Category: Irrigation Engineering Objective Questions

250+ TOP MCQs on Economic Water Rates Vs Prevailing Revenue Rates in India and Answers

Irrigation Engineering written test Questions & Answers on “Economic Water Rates Vs Prevailing Revenue Rates in India”.

1. Depending on how many factors the government generally fixes the irrigation water revenue rates?
a) 3
b) 2
c) 1
d) 4
Answer: b
Clarification: Depending on the factors like water requirement of the crops, and earning capacity of the crop the government generally fixes the revenue rates of irrigation waters.

2. According to which factor the real cost of irrigation water can be worked out?
a) Maintenance Costs
b) Initial Capital
c) Economic Considerations
d) Total Area of Irrigated Land
Answer: c
Clarification: Based on economic considerations like computing the cost of irrigation project, its annual operating and maintenance costs, and recovery of the capital during its lifespan.

3. Economic water rate means the rate worked out in rupees per unit quantity of water.
a) True
b) False
Answer: a
Clarification: To work out the real cost of irrigation waters based on economic considerations, a rate is worked from these considerations. Such a rate, which is worked in terms of rupees per some unit quantity of water, is known as economic water rate.

4. What policy was framed in India to reduce the losses?
a) Irrigation Water Policy
b) Water Law
c) Policy of Water
d) National Water Policy
Answer: d
Clarification: In order to reduce the huge losses and subsidies, given by the government in operations of irrigation water systems and the urgent need to increase the water rates a new policy was framed in India in 1987.

5. Economic water rates are higher than the rates by the government bodies.
a) True
b) False
Answer: a
Clarification: Economic water rates are higher than the rates being charged by the government bodies in the country. For instance in Gujarat state charged a rupees 0.3 per kiloliter, but only rupees 0.18 per kiloliter is charged by State Tube well Corporation. Similar situations are being faced in states like Orissa, Andhra Pradesh etc.

6. What is the main agenda of the National Water Policy?
a) Quality Check of Irrigation Water
b) Recommendations to States
c) Maintenance of the Irrigation Structures
d) Recover Charges
Answer: d
Clarification: The main aim or agenda of the National Water Policy is that to recover the revenue rates to adequate the annual operation and maintenance charges and a part of the fixed capital. In this way, we can also convey the scarcity value of natural resources to the consumers.

7. In the main agenda of National Water Policy what charges can be exempted for time being?
a) Recovery of Annual Operation Charges
b) Recovery of Maintenance Charges
c) Recovery of Fixed Capital
d) Irrigation Water Charges
Answer: c
Clarification: For some time being or in the initial working years of the irrigation structures (say 10 years) the recovery on the fixed capital cost charges as we must first fully rationalize our revenue rates in such a way that we recover annual and maintenance expenditure completely on the irrigation structures.

8. What is the name of the committee appointed by the government on 23-10-1991?
a) Committee on Revenue Collection on Irrigation Water
b) Committee on Pricing of Irrigation Water
c) Committee on Recovery of Charges
d) Committee on Finding the Irrigation Structures
Answer: b
Clarification: With an initiative to rationalize the irrigation water rates in different states, the planning commission of India has appointed a Committee on Pricing of Irrigation Water on 23-10-1991.

9. Which state has completely avoided the irrigation revenue rates?
a) Punjab
b) Maharashtra
c) Andhra Pradesh
d) Gujarat
Answer: a
Clarification: Punjab state abolished water charges for the irrigation water supplies on February 1997, as all most all the whole population of the state is farmers. So, in order to benefit its state farmers and to politically become strong among them the government completely abolished the irrigation revenue rates or charges in the state.

10. On what factors does the economic water rate depend?
a) 4
b) 5
c) 3
d) 6
Answer: b
Clarification: The economic water rate depends upon the initial cost of the irrigation project, its effective life span, rate of interest considered on the capital, annual operation charges, and maintenance charges.

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250+ TOP MCQs on Irrigation Channels Design – Method for Design of Non-Scouring and Answers

Irrigation Engineering Multiple Choice Questions on “Irrigation Channels Design – Method for Design of Non-Scouring”.

1. Whose theory was the first to provide semi-theoretical analysis of the problem of incipient condition of bed motion?
a) Lacey’s theory
b) Kennedy’s theory
c) Shield’s theory
d) Strickler’s equation
Answer: c
Clarification: Shield was the first person who designed non-scouring channels by providing semi-theoretical analysis. He defined that the bed particles need a drag force greater than or equal to the resistance offered by the particle for its movement.

2. Which one is the correct expression for Shield’s entrainment function?
a) Tc / Yw. d. (Gs – 1)
b) Tc.Yw / d. (Gs – 1)
c) Tc / Yw. d
d) Tc .Yw. d / (Gs – 1)
Answer: a
Clarification: Shield introduced a dimensionless number which is called entrainment function (Fs) which is a function of Reynold’s number at a critical stage of bed movement in alluviums. It is based on the experimental work done by the Shield.
Mathematically, Fs = Tc / Yw. d. (Gs – 1).

3. For the design of non-scouring channels in coarse alluviums, the shield’s entrainment function should be ____________
a) Fs > 0.056
b) Fs < 0.056
c) Fs = 0.056
d) Fs = 0
Answer: c
Clarification: According to Shield for coarse alluvium, Fs = 0.056 (d > 6 mm). When the particle Reynold’s number is more than 400, the application of the curve plotted by Shield between Reynold’s number and entrainment function becomes simpler. Also, the value of the entrainment function becomes constant and equal to 0.056.

4. Strickler’s formula is only applicable to the flexible boundary channels.
a) True
b) False
Answer: b
Clarification: 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.

5. 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
Clarification: 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.

6. 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
Clarification: 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.

7. 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
Clarification: The strickler’s formula is n = d1/6/24 where d = 0.05 m.
n = 0.051/6/24 = 0.025 m.

8. 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
Clarification: 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%.

9. 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
Clarification: 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.

10. 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
Clarification: 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.

11. 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
Clarification: 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.

12. 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
Clarification: 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.

250+ TOP MCQs on Water Requirements of Crops – Consumptive Use of Evapotranspiration and Answers

Irrigation Engineering online test on “Water Requirements of Crops – Consumptive Use of Evapotranspiration”.

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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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.

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250+ TOP MCQs on Arch Dams – Types and Answers

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

1. An arch dam looks like a single arch in _________________
a) plan
b) front elevation
c) side elevation
d) both plan and front elevation
Answer: a
Clarification: A solid wall curved in plan standing across the entire width of the river valley in a single span is an arch dam. The dam body is usually made of cement concrete. When multiple or number of arches are used supported between intermediate piers, the dam is known as buttress dam.

2. A non-vertical arch dam is known as a _______________
a) buttress dam
b) double curvature arch dam
c) shell arch dam
d) both double curvature and shell arch dams
Answer: d
Clarification: The economy in dam thickness is further increased by making the dam body not only curved in the plan but also in section. This non-vertical dam is known as the double curvature arch dam or shell-arch dam. These dams are designed as shell structures which are quite complex.

3. Greater is the wall curvature in an arch dam, greater will be the economy in the dam thickness.
a) True
b) False
Answer: a
Clarification: The curved wall will structurally behave as a cantilever retaining wall standing up from its base and partly the load will be transferred to the two ends of the arch span. The load on the cantilever wall reduces as it is distributed to the side walls which in turn reduces the thickness.

4. Which of the following is the most economical type of arch dam?
a) Constant radius type
b) Variable radius type
c) Constant angle type
d) Variable angle type
Answer: c
Clarification: A special type of variable radius arch dam in which the central angles of the horizontal arch rings are of the same magnitude at all elevations is called a constant angle arch dam. Such a dam proves to be economical as the design can be made by adopting the best central angle of 133°-34′.

5. A constant angle arch dam when compared to constant radius arch dam utilizes concrete quality of about ______________
a) 43%
b) 130%
c) 230%
d) 113%
Answer: a
Clarification: The simplest in design as well as construction is a constant radius arch dam but uses the maximum concrete. An intermediate choice is a variable radius arch dam using around 58% of the concrete used by constant radius arch dam. A constant angle arch dam uses about 43% of the concrete used by a constant radius arch dam.

6. Which among the following type of dam section is expected to be the thinnest and the most economical?
a) Constant angle arch dam
b) Shell-arch dam
c) Constant radius arch dam
d) Concrete gravity dam
Answer: b
Clarification: Shell arch dams are much more economical than the constant angle arch dams as their sections are quite thin. The Idduki dam in India is only 45 m thick at its base even with 170.7 m height whereas the famous Hoover dam of the USA which is a constant radius arch dam is 201m thick at its base with only 222 m in height.

7. The most economical central angle of the arch rings of an arch dam can be adopted only at one place preferably at mid-height in an arch dam is of the type __________________
a) constant angle arch dam
b) constant radius arch dam
c) both constant angle and constant radius arch dam
d) variable radius arch dam
Answer: b
Clarification: The most economical central angle is equal to 1330-34’ and such an angle can be adopted only at one place in constant radius arch dam since as there is considerable variation with height due to narrow V-shape of the valley. It is therefore considered to keep the economical angle of 1330-34’ at about mid-height and the angle at the top will be more than this value due to topographical conditions.

8. In an arch dam, the extrados curve refer to the arch rings corresponding to the ____________________
a) the upstream face of the dam
b) downstream face of the dam
c) side face of the dam
d) either upstream or downstream face of the dam
Answer: a
Clarification: The arch rings corresponding to the upstream face is called the extrados curve. The arch rings corresponding to the downstream face are the intrados curve.

9. A V-shaped valley with stronger foundations can suggest the choice of an arch dam is of the type ___________________________
a) constant radius arch dam
b) variable radius arch dam
c) constant angle arch dam
d) variable angle arch dam
Answer: c
Clarification: The variable center arch dam is also preferred for V-shaped valleys as compared to constant radius arch dams. The constant radius arch dam may be preferred for comparatively wider U-shaped valleys.

10. A constant radius arch dam is also sometimes called as _______________
a) constant centre arch dam
b) constant angle arch dam
c) variable angle arch dam
d) variable centre arch dam
Answer: a
Clarification: In a constant radius arch dam, the centres of extrados, intrados as well as the centrelines of the horizontal arch rings lie on a straight vertical line at various elevations that passes through the centre of the horizontal arch ring at the crest. The centre of the arch rings is not at one point but lies along on vertical line at different heights where in variable centre arch dam the centres do not lie on the same vertical line.

11. The type of arch dam which generally requires overhangs at abutments is of ________________________
a) constant radius type
b) variable radius type
c) constant angle type
d) variable angle type
Answer: c
Clarification: The design of a constant angle dam usually involves providing overhangs at abutments which requires stronger foundations. Hence, they are not used if the foundations are weak. Such a dam proves to be most economical out of all other dams.

250+ TOP MCQs on Chute Spillway and Answers

Irrigation Engineering Multiple Choice Questions on “Chute Spillway”.

1. Which of the following is the simplest type of spillway which can be provided independently and at low costs?
a) Ogee spillway
b) Trough spillway
c) Siphon spillway
d) Saddle spillway
Answer: b
Clarification: The trough or chute spillway is adaptable to any type of foundations and is the simplest type of spillway. It can be easily provided independently at low costs but it requires ample room adjacent to the dam.

2. If the spillway is constructed in continuation to the dam at one end, it may be called as ________
a) saddle weir
b) flank weir
c) waste weir
d) temporary weir
Answer: b
Clarification: Saddle weir is the one when the spillway is constructed in a natural saddle in a bank of the river separated from the main dam by a high ridge. Flank weir is the one when the spillway is constructed in continuation to the dam at one end.

3. The spillway which can be adopted with ease on gravity as well as earthen dams is _________________
a) ogee spillway
b) chute spillway
c) both ogee and chute spillway
d) straight drop spillway
Answer: b
Clarification: Straight drop spillway is a low weir and simple vertical fall type structure which may be constructed on thin arch dams or small bunds etc. An ogee spillway is most suitable for concrete gravity dams. Chute spillway can be provided easily on earth and rock-fill dams.

4. The famous Bhakra dam of India is provided with ________________
a) trough spillway
b) ogee spillway
c) shaft spillway
d) siphon spillway
Answer: a
Clarification: Bhakra Dam is a concrete gravity dam on the Sutlej River in Himachal Pradesh of the total height of 226 m approximately. It forms the Gobindsagar Reservoir with controlled overflow chute or trough spillway and four spillway gates for an emergency.

5. The surplus reservoir water after spilling over the crest of the spillway flows on the chute is __________________
a) parallel to the crest in a trough spillway
b) parallel to the crest in a side-channel spillway
c) perpendicular to the crest in a side-channel spillway
d) obliquely to the crest in a chute spillway
Answer: b
Clarification: In chute spillway, after crossing over the crest of the spillway water flows at right angle shoots down a channel or a trough to meet the river channel downstream of the dam. The water after spilling over the crest flows parallel to the crest in a side-channel spillway.

6. Whenever the slope of chute changes from steeper to milder ____________________ curve shall be provided.
a) a concave vertical curve
b) a convex vertical curve
c) a hyperbolic curve
d) a parabolic curve
Answer: a
Clarification: A concave vertical curve is provided whenever the slope changes from steeper to milder. The radius of this curve should be less than 10 times the depth of water (in meters).

7. The side slope of approach channel of chute spillway is ___________________
a) 1:1
b) 1:1 (frac{1}{2})
c) 1:3
d) 1:2.5
Answer: a
Clarification: An entrance channel also called as an approach channel is trapezoidal in shape with side slopes 1:1. It is constructed so as to lead the reservoir water up to the control structure or low ogee weir.

8. The spillway which may sometimes be called as a waste weir is ______________________
a) an ogee spillway
b) a trough spillway
c) a shaft spillway
d) a siphon spillway
Answer: b
Clarification: A chute or trough spillway is sometimes called as a waste weir. It is called as chute spillway because after crossing over the crest of the spillway water flow shoots down a channel or trough to meet the river channel downstream of the dam. The trough is taken straight from the crest to the river.

9. The portion of a chute spillway which is known as its control structure is ________________
a) low ogee weir
b) chute channel
c) approach channel leading the water from the reservoir to the ogee weir
d) silting basin at its bottom
Answer: a
Clarification: Low ogee weir is required as a control structure in the chute spillway. The entire chute spillway is divided into the entrance channel, control structure, Chute channel or discharge channel, and energy dissipation arrangements in the form of silting basin at the bottom.

10. The type of spillway which is provided on narrow valleys where no side flanks are available is ____________________
a) ogee spillway
b) chute spillway
c) side-channel spillway
d) straight drop spillway
Answer: c
Clarification: The side channel spillway is suitable when the valley is too narrow. When there is no room for provision of chute spillway (as side flanks of sufficient width are required), this type is adopted as it requires limited space. The situation required for the chute spillway and side-channel spillway is mostly the same.

11. Which of the following curve is provided when the slope of the chute changes from milder to steeper?
a) Concave vertical curve
b) Hyperbolic curve
c) Convex vertical curve
d) Straight plan
Answer: c
Clarification: A convex vertical curve shall have to be provided when the slope of the chute changes from milder to steeper. The curvature should approximate to a parabolic shape. A concave vertical curve is provided whenever the slope changes from steeper to milder.

12. Calculate the freeboard for the top levels of the side walls if the mean velocity of water in the chute reach is 3.5 m/s and the mean depth of water in the chute reach under consideration is 4.7 m.
a) 0.60 m
b) 1 m
c) 0.85 m
d) 0.55 m
Answer: c
Clarification: The freeboard to be provided above the top nappe of side walls is generally given by the equation:
F.B = 0.61 + 0.04 Vm. dm1/3 where ‘Vm‘ is the mean velocity of water in the chute reach and ‘dm’ is the mean depth of water in the chute reach under consideration.
F.B = 0.61 + 0.04 x 3.5 x 4.71/3 = 0.84 m.

13. The minimum slope of the chute is governed by the condition that _____________
a) the subcritical flow must be maintained
b) the supercritical flow must be maintained
c) a critical flow must be maintained
d) maybe supercritical or subcritical flow is possible
Answer: b
Clarification: The water flows through the chute channel after spilling over the control structure and the minimum slope is governed by the condition that the supercritical flow must be maintained. The slope should be sufficient to meet the flow requirement from the crest without endangering the stability or heavy excavations.

14. What is the approximate percentage of light reinforcement that is provided in the top of the reinforced concrete slabs at the base?
a) 0.50% of the concrete area
b) 0.45% of the concrete area
c) 0.30% of the concrete area
d) 0.25% of the concrete area
Answer: d
Clarification: The base of the channel is usually made of reinforced concrete slabs 25 to 50 cm thick and light reinforcement of about 0.25% of the concrete area is also provided in both the directions. The chute is sometimes narrowed for the economy and then widened near the end to reduce the discharging velocity.

250+ TOP MCQs on Reservoir Losses and Answers

Irrigation Engineering Multiple Choice Questions on “Reservoir Losses”.

1. As the height of a proposed dam is increased, the cost per unit of storage initially increases and then decreases.
a) True
b) False
Answer: b
Clarification: The cost per unit storage of the dam initially decreases reaches to a certain minimum value and then increases when the height of the dam is increased. This lowest point on the curve gives the economical height for which the cost per unit storage is minimum.

2. The economical height of a dam is that height for which the ____________________
a) cost per unit of storage is minimum
b) benefit-cost ratio is maximum
c) net benefits are maximum
d) benefit-cost ratio is minimum
Answer: a
Clarification: The height of the dam corresponding to which the cost of the dam per unit storage is minimum is the economical height of the dam. The cost per unit storage of the dam initially decreases reaches to a certain minimum value and then increases when the height of the dam is increased.

3. If 20% of the reservoir capacity is earmarked for dead storage in a storage reservoir of 30 M.cum and the average annual silt deposition in the reservoir is 0.1 M cum, then the useful life of the reservoir will start reducing after ____________
a) 60 years
b) 120 years
c) 240 years
d) 300 years
Answer: a
Clarification: Given, Reservoir capacity = 30 M.cum
Dead storage is 20% of the reservoir capacity = 0.2 x 30 = 6 M.cum
Time during which 6 M cum is filled by sediment @0.1 M.cum/yr = 6/0.1 = 60 years.

4. According to the Central Water Commission, the average annual evaporation loss is estimated to be _____________
a) 100 cm
b) 125 cm
c) 225 cm
d) 250 cm
Answer: c
Clarification: The Central Water Commission in 1990 has estimated the average annual evaporation loss to be 225 cm on the basis of a review conducted on 130 sample reservoirs. The estimated total water loss from all the existing reservoirs to be 27000 Mm3 per annum.

5. Windbreakers, Integrated operation of reservoirs and Use of underground storage are all methods of ______________________
a) evaporation control
b) estimation of absorption losses
c) estimation of absorption losses
d) estimation of reservoir leakage
Answer: a
Clarification: The following are some of the methods of evaporation control such as Windbreakers, Reduction in the exposed water surface, Use of underground storage rather than surface storage, Integrated operation of reservoirs, and Use of chemical for retarding the evaporation rate from the reservoir surface. Use of Water Evapo-Retardants (WERs) is considered to be an only practical solution for the conservation of freshwater in spite of its limitations and disadvantages.

6. Which of the following is not used as WERs?
a) Hexa decanol
b) Octa decanol
c) Docosanol
d) Cyclohexanol
Answer: d
Clarification: Fatty alcohols like Cetyl alcohol (popularly known as Hexa decanol), Stearyl alcohol (also called Octadecanol), and Behenyl alcohol (also called docosanol) are found to be quite suitable. A mixture of these chemicals have also been generally used and the chemicals should be 99% pure for getting the desired properties of the monolayer.

7. Which of the following dam is situated in the seismologically inactive zone in Peninsular India?
a) Koyna dam
b) Hirakud dam
c) Tehri dam
d) Idukki dam
Answer: a
Clarification: A very strange phenomenon has been observed in several dams of the world, the reservoir basins which were seismologically inactive started showing seismic activities when the reservoir was filled up with water. Koyna dam (Maharashtra) is an important example of such a case.

8. Which of the following measure does not prevent or reduce the intensity of the earthquake?
a) Filling the reservoir up to a limited safe level
b) Draining out water from weaker adjoining rocks
c) Active exploration of the dam site for the absence of inactive faults before selecting the same
d) Increasing the pore pressure
Answer: d
Clarification: Earthquakes occur due to increased pore-pressure in the adjoining rocks which lowers their shearing strength resulting in the release of tectonic strain. Draining out water from weaker adjoining rocks actually reduces pore pressure.

9. Which of the following guideline is incorrect for the selection of a suitable site for a reservoir?
a) The geological formation should be such that it entails minimum leakage
b) The cost of the dam is a controlling factor
c) Too much silt-laden tributaries should be avoided as far as possible
d) The reservoir basin should have a shallow wide opening in the valley
Answer: d
Clarification: The site should be such that a deep reservoir is formed because of lower land cost per unit of capacity, less evaporation loss, and less possibility of weed growth as compared to shallow ones. The basin should have a narrow opening so that the length of the dam is minimum.