Previous Year Paper

BPSC AE Paper 6 (Civil) 25 March 2022 Official Paper (Previous Year Paper)

Held in 2022

50 questions60 minutes
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Full Test (50 questions)

Question 1

Identify the correct sequence in a house water connection.

  1. AOption (a)

    Water meter, Gooseneck, Ferrule, Stopcock, Service pipe

  2. BOption (b)

    Water meter, Service pipe, Stopcock, Gooseneck, Ferrule

  3. COption (c)

    Ferrule, Gooseneck, Service pipe, Stopcock, Water meter

  4. DOption (d)

    Ferrule, Water meter, Gooseneck, Service pipe, Stopcock

View answer & explanation

Correct answer

Answer: (c)C.

Ferrule, Gooseneck, Service pipe, Stopcock, Water meter

Why this answer is correct

Concept:

The house water connection

To get water supply from municipalities connection consists of the following sequence:

  1. Ferrule:
  • A ferrule is a right-angled sleeve made of brass or gunmetal and is joined to a hole drilled in the water main to which it is screwed down with a plug.
  1. Goose Neck:
  • The gooseneck is a small-sized curved pipe made up of flexible material and is about 75 cm in length forming a flexible connection between the water main and the service pipe.
  1. Service Pipe:
  • It is a galvanized iron pipe of size less than 50 mm dia. It should be laid underground in a trench in which no sewage or drainage pipe is laid. The service pipe which supplies water to the building through the municipal mains is connected to the main through gooseneck and ferrule.
  1. Stop Cock:
  • The stop cock is provided before the water enters the water meter in the house. It is placed in a suitable masonry chamber with a removable cover and is fixed in the street close to the boundary wall in an accessible station.
  1. Water Meter:
  • It measures and records the quantity of water consumed in the house.

Question 2

The mass curve method for determination of storage capacity of a reservoir uses

  1. AOption (a)

    census data and cash flow

  2. BOption (b)

    cash flow data and water flow data

  3. COption (c)

    water inflow data and water demand data

  4. DOption (d)

    water demand data and water wastage data

View answer & explanation

Correct answer

Answer: (c)C.

water inflow data and water demand data

Why this answer is correct

Explanation:

Mass curve method:

  • This method is also known as the Rippl mass curve method after the developer of this method. This is a simple method commonly used to estimate the required storage capacity of a reservoir in the project planning stage.
  • The method uses the most critical period of recorded flow to compute storage.
  • The critical period is defined as the duration in which a full reservoir depletes and passes through various states and empties without spilling. In the methods based on the essential period concept, a sequence of streamflows containing a critical period is routed through an initially full reservoir in presence of specified demands.
  • The reservoir capacity is obtained by finding the maximum difference between cumulative inflows and cumulative demand curves.
  • To determine the minimum required storage, the mass curve of inflow and the mass curve of demand are accumulated separately.
  • For a constant draft, the mass curve of demand will be a straight line having a slope equal to the demand rate. Now, at each high point on the mass inflow curve, a line is drawn parallel to the mass curve of demand and is extended until it meets the mass curve of inflows.

Question 3

Identify the wrong empirical formula name for computing the peak drainage discharge.

  1. AOption (a)

    Ven Te Chow formula

  2. BOption (b)

    Nawab Jung Bahadur formula

  3. COption (c)

    Dicken's formula

  4. DOption (d)

    Ryve's formula

View answer & explanation

Correct answer

Answer: (a)A.

Ven Te Chow formula

Why this answer is correct

Explanation:

Ryve’s Method

This formula is used only in Southern India.

QP=CRA23{Q_P} = {C_R}{A^{\frac{2}{3}}}

Rational Method

This method is based on the principle of the relationship between rainfall and runoff. Hence, it can be considered to be similar to the empirical method.

QP=136kPcA1{Q_P} = \frac{1}{{36}}k{P_c}{A^1}

Dicken’s Method

It was earlier adopted only in Northern India but now it can be used in most of the States in India after proper modification of the constant.

QP=CDA34{Q_P} = {C_D}{A^{\frac{3}{4}}}

Nawab jung bahadur formula:

Qp=C2×A0.93−logA14Q_{p}=C_{2}\times A^{0.93-\frac{log A}{14}}

Where, Qp = Discharge, A = Area in km2

Inglis Method

This formula is used only in Maharashtra. Here three different cases are taken into consideration.

QP=124AA+10.4≅123A{Q_P} = \frac{{124A}}{{\sqrt {A + 10.4} }} \cong 123\sqrt A

where

QP = Peak discharge in m3/s.

A = Area in km2.

CR = CD = coefficient applicable in certain region.

Additional Information Van Te chow developed a formula for hydrologic frequency drainage design, a method of backwater curve computation.

Question 4

Which of the following hydraulic formulae is not used in determining flow velocities in sewers and drains?

  1. AOption (a)

    Horton's formula

  2. BOption (b)

    Kutter's formula

  3. COption (c)

    Crimp and Burges' formula

  4. DOption (d)

    All of the above

View answer & explanation

Correct answer

Answer: (a)A.

Horton's formula

Why this answer is correct

Explanation:

Various equations used in sewer design are as follows:

1) Chezy formula

V=C×RSV=C\times \sqrt{RS}

Where, V = Velocity of flow, S = Slope, R = Hydraulic radius

Value of C is obtained from

A) Ganguillet-Kutter’s Formula

C=23+0.00155S+1N1+(23+0.00155S)NRC= \frac{23+\frac{0.00155}{S}+\frac{1}{N}}{1+\left ( 23+\frac{0.00155}{S} \right )\frac{N}{\sqrt{R}}}

B) Bazin formula

C=157.61.81+mRC= \frac{157.6}{1.81+\frac{m}{\sqrt{R}}}

2) Manning formula

V=1nR23S12V = \frac{1}{n}{R^{\frac{2}{3}}}{S^{\frac{1}{2}}}

3) Crimp and burges formula

V=83.47×R2/3×S1/2V= 83.47\times R^{2/3}\times S^{1/2}

Additional Information Horton formula is used for finding the infiltration and is given by

f(t) = fc + (f0 – fc)e - kt

Where, f(t) = infiltration at any time t, f0 = Infiltration at T = 0, fc = Constant infiltration value

Question 5

P. Trickling filter consists of a bed of crushed stone, gravel or slag of relatively large size.

Q. As sewage passes through the filtering media of the trickling filter, an organic film is formed around the particles of filtering media.

Identify the correct option.

  1. AOption (a)

    P is correct and Q is incorrect

  2. BOption (b)

    P is incorrect and Q is correct

  3. COption (c)

    Both are incorrect

  4. DOption (d)

    Both are correct

View answer & explanation

Correct answer

Answer: (d)D.

Both are correct

Why this answer is correct

Concept:

Trickling Filters:

  • A trickling filter is an attached growth process i.e. process in which microorganisms responsible for treatment are attached to inert packing material.
  • The packing material used in attached growth processes includes rock, gravel, slag, sand, redwood, and a wide range of plastic and other synthetic materials.

Figure: Trickling Filter

The process of trickling filter is as follows:

  • The wastewater in the trickling filter is distributed over the top area of a vessel containing non-submerged packing material.
  • Air circulation in the void space, by either natural draft or blowers, provides oxygen for the microorganisms growing as an attached biofilm.
  • During operation, the organic material present in the wastewater is metabolized by the biomass attached to the medium. The biological slime grows in thickness as the organic matter abstracted from the flowing wastewater is synthesized into new cellular material.
  • The thickness of the aerobic layer is limited by the depth of penetration of oxygen into the microbial layer.
  • The micro-organisms near the medium face enter the endogenous phase as the substrate is metabolized before it can reach the micro-organisms near the medium face as a result of increased thickness of the slime layer and lose their ability to cling to the media surface.
  • The liquid then washes the slime off the medium and a new slime layer starts to grow. This phenomenon of losing the slime layer is called sloughing.
  • The sloughed-off film and treated wastewater are collected by an under drainage which also allows circulation of air through the filter.
  • The collected liquid is passed to a settling tank used for solid-liquid separation.

Question 6

Identify the correct statement or statements from practical and technical point of view.

  1. AOption (a)

    In a two-chamber septic tank, the first chamber is a grit chamber and second chamber is anaerobic chamber.

  2. BOption (b)

    In a two-storied septic tank, the top storey is the grit chamber while the other is anaerobic chamber.

  3. COption (c)

    Modern septic tanks possess a self-cleansing property.

  4. DOption (d)

    All are wrong statements

View answer & explanation

Correct answer

Answer: (a)A.

In a two-chamber septic tank, the first chamber is a grit chamber and second chamber is anaerobic chamber.

Why this answer is correct

Explanation:

Septic tank:

  • Septic tank is a tank where sewage is kept for long time for sedimentation of suspended solids by anaerobic microorganisms activity.
  • The tank is rectangular is shape, consisting of rooftop with 2 chambers separated by a baffle wall
  • The first one is called grit chamber where the sewage enters and the rest is called anaerobic unit.
  • Inlet and outlet pipe are fitted in such a manner that, ie while entering or exiting there should not be any unsettled influence in the sewage of anaerobic chamber
  • Sand, grit etc is settled in the grit chamber before the sewage is directed to the anaerobic section. In anaerobic chamber natural solids settle at the base of the tank, where anaerobic bacteria converts complex mixes into simple and stable mixes, which does not create any nuisance when disposed

Question 7

The elementary profile of a concrete gravity dam is

  1. AOption (a)

    right-angled triangle

  2. BOption (b)

    isosceles triangle

  3. COption (c)

    equilateral triangle

  4. DOption (d)

    None of the above

View answer & explanation

Correct answer

Answer: (a)A.

right-angled triangle

Why this answer is correct

Concept-

The elementary profile of a dam subjected only to external water pressure on the upstream side will be a right-angled triangle, having zero width at water level and a base width (B) at the bottom, i.e., the point at which maximum hydrostatic pressure acts.

When reservoir is empty

When the reservoir is empty, only the self-weight of the dam acts at a distance of B/3 from the heel.

This is the maximum possible innermost position of the resultant for no tension to develop.

Hence, such a line of action of W is most ideal as it gives maximum possible stabilizing moment about the toe without causing tension at toe when the reservoir is empty.

When the reservoir is full

The base width is governed by

(i) The resultant of all forces, i.e., P wan U passed through the outermost middle third point.

(ii) The dam is safe in sliding.

Taking moments of all forces about the middle third point, we get

W(B3)−U(B3)−pH3=R×0W\left( {\frac{B}{3}} \right) - U\left( {\frac{B}{3}} \right) - \frac{{pH}}{3} = R \times 0

(W−U)(B3)−pH3=0→(1)\left( {W - U} \right)\left( {\frac{B}{3}} \right) - \frac{{pH}}{3} = 0 \to \left( 1 \right)

ButW=12×B×H×Sc×rωBut{\rm{ }}W = \frac{1}{2} \times B \times H \times {S_c} \times {r_\omega }

Sc = specific gravity

rω = unit weight of water = 9.81 kN / m3

U=12.C.rω.H.BU = \frac{1}{2}.C.{r_\omega }.H.B

If C is 1, B=Hμ(Sc−1)B = \frac{H}{{\mu \left( {{S_c} - 1} \right)}}

If C is 0, that is there is no uplift pressure, B≥Hμ.ScB \ge \frac{H}{{\mu .{S_c}}}

The value of B chosen should be greater of the two values from the above two equations.

The base width of an elementary profile of a concrete gravity dam with no tension criteria and neglecting the uplift pressure is approximately 65% of dam height.

Question 8

What is a culvert?

  1. AOption (a)

    A bridge with span up to 8 m

  2. BOption (b)

    A bridge of span up to 6 m

  3. COption (c)

    A bridge of span up to 3 m

  4. DOption (d)

    None of the above

View answer & explanation

Correct answer

Answer: (b)B.

A bridge of span up to 6 m

Why this answer is correct

Explanation:

According to IRC 5: 1998

Culvert:

  • A culvert is a cross drainage structure having a total length of 6 m or less between the inner faces of dirt walls or extreme vent way boundaries measured at right angles.

Bridge:

  • A bridge is a structure whose total length is above 6 m between the inner faces of the dirt walls for carrying traffic or other moving loads over a depression or obstruction such as channel, road, or railway.
  • These bridges are further classified as minor or major bridges according to their span.
Minor bridgesMinor bridges have a total length(Span) less than 60 m.
Major bridgeMajor bridges have a total span greater than 60 m

Question 9

P. A weir is a hydraulic structure where majority or the entire ponding is achieved by a raised crest and smaller part or nil part by the shutters.

Q. A weir is a hydraulic structure that is made by constructing the width of the channel only.

R. A barrage is a hydraulic structure that is made by creating a choking condition only.

S. A barrage is a hydraulic structure where majority or the entire ponding is achieved by a shutter and smaller part or nil part by the raised crest.

Identify the correct statements in the above.

  1. AOption (a)

    P and R

  2. BOption (b)

    Q and S

  3. COption (c)

    Q and R

  4. DOption (d)

    P and S

View answer & explanation

Correct answer

Answer: (d)D.

P and S

Why this answer is correct

Concept:

Weir:

  • If the major part of the entire ponding of water is achieved by the raised crest and a smaller part or nil part of it is achieved by the shutters, then this barrier is known as a weir.

Barrage:

  • If most of the ponding is done by gates and a smaller or nil part of it is done by the raised crest, then the barrier is know as a barrage.
  • It is a structure similar to a weir with the only difference being that the crest is kept at a low level and ponding of water is accomplished mainly by means of gates. During floods these gates can be raised above HFL and thus enable the high flood to pass with minimum afflux.

Question 10

Silt excluders are constructed ________ while silt ejectors are constructed ________.

  1. AOption (a)

    on the river bed, on the diversion headworks

  2. BOption (b)

    on the river bed, on the bed of the off taken canal

  3. COption (c)

    on the bed of the off taken canal, on the bed of the river

  4. DOption (d)

    on the bed of the off taken canal, on the river bed

View answer & explanation

Correct answer

Answer: (b)B.

on the river bed, on the bed of the off taken canal

Why this answer is correct

Concept:

Silt Control Devices: The entry of silt into a canal, which takes off from a diversion head works, can be reduced by constructing certain special works which are called silt control works or silt control devices. These works may be classified into the following two types:

Silt Excluders:

  • Silt excluders are those works which are constructed on the bed of the river at upstream of the head regulator.
  • The clearer water enters the head regulator and consequently the silted water enters the silt excluder. In this type of works, the silt is, therefore, removed from the water before it enters the canal.
<br>

Silt Ejectors:

  • Silt ejectors, also called silt extractors, are those devices that extract the silt from the canal water after the silted water has travelled a certain distance in the off-take canal.
  • These works are, therefore, constructed on the bed of the canal, and a little distance downstream from the head regulator.

Question 11

Which of the following give a discharge that is relatively constant and fixed within limits?

  1. AOption (a)

    Non-modular modules

  2. BOption (b)

    Semi-modules

  3. COption (c)

    Modular outlets

  4. DOption (d)

    None of the above

View answer & explanation

Correct answer

Answer: (c)C.

Modular outlets

Why this answer is correct

Concept:

Canal Outlets or Module:

A canal outlet is a small structure built at the head of the watercourse so as to connect it with a minor or a distributary channel. It is also called sluice.

Types of Canal Outlets:

Canal outlet can be classified mainly into three classes:

(i) Non-modular outlets:

  • Non-modular outlets are those outlets whose discharge depends on the difference of water levels (i.e. head) between the distributary and the watercourse.
  • Such an outlet is controlled by a shutter at its upstream end.
  • Non-modular outlets are very suitable for low head conditions.
  • The diameter of the pipe varies from 10 to 30 cm.

Example: Submerged pipe outlets and masonry sluices

<br>

(ii) Semi-modular outlets:

  • Semi-modular outlets are those outlets whose discharge depends only on the water level in the distributary and is unaffected by the water level in the watercourse provided that a minimum working head required for working is available.

Example: Pipe outlet, Venturi flume

<br>

(iii) Modular outlets:

  • Modular outlets are those whose discharge is independent of the difference of water levels in the distributary and the watercourse and hence remains constant within the limits
  • These are called Rigid modules.

Example: Gibb's module

Question 12

What are bed bars?

  1. AOption (a)

    They are iron rods placed on the bed of canal to ensure smooth flow in the canal

  2. BOption (b)

    They are small sediment island like deposits formed on the bed of the canal

  3. COption (c)

    They are constructed, at suitable intervals, along an unlined canal so as to serve as permanent reference marks

  4. DOption (d)

    They are very large islands formed on the bed of rivers

View answer & explanation

Correct answer

Answer: (c)C.

They are constructed, at suitable intervals, along an unlined canal so as to serve as permanent reference marks

Why this answer is correct

Explanation:

Bed bars in the canal

  • The maintenance of the canal includes silt clearance and removing aquatic weeds to maintain the desired cross-sectional and longitudinal profiles.
  • Silt clearance consists of removing the silt deposited during the course of running the canal over a period of time.
  • The silt is removed down to the bed bars that are laid at suitable intervals of 200 m for small canals to 500 m for large canals.
  • These bed bars are usually masonry walls constructed at the bottom of the canal such that the top of each bar, is flush with the theoretical bed level and the side of the canal.
  • The bed bars reveal whether the canal is silting or scouring in any section of the canal.

Question 13

Assuming 80% efficiency, 100 m3/sec of design discharge, and 100 m of design head, what will be the approximate electrical power producted?

  1. AOption (a)

    800 megawatts

  2. BOption (b)

    8000 kilowatts

  3. COption (c)

    80000000 watts

  4. DOption (d)

    None of the above

View answer & explanation

Correct answer

Answer: (c)C.

80000000 watts

Why this answer is correct

Concept:

Power developed in a reaction turbine is given by:

P=;ρQgHηo1000P = ;\frac{{\rho QgH{\eta _o}}}{{1000}}

Where,

ρ = density of the water (kg/m3)

Q = Discharge through the turbine (m3/s)

H = Head (m)

ηo = the overall efficiency of the turbine 

Calculation:

ρ = 1000 kg/m3

Q = 100 m3/sec

H = 100 m, Efficiency = 80%

Power developed in a reaction turbine is given by:

P=;1000×100×10×100×0.801000P = ;\frac{{1000 \times 100 \times 10 \times 100 \times 0.80}}{{1000}}

P = 80000 kW = 80000000 W

Question 14

P% of earth dam failures are due to hydraulic failures, Q% of the earth dam failures are due to seepage failures, R% of the earth dam failures are due to structural failures, where

  1. AOption (a)

    P = 40, Q = 33, R = 25

  2. BOption (b)

    P = 25, Q = 40, R = 33

  3. COption (c)

    P = 33, Q = 25, R = 40

  4. DOption (d)

    P = 40, Q = 25, R = 33

View answer & explanation

Correct answer

Answer: (a)A.

P = 40, Q = 33, R = 25

Why this answer is correct

Concept:

Failures of earthen embankment dams are grouped into three categories

a) Hydraulic Failure: They occur from the uncontrolled flow of water over and adjacent to the embankment and due to the erosive action of water on the embankment slopes. 4 out every 10 earthen dams have failed due to hydraulic failures.

Various hydraulic failures are:-

i) Overtopping:

 

(ii) Wave erosion: Notching of the upstream face by wave action that reduces the embankment cross section thickness and weakens embankment material.

(iii) Toe erosion: Erosion of downstream toe of the earth slope caused by misdirected spillway outlet discharge.

(iv) Gullying: Rainfall erosion of embankment slopes. Also caused by traffic from people and vehicles.

b) Seepage Failure: Seepage, if uncontrolled, can erode fine soil material from the downstream slope or foundation and continue moving towards the upstream slope to form a pipe or cavity to the pond or lake often leading to a complete failure of the embankment. This action is known as “piping.”1 out of every 3 failed dam cases has seepage failures.

c) Structural Failure: Sloughs, bulges, cracks or other irregularities in the embankment are examples of structural failure. About 1/4th of earthen dams have failed due to structural failures.

Hence, 40 % of earth dam failures are due to hydraulic failures, 33% of the earth dam failures are due to seepage failures, 25% of the earth dam failures are due to structural failures.

Question 15

Select the correct statement from the following:

  1. AOption (a)

    The canal flows below the drain in aqueduct and syphon aqueduct, and the canal flows above the drain in super passage and syphon.

  2. BOption (b)

    The drain flows below the canal in aqueduct and syphon aqueduct, and the drain flows above the canal in super passage and syphon.

  3. COption (c)

    It is engineer's choise to let the drain flow above or below the canal in aqueduct and syphon aqueduct and also in super passage and syphon.

  4. DOption (d)

    The above statements are wrong

View answer & explanation

Correct answer

Answer: (b)B.

The drain flows below the canal in aqueduct and syphon aqueduct, and the drain flows above the canal in super passage and syphon.

Why this answer is correct

Concept:

A cross drainage work is a structure constructed for carrying a canal across a natural drain or river intercepting the canal to dispose drainage water without mixing with the continuous canal supplies.

Types of Cross Drainage Works:-

Based on the relative bed levels, water levels of the canal and the drain and their relative discharge the Cross Drainage works are of the following types:-

(1) Cross drainage works carrying the Canal over the Natural Drain:

(i) Aqueduct

  • An aqueduct is a hydraulic structure which carries a canal (through a trough) across and above the drainage similar to a bridge in which instead of the road or a railway, a canal is carried over a natural drain.

  • In the case of an aqueduct, HFL (highest flood level) of the drainage should remain lower than the level of the underside of the canal trough.

  • The drain flows at atmospheric pressure under the work.

  • Generally, an inspection road is provided along with the trough

(ii) Syphon aqueduct

  • A syphon aqueduct is a cross drainage structure similar to an aqueduct except that the streambed is depressed locally where it passes under the trough of the canal and the barrels discharges the streamflow under pressure.
  • A syphon aqueduct is constructed where the water surface level of the drain at high flood is higher than the canal bed.
  • The horizontal floor of the barrels is provided with slopes at its ends to join the drain bed on either side.
  • The drain water flows under pressure through the barrels which act as inverted syphons and hence this cross drainage work is known as syphon aqueduct.

 

2) Cross drainage works carrying the natural drain over the canal-

(i) Super passage

A super passage is also similar to a bridge in which the natural drain is carried over the canal.

  • A super passage is reverse of an aqueduct.

  • A super passage is constructed where the bed of the drain is well above the canal F.S.L

  • The canal flows at atmospheric pressure under the work.

  • In this case it is not possible to provide an inspection road along the canal.

(ii) Syphon

  • A syphon is similar to a syphon aqueduct with the difference that in the case of a syphon the canal water is carried through the barrels under the drain.
  • A syphon is constructed where the full supply level of the canal is higher than the bed of the drain.
  • The barrels in this case also act as inverted syphons through which the canal water flows under pressure.

​

  1. Cross drainage works admitting the drain water into the canal-

In this type of cross drainage works, the canal water and the drain water are allowed to intermingle with each other. This may be achieved by the following two types of cross drainage works:-

(i) Level crossing

(ii) Inlet and outlet

Question 16

The flexibility of a module is defined as

  1. AOption (a)

    (outlet discharge)/(distributary channel discharge)

  2. BOption (b)

    (depth of water in the distributary channel)/(head acting on the outlet)

  3. COption (c)

    (depth of water in the distributary channel)m/(head acting on the outlet)n

  4. DOption (d)

    (rate of change of discharge of the outlet)/(rate of change of discharge of the distributary channel)

View answer & explanation

Correct answer

Answer: (d)D.

(rate of change of discharge of the outlet)/(rate of change of discharge of the distributary channel)

Why this answer is correct

1) Flexibility is the ratio of the rate of change of discharge at an outlet to the rate of change of discharge of the distributary channel.

Flexibility, F =(dqq)(dQQ) = \frac{{\left( {\frac{{dq}}{q}} \right)}}{{\left( {\frac{{dQ}}{Q}} \right)}}

where,

q = discharge passing through the outlet = C1yn

Q = flow rates in the distributary channels = C2Hm

y = depth of flow in the channel, H = Head on the outlet

m = outlet index,  n = channel index

C1 and C2 = constant 

F =(dqq)(dQQ) = \frac{{\left( {\frac{{dq}}{q}} \right)}}{{\left( {\frac{{dQ}}{Q}} \right)}}= =mn×yH×dHdy = \frac{m}{n} \times \frac{y}{H} \times \frac{{dH}}{{dy}}

F = 0 for modular outlet

F = =mn×yH = \frac{m}{n} \times \frac{y}{H} for semi-modular channel, dy = dH

2) Sensitivity is the ratio of the rate of change of discharge through an outlet to the rate of change in the water surface level of distributary channel with respect to the normal depth of the flow in the channel.

Sensitivity =(dqq)(dSy) = \frac{{\left( {\frac{{dq}}{q}} \right)}}{{\left( {\frac{{dS}}{y}} \right)}}        where, S = Gauge reading 

3) Proportionality is the property of the channel and an outlet is said to be proportional when the rate of change of outlet discharge equals to the rate of change of channel discharge.

Proportionality =(dqq)(dQQ) = \frac{{\left( {\frac{{dq}}{q}} \right)}}{{\left( {\frac{{dQ}}{Q}} \right)}} = 1 =mn=Hy = \frac{m}{n} = \frac{H}{y}= =Outlet;IndexChannel;Index = \frac{{Outlet;Index}}{{Channel;Index}}

Depending on the value of Flexibility, the outlet can be classified as

  1. Proportional outlet (F = 1)

  2. Hyper Proportional outlet (F > 1)

  3. Sub Proportional outlet (F < 1)

4) Setting is the ratio of the depth of the crest level of the module below the free surface level of the distributary channel to the full supply depth of the distributary channel.

Setting =Hy = \frac{H}{y}

Question 17

Lysimeters are used to measure

  1. AOption (a)

    groundwater recharge rates

  2. BOption (b)

    evapotranspiration

  3. COption (c)

    interception losses

  4. DOption (d)

    lake evaporation

View answer & explanation

Correct answer

Answer: (b)B.

evapotranspiration

Why this answer is correct

Humidity: It is a measure of moisture present in the air.

The amount of moisture present in the air when expressed as mass per unit volume is known as absolute humidity.

Humidity is measured by the psychrometer.

Transpiration: It is the process by which water leaves the body of a living plant and reaches atmosphere as water vapour.

Transpiration is measured by phytometer.

Evaporation: It is the process in which a liquid change to the gaseous state at the free surface below the boiling point through the transfer of heat energy.

Evaporimeter measures evaporation.

Evapotranspiration is measured by lysimeter.

Question 18

What are the lower and upper limits of catchment area for applicability of the use of unit-hydrograph?

  1. AOption (a)

    200 ha and 5000 km2

  2. BOption (b)

    2000 m2 and 5000 km2

  3. COption (c)

    200 ha and no upper limit

  4. DOption (d)

    None of the above

View answer & explanation

Correct answer

Answer: (a)A.

200 ha and 5000 km2

Why this answer is correct

Explanation:

Instantaneous unit hydrograph (IUH):

  • A unit hydrograph of infinitesimal duration (Very small duration) is called an instantaneous unit hydrograph.
  • It mainly used when rainfall duration and intensity are variable because in that case, we need to divide the rainfall into various small durations and then find its response.
  • The advantage of IUH is that it is independent of the duration of effective rainfall hyetograph and has one parameter less than the D-hr unit hydrograph.
  • IUH is independent of rainfall characteristics.
  • It is indicative of catchment storage characteristics.
  • An instantaneous unit hydrograph is also obtained from S- curve. But that is an approximate method.
  • Also called Clark's method

​

Additional Information

Uses and Limitations of unit hydrograph:

Uses:

  • The development of flood hydrograph for extreme rainfall magnitude for use in the design of the hydraulic structure.
  • Extension of flood flow records based on rainfall records.
  • Development of flood records based on rainfall
  • Development of flood forecasting and warning systems based on rainfall.

Limitations:

  • Unit hydrograph cannot be used for catchments are greater than 5000 km2
  • Unit hydrograph cannot be used for catchment less than 200 hec or 2 km2
  • Precipitation must be in the form of rainfall only ie not be in the form of snow.​

Question 19

Which of the following meters is used in river flow measurement?

  1. AOption (a)

    Acoustic Doppler velocimeter

  2. BOption (b)

    Electromagnetic velocimeter

  3. COption (c)

    Pygmy current meter

  4. DOption (d)

    All of the above

View answer & explanation

Correct answer

Answer: (d)D.

All of the above

Why this answer is correct

Explanation:

1) Acoustic doppler velocimeter:

  • The acoustic Doppler velocimeter (ADV) is a new acoustic instrument with the potential to offer accuracy for laboratory devices even under field conditions.
  • An acoustic Doppler velocimeter (ADV) operates by the principle of Doppler shift.
  • The ADV uses this principle to measure the velocity of water in three dimensions. The device sends out a beam of acoustic waves at a fixed frequency from a transmitter probe.

2) Electromagnetic velocimeter:

  • Electromagnetic current meters produce a voltage proportional to the velocity.
  • One advantage of these current meters is the direct analog reading of velocity; counting of revolutions is not necessary.
  • These current meters can also measure crossflow and are directional.
  • Electromagnetic current meters, while still not as reliable as the anemometer type, have improved greatly in recent years. Their use of near metallic objects is still a limitation.

3) Current meter:

  • The Pygmy Current Meter is a versatile instrument with a long history of accuracy, reliability, and durability.
  • The precisely balanced Bucket Wheel mounts on a vertical pivot inside a yoke and rotates by water flow.
  • Each rotation of the Bucket Wheel produces an audible click which users monitor either manually with a headset or with digital equipment.
  • Rotation rate is proportional to water velocity

Question 20

Which of the following is used for channel flow routing?

  1. AOption (a)

    Puls method

  2. BOption (b)

    Modified Puls method

  3. COption (c)

    Goodrich method

  4. DOption (d)

    Muskingum method

View answer & explanation

Correct answer

Answer: (d)D.

Muskingum method

Why this answer is correct

Explanation:

  • A common method of Hydrologic channel routing is Muskingum method, in which the channel storage in a reach is expressed as a function of both inflow and outflow discharge.

S = k [x Im + (1 - x)Qm]

Where, s = Storage in a channel reach

I = Inflow Discharge

Q = Outflow Discharge

k → Coefficient is known as the storage time coefficient. It was unit of time, it is approx. equal to time of travel of flood wave through the channel reach.

x → Weightage factor. Its range series from 0 to 0.5

  • For naturally occurring channels m = 1, Hence the equation becomes

S = k [x I + (1 - x)Q]

  • For x = 0, storage is only a function of outflow discharge. Such reservoirs are also known as Linear Reservoir.

The Modified Puls routing method and Goodrich routing method:

The Modified Puls routing method and Goodrich routing method is a reservoir routing (Lumped routing) which is used in

  • the design of the capacity of spillways and other reservoir outlet structures
  • the location and sizing of the capacity of reservoirs to meet specific requirements

Question 21

Find the delta for a crop when its duty is 1728 hectares/cumec on the field, the base period for the crop is 240 days.

  1. AOption (a)

    360 cm

  2. BOption (b)

    240 cm

  3. COption (c)

    120 cm

  4. DOption (d)

    60 cm

View answer & explanation

Correct answer

Answer: (c)C.

120 cm

Why this answer is correct

Concept: 

 

The relation between Duty (D), Delta (Δ), and Base period (B) is given as:

Δ(metres)=8.64BD{\bf{Δ }}\left( {{\bf{metres}}} \right) = \frac{{8.64B}}{D}

Calculation:

Base period (B) = 240 days

Duty (D) = 1728 hec/cumec

Delta(Δ)=8.64 BD=8.64×2401728Delta\left( Δ \right) = \frac{{8.64\ B}}{{\rm{D }}} = \frac{{8.64 \times 240}}{{1728}}

Δ = 1.2 m = 120 cm

Question 22

Blaney-Criddle formula is used for

  1. AOption (a)

    estimation of groundwater recharge

  2. BOption (b)

    reservoir evaporation

  3. COption (c)

    Both of the above

  4. DOption (d)

    None of the above

View answer & explanation

Correct answer

Answer: (d)D.

None of the above

Why this answer is correct

Horton's Formula:

Infiltration is the process in which water enters into the soil and the rate at which water enters into the soil is known as infiltration rate. Under the ponding condition, the infiltration rate decreases exponentially, and it can be represented by the Horton equation. According to Horton’s equation,

\({\rm{f}} = {{\rm{f}}_{\rm{c}}} + \left( {{{\rm{f}}0} - {{\rm{f}}{\rm{c}}}} \right) \times {{\rm{e}}^{ - {\rm{kt}}}}\)

Where,

f = infiltration capacity at time t, f0 = initial infiltration capacity, fc = constant infiltration capacity after long time, and k= Horton’s decay constant

Lysimeter:

A Lysimeter is a measuring device that can be used to measure the amount of actual Soil moisture lost from the soil column (indirectly actual evapotranspiration) that is released by plants (usually crops or trees). By recording the amount of precipitation that an area receives and the amount lost through the soil, the amount of water lost to evapotranspiration can be calculated. Lysimeters are of two types: weighing and non-weighing.

Blaney-Criddle method: 

It is used to determine evapotranspiration i.e. consumptive use of the crop.

It is one of the oldest methods to measure evapotranspiration (Consumptive use of water).

Pan coefficient:

It is the ratio of lake evaporation and pan evaporation.

In order to convert pan evaporation value into lake evaporation value, a pan coefficient is used.

Lake evaporation = Pan Coefficient (Cp) × Pan Evaporation

Question 23

If the hydraulic radius of a river is 2 m and longitudinal bed slope of the channel is 0.0001, what will be the expected smallest size of the sediment particles on the bed of the river?

  1. AOption (a)

    About 4.5 mm

  2. BOption (b)

    About 2.2 mm

  3. COption (c)

    1.18 mm

  4. DOption (d)

    None of the above

View answer & explanation

Correct answer

Answer: (b)B.

About 2.2 mm

Why this answer is correct

Concept:

The minimum size of sediment that may remain stable in a channel with hydraulic radius R and bottom slope S, is

d ≥ 11 RS

Where R = hydraulic radius and S = bed slope

Calculation:

Given data,

R = 2 m = 2000 mm

S = 0.0001

The expected smallest size of the sediment particles on the bed of the river is

d ≥ 11 RS

d ≥ 11× 2000 × 0.0001

d ≥ 2.2 mm

Question 24

A natural river is conveying a flow of 2500 cumec. What is the expected wetted perimeter of the river cross-section?

  1. AOption (a)

    23.75 m

  2. BOption (b)

    2375 m

  3. COption (c)

    237.5 m

  4. DOption (d)

    327.5 m

View answer & explanation

Correct answer

Answer: (c)C.

237.5 m

Why this answer is correct

Explanation:

According to Lacey's theory, following formulae has been given below:

  1. Hydraulic mean depth, R=52V2fR = \frac{5}{2}\frac{{{V^2}}}{f}
  2. Area of the channel section, A=QVA = \frac{Q}{V}
  3. Wetted perimeter, P=4.75QP = 4.75\sqrt Q
  4. Regime mean velocity, V=(Qf2140)1/6V = {\left( {\frac{{Q{f^2}}}{{140}}} \right)^{1/6}}
  5. Bed Slope, S=f5/33340Q1/6S = \frac{{{f^{5/3}}}}{{3340{Q^{1/6}}}}

Calculation:

Q = 2500 cumec

P=4.75Q=4.752500P = 4.75\sqrt Q =4.75\sqrt {2500 } = 237.5 m

Question 25

The mathematical expression commonly used for spacing design of tile drainage system is called

  1. AOption (a)

    V. T. Chow equation

  2. BOption (b)

    Hooghoudt equation

  3. COption (c)

    Horton equation

  4. DOption (d)

    R. J. Garde equation

View answer & explanation

Correct answer

Answer: (b)B.

Hooghoudt equation

Why this answer is correct

Concept:

Tile drainage is a type of drainage system that removes excess water from soil below the surface. Plants need air as well as moisture in their root zones for their survival, excess irrigation farm water is free to move into the underground tile drains, this water, if not removed, retards the plant growth because It fills the soil voids and restricts proper aeration, Therefore there is a need for Sub-surface drainage or tile drainage.

The spacing of the drains is given by the Hooghoudt equation:

S=4kq(b2−a2)S = \frac{{4k}}{q}\left( {{b^2} - {a^2}} \right)

Where,

k = Coefficient of permeability

q = discharge per unit length in the drain

a = depth of the impervious strata below the drain

b = maximum height of the drained water table above the impervious layer

Question 26

Groundwater flow is analysed by using

  1. AOption (a)

    Darcy's law

  2. BOption (b)

    Reynolds' law

  3. COption (c)

    Newton's law

  4. DOption (d)

    Galerkin's principle

View answer & explanation

Correct answer

Answer: (a)A.

Darcy's law

Why this answer is correct

Concept:

Darcy's law:

Darcy's law states that there is a linear relationship between flow velocity (v) and hydraulic gradient (i) for any given saturated soil under steady laminar flow conditions.

It is used for groundwater flow.

If the rate of flow is q (volume/time) through a cross-sectional area (A) of the soil mass,

Darcy's Law can be expressed as:

v = q/A = k × i

Darcy’s law is valid for fine saturated sands only.

Where

k = permeability of the soil

i = Δh/L

Δh = difference in total heads

L = length of the soil mass

Question 27

Given dmm = average particle size in mm, Lacey's silt factor is calculated as

  1. AOption (a)

    f=1.72dmmf = 1.72\sqrt {{d_{mm}}}

  2. BOption (b)

    f=1.75dmmf = 1.75\sqrt {{d_{mm}}}

  3. COption (c)

    f=1.76dmmf = 1.76\sqrt {{d_{mm}}}

  4. DOption (d)

    None of the above

View answer & explanation

Correct answer

Answer: (c)C.

f=1.76dmmf = 1.76\sqrt {{d_{mm}}}

Why this answer is correct

Explanation:

According to Lacey’s, the design formulas to build canal is as follows:

  1. Silt factor ⇒ f=1.76dmm\rm{f = 1.76\sqrt {{d_{mm}}}}

  2. Velocity of flow ⇒ V=[Qf2140]16\rm{V = {\left[ {\frac{{Q{f^2}}}{{140}}} \right]^{\frac{1}{6}}}}

  3. Hydraulic mean depth ⇒ R=5V22f\rm{R = \frac{{5{V^2}}}{{2f}}}

  4. Wetted perimeter ⇒ P=4.75Q\rm{P = 4.75\sqrt Q}

  5. Bed slope ⇒ S=f533340×Q16\rm{S = \frac{{{f^{\frac{5}{3}}}}}{{3340 \times {Q^{\frac{1}{6}}}}}}

Question 28

Armoring is defined as

  1. AOption (a)

    the process of strengthening the embankment of rivers

  2. BOption (b)

    the process of sand deposition on the river bed

  3. COption (c)

    the process of bank erosion in rivers

  4. DOption (d)

    the process of transport of smaller size particles from the river bed surface leaving only the larger size particles during sediment transport

View answer & explanation

Correct answer

Answer: (d)D.

the process of transport of smaller size particles from the river bed surface leaving only the larger size particles during sediment transport

Why this answer is correct

Explanation:

Armoring

  • The process of transport of smaller size particles from the river bed surface leaves only the larger size particles during sediment transport.

Armoring Process

  • Flow develops shear stresses less than required to move large particles, but large enough to move fines
  • Flow entrains fine particles, winnowing them from the bed surface
  • Coarse layer forms, sheltering fine grains
  • The coarse layer increases resistance to entrainment

Question 29

A general mobile-boundary channel can be considered to have which of the following degrees of freedom?

  1. AOption (a)

    Depth of flow, bed width, side slope and longitudinal slope

  2. BOption (b)

    Depth of flow, bed width, side slope and planiform

  3. COption (c)

    Depth of flow, bed width, longitudinal slope and planiform

  4. DOption (d)

    Depth of flow, bed width, planiform and sediment grading

View answer & explanation

Correct answer

Answer: (c)C.

Depth of flow, bed width, longitudinal slope and planiform

Why this answer is correct

Explanation:

Rigid and Mobile Boundary Channels:

  • Rigid channels are those in which the boundary is not deformable.
  • The shape and roughness magnitudes are not functions of flow parameters.
  • For example, lined canals and nonerodible unlined canals.
  • In Rigid channels, the flow velocity and shear stress distribution will be such that no major scouring, erosion, or deposition will take place in the channel and the channel geometry and roughness are essentially constant with respect to time.
  • When the boundary of the channel is mobile and flow carries considerable amounts of sediment through suspension and is in contact with the bed. Such channels are classified as mobile channels.
  • In the mobile channel, not only depth of flow but also bed width, and longitudinal slope of a channel may undergo changes with space and time depending on the type of flow. The resistance to flow, the quantity of sediment transported and channel geometry all depend on the interaction of flow with channel boundaries.

Question 30

Identify the more correct statement.

  1. AOption (a)

    Alternate depth is related to specific energy equation and sequent depth is related to specific force equation.

  2. BOption (b)

    Alternate depth is related to specific force equation and sequent depth is related to specific energy equation. 

  3. COption (c)

    Alternate depth is related to total energy equation and sequent depth is related to total force equation.

  4. DOption (d)

    Alternate depth is related to total momentum equation and sequent depth is related to total energy equation.

View answer & explanation

Correct answer

Answer: (a)A.

Alternate depth is related to specific energy equation and sequent depth is related to specific force equation.

Why this answer is correct

Explanation:

Critical Depth: The condition of minimum specific energy is known as the critical-flow condition and the corresponding depth (yc) is known as critical depth.

Alternate Depths: The two possible depths (y1, y2­) have the same specific energy are known as alternate depths.

Sequent/Conjugate Depths: The two possible depths (y1, y2­) have the same specific force are known as sequent depths.

Normal depth: The depth of flow in a channel when the slope of the water surface and channel bottom is the same and the water depth remains constant is known as normal depth.

Question 31

Identify the correct statement.

  1. AOption (a)

    At critical flow through a channel section, the specific energy is minimum and the specific force is minimum for a given discharge.

  2. BOption (b)

    At critical flow through a channel section, the alternate depths merge to a single value but the sequent depths differ.

  3. COption (c)

    At critical flow through a channel section, any surface disturbance wave always propagates in both upstream and downstream directions.

  4. DOption (d)

    At critical flow through a channel section, the specific force is equal to the specific energy.

View answer & explanation

Correct answer

Answer: (a)A.

At critical flow through a channel section, the specific energy is minimum and the specific force is minimum for a given discharge.

Why this answer is correct

As can be observed from the graph, specific energy is minimum for a given discharge at a critical state.

As can be observed from the graph, at critical state discharge is maximum for a given specific energy.

At critical state, specific force is minimum for a given discharge.

Question 32

What is the qualification of uniform flow?

  1. AOption (a)

    Slope of the energy line ≠ slope of the water surface = slope of the channel bed

  2. BOption (b)

    Slope of the energy line ≠ ∝ and slope of the water surface = slope of the channel bed

  3. COption (c)

    Slope of the energy line = slope of the water surface = slope of the channel bed = ± a constant

  4. DOption (d)

    Slope of the energy line = slope of the water surface = slope of the channel bed

View answer & explanation

Correct answer

Answer: (d)D.

Slope of the energy line = slope of the water surface = slope of the channel bed

Why this answer is correct

Explanation

In the uniform flow in a channel of a small bed slope, the hydraulic grade line coincides with the water surface line.

Important Points While analyzing the GVF, the following assumptions are made:

  1. Region of high curvature is excluded from analysis and pressure distribution is assumed to be hydrostatic.
  2. Resistance to flow at any depth is given by uniform flow equations such as Chezy’s equation or Manning’s formula by replacing bed slope with energy slope. 3. Slope of the channel is small i.e Hydraulic grade line coincides with the water surface line.

Additional Information

 Energy loss in GVF can be written as

So = dz/dx, ( Bed slope)

Sf = dH/dx (Slope of total energy line)

We know Total Head (H) = z + y + v2/2g = z + E

E is specific energy, z is datum head

On differentiating We get

dH/dx = dE/dx + dz/dx

So, dE/dx = dH/dx - dz/dx = Sf - S0

We know upon interacting the end result we get is (Considering manning substitution)

dydx=So×[1−(y0y)103][1−(ycy)3]\dfrac{{dy}}{{dx}} = {S_o} \times \dfrac{{\left[ {1 - {{\left( {\frac{{{y_0}}}{y}} \right)}^{\frac{{10}}{3}}}} \right]}}{{\left[ {1 - {{\left( {\frac{{{y_c}}}{y}} \right)}^3}} \right]}}

From this 

we can say Energy Loss is directly proportional to Head x (yx)

So, (here head is given in terms of 'd')

dE/dx = Kdx

Where, d is depth of flow and K is constant.

  1. Roughness coefficients C and N are independent of depth of flow.

Question 33

For an open rectangular channel, the Froude number at critical flow is

  1. AOption (a)

    0

  2. BOption (b)

    0.5

  3. COption (c)

    1.0

  4. DOption (d)

    None of the above

View answer & explanation

Correct answer

Answer: (c)C.

1.0

Why this answer is correct

Explanation:

Critical flow

It is defined as the flow at which specific energy is minimum or the flow corresponding to critical depth is defined as critical flow.

Relation of critical velocity with critical depth is:

Vc=g×hc{V_c} = \sqrt {g \times {h_c}}

Vcg×hc=Froude;number=1\frac{{{V_c}}}{{\sqrt {g \times {h_c}} }} = Froude;number = 1

Froude number is 1 for critical flow.

Tranquil flow or streaming flow or sub-critical flow

  • When the depth of a flow in a channel is greater than critical depth (hc), the flow is said to be sub-critical flow.
  • Froude number is less than 1 for sub-critical flow.
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Torrential flow or shooting flow or super-critical flow

  • When the depth of a flow in a channel is less than critical depth (hc), the flow is said to be a super-critical flow.
  • Froude number is greater than 1 for super-critical flow.

Question 34

A rectangular open channel having bed width 2 m is laid at a bed slope - 0.0009 and carries a flow of 10 cumec. What is its normal depth of flow?

  1. AOption (a)

    It needs a calculator for doing the calculations and hence one cannot answer it now

  2. BOption (b)

    Many data are missing in the problem statement

  3. COption (c)

    The problem statement is wrong

  4. DOption (d)

    The uniform flow does not exist for the channel

View answer & explanation

Correct answer

Answer: (d)D.

The uniform flow does not exist for the channel

Why this answer is correct

Concept:

As per Manning's equation, the discharge in an open channel is given by

Q=R23S122×AQ = \frac{{{R^{\frac{2}{3}}}{S^{\frac{1}{2}}}}}{2} \times A

Where A = Area of Flow, R = Hydraulic radius, S = Bed Slope of the channel,

When the slope (S) < 0, So the slope is adverse slope

So, uniform discharge or depth cannot be established when the slope is negative

Classification of the channel:

Channel categorySymbolCharacteristic conditionRemark
Mild slopeMYo > YcSubcritical at normal depth
Steep slopeSYo < YcSupercritical flow at normal depth
Critical slopeCYo > YcCritical depth at normal depth
Horizontal bedHSo = 0Cannot sustain uniform flow
Adverse slopeASo < 0Cannot sustain uniform flow

Question 35

The equation relating the pre-and post-hydraulic jump flow depths for rectangular channel is given by

Assume standard nomenclature of the terms used above.

  1. AOption (a)

    y1y2=−1+1+8Fr122\frac{{{y_1}}}{{{y_2}}} = \frac{{ - 1 + \sqrt {1 + 8Fr_1^2} }}{2}

  2. BOption (b)

    y1y2=+1+1+8Fr122\frac{{{y_1}}}{{{y_2}}} = \frac{{ + 1 + \sqrt {1 + 8Fr_1^2} }}{2}

  3. COption (c)

    y2y1=+1+1+8Fr122\frac{{{y_2}}}{{{y_1}}} = \frac{{ + 1 + \sqrt {1 + 8Fr_1^2} }}{2}

  4. DOption (d)

    y2y1=−1+1+8Fr122\frac{{{y_2}}}{{{y_1}}} = \frac{{ - 1 + \sqrt {1 + 8Fr_1^2} }}{2}

View answer & explanation

Correct answer

Answer: (d)D.

y2y1=−1+1+8Fr122\frac{{{y_2}}}{{{y_1}}} = \frac{{ - 1 + \sqrt {1 + 8Fr_1^2} }}{2}

Why this answer is correct

Concept:

Hydraulic jump: Whenever supercritical flow merges into subcritical flow then to reduce the energy a jump is formed, called a hydraulic jump.

For a rectangular frictionless channel, both depths are related as:

y2y1=12[1+8F12−1]\frac{{{{\rm{y}}_2}}}{{{{\rm{y}}_1}}} = \frac{1}{2}\left[ {\sqrt {1 + 8{{\rm{F}}_1}^2} - 1} \right]

Froude number (F) = Q2TgA3\sqrt {\frac{{{{\rm{Q}}^2}{\rm{T}}}}{{{\rm{g}}{{\rm{A}}^3}}}}

Where,

Y1 = Pre jump depth, Y2 = Post jump depth, F1= Froude Number before jump, Q = Discharge, T = top width, and A = area of flow

Length of hydraulic jump (L) is calculated empirically by: L = 7 × (Y2 - Y1)

Question 36

The most popular form of gradually varied flow equation is expressed as

Assume standard nomenclature of the terms used above.

  1. AOption (a)

    dydx=S0−Sf1−αFr2\frac{{dy}}{{dx}} = \frac{{{S_0} - {S_f}}}{{1 - \alpha F{r^2}}}

  2. BOption (b)

    dydx=Sf−S01−αFr2\frac{{dy}}{{dx}} = \frac{{{S_f} - {S_0}}}{{1 - \alpha F{r^2}}}

  3. COption (c)

    dxdy=S0−Sf1+αFr2\frac{{dx}}{{dy}} = \frac{{{S_0} - {S_f}}}{{1 + \alpha F{r^2}}}

  4. DOption (d)

    dxdy=S0−Sf1−αFr2\frac{{dx}}{{dy}} = \frac{{{S_0} - {S_f}}}{{1 - \alpha F{r^2}}}

View answer & explanation

Correct answer

Answer: (a)A.

dydx=S0−Sf1−αFr2\frac{{dy}}{{dx}} = \frac{{{S_0} - {S_f}}}{{1 - \alpha F{r^2}}}

Why this answer is correct

Explanation:

In gradually varied flow, the total head H of the liquid at any cross-section can be written as

H=zb+y+V22gH = {z_b} + y + \frac{{{V^2}}}{{2g}}.........(1)

Differentiating H with respect to x, we get

dHdx=dzbdx+dydx+ddx(V22g)\frac{{dH}}{{dx}} = \frac{{d{z_b}}}{{dx}} + \frac{{dy}}{{dx}} + \frac{d}{{dx}}\left( {\frac{{{V^2}}}{{2g}}} \right)

dHdx\frac{{dH}}{{dx}} represent the energy slope (Sf). Since total energy always decreases in the direction of motion, therefore

dHdx=−Sf\frac{{dH}}{{dx}} = - {S_f}

dzbdx\frac{{d{z_b}}}{{dx}} denotes the bottom slope. It is common to consider the channel slope with bed elevations decreasing in the downstream direction, therefore

dzbdx=−So\frac{{d{z_b}}}{{dx}} = - {S_o}

dydx\frac{{dy}}{{dx}} denoting water surface slope relative to the bottom of the channel.

ddx(V22g)=ddy(Q22gA2)dydx=−Q2gA3dAdydydx=−−Q2TgA3dydx\frac{d}{{dx}}\left( {\frac{{{V^2}}}{{2g}}} \right) = \frac{d}{{dy}}\left( {\frac{{{Q^2}}}{{2g{A^2}}}} \right)\frac{{dy}}{{dx}} = - \frac{{{Q^2}}}{{g{A^3}}}\frac{{dA}}{{dy}}\frac{{dy}}{{dx}} = - - \frac{{{Q^2}T}}{{g{A^3}}}\frac{{dy}}{{dx}}

Now, equation (1) will become,

−Sf=−So+dydx−Q2TgA3dydx- {S_f} = - {S_o} + \frac{{dy}}{{dx}} - \frac{{{Q^2}T}}{{g{A^3}}}\frac{{dy}}{{dx}}

Fr=Vgy=QAgAT=QTgA3{F_r} = \frac{V}{{\sqrt {gy} }} = \frac{Q}{{A\sqrt {g\frac{A}{T}} }} = \frac{{Q\sqrt T }}{{\sqrt {g{A^3}} }}

So, the equation of gradually varied flow is expressed as

dydx=So−Sf1−Q2TgA3=So−Sf1−Fr2\frac{{dy}}{{dx}} = \frac{{{S_o} - {S_f}}}{{1 - \frac{{{Q^2}T}}{{g{A^3}}}}} = \frac{{{S_o} - {S_f}}}{{1 - F_r^2}}

The same equation, when the velocity correction factor(α) is taken into consideration, can be written as

dydx=So−Sf1−Q2Tα×g×A3=So−Sf1−α×Fr2\frac{{dy}}{{dx}} = \frac{{{S_o} - {S_f}}}{{1 - \frac{{{Q^2}T}}{{α \times g\times {A^3}}}}} = \frac{{{S_o} - {S_f}}}{{1 - α \times F_r^2}}

Question 37

Equation for critical flow by using usual terms is given by

  1. AOption (a)

    Q2TgA3\frac{{{Q^2}T}}{{g{A^3}}} = 1

  2. BOption (b)

    Q2TgP3\frac{{{Q^2}T}}{{g{P^3}}} = 1

  3. COption (c)

    Q2PgA3\frac{{{Q^2}P}}{{g{A^3}}}  = 1

  4. DOption (d)

    Q2TgD3\frac{{{Q^2}T}}{{g{D^3}}}  = 1

View answer & explanation

Correct answer

Answer: (a)A.

Q2TgA3\frac{{{Q^2}T}}{{g{A^3}}} = 1

Why this answer is correct

Concept

For constant specific energy, the discharge will be maximum when the flow is critical and it satisfies the following condition:

Q2TgA3=1\frac{{{Q^2}T}}{{g{A^3}}} = 1

Where,

Q is the discharge

T is the top width of the channel

A is the wetted area of the channel

Specific energy under critical conditions for a given critical depth yc is given as:

Ec=yc+;V22g{E_c} = {y_c} + ;\frac{{{V^2}}}{{2g}}

Where,

V is the velocity of flow

yc is the depth flow at a particular section

Question 38

The most economic cross-section is the one

  1. AOption (a)

    having minimum velocity for a given cross-section

  2. BOption (b)

    having minimum wetted perimeter for a given flow cross-sectional area

  3. COption (c)

    Both of the above

  4. DOption (d)

    None of the above

View answer & explanation

Correct answer

Answer: (b)B.

having minimum wetted perimeter for a given flow cross-sectional area

Why this answer is correct

Concept:

Most economical or most efficient or best section of the channel

  • A channel section is said to be economical when it can pass a maximum discharge for a given cross-sectional area, resistance coefficient, and bottom slope.
  • Discharge, Q=A.1nR2/3So1/2Q = A.\frac{1}{n}{R^{2/3}}S_o^{1/2}
  • Discharge is maximum when Hydraulic radius (R) is maximum and wetted perimeter is minimum (R = A/P)
  • A section of a channel is said to be most economical when the cost of construction of the channel is minimum. But the cost of construction of a channel depends upon the excavation and the lining. To keep the cost down or minimum, the wetted perimeter, for a given discharge should be minimum. This condition is utilized for determining the dimensions of a economical sections of different form of channels.

Question 39

Width of carriageway recommended by the IRC for two-lane road with raised kerbs is

  1. AOption (a)

    7.0 m

  2. BOption (b)

    7.25 m

  3. COption (c)

    7.5 m

  4. DOption (d)

    7.75 m

View answer & explanation

Correct answer

Answer: (c)C.

7.5 m

Why this answer is correct

Explanation:

Width of Pavement or Carriageway:

  • The pavement width depends upon the number of lanes and the width of the single lane required. The lane width determines on the basis of the width of the vehicle and minimum side clearance.
  • The maximum permissible width of the vehicle as per IRC specification is 2.44 m.
Class of RoadWidth of Carriageway
Single lane3.75 m
Two lanes, without raised kerbs7.0 m
Two lanes, with raised kerbs7.5 m
Intermediate carriageway ( except on important road )5.5 m
Multi-lane pavement3.5 m per lane

Important Points

The intermediate carriageway is more than one lane but less than two lanes. This is only provided to allow maneuvering.

Question 40

While aligning a road in a hilly area with a ruling gradient of 6.5%, a horizontal curve of radius 60 m is encountered. Find the compensated gradient at the curve.

  1. AOption (a)

    5%

  2. BOption (b)

    5.25%

  3. COption (c)

    1.5%

  4. DOption (d)

    1.25%

View answer & explanation

Correct answer

Answer: (b)B.

5.25%

Why this answer is correct

Concept:-

IRC Recommendations for grade compensation:

IRC gave the following guidelines for grade compensations:-

  1. Grade compensation is not required for grades flatter than 4% because the loss of tractive force is negligible.
  2. Grade compensation is 30+RR%\frac{{30 + {\rm{R}}}}{{\rm{R}}}{\rm{\% }}, where R – radius of the horizontal curve in m.
  3. The maximum grade compensation is limited to 75R%\frac{{75}}{{\rm{R}}}{\rm{\% }}

Calculation:

Given:

⇒GC=30+6060≤7560\Rightarrow {\rm{GC}} = \frac{{30 + 60}}{{60}} \le \frac{{75}}{{60}}

⇒ GC = 1.25 %

∴ The compensated gradient is 6.5 - 1.25 = 5.25

Check: Compensated Gradient (C.G) should be ≥ 4 % and 5.25 > 4

Hence, okay

Question 41

What is the PCU value of a bullock cart as per recommendation made by IRC?

  1. AOption (a)

    7.0

  2. BOption (b)

    8.0

  3. COption (c)

    3.0

  4. DOption (d)

    4.0

View answer & explanation

Correct answer

Answer: (b)B.

8.0

Why this answer is correct

Concepts:

Passenger car unit (PCU):

  • A Passenger Car Unit is a measure of the impact that a mode of transport has on traffic variables (such as headway, speed, and density) compared to a single standard passenger car. PCU is a vehicle unit used for expressing highway capacity.
  • For instance, one car is considered as a single unit, cycle, the motorcycle is considered a half car unit. Bus and truck cause a lot of inconvenience because of their large size and are considered equivalent to 3.5 cars or 3.5 PCU.
  • The PCU of various vehicles as per IRC is given below:
VehiclePCU
Car or taxi1
Motor Cycle0.5
Bus or truck3.5
Cycle0.2
Bullock Cart6
Bullock Cart (large)8

Question 42

One passage of an overloaded vehicle with 3 times the standard load will cause damage equivalent to N passages of the standard load. Here N =

  1. AOption (a)

    81

  2. BOption (b)

    18

  3. COption (c)

    36

  4. DOption (d)

    63

View answer & explanation

Correct answer

Answer: (a)A.

81

Why this answer is correct

Concepts:

The equivalent axle load factor (EALF) defines the damage per pass to a pavement in relation to the damage caused by a standard axle load. The standard axle load depends on the type of axle.

As per the fourth power law, EALF is determined as:

EALF=(PPS)4EALF = {(\frac{P}{P_S})}^ 4

Where

Ps is standard axle load

P = actual load on pavement

Calculation:

Let Standard load , Ps =  X kN then

Overloaded Vehicle load , P = 3 X ( Given)

EALF=(3XX)4EALF = {(\frac{3X}{X})}^ 4

EALF = 81

This means that one passage of an overloaded vehicle with 3 times the standard load will cause damage equivalent to 81 passages of the standard load.

Question 43

When a circular load of radius a with uniform pressure p is applied in the surface of a homogeneous layer, the vertical stress σ*z* under the centre of the load at depth z is given by

  1. AOption (a)

    σz = p [1−z2(a3+z3)2/3]\left[ {1 - \frac{{{z^2}}}{{{{\left( {{a^3} + {z^3}} \right)}^{2/3}}}}} \right]

  2. BOption (b)

    σz = p[1−z3(a2+z2)3/2]\left[ {1 - \frac{{{z^3}}}{{{{\left( {{a^2} + {z^2}} \right)}^{3/2}}}}} \right]

  3. COption (c)

    σz = p[1−z1.5(a2+z2)3/4]\left[ {1 - \frac{{{z^{1.5}}}}{{{{\left( {{a^2} + {z^2}} \right)}^{3/4}}}}} \right]

  4. DOption (d)

    σz = p[1−z2(a2+z2)]\left[ {1 - \frac{{{z^2}}}{{\left( {{a^2} + {z^2}} \right)}}} \right]

View answer & explanation

Correct answer

Answer: (b)B.

σz = p[1−z3(a2+z2)3/2]\left[ {1 - \frac{{{z^3}}}{{{{\left( {{a^2} + {z^2}} \right)}^{3/2}}}}} \right]

Why this answer is correct

Concepts:

The vertical stress at a point directly under the center due to uniform circular load is given as:

σz = p × ( 1- Cos3θ)

where

 p is the distributed load

and

Cosθ=Z√a2+Z2Cosθ = \frac{Z}{√{a^2 + Z^2}}

The substituting the value of Cosθ in the above equation we get

σz=p[1−Z3(a2+Z2)3/2]σ_z = p [1- \frac{Z^3}{(a^2 + Z^2)^ {3/2}}]

Question 44

  1. Basic capacity
  2. Possible capacity
  3. Traffic flow

R. It is the maximum number of vehicles that can pass a given point on a lane or roadway during one hour under prevailing roadway and traffic conditions.

S. It is the number of vehicles moving in a specified direction on a given lane or roadway that passes a given point or cross-section during specified unit of time.

T. It is the maximum number of passenger cars that can pass a given point on a lane or roadway during one hour under nearly ideal roadway and traffic conditions which can possibly be attained.

Select the correct combination:

  1. AOption (a)

    1-T, 2-S, 3-R.

  2. BOption (b)

    1-R, 2-T, 3-S.

  3. COption (c)

    1-T, 2-R, 3-S.

  4. DOption (d)

    1-R, 2-S, 3-T.

View answer & explanation

Correct answer

Answer: (c)C.

1-T, 2-R, 3-S.

Why this answer is correct

Concepts:

1) Traffic Volume:

  • It is the no. of vehicles that can be passed through a particular section of road during a given period. The period may be hours, minutes, seconds, days or weeks, or any other as per the requirement.
  • Generally, it is expressed in Veh/hr, if the period chosen is hours.

2) Basic Traffic capacity:

  • It is the maximum possible no of vehicles that can be passed through a particular section of road during a given period under most ideal conditions.
  • It is given as:

QB=1000S×VQ_B = \frac{1000}{S} \times V

Where

S is the space headway in ‘m’

QB is the Basic Traffic Capacity in Veh/hr

V is the Traffic Speed in km/hr

3) Design or Practical or Possible Traffic capacity: 

  • It is the maximum possible no of vehicles that can be passed through a particular section of road during a given period during the prevalent conditions on the road.
  • It is always less than or equal to Basic Traffic Capacity.

Question 45

The following gives the penetration grades of the bitumen binder and corresponding equivalent viscosity grades. Select the correct option:

Penetration grades of bitumen binderEquivalent viscosity grade
(A)(B)(C)(D)
80-100VG-10VG-80VG-100VG-90
70-80VG-20VG-70VG-80VG-75
60-70VG-30VG-60VG-70VG-65
30-40VG-40VG-30VG-40VG-35
  1. AOption (a)

    (A)

  2. BOption (b)

    (B)

  3. COption (c)

    (C)

  4. DOption (d)

    (D)

View answer & explanation

Correct answer

Answer: (a)A.

(A)

Why this answer is correct

Concepts:

The penetration grade of bitumen binder, equivalent viscosity grade, and their general application are given below in the tabulated form:

Penetration grade of bitumen binder,Equivalent viscosity gradeApplication
80-100VG 10Used in spraying applications such as surface dressing and paving in very cold climatic region
70-80VG 20Used in cold climatic, high-altitude regions
60-70VG 30Used in paving in normal situations
30-40VG 40Used in highly stressed areas such as those in intersections, near toll booths, and truck parking tolls.

Question 46

The aggregates used for construction of BM base course shall fulfill the following properties. Select the correct option:

PropertyOptions
(A)(B)(C)(D)
Loss Angeles abrasion value:<40%<30%<40%<40%
Aggregate impact value:<40%<40%<25%<30%
Combined flakiness and elongation index:<30%<30%<10%<40%
Water absorption:5% to 3%0%<1%<2%
  1. AOption (a)

    (A)

  2. BOption (b)

    (B)

  3. COption (c)

    (C)

  4. DOption (d)

    (D)

View answer & explanation

Correct answer

Answer: (d)D.

(D)

Why this answer is correct

Concepts:

As per IRC, specification for bituminous macadam are:

1.  The coarse aggregate shall consist of crushed rock, crushed gravel, or other gravel or other hard material retained on a 2.36 mm sieve. The aggregate shall satisfy and physical requirements:

PropertyTestRequirements
CleanlinessGrain size analysisMax 5 % passing through 0.075μ sieve.
Particle ShapeCombined flakiness and elongation IndexMax. 40 %
StrengthLos Angeles Abrasion ValueMax. 40 %
Aggregate Impact valueMax. 30 %
Water AbsorptionWater AbsorptionMax 2 %
StrippingCoating and stripping of Bitumen AggregateMinimum retained coating 95 %2.

 

  1. Fine aggregate shall consist of crushed or naturally occurring mineral material, or a combination of two, passing a 2.36 mm sieve and retained on a 75-micron sieve. 3.  The amount of rounded, natural sand in the total fine aggregate shall be limited to 10% if the BM is used within 100mm from the road surface and to 50% if the BM is used more than 100 mm below the road surface.

Question 47

Given that Ps = saturation population, P0 = population at the starting point, P = population at any time t from the starting point, K = constant, t = time in years after the starting point, the population at any time t from the starting point is computed according to the logistic curve model as

  1. AOption (a)

    loge(Ps−PP)\left( {\frac{{{P_s} - P}}{P}} \right)  – loge  (Ps−P0P0)\left( {\frac{{{P_s} - {P_0}}}{{{P_0}}}} \right)  = – KPs·t

  2. BOption (b)

    loge(Ps−PP)\left( {\frac{{{P_s} - P}}{P}} \right)  + loge (Ps−P0P0)\left( {\frac{{{P_s} - {P_0}}}{{{P_0}}}} \right)  = KPs·t

  3. COption (c)

    loge (Ps−PP)\left( {\frac{{{P_s} - P}}{P}} \right) – loge (Ps−P0P0)=(KPs)t \left( {\frac{{{P_s} - {P_0}}}{{{P_0}}}} \right) = {(KP_s)}^t

  4. DOption (d)

    None of the above

View answer & explanation

Correct answer

Answer: (a)A.

loge(Ps−PP)\left( {\frac{{{P_s} - P}}{P}} \right)  – loge  (Ps−P0P0)\left( {\frac{{{P_s} - {P_0}}}{{{P_0}}}} \right)  = – KPs·t

Why this answer is correct

Concepts:

Logistic curve method:

  • A logistic curve is an S-shaped curve that is used to model population forecasting that increases initially, rapidly then, and slowly later on and levels off or reached a saturation value after a period of time.

  • A logistic curve is the graphical representation of population growth with time which is shown as:

  • This method is used when the growth rate of the population due to births, deaths, and migrations takes place under normal situations and it is not subjected to any extraordinary changes like epidemics, war, earthquakes or natural disasters, etc.

  • Verhulst gives a mathematical solution for a logistic curve which can be represented as:

 loge(Ps−PP)\left( {\frac{{{P_s} - P}}{P}} \right)  – loge  (Ps−P0P0)\left( {\frac{{{P_s} - {P_0}}}{{{P_0}}}} \right)  = – KPs·t

that Ps = saturation population, P0 = population at the starting point, P = population at any time t from the starting point, K = constant, t = time in years after the starting point

Question 48

According to the Indian Standard Code of Practices, the minimum total domestic consumption for a town or a city with full flushing system should be taken in terms of litres per head per day (l/h/d) as

  1. AOption (a)

    135 l/h/d

  2. BOption (b)

    170 l/h/d

  3. COption (c)

    235 l/h/d

  4. DOption (d)

    200 l/h/d

View answer & explanation

Correct answer

Answer: (d)D.

200 l/h/d

Why this answer is correct

Concepts:

Water Supply requirements as per IS 1172:1993:

  1. A minimum of 70 to 100 liters per head per day may be considered adequate for the domestic needs of urban communities, apart from non-domestic needs such as flushing requirements.
  2. For communities having a population up to 20,000 without a flushing system and water is delivered through stand post and water supply through house service connection is    40 lpcd and 70 to 100 lpcd respectively.
  3. For communities having populations between 20,000 to 100,000 together with a full flushing system, the minimum water supply considered is 100 to 150 lpcd/l/h/d.
  4. For communities having a population above 100,000 together with a full flushing system, the minimum water supply considered is 150 to 200 lpcd.  This value may be reduced to 135 lpcd for houses for Low Income Group (LIG) and Economical Weaker Section (EWG) depending upon prevailing conditions.

Under ordinary conditions, the minimum water demand for a town or city with a full flushing system can be taken as  200 lpcd or l/h/d

Question 49

All hand-operated pumps for lifting water from tube wells are

  1. AOption (a)

    handle pump

  2. BOption (b)

    piston pump

  3. COption (c)

    cylinder pump

  4. DOption (d)

    reciprocating pump

View answer & explanation

Correct answer

Answer: (d)D.

reciprocating pump

Why this answer is correct

Concepts:

All hand-operated pumps for lifting water from tube wells are of the reciprocating type.

Reciprocating pumps:

  • They are normally used for drinking water supply and irrigation as well.
  • The main parts of the reciprocating pumps are:

      1) Pump cylinder in which an airtight piston or plunger moves up and down with the help of a pump rod, 

       2) Valves - Handle the operation of the pump.

       3) Pipe and stainer

  • Working: As the plunger rises, water is drawn through a non-return valve at the bottom of the cylinder into the cylinder, and on the downward stroke the water is released to the upper side of the plunger. On the next upward movement of the plunger, water is raised to the pump head and discharged through the spout.
  • They can be operated manually, with animal power and auxiliary power sources.

Pump Cylinder:

When the water level is very deep the pump cylinder has to be lowered close to the water surface to reduce the suction head and then the reciprocating pump is used to lift the water from underground sources like tube well.

Centrifugal Pumps:

  • They are the most commonly used pumps for irrigation in agriculture.
  • They are operated with either an electric motor or an engine.
  • Main Parts:  casing, impeller, high tensile shaft, bearing pedestal, stuffing box, flanges, and coupling.
  • Working: Upon rotation of the impeller, the water enters the eye, which is thrown radially outward to the periphery. Such action causes a vacuum in the eye and thus more water enters the suction pipe to maintain the continuous flow. The impeller accelerates the water to a high velocity and the casing converts this velocity head into a pressure head and water delivers to the desired level due to this high-pressure head.

Question 50

Identify the correct statement from the following:

  1. AOption (a)

    The uniformity coefficient value for a slow sand filter varies from 2.5 to 3.3 while that for a rapid gravity filter it varies from 1.3 to 1.7.

  2. BOption (b)

    The uniformity coefficient value for a slow sand filter varies from 1.8 to 2.5 while that for a rapid gravity filter it varies from 1.3 to 1.7.

  3. COption (c)

    The uniformity coefficient value for a slow sand filter varies from 1.5 to 2.3 while that for a rapid gravity filter it varies from 1.1 to 1.4.

  4. DOption (d)

    The uniformity coefficient value for a slow sand filter varies from 1.8 to 2.5 while that for a rapid gravity filter it varies from 1.1 to 1.4.

View answer & explanation

Correct answer

Answer: (b)B.

The uniformity coefficient value for a slow sand filter varies from 1.8 to 2.5 while that for a rapid gravity filter it varies from 1.3 to 1.7.

Why this answer is correct

Explanation:

The difference between Slow Sand Filter (SSF) and Rapid Sand Filter (RSF) is given below:

ParameterSSFRSF
Rate of filtrationLow - 100-200 I/m2/hHigh - 3000-6000 I/m2/h
Depth of filter Media90 – 110 cm60-90 cm
Coefficient of Uniformity1.8 - 2.5 or 31.2-1.7
Effective Size0.2 – 0.35 mm0.35 – 0.6 mm
Gradation of filter mediaWell GradedPoorly Graded
Base MaterialBase Material is gravel and laid in 4 to 5 layers.Also gravel but laid in 5 to 6 layers.
Base Material thickness30-75 cm thick6- 90 cm thick
Cleaning period of filter mediaAfter 20-30 daysNo Cleaning; backwashing is done.
Surface or plan area of the tank50– 1000 sqm20 – 80 sqm
Depth of tank2.5 – 4 m2.5 – 3.5 m
EfficiencyRemoves 98 to 99 % bacteria80 -90 % removal of bacteria
Post TreatmentDisinfection is not a mustDisinfection is a must.
Turbidity removalUp to 50 ppmUp to 35 ppm

50 questions in the full paper