Category: Irrigation Engineering Objective Questions

250+ TOP MCQs on Shaft Spillway and Answers

Irrigation Engineering Multiple Choice Questions on “Shaft Spillway”.

1. A shaft spillway is located _______________________
a) inside the body of a gravity dam
b) inside the upstream reservoir
c) inside the downstream reservoir
d) on the side flanks of the main dam
Answer: b
Clarification: The shape of the shaft spillway is just like a funnel. The lower end of the funnel is turned at right angles and then taken out below the dam horizontally. Water spills over the circular crest and then enters the vertical shaft and comes out of the dam through the horizontal tunnel.

2. The siphons installed within a gravity dam to spill the surplus reservoir water are known as ______________________
a) hooded type siphon spillway
b) tilted outlet type siphon spillway
c) both hooded and titled type siphon spillway
d) shaft spillway
Answer: b
Clarification: The siphon pipe is installed within the body of the dam and an air vent may be connected with it. The outlet is submerged to avoid the entry of the air in the siphon from the d/s end. When the air enters the siphon through the vent, siphoning action cannot take place and the flow will stop.

3. The siphons installed over overflow dams constitute what are known as __________________
a) hooded type siphon spillway
b) tilted outlet type siphon spillway
c) discharge carriers
d) baby siphons
Answer: a
Clarification: In hooded type spillway, a reinforced concrete hood is constructed over an ordinary overflow section of a gravity dam. The inlet is kept submerged to prevent the entry of debris, ice, etc. A small de-priming hood is kept above the main hood and they are connected by an air vent.

4. The only spillway among the following through which the discharge does not increase as fast as it increases in all others is _________________
a) chute spillway
b) side-channel spillway
c) ogee spillway
d) shaft spillway
Answer: d
Clarification: A shaft spillway is generally adopted when the possibility of an overflow spillway and a trough spillway has been ruled out. The discharge does not increase at such a high rate as in weir type spills. Hence, it may not prove as useful as a weir type spills if unestimated high floods occur.

5. The only spillway among the following through which the discharge is almost at its capacity rate even from the start of its functioning is ________________
a) chute spillway
b) side-channel spillway
c) ogee spillway
d) siphon spillway
Answer: d
Clarification: Since the change in the effective head is small as compared to the corresponding change in head over an ogee spillway. When once the water level has risen above the normal pool level, the discharge through the siphon is always at its capacity. This makes it advantageous in disposing of sudden surges of water.

6. An air vent is provided at FRL to break the siphoning action at that level in a ___________________
a) hooded type of siphon spillway
b) tilted outlet type of siphon spillway
c) saddle siphon spillway
d) volute siphon spillway
Answer: c
Clarification: When the water level falls to FRL an air vent is provided on the crown to break the siphoning action. Air enters the lower limb through the air vent and siphonic action is stopped.

7. Which of the following spillway is designed in India by Ganesh Iyer?
a) Saddle siphon spillway
b) Saddle spillway
c) Glory hole spillway
d) Volute Siphon spillway
Answer: d
Clarification: Volute siphon spillway consists of a vertical shaft circular in the section provided with a cap covering. It has low priming depth and higher discharge efficiency. It can be provided with an earth dam also.

8. The crest of a siphon spillway is fixed at ________________
a) full reservoir level
b) dead storage level
c) maximum water level
d) top of the dam
Answer: a
Clarification: The crest of a siphon spillway is fixed at FRL. When the water level in the reservoir rises above FRL the sheet of water starts spilling over the crest. When the waterfalls to FRL, air enters the vent and siphonic action is stopped.

9. A siphon spillway is sufficiently independent of the water surface elevation of the reservoir.
a) True
b) False
Answer: a
Clarification: The discharge through the siphon is given by the equation-
Q = Cd. A. (2gH)1/2 where Cd is the coefficient of discharge, A is the area of the cross-section and H is the effective head i.e. the difference of the water level in the reservoir and the tailwater level for a submerged outlet. The discharge through the spillway is affected to a very less extent is the water surface in the reservoir rises.

10. The allowable maximum negative head on an average is equal to __________
a) 2.5 m or so
b) 5 m or so
c) 7.5 m or so
d) 10 m or so
Answer: c
Clarification: The vertical distance from the crown of the siphon down to the discharging point or HGL should not exceed a value of 7.5 m under average conditions. This is necessary to avoid cavitation.
Limiting vaccum pressure = Atmospheric pressure – Vapour pressure = 10 – 2.5 = 7.5 m.

11. The structure which is not used in a shaft spillway is ________________
a) tunnel
b) bridge
c) radial gates
d) radial piers
Answer: c
Clarification: The radial gates consists of steel plates with struts which are pinned at the center of the circular plate. When the water level rises above FRL the gates are raised by hoisting arrangement to allow the flood water to discharge freely. Each gate is separately housed between two piers constructed on the crest.

12. Morning glory is the _____________________________
a) special flared inlet of the shaft spillway of a dam of very small height
b) special flared inlet of the shaft spillway of a large dam project
c) a horizontal tunnel constructed in a shaft spillway across the body of a gravity dam to carry surplus reservoir water to the d/s river
d) a horizontal tunnel constructed in a shaft spillway of an earthen dam through its foundation to carry surplus reservoir water to the d/s river
Answer: b
Clarification: In shaft spillway of large heights, RCC may be used and for smaller heights, no special inlet design is necessary. A flared inlet called morning glory is often used in large projects. The horizontal tunnel is either taken through the dam body or below the foundations.

13. For a saddle siphon, the maximum operative head is 4.53 m. The width and height of the throat of the siphon are 5 m and 2.25 m respectively. The coefficient of discharge is 0.90. How many units are required to pass a flood of 350 cumecs? (Assume g = 10 m/s2)
a) One
b) Two
c) Three
d) Four
Answer: d
Clarification: The discharge through the siphon is given by the equation:
Q = Cd. A. (2gH)1/2 Where Cd = 0.9, H = 4.53 m and A = 5 x 2.25 = 11.25 m2
Q = 0.9 x 11.25 x (2 x 9.81 x 4.53)1/2 = 95.45 m3/sec
No. of units of saddle siphon = 350 / 95.45 = 3.66 units. = 4 units.

14. A siphon spillway in a concrete gravity dam is in the shape of ________________
a) U-shape
b) Inverted U-shape
c) Horizontal bend through the abutment
d) Any straight line
Answer: b
Clarification: A siphon spillway is a closed conduit of the shape of an inverted U-tube with unequal legs and works on the principle of siphonic action. It consists of a siphon pipe one end of which is in u/s side and the other end discharges water on the d/s side.

250+ TOP MCQs on Gravity Dam Design – Typical Cross-section and Answers

Advanced Irrigation Engineering Questions and Answers on “Gravity Dam Design – Typical Cross-section”.

1. The vertical component of the earthquake wave which produces adverse effects on the stability of a dam when is acting in ____________________
a) upward direction
b) downward direction
c) both upward and downward direction
d) any direction
Answer: b
Clarification: When the vertical acceleration is acting downward the foundation try to move downward away from the dam body. This reduces the effective weight and the stability of the dam. It is the worst case for design.

2. The horizontal component of an earthquake wave producing instability in a dam is the one which acts __________________
a) towards the reservoir
b) towards the dam
c) away from the reservoir
d) away from the dam
Answer: a
Clarification: Hydrodynamic pressure and horizontal inertia force are caused by the horizontal acceleration of an earthquake wave acting towards the reservoir. As the foundation and dam accelerate towards the reservoir there is an increase in the water pressure which is resisted by water, the extra pressure is hydrodynamic pressure.

3. The vertical downward earthquake acceleration av = 0.1g acting on a gravity dam will ________________________
a) increase the resisting weight of the dam by 10%
b) decrease the resisting weight of the dam by 10%
c) increase the uplift by 10%
d) decrease the uplift by 10%
Answer: b
Clarification: The net effective weight of the dam is given as W [1 – Kv] and vertical acceleration av = Kv.g = 0.1.g i.e 10%
where W is the total weight of the dam and Kv is the fraction of gravity adopted for vertical acceleration.
This is considered as the worst case for design as the foundation tries to move downward away from the dam body when there is downward vertical acceleration. This reduces the effective weight and the stability of the dam.

4. A gravity dam is subjected to hydrodynamic pressure caused by __________
a) the rising waves of the reservoir when a flood wave enters into it
b) the rising waves in the reservoir due to high winds
c) the increase in water pressure momentarily caused by the horizontal earthquake acting towards the reservoir
d) the increase in water pressure momentarily caused by the horizontal earthquake acting towards the dam
Answer: c
Clarification: Hydrodynamic pressure is the extra pressure exerted when the horizontal acceleration of an earthquake wave is acting towards the reservoir. As the foundation and dam accelerate towards the reservoir there is an increase in the water pressure which is nothing but hydrostatic pressure acting at a height of 3H/4π above the base. Mathematically, this value is given by Von-Karman equation.

5. In a concrete gravity dam with a vertical upstream face the stabilising force is provided by the ___________________
a) weight of the dam
b) the water supported against the upstream slope
c) uplift pressure
d) water pressure at the tail end
Answer: a
Clarification: The major resisting force is the weight of the dam body and its foundation. A unit length of the dam is considered in the 2D analysis of a gravity dam. The resultant of all the downward forces will represent the total weight of the dam acting on it.

6. What is the value of horizontal destabilizing force caused by the formation of waves in a storage reservoir having a fetch of 52 km due to high wind of 172 km/h?
a) 30 KN
b) 60 KN
c) 130 KN
d) 180 KN
Answer: d
Clarification: When F > 32 km, the wave height is given by hw = 0.032 (V.F)1/2
hw = 0.032 (172 x 52)1/2 = 3.02 m
The force caused by waves is given by the equation – Pw = 19.62 hw2 KN/m run of the dam
Pw = 19.62 x 3.022 = 179.69 KN.

7. Calculate the value of free-board that should be provided for a reservoir having a wind velocity of 92 km/h and it extends up to 18 km upstream.
a) 1.2 m
b) 1.6 m
c) 2.25 m
d) 2.5 m
Answer: c
Clarification: When F < 32 km, the wave height is given by – Hw = 0.032(V.F)1/2 + 0.763 – 0.271 F1/4
Hw = 0.032(92 x 18)1/2 + 0.763 – 0.271 (18)1/4 = 1.5 m
Free board is generally provided equal to 1.5 Hw = 1.5 x 1.5 = 2.25 m.

8. The upward acceleration of dam due to seismic activity will ________________________
a) increase the base pressure
b) decrease the base pressure
c) not affect the effective weight of the dam
d) increase the horizontal dynamic force
Answer: a
Clarification: Vertical acceleration can be resolved in two parts i.e. upward vertical accelerations and downward vertical accelerations. The effective weight of the dam increases when there is an upward vertical acceleration as it brings the foundation closer to the dam. In downward acceleration, the effective weight reduces and is considered as the worst design case.

9. What is the average value of acceleration that is sufficient for high dams in seismic zones?
a) 0.1g to 0.15g
b) 0.05g to 0.1g
c) 0.3gs
d) 0.15g (where g is the acceleration due to gravity)
Answer: a
Clarification: An average value of acceleration equal to 0.1 to 0.15 g is sufficient for high dams in seismic zones and for areas not subjected to extreme earthquake ah = 0.1 g and av = 0.05 g may be used. These forces are neglected in areas of no earthquake or very less earthquake.

10. What is Von Karman’s formula for hydrodynamic force (Pe)?
a) Pe = 0.726 pe H
b) Pe = 0.424 pe H
c) Pe = 0.555 pe H
d) Pe = 0.555.Kh. Yw H2
Answer: d
Clarification: According to Von-Karman, the hydrodynamic force is given by –
Pe = 0.555. Kh. ϒw H2 where, ϒw is the unit weight of water, H is the depth of water and Kh is a fraction of gravity adapted for horizontal acceleration which acts at the height of 4H/3π above the base. In addition to this, an inertia force is also produced by the horizontal acceleration into the body of the dam.

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250+ TOP MCQs on Ground Water through Wells and Tubewells – Infiltration and Answers

Irrigation Engineering Multiple Choice Questions on “Ground Water through Wells and Tubewells – Infiltration”.

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

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

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

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

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

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

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

8. What is the correct expression of Horton’s equation?
a) ft = fC + (fC – fO) e-K.t
b) ft = fC + (fC – fO) eK.t
c) ft = fC + (fO – fC) eK.t
d) ft = fC + (fO – fC) e-K.t
Answer: d
Clarification: Horton’s equation assumes an infinite water supply at the surface i.e. it assumes the saturation conditions at the soil surface. The infiltration rate at any time‘t’ is given by –
ft = fC + (fO – fC) e-K.t
Where fC = final constant rate of infiltration at saturation, fO = initial rate of infiltration capacity, k is the decay constant depending upon soil and vegetation and t = time from the beginning of the storm.

250+ TOP MCQs on Canal Falls – Types and Answers

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

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

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

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

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

5. What type of fall does diagram represent?
irrigation-engineering-questions-answers-types-falls-q5
a) Simple Vertical Drop Fall
b) Inglis Fall
c) Straight Glacis Fall
d) Ogee Fall
Answer: a
Clarification: The diagram gives clear image that in this fall a high crested fall takes place into the water cushion below. There is no clear hydraulic jump or energy dissipation, as the velocity jet enters the deep pool of downstream. Hence it is a simple vertical drop type fall.

6. What type of fall does the diagram depict?
irrigation-engineering-questions-answers-types-falls-q6
a) Ogee Fall
b) Trapezoidal Notch Fall
c) Simple Vertical Drop Type Fall
d) Straight Glacis Fall
Answer: a
Clarification: The diagram shows that water is gradually led down the convex and concave curves. It also shows the draw down on the upstream side which results in lower depths, higher velocities and subsequent soil erosion. The curves provide smooth surfaces which results in less energy dissipation, which again causes soil erosion. Hence the diagram is ogee fall type.

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

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

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

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

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

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.