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BPSC AE Civil Paper 6 (Civil) Mock Test - 1 - Civil Engineering (CE) MCQ


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

BPSC AE Civil Paper 6 (Civil) Mock Test - 1 for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The BPSC AE Civil Paper 6 (Civil) Mock Test - 1 questions and answers have been prepared according to the Civil Engineering (CE) exam syllabus.The BPSC AE Civil Paper 6 (Civil) Mock Test - 1 MCQs are made for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for BPSC AE Civil Paper 6 (Civil) Mock Test - 1 below.
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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 for BPSC AE Civil Paper 6 (Civil) Mock Test - 1 - 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 for BPSC AE Civil Paper 6 (Civil) Mock Test - 1 - 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

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

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

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 7

Which of the following statements is/are correct?

1. Isochrones are curves of equal pore water pressure

2. Isochrones depict the variation of the pore water pressure along with the depth of the soil sample

3. Isochrones vary with time

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

Isochrones:

Isochrones is a line on the map which connects points having an equal time of travel of the surface runoff to the catchment outlet. These are some properties of Isochrones:

  • Isochrones vary with time, It's most commonly used to depict travel times, such as drawing a 30-minute travel time perimeter around a start location. The isochrone below joins up all points within a 45-minute drive from the origin.
  • Isochrones depict the variation of the pore water pressure along with the depth of the soil sample.
  • Isochrones are mainly used for transport planning, property search, sales territory planning, etc. ​

Important Points

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

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

Detailed Solution for BPSC AE Civil Paper 6 (Civil) Mock Test - 1 - 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

Identify the layout of water distribution system given in image.

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

Dead-end System:

  • It is also called a tree system, it consists of one main supply line, from which originates (generally at right angles) a number of sub-main pipes; Each sub-main is then divided into several branch pipes, called laterals
  • This type of distribution system is suitable for older towns that have developed in a haphazard manner, without properly planned roads e.g. old cities, irregularly grown cities, etc; It is economical and simple

Gridiron System:

  • It is also known as an interlaced system or reticulation system, the mains, sub-mains, and branches are all connected with each other
  • There is no dead-end in the system as looping is provided and is most suitable for well-planned cities.
  • There is equal pressure in all pipes and multiple flow path.

Radial system:

  • If a city or town is having a system of radial roads emerging from different centers, the pipeline can be best laid in a radial method by placing the distribution reservoir at these centers
  • The water from the distribution reservoir is then supplied radially to distribution pipes and thus also known as zonal distribution

Ring System:

  • It is also called a circular system; In this system, a closed ring, either circular or rectangular, of the main pipes, is formed around the area to be served.
  • The distribution area is divided into rectangular or circular blocks, and the main water pipes are laid on the periphery of the blocks
  • This system is very suitable for towns and cities having well-planned roads.
  • It is economical and the pressure at the ends are reasonably equal

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

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

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

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

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 12
Ocean and atmospheric assimilation comes under which stage of hierarchy of options for handling hazardous waste ?
Detailed Solution for BPSC AE Civil Paper 6 (Civil) Mock Test - 1 - Question 12

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 13

Hydrologic flood routing methods make use of:

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

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 14

Expansion joints are provided if the length of concrete structures exceeds _____

Detailed Solution for BPSC AE Civil Paper 6 (Civil) Mock Test - 1 - Question 14
  • According to IS 456 : 1978, the expansion joints are provided in the structures in which marked changes in plane dimension take place abruptly.
  • They are required when the length of structure exceeds 45m
  • Expansion joints shall be so provided that the necessary movements within the minimum resistance at the joints. The structures adjacent to the joints should preferably be supported on separate columns or walls but not necessarily on separate foundations.
BPSC AE Civil Paper 6 (Civil) Mock Test - 1 - Question 15

The first watering before sowing the crop is called as

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

Some times in the initial stages before the crop is sown, the land is very dry.

In such cases the soil is wetted with water, so as to help in sowing of the crops, such an irrigation is known as paleo irrigation.

The 1st watering which is given to the crop when the crop is few centimetres high is called kor watering.

The kor watering must be applied within a fixed limited period called as KOR period.

Crop ratio is defined as the ratio of the land irrigated during the two main crop season Rabi and Kharif.

Root zone depth is the depth within the soil profile roots can effectively extract water and nutrients for its growth.

Mistake Point:

Kor watering - After sowing when the crop is few centimeters high.

Paleo - Before sowing to make the soil wet and easy to work on.

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

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

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

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 17

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

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 18
An RCC column will be considered as slender when the ratio of effective length (in depth direction) to depth ratio is:
Detailed Solution for BPSC AE Civil Paper 6 (Civil) Mock Test - 1 - Question 18

Concept:

Slenderness ratio() of an RC column:

It is the ratio of its effective length to the least lateral dimension of the column.

According to slenderness ration column is classified as follows:

  • ≤ 3 → Pedestal
  • 3 < < 12 → short column
  • 12 ≤ → Long column
BPSC AE Civil Paper 6 (Civil) Mock Test - 1 - Question 19

Identify the FALSE statement from the following, pertaining to the design of concrete structures -

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

Statement 1- False

One way slab is a slab that is supported by beams or walls or both on the two opposite sides to carry the load in one direction.

If it is supported on all four sides but the ratio of longer span (l) to shorter span (b) is equal to or greater than 2 then it is also considered a one-way slab.

The two-way slab is a slab supported by beams on all four sides/edges and the ratio of longer span (l) to shorter span (b) is less than 2.

Statement 2 – True

The assumption made in design for flexure is that plane sections normal to the axis remain plane after bending. It means that the strain at any point in the cross-section is proportional to the distance from the neutral axis. This is valid for both limit state of collapse and the working stress method.

Statement 3 - True

In the case of staircase design, the dead weight of the steps is calculated by treating the step to be an equivalent horizontal slab of thickness equal to half of rising.

Statement 4- True

Torsional Reinforcement provisions as per IS 456:2000:

  • Torsional reinforcement is required to be provided at the corners of simply supported rectangular slabs if the corner is free to lift up.
  • Torsion reinforcement shall be provided at any corners in a two-way slab which is simply supported on both edges meeting at the corners if restrained. It shall consist of top and bottom reinforcement, each with layers of bars placed parallel to the side of the slab and extending from the edges a minimum distance of 1/5th of the shorter span.
  • Torsional reinforcement need not be provided at any corner where both edges are continuous.
BPSC AE Civil Paper 6 (Civil) Mock Test - 1 - Question 20

Which of the following methods is NOT used for determining total hardness of water?

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

The methods used for determining the total hardness of water are as follows:

Versenate method is used to determine the total hardness of fresh water. The method uses a molecule called EDTA (ethylene diamine tetra acetic acid) which forms a complex with calcium and magnesium ions. EDTA is short for. A blue dye called Eriochrome Black T (EBT) is used as the indicator.

Clarks’s Method is the water softening process. It removes the hardness of water by converting bicarbonates into carbonate.

The Hoyer system is generally used for the production on large scale of pre-tensioned structural element of the structure.

Hehner’s method is used to determine the temporary hardness of water. It is determined by finding the alkalinity of water before boiling and that left after boiling.

Hence, Zeolite method is not used for determining total hardness of water.

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

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

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 22

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

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

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 23

The pipe joint commonly used in pumping station

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

Flanged joint:

  • This joint is commonly used for joining pipes in pumping stations, filter plants, hydraulic laboratories, boiler houses, etc.
  • These joints are preferred due to the easy process of assembly and disassembly, however, these connections are costly.
  • These joints can be disassembled and re-assembled when required.
  • A pipe has flanged ends on both sides of the pipe length. Both ends of pipes are joined at a proper level near one another. A hard rubber washer is placed between flanges and bolted. Flanges are generally fixed to the pipe by welding or threading.
  • In certain cases, a flange-type joint is also called a lap joint. It may also be made by forging the process and machining the pipe end. There is no leakage in flanged joints even after rapid temperature fluctuations.

Additional Information

Pipe Joints

Pipes are connected with the help of joints. A variety of joints are used in the assembly of pipes. Connecting two or more pipes together is called a fitting. Various types of joints could be used in a pipe as per the requirement. Joints are also used for multiple pipe connections and are an important component of the plumbing system. Generally, the pipe joint fitted can easily sustain the pressure created in the pipe.

Types of pipe joints Various types of pipe joints are as follows:

  1. Threaded joint
  2. Welded joint (butt welded, socket welded)
  3. Brazed joint
  4. Soldered joint
  5. Grooved joint
  6. Flanged joint
  7. Compression joint
BPSC AE Civil Paper 6 (Civil) Mock Test - 1 - Question 24

In the case of Pelton turbine installed in a hydraulic power plant, the gross head available is the vertical distance between:

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

Gross Head:

  • The gross head is the difference between the water level at the reservoir and the water level at the tailrace.

Additional Information

Net or Effective Head:

  • The head available at the inlet of the turbine is known as the net or effective head.
BPSC AE Civil Paper 6 (Civil) Mock Test - 1 - Question 25

In which one of the following grades of a highway is an emergency escape ramp Provided?

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

Emergency Escape Ramps:

  • For use by heavy vehicles losing control because of brake failure (caused by heating or mechanical failure), an emergency escape ramp is provided on a long, steep downgrade.
  • To decelerate and stop away from the main traffic stream, these vehicles can use the ramp.
  • The emergency escape ramps are of four basic types: sandpile, descending grade, horizontal grade, and ascending grade.
  • The loose sand or an arresting bed of loose gravel provides the rolling resistance on the ramps.
  • A force of gravity opposite to the vehicle movement is provided by the ascending grade ramp, and, therefore, its length can be shorter than the descending and horizontal grade ramps.
  • Depending on a particular topographic situation, the specific ramp type is applicable.
BPSC AE Civil Paper 6 (Civil) Mock Test - 1 - Question 26

Advantages of asphaltic concrete (Bituminous Concrete) are :

(a) Durability
(b) Imperviousness
(c) Load spreading properly
(d) Quickly openable to traffic
(e) Good skid Resistance

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

Asphaltic concrete is a composite material made from a mixture of aggregates, filler, and bitumen or tar as binding material.

Advantages of Asphalt Concrete:

  1. It is less expensive than concrete.
  2. It takes less time to build a road i.e., quickly openable to traffic.
  3. It is long-lasting and only requires surface maintenance.
  4. It is impervious or it can be used as a waterproofing agent in many works.
  5. The load transfer through asphaltic concrete is via grain-to-grain contact.
  6. It provides better riding quality due to its flexibility properties i.e., it has better skid resistance.
  7. It reduces the extent and severity of reflective cracking in pavements.

Disadvantages of Asphalt Concrete:

  1. It becomes softer in hot climates.
  2. When the asphalt is not laid properly, it causes cracks and issues.
  3. The construction method requires heavy equipment to be installed.
BPSC AE Civil Paper 6 (Civil) Mock Test - 1 - Question 27

Acceptable noise level in dB for auditorium is:

Detailed Solution for BPSC AE Civil Paper 6 (Civil) Mock Test - 1 - 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

The percentage compensation ingradient for ruling gradient of 4% and horizontal curve of radius 760 m is

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

Grade compensation:

  • The grade compensation is the reduction in gradient on a horizontal curve.
  • As per IRC, grade compensation is given by (30 + R)/R % and limited to maximum of 75/R % , where R = Radius of the curve in 'm'.
  • Grade compensation is not required for the grades flatter than 4%.

And the given gradient is 4% only, so we can't do any grade compensation to it, as the value will go less than 4 % and it will fail the above criteria

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

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

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