Category: Irrigation Engineering Objective Questions

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 Channel Cross-sections and Answers

Irrigation Engineering Multiple Choice Questions on “Channel Cross-sections”.

1. 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
Clarification: Generally, Triangular and Trapezoidal channel sections are adopted. For smaller discharges, the Triangular channel section is adopted and trapezoidal channel for larger discharges.

2. 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
Clarification: 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.

3. 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
Clarification: 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.

4. 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
Clarification: 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.

5. 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
Clarification: 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.

6. 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
Clarification: 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.

7. 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
Clarification: 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.

8. 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
Clarification: 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.

250+ TOP MCQs on Canal Irrigation System – Alignment and Answers

Irrigation Engineering Multiple Choice Questions on “Canal Irrigation System – Alignment”.

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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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. What type of canal alignment does the diagram represent?

a) Alignment of Side Slope Canal
b) Alignment of Contour Canal
c) Alignment of Watershed Canal
d) Alignment of Field Channel
Answer: c
Clarification: The image indicates an alignment of canal perpendicular to the ground contours. So, therefore the fitting example of this type of alignment is alignment of side slope canal.

9. Which type of alignment does the given figure represent?

a) Alignment of Side Slope Canal
b) Alignment of Contour Canal
c) Alignment of Watershed Canal
d) Alignment of Field Channel
Answer: b
Clarification: The figure shows a river flowing in a hill area, and we can observe that a canal is aligned along the contours of the area. The canal is connected to the river by head works. So, from these points we can conclude or say that it an alignment of contour canal.

10. Which type of canal is the farmer’s responsibility?
a) Contour Canal
b) Side Slope Canal
c) Watershed Canal
d) Field Channel
Answer: d
Clarification: 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.

11. Which type of canal need syphons?
a) Side Slope Canal
b) Contour Canal
c) Watershed Canal
d) Field Channel
Answer: c
Clarification: 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.

12. 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
Clarification: 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.

250+ TOP MCQs on Pressure Conduites and Answers

Irrigation Engineering Multiple Choice Questions on “Pressure Conduites”.

1. The flow through the penstocks and pressure conduits is generally ____________________
a) laminar
b) turbulent
c) both laminar and turbulent
d) unpredictable
Answer: b
Clarification: Penstocks are the huge diameter pipe which carries water under pressure and the structural design is similar to that of pressure pipes. Since there is a possibility of sudden load changes in penstocks which changes the pressure and flow velocity which characterizes the turbulent flow in fluid dynamics.

2. The head loss in the flow of water through a penstock pipe of given length ____________
a) increases with the increase in flow velocity
b) decreases with the increase in flow velocity
c) decreases with the increase in the roughness of the pipe surface
d) decreases with time
Answer: a
Clarification: The head loss (HL) by pipe friction is given by Darcy-Weisbach equation –
HL = f’. V2. L / 2gd where L is the length of pipe in meters, d is the diameter of the pipe, V is the velocity of the pipe and f’ is the friction factor which depends on Reynold’s number and the relative roughness of the pipe.
It is clear from the given equation that the head loss is directly proportional to the length of the pipe hence the head loss increases with the increase in the length of the pipe.

3. Hoop’s reinforcement is provided in cement concrete pressure pipes in order to counteract the _________
a) water hammer pressure
b) internal hammer pressure
c) stresses caused by the external backfills
d) both water hammer pressure and internal hammer pressure
Answer: d
Clarification: Hoop’s tension is the internal pressure exerted on the walls of the pipe by flowing water and the circumferential tensile stress (T1) is given by –
T1 = p1 d/2t where p1 is the internal static pressure, d is the diameter of the pipe and t is the thickness of the pipe shell.
Similarly, the circumferential tensile stress caused by water hammer pressure is given by –
T2 = p2 d/2t where p2 is the maximum water hammer pressure developed in pipelines.

4. The more rapid the closure of the valve, greater is the water hammer pressure developed.
a) True
b) False
Answer: a
Clarification: When a liquid is flowing in a pipeline is abruptly stopped by the closing of the valve, it destroys the momentum and retards the velocity of water column behind. This exerts a thrust on the valve, an additional pressure on the pipe shell and it may be so high as to cause bursting of the pipe shell. The more rapid is the closure of the valve, the more rapid is the change in momentum and hence, greater hammer pressure is developed.

5. Which of the following types of pressure conduits is preferably used for large heads?
a) PCC pipes
b) Cast iron pipes
c) Pre-stressed concrete pipes
d) Asbestos pipes
Answer: c
Clarification: Pre-stressed concrete pipes have more strength than RCC pipes and are more economical. The strength can be achieved by circumferential pre-stressing so as to increase the tensile stress. The main advantage is that they offer a cost advantage over other pipes for large diameter and higher pressures.

6. Which head loss formula is also applicable to turbulent flow in pressure conduits?
a) Darch-Weisbach equation
b) Manning’s formula
c) Hazen-Williams formula
d) Both Darch-Weisbach and Manning’s formula
Answer: b
Clarification: Manning’s formula is also applicable to the turbulent flow in pressure conduits. It yields good results provided the roughness coefficient is accurately determined. The head loss is expressed as –
HL = n2. V2. L / R4/3 where, n = manning’s rugosity coefficient, L is the length of pipe, V is the flow velocity through pipe and R is the hydraulic mean depth of the pipe.

7. __________________ joint is often used for connecting cast iron pipes.
a) Flexible joint
b) Expansion joint
c) Collar joints
d) Bell and spigot joint
Answer: d
Clarification: The pipes which are to be joined are made in such a way that one end is enlarged (also called socket or bell) and the other end is normal (i.e. spigot). The spigot end is inserted into the bell and the remaining space is filled with molten lead which gets solidified and thus making a water-tight joint.

8. Which of the following conduits is used when large diameter pipes of smaller wall thickness are required?
a) Cast iron conduits
b) Galvanized steel pipes
c) Hume steel conduits
d) Centrifugal type RCC pipe
Answer: b
Clarification: Galvanised steel pipes (with circumferential corrugations) are much stronger than ordinary steel pipes. These are usually manufactured in various sizes varying from 20 cm to 2 meters in diameter. They are lighter and can be more easily transported at distances and are widely used where large pipes of smaller wall thickness are required.

9. A special flexible joint called simplex joint is generally used to join _______________
a) vitrified clay conduits
b) galvanised iron pipes
c) asbestos conduits
d) hume steel conduits
Answer: c
Clarification: Asbestos conduits are highly flexible and may permit as much as 120 deflections in laying them around curves. Expansion joints are not required as the coefficient of expansion is low and the joints are also flexible i.e. simplex joint. Its assembly consists of pipe sleeve and two rubber rings which are compressed between the pipe and the interior of the sleeve.

10. Which of the following statement is incorrect?
a) Steel conduits are lighter than cast iron pipes
b) RCC pipes are generally made from 1:2:4 cement concrete with a maximum size of aggregates as 6 mm
c) RCC pipes and Hume steel conduits are heavy and difficult to handle
d) Riveted pipes are smoother and stronger than welded pipes
Answer: d
Clarification: Steel is strong in tension and even large size diameter pipes can be made of thin shells. They are therefore lighter than cast iron pipes. Welded pipes are smoother and stronger than riveted pipes. But steel pipes get rusted easily and are protected on the inside as well as outside by protective coatings.

11. Which of the following pipes is used for carrying hot water in the interior of the building?
a) Wrought iron pipes
b) Copper pipes
c) Galvanised iron pipes
d) Plastic pipes
Answer: b
Clarification: Copper pipes are very costly but they are highly resistant to acidic as well as alkaline waters. They can be bent easily and do not sag due to heat. Hence, they are very useful in carrying hot waters.

12. Wrought iron pipes are heavier than cast iron pipes and cannot be easily fabricated.
a) True
b) False
Answer: b
Clarification: Wrought iron pipes can be more easily cut threaded and worked but are more costly. They are neat in appearance but are less durable and corrode quickly. These are lighter than cast iron pipes and are generally protected by galvanizing with zinc coatings.

250+ TOP MCQs on Irrigation Canal – Maintenance and Answers

Irrigation Engineering Multiple Choice Questions on “Irrigation Canal – Maintenance”.

1. The berms help in checking excessive loss of water due to seepage.
a) True
b) False

Answer: a
Clarification: Berms provide a path for inspection. Excessive loss of water due to seepage is checked. It is so because berms are formed of fine silt which makes berms fairly impervious.

2. Which of the following is not a reason which leads to a canal breach?
a) Due to faulty design and construction of banks
b) Due to leakage or piping
c) Due to the maintenance of service roads
d) Due to intentional cuts made by cultivators

Answer: c
Clarification: Breaches are the gaps created in the canal banks due to breaking up of the banks. The maintenance of service roads includes removal of grass and small bushes, levelling and surfacing of roads. The slopes if damaged should be repaired to facilitate drainage of roads.

3. Which of the following is not a preventive measure for the maintenance of the canal?
a) Strengthening of canal banks
b) Formation of berms by internal silting
c) Provision of Pushta or cover to the saturation line
d) Silt removal

Answer: d
Clarification: Preventive measures are aimed at providing additional stability at various components parts of the canal. Remedial measures are aimed at repairing or removing the causes of trouble like aquatic weed control, silt removal, and maintenance of service roads.

4. Internal silting method and external silting method are the methods of _____
a) strengthening of canal banks
b) formation of berms
c) silt removal
d) formation of breaches

Answer: a
Clarification: Internal silting method is adopted only when new canals are constructed as the section has to be sufficiently wide. In the external silting method, subsidiary banks are constructed to hold the water externally in a confined position. Both of these are methods of strengthening of canal banks.

5. Which of the following statement is wrong about the internal silting method?
a) The extra strength is gained by increasing the section of the banks
b) The additional soil is derived from the natural process of silting internally
c) This method of strengthening the canal is very cheap, simple and efficient
d) To decelerate the silting process low submersible spurs may be constructed

Answer: d
Clarification: The spurs project into the canal section from the banks. They provide high silting rate and spurs may be raised after silting has taken place up to the top of spurs. Thus silting takes place in successive layers and finally, silt deposit reaches a predetermined height which is then raised up to the top of the bank.

6. In the external silting method, subsidiary banks are constructed which run parallel to canal main banks.
a) True
b) False

Answer: a
Clarification: It is very essential to bring the flow of water out of the channel section to accomplish silting of banks externally. Subsidiary banks are required to hold the water externally in a confined position and are constructed parallel to the canal main banks.

7. In Long reach system, only a portion of full supply discharge is taken in the compartment.
a) True
b) False

Answer: b
Clarification: Canal water is taken in the compartment through the inlet. When the length of the compartment is between 1200 to 1500 m it is called long reach the system. The whole canal discharge is taken in the compartment and this is done when the water is not required for irrigation purposes.

8. Which of the following combination is incorrect?
a) Silt removal – Silt ejectors and silt escapes
b) Aquatic weed control – Mechanical removal of plants
c) Breach due to piping – provision of a sand core in the bank
d) Breaches in a big canal – Dumping huge quantity of earth

Answer: d
Clarification: This process cannot be adopted for a bigger canal. The outflow from the canal breach will be large and all the soil dumped for closing the breach may be washed off. For bigger canals, the outflow is reduced first followed by cutting the sides of the gap and then driving the double pile in the opening of the breach.

9. Berm is _____________
a) a horizontal strip at the N.S.L
b) a device to remove silt
c) another name of the canal bank
d) a gap created in the bank of the canal

Answer: a
Clarification: It is a narrow horizontal strip of natural ground left between the canal section and inner toe of the bank. This is provided in such a way that the bed line and the bank line remain parallel.

10. The Remodelling of canal system is done to __________
a) beautify the canal surroundings
b) correct deformations in the canal
c) exhibit the canal in exhibition
d) widen up the channel section

Answer: b
Clarification: Remodelling of canal aims at correcting the distribution system as per existing demand. It includes silt clearing operations, remodelling of outlets, the rectification of channels, modification of faulty hydraulic structures etc.