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BPSC AE Civil Paper 6 (Civil) Mock Test - 1 Free Online Test 2026


Full Mock Test & Solutions: BPSC AE Civil Paper 6 (Civil) Mock Test - 1 (50 Questions)

You can boost your Civil Engineering (CE) 2026 exam preparation with this BPSC AE Civil Paper 6 (Civil) Mock Test - 1 (available with detailed solutions).. This mock test has been designed with the analysis of important topics, recent trends of the exam, and previous year questions of the last 3-years. All the questions have been designed to mirror the official pattern of Civil Engineering (CE) 2026 exam, helping you build speed, accuracy as per the actual exam.

Mock Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 60 minutes
  • - Total Questions: 50
  • - Analysis: Detailed Solutions & Performance Insights

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BPSC AE Civil Paper 6 (Civil) Mock Test - 1 - Question 1

A wastewater sample of 2 ml is made up to 300 ml in a BOD bottle with distilled water. The initial D.O. of the sample is 8 mg/l and after 5 days it is 2mg/l. Its BOD is:

Detailed Solution: Question 1

Concept:

Biochemical Oxygen demand (BOD):

It is the total amount of oxygen required to oxidize the biodegradable organic matter present in wastewater through microbial utilization of organics.

BOD = (Initial Dissolve Oxygen - Final Dissolve Oxygen) × Dilution Factor

where

Calculation:

Given:

DOInitial = 8 mg/L

DOFinal = 2 mg/L

Dilution Factor = 300/2

BOD = (8-2) x (300/2) = 900 mg/L 

BPSC AE Civil Paper 6 (Civil) Mock Test - 1 - Question 2

As per Lacey's theory, the wetted perimeter of a regime-channel for a discharge of 100 m3/s, is

Detailed Solution: Question 2

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

Hydraulic mean depth,

Area of the channel section,

A = Q / V

Wetted perimeter,

P = 4.75√Q

Regime mean velocity,

Bed Slope,


Calculation:

Given data,

Q = 100 m3/s

P = 4.75√Q = 4.75√100 = 47.5m

BPSC AE Civil Paper 6 (Civil) Mock Test - 1 - Question 3

If modulus of elasticity of the subgrade is 25 MPa, then deflection at the surface of flexible pavement due to a wheel load of 40 kN and a tyre pressure of 0.6 MPa will be:

Detailed Solution: Question 3

Concept:

Burmister’s Method:

Assumptions

  • Materials in each layers are isotropic, homogeneous and elastic.
  • Pavement forms a stiffer layer having higher value of E than that of subgrade.
  • Layers are in constant contact.
  • Surface layer is infinite in horizontal direction but finite in vertical direction.

This method gives the following design deflection values for pavements with different layers –

  1. For flexible pavement –

  1. For rigid pavement –

Where, p = contact pressure at road surface due to wheel load (kg/m2)

Es = modulus of elasticity of subgrade (kg/cm2),

a = radius of contact area(cm) and

F2 = deflection factor which depends upon Es/EP and h/a, for single layer system F2 = 1.

Formula used:-

Design deflection for single-layered flexible pavement is given by,

Radius of contact area is given by,

Calculation:

Wheel Load, P = 40 kN

Modulus of elasticity for subgrade and pavement, ES = EP = 25 MPa

Tyre pressure, p = 0.60 MPa

Radius of contact (a)

Design deflection for flexible pavement

BPSC AE Civil Paper 6 (Civil) Mock Test - 1 - Question 4

Which of following statements is/are true for manometers?

l) Manometers are easy to operate.

ll) Manometers do not require frequent calibration.

lll) Manometers are made of steel.

Detailed Solution: Question 4

The correct statements are:-

(i) manometers are easy to operate

(ii) manometers do not require frequent calibration

Key Points :

  • manometers are a device used to measure the pressure of the fluid by balancing of the fluid column
  • They are generally made up of Glass
  • They are generally a tube of Glass in U - Shape
  • They are simple in construction and are easy to operate
  • They do not require frequent calibration
  • They are generally calibrated once a year.

Additional Information

Types of manometer :

  • piezometer is wed to measure very low pressure
  • Differential monometers are used to measure the difference of pressure between two points.

BPSC AE Civil Paper 6 (Civil) Mock Test - 1 - Question 5

In the design of storm sewers 'time of concentration' is relevant to determine the

Detailed Solution: Question 5

Time of concentration:

  • Time of Concentration is the time required by the entire drainage area to contribute to the runoff is called the time of concentration or time required by the most extreme point in the drainage to reach the point of interest.
  • In Other Words, it is the maximum time taken by the rainwater to reach the outlet of the basin.
  • The rainfall intensity corresponding to a duration of Tc and desired period of exceedance, P is found from the rainfall frequency-duration relationship for the given catchment area.
  • This relationship is given by


Where, IC = Rainfall intensity, TC = Time of concentration, T = Return period = 1/P, and K, a, x, n are coefficients corresponding to a particular location

Additional Information

Time of concentration is also given by Kirpich equation

TC = 0.01947 x L0.77 × S-0.385

Where TC = Time of concentration in minutes, L = Maximum length of travel of water in meters, S = Slope of the catchment = Δ H/L, Δ H = Head difference

BPSC AE Civil Paper 6 (Civil) Mock Test - 1 - Question 6

Select which is not the reason to provide extrawidening:

Detailed Solution: Question 6

Extra Widening

The extra width of carriageway that is required on a curved section of a road over and above that is required on a straight alignment is known as Extra Widening.

Extra widening of road is given by,

Where,

Wm = Mechanical widening, Wps = Psychological widening, n = number of lanes, l = length of wheelbase, R = Radius of curve, V = Velocity of vehicle (in kmph)

Note:

Reasons to provide extra widening are:

(i) To avoid off tracking due to rigidity of wheelbase

(ii) To counteract transverse skidding

(iii) To increase the visibility at curves

(iv) To encounter psychological tendency while overtaking operation

Important Point:

IRC recommended values of extra widening for single and two-lane pavement are given below:

1) If R > 300 m, then extra widening is not provided.

2) If R < 50 m, Then extra widening is provided at the inner edge.

3) If 50 < R < 300 m, then extra widening is provided at both edges.

BPSC AE Civil Paper 6 (Civil) Mock Test - 1 - Question 7

A rectangular suppressed weir is a device used to measure stream flow in an open channel. For a rectangular suppressed weir flowing free, the discharge Q is related to the head H over the weir as _________.

Detailed Solution: Question 7

Concept:

NOTE: Length is not given in the question or given options, so we have to calculate discharge per unit length in the given weir

Weir or notch is a physical structure of masonry constructed across the channel width to calculate the discharge of the channel section.

Rectangular (suppressed) Notch:

The discharge through a rectangular notch weir is,

Where, Q = discharge of fluid, Cd = Coefficient of discharge and H = height of water above the notch

Additional Information

Triangular Notch:

A V-notch weir is also called the triangular notch or weir. The discharge over a triangular weir or notch is given by the:

Where, Q = discharge of fluid, Cd = Coefficient of discharge, θ = Notch angle and H = height of water above the notch

Trapezoidal weir (or) Notch:

Where, (θ/2) = weir angle of inclination with the vertical.

Cd1= Coefficient of discharge for rectangular portion.

Cd2= Coefficient of discharge for the triangular portion

BPSC AE Civil Paper 6 (Civil) Mock Test - 1 - Question 8

Depth-Area-Duration curves of precipitation are drawn as:

Detailed Solution: Question 8

Concept:

DAD curve:

  • The areal characteristics of a rain storm are represented by a depth-area-duration curve. Once the sufficient rainfall records for the region are collected the basic or raw data can be analyzed and processed to produce useful information in the form of curves or statistical values for use in the planning of water resources development projects.
  • Many hydrologic problems require an analysis of time as well as areal distribution of storm rainfall. Depth-Area-Duration (DAD) analysis of a storm is done to determine the maximum amounts of rainfall within various durations over areas of various sizes
  • The depth area curve for a particular storm can be calculated with the equation

:

Where P is average depth of rainfall over an Area (in cm)

P0 is the highest amount of rainfall at the storm center (in cm)

A is area (km2)

K and n are constants for a particular region: n is < 1.

With increase in Area P value decreases.

To find out how much of rainfall will occur in an area by converting rainfall data to areal rainfall data, Depth area duration curve is used.

DAD curve expresses graphically the relation between progressively decreasing average depth of rainfall over a progressively increasing area from the centre of the storm outward to its edges for a given duration of rainfall.

For a given area the maximum average depth of rainfall increases with storm duration.

BPSC AE Civil Paper 6 (Civil) Mock Test - 1 - Question 9

When the size of water droplets is under 0.5 mm and its intensity is < 1 mm per hour, what is it called as?

Detailed Solution: Question 9

Concept:

Forms of the precipitation are as follows:

BPSC AE Civil Paper 6 (Civil) Mock Test - 1 - Question 10

If the odour intensity or pO value is 4, then what is the interpretation?

Detailed Solution: Question 10

Concept:
PO value:

  • The Odour in water changes with temperature. the water Odour is tested at 20oC to 25oC
  • The apparatus used to measure Odour in water is known as osmoscope.
  • A commercial osmoscope is graduated with a PO value from 0 - 5.

PO value is 4 then the interpretation is a very distinct Odour.

BPSC AE Civil Paper 6 (Civil) Mock Test - 1 - Question 11

The depth of water required to bring the soil moisture content of a given soil up to its field capacity is known as

Detailed Solution: Question 11

Readily available moisture-

  • It is that portion of the available moisture that is most easily extracted by the plants and is approximately 75 to 80% of the available moisture.

Field Capacity-

  • Immediately after a rain or irrigation water application, when all the gravity water has drained down to the water table, a certain amount of water is retained on the surfaces of soil grains by molecular attraction and by loose chemical bonds. This water cannot be easily drained under gravity and this is called field capacity.

Permanent wilting point-

  • It is the water content at which the plant can no longer extract sufficient water for its growth, and wilts up.

Soil moisture deficiency-

  • The water required to bring the soil moisture content of a given soil to its field capacity is called field moisture deficiency or soil moisture deficiency.

BPSC AE Civil Paper 6 (Civil) Mock Test - 1 - Question 12

How many major crossing conflict points exist on a right-angled two-lane road intersection with two-way traffic on both roads?

Detailed Solution: Question 12

For the intersection of two-lane two-way roads from both sides:

Major conflicting points = 16, Minor conflicting points = 8 and thus total conflicting points = 24

For the intersection of two-lane two-way rod and two-lane one-way road:

Major conflicting points = 7 Minor conflicting points = 4 and thus total conflicting points = 11

For the intersection of a two-lane one-way rod from both the sides:

Major conflicting points = 4 Minor conflicting points = 2 and thus total conflicting points = 6

Important Points:

BPSC AE Civil Paper 6 (Civil) Mock Test - 1 - Question 13

Which of the following is NOT a function of a cross-regulator?

Detailed Solution: Question 13

Canal-regulator:

  • Canal Regulators are required on a channel to regulate the supply of water.
  • For effecting control of flow on the irrigation system some structures are required to be constructed at the barrage and also on the off-taking distributary canals and these structures are : 1. Canal Head Regulator 2. Canal Cross Regulator

​Canal Cross Regulator:

  • It is a control structure constructed across a canal to regulate the level of the water upstream of the cross regulator.
  • It is general practice to provide a cross regulator in the main canal downstream of the off-taking canal.
  • The functions of cross regulator:
    (a) To allow the desired discharge in the off-taking canal that is situated upstream of the cross regulator,
    (b) To release water from the canals by operating along with escapes,
    (c) To maintain the water surface slopes in appropriate conjunction with the flow control over falls so the canals may attain their regime slopes and cross sections,
    (d) To control the discharge released by one canal into another canal or lake.

​Canal Head Regulator:

  • A canal head regulator which is located just upstream of a barrage (or placed appropriately near about a reservoir) is provided to help,
    (a) regulate the discharge flowing into the off-taking canal, and
    (b) control the entry of sediment into the canal.
  • A head regulator is generally aligned at an angle of 90° to 110° to the axis of the barrage in order to minimize the entry of sediment entering into the channel, besides
    preventing backflow and stagnant pools in the under-sluice pocket that lies in the vicinity of the regulator.

BPSC AE Civil Paper 6 (Civil) Mock Test - 1 - Question 14

Calculate conveyance efficiency. The quantity of water pumped into the farm distribution system is 2 cumec. 1.6 cumec is delivered to the turn - out which is at a distance of 1 km from the well.

Detailed Solution: Question 14

Concept:

Where ηc = Conveyance efficiency.

wf = Water delivered to the field.

wr = Water delivered from the reservoir.

Calculation:

Given

Water delivered to the field (wf) = 1.6 cumec

Water delivered from the reservoir (wr ) = 2 cumec

ηc = (1.6/2)  x 100

ηc = 80%

Conveyance efficiency (ηc) = 80%

BPSC AE Civil Paper 6 (Civil) Mock Test - 1 - Question 15

The ratio of quantity of liquid discharged per second from the pump to the quantity of liquid passing per second through the impeller is known as

Detailed Solution: Question 15

Concept:

In the case of a centrifugal pump, the power is transmitted from the shaft of the pump to the impeller and then from the impeller to the water. The following are the important efficiencies of a centrifugal pump:

Manometric Efficiency (ηman): It is the ratio of the manometric head to head imparted by the impeller to the water.

Mechanical Efficiency (ηm): It is the ratio of the power available at the impeller to the power at the shaft of the centrifugal pump.

Overall Efficiency (ηo): It is defined as a ratio of the power output of the pump to the power input to the pump.

ηo = ηman × ηm

Volumetric efficiency: It is the ratio of the quantity of liquid discharged per second from the pump to the quantity of liquid passing per second through the impeller.

BPSC AE Civil Paper 6 (Civil) Mock Test - 1 - Question 16

Ocean and atmospheric assimilation comes under which stage of hierarchy of options for handling hazardous waste ?

Detailed Solution: Question 16

Concept-

Hazardous waste options:

A three-stage hierarchy of options for handling hazardous wastes is:

First stage:

  • The top tier includes plant options such as process manipulation, recycling, and reuse options that reduce the production of hazardous waste in the first place. It also contains the most desirable options.

Middle stage:

The middle stage highlights processes that convert hazardous waste to less hazardous or non-hazardous substances include

  • Incineration
  • Land treatment
  • Ocean and atmospheric assimilation
  • Chemical, physical, and biological treatments
  • Thermal treatments

Last stage:

  • This is the stage that is the least preferred or desirable tier that is perpetual storage the cheapest alternative.
  • A few processes include landfill, underground injection, arid region unsaturated zone, surface impoundments, salt formations, and waste piles.

BPSC AE Civil Paper 6 (Civil) Mock Test - 1 - Question 17

Hydrologic flood routing methods make use of:

Detailed Solution: Question 17

Flood routing: It is the technique of determining the flood hydrograph at a section of a river by utilizing the data of flood flow at one or more upstream sections.

Hydrologic routing methods employ the equation of continuity.

In its simplest form, inflow to river reach is equal to the outflow of the river reach plus the change of storage.

I = O + (ΔS/Δt)

Where,

I = Average inflow to the reach during ∆t,

O = Average outflow to the reach during ∆t, and

ΔS = Change in storage

Note:

In the case of hydraulic routing, the continuity equation together with the equation of momentum is used.

BPSC AE Civil Paper 6 (Civil) Mock Test - 1 - Question 18

The rain waterholes in the parapet or in edging is called as-

Detailed Solution: Question 18

Weep holes are provided in masonry walls, parapets, wing walls and earth retaining structures for safe drainage of rain water

Weep holes serve the purpose in the following ways.

  1. Relieving hydrostatic pressure on the walls.
  2. Reducing water pressure, thus reducing the reinforcement requirements and thickness of walls
  3. Providing ventilation facility

BPSC AE Civil Paper 6 (Civil) Mock Test - 1 - Question 19

Match List-I (Control structures) with List-II (Functions of the control structures) and select the correct answer:

Detailed Solution: Question 19

Canal drop:

It is a structure constructed across a channel to permit lowering down of water level in order to dissipate the surplus energy possessed by the falling water which may otherwise scour the bed and banks of the channel.

Canal Escape:

During floods/rains, when the canal carries water beyond its full capacity, water is released through Canal Escapes.

Canal Cross regulators:

It is provided to admit water into the off taking canal and to regulate the supplies into the canal ( Depth of flow). It also regulated the discharge to be passed into the canal and controls the silt entry into the canal.

Canal outlets:

A canal outlet or a module is a small structure built at the head of the watercourse so as to connect it with a minor or a distributary channel. In other words, canal outlets are devices to regulate the flow of water from a bigger channel into a smaller one.

BPSC AE Civil Paper 6 (Civil) Mock Test - 1 - Question 20

The discharge of water through a rectangular channel of width 6 m is 18 m3/s, when the depth of flow of water is 3 m, then the specific energy of flowing water is

Detailed Solution: Question 20

Specific energy

In open channel flow, specific energy ( E ) is the energy length, or head, relative to the channel bottom. Specific energy is expressed in terms of kinetic, potential, and internal energy.

E = y + v2/2g

Q=A x V ,Q = Discharge in m3/s , V = velocity of water , y = depth

Q = B x y x V

Calculation:

Given data;

Width = 6 m

depth of water = 3 m and

discharge = 18 m3/s

The specific energy ( E ) is

Note:

The most appropriate answer is option 1.

BPSC AE Civil Paper 6 (Civil) Mock Test - 1 - Question 21

Match the following sewer components with their functions.

Detailed Solution: Question 21

Catch basin:

  • Catch basins are provided to stop the entry of heavy debris present in the storm water into the sewers.
  • At the bottom of the basin, space is provided for the accumulation of impurities.
  • The perforated cover is provided at the top of the basin to admit rainwater into the basin.
  • A hood is provided to prevent the escape of sewer gas.

Lamp hole:

  • It is an opening or hole constructed in a sewer for the purpose of lowering a lamp inside it.
  • It consists of stoneware or concrete pipe, which is connected to a sewer line through a T-junction.

Flushing tank:

  • An arrangement to hold and throw water into the sewer for the purpose of flushing.
  • When the gradient of the sewers is flat and the velocity of sewage is low, the suspended matter starts settling and causes clogging of the sewer, at such places, where self-cleansing velocity is not available, a flushing tank is provided to flush the sewer.

Gullies:

  • Gully is a type of drainage inlet of the drainage system. Though it is designed to catch stormwater, road gullies and catch pits are intended to catch stormwater at different locations.

BPSC AE Civil Paper 6 (Civil) Mock Test - 1 - Question 22

​Deep beams are designed for

Detailed Solution: Question 22

Deep beams are structural elements loaded as simple beams in which a significant amount of the load is carried to the supports by a compression force combining the load and the reaction. As a result, the strain distribution is no longer considered linear, and the shear deformations become significant when compared to pure flexure.

In view of ample shear strength, deep beams are primarily recommended as transfer girders. These members transfer loads through-loading face to supports in the transverse direction. The deep horizontal members predominantly fail in shear rather than flexure. These beams are characterized with small span-to-depth ratio. Pile caps, corbel, brackets, foundation walls and off-shore structures are few examples of RC deep beams.

According to IS 456-2000 a beam shall be deemed to be a deep beam when the ratio of effective span-to-overall depth, l/D is less than:

1) 2.0, for simply supported beam; and

2) 2.5, for a continuous beam.

Though different codes define deep beams in different clear span-to-depth ratios, as a general rule deep beams are recognized by their relatively small span-to-depth ratio.

Hence, Deep beams are designed for bending moment and checked for shear.

IS 456:2000 page 51.

BPSC AE Civil Paper 6 (Civil) Mock Test - 1 - Question 23

According to IS: 4111-1986, the spacing of manholes above ______ may be allowed on straight runs for sewers of diameter above 900 to 1500 mm.

Detailed Solution: Question 23

Concept:

According to Clause, 3.2.2.1 of IS: 4111 (Part 1) - 1986:

The spacing of manholes above 90 to 150 m may be allowed on straight runs for sewers of diameter above 900 to 1500 mm and above 150 to 200 m the spacing of manholes may be on straight runs for sewers of 1.5 to 2.0 m dia, which may further be increased up to 300 m for sewers of over 2.0 m diameter. A spacing allowance of 100 m per 1 m dia of sewer is a general rule in the case of very large sewers.

Additional Information:

According to Clause 3.2.1 of IS: 4111 (Part 1) - 1986:

  • Manholes should be built at every change of alignment, gradient, or diameter, at the head of all sewers and branches, and at every junction of two or more sewers.
  • On sewers that are to be cleaned manually which cannot be entered for cleaning or inspection the maximum distance between manholes should be 30 m.

BPSC AE Civil Paper 6 (Civil) Mock Test - 1 - Question 24

The estimated value of elastic modulus (GPa) of concrete of M65 grade as per IS 456 would be approximately:

Detailed Solution: Question 24

Concept:
Elastic modulus of concrete,

Ec = 5000 x √fck

where, fck = grade of concrete

Long-term elastic modulus of concrete

Eθ = Ec / (1 + θ)

θ = creep coefficient

Calculation:

Modulus of elasticity as per IS 456: 2000 is given by

E = 5000√65 = 40311 MPa = 40.311 GPa ≈ 40GPa

BPSC AE Civil Paper 6 (Civil) Mock Test - 1 - Question 25

A canal is constructed with the idea of feeding two or more canals is known as

Detailed Solution: Question 25

Feeder canals: It is constructed with the idea of feeding two or more canals. When the main canal is divided into two canals then it is called a feeder canal. e.g. Lower Chenab canal feeder.

Irrigation canal: A canal aligned along the boundaries of cultivatable areas in order to supply water for the purpose of agriculture is said to be an irrigation canal.

Inundation canals: These are long canals taken off from large rivers. They receive water when the river stage is high especially during the flood. Thus they are used to divert the flood water of rivers.

Perennial Canals: These canals are lined to dams and barrages and receive or carry water throughout the year. They never dry and able to irrigate vast areas.

Carrier canals: It is the one which besides doing irrigation carries water for another canal. It’s a link canal and has an outlet e.g. Upper Chenab Canal in Punjab.

Navigation canal: A canal that is constructed especially for navigational purposes is known as a navigation canal. The water level required in a navigation canal is generally a lot higher to accommodate large ships, vessels, etc.

BPSC AE Civil Paper 6 (Civil) Mock Test - 1 - Question 26

Permanent hardness is removed by :

(a) Lime soda process

(b) Boiling

(c) Demineralisation process

(d) Base exchange process

Detailed Solution: Question 26

The hardness of water:

  • Hardness in water is that characteristic which prevents the formation of sufficient leather or foam when such hard water is mixed with soap.
  • It is usually caused by the presence of calcium and magnesium salts present in water, which form scum by reaction with soap.

Water softening:

  • The process of removal of hardness from water is known as Water softening.

Removal of hardness;

1. Temporary hardness:

  • The temporary or carbonate hardness is caused by carbonates and bicarbonates of Ca and Mg or bivalent cations.

The methods of removing temporary hardness:

  • Boiling
  • Addition of lime.

2. Permanent hardness:

The permanent or non-carbonate hardness, which is caused by the sulphates, chlorides and nitrates of Ca and Mg is removed by special methods.

The methods of removing permanent hardness

  • Lime soda process
  • Base exchange process
  • Demineralization​

Lime Soda Process

  • Chemicals normally used are lime {calcium hydroxide, Ca(OH)2} and soda ash (sodium carbonate, Na2CO3).
  • Lime is used to remove chemicals that cause carbonate hardness.
  • Soda ash is used to remove chemicals that cause non-carbonate hardness.
  • Lime soda softening cannot produce water completely free of hardness because of the solubility of CaCO3 and Mg(OH)2.
  • Thus the minimum calcium hardness that can be achieved is about 30 mg/L as CaCO3, and the magnesium hardness is about 10 mg/L as CaCO3.

Ion-exchange method:

  • ​This method is also called the zeolite/permutit process.
  • In this method, the permanent hardness of water is removed by using resins.
  • Ca++/Mg++ ions are exchanged with Cl-, SO-24 ions are exchanged with anion exchange resin (RNH2OH).
  • Demineralized water is formed in this process.
  • Hydrated sodium aluminium silicate is zeolite/permutit.
  • For the sake of simplicity, sodium aluminium silicate (NaAlSiO4) can be written as NaZ.
  • When this is added to hard water, exchange reactions take place.

Permanent hardness can also be removed by:-

Treatment with washing soda (sodium carbonate):-

  • Washing soda reacts with soluble calcium and magnesium chlorides and sulphates in hard water to form insoluble carbonates.
  • Sodium carbonate is water-soluble in nature and hence, adds a large number of carbonate ions to the water.
  • Because of this, the calcium ions come from hard water and the carbonate ions from sodium carbonate.

Note: Permanent Hardness is not removed by Boiling.

BPSC AE Civil Paper 6 (Civil) Mock Test - 1 - Question 27

Acceptable noise level in dB for auditorium is:

Detailed Solution: Question 27

Acceptable Indoor noise levels for various types of buildings

Important Points

Acceptable outdoor noise levels in residential areas

BPSC AE Civil Paper 6 (Civil) Mock Test - 1 - Question 28

Statement (I): Instantaneous unit hydrograph (IUH) is used in theoretical analysis of rainfall excess-runoff characteristics of a catchment.

Statement (II): For a given catchment, IUH, being independent of rainfall characteristics, is indicative of the catchment storage characteristics.

Detailed Solution: Question 28

The unit hydrograph (UH) of a catchment is defined as the hydrograph resulting from an effective rainfall of 1 cm evenly distributed over the basin during the time T. We can plot any no. of UHs for a given basin by changing the duration of rainfall i.e. we may have 2 hr UH, 3hr UH, 4 hr UH… so on.

Therefore, theoretically an infinite number of unit graphs are possible for a given basin.

Hydrograph is the graphical representation of discharge Vs time at any given point on stream.

The following are some factors which affect the hydrograph:

1. Rainfall characteristics: rain fall duration, its distribution over the basin, rainfall intensity.

2. Catchment characteristics: slope of basin, basin area, shape of basin.

3. Land use pattern

4. Soil moisture conditions

BPSC AE Civil Paper 6 (Civil) Mock Test - 1 - Question 29

________ is the number of vehicles occupying a unit length of a lane of roadway at given instant.

Detailed Solution: Question 29

Traffic Density

  • It can be defined as the number of vehicles per unit length.
  • Unit - Veh/Km

Traffic Capacity

  • the maximum number of vehicles in a lane or a road that can pass a given point in unit time, usually an hour, i.e., vehicles per hour per lane or roadway.
  • Unit - Veh/Hr/lane

Traffic volume:

The number of vehicles moving in a specified direction on a roadway that passes a given point during a specific unit of time is called traffic volume.

The relation between traffic (density, volume, and speed) is given by:
Traffic volume(q) = Speed(v) x Traffic density(k)

Where,

q is in veh/hr, v is in km/hr, k is in veh/km

Important PointsThe relationship between the three different fundamental quantities are depicted below:

BPSC AE Civil Paper 6 (Civil) Mock Test - 1 - Question 30

If the specific energy at the up stream section of a rectangular channel is 3 m and minimum specific energy is 2.5 m, the maximum height of jump without causing afflux will be:

Detailed Solution: Question 30

Flow-through hump:

  • When ΔZ is equal to ΔZm (ΔZm is the minimum height of hump for critical flow or maximum height of hump for which upstream flow is not affected) then point Q corresponding to the depth of flow y2 shift to point C corresponding to the depth of flow yc, E1 = Ec + ΔZm, it means at ΔZm, the flow at downstream becomes critical and at ΔZ < ΔZm, the flow over hump remains subcritical.

For avoiding the afflux at downstream, apply energy equation u/s and d/s, we get

E1 = Ec + ΔZm

Where, E1 = Energy at U/S, Ec = Energy at D/S and ΔZm = Change in hump height

ΔZm = 3 - 2.5 = 0.5 m

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