Category: Irrigation Engineering Objective Questions

250+ TOP MCQs on Behavior of Rivers and Answers

Irrigation Engineering Multiple Choice Questions on “Behavior of Rivers”.

1. The numerous problems faced by the river is due to sediment carried by it.
a) False
b) True
Answer: b
Clarification: The sediment which is being carried by the river poses numerous problems like the flood level gets increased, silting occurs in irrigation and navigation channels, meandering of rivers takes place, and the river gets splitted into number of channels.

2. Meandering of the river causes some serious problems.
a) True
b) False
Answer: a
Clarification: Meandering of a river causes the river to change its original path and forces them to flow in new courses devastating vast areas of land and damaging nearby irrigation structures like dams, weirs, canals, and other structures like railway lines, and roads etc.

3. What is the shape of cross section of a straight reach?
a) Rectangular
b) Circular
c) Trough
d) Trapezoidal
Answer: c
Clarification: The shape of cross section of a straight reach is in the shape of a trough, with high speed velocities of flow in the middle of the section.

4. What type of structure does the diagram represent?
irrigation-engineering-questions-answers-behavior-rivers-q4
a) Cut-Off
b) Meander
c) Bends
d) Straight Reach
Answer: d
Clarification: The diagram shows a structure with cross section shape as a trough and the currents in the diagram shows transverse rotary currents, which in turn gives the information that the high velocity flow is in the middle of the section, and water level is high at the ends than at the middle. So, therefore according to this information interpreted the diagram represents a straight reach.

5. Which type of river tends to develop bends?
a) Alluvial River
b) Flashy River
c) Virgin River
d) Himalayan River
Answer: a
Clarification: Alluvial River tends to form the bends as this type of rivers are always characterised by scouring on the concave side and silting on the convex side. And this scouring and silting action gets continued by the action of centrifugal force in the bend.

6. Where are shoals formed in a bend in a river?
a) Concave Side
b) Convex Side
c) Middle Section of the River
d) At the Bottom of the River Bed
Answer: b
Clarification: Velocities at the bottom of the bend are much less than the velocities at the top, and enough centrifugal force is not present to counteract this tendency of water at the top. Due to this the water dives in from the top at the concave side and moves at the bottom towards the convex side. These rotary currents cause erosion at the concave side and deposition at the convex side, forming shoals on the convex side.

7. The complete formation of which curve leads to the formation of a meander?
a) Horizontal Curve
b) S-Curve
c) Vertical Curve
d) Circular Curve
Answer: b
Clarification: When bend is developed in a river, the process moves downstream by the formation of shoals on the convex side with the help of secondary currents. The formation of shoals on the convex side leads further erosion on the concave side. So, therefore formation of successive bends of reverse order leads to complete formation of an s-curve called meander.

8. What is the name given to the straight reaches between two successive curves in a meander?
a) Cut-Off
b) Meander Belt
c) Crossings
d) Bend
Answer: c
Clarification: The consecutive curves in a meander are connected with short straight reaches called crossings or in other words two consecutive clockwise and anti-clockwise loops in meander are connected with short straight reaches called crossings.

9. How many numbers of variables govern the meandering process?
a) 2
b) 3
c) 5
d) 4
Answer: d
Clarification: Four variables govern the meandering process, namely silt grade and silt charge, discharge, bed and side materials and their action towards erosion, and valley slope. These all variables affect the meandering process, and all of them are interdependent.

10. What was the initial belief for the meandering process?
a) Velocity of the Flow
b) Excessive Bed Slope
c) Discharge
d) Type of Sediment Load Carried
Answer: b
Clarification: The initial belief was that the meandering process is caused by the presence of excessive bed slope in the river that the excess energy is dissipated by increasing the channel length by meandering. But however, this belief was proved to be wrong.

250+ TOP MCQs on Cross-Section of an Irrigation Canal and Answers

Irrigation Engineering Multiple Choice Questions on “Cross-Section of an Irrigation Canal”.

1. The actual capacity of silt laiden water channel is worked out with 1/2 : 1 side slopes.
a) True
b) False
Answer: a
Clarification: Flatter slopes such as 1 : 1 and 1.5 : 1 are also constructed at the time of execution, but actual capacity of channel is worked out in 1/2 : 1 slope because even if we provide 1 : 1 or 1.5 : 1 slope the silt gets deposited on this slope and gives a new side slope 1/2 : 1. Therefore the actual capacity is calculated using 1/2 : 1 slope.

2. What is range of side slopes for a canal in cutting?
a) 1 H : 1 V to 1.5 H : 1 V
b) 1.5 H : 1 V to 1.5 H : 1 V
c) 1.5 H : 1 V to 1 H : 1 V
d) 1 H: 0.5 V to 1 H : 1 V
Answer: a
Clarification: The side slopes of a canal are designed such that they are stable, depending upon the type of soil present in canal area. A canal in cutting is provided with steeper slope than a canal in filing, as the stability of soil is more in case of canal in cutting than in canal in filing.

3. With what equation we can calculate the berm width?
a) (s1 – s2) d1
b) (s2 – s1) d1
c) (s2) d1
d) (s1) d1
Answer: b
Clarification: Berm is the horizontal distance between the toe of the bank and top edge of the cutting. If s2 : 1 is the slope in filling and s1 : 1 is the slope in cutting, and the initial depth of the berm from the bed of the canal is d1 then width of the berm is (s2 – s1) d1. As d1 varies, the width of the berm also varies. As siltation occurs the formula is (s2:1/2) y (y=over all new depth of the berm).

4. Deposition of silt to form the berms completely has lot of advantages.
a) False
b) True
Answer: b
Clarification: The silt which gets deposited is very fine and impervious therefore it acts as a good lining which reduces losses, leakage, breaches etc. This also gives strength to the banks and protects them against erosion and breaches. This deposition also provides scope for future widening of the canal.

5. What is the primary purpose of the banks?
a) Aesthetic View
b) Inspection Paths
c) Retaining Water
d) Means of Communication
Answer: c
Clarification: The main purpose of the banks is to retain water in the canal itself. This retaining action is always needed, in cases of low availability of water or during the period of floods. The other purposes of the banks include a means of communication and inspection paths.

6. What type of roads is provided for inspection purposes?
a) Side Road
b) Dirt Road
c) Service Roads
d) Toll road
Answer: c
Clarification: For inspection purposes of the canal, service roads are provided and they even serve as a means of communication in remote areas. Depending upon the size of the canal these roads are provided at a height of 0.4 m to 1 m above the fixed surface level.

7. What is provided as a measure of safety during driving?
a) Freeboard
b) Banks
c) Service Roads
d) Dowlas
Answer: d
Clarification: A structure of 0.3 m high and 0.3 to 0.6 m wide is provided along the banks, with side slopes of 1.5: 1 to 2: 1. This structure is known as dowlas. They also help in preventing soil erosion during rains.

8. Which structure is provided when the disposition of earth becomes costlier?
a) Borrow pits
b) Spoil Banks
c) Back Berm
d) Freeboard
Answer: b
Clarification: The disposition of earth becomes costlier when the earthwork in excavation exceeds earthwork in filling. To dispose this earth economically, it should be collected on the edge of the bank embankment. The soil is deposited on both banks or only on one bank in the form of heaps. Longitudinal drains are excavated to run along their sides to dispose rain water.

9. What type of structure is provided when earth has to be brought from somewhere else for canal construction?
a) Freeboard
b) Borrow Pits
c) Spoil Banks
d) Berms
Answer: b
Clarification: When the earthwork in filling exceeds the earthwork in excavation then the earth has to be brought from somewhere else to facilitate the construction of the canal. The pits or the places where the earth has been brought are called borrow pits.

10. Which type of borrow pits are preferred more?
a) External Borrow Pits
b) Internal Borrow Pits
c) Borrow pits
d) Spoil Banks
Answer: b
Clarification: The pits which are dug outside the canal are called external borrow pits and which are dug somewhere within the canal are called internal borrow pits. Internal borrow pits are preferred because bringing soil from distances is costlier than finding it near the canal itself. Moreover, there is a problem of mosquito nuisance in the external pits due rainwater collection.

11. Which type of structure does the diagram represent?
irrigation-engineering-questions-answers-cross-section-irrigation-canal-q11
a) Borrow Pit
b) Freeboard
c) Spoil Banks
d) Berm
Answer: a
Clarification: Clearly the diagram represents the excavation of pit of width less than half the width of the canal and the depth of the pit is less than 1 m. So, therefore according to this information we can say that the diagram represents a borrow pit. And more information we can obtain from the diagram is that the borrow pit is an internal borrow pit.

12. What structure does the diagram represent?
irrigation-engineering-questions-answers-cross-section-irrigation-canal-q12
a) Back Berm
b) Berm
c) Freeboard
d) Spoil Banks
Answer: d
Clarification: From the diagram, we can observe that FSL is obtained in the canal and is provided with berms, drains, and service roads. So, therefore this represents a phenomenon where the earthwork of excavation exceeds the earthwork of filling. Therefore to economically dispose the earth we need construction of spoil banks with drains which the clearly represents. Therefore this diagram represents the spoil banks structure.

13. What is the condition for digging of borrow pits?
a) B/2 > b
b) Depth of pit 7 < 1 m
c) Should in the Centre
d) Dig only External Pits
Answer: a
Clarification: For digging up of borrow pits the main condition is that the width of the pit (b) should be less than half the value of width of the canal (B/2). Other secondary conditions include that the pit should be dug in the centre, depth of the pit should be < 1 m, and it should start from a point at a distance of 5m from toe for small canals and 10 m for large canals.

250+ TOP MCQs on Water Requirements of Crops – Estimation of Consumptive Use – 2 and Answers

Irrigation Engineering Interview Questions and Answers on “Water Requirements of Crops – Estimation of Consumptive Use – 2”.

1. Which method involves the use of crop factor?
a) Penman’s equation
b) Hargreaves method
c) Blaney-Criddle Formula
d) Tanks and Lysimeter
Answer: c
Clarification: Blaney-Criddle Formula is expressed as Cu = k.f
Where, f = p/40 [1.8t + 32], Cu = seasonal consumptive use and k = crop factor. Crop factor is determined by experiments under environmental conditions of the particular area for each crop.

2. What are the dimensions of the standard class-A pan?
a) Diameter-1.2 m, Depth – 25 cm
b) Diameter-2 m, Depth – 25 cm
c) Diameter-2 m, Depth – 30 cm
d) Diameter-1.2 m, Depth – 30 cm
Answer: a
Clarification: Class-A pan is used for experimental determination of pan evaporation. This pan has a diameter of 1.2 m, 25 cm deep, and the bottom is raised 15 cm above the ground surface.

3. Penman’s equation for the estimation of PET has been derived by using ___________
a) energy balance approach
b) mass transfer approach
c) combination of energy balance approach and mass transfer approach
d) combination of energy balance approach and energy transfer approach
Answer: c
Clarification: The formula was developed by Penman in 1948 for determining the consumptive use of different areas of the basin. This formula has sound theoretical reasoning and is derived by combining energy balance and mass transfer approaches for the computation of transpiration and evaporation respectively.

4. Albedo or reflection coefficient factor is used in which method?
a) Blaney- Criddle equation
b) Christiansen equation
c) Penman equation
d) Tank and Lysimeter method
Answer: c
Clarification: Albedo is a factor used in Penman’s equation which is used for a different type of areas for calculation of consumptive use. The value of albedo varies for water surface, bare lands and snow.

5. Penman’s equation can also be used to compute evaporation from a water surface like a lake if _________
a) reflection coefficient = 0.05
b) reflection coefficient > 0.05
c) reflection coefficient < 0.05
d) reflection coefficient = 0.45- 0.90
Answer: a
Clarification: The value of reflection factor or albedo varies for water surface, bare lands and snow. When the value is equal to 0.05, it can be used to compute evaporation from a water surface.

6. The pan evaporation can also be determined by Christiansen formula.
a) True
b) False
Answer: a
Clarification: Christiansen formula for pan evaporation is Ep = 0.459 R.Ct.Cw.Ch.Cs.Ce where, R = Extra-terrestrial radiation in cm
Ct = Coefficient of temperature
Cw = Coefficient for wind velocity
Ch = Coefficient for relative humidity
Cs = Coefficient for percent of possible sunshine
Ce = Coefficient of elevation.

7. What is the consumptive use for a crop in the month of April having a consumptive use coefficient equal to 0.80 and pan evaporation is 35 cm?
a) 28 cm
b) 43.75 cm
c) 35.80 cm
d) 40.80 cm
Answer: a
Clarification: The formula is Et = k. Ep
Et = 0.8 x 35 = 28 cm
Hence, the required value of consumptive use = 28 cm.

8. Compute the total consumptive use for a drainage basin by Penman’s formula. The slope of saturation vapour pressure V/s chart at 40°C is 2.95 mm of Hg/°C and the net incoming solar radiation is 5.705 mm of evaporative water/day. Assume, Ea = 18.07 mm/day and Psychrometric constant = 0.49 mm of Hg/°C.
a) 17.01 mm/day
b) 18 mm/day
c) 11.07 mm/day
d) 20.45 mm/day
Answer: a
Clarification: Penman’s formula for Cu is equal to [A.Hn + Ea.Y] / (A + Y)
Cu = [2.95 x 5.705 + 18.07 x 0.49] / (2.95 + 0.49)
Cu = 17.01 mm/day.

9. What is the correct expression of parameter Ea in Penman’s equation?
a) Ea = 0.35 (1 + V2/160) (es – ea)
b) Ea = 0.35 (1 + V2/160) (es + ea)
c) Ea = 0.35 (1 – V2/160) (es – ea)
d) Ea = 0.35 (1 – V2/160) (es + ea)
Answer: a
Clarification: The parameter Ea is estimated by as: Ea = 0.35 (1 + V2/160) (es – ea).
Where, V2 is the wind mean speed at 2m above the ground in km/day, es is Saturation vapour pressure at mean air temperature in mm of Hg, and ea is actual mean vapour pressure of air in mm of Hg.

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250+ TOP MCQs on Energy Dissipation Below Overflow Spillway – 2 and Answers

Irrigation Engineering Quiz on “Energy Dissipation Below Overflow Spillway – 2”.

1. A troublesome and oscillating hydraulic jump is normally met in cases of _____________
a) weirs and barrages
b) overflow spillways of dams
c) weirs
d) large spillways
Answer: a
Clarification: When the Froude number lies in the range of 2.5 to 4.5, the jump is troublesome and oscillating as in case of weirs and barrages. There is an oscillating jet entering the jump bottom to the surface and back again which produces a large wave of irregular period doing unlimited damage.

2. A very steady and stable hydraulic jump is usually formed in the flows involving the approaching Froude number in the range of _____________________
a) less than 2.5
b) 2.5 to 4.5
c) 4.5 to 9.0
d) more than 9.0
Answer: c
Clarification: When the Froude number is in the range of 4.5 to 9.0, the jump performs at its best and is called steady jump. The length of the jump is almost constant and equal to 6 y2 where y2 is the post jump depth. The dissipation of energy ranges from 45 to 70%.

3. Standard USBR stilling basin-II is useful for energy dissipation at the bottom of the overflow structure, if the approaching Froude number is ________________
a) less than 4.5
b) more than 4.5
c) less than 2.5
d) more than 2.5
Answer: b
Clarification: USBR has standardized stilling basins for different Froude numbers. USBR stilling basin- II is recommended for large structures when the Froude number is more than 4.5. The dissipation of energy will be 45 to 85%.

4. When the Froude number is in the range of 2.5 to 4.5, the jump is ____________________
a) weak and energy loss is low
b) troublesome and oscillating
c) steady jump
d) strong jump
Answer: b
Clarification: When the Froude number is in the range of 2.5 to 4.5, the jump is troublesome and oscillating. This gives rise to the heavy waves on the surface and wave suppressors are needed in this range.

5. Which of the following stilling basin is applicable to only rectangular cross-sections?
a) U.S.B.R stilling basin-II
b) U.S.B.R stilling basin-IV
c) U.S.B.R stilling basin-I
d) U.S.B.R stilling basin-III
Answer: b
Clarification: USBR stilling basin-IV is used for Froude number varying between 2.5 and 4.5 which generally occurs in canal weirs, canal falls, diversion dams, etc. They are tried to be controlled by providing large chute blocks since oscillating waves are generated in this range. This is applicable only to rectangular cross-sections.

6. The Froude number of a hydraulic jump is 5.5. The jump can be classified as ____________
a) a weak jump
b) an oscillating jump
c) rough and choppy jump
d) steady jump
Answer: d
Clarification: When the Froude number is in the range of 4.5 to 9.0, the jump is steady. This is generally the case of dams and spillways. The oscillating jump occurs between the range of 2.5 to 4.5 and when Froude number is less than 2.5, the jump is weak and when it is more than 9.0, the jump is rough and choppy.

7. Standard stilling basin is provided at the toe of a dam spillway for energy dissipation is usually provided with auxiliary devices like chute blocks and dentated sills for the basic purpose of reducing the length of the stilling basin from about ________________
a) 6 y2 to 4 y2
b) 4 y2 to 2 y2
c) 6 y2 to 2 y2
d) 3 y2 to y2
Answer: a
Clarification: In the case of dams and spillways, the jump is steady and the length of the jump is almost constant and is equal to 6 y2 where y2 is the post jump depth. The length of the basin is related to Froude number as the economy in the length of the basin up to about 35% i.e. 4.3 y2 is thus obtained for auxiliary devices.

8. In the case of dam spillways, the approaching Froude number usually lies in the range of __________
a) less than 2.5 m
b) 2.5 – 4.5
c) 4.5 – 9.0
d) more than 9.0
Answer: c
Clarification: Froude number varying between 2.5 and 4.5 generally occurs in canal weirs, canal falls, diversion dams, etc. For dams and spillways, the Froude number is in between 4.5 and 9.0 i.e. steady jump and the jump is well balanced.

9. _____________________ are suitable only for low spillways or weirs.
a) Chute blocks
b) Sloping aprons
c) Baffle wall or friction blocks
d) Roller buckets
Answer: c
Clarification: Baffle piers or walls are blocks placed within the basin across the basin floor and they help in breaking the flow and dissipate the energy mostly by the impact. They generally give away under high-velocity jets due to their cavitation effects. It is unsuitable for large works.

10. Which of the following helps in dissipating the residual energy and to reduce the length of the jump or the basin?
a) Dentated sills
b) Chute blocks
c) Roller bucket
d) Baffle piers
Answer: a
Clarification: Chute blocks help in stabilizing the flow and thus improve the jump performance. Baffle piers help in breaking the flow and dissipate energy mostly by the impact. Dentated sills are generally provided at the end of the stilling basin and diffuse the residual portion of high-velocity jet reaching the end of the basin.

Irrigation Engineering for Quizzes,

250+ TOP MCQs on Gravity Method and Answers

Irrigation Engineering Multiple Choice Questions on “Gravity Method”.

1. When the reservoir is empty, the single force acting on it is the self-weight of the dam which acts at a distance of ____________
a) B/2 from the heel
b) B/6 from the heel
c) B/3 from the heel
d) B/4 from the heel
Answer: c
Clarification: The only single force on the dam when the reservoir is empty is the self-weight of the dam acting at a distance of B/3 from the heel. It provides maximum possible stabilizing moment about the toe without causing tension.

2. When the reservoir is empty, the maximum vertical stress equal to ________________
a) At heel = 2W/B and at toe = 0
b) At heel = 0 and at toe = 2W/B
c) At heel = toe = zero
d) At heel = toe = 2W/B
Answer: a
Clarification: The vertical stress distribution at the base when the reservoir is empty is given as –
Pmax/min = V/B [1 + 6e/B] and V/B [1 – 6e/B] where e = B/6 and V = total vertical force = weight W
Pmax = 2W/B and Pmin = 0.
The maximum vertical stress at the heel is equal to 2W/B and at the toe is zero.

3. The two-dimensional stability analysis of gravity dams proves better for U-shaped valleys than for V-shaped valleys.
a) True
b) False
Answer: a
Clarification: The transverse joints in the dam body are generally not grouted in U-shaped valleys but are keyed together in V-shaped valleys. In V-shaped valleys, the entire length of the dam acts monolithically as a single body. The assumption that the dam is considered to made up of a number of cantilevers of unit width each may involve errors here.

4. Calculate the value of minimum base width for an elementary triangular concrete gravity dam supporting 72 m height of reservoir water and full uplift? (Take specific gravity of concrete as 2.4 and coefficient of friction as 0.7)
a) 36.28 m
b) 39.77 m
c) 51.5 m
d) 73.5 m
Answer: d
Clarification: Using formula –
Case 1: B = H / (Sc – c)1/2 (For full uplift c = 1 and specific gravity of concrete = 2.4 )
= 72/ (2.4 – 1)1/2 = 60.85 m
Case 2: B = H/μ (S – 1) where μ = coefficient of friction taken as 0.7
B = 72 / 0.7 x 1.4 = 73.46 m
The highest among the two base width value is to be selected i.e. B = 73.46 m.

5. For usual values of permissible compressive stress and specific gravity of concrete, a high concrete gravity is the one whose height exceeds ______________
a) 48 m
b) 70 m
c) 88 m
d) 98 m
Answer: c
Clarification: The limiting height is – Hmax = f / (Sc + 1) ϒw Permissible strength of concrete = 3000 KN/m2, Sc = specific gravity of concrete = 2.4
Hmax = 3000/[(2.4 + 1) x 9.81] = 89.9 m.

6. For triangular dam section of height H for just no tension under the action of water pressure, self-weight and uplift pressure, the minimum base width required is _____________
a) H / (S-1)
b) H / S1/2
c) H / (S – 1)-1
d) H / (S-1)1/2
Answer: d
Clarification: The minimum base width (B) of a gravity dam having an elementary profile –
B = H / (S – 1)-1 where S is specific gravity of concrete and H is the height of water.
If uplift is not considered – B = H/S1/2.

7. If the eccentricity of the resultant falls outside the middle third, the ultimate failure of the dam occurs by ______________
a) tension
b) crushing
c) sliding
d) overturning
Answer: a
Clarification: When eccentricity is greater than B/6 (eccentricity falls outside the middle third), tension may develop. When tension prevails, cracks develop near the heel and uplift pressure distribution increases reducing the net salinizing force.

8. What is the value of eccentricity for no tension condition in the dam?
a) e < B/6
b) e > B/6
c) e > B/3
d) e < B/3
Answer: a
Clarification: The resultant of all the forces i.e hydrostatic water pressure, uplift pressure and self-weight of the dam should always lie within the middle third of the base for no tension. When e < B/6, the value of stress intensity at toe and heel are positive i.e compression on both sides.

9. What is the formula for limiting height of a gravity dam?
a) Hmax = f / (Sc + 1) γw
b) Hmax = f / (Sc – 1) γw
c) Hmax = f / (Sc + C) γw
d) Hmax = f / (Sc – 1) γw
Answer: a
Clarification: The critical height or limiting height of a dam having elementary profile is –
Hmax = f / (Sc + 1) γw where f = allowable stress of the dam material, Sc = Specific gravity of concrete and γw = unit weight of water.
This limiting height draws a dividing line between a low gravity dam and a high gravity dam.

10. Calculate the top width of the dam if the height of water stored is 84m.
a) 5 m
b) 2.5 m
c) 5.55 m
d) 7.75 m
Answer: a
Clarification: Bligh has given an empirical formula for finding out the thickness of the dam at top.
A = 0.522 H1/2 = 0.522 x 841/2 = 5.05 m.
As per Creager, the economical top width has been found to be equal to 14% of the dam height without considering earthquake forces.

250+ TOP MCQs on Aquifers and Their Type-1 and Answers

Irrigation Engineering Multiple Choice Questions on “Aquifers and Their Type-1”.

1. The geological formation which yields only insignificant quantity of groundwater is _____________
a) aquifer
b) aquifuse
c) aquiclude
d) aquitard
Answer: d
Clarification: The geological formations which are porous and contain a good amount of water but does not yield water freely to wells due to its lesser permeability is called aquitard. The water yield from such a formation is insignificant. Sandy clay is an example of the aquitard.

2. The geological formation which may contain water but does not contain any yield is ____________
a) aquifer
b) aquifuse
c) aquiclude
d) aquitard
Answer: c
Clarification: The geological formations which are porous but have no permeability are termed as an aquiclude. Water cannot be extracted from such formations. A clay layer is an example of aquiclude.

3. Which of the following geological formation contains and readily yields water to our tube wells?
a) Aquifer
b) Aquifuse
c) Aquiclude
d) Aquitard
Answer: a
Clarification: The geological formations which are both porous and permeable hence sufficient quantity of water can be extracted from them. Aquifers vary in depth and thickness and in general, confined and non-confined aquifers are two main categories of the aquifer.

4. Which of the following geological formation does not contain any amount of groundwater?
a) Aquitard
b) Aquifer
c) Aquiclude
d) Aquifuge
Answer: d
Clarification: The geological formations which are very dense and contain no water in voids and are neither porous nor permeable are termed as aquifuge. Granite rock is an example of aquifuge.

5. The quantum of water contained in the soil pores which cannot be extracted by gravity drainage is called _____________
a) pellicular water
b) capillary water
c) hygroscopic water
d) available water
Answer: a
Clarification: When the saturated formations are drained under the action of gravity drainage, the volume of water drained is less than the volume of void space. The water contained in these voids cannot be drained out by force of gravity. The water which is always retained by these interstices due to molecular attraction is called pellicular water.

6. What is the volume of groundwater which can be extracted by gravity drainage from a soil stratum when expressed as percentage fraction of the volume of the soil stratum?
a) Pellicular water
b) Available water
c) Specific yield
d) Field capacity
Answer: c
Clarification: Yield is the volume of groundwater extracted by gravity drainage from a saturated water-bearing material. It is known as the specific yield when it is expressed as the ratio of the volume of the total material drained.
Specific Yield = (Volume of the water obtained by gravity drainage/ total volume of drained material) x 100.

7. Field capacity of a ground aquifer equals _________
a) specific yield
b) 100 – specific yield
c) 100/ specific yield
d) field capacity
Answer: b
Clarification: Field capacity can be expressed as the percentage of the total volume of the material drained. Specific retention or field capacity is given as –
F.C = volume of water held against gravity drainage/total volume of the material drained x 100
The summation of Specific yield and field capacity is equal to the porosity.

8. Specific retention of groundwater is larger in coarse-grained soils.
a) True
b) False
Answer: b
Clarification: Specific retention can be defined as the amount of water held between the grains due to molecular attraction on the walls of the interstices. If the effective size of grain decreases, the surface area between the interstices will increase causing more specific retention and less specific yield. In fine soils like clay, the specific retention would be more and it results in small specific yield.

9. Water wells excavated through confined aquifers are known as ______________
a) artesian wells
b) non-artesian wells
c) gravity wells
d) water table wells
Answer: a
Clarification: The gravity wells which are constructed to tap water from the unconfined aquifer are known as unconfined or non-artesian wells. Such wells are also known as water table wells or gravity wells since the water level in these wells is equal to the level of the water table.

10. In case of a flowing well, the piezometric surface is always _________
a) below the ground level
b) above the ground level
c) at the ground level
d) above or below the ground level
Answer: b
Clarification: A well excavated through confined aquifer yields water that often flows out automatically under the hydrostatic pressure. It may even rise or gush out of the surface for a reasonable height. Flowing well is a type of artesian well where water gushes out automatically.