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30 Questions MCQ Test UPSSSC JE Civil Mock Test Series 2026 - UPSSSC JE Civil Mock Test - 5

UPSSSC JE Civil Mock Test - 5 for Civil Engineering (CE) 2026 is part of UPSSSC JE Civil Mock Test Series 2026 preparation. The UPSSSC JE Civil Mock Test - 5 questions and answers have been prepared according to the Civil Engineering (CE) exam syllabus.The UPSSSC JE Civil Mock Test - 5 MCQs are made for Civil Engineering (CE) 2026 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for UPSSSC JE Civil Mock Test - 5 below.
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UPSSSC JE Civil Mock Test - 5 - Question 1

Select the INCORRECT statement.

Detailed Solution for UPSSSC JE Civil Mock Test - 5 - Question 1

Concept:

Retarding admixtures

  • It slows down the rate of hydration of cement in its initial stage and increases the initial setting time of concrete.
  • These are also called retarders and are used especially in high-temperature zones where concrete will set quickly.
  • The quick setting in some situations may lead to discontinuities in structure, poor bond between the surfaces, creates unnecessary voids in concrete etc.

Air entraining admixtures

  • These are one of the most important inventions in concrete technology.
  • Their primary function is to increase the durability of concrete under freezing and thawing conditions.
  • When added to the concrete mix, these admixtures will form millions of non-coalescing air bubbles throughout the mix and improve the properties of concrete.
  • Air entrainment in concrete will also improve the workability of concrete, prevent segregation and bleeding, lower the unit weight and modulus of elasticity of concrete improve the chemical resistance of concrete and reduction of cement or sand or water content in concrete, etc.

Damp proofing admixtures-

  • Damp-proofing or waterproofing admixtures are used to make the concrete structure impermeable against water and to prevent dampness on the concrete surface.
  • In addition to the waterproof property, they also acts like accelerators in early stage of concrete hardening.
  • Damp proofing admixtures are available in liquid form, powder form, paste form, etc.

Pozzolanic admixtures

  • These are used to prepare dense concrete mix which is bets suitable for water retaining structures like dams, reservoirs, etc.
  • They also reduce the heat of hydration and thermal shrinkage.
  • Best pozzolanic materials in optimum quantity give best results and prevents or reduce many risks such as alkali-aggregate reaction, leaching, sulfate attack etc.
UPSSSC JE Civil Mock Test - 5 - Question 2

Identify the following foundation plan shown in the figure:

Detailed Solution for UPSSSC JE Civil Mock Test - 5 - Question 2

Concept:
Combined footings:
When two or three columns are located very near to each other, we provide a single substructure for them and the footing is known as a combined footing. Combined footings are also usually made of reinforced concrete. A combined footing may be rectangle or trapezoidal in plan.

Other Related PointsStepped footing:


In this type of footings, three concrete cross-sections are stacked upon each other and forms as a step. This is also called as a step foundation. This is mostly used in residential buildings
Sloped footing:


It is a footing in which 45° slope is maintained from all sides. Due to the provision of slope, the cost of footing reduces as concrete and reinforcement required will be less compared to the trapezoidal footing.
Raft or Mat foundation:


When we have to spread the load over a large area with less depth then we have to increase the footing area. If we increase the footing area the footings are overlapped , hence the raft foundation with a solid RCC slab covering the entire area beneath the structure and supporting all the columns is provided.

UPSSSC JE Civil Mock Test - 5 - Question 3

Select the correct option with regard to compression members.
Statement 1: For very short compression steel members, the failure stress will equal the yield stress and buckling will occur.
Statement 2: For long compression steel members, the Euler formula predicts the strength very well, where the axial buckling stress remains below the proportional limit.

Detailed Solution for UPSSSC JE Civil Mock Test - 5 - Question 3

Statement 1: For very short compression steel members, the failure stress will equal the yield stress and buckling will occur.
This statement is incorrect.
For very short compression steel members, the failure stress is typically higher than the yield stress of the material. Short members fail due to material yielding rather than buckling. Buckling occurs when the length of the member is relatively long compared to its cross-sectional dimensions.
Statement 2: For long compression steel members, the Euler formula predicts the strength very well, where the axial buckling stress remains below the proportional limit.
This statement is correct.
For long compression steel members, the Euler formula is commonly used to predict their strength. The Euler formula calculates the critical buckling stress for an idealized, perfectly straight member with pinned ends. The formula assumes that the buckling stress remains below the proportional limit of the material, ensuring that elastic behavior is maintained.

UPSSSC JE Civil Mock Test - 5 - Question 4

If d and t are the effective depth and thickness of a beam respectively and ϵ is the yield stress ratio, the webs shall be checked for shear buckling when ________.

Detailed Solution for UPSSSC JE Civil Mock Test - 5 - Question 4

Transverse Stiffeners are provided to increase buckling resistance of the web due to inclined compressive stress due to shear, It is provided vertically along the span.
Horizontal Stiffener / Longitudinal Stiffener is designed to prevent web buckling due to bending compression.
Both traverse and longitudinal stiffeners are provided to check the web buckling.
End Bearing Stiffeners
are provided at the supports & Load Bearing Stiffeners are provided at the points of concentrated loads.

As per IS 800 resistance to shear buckling must be verified when
> 67 ϵ for unstiffened webImportant Pointsif < 67ϵ ⇒ unstiffened girder can be designed i.e. No girder required.
If 85 ϵ < < 200 ϵ ⇒ Vertical stiffener (C1 and C2) may be provided.
If 200 ϵ < < 250 ϵ ⇒ Vertical stiffener along with longitudinal stiffener at 0.2 d may be provided.
If 250 ϵ < < 345 ϵ ⇒ Vertical stiffeners along with two longitudinal stiffeners at 0.2 d and 0.5 d respectively may be provided.

UPSSSC JE Civil Mock Test - 5 - Question 5
Calculate R-log removal value for 99.0% removal.
Detailed Solution for UPSSSC JE Civil Mock Test - 5 - Question 5

Calculation:

R-log removal =

Where, Nt = percentage of removal after time T= 99

No = Original percentage = 100 %

=

= - Log (1 - 0.99)

= - Log (0.01)

R-Log removal = 2

Important Points As per Watson and Chicks law

Where,

K = Rate constant which depends on the type of microorganism, the type of disinfectant, and the temperature

n = Constant for a particular microorganism and type of disinfectant

C = Disinfectant concentration (weight/volume)

Nt = Number of microorganisms present at any time ‘t’

UPSSSC JE Civil Mock Test - 5 - Question 6

Which of the following properties of a material is to absorb water vapour from the air and depends on the relative humidity, porosity, air temperature etc?

Detailed Solution for UPSSSC JE Civil Mock Test - 5 - Question 6

Hygroscopicity:

  • It should absorb moisture from air or soil.
  • Hygroscopic water means how much water can be held by the soil over its surface by the action adhesion forces.
  • Specific surface is the ratio of surface per unit volume. More surface area implies a more specific surface. Clay has more surface area (more specific surface) as compared to sand since clay has smaller particles, therefore, more water can be held by clay particles over their surface. So, clay exhibits more hygroscopicity.

Water Permeability

  • It is the property of the soil with how much ease water can flow through the soil mass. Higher the easiness of flowing the water, the higher the permeability of the soil sample. Generally, coarse-grained soils like sand have higher permeability than that fine-grained soils like clay.
UPSSSC JE Civil Mock Test - 5 - Question 7

Which types of mortars are intended for architectural or ornamental parts, and application of decorative layers on walls and panels?

Detailed Solution for UPSSSC JE Civil Mock Test - 5 - Question 7

Classification of Mortar based on the nature of the application
Based on the nature of the application, mortar may be classified as follows:

  1. Masonry mortar
  2. Finishing mortar

Masonry mortar

  • These mortars are used for the construction of masonry. Masonry mortars may be either lime mortar, cement mortar, and lime-cement mortar. Composite mortars (i.e. lime cement mortars) are preferred since they possess both plasticity as well as strength.

Finishing mortar

  • Finishing mortars are those which are used for finishing works such as plastering, pointing, ornamental finishing, etc. Such a mortar should have strength, mobility, water retentivity, and resistance to atmospheric actions such as temperature, humidity, rain, wind, dust storm, etc.
UPSSSC JE Civil Mock Test - 5 - Question 8

An offset is laid out 5° from its true direction on the field. Find the resulting displacement of the plotted point on the paper in a direction parallel to the chain line on the paper if the length of offset is 20 m and the scale is 10 m to 1 cm. (Cos 5° = 0.996, Sin 5° = 0.087)

Detailed Solution for UPSSSC JE Civil Mock Test - 5 - Question 8

Concept:
Length of limiting offset is given by:


where
S = scale of the drawing
θ = Maximum allowed error in laying direction of an offset.
L = limiting length of the offset
e = Displacement on paper
Calculation:
Given,
θ = 5°, S = 1 cm to 10 m, L = 20 m
Let AB be the actual length of the offset which was laid out 5° from its true directions. = 5

BB’ is the displacement of the point
Displacement on land = L sin 5°

Now, Displacement on paper,
L = 0.174 cm

UPSSSC JE Civil Mock Test - 5 - Question 9

Which of the following tools is used for lifting and spreading mortar, as well as for cutting bricks?

Detailed Solution for UPSSSC JE Civil Mock Test - 5 - Question 9

Tools for Brickwork
Trowel: It is a tool used for lifting and spreading mortar, as well as for cutting bricks.

Bolster: This tool is used for cutting bricks accurately.
Brick Hammer: It is used for pushing the bricks in the courses and occasionally, for cutting bricks.
Mason Square: In order to check the right-angle Mason’s square is used.

UPSSSC JE Civil Mock Test - 5 - Question 10

Notches are used to measure:-

Detailed Solution for UPSSSC JE Civil Mock Test - 5 - Question 10

UPSSSC JE Civil Mock Test - 5 - Question 11

A cantilever beam with a load acting at end point. The mass 'M' is attached at the end of the beam. The frequency of the beam is given as _____.

Detailed Solution for UPSSSC JE Civil Mock Test - 5 - Question 11
Spring Stiffness (Keq)
  • Spring Stiffnessor spring constant represented by k represents how much resistance it offers to displacement when a force is applied to it. The more stiff the spring the lesser it would deflect ( compression or tension).

The cantilever with a Point Load at its Free End


​Maximum deflection at the free end is given by,
Maximum Deflection
Spring Stiffness (Keq) of the cantilever with a Point Load at its Free End

The frequency is given as,

Additional Information

UPSSSC JE Civil Mock Test - 5 - Question 12

A prismatic bar of cross-sectional area 1 mm2 and E = 200 GPa is fastened between two rigid walls at A and B and are subjected to loads as shown.

The support reactions at B and A (in kN) are

Detailed Solution for UPSSSC JE Civil Mock Test - 5 - Question 12

Concept:
(δl)1 + (δl)2 + (δl)3 = 0

Calculation:
Draw the free body diagram of all the blocks

∵ RA + RB = 40.8 ............. (1)
(δl)1 + (δl)2 + (δl)3 = 0

100 RA + 200 RA – 2720 – 300 RB = 0
300 RA – 300 RB = 2720
RA - RB = 9.067 ----(2)
Solving equation (1) and (2),
∴ RA = 24.93 kN and RB = 15.86 kN

UPSSSC JE Civil Mock Test - 5 - Question 13

In a chain survey work, in order to determine the length across a river of a continuing chain line, the following observations were made as shown in the figure. The points C, A, and B are collinear. Also, the points B, D, and E are collinear. CA = AD = 40 m. The angle CAD is 90°. Distance CE = 75 m, angle ACE = 90°. Determine length AB.

Detailed Solution for UPSSSC JE Civil Mock Test - 5 - Question 13


Similar triangle ABD and FDE
AB/AD = DF/FE
DF = AC and FC = CE - AD
AB/AD = AC/(CE-AD)
AB = AC × AD/ (CE-AD) = 40 × 40 /(75-40) = 45.71

UPSSSC JE Civil Mock Test - 5 - Question 14

Read the following:
A. Polar modulus is defined as the ratio of the polar moment of inertia to the radius of the shaft.
B. Polar modulus is also called as section modulus.
C. Polar modulus of a solid circular shaft is calculated using π/16D3formula.
D. Polar modulus of a hollow circular shaft is calculated using  π/32(D4 - d4)3
Identify the correct statement from the following:

Detailed Solution for UPSSSC JE Civil Mock Test - 5 - Question 14

Explanation:
Polar modulus is defined as the ratio of the polar moment of inertia to the radius of the shaft. It is also called as torsional section modulus. It is denoted by Zp.
Polar section modulus of a solid shaft is given by,

For Hollow solid shaft

UPSSSC JE Civil Mock Test - 5 - Question 15
A raft foundation of 6 m × 9 m is placed at a depth of 3 m in a cohesive soil having c = 120 kN/m2. The net ultimate bearing capacity of the soil using Terzaghi's theory will be.
Detailed Solution for UPSSSC JE Civil Mock Test - 5 - Question 15

Concept:

As per Terzaghi's theory, ultimate bearing capacity of strip footing is given by the equation

qu = cNc + γDfNq + 0.5BγNγ

Modified equation of ultimate bearing capacity becomes for rectangular or raft footing:

Net ultimate bearing is given by, qnu = qu - γDf

Calculation:

Given soil is cohesive, so for cohesive soil, ϕ = 0°

For ϕ = 0°, Nc = 5.7, Nq = 1 and Nγ = 0

B = 6 m, L = 9 m, Df = 3 m and c = 120 kN/m2

Ultimate bearing capacity is given by,

Net ultimate bearing capacity is given by

qnu = qu - γDf

= 820.8 kN/m2

UPSSSC JE Civil Mock Test - 5 - Question 16

Consider the following statements related to the advantages of concrete sleepers :

1. Concrete sleepers can generally be mass-produced using local resources.

2. Concrete sleepers are not suitable for better packing.

3. Concrete sleepers have a very long life span.

4. Concrete sleepers have no scrap value.

Which of the above statements is/are correct?

Detailed Solution for UPSSSC JE Civil Mock Test - 5 - Question 16

Explanation:

Concrete Sleeper:

(i) Concrete sleeper is a kind of railway sleeper made of steel-reinforced concrete. With the characteristic of water-resistant, sun-resistant, and corrosion-resistant.

(ii) Concrete sleeper has been widely used all over the world since 1950.

Advantages of Concrete Sleepers:

  • It is more durable having greater life (up to 50 years)
  • It is economical as compared to wood and steel.
  • Concrete sleepers can generally be mass-produced using local resources.
  • Easy to manufacture.
  • It is not susceptible to vermin attack
  • It is not susceptible to fire
  • Good for track circuited areas

Disadvantages of Concrete Sleepers

  • It is brittle and cracks without warning.
  • It cannot be repaired and required replacement.
  • Fittings required are greater in number.
  • No scrap value

Other Related Points

Advantages of Wooden Sleepers:

  • They are cheap and easy to manufacture
  • They are easy to handle without damage
  • They are more suitable for all types of ballast
  • They absorb shocks and vibrations better than other types of sleepers.
  • Ideal for track circuited sections
  • Fittings are few and simple in design
  • Good resilience
  • Ease of handling
  • Adaptability to non-standard situation
  • Electrical insulation

Disadvantages of Wooden Sleepers:

  • They are easily liable to attack by vermin and weather
  • They are susceptible to fire
  • It is difficult to maintain gauge in the case of wooden sleepers
  • The scrap value is negligible
  • Their useful life is short about 12 to 15 years.

Advantages of Steel Sleepers:

  • It is more durable. Its life is about 35 years
  • Lesser damage during handling and transport
  • It is not susceptible to vermin attack
  • It is not susceptible to fire
  • Its scrap value is very good

Disadvantages of Steel Sleepers:

  • It is liable to corrosion.
  • Not suitable for track circuiting
  • It can be used only for rails for which it is manufactured
  • Cracks at rail seats develop during the service.
  • Fittings required are greater in number

Advantages of Cast Iron Sleepers:

  • Service life is very long
  • Less liable to corrosion
  • Form good track for light traffic up to 110 kmph as they form rigid track subjected to vibrations under moving loads without any damping
  • Scrap value is high

Disadvantages of Cast Iron Sleepers:

  • Gauge maintenance is difficult as tie bars get bent up
  • Not suitable for circuited track
  • Need a large number of fittings
  • Suitable only for stone ballast
  • Heavy traffic and high speeds (>110kmph) will cause loosening of keys and development of high creep
UPSSSC JE Civil Mock Test - 5 - Question 17
Which of the following is a practically impermeable?
Detailed Solution for UPSSSC JE Civil Mock Test - 5 - Question 17

Permeability is defined as the property of a porous material which permits the passage or seepage of water through its interconnecting voids.

Usually, the finer the soil texture, the slower the permeability.

The decreasing order of permeability is:

Gravel > Coarse sand > Sand mixture > Clay

Clay is impermeable, or at least it has a very low permeability. The grains in clay are so fine that the spaces between the grains are extremely small. This makes it difficult for water to squeeze through, so water is more likely to simply run off.
UPSSSC JE Civil Mock Test - 5 - Question 18
Which of following tests is conducted to assess shear strength parameter of the soil.
Detailed Solution for UPSSSC JE Civil Mock Test - 5 - Question 18

Concept-

Shear strength of soil is given by τ’ = C’ + σ’ tan φ

Where τ’ = Effective shear strength, C’ = Effective cohesion, σ’ = Effective stress, φ = Angle of internal friction

Here C’ and φ are the shear strength parameter of the soil.

Triaxial shear strength test on soil measures the mechanical properties of the soil.

In this test, soil sample is subjected to stress, such that the stress resulted in one direction will be different in perpendicular direction.

The material properties of the soil like shear resistance, cohesion and angle of internal friction are determined from this test. The test is most widely used and is suitable for all types of soils.

Vane shear test

It is used to determine the undrained shear strength of soils especially soft clays.

This test can be done in laboratory or in the field directly on the ground. Vane shear test gives accurate results for soils of low shear strength (less than 0.3 kg/cm2).

Compaction test-

Compaction test of soil is carried out using Proctor’s test to understand compaction characteristics of different soils with change in moisture content.

Compaction of soil is the optimal moisture content at which a given soil type becomes most dense and achieve its maximum dry density by removal of air voids.

UPSSSC JE Civil Mock Test - 5 - Question 19

Keeping the instrument height as 1.5 m, length of staff 4 m, the slope of the ground as 1 in 10, the sight on the down-slope, must be less than________m.

Detailed Solution for UPSSSC JE Civil Mock Test - 5 - Question 19

Concept:

From the figure
Tanθ = Ground slope = difference b/w lenght of staff and (HI)/Horizontal distance
Tanθ = 1/10 = 4 - 1.5/x
Calculations:
Given,
Height of instrument (HI) = 1.5 m
Length of staff = 4 m
Ground slope (down slope) = 1/10
Tanθ = 2.5/x = 1/10
x = 25 m.

UPSSSC JE Civil Mock Test - 5 - Question 20
Which situation is not considered in the design of sight distance?
Detailed Solution for UPSSSC JE Civil Mock Test - 5 - Question 20

Explanation:

Sight distance available from a point is the actual distance along the road surface, over which a driver from a specified height above the carriageway has visibility of stationary or moving objects. Three sight distance situations are considered for design:

  • Stopping sight distance (SSD) or the absolute minimum sight distance
  • Intermediate sight distance (ISD) is defined as twice SSD
  • Overtaking sight distance (OSD) for safe overtaking operation
  • Headlight sight distance is the distance visible to a driver during night driving under the illumination of headlights
  • Safe sight distance to enter into an intersection
UPSSSC JE Civil Mock Test - 5 - Question 21
Facultative bacteria are able to work in
Detailed Solution for UPSSSC JE Civil Mock Test - 5 - Question 21

Bacteria: These are the microscopic unicellular plants or organisms. For the study of sanitary engineering, they are divided into three groups are as follows:

Aerobic bacteria: The aerobic bacteria require light and free oxygen for their existence and development.

Anaerobic bacteria: The anaerobic bacteria do not require light and free oxygen for their existence and development.

Facultative bacteria: The facultative bacteria can exist in the presence or absence of oxygen, but they grow in plenty in the absence of air.
UPSSSC JE Civil Mock Test - 5 - Question 22

If a piece of material neither expands nor contracts in volume when subjected to stresses, then the Poisson’s ratio must be

Detailed Solution for UPSSSC JE Civil Mock Test - 5 - Question 22

Explanation:

ϵv = ϵx + ϵy + ϵz
εx = (1/E) [σx - ν (σy + σz)]
εy = (1/E) [σy - ν (σx + σz)]
εz = (1/E) [σz - ν (σx + σy)]
Total strain or volumetric strain is given by:
εv = (1/E) [σx + σy + σz] (1 - 2ν)
There will be no change in volume if volumetric strain is zero.
ϵv = 0 ⇒ ν = 0.5

UPSSSC JE Civil Mock Test - 5 - Question 23

Horizontal distances obtained by tacheometric observations

Detailed Solution for UPSSSC JE Civil Mock Test - 5 - Question 23

Concept:
Slope, temperature, and tension corrections
are required in chain surveying, not in Tacheometric surveying.
In Method of Tacheometry

In this figure, the Line of sight is horizontal and staff is held truly vertical.
Tacheometric distance formula
D = (f/i)s + (f + d)
D = KS + C
Where,
K is (f/i) → called as multiplying constant
C is (f + d) → called as additive constantImportant PointsFor Annallatic Len
K = 100
C = 0

UPSSSC JE Civil Mock Test - 5 - Question 24

According to Indian Road Congress, the roads are classified into:

Detailed Solution for UPSSSC JE Civil Mock Test - 5 - Question 24

IRC(Indian Roads Congress) has classified the roads in the India in the following five categories:
(a) National Highways
(b) State Highways
(c) Major District Roads
(d) Other District Roads
(e) Village Roads

National Highways(NH):

  • National highways are the major arterial roads spanning in the length and breadth of the country and connects the Capital to the various state capitals of the country or with the neighboring countries.

State Highways(SH):

  • State highways are the roads which connect the state capital to other states and to the district headquarters in the state. They have design specifications similar to those of the National Highways because they carry enough traffic.

Major District Roads(MDR):

  • These roads connect the district headquarters to the main town centers in the district, and to the headquarters of the other districts also. They also connect these major town centers to the other state highways of importance. They have lower design specifications as compared to the NH and SH.

Other district roads(ODR) :

  • These roads connect the rural areas town centers to the major district roads of higher importance.They provide the facilities for the transportation of the raw materials or the goods mainly of agricultural products from the rural towns to the higher markets and vice-versa.

Village Roads(VR):

  • These roads connect the rural villages with one another and to the nearest higher level road or to the nearest town center. They have lower design specifications and many of them are not even metaled.
UPSSSC JE Civil Mock Test - 5 - Question 25
Who secured Rank 1 in the International Digital Teachers' Olympiad 2024?
Detailed Solution for UPSSSC JE Civil Mock Test - 5 - Question 25

The correct answer is D. Ms. Shilpi Agarwal

Ms. Shilpi Agarwal secured the top position in the International Digital Teachers' Olympiad 2024, an event designed to showcase and improve digital teaching skills among educators worldwide. This prestigious competition is organized by upEducators and involved a significant number of participants, highlighting the importance of digital pedagogy in education.

Details on Other Options:

  • A: This option is incorrect as Mr. Vijaya Kumar M did not achieve the top rank in the Olympiad, which was won by Ms. Shilpi Agarwal.
  • B: This option is not correct because Ms. Neethu William did not secure the first position; the honor went to Ms. Shilpi Agarwal.
  • C: This option is incorrect since Ms. Sonali Kumar did not rank first in the competition, which was topped by Ms. Shilpi Agarwal.

Conclusion:

Option: D, as the most accurate and relevant option, stands out due to Ms. Shilpi Agarwal's exceptional achievement in the Olympiad, clearly differentiating her from the other candidates. Thus, Ms. Shilpi Agarwal is the correct choice.

UPSSSC JE Civil Mock Test - 5 - Question 26
Which Doab region of Uttar Pradesh was ruled by Nawabs of Farrukhabad?
Detailed Solution for UPSSSC JE Civil Mock Test - 5 - Question 26

The correct answer is Central.

Explanation

  • The Nawabs of Farrukhabad ruled the Central Doab regions.
  • After the death of Aurangzeb, five independent kingdoms were established in Uttar Pradesh.
  • Pathan Sarda Najib Khan ruled Northern Regions of Uttar Pradesh: Bareily and Meerut.
  • Rahmat Khan ruled Rohillakhand (Rohil Pradesh of Meerut and Doab).
  • The Nawabs of Awadh ruled the regions of Faizabad and Lucknow.
  • The Marathas ruled the Bundelkhand region.
UPSSSC JE Civil Mock Test - 5 - Question 27
Kalyan Singh served as the Chief Minister of Uttar Pradesh for the first time during which period?
Detailed Solution for UPSSSC JE Civil Mock Test - 5 - Question 27

The correct answer is 1991-1992.

Explanation

  • Kalyan Singh served two terms as the Chief Minister of Uttar Pradesh, first from June 1991 to December 1992 and then from September 1997 to November 1999.
  • His leadership period is notably remembered for the implementation of the party's core ideologies and development initiatives, despite being marked by controversies.
  • Kalyan Singh is most prominently remembered for his role during the Ayodhya movement.
  • Apart from his political career in Uttar Pradesh, Kalyan Singh also served as the Governor of Rajasthan from 2014 to 2019 and briefly served as the Governor of Himachal Pradesh in 2019.
UPSSSC JE Civil Mock Test - 5 - Question 28
In which city did the municipal corporation issue notices to nine departments for affecting air quality? (May 2024)
Detailed Solution for UPSSSC JE Civil Mock Test - 5 - Question 28

The correct answer is B. Prayagraj

The Prayagraj Municipal Corporation (PMC) has taken action by issuing notices to nine departments, including the Prayagraj Development Authority (PDA), Public Works Department (PWD), North Central Railway (NCR), and Jal Nigam, due to their impact on air quality in Smart City Prayagraj. This initiative aims to address the rising air pollution linked to ongoing construction activities in the city.

Details on Other Options:

  • A: This option is incorrect as Agra has not been reported to have issued notices regarding air quality by its municipal corporation.
  • C: Lucknow is not the correct answer because there are no recent reports of similar notices being issued by its municipal corporation concerning air quality issues.
  • D: Ghaziabad is also incorrect as it has not been highlighted in the context of notices related to air quality affecting departments.

Conclusion:

Option: B, stands out as the most accurate and relevant choice due to the direct actions taken by the Prayagraj Municipal Corporation in response to air quality concerns, clearly differentiating it from the other options. Thus, Prayagraj is the correct choice.

UPSSSC JE Civil Mock Test - 5 - Question 29
Who won the women's 10,000m event at the Portland Track Festival 2024 athletics meet?
Detailed Solution for UPSSSC JE Civil Mock Test - 5 - Question 29

The correct answer is A. Sanjivani Jadhav

Sanjivani Jadhav won the women's 10,000m event at the 2024 Portland Track Festival, finishing with a time of 32:22.77. This victory highlights her impressive performance, showcasing her strength in long-distance running among a competitive field.

Details on Other Options:

  • B: Dutee Chand is primarily known for her sprinting events, and her participation in the 10,000m event is not documented, making this option incorrect.
  • C: Hima Das specializes in shorter distances, particularly the 400m, and did not compete in the 10,000m race, thus this option is not applicable.
  • D: Parul Chaudhary finished third in the women's 3000m steeplechase at the festival, but she did not win the 10,000m event, which makes this option incorrect.

Conclusion:

Option: A, as the most accurate and relevant option, stands out for its representation of the actual winner of the women's 10,000m event, clearly differentiating it from the other options. Thus, Sanjivani Jadhav is the correct choice.

UPSSSC JE Civil Mock Test - 5 - Question 30
Which among the following was the most well developed state amongst all the ancient Kingdoms of Uttar Pradesh in the Vedic times?
Detailed Solution for UPSSSC JE Civil Mock Test - 5 - Question 30

The correct answer is B. Panchala

Panchala was a prominent ancient kingdom in northern India, situated in the Ganges-Yamuna Doab region of the upper Gangetic plain. During the Late Vedic period, it emerged as one of the most powerful states in Ancient India, closely associated with the Kuru Kingdom. By the fifth century BCE, Panchala transformed into an oligarchic confederacy, becoming one of the sixteen Mahajanapadas of the Indian subcontinent. Eventually, it became part of the Mauryan Empire and was later incorporated into the Gupta Empire, marking its historical significance.

Details on Other Options:

  • A: Hastinapur was the capital of the Kuru Empire, located in what is now Meerut district, Uttar Pradesh. While significant, it did not reach the same level of development as Panchala.
  • C: Kannauj, previously known as Kanyakubja, served as the capital of the Amavasu kingdom. Although important, it was not as well developed as Panchala during the Vedic period.
  • D: Kosala was an ancient kingdom in the Awadh region. It was smaller during the late Vedic era and is noted for being the birthplace of Jainism and Buddhism, but it did not match the development of Panchala.

Conclusion:

Option: B, as the most accurate and relevant choice, stands out for its historical prominence and political power during the Vedic period, clearly differentiating it from the other options. Thus, Panchala is the correct choice.

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