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MPPSC AE CE Mock Test - 5 - Civil Engineering (CE) MCQ


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30 Questions MCQ Test - MPPSC AE CE Mock Test - 5

MPPSC AE CE Mock Test - 5 for Civil Engineering (CE) 2025 is part of Civil Engineering (CE) preparation. The MPPSC AE CE Mock Test - 5 questions and answers have been prepared according to the Civil Engineering (CE) exam syllabus.The MPPSC AE CE Mock Test - 5 MCQs are made for Civil Engineering (CE) 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for MPPSC AE CE Mock Test - 5 below.
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MPPSC AE CE Mock Test - 5 - Question 1

How many forest territorial circles has Madhya Pradesh been divided into?

Detailed Solution for MPPSC AE CE Mock Test - 5 - Question 1

The correct answer is 16.
Explanation

  • Madhya Pradesh has been divided into 16 forest territorial circles.
  • Madhya Pradesh has 10 National Parks and 25 sanctuaries.
  • The forest circles comprise 63 forest divisions.
  • Each agro-climatic zone has a research extension unit.

Other Related Points

  • As per the latest estimates of the Forest Survey of India, published in the State of Forest Report (IFSR) 2017, the total forest cover of M.P. is 77,414 km2., which is 25.11% of the total land area.
  • The maximum area of Tropical Dry Deciduous Forests is found in Madhya Pradesh.
  • Mizoram has the highest area under forest cover in terms of percentage.
  • Madhya Pradesh has the highest area under forest cover in terms of area.
MPPSC AE CE Mock Test - 5 - Question 2

The core part of an operating system which is one of the first programs loaded on start-up and manages computer system resources is:

Detailed Solution for MPPSC AE CE Mock Test - 5 - Question 2

The correct answer is Kernel.

Explanation

  • The first program loaded on start-up and manages computer system resources is Kernel.
  • A kernel is the central part of an operating system.
  • It manages the operations of the computer and the hardware, most notably memory and CPU time.
  • There are five types of kernels:
    • A microkernel, which only contains basic functionality
    • A monolithic kernel, which contains many device drivers
    • Hybrid Kernel
    • Exokernel
    • Microkernel

Other Related Points

  • Resources under the Operating System control:
    • The processor.
    • Main memory.
    • Input/Output Device.
    • Secondary storage devices.
    • Communication devices and ports.
MPPSC AE CE Mock Test - 5 - Question 3

Which district in Madhya Pradesh has the maximum population growth rate?

Detailed Solution for MPPSC AE CE Mock Test - 5 - Question 3

The correct answer is Indore.
Explanation

  • Indore has the maximum population growth rate in Madhya Pradesh.
  • As per Population Census 2011, the total population of the Indore district is 3,276,697.
  • The population growth rate of Indore is 32.88 %.
  • The population of Indore in 2011 increased by 32.88% compared to the population in 2001.
  • Indore District recorded an increase of 34.30% to its population compared to 1991.
  • As per Population Census 2011, Madhya Pradesh has a population of approximate 7.27 Crores.
  • The population density of Madhya Pradesh state is 236 per sq km.

Other Related Points

  • The population growth rate of Jabalpur is 14.51 % compared to the 2001 census.
  • The population growth rate of Bhopal is 28.62 % compared to the 2001 census.
  • The population growth rate of Gwalior is 24.50 % compared to the 2001 census.
MPPSC AE CE Mock Test - 5 - Question 4

In which place are the Caves of Mrigendra Nath in Madhya Pradesh located?

Detailed Solution for MPPSC AE CE Mock Test - 5 - Question 4

The correct answer is Raisen.
Explanation

  • Caves of Mrigendra Nath can be found in Raisen district of Madhya Pradesh.
  • These caves are located at Patni village in Raisen district of Madhya Pradesh.
  • They were discovered in 2009 by the Archeological Department of Madhya Pradesh.
  • These caves have striking similarities with the caves of Bhimbetka
  • The caves of Bhimbetka are also located in the Raisen district of Madhya Pradesh.
  • Caves of Bhimbetka were given World Heritage Site status by UNESCO in 2003.

Other Related Points

  • Caves of Udaiyagiri are located in Vidisha district of Madhya Pradesh.
  • Adamgarh caves are located in the Hoshangabad district on the banks of the Narmada river.
  • Bagh caves are located in Dhar district on the banks of Bagh River.
  • Bagh caves are very similar to Ajanta caves in their structural design and planning.
MPPSC AE CE Mock Test - 5 - Question 5

Which historical event significantly shaped the anti-infiltration narrative behind Assam’s eviction drives?

Detailed Solution for MPPSC AE CE Mock Test - 5 - Question 5

The Assam Agitation (1979–1985) and the subsequent Assam Accord were pivotal in shaping the anti-infiltration narrative targeting Bengali-speaking Muslims. This historical context drives the current eviction policies, which aim to protect local land and heritage but raise human rights and regional tensions.

MPPSC AE CE Mock Test - 5 - Question 6

Who among the following are the beneficiaries of ‘Mukhyamantri Kaushalya Yojana’ of Madhya Pradesh Government?

Detailed Solution for MPPSC AE CE Mock Test - 5 - Question 6

The correct answer is Women.

  • Chief Minister Kaushalya Yojana:
    • It is being launched to train the women and make women self-reliant by the state government of Madhya Pradesh. So, option 4 is correct.
    • Objectives of this scheme:
      • Provide the necessary skill for women to get employment and self-employed.
      • Ensure the participation of women by providing skills in non-conventional areas.

Important Points

  • List of Important Centrally Sponsored Schemes by the Indian government.

Other Related Points

  • The state government of Haryana has launched an initiative called ‘Reading Mission- Haryana’.
    • The initiative was launched to encourage reading habit among the young students of the state.
  • ‘Pyaar ka Paudha’ (a plant of love), is a campaign of Bihar state, for promoting the planting of trees.
  • The ‘BhoomiRashi’ Portal is a land acquisition portal launched in March 2018 by the Ministry of the Road Transport and Highways.
MPPSC AE CE Mock Test - 5 - Question 7

Which of the following National Parks is Not located in Madhya Pradesh?

Detailed Solution for MPPSC AE CE Mock Test - 5 - Question 7

The correct answer is Manas National Park.
Explanation

  • National Parks in Madhya Pradesh:

Other Related Points

  • Manas National Park is located in Assam.
  • It is also a UNESCO World Heritage Site.
  • It is home to animals like Wild Buffaloes, Royal Bengal Tiger, Golden Langur, Hispid Hare, etc.
  • It has been named after the river Manas, which is a tributary of the river Brahmaputra.
MPPSC AE CE Mock Test - 5 - Question 8

Which of the following is not a composition of Pandit Makhanlal Chaturvedi?

Detailed Solution for MPPSC AE CE Mock Test - 5 - Question 8

The correct answer is Rasikpriya.
Explanation

  • Makhanlal Chaturvedi
    • He was born at a place called Babai in the Hoshangabad district of Madhya Pradesh.
    • His father's name was Nandlal Chaturvedi who was a teacher in the primary school of the village.
    • He was a famous poet, writer, and journalist of India whose works became very popular.
    • As editor of prestigious papers like Prabha and Karmaveer, he campaigned vigorously against British rule.
    • Along with patriotism, nature and love have also been depicted in his poems.
    • His notable writings include Himkiritani, Bijuri, Himtarangini, Maran Jwar and Kajal

Important Points

  • Kumbha wrote a commentary on Gita Govinda of Jaidev, and this commentary is named Rasikpriya.
MPPSC AE CE Mock Test - 5 - Question 9
In which famous place Jyotirling Mamleshvar is situated?
Detailed Solution for MPPSC AE CE Mock Test - 5 - Question 9

The correct answer is Omkareshvar.

Explanation

  • Mamleshwar Temple:
    • Mamleshwar Temple is built on the south banks of river Narmada.
    • It is situated right opposite the river bank where Omkareshwar is located.
    • It is one of the Jyotirlingas located near Amreshwar temple and right opposite the Omkareshwar temple.
MPPSC AE CE Mock Test - 5 - Question 10

The thickness of the gusset plate for column base should not be less than

Detailed Solution for MPPSC AE CE Mock Test - 5 - Question 10

Concept:
Column base:

  • Column base transmits the column load to the concrete or masonry foundation blocks.
  • The column base spreads the load on a wider area so that the intensity of bearing pressure on the foundation block is within the bearing strength.

There are two types of column bases commonly used in practice:

  • Slab base
  • Gusseted base

Slab Base:

  • These are used in columns carrying small loads.
  • In this type, the column is directly connected to the base plate by welding or through cleat angles.
  • The load is transferred to the base plate through bearing.

Gusseted Base:

  • For columns carrying heavy loads, gusseted bases are used. In a gusseted base, the column is connected to the base plate through gussets.
  • The load is transferred to the base partly through bearing and partly through the gusset.
  • The thickness of the gusset plate in any case should not be less than 12mm.
MPPSC AE CE Mock Test - 5 - Question 11

Match the designated component in List 1 with its purpose at airports in List 2.

Detailed Solution for MPPSC AE CE Mock Test - 5 - Question 11

MPPSC AE CE Mock Test - 5 - Question 12

The velocity of drainage water in the barrels of syphon aqueduct is normally limited to

Detailed Solution for MPPSC AE CE Mock Test - 5 - Question 12

Concept:
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​​

  • 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 velocity of drainage water in the barrels of syphon aqueduct is normally limited to 2 - 3 m/sec
  • 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 passageA 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.


3) 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:

MPPSC AE CE Mock Test - 5 - Question 13

What should be the clear distance between the lateral restraint for continuous beams?

Detailed Solution for MPPSC AE CE Mock Test - 5 - Question 13

Concept:
As per IS 456:2000(Clause 23.3):

Where, d = the effective depth of the beam and b = the breadth of the compression face midway between the lateral restraints.

MPPSC AE CE Mock Test - 5 - Question 14

By what value the bond stress of deformed bars is increased in tension?

Detailed Solution for MPPSC AE CE Mock Test - 5 - Question 14

Explanation:
As per IS 456:2000, clause B 2.1.2, (conforming to IS 1786),
The allowable bond stress in deformed bars may be increased by 60% as compared to the allowable bond stress in plain bars.
Important Points
The value of bond stress is increased by 25% when the bar is in compression.
Other Related Points
The bond stress for different grades of concrete as per different established provisions are as follows:

The bond stresses value tabulated above is for plain bar in tension.

MPPSC AE CE Mock Test - 5 - Question 15

The coefficient of roughness for an earthen canal, in excellent condition, is taken as

Detailed Solution for MPPSC AE CE Mock Test - 5 - Question 15

Explanation:
Manning Equation:

As per Manning’s average velocity of flow in an open channel undergoing uniform flow is given by;

Where,
N = Roughness coefficient that depends upon nature of the boundary
So = Slope of channel bed
R = Hydraulic mean depth of flow

  • Value’s of N generally varies in the range of 0.010 to 0.05050.
  • For different types of channel manning coefficient is as follow’s;

These values are totally based on field observation that’s why Manning equation is used in practice.

MPPSC AE CE Mock Test - 5 - Question 16
According to IS 1893-1962 how many seismic zone’s are there-
Detailed Solution for MPPSC AE CE Mock Test - 5 - Question 16

The correct answer is option 1) i.e. 5.

  • Earthquake-prone areas of the country have been identified on the basis of scientific inputs relating to seismicity, earthquakes occurred in the past and tectonic setup of the region.
  • According to IS 1893-1962, India has been grouped into five seismic zones.
  • The Geological Survey of India (GSI) first published the seismic zoning map of the country in the year 1935.
  • According to IS 1893 - 2016, India has been grouped into Four seismic zones.

Note:

  • An Earthquake is caused by the forces originating in the interior of the earth. These internal forces are called endogenetic forces.
  • The magnitude of an earthquake is measured by the Richter scale.
  • The place of origin of Earthquake is called Focus. The place on the ground surface which is perpendicular to the focus is called Epicentre.
  • Seismology is the special branch of geology that deals with the study of Earthquakes.
  • Shadow zones are zones where earthquakes do not occur or rarely occur.
MPPSC AE CE Mock Test - 5 - Question 17

Limiting value of Poisson's ratio are:

Detailed Solution for MPPSC AE CE Mock Test - 5 - Question 17

Explanation:
Poisson ratio:
Poisson's ratio is the ratio of transverse strain to longitudinal strain.


We know the relation between elastic constants and Poisson's ratio is given by:

  • E = 2G (1 + μ) ...(i)
  • E = 3K (1 - 2μ) ...(ii)

The elastic constants cannot be negative therefore from eq (i) and (ii)

  • (1 + μ) = 0 or μ = -1 and
  • (1 - 2μ) = 0 or μ = 0.5

∴ The range of Poisson's ratio (μ) for any material is -1 < μ < 0.5
The normal value of μ lies between -1 and 0.5 whereas practical value of μ lies between 0 to 0.5
Other Related Points

  • Most materials have Poisson's ratio values ranging between 0.0 and 0.5.
  • A perfectly incompressible material deformed elastically at small strains would have a Poisson's ratio of exactly 0.5.
  • Rubber has a Poisson ratio of nearly 0.5.
  • Cork's Poisson ratio is close to 0, showing very little lateral expansion when compressed.
MPPSC AE CE Mock Test - 5 - Question 18

In plastic theory, among the given shape of the cross section, which the steel member has the largest shape factor?

Detailed Solution for MPPSC AE CE Mock Test - 5 - Question 18

Concept:
Shape Factor for the different section:

Other Related Points
Shape Factor:

  • It represents the reserve of the strength of a section beyond the yield point.
  • It doesn't show the reserve of the strength of the complete structure.
  • It depends upon the geometry of the cross-section only.
  • It is defined as the ratio of the plastic moment capacity of a beam section to its yield moment capacity.

Mathematically,

Where,
fy = Yield stress.
Zp = Plastic modulus of a section.
Z = Elastic modulus of a section.
Zp is calculated about the equal area axis and Z is calculated about the centroidal axis.

MPPSC AE CE Mock Test - 5 - Question 19

In earthwork computations on a longitudinal profile, the diagram prepared to work out the quantity of earthwork is:

Detailed Solution for MPPSC AE CE Mock Test - 5 - Question 19

Explanation:
Mass Haul Curve:

  • This is a curve representing the cumulative volume of earthwork at any point on the curve, the manner in which earth to be removed.
  • It is necessary to plan the movement of excavated soil of worksite from cuts to fill so that haul distance is minimum to reduce the cost of earthwork.
  • The mass haul diagram helps to determine the economy in a better way.
  • The mass haul diagram is a curve plotted on a distance base with the ordinate at any point on the curve representing the algebraic sum of the volume of earthwork up that point.

A haul refers to the transportation of your project’s excavated materials. The haul includes the movement of material from the position where you excavated it to the disposal area or a specified location. A haul is also sometimes referred to as an authorized haul.
Haul = Σ Volume of earthwork × Distance moved.

MPPSC AE CE Mock Test - 5 - Question 20
If the peak ordinate of a 4 h unit hydrograph of a basin is 270 m3/s, then the peak ordinate of a 8 h unit hydrograph of the same basin will be:
Detailed Solution for MPPSC AE CE Mock Test - 5 - Question 20

Concept:

Hydrograph:

  • A hydrograph is a plot between discharge and time at any given section of a river, channel, etc.
  • It is a response of a given catchment to the rainfall input.

Unit hydrograph (UH):

  • A unit hydrograph can be defined as the hydrograph of direct runoff resulting from one unit depth (1 cm) of rainfall excess occurring uniformly over the basin for a specified duration.
  • A unit hydrograph is based on the principle of linearity and the principle of superposition.
  • If a D-hr unit hydrograph is available, a unit hydrograph with other duration such as 2D, 3D, etc. can be obtained easily from the principle of superposition. As the duration of the unit hydrograph increases, the base period also increases, and consequently, the peak ordinate decreases.

Hence, If the peak ordinate of a 4 h unit hydrograph of a basin is 270 m3/s, then the peak ordinate of an 8 h unit hydrograph of the same basin will be Less than 270 m3/s.

MPPSC AE CE Mock Test - 5 - Question 21

Find the hydraulic mean radius of a channel, having mean velocity of 2 m/sec., if the silt factor is 1.0 and channel is designed by Lacey's theory -

Detailed Solution for MPPSC AE CE Mock Test - 5 - Question 21

Concept:
As per Lacey’s theory-
Hydraulic radius (R) is given by,

Where,
V is the flow velocity
f is silt factor
Calculation:
Given,
v = 2 m/s, f = 1
Hydraulic mean radius,

= 10 m
Important Points

Following relations are used in Lacey's Theory

The velocity of flow is given by V =

Where,
f = silt factor = 1.76√d and d = average size of soil particles in mm.
Q is discharge , Area of the section is given by A = Q/V
Perimeter is given by, P = 4.75√Q

Bed slope(S), is given by S =

MPPSC AE CE Mock Test - 5 - Question 22

If the dimension of water tank is greater than 15 m in this case minimum steel requirement of steel is

Detailed Solution for MPPSC AE CE Mock Test - 5 - Question 22

Explanation:
(i) As per CI 8.1.1 of IS 3370:2009, The minimum reinforcement in walls. floors and roofs in each of two directions at right angles. within each surface zone shall not be less than 0.35 percent of the surface zone.
(ii) For high strength deformed bars and not less than 0.64 percent for mild steel reinforcement bars. The minimum reinforcement can be further reduced to 0.24 percent for deformed bars and 0.40 percent for plain round bars for tanks having any dimension not more than 15 m.
(iii) In wall slabs less than 200 mm in thickness, the calculated amount of reinforcement may all be placed in one face. For ground slabs less than 300 mm thick the calculated reinforcement should be placed in one face as near as possible to the upper surface consistent with the nominal cover. Bar spacing should generally not exceed 300 mm or the thickness of the section. whichever is less.
Other Related Points

  • Maximum cement content is 400 kg/m3 to take care of shrinkage effect.
  • Minimum cement content is 320 kg/m3.
  • Minimum grade of concrete is M30.
  • Maximum w/c ratio is 0.45.
  • Minimum nominal cover is 45 mm.
  • Maximum allowed crack width is 0.2 mm in LSM design.
MPPSC AE CE Mock Test - 5 - Question 23

A quick and reasonably accurate method of determining the moisture content of fine aggregate is:

Detailed Solution for MPPSC AE CE Mock Test - 5 - Question 23

Calcium carbide method: (Rapid Moisture Meter)

  • The calcium carbide method is the quickest and reasonably accurate method of determining of the water content of soil using a portable moisture content kit. It usually requires 5 to 7 minutes for determining the water content.
  • The method is based on the principle that when the water in the soil reacts with calcium carbide, acetylene gas is produced and the pressure exerted by the acetylene gas on a diaphragm gives a measure of the water content.
  • The water content obtained from the calcium carbide method is based on the initial weight of wet soil. It should be converted to water content based on the dry weight of the soil.​

Additional Information
Sand bath method:

(i) It is a quick field method for the determination of water content.
(ii) This method is used when an electric oven is not available.
In this method there is no control over heat given to soil sample, due to which it has the following two limitations:
a) Results obtained from this method are inaccurate as such it may also include structural water present in the soil.
b) It is not suitable for organic soil and soil having higher gypsum content.
Pycnometer method:
(i) It is quick method gives result in 10 – 20 minutes.
(ii) This method is used only when the specific gravity of soil solids is known.
(iii) It is used only for cohesionless soils because removal of entrapped gasses is very difficult in the case of cohesive soils.

Oven drying method:
(i) Oven dry method is the most accurate and simplest method for water content determination.
(ii) In this method complete drying of soil sample occurs and water content in the sample is calculated accurately by a maintained temperature in the oven ( 105°C to 110°C) for 24 hours.
(iii) For highly organic soils a low temperature of about 60° C is preferable.
(iv) If Gypsum is present, the temperature should not be more than 80°C but for a long time.

MPPSC AE CE Mock Test - 5 - Question 24

Identify the WRONG diagram. (G-Centroid, s-shear center)

Detailed Solution for MPPSC AE CE Mock Test - 5 - Question 24

Explanation:
Shear centre:
The shear centre is the point through which if the resultant shear force acts then the member is subjected to simple bending without twisting.
Location of shear centre:
(i) Shear centre generally does not coincide with the centroid of section except in special cases when the area is symmetrical bout both axis.
(ii) Shear centre always lies on the axis of symmetry if exists.
(iii) If there are two or more two axis of symmetry exist, then the shear center will coincide with the point of intersection of the axis of symmetry. In this case shear centre of the area will be the same as the centroid of the area.
(iv) If a section is made of two narrow rectangles then the shear center lies on the junction of both rectangles.

MPPSC AE CE Mock Test - 5 - Question 25
Three pipes of lengths 800 m, 500 m and 400 m and of diameter 500 mm, 400 mm and 300 mm respectively are connected in series. These pipes are to be replaced by a single pipe of length. Find the diameter of the single pipe.
Detailed Solution for MPPSC AE CE Mock Test - 5 - Question 25

Explanation:

As the pipes are connected in series, total equivalent length of pipe

Le = L1 + L2 + L3 = 800 + 500 + 400 = 1700m.

Since the pipes are connected in series, the discharge will remain constant hence we can write down,

De5 = 0.007112

therefore De = 0.3718 m = 371.8 mm.

MPPSC AE CE Mock Test - 5 - Question 26
Sedimentation with coagulation treatment of water is effective when water is
Detailed Solution for MPPSC AE CE Mock Test - 5 - Question 26

Explanation:

Sedimentation:

(i) The process of removal of suspended solids from the water is termed sedimentation.

(ii) Sedimentation is a natural process by which solids with higher density than the fluid in which they are suspended, settle under the action of gravity.

(iii) Sedimentation tanks are designed to reduce the velocity of flow to reduce turbulence.

(iv) Efficiency of sedimentation is comparatively less if water consists of fine suspended particles, hence in such cases coagulation aided sedimentation is adopted.

(v) Coagulation aided sedimentation is performed in 3 stages:

  • Coagulation (Fast mixing)
  • Flocculation (slow mixing)
  • Sedimentation

Coagulation:

(i) It is the process in which certain chemical terms such as coagulants are added to the water so as the neutralize negative protective charge over the particle to increase their opportunity to come in contact with each other, thus grow in size and get easily settled in following sedimentation process.

(ii) To neutralize the charge over the suspended impurities minimum energy (threshold energy) is required which is provided by induced fast mixing in this process

(iii) The pH of the water plays an important role when alum is used for coagulation because the solubility of the aluminum species in water is pH dependent. If the pH of the water is between 4 and 5, alum is generally present in the form of positive ions (i.e., Al(OH)2+, Al8(OH)4+, and Al3+). However, optimum coagulation occurs when negatively charged forms of alum predominate, which occurs when the pH is between 6 and 8.

Flocculation:

(i) Flocculation is the process in which neutralize particle is brought in intimate contact with each other to promote their agglomeration resulting in an increased size of flocs, which can get easily settled in the following sedimentation process.

(ii) To increase the opportunity of the neutralized particle to come in contact with each other slow mixing is induced during flocculation.

Hence, Sedimentation with coagulation treatment of water is effective when water is slightly alkaline.

MPPSC AE CE Mock Test - 5 - Question 27
The reduction in gradients at curve is known as grade compensation. In India, the following value of grade compensation have been allowed for broad gauge track
Detailed Solution for MPPSC AE CE Mock Test - 5 - Question 27

Explanation:

Grade compensation (GC):

(i) If there is any curvature along with the gradient then some compensation has to be made in terms of two types of resistance.

(ii) One is the resistance offered by the gradient and the other is the resistance offered by the curvature.

BG → 0.04% per degree of curve or 70/R

MG → 0.03% per degree of curve or 52.5/R

NG → 0.02% per degree of curve or 35/R

MPPSC AE CE Mock Test - 5 - Question 28
The best direction of runway is
Detailed Solution for MPPSC AE CE Mock Test - 5 - Question 28

Concept:

Wind Rose Diagram

  • A wind rose is a graphic tool used by meteorologists to give a succinct view of how wind speed and direction are typically distributed at a particular location.
  • Historically, wind roses were predecessors of the compass rose (found on charts), as there was no differentiation between a cardinal direction and the wind which blew from such a direction.
  • Using a polar coordinate system of gridding, the frequency of winds over a time period is plotted by wind direction, with color bands showing wind speed ranges.
  • The direction of the longest spoke shows the wind direction with the greatest frequency.
  • The wind rose diagrams help you visualize wind patterns at a site. Use them to better inform your design decisions, but be aware of unique microclimates and site considerations that wind rose intensity blows from the respective direction.
  • The procedure for determining the orientation of the runway from this type of wind rose is described below. Draw three equispaced parallel lines on a transparent paper strip in such a way that the distance between the two nearby parallel lines is equal to the permissible crosswind component.
  • The best direction of runway is along the direction of the longest line on the windrose diagram.
MPPSC AE CE Mock Test - 5 - Question 29
A pile group consists of 9 friction piles of 300 mm diameter and 10 m length driven in clay (Cu = 100 kN/m2, = 20 kN/m3), The centre to centre distance of piles is 0.75 m in both direction. Factor of safety = 3, adhesion factor = 0.60, then the safe load for pile group is
Detailed Solution for MPPSC AE CE Mock Test - 5 - Question 29

Concepts:

The ultimate load-carrying capacity of a friction pile in clay is given as:

Where

α is the adhesion factor,As is the skin area, C is the Cohesion, which is taken 9 in general for clay piles

Calculation:

Case 1)

For Individual friction pile:

Skin friction area, As = perimeter of pile × Length of the pile, D is dia of pile

Qu = 9 × (0.6 × 100 × π × 0.3 × 10) = 5089.8 kN........ (1)

Case 2)

When friction pile acts as group action:

Skin friction area, As = perimeter of pile group × Length of pile

Perimeter of pile group = 4 × (2S + D)

End bearing area = (2S + D)2

S is the center-to-center distance between pile and D is dia of pile

Qu = 1 × 100 × 4 × (2 × 0.75 + 0.3) × 10 = 7200 kN ...... (2)

The pile load capacity will be a minimum of 1 and 2 i.e.

Qu = 5089.8 kN

The safe load capacity of the pile is calculated as:

Qs = 1696.33 kN

MPPSC AE CE Mock Test - 5 - Question 30

As per IS : 1893 (Part I) 2016, the design horizontal seismic co-efficient Ah for a structure is determined by

Detailed Solution for MPPSC AE CE Mock Test - 5 - Question 30

Explanation:
As per clause 6.4.2 of IS 1893-1 (2002) page no. 14,
The design horizontal seismic coefficient Ah for a structure shall be determined by the following expression:

  • Provided that for any structure with T < 0.1 s, the value of Ah will not be taken less than Z/2 whatever be the value of I/R

Where,
Z = Zone factor,
I = Importance factor, depending upon the functional use of the structures, characterized by hazardous consequences of its failure, post-earthquake functional needs, historical value, or economic importance,
R = Response reduction factor, depending on the perceived seismic damage performance of the structure, characterized by ductile or brittle deformations. However, the ratio (I/R ) shall not be greater than 1.0,
Sa/g = Average response acceleration coefficient
Important Points
First step to calculating earthquake loads on the structure is to identify the earthquake zone for which the structure needs to be designed.
Earthquake zones of India:

Zone factor:

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