Category: Irrigation Engineering Objective Questions

250+ TOP MCQs on Design Criteria of Earth Dams and Answers

Irrigation Engineering Multiple Choice Questions on “Design Criteria of Earth Dams”.

1. 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
Clarification: 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.

2. 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
Clarification: 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.

3. 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
Clarification: 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.

4. 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
Clarification: 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.

5. When the height of the dam is in 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
Clarification: 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.

6. 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
Clarification: 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.

7. 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
Clarification: 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.

8. 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
Clarification: The provision of berms serves the following purposes –

  1. It behaves like a good lining for reducing losses and leakage.
  2. They provide protection against erosion and breaches due to wave action.
  3. They help the channel to attain regime conditions as they help in providing a wider waterway.
  4. It can be used as borrow pits for excavating soil to be used for filling.

9. 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
Clarification: 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.

10. The upstream face of the earth dam is considered as _____________________
a) equipotential line
b) streamline
c) streak line
d) path line
Answer: a
Clarification: 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.

10. The effects of capillary fringe are on the slightly safer side and are neglected.
a) True
b) False
Answer: a
Clarification: 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.

11. 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
Clarification: 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.

250+ TOP MCQs on Categories of Dams and Answers

Irrigation Engineering Multiple Choice Questions on “Categories of Dams”.

1. Aswan dam on Nile River in Egypt was perhaps the first modern dam of the world. Its construction was completed in the year _________
a) 1664
b) 1798
c) 1902
d) 1931
Answer: c
Clarification: The first modern dam of the world was completed in 1902 and is constructed on the Nile River in Egypt at Aswan. It was designed primarily to control the flooding of the Nile River, to promote irrigation and for navigation along the river.

2. Which of the following is the only modern dam that has failed?
a) Koyna dam in India
b) Vega de Tera dam in Spain
c) Mauvoision dam in Switzerland
d) Trinity dam in California, USA
Answer: b
Clarification: Vega de Tera Dam in Spain was wrecked by heavy rains resulting in the failure of the structure due to faulty design and poor construction. The koyna dam in India was on the verge of failure due to earthquake (in 1968).

3. Earthen dams are _________________
a) rigid dams
b) non-rigid dams
c) overflow dams
d) diversion dams
Answer: b
Clarification: Earthen dams, Rock-fill dam, and combined earth and rockfill dam are all non-rigid dams. Timber dam, Steel dam, Arch dam, Buttress dam, and solid gravity dam are all rigid dams.

4. The prominent modern dam which has changed its name several times is _____________
a) Nagarjuna dam
b) Hoover dam
c) Bhakra dam
d) Idukki dam
Answer: b
Clarification: The name of the Hoover dam was a political issue and it was first of all named as Boulder dam. Later it was changed to Hoover dam when Herbert Hoover became the President of America. In 1933, President Roosevelt changed this name again to Boulder dam for 14 years and then once again the name was changed to Hoover dam by Republicans which still exists today.

5. The famous Roosevelt dam in USA is of the type __________________
a) Rock-fill dam
b) Solid Masonry Gravity Dam
c) RCC arch bridge
d) A hollow masonry gravity dam
Answer: b
Clarification: The Roosevelt dam was completed in 1911 on the Salt River of Arizona (U.S.A). It was constructed with the solid blocks of concrete which were of the type known as Solid Masonry Gravity Dam.

6. A _________________ dam is generally called as a weir or barrage.
a) storage dam
b) detention dam
c) diversion dam
d) rigid dam
Answer: c
Clarification: A small dam constructed only to raise the water level on the upstream side for diverting it to the canals is called a diversion dam. Nangal Dam is an example of a diversion dam. It is actually called a weir or a barrage.

7. Multiple arch dam is an example of _______________
a) Arch dam
b) Shell-arch dams
c) Buttress dams
d) Coffer dam
Answer: c
Clarification: In buttress dams, the load is mainly transferred to the supporting buttresses. The most common types of this dam are Deck slab type and multiple arch types. The multiple arch dams are used for higher dam heights.

8. Which type of dam design gives a wider choice of materials including earth-fill and rock-fill dams?
a) Non-overflow dams
b) Overflow dams
c) Rigid dams
d) Non-rigid dams
Answer: a
Clarification: Non-overflow dams are not designed to be overtopped. Sometimes it is combined together with the overflow dam to form a composite single structure. Overflow dams are designed to pass the surplus water over their crest and are often called as spillways.

9. The highest concrete gravity dam of the world is the Grand Dixene dam in Switzerland whose height is ____________
a) 325 m
b) 317 m
c) 285 m
d) 272 m
Answer: c
Clarification: This dam is a concrete gravity dam and is the highest gravity dam in the World with a height of 285 m. The primary purpose of the dam is hydroelectric power generation.

10. Which of the following is known as a debris dam?
a) Storage dam
b) Detention dam
c) Diversion dam
d) Rigid dam
Answer: b
Clarification: The small dams which are constructed to delay and detain the flow of floodwater are called detention dams. They are called check dams also. It is sometimes constructed to trap sediments and is often called debris dams.

250+ TOP MCQs on Regulation Modules – Non Modular Outlet Types and Answers

Irrigation Engineering Multiple Choice Questions on “Regulation Modules – Non Modular Outlet Types”.

1. Which type of non-modular outlet is used in South India?
a) Open Sluice
b) Pipe Outlet
c) Gibb’s Module
d) Open Flume
Answer: b
Clarification: A non-modular may be in the form of a rectangular opening or open sluice or a simply submerged pipe. Pipe outlet is simple type of non-modular and therefore it is widely used in South India.

2. The discharge through the open sluice is given by suppressed weir formula.
a) True
b) False
Answer: a
Clarification: Using suppressed weir formula the discharge through the opening is given when discharge consists of flow over upper part of section (HL may be considered as free discharge), then the discharge is given by q1 = (2/3) x Cd1 x (sqrt{2g}) x B x HL3/2 and if the flow is through the remaining part (through submerged orifice) the discharge is given by q2 = Cd2 x (sqrt{2gH_L}) x B x d.
Total discharge is given by q = q1 + q2.

3. Why would a submerged pipe outlet be laid inclined?
a) To Provide Proper Exit Gradient
b) To Maintain Proper Hydraulic Jump
c) To Increase Silt Conductivity
d) To Increase the Discharge Capacity
Answer: c
Clarification: Generally the diameter of the pipe varies from 10 to 30 cm. They are generally laid in concrete and fixed horizontally. They are also laid sloping upwards by depressing the upstream end of the pipe so as to increase the silt conductivity.

4. How many types of submerged pipe outlets can be laid?
a) 2
b) 3
c) 4
d) 5
Answer: a
Clarification: The submerged pipe outlets can be laid either horizontally or sloping upwards. Generally, they are embedded in concrete and fixed horizontally at right angles to the direction of flow. They can also be laid sloping upwards to increase the silt conductivity.

5. What is the formula for velocity through the pipe?
a) Total loss of head = Entry loss + Frictional loss
b) Total loss of head = Frictional loss + Velocity head at exit
c) Total loss of head = Entry loss + Velocity head at exit
d) Total loss of head = Entry loss + Frictional loss + Velocity head at exit
Answer: d
Clarification: If the difference in the water level of the distributary and the water course is known then by using the formula total loss of head = entry loss + frictional loss + velocity head at the exit. Mathematically it is represented as HL = 0.5V2/2g (1.5 + f’ l/d) where l = length of pipe, d = diameter of pipe and f’ = coefficient of friction.

6. What is the formula for discharge through the pipe?
a) Q = Cd x A
b) Q = Cd x A x (sqrt{2gH_L})
c) Q = Cd x (sqrt{2gH_L})
d) Q = A x (sqrt{2gH_L})
Answer: b
Clarification: After knowing the velocity through the pipe, we can calculate discharge using the relation discharge = velocity x area. Mathematically q = V x A, therefore q = Cd x A x (sqrt{2gH_L}) where q = discharge, Cd = coefficient of friction, A = area of pipe, HL = difference of head between FSL of distributary and FSL of water course.

7. To ensure equitable distribution of water flexible and rigid modules are preferred.
a) False
b) True
Answer: b
Clarification: Flexible and rigid modules are preferred to non-modular outlets because to ensure the equitable distribution of water in the water courses irrespective of their being at high or low levels.

8. What is the range of modular outlet called?
a) Minimum Modular Head
b) Working Range
c) Minimum Loss of Head
d) Modular Ratio
Answer: b
Clarification: Outlets are termed as modular outlets only when they work in certain range or say certain limits of water level in the distributary channel and the water course. The range over which each module acts as a modular outlet is called a working range or range of modularity.

9. To ensure the modularity of the modular outlet what should be maintained?
a) Minimum Modular Head
b) Efficiency of an Outlet
c) Drawing Ratio
d) Modular Range
Answer: a
Clarification: A minimum difference of water levels on both sides of each module should always be present to ensure its modularity. This minimum difference of levels is known as minimum modular head or minimum loss of head of the modular outlet.

10. What is the value of HL?
a) Water Level of Water course
b) Water Level of Distributary
c) The Water Level of Distributary – Water Level of Water Course
d) Discharge/Area
Answer: c
Clarification: HL is defined as the difference in the water levels between distributary channel and the water course. HL in design is taken as equal to the average of head loss values observed once a fortnight throughout the season. Mathematically it is defined as (H – d).

250+ TOP MCQs on Diversion Head Works and its Components and Answers

Irrigation Engineering Multiple Choice Questions on “Diversion Head Works and its Components”.

1. Into how many components the diversion headwork is divided?
a) 8
b) 5
c) 4
d) 7
Answer: a
Clarification: The diversion head work is generally divided into eight component parts, namely weir, divide wall, fish ladder, pocket or approach ladder, scouring sluices, silt prevention devices, canal head regulator, and river training works.

2. In order to find the proper location for the head works on the river, the river is divided into how many stages?
a) 5
b) 2
c) 3
d) 4
Answer: d
Clarification: In order to find an appropriate location for the head work on the river, the river is divided into four stages. They are mountainous stage, boulder stage, alluvial plain, and delta stage.

3. What does the diagram represent?
irrigation-engineering-questions-answers-diversion-head-works-components-q3
a) Divide Wall
b) Plan of Fish Ladder
c) Scouring Sluices
d) Canal Head Regulator
Answer: b
Clarification: The figure represents the plan of a fish ladder installed in the channel, where the water comes to the channel from end and leaves it at the other end. Baffle walls are also provided in the path to control the silt and sediment load and also the velocity of the flow, thus helping the fish.

4. Divide wall helps in concentrating scouring action.
a) True
b) False
Answer: a
Clarification: If perhaps divide wall is not provided then the currents approach the scouring sluices from all directions and their effectiveness is reduced. Thus the dividing wall helps in concentrating the scouring action of the under sluices from washing out the silt deposited in the pocket.

5. Head regulator helps in controlling the flow in the canal.
a) False
b) True
Answer: b
Clarification: As the name itself suggests (regulator) head regulator helps in regulating the supply of flow easy in a canal, controls silt entry into the canal, and shut river floods.

6. By constructing which structure we can help the fish in their migration?
a) Scouring Sluices
b) Silt Excluder
c) Fish Ladder
d) Divide Wall
Answer: c
Clarification: Before the start of monsoons the fish migrate to the upstream in search of warm water. So, therefore some provision is made to make some space available for them to travel. And this achieved by the construction of fish ladder as they provide the room for movement and also slows down the flow for easy traveling of fish.

7. What device is placed in front of head regulator for silt removal?
a) Weir
b) Silt Extractor
c) Silt Excluder
d) Barrage
Answer: c
Clarification: Silt excluder is placed in front of the head regulator by which the silt is removed from the water even before the water enters the canal. The fundamental principle on which this device acts is the fact that stream carrying silt in suspension, the concentration of silt charge is more in upper layers than in lower layers. Therefore this device is so designed for separating these two layers without disturbance.

8. Which device is used for silt removal after it enters the canal?
a) Silt Excluder
b) Silt Ejector
c) Weir
d) Barrage
Answer: b
Clarification: This device can be called as a failsafe device in case the silt excluder does not work properly or the silt charge is beyond the capacity of the silt excluder. This device removes or ejects or extracts the silt which has entered the canal and is thrown out. This device placement is a curative measure and is constructed at some distance from the head regulator.

9. Depending on how many considerations the capacity of under sluices is fixed?
a) 5
b) 4
c) 2
d) 3
Answer: d
Clarification: The discharging capacity of an under sluice is fixed by the considerations like to ensure proper scouring and its capacity should be double the canal discharge, sluices should of sufficient capacity to discharge winter freshlet, and during floods 10 to 15 percent of maximum flood discharge should be done.

10. How many river training works are needed on the canal head works?
a) 5
b) 3
c) 4
d) 2
Answer: b
Clarification: Three river training works are needed on the canal head works, to prevent the river from outflanking the works due to a change in its course and ensure smooth and an axial flow of water. The works include guide banks, marginal bunds, and spurs. Guide banks force the river into the restricted channel, thus ensuring smooth and axial flow near the weir site. In order to protect the area from submergence due to raise in HFL these marginal bunds are provided. The spurs are the works that protect the marginal bunds.

250+ TOP MCQs on Hard Surface or Rigid Linings and Answers

Irrigation Engineering Multiple Choice Questions on “Hard Surface or Rigid Linings”.

1. Which type of lining is applied under pressure?
a) Shotcrete Lining
b) Boulder Lining
c) Asphaltic Concrete Lining
d) Brick Lining
Answer: a
Clarification: The name given to this lining is to designate that the mortar is applied under pressure with the help of a nozzle on the surface of the canal.

2. Which type of lining is popular in India?
a) Boulder Lining
b) Brick Lining
c) Cement Concrete Lining
d) Shotcrete Lining
Answer: b
Clarification: Brick Lining is very popular in India due to bricks can be laid by ordinary masons, rigid quality control is not needed, expansion joints are not needed, huge employment is provided, isolated damaged parts can be easily repaired, plastering of bricks can be done to increase the canal capacity and its life span also.

3. Which type of lining is mixed and placed under temperature?
a) Cement Concrete Lining
b) Shotcrete Lining
c) Asphaltic Concrete Lining
d) Dry Stone Lining
Answer: c
Clarification: Asphaltic concrete lining has been under evolution. It is used in very few places. This concrete is a careful mixture of asphalt and graded stone aggregate, and is mixed and applied under elevated temperature. This is a flexible lining and readily confirms to subgrade.

4. What is the biggest advantage of boulder lining?
a) Impervious Lining
b) Dry Stone Lining
c) Dhamalis
d) Pervious Lining
Answer: d
Clarification: As the boulder lining is a pervious lining, it allows free flow of water from submerged or saturated subgrade into the canal. Moreover, this lining does not need any drainage arrangements. This lining can be used in areas where water table is higher than FSL of the canal.

5. Which type of lining is only used for lining the side slopes of the canal?
a) Shotcrete Lining
b) Dry Stone Lining
c) Asphaltic Concrete Lining
d) Cement Concrete Lining
Answer: b
Clarification: This type of lining is only used for lining the side slopes of the canal by proper placing and packing of stones, either after laying a filter paper or not, depending on the soil requirement. Therefore this lining does not help in preventing the seepage losses but helps to maintain the shape of the canal, hence reducing the maintenance cost.

6. Shotcrete lining can be used highly for smaller jobs.
a) True
b) False
Answer: a
Clarification: Shotcrete lining can be used for smaller jobs, because this lining requires lighter equipment and small crew to carry out the works. It can also be used in rehabilitation of old canals and repair works. The trimming of the canal section can be avoided by using this lining in irregular canals.

7. Brick lining gives a very satisfactory service.
a) False
b) True
Answer: b
Clarification: If there is settlement of subgrade in a canal where brick lining is present, the mortar joints present between the bricks or tiles provide numerous cracks so that seepage of canal water would be insignificant. The local area can be repaired without any problem even in case of abnormal settlement.

8. Which type of cement grade is used in cement concrete lining?
a) M20
b) M25
c) M10
d) M15
Answer: d
Clarification: The mixture ratio of M15 grade is 1:2:4 (cement, sand, coarse aggregate) and this ratio is a good quality type for construction of cement concrete lining. This lining gives satisfactory service, high durability, hydraulic efficiency, weed free surface, long life, minimum maintenance cost.

9. For how many days curing of concrete lining takes place?
a) 30 days
b) 28 days
c) 27 days
d) 26 days
Answer: b
Clarification: Curing is mainly done because the concrete which is laid for the lining gains hardened properties over time. After laying the concrete lining a period of 24 to 36 hours gap is given before curing starts. Curing process is done for a period of 28 days. On the canal bed it is done by constructing 150 mm deep earthen bunds across, so that water stands regularly on the bed. On the side slopes it is done by constructing masonry weep holes.

10. In the laying of concrete, what paper is laid to make filter blanket effective?
a) Filter Layer
b) Filter Blanket
c) Tar Paper
d) Sandpaper
Answer: c
Clarification: Before laying the concrete the subgrade should be levelled dressed and moistened thoroughly so that the moisture is not taken from the concrete laid. To avoid this inverted filter blanket is laid over the subgrade before laying the concrete. And to make this blanket effective and to prevent the ingression of concrete being laid on the blanket a tar paper is placed over the filter blanket, before the placing of cement concrete.

250+ TOP MCQs on Canal Irrigation System – Curves in Channels and Answers

Irrigation Engineering Multiple Choice Questions on “Canal Irrigation System – Curves in Channels”.

1. A curve in channel results in _____________
a) silting on inside (convex side) and scouring on outside (concave side)
b) scouring on inside (convex side) and scouring on outside (concave side)
c) both silting and scouring on inside
d) both silting and scouring on outside
Answer: a
Clarification: When a channel is aligned the proposed curve should be as gentle as possible. It leads to silting on the convex side and scouring on the concave side due to the disturbance of flow caused by curves. Stone pitching is sometimes proposed to avoid erosion and scouring.

2. If the discharging capacity of the channel is less than 0.5 cumecs then, the minimum recommended value for curve radius is _________
a) 150 m
b) 100 m
c) 300 m
d) 600 m
Answer: b
Clarification: If the discharge is more, the curve should be gentler and should have more radius. When the capacity is less than 0.5 cumecs, the minimum recommended curve radius is 100m. If the discharge lies in the range 0.5-3 cumecs, the value is 150m.

3. The Gross Command Area (G.C.A) represents the geographical area of the Doab.
a) True
b) False
Answer: a
Clarification: The G.C.A includes the cultivated as well as an uncultivated area like ponds, residential areas, roads, etc. An irrigation canal lies in the area between the two drainages and can economically irrigate it. Thus, the drainages fix the boundary of the gross command of a canal system and it becomes uneconomical to use the system across the two drainages i.e boundaries.

4. What is the Annual Intensity of irrigation (AII)?
a) Gross Irrigated Area/Cultivable Command Area
b) Cultivable Command Area/Gross Irrigated Area
c) Net Irrigated Area/Cultivable Command Area
d) Cultivable Command Area/Net Irrigated Area
Answer: a
Clarification: The AII is the sum total of intensities of irrigation of all the seasons of the year. IT can also be defined as the percentage of CCA which may be irrigated annually. It is thus obtained by dividing the gross irrigated area by the CCA.

5. What is Gross Cropped Area?
a) Net irrigated area + Area irrigated more than once during the same year
b) Net cropped area + Area sown more than once during the same year
c) Net irrigated area + Net cropped area
d) Total area irrigated once a year + area irrigated more than once in that year
Answer: b
Clarification: Sometimes two crops in two seasons are grown during a particular year on the same area. Hence this area will be sown more than once during that year. When this area is added to the area which is sown only once then the value obtained is Gross Cropped Area.

6. The ratio of the actual operating period of a distributary to the crop period is called as ____________
a) capacity factor
b) time factor
c) full supply coefficient
d) nominal duty
Answer: b
Clarification: No distributary is allowed to operate on all the days during any crop season to check for the dangers of over-irrigation leading to waterlogging and salinity. This ratio of actual operating period of a distributary to the crop period is called time factor. It is useful in computing the design capacity of a distributary.

7. The number of hectares irrigable per cumec of the canal capacity at its head is known as ____________
a) nominal duty
b) duty on capacity
c) design full supply discharge
d) capacity factor
Answer: b
Clarification: Full supply coefficient is also called duty on capacity. It is the design duty at the head of the canal. It is obtained by dividing the area estimated to be irrigated during the base period with the design full supply discharge at the head of the canal.

8. The cropped area in Rabi season is usually increased.
a) True
b) False
Answer: a
Clarification: The water requirement during Rabi season reduces to about 2/3rd times the full supply, the capacity factor usually varies from 0.60 to 0.70 for rabi season. The cropped area is increased in order to improve this factor.

9. The Gross Command Area for a distributary is 8000 hectares 80% of which is culturable irrigable. The intensity of irrigation for Rabi season is 50% and the average duty at the head of the distributary is 2000 hectares/cumec. Determine the discharge required at the head of the distributary.
a) 1.33 cumec
b) 1.60 cumec
c) 1.40 cumec
d) 1.44 cumec
Answer: b
Clarification: Cultivable Command Area CCA = 8000 x 80/100 = 6400 hectares
The area to be irrigated = CCA x Intensity of irrigation = 6400 x 50/100 = 3200 hectares
The water required at the head of the distributary = 3200/2000 = 1.6 cumec.

10. The CCA of a water course is 1500 hectares. The intensity of sugarcane and wheat crops is 20% and 40% respectively. The duties for the crops at the head of the watercourse are 730 hectares/cumec and 1600 hectares/cumec respectively. Calculate the design discharge at the outlet assuming a time factor equal to 0.8.
a) 0.55 cumec
b) 1.0 cumec
c) 0.72 cumec
d) 0.92 cumec
Answer: c
Clarification: Area to be irrigated under sugarcane = 1500 x 20/100 = 300 hectares
Area to be irrigated under wheat = 1500 x 40/100 = 600 hectares
Discharge required for sugarcane = 300/730 = 0.410 cumec
Discharge required for wheat = 600/1800 = 0.333 cumec
The discharge required at the head of the water course = 0.410 + 0.333 = 0.743 cumec
The actual design discharge at the outlet = 0.743/0.8 = 0.92 cumec.

11. The transplantation of rice takes 18 days and the total depth of water required by the crop is 60 cm on the field. During this transplantation period, rain starts falling and about 10 cm of rain is being utilized to fulfill the rice demand. Find the duty of irrigation water required for rice assuming 25% losses of water in watercourses.
a) 233.28 hectares/cumec
b) 240 hectares/cumec
c) 230 hectares/cumec
d) 244.44 hectares/cumec
Answer: a
Clarification: Extra water depth required = 60 – 10 = 50 cm
Duty of irrigation water = 864 B/D = 864 x 18/50 = 311.04 hectares/cumec
Duty at the head of the watercourse = 311.04 x (1-0.25) = 233.28 hectares/cumec.