Previous Year Paper

BPSC AE Paper 6 (Civil) 11 Nov 2022 Official Paper (Previous Year Paper)

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Full Test (50 questions)

Question 1

When a canal flowing under pressure is carried below a natural drainage such that its FSL does not touch the underside of the supporting structure, the structure so provided, is called 

  1. AOption (a)

    aqueduct

  2. BOption (b)

    superpassage

  3. COption (c)

    siphon-aqueduct

  4. DOption (d)

    siphon

View answer & explanation

Correct answer

Answer: (b)B.

superpassage

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 2

A channel has three reaches P, Q and R connected in series with the following properties:

ReachBed slopeNormal depth (in m)Critical depth (in m)
P0.00052.271.29
Q0.0090.821.29
R0.0051.171.29
<br>

What gradually varied flow profiles will occur?

  1. AOption (a)

    M2, S2, S3

  2. BOption (b)

    M1, S1, S3

  3. COption (c)

    M1, H2, S2

  4. DOption (d)

    M2, S1, S3

View answer & explanation

Correct answer

Answer: (a)A.

M2, S2, S3

Why this answer is correct

Explanation:

  1. Reach P:
  • Slope: Mild (0.0005)
  • Normal depth (yn): 2.27 m
  • Critical depth (yc): 1.29 m
  • Since yn​>yc​and the slope is mild, the flow profile is M2.
  1. Reach Q:
  • Slope: Steep (0.009)
  • Normal depth (yn): 0.82 m
  • Critical depth (yc): 1.29 m
  • Since yn​<yc​ and the slope is steep, the flow profile is S2.
  1. Reach R:
  • Slope: Steep (0.005)
  • Normal depth (yn): 1.17 m
  • Critical depth (yc): 1.29 m
  • Since yn​<yc  and the slope is steep, the flow profile is S3.

Question 3

At a gauging site, if Qr is the discharge at a stage when the water surface is rising and Qf the discharge at the same stage when the water surface is falling, then

  1. AOption (a)

    Qr > Qf

  2. BOption (b)

    Qr = Qf

  3. COption (c)

    Qr/Qf = constant

  4. DOption (d)

    Qr < Qr

View answer & explanation

Correct answer

Answer: (a)A.

Qr > Qf

Why this answer is correct

Explanation

The relationship between the amount of water flowing in a river or stream and stage at any particular point is usually known as the stage-discharge relationship.

At the same stage, more discharge passes through the river during rising stages than in falling ones.

Note:

Stage discharge relation is also known as the rating curve.

Question 4

A river carrying a discharge of 100 cumecs has a fall of 200 m. The power available from the fall using an overall efficiency of 85% is

  1. AOption (a)

    166.6 MW

  2. BOption (b)

    66.67 MW

  3. COption (c)

    166.6 kW

  4. DOption (d)

    266.7 MW

View answer & explanation

Correct answer

Answer: (a)A.

166.6 MW

Why this answer is correct

Concept:

Consider specific weight of Water =9.81 k/m3\rm{=9.81\ k/m^3}

P=WQHη kW\rm {P=WQHη \ kW}

=9.81×100×200×0.85\rm{=9.81×100×200×0.85} 

=166770 kW\rm{=166770\ kW}

=166.77 MW\rm{=166.77\ MW}

Question 5

Power transmitted through pipes will be maximum when

  1. AOption (a)

    the head due to friction = one-fourth the total head at inlet of the pipe

  2. BOption (b)

    the head due to friction = the total head at inlet of the pipe

  3. COption (c)

    the head due to friction = one-third the total head at inlet of the pipe

  4. DOption (d)

    the head due to friction = half the total head at inlet of the pipe

View answer & explanation

Correct answer

Answer: (c)C.

the head due to friction = one-third the total head at inlet of the pipe

Why this answer is correct

Explanation:

The power transmitted through a pipe is maximum when the loss of head due to friction is given by (H = head supplied),

hf=H3{h_f} = \frac{H}{3}

Power transmission through pipes: 

  • Power is transmitted through pipes by flowing water or other liquids through them.
  • Power transmitted through pipes will be dependent over the following factors as mentioned here.
  1. Rate of flow through the pipe.
  2. Total head available at the end of the pipe

Now we will consider a tank with which a pipe AB is connected. Let us consider the following terms from figure

L = Length of the pipe, D = Diameter of the pipe, H = Total head available at the inlet of the pipe, V= Velocity of flow in the pipe, hf = Loss of head due to friction, f = Co-efficient of friction 

Power transmitted at the outlet of the pipe can be determined by:

P = ρQgHnet

P=ρg1000×π4d2×V(H−hf)P = \frac{{ρ g}}{{1000}} × \frac{\pi }{4}{d^2} × V\left( {H - {h_f}} \right)  =  P=ρg1000×π4d2×V(H−4flV22gd)P = \frac{{ρ g}}{{1000}} × \frac{\pi }{4}{d^2} × V\left( {H - \frac{{4fl{V^2}}}{{2gd}}} \right)

Condition for maximum transmission of power: 

Now we will find here the condition for maximum transmission of power and it could be secured by differentiating the equation of power transmitted at the outlet of the pipe.

ddV(P)=0\frac{d}{{dV}}\left( P \right) = 0

ddV[ρg1000×π4d2(HV−4flV32gd)]=0\frac{d}{{dV}}\left[ {\frac{{ρ g}}{{1000}} × \frac{\pi }{4}{d^2}\left( {HV - \frac{{4fl{V^3}}}{{2gd}}} \right)} \right] = 0

ρg1000×π4d2(H−4×3×f×L×V22gd)=0\frac{{ρ g}}{{1000}} × \frac{\pi }{4}{d^2}\left( {H - \frac{{4 × 3 × f × L × {V^2}}}{{2gd}}} \right) = 0

(H−34fLV22gd)=0\left( {H - 3\frac{{4fL{V^2}}}{{2gd}}} \right) = 0

H - 3hf = 0,

hf=H3{h_f} = \frac{H}{3}

Important Points

Head loss due to friction is given by:

hf;=;fLv22gd{\rm{h_f;}} = {\rm{;}}\frac{{{\rm{f}}L{{\rm{v}}^2}}}{{2{\rm{gd}}}}

d = diameter of pipe, f = friction factor = 4 × co-efficient of friction, L = length of pipe, and v = velocity of flow

Question 6

At two points 1 and 2 in a pipeline, the velocities are V and 2V respectively. Both the points are at the same elevation. The fluid density is ρ. The flow can be assumed to be incompressible, inviscid, steady and irrotational. The difference in pressures P1 and P2 at points 1 and 2 is

  1. AOption (a)

    1.5 ρV2

  2. BOption (b)

    2 ρV2

  3. COption (c)

    3 ρV2

  4. DOption (d)

    0.5 ρV2

View answer & explanation

Correct answer

Answer: (a)A.

1.5 ρV2

Why this answer is correct

Explanation:

Given Data:

V1​=V

V2​=2V

The Bernoulli equation for two points in a flow field is given by:

P1+12ρV12=P2+12ρV22P_1 + \frac{1}{2} \rho V_1^2 = P_2 + \frac{1}{2} \rho V_2^2

Where:

P1​ and P2 are the pressures at points 1 and 2.

V1and V2are the velocities at points 1 and 2.

ρ is the fluid density.

Applying Bernoulli's Equation:

P1+12ρV2=P2+12ρ(2V)2P_1 + \frac{1}{2} \rho V^2 = P_2 + \frac{1}{2} \rho (2V)^2

P1−P2=12ρ(4V2)−12ρV2P_1 - P_2 = \frac{1}{2} \rho (4V^2) - \frac{1}{2} \rho V^2

P1−P2=12ρ(4V2−V2)P_1 - P_2 = \frac{1}{2} \rho (4V^2 - V^2)

P1−P2=1.5ρV2P_1 - P_2 = 1.5 \rho V^2

Question 7

Flow net is used for the determination of 

  1. AOption (a)

    quantity of seepage

  2. BOption (b)

    hydrostatic pressure

  3. COption (c)

    exit gradient

  4. DOption (d)

    All of the above

View answer & explanation

Correct answer

Answer: (d)D.

All of the above

Why this answer is correct

Flow Nets:

It is a graphical solution to the Laplace equation for two-dimensional seepage flow. Two orthogonal sets of curves forming a flow net are:

  • Equi potential lines → connecting points of the equal total head
  • Flow lines → indicating the direction of seepage down under a hydraulic gradient.

In the case of isotropic soil, these lines are orthogonal to each other but need not be orthogonal in anisotropic soil.

Application of flow net is as follows:

  1. Determination of seepage
  2. Determination of hydrostatic pressure
  3. Determination of seepage pressure
  4. Determination of exit gradient

Important Points

Various methods of obtaining flow nets are:

  1. Graphical method

  2. Electrical Analogy method

  3. Soil Model

  4. Plastic Model

  5. Solution of Laplace equation/Analytical method.

Sand model is not used to obtain flow nets.

Question 8

Total Kjeldahl nitrogen is a measure of 

  1. AOption (a)

    total organic and ammonia nitrogen 

  2. BOption (b)

    total ammonia nitrogen

  3. COption (c)

    total inorganic and ammonia nitrogen

  4. DOption (d)

    total organic nitrogen

View answer & explanation

Correct answer

Answer: (a)A.

total organic and ammonia nitrogen 

Why this answer is correct

Explanation:

Kjeldahl Nitrogen = Organic nitrogen + free ammonia

Nitrogen in water:

The nitrogen in water indicates organic contamination of water i.e when water is contaminated by sewage then nitrogen compounds are traced in water. We can trace the following compounds,

  1. Ammonia nitrogen

  2. Organic nitrogen or Albuminoid nitrogen

  3. Nitrites

  4. Nitrates

Nitrogen compounda are measured by colorometry

Total Kjeldahl nitrogen (TKN): 

It is the sum of organic nitrogen, ammonia (NH3), and ammonium (NH4+) in the chemical analysis of soil, water and wastewater.

To calculate Total Nitrogen (TN), the concentrations of nitrate-N and nitrite-N are determined and added to the total Kjeldahl nitrogen

Question 9

From amongst the following sewage treatment options, the largest land requirement for a given discharge will be needed for

  1. AOption (a)

    anaerobic pond

  2. BOption (b)

    oxidation ditch

  3. COption (c)

    oxidation pond

  4. DOption (d)

    trickling filter

View answer & explanation

Correct answer

Answer: (c)C.

oxidation pond

Why this answer is correct

Explanation:

Oxidation Pond:

  • Oxidation ponds, also called lagoons or stabilization ponds, are large, shallow ponds designed to treat wastewater through the interaction of sunlight, bacteria, and algae.
  • Algae grow using energy from the sun and carbon dioxide and inorganic compounds released by bacteria in water.
  • During the process of photosynthesis, the algae release oxygen needed by aerobic bacteria. Mechanical aerators are sometimes installed to supply yet more oxygen, thereby reducing the required size of the pond.
  • Sludge deposits in the pond must eventually be removed by dredging. Algae remaining in the pond effluent can be removed by filtration or by a combination of chemical treatment and settling.
  • This method would also have disadvantages that it requires extensive land area, potential odour problem, mosquito menace, little control over the effectiveness of the treatment process, and the main disadvantage is seepage of effluents into the soil which may also lead to groundwater pollution.

Oxidation Ditch:

  • The oxidation ditch (OD) is a sort of equipment used for long-term aeration.
  • It consists of a long channel of an elliptical or circular shape equipped with aeration equipment called a rotor for generating a water flow and stirring water in the channel to supply oxygen.
  • Though it requires a relatively large area, it has a simple structure and can be easily operated as well as being able to remove nitrogen easily. Thus, it has recently been widely used in relatively small wastewater treating plants.
  • It can be said the OD method is particularly suitable for small-scale equipment.

Question 10

The relative stability of a sewage sample whose dissolved oxygen is same as the total oxygen required to satisfy the BOD is

  1. AOption (a)

    100

  2. BOption (b)

    infinite

  3. COption (c)

    zero

  4. DOption (d)

    1

View answer & explanation

Correct answer

Answer: (a)A.

100

Why this answer is correct

Explanation:

Relative stability:

The ratio of oxygen available in the wastewater (as dissolved oxygen) to the total oxygen required to satisfy its first state BOD is called relative stability.

Relative;stability;=;BOD;removedTotal;BOD×100Relative;stability; = ;\frac{{BOD;removed}}{{Total;BOD}} \times 100

Note:

As the DO level of the sewage is equal to the total oxygen required to decompose the organic matter present, total BOD will be removed.

∴ Relative stability = 100%

Question 11

If the duty for a crop is 864 hectares per cumec on the field and base period for this crop is 120 days, then the delta for a crop is

  1. AOption (a)

    864 cm

  2. BOption (b)

    864 m

  3. COption (c)

    120 m

  4. DOption (d)

    120 cm

View answer & explanation

Correct answer

Answer: (d)D.

120 cm

Why this answer is correct

Concept:

Duty: The duty of water is the relationship between the volume of water and the area of the crop it matures. It is defined as the area irrigated per cumec of discharge running for base period B. The duty is generally represented by D.

 D=8.64BΔD = \frac{{8.64B}}{\Delta }

Where,

Δ = Total depth of water required (in meter)

D = Duty in ha/cumec

B = Base period (in days)

Calculation:

Given:

B = 120 days

Duty = 864 hectares/cumec

⇒ Total depth of water required (Δ),

⇒ Δ=8.64×120864=1.2 m\rm \Delta = \frac{{8.64 \times 120}}{{864}} = 1.2 \ m

∴ Delta of crop = 120 cm

Question 12

The maximum permissible limit for fluoride content in drinking water is

  1. AOption (a)

    1.5 mg/1

  2. BOption (b)

    5 mg/1

  3. COption (c)

    10 mg/1

  4. DOption (d)

    0.1 mg/1 

View answer & explanation

Correct answer

Answer: (a)A.

1.5 mg/1

Why this answer is correct

Explanation:

Water quality for different compounds As per l.S.10500 are as follows:

CharacteristicRequirement (Acceptable limit)Permissible limit in the absence of alternate source
Nitrate45 mg/lNo relaxation
Fluoride1.0 mg/l1.5 mg/l
Chlorides250 mg/l1000 mg/l
Free residual Chlorine0.2 mg/l1 mg/l
Total hardness200 mg/l600 mg/l

Question 13

Zero hardness of water is achieved by

  1. AOption (a)

    excess lime treatment

  2. BOption (b)

    ion-exchange treatment

  3. COption (c)

    excess alum and lime treatment

  4. DOption (d)

    lime-soda process

View answer & explanation

Correct answer

Answer: (b)B.

ion-exchange treatment

Why this answer is correct

Explanation:

Water softening: 

  • It is the process of hardness removal from the water. It is caused by multivalent cation and affects water quality.

Lime soda method: 

  • It is a water softening method in which lime and soda ash are added to the water, which causes the precipitation of multivalent cation as CaCO3.
  • Precipitation of CaCO3 occurs only when the pH of water is greater than 9, so in case of less pH alkalinity is added to the water. In this process small amount of Ca2+ and Mg2+ precipitates very late, which will create incrustation in the pipe, so to avoid this recarbonation is done to dissolve back this small amount of cation.
  • Due to this, the method does not give zero hardness.

Ion Exchange Process: 

  • Ion-exchange resin, (zeolite) exchanges one ion from the water being treated for another ion that is in the resin (sodium is one component of softening salt, with chlorine being the other).
  • Zeolite resin exchanges sodium for calcium and magnesium. It can produce water with zero hardness.

Question 14

IRC recommends that longitudinal stresses on the bridges should be considered as

  1. AOption (a)

    30% of the live load

  2. BOption (b)

    20% of the live load

  3. COption (c)

    10% of the live load

  4. DOption (d)

    40% of the live load

View answer & explanation

Correct answer

Answer: (b)B.

20% of the live load

Why this answer is correct

Longitudinal forces in bridge:

Longitudinal forces mean force acting on a bridge along its length or along with the motion of traffic.

The longitudinal forces are caused by braking or accelerating of the vehicle on the bridge. When the vehicle stops suddenly or accelerates suddenly it induces longitudinal forces on the bridge structure, especially on the substructure. IRC recommends 20% of the live load should be considered as a longitudinal force on the bridges.

Dead load is due to the weight of the bridge and it always acts in the direction of gravity i.e. perpendicular to the roadway of the bridge.

Thermal stresses are caused due to temperature. When the temperature is very high or very low they induce stresses in the bridge elements, especially at bearings and deck joints. They induce both vertical and horizontal forces.

Wind load acts in every direction of the bridge. However, their impact as lateral wind force is more as compared to the longitudinal direction.

Other types of forces and load on bridges:

1. Centrifugal forces: If bridge is to be built on horizontal curves, then the movement of the vehicle along curves will cause centrifugal force on to the superstructure. 2. Buoyancy Effect: Buoyancy effect is considered for substructures of large bridges submerged under deep water bodies. 3. Seismic Loads: They induce both vertical and horizontal forces during the earthquake. 4. Erection Stresses: Erection stress are induced by the construction equipment during the bridge construction.

Question 15

In disinfection, which of the following forms of chlorine is most effective in killing the pathogenic bacteria?

  1. AOption (a)

    OCI

  2. BOption (b)

    NH2Cl

  3. COption (c)

    HOCI

  4. DOption (d)

    CI2

View answer & explanation

Correct answer

Answer: (c)C.

HOCI

Why this answer is correct

concept:

  • HOCl (Hypochlorous Acid):
  • Hypochlorous acid is the most effective disinfecting form of chlorine because it is a neutral molecule that easily penetrates the cell walls of microorganisms, including bacteria, viruses, and protozoa. Once inside the cell, it disrupts vital cellular processes, leading to the death of the organism.
  • HOCl is produced when chlorine (Cl2) or sodium hypochlorite (NaOCl) is dissolved in water. It exists in equilibrium with the hypochlorite ion (OCl), but HOCl is more effective because it is more reactive and can diffuse more readily across cell membranes.
  • OCl (Hypochlorite ion):
  • The hypochlorite ion is also a disinfectant, but it is less effective than HOCl because it is negatively charged and does not penetrate cell membranes as efficiently as HOCl.
  • NH2Cl (Chloramine):
  • Chloramine is used as a disinfectant, but it is less effective than free chlorine (HOCl and OCl) in killing bacteria. It is often used in water treatment where a longer-lasting disinfectant is needed, but it is slower-acting compared to HOCl.
  • Cl2 (Chlorine gas):
  • Chlorine gas, when dissolved in water, forms HOCl and OCl^-, but in its gaseous form, it is not directly effective as a disinfectant until it is dissolved and forms these active species.

Question 16

For a waste, the 5-day BOD at 20 °C is found to be 200 mg/1. For the same waste, 5-day BOD at 30 °C will be

  1. AOption (a)

    more than 200 mg/1

  2. BOption (b)

    200 mg/1

  3. COption (c)

    zero, as the bacteria cannot withstand such high temperature

  4. DOption (d)

    less than 200 mg/1

View answer & explanation

Correct answer

Answer: (a)A.

more than 200 mg/1

Why this answer is correct

concept:

As temperature increases from 20°C to 30°C, the growth of microbial organism will increase. So the degradation of organic matter will be more as the growth of organism increases. So that the 5 day BOD at 30°C will be more than that of 5 day BOD at 20°C.

The growth of organism will decrease if the temperature is  further increased above 60°C – 70°C.

from the given data, for the 5-day BOD at 30 °C it will be more than 200 mg/1.

Question 17

In water treatment, slow sand filter, when compared to rapid gravity filters, produces

  1. AOption (a)

    more contaminated effluent

  2. BOption (b)

    equally contaminated effluent

  3. COption (c)

    Uncertain 

  4. DOption (d)

    less contaminated effluent

View answer & explanation

Correct answer

Answer: (d)D.

less contaminated effluent

Why this answer is correct

Explanation:

In water treatment, slow sand filters when compared to rapid sand filters produce Lesser contaminated effluent.

Slow sand filterRapid sand filter
Pre-treatmentEffluents from plain sedimentation coagulation, not req.Coagulation, flocculation, and sedimentation is a must.
Filter SandD10 b/w 0.2 and 0.4 mm Cu ⇒ 1.8 to 3. Gravel -30-75 cm thickD10 = 0.35 mm - 0.55 mm Cu = 1.2 to 1.8 Gravel-60-90 cm thick
Size of each unitLarge such as 30 m × 60 m. An area from 100-2000 m2. Less flexible on shock loadingSmall 5m × 8m. Area < 100m2 More flexible on shock loading
Rate of filtration100 to 200 litres/hr/m23000 to 6000 L/hr/m2
EfficiencyVery efficient in removing bacteria (98-99%) but less efficient in removing colour.Less efficient in bacteria (80-90%), very efficient in colour removal.
Post-treatmentDisinfection is not must. Almost pure water obtained. Turbidity removal up to 50 ppmDisinfection is the must. Turbidity up to 35 ppm.
Method of cleaningScrapping and removing the top 1.5 to 3 cm thick layer and washing down by hoses.Backwashing with or without compressed air.
Qty of wash water req.0.2 to 0.6% of total filtered water is req. in every 1 to 3 monthsA large amount of total filtered water is req. in every 1 to 3 days.
(1-5%)

Question 18

Bitumen of grade 80/100 means

  1. AOption (a)

    its penetration value is 10 mm

  2. BOption (b)

    its penetration value is 8 mm to 10 mm

  3. COption (c)

    its penetration value is 8 cm to 10 cm

  4. DOption (d)

    its penetration value is 8 mm

View answer & explanation

Correct answer

Answer: (b)B.

its penetration value is 8 mm to 10 mm

Why this answer is correct

Explanation:

  • Bitumen is classified on the basis of either its penetration value or its viscosity.
  • The grade of bitumen is decided on the basis of the penetration test.

Penetration test:

  • The hardness of bitumen is obtained by the penetration test.
  • It measures the distance a standard blunt pointed needle will vertically penetrate a sample of material at 25°C, the load being 100 gram,s and the time of application of load being 5 seconds.
  • The unit of penetration is 1/10 mm. Thus 80/100 grade bitumen means 80 and 100 give the idea of the penetration value of the bitumen and its value is equal to 8 - 10 mm penetration.
  • 30/40 and 80/100-grade bitumen are more commonly used, depending upon bitumen construction type and climate conditions.
  • In hot climates, a lower penetration grade bitumen like 30/40 bitumen is preferred.

Question 19

As per IRC recommendations, the maximum spacings of expansion and contraction joint in rigid pavements are respectively

  1. AOption (a)

    140 m, 4.5 m

  2. BOption (b)

    25 m, 40 m

  3. COption (c)

    140 m, 45 m

  4. DOption (d)

    45 cm, 4.5 m

View answer & explanation

Correct answer

Answer: (a)A.

140 m, 4.5 m

Why this answer is correct

Explanation:

Movement joints:

When the concrete is placed, it will contract slightly during the curing process, this is the primary cause of small surface cracks that appear during the curing process. When concrete sets, it will expand and/or contract slightly with ambient temperature.

Hence we need to provide movement joints in the concrete pavements or slab

Various types of movement joints provided in the cement concrete pavement are:-

  1. Expansion joints 2) Contraction joints 3) Construction joints

Expansion joint:

  • These are provided to allow the expansion of pavement due to a rise in temperature with respect to construction temperature.
  • The maximum spacing between the expansion joint is 140 m.

Contraction Joints: 

  • These joints are intended to control the cracks in a slab that are developed due to the stresses caused by moisture-related concrete shrinkage, thermal contraction, temperature warping.
  • They may be oriented transversely (i.e. perpendicular to the direction of traffic flow) or longitudinally (i.e. parallel to the direction of traffic flow).

According to IRC 15: 2002, The maximum spacing of contraction joints is given in the below table:

Categories of slabSlab Thickness (cm)Maximum Contraction Joint Spacing (m)
Unreinforced Slabs104.5
154.5
204.5
Reinforced Slabs107.5
1513.0
2014.0

Question 20

The speed and delay studies on a defined section of a highway are conducted by

  1. AOption (a)

    traffic counters

  2. BOption (b)

    moving car method

  3. COption (c)

    enoscope

  4. DOption (d)

    radar gun

View answer & explanation

Correct answer

Answer: (b)B.

moving car method

Why this answer is correct

Concept:

  1. Fast-moving car method (or Floating Car Method or Moving Observer Method): In this method, a test vehicle is driven a number of times through a pre-selected stretch of a route at an approximately average speed of the stream to find out journey speed, time and location of the delay. In traffic engineering, this method is used in speed and delay studies.
  2. Enoscope: This is a device that is an L-shaped mirror box, open at both ends with a mirror set at 45° angle to the arm of the device. This device is almost a ‘half-periscope’. Using an Enoscope an observer can measure the spot speed of different vehicles. In traffic engineering, this device is used in spot speed studies.
  3. Radar Speed Meter: The radar speed meter is a pre-calibrated device that directs a radar or an audio beam of a certain frequency to moving vehicles to record and display their spot speed. In this device, there is a frequency difference in the signal transmitted and the signal received which is proportional to the speed of the vehicle. In traffic engineering, this device is used in spot speed studies.
  4. Traffic Contours (or Traffic Flow Map): It is the contour map of all traffic volumes plotted for a certain region and for a particular time frame to find out traffic flow patterns and behavior. This traffic contour helps in designing and evaluating the efficiency of different components of a traffic management system. In traffic engineering, this map is used in traffic volume studies.

Question 21

In the revised CBR design method recommended by the IRC for the design of flexible pavement, the total thickness depends upon

  1. AOption (a)

    CBR value of soil and magnitude of wheel load

  2. BOption (b)

    CBR value of soil and number of commercial vehicles per day

  3. COption (c)

    CBR value of soil and cumulative standard axle load

  4. DOption (d)

    CBR value of soil only

View answer & explanation

Correct answer

Answer: (c)C.

CBR value of soil and cumulative standard axle load

Why this answer is correct

concept:

The CBR method of pavement design gives the total thickness requirement of the pavement above a subgrade.

In revised CBR method for the design of flexible pavement IRC has provided charts for different CBR in which relation between pavement thickness and cumulative traffic axle is given.

CBR Value of Soil: This measures the strength of the subgrade soil and is a critical factor in determining the pavement thickness.

Cumulative Standard Axle Load: This considers the traffic load, specifically the cumulative number of standard axles (usually 8.2 tonnes) that the pavement is expected to carry over its design life.

In this method, the chart contains several curves (A, B, C, D, E, F, and G) which represent the different levels of traffic intensities. Based on this we can find out the layers thicknesses.

Question 22

As per IRC, the minimum length of transition curve for a mountainous terrain road with radius of curvature 100 m and design speed of vehicle 100 kmph is

  1. AOption (a)

    200 m

  2. BOption (b)

    100 m

  3. COption (c)

    170 m

  4. DOption (d)

    270 m

View answer & explanation

Correct answer

Answer: (b)B.

100 m

Why this answer is correct

Concept: 

Length of transition curve 

Based on IRC Formula

ls=V2R{l_s} = \frac{{{V^2}}}{R} for Hilly terrain and ls=2.7V2R{l_s} = \frac{{2.7{V^2}}}{R} for Plain and Rolling terrain.

Where

Ls = length of the transition curve (m),  V = speed of the vehicle (km/hour),

 R = radius of the transition (m),

Calculation:

Given,

Design Speed = 100 kmph, Radius (R) = 100 m

iii) By empirical formula

For Mountainous or Hilly Terrain 

V=V2R=1002100=100;m{\rm{V}} = \frac{{{{\rm{V}}^2}}}{{\rm{R}}} = \frac{{ {{100}^2}}}{{100}} = 100{\rm{;m}}

∴ The length of the transition curve-based IRC empirical formula is 100 m

Additional InformationLength of transition curve (Maximum value is to be considered among the below options)

a) Based on the rate of change of centrifugal acceleration (c)

ls=v3c×R{{\rm{l}}_{\rm{s}}} = \frac{{{v^3}}}{{c \times R}}

where c=8075+Vc = \frac{{80}}{{75 + V}} ……….. c value should be between 0.5 to 0.8

b) Based on the rate of change of superelevation (e)

  1. Superelevation attained by rotation about the inner edge or outer edge

ls=eN(w+w×e){l_s} = eN\left( {w + w \times e} \right)

  1. Superelevation attained by rotation about the center

ls=eN(w+w×e)2{l_s} = \frac{{eN\left( {w + w \times e} \right)}}{2}

c) Based on IRC Formula

ls=V2R{l_s} = \frac{{{V^2}}}{R} for Hilly terrain and ls=2.7V2R{l_s} = \frac{{2.7{V^2}}}{R} for Plain and Rolling terrain.

Where

Ls = length of the transition curve (m), v = speed of the vehicle (m/s), V = speed of the vehicle (km/hour),

e = required superelevation, N = rate of change of superelevation to be attained, R = radius of the transition (m), and

w = width of road/pavement

Question 23

The weight of tracked vehicle considered for class 70R loading as per IRC is

  1. AOption (a)

    600 kN

  2. BOption (b)

    700 kN

  3. COption (c)

    800 kN

  4. DOption (d)

    500 kN

View answer & explanation

Correct answer

Answer: (b)B.

700 kN

Why this answer is correct

Concept:

IRC Class AA Loading

  • Two different types of vehicles are specified under this category grouped as tracked and wheeled vehicles.
  • The IRC Class AA tracked vehicle (simulating an army tank) of 700 kN and a wheeled vehicle (heavy duty army truck) of 400 kN
  • All the bridges located on National Highways and State Highways have to be designed for this heavy loading.
  • These loadings are also adopted for bridges located within certain specified municipal localities and along specified highways. Alternatively, another type of loading designated as Class 70 R is specified instead of Class AA loading.

Question 24

The fixed type bearing used in bridges is

  1. AOption (a)

    sliding plate bearing

  2. BOption (b)

    elastomeric bearing

  3. COption (c)

    RC rocker bearing.

  4. DOption (d)

    None of the above.

View answer & explanation

Correct answer

Answer: (c)C.

RC rocker bearing.

Why this answer is correct

  1. Sliding Plate Bearing:
  • Function: Sliding plate bearings are designed to accommodate large longitudinal movements in bridge structures. They consist of two plates, one of which slides over the other. The sliding action is usually facilitated by low-friction materials like Teflon.
  • Use Case: These bearings are not typically used as fixed bearings because they allow horizontal movement. They are commonly used in bridges where there is a need to accommodate expansion and contraction due to temperature changes or other factors.
  1. Elastomeric Bearing:
  • Function: Elastomeric bearings are made of layers of rubber and steel. They allow for rotation and horizontal movements, making them ideal for bridges that need to accommodate thermal expansion and contraction, as well as minor vertical displacements.
  • Use Case: These bearings are flexible and accommodate both rotation and translation. They are not considered fixed bearings because they do not restrict movement entirely, allowing for multi-directional displacement and rotation.
  1. RC Rocker Bearing:
  • Function: Reinforced Concrete (RC) rocker bearings are designed to allow rotation while preventing horizontal movement. This type of bearing is rigid and is typically used as a fixed bearing in bridge structures to provide stability.
  • Use Case: Fixed bearings like the RC rocker bearing are essential in bridges to ensure that one end of the bridge remains in place, controlling longitudinal and transverse movements.

Question 25

The value of the impact factor for the concrete bridge of 3 m span is

  1. AOption (a)

    0.5

  2. BOption (b)

    0.7

  3. COption (c)

    0.9

  4. DOption (d)

    0.3

View answer & explanation

Correct answer

Answer: (a)A.

0.5

Why this answer is correct

concept:

Impact factor is taken to accommodate impact of live load on structures.

As per IRC 112, different types of vehicles have different impact factor.

For Reinforced concrete structures, it is given by:

I.F=4.56+L\text{I}.\text{F}=\frac{4.5}{6+\text{L}}

Where,

L = span of the bridge=3 m.

I.F=4.56+3\text{I}.\text{F}=\frac{4.5}{6+\text{3}}

I.F=0.5\text{I}.\text{F}=0.5

Question 26

The minimum width of carriage-way for a two-lane bridge, other than minor village roads and national highways, is

  1. AOption (a)

    5.5 m

  2. BOption (b)

    7.5 m

  3. COption (c)

    9 m

  4. DOption (d)

    4.25 m

View answer & explanation

Correct answer

Answer: (b)B.

7.5 m

Why this answer is correct

Explanation:

According to IRC 5: 1998, Clause 112

  • The width of the carriageway for a bridge constructed for road traffic only, the width of the carriage should be minimum of 4.25 m for a single lane and 7.5 m for two lane road.
  • It shall increase by 3.5 m for every additional lane of traffic for the multilane bridge.
  • The width of central/verge/median when provided shall not less than 1.2 m.
  • Road bridges shall be provided for one lane, two-lane or multiple of 2 lanes, three-lane bridges with 2 directional traffic shall not be constructed

Question 27

With the increase in capacity-inflow ratio, the trap efficiency of a reservoir

  1. AOption (a)

    decreases

  2. BOption (b)

    remains unchanged

  3. COption (c)

    may increase or decrease depending upon reservoir's, characteristic

  4. DOption (d)

    increases

View answer & explanation

Correct answer

Answer: (d)D.

increases

Why this answer is correct

Explanation:

Trap Efficiency:

  • Trap efficiency of a reservoir is the ratio of total sediments retained by the reservoir to the total sediments entering into the reservoir (inflow sediments).
  • Trap efficiency of the reservoir is the function of the ratio of reservoir capacity and total inflow.
  • The reservoir capacity is inversely related to the number of sediments deposited. As time passes, the rate of silting or the rate of deposition of sediments reduces but the amount of sediments deposited will be increased which in turn reduces the capacity of the reservoir. Hence, trap efficiency is also gets reduced.

Additional InformationOther factors affecting trap efficiency:

  • Size of reservoir and stream: A larger reservoir on a small stream has less capacity inflow ratio and hence has higher trap efficiency.
  • The velocity of flow of stream: Higher velocity less deposition of sediments, lower is the trap efficiency.

Question 28

The critical shear stress, at which the incipient motion of sediment takes place, is proportional to

where d is grain size.

  1. AOption (a)

    d

  2. BOption (b)

    d3/2

  3. COption (c)

    d2

  4. DOption (d)

    √d

View answer & explanation

Correct answer

Answer: (a)A.

d

Why this answer is correct

Concept:

When the velocity through a channel is very low the bed sediment particles do not moves, but if the velocity increases gradually there will be a point of time when the bed sediment particles starts moving  with the flow.

This beginning of the movement of bed sediment particles is called as incipient motion of sediment.

  • At incipient motion of sediment the shear stress applied by the flowing water on the bed sediment particles just exceeds the resistance of those particles and the particles start to move. This shear stress is called as critical shear stress.
  • Incipient motion of sediment is also known as threshold point or critical condition.
<br>

The relationship between critical shear stress and diameter of the particle is best represented by Shield’s formula which is given by,

\({{\rm{F}}{\rm{s}}} = \frac{{{{\rm{\tau }}{\rm{c}}}}}{{{{\rm{\gamma }}{\rm{w}}}{\rm{d}}\left( {{{\rm{S}}{\rm{s}}} - 1} \right)}}\)

Where,

FS = Shield’s Entrainment function (a dimensionless number),

τc = Critical shear stress,

d = Diameter of the particles,

Ss = Specific gravity of the particles, and

γw = Specific weight of water

For a particular type of sediment material Fs is constant.

 ∴ τc ∝ d.

Question 29

Pure clayey soils are generally not preferred for the central impervious cores of zoned type of earthen dams because

  1. AOption (a)

    clays are highly impervious

  2. BOption (b)

    clays are highly pervious

  3. COption (c)

    clays are susceptible to cracking

  4. DOption (d)

    None of the above

View answer & explanation

Correct answer

Answer: (c)C.

clays are susceptible to cracking

Why this answer is correct

Clayey Soils in Central Impervious Cores of Zoned Earthen Dams:

  • Pure clayey soils are generally avoided in the central impervious cores of zoned earthen dams.
  • One of the main reasons for this is that clays are susceptible to cracking under certain conditions, particularly due to drying or when subjected to differential settlement.
  • When cracks develop, they can compromise the impervious nature of the core, leading to potential leakage or failure of the dam.
  • Therefore, clays' susceptibility to cracking makes them a less preferred choice for the central impervious core.

Question 30

The line joining the static water levels in several wells excavated through a confined aquifer is known as

  1. AOption (a)

    piezometric surface

  2. BOption (b)

    perched water table

  3. COption (c)

    hypsometric curve

  4. DOption (d)

    cone of depression

View answer & explanation

Correct answer

Answer: (a)A.

piezometric surface

Why this answer is correct

Concept:

Piezometric surface of the aquifer: 

  • A confined aquifer is also known as an artesian aquifer is confined between two impervious beds such as aquicludes or aquifuges.
  • Confined aquifers are permeable rock units that are usually deeper under the ground than unconfined aquifers.
  • The water in the confined aquifer will be under pressure and hence the piezometric level which is the surface joining the static water levels will be much higher than the top level of the aquifer.
  • The line joining static water level in several wells, excavated through a confined aquifer is known as the piezometric surface.

Water-table surface: 

  • The surface at which pressure head and atmospheric head are equal.

Capillary fringe:

  • It is the layer above the groundwater-surface in which water seeps to some extent due to capillary action.

Cone of depression: 

  • A depressed cone appears when groundwater is pumped out from the well.

Question 31

The relation between transmissibility (T) and permeability (K) for an aquifer of depth (B) is

  1. AOption (a)

    T = KB

  2. BOption (b)

    T = K log B

  3. COption (c)

    T = In (KB)

  4. DOption (d)

    K = TB

View answer & explanation

Correct answer

Answer: (a)A.

T = KB

Why this answer is correct

Concept:

The coefficient of transmissibility (T)

The rate of flow of water through the entire medium of unit width under a unit hydraulic gradient is termed as the coefficient of transmissibility (T)

T = K (A) 

∵ A = d × 1

T=K×dT = K × d m2/sec

or

​T=K×BT = K × B m2/sec​

Where, K = coefficient of permeability, T = the coefficient of transmissibility, B or d =  the aquifer thickness

Question 32

For the head regulator, the most severe condition of uplift pressure on the floor occurs when

  1. AOption (a)

    the canal runs dry and the river flow is at high flood level

  2. BOption (b)

    the canal runs at full supply depth and the river. flow is at pond level

  3. COption (c)

    the canal runs dry and the river flow is at pond level

  4. DOption (d)

    the flow in the level is at flood level and canal is running at full supply depth

View answer & explanation

Correct answer

Answer: (a)A.

the canal runs dry and the river flow is at high flood level

Why this answer is correct

Concept:

Diversion head works like Weir or barrage is constructed across a perennial river to raise water level and to divert the water to canal, is known as diversion head work Flow of water in the canal is controlled by canal head regulator. It controls the entry of silt into canals and provides some poundage by creating small pond.

Components of Diversion headwork are as follows:

  1. Canal head regulator
  • A canal head regulator is provided at the head of each main canal off taking from diversion headwork.
  • It should be so aligned that its axis makes an angle of 90 to 120° with the axis of weir or parallel to the divide wall as shown in the figure.
  • A head sluices or canal head regulator (CHR) is provided at the head of the off-taking canal, and serves the following functions:
  • It controls the entry of silt in the canal.
  • It regulates the supply of water entering the canal.
  • It prevents the river floods from entering the canal.

 The maximum uplift under the worst condition is  that canal runs dry and the river flow is at high flood level.

Question 33

Canal drops are required to

  1. AOption (a)

    dissipate excess energy

  2. BOption (b)

    dissipate inadequate land slope

  3. COption (c)

    dissipate excess land slope

  4. DOption (d)

    None of the above

View answer & explanation

Correct answer

Answer: (a)A.

dissipate excess energy

Why this answer is correct

Explanation:

Canal drop or Fall:

A canal fall or drop is an irrigation structure constructed across a canal to lower down its bed level to maintain the designed slope when there is a change of ground level to maintain the designed slope when there is change of ground level. This falling water at the fall has some surplus energy. The fall is constructed in such a way that it can destroy this surplus energy.

Additional Information

Types of Canal Fall:

  1. Ogee Fall
  2. Rapid Fall
  3. Stepped Fall
  4. Notch Fall
  5. Vertical Drop Fall
  6. Glacis Type Fall
  • Straight Glacis Type
  • Parabolic Glacis Type

Question 34

The soil becomes partially infertile when its pH value is about

  1. AOption (a)

    zero 

  2. BOption (b)

    7

  3. COption (c)

    11

  4. DOption (d)

    None of the above

View answer & explanation

Correct answer

Answer: (c)C.

11

Why this answer is correct

Explanation:

When the soil has pH value greater than 11, it is termed as practically infertile.

Causes of high soil pH:

  • Sources of alkalinity.
  • Role of cations (Na and Ca)
  • Influence of soluble salt levels
  • Influence of carbon dioxide and carbonates.

Important Points

  1. If the pH in the case of saline soil is around 8.5, very good structure
  2. And in the case of saline-sodic soil, it is from 8.5 to 10, good structure.
  3. And in the case of sodic soil, it is generally greater than 10, structureless.

Note:

Why high pH soils are infertile:

The high pH caused due to the presence of sodium carbonate which is highly soluble.

This high pH causes fulvic and humic acid of organic matter to get dissolve and when the water evaporates, it gives the surface a black color.

This sodic soil creates physical disintegration and due to high sodium and low salt concentration, slaking, swelling and dispersion occurs which caused coagulation of soil particles as Vander Vaal forces dominate due to deficiency of hydrated ions. 

Complete dispersion causes soil crusting and as a result of this germination problem due to clogs that restrict soil and air movement and makes soil infertile.

Question 35

Penman's equation is based on 

  1. AOption (a)

    energy budgeting and water budgeting

  2. BOption (b)

    energy budgeting and mass transfer

  3. COption (c)

    water budgeting and mass transfer

  4. DOption (d)

    energy budgeting only

View answer & explanation

Correct answer

Answer: (b)B.

energy budgeting and mass transfer

Why this answer is correct

Explanation:

Evapotranspiration Equation:

There are large number of evapotranspiration equations available, they are purely empirical and not theoretical. The most used equation is Penman's Equation.

Penman's Equation:

Penman's equation is based on sound theoretical reasoning and is obtained by combination of energy and mass transfer approach.

PET=AHn+EaγAγPET = \frac{{A{H_n} + {E_a}γ }}{{Aγ }}

PET = Daily potential evapotranspiration (in mm per day)

A = Slope of the saturation vapour pressure vs. temperature curve at the mean air temperature (in mm of Hg per °C)

Hn = Net radiation

Ea = Parameter including wind velocity and saturation deficit.

γ = Psychromrtic constant = 0.49 mm of Hg/°C

Question 36

The ratio of AET to PET lies between

  1. AOption (a)

    0.6 to 0.9

  2. BOption (b)

    0 to 1.0

  3. COption (c)

    0.9 to 1.0

  4. DOption (d)

    0 to 0.6

View answer & explanation

Correct answer

Answer: (b)B.

0 to 1.0

Why this answer is correct

Concept:

Actual Evapotranspiration:

It is the quantity of water that is actually removed from a surface due to the processes of evaporation and transpiration.

Potential Evapotranspiration:

It is a measure of the ability of the atmosphere to remove water from the surface through the processes of evaporation and transpiration assuming no control on water supply.

The relation between actual evapotranspiration and potential evapotranspiration with different soil moisture conditions is as follows:

  1. If the water supply to the plant is adequate, soil moisture will be at field capacity then, the ratio of AET to PET = 1.
  2. If the water supply to the plant is inadequate, then the ratio of AET to PET is less than 1.
  3. In clayey soils, AET/PET is almost equal to 1.
  4. When the soil moisture approaches the permanent wilting point, AET tends to 0.

Hence, If the moisture content in soil is at the field capacity of the soil, the ratio of Actual Evapotranspiration (AET) to the potential Evapotranspiration (PET) is equal to 1.

Additional InformationField Capacity (FC):

The field capacity is the amount of water held in soil after excess water has been gravity drained.

Permanent Wilting point (PWT):

A permanent wilting point is the moisture content of a soil at which the moisture is no longer available in sufficient quantity to sustain the plants.

Question 37

A through bridge is one with

  1. AOption (a)

    bridge flooring provided at the top of superstructure

  2. BOption (b)

    bridge flooring provided at the bottom of super-structure

  3. COption (c)

    bridge flooring provided at some intermediate level of superstructure

  4. DOption (d)

    None of the above

View answer & explanation

Correct answer

Answer: (b)B.

bridge flooring provided at the bottom of super-structure

Why this answer is correct

Concept:

Bridges:

A bridge is a structure built to span a physical obstacle, such as a body of water, valley, or road, without closing the way underneath. A bridge in which the roadway or track passes between the supporting elements (as trusses, girders, or arches).

Bridges classification based on their floor position:

1). Deck Bridge: The super-structure or floor of the bridge is positioned in between the high flood level and formation level. Therefore no top bracing is required.

2). Through Bridge: The Super-structure of the bridge is completely above the formation level. In the case of through Type Bridge, flooring is supported or suspended at the bottom of the superstructure and the top chord is braced laterally.

<br>

 

3). Semi-Through Bridge: The Super-structure of the bridge is partly above and partly below the formation level. here are also double-deck bridges which are constructed to carry the traffic on both roadways and railways at the same time.

Question 38

In class B train of vehicles loading, the nose to tail distance between successive trains

  1. AOption (a)

    shall be less than 18 m

  2. BOption (b)

    shall not be less than 90.0 m

  3. COption (c)

    shall be less than 90.0 m

  4. DOption (d)

    shall not be less than 18.5 m

View answer & explanation

Correct answer

Answer: (d)D.

shall not be less than 18.5 m

Why this answer is correct

Concept:

IRC Class A Loading:

  • The Indian Roads Congress (IRC) is the Apex Body of Highway Engineers in the country. The IRC was set up in December 1934 on the recommendations of the Indian Road Development Committee best known as Jayakar Committee set up by the Govt. of India with the objective of Road Development in India.
  • The Class A loading was proposed by IRC with the objective of covering the worst combination of axle spacing and axle loads likely to arise from the various types of vehicles that are normally expected to use the road.
  • The loading consists of a train of wheel loads that is composed of a driving vehicle and two trailers of specified axle spacings and loads. This loading in bridge designing is generally adopted on all roads on which permanent bridges and culverts are constructed.
<br> <br>

Explanation:

  • The nose-to-tail distance between successive trains shall not be less than 18.5m.
  • No other live load shall cover any part of the carriageway when a train of the vehicle (or trains of the vehicle in multilane Bridge) is crossing the bridge.
  • The minimum clearance ‘f’ between the outer edge of the wheel and the roadway face of the kerb, and the minimum clearance ‘g’, between the outer edge of passing or crossing vehicles on a multi-lane bridge, shall be as given below.

​

Question 39

The term 'wash load' refers to

  1. AOption (a)

    the suspended load during a flood

  2. BOption (b)

    the part of suspended load comprising of particles not available in the bed material

  3. COption (c)

    the bed load after the fines have been washed out

  4. DOption (d)

    the saltating part of bed material load

View answer & explanation

Correct answer

Answer: (b)B.

the part of suspended load comprising of particles not available in the bed material

Why this answer is correct

Explanation:

Bed load is the term bed load describes particles in flowing fluid that are transported along the bed. Bed load moves by rolling, sliding, and saltating. These heavier particles are usually sands and gravels. 

So, Bed load is the combination of contact load and saltation load.

Suspended load/Wash load consists of particles suspended in the water above the bottom. Suspended load consists of generally finer, smaller particles than bed load (like silt and clay).

Dissolved load is the solids dissolved in the water move along with the water (like dissolved salts).

Question 40

Tortuosity in a meandering river is

  1. AOption (a)

    greater than 1

  2. BOption (b)

    less than 1

  3. COption (c)

    less or greater than 1, depending upon the river

  4. DOption (d)

    1

View answer & explanation

Correct answer

Answer: (a)A.

greater than 1

Why this answer is correct

Concept:

Meandering River:

  • The formation of successive bends of reverse order may lead to the formation of a complete S curve called Meanders.
  • When consecutive curves of reverse order connected with short straight reaches called Crossings are developed in a river reach, the river is stated to be a Meandering river.

 

Causes:

  1. Extra turbulence generated by excess of river sediment during floods. When the silt charge is in excess of the quantity required for stability, the river starts building up its slope by depositing the silt on its bed.
  2. The increase in bed slope tends to increase the width of channel, if the banks are not resistant. The banks are attacked by river water and in the process one bank is likely to be attacked slightly more than the other, causing a slight deviation in flow. This slight deviation from the uniform axial flow, helps in improving more and more flow towards one bank than towards the other. This process continues causing once bank convex and another concave. Hence, increasing slope is also a reason for meandering.

Meander parameters:

Tortuosity: 

  • It is the ratio of length along the channel (actual length) to the direct axial length.

Tortuosity=Curved;length;along;the;river;the;direct;axial;length;of;the;riverTortuosity = \frac{{Curved;length;along;the;river;}}{{the;direct;axial;length;of;the;river}}

As, the curved length of the river is always greater than the direct axial length of the river. So, tortuosity of the river is always greater than 1

Meander Length: 

  • It is the axial length of one mender, i.e. tangential distance between the corresponding points of meander.

Meander Belt: 

  • It is the distance between the outer edges of clockwise and anticlockwise loops.

Meander Ratio:

  • It is the ratio of the meander belt to the meander length.

Crossings or cross-over: 

  • It is the shortest reach of the river, connecting two consecutive clockwise and anticlockwise loops.

Question 41

A person standing on the bank of a canal drops a stone on the water surface. He notices that the disturbance on the water is not travelling upstream. This is because the flow in the canal is

  1. AOption (a)

    supercritical 

  2. BOption (b)

    steady

  3. COption (c)

    uniform 

  4. DOption (d)

    subcritical

View answer & explanation

Correct answer

Answer: (a)A.

supercritical 

Why this answer is correct

Explanation:

The flow in an open channel flow can be divided on the basis of the Froude number

Fr < 1 for the flow to be subcritical

Fr = 1 for the flow to be critical

Fr > 1 for the flow to be supercritical

For the supercritical flow

vgy>1\frac{v}{{\sqrt {gy} }} > 1

v>gyv > \sqrt {gy}

V > C

Therefore, if a person drops a stone on the water surface of a canal, disturbances on the water surface is not travelling upstream because the waves will be washed away, and he is not able to see the waves travelling on the upstream and they can be seen on downstream side. Hence the flow is super-critical.

Question 42

A culvert is designed for a peak flow Qp on the basis of rational formula. If a storm of the same intensity as used in design but of duration twice larger occurs, then the resulting peak discharge will be

  1. AOption (a)

    2Qp

  2. BOption (b)

    Qp/2

  3. COption (c)

    Qp2\rm Q_{p^2}

  4. DOption (d)

    Qp

View answer & explanation

Correct answer

Answer: (d)D.

Qp

Why this answer is correct

Explanation:

Rational method:

According to the Rational method, the stormwater or runoff is defined by

Q=K×A×I×RQ=K\times A\times I\times R

where Q is expressed in liter per second,

'K' i a constant,

'A' is the catchment area in hectares,' I' is the impermeability factor

and 'R' is the intensity of rainfall in mm per hour.'

 

There is no term for duration of rainfall in the rational formula. So peak discharge​ will not change.

Question 43

Which of the following statements is correct as per Lacey's method for design of alluvial channels?

  1. AOption (a)

    Hydraulic radius increases with an increase in silt factor.

  2. BOption (b)

    Wetted perimeter decreases with an increase in design discharge.

  3. COption (c)

    Wetted perimeter increases with an increase in silt factor.

  4. DOption (d)

    Wetted perimeter increases with an increase in design discharge.

View answer & explanation

Correct answer

Answer: (d)D.

Wetted perimeter increases with an increase in design discharge.

Why this answer is correct

Concept:

Lacey gave three independent equation after intensive research in alluvium and gave the idea of regime.

The equations given by Lacey are:

v=√25Rf{\bf{v}} = √ {\frac{2}{5}{\bf{Rf}}}

Af2=140v5{\bf{A}}{{\bf{f}}^2} = 140{{\bf{v}}^5}

v=10.8R23S13{\bf{v}} = 10.8{{\bf{R}}^{\frac{2}{3}}}{{\bf{S}}^{\frac{1}{3}}}

As per lacey, the wetted perimeter P is given as 

P = 4.75 √Q

Hence the wetted perimeter of the alluvial channel increases with an increase in design discharge.

Important Point:

Difference between Lacey and Kennedy theory are:-

Lacey TheoryKennedy’s Theory
Lacey considered a semi-elliptical sectionKennedy considered a trapezoidal section
Lacey considered boundary material and defined the silt factorKennedy did not give any importance to boundary material.
Lacey gave the equation for the bed slopeKennedy didn’t give any such equation
This theory applies to both the alluvial canal and riverThis theory applies to alluvial canal only

Question 44

A stilling basin is the one 

  1. AOption (a)

    which is provided at the foot of a spillway for energy dissipation

  2. BOption (b)

    which is provided in head race for desilting purposes 

  3. COption (c)

    which is used to regulate the water supply to a power plant

  4. DOption (d)

    None of the above

View answer & explanation

Correct answer

Answer: (a)A.

which is provided at the foot of a spillway for energy dissipation

Why this answer is correct

Explanation:

  • According to IS 4997-1968: Stilling basin is a structure in which all or part of energy dissipating action is confined. In stilling basin kinetic energy first causes turbulence and is ultimately lost as heat energy.
  • Stilling basin is located at downstream of the spillway and is used for energy dissipation.

Type of Stilling Basin:

  1. Hydraulic jump-type stilling basin
  • Horizontal apron type;
  • sloping apron type;       2. Jet diffusion and free jet stilling basin
  • Jet diffusion basin;
  • Free jet stilling basin;
  • Hump stilling basin;
  • Impact stilling basin;

​

Question 45

A penstock is 3000 m long. Pressure wave travels in it with a velocity 1500 m/s. If the turbine gates are closed uniformly and completely in 4.5 s, then it is called a

  1. AOption (a)

    slow closure

  2. BOption (b)

    sudden closure

  3. COption (c)

    uniform closure

  4. DOption (d)

    rapid closure

View answer & explanation

Correct answer

Answer: (a)A.

slow closure

Why this answer is correct

Concept:

Critical Time of closure is given by:

tc=2LC{t_c} = \frac{{2L}}{C}

Where 2L/C = Time taken by pressure wave to travel from the valve to the tank and from tank to valve.

L = Length of pipe, C = Velocity of pressure wave

(i) The valve closure is gradual if, tc>2LC{t_c} > \frac{{2L}}{C}

(ii) The valve closure is sudden if, tc<2LC{t_c} < \frac{{2L}}{C}

Calculation:

Given,

L = 3000 m, C = 1500 m/s, tc = 4.5 second

Critical time of closure, tc=2LC{t_c} = \frac{{2L}}{C}

⇒ tc=2 × 30001500=4sec⁡{t_c} = \frac{{2 \ \times\ 3000}}{{1500}} = 4\sec

∵ tc (4.5 sec) > 4 sec, So it is the condition of gradual closure (slow closure).

Question 46

The relationship among specific yield (Sy), specific retention (Sr) and porosity (n) of an aquifer is

  1. AOption (a)

    Sy = Sr - n

  2. BOption (b)

    Sy = n - Sr

  3. COption (c)

    Sy = Sr + 2n

  4. DOption (d)

    Sy = Sr + n

View answer & explanation

Correct answer

Answer: (b)B.

Sy = n - Sr

Why this answer is correct

Following are the properties of aquifer

  1. Porosity (n):

Porosity is measure of storage property of aquifer

n=VvV{\rm{n}} = \frac{{{{\rm{V}}_{\rm{v}}}}}{{\rm{V}}}

n > 25 % → Sufficient amount of groundwater present

5 % < n > 25 % → Moderate amount of groundwater

n < 5 % → Less amount of groundwater

  1. Specific yield (Sy)

It is the ratio of the volume of water that is extracted from a saturated rock (due to gravity) to the total volume of the rock.

Sy=Volume;of;water;extractedVolume;of;aquifer{{\rm{S}}_{\rm{y}}} = \frac{{{\rm{Volume;of;water;extracted}}}}{{{\rm{Volume;of;aquifer}}}}

  1. Specific Retention (Sr)

Specific retention of a rock or soil is the ratio of the volume of water a rock can retain against gravity to the total volume of the rock.

The relation between porosity (n), specific yield (Sy) and surface retention (Sr)

n = Sy + Sr

Question 47

The clogging of well screens and consequent reduction in pump efficiency is indicated by

  1. AOption (a)

    high value of well losses

  2. BOption (b)

    low value of well losses

  3. COption (c)

    variable value of well losses

  4. DOption (d)

    None of the above.

View answer & explanation

Correct answer

Answer: (a)A.

high value of well losses

Why this answer is correct

Explanation:

  • Well Losses: Well losses refer to the additional head loss that occurs as water flows through the well screen and into the pump. This can be due to several factors, including the resistance offered by the well screen, clogging, or other obstructions within the well.
  • Clogging: When a well screen becomes clogged, the resistance to water flow increases, leading to a higher head loss or well loss. This is typically observed as a decrease in the efficiency of the pump, as more energy is required to draw the same amount of water.
  • High Well Losses: A high value of well losses is a direct indicator that there is an obstruction or clogging in the well screen. This condition results in increased energy consumption by the pump and reduced overall efficiency.

Question 48

When water flows over a rectangular suppressed weir, the pressure beneath the nape is

  1. AOption (a)

    slightly above atmospheric

  2. BOption (b)

    just equal to atmospheric

  3. COption (c)

    negative

  4. DOption (d)

    very high

View answer & explanation

Correct answer

Answer: (c)C.

negative

Why this answer is correct

Explanation:

  • Nappe: The nappe is the sheet of water that flows over the weir crest. When the nappe is free-falling, there is often a gap between the underside of the nappe and the top of the weir.
  • Pressure Beneath the Nappe: The pressure beneath the nappe can become negative (less than atmospheric pressure) due to the air entrainment and the free-fall of the water.
  • This creates a suction effect in the area beneath the nappe. This negative pressure can sometimes cause the nappe to cling to the weir, or in other cases, air may enter under the nappe, depending on flow conditions.

Question 49

In a barrage, the crest level is kept

  1. AOption (a)

    high with large gates

  2. BOption (b)

    high with no gates

  3. COption (c)

    low with no gates

  4. DOption (d)

    low with large gates

View answer & explanation

Correct answer

Answer: (d)D.

low with large gates

Why this answer is correct

Concept:

Barrage is a structure similar to weir with the only difference 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.

Regulator are located just u/s of a barrage in order to regulate the discharge flowing into off-taking channel and to control the entry of sediment in the channel.

Escape is a structure constructed on an irrigation channel for the disposal of surplus water from the channel. It is also known as surplus water escape or canal surplus escape. These are also called the safety valves of an irrigation channel system.

Question 50

If the peak load for a power plant equals the plant capacity, then the ratio of capacity factor to load factor will be

  1. AOption (a)

    0

  2. BOption (b)

    >1

  3. COption (c)

    <1

  4. DOption (d)

    1

View answer & explanation

Correct answer

Answer: (d)D.

1

Why this answer is correct

Concept:

Load factor (Lf):

The ratio of average load to the maximum demand during a given period is known as the load factor.

Lf=ALMDL_f = \frac{{{AL}}}{MD}

Where,

AL is average load and MD is maximum demand.

  • The load factor may be the daily load factor, monthly load factor, or annual load factor if the time period considered is a day or month, or year.
  • The load factor is always less than 1 because the average load
  • is smaller than the maximum demand.
  • The load factor plays a key role in determining the overall cost per unit generated.
  • The higher the load factor of the power station, the lesser will be the cost per unit generated.

Plant capacity factor (Pcf):

It is the ratio of actual energy produced to the maximum possible energy that could have been produced during a given period.

Pcf=ALPCP_{cf} = \frac{{{AL}}}{PC}

Where PC is the plant capacity.

Conclusion:

If plant capacity (PC) and maximum demand (MD) are equal,

PC = MD = X

From the above concept,

Load factor will be,

Lf=ALXL_f = \frac{{{AL}}}{X} ... (1)

Plant capacity will be,

Pcf=ALXP_{cf} = \frac{{{AL}}}{X} .... (2)

From equation (1) and (2),

Lf = Pcf

Hence, when maximum demand is equal to the plant capacity then the Load factor and capacity factor are equal.

50 questions in the full paper