ratio increases, and if amount of water decreases the ratio decreases.

8. In transpiration ratio what type of crop is considered to be weighed, for mass of dry matter produced?
a) Marketed Crop
b) Cash Crop
c) Horticulture Crop
d) Plantation Crop
Answer: a
Explanation: Mass of dry matter is taken as the weight of the entire plant including its roots. And also sometimes only marketed crop like wheat, gram etc is weighed.

9. Given information is that total mass of water required for the growth of a plant is 285 kgs, and mass of the marketed crop is 19 kgs. Find the transpiration ratio?
a) 12
b) 16
c) 25
d) 15
Answer: d
Explanation: We have transpiration ratio (TR) = (total mass of water transpired by the plant during its growth / mass of dry matter produced)
= (285/19)
= 15.

10. A crop is being tested in the lab to its loss due to transpiration. Given that initial weight of instrument is 100 kgs, and final mass of instrument is 150 kgs. The amount of water added during full growth of plant is 70 kgs. Find the loss due transpiration when the method used is phytometer method?
a) 20 kgs
b) 15 kgs
c) 25 kgs
d) 10 kgs
Answer: a
Explanation: When phytometer method is used the formula for transpiration loss (T)
= (M1 + M) – (M2)
M1 = initial mass of instrument
M = mass of water added during full growth of plant
M2 = final mass of instrument
Therefore (T) = (100 + 70) – 150 = 170 – 150 = 20 kgs.

11. By what number should the value from the phytometer method be multiplied when it comes to obtain possible field results?
a) 5
b) 6
c) 9
d) A Constant Number
Answer: d
Explanation: Since the phytometer method is tested in lab in artificial conditions, the obtained result is only relevant to the lab due different soil conditions in the open field. So, therefore in order to get or obtain exact field results we need to multiply the value of transpiration loss obtained from the method with a constant number.

12. Which type of soil has less ratio of AET/PET than the other?
a) Clayey Soil
b) Sandy Soil
c) Alluvial Soil
d) Black Soil
Answer: b
Explanation: The ratio AET/PET is directly proportional to available moisture. Sandy soil has less availability of moisture than the other type of soils like clayey soil, alluvial soil, and black soil. So, this gives that sandy soil has less AET/PET ratio.

13. On what factors Potential Evapotranspiration critically depends?
a) Climatological Factors
b) Types of Crop
c) Types of Soils
d) Vegetation
Answer: a
Explanation: Potential Evapotranspiration (PET) involves transpiration process. As this process consists of transpiration losses through leaves of plant, and evaporation losses from surroundings of the plant and this directly dependent on availability of moisture which when is sufficiently available to meet the needs of the vegetation then it is called PET. Therefore this clearly states PET critically depends on climatological conditions.

14. On what factors Actual Evapotranspiration depends?
a) Climatological Factors
b) Types of Crop
c) Types of Soils
d) Characteristics of Soil and Vegetation
Answer: d
Explanation: Actual Evapotranspiration (AET) is actual and real evapotranspiration that occurs in any specific situation of the field. The specific situation of the field directly depends on the characteristics of soil and the type of vegetation present in that particular soil. So, therefore AET indirectly depends on the characteristics of soil and vegetation, in the field.

15. Which type of method is adopted for research studies on crops?
a) Phytometer Method
b) Lysimeter Method
c) Furrow Irrigation Method
d) Drip Irrigation Method
Answer: b
Explanation: Actually lysimeter method is used to determine the AET. This method consists of a tight tanker filled with a block of soil and is installed in a field of growing plants. The conditions to maintain this tanker on par with conditions of the field and the measurement of water added to the tanker to maintain moisture content are time-consuming and costlier field studies.

16. Average yield of a storage reservoir is the arithmetic average of its ________________
a) firm yields over a long period
b) secondary yields over a long period
c) firm and secondary yields over a long period
d) reservoir yield over a long period
Answer: c
Explanation: Firm yield is the yield corresponding to the most critical year on record. Secondary yield is the water available in excess of the firm yield during years of higher inflows. Average yield is the average of both firm yield and secondary yield.

17. While planning a water supply reservoir as compared to an irrigation reservoir the design yield may be kept __________________
a) higher
b) lower
c) equal
d) lower or higher as per designers discretion
Answer: b
Explanation: The dependability percentage value will depend upon the risk which can be absorbed for the proposed use of water. The city water supply projects can absorb lesser risk as compared to the irrigation projects. Hence, higher percentage values are considered.

18. Yield of a reservoir represents ___________________
a) the inflow into the reservoir
b) the capacity of the reservoir
c) the outflow demand on the reservoir
d) the optimum value of catchment yield
Answer: c
Explanation: The yield of the catchment is the long-range runoff from a catchment. It helps in designing the capacity of the reservoir. The outflow from the reservoir (reservoir yield) is represented by the mass demand line.

19. Design yield of a storage reservoir is kept _______________
a) higher than its firm or safe yield
b) lower than its firm or safe yield
c) equal to its firm or safe yield
d) higher or lower than the firm yield depending upon the designer’s intuition
Answer: b
Explanation: An intermediate dependability percentage such as 50% to 75% is used to compute the design yield or dependable yield. This value will depend upon the risk which can be absorbed for the proposed use of water.

20. Who did the first effort to give empirical relations for converting the yearly rainfall value into the yearly runoff for the given catchment?
a) Mr. W L Strange
b) Alexander Binnie
c) Mr. T.G Barlow
d) Sir Inglis
Answer: b
Explanation: The first effort made in India was from Sir Alexander Binnie to connect the long-range rainfall and the yield. The observations were made on two rivers in the central provinces for the entire monsoon period. He worked out certain percentages to connect the monthly rainfall with the monthly yield which was further adjusted by Mr. Garret.

21. In which of the following method, the catchments were categorized as good, bad and average catchments to account for the general characteristics of the catchment?
a) Binnie’s percentages
b) Strange’s tables
c) Barlow’s tables
d) Inglis formula
Answer: b
Explanation: Strange’s tables was an improvement over Binnie’s tables. The catchments prone to producing higher yield were good catchments and that of low yields were bad catchments. Average catchments were the intermediate types.

22. What is the value of Barlow’s runoff percentage (K) for hills and plains with little cultivation?
a) 10
b) 15
c) 20
d) 25
Answer: d
Explanation: For Flat, cultivated, and adsorbent soils, the runoff percentage is taken as 10 and for flat, partly cultivated stiff soils, it is taken as 15. For average catchment, the value is taken as 20 and for hills and plains with little cultivation, the value of the runoff percentage is 25.

23. What is the value of Lacey’s monsoon duration factor for a good year?
a) 0.5
b) 1.2
c) 1.5
d) 1.8
Answer: c
Explanation: The monsoon duration factor (m) for a bad year is 0.5 and that of the normal year the value is 1.2. For a good year, the monsoon duration factor is 1.5.

24. Which of the following yield formulas is used for the catchments of West Maharashtra state of India?
a) Inglis formula
b) Khosla’s formula
c) Lacey’s formula
d) Barlow’s table
Answer: a
Explanation: Inglis formula was given by Sir Inglis derived his formula for catchments of Maharashtra state of India. He divided the areas into ghat areas (Sahyadri ranges) and non-ghat areas depending upon the rainfall intensity and gave the formula to determine the yield.

25. Which of the following formula for calculating yield can be applied to all catchments?
a) Inglis formula
b) Khosla’s formula
c) Lacey’s formula
d) Barlow’s table
Answer: b
Explanation: Khosla formula is based upon the recent research work conducted in the field. It is a simple and useful formula and can be applied to the entire country. The formula is given as-
Yield (Q) = P – 0.48 Tm where Q = the yield in cm, P = rainfall in cm, and Tm = mean annual temperature of the area.

26. According to the Inglis formula, the non-ghat areas are the one where rainfall is _____________
a) 200 cm or more
b) 200 cm or less
c) 100 cm or more
d) 100 cm or less
Answer: b
Explanation: The ghat area is the one where rainfall is 200 cm or more and in non-ghat areas the rainfall is less than 200 cm. The formulas are-
For ghat areas, Yield = (0.85P – 30.48)
For non-ghat areas, Yield = [P (P – 17.78)] / 254 where P is the rainfall in cm.

1. The field measurement of infiltration is done by ____________________
a) potentiometer
b) lysimeter
c) infiltrometer
d) tensiometer
Answer: c
Explanation: Infiltrometer is used for field measurement of infiltration. There are two types of infiltrometer – Single ring infiltrometer and Double ring infiltrometer. Single ring infiltrometer always overestimate because of the lateral movement of water and to overcome this double-ring infiltrometer is used.

2. Which of the following is used for laboratory measurement of infiltration?
a) Infiltrometer
b) Rainfall Simulator
c) Tensiometer
d) Lysimeter
Answer: b
Explanation: Rainfall simulators are used for laboratory measurement of infiltration and Infiltrometer is used for field measurement of infiltration. Lysimeter is used to measure evapotranspiration and tensiometer is used to measure capillary potential.

3. As the temperature increases, the rate of infiltration also increases.
a) True
b) False
Answer: a
Explanation: The rate of infiltration is directly proportional to the temperature. As temperature increases, viscosity decreases and resistance to flow decreases and infiltration increases.

4. Sandy soil has more infiltration capacity as compared to clayey soil.
a) True
b) False
Answer: a
Explanation: Soils having small pore size such as clay have low infiltration capacity than the soils having large pore size such as sandy soil. One exception is when the clay is present in dry conditions; the soil can develop large cracks which lead to higher infiltration capacity.

5. Rate of infiltration determined by infiltrometer is less than the one determined by rainfall simulator.
a) True
b) False
Answer: b
Explanation: The rate of infiltration measured by infiltrometer is more than the infiltration rate determined by the rainfall simulator. The rate of infiltration is directly proportional to the depth of surface retention. As the depth of retention is more in infiltrometer hence the rate of infiltration is more in infiltrometer.

6. Deep vertical movement of water in the ground is called as ____________________
a) infiltration
b) percolation
c) runoff
d) seepage
Answer: b
Explanation: Infiltration is the process by which the water seeps into the surface strata of the earth to meet soil moisture deficiencies. Percolation is the deep vertical movement of water in the ground.

7. Vegetation cover or grass-cover _____________
a) increases the field capacity
b) decreases the field capacity
c) may increase or decrease the field capacity
d) have no effect on field capacity
Answer: a
Explanation: Grass cover or vegetation cover increases the field capacity by trapping water and reducing the effect of raindrop compaction. Vegetation and grass cover also reduces the surface compaction of the soil which again allows for increased infiltration.

9. 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
Explanation: 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.

10. A minor irrigation scheme serves up to ________________
a) 100 hectares
b) 500 hectares
c) 1000 hectares
d) 2000 hectares
Answer: d
Explanation: 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

11. Energy is required in the utilisation of _____________
a) groundwater
b) surface water
c) both groundwater and surface water
d) capillary water
Answer: a
Explanation: 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.

12. 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
Explanation: 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.

13. 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
Explanation: 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.

14. 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
Explanation: 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.

15. 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
Explanation: 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.

16. 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
Explanation: 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.

17. 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
Explanation: 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.

18. 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
Explanation: 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.

19. 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
Explanation: 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.

20. 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
Explanation: 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.

21. The geological formation which yields only insignificant quantity of groundwater is _____________
a) aquifer
b) aquifuse
c) aquiclude
d) aquitard
Answer: d
Explanation: The geological formations which are porous and contain a good amount of water but does not yield water freely to wells due to its lesser permeability is called aquitard. The water yield from such a formation is insignificant. Sandy clay is an example of the aquitard.

22. The geological formation which may contain water but does not contain any yield is ____________
a) aquifer
b) aquifuse
c) aquiclude
d) aquitard
Answer: c
Explanation: The geological formations which are porous but have no permeability are termed as an aquiclude. Water cannot be extracted from such formations. A clay layer is an example of aquiclude.

23. Which of the following geological formation contains and readily yields water to our tube wells?
a) Aquifer
b) Aquifuse
c) Aquiclude
d) Aquitard
Answer: a
Explanation: The geological formations which are both porous and permeable hence sufficient quantity of water can be extracted from them. Aquifers vary in depth and thickness and in general, confined and non-confined aquifers are two main categories of the aquifer.

24. Which of the following geological formation does not contain any amount of groundwater?
a) Aquitard
b) Aquifer
c) Aquiclude
d) Aquifuge
Answer: d
Explanation: The geological formations which are very dense and contain no water in voids and are neither porous nor permeable are termed as aquifuge. Granite rock is an example of aquifuge.

25. The quantum of water contained in the soil pores which cannot be extracted by gravity drainage is called _____________
a) pellicular water
b) capillary water
c) hygroscopic water
d) available water
Answer: a
Explanation: When the saturated formations are drained under the action of gravity drainage, the volume of water drained is less than the volume of void space. The water contained in these voids cannot be drained out by force of gravity. The water which is always retained by these interstices due to molecular attraction is called pellicular water.

26. What is the volume of groundwater which can be extracted by gravity drainage from a soil stratum when expressed as percentage fraction of the volume of the soil stratum?
a) Pellicular water
b) Available water
c) Specific yield
d) Field capacity
Answer: c
Explanation: Yield is the volume of groundwater extracted by gravity drainage from a saturated water-bearing material. It is known as the specific yield when it is expressed as the ratio of the volume of the total material drained.
Specific Yield = (Volume of the water obtained by gravity drainage/ total volume of drained material) x 100.

27. Field capacity of a ground aquifer equals _________
a) specific yield
b) 100 – specific yield
c) 100/ specific yield
d) field capacity
Answer: b
Explanation: Field capacity can be expressed as the percentage of the total volume of the material drained. Specific retention or field capacity is given as –
F.C = volume of water held against gravity drainage/total volume of the material drained x 100
The summation of Specific yield and field capacity is equal to the porosity.

28. Specific retention of groundwater is larger in coarse-grained soils.
a) True
b) False
Answer: b
Explanation: Specific retention can be defined as the amount of water held between the grains due to molecular attraction on the walls of the interstices. If the effective size of grain decreases, the surface area between the interstices will increase causing more specific retention and less specific yield. In fine soils like clay, the specific retention would be more and it results in small specific yield.

29. Water wells excavated through confined aquifers are known as ______________
a) artesian wells
b) non-artesian wells
c) gravity wells
d) water table wells
Answer: a
Explanation: The gravity wells which are constructed to tap water from the unconfined aquifer are known as unconfined or non-artesian wells. Such wells are also known as water table wells or gravity wells since the water level in these wells is equal to the level of the water table.

30. In case of a flowing well, the piezometric surface is always _________
a) below the ground level
b) above the ground level
c) at the ground level
d) above or below the ground level
Answer: b
Explanation: A well excavated through confined aquifer yields water that often flows out automatically under the hydrostatic pressure. It may even rise or gush out of the surface for a reasonable height. Flowing well is a type of artesian well where water gushes out automatically.

31. Which of the following test is commonly adopted for determining soil permeability of soil formations?
a) Pumping-in test and Pumping-out test
b) Horizontal capillary test
c) None of the mentioned
d) All of the mentioned
Answer: a
Explanation: Pumping-out test is the frequently used field method, as it involves discharging water from the ground.

32. The value of specific yield SY, of an aquifer depends on ___________
a) Grain size
b) Compaction of stratum
c) Grain shape
d) All of the mentioned
Answer: d
Explanation: Since specific yield is the volume of water drained, the drainage of water depend on grain size, shape and compaction of stratum in the soil.

33. Specific yield of unconfined aquifer indicates __________
a) Water capacity
b) Volume of water
c) Water retained
d) All of the mentioned
Answer: a
Explanation: Specific yield is an indication of the water yielding capacity of an unconfined aquifer.

34. Dupuit’s theory was later modified by _________
a) Thiem
b) Darcy
c) Alam Singh
d) Louden
Answer: a
Explanation: The theory proposed by Dupuit’s on radial flow was modified by Thiem in 1906.

35. The parabolic depression in the aquifer is called as __________
a) Cone of depression and Drawdown curve
b) Parabolic curve
c) None of the mentioned
d) All of the mentioned
Answer: a
Explanation: The parabolic depression formed when water is pumped from the aquifer through well can be called as cone of depression or the drawdown surface.

36. In most of the confined aquifers,the value of storage coefficient ranges between __________
a) 0.05 to 0.0005
b) 0.5 to 0.005
c) 0.00005 to 0.005
d) 0 to 5
Answer: c
Explanation: By pumping test method, the value of storage coefficient ranges was found to be in between 0.00005 to 0.005.

37. Which of the following is depicted by Dupuit’s theory?
a) The velocity of flow
b) Co-efficient of transmissibility
c) Radial flow of water
d) Volume of aquifer
Answer: c
Explanation: The analysis of the radial flow of water to a well was originally proposed by Dupuit’s theory.

38. The co-efficient of transmissibility (T) of an aquifer is given by which of the following equation?
a) T=b k
b) T=b/k
c) T=(b k)2
d) T=k/b
Answer: a
Explanation: The coefficient of transmissibility T is equals the field of co-efficient of permeability K multiplied by the aquifer thickness b:
T=b k.

39. According to Dupuit’s theory,the velocity of flow is proportional to__________
a) Sine of hydraulic gradient
b) Tangent of hydraulic gradient
c) None of the mentioned
d) All of the mentioned
Answer: b
Explanation: Based on the assumption proposed by Dupuit’s theory, the velocity of flow is proportional to the tangent of the hydraulic agent instead of its sine.

40. Dupuit’s theory states that, the Darcy’s law equation is not valid near_________
a) Well face
b) Top Surface of well
c) Sides of the well
d) All of the mentioned
Answer: a
Explanation: Since the flow of water is not horizontal near wall and the flow no longer remains laminar. Thus, Darcy’s law equation is not valid near the well face.

41. The rate of water contribution to the well increases with the increase in the depression head.
a) True
b) False
Answer: a
Explanation: The velocity of the percolating water depends on the depression head. If the amount of water withdrawn from the well is more, increasing the depression head, higher flow velocities will prevail in the vicinity of the well. At a certain stage, depression head may become so great that the soil particles also start coming with the percolating water.

42. The performance of a well is measured by its ________
a) specific capacity
b) specific yield
c) storage coefficient
d) permeability coefficient
Answer: a
Explanation: Specific capacity is the measure of the well performance indicating the rate of water percolation into a well. It is also defined as the yield of a well under the head of one meter. The specific capacity of the well is not constant but decreases as the discharge increases.

43. The safe depression head for open wells is generally taken to be x times the critical depression head when x is ______
a) 1/6
b) 1/3
c) 1/2
d) 3/4
Answer: b
Explanation: The depression head is kept equal to one-third of the critical depression head. The maximum or critical yield corresponds to the critical depression head and the maximum safe yield corresponds to the working head.

44. The yield of a well depends on ________________
a) permeability of soil
b) area of aquifer opening into the wells
c) actual flow velocity
d) permeability and actual flow velocity
Answer: a
Explanation: Yield of a well is the rate at which water percolates into the well under the safe maximum depression head or the critical depression head. It depends on the position of the water table, permeability, and porosity of the soil, rate of water withdrawal from the well, and amount of water storage in the well.

45. Which type of open well is suitable when the sub-soil is formed of gravel or coarse sand deposits?
a) Unlined wells
b) Wells with pervious lining
c) Wells with impervious lining
d) Temporary wells
Answer: b
Explanation: Permanent lining or wells with impervious lining is sunk in regions with alluvial soil formation. Unlined wells or temporary wells are of very shallow depths and can be constructed only when the water table is very near to the ground. Wells with pervious lining is generally suitable in the strata of gravel or coarse sand.

46. Depending upon the type of sub-soil formation, the construction of well is done in two ways i.e. Dug wells and Sunk wells. Dug wells are constructed in soft formations.
a) True
b) False
Answer: b
Explanation: Sunk wells are constructed in soft formations and the subsoil formation may be sandy or clayey. The lining is first constructed above the ground level and then it sunk in the sub-soil formation by putting a load on the lining, therefore, it is called sunk well.

47. A well 3 meters in diameter has its normal water level 3 meters below the ground level. By pumping water level in the well is depressed to 10 meters below the ground level. In 4 hours the water rises by 5 meters. Calculate the specific yield of the well.
a) 2.213 m3/hr
b) 5 m3/hr
c) 1.242 m3/hr
d) 3.224 m3/hr
Answer: a
Explanation: Specific yield (K) is calculated from the formula, K = 2.303 (A/T) log (H1/H2) where A is the area of the well, T is the total time of recuperation to bring the water level from depth H1 to H2.
Given values, A = 3.14/4 x 9 = 7.065 sq. m, T = 4 hours, H1 = 7m, H2 = 2m
K = 2.303 x 7.065 x 1/4 x log (7/2) = 2.213 m3/hr under a head of one meter.

48. An open well is called shallow well when ____________
a) the depth of well is small
b) the water table is high
c) it does not encounter mota formation
d) it finds the foundation in the mota formation
Answer: c
Explanation: A shallow well is that well which does not encounter mota formation. A deep well is that well which goes below the water table and finally finds its foundation in the mota formation.

49. Open well has big diameter than tube well because ____________
i. Open well has to irrigate more area.
ii. Water contribution to the well is natural and therefore, the percolation area has to be more.
iii. Storage of water has to be made before irrigation is done.
a) i and ii
b) i and iii
c) ii and iii
d) i, ii and iii
Answer: c
Explanation: Open wells have comparatively bigger diameter varying from 2-9 m and are suitable for low discharges in the order of 1-5 liters per second. The yield of the well is limited as the groundwater storage is limited. An open well irrigates small area and its cost per hectare is generally more than that of a tube well.

50. If the sub-soil formation is of fine sand, the rough value of the specific yield of well per unit area of the well (K/A) given by Marriot is ____________
a) 0.25
b) 0.50
c) 0.75
d) 1.0
Answer: b
Explanation: When the limit of critical velocity is not exceeded it may be rightly assumed that K/A is constant for a well. K/A is the specific yield of well per unit area of the well. The value for K/A given by Marriot for clay sub-soil is 0.25, for fine sand = 0.50 and for coarse sand, it is 1.0.

51. Which of the following method of lifting water is very suitable when the depth of well is from 12 to 20 meters?
a) Denkli
b) Churus
c) Persian wheel
d) Windmill
Answer: c
Explanation: Denkli is generally provided on the unlined wells which are supposed to irrigate only small fields. Churus is suitable for lined wells and can be conveniently used deep wells up to a maximum depth of 30 m. Persian wheel (also known as Rahat in some parts of India) is suitable when the depth of well is 12-30 m. The windmill is not very reliable.

52. Mota layer sometimes is also known as Marbarwa or Magasan is an impervious layer.
a) True
b) False
Answer: a
Explanation: Mota is an impervious patch of small thickness situated in the pervious soil mass. It gives structural support to the open well and is useful for unlined and partly lined wells. It can either be continuous or may be localized.

53. A well is to be constructed in a fine sandy sub-soil formation. The discharge of the well under the depression head of 4 m is 0.004 m3/sec. Calculate the diameter of the well.
a) 2 m
b) 2.5 m
c) 3 m
d) 3.5 m
Answer: c
Explanation: Given, Yield of well = 0.004 x 3600 = 14.4 m3/hour
Q or yield = K x H
14.4 = K x 4 or K = 3.6 m3/hour under head of 1 meter.
From Marriot’s table, the value of K/A for sandy sub-soil formation = 0.5
3.6/A = 0.5 or A = 7.2 m2
Diameter of well = (A x 4/3.14)1/2 = 3.07 m.

54. The most widely used type of a deep state tube well in India is __________
a) cavity well
b) strainer tube well
c) slotted pipe gravel packed tube well
d) both cavity and strainer tube well
Answer: b
Explanation: Strainer type tubewell is a bored hole in which a metal pipe with suitable perforations is inserted. A strainer type tubewell is suitable when water-bearing stratum is available at less depth and there is a possibility of tapping water through sides. It is generally unsuitable for fine sandy strata.

55. The general average yield from standard tube wells is of order of____________
a) 5 L/sec
b) 50 L/sec
c) 500 L/sec
d) 5000 L/sec
Answer: b
Explanation: Tube well irrigation is very suitable where the subsoil formation is suitable for storing water. The general average yield from the standard deep tube-wells is of the order 40 to 50 liters/sec. Shallow tube-wells may yield as high as 15-20 liters/sec if located at proper places.

56. Which of the following tubewell is suitable when a deep bearing stratum lies below an impervious layer and water contribution can take place through bottom only?
a) Strainer type tubewell
b) Abyssinian tubewell
c) Cavity type tubewell
d) Slotted type tubewell
Answer: c
Explanation: A strainer type tubewell is suitable when water-bearing stratum is available at less depth and there is a possibility of tapping water through sides. The slotted type tubewell is intermediate to the strainer and cavity type tubewell and is used where both the above cannot be used.

57. In cavity tubewell, there is a possibility of water tapping through sides and the flow is radial.
a) True
b) False
Answer: b
Explanation: In cavity-type tubewell, water contribution to the bored hole takes place through the bottom layer only as is done by screen well. The flow in a cavity well is essentially spherical. In strainer type tubewell, water enters into the well from the sides and the flow is radial.

58. What is the measure of the fineness of an aquifer?
a) Average grain size
b) Effective diameter
c) Mean particle size
d) Uniformity coefficient
Answer: b
Explanation: Effective diameter is an index of the measure of the fineness of an aquifer. It is used for designing various intake components of a tubewell. For permeability D10 (i.e. 90% grains retained on the sieve) is taken as the effective size.

59. What is Pack Aquifer Ratio?
a) D50 of the gravel / D50 of the aquifer
b) D50 of the aquifer / D50 of the gravel
c) D60 of the aquifer material / D10 of the aquifer material
d) D60 of the gravel / D10 of the aquifer
Answer: a
Explanation: The gravel pack is usually designed on the basis of Pack Aquifer (PA) ratio. It is usually defined as the ratio of D50 size of the gravel pack material to the D50 size of the aquifer material. It should be designed before designing the size of the slots to be made in the well-screens.

60. According to the Central Board of Irrigation and Power, the PA ratio for graded aquifer should be _______________
a) 9 – 12.5
b) 12 – 15.5
c) > 15.5
d) < 9
Answer: b
Explanation: The Central Board of Irrigation and Power suggested the following criteria for PA ratio.
i. For uniform aquifers (having Cu < or = 2.0), the PA ratio should be in between 9 – 15.5
ii. For graded aquifers (having Cu > 2.0), the PA ratio should be in between 12 – 15.5.

61. For good design, the uniformity coefficient (Cu) of the gravel material should preferably be ______
a) > 2.0
b) < 2.0
c) > 2.5
d) < 2.5
Answer: d
Explanation: The uniformity coefficient was proposed by Hazen. The value of the coefficient of the gravel material is preferably kept to be equal to or less than 2.5. A higher value may cause segregation of the gravel during pouring which will result in poor efficiency of the well.

62. The efficiency of the centrifugal pumps commonly used for lifting water from wells may be of the order of _____________
a) 30%
b) 65%
c) 90%
d) 95%
Answer: b
Explanation: Jet pumps have low efficiency of the order of 35% as compared to 65-85% for other centrifugal pumps. It is not used for irrigation tube-wells.

63. A tube well is suitable when the subsoil formation is made up of _____________________
a) cracked and faulted rock
b) alluvium
c) alluvium and various layers of sandy soil, clayey soil, and gravel
d) clayey soil
Answer: c
Explanation: In India most of the land especially the Indo-Gangetic plains, Narmada valley consists of deep alluvial soils. The sub-soil water slowly penetrates and is stored in porous sand and gravel beds which are extensively found in India. Tube wells can be easily installed in such lands and are useful for irrigation.

64. Bored tube wells in rocky consolidated formations as are encountered in the South Indian States of our country are usually drilled by _____________
a) rotary drilling rigs
b) percussion drilling rigs
c) down the hole hammer (DTH) rigs
d) cable method of drilling
Answer: c
Explanation: In rocky areas, down the hole hammer rigs (DTH rigs) are used to drill isolated holes of 10 to 15 cm in diameter. They are usually in the depth range of 100 m and mostly been constructed in southern states of India. This type of wells usually depends on joints and fissures in the rocks for their water supply.

65. Which of the following drilling method is unsuitable in loose formations such as unconsolidated sand and gravel or quicksand?
a) Percussion drilling
b) Wash boring method
c) Rotary boring method
d) Reverse rotary method
Answer: a
Explanation: Percussion method of drilling the well hole is also known as the cable method of drilling. It is ineffective in loose material because of slumping and caving of the material around the drilling bit. This method is useful for cutting consolidated rocks from soft clay to hardest rocks.

66. Which drilling method is best suited for underground formations made of gravel, sand and clayey deposits?
a) Percussion drilling
b) Wash boring method
c) Rotary boring method
d) Reverse rotary method
Answer: b
Explanation: Percussion method is suitable for soft and fissured rock formations. The rotary boring method can be successfully used for rock as well as unconsolidated formations. The reverse rotary method is used for soft unconsolidated formations made of sand, silt or soft clay.

67. Which is the fastest method of drilling and especially useful in unconsolidated formations?
a) Cable tool method
b) Water-jet boring method
c) Hydraulic Rotary method
d) Reverse Rotary method
Answer: c
Explanation: Hydraulic rotary method can handle hard and soft foundations with ease and the danger of accidents is lesser. The rate of drilling by this method for consolidated rock formation may vary from 10 to 15 m per day and in unconsolidated formations 100 to 150 m per day.

1. A dam reservoir which is not provided with gate controls on its spillway and other sluices is called ________________
a) detention dam
b) storage reservoir
c) retarding basin
d) flood control reservoir
Answer: c
Explanation: Storage reservoir is the one having gates and valves installed at its spillway and at its sluice outlets. The retarding basin is the one with uncontrolled and ungated outlets. The cost of gate installation is saved and there are no gates hence, the possibility of human error and negligence in their operation is eliminated.

2. A hydel power project has been envisaged to serve the water supply and irrigation needs of the area at its inception stage. The dam reservoir so constructed is known as __________________
a) multipurpose reservoir
b) single-purpose reservoir
c) distribution reservoir
d) retarding reservoir
Answer: a
Explanation: A reservoir planned and constructed to serve various purposes together is a multipurpose reservoir. It is designed to protect the downstream areas from floods, to conserve water, irrigation, industrial needs, hydroelectric purposes, etc. Bhakra dam and Nagarjuna Sagar dam are important multipurpose dams.

3. A dam reservoir catering to flood control, irrigation, and water supply basically designed for irrigation alone is a _______________-
a) multipurpose reservoir
b) single-purpose reservoir
c) distribution reservoir
d) retarding basins
Answer: b
Explanation: A reservoir planned, designed and constructed for one purpose is called a single purpose reservoir whereas a reservoir planned and constructed to serve various purposes together is a multipurpose reservoir. A small storage reservoir constructed within a city water supply system is called distribution reservoir

4. Which reservoir is also known as Mitigation reservoir?
a) Conservation reservoir
b) Flood control reservoir
c) Multipurpose dam
d) Storage reservoir
Answer: b
Explanation: A flood control reservoir protects the downstream areas by storing a portion of the flood flows to minimize the flood peaks. The entire inflow entering the reservoir is discharged or gradually released to recover the capacity for the next flood. It is also called as a mitigation reservoir.

5. A reservoir having gates and valves installation at its spillway and at its sluice outlets ___________
a) storage reservoir
b) retarding basin
c) both storage and retarding reservoir
d) distribution reservoir
Answer: a
Explanation: Storage reservoir is the one having gates and valves installed at its spillway and at its sluice outlets whereas retarding basin is the one with uncontrolled and ungated outlets. It provides more flexibility in operation and better control.

6. Which reservoir is helpful in permitting the pumps or the water treatment plants to work at a uniform rate?
a) Storage reservoir
b) Detention reservoir
c) Multipurpose reservoir
d) Distribution reservoir
Answer: d
Explanation: A small storage reservoir constructed within a city water supply system is called a distribution reservoir. It stores water during hours of no demand or less demand and supply water from their storage during the critical time of maximum demand.

7. In which of the following reservoir the flood crest downstream can be better controlled and regulated properly?
a) Distribution reservoir
b) Multipurpose reservoir
c) Storage reservoir
d) Retarding reservoir
Answer: c
Explanation: Storage reservoirs are preferred on large rivers and require better control. It is provided with gated spillway and sluiceways for more flexibility of operation, better control and to increase the usefulness of the reservoir.

8. What are the types of flood control reservoirs?
a) Multipurpose reservoir and Single purpose reservoir
b) Storage reservoir and retarding reservoirs
c) Distribution reservoir and Storage reservoir
d) Distribution reservoir and multipurpose reservoir
Answer: b
Explanation: There are two basic types of flood-mitigation reservoirs i.e. storage reservoir and retarding reservoirs. Storage reservoir is the one having gates and valves installed at its spillway and at its sluice outlets whereas the retarding basin is the one with uncontrolled and ungated outlets.

9. As the reservoir elevation increases, the outflow discharge increases.
a) True
b) False
Answer: a
Explanation: When floods occur the reservoir gets filled and discharges through sluiceways and the water level goes on rising until the flood has subsided. The inflow becomes equal or less than the outflow. The water gets completely withdrawn until the stored water is completely discharged.

10. The maximum discharging capacity of a retarding reservoir should not be equal to the maximum safe carrying capacity of the channel downstream.
a) True
b) False
Answer: b
Explanation: Since the retarding reservoir is not always filled much of the land below the maximum reservoir level will be submerged only temporarily and occasionally. The automatic regulation of outflow depends upon the availability of water. The maximum discharging capacity should be equal to the maximum safe capacity of the channel d/s.

11. What is the most important physical characteristic of a reservoir?
a) Storage capacity
b) Annual Yield
c) Average yield
d) Reservoir water level
Answer: a
Explanation: The main function of a reservoir is to store water and thus to stabilize the flow of water therefore, the most important physical characteristic is its storage capacity. It is determined from the contour maps of the area. The planimeter is used to measure the area enclosed within each contour.

12. In general practice adopted for capacity computations is to actually survey the site contours only at vertical distances of __________________
a) 5 m or more
b) Less than 5 m
c) 0.5 m or more
d) Less than 0.5 m
Answer: a
Explanation: The area of the intervening contours at smaller intervals of 0.5 m or so and at vertical distances of 5 m or so. It is then interpolated by taking the square root of the surveyed contours. It is also assumed that the square root of the interpolated ones varies in exact proportion to their vertical distances.

13. The area-elevation curve when integrated will yield the capacity-elevation curve.
a) True
b) False
Answer: a
Explanation: The surveyed area at large intervals is plotted on a graph and area-elevation curve is drawn. The equation of this curve can be obtained by statistical methods. It can be integrated to obtain the equation of the capacity-elevation curve to obtain storage capacity.

14. Which of the following method does not give more than 3% error in the determination of the capacity of a reservoir?
a) Prismoidal formula
b) Trapezoidal method
c) Average formula
d) Integration method
Answer: d
Explanation: The capacity of the reservoir can be determined by surveying only a few contours by using this method. When it is cross-checked with the capacity worked out by surveying a large number of contours this method gives less than 3% error. Since the areas of contours are not very precise figures this error is not considered much.

15. The surcharge storage in a dam reservoir is the volume of water stored ________________
a) between the minimum and maximum reservoir levels
b) between the minimum and normal reservoir levels
c) between normal and maximum reservoir levels
d) below the minimum pool level
Answer: c
Explanation: Surcharge storage is uncontrolled storage which is the volume of water stored between the normal pool level and the maximum pool level. It is that portion above the crest of the dam’s spillway which cannot be regulated.

16. The useful storage in a dam reservoir is the volume of water stored __________________
a) between the minimum and maximum reservoir levels
b) between the minimum and normal reservoir levels
c) between normal and maximum reservoir levels
d) below the minimum pool level
Answer: b
Explanation: Useful storage can be defined as the volume of water stored in a reservoir between the minimum pool and normal pool level. Conservation storage and flood-mitigation storage are the two sub-divisions of useful storage in a multi-purpose reservoir.

17. The dead storage in a dam reservoir is the available volume for collection of silt and sediment between _______________
a) bed level of the reservoir and minimum reservoir level
b) bed level of the reservoir and the silt level in the reservoir
c) bed level of the reservoir and the normal pool level
d) above the minimum pool level
Answer: a
Explanation: Useful storage can be defined as the volume of water stored in a reservoir between the minimum pool and normal pool level. Dead storage is the water stored in the reservoir below the minimum pool level. It is not of much use in the operation of reservoirs.

18. Bank storage in a dam reservoir _________________
a) increases the compound reservoir capacity
b) decreases the computed reservoir capacity
c) sometimes increases and sometimes decreases the computed reservoir capacity
d) has no effect on computed reservoir capacity
Answer: a
Explanation: The certain amount of water that seeps into the permeable reservoir banks when the reservoir is filled up is known as bank storage. This water comes out as soon as the reservoir gets depleted and it effectively increases the capacity of the reservoir.

19. The valley storage reduces the effective storage capacity of a reservoir.
a) True
b) False
Answer: a
Explanation: The effective storage for flood mitigation is equal to the useful storage plus surcharge storage minus valley storage i.e. storage corresponding to the rate of inflow in the reservoir. The actual net increase in the storage is equal to the storage capacity of the reservoir minus the natural valley storage. Thus, the available storage is reduced for flood mitigation reservoir.

20. The maximum level to which the water rises during the worst design flood is known as _________________________
a) full Reservoir level
b) maximum conservation level
c) minimum pool level
d) surcharge storage
Answer: a
Explanation: Normal pool level or maximum conservation level is the maximum elevation to which the reservoir water rises during normal operating conditions. Minimum pool level is the lowest water surface elevation which has to be kept under normal operating conditions. Surcharge storage can be defined as the volume of water stored between the normal pool level and the maximum pool level.

21. Which of the following formula can be used to determine the storage when three consecutive sections at equal height are taken?
a) Prismoidal formula
b) Trapezoidal method
c) Average formula
d) Integration method
Answer: a
Explanation: According to the Prismoidal formula,
Storage = h/6 [A1 + 4A2 + A3] where A1, A2, and A3 are the areas of succeeding contours and h is the vertical distance between two alternate contours.
It can be preferably used where three consecutive sections at equal height are taken.

22. The factor of safety against overturning generally varies between ___________
a) 2 to 3
b) 1.5 to 2
c) 0.5 to 1.5
d) 1 to 2
Answer: a
Explanation: Factor of safety against overturning can be determined by the ratio of righting moments about the toe to the overturning moments about the toe. The value generally varies between 2 to 3.

23. What is the maximum permissible tensile stress for high concrete gravity dam under worst conditions?
a) 500 KN/m2
b) 500 kg/cm2
c) 5 kg/m2
d) 50 KN/m2
Answer: a
Explanation: The masonry and concrete gravity dams are usually designed in such a way that no tension is developed anywhere in the structure. The maximum permissible tensile stress for high gravity dams is taken as 500 KN/m2 under worst conditions. If subjected to such tensile stresses crack develops near the heel.

24. Which failure occurs when the net horizontal force above any plane in the dam or at the base of the dam exceeds the frictional resistance developed at that level?
a) Overturning
b) Crushing
c) Sliding
d) BY development of tension
Answer: c
Explanation: Sliding should always be fully resisted. At any horizontal section of the dam, the factor of safety against sliding is –
FOS = u Ph / Pv where u = coefficient of friction, Ph = Sum of horizontal forces causing sliding and Pv = Algebraic sum of vertical forces.

25. Which failure occurs when the minimum stress exceeds the allowable compressive stress of the dam material?
a) Overturning
b) Crushing
c) Sliding
d) By development of tension
Answer: b
Explanation: The compressive stress produced if exceeds the allowable stresses then the dam material may get crushed, a dam may fail by the failure of its own material. The allowable compressive stress of concrete is generally taken as 3000 KN/m2.

26. Tension cracks in the dam may sometimes lead to the failure of the structure by?
a) Sliding of the dam at the cracked section
b) Overturning about the toe
c) Crushing of concrete starting from the toe
d) Both overturning and crushing
Answer: c
Explanation: When tension prevails, cracks develop near the heel and uplift pressure increases, reducing the net salinizing force. This crack by itself does not fail the structure but it leads to failure of the structure by producing excessive compressive stresses.

27. The major principal stress at the toe of a gravity dam under full reservoir condition neglecting the tailwater effect is given by ____________________
a) Pv
b) Pv tanQ2
c) Pv secQ2
d) Pv sinQ2
Answer: c
Explanation: When there is no tailwater, the principal stress in such a case is Pv secQ2 where Pv is the intensity of vertical pressure. This value of principal stress should not be allowed to exceed the maximum allowable compressive stress of dam material.

28. Which of the following criteria has to be satisfied for no tension at any point on a gravity dam?
a) The resultant of all the forces must always pass through the mid-point of the base of the dam
b) The resultant force for all conditions of loading must pass through the middle third of the base
c) The resultant of all the forces must pass through the upstream extremity of the middle third of the base
d) The resultant of all the forces must pass through the downstream extremity of the middle third of the base
Answer: b
Explanation: The minimum vertical stress Pmin is equal to zero in order to ensure that no tension is developed anywhere. If Pmin = 0, e = B/6 i.e. the maximum value of eccentricity that can be permitted on either side of the center is equal to B/6. This concludes the fact that the resultant of all forces must lie within the middle third of the joint width.

29. The bottom portion of concrete or a masonry gravity dam is usually stepped in order to _______
a) increase the overturning resistance of the dam
b) increase the shear strength
c) decrease the shear strength
d) increase the frictional resistance
Answer: b
Explanation: The foundation is stepped at the base to increase the shear strength at the base and at other joints and measures is taken to ensure a better bond between the dam and the rock foundation. By ensuring a better bond between the surfaces the shear strength of these joints should be made as good as possible.

30. The governing compressive stress in a concrete gravity dam which should not be allowed to exceed the permissible value of about 3000 KN/m2 while analyzing full reservoir case is ____________________
a) the vertical maximum stress at the toe
b) the major principal stress at toe
c) the shear stress at the toe
d) the principal stress at the heel
Answer: b
Explanation: In reservoir full case, the resultant is nearer to the toe and hence, maximum compressive stress is produced at the toe. The vertical direct stress distribution at the base is the sum of the direct stress and the bending stress and is given by the equation –
Pmax = V/B [1 + 6e/B] where V is the total vertical force, e is the eccentricity of the resultant force from the center of the base and B is the base width.

31. If the uplift increases and the net effective downward force reduces, the resultant will shift towards the toe.
a) True
b) False
Answer: a
Explanation: The resultant shifts towards the toe if the uplift increases and the net effective downward force reduces. This further increases the compressive stress at the toe and further lengthening the crack due to the development of tension. It finally leads to the failure of the toe by direct compression.

32. In high dams, the safety against sliding should be checked only for friction.
a) True
b) False
Answer: b
Explanation: The safety against sliding should be checked only for friction in case of low dams and in high dams, the shear strength of the joint (i.e. an additional shear resistance) is also considered for economical design. The dam section is given an extra slope or batter on the U/s or D/s side as per requirements for achieving stability.

33. For full reservoir condition in a gravity dam, the critical combination of vertical and horizontal earthquake accelerations to be considered for checking the stability is ________________________
a) vertically upward and horizontally downstream
b) vertically downward and horizontally downstream
c) vertically upward and horizontally upstream
d) vertically downward and horizontally upstream
Answer: b
Explanation: Horizontal inertia force should be considered to be acting at the center of the gravity of the mass regardless of the shape of cross-section and it acts horizontally downstream in worst cases under full reservoir case. This force would produce the worst results is it is additive to the hydrostatic water pressure (acting towards the downstream).

34. The base width of a solid gravity dam is 35 m and the specific gravity of dam material is 2.45. What is the approximate allowable height of the dam having an elementary profile without considering the uplift?
a) 64.68 m
b) 54.80 m
c) 164 m
d) 80 m
Answer: b
Explanation: The base width at bottom is given by B = H/Sc1/2 (c = 0 since uplift is not considered).
B = 35 m and Sc = 2.45
Allowable height of the dam H = 35 x 2.451/2 = 54.8 m.

35. A low gravity dam of elementary profile made up of concrete of relative density 2.57 and safe allowable stress of foundation material 4.2 MPa. What is the maximum height of the dam without considering the uplift force?
a) 120 m
b) 217 m
c) 279 m
d) 325 m
Answer: a
Explanation: The maximum possible height of low gravity dam is H = f / ϒw (Sc + 1) where f = allowable stress of dam material = 4.2 MPa, Sc = 2.57, and ϒw = 9.81 KN/m2.
H = [4.2 / (9.81 x 3.57)] x 1000 = 119.92 m.

36. The vertical stress at the toe was found to be 3.44 MPa at the base of the gravity dam section. If the downstream face of the dam has a slope of 0.617 horizontal: 1 vertical, the maximum principal stress at the toe of the dam when there is no tailwater is _______________
a) 1.7 MPa
b) 2.4 MPa
c) 3.6 MPa
d) 4.8 MPa
Answer: d
Explanation: The principal stress at the toe is given by Pat toe = Pv. secΦ2 (without considering the tailwater) where Pv = 3.44 MPa and tan Φ = 0.617/1 i.e.Φ = 31.67°
Pat toe = 3.44 x sec(31.67°)2 = 4.74 MPa.

37. What is the recommended value of shear friction factor against sliding?
a) More than unity
b) Less than unity
c) More than 3 to 5
d) Less than 3
Answer: c
Explanation: Shear Friction Factor is given by –
SFF = sliding factor (SF) + B.q / ∑Ph where B = width of joint or section area = B x 1, q is the shear strength of the joint, and Ph is the sum of horizontal force causing sliding. SF must be greater than 1 and SFF must be greater than 3 to 5. This analysis is carried out for a full reservoir case as well as an empty case.

38. The small openings made in the huge body of a concrete gravity dam such as sluices and inspection galleries can be assumed to be causing only local effects without any appreciable effect on the distribution of stresses as per the principle of_____________________
a) Laplace
b) St. Venant
c) Reynold
d) St. Francis
Answer: b
Explanation: Small openings made in the dam only produce local effects as per St. Venant’s principle. They do not affect the general distribution of stresses. This is one of the most important assumptions made in the two-dimensional analysis of gravity dams.

39. A concrete gravity dam having a maximum reservoir level at 200 m and the RL of the bottom of the dam 100 m. The maximum allowable compressive stress in concrete is 3000 KN/m2 and the specific gravity of concrete is 2.4. Calculate the height of the dam and check whether it is a high dam or low dam.
a) H = 90 m High gravity dam
b) H = 90 m Low gravity dam
c) H = 214.2 m High gravity dam
d) H = 214.2 m Low gravity dam
Answer: a
Explanation: The limiting height of the dam is given by-
H = f / ϒw (Sc + 1) where f = allowable stress of dam material = 3000 KN/m2, Sc = 2.4, and ϒw = 9.81 KN/m2.
H = 3000 / 9.81 x 3.4 = 89.9 m.
This value is more than the height of the dam so it is a high gravity dam.

40. The axis of a gravity dam is the ______________________
a) line of the crown of the dam on the downstream side
b) line of the crown of the dam on the upstream side
c) centre-line of the top width of the dam
d) line joining mid-points of the base
Answer: b
Explanation: The axis of the dam is taken as the reference line which is defined separately in the plan and in the cross-section of the dam. In plan, it is the horizontal trace of the U/s edge of the top of the dam. In the cross-section, the vertical line passing through the U/s edge of the top of the dam is considered as the axis of the dam.

41. Presence of tail-water in a gravity dam ____________________
a) increases the principal stress and decreases the shear stress
b) increases both the principal stress and the shear stress
c) decreases the principal stress and increases the shear stress
d) decreases both the principal stress and the shear stress
Answer: d
Explanation: The principal stress is given by the formula –
P = Pv sec(Φ)2 – p’ tan(Φ)2 where Pv is the intensity of vertical pressure and p’ is the tail-water pressure
The shear stress on the horizontal plane near the toe is given by –
S = (Pv – p’) tan(Φ)
From both the equations, it is clear that the tail-water pressure is opposite in nature and it reduces the value of principal stress and shear stress.

42. Which of the following earth dam is suitable only on impervious foundation?
a) Zoned embankment type
b) Homogenous embankment type
c) Non-homogenous type
d) Diaphragm type
Answer: b
Explanation: Homogenous earth dam may be constructed of uniform and homogenous material when locally available. The percolation of water is not checked in such dams. It is suitable only on impervious foundations.

43. Which of the following zone in Zoned type embankment prevents piping through cracks?
a) Central core
b) Transition zone
c) Outer zone
d) Core wall
Answer: b
Explanation: The outer zone is made of fairly pervious material like grit, gravel, and murrum. The innermost zone is done with fairly impervious material. The central core is covered by comparatively pervious transition zone which prevents piping through the cracks which may develop in the core.

44. Which of the following soil material is most preferred for the central core material of zoned embankment type dam?
a) Highly impervious clay
b) Sands or silty clays
c) Coarse sands
d) Gravels
Answer: b
Explanation: Clay shrinks and swells and is highly impervious is not suitable. It is sometimes mixed with fine sand or fine gravel so as to make it a suitable material for the impervious central core. Coarse sands and gravels are used in the outer shell.

45. Which embankment has the thickness of the diaphragm at an elevation less than 10 meters or the height of the embankment above the corresponding elevation?
a) Diaphragm type dam
b) Zoned type dam
c) Non-homogenous earth dam
d) Homogenous earth dam
Answer: a
Explanation: The thickness of the core differentiates the diaphragm embankment from the zoned type embankment.
Embankment type The thickness of core or diaphragm

46. Diaphragm-type embankment < 10 m or
< the height of embankment
2. Zoned-type embankment ≥ 10 m
5. The blanket in earth dam is provided ___________________
a) at the ground level on u/s side
b) at the ground level on the d/s side
c) at the ground level of the D/s side of the dam
d) on the D/s slope
Answer: a
Explanation: A blanket of relatively impervious material may be placed on the upstream face. Since homogenous earth dam poses the problems of seepage, a homogenous section is generally added with an internal drainage system. This keeps the seepage line well within the body of the dam.

47. Which of the following statement is correct with reference to earthen dams?
a) These dams are very costly as compared to other types
b) Gravity dams are less susceptible to failure as compared to rigid dams
c) These dams are suitable for construction on almost every type of foundation
d) Highly skilled labor is generally not required
Answer: c
Explanation: Earthen dams are suitable for any type of foundation but there has to be a separate provision for the spillway location. The mode of failure is sudden and requires high maintenance and the overall life is not so long. It requires low to moderately skilled labors and moderate capital cost.

48. During the construction of an earthen dam by hydraulic fill method, development of pore pressure becomes important in the __________________
a) central impervious core
b) pervious outer shell
c) transition zone
d) both central core and outer shell
Answer: a
Explanation: In this method, an embankment is created by water and earth slurry which is discharged inwards to form a central pool at site. When water is pumped to the site it spreads the coarser material drops out first and the finest at last. Thus the core material settles within the central pool forming a zoned embankment and high pore pressures develop in the core.

49. The process of laying and compacting earth in layers by power rollers under OMC for construction of earthen dams is known as ______________________
a) Rolled fill method
b) Hydraulic fill method
c) OMC method
d) Compaction
Answer: a
Explanation: In the rolled-fill method, the embankment is constructed by placing suitable soil materials in thin layers and compacting the layers with rollers. Power-operated rollers are used for dams and ordinary road rollers can be used for low embankments. The best compaction can be obtained at the optimum moisture content (OMC).

50. The central core of the zoned embankment type earth dam ________________________
a) checks the seepage
b) prevents piping
c) gives stability to the central impervious fill
d) distribute the load over a large area
Answer: a
Explanation: The central core or hearting is done with fairly impervious material and it checks the seepage. The transition zone of mediocre permeability prevents piping through cracks. The outer zone gives stability to the central fill.

51. Which type of dam is suitable on shallow pervious foundations?
a) Zoned embankment type
b) Homogenous embankment type
c) Both Non-homogenous type and homogenous type
d) Diaphragm type
Answer: a
Explanation: Zoned earth dam is also called a non-homogenous or heterogeneous earth dam and it is suitable on shallow pervious foundations. This dam is widely constructed and the materials of the zone are selected depending upon the availabilities.

52. What is the U.S.B.R recommended value for freeboard when the height of the dam is more than 60 m?
a) 2 m to 3 m
b) 2.5 m above the top of gates
c) 3 m above the top of gates
d) More than 3 m
Answer: c
Explanation: For controlled spillway, if the height of the dam is less than 60 m, the minimum recommended freeboard value is 2.5 m above the top of gates. If the height of the dam is more than 60 m, the minimum freeboard value is 3 m.

53. What is the recommended formula for top width of a very low dam?
a) H + 3
b) 0.2H + 3
c) 0.2 H
d) H + 5
Answer: b
Explanation: The top width (A) is generally governed by minimum roadway width requirements in case of small dams. The top width of the earth dam for very low dams is given by –
A = H/5 + 3 where H is the height of the dam.

54. What is the Terzaghi’s recommended value of U/s side slope for earth dam of height less than 15m of homogenous silty clay?
a) 2: 1
b) 2.5: 1
c) 3: 1
d) 3.5: 1
Answer: b
Explanation: For earth dams of homogenous silty clay of height less than 15 m, the recommended u/s slope is 2.5: 1 and d/s slope is 2: 1. For height more than 15 m, u/s slope value is 3: 1 and d/s slope is 2.5: 1.

55. If the height of the dam is 10 m, then the value of top width (A) according to Strange’s recommendations is _________________
a) 1.85 m
b) 2.5 m
c) 3.0 m
d) 4.0 m
Answer: b
Explanation: If the height of the dam is up to 7.5 m then the top width value according to Strange’s recommendation is 1.85 and if the height is in between 7.5 to 15 m, then the recommended value is 2.5. If the height is in between 15 to 22.5 m then the top width is 3.0 m.

56. When the height of the dam is between 7.5 to 15, the Strange’s recommended value for maximum freeboard of low earth dams is ______________
a) 1.2 to 1.5
b) 1.5 to 1.8
c) 1.85
d) 2.1
Answer: c
Explanation: For the height of the dam up to 4.5 m, the maximum freeboard is 1.2 to 1.5 m and for the height range 4.5 to 7.5 m, the maximum freeboard is 1.5 to 1.8 m. If the height of the dam is in between 7.5 to 15 m, the maximum value is 1.85 m.

57. Calculate the top width (A) of the earth dam of height 50 m.
a) 5.0 m
b) 4.75 m
c) 6.10 m
d) 3 m
Answer: c
Explanation: For dams higher than 30 m, the top width (A) is given by the following formula-
A = 1.65 (H + 1.5)1/3 where H is the height of the dam.
A = 1.65 (50 + 1.5)1/3 = 6.13 m.

58. A phreatic line in seepage analysis is defined as the line on which pressure is _______________
a) equal to the atmosphere
b) greater than atmosphere
c) lower than atmosphere
d) varying
Answer: a
Explanation: The line which joins the points in a dam section at which pressure is equal to the atmospheric pressure is called phreatic line. It is also called as line of seepage or saturation line. There is capillary fringe i.e zone of capillary which has negative hydrostatic pressure above this line.

59. Provision of horizontal berms at suitable vertical intervals may be provided in the downstream face of an earthen dam in order to _________________
a) allow the movement of cattle
b) allow the inspection of vehicles to move
c) reduce the erosion caused by the flowing rainwater
d) increase the erosion
Answer: c
Explanation: The provision of berms serves the following purposes –
It behaves like a good lining for reducing losses and leakage.
They provide protection against erosion and breaches due to wave action.
They help the channel to attain regime conditions as they help in providing a wider waterway.
It can be used as borrow pits for excavating soil to be used for filling.

60. During seepage through an earthen mass, the direction of seepage is ________________ to the equipotential lines.
a) perpendicular
b) parallel
c) not defined
d) diagonal
Answer: a
Explanation: The seepage through a pervious soil material for 2-D flow is given by Laplacian equation and the graphical solution suggests that the flow through the soil can be represented by flow-net. It consists of 2 sets of curves equipotential line and streamlines which is mutually perpendicular to each other.

61. The upstream face of the earth dam is considered as _____________________
a) equipotential line
b) streamline
c) streak line
d) path line
Answer: a
Explanation: Equipotential lines are the lines of equal energy. Every point on the upstream face of an earth dam will be under equal total energy, this line acts as an equipotential line. Similarly, the upstream floor of a weir acts as an equipotential line.

62. The effects of capillary fringe are on the slightly safer side and are neglected.
a) True
b) False
Answer: a
Explanation: When there is appreciable flow through the dam body below the phreatic line, it reduces the effective weight of the soil and also the shear strength of the soil due to pore pressure. The capillary tension in water leads to increased intergranular pressure as the insignificant flow through the fringe leads to greater shear strength. Hence, the effects are neglected.

63. Which of the following line acts as a dividing line between dry (or moist) and submerged line?
a) Equipotential line
b) Path line
c) Seepage line
d) Streak line
Answer: c
Explanation: The soil above the phreatic line or seepage line will be taken as dry and the soil below it shall be taken as submerged for computation of shear strength of the soil. This helps in drawing the flow net.

1. Kor demand of the crop should be taken into account while fixing the capacity of a canal.
a) Tue
b) False
Answer: a
Explanation: Suppose take wheat crop for example, let us say it requires a 60 cm of water in a total period of about 240 days (for a given area of field) thus giving an average outlet factor of 3456 hectares/cumec. But the kor depth of wheat is 14 cm in about 4 weeks which gives an outlet of 1728 hectares/cumec. Therefore it is clear that for a given area (A) outlet factor of 1728 is very much more than 3456. Hence kor demand should be given importance while designing irrigation canal.

2. What is the most important point to be considered while fixing the canal capacity?
a) Keenest Demand
b) Average Demand
c) Water Demand
d) Demand
Answer: a
Explanation: While fixing the capacity of a canal the main importance that should be kept in mind is the keenest demand, but not the average demand. For instance, let rice require 160 cm of water during 320 days which gives an outlet factor of 1728 hectares/cumec. (i.e D = 864B/∆ = 864 x 320 / 160 = 1728).

3. Let us consider in a given area the plantation of a certain crop takes 20 days, and the total water depth required by this crop is 80 cm on the field. Find the duty of irrigation water required for the crop during this period. (a) Assuming 20% losses of water in the watercourses, find duty at the head of the course. (b) Find the duty of the water at the head of the distributary, assuming 10% losses from the distributary head.
a) 182.9 hectares/cumec, 194.4 hectares/cumec
b) 172.8 hectares/cumec, 185.4 hectares/cumec
c) 172.8 hectares/cumec, 194.4 hectares/cumec
d) 185.6 hectares/cumec, 184.6 hectares/cumec
Answer: c
Explanation: Total depth of water needed = 80 cm
Period for which water is needed = 20 days
Duty of irrigation water = (864B/d) = (864 x 20 / 80) = 216 hectares/cumec
(a) Given 20% losses in the watercourse
= 216 x 0.8 = 172.8 hectares/cumec
(b) Given 10% losses in the head of the distributary
= 216 x 0.9 = 194.4 hectares/cumec.

4. A pump was installed in a field to supply water to the crops. The duty for this crop is 432 hectares/cumec on the field and the efficiency of pump is 50%. The sown area of the field is 5 hectares. Determine the maximum output required (H.P) of the pump, if the highest water level is 4 meters below the highest portion of the field. Assume negligible field channel losses.
a) 0.77H.P
b) 0.85H.P
c) 0.80H.P
d) 0.31H.P
Answer: d
Explanation: Area of field to be irrigated = 5 hectares
Duty of water for crop = 432 hectares/cumec
Discharge required for the crop is = (5 / 432) = 1/86.4 cumec
Volume of water lifted per second = 1 / 86.4 cumec
Therefore, weight of water lifted per second = (1 / 86.4) x 9.81 = 0.1135 KN/sec
(unit wt. of water = 9.81 KN/m3)
Maximum static lift of pump = 4 metres
Work done by the pump in lifting water = 0.1135 x 4 = 0.454 KWatt
The input of the pump = (0.454 / 0.735) = 0.62
(1 metric H.P = 0.735 KWatt)
Output H.P of the pump = (input/η) = (0.62 / 0.5) = 0.31 H.P.

5. At a certain place, the transplantation of a crop takes 15 days, and the total depth of water required by the crop is 90 cm on the field. During the plantation rain falls and about 20 cm is utilised to fulfil the demand. Now determine the duty of irrigation water required during the plantation.
a) 185.14hectares/cumec
b) 186.25hectares/cumec
c) 187.45hectares/cumec
d) 184.89hectares/cumec
Answer: a
Explanation: Total depth of water required = 90 cm
Useful rainfall = 20 cm
Extra water depth needed after useful rainfall = 90 – 20 = 70 cm
Period for which water is needed = 15 days
Duty of irrigated water (∆) = (864B/d) = (864 x 15 / 70) = 185.14 hectares/cumec.

6. Let us take the gross commanded area of a watercourse is 2000 hectares, 80% of which is culturable irrigable. Intensities of sugarcane and paddy are 30% and 50%. The duties of the crops at the head of the watercourse are 850 hectares/cumec and 1900 hectares/cumec. Find the discharge required at the head of the watercourse?
a) 0.98 cumec
b) 0.97 cumec
c) 0.9 cumec
d) 0.986 cumec
Answer: d
Explanation: Given G.C.A = 2000 hectares
Now the C.C.A = 2000 x 80 / 100 = 1600 hectares
Intensity of irrigation for sugarcane = 30%
Area to be irrigated under sugarcane = 1600 x 30/100 = 480 hectares
Intensity of irrigation of paddy = 50%
Area to be irrigated under paddy = 1600 x 50/100 = 800 hectares
Duty of sugarcane = 850 hectares/cumec
Duty of paddy = 1900 hectares/cumec
Discharge for sugarcane = 480 / 850 = 0.565 cumec
Discharge for paddy = 800 / 1900 = 0.421 cumec
Now the total canal capacity at the head of watercourse is = 0.565 + 0.421 = 0.986 cumec.

7. We can find out monthly or fortnightly water requirements of various crops using intervals.
a) True
b) False
Answer: a
Explanation: The water depth will be multiplied by crop area which is required in this interval, so as to give the volume of water required in this interval. By dividing the volume by interval, we can find out the discharge needed for each interval. The summation of this will give the discharge required for all the crops in each interval. The canal can then be designed for the maximum of these values.

8. By how much percentage canal capacity is increased to meet peak demands?
a) 30 to 35%
b) 20 to 25%
c) 15 to 20%
d) 20 to 30%
Answer: b
Explanation: In order to be more precise, the intervals are kept as small as possible. Monthly water requirements studies are conducted to provide a sufficient quantity of water for irrigation. So, the canal capacity is increased by 20 to 25% to meet the peak demands in a month.

9. Suppose the culturable commanded for a distributary is 8000 hectares, and the intensity of irrigation for Kharif season is 60% and that of rabi season is 30%. The average duty at the head of a distributary is 4000 hectares/cumec for Kharif season and for Rabi season it is 1800 hectares/cumec, find out discharge required at the head of the distributary?
a) 1.35 cumec
b) 1.32 cumec
c) 1.33 cumec
d) 1.3 cumec
Answer: c
Explanation: Area to be irrigated in Kharif season = 8000 x 60/100 = 4800 hectares
Area to be irrigated in Rabi season = 8000 x 30/100 = 2400 hectares
Water required at the head of distributary to irrigate Kharif area = 4800 / 4000 = 1.2 cumec
Water required at the head of distributary to irrigate Rabi area = 2400 / 1800 = 1.33 cumec
Therefore, the required discharge is maximum of the two = 1.33 cumec.

10. Determine the discharge required at the head of the distributary in a canal where Kharif area to be irrigated is 2400 hectares and Rabi area to be irrigated is 1800 hectares, for fulfilling maximum crop requirement. Assume suitable values for kor depth and kor period.
a) 2.08 cumec
b) 2 cumec
c) 2.1 cumec
d) 2.2 cumec
Answer: a
Explanation: Now let us assume kor period of 4 weeks for Rabi and 3 weeks for Kharif crop. Also, assume kor depth of 12.5 cm for Rabi and 21 cm for Kharif crop.
Outlet factor for rabi = (864B/∆) = 864 x 4 x 7 / 12.5 = 1935.36 hectares/cumec
Outlet factor for kharif = (864B/∆) = 864 x 3 x 7 / 21 = 864 hectares/cumec
Area to be irrigated for Rabi season = 2400 hectares
Area to be irrigated for Kharif season = 1800 hectares
Water required for Kharif season at the distributary = 1800 / 864 = 2.08 cumec
Water required for Rabi season at the distributary = 2400 / 1935.36 = 1.24 cumec
Therefore, the canal capacity at the head of the distributary is the maximum of the two, 2.08 cumec.

11. What are the types of sediment load?
a) Bedload and Suspended load
b) Bedload and Dissolved load
c) Bedload, Dissolved load and Suspended load
d) Suspended load and Dissolved load
Answer: c
Explanation: Bedload is that portion of sediment load that is transported along the bed by sliding, rolling or hopping. The dissolved load is the one that is chemically carried by water in the form of ions. When the material is maintained in suspension due to the turbulence of flowing water, it is called a suspended load.

12. Increased dissolved load gives the flooded stream its muddy color.
a) True
b) False
Answer: b
Explanation: Due to unusual high velocity in the flooded streams, there is heavy sediment transportation and deposition downstream. The increased suspended load may be visible giving the stream a muddy color.

13. At low velocity, the bed does not move at all but as velocity increases, it goes on assuming different shapes.
a) True
b) False
Answer: a
Explanation: Depending upon the discharge or the velocity of the water, the channel bed may be distorted into various shapes by the moving water. Ripples are superimposed on the bed when they first appear. At high velocities, ripples disappear and only dunes are left.

14. On increasing velocity beyond the threshold stage of motion, the bed develops ____________ shape.
a) dunes with ripples
b) saw-tooth ripples
c) anti-dunes
d) flat surface
Answer: b
Explanation: A channel bed made of fine sand less than 2 mm diameter develops the shape of ripples of saw-tooth type when the velocity is increased beyond the threshold stage of motion. If the velocity is increased further, large periodic irregularities appear that are called dunes.

15. Which of the following statement is wrong?
a) Ripples do not occur if the size of the bed particles is coarser than 0.6 mm
b) Dunes are much larger in length and height than ripples
c) Crests of both ripples and dunes extend across the entire width of the stream
d) Ripples are less rounded than Dunes
Answer: c
Explanation: Both formations of ripples and dunes tend to occur in the form of short-crested waves. Sub-critical flow conditions exist in both these regimes at this stage i.e. Froude’s number is less than unity. Crests of both do not extend across the entire width of the stream.

16. When the velocity is further increased beyond the stage where sand waves are formed in association with the surface waves, the waves are then called as ____________
a) dunes
b) ripples
c) flat surface
d) antidunes
Answer: d
Explanation: As the velocity is further increased, the surface waves become so steep that they move upstream and break intermittently and the sediment particles keep on moving downstream only. The flow becomes super-critical and sand waves are then called antidunes.

17. In the case of canals and natural streams, anti-dunes rarely occur.
a) True
b) False
Answer: a
Explanation: As compared to the ripples and dune regime, the resistance to flow is very small. The sediment transport rate is exceptionally high in this regime. Hence, it rarely occurs in natural streams.

18. On which factor does the movement of bedload depends?
a) Velocity of Flow
b) Type of Flow
c) Depth of Flow
d) Width of the River
Answer: a
Explanation: Bedload is a sediment load, and it moves by the actions like rolling, sliding and hopping which in turn depend on the velocity of flow. The bed load remains in the bottom layers of the flow.

19. Commonly the bed load is 10% of total suspended load.
a) True
b) False
Answer: a
Explanation: Sometimes bed load is estimated to be in between 3 to 25% of the total suspended load, depending upon the reflexes of the sediment to forces from the sectors like physical, chemical, and biological. In between this percentage we commonly take bed load is 10% of total suspended load.

20. What type of force is completely responsible for the bedload movement?
a) Forces of Turbulence
b) Drag Force
c) Capillary Force
d) Gravity Force
Answer: b
Explanation: Generally we know that bedload is a type of sediment which moves adjacent to the bed of the channel. Shear stress (τo) is responsible for this movement of bedload along the bed if the channel, which is developed by the flowing water along the channel bed. This shear force is called drag force or tractive force.

21. Which type of force is needed for suspension of suspended load in flowing water?
a) Capillary Force
b) Drag Force
c) Forces of Turbulence
d) Gravity Force
Answer: c
Explanation: Suspended load is a sediment load, but remains in suspension in the flowing water of the channel. The forces which cause this suspension are the forces of turbulence, which are generated by the flow of the channel itself.

22. A part of tractive force does not have any part in transporting the bed material, i. e bed load.
a) True
b) False
Answer: a
Explanation: In order to transport the bed load the tractive force (τo) should exceed the critical tractive force (τc), therefore the rate of movement of bedload becomes a function of (τo – τc). But, as soon as the sediment starts moving it is opposed by the ripples generated by the channel bed, and to overcome these ripples a large part of tractive force is lost and cannot be used again. So, this part of the force does not have any part in transporting the bed load.

23. Given that the bed slope of a channel is 1 in 2000 and the discharge is 60 cumecs. The depth of the channel is fixed and is given as 2 m. The tractive force needed for the movement of bedload is? Take critical velocity ratio as 1.1.
a) 8.89 N/m3
b) 8.87 N/m3
c) 8.85 N/m3
d) 8.83 N/m3
Answer: d
Explanation: Critical velocity (Vo) = 0.55my0.64
= 0.55 x 1.1 x (2)0.64
= 0.943 m/sec
Area = Q / Vo = 60 / 0.943 = 63.63 m2 (Given Q = 60 cumecs)
A = y (b + y (1/2)) (y = depth, b = base width)
63.63 = 2(b + 1) (for side slope as (1/2:1/2H:V))
b = 30.82 m
Perimeter (P) = b + √5y = 30.82 + 2 x √5
P = 35.3 m
R = A/P = 63.63 / 35.3 = 1.8 m (R = hydraulic mean depth)
Now, tractive force (τo) = γwRS = 9.81 x 103 x 1.8 x (1/2000) (γw = 9.81 x 103 N/m3)
= 8.83 N/m3.

24. What cause is prime responsible for the heavy movement of water from main canal to branch canal?
a) Existence of Favorable Gradient
b) Velocity of Flow
c) Type of Flow
d) Depth of the Canal
Answer: a
Explanation: When a branch canal is connected to the main canal, the branch canal starts taking water discharge which reduces the discharge in main canal thereby reduces sediment load capacity. This leads to the deposition of sediment load which further diverts the flow towards branch canal. This process continues till all the water in the main canal gets discharged into branch canal. The prime cause of this phenomenon is due to the existence of favorable gradient in the branch canal.

25. Water in which condition or state carry a maximum amount of sediment?
a) Uniform State
b) Vapor State
c) Floods
d) Ice
Answer: c
Explanation: According to Einstein’s equation (qb/q≼qS2/d3/2), the sediment carrying capacity of the channel depends upon the discharge per unit width. Therefore, this gives that floods can carry more amount of sediment than any other state of water because the discharge per unit width of channel is maximum in case of floods. Most of the annual sediment load is done by floods.

26. What minimum value of shear stress is needed to move the sediment?
a) Critical Velocity Ratio
b) Critical Shear Stress
c) Critical Velocity
d) Drag Force
Answer: b
Explanation: Generally in nature, any two moving bodies oppose each other due to internal presence of friction between them. Here as the sediment load moves along the bed of the channel, then internal friction is developed between them depending upon the soil. Therefore to overcome this friction a minimum value of shear stress is needed. This minimum value is called critical shear stress (τc).

27. Design a channel carrying a 30 cumecs. The median grain diameter is taken as 0.5 mm. The bedload concentration is 60 p.p.m by weight. Use Lacey’s Regime perimeter and Meyer-peter’s formulas.
a) B = 26 m, S = 1/5600, y = 1.25 m
b) B = 22 m, S = 1/5800, y = 1.55 m
c) B = 24 m, S = 1/5500, y = 1.45 m
d) B = 25 m, S = 1/5700, y = 1.35 m
Answer: d
Explanation: Quantity of bed load transported by weight = 40/106
Quantity of bed load transported per second = 40/106 (30 x 9.81 x 1000) = 11.8 N/sec
Lacey’s Regime perimeter = 4.75 x √Q = 26.03 m
Let us take channel bed width (B) as 25 m
Bed load per unit width = gb = 11.8 / 25 = 0.472 N/m/sec
Meyer Peter equation –
gb = 0.417 x [τo(n’/n) – τc]3/2
n’ = (1/24) x (0.5)1/6
= 0.011
n = 0.02
n’/n = 0.55
τc = 0.687 x da = 0.687 x 0.5 = 0.3435 N/m2
gb = 0.417 [9.81 x 1000 x RS x (0.55)3/2 – 0.3435]3/2
RS = 0.0002
Manning’s equation –
Q = 1/n x R2/3 x S1/2
R2/3 x S1/2 = 0.6
S = 0.0002/R
S = 1/5700
R = 1.15 m
Now P = 25+y√5, A = 25y + y2/2 for trapezoidal channel of 1/2:1 slopes
R = A/P = (25+√5y)/ (25+y2/2) = 1.15
From this y = 1.35 m
Therefore B = 25 m, y = 1.35 m, S = 1/5700.

28. Strickler’s formula is only applicable to the flexible boundary channels.
a) True
b) False
Answer: b
Explanation: Strickler’s formula is not used for moveable boundary channels as it does not account for the roughness due to undulations in the bed channel. It is used to calculate Manning’s rugosity coefficient and is valid for rivers with the bed of coarse material i.e. rigid boundary channels.

29. What is the minimum size of the bed material that will remain at rest in a channel?
a) d > or = 11 R.S
b) d < or = 11 R.S
c) d = 11 R.S
d) d > 11 R.S
Answer: a
Explanation: For no movement of the sediment particles, Tc > or = T0.
0.056. Yw. d. (Gs – 1) >or = Yw. R. S where, for sand of gravel Gs = 2.65
d >or = 10.82 R.S
d >or = 11 R.S.

30. What is the limitation of the Shield’s expression?
a) It can be used when the diameter of a particle (d) is < 6mm and Reynold’s number > 400
b) It can be used only when Reynold’s number > 400
c) It can be used only when the diameter of the particle is < 6 mm
d) It can be used when the diameter of a particle is > 6mm and Reynold’s number > 400
Answer: d
Explanation: The curve plotted by Shield between Reynold’s number and Entrainment function forms a suitable basis for the design of channels where it is required to prevent bed movement. When the particle size exceeds 6 mm such as for coarse alluvium soils, the particle Reynold’s number has found to be more than 400 representing roughness.

31. Calculate the Manning’s rugosity coefficient in a coarse alluvium gravel with D-75 size of 5 cm.
a) 0.025 m
b) 0.035 m
c) 0.055 m
d) 0.1 m
Answer: a
Explanation: The strickler’s formula is n = d1/6/24 where d = 0.05 m.
n = 0.051/6/24 = 0.025 m.

32. The manning’s coefficient for a lined trapezoidal channel with a bed slope of 1 in 4000 is 0.014 and it will be 0.028 if the channel is unlined. The area in the case of the lined section is 19.04 m2 and for the unlined section is 29.09 m2. What percentage of earthwork is saved in a lined section relative to an unlined section, when a hydraulically efficient section is used in both the cases?
a) 24.44 %
b) 34.55%
c) 50%
d) 37.66%
Answer: b
Explanation: The percentage of earthwork saving due to the lining is (A2 – A1)/A2 x 100.
= (29.09 – 19.04) / 29.09 x 100
= 34.55%.

33. Calculate the critical tractive stress if the median diameter of the sand bed is 1.0 mm.
a) 0.53 N/m2
b) 0.61 N/m2
c) 0.73 N/m2
d) 1.61 N/m2
Answer: a
Explanation: The given size of the particle is less than 6 mm so; shield’s equation cannot be used. The general relation is given by Mittal and Swamee which gives results within +5% of the values of Shield’s curve for all particle sizes.
Tc = 0.155 + [0.409 d2/(1 + 0.177 d2)0.5]
= 0.155 + [0.409 x 12/(1 + 0.177 x 12)0.5] = 0.53 N/m2.

34. Calculate the ratio of the tractive critical stress to the average shear stress if the water flows at a depth of 0.8 m in a wide stream having a bed slope of 1 in 3000. The median diameter of the sand bed is 2 mm.
a) 0.53
b) 1.86
c) 0.86
d) 1.53
Answer: a
Explanation: The critical tractive stress is given by (d < 6 mm) –
Tc = 0.155 + [0.409 d2/(1 + 0.177 d2)0.5]
= 0.155 + [0.409 x 22/(1 + 0.177 x 22)0.5] = 1.40 N/m2.
The average shear stress = Yw. RS = 9.81 x 0.8 x 1/3000 = 2.616 N/m2
Ratio = 1.40/2.616 = 0.53.

35. Determine the shear stress required to move the single grain on the side slopes, if the critical shear stress required moving the similar grain on the horizontal bed is 2.91 N/m2. Consider the angle of the side slope with the horizontal as 30° and the angle of repose of soil as 37°.
a) 1.61 N/m2
b) 2.61 N/m2
c) 2.91 N/m2
d) 1.00 N/m2
Answer: a
Explanation: The equation required is Tc’ = (1 – Sin2Q/Sin2R)1/2.Tc where Q = 30°, R = 37° and Tc = 2.91 N/m2.
Tc’ = (1 – Sin230°/Sin237°)1/2x 2.91 = 1.61 N/m2.

36. The shear stress required to move grain on the side slopes is less than the shear stress required to move the grain on the canal bed.
a) True
b) False
Answer: a
Explanation: The actual shear stress on the channel bed is Yw.RS while on slopes this value is given by 0.75 Yw. R.S. The following equation shows that Tc’ < Tc.
Tc’ = {(1 – Sin2Q/Sin2R)1/2.Tc } where Tc’ = shear stress on the side slopes, Tc = shear stress on the horizontal bed, Q = Angle of side slope with the horizontal, and R = angle of repose of the soil.

37. What types of channel sections are usually adopted?
a) Triangular and Circular channel
b) Triangular and Trapezoidal channel
c) Rectangular and Trapezoidal channel
d) Rectangular and Triangular channel
Answer: b
Explanation: Generally, Triangular and Trapezoidal channel sections are adopted. For smaller discharges, the Triangular channel section is adopted and trapezoidal channel for larger discharges.

38. What is the maximum permissible velocity in Cement concrete lining (Unreinforced) as per Indian Standards?
a) 2 to 2.5 m/sec
b) 1.5 to 2 m/sec
c) 1.2 to 1.8 m/sec
d) 1.5 to 2.5 m/sec
Answer: a
Explanation: The maximum permissible velocity for unreinforced cement concrete lining is 2.0 to 2.5 m/sec. For Boulder lining, the maximum permissible velocity is 1.5 m/sec.

39. What is the minimum value of free-board for Main and Branch Lined canals if the discharge is more than 10 cumecs as per specified by BIS code?
a) 0.60
b) 0.50
c) 0.75
d) 0.30
Answer: c
Explanation: In main and branch canals for discharge more than 10 cumecs, the minimum value of freeboard is 0.75 m. For branch canals and distributaries in which discharge is less than 10 cumecs the value is 0.60 m.

40. The distance measured above the F.S.L and to the top of the lining is known as free-board.
a) True
b) False
Answer: a
Explanation: The vertical distance between the full supply level (F.S.L) and the top of the bank is called freeboard. It depends on the size of the canal, location of the canal and water level fluctuations.

41. What is the minimum value of inspection bank width recommended by Indian Standards if the discharge is more than 30 cumecs?
a) 8.0
b) 5.0
c) 6.0
d) 7.0
Answer: a
Explanation: The minimum value of the top width of the bank depends on the value of discharge. For discharge greater than 30 cumecs, the minimum value of the top width of the inspection bank is 8.0 m and for Non-inspection bank is 5.0 m.

42. When the concrete lining is not reinforced in a channel, velocities up to 2.5 m/sec are permitted.
a) True
b) False
Answer: a
Explanation: Higher velocities can be safely used in lined canals. The maximum permissible velocities for concrete linings when the lining is not reinforced are up to 2.5 m/sec. The lining can be reinforced if still higher velocities are desired.

43. Calculate hydraulic mean depth for concrete-lined channel to carry a discharge of 350 cumecs at a slope of 1 in 5000. The value of manning’s constant for the lining is 0.015. The side slopes of a channel may be taken as 1.5:1. Assume limiting velocity in the channel as 2m/sec.
a) 2.20 m
b) 2.40 m
c) 2.60 m
d) 2.80 m
Answer: d
Explanation: Using Manning’s equation: V = 1/n.R2/3.S1/2
2 = 1/0.015 x R2/3 x 1/50001/2
R = 2.80 m.

44. Calculate the central depth of a triangular channel section to carry a discharge of 15 cumecs. Consider the available slope as 1 in 9000. Assume the side slopes of the channel be 1.25:1 and manning’s constant is 0.015 for good brick work in lining.
a) 2.94 m
b) 3.14 m
c) 2.25 m
d) 2.77 m
Answer: a
Explanation: Given data- tanQ = 1/1.25; Q = 0.675 radian
A = y2 (Q + cotQ) = 1.925y2 and R = y/2
Q = 1/n.A.R2/3.S1/2
15 = 1/0.015 x 1.925y2 x (0.5y)2/3x 1/90001/2
y = 2.94 m.

1. Which type of canal does not need cross drainage structures?
a) Side Slope Canal
b) Contour Canal
c) Watershed Canal
d) Field Channel
Answer: a
Explanation: Side slope canal is that type of canal which runs perpendicular to the ground contours, i.e they run parallel to natural drainage flow and do not intercept the drainage channels and therefore avoiding the construction of cross drainage structures.

2. What is the name given to the junction of two streams?
a) Ridge
b) Area of Mixture
c) Merging
d) Area of Mingling
Answer: a
Explanation: The dividing line between the catchment areas of two streams is called ridge or watershed. The main watershed between two streams divides the drainage area of the two streams.

3. On flatlands what type of canal alignment is used?
a) Side Slope Canal
b) Contour Canal
c) Watershed Canal
d) Field Channel
Answer: c
Explanation: For flat lands, the slopes are relatively flat and uniform and it is very easy, and advantageous to align canals along watershed. Therefore the name of the alignment is watershed canal.

4. A canal aligned on the watershed saves the cost of constructed cross drainage works.
a) True
b) False
Answer: a
Explanation: By aligning a canal on the ridge, helps to irrigate the land on both sides of the canal. Moreover, the drainage flows away from the ridge, this gives an advantage in a way that the drainage does not cross a canal aligned on the ridge. Therefore the cost of construction of cross drainage works is reduced.

5. Which type of canal is most useful in hilly areas?
a) Side Slope Canal
b) Contour Canal
c) Watershed Canal
d) Field Channel
Answer: b
Explanation: The idea of watershed canal in hilly areas is does not work because it is highly uneconomical, since the river flows in valley below the ridge. And the watershed may be hundreds of meters above the river.

6. Contour canal can irrigate only on one side of the canal.
a) True
b) False
Answer: a
Explanation: In this canal the river slope is much steeper than the canal bed slope so therefore the canal encloses more and more area between itself and the river. It should be noted that more fertile lands are located at the lower levels. In other words we can say contour canal irrigates only on one side as the other side is higher.

7. What type of canal necessitates construction of cross drainage works more than any other types?
a) Side Slope Canal
b) Contour Canal
c) Watershed Canal
d) Field Channel
Answer: b
Explanation: In this contour canal the drainage of the river flow is always perpendicular to the ground contours, and this would certainly require crossing of natural drains and streams, which necessitates the construction of cross drainage structures.

8. Which type of canal is the farmer’s responsibility?
a) Contour Canal
b) Side Slope Canal
c) Watershed Canal
d) Field Channel
Answer: d
Explanation: The maintenance of field channel is the responsibility of the farmers because this channel is laid along the field boundaries and supply water to the fields to meet their requirements.

9. Which type of canal need syphons?
a) Side Slope Canal
b) Contour Canal
c) Watershed Canal
d) Field Channel
Answer: c
Explanation: The alignment of watershed canal should run in areas where there homes, sacred places etc, and does not run in loops of the ridge line but run in straight. So, for these depressions in the ridge line necessitates construction of viaducts or syphons to maintain the canal fixed supply level.

10. If a looping is present in the ridge line they how can that area be irrigated with?
a) Distributary
b) Tributary
c) Weir
d) Canal
Answer: a
Explanation: The alignment of watershed canal does not run along the loops of the ridge line, but instead they run in straight line leaving a small area between the alignment and the looped ridge line. This area can be irrigated with help of a distributary which starts from the starting of the loop of the ridge line and runs along the ridge line and finally joins the alignment at the end of the loop.

11. Loss of canal discharge occurs mainly due to?
a) Seepage and Percolation
b) Percolation and Absorption
c) Seepage and Evaporation
d) Seepage and Absorption
Answer: c
Explanation: Seepage and Evaporation are the two main reasons for loss of canal discharge. Out of these, seepage loss is most significant in the initial stages (about 40%). Evaporation loss may range from 2 to 3% of the total canal discharge.

12. Which of the following factor do not affect seepage loss?
a) Underground water-table condition
b) The porosity of the soil
c) Physical properties of the canal water
d) Prevailing wind velocity in the region
Answer: d
Explanation: The loss due to seepage is the most significant so far as irrigation loss from canals is considered. The porosity of the soil, underground water-table condition, condition of the canal system, and physical properties of the canal water are all factors affecting seepage loss.

13. In which of the following phenomenon there is a continuous direct flow between the canal and the underground water table?
a) Absorption
b) Seepage
c) Percolation
d) Both Absorption and Percolation
Answer: c
Explanation: When the ground water-table is nearer to the natural surface, the water which has entered the subsoil may join the saturation zone or underground reservoir to maintain a continuous direct flow. Due to the difference of level, the water directly flows from the canal to the underground reservoir through the soil pores.

14. Which of the following factor have no influence on evaporation loss?
a) The temperature of the region
b) Prevailing wind velocity in the region
c) Area of water surface exposed to the atmosphere
d) The porosity of the soil
Answer: d
Explanation: Evaporation loss is not that significant and may range from 2-3% of the total canal discharge. The rate of loss depends mainly on temperature and humidity of the region, prevailing wind velocity and the area of water surface exposed to the atmosphere.

15. In the phenomenon of Absorption, the extent of saturation goes on increasing from the ground level below the soil with depth.
a) True
b) False
Answer: b
Explanation: The soil which is in immediate contact with the channel section forms a bulb of saturated soil. The soil layer below the saturated bulb is partially saturated. Thus the extent of saturation goes on decreasing from the ground level and a zone of unsaturation exists between them.

16. According to Etcheverry and Harding, the range of conveyance loss is highest in which soil?
a) Sandy Loam
b) Loose sandy soil
c) Clay Loam
d) Rocks
Answer: b
Explanation: The range of water lost in conveyance for loose sandy soil is 5.18 to 6.10.
For Sandy loam = 3.66 to 5.18
For Clay loam = 1.83 to 3.66
For rocks = about 0.9.

17. Which of the following empirical formula is used in Punjab for calculating channel losses?
a) Q’ = 1.9 Q1/6
b) Q’ = 1/200 x (B+D)2/3
c) Q’ = 1.9 Q
d) Q’ = 1/200 x (B-D)2/3
Answer: a
Explanation: The formula for channel losses used in Punjab and Haryana states is Q’ = 1.9 Q1/6.
Where Q’ = Losses in cumecs per million sq.m of the wetted perimeter, Q = Discharge in cumecs.
The formula is used in U.P is; Q’ = 1/200 x (B + D)2/3 where B is the bed width of channel and D is the depth of water in the channel in meters.

18. Inflow and Outflow method is generally used to measure ____________
a) transpiration loss
b) seepage loss
c) evaporation loss
d) percolation loss
Answer: b
Explanation: Seepage loss is generally measured by Inflow and Outflow method. Regular and simultaneous discharge measurement is done at the entrance and end of the selected reach. Same gauge and discharge are maintained at the entrance during the period of observations. The difference between the inflow and outflow over the reach gives the quantity of water lost.

19. Which of the following statement is incorrect?
a) The seepage loss is very high in black cotton soils and heavy clay as compared to sandy porous soil
b) The seepage through a new canal is more than from a silted canal
c) The more the silt, the lesser are the losses
d) The more the velocity, the lesser will be the losses
Answer: a
Explanation: The seepage losses depend upon all these factors. The seepage loss is very high in sandy or porous soil as compared to black cotton soil or clays. The sandy soil has high permeability so the soil structure allows more water to penetrate and hence there is more seepage.

20. Which soil has the least range of conveyance loss according to Etcheverry and Harding?
a) Sandy Loam
b) Clay Loam
c) Loose sandy soil
d) Impervious clay loam
Answer: d
Explanation: The loss is generally expressed in cubic meters per sec per million square meters of the wetted perimeter. For Impervious clay loam, the range is between 1.02 to 1.22.

21. The quantity of water that can be diverted from the river into the main canal has no dependency on which of the following factor?
a) The water available in the river
b) The capacity of the main canal
c) Water demand of the branch canals
d) The capacity of the branch canal
Answer: d
Explanation: This method of distribution of available water into various canals is called canal regulation and the flow from the main canal is followed by the branch canals and then to the distributaries. The flow distribution depends on the demand for water on various canals.

22. In case of high demand but insufficient supplies, all the smaller channels are made to run simultaneously and continuously with reduced supplies.
a) True
b) False
Answer: b
Explanation: This alternative is generally not preferred as it causes siltation, weed growth, seepage, and water-logging. To compensate for these disadvantages and provide time for sufficient inspections and repairs of the channels, distributaries are made to run turn by turn with full supplies.

23. Which of the following combination is wrong?
a) Roster – Indication of allotted supplies to different channels
b) Flexible regulation – Allocation of supplies with anticipated demand
c) Head regulator – Regulation of discharge in the canal
d) Chak – Command area of an inlet
Answer: d
Explanation: Chak is the command area of an outlet. It is a function of operating characteristics of the system, peak design requirement of an outlet and field application rate.

24. Which of the following characteristics is wrong about extensive irrigation?
a) The irrigation extends to a large area with the lowest available supply
b) Agricultural production and protection against famine will be at optimum levels
c) The crop production will be minimal per unit of available water
d) It creates a perpetual scarcity of water
Answer: c
Explanation: In spite of the fact that this method requires good control and monitoring on the release of water, it is usually adopted in India. The production would be the maximum per unit of available water, though it may not be optimum of the land covered.

25. What are the categories of the entire water conveyance system?
a) The primary distribution system and secondary distribution system
b) The primary distribution system, the secondary distribution system, and Water allowance
c) The primary distribution system, secondary distribution system, and tertiary distribution system
d) The primary distribution system, secondary distribution system, tertiary distribution system, and Water allowance
Answer: c
Explanation: An integral management system designed to deliver a constant flow of water among irrigators along a tertiary canal is called Warabandi. This entire water conveyance system is divided into Primary, Secondary and Tertiary distribution system. Water allowance is just a certain rate of flow which is allocated to each unit of C.C.A in this warabandi system.

26. Which of the following statement is wrong?
a) The primary and secondary distribution system is fully controlled by the State Irrigation Department
b) The tertiary system is managed by farmers
c) The distribution of water managed by farmers is done on a seven-day rotation basis
d) Water allowance is not a compromise between the likely demand and the supply for a given project
Answer: d
Explanation: The value of water allowance decides the carrying capacities of the water-courses and the distributaries. Hence, it is generally a compromise between the likely demand and the supply for a given project.

27. What is the correct formula for Flow Time in an hour for an individual farmer?
a) (FT for the unit area) x (area of farmer’s fields) + (his bharai – his jharai)
b) (FT for the unit area) x (area of farmer’s fields) – (his bharai – his jharai)
c) FT for the unit area + (his bharai – his jharai)
d) FT for the unit area – (his bharai – his jharai)
Answer: a
Explanation: The formula for the flow time an hour for an individual farmer after the allowance of Bharai and Jharai timings is –
(FT for the unit area) x (area of farmer’s fields) + (his bharai – his jharai)
Bharai is generally zero in case of the tail (i.e. the last farmer) and Jharai is usually zero for all except for the tail.

28. What is the formula for the Flow Time for the unit area in hours?
a) 168 – (total Bharai – total Jharai) / Total area to be irrigated by the water-course
b) 168 + (total Bharai – total Jharai) / Total area to be irrigated by the water-course
c) 168 – (total Bharai – total Jharai)
d) 168 + (total Bharai – total Jharai)
Answer: a
Explanation: The distribution is done on the basis of seven-day rotation i.e. 24 x 7 = 168 hours.
The bharai time is debited from the common pool time of 168 hours and that value of bharai which is not efficient for field applications, the tail end farmer is compensated and allowed a certain recovery of the bharai time is called jharai.
Hence, the formula for Flow Time for a unit area of a land holding is given as –
FT = 168 – (total Bharai – total Jharai) / Total area to be irrigated by the water-course.

29. What is the safe limiting velocity for cement concrete lining?
a) 1.5 m/s
b) 2.2 m/s
c) 2.7 m/s
d) 1.8 m/s
Answer: c
Explanation: For cement concrete lining, the safe limiting velocity is 2.7 m/s. For boulder lining, it is 1.5 m/s and for burnt clay tile lining it is approximately 1.8 m/s. This serves as a guide for selecting canal slopes and alignments.

30. The most economical type of lining is the one which shows ____________
a) minimum benefit-cost ratio
b) maximum benefit-cost ratio
c) zero benefit-cost ratio
d) benefit-cost ratio = 1
Answer: b
Explanation: The one which shows the maximum annual benefit-cost ratio is the most economical lining. It may have a higher cost initially but longer life than another type of lining having a lesser value of the annual benefit-cost ratio.

31. Which of the following guideline is not recommended for the choice of lining when the bed width of the canal is up to 3 m?
a) Single burnt clay tile lining or brick lining
b) PCC slab lining
c) Flexible membrane lining with adequate earth/tile cover
d) In-situ cement concrete lining in bed as well as on sides
Answer: d
Explanation: The choice of suitable lining for different size of a canal depends on canal slopes and alignments, size and importance of canal, the climate of the area, initial expenditure, etc. In-situ cement concrete lining in bed and on sides is generally preferred for canals when bed width is greater than 8 m.

32. Which of the following is not a cause for the hydrostatic pressure on the lining?
a) Seeping of the rainwater in the backfill
b) When the water table remains below the canal bed
c) The backfill is of low drainage
d) The backfill is of high permeability (i.e. > 3 cm/sec)
Answer: d
Explanation: If the subgrade is made of clear gravel or sand of good permeability (> 3 cm/sec) and the water table is not likely to go above the canal bed no drainage arrangements may be necessary. There will be no appreciable time lag in the dissipation of drawdown pore pressures in the subgrade for such case.

33. Drainage relief pockets are provided ___________
a) at isolated locations in the bed as well as sides of the lined canal
b) in the bed of the canal
c) in the sides of the canal
d) anywhere in the canal
Answer: a
Explanation: Drainage relief pockets are provided with pressure relief valves filled with a graded filter containing gravel, coarse sand, and fine sand. They are provided at isolated locations in the bed as well as on the sides below the lining at suitable spacing (@ 15 to 20 m intervals).

34. The permeability of lining is also governed by the depth of water in the canal and type of subgrade soil.
a) True
b) False
Answer: a
Explanation: The permeability of lining may decide the quantum of seepage loss from a canal. The seepage losses from a canal for a particular area will depend on the local conditions such as the values of land and water, population intensity, etc.

35. Pressure relief valves may help in ___________
a) releasing the hydrostatic pressure
b) holding the hydrostatic pressure
c) increasing the hydrostatic pressure
d) may increase or decrease the hydrostatic pressure
Answer: a
Explanation: Pressure relief valve is a flap valve opening upwards into the canal. It helps in releasing the hydrostatic pressure as soon as the differential head exceeds the safe pressure. These valves open out as soon as the differential pressure goes beyond 10 cm.

36. Pipe drains run _______________
a) longitudinally on the bed and transverse to the length of canal on the side slopes
b) longitudinally to the length of the canal on the side slopes and transverse to the bed
c) longitudinally to the length of the canal
d) transverse to the bed
Answer: a
Explanation: Provision of open-jointed pipe drains is one of the methods of drainage arrangements. It will run longitudinally on a trench excavated below the lining on the canal bed along the length of the canal and transverse to the length of the canal on the side slopes.

37. Which of the following statement is wrong about the requirement of good lining?
a) The lining should be able to withstand the differential sub-soil pressure
b) Brick lining, concrete lining or precast slab lining can be easily repaired as compared to cast-in-situ concrete lining
c) The hydraulic efficiency generally reduces with time
d) Brick tile lining may provide better abrasion resistance than cement concrete and boulder lining
Answer: d
Explanation: A canal may have to transport a considerable amount of sediment load which damages the lining by abrasion. Cement concrete and stone boulder linings provide better abrasion resistance as compared to the brick tile lining.

38. Which type of lining is adopted when the channels have become stable and no danger of scouring is expected?
a) Brick lining
b) Single burnt clay tile lining
c) In-situ cement concrete lining
d) Flexible membrane lining in the bed and rigid lining on the sides
Answer: d
Explanation: The burnt clay tile lining is to be adopted at places where aggregates for the manufacture of concrete are not available economically. The brick lining is used where seepage considerations are important. In-situ cement concrete lining is provided when the canal has a bed width greater than 8 m.

39. What number of major defects does the ogee fall have?
a) 1
b) 2
c) 3
d) 4
Answer: b
Explanation: Ogee fall has two major defects. Heavy drawdown occurs on the upstream side, which results in lower depths, higher velocities and bed erosion. The kinetic energy of the flow will not be dissipated due to smooth transition, causing erosion on downstream bed and banks.

40. In which fall the depth discharge relationship is unaffected?
a) Trapezoidal Notch Fall
b) Ogee Fall
c) Rapids
d) Straight Glacis Fall
Answer: a
Explanation: In this fall the notches could be designed to maintain the normal water depth in the upstream channel at any two discharges, as the intermediate values do not vary much. Therefore the depth discharge relationship of the channel is not affected by the introduction of the fall.

41. Which type of fall is suitable for 60 cumecs discharge and 1.5 m drop?
a) Montague Type Fall
b) Rapids
c) Straight Glacis Fall
d) Baffle Fall
Answer: c
Explanation: In this type of modern fall, a straight glacis is provided after the raised crest. The hydraulic jump is made to happen on this glacis causing sufficient energy dissipation. If not flumed this fall gives good performance. This fall is suitable for 60 cumecs discharge and 1.5 m drop.

42. What is the reason for the construction of baffle wall in baffle fall?
a) To Maintain Uniform Velocity Flow
b) To Ensure Formation of Jump
c) To Reduce Soil Erosion
d) To Ensure Uniform Discharge of the Flow
Answer: b
Explanation: The baffle wall is provided at a calculated height and a calculated distance from the toe of the glacis to ensure proper formation of the jump on the baffle platform. This type of fall is suitable for all discharges and for drops which are more than 1.5 m.

43. Which type of fall is not adopted in India?
a) Ogee Fall
b) Rapids
c) Inglis Fall
d) Montague Type Fall
Answer: d
Explanation: In this type of fall the energy dissipation is incomplete on a straight glacis due to the vertical component of velocity remains unaffected. So, therefore due to this reason the straight glacis is replaced by a parabolic curve known as Montague profile. This curved glacis is difficult to construct and therefore is costlier. Hence it is not adopted in India.

44. Canal drop or canal fall is needed construction in the canal bed.
a) False
b) True
Answer: b
Explanation: If the available natural slope is steeper than the bed slope of the canal, then this is adjusted by the construction of vertical falls or drops in the canal bed at suitable intervals. But such a drop will not be stable, therefore in order to maintain this drop a masonry structure is constructed. This is known as canal fall or canal drop.

45. On what term does the construction of a fall in the case of the main canal depend?
a) Discharge Capacity of the Canal
b) Length of the Canal
c) Topography
d) Cost of Excavation and Filling versus Cost of fall
Answer: d
Explanation: The main does not irrigate any area directly, so therefore the site of the fall is based on the considerations of economy in cost of excavation and filling versus cost of fall. The excavation and filling on both sides of the fall should be balanced because unbalanced work causes extra cost.

46. On what factor in case of branch canals, the construction site for a fall depends?
a) Cost of Excavation
b) Topography
c) Commanded Area
d) Cost of fall
Answer: c
Explanation: By considering the commanded area of a branch canal or a distributary canal the location for the falls is decided. The procedure is to fix the FSL needed at the head of the off taking channels and outlets. Thus the FSL can be marked at all commanded points and hence deciding appropriate locations for the falls in canal FSL and therefore in canal beds.

47. How many types of falls are present?
a) 4
b) 5
c) 7
d) 8
Answer: d
Explanation: Since the idea of the construction of falls various types of falls have been designed. Some important falls are ogee falls, rapids, trapezoidal notch falls, simple vertical drop type and sarda type falls, straight glacis falls, Montague type falls and inglis falls.