Category: Irrigation Engineering Objective Questions

250+ TOP MCQs on Canal Falls – Design of Syphon Well Drop and Answers

Irrigation Engineering Multiple Choice Questions on “Canal Falls – Design of Syphon Well Drop”.

1. ______________ fall is adopted for smaller discharges and larger drops.
a) Ogee fall
b) Sarda fall
c) Glacis fall
d) Well-type fall
Answer: d
Clarification: A Cylinder fall popularly known as Syphon or Well-Type fall is useful for affecting larger drops and for small discharges. They are commonly used where high levelled small drains do outfall into low levelled bigger drains and also as tail escapes for small canals.

2. Consider the following statements.
I. Syphon falls are commonly used where high levelled small drains do out-fall into a low levelled bigger drain.
II. The D/s well is necessary in case of falls greater than 1.8 m and for discharges greater than 0.29 cumecs.
III. The design discharge is determined based on the fact that V-notch is used.
Which of the following statement is correct?
a) I only
b) II only
c) I and II
d) I, II and III
Answer: c
Clarification: The design discharge of a notch fall is obtained by adding the discharge of rectangular and V-notch i.e. trapezoidal notch taking the value of the coefficient of the discharge as 0.75. The notches are designed by taking into consideration full supply discharge and half supply discharge w.r.t normal water depths in the channel in both cases.

3. Calculate the velocity over the notch (V1) and the velocity through the pipe (V3) if the full supply discharge is 2 cumecs. The area of flow over the trapezoidal notch is 0.58 m2. Assume the value of diameter of the pipe used as 1 m.
a) V1 = 3.44 m/sec and V3 = 2.54 m/sec
b) V1 = 2.54 m/sec and V3 = 3.44 m/sec
c) V1 = 3.44 m/sec and V3 = 3.44 m/sec
d) V1 = 0.44 m/sec and V3 = 2.04 m/sec
Answer: a
Clarification: The velocity over the notch (V1) = full supply discharge / Area over the notch and the velocity through the pipe (V3) = Full supply discharge / Area of pipe.
V1 = 2/0.58 = 3.44 m/sec
V3 = 2/0.785 = 2.54 m/sec (Area of pipe = 0.785 m2).

4. What is the height of the centre of pressure above the water level in the inlet well if the R.L of the centre of pressure is 150.83 m and the R.L of water level in inlet well is 148.30 m?
a) 2.53 m
b) 1.78 m
c) 3 m
d) 1.99 m
Answer: a
Clarification: The height of the centre of pressure (Y) above the water level in the inlet well is given by –
Y = R.L of C.P – R.L of water level in inlet well
Y = 150.83 – 148.30
Y = 2.53 m.

5. Calculate the diameter of the inlet well if the X and Y coordinate of the jet (issuing from the centre of pressure) w.r.t the water surface in the inlet well is 1.05 m and 1.88 m respectively.
a) 1.575 m
b) 2.82 m
c) 1.79 m
d) 0.83 m
Answer: a
Clarification: The diameter of the inlet well is kept as 1.5 times the value of X.
Given, X = 1.05 and Y = 1.88
Diameter = 1.5 x 1.05 = 1.575 m.

6. Which of the following loss is not considered while calculating the total head loss between the inlet well and D/s FSL?
a) Entry and exit losses
b) Loss due to friction
c) Loss due to enlargement of the section
d) Loss due to contraction of section
Answer: d
Clarification: The head loss between the inlet well and the D/s FSL is given by adding the losses due to entry, exit, loss due to sudden enlargement and loss due to friction in the pipe.
HL = 0.5 V22/2g (Entry loss) + (V2 – V3)2/2g (sudden enlargement loss) + fLV32/2gd (Friction loss) + V32/2g (Exit loss)
Where, V2 = Velocity of entry in pipe, V3 = Velocity through the pipe, L = assumed pipe length, d = diameter of the pipe, f = Darcy’s coefficient of friction = 0.012 and g = acceleration due to gravity.

7. The discharge formula used in the design of the syphon well drop is ____________
a) 8/15 Cd. (2g)1/2. tan (Q/2). H5/2
b) 2/3 Cd. (2g)1/2. L. H3/2
c) 8/15 Cd. (2g)1/2. tan (Q/2). H3/2
d) 2.22H3/2 [L + 0.4nH]
Answer: d
Clarification: The discharge passing through a trapezoidal notch is given by adding the discharge of a rectangular notch and V-notch which is given by-
Q = 2/3 Cd. (2g)1/2. L. H3/2 + 8/15 Cd. (2g)1/2. tan (Q/2). H5/2
Q = 2.22H3/2 [L + 0.4nH]
(Where n = 2 tan (Q/2) and coefficient of discharge = 0.75).

8. Which of the following statement is wrong?
a) In a trapezoidal notch fall, the top width of the notch is kept between 3/4th to full water depth above the sill of the notch
b) The syphon fall is designed to maintain a fixed supply level in the channel
c) A cylinder fall is popularly known as syphon well drop
d) Energy dissipators are provided for small discharges
Answer: d
Clarification: In syphon well drops, the downstream well is provided for discharges greater than 0.29 cumecs. A trapezoidal notch is constructed in the steining of the well, the water falls into the inlet well from where it emerges near the bottom dissipating its energy in turbulence inside the well.

250+ TOP MCQs on Indian Rivers and Their Classification and Answers

Irrigation Engineering Multiple Choice Questions on “Indian Rivers and Their Classification”.

1. In to how groups rivers in India are divided?
a) 3
b) 2
c) 4
d) 5
Answer: b
Clarification: Rivers in India are broadly classified into two types, namely Himalayan Rivers and Non-Himalayan Rivers. This classification is based on the basis of the origin of the rivers.

2. Which type of rivers can derive water from all seasons?
a) Tidal Rivers
b) Non-Himalayan Rivers
c) Himalayan Rivers
d) Deltaic Rivers
Answer: c
Clarification: These types of rivers can derive water from rains in monsoon and winter seasons, but in summer they can derive water from the melting snow or ice from the mountains as these rivers take off from the Himalayas.

3. Himalayan rivers carry huge amounts of sediment.
a) True
b) False
Answer: a
Clarification: The Himalayan Rivers carry huge amounts of sediment because they flow through Himalayan rocks which are soft and friable. The reason is that the zone in which these rivers flow is very much prone to earthquake disturbances causing landslides and increased rock sediment.

4. What is the example of Himalayan Rivers?
a) Cauvery
b) Godavari
c) Tapti
d) Ganga
Answer: d
Clarification: Ganga is the Himalayan River because it has its origin in Himalayan mountain glacier called Ganges, it flows through the alluvial plains in Northern India, and it is a perennial river.

5. The North Indian rivers rise in high floods.
a) True
b) False
Answer: a
Clarification: In months of August and July the rainfall is more than 3/4 of the average rainfall. So, due to this reason the North Indian Rivers rise in high floods.

6. What is the source of water for the Non-Himalayan Rivers?
a) Summer Season
b) Rainy Season
c) Winter Season
d) Himalayan Rivers
Answer: c
Clarification: The Non-Himalayan Rivers are not perennial rivers. They receive most part of the water from the rainy seasons. And for the rest of the year they receive water from the groundwater as base flow.

7. In which parts of India do non perennial rivers flow?
a) North India
b) Central and South India
c) Eastern India
d) West India
Answer: b
Clarification: These rivers flow in Central and Southern India, as these regions are full of plateaus, valleys which is not the case in Northern India where the land terrain is of plains. Here the rivers do not have a constant water source as in case of perennial rivers, which have Himalayan mountain ranges ice as their source.

8. Which river is the example of Non-Himalayan Rivers?
a) Chenab
b) Beas
c) Godavari
d) Jhelum
Answer: c
Clarification: Godavari is the best example of non-perennial rivers. It originates from Maharashtra (Central India) and merges into the Bay of Bengal in Andhra Pradesh (South India).

9. Which type of rivers is more stable?
a) Himalayan Rivers
b) Non-Himalayan Rivers
c) Virgin Rivers
d) Flashy Rivers
Answer: b
Clarification: The Non-Himalayan Rivers are more stable compared to Himalayan Rivers as they flow through non-alluvial rivers. They also pose no risk of a rise in high floods during rainy seasons because these rivers draw their waters from these seasons itself.

10. The flood discharge in Himalayan Rivers can be great as how times the normal winter flow?
a) 25 to 75
b) 25 to 50
c) 75 to 100
d) 50 to 100
Answer: d
Clarification: Since the Himalayan Rivers are perennial rivers, there is a risk of a rise in high floods during rainy seasons. This leads to flood discharge as great as 50 to 100 times the normal winter flow. The surface runoffs are also unpredictable.

11. According to what variations the Himalayan Rivers are very complex?
a) Velocity
b) Discharge and Sediment Load
c) Type of Bed Soil
d) Terrain of the River Basin
Answer: b
Clarification: As there is a risk of high floods in these rivers, the flood discharge and surface runoffs are unpredictable and this makes it tough to control them. Moreover, the sections which are required for the passing of these flood discharges are vastly out of proportion to the sections required for normal winter flows. So, therefore these variations in discharge and sediment load make the hydraulics of these rivers very complicated and cause them to meander.

250+ TOP MCQs on Sediment Transport – Suspended Load and its Measurement and Answers

Irrigation Engineering Questions & Answers for Exams on “Sediment Transport – Suspended Load and its Measurement”.

1. Whose equation was the first on the rate of bed load transport?
a) DU-Bois empirical formula
b) Shield’s formula
c) Meyer-Peter’s formula
d) Einstein’s formula
Answer: a
Clarification: The first equation was proposed by DU-Bois on the rate of bedload transport. He assumed that the bed load transportation rate was proportional to the excess of existing tractive force over the critical value required to initiate the movement.

2. Which of the following statement is not correct about Einstein’s theory?
a) He put forward a mathematical approach to the problem of bed load transport
b) He further postulated probability of the grain being dislodged is directly proportional to the lift force which the flow can exert on the grain
c) He introduced a factor called Einstein’s bedload function
d) He gave a curve graph which can be used to compute the bedload transport rate of a given channel
Answer: a
Clarification: He put forward a semi-theoretical approach to the problem of bed load transport. It was assumed that every particle travels a certain minimum distance before coming to rest after it is dislodged from the bed.

3. If the width of a river increases, the discharge per unit width will decrease and therefore, sediment carrying capacity will increase.
a) True
b) False
Answer: b
Clarification: The sediment carrying capacity will reduce if the width of the river increases as the discharge per unit width will decrease. The deposition of the sediment will start which will increase the bed slope.

4. Which of the following statement is not correct about sediment load phenomenon and its measurement?
a) The material is kept in suspension by the turbulence or by the generation of eddies
b) In laminar flow, the shear stress is caused due to the difference of the velocities at the top and the bottom
c) In turbulent flow, momentum transfer is not very significant
d) Due to the formation of eddies, the sediment transfer from high concentration regions to the low concentration regions takes place
Answer: c
Clarification: In turbulent flow, momentum transfer or mass exchange takes place due to the jumping of particles from higher velocity region to lower velocity region. Due to this transfer of momentum between two adjacent fluid layers, effective shear stress is caused at the interface between the layers.

5. The sediment confined along and above the bed up to a depth ‘2d’ (d being grain size) is treated as bed load.
a) True
b) False
Answer: a
Clarification: The sediment moves only on the top layers of the bed. Hence, the lower limit can be considered as equal to the thickness of the bed layer which is approximately equal to 2d and the upper limit as the total water depth. The sediment concentration at this depth can be considered equal to the concentration of bed load.

6. In a wide stream, a suspended load sample taken at a height of 0.30 m from the bed. The stream is 5.0 m deep and has a bed slope of 1/4000. The bed material can be of uniform size. Estimate the shear friction velocity.
a) 0.111 m/s
b) 0.211 m/s
c) 0.311 m/s
d) 0.711 m/s
Answer: a
Clarification: V*= Shear friction velocity = (S0/Dw)1/2 where Dw is the density of water and So (Yw.R.S) is shear stress at the bottom.
V* = [(Dw. g.R.S)/Dw]1/2 = (g.R.S)1/2
V* = (9.81 x 5 x 1/4000)1/2 = 0.111 m/s.

7. For the usual turbulent flow, calculate the critical shear stress if the mean diameter of the grain particle of bed material is 0.3 mm.
a) 0.300 N/m2
b) 0.106 N/m2
c) 0.206 N/m2
d) 0.116 N/m2
Answer: c
Clarification: For usual turbulent flow, the value of critical shear stress is given by 0.687d where d is the arithmetic average diameter of sediment in mm.
S0 = 0.687 d
S0 = 0.687 x 0.3 = 0.206 N/m2.

8. Calculate the corresponding hydraulic mean depth that would exist in the channel if the bed was unrippled. The rugosity coefficient in an unrippled channel is 0.015 and the rugosity coefficient actually observed by experiments on the rippled bed of channel is 0.020. Consider the value of hydraulic mean depth of the channel as 1.5 m.
a) 1.5 m
b) 2.5 m
c) 0.97 m
d) 0.77 m
Answer: c
Clarification: The formula that is used to take into account the effect of bed ripples is-
R’ = R (n’/n) 3/2 where, R = 1.5 m, n’ = 0.015, and n = 0.020
R’ = 1.5 x (0.015/0.020)3/2 = 0.97 m.

9. Calculate the quantity of bed load transport by using Meyer and Peter formula if the effective tractive force that causes bed load transportation is 2.5 N/m2.
a) 1.04 N/m/second
b) 1.5 N/m/second
c) 2.5 N/m/second
d) 0.476 N/m/second
Answer: a
Clarification: Meyer and Peter has suggested the following formula for calculating the quantity of bed load transport –
Gb = 0.417 x Teff where Gb is the rate of bed transport by weight in N/m/second and Teff is an effective unit tractive force.
Gb = 0.417 x 2.5 = 1.04 N/m/second.

10. Calculate the effective unit tractive force that causes bed load transportation if the unit tractive force produced by the flowing water is 3.20 N/m2. The Manning’s coefficient on an unrippled bed is 0.0108 and the actual observed value of Manning’s coefficient on ripped channels is 0.0222. Take the value of critical shear stress that is required to move the grain particle as 0.5 N/m2.
a) 1.58 N/m2
b) 0.222 N/m2
c) 1.08 N/m2
d) 0.58 N/m2
Answer: d
Clarification: The unit tractive force causing bed load to move is reduced by ripples in the ratio of
S0‘= S0 (n’/n) 3/2 = 3.2 (0.0108/0.0222)3/2 = 1.08 N/m2.
The effective unit tractive force = S0’ – Sc = 1.08 – 0.5 = 0.58 N/m2.

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250+ TOP MCQs on Water Requirements of Crops – Estimation of Consumptive Use – 1 and Answers

Irrigation Engineering Multiple Choice Questions on “Water Requirements of Crops – Estimation of Consumptive Use – 1”.

1. Which formula is extensively used for estimating seasonal water requirements?
a) Blaney-Criddle formula
b) Hargreaves pan evaporation method
c) Penman’s equation
d) Christiansen formula
Answer: a
Clarification: This formula is also used for estimating monthly consumptive use.
Blaney-Criddle formula is written as Cu = k.f where, f = p/40[1.8t + 32].
Where Cu = monthly/seasonal consumptive use, k = crop factor, p = monthly % of annual daylight hours in that period, and t = mean monthly temperature

2. Which method is time-consuming and expensive?
a) Tanks and Lysimeter
b) Vapour transfer method
c) Field plot method
d) Integration method
Answer: a
Clarification: In the Lysimeter method, the plants are grown in the tank with a sand layer at the bottom and a pan for collecting the surplus water. The consumptive use is measured by the amount of water required for the satisfactory growth of plants within tanks. Other methods are more reliable than this method.

3. Which of the following is not an empirical method of determining consumptive use?
a) Lowry Johnson method
b) Penman’s equation
c) Hargreaves method
d) Inflow-outflow method
Answer: d
Clarification: Inflow-outflow method is a direct method in which field observations are made and physical models are used for the purpose. Lowry Johnson, Penman, and Hargreaves method are all empirical methods.

4. The evapotranspiration rate is higher in light green vegetation than in dark green vegetation.
a) True
b) False
Answer: b
Clarification: The production of the rate of absorption of solar energy is high in dark green vegetation hence; its evapotranspiration rate will be higher than that of light green vegetation. The plant diseases causing yellowing of the leaves of the plants greatly reduce evapotranspiration.

5. The monthly consumptive use values for paddy are 26.69, 8.76, 14.38, 22.73, 21.29, 25.50 and 15.06 for the duration of month from June to July. What is the average daily consumptive use?
a) 7.0 mm
b) 8.0 mm
c) 7.5 mm
d) 8.5 mm
Answer: c
Clarification: The total consumptive use for paddy = 29.69 + 8.76 + 14.38 + 22.73 + 21.29 + 25.50 + 15.06 = 137.41 cm
Average daily consumptive use = 137.41 / period of growth in days
= 137.41 / (30+31+31+30+31+30) = 7.5 mm.

6. Which of the following statement is wrong about consumptive use coefficient?
a) It is different for different crops
b) It varies with the crop growth
c) It is different for the same crop at different places
d) It is independent of the crop type and its growth
Answer: d
Clarification: Consumptive use coefficient (k) = evapotranspiration (Cu or Et) / pan evaporation (Ep). It is different for different crops, varies with crop growth and is different at different crop stages for the same crop.

7. What is the range of reflection coefficient for close-grained crops?
a) 0.15-0.25
b) 0.05-0.45
c) 0.05
d) 0.45-0.90
Answer: a
Clarification: The range of reflection coefficient for different surfaces are-

Close-grained crops 0.15-0.25
Bare lands 0.05-0.45
Water surface 0.05
Snow 0.45-0.90

8. Which method is widely used in India for the computation of consumptive use?
a) Penman’s equation
b) Hargreaves – Christiansen equation
c) Blaney-Criddle equation
d) Tanks and lysimeter
Answer: a
Clarification: The value obtained from Penman’s equation is almost equal to the values obtained from the actual field observations made in pan evaporation method. The values obtained from the Blaney-Criddle equation were on the much higher side (about 30%) and the values obtained from the Hargreaves method is on the lower side (about 15%-20%). Lysimeter method is time-consuming and expensive.

9. The average value of Et/Ep for citrus crops?
a) 0.90
b) 0.7-1.10
c) 0.60
d) 0.66-1
Answer: c
Clarification: Et/Ep ratio is consumptive use coefficient (k). Group F includes citrus fruits like oranges, grapefruit, etc. the Et/Ep ratio is fairly constant throughout the year and average to a value of about 0.60.

10. The monthly consumptive use is given by _____________
a) Cu = k.p/40 [1.8t + 32]
b) Cu = k.p/40 [1.8t – 32]
c) Cu = k.p/40 [t + 32]
d) Cu = k.p/40 [t – 32]
Answer: a
Clarification: BLaney – Criddle formula is used to calculate monthly consumptive use.
Cu = k.p/40 [1.8t + 32]
Where, Cu is monthly consumptive use in cm, K = crop factor, T = mean monthly temperature in °C, and P = monthly % of annual day light hours that occur during the period.

11. Determine the pan evaporation from the following data using the Christiansen method.
Extra-terrestrial radiation in cm = 47.3cm
Coefficient of temperature(Ct) = 1.403
Coefficient for wind velocity (Cw)= 1.2
Coefficient for relative humidity (Ch) = 1
Coefficient for percent of possible sunshine(Cs) = 1.073
Coefficient of elevation(Ce) = 1.014
a) 40 cm
b) 30.44 cm
c) 39.8 cm
d) 37.98 cm
Answer: c
Clarification: Christiansen formula for pan evaporation is Ep = 0.459 R.Ct.Cw.Ch.Cs.Ce
Ep = 0.459 x 47.3 x 1.403 x 1.200 x 1.000 x 1.073 x 1.014
= 39.8 cm.

250+ TOP MCQs on Energy Dissipation Below Overflow Spillway – 1 and Answers

Irrigation Engineering Multiple Choice Questions on “Energy Dissipation Below Overflow Spillway – 1”.

1. The energy dissipation at the toe of the spillway is affected basically by the use of hydraulic jump in _______________________
a) roller bucket
b) a ski-jump bucket
c) a sloping apron below the downstream river bed
d) both roller and ski-jump bucket
Answer: c
Clarification: Most of the kinetic energy is destroyed by creating a condition suitable for the formation of a hydraulic jump. Sometimes the depth of tail-water may be more than that necessary to create the hydraulic jump. The depth of water can be reduced to create a hydraulic jump by providing a sloping apron.

2. The most ideal condition for energy dissipation in the design of spillways is the one when the tail-water rating curve coincides with the jump rating curve at all discharge.
a) True
b) False
Answer: a
Clarification: The most ideal condition for jump formation is when TWC coincides with JHC at all discharge. To ensure protection in the region of a hydraulic jump, a simple concrete apron of apron length 5 (y2 – y1) is provided.

3. When the tail-water depths in the river downstream of a spillway are quite low such that the tail-water curve at all discharges lies below the post jump depth curve, then the energy dissipation can be affected best by ___________________
a) a roller bucket
b) a ski-jump bucket
c) either roller or ski-jump bucket
d) a sloping apron
Answer: b
Clarification: Energy dissipation bucket called ski-jump bucket is used when the tail-water depth is insufficient or low at all discharge. It requires sound and rocky river bed. Water may shoot up out of the bucket and fail harmlessly into the river at some distance downstream of the bucket.

4. The device which does not help in energy dissipation at the bottom of a hydraulic structure over which water spills is ________________
a) chute block
b) dentated sill
c) morning glory
d) baffle piers
Answer: c
Clarification: 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. Chute blocks, dentated sills and baffle piers are all auxiliary devices which help in energy dissipation.

5. The formation of hydraulic jump at the foot of a spillway is one of the common methods of energy dissipation because ______________________
a) it destroys more than 90% of total energy by the turbulence produced in the jump
b) it reduces the kinetic energy by increasing the depth of flow
c) its action is not understood
d) it reduces the kinetic energy by decreasing the depth of flow
Answer: a
Clarification: Hydraulic jump is generally accompanied by large scale turbulence dissipating most of the kinetic energy of the super-critical flow. It is the most suitable method because the energy is lost in the impact of the water against water. Most of the kinetic energy is destroyed by creating a condition suitable for the hydraulic jump.

6. A ski-jump bucket is also known as _____________________
a) flip bucket
b) solid roller bucket
c) slotted roller bucket
d) flexible bucket
Answer: a
Clarification: A ski-jump bucket is also called flip bucket is used for energy dissipation when tail-water depth is insufficient or low at all discharge. A part of energy dissipation takes place by impact and some of the energy is dissipated in the air by diffusion or aeration.

7. The percentage of energy dissipation in a hydraulic jump increases with the increase in the Froude number.
a) True
b) False
Answer: a
Clarification: The energy dissipation in the jump depends upon the Froude number, if this Froude number is higher, the greater energy dissipation can take place.

S NO. Froude number % loss in energy
1. 2.5 17
2. 4.5 45
3. 9.0 70

8. Which of the following stilling basin help in stabilizing the flow and improve the jump performance?
a) dentated sills
b) chute blocks
c) baffle piers
d) friction blocks
Answer: b
Clarification: Chute blocks are a row of small projections like teeth of saw and are provided at the entrance of the silting basin. It produces a shorter length of jump and stabilizes the flow. Hence, they improve jump performance.

9. What is the expected solution for the case when the T.W.C is lying above the J.H.C curve at all discharges?
a) By providing a simple concrete apron of length 5(Y1 – Y2)
b) By providing a sloping apron above the river bed
c) By providing a sloping apron below the river bed
d) Provision of a ski-jump bucket
Answer: b
Clarification: When the TWC is lying above the JHC at all discharges, the problem can be solved by-
1) By constructing a sloping apron above the river bed
2) By providing a roller bucket type of energy dissipator.
In this case, the jump is formed at the toe will be drowned by the tail-water and little energy will be dissipated.

10. A sloping apron is provided partly above the river bed and partly below the river bed in case of ____________________________
a) when TWC coincides with the JHC at all discharges
b) when TWC lies above the JHC at all discharges
c) when TWC lies below the JHC at all discharges
d) when TWC lies above the JHC at low discharges and below the JHC at high discharges
Answer: d
Clarification: At low discharges, the jump will be drowned and at high discharges tail-water depth is insufficient. When TWC lies above the JHC at low discharges and below the JHC at high discharges, the solution is the provision of sloping apron partly above and partly below the river bed. The horizontal apron and end-sill are also provided.

11. When the TWC lies below the JHC at all discharges, the problem can be solved by which of the following provisions?
i. Constructing a sloping apron above the river bed
ii. Provision of roller bucket type of energy dissipator
iii. Provision of a ski-jump bucket
iv. A sloping apron below the river bed
v. Construction of a subsidiary dam
vi. A sloping apron partly above and partly below the river bed
a) i, iii and v
b) i, ii and vi
c) iii, iv and v
d) i, iii, iv and v
Answer: c
Clarification: When TWC lies below the JHC at all discharges, the expected solution is –
i. Provision of a ski-jump bucket
ii. A sloping apron below the river bed of length 5 (y2 – y1)
iii. Construction of a subsidiary dam below the main dam.

250+ TOP MCQs on Gravity Dams – Stability and Answers

Irrigation Engineering Multiple Choice Questions on “Gravity Dams – Stability”.

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

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

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

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

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

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

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

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

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

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