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UPSSSC JE CIVIL
SOLVED PAPERS
1.UPSSSC JE CIVIL 27/12/2015
2.UPSSSC MANDI PARISHAD DRAFTSMAN 27/12/2015
3.UPSSSC JE 31/07/2016
उत्तर प्रदेश अधीनस्थ सेवा चयन आयोग

4.UPSSSC JE 2016- 19/12/2021

5.UPSSSC JE 2018- 16/04/2022
6.UP HOUSING & DEVELOPMENT BOARD JE 11/09/2022
7.UPSSSC JE MANDI PARISHAD DRAFTSMAN 22/05/2022
8.UPSSSC RAJ INSTRUCTOR CIVIL WORK 2016: 26/03/2023
9. UPSSSC JE 2018 RE-EXAM- 27/05/2023

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9255624029 ANUP SINGH
CIVIL KI GOLI S. GUPTA
 UPSSSC JE CIVIL
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 INDEX
S. No. Of
Description Page
no. QUESTIONS
(i) Subject-Wise Weightage Analysis 3

(ii) Exam Pattern & Syllabus 4-5

1. UPSSSC JE CIVIL 27/12/2015 6-33 125

UPSSSC MANDI PARISHAD DRAFTSMAN 125

2. 34-59
27/12/2015
3. UPSSSC JE 31/07/2016 60-85 125

4. UPSSSC JE 2016- 19/12/2021 86-120 125

5. UPSSSC JE 2018- 16/04/2022 121-154 150

6. UP Housing & Development Board JE 11/09/2022 155-177 100

UPSSSC JE MANDI PARISHAD DRAFTSMAN 40
7. 178-189
22/05/2022
UPSSSC RAJ INSTRUCTOR CIVIL WORK 2016: 35
8. 190-197
26/03/2023
9. UPSSSC JE 2018 RE-EXAM- 27/05/2023 198-236 150

Total QUESTIONS 975

For Watch All Papers Detailed Video Solution

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E-Book Pdf Of This Book Available on Civil Ki Goli App & Website
PREVIOUS YEAR Papers Weightage: -
JE Raj
AVAS

Civil Ki
J.E. J.E. J.E. 2016 J.E. 2018 Instructor Draftsman Draftsman
Subject VIKAS JE
2015 2016 (19/12/21) 2018 Re- Work 2015 2022
2022
Exam 2016
BMC 6 2 14 9 12 10 6 15
Soil 14 14 13 15 19 15 3 1
FM 9 16 9 16 12 4 1
HM 1 2
OCF 2 1 1
CT 5 5 9 12 7 4 1 6

Goli
SOM 5 22 1 12 12 6
Irrigation - 2 0 - 3
Hydrology 2 1
Surveying 15 1 37 21 15 10 4 5 10
Mechanics - 5 1 -
Steel 8 13 2 2 8 1
S.A. 6
RCC 21 27 8 17 16 13 4 10
BCME 1 2 8 3 7 1 7 66 11
ENVIRONMENT 17 3 7 13 14 11 2 7 7
Estimation 10 2 13 20 10 8
Highway 14 11 3 10 10 6 2 2
Railway 5 3
Bridge 1
Earthquake 1
AutoCad 1 8
CPM-PERT 2 1
Drawing 9
 Exam Pattern & Syllabus

Subject QUESTIONS Marks Time

Civil Engineering 65 65
Computer 15 15 2 Hours
UP GK 20 20

𝟏
NEGATIVE = 𝟒𝐭𝐡 =0.25 Marks

CIVIL ENGINEERING SUBJECT-WISE QUESTIONS:-

Subject Number of Questions

Strength of Materials & Theory of Structures 5
Design of Reinforced Concrete Structures 10
Steel & Machinery Structures 5
Estimation, Costing & Valuation 5
Soil Mechanics and Foundation Engineering 5
Surveying 5
Transportation Engineering 5
Environment and Public Health Engineering 5
Building Materials and Construction 10
Irrigation Engineering 5
Fluid Mechanics 5

Syllabus

1. Strength of Materials & Theory of Structures :- Effect of a force, tension and compression, free body
diagram, virtual work, force distribution system. Principle of energy, force conservation of energy and
momentum, rotation of rigid bodies about fixed axis, mass moment of inertia. Stresses and strains,
types of stresses and strains, definition of tension, simple compression, shear, bending, torsion,
volumetric and lateral strain, Poisson's ratio, Hooke's law. Bending moment and shear force, types of
beam, simply supported, cantilever, fixed, overhanging and continuous beams. Analysis of trusses, slope
and deflection of beams. Long columns, short columns & struts, slenderness ratio. Torsion in circular
shaft, combined bending, torsion and axial thrust, strength of hollow and solid shaft.
2. Design of Reinforced Concrete Structures:- Design based on working stress method, flexural strength,
shear strength and bond strength of a singly reinforced RCC beam, T-beams. Simply supported and
cantilever beams, determinate and indeterminate structures, design of lintels, design of a cantilever
beam and slab, design of doubly reinforced concrete beam. Design of RCC slab, design of one way and
two-way slab, stair cases. Design of reinforced brick beams, slab & lintels, design of T-beams. Design of
columns and column footings, cantilever retaining walls, components of overhead water tank and
multistoried framed structures. Introduction to design based on limit state method. Prestressed
concrete.

3. Steel & Machinery Structures:- Tension and compression members in steel. Design of steel beams.
Design of simple column & bases. Design of simple trusses, purlins, plates and girder.

4. Estimation, Costing & Valuation:- Estimating, costing and method of valuation, analysis of rates.
Method and unit of measurement. Value and cost, scrap value, salvage value, depreciation value.
Simpson's rule, centerline method, mid-section formula for earthwork calculation. Construction
management, account & entrepreneurship development.

5. Soil Mechanics and Foundation Engineering:- Fundamental terms and their relationships.
Classification and identification of soil. Phase relationship index property, laboratory determination.
Capillary phenomenon permeability, factors affecting permeability. Compaction, methods of
compaction, consolidation, difference compaction and consolidation, stresses in soil, shear strength,
coulomb's equation, unconfirmed compression test. Earth pressure retaining structures- dams and
retaining walls. Concept of slope establishments. Shallow and deep foundations, classifications of piles,
piles and pile caps. Stablisation of soils by lime and cement. Sub-surface exploration & standard tests,
effective stress, active and passive earth pressures, dams and retaining wall.

6. Surveying :- General principles of surveying- Surveying, principle of surveying, measurement of

distance, working of prismatic compass, chain surveying, compass traversing, bearings, local attractions,
types of traversing, traverse computations, corrections and missing readings. Leveling- theory,
Temporary & permanent adjustment of levels, types of leveling, reciprocal leveling, L section and cross
section, retraction and curvature corrections. Contouring- characteristics, method of contouring, uses
and plotting of contours. Plane table surveying orientation, plotting methods, two & three- point
problems. Lehmann's rule errors and precautions. Theodolite- adjustment (temporary & permanent)
measurement of angles, curves, horizontal and vertical curves, their design and layout transition curves.
Surveying equipment, minor instruments Abney level, tangent clinometer, Ceylon ghat tracer,
pantograph and planimeter, temporary & permanent adjustment of dumpy level.

7. Transportation Engineering:- Highway engineering- cross sectional elements, geometric design, types
of pavements, pavement materials aggregates and bitumen, different tests, design of flexible and rigid
pavements. Water Bound Macadam (WBM) and Wet Mix Macadam (WMM), gravel road, bituminous
construction, rigid pavement joint, pavement maintenance, highway drainage. Railway engineering-
components of permanent way-sleepers, ballast, fixtures and fastening, track geometry, points and
crossings, track junction, stations and yards. Traffic engineering- different traffic survey, speed-flow-
density and their interrelationships, intersections and interchanges, traffic signals, traffic operation,
traffic signs and markings, road safety.
8. Environmental and Public health Engineering:- Sources of water, quality of water supply, water
treatment, water distribution, laying of pipes, building water supply & maintenance. Quantity of
sewage, sewerage systems, sanitation and drainage, disposal of rainfall and domestic wastes, waste
water and garbage, plumbing for buildings, septic tanks and soak pit, sewage treatment, circular sewer,
oval sewer, sewer appurtenances. Solid waste management types and effects, types of pollutions,
containment, effects, control. Pollutants, causes, effects, control.

9. Building Materials and Construction:- Bricks, their classification and characteristics, building stones
source, quarrying, classification and properties, lime properties, cement types, properties and tests,
storage of cement. Timber and wood-based products, types, properties and uses. Types of laminates
and its uses, paints, varnishes and distempers, glass and plaster, shuttering etc. Lime concrete, cement
concretes, ingredients, grading of aggregates, workability, importance of water quality, water cement
ratio, batching, mixing, laying, compaction and curing. Building constructions, detailing of walls, floors,
roots, stair cases, doors and windows, finishing of building, plastering, pointing, damp proofing etc.
Ventilation and air conditioning, firefighting.

10. Irrigation Engineering:- Introduction, definition of irrigation, necessity of irrigation, types of

irrigation, sources of irrigation, rainfall & run-off, catchment area, Dicken's & Ryve's formula, types of
rain gauges, stream gauging. Water Requirement of crops, crop season, duty, delta and base period,
their relationship, irrigation methods & efficiencies. Gross command area, culturable command area,
intensity of irrigation, irrigable area. Types of wells, aquifer, types of ground water flow, construction of
open wells and tube wells. Yield of an open/tube well and problems, wind mills, lift canals and their
design, construction and water scheduling. Irrigation canals, perennial irrigation, different parts of
irrigation canals and their functions, canal cross sections, classification of canals, design of irrigation
canals Chezy's formula, Manning's formula, Kennedy's and Lacey's silt theory and equations, critical
velocity ratio, various types of canal lining and its advantages. Canal head works, layout and functions of
different part, difference between weir and barrage. Regulatory works, cross head regulators, falls,
energy dissipates, outlets and escapes. Cross drainage works, aqueduct, syphon, super passage, level
crossing, inlet and outlet. Dams, classification of dams, labelled cross section, spillways. Water logging,
causes and effect of water logging, detection, prevention and remedies, surface and sub-surface
drainage and their layout, field drainage, salinity controlling measures, groundwater recharging
measures. flood protection, estimation of flood discharges, systems of flood warning, river behaviors,
training works and control, marginal embankments, their design, causes of failure, spurs & dykes,
attracting & repelling types. flood management, relief & rehabilitation measures.

11. Fluid Mechanics:- Properties of fluids, hydrostatic pressure, measurement of pressure, kinematics of
fluid flow. Dynamics of fluid flow, Bernoulli's theorem, measurement of flow, pitot tube, piezometer,
orifices, venturi meter, current meter. Flow through pipes, water hammer, Reynold's number. Flow
through open channels, uniform flow, Chezy's and Manning's formula, most economical section.
Measurement of discharges by weirs and notch. Hydraulic pumps (reciprocating and centrifugal),
turbines (impulse & reaction).
(Knowledge of Concepts of Computer and Information Technology and Contemporary Technological
Development and Innovation in this field) :-

• History, Introduction and Application of Computer, Information Technology, Internet and World Wide
Web (WWW).

• General Knowledge related to:

1. Hardware and Software.

2. Input and Output.

3. Internet Protocol/IP Address.

4. IT gadgets and their application.

5. Creation of e-mail ID and use/operation of e-mail.

6. Operation of Printer, Tablet and Mobile.

7. Important elements of Word Processing (MS-Word) and Excel Processing (MS-Excel).

8. Operating System, Social Networking, e-Governance.

• Digital Financial Tools and Applications.

• Future Skills and Cyber Security.

• Technological Development and Innovation in the field of Computer and Information Technology
(Artificial Intelligence, Big Data Processing, Deep Learning, Machine Learning, Internet of Things) and
India's achievements in this field etc.

(General Information related to The State of Uttar Pradesh):-

In this part of the question paper, questions based on History, Culture, Art, Architecture, Festivals, Folk
Dance, Literature, Regional Languages, Heritage, Social Customs and Tourism, Geographical Landscape
and Environment, Natural Resources, Climate, Soil, Forest, Wildlife, Mines and Minerals, Economy,
Agriculture, Industry, Business and Employment, Polity, Administration of Uttar Pradesh and Current
Events and Achievements of Uttar Pradesh State in various fields etc. will be asked from the candidates.
 Q.03. Pressure energy of fluid can be increased by
UPSSSC JE Civil 2015 using:
(a) Turbine (b) Pump
❖ FLUID MECHANICS
(c) Viscometer (d) Hydraulic ram
Q.01. When various fluid particles move in Zig-Zag
paths, flow is called: Ans: (b) Centrifugal pump: A centrifugal pump is a
mechanical device designed to move fluid by means of
(a) Laminar flow (b) Turbulent flow
the transfer of rotational energy from one or more
(c) Uniform flow (d) None of above driven rotors, called impellers.
Ans : (b) Turbulent flow: It is that type of flow in which • Fluid enters the rapidly rotating impeller along it’s axis
fluid particles move in a zig-zag way. Due to the zig-zag and is cast out by centrifugal force along it’s
movement of fluid-particle eddies formation take place circumference through the impeller’s vane tips.
which is responsible for high energy loss. • The action of the impeller increases the fluid’s velocity
and pressure and also directs it towards the pump
outlet. The pump casing is specially designed to
constrict the fluid from the pump inlet, direct it into
the impeller and then slow and control the fluid before
discharge.
Laminar flow: It is a type of flow in which the fluid
Q.04. The ratio between inertia force and elastic force
particles move along well-defined paths or streamline
is called:
and all the streamlines are straight and parallel. It is also
called streamline flow or viscous flow. (a) Cauchy’s number (b) Mach’s number
(c) Weber number (d) Reynold’s number
Ans:(a) Cauchy Number : Cauchy number is defined as
the ratio of inertia force to elastic force
𝐈𝐧𝐞𝐫𝐭𝐢𝐚 𝐟𝐨𝐫𝐜𝐞 𝛒𝐋𝟐 𝐕𝟐 𝛒𝐕𝟐
𝐂𝐧 = = =
𝐄𝐥𝐚𝐬𝐭𝐢𝐜 𝐟𝐨𝐫𝐜𝐞 𝐊𝐋𝟐 𝐊
Mach number :- Mach number is defined as the ratio of
Q.02. Rotameter is used to measure inertia force to Compressibility force.
(a) Velocity of fluid (b) Kinetic energy of fluid
𝐈𝐧𝐞𝐫𝐭𝐢𝐚 𝐟𝐨𝐫𝐜𝐞 𝛒𝐀𝐕 𝟐 𝐕𝟐 𝐕 𝐕
(c) Viscosity of fluid (d) Flow of fluid 𝐌=√ =√ √ 𝐊 = =
𝐂𝐨𝐦𝐩𝐫𝐞𝐬𝐬𝐢𝐛𝐢𝐥𝐢𝐭𝐲 𝐟𝐨𝐫𝐜𝐞 𝐊𝐀 𝐊
√𝛒 𝐂
𝛒

Ans : (d) Rotameter : A rotameter is a device that

𝑲
measures the flow rate of fluid in a closed tube. {√ = 𝑪 = Velocity of sound
𝝆
• It is a variable area type of flow meter. In this, a float 𝐕𝐞𝐥𝐨𝐜𝐢𝐭𝐲 𝐨𝐟 𝐛𝐨𝐝𝐲 𝐦𝐨𝐯𝐢𝐧𝐠 𝐢𝐧 𝐟𝐥𝐮𝐢𝐝
rises inside a conically shaped glass tube, as the flow 𝐌=
𝐯𝐞𝐥𝐨𝐜𝐢𝐭𝐲 𝐨𝐟 𝐬𝐨𝐮𝐧𝐝 𝐢𝐧 𝐟𝐥𝐮𝐢𝐝
increases, its position on a scale can be read off as the • For the compressible fluid flow, Mach number is an
flow rate. important dimensionless parameter. On the basis of
• The construction is as shown: the Mach number, the flow is defined.
Mach Number Type of flow
M < 0.8 Sub-sonic flow
0.8<M<1.3 Trans-sonic
M=1 Sonic flow
1.3<M<5 Super-sonic flow
M>6 Hypersonic flow

Q.05. The most familiar form of Bernoulli’s equation is

𝐏𝟏 𝐕𝟐 𝐏𝟐 𝐕𝟐
(a)
𝐖
+ 𝐙𝟏 + 𝟐𝐠𝟏 = 𝐖
+ 𝐙𝟐 + 𝟐𝐠𝟐
𝐝𝐩
(b) + 𝐠. 𝐝𝐳 + 𝐯. 𝐝𝐯 = 𝟎
Limitations : - 𝛒
• Cannot be placed horizontally without amendments. 𝑷 𝑽𝟐
(c) [ + 𝒁 + 𝟐𝒈] any section=constant head
• Cannot bear high pressure and temperature. 𝑾
• Only for measuring small flow rates. (d) None of these
• Free from solid impurities.

UPSSSC JE 2015 6 Civil KI Goli

Ans : (c) Bernoulli's principle :- For a streamlined flow of (a) In stable equilibrium (b) In unstable equilibrium
an ideal liquid in a varying cross-section tube the total (c) In neutral equilibrium (d) in Real equilibrium
energy per unit volume remains constant throughout Ans : (c) Meta Centre:- It is defined as the point about
the fluid. which a body starts oscillating when the body is tilted by
• This means that in steady flow the sum of all forms of a small angle. The meta centre may also be defined as
mechanical energy in a fluid along a streamline is the the point at which the line of action of the force of
same at all points on that streamline. buoyancy will meet the normal axis of the body when
the body is given a small angular displacement.

• For the body shown in the figure, M is above G, and

the couple acting on the body in its displaced position is
a restoring couple which tends to turn the body to its
original position
From Bernoulli's principle : - • If M were below G, the couple would be an overturning
𝐏𝟏 𝟏 𝐏𝟐 𝟏 couple and the original equilibrium would have been
+ 𝐠𝐡𝟏 + 𝐕𝟐 = + 𝐠𝐡𝟐 + 𝐕𝟐
𝝆 𝟐 𝟏 𝛒 𝟐 𝟐 unstable
𝐏 𝟏
+ 𝐠𝐡 + 𝐕 𝟐 = constant • When M coincides with G, the body will assume its new
𝛒 𝟐
position without any further movement and thus will be
𝐏 𝐕𝟐
𝐖
+ 𝟐𝐠 + 𝐙 = constant in neutral equilibrium
Where, W = ρ g Q.08. Flow in a pipe is laminar if the Reynold’s number
• In fluid dynamics, Bernoulli’s theorem gives the is
relation among the elevation, velocity, and pressure in a (a) Less than 2000
moving fluid, for which, the viscosity and compressibility (b) Between 2000 and 4000
are negligible and the flow is laminar or steady. (c) Between 4000 and 6000
• Thus the theorem is applicable for incompressible (d) Equal to 10000
flows.
Ans : (a) Laminar flow:- A flow is said to be laminar when
Q.06. If salt is added in water, the surface tension of various fluid particles move in layer or laminae, or well-
water will: defined paths, with one layer of fluid, smoothly sliding
(a) Increase (b) Decrease over an adjacent layer.
(c) Will not change (d) None of the above Reynolds number:- It is a dimensionless value that is
Ans : (a) Surface tension:- Surface tension could be applied in fluid mechanics to represent whether the fluid
defined as the property of the surface of a liquid that flow in a duct or pat a body is laminar or turbulent. The
allows it to resist an external force, due to the cohesive value is obtained by comparing the inertial forces with
nature of the water molecules. The surface tension of the viscous force.
𝐈𝐧𝐞𝐫𝐭𝐢𝐚𝐥 𝐅𝐨𝐫𝐜𝐞
water provides the necessary wall tension for the Reynolds Number =
𝐕𝐢𝐬𝐜𝐨𝐮𝐬 𝐅𝐨𝐫𝐜𝐞
formation of bubbles with water. The tendency to Reynolds number for flow in channels is given as,
minimize that wall tension pulls the bubbles into 𝛒𝐕𝐑
spherical shapes. Re =
𝛍
The surface tension of the liquid increases after we add Where V is the mean velocity of the flow, ρ is the mass
salt to the water because the ions from the salt interact density, μ is the dynamic viscosity and D is known
with the water molecules, leading to a stronger diameter of the pipe.
attraction between the molecules at the surface. Surface The Kind of flow is based on the value of Re:
tension arises due to the cohesive nature of a liquid
If Re < 2000, the flow is called laminar
which resists an external force on the liquid surface.
If Re > 4000, the flow is called turbulent
Q.07. When the metacenter and center of gravity of any
If 2000 < Re < 4000, the flow is called transition.
floating body coincide, the floating body will be:

UPSSSC JE 2015 7 Civil KI Goli

 ❖ HIGHWAY therefore to check the tendency of the vehicle to
overturn or skid.
Q.09. The color of the upper part of kilometer stone on
The value of super-elevation e is given by,
the roadside in case of state highway is
𝐯𝟐
(a) Green (b) Yellow 𝐞+𝐟=
𝐠𝐑
(c) Brown (d) Red Where,
V = speed of the vehicle,
Ans : (a) The various type of milestone color codes being
R = radius of the horizontal curve,
used in India for highways are specified in the below
g = acceleration due to gravity,
sstabulated form : -
f = Lateral friction.
Type of Highway Milestone Color Code According to IRC, the maximum superelevation is given
National Highway Yellow & White Strips by:
State Highway Green & White Strips Type of Terrain emax
City or Main District Blue or Black & White Plain and roling 7% (1 in 16)
Roads strips Hilly road 10%
Village Roads Orange & White strips Urban road 4%
From the above table, We can say that the color of the Q.12. The specific gravity of bitumen is
upper part of the kilometer stone on the roadside in the (a) 1.78 (b) 1.02 (c) 1.30 (d) 0.85
state highway is Green.
Ans : (b) Specific gravity of bituminous material:
• It is defined as the ratio of the mass of a given volume
of a material to the equal volume of water, the
temperature of both being 27°C.
• The specific gravity of pure bitumen ranges from 0.97
to 1.02.
• According to Indian Standard (BIS), the minimum
specific gravity of paving bitumen at 27°C shall be 0.99
Q.10. The value of ruling gradient in plains as per IRC is for grades A25, A35, A45, A65, S35, S45 and S65, 0.98
for A90 and S90 and 0.97 for A200 and S200.
(a) 1 in 10 (b) 1 in 15
(c) 1 in 30 (d) 1 in 40 Q.13. The highest point on a carriage way is known as
(a) Super elevation (b) Crown
Ans: (c) Ruling gradient:- Ruling gradient is the
maximum gradient within which the designer attempts (c) Camber (d) Gradient
to design the vertical profile of a road. Ans : (b) Crown: The highest point of the camber is called
• Gradient up to the ruling gradient are adopted as a the crown of the road.
normal course in design of vertical alignment and Camber: It is the slope provided to the road surface in
accordingly the quantities of cut and fill are decided. the transverse direction to drain off the rainwater from
Hence ruling gradient is also known as design gradient. the road surface. It is also known as the cross slope of
The IRC has recommended ruling gradient values of: the road. Camber can be written as 1 in n or x%.
1. 1 in 30 on plain and rolling terrain. There are generally three types of the cambers :
2. 1 in 20 on mountainous (hilly) terrain. (a) Straight Camber (b) Parabolic Camber (c) Mixed
3. 1 in 16.7 on steep terrain. Camber.
Gradient: It is the slope provided to the surface of the
Terrain Ruling Limitings Exceptional road in the longitudinal direction for the vertical
Plain/Rolling 3.3 5.0 6.7 alignment of the road. There are three kinds of
Hily 5.0 6.0 7.0 gradients:
Steep 6.0 7.0 8.0 (a) Ruling Gradient (b) Limiting Gradient (c) Exceptional
Gradient
Q.11. The maximum allowable superelevation is: Super elevation:- It is the amount by which the outer
(a) 1 in 12 (b) 1 in 18 edge of a curve on a road or railway is raised above the
(c) 1 in 15 (d) 1 in 30 inner edge to counteract the effect of centrifugal force.
Ans : (c) Super-elevation: It is the inward transverse
slope provided throughout the length of the horizontal
curves to counteract the centrifugal force which tends to
pull out the vehicle in an outward direction and

UPSSSC JE 2015 8 Civil KI Goli

Q.14. Penetration test on bitumen is used for Since the bituminous macadam is an open-graded
determining it’s: mixture there is a potential that it may trap water or
(a) Grade (b) Viscosity moisture within the pavement system. Therefore,
(c) Ductility (d) Temperature susceptibility providing a proper drainage outlet to the BM layer
should be considered to prevent moisture-induced
Ans : (a) Penetration test : - Hardness of bitumen is damage to the BM and adjacent bituminous layers.
obtained by the penetration test.
Q.17. One degree of curve of a roadway is equivalent
• It measures the distance a standard blunt pointed
to
needle will vertically penetrate a sample of material at
25°C, the load being 100 gram and time of application (a) 1600/R (b) 1720/R
of load being 5 seconds. (c) 1750/R (d) 1800/R
• The unit of penetration is 1/10 mm. Thus 80/10 grade Ans : (b) Arc definition of degree of curve:- According to
bitumen means 8 - 10 mm penetration. this, the degree of curve is the central angle subtended
• 30/40 and 80/100 grade bitumen are more commonly by an arc of length 20 m or 30 m depending on the case.
used, depending upon bitumen construction type and Chord definition of degree of curve:- According to this,
climate conditions. the degree of curve is the central angle subtended by a
chord of length 20 m or 30 m depending on the case.

For 20 m arc or chord length:

R = Radius of the curve
D = degree of the curve as per arc definition for 20 m
arc length i.e 20 m arc length subtends an angle of D° at
the center.
𝛑
D° = 𝐃 radians
𝟏𝟖𝟎
𝐀𝐫𝐜 𝟐𝟎
θ(radians) = =
𝐑𝐚𝐝𝐢𝐮𝐬 𝐑
𝐑𝐃𝛑
= 𝟐𝟎
𝟏𝟖𝟎
𝟏𝟏𝟒𝟔
𝐑=
Q.15. Slope of village roads should be usually less than 𝐃

(a) 1 in 12 (b) 1 in 10 For 30 m arc or chord length:

(c) 1 in 4 (d) 1 in 3 R = Radius of the curve
D = degree of the curve as per arc definition for 30 m
Ans : (a) The slope of village roads should be usually
arc length i.e 30 m arc length subtends an angle of D° at
taken 1 in 12.
the center.
Q.16. The Thickness of ‘Domer’ laid (Roads) layer in 𝛑
D° = 𝐃 radians
‘Bitumen’ Roads is - 𝟏𝟖𝟎
𝐀𝐫𝐜 𝟑𝟎
(a) 5 to 10 cm (b) 2 to 3 cm θ(radians) = =
𝐑𝐚𝐝𝐢𝐮𝐬 𝐑
(c) 1 to 2 cm (d) 0.5 to 1.0 cm 𝐑𝐃𝛑
= 𝟑𝟎
𝟏𝟖𝟎
Ans: (a) Bituminous Macadam (BM) shall consist of 𝟏𝟕𝟏𝟖.𝟖
𝐑=
mineral aggregate and appropriate binder, mixed in a hot 𝐃

mix plant and laid with a mechanized paver. It is an open- As per Given options most appropriate answer is
graded mixture suitable for base course. It is laid in a 1720/R
single course or in multiple layers on a previously Q.18. Los Angeles machines is used to test the
prepared base. The thickness of the single layer shall be aggregate for _____.
50 mm to 100 mm. i.e (5 cm to 10 cm) (a) crushing strength (b) impact value
(c) abrasion resistance (d) water absorption

UPSSSC JE 2015 9 Civil KI Goli

Ans : (c) Various Tests for Aggregates with IS codes:
Property of
Type of Test Test Method
aggregate
Crushing IS : 2386 (part
Crushing test
strength 4) -1963
Los Angeles abrasion IS : 2386 (Part
Hardness
test 5)-1963
IS : 2386 (Part
Toughness Aggregate impact test
4)-1963
Soundness test-
IS : 2386 (Part
Durability accelerated durability ❖ RAILWAY
5)-1963
test
Q.20. The width of the broad gauge is
Shape IS : 2386 (Part
Shape test (a) 1.576 m (b) 1.676 m
factors 1)-1963
Specific Specific gravity test (c) 1.776 m (d) 1.67 m
IS : 2386 (Part
gravity and and water absorption Ans:(b) The standard width of broad gauge (BG) in
3)-1963
porosity test railway systems is 1.676 meters.
Adhesion to Stripping value of According to Indian railways, the clear distance between
IS: 6241-1971
bitumen aggregate the inner faces of two rails for the different type of
Q.19. Bottom most layer of Pavement is known as - gauges are:
(a) Wearing course (b) Base course Gauge Distance Between Rails
(c) Sub-base course (d) Sub grade Broad Gauge 1.676 m
Ans : (d) The bottom most layer of a pavement structure Meter Gauge 1.0 m
is known as the subgrade. The subgrade is the native soil
Narrow Gauge 0.762 m
(or improved soil), usually compacted, that serves as the
foundation for the pavement structure. Q.21. The number of fish bolts per fish plate joining
Types of pavement structure: Pavements are generally rails is
classified into two categories based on structural (a) 2 (b) 4 (c) 6 (d) 8
behavior: Ans : (b) The name ‘fish plate’ derives from the fish-
(i) Flexible pavement shaped section of this fitting.
(ii) Rigid pavement • The function of a fish plate is to hold two rails together
Flexible pavement : in both the horizontal and vertical planes. Fish plates are
(i) The pavements which have very less flexural strength manufactured using a special type of steel (Indian
are called flexible pavements. Railways specification T-1/57) with the composition
given below:
(ii) This type of pavements transmit the load to the lower
layer by grain to grain transfer. (a) Carbon: 0.30–0.42%
(iii) A typical flexible pavement consists of four (b) Manganese: not more than 0.6%
components as shown in the figure below (c) Silicon: not more than 0.15%
•Soil subgrade •Sub-base course (d) Sulphur and phosphorous: not more than 0.06%
•Base course •Surface course • The number of bolts per fish plate is 4
• The steel used for fish plates should have a minimum
tensile strength of 5.58 to 6.51 t/cm2 with a minimum
elongation of 20%.
• Fish plates are designed to have roughly the same
strength as the rail section, and as such the section area
of two fish plates connecting the rail ends is kept about
the same as that of the rail section.

Rigid pavement:
(i) Rigid pavement are those which possess worthy
flexural strength.
(ii) The rigid pavement transmits the wheel load stresses
through a wider area below by the slab action.

UPSSSC JE 2015 10 Civil KI Goli

Aquatic Vegetation: Aquatic vegetation and algae Concentration in ambient
directly release oxygen into the water during air
photosynthesis (during the day). At night, plants actually
Time- Ecologically,
use oxygen for their metabolism.

S. No.
Thus, the dissolved oxygen concentration decreases Pollutant weighted Industrial, sensitive
with the increase in the temperature of the water. Average Residential, Area
Rural and (notified by
Q.26. As per Indian standard (IS 10500: 2012) of other areas central
drinking water specification, the concentration of iron Government)
in drinking water should not exceed:
Sulphur Annual 50 20
(a) 0.5 mg/L (b) 0.4 mg/L 1 dioxide
(c) 0.3 mg/L (d) 0.2 mg/L (SO2) µg/m3 24 hrs 80 80
Ans : (c) As per IS 10500: 2012. The permissible limits Nitrogen Annual 40 30
of various compounds are as follows: dioxide
2
Permissible Cause for (NO2), 24 hrs 80 80
Parameter
Limit Rejection µg/m3
Total suspended
500 2000 Particulate Annual 60 60
solids (mg/L)
Matter (size
Turbidity (NTU) 1 5 less than 10
3
Colour (TCU) 5 15 µg/m3) or 24 hrs 100 100
Taste & Odour PM 10,
1 3 µg/m3
(TON)
Total dissolved Particular Annual 40 40
500 2000
solids (mg/L) Matter (size
Alkalinity (mg/L as less than
200 600 4.
CaCO3) 2.5 µm) 24 hrs 60 60
pH 6.5-8.5 No Relaxation or PM 2.5,
Hardness (mg/L as µg/m3
200 600
CaCO3) Ozone (O3), Annual 100 100
Chloride content 5
250 1000 µg/m3 24 hrs 180 180
(mg/L)
Free ammonia Lead (Pb), Annual 0.50 0.50
0.15 No Relaxation 6
(mg/L) µg/m3 24 hrs 1.0 1.0
Nitrite (mg/L) 0 0 Carbon 8 hrs 02 02
Nitrate (mg/L) 45 No Relaxation Monoxide
7
Fluoride content (CO), 1 hr 04 04
1 1.5 mg/m3
(mg/L)
Iron as Fe (mg/L) 1(Old=0.3) No Relaxation Ammonia Annual 100 100
Sulphate (mg/L) 200 400 8 (NH3),
µg/m3 24 hrs 400 400
Q.27. As per National Ambient Air Quality standards,
maximum permissible concentration of NO2 (24 hr. Benzene
average) in ambient air in residential area is 9 (C6H6), Annual 100 100
(a) 100 μg/m3 µg/m3
(b) 80 μg/m3 Benzo (a)
(c) 60 μg/m3 Pyrene
(BaP)-
(d) 40 μg/m3 10 Annual 01 01
particular
Ans : (b) According to the National Ambient Air Quality phase only,
Standards (NAAQS), the permissible limit of NO2 (24 hr. ng/m3
average) in ambient air in a residential area is 80 µg/m3.
Arsenic,
As per National Ambient Air Quality standards 11 06 06
ng/m3

UPSSSC JE 2015 12 Civil KI Goli

Q.28. To control vehicular pollution from exhaust of Ans : (b) Cowl is provided at upper end of ventilating
vehicles, following is fitted: column to prevent blockage by nesting birds , Also it
(a) Electrostatic precipitator (b) Catalytic converter helps to escape out foul gases.s
(c) Bag filter (d) Cyclone separator
Ans : (b) Catalytic Convertor is a vehicular pollution
control device used in automobiles, which converts the
toxic gases pollutants present in engine exhaust to less
toxic pollutants by filtering or absorbing by a catalyst &
also a chemical reaction.
• It converts nitrogen oxide, carbon monoxide &
hydrocarbon like harmful products into carbon dioxide,
water vapour & nitrogen gas like less harmful products.
Electrostatic precipitator: It is a device that removes
suspended dust particles from a gas or exhaust by
applying a high-voltage electrostatic charge and
collecting the particles on charged plates.
• Cyclone separators are separation devices (dry
scrubbers) that use the principle of inertia to remove
particulate matter from flue gases.
• A bag filter is an air pollution control device and dust
collector that removes particulates out of air or gas Q.31. As per Indian standard (IS 10500:2012) of
released from commercial processes or combustion for drinking water specification, pH value should be in the
electricity generation. range of:
Q.29. For a sewer line of 50 cm diameter, the spacing of (a) 6.5 - 8.0 (b) 6.5 - 8.5
manhole along straight run would be approximately. (c) 6.0 - 8.5 (d) 7.0
(a) 10 m (b) 100 m Ans : (b) Permissible values of pH for public supplies may
(c) 500 m (d) 1000 m range between 6.5 to 8.5.
Ans : (b) As per IS 1742:1960: As per IS 10500: 2012. The permissible limits of various
compounds are as follows:
Size of Sewer spacing of manholes on
Parameter Permissible Cause for
straight reaches of
Limit Rejection
sewer lines
Total suspended solids 500 2000
Diameter up to 0.3 m 45 m
(mg/L)
Diameter up to 0.6 m 75 m Turbidity (NTU) 1 5
Diameter up to 0.9 m 90 m Colour (TCU) 5 15
Diameter up to 1.2 m 120 m Taste & Odour (TON) 1 3
Diameter up to 1.5 m 250 m Total dissolved solids 500 2000
(mg/L)
Diameter greater than 300 m
1.5 m Alkalinity (mg/L as 200 600
CaCO3)
So as per the given options most appropriate answer, for pH 6.5-8.5 No Relaxation
50 cm diameter of sewer spacing of manhole is 100 m.
Hardness (mg/L as 200 600
Q.30. “Cowl” is provided at:
CaCO3)
(a) Lower end of the ventilating column
(b) upper end of ventilating column Chloride content 250 1000
(c) Upper end of the manhole (mg/L)
(d) Lower end of manhole Free ammonia (mg/L) 0.15 No Relaxation

UPSSSC JE 2015 13 Civil KI Goli

Dimension of T-beam : - The effective width of the flange • Prestressed concrete requires high-quality dense
is adopted as the minimum of c/c distance of the nearby concrete of high strength. Perfect quality concrete in
ribs or beams. production, placement and compaction is required.
• The overall thickness of the slab crossing over the • It requires high tensile steel, which is 2.5 to 3.5 times
beam is taken as flange thickness. costlier than mild steel.
• The breadth of the rib is taken down earth ground. It • Prestressing process requires complicated tensioning
should be adequate to hold the steel zone in it, equipment and anchoring devices.
effectively. It might be taken as between 1/3 to 2/3 of • Prestressed concrete needs skilled labours.
the rib depth of the beam.
Q.39. If σs is the stress in the bar and 𝜼𝒅 is the design
• The depth of T-beam is taken between 1/10 to 1/20 of
bond stress then the development length of a bar of
the span. diameter ϕ is given by:
Q.37. For a slab spanning in two directions, the ratio of 𝟒𝝓×𝝈𝜹 𝝓×𝝈𝒔
(a) (b)
span to the depth of slab should not exceed 𝜼𝒅 𝟒×𝜼𝒅
(a) 10 (b) 20 (c) 35 (d) 50 𝟐𝝓×𝝈𝒔 𝟑𝝓×𝝈𝒔
(c) (d)
𝟑×𝜼𝒅 𝜼𝒅
Ans : (c) Deflection check for two way slab:
Ans : (b) Development Length : - A development length
Support Span/overall depth
is the amount of rebar length needed to be embedded
Condition Mild steel Fe 415/Fe 500
or projected into concrete to create desired bond
Simply strength between the two materials and develop
35 28
supported required stress in steel at that section.
Continuous 40 32
As per IS 456: 2000, clause 26.2.1,
So, From the above table It is clear that Span/depth ratio
The development length of bars is given by,
for two way slab should not be greater than 40. But From
𝝓×𝝈𝒔
the given options, 35 is the closest one. 𝑳𝒅 = 𝟒×𝜼𝒅
Q.38. In a pre-stressed concrete structure where,
(a) dead load of structure is reduced ϕ = nominal diameter of the bar
(b) cracking of concrete is avoided σs = stress in the bar at the section considered at design
(c) the cost of supporting structure and foundation is load
reduced 𝜼𝒅 = design bond stress
(d) all of the above • The value of bond stress is increased by 60% for a
Ans : (d) Prestressed Concrete : - Concrete in which deformed bar in tension and a further increase of 25% is
reinforcing steel bars are stretched and anchored to made for bars in compression.
compress it and thus increase its resistance to stress. In the case of bundled bars in contact, the development
Advantages: length is increased than that for the individual bars by
• It is suitable for long-span structures and liquid • 10% for two bars in contact
retaining structures. • 20% for three bars in contact
• The load-carrying capacity of the beam is also higher. • 33% for four bars in content
• It reduces the deflection and hence higher stiffness. Q.40. In the design of two-way slab restrained at all
• Sections are considered uncracked while analysis edges, torsional reinforcement required is
because the members are free from tensile stresses, (a) 0.75 times the area of steel provided at mid-span in
and thus protects the steel from corrosion. the same direction
• The prestressing force induces compression on the
(b) 0.375 times the area of steel provided at mid-span
section which increases the shear capacity of the
in the same direction
section.
• It has high resistance against fatigue, impact, cracks, (c) 0.375 times the area of steel provided in the shorter
and corrosion. span
(d) Nil
Practically, dead loads are neutral in a prestressed
concrete beam. Consequently, the dead load weight of Ans : (a) Torsional Reinforcement detailing in Slab :
the structure is decreased which results in reduced • Torsional reinforcement is provided in the form of a
consumption of materials. grid or mesh both at the top and bottom of the slab.
 Disadvantages of prestressing: • IS 456: 2000 recommends that the torsional
reinforcement grid should extend beyond the edge for
a distance not less than 20% of a shorter span. The

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 total area of torsional steel provided in each of the four Q.43. In a single reinforced beam, if the permissible
layers should not be less than:- stress in concrete reaches earlier than that in steel, the
1. 0.75Astx if both the meeting edges are restrained. beam section is called __________.
2. 0.375Astx if one of the two meeting edges, one in (a) under-reinforced section (b) over reinforced section
continuous and other discontinuous (c) economic section (d) critical section
Here, Astx denotes are of flexural steel required for Ans : (b) In under reinforced sections, steel is less so it
maximum mid-span moment in the short direction. starts to yield first and then cracks starts appearing on
No torsional reinforcement is required if both the concrete and structure fail slowly i.e. ductile failure. It
meeting edges are continuous. gives us warning and time to escape before collapse any
Q.41. In double reinforced sections, total other vulnerable conditions like or earthquake.
reinforcement percentage should not exceed: • In over reinforced sections, concrete breaks first and it
(a) 4 (b) 6 (c) 8 (d) 10 is a brittle failure, and structure will collapse instantly.
So we have no time to save our life or escape from
Ans : (c) For Doubly Reinforced Beams: building before collapse.
• Maximum total reinforcement percentage = 8% of • In balanced section, both steel and concrete will reach
gross area their limiting values simultaneously.
• Maximum compression reinforcement = 4% of gross Under Over
area Balanced section
reinforced reinforced
• Maximum tension reinforcement = 4% of gross area
Xu < Xulim Xu = Xulim Xu > Xulim
Reinforcement of different sections:
For Beams: Ast < Astlim Ast = Astlim Ast > Astlim
• Maximum reinforcement = 4% of gross area MOR < MOR >
MOR = MORlim
• Minimum reinforcement = 0.85bd/fy MORlim MORlim
For column: Z > Zlim Z = Zlim Z < Zlim
• Maximum reinforcement = 6% of gross area Steel yields Concrete
• Minimum reinforcement = 0.8% of gross area first then Both steel yields reaches
concrete and concrete maximum
For slab:
reaches reaches maximum strain of
• Minimum reinforcement = 0.15% of gross area(For maximum strain of 0.0035 0.0035
mild steel) strain of simultaneously before steel
• Minimum reinforcement = 0.12% of gross area(For 0.0035 yields
HYSD bars)
Q.44. Tension bars in a cantilever beam must be
Q.42. A reinforced concrete beam, supported on enclosed in the support up to:
columns at ends, has a clear span of 5 m and 0.5 m Where, d: Effective depth of beam, Ld: Development
effective depth. It carries a total uniformly distributed length
load of 100 kN/m. The design shear force for the beam
(a) Ld + 10d (b) Ld /3
is
(c) 12ϕ (d) Ld
(a) 250 kN (b) 200 kN
(c) 175 kN (d) 150 kN Ans : (d) Development Length: A development length is
the amount of rebar length that is needed to be
Ans : (b) Given, embedded or projected into concrete to create desired
Clear span = 5 m, Effective depth = 0.5 m bond strength between the two materials and also to
Uniformly distributed load(W) = 100 KN/m develop required stress in steel at that section.
The shear force of the beam in case of uniformly As per IS 456: 2000, clause 26.2.1,
distributed load, The development length is given by:
𝐖×𝐋 𝟏𝟎𝟎×𝟓
𝐕= = = 𝟐𝟓𝟎 𝐊𝐍 𝝓𝝈𝒔
𝟐 𝟐 𝑳𝒅 = 𝟒𝝉𝒅
The location of critical section for shear design is
where
determined based on the conditions at the supports. The
location of critical shear is at a distance of effective ϕ = nominal diameter of the bar
depth d. σs = stress in the bar at the section considered at design
Design shear force for the beam: load
Vu = 250 - 100 × 0.5 = 200 KN τbd = design bond stress
And they should be changed as following-

UPSSSC JE 2015 16 Civil KI Goli

• The design bond stress of concrete should be Ans: (d) Excessive reinforcement leads to congestion of
increased by 60% for deformed bars bars which adversely affects the placement and
• For bars in compression, the values of bond stress compaction of concrete.
should be increased by 25% than the bond stress in CI. 26.5.1 of IS 456:2000 limits the maximum area of
tension. steel reinforcement in tension (Ast max) and compression
Q.45. For the deflection of a simply supported beam of (Asc max) to 4% of gross area i.e. 0.04bD.
8 m spans to be within permissible limits, the ratio of Q.48. The minimum size of one side or diameter of
its span to effective depth should not exceed column in reinforced cement concrete structure to
(a) 7 (b) 20 make it earthquake resistant should not be less than:
(c) 26 (d) 8 (a) 200 mm (b) 250 mm (c) 300 mm (d) 350 mm
Ans : (b) Clause 23.3 of in IS 456: 2000, the basic value Ans : (c) The minimum dimension of column shall not
of effective span to depth ratio for span upto 10 m for be less than
different support conditions specified are: • 15 times the largest beam bar diameter of the
longitudinal reinforcement in the beam passing through
Type of beam Span/Depth or anchoring into the column joint, and
Cantilever 7 • 300 mm.
Q.49. The spacing between two successive stirrups
Simply supported 20 (hoops) in reinforced cement concrete beam shall not
be less than:
Continuous 26
(a) 100 mm (b) 125 mm
For spans > 10 m, a Modification factor of 10/(span in (c) 150 mm (d) 175 mm
meter) is to be multiplied with basic values for simply
Ans : (a) Stirrups (hoops) are provided in reinforced
supported and continuous beams. cement concrete beams to resist the shear forces and
Q.46. The diameter of longitudinal reinforcement bars torsional forces. The spacing between two successive
of a RCC column should never be less than: stirrups is an important parameter that affects the
(a) 6 mm (b) 8 mm (c) 10 mm (d) 12 mm strength and stability of the beam. The correct spacing
of stirrups in RCC beams is essential to ensure the safety
Ans : (d) As per IS 456: 2000, clause 26.5.3.1, and durability of the structure.
Longitudinal reinforcement in a column : The spacing between two successive stirrups in
The minimum diameter of longitudinal reinforcement reinforced cement concrete beam shall not be less than
bars: 100 mm. This means that the distance between two
• The main longitudinal reinforcement bars used in the consecutive stirrups should be at least 100 mm.
column shall not be less than 12 mm in diameter. Q.50. The concentration of organic solids in water to be
The minimum number of longitudinal bars: used in reinforced cement concrete contraction should
• The minimum number of the longitudinal bar provided not be more than:
in the column shall be four in rectangular columns six (a) 50 mg/L (b) 100 mg/L
in circular columns (c) 150 mg/L (d) 200 mg/L
Percentage of steel: - The cross-sectional area of
Ans:(d) As per clause 5.4 of IS 456: 2000
longitudinal reinforcement shall be not less than 0.8
percent nor more than 6 percent of the gross cross- Potable water is considered satisfactory for mixing
sectional area of the column. Concrete and the permissible limits for solids is shown in
table below:
Helical reinforcement: - A reinforced concrete column
having helical reinforcement shall have at least six bar of Type of solid Max. Permissible limit
longitudinal reinforcement within the helical Organic 200 mg/l
reinforcement.
Inorganic 3000 mg/l
Spacing : - The spacing of longitudinal bars measured
along the periphery of the column shall not exceed 300 Sulphates 400 mg/l
mm. Chlorides 2000 mg/l
Q.47. A reinforced cement concrete beam can have (concrete)
maximum area of tension reinforcement as Chlorides (RCC) 500 mg/l
(a) 0.06bD (b) 0.02bD
Suspended matter 2000 mg/l
(c) 0.05bD (d) 0.04bD

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 𝐋 Axial
Maximum deflection At centre, x = Shear Bearing
𝟐 Types of fastener tension
𝐏 𝐋𝟑 (Mpa) (Mpa)
δ= (Mpa)
𝟒𝟖 𝐄𝐈
(i) Power -driven
Where, P = Applied load; E = Young's modulus; I
Shop river Field 100 100 300
= Moment of Inertia; L = length of the beam Rivet 90 90 270
(ii) Hand driven
❖ STEEL STRUCTURE 80 80 250
rivets
Q.65. If, the diameter of rivet is 25 mm or less; diameter
(iii) Closed
of rivet hole will be diameter rivet.
tolerance and 120 100 300
(a) more than 1.5 mm (b) less than 1.5 mm turned bolts
(c) more than 2.0 mm (d) Equal (iv) Bolt in
120 80 250
Ans : (a) Gross diameter is the diameter of rivet when it clearance holes
is fully inserted in the rivet hole or body. This diameter is So from the table, it is clear that For field rivets the
more than the diameter of shank (when it is cold or maximum permissible stresses in rivets and bolts as
before insertion i.e. nominal diameter. given in codes are reduced by 10%.
Due to the heating effect, the size of rivets gets Q.68. The effective length of a steel column, effectively
expanded which upon cooling gets reduced (called held in position and restrained against rotation at both
shank diameter). ends is:
Rivet hole diameter is calculated as: (a) 0.80 L (b) 1.0 L (c) 0.65 L (d) 0.5 L
Rivet size, d (mm) Rivet Hole size d0 (mm) Ans : (c) As per IS code effective length for different end
conditions:
≤ 25 d + 1.5 mm
Recommended
> 25 d + 2 mm Degree of end restraint of compression
value of
member
Q.66. Tensile strength of mild steel is- effective length
Effectively held in position and
(a) 1400 to 1800 kg/cm2 (b) 1800 to 2500 kg/cm2 0.65 L
restrained against rotation at both ends
(c) 4200 to 5400 kg/cm2 (d) 5500 to 5700 kg/cm2 Effectively held in position at both ends
0.80 L
Ans : (c) Low-carbon steel, also known as mild steel is restrained against rotation at one end
now the most common form of steel because its price is Effectively held in positions at both
ends, but not restrained against 1.00 L
relatively low while it provides material properties that
rotation
are acceptable for many applications.
Effectively held in position and
• Low-carbon steel contains approximately 0.05 - 0.25% restrained against rotation at one end,
carbon making it malleable and ductile. Mild steel has a 1.20 L
and at the other end restrained against
relatively low tensile strength, but it is chseap and easy rotation but not held in position
to form; surface hardness can be increased through Effectively held in positions and
carburizing restrained against rotation at one end
and at the other end partially restrained 1.50 L
Properties of Mild steel:
against rotation but not held in
Properties Value position.
Effectively held in position at one end
Density 7850 kg/m3
but not restrained against rotation, and
2.00 L
ultimate tensile 400 - 550 MPa (4000 - 5500 at the other end restrained against
strength kg/cm2) rotation but not held in position
Effectively held in position and
Yield strength 250 MPa restrained at one end but not held in
2.00 L
Young's modulus of 200 GPa position nor restrained against rotation
Elasticity at the end

Melting point 1450°c Q.69. Rivets are generally specified by:

(a) Thickness of plates to be joined (b) Overall length
Q.67. For field rivets the maximum permissible stresses
(c) Shank diameter (d) Plate thickness
in rivets and bolts as given in codes are reduced by:
(a) 5% (b) 10% (c) 15% (d) 20% Ans : (c) Rivet : - Rivets are used as fasteners for making
permanent joints of two or more pieces of metals.
Ans : (b) As per IS 800:1984 Recommendations,
• Rivets are made of mild steel or wrought iron and
Maximum permissible stress in rivets and bolts
comprise head, tail, and shank.

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• The rivet is specified by the diameter of its shank. saturation factor (LSF) and it should not be less than
0.66 and it should not be greater than 1.02, when
calculated by the following formula:
𝐂𝐚𝐎
𝐋𝐒𝐅 =
𝟐.𝟖𝐒𝐢𝐎𝟐+𝟏.𝟐𝐀𝐥𝟐𝐎𝟑+𝟎.𝟔𝟓𝐅𝐞𝟐𝐎𝟑
Total loss on ignition: - This should not be greater than
4 percent.
Total Sulphur content: - The Sulphur content is
calculated as SO3 and it should not be greater than
2.75%.
Weight of insoluble residue: - This should not be greater
than 1.5%.
Q.70. In general the depth of Plate girder is kept as Weight of magnesia: -This should not exceed 5%.
________ of span. Q.73. As per IS 269-1975 initial setting time of ordinary
(a) 1/5 to 1/8 (b) 1/8 to 1/10 cement is -
(c) 1/10 to 1/12 (d) 1/12 to 1/16 (a) 15 minute (b) 30 minute
Ans: (c) The depth between the outer surfaces of the (c) 60 minute (d) 75 minute
flanges is termed as overall depth or depth of the plate Ans : (b) Setting time of the cement is tested in order to
girder. check its deterioration due to storage. Setting time of
• In general, depth of the plate girder is kept 1/10th to cement is further referred to as initial setting time and
1/12th of the span. The distance between C.G of final setting time
compression flange or C.G. of tension flange is known Initial setting time : - Initial setting time is defined as the
as effective depth of plate girder. time that is measured from the instant water is added
• The distance between vertical legs of flange angles at into the cement up to the time it starts losing its
the top and at the bottom is known as clear depth of plasticity.
plate girder.
Final setting time : - Final setting time is referred to as
• When the depth of plate girder is less than 750mm,
the time which is measured from the instant of water
then such girders are known as ‘Shallow Plate Girders’.
added into the cement up to the time it completely loses
● When the depth of plate girder is more than 750mm,
its plasticity and attains sufficient firmness so as to resist
then such girders are known as ‘Deep Plate Girders’. definite pressure.
❖ BMC Cement
Initial setting Final setting
Q.71. Plaster of Paris is obtained by calcining ______ time time
(a) bauxite (b) gypsum Ordinary Portland
30 minute 600 minute
(c) kankar (d) lime stone cement
Ans : (b) Plaster of Paris is made by heating the mineral Rapid hardening
30 minute 600 minute
gypsum. When gypsum is heated to about 150oC it losses cement
water and produces the powder, plaster of Paris. High alumina
Hemihydrate Calcium Sulphate (2CaSO4.H2O) is 3.5 hour 5.5 hour
cement
commonly known as plaster of paris. It is formed by
Low heat cement 1 hour 600 minute
heating gypsum.
Q.72. The insoluble residue in cement should be Q.74. The size of modular brick is
(a) Between 20% to 25% (b) Less than 20% (a) 10 × 10 × 9 cm (b) 19 × 9 × 9 cm
(c) Between 10% to 20% (d) Less than 1.5% (c) 22.5 × 10 × 8.5 cm (d) 22.5 × 8.0 × 9 cm

Ans : (d) Chemical composition : - The various tests are Ans : (b) Modular Bricks :
carried out to determine the chemical constituents of Standard size of brick = 19 cm × 9 cm × 9 cm
cement. Nominal size of a brick with mortar = 20 cm × 10 cm × 10
Following are the chemical requirements of ordinary cm
cement as per IS: 269- 1998: Non-Modular Bricks: - Conventional size of brick = 22.4
• The ratio of the percentage of alumina to that of iron cm × 11.4 cm × 7.6 cm
oxide: This ratio should not be less than 0.66. ∴ The actual size of the standard modular brick as per
• Ratio of percentage of lime to those of alumina, iron Indian Standards is 19 cm × 9 cm × 9 cm.
oxide, and silica: This ratio is known as the lime
UPSSSC JE 2015 22 Civil KI Goli
Q.75. The moisture content in a well seasoned timber 760 to 840 mm (800 ± 40 mm)
is (For 40 mm high bricks)
(a) 4% to 6% (b) 10% to 12% (b) For non - modular size
(c) 15% to 20% (d) 100% Length 4520 to 4680 mm (4600 ± 80 mm)
Ans: (b) Seasoning of Timber (IS 1141) : - Seasoning of Width 2240 to 2160 mm (2200 ± 40 mm)
timber is the process by which moisture content in the Height 1440 to 1360 mm (1400 ± 40 mm)
timber is reduced to required level. (For 70 mm high bricks)
• By reducing moisture content, the strength, elasticity 640 to 560 mm (600 ± 40 min)
and durability properties are developed.
(For 30 mm high bricks)
• A well-seasoned timber has 10-12% moisture content
in it. Q.78. Thickness of plastering is usually:
Q.76. Putty is made up of (a) 6 mm (b) 12 mm (c) 25 mm (d) 40 mm
(a) White lead and turpentine Ans : (b) The measurement for different works are as
(b) Powdered chalk and raw linseed oil follows:
(c) Red lead and linseed oil 1. Plastering usually 12mm thick is calculated in square
meter.
(d) Zinc oxide and boiled linseed oil
2. Pointing in-wall calculated in square meter.
Ans: (b) Putty : - Putty is a material with high plasticity, 3. Ornamental and large cornice is measured in running
similar in texture to clay or dough. meters or running feet.
It is made up of powdered chalk and raw linseed oil. 4. Pillars are taken separately in cubic meters for their
There are different types of putty mention bellow; net volume.
• Wood filler putty All work shall be measured subjected to following
• Epoxy putty tolerance unless otherwise stated:
• Painter's putty (a) Dimension shall be measured to the nearest 0.01 m.
• Wall putty (b) The area shall be worked out to the nearest 0.01 sq.
m.
Q.77. For checking the length of bricks as per Indian
(c) Cubic content shall be calculated to the nearest 0.01
standards how many bricks are to be taken:
cu.m.
(a) 10 (b) 15 (c) 20 (d) 25 (d) Weight shall be worked out to the nearest 0.001
Ans : (c) Tests for the acceptance of bricks for building tonnes.
construction are : (e) (e). The thickness of the RCC slab shall be measured
• Dimension and tolerance test to the nearest 0.005 m.
• Compressive strength test (f) The thickness of woodwork shall be measured to the
• Water absorption test nearest to 0.002 m.
• Efflorescence test (g) The thickness of steelwork shall be the measure to
the nearest 0.001 m.
Dimension and tolerance test : Take 20 bricks out of the
(h) The length of the steel reinforcement bar shall be
given sample. The dimensions of 20 bricks should be
measured to the nearest 0.005 m and its diameter to
within the following limits.
the nearest 0.0001 m.
Modular Bricks : (i) The area shall be measured to the nearest 0.01
Standard size of brick = 19 cm × 9 cm × 9 cm sq.m., but the area of steel shall be measured to the
Nominal size of a brick with mortar = 20 cm × 10 cm × 10 nearest 0.0001 sq.m.
cm (j). Volume shall be measured to the nearest 0.01 cu.m.
Non Modular Bricks : But the volume of wood shall be measured to the
The size of non-modular brick size is 230 × 110 × 70 mm nearest 0.001 cu.m.
Tolerance :
❖ ESTIMATION COSTING
The dimensions of bricks when tested in accordance
with shall be within the following limits per 20 bricks: Q.79. The plinth area rate for construction of first and
third floor of the building as per UPPWD has the
(a) For modular size
following relationship:
Length 3720 to 3880 mm (3800 ± 80 mm)
(a) Rate for first floor is higher than rate for third floor
Width 1760 to 1840 mm (1800 ± 40 mm)
(b) Rate for first floor is lesser than rate for third floor
Height 1760 to 1840 mm (1800 ± 40 mm)
(c) Both rates are same
(For 90 mm high bricks)
(d) None of above
UPSSSC JE 2015 23 Civil KI Goli
Ans : (c) Plinth Area: A covered built-up area measured various stages of construction, including excavation,
at the floor level of any buildings storey or (at the floor foundation, structural work, finishes, and other related
level of the building's basement) is called the plinth area. activities.
It is the measure of a buildings useable area. • In booking dimensions the order shall be in the
• It is also known as the built-up area that is the whole sequence of length, breadth, and height or depth or
area occupied by the building along with external and thickness.
internal walls. It is usually 10% to 20% higher than the Q.82. Unit of measurement of D.P.C. is
carpet area. (a) Cu. M (b) Sq. m (c) Meter (d) Kg
• It should be computed for the enclosed area by
Ans : (b) Damp-proof course (DPC) : - It is generally
measuring the dimensions of the external building at the
applied at basement levels/plinth level, which restricts
floor level. The courtyard and other areas will not
moisture movement through walls and floors.
include in the plinth area.
• A cement concrete layer in the proportion 1 : 2 : 4 is
Plinth Area Estimate : It is prepared based on the
generally provided at the plinth level.
buildings plinth area. It is a rough or approximate cost
• The depth of the cement concrete layer varies from 40
estimate in which the buildings plinth area is multiplied
mm to 150 mm.
by the plinth area to get the buildings cost.
• Estimation of D.P.C work is performed/measured in
• For storeyed buildings, we prepare plinth area
square meters of area.
estimates separately for each storey.
• As per UPPWD Plinth area rate for the first floor is same Q.83. Due to end of circulation of shape or design
as the rate for the third floor. property is known as ________.
(a) Uncirculation (b) Distress value
Q.80. For calculating the total quantity of paint on both
sides of flush doors, the outer dimensions of the door (c) Salvage value (d) Scrap value
is multiplied by: Ans : (a) Uncirculation : - It is the value of material due
(a) 1.25 (b) 2.25 to end of circulation of shape or design property is
(c) 1 (d) 2 known as Uncirculation.
Scrap value : - It is the value of dismantled materials for
Ans : (b) The coefficient of Painting for woodworks
a building when its life is over. The scrap value of a
(Doors and windows) used in different types of work
building may be about 10% of its total cost of
are as follows:
construction.
Coefficient of Salvage value : It is the value at the end of the utility
S. No

Description of work painting(for period without being dismantled. It may be zero,

Each side) positive or negative.
1. Flush doors 1.20 It also may be defined as Due to end of circulation of
Panelled or framed and braced shape or design property is known as salvage value.
2. 1.30 Distress value : - Some time due to fear of war or riot
doors, windows, etc.
the value of a property cannot fetch the full market
Fully glassed or gauged doors,
3. 0.80 value. Then this value of the property is called distress
windows, etc.
value.
Fully venetioned or louvered
4. 1.80 Q.84. In masonry works, while doing ‘Rate Analysis’,
doors, windows, etc.
water-surcharge is taken as
5. Collapsible gates 1.50 (a) 10% (b) 5%s (c) 3% (d) 1.5%
Hence, For both side, it should be 2 × 1.2 = 2.4 Ans : (d) The analysis of rates is worked out for the unit
The closest answer is option B. payment of the particular item of work under two heads:
Q.81. The order of booking dimensions in the standard Materials and Labour.
measurement book is • The cost of items of work = Material cost + Labour cost
(a) Length, breadth, height (b) Breadth, length, height Other costs included to the above cost of items of work
(c) Height, breadth, length (d) Height, length, breadth are:
• Tools and Plants (T&P) = 2.5 to 3% of the labour cost •
Ans : (a) A Measurement Book, also known as an MB, is
Transportation cost (if conveyance more than 8 km is
a document used in construction and engineering
considered.)
projects to record measurements, quantities, and other
relevant information related to on-site work. It serves as • Water charges = 1.5 to 2% of total cost
a detailed record of the measurements taken during • Contractor's profit = 10%

UPSSSC JE 2015 24 Civil KI Goli

Q.85. In Building construction, for one-time surcharge provides a finished surface over the masonry that is firm
for Electricity and water is added for the cost of and smooth hence it enhances the appearance of the
______% total cost of building. building. Plastering usually 12mm thick is calculated in a
(a) 12% (b) 10% (c) 8% (d) 5% square meter.
Ans : (c) While preparing an estimate, it is not possible Deductions related to plastering are as follows : -
to work out in detail in case of petty items. Items are • No deduction is made for the end of beams, posts,
other than civil engineering items are called lumpsum rafters, etc.
items. • For small opening up to 0.5 m2 area, no deduction in
The following are some of the lumpsum items in the plastering is made.
estimate. • For the opening of size 0.5 m2 to 3 m2 area, the
deduction is made on one face of the wall.
Work Percentage of ● For openings of size above 3 m2, the deduction is
estimate made on both faces of the wall, but the area of the
Water supply and sanitary 8 sill, jamb, and soffits of the opening is added.
arrangements
❖ SURVEY
Electrical installation 8 Q.89. Theodolite measures:
Contingencies and 3-5 (a) Horizontal angle only
unforeseen items (b) Vertical angle only
Work charge establishments 1.5-2 (c) Horizontal and vertical angles
Q.86. Weight of one bag of cement is: (d) Horizontal distance
(a) 70 kg (b) 50 kg (c) 60 kg (d) 65 kg Ans : (c) Theodolite : - A theodolite is a precision optical
Ans : (b) Cement is used as a binding material in mortar instrument for measuring angles between designated
and concrete. visible points in the horizontal and vertical planes.
It consists of a moveable telescope mounted so it can
• Mass of one bag = 50 kg
rotate around horizontal and vertical axis and provide
• Density of cement = 1400 kg/m3 angular readouts.
• Volume of one bag of cement = 50 / 1400 = 0.035714
m3 =35 litres
• No. of bags in 1 m3 = 1 / 0.035714 = 28 bags
• Specific gravity of cement = 3.15
Q.87. In analysis of rates, contractor profit is taken at
the rate of
(a) 1% (b) 5% (c) 10% (d) 20%
Ans : (c) The analysis of rates is worked out for the unit
payment of the particular item of work under two
heads : - Materials and Labour.
Types of theodolites:
• The cost of items of work = Material cost + Labour cost
1. Transit theodolite: - A theodolite is called a transit
Other costs included to the above cost of items of work theodolite when its telescope can be transited i.e
are: revolved through a complete revolution about its
• Tools and Plants (T&P) = 2.5 to 3% of the labour cost horizontal axis in the vertical plane.
Transportation cost (if conveyance more than 8 km is 2. Non-transit theodolite: - A theodolite is called a
considered.) transit theodolite when its telescope cannot be
• Water charges = 1.5 to 2% of total cost transited. They are inferior in utility and have now
• Contractor's profit = 10% become obsolete.
Q.88. No deduction is made while plaster Uses of theodolite: →
measurement in case of small openings up to • Measuring horizontal and vertical angles.
(a) 0.1 sq. m (b) 0.3 sq. m • Locating points on a line
(c) 0.5 sq. m (d) 0.7 sq. m • Prolonging survey lines
• Finding difference of level
Ans : (c) Plastering: Plaster is a thin layer of mortar • Setting out grades
applied over the masonry surface and it acts as a damp-
• Ranging curves
proof coat over the brick masonry work. Plastering also
• Tacheometric Survey

UPSSSC JE 2015 25 Civil KI Goli

Q.90. The series of closed contours having higher • Two vanes are hinged at their two ends i.e Object vane
contour values inwards represent and sight vane.
(a) Flat plane (b) Valley • A scale is attached to the fiducial edge so as to plot
(c) Hill (d) Lake distances to the needle.

Ans : (c) Hill :- Closed contour lines with higher values

inside indicate hill and lower values inside indicate pond
or depression.
Steep slope:- Contour lines are closely spaced and for
mild slope, contour lines are apart from each - other.
Overhanging cliff:- Two contours of different elevations
cannot cross each other except in the case of a vertical
cliff. Alidades are of two types:
Vertical cliff :- Contour lines of different elevations unite 1. Plain Alidade: It is very simple and is used for ordinary
to form one line. work.
Uniform slope:- Parallel contour lines
2. Telescopic Alidade: It is used to take inclined sights.
Q.92. The longest chain line passing through the center
of the survey area is known as
(a) Base line (b) Tie line
(c) Check line (d) None of above
Ans : (a) In chain surveying, the following lines are used
Main line: The line joining main survey stations are
known as Main survey lines.
Base line: The longest line that will divide the total area
into two parts is known as Base line. This line is the main
reference line for fixing the positions of various stations
and also to fix the direction of other lines.
This should be carefully measured and laid as the
accuracy of the entire triangulation critically depends on
this measurement.
Check line: The lines which are run in the field to check
the accuracy of the work are known as Check line. They
are also known as Proof lines.
Q.93. The survey, in which the curvature of the earth is
considered, is called
(a) Geodetic surveying (b) Plane surveying
(c) Geographical surveying (d) Land surveying
Ans: (a) Plane Surveying: It is a type of surveying in
which the mean surface of the earth is considered as a
plane and the spheroidal shape is neglected. All triangles
formed by the survey lines are considered as a plane
triangle. The level line is considered as straight.
Geodetic Surveying: It is a type of surveying in which the
Q.91. Alidade is used in mean surface of the earth is taken into account. All lines
(a) Chain surveying (b) Levelling lying on the earth's surface are curved and all triangles
(c) Plane table surveying (d) Compass surveying formed by the survey lines are spherical triangle.
Ans : (c) An Alidade is an instrument used for sighting Q.94. If the fore bearing of a line AB is 35° and that of
the point in plane table surveying. line BC is 15°, then the included angle between the line
• One edge of the ruler is bevelled or Fiducial and the is
other is graduated (used for drawing line of sight). (a) 20° (b) 50° (c) 160° (d) 230°
• It is a wooden or brass ruler of 50-60 cm in length. Ans : (c) Given
Fore bearing of line AB = 35°

UPSSSC JE 2015 26 Civil KI Goli

Fore bearing of line BC = 15°
Included angle = F.B. of next line
B.B of previous line
B.B. of previous line = F.B ± 180°
B.B. of previous line = 35° + 180°
B.B. of previous line = 215°
Included angle = 15° - 215°
Included angle = - 200°
We got the negative value of included angle so we add
360°
Included angle = 360° - 200
Included angle = 160°
Least count : - The smallest value that can be measured
Q.95. As applied to staff readings, the corrections for
by the measuring instrument is called its Least Count.
curvature and refraction are respectively
• Least count of Prismatic Compass = 30 minutes = 30’
(a) + and - (b) - and + (c) + and + (d) - and -
• Least Count of surveyor’s Compass = 15 minutes = 15’
Ans : (b) In staff readings, the corrections for curvature
and refraction are respectively - and +. This means that Q.98. Principle of surveying followed to prevent
the curvature correction is always positive and added to accumulation of errors is:
the staff reading, while the refraction correction is (a) To work from whole to part
always negative and subtracted from the staff reading. (b) To work from part to whole
• Correction due to curvature = -0.0785D2 (c) Both (1) & (2)
• Correction due to refraction = 0.0112D2 (d) None of above
• Combined correction = -0.0673D2 Ans : (a) Principle of surveying:
Q.96. The suitable contour interval for a map with scale (1) Whole to part : In working from whole to part, the
1: 10,000 is error will localise and prevent the accumulation of error
(a) 2 m (b) 5 m (c) 10 m (d) 20 m while working from part to whole (figure A), the error
𝟐𝟓 will accumulate. Hence more error in working from part
Ans : (a*) Contour Interval = ( meters) to whole.
𝐍𝐨.𝐨𝐟 𝐜𝐦 𝐩𝐞𝐫 𝐤𝐦
𝟓𝟎
Also Contour Interval = (feet)
𝐍𝐨.𝐨𝐟 𝐢𝐧𝐜𝐡𝐞𝐬 𝐩𝐞𝐫 𝐦𝐢𝐥𝐞
For the scale of 1: 10000.
1 cm represents 10000 cm
∴ 1 cm represents 10000 × 10-5 km = 0.1 km
∴ 1 cm represents 0.1 km.
So 1 km represent 10 cm (2) Location w.r.t at least 2 well-defined control point:
𝟐𝟓 Location of a point should be respected to at least 2 well-
∴ Contour interval = = 2.5 meters
𝟏𝟎
defined control points.s
*Hence, the most appropriate answer is option (a).
To establish a station/point, minimum one angular and
Q.97. The least count of prismatic compass is: one linear or two linear or two angular measurements
(a) 1° (b) 30' (c) 15' (d) 20" are necessary. The following different conditions can be
Ans : (b) Prismatic compass : - The prismatic compass observed in the figure below.
consists of a circular box about 85 to 110 mm in
diameter.
• At the center of the metal box, a needle and pivot are
provided.
• The graduations are in degrees to 30 minutes and from
0o to 360o in the clock wise direction.
• A glass cover is fitted over the box to protect the
needle from dust. The compass is fitted to a tripod
stand.

UPSSSC JE 2015 27 Civil KI Goli

Q.99. As per Indian Standard, the length of one link in (3) Traversing method: The traverse is directly plotted on
the 30 m chain should be the paper by drawing the traverse lines with the alidade.
(a) 20 cm (b) 30 cm (c) 40 cm (d) 10 cm No angle measuring instrument is used.
Ans:(a) The 30 m chain is divided into 150 links. So, each (4) Resection method: Resection is a general term used
link is of 0.2 m. or the process of determining the location of the
instrument station. The principle of resection is opposite
The tallies are provided after every 25 links (5 metres). A
to that of the intersection method.
round brass ring is fixed after every metre as can be
observed in the figure above. Q.102. Line joining points of equal elevations on earth
surface is called
(a) Contour surface (b) Contour gradient
(c) Contour line (d) All of above
Ans : (c) An imaginary line joining the points of equal
elevation on the surface of the Earth represents the
contour line.
A line laying on the ground which maintains a constant
Q.100. If reduced bearing of a line is N87°W , its whole inclination to the horizontal is known as contour
circle bearing will be: gradient. It is found out by instrument clinometer.
(a) 87° (b) 93° (c) 3° (d) 273° Characteristics of the contour are : -
• The variation of vertical distance between any two
Ans : (d) Given
contour lines is assumed to be uniform.
Reduced bearing of line = N 87° W • The horizontal distance between any two contour lines
For reduced bearing in NW quadrant indicates the amount of slope and varies inversely on
RB s360°WCB the amount of slope.
WCB = 360° - 87° • The steepest slope of terrain at any point on a contour
WCB = 273° is represented along the normal of the contour at that
point. They are perpendicular to ridge and valley lines
Conversion table for WCB and QB : -
where they cross such lines.
S. Whole Circle Reduced Bearing Quadrant • Contours do not pass through permanent structures
No. Bearing (WCB) (RB) such as buildings
1 0° - 90° WCB NE • Contours of different elevations cannot cross each
other (caves and overhanging cliffs are the exceptions).
2 90° - 180° 180° - WCB SE
● Contours of different elevations cannot unite to form
3 180° - 270° WCB - 180° SW one contour (vertical cliff is an exception).

4 270° - 360° 360° - WCB NW ❖ SOIL

Q.101. In a plane table survey, the plotting of Q.103. For a given degree of compaction, a graph
inaccessible points can be conveniently done by between the dry unit weight of soil and water content
in soil is called
(a) Method of resection (b) Method of radiation
(a) Compression graph (b) Moisture-density graph
(c) Method of traversing (d) Method of intersection
(c) Void ratio graph (d) Porosity graph
Ans : (d) Plane table surveying : - It is a graphical method
of surveying in which the fieldwork and plotting are Ans : (b) Compaction Curve of Soil : - The compaction
carried out simultaneously. curve is the curve drawn between the water content (X-
axis) and the Respective dry density (Y-axis). So, It is also
Principle : - The rays drawn from different points should
known as Moisture-density graph.
pass through a single point i.e position of station point.
• The observation will be initially an Increased dry
Methods:
density with an increase in the water content. Once it
(1) Radiation method: It is adopted when the points to reaches a particular point a decrease in dry density is
be plotted are intervisible and accessible from a single observed.
instrument station. Orientation is not required because
• The maximum peak point of the soil compaction curve
only one instrument station is required.
obtained is called the Maximum Dry density value. The
(2) Intersection method: It is adopted when the points water content corresponds to this point is called the
to be plotted are intervisible but inaccessible. Optimum water content (O.W.C) or Optimum Moisture
Orientation is required Content (O.M.C)

UPSSSC JE 2015 28 Civil KI Goli

 Q.106. A soil has 30% air voids. It has a porosity of 0.4,
the air content of that soil shall be:
(a) 0.75 (b) 0.12 (c) 1.33 (d) 0.77
Ans : (a) Given
Percentage air voids(na) = 30% = 0.30,
Porosity(n) = 0.4
na = ac x n
0.3 = ac x 0.4
ac = 0.75.
Hence, the Air content of the soil is 0.75
Q.107. Due to rise in water table, the effective stress in
• The Graph shown in fig. is the compaction curve. soil
initially for a water content lesser than ( O.M.C ) the (a) Increases (b) Decreases
soil is rather stiffer in nature that will have lots of void (c) Does not change (d) May increase or decrease
space and Porosity. This is the reason for lower dry
Ans: (b) Effective stress is the ratio of force at the contact
density attainment.
of particles of soil to the total area. It cannot be obtained
• The graph represents a zero-air Void or 100%
practically but we can calculate the effective stress by
Saturation line. This is based on the theoretical
measuring total stress and pore water pressure as : -
maximum dry density where it occurs when there is
100% saturation. As the condition of zero voids in the Total stress (σ) = effective stress (σ’) + pore water
soil is not real and a hypothetical assumption, the soil pressure (u)
can never become 100 % Saturated. Effective stress (σ') = Total stress (σ) - pore water
pressure (u)
Q.104. The effective size of soil is
• If we increase the ground water table then value of
(a) D10 (b) D20 (c) D30 (d) D60
pore water pressure increases and effective stress
Ans : (a) Effective size(D10):- The diameter in the decreases.
particle-size distribution curve corresponding to 10% • If we lowering the groundwater table below ground
finer is defined as the effective size or D10. then the value of pore water pressure decreases and
• To obtain the grading characteristics, three points are effective stress increases.
located on the grading curve. Q.108. If void ratio is 0.67, water content = 0.188 and
D60 = size at 60% finer by weight sp. Gravity = 2.68, the degree of saturation of soil is:
D30 = size at 30% finer by weight (a) 25% (b) 40% (c) 75% (d) 60%
D10 = size at 10% finer by weight Ans : (c) Given, Void ratio (e) = 0.67,
The other parameter where generally D10 is involved in Water content (w) = 0.188,
the calculations are
Specific gravity (G) = 2.68
𝑫𝟔𝟎
Uniformity coefficient, = 𝑪𝒖 = We know, S e = w G
𝑫𝟏𝟎
𝑫𝟐𝟑𝟎 S × 0.67 = 0.188 × 2.68
Curvature coefficient, = 𝑪𝒖 =
𝑫𝟏𝟎 ×𝑫𝟔𝟎 S = 0.752 = 75.2%
Q.105. Which of the following soils has the finest Q.109. When the plastic limit of a soil is greater than
grains? the liquid limit, then the plasticity index is reported as:
(a) Silts (b) Clays (a) 1 (b) Negative
(c) Sands (d) Fine sands (c) Zero (d) Non-Plastic (NP)
Ans : (b) The following table gives the particle size for Ans : (c) Plasticity index (PI) is the range of water content
different types of soil : - over which the soil remains in the plastic state.
Plasticity Index = Liquid Limit – Plastic Limit
Type of soil Size of particle
• This parameter cannot be negative if plastic limit, in
Sand 0.075 mm to 4.75 mm some exceptions, is larger than the liquid limit, it is
considered to be zero and soil is considered non-
Silt 0.002 mm to 0.075 mm plastic.
Clay <0.002 mm • The plasticity of a soil is its ability to undergo
deformation without cracking. It is an important index

UPSSSC JE 2015 29 Civil KI Goli

 property of fine-grained soil, especially for clayey soils. Q.113. Plasticity index of soil is equal to
The adsorbed water in clayey soils leads to the (a) Liquid limit – Plastic limit
plasticity of the soil. (b) Liquid limit – Elastic limit
• Generally, sands do not possess any plasticity and their
(c) Elastic limit – Plastic limit
plasticity index is assumed to be zero.
(d) Elastic limit – Consistency limit
Q.110. The minimum depth of foundation in clayey soil
is Ans : (a) Plasticity index (PI) is the range of water
content over which the soil remains in the plastic state.
(a) 0.5 m (b) 0.7 m
Mathematically defined as,
(c) 0.9 m (d) 1.20 m
Plasticity Index = Liquid Limit – Plastic Limit.
Ans : (c) Foundation depth in clay soil : Foundation The plasticity of a soil is its ability to undergo
depth in clay soil depend on the plasticity nature of deformation without cracking. It is an important index
volumetric expansion and contracting, depth of the property of fine-grained soil, especially for clayey soils.
foundation of footing in clay soil of higher plasticity, it
∴ Plasticity index is maximum for clayey soil.
could be 1 m (1000 mm) deep, of medium plasticity, it
could be 0.9 m (900 mm) deep and for low plasticity, it Based on the plasticity index, the soil may be classified
could be 0.75 m (750 mm) deep beneath the soil where as followed:
hard strata of soil have been found. Plasticity Index Soil Description
Minimum depth of foundation in clay soil : According to 0 Non-Plastic
volumetric changes expanding and shrinkage nature and
plasticity condition, minimum depth of the foundation 1-5 Slight Plastic
of footing in clay soil is kept around 0.9 m(900 mm) deep 5-10 Low Plastic
beneath the soil where hard strata of soil have been
10-20 Medium Plastic
found.
Maximum depth of foundation in clay soil : According 20-40 Highly Plastic
to volumetric changes expanding and shrinkage nature >40 Very Highly Plastic
and plasticity condition, maximum depth of the Q.114. Uniformity coefficient of filter sand is
foundation of footing in clay soil is kept around 1 m(1000 represented by
mm) deep beneath the soil where hard strata of soil have
(a) D50/D5 (b) D50/D10
been found.
(c) D60/D5 (d) D60/D10
Q.111. If angle of internal friction of soil is 30°, the
coefficient of active earth pressure will be: Ans: (d) Coefficient of Uniformity / Uniformity
Coefficient: It is defined as the ratio of D60 and D10 sieve
(a) 1/2 (b) 1/3 (c) 1/4 (d) 2/3
sizes in sieve analysis of granular material.
Ans : (b) Given Higher is the value of Cu larger is the range of the particle
Internal friction of soil = 30° size.
𝟏−𝒔𝒊𝒏 𝝓 𝑫𝟔𝟎
Coefficient of active earth pressure = Cu =
𝟏+𝒔𝒊𝒏 𝝓
𝑫𝟏𝟎
𝟏−𝒔𝒊𝒏 𝟑𝟎° 𝟏
=
𝟏+𝒔𝒊𝒏 𝟑𝟎°
=
𝟑 • For uniformly graded soil, Cu = 1
• For well graded sand, Cu > 6
Q.112. The angle of internal friction ϕ for cohesive soil
• For well graded gravel Cu > 4
is equal to
Coefficient of Curvature / Curvature Coefficient:
(a) 0° (b) 30° (c) 45° (d) 15°
𝑫𝟐𝟑𝟎
Ans:(a) The shear strength of soil specimen as per Mohr Cc =
𝑫𝟔𝟎 ×𝑫𝟏𝟎
and Columbus Criterion is given by:
• For well graded soil, 1 < Cc < 3
τ= C + σ tan ϕ • For gap graded soil, 1 < Cc or Cc > 3
Where where
C = cohesion and ϕ is the angle of internal friction and D60 = size at 60% finer by weight
these both are called shear strength parameters. These
D30 = size at 30% finer by weight
sear strength parameters can only be determined
experimentally using the triaxial test. D10 = size at 10% finer by weight = Effective size
σ is total stress at the given point. Q.115. The maximum size of clay particle is:
• For pure sand, C = 0 and ϕ ≠ 0 (a) 0.1 mm (b) 0.03 mm
• For pure clay, C ≠ 0 and ϕ = 0. (c) 0.002 mm (d) 0.0002 mm

UPSSSC JE 2015 30 Civil KI Goli

Ans : (c) The Size of clay particles is less than 0.002 mm. Q.117. The sloped roof which is sloped in four sides is
In the Indian Standard Soil Classification System (ISSCS) called as:
or BIS, soils are classified according to their grain size as (a) Shed roof (b) Gable roof
boulder, cobble, gravel, sand, silt, or clay, as shown (c) Hip roof (d) Mansard roof
below in the tabulated form. Ans : (c) Hipped roof : - It has slopes on all four sides.
Soil classification based on the particle size range The sides are all equal length and come together at the
Type of Sub top to form a ridge.
Soil Group Size Range
Soil Group Shed roof : - It is also referred to as a skillion. It is a single,
sloping roof, usually attached to a taller wall.
Very Coarse Boulder > 300 mm
Gable end roof : - It is also known as pitched or peaked
Soils Cobble 80-300 mm roof and is recognized by triangular shape.
Coarse 20-80 mm Hipped roof : - It has slopes on all four sides. The sides
Gravel 4.75-20 are all equal length and come together at the top to form
Fine a ridge.
mm
Coarse 2-4.75 mm Gambrel roof : - It is also known as a barn roof. It is
Coarse Soils similar to mansard, the difference between the two is
0.425-2 that the Gambrel only has two sides, while the mansard
Medium
Sand mm has four.
0.075- The lower side of the Gambrel roof has an almost
Fine
0.425 mm vertical, steep slope, while the upper slope is much
0.002- lower.
Silt Different Types of roofs is depicted in the diagram
Fine Soils 0.075 mm
below:
Clay < 0.002 mm

❖ BCME
Q.116. The member of roof truss which supports the
purlins is called as:
(a) Sag rod (b) Main strut
(c) Principal rafter (d) Principal tie
Ans : (c) Purlins: These are the members which are
spanning on the roof frames to support the roof
coverings and runs parallel to ridge to connect different
trusses situated in the longitudinal direction.
Rafters: These are a series of sloped structural members
Q.118. In design of cinema hall width of Tread is taken
that extend from the ridge to the downslope perimeter
as
or to the bottom chords and are designed to support the
roof deck and its associated loads. (a) 25 to 30 cm (b) 30 to 50 cm
(c) 50 to 75 cm (d) 90 to 110 cm
The principal rafter is the top chord member of the truss
and is subjected to compressive forces from loads Ans : (a) As per the Indian National building code
transferred by purlins at the nodes. The rafters act as Staircase Requirements :
simply supported beams between the purlins The minimum clear width, minimum tread width, and
maximum riser of staircases for buildings-
Minimum width - The minimum width of the staircase
shall be as follows:
(a) Residential buildups (dwellings) 1.0 m
NOTE— For row housing with 2 storeys, the minimum
width shall be 0.75 m.
(b) Residential hotel buildings 1.5 m
(c) Assembly buildings (like auditorium, theatres, and
cinemas) 2.0 m
(d) Educational building 1.5 m
(e) Institutional buildings 2.0 m
(f) All other buildings 1.5 m
UPSSSC JE 2015 31 Civil KI Goli
 (a) Dead shores (b) Flying shores
UPSSSC MANDI PARISHAD (c) Live shores (d) Raking shores
DRAFTSMAN 2015 Ans : (c) Shoring is the temporary support which is
provided to improve the lateral strength of walls during
❖ BCME repairs.
Q.01. Which of the following are not the types of stone
masonry?
(a) Uncoursed rubble masonry
(b) Random rubble masonry
(c) Freedom rubble masonry
(d) Dry rubble masonry
Ans : (c) STONE MASONRY:- Masonry may be defined as
the construction of building units bounded together
with mortar. When stones are used as the building units
or building blocks, we have stone masonry. In stone
masonry, the stones are placed in position such that the
natural bedding plane normal to the direction of
pressure they carry.
CLASSIFICATION OF STONE MASONRY :
Stone masonry can be classified as follows:
1. Rubble Masonry :- In this category, the stones used
are either undressed or roughly dressed having wider
joints. This can be further subdivided as uncoursed,
coursed, random, dry, polygonal and flint.
It is further sub-divided in the following categories:
(a) Uncoursed rubble masonry
(b) Random Rubble masonry
(c) Coursed rubble masonry
Types of shoring:
(d) Dry rubble masonry
1. Raking Shores: In this method, inclined members,
2. Ashlar masonry :- Ashlar masonry consists of blocks
called rakers are used to give lateral support to the wall.
of accurately dressed stone with extremely fine bed and
joints. The blocks may be either square and rectangular 2. Flying Shores: Flying shores are a support system
shape. The height of stone varies from 25 to 30 cm. which provides horizontal support to two parallel party
walls when removal or collapse of the intermediate
Ashlar masonry may be subdivided into the following
building takes place.
categories:
3. Dead Shores: Dead shore is temporary structure
(a) Ashlar fine
erected to increase the size of opening in an existing wall
(b) Ashlar rough
Q.03. At which level Damp Proof Course (D.P.C) is
(c) Ashlar rock, rustic or quarry faced
provided for the construction of the building?
(d) Ashlar chamfered
(a) At ground level (b) At lintel level
(e) Ashlar facing
(c) At slab level (d) At plinth level
Ans :(d) Damp Proof Course or DPC:
• It is the protective layer applied to prevent the rising
of moisture to the walls from the ground due to
capillary action.
• It is usually 2.5 cm thick with a ratio of rich cement
concrete 1:1.5:3 or 2 cm thick with cement mortar 1:2
mixed with standard waterproofing material.
• It is provided from the plinth level to the entire width
Rubble masonry Ashlar masonry of the plinth wall, and the quantities are calculated in
Q.02. Which of the following temporary structure square meters (length x breadth).
(shoring) are not used for the construction of building?

UPSSSC DRAFTSMAN 2015 34 Civil Ki Goli

 the shallow foundation cannot support the load of the
structure.
Types of Deep Foundation
1. Pile Foundation
2. Pier Foundation
3. Well (Caissons) Foundation
Q.06. Which of the following joint is not used for the
framing of wood?
(a) Bearing joint (b) Framing joint
(c) Expansion joint (d) Longitudinal joint
Ans : (c) Expansion joints are typically used in
Q.04. Which of the following main plan is required for construction to allow for the expansion and contraction
designing and planning of new Districts /city? of materials due to temperature changes or other
(a) Electrical plan (b) Site plan factors. They are not specific to wood framing but are
(c) Master plan (d) Building plan used in various types of construction to prevent cracking
or structural damage from thermal movement.
Ans: (c) Master plan:- A plan made to design a city is
called a master plan. A master plan includes housing Q.07. Which of the following foundation is a deep
areas, schools, market, industrial areas, office areas, foundation?
parks, and recreational areas. Based on the master plan, (a) Combined footing (b) Pile foundation
city planners decide the types of road, amount of water (c) Eccentrically loaded footing (d) Mat foundation
or electricity required, how to dispose of waste, how to
Ans : (b) Pile Foundation :- Pile is a slender member with
clean the sewage and all other details for a city.
small area of cross-section relative to its length. They
Site plans:- Site plans are used to locate the position of can transfer load either by friction or by bearing.
buildings in relation to setting out points, means of
Pile foundation are used when :
access, and the general layout of the site. These plans
may also contain information on services, drainage • The load is to be transferred to stronger or less
networks, etc. compressible stratum, preferably rock. The granular
soils need to be compacted.
Q.05. Which of the following is not a type of shallow • The horizontal and the inclined forces need to be
foundation? carried from the bridge abutments and the retaining
(a) Spread footing (b) Grilling foundation walls
(c) Raft foundation (d) Caisson
Ans:(d) Foundation :- It is defined as that part of the
structure that connects and transmits the load from the
structure to the ground soil.
Types of Foundation :
A. Shallow Foundation:- Shallow Foundations are those
foundations in which the depth is placed less than the
width of the foundation (D < B). Shallow foundations are
generally termed as spread footing as they transmit the
Q.08. Which of following is not a type of scaffolding ?
load of the super structure laterally into the ground.
(a) Steel scaffolding (b) Wooden scaffolding
Types of Shallow Foundation:
(c) Needle scaffolding (d) Paper scaffolding
1. Wall Footing
Ans : (d) Types of Scaffolding:
2. Isolated column/Column Footing
1. Single / putlog scaffolding: In case of brick masonry
3. Combined Footing
all the standards, ledgers, putlogs are arranged parallel
4. Cantilever (Strap) Footing to the wall at distance of 1.20 mts.
5. Mat (Raft) Foundation 2. Double scaffolding: In case of stone masonry works,
B. Deep Foundation:- Deep Foundation are those it is difficult to provide holes in the walls to support
foundations in which the depth of the foundation is putlogs.
greater than its width (D>B). The D/B ratio is usually 4-5 3. Cantilever/needle Scaffolding: where ground is weak
for deep foundation. Deep foundations are used when to support the standards.

UPSSSC DRAFTSMAN 2015 35 Civil Ki Goli

Ans : (d) Scaffolding:- It is a temporary structure to An unbroken series of steps between
Flight
support a platform over which the workman can stand landings.
or sit to execute the construction work in case of height The vertical distance between two
Rise
of the wall exceed 1.5 m. successive tread faces.
• Ordinary Scaffolding consists of standards, ledgers, The horizontal distance between two
Going
putlogs, guardrails, free boards, and braces. successive riser faces.
Shoring:- The process of construction of temporary The projecting part of the tread
Nosing
structures to support unsafe structures temporarily. beyond the face of the riser.
Shoring can be used when walls bulge out, or when walls A molding provided under the nosing
Scotia
crack due to unequal settlement of the foundation. to provide strength to the nosing.
Shuttering:- Shuttering, also called formwork, is a Soffit The underside of a stair.
temporary support used to mold concrete into the Pitch or The angle which the line of the nosing
desired shape, in which the concrete gains initial Slope of the stair makes with the horizontal.
strength, hardens, and matures. As we produce PCC, These are the sloping members which
Strings
RCC construction elements in the form of shuttering for support the steps in a stair.
buildings, bridges, tunnels, hydroelectric dams, sanitary A vertical member which is placed at
pipelines, and more, shuttering comes in different Newel post the ends of flight to connects the ends
shapes and sizes. of strings and handrail.
Q.15. Which of the following technical term is not used A vertical member of wood or metal,
Baluster
for the construction of staircase in the building? supporting the handrail.
The clear vertical distance between
(a) Tread (b) Winder Headroom
the tread and overload structure.
(c) Nosing (d) Batten
A step that is narrower at one side
Ans : (d) Staircase: Stairs are a set of steps that give Winder than the other, used to change the
access from floor to floor. direction of a staircase.
• The room or enclosure of the building, in which stair is Q.16. Which of the following type of flooring is most
located is known as the staircase. suitable for the dancing room?
• A staircase provides access & communication between (a) Cement concrete flooring (b) Wooden flooring
floors in multi-story buildings.
(c) Tiles flooring (d) Mosaic flooring
Ans : (b) Wood Flooring :- One of the most common
floorings in residential houses is flooring made with
wood or timber. It is preferred in cold climates and at
places where wood is cheap. Apart from residential
houses, other places with suitable wooden floors
include dance floors, auditoriums, etc. While placing
these floors, a damp-proof course is necessitated under
the floor.
Cement Concrete Flooring:- Concrete is the most
common type of flooring because it can be used in any
type of building and is cheaper and more durable than
other options. As a base course, you can use a concrete
The following are the components of the staircase mix of 1:3:6 to 1:5:10 or lime concrete with 40% of 1:2
Component Definition lime sand mortar and 60% of coarse aggregate. As a
A portion of the stair which permits topping, a layer of 1:2:4 cement concrete mix 40 mm
Step ascent or descent. A stair is composed thick is placed on top.
of a set of steps. Mosaic flooring:- It consists of Concrete layer. A layer of
The upper horizontal portion of a step cementing material of about 3 mm thick (consisting
Tread upon which the foot is placed while 2:1:1 ratio of lime and marble and pozzolana material),
ascending or descending. marble pieces or tiles.
The vertical portion of a step providing Q.17. Which of the following is the type of staircase?
Riser
support to the tread. (a) Dog-legged stair (b) Circular stair
A level platform at the top or bottom (c) Bifurcated stair (d) All of above
Landing
of a flight between floors.
Ans : (d) Classification of staircase :-

UPSSSC DRAFTSMAN 2015 38 Civil Ki Goli

1. Straight Stairs
2. Turning Stairs
(a) Quarter-Turn Stairs
(b) Half-Turn Stairs
(i) Dog-legged Stairs
(ii) Open Newel Stair
(c) Three-Quarter Turn Stairs
(d) Bifurcated Staircase
3. Continuous Staircase
(a) Circular Staircase
(b) Spiral Staircase
(c) Helical Staircase
DOG-LEGGED STAIRCASE:- It consists of two straight Q.18. Maximum number of steps constructed in one
flights with 180° turn between the two. flight for the construction of staircase followed as per
• They are very commonly used to give access from floor the standards.
to floor. (a) 8 Nos (b) 10 Nos
(c) 14 Nos (d) 20 Nos
Ans. (c) The following are some of the general
guidelines to be considered while planning a staircase:
1. The respective dimensions of tread and riser for all the
parallel steps should be the same on the consecutive
floor of a building.
2. The minimum width of stairs should be 850 mm,
though it is desirable to have a width between 1.1 to 1.6
m. In a public building, cinema halls, etc. large widths of
Bifurcated staircase:- Apart from dog legged and open the stair should be provided.
newel type turns, stairs may turn in various forms. 3. It is not desirable to provide a flight with more than
• They depend upon the available space for stairs. 12 or at the most 15 steps and less than 3 steps. Suitable
Quarter turned, half turned with few steps in between landings should be provided to give comfort and safety
and bifurcated stairs are some of such turned stairs. to the users of the stair.
4. The minimum vertical headroom above any step
should be 2 m.
5. The size of the tread generally ranges from 270 mm
for residential buildings and 300 mm for public buildings.
6. The size of the riser generally ranges from 150 mm for
public buildings to 190 mm for residential buildings.
7.The minimum width of landing shall be equal to the
staircase width.
Q.19 . The measurement of tread & riser for the
staircase construction used normally in public building.
(a) 30 cm and 15 cm (b) 25 cm and 12 cm
(c) 30 cm and 12 cm (d) 25 cm and 16 cm
Ans : (a) (i) For residential building, common size of
steps is (16 x 25) cm and Public buildings: (27 cm x 15
cm) to (30 x 14 cm)
(ii) Following thumb rules are commonly used for
obtaining a satisfactory proportion of the tread and riser
Circular Staircase:- It is known as spiral stair. of a step:
• When viewed from top it appears to follow a circle • (Rise in cm) + (Tread in cm) = 40 to 45
with a single centre of curvature. • (Rise in cm) x (Tread in cm) = 400 to 450
• The spiral stairs are provided where space available is • (2 x Rise in cm) + (Tread in cm) = 60
limited and traffic is low. • Take rise 140 mm and going = 300 mm as standard.

UPSSSC DRAFTSMAN 2015 39 Civil Ki Goli

Q.20. Which of the following is not the type of pitched • Elevation of Stairs
roof ? • Section of stairs
(a) Lean to roof (b) Flat roof Q.22. Which of the following is not the type of roof ?
(c) Coupled roof (d) Collar roof (a) Flat roof (b) Pitched roof
Ans : (b) Roof is a covering supported on the top of walls (c) Sloppy roof (d) Coffer roof
and pillars. The main function of a roof is to enclose the
Ans: (d) Flat roof: (i) Flat roofs are mainly used in plains,
space or building and to protect the same from the
where there is low to moderate rainfall and temperature
damaging effects of weather elements such as rain,
is high. This type of roof is generally horizontal but
wind, heat, snow etc. A good roof also increases the life
having a slope of not more than 10° to drain rainwater.
of the building.
It is also called a terraced roof.
Types of Roofs:
(ii) The roof must slope in one direction or the other to
1. Pitched or Sloping Roofs cause rainwater to flow off rapidly and easily.
(a) Single Roofs (iii) The construction of the flat roof is the same as that
• Lean to roof/ varandah roof • Couple Roof of floors except that the top surface is made slightly
• Couple Close Roof • Collar-Beam Roof sloping in the case of flat roofs.
(b) Double or Purlin Roofs Pitched and sloping roof: A sloping roof is known as a
(c) Triple-membered or Framed or Trussed Roofs pitched roof. These are suitable in those areas where
• King Post Roof Truss rainfall/ snowfall is very heavy
• Queen Post Roof Truss Q.23. Which of the following answer is correct in
• Combination of King-Post and Queen Post Trusses respect of cantilever construction in building?
• Mansard Roof Trusses (a) One end is resting on the column or wall and other
end is free
• Truncated Roof Truss
(b) One end is resting on wall and other end is resting
• Belfast Roof Truss or Lattice Roof Truss on column
• Composite Roof Trusses (c) Both ends are resting on the wall
• Steel Sloping Roof Trusses (d) Both ends are resting on the column
2. Flat Roofs or Terraced Roofs Ans : (a) Cantilever Construction:- A cantilever is a rigid
• Mud-Terrace Roofing structural element that extends horizontally and is
• Brick Jelly or Madras Terrace Roofing supported at only one end. The unsupported end is free
• Bengal Terrace Roofing to move and often carries loads.
4. Curved Roofs: Q.24. Which of the following term (technical) is not
related to the pitched roof construction?
(a) Eaves (b) Valley
(c) Hip (d) Nosing
Ans : (d) Pitched Roof Components :
The pitched roof components are explained below:-
Span-:-The span is the horizontal distance between the
internal faces of walls or supports.
Rise:-Rise is the vertical distance between the wall plate
and the top of the ridge.
Pitch:- Pitch is the angle at which the sides of a roof are
inclined to the horizontal plane.
Ridge:- The ridge is the apex line of a sloping roof.
Hip:-Hip is the ridge formed by the intersection of two
Q.21. Which of the following technical term is not used sloped surfaces with an exterior angle greater than 180°.
for the construction of Staircases in any building? Eaves:- The eaves are the areas where a roof extends a
(a) Plan of stairs (b) Stairs well short distance past a building's wall. This is an important
(c) Section of stairs (d) Elevation of Stairs part of the roofing because it must be properly finished
to prevent animals from nesting and potential water
Ans : (b) following technical terms are used for the
ingress.
construction of Staircases in any building : -
• Plan of stairs

UPSSSC DRAFTSMAN 2015 40 Civil Ki Goli

(a) Critical Path Method (b) Fabricated Door & Windows:- Fabricated doors and
(b) Cement Plastic Method windows are pre-made units that can be easily installed
(c) Critical Planning Method into the structure. These are common in prefabricated
construction because they ensure uniformity and quality
(d) Cement Plaster Method
control, and they simplify the installation process.
Ans : (a) CPM (Critical Path Method): (c) Fabricated Frame:- Fabricated frames, which include
1. CPM is Activity Oriented. structural components like steel or wooden beams and
2. CPM involves the critical path which is the largest path columns, are crucial in prefabricated construction.
in the network from starting to ending event and These frames form the skeleton of the building and are
defines the minimum time required to complete the essential for ensuring structural integrity.
project. Note:- Fabricated tiles are not typically part of the
3. So it is deterministic. structural elements in fabricated buildings. Tiles,
whether for flooring or wall coverings, are usually
Q.43. The measurement of a brick is 230 mm × 115 mm considered finishing materials. They are often installed
× 75 mm, which of the following is the measure of its on-site after the main structure is assembled.
stretcher?
Q.45. Which of the following types of wall are not
(a) 115 mm × 75 mm (b) 230 mm × 115 m
constructed in the building?
(c) 230 mm × 75 mm (d) None of is the above
(a) Load bearing wall (b) Retaining wall
Ans : (c) Stretcher Face: This is the longest face of the (c) Non load bearing wall (d) Boundary wall
brick, measuring the length and the height. For a brick
with dimensions 230 mm × 115 mm × 75 mm, the Ans : (b) Based on its functions walls are divided into
stretcher face would be 230 mm × 75 mm. This face is the following types for the building construction:-
typically visible in a stretcher bond, where bricks are laid • Load Bearing Walls
lengthwise along the wall. • Non-Load Bearing Walls
Header Face: This is the face of the brick that measures • Cavity Walls
the width and the height. For the same brick dimensions, • Shear Walls
the header face would be 115 mm × 75 mm. This face is • Partition Walls
visible when bricks are laid end-to-end (width-wise) • Panel Walls
along the wall, as seen in a header bond. • Veneered Walls
• Faced Walls
Note:- Retaining wall is also type of wall but it is used as
earthen wall to retain soil pressure, not in building
construction.
Q.46. For the construction of load bearing wall
foundation first layer provided is called L.C. What is the
full form L.C ?
Q.44. For the construction of fabricated structure
(a) Lime Concrete (b) Lead Cement
which of the following is not used?
(c) Lean Concrete (d) Lower Compound
(a) Fabricated wall panel
(b) Fabricated door & windows Ans : (c) Load-bearing wall :
• A wall that is constructed to support the above slab or
(c) Fabricated frame
other building elements in a structure is called a load-
(d) Fabricated tiles
bearing wall.
Ans : (d) Fabricated structures generally refer to • For the construction of a load-bearing wall foundation,
buildings or components that are manufactured off-site the first layer provided is typically a layer of lean
and then assembled on-site. This approach is often used concrete to provide a stable base and distribute the
to save time, reduce labour costs, and ensure higher load evenly.
quality control. Here’s an explanation of the options • It carries the weight of a house from the roof and
provided: upper floors.
(a) Fabricated Wall Panel:- Fabricated wall panels are • Load-bearing walls transfer loads all the way to the
pre-manufactured sections of walls that are transported foundation or other suitable frame members.
to the construction site and assembled. These panels • It can support structural members like beams (sturdy
can include insulation, electrical wiring, and other pieces of wood or metal), slab, and walls on the above
features, allowing for quick and efficient construction. floors above.
• A wall directly above the beam is called a load-bearing
wall if it is designed to carry the vertical load.

UPSSSC DRAFTSMAN 2015 45 Civil Ki Goli

Ans : (c) The primary compounds in Portland cement are Q.73. Which of the following material is not used as a
lime (CaO), silica (SiO₂), and alumina (Al₂O₃), which damp proofing material for building?
together contribute to the cement's strength and (a) Tapecrete (b) Rubber solution
durability. (c) DPC powder (d) Fair mate
Chemical Percentage Ans : (b) A damp proof course is one of the most
S.No Ingredient
Form (%) important elements of a property. The DPC protects the
property against moisture rising from the ground.
1 Lime CaO (62-65)
Estimation of DPC is measured in square meters of area.
2 Silica SiO2 (17-25) Most commonly used material for damp proofing other
than concrete is Bitumen. Other materials are
3 Alumina Al2O3 (3-8)
Tapecrete, DPC powder, Fair mate etc.
Calcium Q.74. Which of the following is the right answer for the
4 CaSO4 (3-4)
Sulphate English bonded wall?
5 Iron Oxide Fe2O3 (3-4) (a) Alternate header & stretcher
(b) One course of stretcher & then other course of
6 Magnesia MgO (1-3) header in same wall
7 Sulphur S (1-3) (c) Only stretcher are provided in every course
(d) Header is provided vertically in each course
8 Alkalies Na2O, K2O (0.2-1) Ans : (b) Different type of bonding in Brick masonry:
Q.70. How many days of curing are necessary for English Bond : It is the arrangement of bonding that
concrete work in the building in winter moist season? consists of alternate courses of stretcher and header
(a) 6 days (b) 14 days placed one over other.
(c) 20 days (d) 8 days In order to break the alignment of vertical joints to be in
the same straight line queen closer is placed next to the
Ans : (b) Concrete curing is essential to prevent the
quoin header in each alternate course. Header courses
concrete from drying out too quickly, which can cause
must not start from queen closer as it is liable to get
surface cracks and reduce the strength and durability of
displace off.
the concrete. In cooler and moist conditions, curing
times are often extended to ensure that the concrete A lap of one-fourth brick is available for each stretcher
reaches its desired properties. Therefore, 14 days is a over the header in the course below it.
standard recommendation for curing concrete under For the wall having the thickness in the odd multiples of
such conditions. half brick thick, each course shows a stretcher on one
face and a header on the other face.
Q.71. Which of the following is not dry building
material? Flemish Bond: It is the arrangement of bonding that
consists of an alternate header and stretcher in each
(a) marble (b) Brick
course.
(c) Paint (d) Cement
The header in each alternate course is centered over the
Ans : (c) Paint is not considered a dry building material stretcher in the course below it.
because it is applied in a liquid form and requires a
In order to break the alignment of the vertical joint to be
drying or curing period after application. In contrast,
in the same straight line queen closer is placed next to
materials like marble, brick, and cement are typically the quoin header.
used in their dry states during construction.
For walls having a thickness greater than 1.5 brick thick,
Q.72. Which of the following word is not related to an English bond is found to be stronger than a Flemish
brick? bond, however, a Flemish bond renders a higher
(a) Stretcher (b) Queen closer aesthetic appearance than an English bond.
(c) Bat (d) Ashlar Flemish bond is more economical than the English bond.
Ans : (d) Stretcher: A brick laid flat with its longest side Stretcher bond: It is the arrangement of bonding that
exposed. consists of a stretcher in each course.
Queen closer: A brick cut longitudinally in such a way In order to break the alignment of the vertical joint to be
that it is one quarter of a full brick in size. in the same straight line each alternate course started
Bat: A partially brick made by cutting a full-size brick into with the half-bat.
smaller pieces.
Note:-Ashlar is not related to brick

UPSSSC DRAFTSMAN 2015 49 Civil Ki Goli

of walls in double lines and can handle different levels of Description Unit
a building. Earthwork, Stone/Brick Work, Wood
m³
Q.97. On the drawing of map in CAD software to make Work/Sunshade, R.C.C.
computer aided drafting. Which of the following scale Surface/Shallow Excavation, shutter,
maps are prepared in computer screen m²
panal, batten
(a) 1 : 200 (b) 1 : 2 Pointing, Soling, DPC, Plastering, door,
m²
(c) 1 : 100 (d) 1 : 1 window
Ans : (d) In CAD software, the drawings are usually Steel/Iron Work kg/Quintal
created at a full scale (1:1), meaning that the dimensions Dressing of stone/Half Brick
m²
are represented at their actual size. This approach wall/partition wall
allows for precise and accurate design work. Painting Work/Distemper/Colour
m²
Washing/Jali Work
Q.98. On drafting of drawing in CAD, print from
printer/plotter is taken. Which of the following is the Q.102. Which of the following is the full form of F.A.R.
largest size of paper? for the designing and construction of building?
(a) A₁-size (b) A3-size (a) Fire Air Ratio
(c) A2-size (d) Ao-size (b) Floor Area Ratio
Ans : (d) The A0 size is the largest (c) Front Area Ratio
(d) Function Area Ratio
• A0-size: 841 x 1189 mm
• A1-size: 594 x 841 mm Ans : (b) Floor Area Ratio: The FAR of a project is the
• A2-size: 420 x 594 mm total floor area of the building (including the space
• A3-size: 297 x 420 mm covered by all the floors in the building) divided by the
area of land on which the project is being constructed.
Q.99. On the drafting of drawing from auto CAD, offset
command is used mainly for which of the following Q.103. Give the area of plot is 500 sq.m. and total
work ? constructed area of all the floors for building is 400
sq.m. Find out the F.A.R.
(a) Making perpendicular line
(a) 1.25 (b) 2.50
(b) Draw the parallel line & shape
(c) 2.25 (d) 1.50
(c) Curve line draws with given line
(d) Making an angle Ans : (*)
Given:
Ans : (b) The offset command in AutoCAD is used
to create parallel lines, concentric circles, and parallel Plot area = 500 sq.m.
curves. We can offset any object through a point or at a Total constructed area = 400 sq.m.
specified distance. We can create as many parallel lines 𝐓𝐨𝐭𝐚𝐥 𝐂𝐨𝐧𝐬𝐭𝐫𝐮𝐜𝐭𝐞𝐝 𝐀𝐫𝐞𝐚
FAR =
𝐩𝐥𝐨𝐭 𝐚𝐫𝐞𝐚
and curves with the help of the offset command.
𝟒𝟎𝟎
Q.100. On drafting of drawing from the computer by = = 0.8
𝟓𝟎𝟎
auto CAD software if we want to use erase command Options are not correct.
directly from key board then which of the following Q.104. A wall 4m long, 3m high and 30 cm wide is
method is adopted ? constructed and plastered (both side of wall) If cost of
(a) O ↵ (Enter) (b) E ↵ (Enter) the construction of wall is Rs. 150 per cubic meter and
(c) Ex ↵ (Enter) (d) L-Control rate of the plaster are Rs. 10 per sq. m then the total
Ans : (b) The " E ↵" command in AutoCAD is the shortcut estimated cost of the above work is
for the "ERASE" command. Typing "E" and then pressing (a) Rs. 780/- (b) Rs. 925/-
Enter activates the erase command, allowing you to (c) Rs. 670/- (d) Rs. 720/-
select and delete objects. Ans : (a) Given , Length of wall = 4m
❖ ESTIMATION Width of wall = 30cm = 0.3 m
Q.101. In estimating and costing the earthwork Height of wall = 3m
excavation of foundation is measured in which unit? Volume of construction of wall = 4 × 3 × 0.3 = 3.6 m3
(a) Running meter (b) Square meter Cost of construction = 3.6 × 150 = 540 rs
(c) Kilogram (d) Cubic meter Area of plaster = 2 × 3 × 4 = 24 m2
Ans : (d) Cost of plaster = 24 × 10 = 240 rs
Total cost = 540 + 240 = 780 rs

UPSSSC DRAFTSMAN 2015 54 Civil Ki Goli

Q.105. Which of the following unit is used to measure Supplementary estimate- It is required when further
Reinforced Cement Concrete (R.C.C.) in building development is required during the progress of original
estimate. work.
(a) Sq.metre (b) Running meter Revised estimate- It is prepared when Original sanction
(c) Cubic meter (d) Quintal amount or estimate exceed more than 5% or The
Ans : (c) expenditure of the work exceeds more than 10% of the
administrative approval.
Description Unit
Q.107. 1.5 sq.m. Granite is required for the top of the
Earthwork, Stone/Brick Work, Wood m³
kitchen platform. Given the cost of the granite is Rs.
Work/Sunshade, R.C.C.
3000 per sq. m. Then find the estimated cost of the
Surface/Shallow Excavation, shutter, m²
above work.
panal, batten
(a) Rs. 4550/- (b) Rs. 4500/-
Pointing, Soling, DPC, Plastering, door, m²
window (c) Rs. 5000/- (d) Rs. 4450/-
Steel/Iron Work kg/Quintal Ans:(b) Total Cost = Area × cost per sq.m
Dressing of stone/Half Brick m² =1.5 × 3000 = 4500 Rs
wall/partition wall Q.108. A room is 4m long and 3m wide, the tile flooring
Painting Work/Distemper/Colour m² is required for the floor of the room. The cost of 1 sq.m.
Washing/Jali Work Tile is Rs. 750/-. Then find out the total estimated cost
Q.106. Which of the following term is not related with of the above work
the building estimate? (a) Rs. 9000/- (b) Rs. 8750/-
(a) Lump-sum estimate (b) Item wise estimate (c) Rs. 9050/- (d) Rs. 8975/-
(c) Group estimate (d) Detailed estimate Ans:(a) Given length =4m , width= 3m , cost of 1 sq.m.
Ans :(c) Group estimate does not exist in the standard Tile = 750 rs
categories of estimating methods used in building and Area of the floor: length × width
construction projects. = 4m×3m = 12 m2
Estimation- It is the process of working out the total cost: Area × cost per sq.m
approximate cost of an engineering project before =12 ×750 = 9000 Rs
execution of the work. ❖ RCC
Type of Estimate: Q.109. What is the full form of B.I.S. in Indian standard
Preliminary or Approximate Estimate or Abstract (I.S.) code of practice for Plain and Reinforced
Estimate: It is required for preliminary studies of various Concrete?
aspect of a project or work, to decide the financial (a) Bureau of Indian Investigation
position & policy for administrative action by the (b) Board of Indian Standards
competent authority. In it various quantities are worked (c) Building important service
out with the help of many short cuts.
(d) Bureau of Indian Standards
Plinth Area Estimate: Square Meter Method
Ans : (d) Bureau of Indian Standards (BIS) is the official
This is prepared on the basis of plinth area (BL) of
body responsible for the standardization and quality
building.
control of various products and processes in India.
Cube Rate Estimate: It is prepared on the basis of the
BIS develops and publishes standards, including those
cubical contents (LBH) of the building.
related to civil engineering and construction, ensuring
Detailed Estimate: It is an most accurate & reliable consistency and safety in practices and materials used.
method.It consists of working out the quantities of each
item of works and working the cost. Q.110. Which of the following is not the type of slab?
Lump-sum estimate: This type of estimate provides a (a) One way slab (b) Three way slab
single total cost for the entire project without breaking (c) Two way slab (d) R.B. slab
it down into individual components. It is used when a Ans:(b) Slabs are considered as plane element or plate
rough or approximate cost is sufficient. elements which are classified based on aspect ratio.
Item-wise estimate: This type of estimate details the Aspect ratio: It is defined as ratio between longer span
cost for each item or component of the project (Ly) of the slab to shorter span (Lx) of the slab.
separately. It allows for a more precise and accurate One way slab: Slab spanning in one direction. It is
calculation of the total project cost. supported only on the two opposite sides.

UPSSSC DRAFTSMAN 2015 55 Civil Ki Goli

 𝑳𝒀 The different types of beam are :
≥𝟐
𝑳𝑿 Cantilever beam: A beam fixed at one end and free at
the other end is known as a cantilever beam.
Simply Supported beam : A beam supported at its both
ends is known as a simply supported beam.
Fixed beam: A beam whose both ends are fixed, is
known as a fixed beam.
Continuous beam: A beam supported on more than two
supports is known as a continuous beam.
Two way slab: Slab spanning in both the direction. slab Overhanging beam: A beam having its end portion
supported by beams on all the four sides and the loads extended beyond the support, is known as overhanging
are carried by the support along with both directions. beam. A beam may be overhanging on one side or on
𝑳𝒀 both sides.
≤𝟐
𝑳𝑿

Ribbed slab (R.B. slab): Ribbed slab is a structural

component which is plain on its top and contains grid
like system on its bottom surface. The top of ribbed slab
is normally thin and the bottom grid lines are generally
ribs which are laid perpendicular to each other with
equal depth.
•It is special type slab and beam system in which slab,
which is called topping is very thin (50 mm to 100 mm)
and the beam, which is called rib are very slender and
closely spaced (less than 1.5 m apart)
•In ribbed slab system, the thickness of floor slab Q.112. Which type of building is specially constructed
generally varies from 100 mm - 200 mm. with the help of Reinforced Cement Concrete (R.C.C.)?
Characteristics of Ribbed Slabs : (a) Earthquake resistant (b) Normal building
•Suitable for flat areas (c) Green bhawan (d) Load bearing building
•Volume of concrete used is very less compared to Ans : (d) Earthquake Resistant Buildings : R.C.C.
others structures are designed to withstand seismic forces,
•Reinforcement provided is in the form of mesh or making them suitable for earthquake-resistant
individual bars buildings.
•The bottom surface of slab is looks like waffle which is •The steel reinforcement within the concrete helps to
obtained by using cardboard panels or pods etc. absorb and dissipate the energy generated during an
earthquake, reducing the risk of structural failure.
•These buildings are specially designed to absorb and
dissipate seismic energy, ensuring they remain standing
and functional after an earthquake.
Q.113. For the pure construction of Reinforced Cement
Concrete (R.C.C.) Structure, which of the following
Three-way slab is not a type of slab. Slabs typically carry building material is used minimum ?
loads in one or two directions.
(a) Concrete (b) Brick
Q.111. Which of the following is not the type of beam? (c) Steel (d) Cement and coarse sand
(a) T-beam (b) L-beam
Ans:(b) Bricks are primarily used for non-structural
(c) Simply supported beam (d) P-beam purposes such as partition walls or facades in R.C.C.
Ans:(d) P- beam is not the type of beam. constructions. The main structural materials are

UPSSSC DRAFTSMAN 2015 56 Civil Ki Goli

concrete (comprising cement, sand, and aggregates) and One way slab: Slab spanning in one direction. It is
steel, which form the R.C.C. elements.Therefore, bricks supported only on the two opposite sides.
are used minimally in the context of R.C.C. structures. 𝑳𝒀
>𝟐
Q.114. l/d ratio of simply supported beam is: 𝑳𝑿
(a) 25 (b) 24 (c) 22 (d) 20
Ans : (d) Basic values of span to effective depth ratios for
spans up to 10 m:
Type of beam Span/depth ratio
Cantilever beam 7
Simply supported beam 20
Two way slab: Slab spanning in both the direction. slab
Continuous beam 26 supported by beams on all the four sides and the loads
are carried by the support along with both directions.
𝐋𝐘
Type of reinforcement ≤𝟐
Type of slab 𝐋𝐗
Mild steel Fe 415
Simply supported 35 28
Continuous 40 32
𝟏𝟎
Where as factor is neglected.
𝐒𝐩𝐚𝐧
𝟏𝟎
• If span > 10 m, then factor is also multiplied (But
𝐒𝐩𝐚𝐧
if cantilever beam > 10 m, then actual deflection
calculations should be made) Q.117. Minimum reinforcement is provided in R.C.C.
slab:
Q.115. M-20 concrete mix is
(a) Central strip (b) Side strip
(a) 1 : 2 : 4 (b) 1 : 3 : 6
𝟏 (c) Middle strip (d) Edge strip
(c) 1:1 ∶ 𝟑 (d) 1 : 4 : 8
𝟐 Ans:(d) Minimum reinforcement is provided at the edge
Ans : (c) In M 20 grade concrete, strips to control cracking due to shrinkage and
𝟏 temperature variations in R.C.C. Slab and maximum
the ratio of cement: sand: ballast will be 1:1 ∶ 𝟑
𝟐
reinforcement at middle strip.
M20 signifies a mixture of cement sand and aggregate
The minimum reinforcement in edge strip, parallel to
which are prepared in such a manner that a cement
that edge shall be 0.15% for mild steel and 0.12% for
concrete cube of size 15 cm x 15 cm x 15 cm is formed
HYSD bars, of the cross-section area of concrete.
with characteristic strength (fck) of 20 MPa while
examining it after being cured for 28 days.
Strength Nominal Mix Design
M5 1:5:10
M7.5 1:4:8
M10 1:3:6
M15 1:2:4 Q.118. The minimum number of bars in circular
M20 1:1.5:3 columns is:
(a) 4 (b) 5 (c) 6 (d) 8
M25 1:1:2
𝐋𝐘
Ans : (c) Minimum Number of bars
Q.116. The ratio for a two way slab is: • For rectangular columns = 4
𝐋𝐗
(a) = 2.0 (b) ≤ 2.0 • For circular columns = 6
(c) = 1 (d) > 2.0
Ans : (b) Slabs are considered as plane element or plate
❖ ENVIRONMENT
elements which are classified based on aspect ratio. Q.119. Which of the following is not the source of
surface water?
Aspect ratio: It is defined as ratio between longer span
(Ly) of the slab to shorter span (Lx) of the slab. (a) Lakes (b) Reservoir
(c) Springs (d) Rivers

UPSSSC DRAFTSMAN 2015 57 Civil Ki Goli

Ans : (c) Springs: Points where groundwater naturally Q.122. Which of the following trap is not used in the
flows to the surface, often forming small streams or sewage drainage system of residences?
pools. Springs are primarily a source of groundwater, (a) Water seal trap (b) p-trap
not surface water. (c) Nahani trap (d) S-trap
Q.120. As per I.S. 1172-1983 (3rd REVISED) the required Ans:(a) All the traps mentioned (p-trap, nahani trap, and
water per person per day in domestic residential use is- s-trap) are commonly used in the sewage drainage
(a) 85 litres (b) 100 litres systems of residences. However, "water seal trap" is not
(c) 115 litres (d) 135 litres a specific type of trap.
Ans : (d) Traps: Traps simply trap the water and sewer gas. It is a
Demand in l / h / d plumbing device used to prevent smell, bacteria, also
insects from entering your home. Different types of trap
With full
Use Without full mention bellow,
flushing
flushing system Gully Trap: A gully trap is provided outside the building
system
Domestic use 200 135 to connect waste pipe to the external sewerage line. It
Industrial use 50 40 collects wastewater from the kitchen sink, wash basins,
Commercial use 20 25 bathroom, and other wash areas.
Civic or Public
10 15
use
Wastes and
55 55
thefts, etc.
Total per capita
335 270
demand (q)
Q.121. Which of the following is NOT a type of
sewerage system?
(a) Separate system (b) Sullage system
(c) Combined system (d) Partially combined system Nahani trap: This trap is generally used to admit sullage
from the floors of rooms, bathrooms, kitchen, etc. in to
Ans : (b) Sullage refers to wastewater from household
the sullage pipe.
activities like bathing, washing, and cooking, excluding
toilet waste. There isn't a specific "sullage system" as a
type of sewerage system.
Sewerage system: The entire system developed for the
collection, treating, and disposing of the sewage safely
is called the sewerage system. Modern sewerage
systems fall under mainly three categories-
(1). Combined sewerage system: A combined sewer
system is a sewer that accepts stormwater, sanitary
water(sewage), then the sewage is treated in the
sewerage treatment plant. This system is mainly used in Intercepting Trap: Intercepting trap is provided at the
towns where streets are narrow and rainfall is less than intercept of building sewer and Public sewer.
moderate. Intercepting trap is provided to prevent the foul gases
(2). Separate sewerage system: In this system, the from public sewers entering into the building sewer by
sanitary sewage and stormwater are carried separately providing water seal.
in two sets of sewers. The sewage is conveyed to the
wastewater treatment plant and the stormwater is
discharged into rivers without treatment.
(3). Partially Separate sewerage system: A partially
separate system is a combination of a combined
sewerage system and separate sewerage systems. This
type of sewerage system helps decrease the load from a P, Q, and S -Trap: P, Q, and S traps are classified
combined sewerage system because only the water according to their shape. They essentially consist of U –
from initial rainfalls (water from acid rain) is added to Tube which retains water acting as a seal between the
sewage water and after that this system work as a foul gas and atmosphere. They connect the water closet
separate system. to the soil pipe.

UPSSSC DRAFTSMAN 2015 58 Civil Ki Goli

Q.123. What is the full form of S.T.P. are related to
house drainage and sanitary engineering?
(a) Sewage Treatment Plant
(b) Society Treatment Plant
(c) Service Treatment Plant
(d) System Treatment Plant
Ans:(a) Sewage Treatment Plant (STP) : An STP is a
facility designed to treat and process sewage and
wastewater generated by households, commercial
establishments, and industries. The main goal is to Search Civil Ki Goli On Google Play
remove contaminants and produce environmentally
safe treated effluent, which can be safely discharged Store For app. And you can
into the environment or reused. The treatment process attempt test series and Ebooks.
typically involves physical, chemical, and biological
processes to remove solids, organic matter, and
nutrients from the sewage.
Anti-siphon trap: is used to prevent siphoning action.
Q.124. Which of the following are used for the storage
of the clean water in the building?
(a) Soak pit (b) Water tank
(c) Septic tank (d) Inspection chamber
Ans:(b) A water tank is specifically designed for storing
clean water for various uses within a building, such as
drinking, cooking, bathing, and cleaning. Water tanks
can be placed at ground level, underground, or on
rooftops.
Q.125. Which of the following mainly water is not
stored in following water supply and sanitary
engineering of houses:
(a) Water tank
(b) Flushing cistern
(c) Fixed water purifier or aquaguard
(d) Geyser
Ans : (d) A geyser is an instant water heater, not a
storage unit. It heats water as it passes through the
device, providing hot water on demand.

UPSSSC DRAFTSMAN 2015 59 Civil Ki Goli

 UPSSSC JE Civil ( 31-07-2016)
❖ SOM
Q.01. A simply supported beam carrying UDI on the
entire span. The value of deflection will be
(a) WL3/48EI (b) 5WL4/384EI
(c) WL2/24EI (d) WL3/3EI 𝑭𝟐𝒓 = 𝑭𝟐𝟏 + 𝑭𝟐𝟐 + 𝟐𝑭𝟏 𝑭𝟐 𝒄𝒐𝒔 𝜽
Ans : (c) Deflection of the simply supported beam
𝑭𝒓 = √𝑭𝟐𝟏 + 𝑭𝟐𝟐 + 𝟐𝑭𝟏 𝑭𝟐 𝒄𝒐𝒔 𝜽
𝟓 𝒘𝑳𝟒
carrying UDI on the entire span is yc =
𝟑𝟖𝟒 𝑬𝑰 Here, F1 = P, F2 = Q
Where, y = Deflection of beam  = Slope of beam Resultant force is given by,
𝑷𝑳𝟑 𝑷𝑳𝟐
𝒚𝑩 = 𝜽𝑩 = 𝑭𝒓 = √𝑭𝟐𝟏 + 𝑭𝟐𝟐 + 𝟐𝑭𝟏 𝑭𝟐 𝒄𝒐𝒔 𝜽
𝟑𝑬𝑰 𝟐𝑬𝑰
𝒘𝑳𝟒 𝒘𝑳𝟑 𝑭𝒓 = √𝑷𝟐 + 𝑸𝟐 + 𝟐𝑷𝑸 𝒄𝒐𝒔 𝜽
𝒚𝑩 = 𝜽𝑩 =
𝟖𝑬𝑰 𝟔𝑬𝑰
Q.03. The Lami’s theorem is applicable only for-
𝑴𝑳𝟐 𝑴𝑳 (a) Coplanar forces
𝒚𝑩 = 𝜽𝑩 =
𝟐𝑬𝑰 𝑬𝑰 (b) Concurrent forces
𝒘𝑳𝟒 𝒘𝑳𝟑 (c) Coplanar & Concurrent forces
𝒚𝑩 = 𝜽𝑩 = (d) Any type of forces
𝟑𝟎𝑬𝑰 𝟑𝟒𝑬𝑰
Ans : (c) Lami's theorem is an equation relating the
𝑷𝑳𝟑 𝒘𝑳𝟐 magnitudes of three coplanar, concurrent and non-
𝒚𝒄 = 𝜽𝑩 = collinear forces, which keeps an object in static
𝟒𝟖𝑬𝑰 𝟏𝟔𝑬𝑰
equilibrium, with the angles directly opposite to the
𝟓 𝒘𝑳𝟒 𝒘𝑳𝟑 corresponding forces.
𝒚𝒄 = 𝜽𝑩 = According to the theorem:
𝟑𝟖𝟒 𝑬𝑰 𝟐𝟒𝑬𝑰
𝑴𝑳
yc = 0 𝜽𝑩 =
𝟐𝟒𝑬𝑰

𝑷𝑳𝟑 𝜽𝑨 = 𝜽𝑩
𝒚𝒄 = = 𝜽𝑪 = 𝟎
𝟏𝟗𝟐𝑬𝑰

𝒘𝑳𝟒 𝜽𝑨 = 𝜽𝑩
𝒚𝒄 = = 𝜽𝑪 = 𝟎
𝟑𝟖𝟒𝑬𝑰
Q.02. The resultant of two forces P & Q are acting at
an angle θ is equal to-
(a) √𝐏 𝟐 + 𝐐𝟐 + 𝟐𝐏𝐐 𝐬𝐢𝐧 𝛉
𝐀 𝐁 𝐂
(b) √𝐏 𝟐 + 𝐐𝟐 + 𝟐𝐏𝐐 𝐜𝐨𝐬 𝛉 = =
𝐬𝐢𝐧 𝛂 𝐬𝐢𝐧 𝛃 𝐬𝐢𝐧 𝛄
(c) √𝐏 𝟐 + 𝐐𝟐 − 𝟐𝐏𝐐 𝐬𝐢𝐧 𝛉 Q.04. The efficiency of screw jack may be increased
(d) √𝐏 𝟐 + 𝐐𝟐 − 𝟐𝐏𝐐 𝐜𝐨𝐬 𝛉 by-
(a) Increasing its pitch
Ans : (b) Parallelogram Law of Forces : If two forces,
acting at a point be represented in magnitude and (b) Decreasing its pitch
direction by the two adjacent sides of a parallelogram, (c) Increasing the load to be lifted
then their resultant is represented in magnitude and (d) Decreasing the load to be lifted
direction by the diagonal of the parallelogram passing Ans : (a): Screw jack: A screw jack is a portable device
through that point. consisting of a screw mechanism used to raise or lower
Resultant force Fr, of any two forces F1 and F2 with an the load. There are two types of jack Hydraulic and
angle θ between them, can be given by vector addition mechanical
as

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• A Hydraulic jack consists of a cylinder and piston 𝑹 = √𝟐𝑷 + 𝟐𝑷𝟐 𝒄𝒐𝒔 𝜽
mechanism. The movement of piston rod is used to
𝑹 = √𝟐𝑷𝟐 (𝟏 + 𝒄𝒐𝒔 𝜽)
raise or lower the load.
We know that,
• Mechanical jack can be either hand operated or
power driven 𝟏 + 𝒄𝒐𝒔 𝜽 = 𝟐 𝒄𝒐𝒔𝟐 ( 𝜽/𝟐)
Screw used in screw jack: The screw jack is 𝑹 = √𝟐𝑷𝟐 (𝟐 𝒄𝒐𝒔𝟐 ( 𝜽/𝟐)]
intermittently used device and wear of thread is not an R = 2P cos (𝜽/𝟐)
important consideration. Therefore instead of Q.07. If two forces 3 kg & 4 kg act at right angle to
trapezoidal or V threads, the screw is provided with each other resultant force will be equal to-
square threads.
(a) 7 kg (b) 5 kg (c) 1 kg (d) None of these
• Square threads have higher efficiency and provision
can be made for self-locking arrangement. Ans : (b) Given
• When the condition of self-locking is fulfilled, the load P = 3 kg Q = 4 kg 𝜃 = 900
itself will not turn the screw and descend down, unless R = √𝑷𝟐 + 𝑸𝟐 + 𝟐𝑷𝑸 𝒄𝒐𝒔 𝜽
the handle is rotated in reverse direction with some R = √𝟑𝟐 + 𝟒𝟐 + 𝟐 × 𝟑 × 𝒄𝒐𝒔 𝟗𝟎
effort.
R = 5 kg
• Maximum efficiency of screw jack increases with
increasing its pitch. Q.08. The centre of gravity of a semi circular of radius
10 cm lies at following distance above base
The maximum efficiency of screw jack is
𝟏−𝒔𝒊𝒏 𝝓 (a) 0.23 cm (b) 2.39 cm (c) 4.24 cm (d) 1.33 cm
𝜼=
𝟏+𝒔𝒊𝒏 𝝓 Ans : (c) The centre of gravity of a semi-circle lies at a
Q.05. Two forces act at angle of 120°. If the greater distance of 4r/3π measured along the vertical radius.
forces is 50 kg then their resultant is perpendicular to The semi-circle of radius ‘r’
the smaller forces. Smaller force is
(a) 20 kg (b) 25 kg (c) 30 kg (d) 35 kg
Ans : (c) Given,
Greater force F1= 50kg smaller force F₂ = ?
From Lami's theorem
𝟒𝐫 𝟒×𝟏𝟎
𝑭𝑨 𝑭𝑩 𝑭𝑪 ̅ =
𝐘 = = 𝟒. 𝟐𝟒 𝐜𝐦
= = 𝟑𝛑 𝟑𝛑
𝒔𝒊𝒏 𝜶 𝒔𝒊𝒏 𝜷 𝒔𝒊𝒏 𝜸
Q.09. Moment of inertial of a lamina is also called as
According to question (a) First moment of area (b) Second moment of area
(c) Third moment of area (d) None of these
Ans : (b) Moment of inertia of an area or Second
moment of area (M.I.): M.I. of a body about any axis is
defined as the summation of the second moment of all
elementary areas about the axis.
From Lami's theorem I = Σ(A × d2)
𝐅𝟏
=
𝐅𝟐
=
𝐑 Unit: m or mm4 or cm4
4

𝐬𝐢𝐧 𝟗𝟎 𝐬𝐢𝐧 𝟏𝟓𝟎 𝐬𝐢𝐧 𝟏𝟐𝟎

𝐅𝟏
**This Question repeat 2 times in paper
F2 = × sin 30 = 25 kg
𝐬𝐢𝐧 𝟗𝟎 Q.10. A steel rod of 2 cm diameter and 5 m long is
Q.06. If two equal forces of magnitude P act on angle subjected to an axial pull of 3000 kg. if E = 2.1 × 10 6
of ° their resultant will be- kg/cm2 the elongation of rod will be-
𝛉 𝛉 (a) 2.275 cm (b) 0.2275 cm
(a) 𝐏 𝐜𝐨𝐬 (b) 𝟐𝐏 𝐬𝐢𝐧
𝟐 𝟐 (c) 0.02275 cm (d) 2.02275 cm
𝛉 𝛉
(c) 𝐏 𝐭𝐚𝐧 (d) 𝟐𝐏 𝐜𝐨𝐬 Ans : (b) Given,
𝟐 𝟐
Ans : (d) When two force making an angle , the P = 3000 kg Diameter of rod (d) = 2 cm
resultant (R) of the two force is given by: L=5m E = 2.1 x 106 kg/cm²
𝜫
R = √𝐏 𝟐 + 𝐐𝟐 + 𝟐𝐏𝐐 𝐜𝐨𝐬 𝛉 Cross-section of area of rod(A) = × 𝒅𝟐
𝟒
Since the two forces are equal: =
𝜫
× 𝟐𝟐 = 3.14 cm2
𝟒
R= √𝐏 𝟐 + 𝐏𝟐 + 𝟐𝐏 × 𝐏 𝐜𝐨𝐬 𝛉
We know that

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 𝑷𝑳 𝝈 𝑴
Elongation of rod (Δ) = ∴ =
𝑨𝑬 𝒚 𝑰
𝟑𝟎𝟎𝟎×𝟓×𝟏𝟎𝟎
= 𝑴×𝒚
𝟑⋅𝟏𝟒×𝟐⋅𝟏×𝟏𝟎𝟔 𝝈=
𝒃𝒅𝟐 /𝟏𝟐
= 0.2275 cm
𝐌 × ⅆ×𝟏𝟐 𝒅
Q.11. The bending moment on a section is maximum 𝛔= (∵ 𝒚 = )
𝐛 𝐱 ⅆ𝟑 ×𝟐 𝟐
where shearing force is 𝟔𝐌
(a) Minimum (b) Maximum 𝛔=
𝐛 𝐱 ⅆ𝟐
(c) Equal (d) Changing sign 𝟔𝑴
𝒃𝒙 =
Ans : (d) Let w be load intensity, V be the shear force 𝝈𝒅𝟐

and M be the bending moment. 𝒃𝒙 ∝ 𝑴

𝒅𝑽
𝒘= Q.13. The ratio of the moment of inertia of a circular
𝒅𝒙
𝒅𝑴 plate to that of a square plate for equal depth is-
𝑽=
𝒅𝒙 (a) Less than one (b) Equal to one
For finding the maximum value of Bending Moment: (c) More than one (d) Equal to 6 π
𝒅𝑴 𝜋𝑑 4
=𝟎=𝑽=𝟎 Ans : (a) Moment of inertia of circular plate =
64
𝒅𝒙
𝑎4
Bending moment is maximum where shear force is zero Moment of inertia of Square plate =
12
or its changes sign (positive to negative or vice-versa). Ratio of Moment of inertia of circular plate and Square
𝛑 ⅆ𝟒
𝟔𝟒
plate = 𝐚𝟒
= 0.589
𝟏𝟐
In the given question depth is equal
So, d = a
So , ratio is less than 1
Q.14. The moment diagram for a cantilever beam
whose free end is subjected to a bending moment
(a) Triangle (b) Rectangle
(c) Parabola (d) Cubic parabola
Ans : (b) Vz-x = 0; Mx-z = -M
The moment diagram for a cantilever beam whose free
end is subjects to a bending moment will be a
rectangle.

Q.12. For a beam of uniform strength keeping its

depth constant the width will vary in proportion to-
(a) Bending moment (M) (b) √𝑴
(c) M2 (d) None of the above
Ans : (a)

Q.15. Hook’s law states that stress and strain are-

(a) Directly proportional (b) Inversely proportional
(c) Curvilinearly related (d) None of the above
Ans : (a) Hooke's law: It States that under direct
We have the bending equation for a beam, as loading, within proportionality limit, stress is directly
𝝈 𝑴 𝑬
= =𝑹 proportional to strain. Hooke's law helps us understand
𝒚 𝑰

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how a stretchy object will behave when it is stretched Concurrent forces: The forces, which meet at one
or compacted. point, are known as concurrent forces. The concurrent
Strain ∝ Stress forces may or may not be colinear.
Coplanar concurrent forces: The forces, which meet at
one point and their lines of action also lie on the same
plane, are known as coplanar concurrent forces.
Coplanar non-concurrent forces: The forces, which do
not meet at one point, but their lines of action lie on
the same plane, are known as coplanar non-concurrent
forces.
Q.16. A simply supported beam of span L carrying a Non-coplanar concurrent forces: The forces, which
point load w at the centre of the beam. The maximum meet at one point, but their lines of action do not lie on
bending moment is the same plane, are known as non-coplanar concurrent
(a) WL (b) WL/2 (c) WL/4 (d) WL/6 forces.
𝚺𝐅𝐱 = 𝚺𝐅𝐲 = 𝚺𝐅𝐳 = 𝟎
Ans : (c)
We have, Non-coplanar non-concurrent forces: The forces, which
do not meet at one point and their lines of action do
not lie on the same plane, are called non-coplanar non-
concurrent forces.

Simply supported beam of span L carrying a point load

w at the center of the beam.
Support reaction = R = W/2
Maximum Bending Moment at x = l/2
𝐖 𝐋 𝐖𝐋
(𝐁. 𝐌)𝐱=𝐋 = 𝟐
×𝟐= 𝟒
𝟐
Hence the maximum bending moment i.e WL/4
Q.17. The forces which meet at one point, but their
lines of action do not lie in a plane are called- Q.18. The centre of gravity of a plane lamina will not
(a) Coplanar non concurrent forces be at its geometrical centre if it is a-
(b) Non-coplanar concurrent forces (a) Circle (b) Equilateral triangle
(c) Non coplanar non concurrent forces (c) Rectangle (d) Right angle triangle
(d) Intersecting forces Ans : (d) Circle, square, and rectangle are symmetrical
Ans : (b) When two or more forces act on a body, they plane lamina.
are called to form a system of forces. • Therefore the center of the gravity of these lamina
Coplanar forces: The forces, whose lines of action lie on lies in the geometrical center.
the same plane, are known as coplanar forces. • But the right angle triangle is unsymmetrical
Colinear forces: The forces, whose lines of action lie on therefore it's center of gravity does not lie on its
the same line, are known as colinear forces. geometrical center.

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Diagram Center of gravity • The property, by virtue of which certain materials
return back to their original position after the removal
of the external force, is called elasticity .
𝒂 Q.21. If a steel rod of 20 mm diameter and 5 m long
𝟐 elongates by 2.275 mm, when subjected to an axial
pull of 3000 kg the stress developed is-
(a) 9.5541 kg/cm2 (b) 95.541 kg/cm2
2
(c) 955.40 kg/cm (d) 9554.1 kg/cm2
𝑳 Ans : (c) Given.
𝟐 Diameter of rod (d) = 20 mm = 2 cm.
Length of rod (L) = 5 m = 500 cm
Elongation of rod (𝛿𝑙) = 2.275 mm = 0.2275 cm
Axial Pull (P) = 3000 kg
Elongation in the rod when an axial force is applied,
𝑫 𝐏𝐋
𝛅𝐋 =
𝐀𝐄
𝟐 𝐏𝐋
E=
𝐀𝛅𝐋
𝟑𝟎𝟎𝟎×𝟓𝟎𝟎
𝑬=𝜫
×𝟐𝟐 ×𝟎.𝟐𝟐𝟕𝟓
𝟒
E = 2098746.502 kg/cm²
Hook's Law:
𝒃 𝝈=𝑬×
𝜹𝒍
𝒍
𝟑 𝟎⋅𝟐𝟐𝟕𝟓
𝝈 = 𝟐𝟎𝟗𝟖𝟕𝟒𝟔. 𝟓𝟎𝟐 ×
𝟓𝟎𝟎
= 954.929 kg/cm²
Q.19. What is the unit of the radius of gyration? Q.22. In a continuous curve of bending moment, the
(a) m3 (b) m (c) m-1 (d) m2 point of zero bending moment, where it changes sign,
is called-
Ans : (b) We can define the radius of gyration as the
(a) The point of inflexion
imaginary distance from the centroid at which the area
(b) The point of contra flexure
of cross-section is imagined to be focused at a point in
order to obtain the same moment of inertia. It is (c) The point of a virtual hinge
denoted by k. (d) All of the above
The radius of gyration is given as follows Ans : (b) Points of zero bending moment : The points
𝑰 of contra flexure are points of zero bending moment,
𝒌=√ i.e. where the beam changes its curvature from hogging
𝑴

Where I is the moment of inertia and and M is the mass to sagging.

of the body • In a bending beam, a point of contra flexure is a
• Radius of gyration is the distance from the axis of location where the bending moment is zero (changes
rotation so its unit is meter (m). its sign).
• In a bending moment diagram, it is the point at which
Q.20. The property of a material by virtue of which a
the bending moment curve intersects with the zero
body returns to its original shape after removal of the
lines.
load is called ________.
Point of inflexion : Point of inflexion is the point where
(a) Plasticity (b) Elasticity
the elastic deflection curve changes its curvature not
(c) Ductility (d) Malleability bending moment diagram
Ans : (b) When an external force acts on a body, the Q.23. Shear force diagram for a cantilever beam
body tends to undergo some deformation carrying a uniformly distributed load over its whole
• If the external force is removed and the body comes length is a -
back to its original shape and size, the body is known as (a) Rectangle (b) Triangle
an elastic body
(c) Parabola (d) Cubic parabola

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Ans : (b) For a cantilever beam carrying a uniformly Cantilever beam: When a cantilever beam is subjected
distributed load (UDL) over its entire length, the shear to a hogging bending moment. Then cross-section
force diagram (SFD) is a triangle. above the neutral axis (NA) is subjected to tensile
stresses and the cross-section below the neutral axis
Loading SFD BMD
(NA) is subjected to compressive stress.

Q.24. The bending moment diagram for a cantilever

Q.26. The deflection of any rectangular beam simply
which is subjected to a uniformly distributed load will
supported is:
be a-
(a) Directly proportional to its weight
(a) Triangle (b) Rectangle
(b) Inversely proportional to its weight
(c) Parabola (d) Cubic parabola
(c) Directly proportional to the cube of its depth
Ans : (c) Cantilever beam carries a uniformly
(d) None of the above
distributed load:
Shear force: Shear force diagram is a triangle with zero Ans : (a) As there is no loading given in the question. So
at the free end and WL at the fixed end, ∴ it is linear. deflection due to self-weight will be considered and it
will be a uniform distributed load.
Bending Moment: Bending moment at any section is
proportional to the square of the distance of the
section from the free end. This follows a parabolic
shape.

Deflection of a rectangular beam simply supported

carrying UDL is:
𝟓 𝒘𝒍𝟒
𝜟=
𝟑𝟖𝟒 𝑬𝑰
𝒃𝒅𝟑
𝑰=
𝟏𝟐
Here, w is the weight per unit length of the beam.
So, from this, it is concluded that,
Deflection for the above beam is:
• Directly proportional to its weight
• Inversely proportional to the cube of its depth
Q.25. When a rectangular beam is loaded transversely,
• Inversely proportional to the width of the beam
the maximum compressive stress develops on-
(a) Bottom fiber (b) Top fiber Q.27. When direct stress is greater than bending stress
there will be-
(c) Neutral of fiber (d) None of these
(a) Compression (b) Tension
Ans : (b) Nature of Bending stress:
(c) Compression & tension both (d) None of these
Simply supported beam: When a simply supported
beam is subjected to a sagging bending moment, the Ans : (a) Stress (σ): Stress is the resistance offered by
cross-section above NA is subjected to compressive material against deformation.
stresses, and the cross-section below the neutral axis is Depending upon the nature of stress, stress may be of
subjected to tensile stress. the following two types:

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 𝝅𝒅𝟒 Ans : (d) Stretching bond: Consists entirely of
𝑱=
𝟑𝟐
stretchers. This bond is suitable for thinner walls, such
From torsion equation
as a 10 cm thick wall, because it uses the long face of
𝑻 𝑮𝜽
= the brick along the length of the wall, providing the
𝑱 𝑳
𝑻 ∝ 𝜽 necessary thickness with a single row of bricks.
• Angle of twist is directly proportional to twisting
moment

❖ STRUCTURE ANALYSIS
Q.31. The space diagram of frame structure must have
(a) Loads (b) Reactions
(c) Both (A) & (B) (d) None of these
Ans : (c) A Space Frame structure can be defined as a ❖ RCC
rigid, lightweight, truss-like structure. Q.34. For 1 cubic meter of R.C.C. beam approximate
• The Space frames can be used efficiently to cover quantity of steel reinforcement may be taken as-
huge areas with minimum interior supports. (a) 1% - 2% (b) 0.7% - 1%
• A space frame structure's robustness is due to its (c) 1% - 2.5% (d) 0.5% - 0.8%
rigidity of the triangle and flexing loads that are
Ans : (a) Thumb Rule for Steel in percentage:
transmitted as tension and compression loads along the
length of each strut. • Thumb rule for Steel in general RCC structure is 1% -
4% of wet volume of concrete
• Loads and reactions both are necessary for the frame
• Thumb Rule for Steel in column is 2% - 4% of wet
structures because when the load is applied to the
volume of concrete
structure then it is restricted by the support by
• Thumb Rule for Steel in beam is 1% - 2% of wet
providing the reactions.
volume of concrete
• Space frames are typically designed using a rigidity
• Thumb Rule for Steel in footing is 0.5% - 1% of wet
matrix.
volume of concrete
Q.32. If n & j are the numbers of members and joints • Thumb Rule for Steel in slab is 1% - 1.5% of wet
respectively in a frame, is called redundant- volume of concrete.
(a) n = 2j - 3 (b) n = 3j - 2 Q.35. Under limit state method the value of partial
(c) n < 2j - 3 (d) n > 2j - 3 safety factor for steel is-
Ans : (d) A redundant frame is a frame having more (a) 1.8 (b) 1.6 (c) 1.15 (d) 1.5
number of members than required such that there is Ans : (c) Partial factor of safety for concrete and steel
always a member in which force cannot be computed should be taken as 1.5 and 1.15, respectively when
using equations of static equilibrium assessing the strength of the structures or structural
n > 2j – 3 members employing limit state of collapse.
Perfect frame have the number of members is just
In Limit state method , partially safety factor
sufficient to keep the frame in equilibrium without a
change in its shape under the action of external load. Material Collapse Deflection Cracking
The perfect frame satisfies the following equation
Concrete 1.5 1 1.3
n = 2j - 3
A deficient frame is a frame having less number of Steel 1.15 1 1
members than required for a perfect frame
n < 2j-3
where Q.36. The coefficient of thermal expansion of concrete
n = number of members in the frame is nearly equal to-
j = number of joints in the frame (a) Wood (b) Steel (c) Aluminium (d) Copper
Ans : (b) Concrete and steel have similar coefficients of
❖ BCME thermal expansion, which is why they work well
Q.33. Which type of bond can be made for 10 cm thick together in reinforced concrete structures, expanding
wall- and contracting at similar rates when subjected to
(a) English bond (b) Flemish bond temperature changes.
(c) Heading bond (d) Stretching bond As per Indian standard:

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• Thermal coefficient of expansion of concrete is taken This code uses the limit state design approach as well
to be 10 × 10-6 °C as the working stress design approach. However, the
• Thermal coefficient of expansion of steel is taken to Code recommends the use of the limit state design
be 12 × 10-6 °C. approach.
• Thermal coefficient of expansion of copper is taken to • It is written for use in India. It gives extensive
be 16.7 × 10-6 °C information on the various aspects of concrete.
• Thermal coefficient of expansion of Aluminium is Q.40. If the depth of actual neutral axis in a beam is
taken to be 23 × 10-6 °C more than the depth of critical neutral axis, the beam
Q.37. The spacing of stirrups near the support is- is called:
(a) Minimum (a) Balanced beam (b) Under reinforced beam
(b) Maximum (c) Over reinforced beam (d) None of the above
(c) Between minimum & maximum Ans : (c)
(d) Zero
Ans : (a) The shear reinforcement increases as the
spacing among the stirrups decreases and vice versa.
As shear strength requirement is more at the supports
than the center, the spacing of stirrups increases
towards the center of the beam.

• In under reinforced sections, steel is less so it starts to

yield first and then cracks starts appearing on
concrete and structure fail slowly i.e. ductile failure. It
gives us warning and time to escape before collapse
any other vulnerable conditions like or earthquake.
• In over reinforced sections, concrete breaks first and
Q.38. The reinforcement in RCC takes- it is a brittle failure, and structure will collapse
(a) Tensile stress (b) Compressive stress instantly. So we have no time to save our life or
(c) Shear stress (d) Torsional stress escape from building before collapse.
• In balanced section, both steel and concrete will
Ans : (a) Steel reinforcing rods are used in concrete
reach their limiting values simultaneously.
beams to make it carry tension.
• As concrete is strong in compression but weak in Under Over
Balanced section
tension and its tensile strength is approx 1/10th of reinforced reinforced
compressive strength. To improve its tensile strength, Xu < Xulim Xu = Xulim Xu > Xulim
reinforcement (steel reinforcing rods) is embedded in it.
Ast < Astlim Ast = Astlim Ast > Astlim
• As per IS 456 maximum area of tension reinforcement
in the beam shall not exceed 4% of the total area i.e. MOR < MORlim MOR = MORlim MOR > MORlim
0.04Bd.
Z > Zlim Z = Zlim Z < Zlim
Q.39. IS:456-2000 recommends-
(a) Working stress method of design Steel yields Concrete
Both steel yields and
first then reaches
(b) Ultimate load method of design concrete reaches
concrete maximum strain
(c) Limit state method of design maximum strain of
reaches of 0.0035
(d) Both (A) & (C) 0.0035
maximum before steel
simultaneously
Ans : (c) IS 456-2000 Plain and Reinforced Concrete - strain of 0.0035 yields
Code of Practice is an Indian Standard code of practice Q.41. Value of partial factor of safety for concrete is-
for general structural use of plain and reinforced (a) 2.0 (b) 1.5 (c) 3.0 (d) 5.0
concrete.
Ans : (b) Partial factor of safety for concrete and steel
• The latest revision of this standard was done in the
should be taken as 1.5 and 1.15, respectively when
year 2000 and reaffirmed in 2021.
assessing the strength of the structures or structural
members employing limit state of collapse.
UPSSSC JE 31/07/2016 68 Civil KI Goli
Stress in Steel = 140 N/mm2 Ans : (a) Assumption in limit state method:
For the given condition: • Plane sections normal to the axis remain plane even
σcbc = 4N/mm2 σst = 140 N/mm2 after bending. It means the strain distribution across
𝒎𝝈𝒄𝒃 ×𝒅 the depth of the cross-section is linear.
depth of Neutral axis (x) =
𝒎𝝈𝒄𝒃 +𝝈𝒔𝒕 • At Limiting state, the maximum strain in concrete,
𝟏𝟗×𝟓×𝟓𝟎𝟎 which occurs at
=
𝟏𝟗×𝟓+𝟏𝟒𝟎
𝟒𝟕𝟓𝟎𝟎 outermost compression fibre is 0.0035.
= 𝟐𝟑𝟓 • The stress-strain curve for concrete is having
= 202 parabolic shape up to 0.002 strain and the constant up
Q.52. What tests are done to ensure the acceptance to limit state of 0.0035.
criteria of concrete? The tensile strength of concrete is ignored.
1. Burst modulus test after 7 days • The stress in reinforcement is derived from the
2. Compression strength test after 7 days representative stress- strain curve for the type of steel
used. For design purpose a partial safety factor of 1.15
3. Compression strength test after 28 days
is used. Hence maximum stress in steel is limited to
Choose the correct answer from the options given f/1.15 0.87 x fy.
below:
• The maximum strain in steel at failure shall not be less
(a) 1 and 3 (b) 1 and 2 (c) 1, 2 and 3 (d) only 3 than fy/1.15E, + 0.002.
Ans : (d) Compressive strength
• Three test specimens shall be made for each sample
for testing at 28 days.
• In all the cases 28 days' strength shall alone be the
criterion for acceptance or rejection of the concrete.
• The test results of the sample shall be the average of
the strength of three specimens.
• The individual variation should not be more than +15
percent of the average. If more, the test results of the
sample are invalid.
Q.53. The number of specimens taken for testing
concrete depend upon
(a) the grade of concrete
(b) estimated strength of concrete Q.55. In a beam of 300 mm wide and 500 mm
(c) the amount of concrete used in work effective depth, the value of the design shearing force
(d) all of the above at the critical section is 270 kN. The permissible shear
stress in concrete is 0.4 N/mm2 depending on the
Ans : (c) The minimum number of samples depend on
percentage of steel content. Which of the following
the quantity of concrete work. The details are given
statement is true?
according to Clause 15.2.2 of IS 456:2000
(a) The shear reinforcement will be designed for a
Quantity of
Number of samples shear force of 30 kN
Concrete in m3
1-5 1 (b) The shear reinforcement will be designed for a
6-15 2 shear force of 210 kN
16-30 3 (c) The shear reinforcement will be designed for a
31-50 4 shear force of 270 kN
4 + one additional samples for (d) The section must be redesigned
51 and above
each 50 m3 of work Ans : (b) Given,
Q.54. What is correct with respect to stress strain b = 300 mm d = 500 mm
curve of concrete in limit state method of design as Vu = 270 kN = 270,000 N τc = 0.4 MPa
per IS 456?
we know that
(a) The code allows the use of any appropriate curve
Design shear force (Vus) = Vu - τc × b x d
(b) The curve is parabolic up to a strain of 0.0035
Vus = 270,000 - (0.4 x 300 x 500)
(c) The curve is a straight line up to a strain of 0.002
Vus = 210,000 N = 210 kN
(d) None of these
UPSSSC JE 31/07/2016 71 Civil KI Goli
Q.56. What will be the total strain in Fe 415 grade Q.59. The mixture of different ingredients of cement is
steel corresponding to stress of 0.87fy? burnt at:
(a) 0.0035 (b) 0.0038 (c) 0.002 (d) 0.004 (a) 1000°C (b) 1200°C (c) 1400°C (d) 1600°C
Ans : (b) Given Ans : (c) The mixture of different ingredients of cement
Grade of steel = Fe415 ⇒ fy = 415 N/mm2 is burnt about 1400° to 1500° Celsius where they get
Esteel = 2.0 × 105 N/mm2 fused together and are known as clinkers. The size of
𝟒𝟏𝟓 the clinkers varies from 3 mm to 20 mm.
Strain in Fe415 is = 𝟎. 𝟎𝟎𝟐 + = 𝟎. 𝟎𝟎𝟑𝟖𝟎
𝟏.𝟏𝟓×𝟐.𝟎×𝟏𝟎𝟓 Q.60. Vicat apparatus is used for-
Q.57. Polygonal rings are used for lateral (a) Fineness test (b) Consistency test
reinforcement in the columns. What is the maximum (c) Soundness test (d) None of the above
number of sides this polygonal ring can have?
Ans : (b) Standard consistency test: This test
(a) 1 (b) 10 (c) 8 (d) 12
determines the percentage of water required to make
Ans : (c) Transverse reinforcing bars are provided in workable cement paste.
forms of circular rings, polygonal links (lateral ties) with • Vicat's apparatus is used to perform this test.
internal angles not exceeding 135o or helical • Temperature during test = 27 ± 2° C
reinforcement.
• Relative humidity = 90%
• The transverse reinforcing bars are provided to ensure
• As per Vicat's test 'the percentage of water added to
that every longitudinal bar nearest to the compression
the cement at which the needle can not penetrate 5-7
face has effective lateral support against buckling.
mm from bottom of the mould is called consistency.
Clause 26.5.3.2 stipulates the guidelines of the
• For OPC consistency is around 30%
arrangement of transverse reinforcement. The salient
points are: • In order to make a cement paste of normal
consistency the percentage of water varies from 25 to
35%.

❖ EARTHQUAKE
Q.61. Point area from which the seismic waves
develop is-
(a) epicentre (b) focus
(c) focal depth (d) None of these
• Transverse reinforcement shall only go round corner
Ans : (b) The point within the earth's crust where an
and alternate bars if the longitudinal bars are not
earthquake originates is called the focus. It is also
spaced more than 75 mm on either side
referred to as seismic focus. It generally lies within the
• The maximum number of sides this polygonal ring can depth of 60 kilometers in the earth crust.
have is 8

❖ BUILDING MATERIAL
Q.58. The minimum compressive strength of 43 grade
OPC cement is-
(a) 43 N/m2 (b) 43 kg/m2
2
(c) 43 N/mm (d) 43 kg/mm2
Ans : (c) 43-grade cement stands for an Ordinary Q.62. According to IS 1893-1962 how many seismic
Portland Cement (OPC) having a minimum compressive zone’s are there-
strength of 43 MPa or 43 N/mm² or 430 kg/cm² after 28 (a) Five (b) Four (c) Six (d) Three
days.
Ans : (a) Earthquake-prone areas of the country have
Grade of Cement 28 days been identified on the basis of scientific inputs relating
compressive strength to seismicity, earthquakes occurred in the past and
tectonic setup of the region.
33 grade 33 MPa • According to IS 1893-1962, India has been grouped
43 grade 43 MPa into five seismic zones.
53 grade 53 MPa • The Geological Survey of India (GSI) first published
the seismic zoning map of the country in the year 1935.

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Work Percentage of estimate • It is the total floor area minus the circulation area,
Water supply and 8 verandahs, corridors, passages, staircase, lifts, entrance
sanitary arrangements hall, etc., and minus other non-useable areas as
Electrical installation 8 sanitary accommodations, air conditioning room, etc.
Contingencies and 3-5 Circulation area:
unforeseen items 1. Vertical circulation area of the building is the area or
Work charge 1.5-2 space occupied by staircases, lifts, and the entrance
establishments halls adjacent to them which are required for the
Q.68. One micron is equal to- vertical movement of the users of the building.
(a) 10-6 mm (b) 10-8 mm • It may be taken as (4 to 5) % of the plinth area of the
-3
(c) 10 mm (d) 10-10 mm building.
2. Horizontal circulation area of a building in the area of
Ans : (c) One micron length = 1 micrometer
the verandahs, passages, corridors, balconies, porches,
= 1 µm = 1 × 10-6 m
etc. which is required for the horizontal movement of
We know, the users of the building.
1 m = 1000 mm • It may be taken as (10 to 15)% of the plinth area of
𝟏𝟎𝟎𝟎 𝐦𝐦
∴ 𝟏 × 𝟏𝟎−𝟔 𝐦 = × 𝟏𝟎−𝟔 𝐦 = 𝟏𝟎−𝟑 𝐦𝐦 the building.
𝟏𝐦
Q.70. Unit of measurement for payment of door and
A few common metric prefixes used are as below:
window's shutter is _______
Metric
Symbol Value (a) Cubic meter (b) Meter
Prefix 2
peta P 1015 (c) Meter (d) None of these
tera T 1012 Ans : (c) As the unit of measurement of Doors and
giga G 109 windows is done in square meters, Hence its payment is
mega M 106 done in per square Meters only.
kilo k 103 But the Doors and windows fittings are measured in
hecto h 102 Numbers and its payment is done in per Numbers only.
deka da 101
deci d 10-1 ❖ F.M.
centi c 10-2 Q.71. Unit of water discharge is-
milli m 10-3 (a) m/sec (b) m2/sec
micro µ 10-6 (c) m3/sec (d) None of these
nano n 10-9 Ans : (c) A discharge is a measure of the quantity of any
Angstrom Å 10-10 fluid flow over unit time. The quantity may be either
Pico P 10-12 volume or mass.
Q.69. The area between the walls is called Discharge is denoted as Q which is measured in m 3/sec.
(a) carpet area (b) circulation area Q.72. A 90° degree triangular notch has a discharge
(c) plinth area (d) floor area coefficient of 0.6. If the formula for discharge from the
Ans : (d) Different types of areas are as follows: notch is Q = K × Hn, then what will be the value of K?
Floor area: The floor area of the building is the total Take g = 9.8 m/sec2 and √𝟏𝟎 = 𝟑. 𝟐
area of the floor in between walls and consists of the (a) 1.2 (b) 1.4
floor of all rooms verandas, corridors, etc. (c) 1.8 (d) Data is insufficient
• The floor area is equal to the plinth area minus the
Ans : (b) Given, Cd = 0.6, θ = 90°
area occupied by walls.
Formula for discharge from the notch is
Plinth area: It is the covered built-up area measured at
the floor level of any storey or at the floor level of the Q = K x H …………(i)
basement. Discharge over a triangular weir or notch is given by:
𝟖 𝜽
• The plinth area is also called a built-up area and is the Q= 𝑪𝒅 𝒕𝒂𝒏 √𝟐𝒈𝑯𝟓/𝟐
𝟏𝟓 𝟐
entire area occupied by the building including internal 𝟖 𝟗𝟎
= × 𝟎. 𝟔 × 𝒕𝒂𝒏 √𝟐 × 𝟗. 𝟖𝑯𝟓/𝟐
and external walls. It is generally 10-20% more than the 𝟏𝟓 𝟐
carpet area. = 1.417 x H 5/2………….. (ii)
Carpet area: It is a useful area or liveable area or Comparing both equation : K = 1.417
lettable area.

UPSSSC JE 31/07/2016 74 Civil KI Goli

moisture by evaporation is known as shrinkage of ↓

Environment
concrete. Minimum
Minimum
• For a given humidity and temperature, the total Cement
Exposure Conditions Grade of
shrinkage of concrete is most influenced by the total Quantity
concrete
amount of water present in the concrete at the time (kg/m3)
of mixing and, by the cement content.
1. Concrete surfaces
• Shrinkage of concrete is directly proportional to the
protected against
cement content and as the grade of concrete
weather or
increases the specific surface area increases thus

Mild
aggressive M20 300
shrinkage also increases
conditions, except
Q.98. With regard to the characteristic strength of those situated In
concrete, consider the following statement: coastal areas.
“Test strength of samples means the average strength 2. Concrete exposed to
of X specimen. The strength of a sample should not condensation and

Moderate
vary by more than ± Y% of the average strength. What rain
will be the values of X and Y here? M25 300
Concrete
(a) 5, 15 respectively (b) 5, 5 respectively continuously under
(c) 3, 5 respectively (d) 3, 15 respectively water
Ans : (d) Characteristic strength of concrete (fck): The 3. Concrete completely
compressive strength of concrete is given in terms of immersed in

Severe
the characteristic compressive strength of 150 mm size seawater
M30 320
cubes tested after 28 days Concrete exposed to
• The characteristics strength is defined as the strength a coastal
of concrete below which not more than 5% of the test environment
results are expected to fall 4. Concrete in contact
very Severe

• The test results of the sample shall be the average of with or buried under
the strength of three specimens. The individual M35 340
aggressive
variation should not be more than ±15 percent of the subsoil/groundwater.
average. If more, the test results of the sample are
5. The surface of the
invalid.
member in the tidal
Flexural strength of concrete: - The theoretical zone.
maximum flexural tensile stress occurring in the Members in direct
extreme fibers of RC beam, which causes cracking is M40 360
contact with
Extreme

referred to as the modulus of rupture (fCr). liquid/solid

Q.99. Match the following: aggressive
Exposure chemicals.
Concrete environment
Condition ❖ HIGHWAY
A. Concrete continuously Q.100. On which road pattern were the formulas of
1. Very Severe
underwater the Nagpur Road Plan based?
B. Concrete completely 2. Extreme (a) rectangular and block pattern (b) star and grid
immersed in seawater Severe pattern
(c) star and block pattern (d) star and circular pattern
C. Concrete surface exposed to
3. Severe Ans : (b) Nagpur road plan or First 20 year road plan
sea water spray
1943-63 : - In this plan road network in the country
D. surface of concrete blocks in was classified in to:
4. Moderate
tidal zones (i) National Highways (ii) State highways
(a) A – 3, B – 2, C – 1, D - 4 (iii) Major District roads (iv) Other District roads
(b) A – 4, B – 2, C – 1, D - 3 (v) Village roads
(c) A – 3, B – 2, C – 4, D - 1 In the Nagpur conference, recommendations were
made for the geometric standards of roads, bridge
(d) A – 4, B – 3, C – 1, D - 2
specifications and highway organisations and two plan
Ans : (d) As per IS 456:2000 formulae were finalised. This two plan formulae
assumed star and Grid pattern of road network.

UPSSSC JE 31/07/2016 80 Civil KI Goli

Q.101. What is true about Lucknow road plan? Ans : (c) Two types of friction generally considered in
1. Availability of expressway for fast movement should highway design are as follows:
be 2000 km 1. Longitudinal/Rolling friction: It acts in the rolling
2. The total length of the national highway should be direction of tyres of the vehicle.
66000 km • Its value varies between 0.35 to 0.40.
3. The total length of the state highway should be
Recommended value of longitudinal friction by Indian
145000 km. Road Congress are as follows:
Choose the correct answer from the options given
below :
(a) only 1 (b) 1 and 3
(c) 2 and 3 (d) 1, 2 and 3 2. Lateral/Side/Traverse friction:
Ans : (a) Highway Development • It acts to counter the centrifugal force acting on the
tyre in the lateral direction.
1st 20 Year 2nd 20 Year 3rd 20 Year
Specification • Recommended value of lateral friction coefficient by
Plan Plan Plan
the Indian Road Congress is 0.15.
Nagpur Bombay Lucknow • However, in the case of expressways, it can be
Name of Plan
road plan road plan Road Plan reduced even to 0.1 with a design speed of 120 kmph.
1943-1963 Q.103. For the geometric design of highways, which
Duration Completed 1961-1981 1981-2001 speed has been taken as the design speed?
-1961 (a) 85th percentile speed (b) 90th percentile speed
th
(c) 95 percentile speed (d) 98th percentile speed
Target 16 km/100 32 km/100 82 km/100
Density km² km² km² Ans : (d) The 98th percentile speed is adopted for
geometric design of highway.
27 Lakhs
Total Target 532700 km 10 Lakhs km
km
2702000 km
Achieved NH-66000
709122 km 1502697 km
Target km, SH-
145000 km
Star and Square and
Road Pattern -
Grid Block
Provision of No
1600 km 3200 km
Expressway provision
• Primary
Expressway Geometrical design speed = 98th percentile speed
1. NH NH Safe speed limit = 85th percentile speed
2. SH NH, SH, • Secondary Minimum speed limit = 15th percentile speed
Road MDR, ODR,
3. MDR SH Q.104. What is the value of maximum deviation in
Classification VR and
4. ODR Expressway MDR CBR for values up to 10% of CBR?
5. VR • Tertiary (a) 1.0% (b) 3.0% (c) 5.0% (d) 10.0%
ODR Ans : (b) Following are the some of the important
VR points recommended by the IRC for CBR method of
design (IRC: 37-1970)
Development
15% 5% Nil (a) CBR tests should be performed on remolded soils in
allowances
the laboratory. In situ tests are not recommended for
Q.102. What is the range of value of the coefficient of design purposes.
friction fixed by the Indian Road Congress for (b) For the design of new roads, the subgrade soil
calculating the stopping sight distance? sample should be compacted to OMC to proctor
(a) 0.40 to 0.25 (b) 0.40 to 0.15 density whenever suitable compaction equipment is
(c) 0.40 to 0.35 (d) 0.25 to 0.15 available to achieve that density in the field, otherwise

UPSSSC JE 31/07/2016 81 Civil KI Goli

the soil sample may be compacted to the dry density One part of the rainwater flows on the ground or road
expected to be achieved in the field. surface, while the other part percolates into the ground
(c) In new constructions the CBR test samples may be and reaches the groundwater table, raising its level.
soaked in water for four days period before testing. The subgrade soil above the groundwater table may
(d) At least 3 samples should be tested on each type of raise through the soil pores due to the phenomenon of
soil at the same density and moisture content. capillarity.
(e) If the maximum variation in CBR values of the three Requirements of a poor Drainage System:
specimens exceeds the specified limits, the design CBR • Reduction of the strength of the pavement.
average should be average of at least six samples. • Spoiling the pavement surface by the formation of
(f) The specified limits of maximum variation in CBR are pot-holes and ruts.
3% for values up to 10% of CBR & 10 % for values of 30 • Seeping of surface water through the pavement
- 60 %. layers, shoulders, and sides into the subgrade.
Q.105. Which statement is true for the spacing of • Reduction in the bearing power of the subgrade
contraction joints in cement concrete pavement? through the continued presence of water.
(a) In unreinforced slabs, the maximum spacing of Moisture changes in the subgrade occur due to the
contraction joint can be 4.5 m percolation of rainwater and seepage flow, as also due
to the phenomenon of capillary rise.
(b) In reinforced slab, the spacing may be 13 m for 15
cm thick slab with steel reinforcement of 3.8 kg/cm 2 The aim of subsurface drainage is to keep the
groundwater table (GWT) sufficiently below the level of
(c) In reinforced slab, the spacing may be 14 m for 20
the subgrade – at least 1.2 m.
cm thick slab with steel reinforcement of 2.7 kg/cm 2
If the soil is relatively less permeable, longitudinal as
(d) All of the above
well as transverse drains may be needed to lower the
Ans : (d) Contraction Joints: Contraction joints are groundwater table
provided in cement concrete road to allow for
Q.107. What should be the actual ruling gradient if the
contraction of the slab due to falling in the slab
ruling gradient of 1 in 200 is accompanied by a curve
temperature.
of 3° at B. G. section?
• As per IS 6509: 1985, Table 1, Cl: 5.3.1.2, is shown
(a) 0.5% (b) 0.44% (c) 0.38% (d) 0.41%
below in the tabulated form:
Type Spacing for Spacing for Ans : (c) Given,
Width of Degree of curve D = 3°
of Expansion Contraction
Slab (m)
Slab joint Joint Ruling Gradient = 1/200 × 100 = 0.5 %
0.25 51 m 17 m So for 3° curve, compensation,
0.20 46 m 14 m = 0.04 × 3 = 0.12%
RCC
0.16 36 m 13 m Compensated gradient = 0.5 - 0.12 = 0.38 %
0.10 30 m 7.5
0.20 and ❖ RAILWAY
36 4.5 m
above Q.108. In India, what is the ratio between the weight
PCC
0.15 27 4.5 m of the rail and the axle weight of the engine (both in
0.10 27 4.5 m tons)?
Q.106. Which statement is true for underground (a) 1 : 510 (b) 1 : 410 (c) 3 (d) 1 : 210
drainage in highways? Ans : (a) A rail is designated by its weight per unit
1) Ground water surface should be at least 1.5 m length. The various important factors to be considered
below sub grade in deciding the weight of train are:
2) Transverse drains are also used in low permeability 1. Speed of Train 2. The axle load
soils
3. The Gauge of Track 4. Type of rails
3) Transverse drains are used in all types of soils
5. Spacing of sleepers.
Select the correct answer from the codes given below
For India the ratio of the weight of rail in tonnes to the
(a) only 1 (b) 1 & 3
locomotive axle load in tonnes is fixed at 1: 510.
(c) only 2 (d) 1 & 2
Q.109. How many sleepers will be required for 1280
Ans : (c) Highway Drainage: Highway drainage consists meter long broad gauge track? Take the sleeper
of removing or controlling surface water and subsurface density as M+5 and the length of the rail as 12.8 m.
water away from the road surface and the subgrade (All terms are used in their normal sense)
supporting it.
(a) 1800 (b) 1000 (c) 1200 (d) 1500
UPSSSC JE 31/07/2016 82 Civil KI Goli
Ans : (a) Given: Sleeper Density = M + 5 • It is a web member of a roof truss or brace frame
No of sleepers in 12.8 m rail length = 12.8 + 5 subjected to light axial compressive loads.
= 17.8 ≈ 18 sleepers. Principal strut: It is a top chord member of a roof truss
or a truss girder subjected to heavy axial compression
So, the total number of sleepers (T) is given by
Boom: It is a principle compression member of a crane.
𝐓𝐨𝐭𝐚𝐥 𝐫𝐚𝐢𝐥 𝐥𝐞𝐧𝐠𝐭𝐡 × 𝐍𝐨 𝐨𝐟 𝐬𝐥𝐞𝐞𝐩𝐞𝐫 𝐩𝐞𝐫 𝐮𝐧𝐢𝐭 𝐫𝐚𝐢𝐥 𝐥𝐞𝐧𝐠𝐭𝐡
𝐓=
𝐫𝐚𝐢𝐥 𝐥𝐞𝐧𝐠𝐭𝐡
𝟏𝟐𝟖𝟎×𝟏𝟖
𝐓= = 𝟏𝟖𝟎𝟎
𝟏𝟐.𝟖

Q.110. What is the limit of reduction in cant for a

speed of 120 kmph on broad gauge track?
(a) 7.6 cm (b) 8.6 cm (c) 8.0 cm (d) 10.0 cm
Ans : (d) Given,
Speed = 120 kmph > 100 kmph
∴ Cant Deficiency for BG Track for V > 100 kmph is 100
mm = 10 cm
**This Question repeat 2 times in paper
Maximum cant deficiency : -
Q.113. The ratio of the effective length of column and
For speed the minimum radius of gyration of its cross-sectional
For speed >100
Gauge up to 100 area is known-
kmph
kmph
(a) Buckling factor (b) Slenderness ratio
B.G 76 mm 100 mm
M.G 50 mm - (c) Crippling ratio (d) None of the above
N.G 40 mm - Ans : (b) Slenderness ratio: It is the ratio of the
Q.111. Which is the most commonly used crossing on effective length of a column and the least radius of
broad gauge track in India? gyration of its cross-section.
(a) 1 in 6 and 1 in 8 (b) 1 in 12 and 1 in 16 • It is used extensively for finding out the design load as
(c) 1 in 8.5 and 1 in 16 (d) 1 in 8.5 and 1 in 12 well as in classifying various columns in
Ans : (d) Crossing: A crossing is a device introduced at short/intermediate/long.
the junction where two rails cross each other to permit L
the wheel flange of a railway vehicle to pass from one
λ = k eff
min
track to another. 𝐈
Most commonly used crossing on broad gauge track is 1 𝐤=√
𝐀
in 8.5 and 1 in 12. Leffective = Effective length of the column
❖ STEEL kmin = minimum radius of gyration
Q.112. A long vertical member subjected to an axial I = Moment of Inertia
compressive load is called- A = Area of the cross-section
(a) A column (b) A strut Q.114. A column is known as medium size, if its
(c) A tie (d) All of the above slenderness ratio lies between-
Ans : (a) Compression members: The members (a) 20-32 (b) 32-120
subjected to axial compressive force in a direction (c) 120-160 (d) 160-180
parallel to the longitudinal axis are called compression Ans : (b) Slenderness ratio: It is the ratio of the
members. effective length of a column and the least radius of
Following are some examples of compression gyration of its cross-section.
members 𝐋
𝛌 = 𝐤 𝐞𝐟𝐟
Column or post or stanchion: It is a vertical member of 𝐦𝐢𝐧

a steel building frame that is used to support a floor

girder or floor subjected to heavy axial compressive Type of Column Slenderness Ratio
loads. Short Less than 32
Strut: A structural member subjected to an axial Medium 32-120
compressive force is called a strut. As per definition
Long Greater than 120
strut may be horizontal, inclined or even vertical. A
vertical strut is called a column.
UPSSSC JE 31/07/2016 83 Civil KI Goli
Q.115. The equivalent length of a column having both Side = 10 cm, Le = 500 cm
ends fixed is 𝐛𝟒
𝟏𝟎 𝟒
𝐈 = = = 𝟖𝟑𝟑. 𝟑𝟑𝟑 𝐜𝐦𝟐
(a) L/2 (b) L/4 𝟏𝟐 𝟏𝟐
(c) 2L (d) L 𝐀 = 𝐛. 𝐛 = 𝟏𝟎 × 𝟏𝟎 = 𝟏𝟎𝟎 𝐜𝐦𝟐
Ans : (a) Various column conditions and their
We know that,
respective effective length are tabulated below:
Condition (distance between Effective Slenderness Ratio:
supports is L) length 𝐋𝐞𝐟𝐟
Both ends Hinged (pinned) L 𝛌=
𝐤𝐦𝐢𝐧
One end hinged, another fixed L/√𝟐
𝐈
Both ends fixed L/2 The radius of gyration = 𝑲 = √ 𝐦𝐢𝐧 𝐢𝐦𝐮𝐦
𝐀
One end fixed and another end is 2L
free 𝟖𝟑𝟑.𝟑𝟑𝟑
𝑲 = √ = 𝟐. 𝟖𝟖
Q.116. The equivalent length of a column having 𝟏𝟎𝟎
length ‘l’ fixed at one end and free at the other end is 𝟓𝟎𝟎
(a) 0.5 l (b) 0.7 l So, 𝝀= = 𝟏𝟕𝟑. 𝟐
𝟐.𝟖𝟖
(c) l (d) 2l
Q.119. A twisted bar has about _______ more yield
Ans : (d) Various column conditions and their respective stress than ordinary mild steel bar.
effective length are tabulated below: (a) 10% (b) 20%
Condition (distance between Effective (c) 35% (d) 50%
supports is L) length
Both ends Hinged (pinned) L Ans : (d) Twisted bars: Bond may also be improved by
using twisted bars which are cold worked.
One end hinged, another fixed L/√𝟐
Both ends fixed L/2 • Cold twisted bars in­crease the yield stress by about
One end fixed and another end is 50% and thus save the reinforcing material by 33%.
free
2L • The increase in stresses is due to the path of twist.
The most suitable pitch is 9 cm to 12 cm.
Q.117. The slenderness ration of a vertical column of
• If the pitch exceeds 12 cm, the stress reduces.
square cross-section of 2.5 cm sides and 300 cm
• In twisted bars, hooks are generally not provided and
effective length, is
thus saving in steel and labor.
(a) 200 (b) 360 (c) 240 (d) 416
• On account of increased bond strength, the
Ans : (d) Given, struc­tures do not experience cracks. This is why
Side = 2.5 cm, Le = 300 cm twisted reinforce­ment is very useful for water
𝐛𝟒 𝟐.𝟓𝟒 retaining structures.
𝐈 = = 𝟑. 𝟐𝟓𝟓 𝐜𝐦𝟒
𝟏𝟐 𝟏𝟐
Q.120. For the same depth, the heavier section is-
𝐀 = 𝐛. 𝐛 = 𝟐. 𝟓 × 𝟐. 𝟓 = 𝟔. 𝟐𝟓 𝐜𝐦𝟐
(a) ISLB (b) ISMB
We know that,
(c) ISHB (d) ISWB
Slenderness Ratio:
Ans : (c) Beam, column, channel, and angle sections
𝐋𝐞𝐟𝐟
𝛌= are classified as follows:
𝐤𝐦𝐢𝐧
Beams :
𝐈
The radius of gyration = 𝑲 = √ 𝐦𝐢𝐧 𝐢𝐦𝐮𝐦 • Indian Standard junior beams (ISJB)
𝐀
• Indian Standard light weight beams (ISLB)
𝟑.𝟐𝟓𝟓
𝑲 = √
𝟔.𝟐𝟓
= 𝟎. 𝟕𝟐𝟏 • Indian Standard medium weight beams (ISMB)
𝟑𝟎𝟎
• Indian Standard wide flange beams (ISWB)
So, 𝝀 = = 𝟒𝟏𝟔 Columns/Heavy Weight Beams :
𝟎.𝟕𝟐𝟏

Q.118. The slenderness ratio of a vertical column of • Indian Standard column sections (ISSC)
square cross section of 10 cm side and 500 cm length • Indian Standard heavyweight beam (ISHB)
is Channels :
(a) 117.2 (b) 17.3 • Indian Standard junior channels (ISJC)
(c) 173.2 (d) 137.2 • Indian Standard lightweight channels (ISLC)
Ans : (c) Given, • Indian Standard medium weight channels (ISMC)

UPSSSC JE 31/07/2016 84 Civil KI Goli

• Indian Standard medium weight parallel flange End conditions Le Buckling load
channels (ISMCP) 𝝅𝟐 𝑬𝑰
Both ends hinged Le = L 𝑷𝒃 = 𝟐
Angles : 𝑳
• Indian Standard equal leg angles (ISA) 𝑳 𝟒𝝅𝟐 𝑬𝑰
Both ends fixed 𝑳𝒆 = 𝑷𝒃 =
• Indian Standard 𝟐 𝑳𝟐
One end fixed and 𝝅𝟐 𝑬𝑰
Q.121. If the slenderness ratio of a column is less than Le = 2L 𝑷𝒃 =
another end is free 𝟒𝑳𝟐
80 then column is- One end fixed and 𝑳 𝟐𝝅𝟐 𝑬𝑰
(a) Long column (b) Short column another end is 𝑳𝒆 = 𝑷𝒃 =
(c) Medium column (d) None of these hinged √𝟐 𝑳𝟐
Ans : (c) Slenderness ratio: Slenderness ratio is the Note : - Question NO . 9 And 112 repeat same 2 times
ratio of the length of a column and the least radius of in exam.
gyration of its cross-section. It is used extensively for
finding out the design load as well as in classifying
various columns in short/intermediate/long.
𝐥𝐞
𝛌=𝐤
𝐦𝐢𝐧
Type of Column Slenderness Ratio
Short Less than 32
Medium 32-120
Long Greater than 120
Q.122. Rankine Gordon's formula used to find out
buckling load of column-
(a) Long column (b) Medium column
(c) Short column
(d) All of the above
Ans : (d) Rankine-Gordon's Formula : Rankine-
Gordon's empirical formula is known as Rankine's
formula.
• The Rankine formula is therefore valid for extreme
values of 1/k (means for short and long columns). It is
also found to be fairly accurate for the intermediate
values in the range under consideration.
• Crushing load is responsible for the failure of short
columns.
• Buckling load is responsible for the failure of long
columns.
Q.123. The equivalent length of the column when
both the ends are fixed is ________.
(a) L (b) L/2
(c) L/4 (d) 2L
Ans : (b) Buckling load: the load at which column FOR Revise Complete Civil ENGINEERING All
buckle is termed as Buckling load.
Important data , formulas read ‘Civil Booster
Buckling load is given by:
𝛑𝟐 𝐄𝐈
Handbook ‘
𝐏𝐛 =
𝐋𝟐𝐞 Book Available On Shops, Online.
where
E = Young’s modulus of elasticity
Imin = Minimum moment of inertia
Le = Effective length

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 2. Infrared Wave Instruments: In this instrument
UPSSSC JE Civil 2016 (Exam amplitude modulated infrared waves are used. Prism
Date:19/12/2021) reflectors are used at the end of line to be measured.
• These instruments are light and economical and can be
❖ SURVEY mounted on theodolite With these instruments accuracy
Q.01. Fill in the blank with the correct option: ___ is achieved is ± 10 mm.
used to measure horizontal and vertical angles. • The range of these instruments is up to 3 km.
(a) Prism (b) Compass 3. Visible Light Wave Instruments: These instruments
(c) Theodolite (d) Plum bob rely on propagation of modulated light waves. This type
of instrument was first developed in Sweden and was
Ans : (c) Theodolite: It is an instrument used for named as Geodimeter.
measuring horizontal & vertical angles in surveying.
• During night its range is up to 2.5 km while in day its
• A simple circular curve can be set out using two range is up to 3 km.
theodolite method.
• In this method, only angular measurements are taken Q.03. Velocity correction is applied to account
for__________
with the help of two theodolite instruments.
• Size of theodolite is defined by lower graduated circle. (a) atmospheric errors
Non-transit theodolite can not be rotated through (b) instrument operation errors
180o in vertical plane. (c) standardization errors arising in the instrument
(d) satellite error
Ans: (a) Sources of Errors in Global Positioning
System(GPS):
1. Satellite Errors: Slight inaccuracies in timekeeping by
the satellites can cause errors in calculating positions.
Satellites drift slightly from their predicted orbits which
contributes to errors.
2. Satellite Orbits: Slight shifts of the orbits are possible
due to gravitation forces.
Sun and moon have a weak influence on the orbits.
The resulting error is not more than 2 m.
3. Multi-path error: As the GPS signal finally arrives at
the earth’s surface, it may be reflected by local
obstructions before it gets to the receiver’s antenna. This
is called a multi-path error.
Q.02. Fill in the blank(s) with the correct answer:
4. Atmospheric Effects : The GPS signals have to travel
_______ EDMs can be used up to a range of 100 km
through charged particles and water vapors in the
while ______ type is used with a range up to 3 km.
atmosphere which delays its transmission. Since the
(a) Infrared, microwave (b) Infrared, visible range atmosphere varies at different places and at different
(c) Visible range, ultraviolet (d) Microwave, infrared times, it is not possible to accurately compensate for the
Ans : (d) Types of EDM Instruments: Depending on the delays that occur.
types of carrier wave employed, EDM instruments can • While radio signals travel with the velocity of light in
be classified under the following three heads : outer space, their propagation in the ionosphere and
1. Microwave Instruments troposphere is slower. In the ionosphere (consisting of
2. Infrared Instruments layers) at a height of 80 – 400 km, a large number of
electrons and positively charged ions are formed by the
3. Visible Light Instruments
ionizing force of the sun.
1. Microwave Instruments: These instruments make use
• The layers refract the electromagnetic waves from the
of microwaves. Such instruments were invented as early
satellites, resulting in an elongated runtime of the
as 1950 in South Africa by Dr. T.L. Wadley and named
signals.
them as Tellurometers.
5. Receiver Error: Since the receivers are also not
• The instrument needs only 12 to 24 V batteries. Hence
perfect, they can introduce their own errors which
they are light and highly portable.
usually occur from their clocks or internal noise.
• Tellurometers can be used in day as well as in night.
The range of these instruments is up to 100 km.
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Q.10. Fill in the blank with the correct option: b. Surveyor's 2. Length is 100 feet and has 100 links.
Prismatic compass is mainly used ___________ . chain Brass tags are installed at every 10 feet.
(a) to measure horizontal angle to measure vertical c. Engineer's 3. Length is 33 feet and has 18 links.
angle to measure slope distance. chain
(b) to set out right angle from a survey line.
d. Revenue 4. Widely used and available in lengths
(c) to find the bearing of traversing to find included chain of 5, 10, 20 and 30 meters.
angles.
(a) a - 1, b - 4, c - 2, d - 3
(d) to mark the position of stations to sight the stations
to range straight lines. (b) a - 4, b - 1, c - 3, d - 2
(c) a - 4, b - 1, c - 2, d - 3
Ans : (c) A prismatic compass is a navigation and
surveying instrument which is extensively used to find (d) a - 2, b - 1, c - 4, d - 3
out the bearing of the traversing. It is the most Ans : (c) Chain surveying: It is the branch of surveying
convenient and portable form of magnetic compass that in which the distances are measured with a chain and
can either be used as a hand instrument or fitted on a tape and the operation is called chaining. All the
tripod. distances measured should be horizontal.
Difference between Prismatic compass and Surveyor Chains :- The chains used in surveying are generally of
compass : - the following types.
Prismatic Compass Surveyor Compass 1. Revenue chain: The revenue chain is 33 ft long.
• It is used by the Revenue Department of India.
Smaller in size (85 - 110 Bigger (circular box of
mm diameter) size 150 mm diameter) • The number of links is 16, each link being 2 12 ft.
2. Gunter's chain: It is also known as the surveyor’s
Edge bar type magnetic
Broad Needle with chain.
needle is used, which
aluminium ring is used • It comes in standard 66 ft. this chain consists of 100
also acts as index
links, each link being 0.66 ft or 7.92 inches.
First Object is sighted &
Sighting of the object & • The length 66 ft is selected because it is convenient in
then reading of bearing
reading of the bearing land measurements.
is taken by moving
done simultaneously 3. Engineer's chain:-This chain comes in 100 ft length.
around the box
• Its consists of 100 links each link being 1ft long.
Temporary Adjustment
Temporary Adjustment • At every 10 links, a brass ring or tags are provided for
Centering , Levelling,
Centering & Levelling indication of 10 links. Readings are taken in feet and
focusing
decimal.
Tripod is not essential Tripod is essential
4. Metric chain: Metric chains are the most commonly
Graduation is inverted used chain in India. These types of chains come in many
Graduation is erect as
because we have to see lengths such as 5, 10, 20, and 30 meters.
we can see from the top
them through prism • The most commonly used is a 20m chain. Tallies are
Readings are in W.C.B., provided at every 2m of the chain for quick reading.
having 0° at South, 90° It has 0° at N & S, 90° at Every link of this type of chain is 0.2m. The total length
at West, 180° at North & E & W (Q.B. System) of the chain is marked on the brass handle at the ends.
270° at East • Metric chain - 30 m (150 links) or 20 m (100 links)
Q.12. Consider the following errors associated with
EDM:
(1) Precise centering at the master and slave station.
(2) Error due to variation in meteorological conditions
like temperature, pressure, humidity.
(3) Errors associated with pointing/sighting of a
reflector.
Q.11. Match the following in the context to surveying Which of the aforementioned errors can be classified as
chains. atmospheric errors?
a. Metric 1. Length is 66 feet and 100 links Widely (a) Only 1 and 2 (b) Only 2
chain used for land measurement as 10 (c) Only 1 and 3 (d) Only 3
square chains make 1 acre.

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Ans: (b) Errors in Electronic Distance Measurement • The relative humidity in the air is inversely
Instruments are as follows proportional to the strength of radio waves.
Error in EDM Details • The propagation of electromagnetic waves through
the atmosphere has an important influence on the
Inaccuracy in initial setups of EDMs performance of numerous communication systems
and the reflectors over the like radar, mobile communication, Radio, Wi-Fi, etc.
preferred stations • Wind directly does not affect radio waves but it does
Personal
Instrument and reflector affect the refraction (bending of the waves) capabilities
Errors
measurements going wrong of the medium which is what leads to aberration in
Atmospheric pressures and radio propagation.
temperature determination errors Hence, electromagnetic waves are least affected by
wind speed.
Calibration errors
Q.15. Identify the technique generally preferred for
Instrumental Chances of getting maladjusted time
contouring rough country where ordinary levelling is
error to time generating frequent errors tedious and chaining is slow and inaccurate.
Errors are shown by the reflectors (a) Compass surveying (b) Tachometry surveying
Atmospheric variations in (c) Plane table surveying (d) Levelling
temperature, pressure as well as Ans : (b) Tacheometric surveying: - Tacheometric is a
Natural humidity. Microwave EDM branch of surveying in which horizontal and vertical
Errors instruments are more susceptible to distances are determined by taking angular observation
these. with an instrument known as a tachometer.
Multiple refractions of the signals. • Tacheometric surveying is adopted in rough in rough
Q.13. Which of the following options is not a variation and difficult terrain where direct leveling and chaining
of magnetic declination? are either not possible or very tedious.
(a) Annual variation (b) Diurnal variation • Tacheometry is mainly used while preparing contour
plans and traversing and is also suitable for
(c) Seasonal variation (d) Irregular variation
hydrographic surveys, location surveys of roads,
Ans : (c) Variation in magnetic declination : railways, etc.
Variation of magnetic declination • The main principle of Tacheometry is based on the
Secular Annual Diurnal Irregular basic principle of isosceles triangle.
variation variation variation variation Q.16. In a Tachometer equipped. with an anallatic lens
Due to Due to Due to Due to which is used to measure distances using principles of
over a revolution rotation magnetic angular Surveying without using a chain or measuring
long of earth of earth disturbances tape, the additive and multiplying constants
period of around the about its in earth’s respectively are:
time due sun own axis. magnetic (a) 0, 100 (b) 100, 0
to gradual (yearly) It’s more field (c) 0, 0 (d) 100, 100
shift in at pole, in
Ans : (a) Major Characteristics of a Tacheometer:
earth’s day &
magnetic summer (i) The multiplying constant (k) of the tacheometer is
field time. usually a round figure and mostly it is 100.
(ii) The additive constant (C) of the tacheometer is kept
Q.14. Electromagnetic waves are least affected by
very small and mostly it is kept zero.
which of the following parameters?
(iii) An additional convex lens is provided between
(a) Air temperature (b) Atmospheric pressure
eyepiece and object glass at a fixed distance and known
(c) Vapour pressure (d) Wind speed as anallactic lens to make additive constant zero.
Ans : (d) Electromagnetic waves : - Electromagnetic (iv) The magnifying power of the eyepiece is kept high to
waves are formed when an electric field comes in make the staff graduations clearly visible even at a large
contact with a magnetic field. They are hence known as distance.
‘electromagnetic’ waves. (v) The aperture of the objective is kept usually 35 mm
Factors affecting the Electromagnetic waves:- to 45 mm to make the image sharp.
• The temperature is inversely proportional to the
strength of radio waves.

UPSSSC JE 19/12/2021 90 Civil KI Goli

 (a) Dioptra (b) Theodolite
Q.17. In a closed traverse with five sides, the error (c) Graphometer (d) Geodimeter
found from the fore bearing and back bearing of the Ans : (b) The Given image represents Electronic Digital
last line is + 1°. The correction to the third line will be: Theodolite.
(a) 0°24' (b) 0°36' (c) 0°48' (d) 0°12'
Ans : (b) Given data
Number of sides of closed traverse(n) = 5
Error (e) = +1o
We know 1o = 60’
Now the correction on sides (or lines) of closed traverse
as follows :
𝟔𝟎’ Theodolite: - A Theodolite is a measuring instrument
Correction on each sides = = 12’
𝟓 used to measure the horizontal and vertical angles are
Correction on second side = 12’ × 2 = 24’ determined with great precision.
Correction on third side = 12’ × 3 = 36’ Q.20. Two points A and B are 1530 m apart across a
Q.18. From the given options, identify the tool used by river. The reciprocal levels measured are
engineers to get constant levelling each time. Level at Readings on (in m)
(a) Tribrach (b) Auto Level
A B
(c) Theodolite (d) Abney Level
A 2.165 3.810
Ans : (b) Auto level: - Auto level is an optical instrument
used to establish or verify points in the same horizontal B 0.910 2.355
plane. It is used in surveying and building with a vertical The true difference in level between A and B would be
staff to measure height differences and to transfer, (a) 1.255 m (b) 1.355 m (c) 1.545 m (d) 1.645 m
measure and set heights.
Ans : (c) In reciprocal leveling, the difference in level
• An auto level is a tool that's being used by contractors,
between A and B is given below
builders, land surveyors, and engineers so they can get
(𝐡𝐛 −𝐡𝐚 )+(𝐡′𝐛 −𝐡′𝐚 )
constant and consistent leveling every single time. H=
𝟐
• It's easy to use and can be set up quickly, which can Where,
save both time and money. H = True difference of R.L. between A and B.
Tribrach:- A tribrach is an attachment plate used to When instrument is set up at A
attach a surveying instrument, for example, a theodolite,
ha = Reading on staff at A
total station, GNSS antenna, or target to a tripod.
hb = Reading on staff at B
Theodolite:- it is used to measure horizontal and vertical
angles. When instrument is set up at B
In its modern form, it consists of a telescope mounted to h'a = Reading on staff at A
swivel both horizontally and vertically. h'b = Reading on staff at B
Abney level:- Abney level is a kind of clinometer Given
consisting of a sighting tube, spirit level, and graduated ha = 2.165 m hb = 3.810 m
scale. h'a = 0.910 m h'b = 2.355 m
Q.19. Identify the instrument shown in the image (𝟑.𝟖𝟏𝟎−𝟐.𝟏𝟔𝟓)+(𝟐.𝟑𝟓𝟓−𝟎.𝟗𝟏𝟎)
H= = 1.545 m
𝟐
below:

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Q.21. The triangulation technique is classified as Now the given ordinates are even so we find the area in
primary secondary and tertiary triangulation two-component.
depending on the area it covers. The correct order of A1 = by Simpson's rule from O1 to O3
length of the baseline is given as: A2 = Area of last strip by trapezoidal rule.
(a) Primary > Secondary > Tertiary The traverse area by using Simpson's rule is given from
(b) Secondary > Primary > Tertiary O1 to O3.
(c) Tertiary> Secondary> Primary 𝐝
𝐀 = [(𝐎𝟏 + 𝐎𝟑 ) + 𝟒(𝐎𝟐 )]
𝟑
(d) No relation can be established b/w baseline length 𝟏𝟓
𝐀= (𝟏. 𝟓 + 𝟑. 𝟑) + 𝟒(𝟐. 𝟒)] = 𝟕𝟐 𝐦𝟐
and triangulation class 𝟑
The traverse area by using a trapezoidal rule for the last
Ans : (a) TRIANGULATION :- It is process of measuring
strip is given from O3 to O4.
the angles of a chain or a network or triangles formed by
𝒅
stations marked on the surface of the earth. 𝑨𝟐 = (𝑶𝟑 + 𝑶𝟒 )
𝟐
𝟏𝟓
• The side of the first triangle, whose length is 𝑨𝟐 = (𝟑. 𝟑 + 𝟒. 𝟒) = 𝟔𝟕. 𝟕𝟓 𝒎𝟐
𝟐
predetermined is called the base line and vertices of the
Hence, the traverse area by using Simpson's rule given is
individual triangles are known as triangulation stations.
A = A1 + A2 = 72 + 67.75 = 129.76 m2.
CLASSIFICATION OF TRIANGULATIONS : -
On the basis of quality, accuracy and purpose, Q.23. Which of the following options correctly define
triangulations are classified as : the subsidiary station?
1. Primary Triangulation:- It is the highest grade of (a) Station where observations are not made, but the
triangulation system which is employed either for the angles at the station are used in triangulation series
determination of the shape and size of the earth’s (b) Control points of triangulation network
surface or for providing precise planimetric control (c) Points to provide additional rays to intersected
points on which subsidiary triangulations are connected. points
2. Secondary Triangulation:- It is triangulation system (d) Points close to main stations to avoid intervening
which is employed to connect two primary series and obstructions
thus to provide control points closer together than those Ans : (c) Terms Used in Large Survey Area : -
of primary triangulation.
3. Tertiary Triangulation:- It is the triangulation system
which is employed to provide control points between
stations of primary and second order series.
Triangulation of
Requirement IInd IIIrd
Ist order
order order
30- 1.5-
Length of sides of triangles 8-70km
160km 10km
Length of base line 5-20km 2-5km 0.5-3km

Average triangular closure 1" 3" 6”

(i) Main station : It is a point in chain survey where two
Maximum triangular closure 3" 8" 12”
sides of triangle meet. In above figure, A, B, C, D are main
Limiting strength of figure 25 40 50 stations.
Sets of observations with (ii) Base line : It is the longest survey line from which
16 8 3 direction of all other survey lines are fixed. In figure, AB
theodolite
is the base line. Colby apparatus is used for the
So correct order is Primary > Secondary > Tertiary
measurement of Base lines.
Q.22. Consider an area having ordinates 1.5 m, 2.4 m,
(iii) Main survey line: Chain line joining two main survey
3.3 m, 4.4 m. What would be the traverse area by using
stations. In figure, AD, DB, BC, CA, BA are main survey
Simpson's rule given that d = 15 m?
lines.
(a) 580.5 sq m (b) 129.75 sq m
(iv) Tie station or subsidiary station: Station on survey
(c) 193.5 sq m (d) 200 sq m line joining main stations. It is helpful for locating interior
Ans : (b) Given data, O1 = 1.5 m, O2 = 2.4 m, O3 = 3.3 m, details. In figure, e, f, g, h, i are the tie stations.
O4 = 4.4 m, and d = 15 m

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(v) Proof line or check line: It is provided to check the (a) The reflected energy generated by electromagnetic
accuracy of the field work. In figure, Ck & Dj are check waves
lines (b) Total time taken by electromagnetic waves in
(vi) Offsets: These are lateral measurement made w.r.t. travelling the distance
chain line which may be oblique or perpendicular in (c) The change in the frequency of the electromagnetic
nature. waves
(vii) Chainage: It is the distance measured along the (d) The phase difference between the transmitted and
main survey line in direction of progress of work. the reflected electromagnetic waves
Q.24. Which of the following is typically determined by Ans : (d) Electronic distance measurement (EDM):-
a subtense bar?
Electronic distance measurement (EDM) is a method of
(a) Long distance of up to a km determining the length between two points using
(b) Short distance of up to 200 m electromagnetic waves.
(c) Very short distances of up to a few cm • EDM instruments are highly reliable and convenient
(d) Very long distances of up to 100 km pieces of surveying equipment and can be used to
Ans : (b) Subtense bar is a bar made up of a fixed length measure distances of up to 100 km.
with targets at the ends and with a spirit level at the • They measure the length between two points, using
center for leveling the bar on it stand or tripod. phase changes that occur as electromagnetic energy
waves travels from one end of the line to the other
Subtense bar can measure horizontal and vertical
end.
distance in places where chaining is impossible because
• Modern EDM instruments work on the principle of
of undulations and rough topography. But it can measure
measuring the phase difference between the
short distance up to 200m.
transmitted and the reflected electromagnetic waves
Q.25. Which among the following is not used for setting • They have accuracy up to 1 in 105.
out right angles?
Q.27. Identify the surveying instrument shown in the
(a) Offset rods (b) Cross staff
image:
(c) Optical square (d) Prism square
Ans : (a) Following are the list of instrument used in
surveying along with their usage:
Survey
Purpose
Instrument
To mark survey stations and end
Pegs
points of survey lines on the ground.
To mark the position of the end of (a) Pegs (b) Offset rods (c) Ranging rod (d) Arrows
Arrows
the chain or tape on the ground.
Ans : (d) Arrows :- These are also known as chaining pins
Ranging For locating a number of points on a and are used to mark the end of each chain during the
Rods long survey line. chaining process.
These rods are also similar to • These are made of hardened and tempered steel wire
Offset Rods
ranging rods and they are 3 m long. 4 mm in diameter.
Prism Advanced version of Optical square • The length of arrow is kept 400 mm. These are pointed
Square and used to set out right angles at one end whereas a circular ring is formed at its other
end to facilitate carrying from one station to another.
To indicate whether the line is
Plumb Bob • As the arrows are placed in the ground after every
vertical or not.
chain length, the number of arrows held by the follower
Optical indicates the number of chains that have been
To set out right angles
Square measured.
It is used for setting out right angles
and French cross-staff is an
Cross-staff
advanced version of cross-staff and
can set out 45-degree angles also.
Q.26. Modern EDM instruments work on the principle
of measuring which of the following?

UPSSSC JE 19/12/2021 93 Civil KI Goli

Q.28. Fill in the blank with the correct option: Eye Valve: The eye valve is fine silt with an eye hole in
Instrument error that is independent of distance the bottom to bend the object out of the silt.
measured in EDM is called _________ . Graduation Circle: This is an aluminum graduated ring
(a) Scale Error (b) Zero Error labeled of 0° to 360° to calculate all potential line
(c) Cyclic Error (d) Non-scale Error bearings and connected with a magnetic needle.
Lifting Pin and Lifting Lever: Below the viewing vane. Lift
Ans : (c) The three distinct systematic errors that may
pin pressed while the sight vane is folded. The magnetic
occur in EDM instruments are:
needle was raised from the pivot point with the aid of
Cyclic Error (or Short Periodic Error): Cyclic error is the lever raise.
caused by the non-linearity in amplitude modulation of
Prism: Prism is being used to recognize the graduation
the carrier wave and phase measurement.
on the ring and also to read the same reading by
This cyclic error varies across the modulated wavelength. compass. It’s positioned in the opposite direction of the
For an instrument in good adjustment, this error is object vane.
normally small.
Spring Break: In order to dampen the vibration of the
However, its presence must be determined as an needle once obtaining a measurement to putting it to
indication of the instrument's adjustment. resting easily.
Cyclic error is usually sinusoidal in nature with a Object Vane: The object vane holding horsehair or black
wavelength equal to the unit length of the EDM. thin wire to see the object in line with the sight of the
Additive Constant (correction for zero or index error): object.
All distances measured by a particular EDM/reflector Reflecting Mirror: It is used to obtain a picture of an
combination are subject to a constant error. It is caused object positioned higher or lower surface of the device
by three factors: electrical delays, geometric detours, when bisecting. .
and eccentricities in the EDM; Pivot: The pivot is offered at the middle of the compass
Scale Error: Scale error is proportional to the length of which supports a loosely attached magnetic needle.
the line measured and is caused by: Magnetic Needle: The magnetic needle is the core of the
the drift in frequency of the quartz crystal oscillator in device. That needle determines the angle of the line
the instrument; errors in the measured temperature, from either the magnetic meridian since the needle is
pressure, and humidity which affect the velocity of the often pointing to the north-south pole at the opposite
propagation; and non-homogeneous edges of the needle while freely placed on some
emission/reception patterns from the emitting and support.
receiving diodes (phase inhomogeneities). The scale
Q.30. Consider the given statements, identify the ones
frequency can be checked by:
that cannot be executed with the help of a typical
direct comparison against frequency testing apparatus; theodolite in surveying.
and measurement over a base of known distance.
i) Measuring horizontal and vertical angles
Q.29. Match the following in context of the parts of ii) Measuring horizontal distances only
prismatic compass.
iii) Determining the area of ground
a. Graduation 1. To damp the oscillation of the needle
iv) Locating points on lines
circle before taking a reading and to bring
it to rest quickly the light (a) ii and iv (b) i and iv (c) i and iii (d) ii and iii
b. Prism 2. An aluminium graduated ring Ans : (d) Theodolite is a precision instrument for
marked with 0 degrees to 360 measuring angles in the horizontal and vertical planes.
degrees to measure all possible Modern theodolite consists of a movable telescope
bearing of lines mounted within two perpendicular axes the horizontal
c. Object vane 3. Used to read graduation on the ring or trunnion axis, and the vertical axis.
and to take an exact reading by
compass
d. Spring break 4. Carrying horsehair or black thin wire
to sight the object in line with the
object sight
(a) a - 2, b - 3, c - 4, d - 1 (b) a - 2, b - 3, c - 1, d - 4
(c) a - 3, b - 2, c - 4, d - 1 (d) a - 2, b - 4, c - 3, d - 1
Ans : (a) Parts of Prismatic Compass :

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1. The size of a theodolite is defined by the diameter of Q.32. Which of the following options correctly depicts
the graduated circle of the lower plate. For example, a the deflection angle in traverse?
25 cm theodolite means the diameter of the graduated (a) 180° - included angle (b) 360° - included angle
circle of the lower plate is 25 cm. The size of the (c) Included angle + 180° (d) Included angle + 360°
theodolite varies from 8-25 cm.
Ans : (c) Deflection angle method : - Deflection angle
2. The process of establishing intermediate points, on a
method is used for open traverse, in which traverse line
given straight line whose ends are intervisible is done
make small deflection like railway, canal, sewer lines,
with a theodolite is called lining in.
pipe lines etc.
3. Removal of parallax may be achieved by refocusing
Deflection angle = 180° - included angle
the eyepiece and the objective.
4. Line of collimation: It is an imaginary line passing
through the intersection of the crosshairs at the
diaphragm and the optical center of the object-glass and
its continuation.
5. Axis of Bubble tube: It is an imaginary line tangential
to the longitudinal curve of the bubble tube at its
midpoint.
Q.33. Two points P and Q located on a map have the
6. Axis of a telescope: This axis is an imaginary passing following coordinates:
through the optical center of the object-glass and the
optical center of the eye-piece. Latitude +30 m 10 m

Q.31. Which unit in total station is used for the Departure +40 m -10 m
computation of data collected? Determine the length of PQ.
(a) Data collector (b) EDM (a) 60 m (b) 50 m (c) 34.89 m (d) 53.85 m
(c) Storage system (d) Microprocessor Ans : (d) Given data
Ans : (d) A total station, also known as an electronic Two points P and Q located on a map have the following
tachometer, is an optical instrument. coordinates:
●It is a combination of an electronic theodolite for
measuring horizontal and vertical angles, an
electromagnetic distance measurement (EDM) device
The length of the line PQ is
for measurement of slop distances, and onboard
software to convert the raw observed data to three- = √𝟒𝟎 − (−𝟏𝟎)]𝟐 + (𝟑𝟎 − 𝟏𝟎)]𝟐 = √𝟐𝟗𝟎𝟎 = 𝟓𝟑. 𝟖𝟓 𝐦
dimensional coordinates. Q.34. Fill in the blank with the correct option:
●Thus with a total station, one may determine the actual
…………………..is a preliminary inspection of the area to
positions (X, Y, and Z or northing, easting, and elevation
survey in theodolite traversing.
of surveyed points or the position of the instrument
from known points in absolute terms. Further, the EDM (a) Selection and marking of stations
that measures the slop distance can calculate and (b) Measurement of traverse legs
display horizontal distance and difference in level. (c) Computation
●This is accomplished with the help of a microprocessor (d) Reconnaissance
normally working concentric with the telescope Ans (d) Traversing: The traversing in which the traverse
eyepiece and generally housed in a casting that forms legs are measured by direct chaining on the ground and
part of the telescope. the traverse angle at every traverse station is measured
with a theodolite is known as theodolite traversing.
The procedure of Traverse:
The following steps are required to establish a traverse:
1. Reconnaissance:- Reconnaissance is the preliminary
field inspection of the entire area to be surveyed.
• During reconnaissance surveyor goes to the field,
checks the entire area, and decides the best plan of
working. He has to see the suitability and intervisibility
of traverse stations. Also, he has to decide which method
of traversing will be employed and accordingly. the
instruments and accessories that may be necessary.

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2. Selection of Traverse Stations:- The basic principle of for measurement of slop distances, and onboard
a survey, i.e.. working from whole to the part should be software to convert the raw observed data to three-
adopted. A minimum number of stations should be dimensional coordinates.
selected. Q.37. The magnetic bearing of a line AB is S25°E. If the
• The length of traverse lines is as long as possible to declination is 5° West, then what is the true bearing?
reduce the time and centering effect of the stations. It is (a) S30°E (b) N30°W (c) S20°E (d) N20°W
better to select stations on a level and firm ground. After
Ans : (a) Given,
the selection of stations, they are marked with pegs.
Magnetic bearing of a line AB = S25°E = 155°
3. Linear and Angular Measurements:- Distances
between stations are measured by tape or chain, or Magnetic Declination = 5° West
tacheometric method or electric distance measuring True bearing = Magnetic bearing ± Magnetic Declination
(EDM) instruments like geodimeter. True bearing = 155° - 5° ( western declination “-“ )
• Angular measurements are done by prismatic compass = 150°
or theodolite. = S30°E
4. Plotting of Traverse Survey
Q.35. The whole circle bearings of lines AB and BC are
20°15' and 100°30'. What is the included angle ABC
between the lines AB and BC?
(a) 90°15' (b) 99°45' (c) 229°45' (d) 100°15'
Ans: (b) We know the relationship between FB and BB
Back bearing (BB) = Fore bearing (FB) ± 180°.
+ve sign is used if FB is less than 180° and -ve sign is used
if FB is more than 180°.
Q.38. Match the following in the context of the leveling
staff.
Self-reading The smallest division is 5mm. Its length is
a. 1.
staff about 3 meters.

It has a moving target against which

b. Solid Staff 2. readings are taken by the staff man. The
sliding target comes along with the varnier.

The readings are always taken from the

Telescopic telescope hence appear inverted.
Given. c. 3.
Staff Therefore, readings are taken from
F.B.AB = 20°15' downwards.
F.BBC = 100°30'
It has three telescopic lengths of 1.5, 1.5
Included angle (ABC) = 20°15'+ (180° - 100°30') d. Target staff 4. and 2 meters when stretched out fully.
Included angle(ABC) = 99°45' Normally, its total length is 5 meters.
Q.36. Total Station can directly be used to calculate: (a) a-1, b-3, c-4, d-2 (b) a-3, b-1, c-4, d-2
(1) Horizontal angles (c) a-3, b-1, c-2, d-4 (d) a-3, b-4, c-1, d-2
(2) Vertical angles Ans (a) Leveling Staff: Along with a level, a leveling staff
(3) pH of the water is also required for leveling. The leveling staff is a
(4) Sloping distances rectangular rod having graduations. The staff is provided
(5) Level of PM2.5 in air with a metal shoe at its bottom to resist wear and tear.
Which of the above statements are correct? The foot of the shoe represents zero reading.
(a) 1, 2 and 5 only (b) 1, 2, and 3 only Leveling staff may be divided into two groups:
(c) 2, 3 and 4 only (d) 1, 2 and 4 only A. Self reading staff: This staff reading is directly read by
the instrument man through a telescope.
Ans : (d) A total station, also known as an electronic
• In a metric system staff, one-meter length is divided
tachometer, is an optical instrument.
into 200 subdivisions, each of uniform thickness of 5
● It is a combination of an electronic theodolite for mm. All divisions are marked with black on a white
measuring horizontal and vertical angles, an background. Meters and decimeter are written in red
electromagnetic distance measurement (EDM) device colour.
UPSSSC JE 19/12/2021 96 Civil KI Goli
The portion of brick cut along its length is referred as • The use of silica sand as industrial abrasives, water
closer. Brick cut in such a manner that its one long face filtration, glass-making, ceramics & refractories,
remains uncut. construction materials, etc.
•It is of the following types: Queen Closer (half), Queen Examples of Colour Variation in Quartz Crystals:
Closer (Quarter), King closer, Mitred closer and Bevelled Rose Quartz:-
Closer.
Bat : The potion of brick cut along its width is called bat.
• It is of the following types: Half bat, Quarter bat,
Bevelled Bat.
Q.47. Which among the following is a constituent of a Smoky Quartz:
good brick?
(a) Alumina (b) Steel
(c) Gold (d) Copper
Ans : (a) Good Brick Earth Constituents table: Amethyst:
Properties Excess
Constituents Percentage
Impart causes
Shrinking
Alumina 20 - 30 Plasticity
and warping
Uniform Makes brick Citrine:
Silica 50 - 60
shape Brittle
Prevent Causes the
Lime 4-5 cracking and bricks to
shrinkage melts.
Rock Crystal:
Makes
Imparts red-
Oxide of bricks dark
5-6 brown color
Iron blue and
and strength
blackish.
Yellow tint
Decay of Q.49. In the process of metallizing, which metal is used
Magnesia <1 color to
brick. in present practice of painting of bridges in Indian
brick.
Railways?
Q.48. Consider the given image, identify the one that is
(a) Zinc (b) Bronze (c) Copper (d) Iron
composed mostly of quartz sand.
Ans : (a) Metallizing is a process that involves coating a
surface with a thin layer of metal to protect it from
corrosion, wear, and other forms of degradation. The
process is commonly used on steel and other metal
(a) (b) surfaces, such as bridges, to extend their lifespan and
improve their durability.
• Zinc is commonly used in the metallizing process for
painting bridges in Indian Railways due to its excellent
corrosion resistance properties.
Q.50. Fill in the blank with the correct option:
The process to remove rust, mill scale (oxidization) and
old paints with the use of high velocity impact of
abrasives (sand & grit) is known as _________ .
(c) (d) (a) Surface washing (b) Blast cleaning
Ans : (b) Quartz sand: It is also known as silica sand, (c) Manual hand cleaning (d) Flame cleaning
white sand, or industrial sand.
Ans : (b) Surface preparation method for Painting :
• It is made up of two main elements: silica and oxygen.
Surface washing: Surface washing refers to the removal
Specifically, silica sand is made up of silicon dioxide
of dirt, dust, varnish, and other debris from the surface
(SiO2).
of an object.

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(A) Fat lime: - This lime slakes vigorously due to which Ans : (b) Scraper : - It is a machine for moving earth over
its volume increases by 2 to 2.5 times than its original short distances (up to about two miles) over relatively
volume. Hence it is referred to as fat lime. smooth areas.
• It possesses a perfectly white color, hence also termed • Either self-propelled or towed, it consists of a wagon
as white lime. with a gate having a bladed bottom.
• It is obtained by the calcination of limestone of • The blade scrapes up the earth as the wagon pushes
approximate 95% purity, hence it is also referred to as forward and forces the excavated material into the
pure lime, rich lime, or highly caustic lime. wagon.
• Fat lime is mostly used for plastering and Jackhammer drill: - A Jackhammer drill is a pneumatic or
whitewashing. electro-mechanical tool that combines a hammer
(B) Hydraulic lime: It is also referred to as water lime as directly with a chisel.
it is capable of setting in water and in damp locations. • Jackhammers are used to demolish old concrete,
• It is obtained by the calcination of limestone having remove pavement, and demolish many other surfaces in
purity in the range of 70 - 92%. projects.
• It is insoluble in water and possesses off-white color. Shuttle car: - A vehicle on rubber tires or continuous
• Its hardness is comparatively faster than fat lime, treads to transfer raw materials, such as coal and ore,
hence is used in engineering works where strength is from loading machines in trackless areas of a mine to the
required. main transportation system.
Ex: → Brick masonry and stone masonry. Shredder: - A shredder is a machine or equipment used
(C) Poor lime: It is also referred to as impure lime or lean for shredding. Shredding systems are used to reduce the
lime as it is obtained from the calcination of limestone size of a given material.
having a purity of less than 70%. Q.56. Fill in the blank with the correct option:
• This lime does not undergo slaking, hardness very As per IS10719557 & 1970 code specifications, bricks
slowly, possess muddy white color, hence is used in with compressive strength not less than __________
Engineering works of minor importance. Ex: Brickwork kg/cm2 are termed as First-class bricks.
around the foundation.
(a) 140 (b) 105
Q.54. In which of the following methods of grading of (c) 70 (d) 100
aggregates are the samples of fine and coarse
aggregates passed through nine standard sieves? Ans : (b) According to IS: 10719557-1970, following
classification has been made for bricks on the basis of
(a) By trail
compressive strength:
(b) By minimum voids method
(c) By finesse modules method
(d) By arbitrary standards Classification Compressive Strength

Ans:(c) Grading of aggregates : The particle size Grade A-A > 140 kg/cm2
distribution of an aggregate as determined by sieve
analysis is termed the gradation of aggregates. Grade A > 105 kg/cm2
Finesse modules method: The method can be used to
determine the fineness modulus of coarse aggregates, Grade B > 70 kg/cm2
fine aggregates, and all-in aggregates, i.e., mixed
aggregates. Grade C > 35 kg/cm2
• In this method, a convenient weight of the sample is
taken and sieved through a set of sieves one after • Grade A brick is also known as First Class Brick
another. • Grade B brick is also known as Second Class Brick
• For grading of coarse aggregates the sieve(nine sieves)
• Grade C brick is also known as Third Class Brick
is taken as 80 mm, 63 mm, 40 mm, 20 mm, 16 mm, 12.5
mm, 10 mm, 4.75 mm, and 2.36 mm. Q.57. You wish to check the quality of timber with the
• For grading of fine aggregates, the sieve(seven sieves) help of sound. Which of the following way should you
is taken as 10 mm, 4.75 mm, 2.36 mm, 1.18 mm, 600 use for such a scenario?
microns, 300 microns, and 150 microns. (a) Timber struck by hammer
Q.55. What of the following machines can be used for (b) Two timber pieces struck together
the top-soil removal? (c) Timber tapped by steel square
(a) Jackhammer drill (b) Scraper (d) Timber knocked by chisel
(c) Shuttle car (d) Shredder Ans : (b) Sound characteristic of timber:
UPSSSC JE 19/12/2021 101 Civil KI Goli
• The velocity of sound in timber (hardwood) is Q.59. Match the following classification of glasses with
approximately 12 times more than that in the air. their specific uses:
• A good timber should produce a clear ringing sound Used in the manufacturing of glass
Soda lime
when struck by each other. A dull heavy sound a. 1. articles which have to withstand high
glass
designates decayed timber. temperatures.
• This is the reason why many musical instruments are Used in the manufacturing of
Potash
made of wood. b. 2. artificial gems, electric bulbs, lenses,
lime glass
Prisms, etc.
Q.58. Match the following in the context of paints.
Used in the manufacturing of glass
a. Bleeding 1. It is the swelling of the Potash
c. 3. tubes, laboratory apparatus, plate
paint film generally caused lead glass
glass, window panels, etc.
by moisture or volatile
Common Used in the manufacturing of
substances d. 4.
glass medicine bottles
b. Bittiness 2. The diffusion of coloured (a) a - 1, b - 3, c - 2, d - 4 (b) a - 3, b - 1, c - 2, d - 4
material into the upper
(c) a - 3, b - 1, c - 4, d - 2 (d) a - 1, b - 3, c - 4, d – 2
coat from the under coat
Ans : (b) Different types of glasses, their constituents,
c. Blistering 3. The appearance of a and applications are as follows:
whitish substance on a Type Constituents Applications
surface of varnish or Potash lead Fusion of Silica, Electric glasses and
enamel. glass Potash, and Lead lenses
d. Blooming 4. A condition in which small (Flint glass) Cut glass Prisms
Soda-lime
particles project above the
glass
surfaces of the film of Fusion of Silica,
(Soda ash Doors
paint or varnish yielding a Lime, Soda, and
glass/ Windows
rough surface. Alumina
Window
(a) a - 4, b - 2, c - 1, d - 3 (b) a - 2, b - 4, c - 3, d - 1 glass)
Bottle glass Sodium silicate,
(c) a - 1, b - 4, c - 2, d - 3 (d) a - 2, b - 4, c - 1, d - 3 Household bottles
(Common Calcium silicate,
Medicine bottles
Ans : (d) The common defects in paints are: glass) and Iron silicate
Used in the
1. Blistering and Peeling: Blistering and peeling are
manufacturing of glass
defects in which swelling of the paint film occurs caused
Silicates of articles that have to
by the formation of an air bubble under the paint film Potash lime
potassium and withstand high
due to the presence of moisture or oil or grease matter. glass
calcium temperatures
If it is due to moisture then it is called peeling and if it is (hard glass)
Also used for Hollow
due to oil and grease matter then it is called blistering. equipment for
2. Fading: Fading is the discoloration of the paint surface laboratory
Phosphorus,
due to atmospheric agencies such as sunlight, and Optical glass Lenses
lead silicate
moisture.
Borosilicate Fusion of Silica,
3. Grinning: The visibility of the background due to bottles Borax, Lime, and Laboratory equipment
insuffici ent opacity of paint film even after the final coat (Pyrex glass) Feldspar
is called grinning. Q.60. Which among the following is an oxygenated
4. Chalking: Chalking is the formation of powder on the solvent used in the manufacturing of coatings?
painted surface. This is due to the use of insufficient oil (a) Esters (b) Nitroparaffins
in the primer. (c) Turpentine (d) Cresols
5. Flaking: The detachment of paint film from the Ans : (a) Solvent: - A substance (usually a liquid) that can
surface is called flaking. It occurs when the bond be used to dissolve other materials without chemically
between surface and paint film is poor. changing them.
6. Blooming: The formation of dull patches on a painted The chemical classification of a solvent is based on its
surface is called blooming It is due to improper chemical structure is as follows:
ventilation, weathering, defective paint, etc.
1. Hydrocarbon solvents:- They are classified into three
7. Bittiness: Bittiness is a condition in which small sub-groups based on the type of “carbon skeleton” of
particles project above their molecules, giving us the aliphatic, aromatic, and
paraffinic solvents families.

UPSSSC JE 19/12/2021 102 Civil KI Goli

Paint thinner is a common example of a hydrocarbon (C) Heating: Heating is most suitable for quarrying small,
solvent. thin, and regular blocks of stones from rocks, such as
2. Oxygenated solvents:- They are produced through granite and gneiss.
chemical reactions from olefins (derived from oil or • A heap of fuel is piled and Fred on the surface of the
natural gas), giving us the following sub-groups: rock in a small area.
alcohols, ketones, esters, ethers, glycol ethers, and glycol • The two consecutive lavers of the rocks separate
ether esters. because of the uneven expansion of the two layers.
3. Halogenated solvents :-They are solvents that contain • The loosened rock portions arc broken into pieces of
a halogen such as chlorine, bromine, or iodine. the desired size and are removed with the help of pick-
Q.61. Fill in the blank with the correct option: axes and crowbars.
_________ is most suitable for quarrying small, thin • Stone blocks so obtained are very suitable for coarse
and regular blocks of stones from ‘rocks, such as granite rubble masonry.
and gneiss.
(a) Wedging (b) Boring
(c) Heating (d) Channeling machines
Ans : (c) The methods of quarrying the stone are as
follows:
(A) Blasting: in this method, explosives are used to
convert rocks into small pieces of stones. (D) Wedging: This method is mainly used for the rock of
• This method is used when the stone to be excavated is sedimentary type, which is comparatively soft, such as
of very hard variety and has no cracks or fissures. sandstone, limestone, marble, slate, laterite.
• Moreover, if a stone is to be excavated on a very large • In this method, first of all naturally occurring cracks or
scale, the blasting method will have to be adopted. fissures are located in the rocks, to be excavated.
• After blasting, the excavated stone is sorted out into • The steel wedges or points are then driven with the
different sizes and categories. help of a hammer, in hammer fissures or cracks and
stones are detached.
• Explosives such as blasting powder, blasting cotton,
dynamite, and cordite are used. • The split-out blocks of stone can be converted into
marketable forms and supplied to users.

(E) Channeling Machines: In this method, the

(B) Digging or Excavating: Stones buried in the earth or
channeling machines are driven by steam, compressed
under loose overburden are excavated with pickaxes,
air, or electricity are used to make vertical or oblique
crowbars, chisels, hammers, etc.
grooves or channels on the rock mass and make rapidly
the grooves.
• This process of separation of stone from the rock mass
is almost invariably employed in the case of limestones,
marbles, and other soft sandstones.
• It is possible to separate very large blocks of stones
from the rocks by the application of this method.

UPSSSC JE 19/12/2021 103 Civil KI Goli

 ❖ BCME
Q.62. Fill in the blank with the correct option: Mansard
Truss is a combination of _________ and _________
trusses.
(a) Purlin roof, Queen post
(b) Sloped king post, Sloped queen post
(c) Strutting, King post
(d) King post, Queen post
Ans : (d) Mansard Trusses :- The Mansard Truss is a two-
storey truss with upper portion consisting of the king- Couple Close roof: This type is similar to a couple roof
post truss and the lower portion of queen-post truss. It except that the legs of the common rafters are closed by
is thus a combination of the king-post and queen-post a horizontal tie known as tie beam.
trusses. • This tie beam is connected at the feet of the common
• The Mansard Truss has two pitches. The upper pitch rafters of overturning.
(king-post truss) varies from 30° to 40° and the lower • The tie maybe a Piece of a wood or a steel rod in
pitch (queen-post truss) varies from 60o to 70°, tension.
• The use of Mansard truss results in the economy of • This type of roof can be used for a maximum span of
space and a room may be provided in the roof between 4.5 m.
the two queen-posts. However, this truss has become
obsolete at present mainly because of its odd and ugly
appearance and also due to the introduction of steel
trusses.

Q.64. Choose the option that best describes ‘Gable’.

(a) The triangular upper part of a wall formed at the
end of a pitched roof
(b) The angle formed at the intersection of two roof
slopes is known as hip
(c) The wooden pieces which are placed horizontally on
principal rafters to carry the common rafters
Q.63 Identify the type of roof from the image shown (d) The lower edge of a roof which are resting upon or
below: projecting beyond the supporting walls
Ans : (a) Elements of the pitched roof are as follows :
1. Ridge: - It is defined as the apex line of a sloping roof.
2. Eaves: - The lower edge of the inclined roof is called
Eaves. From eaves, the rainwater from the roof surface
drops down.
3. Hip: - It is the ridge formed by the intersection of two
(a) King post roof (b) Purlin roof sloping surfaces where the exterior angle is greater than
(c) Couple Close roof (d) Queen post roof 180°.
Ans : (c) Couple Roof: This is simple roof, having slope 4. Hipped end: - The slopped triangular surface formed
on both the sides. at the end of the roof.
• It consists of two rafters abutting against each other, 5. Verge: - The edge of a gable, running between the
with the help of a horizontal beam called ridge. eaves and ridge is known as a verge.
• The lower end of the rafter is nailed to the horizontal 6. Cleats: - These are short sections of wood or steel
beams called wall plates. placed of the walls. (angle iron) which is fixed on the principal rafters of
• The rafters are placed suitably and over the wall. trusses to support the purlins.
• The purlins or battens are laid to support the roof 7. Valley: - It is a reverse of a hip. It is formed by the
covering. This type of roof is used up to a span of 3.5 intersection of two roof surfaces, making an external
m. angle less than 180o.

UPSSSC JE 19/12/2021 104 Civil KI Goli

8. Gable: - It is a triangular portion of the roof wall at the Q.66. Specify the name of the roof shown in the image:
end of a ridged roof.

Figure: Elements of pitched roof

(a) Couple roof (b) Lean-to roof
Q.65. Specify the name of the brick used for the image
(c) Single roof (d) Collar beam roof
shown:
Ans : (b) Lean-to roof or Verandah Roof:- This is the
simplest type of pitched roof consisting of rafters sloping
on one side only.
• At their upper ends the rafters are nailed to the
wooden wall plate, placed on corbel which may be of
steel, stone of brick.
• At the lower end the rafters are notched and nailed to
(a) Splay header (b) Splay stretcher
the wooden post plate.
(c) Dog leg (d) Bevelled bat
• The post plate consists of a timber section running
Ans : (a) Splays: These are special moulded bricks which parallel to the wall and supported by posts (which may
are often used to form plinth. Its different types are Splay be of wood, brick or stone) at intermediate points.
stretcher (plinth stretcher) and splay header (plinth • Battens spaced at 15 cm, centre to centre are nailed to
header). the inclined after and upon this the covering material is
laid .
• This type of roof is considered suitable for maximum
span of 2.4 metres.
• Lean to roof is commonly used for covering verandahs,
Splay stretcher Splay header Use of Splay header
out houses, sheds etc.
Dogleg or angle: It is also special form of moulded bricks
which are used to ensure a satisfactory bond at quoins
which are at an angle other than right angle. The angle
and lengths of the faces forming the dogleg vary
according to requirements. These are preferred to
mitred closer.

Q.67. Identify the most accurate brick bond shown in

Bevelled closer: It is a form of king closer in which the the image:
whole length of the brick (i.e. stretcher face) is
chamfered or beveled in such a way that half-width is
maintained at one end and full width is maintained at
the other end.

(a) English bond

(b) One and a half brick Flemish bond
(c) One and a half brick English bond
(d) Flemish bond

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Ans: (c) Important types of bond in brick masonry : - Q.68. Identify the brick bond shown in the image:
Stretcher Bond: All the bricks are laid as stretchers or full
brick which is laid with its length parallel to face the wall.
• Used-in half brick thick walls.
Header Bond: All the bricks are laid as headers or full
brick which is laid with its width parallel to face the wall.
(a) Garden wall (b) Dutch
• The elevation of the wall laid in Header bond.
• Used for curved surfaces in work such as well lining or (c) Raking (d) Zigzag
well foundation etc. Ans : (c*)
English Bond:-Alternate courses of headers and
stretchers.
• Most common and popular bond.
• One of the strongest bonds but requires more facing
bricks than other bonds.
• It is costly as compared to other bond.
Flemish Bond:- Each course consists of a header and
structures alternately arranged.
• Uses a greater number of brickbats and hence it is
economical.
• It is of two types : Single Flemish and Double Flemish.
• Better in appearance than the English bond.
Facing :- This bond is used where the bricks of specific Raking Bond:- Bricks are laid at some inclination to the
thickness are to be used in the facing and backing of the face of the wall.
wall. •When walls are more than two brick thick becomes
• A header course is provided after several stretcher weaker in longitudinal strength as the headers being
courses. used in the interior of the wall to increase the transverse
strength
•This defect is removed by using a raking bond (rake
means inclination)
•Also used for laying bricks on the floor.
•Two types of Raking bond: Diagonal bond and
Herringbone bond.
*Question DELETED by exam commission.

❖ CT
Q.69. Fine aggregate can pass through I.S. sieve
of………………
(a) 13 mm (b) 10 mm (c) 4.75 mm (d) 15 mm
Ans : (c) Aggregates are the inert materials basically used
as fillers with binding material in the production of
mortar & concrete. They give body to the concrete &
occupy 70 to 80 % of volume of concrete.
• If aggregate remain on 4.75 mm sieve then they are
called coarse aggregates.
i.e. Coarse aggregate > 4.75 mm
• If aggregates pass through 4.745 mm grade and retain
on a 75-micron sieve then they are called fine
aggregates.
i.e. Fine aggregate < 4.75 mm
Note: For RCC work, the maximum size of aggregate is
𝟏 limited to 20 - 25 mm
The given diagram is 𝟏 brick wall of English bond.
𝟐

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Q.70. Calculate the water required for the concrete mix Design
M 40
(in Gallon) if the water cement ratio is 0.6 litres/kg and Mix
70 kg cement has been added. 1 US gallon = 3.78 litres Design
M 45
(a) 11.11 (b) 11.50 (c) 9.70 (d) 13.33 Mix
For pre-stressed concrete work
Ans : (a) Given data Design
M 50
Cement quantity = 70 kg Mix
𝐖𝐚𝐭𝐞𝐫 (𝐥𝐢𝐭𝐫𝐞𝐬) Design
𝐂𝐞𝐦𝐞𝐧𝐭 (𝐤𝐠)
= 0.6 litres/kg M 55
Mix
Now using the w/c ratio, find the quantity of water in M 60 Design For high compressive strength
liters and above Mix concrete work
𝐖𝐚𝐭𝐞𝐫 (𝐥𝐢𝐭𝐫𝐞𝐬)
= 0.6 litres/kg Q.72. Fill in the blank with the Correct option:
𝟕𝟎 (𝐤𝐠)
Water quantity = 42 liters Typically, heavy mortar should have a bulk density >
1 US gallon = 3.78 litres __________ .
Hence, the water required to the concrete mix (In (a) 800 kg/m3 (b) 1500 kg/m3
Gallon) is 42/3.78 = 11.11 Gallon (c) 1000 kg/m3 (d) 600 kg/m3
Q.71. Match the following ratios (cement : sand : Ans : (b) Classifications of Mortar on various factors –
aggregates) of concrete with their respective uses: (a) Type of material –
a: 1 : 1 : 2 1. For heavy walls, foundation footings, etc. • Gauged Mortar • Cement Mortar
2. General work of RCC such as stairs, beams, • Gypsum Mortar • Lime Mortar
b: 1 : 15 : 3
columns, etc. (b) Bulk density based –
c: 1 : 2 : 4 3. For water tanks, bridges, sewers, etc. • Light density (  < 15 kN/m3 )
4. For heavy loaded RCC columns and RCC • Heavy density (  > 15 kN/m3 )
d: 1 : 5 : 10
arches of long spans. (c) Special types of mortar –
(a) a-4, b-3, c-1, d-2 (b) a-4, b-1, c-2, d-3 Fire resistants, Sound absorbing, Light weight etc
(c) a-1, b-2, c-3, d-4 (d) a-4, b-3, c-2, d-1 Q.73. Fill in the blanks with an appropriate option:
Ans : (d) 1:1:2 → For heavily loaded RCC columns and The water cement ratio directly affects the strength of
RCC arches of long spans the concrete mix. The typical w/c ratio for different
1:1.5:3 → For water tanks, bridges, sewers, etc. grades of concrete varies between ________ to
_________ .
1:2:4 → General work of RCC such as stairs, beams,
columns, etc (a) 0.4 - 0.6 (b) 0.9 - 1.2
1:5:10 → For heavy walls, foundation footings, etc. (c) 1.2 - 1.5 (d) 1.5 - 2.0
According to IS 456-2000 : - The mixes of grades with Ans : (a) Water cement ratio is the ratio of weight of
their ratios are as follows : - water to the weightof cement. Its normal value lies
between 0.4 to 0.6 as per IS code 10262 (2009) for
Nominal
nominal mixes.
Strength Mix Uses of Mix
Design • However, maximum strength is derived at W/C = 0.4 at
which minimum capillary cavities are expected to form.
M5 1:5:10
For PCC work such as bedding • When it decreased to less than 0.4, there is improper
M 7.5 1:4:8
for footing, etc. consistency and resulting in honeycombed structure.
M10 1:3:6 • If W/C is more than 0.6, the increase in volume of
General work of RCC such as hydrated products will not be able to occupy the space
M15 1:2:4
stairs, beams, columns, etc already filled with water. Hence porosity increases and
For RCC work such as slab, strength decreases.
M20 1:1.5:3 beams, columns, footings, Q.74. Heavy loaded R.C.C columns and R.C.C arches of
water tank, etc.
long spans use concrete composing of cement, sand &
M25 1:1:2 aggregate in which of the following ratios?
For heavy loaded RCC work
Design (a) 1 : 2 : 4 (b) 1 : 4 : 8 (c) 1 : 5 : 10 (d) 1 : 6 : 12
M 30 such as foundations, footing
Mix
columns, beams, slabs, arches, Ans : (a) Heavy loaded R.C.C columns and R.C.C arches
Design etc. of long spans use concrete the ratio is taken as 1:2:4
M 35
Mix

UPSSSC JE 19/12/2021 107 Civil KI Goli

According to IS 456-2000 : The mixes of grades with their (a) a, b, c, d (b) d, c, b, a
ratios are as follows : (c) c, d, b, a (d) c, a, b, d
Nominal Ans: (d) Shuttering: Formwork or shuttering is a
Strength Mix Uses of Mix temporary mold-like arrangement of plates of different
Design shapes, sizes, and materials used for the construction of
M5 1:5:10 different concrete elements like columns, beams, walls,
For PCC work such as bedding for culverts, etc.
M 7.5 1:4:8
footing, etc. Formwork Stripping Time (When Ordinary Portland
M 10 1:3:6 Cement is used):
M15 1:2:4 For heavy loaded RCC work such as Type of Formwork Formwork Removal Time
slab, beams, arches, columns,
M 20 1:1.5:3 footings, etc. Sides of Walls, 24 hours to 48 hours (as per
M 25 1:1:2 Columns and engineer’s decision)
For heavy loaded RCC work such as
Vertical faces of the
Design foundations, footing columns,
M 35 beams, slabs, etc.
beam
Mix
Slabs (props left 3 days
Design
M 45 under)
Mix
Beam soffits (props 7 days
Design
M 50 For pre-stressed concrete work left under)
Mix
Removal of Props of Slabs : -
Design
M 55 (i) Slabs spanning 7 days
Mix
up to 4.5m
M 60
Design For high compressive strength
and (ii) Slabs spanning 14 days
Mix concrete work
above over 4.5m
Q.75. Fill in the blank with the correct option: Removal of props for beams and arches : -
Shuttering of beams spanning over 6 m should be (i) Span up to 6m 14 days
removed in _______
(ii) Span over 6m 21 days
(a) 10 days (b) 21 days (c) 5 days (d) 7 days
Hence, the ascending order of removal of shuttering is
Ans : (b) As per IS 456:2000, the minimum period
as follows : -
before striking formwork is as follows:
First - Shuttering forming the vertical faces of walls
Minimum Period Before
Type of Formwork Second - Shuttering forming soffit of slab
Striking Formwork
Third - Shuttering forming soffit of beam, girders
Vertical formwork to
16-24 hours (1 day) Fourth - Heavily loaded shuttering
columns, walls, beams
Soffit formwork to slabs 3 days Q.77. Consider a 12 inches high slump cone having 8
inches in diameter at the top. How many layers of
Soffit formwork to beams 7 days
concrete should you place to fill this cone?
Props to slabs (spanning up
7 days (a) 3 equal layers by height
to 4.5 m)
(b) 4 equal layers by volume
Props to slabs (spanning
14 days (c) 3 equal layers by volume
over 4.5 m)
Props to beams and arches
(d) 8 equal layers by height
14 days
(spanning up to 6 m) Ans : (a) Slump Test: - The Slump test result is a measure
Props to beams and arches of the behaviour of a compacted inverted cone of
21 days concrete under the action of gravity. It measures the
(spanning over 6 m)
consistency or the wetness of concrete which then gives
Q.76. Arrange the following Steps in ascending order of
an idea about the workability condition of the concrete
removal of shuttering:
mix.
a- Shuttering forming soffit of slab
The procedure of the Concrete Slump test: (As per IS
b- Shuttering forming soffit of beam, girders 1199: 2018)
c- Shuttering forming the vertical faces of walls • Preparation: The mold, typically a truncated cone with
d- Heavily loaded shuttering a base diameter of 200 mm, a top diameter of 100 mm,

UPSSSC JE 19/12/2021 108 Civil KI Goli

and a height of 300 mm, is cleaned and placed on a flat,
non-absorbent surface.
• Filling the Mold: The cone is filled with fresh concrete
in three equal layers. Each layer is tamped 25 times with
a steel rod to ensure uniform compaction and eliminate
air pockets.
• Lifting the Mold: Once the cone is filled and
Q.80. Which of the following grade of concrete falls
compacted, it is lifted vertically without any lateral or
under the category of standard or medium strength
torsional movement. This allows the concrete to settle
concrete?
or slump under its own weight.
(a) M 40 (b) M 15 (c) M 20 (d) M 65
• Measuring the Slump: The difference in height
between the top of the mold and the highest point of Ans : (a) Ordinary concrete grades include grades of
the slumped concrete is measured. This difference is M10, M15, and M20.
called the slump. •Standard grades include M25, M30, M35, M40, M45,
IS 456 has given the following values of slumps for M50, M55 and M60.
different workability of concrete: •High strength concrete grades include greater than
Degree of Workability Slump Value (mm) M60.
Extremely low 0 Q.81. Match the following types of formworks with
their minimum required striking period.
Very low 0-25
a. 16-24 1. Soffit formwork to slabs
Low 25-50
hours
Medium 50-100
b. 7 days 2. Props to beams and arches
High 100-500 spanning up to 6 m
Very High More than 150 c. 14 days 3. Vertical formwork to columns,
walls, and beams
d. 3 days 4. Props for slabs spanning up to 4.5
Q.78. Raw material for cement manufacturing is
m
quarried from which stone quarry among the
following? (a) a - 3, b - 2, c - 4, d - 1
(a) Granite (b) Marble (b) a - 4, b - 3, c - 2, d - 1
(c) Basalt (d) Limestone (c) a - 3, b - 4, c - 2, d - 1
Ans : (d) Cement basically constitutes of calcareous and (d) a - 3, b - 4, c - 1, d - 2
argillaceous materials. Ans : (c) As per IS 456:2000, the minimum period before
Classification Type striking formwork is as follows:
Minimum Period Before
Blast Furnace slag, Slate, Shale and Type of Formwork
Argillaceous Striking Formwork
Clay
Vertical formwork to columns,
16 - 24 h (1 Day)
Calcareous Lime stone, Marl, Cement Rock, Chalk walls, beams.

Q.79. To determine the compacting factor of concrete, Soffit formwork to slabs 3 days
which among the following formula is used during Soffit formwork to beams 7 days
standard tests ? Props to slabs
7 days
lf Weight of partially compacted concrete - A & Weight a. Spanning up to 4.5 m
of fully compacted concrete - B 14 days
b. Spanning over 4.5 m
(a) A × B (b) A / B (c) A + B (d) A - B Props to beams and arches
14 days
Ans : (b) The compacting factor test is a laboratory test a. Spanning up to 6 m
21 days
for measurement of workability of concrete . b. Spanning over 6 m
The compacting factor is defined as the ratio of the
❖ SOIL
weight of partially compacted concrete to the weight of
fully compacted concrete. Q.82. Which of the following foundations is adopted
𝐖𝐭.𝐨𝐟 𝐩𝐚𝐫𝐭𝐢𝐚𝐥𝐥𝐲 𝐜𝐨𝐦𝐩𝐚𝐜𝐭𝐞𝐝 𝐜𝐨𝐧𝐜𝐫𝐞𝐭𝐞 𝐀
where the loose soil extends to a great depth?
Compacting Factor = = (a) Raft foundation (b) Grillage footing
𝐖𝐭.𝐨𝐟 𝐟𝐮𝐥𝐥𝐲 𝐜𝐨𝐦𝐩𝐚𝐜𝐭𝐞𝐝 𝐜𝐨𝐧𝐜𝐫𝐞𝐭𝐞 𝐁
The relationship between workability and compacting (c) Inverted arch footing (d) Pile foundation
factor and slump is given below : -
UPSSSC JE 19/12/2021 109 Civil KI Goli
Ans : (d) Pile Foundations:- p = Stresses at mid-depth of the layer due to added
It is used in the following situations: loads
1. They are constructed where excessive settlement is to H = Total thickness of the layer
be eliminated and where the load is to be transferred Case-For a thick layer, the layer may be divided to the
through soft soil stratum. number of sub-layers, each of thickness H1 and the stress
2. When it is not economical to provide spread at mid-depth of each sub layer is pi.
foundations and hard soil is at a greater depth.
3. When it is very expensive to provide raft or grillage
foundations.
4. When heavy concentrated loads are to be taken up by
the foundations.
5. When the top soil is of compressible nature.
6. When there is a chance of construction of irrigation
canals in the nearby area.
7. In the case of bridges when the scouring is more in the
river bed.
8. In marshy places.
Total settlement(S) is given by
Q.83. A soil layer of depth H is divided into sub layer of
𝑺 = 𝜮𝒎𝒗𝒊 𝜟𝒑𝒊 𝑯𝒊
thickness ‘t' to ease out the calculation of settlement of
soil layer. You wish to calculate the final settlement by Q.84. Match the soil class with their respective
performing some operations on each thin layer. Which descriptions according to VOB/C, DIN 18300(DIN,
of the following option will help you to find the total 2009):
settlement of the layer? a. Class 7 1. Moderately difficult to excavate
(a) The difference of individual settlements of the b. Class 4 2. Easy to excavate rock and similar
various thin layers soils
(b) The square of individual settlements of the various c. Class 1 3. Hard to excavate rock
thin layers
d. Class 6 4. Topsoil
(c) The product of individual settlements of the various
thin layers (a) a - 1, b - 4, c - 1, d - 2 (b) a - 3, b - 1, c - 4, d - 2
(d) The sum of individual settlements of the various (c) a - 2, b - 4, c - 1, d - 3 (d) a - 2, b - 1, c - 4, d - 3
thin layers Ans : (b) Soil and rock classification (according to VOB –
Ans : (d) Settlement of soils (Cohesive) : Part C: General technical specifications in construction
contracts (ATV) – earthworks, ATV DIN 18300
For a thin layer, Δp at mid-depth is considered as average
stress within the layer. Soil and rock classification : -
Class 1 - Top Soil
Class 2: Flowing Soil
Class 3: Easily loosened soil types
Class 4: Moderately hard-to-loosen soil types
Class 5: Hard-to-loosen soil types
Class 6: Easily loosened rocks and comparable soil types
Class 7: Hard-to-loosen rock
Q.85. Which of the following lines connects the liquid
𝑫𝑯 𝑫𝑽
= state with liquid and solid state mixture?
𝑯 𝑽
𝜟𝒆 (a) Solidus (b) Liquidus (c) Solvus (d) Tie line
DH = S = ×𝑯
(𝟏+𝒆𝟎 )
𝒂𝟎
Ans : (b) Liquidus: - It is represented by a line on a phase
𝑺= 𝜟 ×𝑯 diagram that separates a liquid phase from a solid +
(𝟏 + 𝒆𝟎 ) 𝒑
liquid phase region. A system must be heated above the
𝑺 = 𝒎𝒗 𝜟𝒑𝑯
liquidus temperature to become completely liquid.
Where
The liquid system begins to solidify when the
S = settlement temperature cools to the liquidus temperature.
mv = coefficient of volume compressibility

UPSSSC JE 19/12/2021 110 Civil KI Goli

Solidus: It is represented by a line on a phase diagram Q.87. What absorbent Is used to determine water
that separates a solid phase from a solid + liquid phase content in soil as per IS: 2720 (Part Il) - 1973?
region. (a) Potassium carbonate (b) Potassium acetate
The system is not completely solid until it cools below (c) Calcium carbide (d) Potassium citrate
the solidus temperature.
Ans : (c) Calcium carbide Method: This test is done to
Figure: Eutectic binary phase diagram(α = solid 1, β = determine the water content in soil by the calcium
solid 2, L = liquid) carbide method as per IS: 2720 (Part II) – 1973.
Solvus: It is represented by a line on a phase diagram • It is a method for rapid determination of water content
that separates a solid phase from a solid1 + solid2 phase, from the gas pressure developed by the reaction of
where solid1 and solid2 are different microstructures. calcium carbide with the free water of the soil. From the
calibrated scale of the pressure gauge, the percentage of
water on the total mass of wet soil is obtained and the
same is converted to water content on dry mass of soil.
• It is very quick method, takes only 5 to 7 minutes but
does not give an accurate result.
• Calcium carbide powder (CaC2) is added on the moist
soil sample which reacts with the water and as a result
acetylene gas is removed which exerts pressure.
CaC2 + 2H2O → Ca(OH)2 + C2H2 ↑
Q.88. Choose the correct order of materials to fill in the
Eutectic: - It is represented by the horizontal line in a blanks:
eutectic binary phase diagram, connecting the According to Engineering science, particles with
intersections of the solidus and solvus lines from both diameter smaller than 0.002 mm are termed as
sides. ________ while those with diameter between 0.002
The eutectic temperature also is where the liquidus lines and 0.006 mm are called ________ . Those between
for both components meet. 0.075 and 2 mm are called _______ and particles with
diameter more than 4.75 mm are termed as _________
Q.86. Identify the soil that gets accumulated at its place (a) Silt, clay, gravel, sand (b) Clay, silt, sand, gravel
of origin only.
(c) Silt, clay, sand, gravel (d) Clay, sand, silt, gravel
(a) Glacial soil (b) Residual soil
Ans : (b) In the Indian Standard Soil Classification System
(c) Transported soil (d) Aeolian soil (ISSCS) or BIS, soils are classified according to their grain
Ans : (b) All the soils are mainly divided into two parts size as boulder, cobble, gravel, sand, silt, or clay, as
(1). Residual Soil: If Disintegrated materials remain over shown below in the tabulated form.
the parent rock then the soil is called Residual soil. Type of Sub -
Soil Group Size Range
Characteristics of residual soil depend on climatic Soil Group
conditions, natural drainage pattern, form, and extent of
Very Coarse Boulder > 300 mm
vegetation cover.
Soils Cobble 80 – 300 mm
Example - Bentonite is a type of chemically weathered
volcanic ash that is present on the parent rock from Coarse Soils Gravel Coarse 20 – 80 mm
which it is formed. Fine 4.75 – 20 mm
(2) Transported soil: If sediments are transported and
Sand Coarse 2- 4.75 mm
deposited at other places then it is called transported
soil. Medium 0.425- 2 mm
Sediments are transported by winds, water, glacier, etc. Fine 0.075 – 0.425
Some transported soils are- mm
Alluvial – transported in running water (rivers) Fine Soils Silt 0.002 – 0.075
Lacustrine – Deposited by still water (lakes) mm
Marine – Deposited by seawater
Clay < 0.002 mm
Aeolian – Transported by wind
Glacial – Transported by ice
Colluvial – Transported by gravity

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2. No deduction shall be made for openings not Q.112. Choose the correct option that defines
exceeding 0.5 m2 each, and no addition shall be made Obsolescence.
for painting to beading, moulding, edges, jambs, soffits, a. The value at the end of the utility period of the
sills, etc, of such openings. structure without being dismantled.
3. Painting work up to 10 cm in width or in girth and not b. The value of dismantled materials.
in conjunction with similar painted work shall be c. The value which can be obtained at any particular
measured in running meters . time from the open market if put for sale.
4. Areas of uneven surfaces being converted into d. The decrease in value of property when it goes out
equivalent plain areas by multiplying the flat measured of date in style or design.
area by a multiplying factor as specified in Table 1 of IS
(a) a (b) b (c) c (d) d
1200, part 15 code.
5. Painting on eaves-gutters, rain-water pipes, soil and Ans : (d) Obsolescence is the loss in the value of the
ventilating pipes and steel poles shall be measured in property due to changes in fashions, in designs, in
running meters. structures, etc. The obsolescence is due to the changes
in fashions, changes in planning ideas, new inventions,
Manholes are classified according to depth as:
improvements in design techniques, etc.
(i) Shallow manholes: Shallow manholes are those which
are about 0.75 to 0.90 m in depth. Q.113. Consider a 2000 m long cement concrete road
that is 4 m wide and 5 cm thick covering the sub-base
(ii) Normal manholes: Normal manholes (or medium
of 20 cm deep gravel. What would be the value of box
manholes) are those which have depth more than 0.9 m
cutting in road crust?
and up to 2 m.
(a) 1000 m3 (b) 500 m3 (c) 2000 m3 (d) 1500 m3
(iii) Deep manholes: Deep manholes are those having
depth more than 2 m. Ans : (c) Given , Road length (L) = 2000 m
Q.110. An overhead tank of capacity 50,000 litres is to Road width (B) = 4 m
be built in a society. What will be the preliminary Road thickness over sub base = 5 cm
estimate of the project if the per litre cost is Rs. 3? Sub base thickness or depth = 20 cm
(a) Rs. 15,00,000 (b) Rs. 100,000 Total thickness or depth for road for box cutting =
(c) Rs. 75000 (d) Rs. 1,50,000 5 + 20 = 25 cm = 0.25 m
Ans : (d) Given data The value of box cutting in road crust is
The capacity of overhead tank = 50000 liter = Length × width × total depth(for box cutting)
Per liter cost (Rs.) = 3 = 2000 × 4 × 0.25 = 2000 m3
The preliminary estimate(Rs.) of the project is ❖ STEEL
= The capacity × Per liter cost Q.114. Which among the following is true in case of
= 50000 × 3 = 150000 Rs timber and steel shuttering?
Q.111. Which of the following units is used in (a) Easier to dismantle timber forms compared to steel.
estimating R.C.C. & R.B. work on site? (b) Steel forms absorb moisture
(a) sq. m. (b) cu. m. (c) sq. ft. (d) ft (c) Steel forms are stronger, durable, and have a longer
Ans:(b) Based on IS 1200, the units for measurement life than timber formwork and their reuses are more in
for various civil work is given as: number.
(d) Steel does not shrink and expand.
Description Unit
Ans:(c) Shuttering: - Shuttering or formwork is the term
Earthwork, Stone/Brick Work, Wood used for temporary timber, plywood, metal, or other
m³
Work/Sunshade, R.C.C.
material used to provide support to wet concrete mix till
Surface/Shallow Excavation, shutter, panal, it gets strength for self-support.
m²
batten Types of Shuttering : -
Pointing, Soling, DPC, Plastering, door, window m² • Timber Formwork.
• Steel Formwork.
Steel/Iron Work kg/Quintal • Aluminum Formwork.
Dressing of stone/Half Brick wall/partition wall m² • Plywood Formwork.
• Fabric Formwork.
Painting Work/Distemper/Colour Washing/Jali • Plastic Formwork.
m²
Work
Advantages of steel formwork over timber form work:

UPSSSC JE 19/12/2021 116 Civil KI Goli

the suspended particles so as to increase the Scour valve: This valve is used to drain the water out of
opportunity of the contact between the neutralized the pipe system.
particles and a sticker precipitate resulting in the Sluice valve: These valves are used to regulate the flow
formation of a bigger size particle which results in the of water in the pipe system by dividing into a number of
formation of a bigger size particle which can get easily sections.
settled in the following sedimentation process.
Q.121. Match the following air pollutants with their
Flocculation: It is a process in which neutralize particles respective sources.
are gotten intimate contact with each other and with a
sticky precipitate resulting in the formation of bigger size a. Oxides of 1. Pesticides, pints, laboratory.
particles that can get easily settle in the following sulphur
sedimentation tank. b. Mercury 2. Petroleum industry wastewater
• In order to increase the opportunity to combine treatment, oil refineries, etc.
between the particles, slow mixing is also induced in this c. Hydrogen 3. Glass and ceramics, cement
process. Sulphide factories, aluminium industry,
Sedimentation: It is the process of removal of fertilizer industry etc.
suspended particles from the water.
d. Fluorides 4. Power house, smelters, coal and
• The entire theory of sedimentation is based upon a other fossil fuel combustion, etc.
single parameter termed Specific gravity.
(a) a - 4, b - 1, c - 2, d - 3
• In this process, particles are allowed to settle in the
tank due to their own weight and then get removed. (b) a-1, b-4, c-2, d-3
Filtration : (c) a - 4, b - 1, c - 3, d - 2
• Filtration is most often a policing step to remove flocks (d) a - 1, b - 4, c - 3, d - 2
or smaller un- sedimented particles. Ans : (a) Air pollutants with their respective sources are
• Filtration also removes dissolved organic matter, as follows:
dissolved minerals, and microorganisms. Pollutant Common Sources
Disinfection: Chlorination is the most popular method of carbon
automobile emissions, fires, industrial
disinfection in which disinfection is done with the help monoxide
processes
of chlorine. (CO)
Q.120. Which of the following valves should you use nitrogen
automobile emissions, electricity
when you want the flow of water to be in one direction oxides (NO
generation, industrial processes
and NO2)
considering, it does not flow in the reverse direction as
well? Electricity generation, fossil-fuel
sulfur
combustion, industrial processes,
(a) Pressure relief valve (b) Sluice valve dioxide (SO2)
automobile emissions
(c) Reflux valve (d) Air relief valve
nitrogen oxides (NOx) and volatile
Ans : (c) Check valves or reflux valves: It prevent water organic compounds (VOCs) from
to flow back to opposite directions . ozone (O3) industrial and automobile emissions,
They are provided on the delivery side of the pumping gasoline vapors, chemical solvents, and
set and at interconnections between a polluted water electrical utilities
system and a potable water system. These are generally Hydrogen Petroleum industry wastewater
provided on the delivery side of the pump. These valves Sulfide treatment, oil refineries, etc.
work automatically. sources of primary particles include fires,
Pressure reducing valve: In hydraulics, a pressure smokestacks, construction sites, and
reducing valve serves the same purpose as a "pressure particulate unpaved roads; sources of secondary
regulator" valve in a compressed air system. It is one of matter particles include reactions between
gaseous chemicals emitted by power
a variety of pressure control valves available for
plants and automobiles
hydraulic circuits. It is always used in a branch circuit and
never in the full pump flow line. Glass and ceramics, cement factories,
Fluorides aluminum industry, fertilizer industry,
Air Relief Valve: it is a type of safety valve used to control etc.
or limit the pressure in a system, pressure might
otherwise build up and create a process upset, metal processing, waste incineration,
lead (Pb)
fossil-fuel combustion
instrument or equipment failure, or fire. The pressure is
relieved by allowing the pressurized fluid to flow from an Mercury Pesticides, pints, laboratory
auxiliary passage out of the system.

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Q.122. Arrange the following process in order with Ans : (b) Surge arrester: A surge arrester is a protective
context to water purification: device for limiting voltage on equipment by discharging
a- Disinfection or bypassing surge current.
b- Filtration • It is used to control of water hammer in pipe lines.
c- Coagulation Check valve or Reflux valve: These valves are also known
d- Sedimentation as non-return valves or reflux valves.
e- Flocculation • They allow the water to flow in one direction only.
(a) c, e, d, b, a (b) e, c, d, b, a • They may be installed on the delivery side of the
(c) a, b, c, d, e (d) c, e, b, d, a pumping set so as to prevent the backflow of stored or
pumped water when the pump is stopped.
Ans : (a) The sequence of the process in the water
• When the pump is operated, the valve is opened, but
treatment plant :
when the pump is suddenly stopped the valve is
Coagulation → Flocculation→ Sedimentation→
automatically closed. This prevents the backward flow of
Filtration→ Disinfection
water to the pump.
Coagulation: Coagulation is the process in which certain
chemical termed coagulants are added to neutralize the Air Relief Valve: It is a type of safety valve used to
negative protective charge present over the suspended control or limit the pressure in a system, pressure might
particles so as to increase the opportunity of the contact otherwise build up and create a process upset,
between the neutralized particles and a sticker instrument or equipment failure, or fire.
precipitate resulting in the formation of a bigger size • The pressure is relieved by allowing the pressurized
particle which results in the formation of a bigger size fluid to flow from an auxiliary passage out of the system.
particle which can get easily settled in the following Scour valve or Blow off valve or Drain valve: These
sedimentation process. valves are also known as wash-out valves.
Flocculation: It is a process in which neutralize particles • To remove the entire water from the pipe after closing
are gotten intimate contact with each other and with a the supply, small gate valves are provided at a low point.
sticky precipitate resulting in the formation of bigger size • These valves are necessary at low-level points for
particles that can get easily settle in the following completely emptying the pipe for inspection, repair, etc.
sedimentation tank.
• In order to increase the opportunity to combine ❖ HYDRAULIC MACHINE
between the particles, slow mixing is also induced in this Q.124. You have to transform kinetic energy to
process. hydrodynamic energy. Which of the following should
Sedimentation: It is the process of removal of you use to perform the above operation?
suspended particles from the water. (a) Displacement pump
• The entire theory of sedimentation is based upon a (b) Mono pump
single parameter termed Specific gravity.
(c) Reciprocating pump
• In this process, particles are allowed to settle in the
tank due to their own weight and then get removed. (d) Centrifugal pump
Filtration: Ans : (d) Centrifugal pump: Centrifugal pumps transport
• Filtration is most often a policing step to remove flocks fluids by converting rotational Kinetic energy to
or smaller un-sedimented particles Filtration also hydrodynamic energy.
removes dissolved organic matter, dissolved minerals, The rotational kinetic energy comes from an engine or
and microorganisms. an electrical motor
Disinfection: Chlorination is the most popular method of
disinfection in which disinfection is done with the help
of chlorine.
Q.123. Consider the given pairs:
Identify the one that is matched correctly.
(a) Check valve - To maintain the flow of water in pipes,
in all directions.
(b) Surge arrester - Control of water hammer in pipe
lines.
(c) Air valve - To drain or empty the pipe line section.
(d) Scour valve - To release the accumulated air in
pipelines.

UPSSSC JE 19/12/2021 119 Civil KI Goli

Q.125. From the given options, identify the statement
that accurately describes the total efficiency of a
centrifugal pump.
(a) The ratio of manometric head to the energy
supplied by the impeller per Newton of water
(b) The ratio of energy supplied to the pump to the
energy available at the impeller
(c) The ratio of energy is available at the impeller to the
energy lost
(d) The ratio of actual work done by the pump to the
energy supplied to the pump by the prime mover
Ans : (d) The efficiency of a centrifugal pump is three
types.
1. Manometric efficiency ( ηman ) : The ratio of the power
given to water at the outlet of the pump to the power
available at the impeller is known as manometric
efficiency.
Manometric head
𝜼𝒎𝒂𝒏 =
Head imparted by impeller to water
2. Mechanical efficiency(ηm ):The mechanical efficiency
of a centrifugal pump is the ratio of the power available
at the impeller to the power at the shaft of the
centrifugal pump.
Power at the impeller
𝜼𝒎 =
Power at the shaft
3. Overall efficiency ( ηo ): The ratio of actual work done
by the pump to the energy supplied to the pump by the
prime mover.
It is also defined as the ratio of the power output of the
pump to the power input to the pump.
η0 = ηman × ηm

UPSSSC JE 19/12/2021 120 Civil KI Goli

 UPSSSC JE CIVIL 2018 PAPER
(EXAM DATE: 16 JULY 2022)
❖ SURVEY
Q.01. Identify the equipment that is used for marking
and is made up of hardened and tempered steel wire
of good quality. Pegs: - Pegs are used to mark the position of main
station or terminal point of a survey line on the ground.
Its 2/3 length should be under the surface and rest over
the surface.
• Cross section- 25 mm × 25 mm or 30 mm × 30 mm.
(a) (b) • Length - 150 mm

(c) (d)
Ans : (c) Arrows or marking pin:- Arrows are made of
galvanized mild steel used to mark the end point of Q.02. Which of the following statements are
chain or intermediate station. The one pointed part of advantages of using Plane table surveying?
an arrow is inserted into the ground and the other is 1. Suitable for large and precise work
attached with the ring.
2. More rapid and less costly
• Loop dia. - 50 mm • Length- 400 ± 5 mm
3. Chances of getting omission are less
(a) Only 1, 2 statements are correct
(b) Only 1, 3 statements are correct
(c) Only 2, 3 statements are correct
(d) 1, 2, 3 statements are correct
Ans: (c) Plane Table Surveying: - It is a graphical
method of surveying in which field work and plotting
are done simultaneously in the field. It is most suitable
Plum Bob: - A plumb bob is a pointed weight attached for small and medium scale survey.
to the end of the string and is used to find a vertical Advantages of Plane Table Survey: -
reference line called a plumb. Plumb is the vertical • The observations and plotting are done
equivalent to a spirit level. simultaneously, hence there is no risk of omitting
necessary details.
• The errors and mistakes in plotting can be checked by
drawing check lines.
• Irregular objects can be plotted accurately as the lay
of the land is in view.
• It is most rapid and useful for filling in details. No
great skill is required.
• Plane table surveying is less costly than theodolite
survey.
Ranging poles or rod: - It is used to ranging between
two station point. It is also known as flag pole, lining • It is advantageous in magnetic areas where the
rod. compass survey is not reliable.
It is painted with 20 cm wide bands of red and white Disadvantages of Plane Table Survey: -
alternate . • Plane table surveying is not suitable for work in a wet
According to IS : 2288 – 1963- climate and in a densely wooded country.
• The absence of measurements (field notes) is
• Length- 2-3 m • Dia. - 30mm
inconvenient if the survey is to be replotted to some
• Visible up to - 200m different scale.

UPSSSC JE 2018 121 Civil KI Goli

• The graduations are in degrees to 30 minutes and line from either the magnetic meridian since the needle
from 00 to 3600 in the clock wise direction. is often pointing to the north-south pole at the
• A glass cover is fitted over the box to protect the opposite edges of the needle while freely placed on
needle from dust. The compass is fitted to a tripod some support.
stand. Q.15. Match the following parts of prismatic compass
with their functionalities.
1. Metal box a. Heart of the compass
2. Pivot b. Protective casing against dust
3. Lifting pin c. Centrally located part
4. Magnetic d. Provided right below the sight
needle vane
(a) 1-c, 2-a, 3-d, 4-d (b) 1-c, 2-d, 3-a, 4-b
(c) 1-b, 2-c, 3-d, 4-a (d) 1-b, 2-a, 3-d, 4-c
Ans : (c) Parts of Prismatic Compass:
Cylindrical Metal Box:- The cylindrical box with a size of
8 to 12 cm covers the compass and then the whole
casting or body of the compass. It protects casing
against dust.
Eye Valve:- The eye valve is fine silt with an eye hole in
Parts of Prismatic Compass: the bottom to bend the object out of the silt.
Cylindrical Metal Box:- The cylindrical box with a size of Prism:- Prism is being used to recognize the graduation
8 to 12 cm covers the compass and then the whole on the ring and also to read the same reading by
casting or body of the compass. compass. It’s positioned in the opposite direction of the
Eye Valve:- The eye valve is fine silt with an eye hole in object vane. The prism hole is covered by a prism cap
the bottom to bend the object out of the silt. to shield it from dust and moisture.
Prism:- Prism is being used to recognize the graduation Graduation Circle:- This is an aluminum graduated ring
on the ring and also to read the same reading by labeled from 0° to 360° to calculate all potential line
compass. It’s positioned in the opposite direction of the bearings and connected with a magnetic needle.
object vane. The prism hole is covered by a prism cap Lifting Pin and Lifting Lever:- Below the viewing vane.
to shield it from dust and moisture. Lift pin pressed while the sight vane is folded. The
Graduation Circle:- This is an aluminum graduated ring magnetic needle was raised from the pivot point with
labeled from 0° to 360° to calculate all potential line the lever raise.
bearings and connected with a magnetic needle. Spring Break: - In order to dampen the vibration of the
Lifting Pin and Lifting Lever:- Below the viewing vane. needle once obtaining a measurement to putting it to
Lift pin pressed while the sight vane is folded. The resting easily.
magnetic needle was raised from the pivot point with Object Vane:- The object vane holds horsehair or black
the lever raise. thin wire to see the object in line with the sight of the
Spring Break: - In order to dampen the vibration of the object.
needle once obtaining a measurement to putting it to Reflecting Mirror:- It is used to obtain a picture of an
resting easily. object positioned higher or lower surface of the device
Object Vane:- The object vane holds horsehair or black when bisecting.
thin wire to see the object in line with the sight of the Pivot:- The pivot is offered at the middle of the
object. compass which supports a loosely attached magnetic
Reflecting Mirror:- It is used to obtain a picture of an needle.
object positioned higher or lower surface of the device Magnetic Needle:- The magnetic needle is the core of
when bisecting. the device. That needle determines the angle of the
Pivot:- The pivot is offered at the middle of the line from either the magnetic meridian since the needle
compass which supports a loosely attached magnetic is often pointing to the north-south pole at the
needle. opposite edges of the needle while freely placed on
some support.
Magnetic Needle:- The magnetic needle is the core of
the device. That needle determines the angle of the

UPSSSC JE 2018 125 Civil KI Goli

general, there are four main stages of transportation Grid Block
planning:
Provision of No 1600 km 3200 km
1st • Transportation Survey, Data Collection, and
Expressway provision
Stage Analysis
2nd •Build a transportation model by using data Road 1. NH NH, SH, • Primary
Stage collected. Classification 2. SH MDR, ODR, Expressway
•This transportation model predicts future 3. MDR VR and NH
travel demands and network needs. Expressway
4. ODR • Secondary
•The transportation model building stage is 5. VR SH
also sub-divided into four stages MDR
Trip generation: This is the first stage of the • Tertiary
model building process.
ODR
•Estimates the total number of trips
VR
originating in the survey area at one or more
future dates. Development 15% 5% Nil
Trip distribution: This stage involves the allowances
analysis of trips between zones (like public Q.24. You have to calculate the super elevation(e) for
and private transportation systems). a highway having a design speed as v (in km/h). Which
•Estimates the number of trips that will occur of the following options gives the CORRECT relation to
between each origin and destination zone. find the design speed such that radius of the curve is r
Traffic assignment: To predict the traffic flow (in m) and the coefficient of friction is f ?
of trips on a network.
(a) e = (v2 /127r) – f (b) e = v2 f/127r
Model split: This model split is the phase
where transportation engineers incorporate (c) e = (v2 /127r) + f (d) e = v2 r /127f
the mode of travel into their models. Ans : (a) Superelevation: It is the transverse slope to
•The main purpose of the model-split stage is counteract the centrifugal force and reduces the
to determine the trip shares of public, as tendency of the vehicle to overturn or skid.
against private, transport. • It is the rise of the outer edge of the pavement w.r.t
3rd • Forecast of future land use and Strategies inner edge on a horizontal curve thus providing a
Stage for Alternative Policies transverse slope throughout the length of the
4th • Evaluation of Policy horizontal curve
Stage • It is also known as cant or banking.
Q.23. Fill in the blank with the CORRECT option: General equation
The second twenty-year road plan in India was for the 𝐯𝟐
ⅇ= −𝐟
years _______ 𝟏𝟐𝟕𝐫

(a) 1935-55 (b) 1961-81 Q.25. Which among the following is the requirement
(c) 1940-60 (d) 1931-51 for highway drainage system?
Ans : (b) Highway Development (a) Water can percolate through subgrade
(b) Adjoining landed should be in level with road
Specification 1st 20 Year 2nd 20 Year 3rd 20 Year
Plan Plan Plan irrespective of water entering the road
(c) No special adjustments are needed for water
Name of Plan Nagpur Bombay road Lucknow Road logged areas
road plan plan Plan
(d) The side drain should have sufficient capacity and
Duration 1943-1963 1961-1981 1981-2001 longitudinal shape to carry away all the surface water
Completed
Ans : (d) Highway Drainage:- It is the process of
-1961
removing and controlling excess surface and sub-
Target Density 16 km/100 32 km/100 82 km/100 surface water within the right of way which includes
km² km² km² interception and diversion of water from the road
Total Target 532700 km 10 Lakhs km 27 Lakhs km
surface and subgrade.
Requirements of Highway Drainage System:-
Achieved 709122 km 1502697 km 2702000 km
• The surface water from the carriageway and shoulder
Target NH-66000 km,
should effectively be drained off without percolating
SH-145000 km
to subgrade.
Road Pattern Star and - Square and

UPSSSC JE 2018 127 Civil KI Goli

• Adjoining land should be prevented from entering the guidelines follows analytical designs and developed
roadway new set of designs up to 150 msa in IRC:37-2001.
• The side drain should have sufficient capacity and
longitudinal slope to carry away all the surface water ❖ ESTIMATION COSTING
• Special precautions should be taken in waterlogged Q.28. Choose the appropriate service unit for school
areas. building.
• Surface water flow across the road, shoulders and (a) Per seat (b) Per student
along the slope should not cause erosion or cross (c) Per classroom (d) Per meter of span
ruts.
• Seepage and other sources of underground water Ans: (c) Service unit method: - In the Service Unit
should be drained off by sub-surface drainage system. Method, the Service Unit indicates the most important
• Maximum ground water table level should be kept unit in a structure. The entire structure is divided into a
well below the level of subgrade, by at least 1.2 m. number of service units.
• The approximate estimate of the school building is
Q.26. You have to construct a highway having the prepared on the basis of the service unit and the
width of formation 'w' and height of embankment 'h'. service unit for the school building is the classroom.
The slope relative to the side is Y : 1 having no The service units for various structures are given in the
transverse slope. Which of the following options gives below table.
the CORRECT equation for the area of cross-section ? Sr. Types of
(a) wh + Yh (b) wh + Yh Service unit
No. Construction
(c) wh + Yh2 (d) 1/2 (wh + Yh2 ) 1 School, College Classroom
Ans : (c) Given. 2 Hospitals Bed
w = Width of the embankment 3 Hotel Room
4 Hostel Students
h = Height of the embankment Y:1
5 Theatre Seat
= Side slope of the embankment.
6 Stadium Seat
7 Jail Kotani
8 Stable Animal
9 water tank Litre
10 Dam Hectare meter
11 Water supply person
scheme
12 road kilometer
Area of the embankment (A) = Area of the trapezium 13 Bridge Meter
𝟏
A = x (sum of the length of parallel side) x h 14 Power station Kilowatt
𝟐
𝟏 15 Canal Kilometer
A = x [w + (w + 2Yh)] x h
𝟐
𝟏
Q.29. Choose a suitable service unit for the hospital.
A= x [2w+2Yh]x h (a) Per bed (b) Per room
𝟐
A = wh + Yh² (c) Per patient (d) Per meter of span
Q.27. According to IRC : 37-2001, the design Ans : (a) Service unit method : - In the Service Unit
procedures for flexible pavements that are based on Method, the Service Unit indicates the most important
CBR values used for measurement of traffic on the unit in a structure. The entire structure is divided into a
road are up to which of the following? number of service units.
(a) 30 Million Standard Axles The approximate estimate of the school building is
(b) 120 Million Standard Axles prepared on the basis of the service unit and the
(c) 160 Million Standard Axles service unit for the hospital is the bed.
(d) 150 Million Standard Axles Q.30. Prepare an approximate estimate of a factory
building proposed to made up of RCC with the
Ans : (d) Indian roads congress has specified the following data points:
design procedures for flexible pavements based on CBR The total plinth area of the entire building is 120 m 2
values. The Pavement designs given in the previous (built-up area).
edition IRC:37-1984 were applicable to design traffic Plinth area rate for the RCC building is Rs. 6000/-.
upto only 30 million standard axles (msa). The earlier (a) Rs. 7,20,000/- (b) Rs. 14,40,000/-
code is empirical in nature which has limitations
(c) Rs. 9,60,000/- (d) Rs. 8,60,000/-
regarding applicability and extrapolation. This

UPSSSC JE 2018 128 Civil KI Goli

• Books are considered very important account records 1m 3.2808 feet
and maintained very carefully and accurately, as they
may have to be produced in the court as evidence, if 1 mile 1760 yards
and when required .
1 mile 1.609 km
Q.36. You have to construct a new rail line on a valley
surface (having a wavy surface) by performing cut and 1 yard 0.9144 m
fill with earthwork. Which of the following situations
1 foot 0.3048 m
should you try to achieve so that construction cost is
minimized? 1 inch 0.0254 m
(a) Quantity of cut < Quantity of fill Q.38. Analyze the given image and identify the
(b) Quantity of cut > Quantity of fill CORRECT statement.
(c) Quantity of cut = Quantity of fill
(d) Construction cost is independent of quantity of cut
and fill
Ans:(c) To minimize construction costs when
performing cut and fill earthwork for a new rail line on
a valley surface, you should aim to balance the
quantities of cut and fill as closely as possible. This
approach minimizes the need to transport excess
(a) Length of the long wall out-to-out is 6.80 m
material off-site or bring in additional material from
(b) Length of short walls in-to-in is 2.40 m
elsewhere, both of which can significantly increase
costs. (c) Total length of four walls is 24 m, and the length of
the short wall is 8.6 m
Balancing Depth :- Balancing Depth comes when the
canal is in partially embankment and partially in (d) The total length of the centre line of four walls is
cutting. 24 m
• It is the depth of the canal (H) that gives an equal Ans : (d)
amount of filling (i.e. Earth required for the formation
of banks) and cutting (earth from digging).
• For a given cross-section of the canal, It has only one
balancing depth.
• For this depth, the canal will be economical.
• For Balancing depth, Area in cutting = Area in
embankment.
Q.37. Match the following units with their respective
conversions : C/C length of long wall (A to D) and (B to C)
1. 1 INCH a. 1.6093 KILOMETER = 8.60 - 0.3 - 0.3 = 8 m
C/C length of Short wall (D to C) and (A to B)
2. 1 MILE b. 0.9144 METRE
= 4.60 - 0.3 - 0.3 = 4 m
3. 1 METRE c. 25.4 MILLIMETER
The total length of the Centre line of four walls
4. 1 YARD d. 0.4047 HECTARE = 8 + 4 + 8 + 4 = 24 m
5. 1 ACRE e. 3.2808 FOOT Length of the long wall (outer to outer)= 8.60 m
(a) 1-c, 2-a, 3-e, 4-b, 5-d (b) 1-c, 2-d, 3-a, 4-e, 5-b Length of the short wall (in to in)= 4.60 - 0.6 - 0.6 = 3.4
(c) 1-b, 2-c, 3-d, 4-e, 5-a (d) 1-b, 2-e, 3-d, 4-c, 5- m
Ans : (a) Different unit's conversion of length are as Q.39. Consider the given relationships concerning the
follows: formula for the midsection, and identify the CORRECT
statement(s).
Length Equal to
I) Earthwork volume = mid-segment area × (period
1 kilometer (km) 1000 meters (m) within two primary sections)
II) Mean depth is used to determine the area of mid-
1 km 0.6214 miles
segments
1m 1.0936 yards (a) Only I (b) I and II
(c) Only II (d) Neither I or II

UPSSSC JE 2018 130 Civil KI Goli

Ans : (b) Mid Sectional Method and external walls. It is generally 10-20% more than the
Consider a midsection in between 1st and 2nd sections. carpet area.
dm = depth of mid-section = mean depth Following areas are included during measurement of
𝐝𝟏 +𝐝𝟐 plinth area:
𝐝𝐦 = 𝟐 • Area of the wall at the floor level, excluding plinth
Quantity of Earthwork or Earthwork Volume offsets. Areas of the internal shaft for sanitary
= Cross-sectional area of Mid-Section × L installations and garbage chute, electrical, telecom,
𝟏 and firefighting services
= (𝐁𝐝𝐦 + × 𝐒𝐝𝐦 × 𝐝𝐦 × 𝟐) × 𝑳 • Vertical duct for air conditioning and lift well including
𝟐
= (𝐁𝐝𝐦 + 𝐒𝐝𝟐𝐦 )𝐋 landing Staircase room or headroom other than
terrace level , Area of barsati at terrace level
Q.40. Choose the CORRECT measurement mode for
the door and window shutter. Following areas are not included during measurement
of plinth area:
(a) cu.m. (b) sq.m (c) RMT (d) Meter
• In the plinth area Estimate. Open areas. courtyard.
Ans : (b) The correct measurement mode for door and
and balconies should not be included in the plinth
window shutters is square meters (sq.m).
area valuation. This method is also known as an
Description Unit approximate estimate
• Additional floor for seating in assembly buildings,
Earthwork, Stone/Brick Work, Wood m³ theatres, auditoriums
Work/Sunshade
• Cantilevered porch, Balcony, Area of loft, Internal
Surface/Shallow Excavation, shutter, m² sanitary shaft, and garbage shaft
panal, batten • Area of the architectural band, cornice, Open
platform, etc.
Pointing, Soling, DPC, Plastering, door, m² • Towers, turrets, domes projecting above the terrace
window level at the terrace
Steel/Iron Work kg/Quintal Q.43. Consider an asset for which the cost is given as
100 units with the total useful life as 3 years and the
Dressing of stone/Half Brick wall/partition m² depreciation calculated as 20 units. Calculate the scrap
wall value of the asset :
Painting Work/Distemper/Colour m² (a) 60 units (b) 123 units
Washing/Jali Work (c) 8 units (d) 40 units
Q.41. Which of the following is the measurement unit Ans : (d) Given data;
for Weathering Course ? Cost = 100 units. D= 20 units, Life =3 year
(a) SQM (b) RMT (c) KG (d) CUM We know that,
Ans : (a) Weathering Course, often used in construction Scrap Value = Cost of Asset (D x Useful Life)
and civil engineering, typically refers to a layer of =100 (20 × 3) = 40 units
material applied to surfaces such as roofs or pavements Q.44. The plinth area of the proposed building is 340
to protect them from weather-related damage, m2. The estimated cost of building a similar structure
including erosion, water infiltration, and temperature is Rs. 19,35,000 that was having a plinth area of 315
variations. he Weathering Course is generally applied as m2. Calculate the approximate cost of the proposed
a surface layer, so its extent is measured in terms of building.
area covered. This makes square meters (SQM) the
(a) Rs. 22,12,000/- (b) Rs. 35,16,000/-
most appropriate unit of measurement.
(c) Rs. 20,88,569/- (d) Rs. 34,12,000/-
Q.42. You have to estimate the plinth area of an
apartment. Which of the following should you NOT Ans : (c) Given:
include in the calculation ? Plinth area of the proposed building = 340 m2
(a) Verandah area (b) Courtyard area Estimated cost of 315 m2 plinth area of similar structure
(c) Room area (d) Wall thickness = Rs. 19,35,000/-
𝐑𝐬.𝟏𝟗,𝟏𝟑𝟓,𝟎𝟎𝟎
Ans : (b) Plinth area: It is the covered built-up area Cost of 1 m2 plinth area =
𝟑𝟏𝟓
measured at the floor level of any storey or at the floor = Rs. 6142.85 /-
level of the basement. Cost of Proposed building = 340 × 6142.
The plinth area is also called a built-up area and is the = Rs. 20,88,569 /-
entire area occupied by the building including internal
UPSSSC JE 2018 131 Civil KI Goli
electrodeposited coatings that are applied to iron, a primary contributor to air pollution and the cause of
steel, zinc alloys, copper and copper alloys, and harmful corrosion in the engine.
aluminum and aluminum alloys to provide an attractive Q.54. Which type of pollution is primarily responsible
appearance and corrosion resistance. for causing headache and high blood pressure
Q.51. Which of the following is NOT a greenhouse gas? problems?
(a) Nitrous Oxide (b) Methane (a) Land pollution (b) Water pollution
(c) Nitrogen (d) Carbon dioxide (c) Soil pollution (d) Noise pollution
Ans : (c) Greenhouse Effect: The solar energy trapped Ans : (d) Pollution:- Pollution is the introduction of
by the earth atmosphere and radiate it slowly so to harmful substances into the environment which causes
cover our earth with a warm blanket. This is the natural adverse impacts on living beings as well as the
process of the greenhouse effect on earth to maintain environment.
its temperature and makes the earth perfect for life. The substances which cause pollution are known as
The major greenhouse gases are: pollutants.
• Water vapour causes about 36% 70% of the Types of Causes Consequences
greenhouse effect Pollution
• Carbon dioxide (CO2) which causes 9-26%
Air Burning fossil fuels Increase risk of
• Methane (CH) causes 49%
Pollution Exhaust gases respiratory illness and
• Ozone (03)
from industries cardiovascular
• Nitrous Oxide
and factories problems.
• Chlorofluorocarbons (CFCs)
Global warming
Q.52. Which one of the following requirements is
Ozone layer depletion
undesirable for a good trap?
(a) Simple in construction Water Disposing of Increases in toxic
(b) Possess adequate water seal Pollution untreated chemicals (such as
industrial sewage mercury) in water
(c) Internal and external surfaces are rough
into water bodies bodies
(d) Capable of being easily cleaned
Agricultural runoff Increased risk of
Ans : (c) Trap : - The devices which are used to stop the containing water-borne diseases
escape of foul gases inside or outside the houses are pesticides and (diarrhea, skin
known as traps. fertilizers diseases, stomach
• A good trap should possess the following quality Human and problems, etc.,)
• It should provide a sufficient water seal. animal wastes Disruption of the
• It should be capable of being easily cleaned. ecosystem
• It should be easily fixable with the drain or pipes. Soil Intensive farming Dust from toxic soil
• Its internal and external surfaces should be smooth. Pollution and use of may cause respiratory
• It should possess self-cleaning property. excessive chemical problems or even
fertilizer. lung cancer.
• It should be free from any inside projection.
Dumping of Loss of soil nutrients
• It should be simple in construction.
industrial waste Impacts the natural
• It should be made of some non-absorbent material
Mining activities flora and fauna
• It should be provided with an access door for
cleaning. Noise It is caused due to High blood pressure
Pollution unwanted loud Headaches
Q.53. Which among the following is false for onsite sounds in our
reduction of air pollution? Heart diseases
surroundings.
(a) Never burn waste materials Sleep irregularities
(b) Adopt hybrid technology Stress
(c) Use high Sulphur diesel Hearing loss.
(d) Use renewable or sustainable materials Q.55. In a water treatment plant, you have been told
Ans : (c) Sulphur diesel: - Diesel fuel contains sulfur to check the fecal coliform levels. Which of the
which derives from the original crude oil source and following options gives the CORRECT reason behind
monitoring these levels?
can still be present after refining. After combustion in
the engine, the sulfur in fuel forms particulates that are (a) They cause diseases in humans

UPSSSC JE 2018 133 Civil KI Goli

Q.63. Which of the following factors will NOT cause M 20 20
hot weather concreting?
(a) High ambient temperature M 25 25

(b) Reducing setting time of concrete M 30 30

(c) High relative humidity
M 35 35
(d) High speed of the wind
Standard
M 40 40
Ans:(c) Hot weather concreting :- Hot weather Concrete
concreting is defined by ACI as “one or a combination of M 45 45
the following conditions that tend to impair the quality
of freshly mixed or hardened concrete by accelerating M 50 50
the rate of moisture loss and rate of cement hydration, M 55 55
or otherwise causing detrimental results
• High ambient temperature (more than 35oC) M 60 60
• High concrete temperature M 65 65
• Low relative humidity High
• High wind speed Strength M 70 70
Concrete
Q.64. Fill in the blank with a CORRECT option: M 75 75
Using a desirable water-cement ratio effective, M 80 80
economical, and appropriate _______ structures can
be achieved. Q.66. In which of the following methods, the
(a) Strength of concrete (b) Strength of cement ingredients of concrete are estimated and mixed with
both weight and/or volume according to the mix
(c) Strength of aggregate (d) Strength of admixtures
design?
Ans : (a) The water-cement ratio is the ratio of the (a) Batching of concrete (b) Randomizing of concrete
weight of water to the weight of cement used in a
(c) Sizing of concrete (d) Washing of concrete
concrete mix.
• A lower ratio leads to higher strength and durability Ans : (a) Batching : - Process of measuring the required
but may make the mix difficult to work with and form. quantities of all the ingredients of concrete in order to
• Workability can be resolved with the use of obtain the uniform proportion and grading of aggregate
plasticizers or super-plasticizers. is called batching.
• Compressive strength of concrete mixes was found to 1. Volume Batching : - Volume batching is not good for
increase with decrease in water-cement ratio. Any the proportioning the material because of volume of
decrease in water-cement ratio leads to increase in moist sand in a loose condition weights much less than
compressive strength. the same volume of dry compacted sand.
• water-cement ratio plays a significant role in • In this method of batching a wooden or steel gauge
determining the compressive strength of a concrete box is used whose volume is equal to one bag of
mix and is also an important factor for preparing cement i.e. 35 litres.
design mixes. • There is not easy to measure the granular materials in
• By considering a suitable water-cement ratio, terms of volume in volume batching method.
effective, economical, and appropriate strength of • In the method of volume batching cement is always
concrete structures can be achieved. measured by weight and aggregates in volume.
• During filling for mass the material should be filled
Q.65. Choose appropriate Mixed design Grades (as per
loosely and no any compaction is done and after
IS 456 : 2000) for ordinary concrete.
filling the top surface of gauge box is levelled .
(a) M20, M25, M30 (b) M35, M45, M55 • In terms of easiness volume batching is done for small
(c) M10, M15, M20 (d) M45, M50, M55 projects.
Ans : (c) As per IS 456: 2000 • Accuracy of volume batching is not good over the
Group Grade Specified Characteristic
weight batch .
Designation Compressive Strength of 150 mm 2. Weight Batching : - This method of batching is more
Cube at 28 Days in N/mm² accurate and all ingredients are measured in terms of
kg.
M 10 10
Ordinary • Mix. proportion are based on 50 kg bag of cement.
Concrete M 15 15 • This methods are used for a large and all important
concreting works.

UPSSSC JE 2018 136 Civil KI Goli

• Volume batching is adopted in India for all ❖ SOIL MECHANICS
construction work. Q.69. Match the following pairs CORRECTLY:
Q.67. Choose the CORRECT option(s) that talks about A. Gravel Size 1. 80-300 mm
properties of hardened concrete.
B. Sand size 2. 4.72-80 mm
1. Strength of normal concrete varies between 25 MPa
and 40 MPa C. Silt size 3. 0.075-2 mm
2. The denser the concrete the lesser its performance 4. 0.002 – 0.075 mm D. Cobble size
and greater the durability
(a) (A) - (4), (B) - (2), (C) - (1), (D) - (3)
3. Extremely resistant to physio-chemical attack
(b) (A) - (2), (B) - (3), (C) - (4), (D) - (1)
(a) 1 & 2 statements (b) 2 & 3 statements
(c) (A) - (1), (B) - (3), (C) - (2), (D) - (4)
(c) 1 & 3 statements (d) 1, 2 & 3 statements
(d) (A) - (4), (B) - (1), (C) - (3), (D) - (2)
Ans : (c) Properties of Harden Concrete: Ans : (b) Indian standard soil classification system: -
Strength:- The strength of normal concrete varies
Soil Type of soil Subgroup Size range
between 25 MPa and 40 MPa. However, there is high-
group (mm)
performance concrete, which has strength above 50
MPa. Very Boulder >300
Durability: - Extremely resistant to physio-chemical coarse Cobble 80-300
attacks, like rain, frost, pollution, etc. Well-suited for soils
the structures exposed to extreme conditions. Coarse 20-80
Density and Porosity: - The denser the concrete the Gravel
Fine 4.75-20
lesser the pores and the higher will be the strength.
Hence, the performance of concrete will be higher as Coarse Coarse 2-4.75
well as durability (as durability of any material is Sand Medium 0.425-2
directly proportional to its strength and performance). Fine 0.0075-0.425
Fire Resistance : - Acoustic and thermal insulation
properties Resistance to impact. Fine Silt 0.002-0.075
soils
Q.68. Identify the mixer used in the mixing process, in
which the axis of the mixer is always horizontal, and Clay <0.002
discharge takes place by inserting a chute into the Q.70. You are given 1 m³ of wet soil that weights 30
drum or by reversing the direction of rotation of drum. kN. Moreover its dry weight is 25 kN. You have to
(a) Lot mixer (b) Tilting drum mixer determine the water content of the given soil sample.
(c) Non-tilting drum mixer (d) Dual drum mixer Which of the following options represent the
CORRECT answer?
Ans : (c) Non-tilting drum mixer :-
(a) 30% (b) 5%
• A reversing drum mixer (also commonly called a non-
tilting mixer) is a type of concrete mixer that produces (c) 20% (d) 16.7%
concrete in single batches. Ans : (c) Given: Total weight of soil sample= 30kN
• The entire drum rotates around its axis as materials Dry weight of soil sample = 25 kN
are loaded through a charge chute at one end of the Weight of water in soil sample =
drum and exit through a discharge chute at the
Total weight – Dry weight
opposite end of the drum.
Mw = 30 - 25 = 5kN,
Tilting drum mixer :-
Ms = 25kN
• Tilting drum mixer means the drum will discharge
𝐦𝐚𝐬𝐬 𝐨𝐟 𝐰𝐚𝐭ⅇ𝐫
concrete by tilting downwards. Water content =
𝐦𝐚𝐬𝐬 𝐨𝐟 𝐬𝐨𝐥𝐢𝐝
• It is a rapid discharge process and is used for larger 𝟓
= × 100 = 20 %
projects. 𝟐𝟓
• Rapid means it delivers concrete by gravity that is Q.71. In a three-phase diagram for soil, the volume of
tilting the drum downwards because of this the the void is given as:
concrete mix obtained will not be subjected to Vv = V - VS, where V = total volume and VS = volume of
segregation. the solids According to this relation, identify the
• Low workable concrete containing large-sized porosity(n) of soil?
aggregates greater than 7.5cm is mixed efficiently (a) n = V/Vs (b) n = Vv/V
with this tilting type mixer.
(c) n = Vs/Vv (d) n = Vv/Vs

UPSSSC JE 2018 137 Civil KI Goli

Ans : (b) Void ratio (e):- Void ratio is usually defined as Q.75. You are given two fluids f1, and f2, having
the ratio of the volume of voids to the total volume of densities d1 and d2. Which of the following options
soil solid. represent the CORRECT relation between their specific
𝑽 volumes Sv1 and Sv2 consideration d1 > d2?
𝒆 = 𝑽𝒗
𝒔 (a) Sv1 < Sv2 (b) Sv1 > Sv2
Porosity (n): Porosity is defined as the ratio of the (c) Sv1 = Sv2
volume of voids to the total volume of the soil. (d) Cannot be determined due to insufficient
𝑽𝒗
𝒏= information
𝑽
The relationship between void ratio and porosity is as Ans : (a) Given , Densities as d1 and d2
𝐯𝐨𝐥𝐮𝐦ⅇ
follows Specific volume =
𝐧 ⅇ 𝐦𝐚𝐬𝐬
ⅇ = 𝟏−𝐧 n= 𝑽
𝟏+ⅇ S=
𝑴
3 𝐌𝐚𝐬𝐬 𝑴
Q.72. The mass density of oil is 530 kg/m . What is its Mass density = =
𝐕𝐨𝐥𝐮𝐦ⅇ 𝑽
relative density? 𝟏 𝟏
(a) 0.53 (b) 0.23 (c) 0.34 (d) 0.95 Specific volume = =
𝑫𝒆𝒏𝒔𝒊𝒕𝒚 𝑫
𝟏 𝟏
Ans : (a) Given , 𝝆 = 530 kg/m³ and 𝝆w = 1000 kg/m3 So, Sv1 = Sv2 =
𝒅𝟏 𝒅𝟐
We know that
𝐃ⅇ𝐧𝐬𝐢𝐭𝐲 𝐨𝐟 𝐬𝐮𝐛𝐬𝐭𝐚𝐧𝐜ⅇ
According to question
Relative density =
𝐃ⅇ𝐧𝐬𝐢𝐭𝐲 𝐨𝐟 𝐰𝐚𝐭ⅇ𝐫 d1 > d 2
𝟓𝟑𝟎
= = 0.53 𝟏
<
𝟏
𝟏𝟎𝟎𝟎
𝒅𝟏 𝒅𝟐
Q.73. Fill in the blank with an appropriate word:
So Sv1 < Sv2
If the liquid limit of the soil is more than 50%, the soil
is called to have ______. Q.76. Match the following types of soil with their soil
(a) high plasticity (b) low plasticity name.
(c) high elasticity (d) low porosity 1. Talus A. Lake

Ans : (a) Property of soil that allows it to be moulded 2. Loess B. Glacial

into desired shape without changing its volume and 3. Lacustrine C. Wind
without cracks called plasticity. 4. Stratified Drift. D. Gravity
Plasticity Index :- It is the indicative of the range of (a) (1) - (D), (2) - (C), (3) - (A), (4) - (B)
water content over which the soil remains in plastic
(b) (1) - (B), (2) - (C), (3) - (D), (4) - (A)
state.
(c) (1) - (B), (2) - (D), (3) - (A), (4) - (C)
Ip = WL – WP
(d) (1) - (D), (2) - (B), (3) - (A), (4) - (C)
Liquid limit (WL) Plasticity
Ans : (a) Different types of soil deposit according to
< 35 % Low Plasticity transportation agency and method of deposition:
35 – 50 % Medium Plasticity 1. Lacustrine soil: - Soil which is deposited from
> 50 % High Plasticity suspension in fresh still water of the lakes is known as
Lacustrine soil.
Q.74. If sandy soil is tested in a directed shear box in
2. Glacial deposit: - It is the deposits that have been
the saturated condition, what will be the lateral force
transported by ice.
at failure if ϕ = 25° and the normal load is 26 N?
Drift is a general term used for the deposits made by
Given : Value of c for sandy soil = 0 and tan25° = 0.466
glaciers directly or indirectly.
(a) 25.20 N (b) 12.12 N
3. Aeoline Soil: - Soil which is formed due to
(c) 14.17 N (d) 20.22 N transportation by wind is known as Aeoline soil. It is
Ans : (b) Given: For sandy soil, c = 0, also a transported soil.
Normal Load = 26 N, tan25° = 0.4666 Loess is the wind deposited of uniform particles of silt.
Now, P = C + σ tan ϕ 4. Colluvial deposit: - Soil transported under the action
= 0 + 26 + tan 25 of gravity is called colluvial soil.
= 12.12 N Talus is the soil transported by gravity.
5. Marine soil: - Soil which is deposited from the
suspension in sea water is known as marine soil.
UPSSSC JE 2018 138 Civil KI Goli
Q.77. Identify the option that has the CORRECT order
of the geological cycle for the disposition of soil.
(a) transportation-deposition-weathering- upheaval
(b) weathering-transportation-deposition- upheaval
(c) transportation-upheaval-deposition- weathering
(d) weathering-upheaval-transportation- deposition
Ans : (b) cycle of soil formation is : -
Weathering- transportation- deposition - upheaval

Q.78. Identify the image representing combined

footing. Q.79. The critical angle of repose is the steepest angle
of dip, which the exposed face of the soil makes
relative with which of the following ?
(a) Vertical plane
(b) Horizontal plane
(c) Perpendicular to the inclined plane of the soil/
(d) No fixed convention is there
Ans : (b) Angle of repose: The angle of repose of a soil
(a) is the maximum angle that the outer face of the soil
mass makes. The angle of repose or the critical angle of
repose is the steepest angle of dip relative to the
horizontal plane to which a material can be pulled
without slumping.
• At this angle, the material on the slope face is on the
(b) verge of sliding. The angle of repose can range from 0°
to 90°. The morphology of the material affects the
angle of repose.
Q.80. Consider the weight of soil sample and weight
of water 30 kg and 10 kg respectively. What will be the
(c) value of weight of solids (in kg) on considering weight
of air as null or 0?
(a) 40 (b) 300 (c) 20 (d) 3
Ans : (c) Given , weight of soil sample = 30 kg
weight of water = 10 kg
(d) We know that
Ans : (d) Whenever two or more columns in a straight Weight of soil sample = Weight of soil solid + Weight of
line are carried on a single spread footing, it is called a water + Weight of air
combined footing. weight of soil solid = 30 - 10 = 20 kg
Combined footings are provided only when: Q.81. Fill in the blank(s) with the appropriate option.
(i) When two columns are close together, causing It is a fine-grained soil when the percentage of soil
overlap of adjacent isolated footings retained on 75-micron sieve is ______ than 50% and is
(ii) Where soil bearing capacity is low, causing overlap coarse grained when the percentage of soil retained
of adjacent isolated footings on the 75-micron sieve is _______ than 50%.
(iii) When the end column is near a property line so (a) Greater, Greater (b) Less, Greater
that its footing cannot spread in that direction. (c) Less, Less (d) Greater, Less

UPSSSC JE 2018 139 Civil KI Goli

Ans : (b) Group Symbol Classification
Unified Soil Classification Coarse soils
Soil Sample Passing 75μ GW Well-graded GRAVEL
(0.075mm sieve)
GP Poorly-graded GRAVEL
GM Silty GRAVEL
Coarse grained soil Fine grained soil GC Clayey GRAVEL
Retained On 75µ Sieve Retained On 75µ SW Well-graded SAND
Passing from 4.75 mm Sieve Sieve < 50% SP Poorly-graded SAND
Q.82. You have given a saturated sand sample and you SM Silty SAND
have to measure the undrained shear strength. Which SC Clayey SAND
of the following type of tests should you perform on
Fine soils
such a sample?
(a) Vane shear (b) Triaxial shear ML SILT of low plasticity
(c) Direct shear (d) Unconfined compression MI SILT of intermediate plasticity
Ans : (b) Triaxial Test:- This is the most widely used MH SILT of high plasticity
shear strength test and is suitable for all types of soil. CL CLAY of low plasticity
• Drainage can be controlled, whatever be the type of
Cl CLAY of intermediate plasticity
soil i.e. sand can be tested under undrained
conditions and clay can be tested under drained CH CLAY of high plasticity
condition. OL Organic soil of low plasticity
• Pore water pressure can be measured.
Ol Organic soil of intermediate plasticity
• Volume changes can also be measured.
OH Organic soil of high plasticity
• Failure plane is not predetermined.
• There is no rotation of principal stresses during test. Pt Peat
• Stress distribution on failure plane is fairly uniform. Q.84. Which of the following options give the
• The triaxial cell is filled with water and specimen is CORRECT relation of the effective stress(e) with total
sealed inside a rubber membrane. Cell pressures stress(s) and pore water pressure (u)?
applied (called confining pressure) . (a) e = u/s (b) e = su
• With cell pressure held constant, additional axial (c) e = s/u (d) e = s - u
stress is applied gradually until sample fails at
Ans : (d) Effective stress:- It is the part of the total
additional axial stress .
stress that is resisted by soil particles by grain to grain
• Application of additional axial stress, also called
interaction.
deviator stress, produce shear stresses with in soil
mass on all planes except horizontal & vertical planes. • It is also known as intergranular stress.
• Concept of effective stress is developed by Terzaghi.
Q.83. Match the Pair CORRECTLY:
• The difference between the total stress and the pore
A. Well graded Gravel 1. SP pressure is called effective stress.
B. Poorly graded Gravel 2. GW e=s−u
C. Well graded Sand 3. SW where e = effective stress, s = total stress, u = pore
D. Poorly graded Sand 4. GP water pressure
(a) (A) - (2), (B) - (4), (C) - (3), (D) - (1) Q.85. Consider the list of the varieties of soils:
(b) (A) - (3), (B) - (2), (C) - (1), (D) - (4) i) Under-consolidated
(c) (A) - (1), (B) - (4), (C) - (2), (D) - (3) ii) Normally consolidated
(d) (A) - (3), (B) - (2), (C) - (4), (D) - (1) iii) Pre-consolidated
Ans : (a) According to IS Classification system, the soils Which of these is/are completely consolidated under
can be classified into 18 groups. the overburden stress?
Soil classification using group symbols is as follows: (a) i and ii (b) ii and iii
(c) Only ii (d) Only iii

UPSSSC JE 2018 140 Civil KI Goli

Ans : (c) Normally consolidated :- A soil is said to be Type of beam Support conditions
normally consolidated If the effective overburden
Fixed Beam Both the end are fixed with a wall or
pressure that it is currently experiencing is the
column.
maximum it has ever experienced in its history.
OCR for this type of soil is less than 1 Cantilever One end is fixed and another end is
Beam free
Pre-consolidated or over consolidated soil:- A soil is
said to be over-consolidated if the present overburden Propped One end is fixed and another end is
pressure is less than the effective overburden pressure Cantilever roller or hinged supported
it has experienced in the past. Beam
OCR for this type of soil is greater than 1 Simply One end is hinged and another end
Under-consolidated : - Soil is under consolidated when Supported is roller supported.
it is not completely consolidated under the present Beam (SSB)
vertical overburden pressure Continuous when the beam is Continuously
Beam supported by 2 or more support.
❖ STRENGTH OF MATERIAL
Q.86. Which Beam is used for a beam that has more Q.88. Which stress is calculated by dividing the instant
than two spans and more than three supports along load to the reduced cross-sectional area at that point?
its length in a straight line? (a) True stresses (b) Nominal stress
(a) Semi-continuous beam (b) Cantilever beam (c) Working stresses (d) Factor of safety
(c) Simply supported beam (d) Continuous beam Ans : (c) True stress: - It is the stress determined by the
Ans : (d) The different types of beam are: instantaneous load acting on the instantaneous cross-
sectional area.
Continuous beam: - A beam supported on more than
𝑷
two supports is known as a continuous beam. σt=
𝑨
Cantilever beam: - A beam fixed at one end and free at Where, P = Instant load A = Area of cross-section
the other end is known as a cantilever beam. Nominal stress: - Nominal stress or engineering stress
Simply Supported beam: -A beam supported at its both is defined as the force ratio per initial cross-sectional
ends is known as a simply supported beam. area (original cross-section area).
Fixed beam: A beam whose both ends are fixed, is Working stress: - Working stress is the safe stress taken
known as a fixed beam. within the elastic range of the material.
Overhanging beam: A beam having its end portion For brittle materials, it is taken equal to the vided by a
extended beyond the support, is known as overhanging suitable factor of safety. However, for materials
beam. A beam may be overhanging on one side or on possessing a well-defined yield point, it is equal to yield
both sides. stress divided by a factor of safety.
The factor of safety: - The factor of safety is a number
used to determine working stress. It is fixed based on
the experimental works on the material.
Q.89. Assume that the applied force on an object is 16
N when failure occurs and the lap area of the object is
4 m2. What will be the Average Shear Stress (ASS) (in
N/m2) value?
(a) 8 (b) 4 (c) 2 (d) 1
Ans : (b) Given data:
Force F = 16N Area A = 4m²
Average shear stress is to be calculated as we know
Q.87. Which type of beam is fixed at both the ends of 𝑭 𝟏𝟔
the column or wall? ASS = = = 4 N/ m2
𝑨 𝟒
(a) Simply supported beam Q.90. Consider a situation where a failure occurs for
(b) Overhanging beam an object with lap area of 6 m2 and the average shear
(c) Fixed beam strength value (ASS) as 2. What will be applied force
(d) Staggered beam on the object (in N)?
(a) 18 (b) 24 (c) 12 (d) 8
Ans : (c) Beam: Beam is a structural member which is
subjected to transverse loading to its axis. Ans : (c) Given, Area (a) = 6 m2

UPSSSC JE 2018 141 Civil KI Goli

Average shear strength value (ASS) = 2 • Add stability to a structure and used in
Average shear stress is to be calculated as we know areas, like floors and ceilings, to
𝑭 withstand large amounts of stress.
Ass =
𝑨
𝑭 • Made of 2 materials with different
𝟐=
𝟔 properties.
F= 2 × 6 =12N • Constructed from more than one
Q.91. The average shear stress for a rectangular material to increase strength, stiffness
Composite
section is 20 MPa. Calculate the maximum shear and to reduce cost
Beam
stress. • Example of common Composite Beam
(a) 30 MPa (b) 33.2 MPa i.e., I-beams where the web is plywood
and the flanges are solid wood
(c) 35 MPa (d) 30.2 MPa
members
Ans : (a) Given , τavg = 20 MPa
• The beam that rests on two supports
τmax=1.5×τavg and moves freely in the horizontal
τmax = 1.5 × 20 = 30 MPa direction and is subjected to bending
Simply
Q.92. Fill in the blank with the CORRECT option: and shearing.
Supported
The magnitudes of both slope and deflection are zero • Typical practical applications of simply
Beam
at ______ support. supported beams with point loadings
include bridges, beams in buildings,
(a) Free end (b) Fixed support
and beds of machine tools.
(c) Simple support (d) Fringe support
• The beam supported between two
Ans : (b) Fixed beam : - A beam whose both ends are fixed ends is known as a fixed beam or
fixed is known as a fixed beam. Fixed restrained beam.
The slope and deflection are zero at the support which Beam • The rotation and deflection at the ends
can be shown by the deflected shape of the beam as is 0, So there is no degree of freedom
below at the ends.

❖ STEEL
Q.94. A mild steel flat 175 mm wide, 10 mm thick and
4.5 m long and is carrying an axial load of 23.5 kN.
Determine stress developed in this case.
(a) 12.65 N/mm2 (b) 20.30 N/mm2
Q.93. Match the following types of beams with their (c) 13.42 N/mm2 (d) 14.89 N/mm2
respective functionalities.
Ans : (c) Given: 175 mm wide and 10 m thick plate
Reinforced Experiences bending
1. a. Cross-Sectional Area, A = 175 × 10 = 1750 mm2
concrete beam and shearing
To carry transverse Applied axial Load, P = 23.5 kN = 23500 N
2. Composite beam b. We know that
external loads
Simply supported No freedom to rotate at 𝑷
3. c. σ=
𝑨
beam ends 𝟐𝟑𝟓𝟎𝟎
Two materials joined as σ=
4. Fixed beam d. 𝟏𝟕𝟓𝟎
a unit σ = 13.42 N/mm2
(a) (1) - (b), (2) - (d), (3) - (a), (4) - (c)
Q.95 If the value of design wind speed is 35.2 m/s,
(b) (1) - (a), (2) - (d), (3) - (c), (4) - (b)
then determine the value of the design wind pressure.
(c) (1) - (c), (2) - (d), (3) - (b), (4) - (a) (a) 750 N/m2 (b) 743.42 N/m2
(d) (1) - (d), (2) - (b), (3) - (a), (4) - (c) (c) 734.24 N/m2 (d) 744 N/m2
Ans : (a) Various type of the beam and their purpose :
Ans : (b) Given: Vz = Design wind speed = 35.2 m/sec
• It is a concrete beam reinforced by We know that
steel
Reinforced Pz = 0.6 x V2
• Supports large vertical or transverse
Beam loads Pz =0.6 x (35.2)2
• Generally used in large buildings for Pz = 743.424 N/m²
longitudinal support

UPSSSC JE 2018 142 Civil KI Goli

Ans : (d) Partial factor of safety for concrete and steel not less than O-16 percent where high-strength
should be taken as 1.5 and 1.15, respectively when deformed bars are used.
assessing the strength of the structures or structural • The pitch of the distribution bars shall not be more
members employing limit state of collapse. than five times the effective depth or 450 mm
In Limit state method , partially safety factor - whichever is smaller.
• The horizontal distance between two parallel main
Material Collapse Deflection Cracking reinforcements shall not be more than three times
Concrete 1.5 1 1.3 the effective depth of the slab or 450 mm whichever
is smaller.
Steel 1.15 1 1 • Reinforcements shall be so placed that they do not
Q.111. In which of the following types of concrete, the touch bricks at any point.
strength is aided by placing the cables, steel rods, and • A minimum cover of 25 mm shall be provided all-
wires in the concrete before it sets? round the reinforcement.
(a) Prestressed concrete (b) Reinforced concrete • The formwork for the RB and RBC floor or roof shall
not be removed before 14 days of laying.
(c) Ordinary/Plain concrete (d) Precast concrete
• The bricks shall be laid with cement mortar 1:3.
Ans : (a) Prestressed Concrete: - This type of concrete
involves placing steel tendons, rods, or wires in the
concrete before it sets, and then tensioning them either Q.113. Identify the machinery used in the process of
before the concrete sets (pre-tensioning) or after it has Quarrying.
hardened (post-tensioning).
• The tension in the steel creates a compressive stress
in the concrete, counteracting the tensile stresses
that the concrete will be subjected to in service.
• This helps in increasing the strength, especially in
flexural members like beams, slabs, and bridges.
❖ BCME
(a) Dozer (b) Side Discharge loader
Q.112. Which of the following statements are
(c) Backhoe (d) Dragline
CORRECT with respect to Reinforced brickwork,
according to the IS 10440 (1983) : Code of practice for Ans : (c) Backhoe: - A backhoe is an excavating tool that
construction of RB and RBC floors and roofs. consists of a metal plate attached to a long handle with
a. All bricks shall be thoroughly saturated by an acute angle.
submerging them in clear water for at least four hours The plate having a sharp edge is used to excavate the
before use soil. For small work of excavation, it is a widely
b. The formwork for the RB and RBC floor or roof shall preferred tool.
NOT be removed before 14 days after laying
c. The mortar shall be a mix of 1 ∶ 1, 1 part of fresh
Portland cement to 1 parts of coarse sand
d. The rods should be kept in touch with bricks
(a) a & b only (b) a & c only
(c) b & c only (d) a & d only
Ans : (a) As per IS 10440- 1983, the following points
should be remembered: Dragline : - Since the basic character of the machine is
• Bricks shall be kept immersed in water for 4 to 6 dragging the bucket against the material to be dug, it is
hours and removed about 15 to 20 minutes before called as DRAGLINE.
they are used so that their skin is dry.
• One way RB slabs with freely supported ends, shall be
designed to resist a bending moment near midspan of
WL/8, where W is the total uniformly distributed load
over the span and L is the effective span.
• The reinforcement in either direction shall however
not be less than 0.20 percent of the cross-sectional
area of the slab where plain steel bars are used and

UPSSSC JE 2018 146 Civil KI Goli

Side discharge loader : - It is a track-mounted electro- Q.115. Identify the image depicting shuttering.
hydraulically operated side discharging bucket loader
with a smooth control system and excellent
maneuverability. It is a low-profile high-output
machine.

(a) (b)

Dozer: The meaning of dozer is a tractor-driven (c) (d)

machine usually having a broad horizontal blade for Ans : (d) Shuttering: Shuttering, also called formwork,
moving earth. is a temporary support used to mold concrete into the
desired shape, in which the concrete gains initial
strength, hardens, and matures. As we produce PCC,
RCC construction elements in the form of shuttering for
buildings, bridges, tunnels, hydroelectric dams, sanitary
pipelines, and more, shuttering comes in different
shapes and sizes.

Q.114. Consider the following statements.

a. Cracks caused due to shrinkage in masonry walls
can be minimized by excess use of rich cement mortar
in masonry
b. Cracks in masonry can also be reduced by delaying
plasterwork until masonry has dried after proper
curing has undergone most of its initial shrinkage
(a) Only a is true
(b) Only b is true
(c) Both are true
(d) Neither of the two are true
Ans : (b) Shrinkage due to drying out of moisture
content in any building materials such as cement, or
concrete is one of the main causes of cracks in
structures.
• Cracks caused due to shrinkage in masonry walls can Q.116. According to IS : 10440(1983), the formwork
be minimized by avoiding the use of rich cement for the RB and RBC floor or roof shall NOT be removed
mortar in masonry. This is because rich mortar before _____ after laying
contains a very high amount of cement which in turn (a) 28 (b) 7 (c) 14 (d) 21
demands a high amount of water consequently, due Ans : (c) As per IS 10440: 1983, Clause 7.6 :
to this water requirement, when the water dries out The formwork for the RB and RBC floor or roof shall not
quickly then shrinkage cracks appear on the wall. be removed before 14 days after laying.
• It can also be reduced by delaying plasterwork until
As per IS 456 2000, minimum period before striking
masonry has dried after proper curing has undergone
formwork for various components are as follows : -
most of its initial shrinkage.

UPSSSC JE 2018 147 Civil KI Goli

 Minimum Bar 5 60,00
Type of Formwork Period Before 0
Striking
Vertical formwork to columns, walls, 16-24 hours < 0.0015 0.0025
beams 60,00
Soffit formwork to slabs (Props to be 3 days 0
re-fixed immediately after removal Any 0.0015 0.0025
of formwork)
Soffit formwork to beams (Props to 7 days Welded Any Any 0.0012 0.0020
be re-fixed immediately after wire
removal of formwork) Q.119. Fill in the blank with the CORRECT option with
Props to slabs: 1) Spanning up to 4.5 7 days respect to staircase:
m
The horizontal projection between the first and the
Props to slabs: 2) Spanning over 4.5 14 days
last riser of an inclined flight is termed as
m
(a) Pitch (b) Going
14 days
Props to beams and arches: 1) (c) Nosing (d) Landing
Spanning up to 6 m
21 days Ans: (a) Pitch:- It is the horizontal projection between
Props to beams and arches: 2)
the first and the last riser of an inclined flight or It is the
Spanning over 6 m
angle at which the line of the nosing of the stairs makes
Q.117. Choose a suitable option. with the horizontal.
Rubber bitumen mastic is used for which of the The maximum pitch for a domestic building should not
following purpose(s)? exceed 42° and for a public building, it should not
1. Waterproofing exceed 33°.
2. Repairing Q.120. "Cleaning drainage gutters, servicing lifts, and
3. Adhesion other items that should be dealt with regularly." This
is an example of:
(a) Only 1 (b) Only 2
(a) Planned or preventive maintenance
(c) Both 1 and 2 (d) All 1, 2 and 3
(b) Reactive maintenance
Ans : (d) Rubber Bitumen Mastic: - Bitumen-rubber-
(c) Cyclical maintenance
based, elastic mastic is used for waterproofing,
repairing, and adhesion purposes. (d) Corrective maintenance
• The composition contains a crumb of rubber. Ans : (c) Types of Maintenance in Building : -
• It has decent anti-corrosion properties. Cyclical maintenance: - It is working to maintain the
• Used to cover metal structures. general condition of the property and some communal
fittings.
Q.118. According to Building code requirement for the
It includes external & internal re-decoration of areas,
structural concrete (ACI 318-19), minimum transverse
cleaning, checking, repairing, and replacing guttering
reinforcement for Cast-in-place walls using deformed
and downpipes, servicing lifts, etc.
bars is
Corrective (Remedial) maintenance: It is maintenance
(a) 0.05%-0.10% (b) 0.45%-0.60%
that is done after the defects or damages occur in the
(c) 0.75%-0.90% (d) 0.20%-0.25% structure. It is used to rectify and repair the damaged
Ans : (a) Minimum transverse reinforcement for all structure. It deals with clogged sewer lines or
conditions ie when bars are less than and more than 5 overflowing manholes or backing up of sewage into a
varies between 0.0020-0.0025 OR 0.2-0.25% house or structural failure of the system.
The minimum longitudinal and transverse Routine maintenance: - It is also known as service
reinforcement as per ACI 318-19 for Cast-in-place maintenance which is done from time to time to avoid
walls is as follows : deterioration of the structure.
Type of Bar/ Fy, psi Minimum Minimum Preventive maintenance: - This is a type of
Reinforcem wire Longitudinal Transverse maintenance that is done before any defects or
ent Reinforceme Reinforceme damages occur in a structure.
Size
nt nt This helps to prevent any further damage to Building.

Deformed ≤ No ≥ 0.0012 0.002

UPSSSC JE 2018 148 Civil KI Goli

II. The aim is that the amount of earth cutting Ans : (a) These compounds are known as Bogue’s
practically equals the amount in the embankment at a Compounds.
given location Tricalcium silicate (C3S): - This is also called Alite. This is
III. The additional surplus amount of earth is used to also responsible for the early strength of the concrete.
form spoil banks The cement that has more C3S content is good for cold
(a) II and III (b) I and III weather concreting.
(c) I and II (d) I, II and III The heat of hydration is 500 J/Cal. Its proportion is 25 -
Ans : (d) Cutting/Filling of the canal: - Generally, a 50%.
canal with a given terrain can either be in full-filling, full Dicalcium Silicate (C2S): - This compound will undergo a
cutting or partly cutting and filling. reaction slowly. It is responsible for the ultimate
• When the natural surface level (NSL) is above the top strength of concrete.
of the bank of the canal, the entire canal section will This is also called Belite. The heat of hydration is 260
have to be in cutting. This situation is known as Canal J/Cal. Its proportion is 25 - 40%.
in Cutting. Tricalcium aluminate (C3A): - Celite is the quickest one
• When NSL is lower than the bed level of the canal, to react when water is added to the cement. It is
the entire canal section will have to be in filling. This responsible for the flash setting.
situation is known as Canal in filling. The increase of this content will help in the
Balancing depth or economical depth:- A canal section manufacture of Quick Setting Cement. The heat of
is called of balancing depth if, in canal section hydration is 865 J/Cal. Its proportion is 5 - 11%.
"Quantity of earth cutting = Quantity of earth filling" Tetra calcium Alumino ferrite (C4AF): - This is called
• It is only possible when section is partially in cutting Felite. The heat of hydration is 420 J/Cal.
and partially in filling. It has the poorest cementing value but it is responsible
Spoil Banks: - Additional surplus amount of excavated for the long-term gain of strength of the cement. Its
earth is used to make spoil bank. proportion is 8 - 14%.
• The banks constructed from surplus excavated earth Q.140. Choose the CORRECT percentage of 'silica'
on the side cutting parallel to the road alignment in added to a brick to make an exceptional class brick
trapezoidal form is called spoil bank. without cracks.
• The aim of canal excavation is that the amount of eart (a) 50%-60% (b) 80%-90%
h work in cutting is equal to the amount in the emb- (c) 5%-10% (d) 20%-30%
ankment at a given location. Ans : (a) Good Brick Earth Constituents table:
❖ STRUCTURAL ANALYSIS Constituents Percentage Properties Impart
Q.138. What is the carry-over factor for a beam hinged
Alumina 20-30 Plasticity
at both ends?
(a) 1.5 (b) 1 (c) 0.5 (d) 0 Silica 50-60 Uniform shape
Ans : (d) Carry-over factor (COF): It is the ratio of Lime 4-5 Prevent cracking and
moment produced at the far end to the applied shrinkage
moment at that support end.
Oxide of 5-6 Imparts red-brown color
Far End Condition COM COF Iron and strength
Far end is fixed M/2 1/2
Magnesia <1 Yellow tint color to brick
Far end is hinged 0 0
Q.141. Fill in the blank with the CORRECT option:
For cantilever beam -M -1
_______ and _______ are the two major natural forces
For guided roller support -M -1 responsible for causing defects in timber.
❖ BUILDING MATERIAL (a) insects, rupture of tissues
Q.139. Out of the following options, which majorly (b) abnormal growth, rupture of tissues
contribute to the ultimate or final strength of the (c) fungi, abnormal growth
cement? (d) fungi, insects
(a) Dicalcium silicate (b) Tricalcium aluminate Ans : (b) Defects in timber: - Abnormal growth and
(c) Tetra calcium aluminoferrite (d) Gypsum rupture of tissues are the two major natural forces
responsible for causing defects in timber

UPSSSC JE 2018 152 Civil KI Goli

Various types of defects in Timber are given below: 1. Concrete 2. Bricks 3. Clay
(1). Knots: - These are the bases of branches or limbs (a) Only 1 and 2 (b) Only 2 and 3
which are broken or cut off from the tree. (c) Only 1 and 3 (d) 1, 2 and 3
It is formed in a portion from which the branch is Ans : (d) Building material : - Building material is any
removed and receives nourishment from the stem for a material used for construction purposes such as
pretty long time. materials for house building.
(2). Rind galls: - The rind means bark and the gall Classification of building material:-
indicates abnormal growth. Hence peculiar curved
(a) Natural building material :- Natural materials can
swellings found on the body of a tree are known as the
be defined as building materials that are easily
rind galls
accessible in the natural world and can be utilised as
(3). Shakes: These are cracks that partly or completely building materials with little modification. Wood, mud,
separate the fibers of the wood. stone, and sand are some typical examples of natural
(4). Bow: - This defect is indicated by the curvature materials.
formed in the direction of the length of timber. (b) Artificial building Material :- These materials are
(5). Case-Hardening: - The exposed surface of timber made from different types of raw materials through
dries very rapidly. It therefore shrinks and is under rigorous physical and chemical procedures. The process
compression. of manufacturing synthetic building materials results in
The interior surface which has not completely dried is significant alterations to the original material and the
under tension. final product shows a remarkable variation from the
(6). Check: - A check is a crack that separates fibers of originally used raw materials. Concrete ,Brick, cement,
the wood. It does not extend from one end to the bricks, steel, glass, etc. are typical artificial materials.
other. Q.144. Specify the statements that are CORRECT with
(7). Honey-Combing:-Due to stresses developed during respect to carrying out brickwork on site:
drying, the various radial and circular cracks develop in a. Bricks should be completely dry while creating a
the interior portion of timber. wall
• The timber thus assumes the honey-comb texture b. No brickwork shall be carried out during rains
and the defect so developed is known as the honey- unless special arrangements are made to protect
combing. the brickwork from rain for 24 hours
Q.142. Which of the following processes is used to c. Fresh cement mortar should be used within 24
extract or take out stones from the natural rock bed? hours of mixing
(a) Quarrying (b) Placer mining (a) a and c (b) b only (c) c only (d) b and c
(c) Fertilization (d) Building Ans : (b): Brickwork: - Brick to be used in brickwork
with mortar joints shall be moistened with water from
Ans : (a) Quarrying is the multistage process by which
three to four hours before they are used to ensure a
rock is extracted from the ground and crushed to
proper adhesion with mortar.
produce aggregate, which is then screened into the
sizes required for immediate use, or for further • In the case of cement mortar, the time limit is 90
processing, such as coating with bitumen to make minutes after adding water but when surkhi or
bituminous macadam or asphalt. cinders are mixed with mortar as an ingredient, it
shall be used within twenty-four (24) hours of
Methods of Quarrying :-
grinding.
Method Suitability Example • Do not lay bricks when it’s raining to minimize the risk
Sand stone, Lime
Costly, soft and of efflorescence as much as possible. If it starts to rain
Wedging stone, Laterite, Marble
stratified rock. while you’re working, cover the fresh brickwork over
and Slate etc.
Those rock whose with a material such as plastic.
Heating thermal expansion Granite, gneiss etc. • A systematic bond must be maintained throughout
is very low. the brickwork. Vertical joints shouldn’t be continuous
To get stone at a Serpentine, Gypsum, but staggered.
Digging
small scale. Aterite • The joint thickness shouldn’t exceed 1 cm. It should
Obtaining stone in be thoroughly filled with the cement mortar from 1:4
Channeling -
the form of block. to 1:6.
To obtain stone at a • The cover is applied over or the geometrical form is
Blasting -
large scale. given to a part of the structure to enable it to shed
Q.143. Choose the materials which are considered as rainwater.
building materials.

UPSSSC JE 2018 153 Civil KI Goli

• All bricks should be placed on their bed with frogs on Q.148. What is the name of the building material
top. which is used for construction that hardens, sets and
Q.145. Which type of material is mainly used in adheres to other materials to bind them together?
making insulating tubes, switch blocks and lamps (a) Structural Steel (b) Cement
sockets ? (c) Reinforcement Steel (d) Binding Wires
(a) Earthenware (b) Terracotta Ans : (b) Cement : - A cement is a binder, a substance
(c) Porcelain (d) Polished terra cotta used for construction that sets, hardens, and adheres
Ans : (c) Porcelain: Porcelain is a ceramic material to other materials to bind them together.
made by heating substances, generally including Q.149. Match the following type of cement
materials such as kaolinite, in a kiln to temperatures conforming to their respective IS codes.
between 1200°C and 1400 °C. a. OPC 53 Grade 1. IS 269
Porcelain is used for tableware, decorative objects,
b. OPC 33 Grade 2. IS 12269
laboratory equipment, and electrical insulators like
switches, sockets, tubes, etc. c. OPC 43 Grade 3. IS 8112
Terracotta: The most common uses for earthenware (A) a - 1, b - 3, c - 2 (B) a - 3, b - 2, c - 1
are roofing tiles and terracotta pots and dinnerware. (C) a - 2, b - 1, c - 3 (D) a - 1, b - 2, c - 3
Terracotta is used in the manufacturing of water and Ans : (c) The different IS Codes for the production of
waste water pipes, roofing tiles, bricks, and surface different types of cement are tabulated below :
embellishment in building construction.
IS Code Type of Cement
Q.146. Consider the given statements, identify ratio(s)
Specifications for PPC (Portland pozzolana
that describe the CORRECT relation for water and IS 1489
cement).
cement ratio respectively
IS 6909 Specifications for SSC (super-sulfated cement).
i. Volume of concrete to that of water
Specifications for RHPC (Rapid Hardening
ii. Weight of concrete to that of water IS 8041
Portland cement)
iii. Volume of water to that of cement
IS Specifications for SRPC (sulphate resistant
iv. Weight of water to that of cement 12330 portland cement).
(a) Only iv (b) Only iii
Specifications for HAC for structural use (high
(c) iii and iv (d) i and ii IS 6452
alumina cement).
Ans : (c) Water cement ratio: The water-cement ratio is IS
Specifications for 53 grade OPC
defined as the ratio of the weight of water to the 12269
weight of cement used in a concrete mix. IS 8112 Specifications for 43 grade OPC
𝒘𝝎 (𝒎𝒈)𝝎 𝑽𝝎
Water-cement ratio = = (𝒎𝒈)𝒄
= IS 269 Specifications for 33 grade OPC
𝒘𝒄 𝑽𝒄
Where, Q.150. Bogue's compound composition of cement
Ww = Weight of water. W = Weight of cement. consists of
Vw = Volume of Water, V = Volume of Cement (a) C3S and C2S only (b) C3S and C₂A only
From the above equation, we can say that w/c is the (c) C2S, C3S and C4AF (d) C2S, C3S, C3A and C4AF
weight of water to cement, and also the volume of Ans : (d) When these raw materials are put in kiln.
water to cement. Then it fuses & following four major Compounds are
Q.147. Assume the water-cement ratio of a mix to be formed:-
0.6 and the volume of cement is given as 80 kg. What Principal Mineral Compound Formula Avg.
will be the value of required amount of water in liters
Tri calcium Silicate
per 80 kg cement bag ? 3CaO.SiO2 C3S 40%
(Alite)
(a) 30 (b) 48 (c) 80 (d) 120
Dicalcium Silicate
Ans : (b) The water to cement ratio is given as 0.6 2CaO.SiO2 C2S 32%
(Belite)
𝐖ⅇ𝐢𝐠𝐡𝐭 𝐨𝐟 𝐖𝐚𝐭ⅇ𝐫
= 0.6 Tri calcium
𝐖ⅇ𝐢𝐠𝐡𝐭 𝐨𝐟 𝐂ⅇ𝐦ⅇ𝐧𝐭 3CaO.Al2O3 C3A 10%
Aluminate (Celite)
𝐖ⅇ𝐢𝐠𝐡𝐭 𝐨𝐟 𝐖𝐚𝐭ⅇ𝐫
= 0.6 Tetra calcium
𝟖𝟎
Alumina Ferrite 4CaO.Al2O3.Fe2O3 C4AF 8%
Weight of water = 0.6 × 80 = 48 lit. (Felite)

UPSSSC JE 2018 154 Civil KI Goli

 Ans : (c) Topographic map : - Topographic maps are
U.P. HOUSING DEVELOPMENT detailed representations of the Earth's surface that
COUNCIL JE 11-09-2022 include:
Natural features: Such as mountains, valleys, plains,
❖ SURVEY rivers, lakes, and forests.
Q.01. Alidade used in Plane Table surveying is used for- Man-made features: Such as roads, buildings, bridges,
(a) Centring (b) Levelling and other structures.
(c) Approximate orientation (d) Sighting the objects Q.03. Which of the statements is incorrect about the
Ans : (d) Alidade:- It is used to bisect the object, draw Contours?
rays, direction lines etc. (a) Contours are the lines joining the points of equal
• There is one side of alidade is in bevelled form that is elevation
used to draw the line of sight on the sheet. (b) Two contours usually don't intersect each other
• This edge or side is known as fiducial edge or working (c) Contour lines are drawn using the Compass
edge. (d) Contour lines are usually represented on
topographical maps
• Fiducial edge is graduated in mm and cm.
• An ideal position of alidade to keep the left side of Ans : (c) Contours :- Contour lines are the lines joining
station pin. the points of equal elevation on the ground surface, they
There is two types of alidade : form closed loops and never intersect each other except
in case of over hanging cliff. These lines are drawn using
Plain Alidade - It consists of a metal (brass) or a wooden
surveying techniques and instruments such as levels and
ruler (50 - 60 cm length) with two vanes at the ends. Eye
theodolites. They used to represent topographical
vane has a narrow slit while objective vane is in open
maps.
form and carries a hair or thin wire.
Characteristics of Contour line:
• Closed contour lines with higher values inside show a
hill.
• The variation of the vertical distance between any two
contour lines is assumed to be uniform.
• To contour lines having the same elevations cannot
Telescopic Alidade: - These are used when it is required unite and continue as one line. Similarly, a single
to take inclined sight. It essentially consists of a small contour cannot be split into two lines
telescope with a level tube and graduated ring on the • The horizontal distance between any two contour lines
horizontal axis. It provides quick and accurate result. indicates the amount of slope and varies inversely on
the amount of slope.
• It gives higher accuracy and more range of sights.
• The steepest slope of terrain at any point on a contour
• For undulating areas it can be used very efficiently.
is represented along the normal of the contour at that
• Stadia telescope or tacheometric telescope are used
point. They are perpendicular to ridge and valley lines
to exact sighting and to measure directly horizontal
where they cross such lines.
and vertical distance.
• Contours do not pass through permanent structures
• The inclination of the line of sight can be read on the such as buildings.
vertical circle. • Contours of different elevations cannot cross each
other (caves and overhanging cliffs are the
exceptions).
• Contours of different elevations cannot unite to form
one contour (a vertical cliff is an exception).
• Equally spaced contour represents a uniform slope
and contours that are well apart indicate a gentle
slope.
• Closely spaced contours indicate a steep slope.
• Whereas in the case of depressions, lakes, etc the
Q.02. Topographic map represents – higher figures are outside and the lower figure is
inside.
(a) Only natural features of the earth
(b) Only man-made features of the earth Q.04. Total Station can be used for the measurement of -
(c) Both natural and man-made features of the earth (a) Horizontal and vertical angles (b) Elevation
(d) Only hills in a particular geographical area (c) Horizontal distance (d) All of the above

UP AVAS VIKAS JE-2016 155 Civil Ki Goli

 Prismatic Compass Surveyor Compass Ans : (c) Vicat apparatus have to conforming the IS: 5513
Smaller in size (85 - 110 Bigger (circular box of (1996).
mm diameter) size 150 mm diameter) • Vicat apparatus is a penetration device used in testing
Broad Needle with Edge bar type magnetic of hydraulic cements and similar materials to
aluminium ring is used needle is used, which determine their normal or standard consistency. It is
also acts as index also used to determine the initial setting time and final
Sighting of the object & First Object is sighted & setting time of hydraulic cements and similar
reading of the bearing then reading of bearing materials.
done simultaneously is taken by moving • The Vicat's apparatus for consistency test of cement
around the box paste consists of a frame having a movable rod with a
Temporary Adjustment - Temporary Adjustment - cap at one end and at the other end any one of the
Centering , Levelling, Centering & Levelling following attachment, which are interchangeable :
focusing • Needle for determining the initial setting time
Tripod is not essential Tripod is essential • Needle for determining the final setting time
Graduation is inverted Graduation is erect as • Plunger for determining the standard consistency
because we have to see we can see from the top The other two needles are made to freely fall into a Vicat
them through prism mould filled with the cement paste and the amount of
Readings are in W.C.B., It has 0° at N & S, 90° at penetration of the needles of plunger of 10 mm
having 0° at South, 90° E & W (Q.B. System) diameter can be noted using the vertical graduations
at West, 180° at North & from 0 mm to 50 mm.
270° at East

Q.10. Consider the following statements about Vernier

theodolite and choose the correct option
(i) The least count of the Vernier Scale is 20 second
(ii) Method of repetition is used for measurement of
vertical angles.
Q.12 The standard sand used in compressive strength
(a) Only statement (i) is correct
test of cement is also known as-
(b) Only statement (ii) is correct
(c) Both the statement are correct (a) Fine Sand (b) Coarse Sand
(d) None of the statement is correct (c) Portland Sand (d) Ennore Sand
Ans : (a) The Vernier theodolite is also known as a transit Ans :(d) Standard sand :- It is known as Ennore sand and
theodolite. A theodolite is named a transit theodolite as it is obtained from Tamil Nadu. It is used for the
its telescope is transited i.e. rotated through a whole preparation of cement mortar.
revolution regarding its horizontal axis within the • The standard sand shall be of quartz, light grey or
vertical plane. Its least count varies from 10" to 20" whitish variety and shall be free from silt.
depending on the type of Vernier used in it. However, • The standard sand shall be free from organic
most generally it is 20". impurities.
Note:- A horizontal angle is measured either by the • It is used in compressive strength test of cement.
method of repetition or by reiteration. • For the preparation of cement mortar, the cement-
❖ BM+CT sand ratio is 1:3 in which one part is cement and 3 part
Q.11. Vicat's apparatus is used for is sand is used.
(a) Determination of Normal Consistency of cement • The compressive strength of cement is determined by
(b) Determination of Setting times of cement cube test on cement mortar cubes compacted by means
(c) Both of the above of a vibrating machine with vibration frequency should
(d) None of the above be 12000 ± 400 vibrations per minute and vibration time
is of 2 minutes.

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Q.13. The maximum value of expansion allowed in the Item Quantity
Soundness Test of cement is –
1. Bricks 500 Nos. per cu.m of
(a) 5 mm (b) 10 mm (c) 20 mm (d) 25 mm
(20 cm x 10 cm x 10 cm) brickwork
Ans : (b) The soundness of cement is its ability to resist
volume change. According to the Le Chatelier Method 2. Dry mortar for 33% of the volume of
for testing the soundness of cement. Portland pozzolana brickwork brickwork
cement should not have an expansion of more than 10 3. Dry mortar for 42% of the volume of
mm (IS 1489-11991, 7.2) rubble stone masonry stone masonry
Standard specifications : - 1.54 times the volume of
4. Dry concrete
Name of Cement Maximum Expansion wet concrete
OPC 10 mm 1.25 times the volume of
5. Dry mortar
wet mortar
Rapid hardening 10 mm
6. Dry mortar for 12 2 cu.m per 100 sq.m of
Low heat cement 10 mm
mm plastering plastered area
Super sulphated 5 mm
Q.16. The Class 7.5 designated bricks as per BIS have
Portland pozzolana 10 mm minimum average compressive strength-
PSC 10 mm (a) 7.5 N/mm² (b) 75 N/mm²
High alumina cement 5 mm (c) 0.75 N/mm² (d) 750 N/mm²
Q.14. The frog of the brick in the brick masonry is Ans : (a) Classification of on the basis of BIS
generally kept on- Class Average compressive strength
(a) Bottom face (b) Top face designation (N/mm²)
(c) Both of the above (d) None of the above 35 35
Ans : (b) Frog - It is the depression on the top face of the
brick made by stock board. 30 30
• Frog of brick is kept on the top while constructing a 25 25
wall so that mortar is filled properly in it.
• Binding and shear strength of walls are increased due 20 20
to frog. 17.5 17.5
• Object of frog is forming a key between two course of
brick wall 15 15
• Size of frog = 100mm x 40mm x 10mm
12.5 12.5
• Depth of frog = 10mm to 20mm
10 10
7.5 7.5
5 5
3.5 3.5
Q.17. Minimum curing period for OPC in ordinary
conditions is -
(a) 03 days (b) 07 days (c) 14 days (d) 28 days
Q.15. Number of bricks required for one cubic metre of
brick masonry is – Ans : (b) Curing: It is the process of hardening the
(a) 400 (b) 450 (c) 500 (d) 550 concrete mixes by keeping its surface moist for a certain
period, in order to enable the concrete to gain more
Ans : (c) Modular bricks (bricks with mortar) are of strength.
nominal size 20 cm x 10 cm x 10 cm
• Curing plays an important role in the strength
No. of Brinks required in 1 m³ of brickwork development and durability of concrete.
𝟏𝐦𝟑 • Curing takes place immediately after concrete placing
N= = 𝟓𝟎𝟎
𝟎.𝟐𝟎𝐦×𝟎.𝟏𝟎𝐦×𝟎.𝟏𝟎𝐦
and finishing. and involves maintenance of desired
500 bricks are required for 1 cubic meter of brick
moisture and temperature conditions, both at depth
masonry work.
and near the surface, for extended periods of time.

UP AVAS VIKAS JE-2016 158 Civil Ki Goli

As per Clause 13.5.1 of IS 456:2000, The period of Uses: For unimportant and temporary structures and at
curing of concrete depends on: places where rainfall is not heavy.
• Curing should be continued for 7 days (with the These have rough surface with irregular and distorted
temperature being maintained above 10-degree edges.
Celsius) for Ordinary Portland Cement. (D) Fourth Class Bricks:- These bricks are burnt and
• Curing should be continued for 10 days (with a badly distorted in shape and sizes, and are brittle in
recommendation to extend it to 14 days) where nature.
mineral admixtures or blended cement (such as fly-ash Uses: Ballast of such bricks is used for foundations and
mixed) are used or when concrete is exposed to hot floors in lime concrete and road metal.
weather conditions.
• Curing should be continued at least for 14 days for Q.19. The minimum and maximum setting time of
mass concrete work, heavy footings, large piers. Ordinary Portland Cement (OPC) is –
(a) 30 and 60 minutes respectively
Q.18. The maximum water absorption for first class
(b) 300 and 600 minutes respectively
bricks should not be more than the following-
(c) 300 and 60 minutes respectively
(a) 10% (b) 12.5% (c) 15% (d) 25% (d) 60 and 30 minutes respectively
Ans : (c) Classification of on the basis of Field Practice Ans:(a) Characteristics of an Ordinary Portland cement
(a) First Class Bricks :- These are thoroughly burnt and (OPC):
are of deep red, cherry or copper colour. 1. It is available in 3 grades - OPC 33, OPC-43 and OPC-
• Surface should be smooth and rectangular with 53 and the number 33, 43, 53 corresponds to 28 days
parallel, sharp and straight edges and square corners characteristic compressive strength of cement as
• These bricks are table moulded and burnt in kilns. obtained from standard test on cement.
• It should be free from flaws, cracks and stones. 2. Initial setting time shall not be less than 30 minutes
• It should have uniform texture and final setting time should not be more than 10 hours.
• No impression should be left on the bricks when a 3. Soundness lies from 5 to 10 mm when calculated from
scratch is made by finger nail. Le-Chatelier's method.
• Fracture surface of the brick should not show lumps of Q.20. Bogue's compound composition of cement
lime. consists of
• Metallic ringing sound should come when two bricks (a) C3S and C2S only (b) C3S and C₂A only
are struck against each other. (c) C2S, C3S and C4AF (d) C2S, C3S, C3A and C4AF
• On allowing immersed in cold water for 24 hours, Ans :(d) When these raw materials are put in kiln. Then
water absorption should be 12 - 15% of its dry weight it fuses & following four major Compounds are
• Compressive strength ≥10.5 N/mm2 formed:-
Uses: First class brick are recommended for pointing, Principal Mineral Compound Formula Avg.
exposed face work in masonry structures, flooring and
reinforced brick work. Tri calcium Silicate
3CaO.SiO2 C3S 40%
(b) Second Class Bricks:- These bricks are ground (Alite)
moulded and they are burnt in kilns. Dicalcium Silicate
These bricks have same requirements as first class 2CaO.SiO2 C2S 32%
(Belite)
bricks except the following:
(i) Small cracks and distortions are permitted. Tri calcium
3CaO.Al2O3 C3A 10%
Aluminate (Celite)
(ii) Water absorption 16-20% of its dry weight.
(iii) Compressive strength ≥7N / mm2 Tetra calcium
Uses: Second class bricks are recommended for all Alumina Ferrite 4CaO.Al2O3.Fe2O3 C4AF 8%
important or unimportant hidden masonry works and (Felite)
centering and reinforced bricks and reinforced cement Q.21. If the water cement ratio of a concrete mix
concrete structures. increases significantly
(c) Third Class Bricks:- These bricks are ground moulded (a) Strength of the concrete mix will increase
and they are burnt in clamps. (b) Workability of the concrete mix will increase
• These bricks are soft and reddish yellow coloured. (c) Setting time of the concrete mix will decrease
• Produce a dull sound when struck with each other. (d) All of the above are correct
• Water absorption is about 25% of dry weight. Ans: (b) Water-cement ratio: It is the ratio of the weight
of water to the weight of cement used in a concrete mix.

UP AVAS VIKAS JE-2016 159 Civil Ki Goli

• The ratio of water-cement is generally lying in For concrete of very low workability of the order of 0.7
between 0.35 to 0.65. or below, the test is not suitable, because this concrete
Effect of water-cement ratio on concrete : cannot be fully compacted.
• The higher the water content per cubic meter of Compacting Factor
concrete, the higher will be the fluidity of concrete. 𝐖𝐞𝐢𝐠𝐡𝐭 𝐨𝐟 𝐏𝐚𝐫𝐭𝐢𝐚𝐥𝐥𝐲 𝐂𝐨𝐦𝐩𝐚𝐜𝐭𝐞𝐝 𝐂𝐨𝐧𝐜𝐫𝐞𝐭𝐞 (𝐖𝟏)
=
𝐖𝐞𝐢𝐠𝐡𝐭 𝐨𝐟 𝐟𝐮𝐥𝐥𝐲 𝐂𝐨𝐦𝐩𝐚𝐜𝐭𝐞𝐝 𝐂𝐨𝐧𝐜𝐫𝐞𝐭𝐞 (𝐖𝟐)
However, to increase the workability, the water-
cement ratio is to be increased and it can be done
either by increasing the water for the same cement
content or reducing the cement content for the given
water content.
• According to Abram's law, the strength of a concrete
mix is inversely related to the weight ratio of water to
cement. Higher the W/C ratio, the lower will be the
strength.
• A particular relationship between the strength of
concrete and the W/C ratio is given as:

Water Cement Ratio Q.24. Without affecting the compressive strength of

concrete, its workability can be increased by-
Q.22. To produce low heat cement-
(a) Increasing the quantity of water
(a) Gypsum content is increased (b) Decreasing the quantity of water
(b) C3S content is increased (c) Increasing cement content
(c) C3A content is reduced (d) Adding superplasticizer
(d) C3A content is increased
Ans:(d) Super-plasticizers:- They are admixtures that
Ans: (c) Low Heat Cement : - This cement is work on surfactant property, in which they disperse and
manufactured by reducing the proportion of C3A and C3S deflocculate cement particles thus making concrete
and increasing the proportion of C2S. This cement shows flowing, pourable, and easily placed.
a low rate of development of strength (Due to increasing
Examples: Sulfonated Melamine formaldehyde resin,
C2S). This cement is used in mass concreting.
sulfonated naphthalene- formaldehyde resin, Mixtures
Example - Hydraulic structures. of saccharates, and acid amides.
• The low heat cement consists of a lower percentage of • Superplasticizer can reduce the water content as the
C3A among all bogue compounds. water/cement ratio remain the same and can increase
This cement contains a low percentage (5%) of workability at the same water/cement ratio reduce
tricalcium aluminate (C3A) and a higher percentage cement and provide early strength.
(46%) of dicalcium silicate (C2S).
Q.25. Which of the following materials is pozzolana ?
Q.23. The compacting factor test of concrete
(a) Fly ash (b) Rice Husk Ash
determines its-
(c) Both of the above (d) None of the above
(a) Strength (b) Porosity
(c) Workability (d) Density Ans : (c) Pozzolana: Pozzolanic material is essentially
silicious or aluminium compounds that in itself do not
Ans : (c) The compaction factor test determines possess any binding property but when finely grinded
workability by measuring the compaction achieved reacts with lime released during the hydration of
when concrete falls freely from a hopper into a cylinder. cement and results in the formation of a compound
This test is more precise and sensitive than the slump possessing binding property.
test. It is particularly useful for concrete mixes of
Pozzolanas are classified as natural and artificial.
medium and low workabilities as are normally used
when concrete is to be compacted by vibration; such dry Natural Pozzolanas:- All pozzolanas are rich in silica and
concretes are insensitive to slump test. alumina and contain only a small quantity of alkalis.
Following are some of the naturally occurring
pozzolanas:

UP AVAS VIKAS JE-2016 160 Civil Ki Goli

1. Clays and shales which must be calcined to become ,Fe2+ , Mn2+ , Al3+ also cause hardness, but their quantity
active. is very little water. So, for all purpose hardness may be
2. Diatomaceous earth and opaline cherts and shales assumed due to Ca2+ and Mg2+.
which may or may not need calcination(most active). • Hardness is the property of water which does not
3. Volcanic tuffs and pumicites. Fine grained ashes form produce lather or foam with soap and which produces
better pozzolana. However, tuffs- solidified volcanic ash- scale in hot water-pipe, heater, boilers.
may be ground to desired fineness for use. • Due to hard water food becomes tasteless.
4. Rhenish and Bavarian trass. • Ca-hardness does not have any health effect but mg-
hardness with SO42- has laxative effect.
Artificial Pozzolanas : - Some of the examples of
• Ground water has greater hardness than surface
artificial pozzolanas are:
water.
1. Fly ash • Hard water taste sweet and is often used for drinking
2. Ground blast-furnace slag purposes. However, if the level of hardness of the
3. Silica fume water is not taken into consideration, it might have
4. Surkhi hazardous effect. Soft water is generally bitter or
5. Rice husk ash tasteless.
• Hard water damages both industrial and domestic
❖ ENVIRONMENT appliances and therefore is dangerous. It also prevents
Q.26. The pH value of drinking water should be- the formation of soap lather. It can be said that it
(a) 7.0 (b) 4.5 (c) 6.5-8.5 (d) Greater than 9.0 reduces both the efficiency of soap and appliances.
Ans (c) As per IS 10500: 1991 The permissible limits of Q.28. Which of the following method is most suitable
various compounds are as follows: for population forecasting of a rapidly growing young
Permissible Cause for city?
Parameter
Limit Rejection (a) Arithmetic Increase Method
Total suspended (b) Geometric Increase Method
500 2000
solids (mg/L) (c) Both of the above
Turbidity (NTU) 1 5 (d) None of the above
Colour (TCU) 5 15 Ans : (b )
Taste & Odour Population
1 3 Applicable
(TON) forecast method
Total dissolved Large and established cities
500 2000 Arithmetical
solids (mg/L) where there is limited scope of
increase method
Alkalinity (mg/L as expansion
200 600
CaCO3) Applied to young and rapidly
Geometric
pH 6.5-8.5 No Relaxation developing cities with a large
increase method
Hardness (mg/L as scope of expansion
200 600
CaCO3) Incremental Adopted for any city whether old
Chloride content increase method or new
250 1000
(mg/L)
If population is reaching towards
Free ammonia Decreasing rate of
0.15 No Relaxation saturation, and growth rate is
(mg/L) growth method
decreasing
Nitrite (mg/L) 0 0
Q.29. Methane as is released after which type of
Q. 27. Which of the following is a result of using hard decomposition of biodegradable organic solid waste?
water ? (a) Aerobic (b) Anaerobic
(a) Increased soap consumption (c) Both of the above (d) None of the above
(b) Formation of scale at the bottom of steam boilers
in thermal power plants Ans : (b) Biodegradable waste can be transformed into
(c) Both of the above compost by a process called composting. This process
(d) None of the above involves the decomposition of organic material by
microbes, which break down this material into nutrient-
Ans : (c) Hardness : - It is caused by concentration of rich soil conditioner.
multivalent metallic cations in solution. Multivalent
There are two main types of composting: aerobic and
metallic cation which are more abundant in natural
anaerobic.
water are calcium ( Ca2+ ) and magnesium ( Mg2+ ). Sr2+

UP AVAS VIKAS JE-2016 161 Civil Ki Goli

Aerobic composting: This is the most common method • PVC pipes are resistant to corrosion, chemical
where the organic waste is broken down by bacteria and reactions, and scaling, ensuring a long lifespan.
other microorganisms in the presence of oxygen. The • They have good pressure-bearing capacity and can
waste is often turned to ensure even decomposition and handle both cold and hot water supply lines.
to keep the pile aerated. PEX Pipes:- PEX (cross-linked polyethylene) pipes are a
Anaerobic composting: In this method, organic waste is type of flexible plastic piping commonly used in
broken down by microorganisms in an environment plumbing systems.
without oxygen. This method takes longer than aerobic • They also have excellent freeze resistance, as they can
composting and can sometimes produce a foul smell due expand and contract without bursting, making them
to the production of gases like methane and hydrogen suitable for cold climates.
sulphide. • PEX pipes are compatible with both cold and hot water
Q.30. In a BOD test, 10 ml of raw sewage was diluted supply systems, and they have good temperature and
to 100 ml and the dissolved oxygen concentration of pressure resistance.
the diluted sample at the beginning was 8.0 mg/l and it Galvanized Steel:- Pipes Galvanized steel pipes are steel
was 4.0 mg/l at the end of 5 day incubation at 20°C. The pipes that have been coated with a layer of zinc to
BOD of raw sewage will be - protect against corrosion. The process of galvanization
(a) 80 mg/l (b) 10 mg/l involves immersing the steel pipes in a molten zinc bath,
(c) 40 mg/l (d) 120 mg/l allowing the zinc to form a protective layer on the
surface of the pipes. They can withstand high water
Ans : (C) The BOD of the Raw Sewage at the end of 5 pressures and are suitable for both cold and hot water
days incubation period is calculated using the following supply systems.
formula BOD = (Di – Df) × DF
𝟏𝟎𝟎
CPVC Pipes:- CPVC (chlorinated polyvinyl chloride) pipes
= (8-4)× = 40 mg/l are a type of thermoplastic piping commonly used for
𝟏𝟎
Q.31. The hourly variation factor is usually taken as hot and cold water supply systems.
(a) 1.5 (b) 1.8 (c) 2.5 (d) 2.7 • They are an alternative to traditional PVC pipes,
specifically designed to handle higher temperatures
Ans : (a) Variation in rate of demand:-
and pressures.
Description Formula • CPVC pipes are known for their durability, chemical
Maximum Hourly 1.5 x Average Hourly resistance, and fire-retardant properties.
Demand Demand of Maximum Day • They can withstand higher temperatures than PVC
Maximum Daily pipes, making them suitable for hot water
1.8 x Average Daily Demand applications.
Demand
Maximum Weekly 1.48 x Average Weekly • CPVC pipes are also resistant to corrosion, scale, and
Demand Demand chemical reactions, ensuring a longer lifespan and
Maximum Monthly 1.28 x Average Monthly minimizing maintenance needs.
Demand Demand • They are relatively easy to install, as they can be joined
Maximum Seasonal 1.3 x Average Seasonal using solvent cement or mechanical fittings.
Demand Demand • CPVC pipes are commonly used in residential and
commercial buildings for water supply lines, sprinkler
Q.32. Which of the following pipe material is suitable
systems, and other applications where reliable, high-
for hot water supply
temperature water distribution is required.
(a) UPVC (b) CPVC (c) Cast Iron (d) Steel
HDPE Pipes:- HDPE (high-density polyethylene) pipes
Ans : (b) Types of Water Conveyance Pipes are a popular choice for various water supply and
Copper Pipes:- Copper pipes have long been used in distribution applications.
plumbing systems for their excellent durability and • Made from a thermoplastic polymer, HDPE pipes offer
corrosion resistance. numerous advantages that make them desirable in
•They are known for their superior heat conductivity plumbing and infrastructure projects.
and ability to withstand high temperatures, making Concrete Pipes:- R.C.C. (Reinforced Cement Concrete)
them ideal for both hot and cold water supply lines. pipes are commonly manufactured using a mixture of 1
Copper pipes are also resistant to UV rays, making them part cement, 2 parts sand, and 2 parts aggregate.
suitable for outdoor applications. • The aggregates used have a maximum size of 7 mm.
PVC Pipes:- PVC (polyvinyl chloride) pipes are widely • To resist tensile stress, circumferential reinforcement
used in plumbing systems for various water supply and is incorporated into the pipes. Additionally, 0.25%
drainage applications. longitudinal reinforcement is provided. The thickness
of R.C.C. pipes ranges from 25 mm to 65 mm,

UP AVAS VIKAS JE-2016 162 Civil Ki Goli

Q.38. Limit of proportionality depends upon which of R = Radius of curvature.
the following? The relationship between the radius of curvature 'R'
(a) Area of cross-section (b) Type of loading bending moment 'M' and flexural rigidity El is given by
(c) Type of material (d) None of the above 𝐌
=
𝐄
𝐈 𝐑
Ans:(c) Hooke’s Law states that the strain in a solid body
is directly proportional to the applied stress and this Q.42 Buckling load for a given column depends upon
condition is valid upto the limit of proportionality. which of the following?
Limit of proportionality is the stress at which the stress- (a) Length of column only
strain curve ceases to be a straight line. It is the stress at (b) Least lateral dimension only
which extension ceases to be proportional to strain. This (c) Both of the above
limit primarily depends on the material's properties. (d) None of the above

Q.39. The relationship between Young's modulus of Ans: (c) Column: If a beam element is under a
elasticity E, bulk modulus K and Poisson's ratio u is compressive load and its length is an order of magnitude
given by- larger than either of its other dimensions such a beam is
called a column.
(a) E = 2K (1-2μ) (b) E = 3K (1+μ)
Due to its size, its axial displacement is going to be very
(c) E = 3K (1-2μ) (d) Ε = 2K (1+μ)
small compared to its lateral deflection called buckling.
Ans:(c) Relation Between Elastic Modulus - Euler's Buckling Load:
𝚷𝟐 𝐄𝐈𝐦𝐢𝐧
Pcr =
𝐋𝟐𝐞
where, Pcr = critical load for buckling
E = Young's Modulus (GPa)
Q.40. Shear stress on principal planes is- Imin = Area moment of inertia,
(a) Zero (b) Maximum Le = effective length;
(c) Minimum (d) None of the above Slenderness Ratio (S): The ratio between the length and
least radius of gyration.
Ans:(a) Principal Plane:- The planes which have no shear 𝐋𝐞
stress are known as principal planes or the planes of zero 𝑺=
𝐤𝐦𝐢𝐧
shear stress. Kmin = least radius of gyration.
• These planes are subjected to only normal stress. At In conclusion, the buckling load of a given column
this plane normal stress is max./min. depends on the area of cross-section, length and least
• Angle b/w principal plane and max. shear plane is 45° radius of gyration, and modulus of elasticity for the
& 135°. material of the column.
Principal stress:- Normal stress, acting on a principal
❖ STRUCTURE
plane is known as principle stress. Maximum normal
stress is called a major principal stress and minimum Q.43. The number of unknown internal forces in each
normal stress is called minor principle stress. member of a rigid jointed plane frame is –
Oblique plane:- In realistic approach stresses does not (a) 1 (b) 4 (c) 3 (d) 6
act in normal direction but rather in inclined or oblique Ans : (c) In a rigid jointed plane frame, each member can
planes. have the unknown internal forces: axial force, shear
Q.41. Which of the following represents the correct force, and bending moment. Therefore, the number of
relationship between the radius of curvature R, unknown internal forces in each member is 3.
bending moment M and flexural rigidity EI?
(a) M/I = E/R (b) M/R = E/I
(c) M/E = R/I (d) None of the above
Ans : (a) The equation for simple bending is given by:
𝐌 𝐄 𝐅
= =
𝐈 𝐑 𝐘
where I = Moment of Inertia For Rigid jointed space frame number of unknown
E = Modulus of Elasticity internal forces =6[i.e. Mx, My, Mz, Fx, Fy, Fz]
F = Stress at any fibre at a distance of y from the neutral Q.44. A pin-jointed plane frame is unstable if –
axis, (a) (m+r) <2j (b) (m+r) = 2j
M = Bending moment and (c) (m+r) > 2j (d) None of the above

UP AVAS VIKAS JE-2016 164 Civil Ki Goli

Where m is number of members, r is reaction (2) Prismatic member (beam) whose one end is hinged
components and j is number of joints and the other is roller.
Ans:(a) In general, let a frame have j joints ,m members
and r reaction
If m + r = 2j then the frame is perfect frame.
If m + r < 2j then the frame is deficient frame.
If m + r > 2j then the frame is redundant frame.
COF=0
A perfect frame can always be analysed by the condition (3) Prismatic member (beam) whose one end is fixed
of equilibrium. While a redundant frame cannot be fully and the other is roller and also provide hinge in
analysed by the condition of equilibrium. between both support.
Hence, If in a pin-jointed plane frame (m + r) > 2 j, then
the frame is stable and statically indeterminate.
If in a pin-jointed plane frame (m + r) < 2 j, then the
frame is unstable and statically indeterminate.
Q.45. Which of the following is a displacement method
of structural analysis?
(a) Slope Deflection Method COF=a/b
(b) Moment Distribution Method Q.47. Castigliano's first theorem is applicable :
(c) Direct Stiffness Method
(a) For statically determinate structures
(d) All of the above
(b) When the system behaves elastically
Ans : (d) Difference between Force & Displacement (c) Only when principle of superposition is valid
Methods (d) None of the above
Force Methods Displacement Methods
Ans : (b) Castigliano's first theorem states that the
Types of indeterminacy: Types of indeterminacy:
partial derivative of the total strain energy in a structure
Static Indeterminacy Kinematic Indeterminacy
with respect to an applied force gives the displacement
Governing equation: Governing equations:
in the direction of that force. This theorem is applicable
Compatibility Equations Equilibrium Equations
under the conditions that the material is linearly elastic
Force displacement Force displacement and the displacements are small, which implies that the
relations: Flexibility matrix relations: Stiffness matrix
structure behaves elastically and the principle of
Example : - Example : - superposition is valid.
1. Method of consistent 1. Slope deflection method
deformation 2. Moment distribution Q.48. Simple cantilever beam is
2. Theorem of least work method (a) Statically determinate
3. Column analogy method 3. Kani's method (b) Statically Indeterminate
4. Flexibility matrix 4. Stiffness matrix method (c) Unstable structure
method 5. Castigliano's Theorem- I (d) All of the above
5. Castigliano's Theorem- II
Ans:(a) A two-dimensional structure in general is
classified as a statically indeterminate structure if it
Q.46. The carry-over factor in a prismatic member cannot be analysed by conditions of static equilibrium.
whose far end is fixed is-
The conditions of equilibrium for 2D structures are:
(a) 0 (b) ½ (c) ¾ (d) 1 1. The Sum of vertical forces is zero (∑Fy = 0).
Ans : (b) Standard cases of carry-over factor (COF) : 2. The Sum of horizontal forces is zero (Fx = 0).
(1) Prismatic member (beam) whose one end is fixed The Sum of moments of all the forces about any point in
and the other is roller. the plane is zero (ΣΜz = 0).
Simply supported beam: Number of unknowns = 3
Degree of static indeterminacy = 3 – 3 = 0. Hence it is
statically determinate
Cantilever beam: Number of unknowns = 3
Degree of static indeterminacy = 3 – 3 = 0. Hence it is
COF= 1/2 statically determinate.
The carryover factor of a prismatic beam whose one end Three hinged arches: Number of unknown = 4
is fixed and other is roller is =1/2.

UP AVAS VIKAS JE-2016 165 Civil Ki Goli

Degree of static indeterminacy 4 – 3 – 1 = 0. (Additional strength of mix in N/mm². The mixes of grades are as
equation due to internal hinge: B.M = 0) follows:
Hence it is statically determinate. Grade of Mix Ratio
Two hinged arches: Number of unknown = 4 Concrete (Cement:Sand:Aggregate)
Degree of static indeterminacy = 4 - 3 = 1. M5 01:05:10
Hence it is statically indeterminate.
M7.5 01:04:08
❖ RCC M10 01:03:06
Q.49. The maximum permissible free fall on concrete
M15 01:02:04
as per IS 456:2000 is
(a) Less than 1.0 m (b) 1.5 m M20 1:1.5:3
(c) 2.0 m (d) More than 5.0 m Q.52. Vertical formwork of column walls is generally
Ans:(b) Segregation: Segregation can be defined as the removed after
separation of the constituent materials of concrete. (a) 16-24 hours (b) 03 days
• Insufficiently mixed concrete with excess water (c) 07 days (d) 28 days
content shows a higher tendency for segregation. Ans : (a) As per IS 456 2000, minimum period before
• Dropping of concrete from heights as in the case of striking formwork for various components are as
placing concrete in column concreting will result in follows
segregation. Minimum Period
Type of Formwork
• As per clause 13.2 of IS 456: 2000, the maximum Before Striking
permissible free fall of concrete to avoid segregation Vertical formwork to columns, walls,
16-24 hours
may be taken as 1.5 m or 150 cm. beams
Soffit formwork to slabs (Props to be
Q.50. As per IS 456:2000, pH of water used for mixing re-fixed immediately after removal 3 days
and curing the concrete should be : of formwork)
(a) 6.5 to 8.5 (b) Not less than 6.0 Soffit formwork to beams (Props to
(c) 7.0 (d) Highly acidic be re-fixed immediately after 7 days
removal of formwork)
Ans: (b) Important properties of water for concrete:
Props to slabs: 1) Spanning up to 4.5
(1) The pH value must not be less than 6. m
7 days
(2) Presence of salts of manganese, tin, zinc, copper and Props to slabs: 2) Spanning over 4.5
14 days
lead causes a reduction in strength. m
(3) Presence of sugar causes retardation in the setting
Props to beams and arches: 1)
As per clause 5.4 of IS 456: 2000 : Potable water is 14 days
Spanning up to 6 m
considered satisfactory for mixing Concrete and the Props to beams and arches: 2)
21 days
permissible limits for solids is shown in table below: Spanning over 6 m
Type of solid Max. Permissible limit Q.53. If the volume of concreting done in a day lines
Organic 200 mg/l between 1-5m³ how many test samples will be
collected for testing compressive strength at 28 days?
Inorganic 3000 mg/l
(a) 01 (b) 02 (c) 03 (d) 04
Sulphates 400 mg/l
Ans: (a) The minimum number of samples depend on
Chlorides the quantity of concrete work.
2000 mg/l
(concrete) The details are given according to Clause 15.2.2 of IS
Chlorides (RCC) 500 mg/l 456:2000
Suspended matter 2000 mg/l Quantity of
Number of Samples
Concrete in m3
Q.51. Nominal mix proportion for M 20 grade concrete
is – 1 1-5
(a) 1:1.5:3 (b) 1:2:4 (c) 1:3:6 (d) 1:4:8 2 6-15
Ans:(a) IS 456-2000 has designated the concrete mixes
3 16-30
into a number of grades as M10, M15, M20, M25, M30, 4 31-50
M35, and M40. In this designation, the letter M refers to 4+ one additional samples
the mix and the number to the specified 28-day cube 51 and above
for leach 50 m³ of work

UP AVAS VIKAS JE-2016 166 Civil Ki Goli

Q.54. In Limit State, Method (LSM) of design of IS:1489 Part-II – 2015- Portland Pozzolana cement
structures, the values of partial safely factors for (Calcined clay based)
concrete and steel respectively are- 383 :- 2016 Coarse & fine aggregates for concrete
(a) 1.5 and 1.15 (b) 1.15 and 1.5 516:- Strength of concrete tests 650 Specification for
(c) 1.5 and 2.7 (d) 2.7 and 1.5 standard sand (Ennore) for testing
Ans:(a) Partial factor of safety for concrete and steel 1642:2013- Fire safety of buildings(general ) : Details of
should be taken as 1.5 and 1.15, respectively when construction
assessing the strength of the structures or structural 2386- Test for Aggregate (1–8 Parts)
members employing limit state of collapse. 2430- Sampling of aggregate for concrete
In Limit state method , partially safety factor 5816- Splitting tensile strength of concrete
Material Collapse Deflection Cracking 6461- Glossary of terms related to cement concrete (Part
1–12)
Concrete 1.5 1 1.3
7320- Specification of concrete slump test apparatus
Steel 1.15 1 1 13311 Part – 1 Ultrasonic pulse velocity test
Q.55. Which of the following structure is designed Part – 2 Rebound hammer Non-Destructive
using Working Stress Method? testing of concrete
(a) R.C.C. Water Tank (b) Building 875- Design loads (other than earthquakes) for building
& structures
(c) Road (d) Canal
Part I : Dead load.
Ans: (a) Working stress method :- This method is based
Part II: Live load
upon linear elastic theory or depends on the classical
elastic theory. Part III (2015) : Wind load,
• This method ensured adequate safety by suitably Part IV : Snow load
restricting the stress in the materials induced by the Part V : Special loads & load combinations
expected working leads on the structures. 1893- Earthquake resistant design for structures
• The basic assumption of linear elastic behaviour is 456- Plain and reinforced concrete
considered justifiable since the specified permissible Q.57. In a reinforced concrete member, the best way of
stresses are kept well below the ultimate strength of the ensure adequate bond is-
material.
(a) To provide less number of large diameter of
• The ratio of the yield stress of the steel reinforcement reinforcement bars
or the concrete cube strength to the corresponding (b) To provide large number of smaller diameter bars
permissible or working stress value is usually called a (c) To increase the cover for reinforcement
factor of safety. (d) All of the above
Note :- Structures like water tanks, chimneys and RC
Ans : (b) To increase bond strength we should use
bridges, working stress method is generally adopted.
smaller diameter bars in correspondingly large numbers
Working stress method is adopted as it does not account
keeping the same area of reinforcement.
for the variation of loads and the design is mostly
conservative. Water tanks are more vulnerable. Q.58. Minimum clear cover (in mm) to the main steel
bars in slab is taken as (as per IS 456:2000)
Q.56. IS Code used for concrete mix proportioning is –
(a) 20 mm (b) 30 mm (c) 40 mm (d) 75 mm
(a) IS 456:2000 (b) IS 800:2007
(c) IS 875:1987 (d) IS 10262:2009 Ans : (a) Minimum clear cover requirement :
1. Slab = 15 mm
Ans:(d) 10262 - Guidelines for concrete mixed design
2. Beam = 25 mm
269- Specification of OPC 33 grade
3. Column = 40 mm
8112- Specification of OPC 43 grade
4. Footings = 50 mm
12269- Specification of OPC 53 grade
For Slab: For Mild exposure – 15 mm, and For Moderate
8041- Rapid hardening Portland cement
exposure – 30 mm
8042- White Portland cement
However, if the diameter of the bar does not exceed 12
8043- Hydrophobic Portland cement mm, or cover may be reduced by 5 mm. Thus for main
IS:6452- High Alumina cement reinforcement up to 12 mm diameter bar and for mild
IS:1489 Part-I – 2015- Portland Pozzolana cement (fly exposure, the nominal cover is 15 mm.
ash based)

UP AVAS VIKAS JE-2016 167 Civil Ki Goli

For Column: For a longitudinal reinforcing bar in a • not more than 6 per cent for unlapped bars of the
column, concrete cover not less than 40 mm not less gross cross-sectional area of the column
than the diameter of such a bar should be provided. In • not more than 4 per cent for lapped bars of the gross
the case of columns of the minimum dimension of 20 cm cross-sectional area of the column
or under, whose reinforcing bars do not exceed 12 mm,
Helical reinforcement: A reinforced concrete column
the concrete cover of 25 mm to be used for
having helical reinforcement shall have at least
reinforcement
six bar of longitudinal reinforcement within the helical
Q.59. As per IS: 456-2000, the maximum area of reinforcement.
compression reinforcement in a beam is- IS : 456-2000 Spacing: The spacing of longitudinal bars
(a) 1% (b) 2% (c) 3% (d) 4% measured along the periphery of the column shall not
Ans : (d) As per IS:456-2000 the maximum area of exceed 300 mm.
reinforcement in compression and tension for an RC Q.61. Effective length of a column fixed at one end and
beam is 4% of the gross area of the section. It is hinged at other end is-
important to adhere to these limits to ensure the safety (a) l/2 (b) l/√𝟐 (c) 2 (d) 2/3
and durability of the structure
Ans : (b) Actual length- Distance between both ends of
Reinforcement of different sections:
column is known as actual length.
For Beams:
Effective length (leff)- Distance between the point of zero
Maximum reinforcement = 4% of gross area moment or point of contra flexure is called effective
Minimum reinforcement = 0.85bd/fy length of column. The effective length of the column
For column: depends upon its end conditions.
Maximum reinforcement = 6% of gross area Effective length of column for different end conditions:-
Minimum reinforcement = 0.8% of gross area
For slab:
Minimum reinforcement = 0.15% of gross area (For mild
steel)
Minimum reinforcement = 0.12% of gross area (For
HYSD bars)
Q.60. Consider the following statements about
rectangular columns and choose the correct option-
(i) The minimum number of longitudinal bars shall be
four
(ii) The bars shall not be less than 12 mm in diameter
(a) Only statement (i) is correct
(b) Only statement (ii) is correct
(c) Both the statement are correct
(d) None of the statement is correct
Ans : (c) As per IS 456: 2000, clause 26.5.3.1,
Longitudinal reinforcement in a column:
The minimum number of longitudinal bar: The ❖ FM
minimum number of the longitudinal bar provided in Q.62. The discharge through V-notch varies as- V-
column shall be (a) H1/2 (b) H3/2 (c) H5/2 (d) H5/4
• four in rectangular columns Ans :(c) Weir or notch is a physical structure of masonry
• six in circular columns constructed across the channel width to calculate
The minimum diameter of longitudinal the discharge of the channel section.
reinforcement bars : - The main longitudinal Rectangular Notch:- The discharge through a
reinforcement bars used in column shall not be less rectangular notch weir is,
𝟑
than 12 mm in diameter. 𝟐
Q= CdL√𝟐𝐠𝐇 𝟐
𝟑
Percentage of steel: The cross-sectional area of
longitudinal reinforcement shall be Where, Q = discharge of fluid,
• not less than 0.8 per cent of the gross cross-sectional Cd = Coefficient of discharge and
area of the column H = height of water above the notch

UP AVAS VIKAS JE-2016 168 Civil Ki Goli

It depends on the percentage of moisture present in the Plasticity Soil Degree of Degree of
sand & it's fineness. index type plasticity cohesiveness
Effect of Moisture Content on Bulking of Sand 0 Gravel Non- Non-cohesive
• The maximum bulking of sand is 20-40% at 4 to 6% and plastic
water content. sand
• At 12% water content regain its original volume. <7 Silty Low Partly
• Moisture content less than 5% should be preferred for plastic cohesive
construction purpose 7-17 Silt clay Medium Cohesive
• Bulking of Sand is inversely proportional to the plastic
fineness modulus. > 17 Clay High Cohesive
• Note- Size of particle ↑, Fineness modulus ↑, Bulking plastic
decreases. Q.69. Sand has particles size-
Q.67. With increase in moisture content, the bulking of (a) 80-300 mm (b) 20-40 mm
sand (c) 10-20 mm (d) 0.075-4.75 mm
(a) First decreases and then increases Ans : (d) Particle Size classification: This classification of
(b) First increases and then decreases soils is done on the basis of particle size composition.
(c) Decreases According to which soil may be termed as clay, silt, sand,
(d) increases gravel, cobbles and boulders.
I.S. Classification (Grain size distribution):
Ans : (b) Bulking of Sand: The increase in the volume of
sand due to an increase in moisture content is known as (i) Clay = d < 2μ
the bulking of sand. A film of water is created around the (ii) Silt = 2µ < d < 75μ
sand particles which forces the particles to get aside Fine silt = 2μ < d < 10μ
from each other and thus the volume is increased. • Medium silt = 10µ < d < 20μ
•The increase in moisture in sand increases the volume Coarse silt = 20μ < α < 75μ
of sand. The volume increase in dry sand is known as the (iii) Sand = 75μ < d < 4.75 mm
bulking of sand. Bulking of sand depends on the quantity
Fine sand = 0.075 mm < d < 0.425 mm
of moisture in the sand and also the size of the particles.
Five to eight percent of the increase in moisture in the Medium sand = 0.425 mm < d < 2 mm.
sand can increase the volume of sand up to 20 to 40 Coarse sand = 2 mm < d < 4.75 mm
percent. Again the finer the sand is more will be the (iv) Gravel = 4.75mm < d < 80mm
increase in volume and the increase in volume will be Fine gravel = 4.75mm < d < 20mm
relatively less for coarser sand. Coarse gravel = 20mm < d < 80mm
(v) Cobbles = 80 mm < d < 300 mm
(vi) Boulders = d > 300 mm
Q.70. Compaction of clayey soils is done by
(a) Ordinary Roller (b) Sheep-foot roller
(c) Both of the above (d) None of the above
Ans : (b) Mostly four types of rollers used are :
1. Pneumatic tired roller:- Pneumatic tired roller has a
number of rubber tires at the front and at the rear end.
A pneumatic tired roller can be used for highways,
construction of dams, and for both fine-grained and
non-cohesive soils.
2. Tamping roller/sheep foot roller:- Sheep foot roller
also named tamping roller. The front steel drum of the
sheep foot roller consists of many rectangular-shaped
Q.68. If the plasticity index of a soil mass is zero the soil boots of equal sizes fixed in a hexagonal pattern. In
is- sheep foot, roller compaction is by static weight and
(a) Sand (b) Silt (c) Clay (d) None of these kneading of the respective layer. Sheep-foot rollers are
used for compacting fine grained soils such as heavy
Ans: (a) Plasticity index (PI) is the range of water content
clays and silty clays.
over which the soil remains in the plastic state.
Plasticity Index = Liquid Limit – Plastic Limit.

UP AVAS VIKAS JE-2016 170 Civil Ki Goli

3. Smooth wheel rollers:- Smooth wheel roller and Q.73. Which of the following has highest permeability?
vibratory rollers are the same. Both have the same (a) Clay
characteristics. Only the difference in both is vibratory (b) Fine Silt
equipment. The smooth wheel roller has no vibrator
(c) Coarse Sand
attached to the drum. This makes smooth wheel roller
best suited for rolling of weaker aggregates, proof rolling (d) All the above have same permeability
of subgrades, and compacting asphalt pavements. Ans : (c) Permeability: It is the property by which water
4. Vibratory Roller:- Vibratory-type rollers have two can flow through any medium. It is also called hydraulic
smooth wheels/drums plus the vibrators. One is fixed at conductivity.
the front and the other one is on the rear side of the Coefficient of Drainage
vibratory roller. Vibration is to reduce the air voids and Soil Type
Permeability (cm/sec) Properties
to cause densification of granular soils.
Gravel 100-1.0 Very pervious
Q.71. Which of the following is correct for cohesion less
soil? Coarse
1.0-0.01 Pervious
Sand
(a) Angle of internal friction of zero
(b) Apparent cohesion is zero Fine Poorly
0.01-0.001
(c) Both of the above Sand pervious
(d) None of the above Silty Clay 0.001-0.00001 Impervious
Ans : (d) Cohesionless soils are having no cohesion i.e. Clay < 0.000001 Impervious
no intermolecular force of attraction when water is
Q.74. A Pycnometer is used to determine-
added into them. The shear strength of these type of
soils depends on friction between particles and also (a) Water content (b) Void ratio
normal stress acting on them. (c) Dry density (d) Mineral content
Examples: Sand and Gravel. Ans : (a) Pycnometer test is used to determine the
Q.72. Which of following is not a type of shallow specific gravity of cohesion less soils and water content.
foundation? Dry density or in-situ unit weight is determined by
(a) Strip Footing (b) Isolated Footing using the following methods:
(c) Raft foundation (d) Pile foundation 1. Sand Replacement
2. Core-Cutter
Ans : (d) Foundation:- It is defined as that part of the
structure that connects and transmits the load from the 3. Water Displacement
structure to the ground soil. Other methods to determine the water content are:
Types of Foundation: 1. Oven Drying Method
A. Shallow Foundation:- Shallow Foundations are those 2. Calcium Carbide/Rapid Moisture
foundations which the depth is placed less than the 3. Sand Bath Method
width of the foundation (D < B). Shallow foundations are 4. Radiation method
generally termed as spread footing as they transmit the 5. Torsion Balance Moisture Meter
load of the super structure laterally into the ground.
Q.75. A soil has a bulk density of 22 kN/m3 and water
Types of Shallow Foundation
content 10%. What will be the value of dry density of
1. Wall Footing the soil ?
2. Isolated column/Column Footing (a) 18.6 kN/m³ (b) 20.0 kN/m³
3. Combined Footing (c) 22.0 kN/m³ (d) 23.2 kN/m³
4. Cantilever (Strap) Footing 𝛄𝒃
Ans : (b) Dry unit weight, γd=
5. Mat (Raft) Foundation 𝟏+𝒘
𝟐𝟐
B. Deep Foundation:- Deep Foundation are those γ d= = 20 kN/m3
𝟏+𝟎.𝟏
foundations in which the depth of the foundation is γb = Bulk unit weight, w = water constant
greater than its width (D>B). The D/B ratio is usually 4-5
Q.76. The ratio of volume of voids to the total volume
for deep foundation. Deep foundations are used when
of soil mass is called-
the shallow foundation cannot support the load of the
structure. (a) Air content
Types of Deep Foundation (b) Porosity
1. Pile Foundation 2. Pier Foundation (c) Percentage air voids
3. Well (Caissons) Foundation (d) Void ratio

UP AVAS VIKAS JE-2016 171 Civil Ki Goli

Ans : (b) Void ratio (e): Void ratio is usually defined as • The basic unit is also a two-layer sheet like that of
the ratio of the volume of voids to the total volume kaolinite except for the presence of water between
of soil solid. the sheets.
Porosity (n): Porosity is defined as the ratio of the Illite Minerals : - Illite consists of the basic
volume of voids to the total volume of the soil. montmorillonite units but is bonded by potassium ions.
The relationship between void ratio and porosity are as There is about 20% replacement of aluminium with
follows: silicon in the gibbsite sheet due to isomorphous
𝒏 𝒆
e= & n= substitution. This mineral is very stable and does not
𝟏−𝒏 𝟏+𝒆
Water content ratio (w): The water content ratio of the swell or shrinks largely.
soil is defined as the ratio of weight of water to the During dry weather, there will be shrinkage.
weight of solids in a given soil mass.
1. With reference to grain size:
Degree of saturation: It is the ratio of the volume of
water occupied in the soil to the volume of voids of the Kaolinite > Illite > Montmorillonite
given soil sample. 2. With reference to swelling and shrinkage behavior
Q.77. Which of the following relationship is correct? Montmorillonite > Illite > Kaolinite
(a) n=e/(1+e) (b) n=e/(1-e) 3. With reference to Plasticity Index
(c) n=2e/(1+e) (d) n=2e/(1-e) Montmorillonite > Illite > Kaolinite
Ans : (a) Void ratio (e): Void ratio is usually defined as 4. With reference to Dry Strength
the ratio of the volume of voids to the total volume Montmorillonite > Illite > Kaolinite
of soil solid.
Porosity (n): Porosity is defined as the ratio of the Q.79. Which of the following has high porosity as well
volume of voids to the total volume of the soil. The as high permeability?
relationship between void ratio and porosity are as (a) Aquifer (b) Aquiclude
follows: (c) Aquitard (d) Aquifuse
𝒏 𝒆
e= & n= Ans : (a) Aquifer: It is a saturated area beneath the
𝟏−𝒏 𝟏+𝒆
Water content ratio (w): The water content ratio of the water table.
soil is defined as the ratio of weight of water to the • It is basically an underground layer of water-bearing
weight of solids in a given soil mass. permeable rock, rock fractures, or unconsolidated
Degree of saturation: It is the ratio of the volume of materials.
water occupied in the soil to the volume of voids of the • They are huge storehouses of water.
given soil sample. Aquifers are of two types :
Q.78. Which of the following clay mineral shows largest Confined Aquifer: It is an aquifer below the land surface
swelling and shrinkage characteristics? that is saturated with water.
(a) Kaolinite (b) Illite Unconfined Aquifer: It is an aquifer where
(c) Montmorillonite (d) None of the above the groundwater is in direct contact with the
atmosphere through the open pore spaces of the
Ans : (c) Montmorillonite Mineral:- The bonding
between the three-layer units is by van der Waals forces. overlying soil or rock.
This bonding is very weak, and water can enter easily.
Thus, this mineral can imbibe a large quantity of water
causing swelling.
Kaolinite Mineral : - A basic kaolinite unit is a two-layer
unit that is formed by stacking a gibbsite sheet on a silica
sheet These basic units are then stacked one on top of
the other to form a lattice of the mineral.
The units are held together by hydrogen bonds.
The strong bonding does not permit water to enter the
lattice.
Thus, kaolinite minerals are stable and do not expand
under saturation.
• Halloysite mineral is like Kaolinite.

UP AVAS VIKAS JE-2016 172 Civil Ki Goli

 ❖ HIGHWAY Q.82. The ruling design speed on a national highway in
Q.80. The stopping sight distance (SSD) depends upon plain terrain as per IRC recommendation is- IRC
which of the following factors ? (a) 60 kmph (b) 80 kmph
(a) Total reaction time of the driver (b) Speed of vehicle (c) 100 kmph (d) 120 kmph
(c) Efficiency of brakes (d) All of the above Ans : (c) Design speed :- Design speed is the maximum
Ans: (d) Stopping Sight Distance:- Stopping sight speed at which vehicles can continuously run safely
distance (SSD) is the minimum sight distance available under the favourable conditions.
on a highway at any spot having sufficient length to As per IRC :-
enable the driver to stop a vehicle travelling at design Road
Plain Rolling Mountainous Steep
safely without collision with any other obstruction. Classification
• Stopping sight distance = lag distance + braking Expressway
120 100 80 60
(Ruling)
distance.
𝐯𝟐
Expressway
100 80 60 40
• Stopping sight distance, SSD = vt + (Minimum)
𝟐𝐠(𝛈±𝐟)
Where v = Speed of vehicle NH and SH 100 80 65 50
t = reaction time of the driver MDR 80 65 50 40
η = efficiency of the brake ODR 65 50 40 30
f = friction co-efficient on road Village Road 50 40 35 25
N = gradient on road
Q.83. For highway geometric design purposes, which of
So the stopping sight distance depends upon the total the following speed is used?
reaction time, the speed of the vehicle, efficiency of (a) 15th percentile (b) 50th percentile
brakes and the gradient of the road.
(c) 85th percentile (d) 98th percentile
Q.81. Camber in the road is provided for which of the Ans:(d) The 98th percentile speed is adopted for
following purpose? geometric design of highway.
(a) Effective drainage
(b) Counteracting the centrifugal force
(c) Having proper sight distance
(d) None of the above
Ans : (a) Camber: It is the slope provided to the road
surface in the transverse direction to drain off the
rainwater from the road surface.
It is also known as the cross slope of the road.
Camber can be written as 1 in n or x %.
The objectives of providing camber are as follows :
• Surface protection especially for gravel and
• Geometrical design speed = 98th percentile speed
bituminous roads and hence making it more durable
• Safe speed limit = 85th percentile speed
• Sub-grade protection by proper drainage and hence to
• Minimum speed limit = 15th percentile speed
make the surface of pavement more impervious
• Quick-drying of pavement which in turn increases Q.84. In CBR test, the value of CBR is calculated at
safety (a) 2.5 mm penetration only
The main objective of providing a camber will be to drain (b) 5.0 mm penetration only
off rainwater from the road surface, as quickly as (c) Both of the above
possible. As rest options are the ultimate result of the
(d) None of these
quick drainage of water from the pavement.
Ans : (c) California Bearing Ratio (CBR) : -
• The California bearing ratio test is used to evaluate the
suitability of subgrade and the material used in sub-
base and base course.
• It is measured to indicate the relative strength of
paving materials and not the absolute strength.
• CBR test is a strength test conducted on the soil by
introducing surcharge load at the compaction rate of

UP AVAS VIKAS JE-2016 173 Civil Ki Goli

 1.25 mm per minute on a completely soaked soil Ans : (a) Los Angeles abrasion test :- Los Angeles
sample passing through 20 mm sieve size. machine or Los Angeles abrasion test on aggregates is
• The test results have been correlated with the used to measure the aggregate toughness and abrasion
thickness of the various materials required for the resistance such as crushing, degradation and
flexible pavement. disintegration.
• The pressure upto a penetration of 2.5 mm & its ratio • The principle of the Los Angeles abrasion test is to
to the bearing value of Standard Crushed Rock is produce abrasive action by use of standard steel balls
termed as CBR. which when mixed with aggregates and rotated in a
• Average value of 3 test specimens is reported as CBR. drum for a specific number of revolutions also cause the
• 4 days soaked remoulded sample is used. impact on aggregates.
• The maximum abrasion value for the wearing course is
30% and for the base course in WBM is 50%.

𝐥𝐨𝐚𝐝 𝐜𝐚𝐫𝐫𝐢𝐞𝐝 𝐛𝐲 𝐬𝐩𝐞𝐜𝐢𝐦𝐞𝐧

CBR= ×100
𝐥𝐨𝐚𝐝 𝐜𝐚𝐫𝐫𝐢𝐞𝐝 𝐛𝐲 𝐬𝐭𝐚𝐧𝐝𝐚𝐫𝐝 𝐬𝐩𝐞𝐜𝐢𝐦𝐞𝐧 Some of the important aggregate tests are:–
The load values on standard crushed stone are 1370 kg (1) Crushing Test: Resistance to crushing under
(70 kg/cm2) and 2055 kg (105 kg/cm2) at 2.5 mm and 5.0 gradually applied crushing load.
mm penetrations respectively. (2) Impact Test: To determine the toughness of
• If value of CBR at 2.5 mm > value of CBR at 5 mm, adopt aggregate.
CBR value of 2.5 while if value of CBR at 2.5 mm < value (3) Soundness Test: To determine the resistance of
of CBR at 5 mm. Then repeat the test and after it, if aggregates to weathering action.
value of CBR at 5 mm comes greater, finally adopt the (4) Abrasion Test: To determine the hardness of
greater value. aggregates
• CBR pavement design curve give thickness of (5) Shape Test: Flaky index, Elongation index, and
pavement cover to be provided. Angularity number are determined.
(6) Smith Test: To find soluble matter in stone
aggregates.
(7) Specific gravity and Water Adsorption test
(8) Stripping value test: Adhesion of Bitumen
with aggregates.

❖ RAILWAY
Q.86. What is the value of broad gauge used in railway
tracks?
(a) 1.000 m (b) 1.676 m (c) 0.762 m (d) 1.435 m
Ans : (b) The gauge of a railway track is defined as the
clear distance between the inner or running faces of
two-track rails.
The distance between the inner faces of a pair of wheels
is called the wheel gauge.
• The following table gives the length of the rail for
Pedestal of loading machine different types of gauges.
Q.85. Los Angeles testing machine is used to conduct Type of gauge Length of Rail
which of the following test?
B.G 12.8 m
(a) Abrasion test (b) Impact test
(c) Flakiness test (d) Crushing strength test M.G 11.89 m
N.G 11.89 m

UP AVAS VIKAS JE-2016 174 Civil Ki Goli

Indian Railways has standardized a rail length of 13 m • The narrow-gauge railway is the railway track, in which
(previously 12.8 m) for broad gauge and 12 m the distance between two tracks is 2 ft 6 in (762 mm)
(previously 11.8 m) for MG and NG tracks. and 2 ft (610 mm
Q.87. Composite Sleeper Index (CSI) is used for which
❖ IRRIGATION
of the following types of sleepers?
Q.89. The relationship between duty D
(a) Wooden sleepers (b) Cast Iron sleepers
(hectares/cumec), depth of water  (meters) and base
(c) Concrete sleepers (d) Steel sleepers period B (days) is given by –
Ans : (a) This is an index to determine the suitability of a (a)  = 6.48B/D (b) = 4.68B/D
particular timber for use as a wooden sleeper.
(c)  = 8.64B/D (d) = 4.86B/D
Note: This index measures the mechanical strength of
Ans: (c) Duty : It is the number of hectares of land
timber, derived from its composite properties of
strength and hardness. irrigated for full growth of a given crop by a supply of 1
cumec of water continuously during the entire base
The C.S.I. is calculated from the relation.
period of that crop.
𝐒+𝟏𝟎𝐇
C.S.I = Delta: The total water depth required by a crop to attain
𝟐𝟎
Where, its full maturity in its base period.
S = Strength Index, both for green and dry timber at 12% Base Period: The time period that elapsed for the instant
moisture content. of the showing of crop to the instant of its harvesting is
H = Hardness Index, both green and dry timber at 12% called base period or crop period.
moisture content. The relation between Duty (D), Delta (Δ) and Base period
(B) is given as:
𝟖.𝟔𝟒𝐁
=
𝐃
Q.90. The water utilised by the plants is available in
soils mainly in the form of -
(a) Gravity water (b) Free Water
Q.88. Which of the following gauge is generally used in
Metro Rail systems? (c) Chemical Water (d) Capillary Water
(a) Broad gauge (b) Meter gauge Ans:(d) Capillary water : Capillary water is held in the
(c) Narrow gauge (d) Standard gauge capillary pores (micropores). Capillary water is retained
on the soil particles by surface forces. It is held so
Ans: (d) Standard Gauge: - The distance between the strongly that gravity cannot remove it from the soil
two tracks in this railway gauge is 1435 mm (4 ft 81/2 in) particles. The molecules of capillary water are free and
• In India, standard gauge is used only for urban rail mobile and are present in a liquid state. Due to this
transit systems like Metro, Monorail, and Tram. reason, it evaporates easily at ordinary temperature
• Till 2010, the only standard gauge line in India was the though it is held firmly by the soil particle; plant roots
Kolkata (Calcutta) tram system. are able to absorb it.
• All metro lines coming in urban areas will be made only Gravitational water : Gravitational water occupies the
in the standard gauge because it is easy to get rolling larger soil pores (macropores) and moves down readily
stock for the standard gauge as compared to the Indian under the force of gravity. Water in excess of the field
gauge. capacity is termed gravitational water. Gravitational
• By 2016, the lines that are in operation are Delhi water is of no use to plants because it occupies the
Metro. Rapid Metro Rail Gurgaon, Bangalore Metro, and larger pores.
Mumbai Metro. All these are operated separately from Hygroscopic water : The water that held tightly on the
Indian Railways. surface of the soil colloidal particle is known as
Broad Gauge:- Broad gauge is also called wide gauge or hygroscopic water. It is essentially non-liquid and moves
large line. The distance between the two tracks in these primarily in the vapour form. Hygroscopic water held so
railway gauges is 1676 mm (5 ft 6 in). tenaciously by soil particles that plants cannot absorb it.
Metre Gauge:- The distance between the two tracks is Q.91. As per Lacey's theory, the silt factor is -
1,000 mm (3 ft 3 3/8 in). (a) Directly proportional to average particle size
• The meter gauge lines were made to reduce the cost. (b) Inversely proportional to average particle size
Narrow Gauge (c) Directly proportional to the square root of average
• The small gauge is called a Narrow gauge or a small particle size
line. (d) Independent of particle size

UP AVAS VIKAS JE-2016 175 Civil Ki Goli

Ans (c) According to Lacey’s, the design formulas to Detailed analysis of hydrographs is usually important in
build canal is as follows: flood damage mitigation, flood forecasting, or
establishing design flows for structures that convey
(1) Silt factor f = 1.76 √𝐝𝐦𝐦
floodwaters.
Hence Silt factor is directly proportional to the square
Factors that influence the hydrograph shape and
root of average particle size
volume
𝑸𝒇𝟐 𝟏
(2) Velocity of flow V= [ ]𝟔 i. Meteorological factors
𝟏𝟒𝟎
𝟓𝑽𝟐 ii. Physiographic or watershed factors and
(3) Hydraulic mean depth R=
𝟐𝒇 iii. Human factors
(4) Wetted perimeter P= 4.75√𝑸
𝟓
𝒇𝟑
(5) Bed slope S= 𝟏
𝟑𝟑𝟒𝟎×𝑸𝟔

❖ BCME
Q.92. The ratio of cement sand in mortar for full brick
wall masonry is generally taken as –
(a) 1:2 (b) 1:3 (c) 1:6 (d) 1:8
Ans :(c) The typical brick work mortar ratio for building a
brick wall in a typical residential home is
(cement:sand) 1:4 and 1:6.
A typical 4 inch brick work requires a 1:4 mortar ratio
and a 9 inch brick work requires a 1:6 mortar ratio.
❖ HYDRALIC MACHINE
❖ HYDROLOGY Q.95. Which of the following is an impulse turbine ?
Q.93. A rain gauge should preferably the fixed (a) Pelton (b) Francis
(a) Near the building (b) Under the tree (c) Kaplan (d) All of the above
(c) In a closed space (d) In an open space Ans : (a)
Ans:(d) While selecting the site for rain gauge stations Type of Energy available at Example
the following points should be considered: Turbine inlet
The site should be on level ground and in open space. Impulse Only Kinetic Pelton Turbine,
It should never be on sloping ground. Turbine Energy Turgo Cross flow
The site should be such that the distance between the Turbine
gauge station and the objects (like a tree, building, etc) Reaction Both Kinetic and Francis, Kaplan,
should be at least twice the height of the objects. Turbine pressure Energy Girard,
In the hilly area, where level ground is not available, the Fourneyron
site should be so selected that the station may be well propeller
shielded from high wind. Q.96. A Francis turbine under a head of 25 m products
• The site should be easily accessible to the observer. 2000 kW at a speed of 250 rpm. What is the value of its
specific speed?
• The site should be well protected from cattle by wire
fencing (a) 100 (b) 150 (c) 200 (d) 250
Q.94. Hydrograph is the graphical representation of Ans : (c) Given data-
(a) Rainfall intensity and flow velocity Head (H) =25 m N = 250 rpm P= 2500kW
(b) Runoff and time Specific speed of Turbine(NS)=
(c) Ground water flow and time Ns =
𝐍√𝐏
=
𝟐𝟓𝟎√𝟐𝟎𝟎𝟎
= 200
𝟓 𝟓
(d) None of the above 𝐇𝟒 𝟐𝟓𝟒
Ans: (b) Hydrograph: It is a continuous plot of
instantaneous discharge (runoff) v/s time.
❖ CPM & PERT
Q.97. Critical path………………
It results from a combination of physiographic and
meteorological conditions in a watershed and (a) Is always the longest (b) Is always the shortest
represents the integrated effects of climate. hydrologic (c) May be the longest (d) May be the shortest
losses, surface runoff, interflow, and groundwater flow.

UP AVAS VIKAS JE-2016 176 Civil Ki Goli

 * in this question there is mistake in option, as per board
UPSSSC MANDI PARISHAD C is correct.
DRAFTSMAN 22/5/2022 PAPER Q.03. Detailed cross sections of highway are generally
drawn to a natural scale of
❖ SURVEY (a) 1 cm = 2 to 2.5 m (b) 1 cm = 5 to 5.5 m
Q.01. To set a right angle on a survey line, the (c) 1 cm = 8 to 8.5 m (d) 1 cm = 500 to 600 m
instrument is held on
Ans : (a) Preliminary survey plans showing details of the
(a) The tangent of the line with its centre at the mid various alternative alignments and all information
distance and parallel to the right angle. collected should be normally drawn to scale of 10 cm = 1
(b) The line with its centre on the point at which km i.e. 1 cm = 100 m to 25 cm = 1 km.
perpendicular is erected. Detailed cross-sections are generally drawn to the
(c) The main station of survey line and away from the natural scale of 1 cm = 2.0 to 2.5 m. Cross-section should
point at an intervisible distance. be drawn every 100 m or where there are abrupt changes
(d) The nearest tie station and placed at an in level.
intervisible distance. Land acquisition plans and schedules are usually
Ans : (b) To set a right angle on a survey line prepared from the survey drawings for land acquisition
• Let AB be the survey line, and D be a point from which details. These plans show all general details such as
a perpendicular is to be drawn on the zigzag line. buildings, wells, nature of gradients and other details
required for assessing the values. The scale adopted
• Stand at point D and open the tape to sufficient length
maybe 1 cm = 40 m or less.
and draw an arc which cuts any two points of the survey
line AB. Suppose these points are E and F. Bisect line EF. Detailed design for cross drainage and masonry
structure is usually drawn to scale of 1 cm = 1 m. For
• Its midpoint will be C. Match DC.
details of any complicated portion of the structure
• This line will be perpendicular to the survey line AB. enlarged scales up to 8 cm = 1 m or up to half full size may
be employed.
Q.04. The site plan of the town showing the location of
the scheme and the area to be served is prepared on a
scale of
(a) R.F. = 1/800 or so (b) R.F. = 1/900 or so
(c) R.F. = 1/500 or so (d) R.F. = 1/1200 or so
Ans : (c) A site plan of the town showing the location of
the scheme and the area to be served by the scheme
should be prepared. A scale of 1:500 may be used.
Purpose of Survey Scale Interval
(meters)
1. Building sites 1 cm = 10 0.2 to 0.5
Q.02. In India, topographic maps are available with m or less
contour intervals of 2. Town planning 1 cm = 50 0.5 to 2
(a) 0.5 to 1 metre (b) 1 to 2 metre schemes, reservoirs, m to 100 m
etc.
(c) 15 or 30 metre (d) 5 to 10 metre
3. Location surveys 1 cm = 50 2 to 3
Ans : (c)* Contour Interval:- The vertical distance m to 200 m
between consecutive contours is termed as contour
interval. Q.05. The area of any irregular figure of the plotted map
• It is desirable to have a constant contour interval is measured with the help of
throughout the map. (a) Pantagraph (b) Sextant
• However, in special cases, a variable contour interval
(c) Clinometer (d) Planimeter
may also be provided.
• A variable contour interval is, as far as possible Ans : (d) Planimeter: It is an instrument used in surveying
avoided since it gives a false impression of the relative to compute the area of any given plan. Planimeter only
steepness of the ground in different parts of the map. needs plan drawn on the sheet to calculate area.
• Usually contour intervals are taken as 1 to 15 m.

UPSSSC DRAFTSMAN 2022 178 CIVIL KI GOLI

• Generally, it is very difficult to determine the area of known as the ellipsoidal height which cannot be used
irregular plot. So, by using planimeter we can easily directly for practical surveying but needs to be
calculate the area of any shape. transformed into orthometric heights, being the distance
• It is used for measurement of the area of cross- measured along the plumb line between the geoid and a
sections for the highways and railways, and checking point on the Earth’s surface
computed areas in property surveys and property
divisions.
• The area determined with a planimeter is more
accurate when the area is larger and particularly when
the plotting scale is large.

Q.08. _______field book is used for a comparatively

large scale and most detailed dimension work.
Clinometer: It is an instrument for measuring angles of (a) Dotted line (b) Single line
slope, elevation or depression of an object with respect (c) Triple line (d) Double line
to the ground. While the spirit level is restricted to Ans : (b) Field Book - A book in which a surveyor writes
relatively small angles, clinometers can be used for down the chain or tape measurements and other
much larger angles. Clinometers are used to determine technical notes taken in the field.
straightness and flatness of surfaces. There are two main types of field-book :
Optical square: It is small and compact hand
instrument and works on the principle of reflection. It Single line Double line
is used for setting out perpendiculars to a point on a Every page of this kind It is similar to the single-
survey time. A metal cover is also provided to protect it of field book has a line field book but
from dust, moisture etc. single red line ruled instead of a single red
Pantagraph: It is used for enlarging, reducing or down the middle. line in the centre, two
reproducing the plans. Its working is based on the blue or red lines about
principle of similar triangles. 1.5 cm apart are ruled
Q.06. Find the correct climatic condition in which large down in the middle of
area such as a country or a region is covered. each page.
(a) Micro climate (b) Macro climate The space on either The space between these
side of the line is lines represents the chain
(c) Nano climate (d) Meso climate
available for sketching line and is reserved for
Ans (b) Climate : -The climate of the Earth consists of a the various features entering changes, which
series of interlinked physical systems powered by the along the chain line are thus kept entirely
sun. In the built environment we are generally and for writing the separated from the other
concerned with local climatic systems in particular. offset distances. dimensions.
Macro Climate - the climate of a larger area such as a Convenient for large A double line field-book is
region or a country scale and much commonly used.
Micro Climate - the variations in localized climate detailed dimension-
around a building work.
Q.07. The height deduced from GPS observations is
GPS Q.09. Which of the following is not a segment of Global
(a) The ellipsoidal height (b) The sensitive height Positioning System ?
(c) The polymetric height (d) The variable height (a) User segment (b) Independent segment
(c) Space segment (d) Control segment
Ans : (a) To obtain accurate elevations with the Global
Positioning System (GPS), the geoidal heights in the Ans: (b) Global Positioning System (G.P.S.) - GPS is a
area must be known and applied. The height obtained space based all weather radio navigation system that
through Global Positioning System (GPS) observation is provides quickly, accurately and inexpensively the time,

UPSSSC DRAFTSMAN 2022 179 CIVIL KI GOLI

position and velocity of the object anywhere on the ❖ HIGHWAY
globe at any time.
Q.11. The width of pavement or carriageway depends
• Current GPS is based on accurate ephemeris data on on.
the real time location on each satellite and on a (a) Width of traffic lane
precisely kept time.
(b) Numbers of lanes
• It uses the satellite signals, accurate time and
sophisticated algorithms to generate distances in (c) Both width of traffic lane and numbers of lanes
order to triangulate positions anywhere on earth. (d) Type of soil
• Total 24 satellites are used in GPS survey in 6 orbits. Ans: (c) Width of Pavement or Carriageway :
• Min. 4 satellites are required for the GPS to determine The pavement width depends upon the number of lanes
precise position of object and the width of the single-lane required. The lane width
There is three segments comprised in GPS : determines on the basis of the width of the vehicle and
1. Satellite constellations or space segment- It consists minimum side clearance.
of the nominal 24- satellite constellation. Class of Road Width of
2. Ground control or monitoring network (operational Carriageway
control segment) - It is responsible for maintaining the Single lane 3.75 m
satellite and their proper functioning. Two lanes, without raised 7.0 m
3. User receiving equipment or user segment - It kerbs
receives and process the L-band signals transmitted Two lanes, with raised kerbs 7.5 m
from the satellite to determine users position, velocity Intermediate carriageway 5.5 m
and time. Multi-lane pavements 3.5 m per each lane
Q.10. _________is used to locate the boundaries of a Q.12. Control points governing the alignment of the
field and to determine its area. highways are called
(a) Cross staff survey (b) Levelling staff survey (a) Topographic points (b) Obligatory points
(c) Secondary area survey (d) Near distance survey (c) Alignment points (d) Road line points
Ans: (a) A cross-staff survey is a type of surveying Ans : (b) Factors affecting highway alignment are:
technique used to locate boundaries of a field and 1. Obligatory points
determine its area. It involves the use of a cross-staff, 2. Traffic.
which is a simple surveying instrument consisting of a
3. Geometry design.
wooden staff with perpendicular arms. The cross- staff
is used to measure angles and distances in the field. 4. Economy.
Advantages of Cross-staff Survey: 5. Other consideration.
1. Simple and inexpensive: The cross-staff survey Obligatory points : Obligatory points in highway
requires minimal equipment and is cost-effective engineering are controls points governing the alignment
compared to other surveying techniques. of highways.
2. Quick and efficient: The survey can be carried out • Obligatory points are through which Alignment is to
relatively quickly, making it suitable for smaller fields or pass. Example- Bridge site, intermediate town,
areas. Mountain pass, etc. so to connect obligatory point the
alignment should be changed.
3. Accurate for regular-shaped fields: The cross-staff
survey is accurate for fields with regular shapes, such as • Obligatory points through which alignment should not
rectangular or square fields. pass. Example-Religious places, costly structure, etc.
Limitations of Cross-staff Survey : ❖ BCME
1. Not suitable for irregular-shaped fields: The cross- Q.13. In case of box sheeting, the vertical sheets are
staff survey may not provide accurate results for fields kept in position by
with irregular shapes, as it assumes a regular boundary.
(a) Longitudinal rows of spade
2. Limited precision: The accuracy of the cross-staff
(b) Longitudinal rows of wedge
survey depends on the skill of the surveyor and the
instrument used. It may not be as precise as more (c) Longitudinal rows of wales
advanced surveying techniques. (d) Longitudinal rows of shrub
3. Relies on visual alignment: The surveyor needs to Ans : (c) Timbering In Trenches : - When the depth of
visually align the cross-staff with the reference points, trench is large, or when the sub-soil is loose, the sides of
which may introduce human error. the trench may cave in. The problem can be solved by
adopting a suitable method of timbering. Timbering of

UPSSSC DRAFTSMAN 2022 180 CIVIL KI GOLI

trenches, sometimes also known as shoring consists of
providing timber planks or boards and struts to give
temporary support to the sides of the trench.
Timbering of deep trenches can be done with the help
of the following methods:
1. Stay bracing. 2. Box sheeting 3. Vertical sheeting
4. Runner system 5. Sheet piling.
1. Stay bracing:- This method is used for supporting the
sides or a bench excavated in fairly firm soil, when the
depth of excavation does not exceed about 2 metres.
The method consists of placing vertical sheets (called
sheathing) or polling boards opposite each other
against the two walls of the trench and holding them in
position by one or two rows of struts.
2. Box sheeting:- This method is adopted in loose soils,
when the depth of excavation does not exceed 4
meters .The box like structure, consisting of vertical
sheets placed very near to each other (some times Following are the members of door frame
touching each other) and keeping them in position by
(1) Head of Door Frame:- The head is the top horizontal
longitudinal rows (usually two) of Wales. Struts are
member of the frame.
then provided across the Wales.
(2) The sill of Door Frame:- The sill is the bottom
3. Vertical sheeting:- This system is adopted for deep
horizontal member of the frame which may or may not
trenches (up to 10 m depth) in soft ground. The method
be provided. The main reason for providing the sill frame
is similar to the box sheeting except that the excavation
is that the floor acts as a sill itself
is carried out in stages and at the end of each stage, an
offset is provided, so that the width of the trench goes (3) Horn of Door Frame:- It is a horizontal projection of
on decreasing as the depth increases. For each stage, the top and bottom members of the frame to enable the
separate vertical sheeting, supported by horizontal fixing of the frame in the wall opening.
wailings and struts are provided. (4) Jamb of Door Frame:- The door jamb is the vertical
4. Runner system:- This system is used in extremely wall face of a door opening that supports the door frame.
loose and soft ground, which needs immediate support In the door frame, the jamb is the vertical portion of the
as excavation progresses. The system is similar to door frame in which a door is secured. If you want to
vertical sheeting of box system, except that in the place open and lock your door properly, door jamb is
of vertical sheeting, runners, made of long thick important.
wooden sheets or planks with iron shoe at the ends, are (5) Holdfasts of a door frame:- These are the mild steel
provided. flat bars that are used to hold the frame in the position.
5. Sheet piling:- This method is adopted when These bars are generally bent into Z-shape and fixed on
one end to the frame and the other end is inserted in the
(i) soil to be excavated is soft or loose
wall.
(ii) depth of excavation is large
(6) Rebate of a door frame:- It is a depression made all
(iii) width of trench is also large and around the door frame, on one side of which, the door
(iv) there is sub-soil water. shutter is fixed by the means of hinges.
Sheet piles are designed to resist lateral earth pressure. (7) A threshold of a door frame:- It is the cross wooden
These are driven in the ground by mechanical means piece fixed to door under a door frame which forms the
(pile driving equipment). They can be used for sill and may be provided if required.
excavating to a very large depth. (8) The transom of a door frame:- It is a horizontal
Q.14. The bottom surface of door or window opening member of a frame, which is provided to sub-divide a
is known as door opening horizontally.
(a) Lintel (b) Sill (c) Reveals (d) Jambs (9) Mullion of a door frame: It is a vertical member of the
frame, which is provided to sub-divide a door vertically.
Ans: (b) Door frame: The door frame consists of an
assembly of horizontal and vertical members that are Q.15. To make raking shores more effective, rakers
placed at the top, bottom, and sides of an opening to should be inclined to the ground by
form an enclosure providing support for a door. (a) 45o (b) 60o
Generally, the door frame is made of wood. (c) 70 o
(d) 90o

UPSSSC DRAFTSMAN 2022 181 CIVIL KI GOLI

Ans: (a) Raking Shore or Inclined Shore: A Raking shore structure needs pulling down when openings must be
or Inclined shore consists of a wall plate, needles, newly enlarged or made into a wall.
plates, cleats, bracing, and sole plate. • In actual • Its support requirement depends on the types of soil
practice, the angle of inclination in Raking or Inclined and when an excavation depth is at least 1.20-meter
shore may vary from 45° to 75°. difference in levels from ground level.
• The top raker should not be inclined steeper than 75° Q.17. A suspended scaffolding is used for
with the horizontal. (a) Stone masonry (b) Pointing and painting
• The wall plate is placed against the wall and is secured (c) Brick masonry (d) Slab casting
by the means of needles that penetrate into the wall
for a distance of about 150 mm. Ans: (b) The various types of scaffolding and its
application are as follows:
• The wall plate distributes the pressure evenly.
Single scaffolding:- Single scaffolding is generally used
Q.16. In which of the following circumstances the for brick masonry and is also called as brick layer’s
shoring is used ? scaffolding.
(a) When foundation settled down.
• Single scaffolding consists of standards, ledgers,
(b) When an earthquake prone area as a permanent putlogs etc., which is parallel to the wall at a distance
supporting structure. of about 1.2 m.
(c) When a wall shows signs of bulging out due to bad • Distance between the standards is about 2 to 2.5 m.
workmanship Ledgers connect the standards at vertical interval of 1.2
(d) For supporting the structure against snow load. to 1.5 m.
Ans : (c) SHORING : - Shoring is a temporary structure • Putlogs are taken out from the hole left in the wall to
used to prevent the collapse of the main under- one end of the ledgers. Putlogs are placed at an interval
construction structure. of 1.2 to 1.5 m.
Steel scaffolding:- Steel scaffolding is constructed by
steel tubes which are fixed together by steel couplers or
fittings.
• It is very easy to construct or dismantle. It has greater
strength, greater durability and higher fire resistance.
• It is not economical but will give more safety for
workers. So, it is used extensively nowadays.
Trestle scaffolding:- In Trestle scaffolding, the working
platform is supported on movable tripods or ladders.
• This is generally used for work inside the room, such as
paintings, repairs etc., up to a height of 5m.
Suspended scaffolding :- In suspended scaffolding, the
working platform is suspended from roofs with the help
of wire ropes or chains etc., it can be raised or lowered to
our required level.
• This type of scaffolding is used for repair works,
pointing, paintings etc.
Q.18. Choose the correct statement in terms of
disadvantage of a flat roof.
(a) They can be easily made fire proof, in comparison
to pitched roof.
• The most commonly shoring support is required
(b) Flat roof have better insulation property.
during the early stage of construction which is
excavation. (c) They are unsuitable at places of heavy rainfall.
• It is a momentary support, which is used during the (d) Flat roof are proved to be overall economic.
repair or original construction of buildings and in Ans: (c) Flat roofs or terraced roofing : Flat roofs are
excavations. mainly used in plains, where there is low to moderate
• It can be utilized when walls bulge out or cracks due rainfall and temperature is high.
to unequal settlement of foundation and repairs must • This type of roof is generally horizontal but having a
be carried out to the cracked wall, when an adjacent slope of not more than 10° to drain rainwater. It is also
called a terraced roof.

UPSSSC DRAFTSMAN 2022 182 CIVIL KI GOLI

Advantages and disadvantages of flat roofs:
Advantages : 1. The construction of flat roof is
simplified and maintenance is easy.
2. The roof can be used as terrace for playing,
gardening, sleeping and for celebrating functions
3. It is easier to make the flat roof fire-proof than a
sloping roof.
4. They avoid the enclosure of the triangular space. Due Queen Post Truss:- A queen post is a tension member in
to this, the architectural appearance of the building is a truss that can span longer openings than a king post
very much improved. truss.
5. Flat roofs have better insulating properties and they • Span length is in between 8 to 12 meters.
are more stable against high winds. Pratt Truss - Span length is in between 6 to 10 meters.
6. The construction work of upper floors can be easily Howe Truss - Span length is in between 6 to 30 meters.
started. In case of a pitched roof, the entire roof is to
Q.20. To prevent lateral movement of stones, which of
be removed and is to be replaced by a new roof under
the following joints is used?
such circumstances. It is therefore considered to be the
best choice for multi-storeyed buildings. (a) Butt or square joint (b) Tongued or joggle joint
7. They do not require false ceiling, which is essential in (c) Rebated or lapped joint (d) Tabled or bed joint
pitched roofs. Ans : (d) TABLED OR BED JOINT:- This joint is used to
8. Flat roofs are proved to be overall economical. prevent lateral movement of stones such as in sea walls
Disadvantages: 1. The span of flat roof is restricted, where the lateral pressure is heavy.
unless intermediate columns are introduced. Pitched • The joint is made by forming a joggle in the bed of the
roofs can be used over large spans without any stone.
intermediate columns. • The height of the projection is kept about 30-40mm
2 . The self weight of flat roof is very high. Due to this, while the width is kept equal to above 1/3 the breadth
the sizes of beams, columns, foundations of the stone.
3. Flat roofs are exposed to sun and are subjected to BUTT JOINT OR SQUARE JOINT:- This is the most
violent temperature changes, which may lead to cracks commonly used joint in stone masonry.
in the surface of the roof and other structural members • The dressed edges of two adjacent stones are placed
are heavy. side by side.
4. The pockets of water are formed on the surface of TONGUE AND GROOVED JOINT OR JOGGLE JOINT:
the roof, if slope is not sufficient. This leads to the • This type of joint is provided to prevent sliding along the
leakage of the roof and it sometimes proves to be side joints.
difficult to exactly locate the position of leakage on the • The joint is made by providing projection or tongue in
roof. one stone and a corresponding groove or sinking on the
5. They are unsuitable at places of heavy rainfall, hilly adjacent stone.
areas or areas where there is heavy snowfall REBATED OR LAPPED JOINT: This type of joint is provided
6. The dead weight of flat roof is considerable and in arches, gables, copings to prevent the possible
hence it prove to be more expensive. Its initial cost is movement of the stones.
higher than a pitched roof. • The length of the lap or rebate depends upon the
7. The initial cost of flat roof is also more than the nature of the work, but it should not be less than
pitched roof. 70mm.
Q.19. King post truss is suitable for the span of Q.21._______is a vertical member which is placed at the
(a) 5 to 8 m (b) 12 to 15 m ends of flights to connect the ends of strings and hand
(c) 18 to 21 m (d) 24 to 27 m rail
Ans : (a) King Post Truss:- A king post uses one central (a) Scotia (b) Baluster
supporting post, whereas the queen post truss uses (c) Balustrade (d) Newel Post
two. Even though it is a tension member, rather than a Ans: (d) Staircase: Stairs are a set of steps that give access
compression member, it is commonly still called a post. from floor to floor.
• Span length is in the range of 5 to 8 meters • The room or enclosure of the building, in which stair is
located is known as the staircase.

UPSSSC DRAFTSMAN 2022 183 CIVIL KI GOLI

• A staircase provides access & communication Ans : (b) Segments of Arch:
between floors in multi-story buildings. Spandrel: If two arches are constructed side by side, then
a curved triangular space is formed between the extrados
with the base as the horizontal line through the crown.
This space is called a spandrel.
Springing line: The imaginary line joining the springing
points of either end is called a springing line.
Haunch: The lower half of the arch between the crown
and skewback is called haunch.
Skew back: This is an inclined surface or splayed surface
on the abutment, from which the arch curve starts or
ends.
The following are the components of the staircase
Springer: The first voussoir at the springing level which is
Component Definition immediately adjacent to the skewback is called as
springer.
Step A portion of the stair which permits
ascent or descent. A stair is composed Voussoirs: The wedge-shaped units of masonry which are
of a set of steps. forming an arch is called as voussoirs.
Soffit: The inner surface of an arch is called soffit. Soffit
Tread The upper horizontal portion of a step
and intrados are used synonymously.
upon which the foot is placed while
ascending or descending. Key: The voussoir is located at the crown of the arch. Also
called the keystone.
Riser The vertical portion of a step providing
support to the tread.
Landing A level platform at the top or bottom of
a flight between floors.
Flight An unbroken series of steps between
landings.
Rise The vertical distance between two
successive tread faces.
Going The horizontal distance between two
successive riser faces.
Nosing The projecting part of the tread beyond
the face of the riser.
Q.23. A............tool is used for cutting soft bricks and
Scotia A molding provided under the nosing to dressing out surface.
provide strength to the nosing. (a) Bolster (b) Scutch
Soffit The underside of a stair. (c) Trowel (d) Brick hammer
Pitch or The angle which the line of the nosing Ans : (b) Tools that are used in brick masonry
Slope of the stair makes with the horizontal. construction:
Strings These are the sloping members which Tools Use in Construction
support the steps in a stair. Brick Hammer Used for cutting bricks, also for
Newel post A vertical member which is placed at pushing the bricks in courses.
the ends of flight to connects the ends Trowel Used for lifting and spreading
of strings and handrail. mortar; also for cutting bricks
Spirit Level Used, with a straight edge, for
Baluster A vertical member of wood or metal,
getting horizontal surface: also
supporting the handrail.
used for levelling
Headroom The clear vertical distance between the Plumb rule Used for checking verticality of
tread and overload structure. brick walls.
Q.22. The lower half of the arch between the crown Scutch Used for cutting soft bricks and
and skew back is called dressing out surfaces.
(a) Spandrel (b) Haunch Bolster Used for accurate cutting of bricks.
(c) Springer (d) Voussoirs

UPSSSC DRAFTSMAN 2022 184 CIVIL KI GOLI

The effectiveness of this system depends entirely on
the depth of the water seal. No water seal should be
less than 75 mm in depth.

4. One-pipe Partially Ventilated System:- This system is

via media between the first and second one. There is only
one soil pipe into which all WCs, baths, sinks, and
washbasins discharge. In addition, there is a relief vent
2. One Pipe System:- In this system of plumbing, the pipe that ventilates only the traps of WCs and urinals
waste connections from sinks, baths, washbasins, and
the soil pipe which is connected directly to the drainage
system.
Gully traps and waste pipes are completely dispensed
with. But all the traps of WCs basins, etc, are completely
ventilated to preserve the water seal by a separate vent
pipe.

Q.29. _______is used for fixing the pipe in position while

cutting or threading the pipe.
(a) Hacksaw (b) Pipe cutter
(c) Pipe wrench (d) Pipe vice
Ans : (d) Pipe vice :- A pipe vice is another holding tool
that you can use in the process of assembly, threading,
cutting and disassembly. Open and fixed sides are two
types of pipe vice that you can use in the assembly of
pipeline components. You can also use pipe vices for
welding pipes.
3.Two Pipe System:- In this plumbing system, two pipes
are installed. WCs and urinals are connected to vertical
soil pipe baths, kitchens, basins, etc are connected to
another separate vertical waste pipe. Soil pipes and
waste pipes are provided with separate vent pipes.
This system thus requires four pipes and hence proves
very costly. The soil pipe is connected to the drain
directly but the waste pipe should be connected
through a trapped gully.

UPSSSC DRAFTSMAN 2022 186 CIVIL KI GOLI

horizon line. Both horizontal axis (width and depth) Ans : (d) 2H pencils are hard and produce light lines,
recede towards their respective vanishing points on the which are ideal for precise and fine lines, such as center
horizon line. The vertical axis recedes towards the third lines.
vanishing point. • H pencils are slightly softer than 2H and produce
somewhat darker lines, which are suitable for boundary
lines.
• HB pencils are of medium hardness and produce
medium-dark lines, making them suitable for border
lines.
Q.35. The final size of the folded print in method-1 will
be
(a) 290 mm × 190 mm (b) 297 mm × 210 mm
(c) 290 mm × 210 mm (d) 297 mm × 190 mm
Ans : (d) Dimensions for folding of various sizes of
drawing sheets by the two methods are given below :
Method 1 Method 2
Q.32. The angle which can't make using both the set- She Horizont Vertical No. Horizont Vertical No.
square is_________ et al Dimensi of al Dimensi of
(a) 15o (b) 105o (c) 125o (d) 165o Size Dimensi ons from Fol Dimensi ons from Fol
ons from Bottom ds ons from Bottom ds
Ans: (c) 15o can be made by keeping 45o and Left (mm) Left (alt) (alt) (alt
30o adjacent to each other on the line perpendicular to (mm) (mm) (mm) )
the line for which 15o is made, Likewise for 105o and A0 130 + 297 × 2 + 5 139 + 297 + 7
165o also if we just change the alignment with the 109 + 247 210 × 5 297 +
190 × 5 247
required line it possible. But to make 125o there is no
such combination available for Set-squares. A1 146 + 297 + 6 211 + 297 + 4
125 + 297 210 × 3 297
Q.33. Gothic letters are mainly used for 190 × 3
(a) Main-titles of ink-drawings A2 116 + 96 297 1 174 + 297 + 3
× 3 + 190 210 × 2 123
(b) Sub-titles of ink-drawings
A3 125 + 297 1 210 + 297 1
(c) Numerals 105 + 210
(d) Notes, dimensions figures of ink-drawing 190
Ans : (a) Gothic letters: Gothic letters are formed by The final size of the folded print in method I will be 297
thickening the stems of single-stroke letters. These are mm x 190 mm, while that in method II will be 297mm ×
mostly used for main titles of ink-drawings. The outlines 210 mm. In either case the title block is visible in the top
of the letters are first drawn with the aid of instruments part of the folded print.
and then filled-in with ink. The thickness of the stem Q.36. Initial work and construction lines are drawn
may vary from 1/5th to 1/10th of the height of the using_pencil.
letters. Figure 3.9 shows the alphabets and figures in
(a) 3H (b) 4H (c) H (d) 2H
gothic with thickness equal to 1/7 of the height.
Ans : (c) Initial work and construction lines are drawn
using H pencil. 2H pencil is used for outlines, dotted lines,
dimension lines and arrowheads. 3H, 4H are used for
centre lines and section lines.
Q.37. Which of the followings is not a type of
polyhedran ?
Q.34. Match the following grades of pencils with their (a) Tetrahedron (b) Cube
correct uses : (c) Square pyramid (d) Cone
1. 2H i. Boundary lines Ans : (d) Polyhedron: - A three-dimensional shape with
2. H ii. Border lines flat polygonal faces, straight edges, and sharp corners or
3. HB iii. Centre lines vertices is called a polyhedron.
(a) 1. iii, 2. ii, 3. i (b) 1. ii, 2. iii, 3. i Common examples are cubes, prisms, pyramids.
(c) 1. i, 2. iii, 3. ii (d) 1. iii, 2. i, 3. ii However, cones, and spheres are not polyhedrons since
they do not have polygonal faces.

UPSSSC DRAFTSMAN 2022 188 CIVIL KI GOLI

Q.38. If a cone is cut by a section plane parallel to its 3. Well (Caissons) Foundation
base and the portion containing the apex or vertex is Pile Foundation:- Pile is a slender member with small
removed, the remaining portion is area of cross-section relative to its length. They can
(a) Truncated of cone (b) Frustum of cone transfer load either by friction or by bearing. Pile
(c) Solids of revolution (d) Generator of cone foundation are used when:
Ans : (b) A cone is cut through a plane parallel to its • The load is to be transferred to stronger or less
base and then the cone that is formed on one side of compressible stratum, preferably rock.
that plane is removed. The new part that is left over on • The granular soils need to be compacted.
the other side of the plane is called a frustum of cone. • The horizontal and the inclined forces need to be
carried from the bridge abutments and the retaining
walls.

Q.39. Which of the following symbols represents a

connecting wire ?

Ans : (c)
Element of circuit Diagram of circuit
Connecting wire
Open switch
Closed switch CIVIL KI GOLI 9255624029
Light bulb

❖ SOIL
Q.40. A foundation is to transmit loads to lower level
of ground by a combination of friction and end bearing
is known as
(a) Pile foundation (b) Raft foundation
(c) Grillage foundation (d) Wall foundation
Ans: (a) Foundation:- is defined as that part of the
structure that connects and transmits the load from the
structure to the ground soil.
Types of Foundation:
A. Shallow Foundation:
1. Wall Footing
2. Isolated column/Column Footing
3. Combined Footing
4. Cantilever (Strap) Footing
5. Mat (Raft) Foundation
B. Deep Foundation:
1. Pile Foundation
2. Pier Foundation

UPSSSC DRAFTSMAN 2022 189 CIVIL KI GOLI

 UPSSSC RAJ INSTRUCTOR CIVIL
WORK 26 MARCH 2023 PAPER
❖ R.C.C.
Q.01. The maximum compressive strain in concrete in
axial compression is taken as :
(a) 0.0035 (b) 0.002 Cylindrical Test Cube Test
𝟎.𝟖𝟕𝐟𝐲 𝟎.𝟖𝟕𝐟𝐲 Q.04. What is the effective length of column when both
(c) + 0.0035 (d) + 0.002
𝐄𝐬 𝐄𝐬
the ends are restrained against both translation and
Ans : (b) Following assumptions have been made in IS rotation?
456:200 regarding maximum strain in compression (a) 1.0 L (b) 1.2 L (c) 0.8 L (d) 0.65 L
members:
Ans : (d) As per IS code effective length for different end
• Maximum compressive strain in case of direct condition : -
compression in case member subjected to axial
Degree of end restraint of Recommended
compression = 0.002
compression member value of
• Maximum compressive strain in case of bending effective length
compression = 0.0035
Effectively held in position and restrained 0.65 L
• The maximum strain at highly compressed extreme against rotation at both ends
fiber in a section subjected to axial compression and
Effectively held in position at both ends 0.80 L
bending and when there is no tension in the section is = restrained against rotation at one end
0.0035-0.75 x strain at least compressed extreme fiber
Effectively held in positions at both ends, 1.00 L
Q.02. The value of design bond stress for bars in tension but not restrained against rotation
shall be increased by compression for bars in
Effectively held in position and restrained 1.20 L
(a) 60% (b) 35% against rotation at one end, and at the
(c) 25% (d) 20% other end restrained against rotation but
not held in position
Ans : (c) As per IS 456: 2000, Clause 26.2.1.1,
Effectively held in positions and restrained 1.50 L
In limit state method of design, the design bond
against rotation at one end and at the
strength for plain bars in tension are given below in the other end partially restrained against
table rotation but not held in position
Grade of M40 And Effectively held in position at one end but 2.00 L
M20 M25 M30 M35
Concrete above not restrained against rotation, and at the
Design Bond other end restrained against rotation but
Stress 1.2 1.4 1.5 1.7 1.9 not held in position
Tbd N/mm²
Effectively held in position and restrained 2.00 L
• For bars in compression, these values of bond stress at one end but not held in position nor
for bars in tension shall be increased by 25 percent. restrained against rotation at the end
• For deformed bars (e.g. HYSD bars) conforming to IS: Q.05. Which proportion of concrete is suitable for plain
1786, these values shall be increased by 60 percent. concrete?
Q.03. The size of cylindrical test specimen used in (a) 1:1:2 (b) 1:2:4 (c) 1:3:6 (d) 1:1.5:3
compression test of concrete is: Ans : (b) The following proportion of the ingredients of
(a) 150 mm x 300 mm (b) 150 mm x 200 mm concrete mix is used in arbitrary method of
(c) 120 mm x 300 mm (d) 100 mm x 150 mm proportioning
Ans :(a) Compression test of concrete is one of the most Grade Mix Proportion Used In
common and widely used test worldwide. There can be M5 01:05:10 Foundation PCC
two types of specimen
M 7.5 01:04:08 Foundation and Flooring
(1) Cubes of size 150 mm x 150 mm x 150 mm
(2) Cylindrical of 150 mm diameter and 300 mm high. M 10 01:03:06 Flooring
M 15 01:02:04 Other PCC works
M20 1:1.5:3 RCC works

UPSSSC RAJ INSTRUCTOR 2023 190 Civil KI Goli

 ❖ STEEL STRUCTURE • It can be defined as the difference between the height
Q.06. The effective slenderness ratio of laced column is of the concrete before removing slump cone (mould)
increased by__________percent of the actual and height of the concrete after removing of slump cone
maximum slenderness ratio. as measured during concrete slump test.
(a) 5 (b) 10 (c) 1.10 (d) 1.05 Different Types of Slump :
Collapse slump: In this case, fresh concrete collapse
Ans : (a) Slenderness ratio: The slenderness ratio of a
completely. The mix is too wet or high workability mix,
column is defined as the ratio of effective length to the
slump test isn't appropriate for such mix.
corresponding gyration of section. Thus
𝑳𝒆 Shear Slump: If one-half of the cone slides down in an
Slenderness ratio = inclined plane, it is called a shear slump. It is an
𝑹
Effective slenderness ratio: As per IS 800: 2007 - indication of the lack of cohesion of the mix. Again
it's in order to account for the shear deformation effect. perform the experiment to avoid a shear slump.
The effective slenderness ratio of battened columns shall True Slump: Mix has high stiff consistency. In a true
be taken as 1.1 times the maximum actual slenderness slump concrete just subsides shortly and more or less
ratio of the column. maintain the mould shape. This type of slump is most
The effective slenderness ratio of the laced column shall desirable.
be taken as 1.05 times the actual maximum slenderness Zero Slump: If concrete maintains the actual shape of
ratio. the mould , it is called zero slumps which represent stiff,
consistent and almost no workability.
❖ BUILDING MATERIAL
Q.07. A cement with low_____content is found
effective to prevent concrete from sulphate attack.
(a) C3S (b) C2S (C) C3A (d) C4AF
Ans : (c) Sulphate attack: The calcium hydroxide also Q.09. Honeycomb brick work shall be measured in:
reacts with sulphates present in soil or water to form (a) cum (b) sq m (c) m (d) quintal
calcium sulphate which further reacts with C 3A and
Ans : (b) Clearing of shrubs, Partition wall, Honeycomb
cause deterioration of concrete. This is known as
brickwork, Brick Flat Soling, Formwork, DPC measured in
sulphate attack.
square meter.
• Waterborne sulfates react with hydration products of
the tri-calcium aluminate (C3A) phase of portland Q.10. How many bricks are required for 50 sq m half
cement, and with calcium hydroxide (Ca(OH)2) to form brick wall?
an expansive crystalline product called ettringite (a) 5000 (b) 4000 (c) 500 (d) 2500
(calcium sulphoaluminate hydrate). Ans :(d) Modular size of brick with mortar = 20× 10 ×
• Expansion due to ettringite formation causes tensile 10 cm
stresses to develop in the concrete. Area of one brick with mortar = 0.2 × 𝟎. 𝟏 = 0.02 m2
•When these stresses become greater than concrete's 𝟓𝟎
Bricks required for 50 m2 wall = = 2500
tensile capacity, the concrete begins to crack. 𝟎.𝟎𝟐

•These cracks allow easy ingress for more sulfates into Q.11….is used in brick masonry for lifting and spreading
the concrete and the deterioration accelerates. mortar, cutting of brick and construction of joints.
• Sulfates also cause the chemical disintegration of some (a) Brick hammer (b) Trowel
of the cement hydration products. (c) Plumb bob (d) Steel tape
• To remedy the sulphate attack, the use of cement with Ans:(b) Tools that are used in brick masonry
low C3A content is found to be effective. construction:
Q.08. Identify the type of slump when one half of the Brick Hammer: Used for cutting bricks, also for pushing
cone slides down while performing the workability the bricks in courses.
test. Trowel: Used for lifting and spreading mortar, also for
(a) True slump (b) Shear slump cutting bricks.
(c) Collapse slump (d) No slump Spirit Level: Used with a straight edge, for getting
Ans :(b) A slump test is the simplest test to determine horizontal surface; also used for leveling.
the workability of concrete, involves low cost and Plumb bob: Used for checking verticality of brick walls.
provides immediate results. Line and Pins: Used for maintaining alignment of
• It is used to measure the consistency/plasticity of the courses.
concrete mix.
UPSSSC RAJ INSTRUCTOR 2023 191 Civil KI Goli
Scutch: Used for cutting soft bricks and dressing out Q.14. A plate load test is carried out by using bearing
surfaces. plate made of mild steel of thickness not less
Bolster: Used for accurate cutting of bricks. than_________
Jointer: Used for pointing the joints. (a) 3.0 cm (b) 2.5 cm (c) 1.5 cm (d) 7.5 cm
Ans:(b) Plate load test: Plate load test is performed to
determine soil bearing capacity and settlement.
Shape and size of the plate load test are given below:
• Square or circular in shape
• Thickness > = 25 mm
• Size may vary from 300 to 750 mm. therefore, the
maximum size is 750 mm
• For clayey and silty soils and for loose to medium
dense sandy soils with N<15, 450 mm square plate is
used
• For dense sandy or gravelly soils (15 < N < 30) three
plates of sizes 300 mm to 750 mm is used depending
on reaction loading and maximum grain size
Q.12. A lower part of plastered wall where special • Side of plate = 4 x size of particles
treatment is given to make it better resistant is known Q.15. A factor of_________safety of is considered for
as: sites of heavy buildings or structure to determine safe
(a) Dubbing coat (b) Finishing coat bearing capacity of soil.
(c) Under coat (d) Dado (a) 2 (b) 4 (c) 3.5 (d) 2.5 to 3
Ans :(d) Dado: This is lower part of plastered wall, where Ans:(d) Safe Bearing Capacity :- The safe bearing
special treatment is given to make it better resistance. capacity of soil refers to the maximum load that can be
To make the surface of the wall perfectly sealed, wood applied to the soil without failure or excessive
and metal surface require 3 coats of plastering . settlement. The safe bearing capacity is determined by
• The first coat is known as rendering coat. conducting soil tests and analyzing the soil properties.
• The Second coat is known as floating coat. The safe bearing capacity is an important factor in the
• The third coat is known as setting coat or finishing coat. design of foundations and structures.
The value of the factor of safety used for finding the safe
❖ SOIL MECHANICS bearing capacity depends on the type of soil, the load
Q.13. Identify the correct option in context of uniformly that is applied and the level of risk that is acceptable.
graded soil. Generally, a factor of safety of 2 is used for normal soil
(a) D60 = D10 (b) D60 > D10 conditions. However, for more critical structures, a factor
(c) D60 <D10 (d) D60 = 2D10 of safety of 2.5 or 3 may be used.

Ans:(a) Coefficient of Uniformity / Uniformity ❖ ENVIRONMENT ENGINEERING

Coefficient: It is defined as the ratio of D60 and D10 sieve Q.16. The distance between two manholes on straight
sizes in the sieve analysis of granular material. alignment for diameter of sewer above 150 cm is about
Higher is the value of Cu. larger is the range of the (a) 120 m (b) 150 m (c) 75 m (d) 300 m
particle size. Ans :(d) As per IS 1742:1960 : The recommended
𝑫𝟔𝟎
Cu = spacing of manholes on straight reaches of sewer lines
𝑫𝟏𝟎
For uniformly graded soil, Cu = 1 Size of Sewer Recommended Spacing
For well graded sand, Cu > 6
Diameter up to 0.3 m 45 m
For well graded gravel Cu > 4
Coefficient of Curvature / Curvature Coefficient: Diameter up to 0.6 m 75 m
𝑫𝟐
𝟑𝟎 Diameter up to 0.9 m 90 m
CC =
𝑫𝟔𝟎 ×𝑫𝟏𝟎
Diameter up to 1.2 m 120 m
For well graded soil, 1 ≤Cc ≤ 3
Diameter up to 1.5 m 250 m
For gap graded soil, 1 <CC or Cc > 3
Diameter greater than
300 m
1.5 m

UPSSSC RAJ INSTRUCTOR 2023 192 Civil KI Goli

Q.17. The removal of colloidal and dissolved organic 2. Warning Signs: Alert drivers and pedestrians to
matter present in waste water is done by which of the potential hazards or dangers ahead, prompting them to
following treatment process? take necessary precautions. Warning signs are also
(a) Disinfection (b) Chemical flocculation known as cautionary signs.
(c) Biological growth (d) Skimming tank Shape: Triangular with a red border and yellow
background, featuring black symbols or text.
Ans : (c) Colloidal and dissolved organic matter :
Examples: Intersection warning signs, Curve ahead
• Colloidal and dissolved organic matter are small
signs, falling rocks signs, slippery road signs, pedestrian
particles and dissolved substances, respectively, that are
crossing signs, etc.
not effectively removed during primary wastewater
treatment processes such as sedimentation or 3.Informative Signs: Provide helpful information to
screening. drivers, including route guidance, directions, location
services, and other relevant details.
• However, secondary treatment processes, such as
biological treatment or chemical treatment, can Shape: Usually rectangular with a blue background and
effectively remove colloidal and dissolved organic matter white symbols or text.
from wastewater. Examples: Town name signs, Exit signs, directional
• During biological treatment, micro-organisms such as arrows, distance markers, rest area signs, etc.
bacteria, fungi, and algae consume and break down Q.19. What is the full form of M.D.R.?
organic matter, including colloidal and dissolved organic (a) Main Domestic Road
matter. (b) Minor District Road
• In chemical treatment, coagulants and flocculants are (c) Main District Road
added to wastewater to bind to colloidal and dissolved
(d) Major District Road
organic matter and form larger particles that can be
more easily removed through sedimentation or Ans:(d) Classification of Roads Based on Location and
filtration. Function of Roads
This classification of roads in India was proposed by
❖ TRANSPORTATION ENGINEERING Indian Road Congress (IRC)
Q.18. Identify the correct mandatory sign. • National highways (NH)
• State highways (SH)
(a) (b) • Major district roads (MDR)
• Other district roads (ODR)
• Village roads (VR)
(c) (d) National Highways:- These are the main highways
running through the length and breadth of the country,
Ans : (c) Types of traffic signs: connecting major ports, foreign highways and capitals of
state/ union territories and also large industrial and
Classification of traffic signs : -
tourist centers.
1. Mandatory/Regulatory sign: The regulatory or
State Highways:- These are arterial roads of a state
mandatory signs are used to inform the road Users of
linking district headquarters and important cities within
certain laws and regulations to provide safety and free
the state and connecting them with national highways
flow to traffic.
and highways of neighbouring states.
Shape: Typically circular with a red border and black
Major District Roads:- These are important roads within
symbol or white background with black legend.
a district serving areas of production and markets and
Examples: Stop signs, speed limit signs, no parking signs, connecting them with each other or the main highways.
yield signs, one-way signs, etc.
Other District Roads:- These are roads serving rural
areas of production and providing them with outlets to
market centers, block development headquarters or
other main roads.
Village Roads:- These are the roads connecting villages
or groups of villages with each other and to the nearest
road of higher quality.

UPSSSC RAJ INSTRUCTOR 2023 193 Civil KI Goli

 ❖ SURVEY Ans:(b) Traversing : Traversing is the type of survey in
Q.20. When chain is stretched out of the survey line which a number of connected survey lines form the
during chaining, the measured distance is. framework and the directions and lengths of the survey
(a) always be less than actual distance lines are measured with the help of an angle measuring
(b) always be more than actual distance instrument and tape or chain respectively.
(c) same as actual distance Types of traverse : - There are two types of traverse
(d) half the actual distance surveying
Ans :(b) Error due to incorrect chain: If the length of the Closed traverse: When the lines form a circuit that ends
chain used in measuring the length of the line is not at the starting point, it is known as a closed traverse.
equal to the true length or the designated length, the The closed traverse is suitable for locating the
measured length of the line will nor be correct and boundaries of lakes, woods, etc and for a survey of large
suitable correction will have to be applied. areas.
• If the chain is too long, the measured distance will be Open traverse: When the lines form a circuit that ends
less. The error will, therefore, be negative and the elsewhere except the starting point, it is said to be an
correction is positive. open traverse.
• Similarly, if the chain is too short, the measured the open traverse is suitable for surveying a long narrow
distance will be more, the error will be positive and the strip of land as required for a road of the canal or the
correction will be negative. coastline.
Q.21. If the survey is intended for detailed design work
or for accurate earth work calculations
then___________________ contour interval should be
used.
(a) small (b) large
(c) too large (d) none of these
Ans:(a) Contour Interval: A contour interval is a vertical
distance or difference in elevation between contour Closed traverse Open traverse
lines. It is always kept the same or constant for a map. Q.23. While performing survey by using theodolite, the
The contour interval depends upon the following theodolite is supported on.
factors: (a) compass (b) alidade
1.The scale of the map: The contour interval normally (c) tripod (d) line ranger
varies inversely to the scale of the map i.e.. if the scale
of the map is large, the contour interval is considered to Ans:(c) Theodolite: A theodolite is a precision optical
be small and vice versa. instrument for measuring angles between designated
visible points in the horizontal and vertical planes.
2.Purpose of the map: For a very important and detailed
design, contour intervals should be kept small. It consists of a moveable telescope mounted so it can
rotate around horizontal and vertical axis and provide
3.Nature of ground: For flat ground, a small contour
angular readouts.
interval is chosen whereas, for undulating and broken
ground, a greater contour interval is adopted. While performing survey by using theodolite, the
theodolite is supported on tripod.
4.Time & Cost: If the contour interval is small, greater
time and funds will be required in the field survey, in
reduction, and in plotting the map. If the time and funds
available are limited, the contour interval may be kept
large.
5.Slope: Contour interval ki directly proportional to the
slope. Contour intervals for flat countries are generally
small, e.g. 0.25 m. 0.5 m, 0.75 m. etc. Contour interval in
hilly areas is generally greater, e.g. 5 m, 10 m, 15 m, etc.
Q.22. The open traverse is suitable for.
(a) locating the boundaries of lakes
(b) surveying a long narrow strip of land
(c) surveying of large areas
(d) locating the boundaries of plot

UPSSSC RAJ INSTRUCTOR 2023 194 Civil KI Goli

In turn, this fresh air helps force the warm, dirty air Q.30. Which is the given options is incorrect with
inside of the building out through the opening in the reference to the factor influencing sanitation of a house
roof. (a) Lighting (b) Ventilation
• Natural ventilation is generally considered suitable for (c) Roominess (d) Cleanliness
houses & flats (i.e. for small buildings) and not suitable
Ans :(c) Appropriate number of windows, door - lights
for big offices, large factory or workshop. Doors and
(fan-lights) etc. should be installed in all the rooms of the
windows should be located to provide the maximum
building for light and fresh air. Floors of bathrooms and
inflow of air.
toilets should be made of seal-proof materials and
MECHANICAL OR ARTIFICIAL VENTILATION:- In this arrangements for drainage of contaminated water
system of ventilation some mechanical arrangement is should be made through paved underground drains
adopted to provide enough ventilation to room. whose slope should not be less than 1 in 40.
• The outside air is supplied into building by positive The factors affecting the cleanliness of the house are
ventilation. lighting, ventilation and cleanliness etc.
• If the supply of outside air is by means of mechanical Roominess: It is concerned with making an impression
devices such as fan then it is termed as positive of spaciousness. It can be obtained by getting the
ventilation. And for this purpose centrally located supply maximum benefit from the minimum dimension of a
fan of centrifugal type is used. room.
• The process of removal of the air & its disposal outside
by means of device is termed as exhaust of air. and for ❖ AUTOCAD
this purpose wall or roof located exhaust fans of Q.31. Select the input device of computer from the
propeller type are used. given options.
Q.29. __________is the exposed vertical surface left (a) Pen plotter (b) Hard copier
on the sides of an opening after the door or window (c) Mouse (d) Inkjet Printer
frame has been fitted in position. Ans : (c)
(a) Jamb (b) Corbel Input Devices Output Devices
(c) Soffit (d) Reveal They accept and convert They accept data from the
data into a machine- processor and convert them
Ans :(d) JAMBS: This is the vertical cross wall face of a readable format and into the required output
window or door opening which supports the frame and transmit data to the format. They display the
against which the shutters rest when they approach. processor unit. result of input data.
REVEALS: This is the external jamb of a door opening is It involves digital to analogue
at right angles to the face of the wall. conversion.
It usually involves It involves digital to analogue
SOFFITS: A soffit is an exterior or interior architectural
analogue to digital conversion
feature, generally the horizontal, aloft underside of any conversion
construction element. Examples of Input Examples of Output Devices:
LINTELS: Horizontal support of timber stone, concrete, Devices: • Monitor: It displays all the
or steel across the top of a door or window. • Keyboard: It used for programs which are running
entering data and on the computer, known as a
information into the soft copy.
computer system. It is a Printer: It will print out the
primary device for output information which is
inputting text by saved in your computer.
pressing a set of keys. known as a hard copy.
• Mouse: It is a pointing Plotter: It is a device that
device used to input draws pictures on a page
data into the computer as output, after receiving a
system by pointing on it. print command from the
• Joystick: It is another computer. It is also called
pointing device which is graph plotter.
a hand-held stick that Speaker: It is a device
pivots on a base and is connected to a computer's
used for controlling the sound card that outputs
action in video games. sounds generated by the
Digitiser: It allows one to computer. It can be used for
hand-draw images and various sounds meant to
graphics, similar to the alert the user, as well as
music and spoken text

UPSSSC RAJ INSTRUCTOR 2023 196 Civil KI Goli

 way one draws images • Projector: It can take the Q.34. Hydraulic gradient line is the sum of:
with a pencil and paper. display of a computer screen (a) Pressure head and kinetic head
Microphone: It is used to and project a large version of
(b) Datum head and kinetic head
record our voice or any it onto a flat surface
other sound on the
(c) Pressure head and datum head
computer. (d) Pressure head, kinetic head and datum head
Ans : (c) Total energy line (T.E.L) - Line representing the
❖ CPM PERT sum of pressure head, datum head, and velocity head.
Q.32. ___________is the minimum possible time Hydraulic gradient Line (H.G.L) - Line representing the
required to accomplish an activity. sum of pressure head and datum head.
(a) Optimistic time (b) Pessimistic time
(c) Expected time (d) Most likely time
Ans:(a) The PERT (Project Evaluation and Review
Technique) technique is used, when activity time
estimates are stochastic in nature. For each activity.
three values of time (optimistic, most likely, pessimistic)
are estimated.
• Optimistic time (to) estimate is the shortest possible
time require for the completion of the activity
• The most likely time (tm) estimate is the time required
for the completion of activity under normal ❖ HYDROLOGY
circumstances Q.35. Aquifuge is that geological formation which
• Pessimistic time (tp) estimate is the longest possible (a) Neither contains nor yields water
time required for the completion of the activity (b) Contains but not yeild water
❖ FLUID MECHANICS (c) Containing large quantity of water
Q.33. Which among the following fluid possesses (d) None of these
viscosity? Ans:(a) Aquifuge is neither porous, not permeable,
(a) Real fluid (b) Ideal fluid these are not inter-connected opening, so it does not
(c) Ideal plastic fluid (d) Newtonian fluid contain water and does not allow it to pass through it.
• A good example of it is a massive compact rock without
Ans : (a)
any fracture.
Aquifer: An aquifer is a saturated formation of earth
material from which water is yield in sufficient quantity,
due to the high permeability of earth material.
Unconsolidated deposits of sand and gravel are good for
aquifer formation.
Groundwater is generally extracted from aquifers, so it is
of much importance to us.
These are generally of 3 types
• Unconfined aquifer
Newtonian The fluid which obeys Newton's law of
• Confined aquifer
Fluid viscosity (shear stress is proportional to the
rate of shear strain), is known as a • Perched aquifer
Newtonian fluid Aquiclude: It contains a large amount of water in pores,
Non- The fluid which doesn't obey Newton's law but extraction of water is very difficult. It may be
Newtonian of viscosity is known as a Non-Newtonian considered as close to water movement.
Fluid fluid.
• A good example of aquiclude is clay.
Ideal Fluid A fluid is said to be ideal when it is
incompressible and non-viscous. It is an Aquitard: Aquitard form by that material through which
imaginary fluid that doesn't exist in reality. the only seepage is possible but extraction of water is
Real Fluid All the fluids are real as all the fluids not so easy as in aquifer.
possess viscosity. • A good example of aquitard is silty clay material.
Ideal Plastic When the shear stress is proportional to
Fluid the velocity gradient and shear stress is
more than the yield value, it is known as
ideal plastic fluid.
UPSSSC RAJ INSTRUCTOR 2023 197 Civil KI Goli
 UPSSSC JE 2018 RE-EXAM
27/05/2023
❖ STEEL
Bearing/ Crushing failure of rivet: If bearing stress on
Q.01. What is the Imperfection factor for buckling class
the rivet is too large so that contact surface between
"A & B"
rivet and plate may get damaged this is called crushing
(a) 0.54 & 0.67 (b) 0.32 & 0.45
or bearing failure of rivet.
(c) 0.21 & 0.34 (d) 0.43 & 0.56
Ans. (c) Imperfection Factor : It takes into account, the
imperfection that may occur while load transferring,
fabrication, or installation.
• It depends upon the shape of the column cross-section Q.04. Maximum Slenderness ratio of a compression
under consideration, the direction in which buckling member carrying loads resulting from wind or seismic
can occur, and the fabrication process (hot-rolled, load provided the deformation of such members does
welded, or cold-formed). not adversely effect the stress in any part of member-
Classification of different sections under different (a) 250 (b) 120 (c) 350 (d) 180
buckling class i.e. a, b, c, or d used for the design of axial
Ans : (a)
compression members.
As per IS 800: 2007, Table 7 Type of member Maximum
slenderness
ratio
A member carrying compressive 180
Q.02. Calculate the value of minimum edge distance for loads resulting from the dead load
20 mm diameter bolt using machine flame cut. and superimposed load
(a) 25 mm (b) 22 mm (c) 20 mm (d) 33 mm A member subjected to compressive 250
loads resulting from
Ans. (d) Given, wind/earthquake forces provided
Diameter of bolt = 20 mm, the deformation of such members
as the bolt diameter is in the range of (16 - 24) mm does not adversely affect the stress
Hole Dia = Bolt dia + 2 mm = 20 + 2 = 22 mm in any part of the structure
Minimum edge Distance for machine flame cut A member normally carries tension 350
= 1.5 × diameter of the bolt hole but is subject to the reversal of
stress due to wind earthquake
= 1.5 × 22 = 33 mm
force.
Q.03. Rivet value is defined as - Q.05. According to IS 800,2007 ,in the design of a
(a) strength of joint tension member using bolted connections, the net area
(b) minimum of shearing and bearing strength of rivet required to carry the design load t is given by equation..
(c) bearing strength of rivet
Where
(d) shearing strength of rivet
An = net cross sectional area required
Ans : (b) Rivet value is defined as minimum of shearing fu = yield stress in steel
and bearing strength of rivet.
(a) An = Tu/(fu/1.5) (b) An = Tu/(fu/1.25)
Failure in a riveted joint may be occurred due to:
(c) An = Tu × fu/1.5 (d) An = Tu × fu/1.25
(i) Shear failure of the rivet
Ans : (b) As per Clause 6.3 Design Strength Due to
(ii) Bearing failure of the rivet
Rupture of Critical Section of IS 800: 2007, design
(iii) Tearing failure of the rivet
strength is given for different members separately as
(iv) Bearing failure of the plate follows.
(v) Tearing failure of the plate 𝛂 𝐀𝐧 𝐟𝐮
𝐓𝐮 =
(vi) Block shear failure 𝛄𝐦𝟏

Shearing of the rivet: When shear stress in the rivets When α = 1, above equation can be written as
exceeds the maximum allowable shearing stress then 𝐀𝐧 𝐟𝐮
𝐓𝐮 =
𝛄𝐦𝟏
the shear failure of rivets takes place.

UPSSSC JE 2018 RE-EXAM 27/05/2023 198 Civil Ki Goli

Where 𝛄𝐦𝟏 - partial safety factor for failure at ultimate
stress = 1.25 (refer to Table 5 of IS 800: 2007)
fu - ultimate stress of the material
An - the net effective area of the member
𝐀𝐧 =
𝐓𝐮
𝐟𝐮 /𝟏.𝟐𝟓
❖ RCC
Q.09. What is minimum cement content required for
Q.06. Which of the following assumptions is used for RC work which is exposed to aggressive sub-soil or
the design of an axially loaded compression member ground water?
(steel)? (a) 300 kg/m3 (b) 320 kg/m3
(a) The modulus of elasticity is not assumed to be (c) 60 kg/m 3
(d) 340 kg/m3
constant. Ans. (d)
(b) Secondary stresses are neglected. Minimu Minimu
Exposur maximu maximu
(c) 25% to 40% of secondary stresses are taken into m m
e m water- m water- Exposure
cement cement
account. conditio
content
cement
content
cement condition
n ratio ratio
(d) The ideal column is not straight and has (kg/m3) (kg/m3)
crookedness. PCC RCC
Concrete
Ans : (b) Compression member : A compression exposed to
Mild 220 0.6 300 0.55
member is a structural member which is straight and coastal
subjected to two equal and opposite compressive forces area
Concrete
applied at its ends. continuous
Moderat
Assumptions made while designing a compression e
240 0.6 300 0.5 ly
member (or column): underwate
r
• The ideal column is assumed to be absolutely straight Concrete
having no crookedness, which never occurs in practice. Severe 250 0.5 320 0.45
immersed
• The modulus of elasticity is assumed to be constant in under
seawater
a built-up column. Concrete
• Secondary stresses(which may be of the order of even buried
Very
25%-40% of primary stresses) are neglected. severe
260 0.45 340 0.45 under
aggressive
Q.07. In case of web crippling, the dispersion of load subsoil
from bearing plate takes place at: Concrete
(a) 30° (b) 60° (c) 45° (d) 10° Extreme 280 0.40 360 0.4 under tidal
zone
Ans : (a) Buckling strength: A certain portion of the Q.10. What is the minimum area of distribution bars
beam at supports acts as a column to transfer the load required for slabs where steel Fe415 is used?
from the beam to the support. Hence, under this (a) 0.12% of gross area of slab
compressive force, the web may buckle. This may (b) 4% of gross area of slab
happen under a concentrated load on the beam also. (c) 6% of gross area of slab
The load dispersion angle is taken as 45° with the (d) 0.15% of gross area of slab
horizontal.
Ans. (a) As per IS 456 : 2000
Q.08. Imperfection factor for buckling class c. Minimum percentage of reinforcement:
(a) 0.21 (b) 0.34 (c) 0.49 (d) 0.76 • 0.15 % of the gross area if a mild steel bar is used.
Ans : (c) Imperfection Factor:It takes into account, the • 0.12 % of the gross area if the HYSD bar (Fe415) is
imperfection that may occur while load transferring, used.
fabrication, or installation. Q.11. The analysis of slab spanning in one direction is
• It depends upon the shape of the column cross-section done by assuming it to be a beam of
under consideration, the direction in which buckling (a) 1m length (b) 1m2 area
can occur, and the fabrication process (hot rolled, (c) 1m width (d) Least thickness
welded, or cold-formed).
Ans : (c) Analysis of slab spanning in one direction: The
• Classification of different sections under different
analysis of slab spanning in one direction is done by
buckling classes i.e. a, b, c, or d used for the design of
assuming it to be a beam of 1 m width, though the slab
axial compression members.
width is continuous and is not composed of individual
As per IS 800: 2007, Table 7 beams of 1 m width. The reinforcement are calculated
for 1 m width and the bars are distributed accordingly.

UPSSSC JE 2018 RE-EXAM 27/05/2023 199 Civil Ki Goli

Q.16. According to IS: 456 ∶ 2000, if 46 cubic metres • Nominal cover can be decreased by 5 mm if the
quantity of M25 grade of concrete is required at a site, concrete grade is M35 or higher.
the minimum number of cube samples that should be Q.19. In singly reinforced sections, when the section is
taken from the site are _______. under-reinforced, the relation between depth of
(a) 9 (b) 8 (c) 4 (d) 7 neutral axis (xu) and the limiting value of depth of
Ans : (c) By Clause 15.2 of IS 456-2000 neutral axis (xu' max) is:
The minimum frequency of sampling of concrete of each (a) xu = xu' max (b) xu < xu' max
grade shall be in accordance with the following: (c) xu > xu' max (d) none of the above
Quantity of Ans : (b) The characteristics of the balanced section
Concrete in Number of Samples under-reinforced section and the over-reinforced
the Work, m3 section are given below in the tabulated form.
1 -5 1 Under Over
6 -15 2 Balanced section
reinforced reinforced
16 - 30 3
Xu < Xulim Xu = Xulim Xu > Xulim
31 - 50 4
51 and above 4 plus one additional sample Ast < Astlim Ast = Astlim Ast > Astlim
for each additional 50 m3 or
part there of MOR < MORlim MOR = MORlim MOR > MORlim
Q.17. The development length Ld is given by ______, Z > Zlim Z = Zlim Z < Zlim
where ϕ = nominal diameter of the bar, σs = stress in
Steel yields first Both steel yields and Concrete
the bar at the section considered at design load, and τbd
then concrete concrete reaches reaches
= design bond stress.
reaches maximum strain of maximum strain
𝛟𝝈𝑺 𝛟𝝈𝑺 𝟒𝛟𝝈𝑺 𝝈𝑺
(a) (b) (c) (d) maximum 0.0035 of 0.0035 before
𝟒𝝉𝒃𝒅 𝝉𝒃𝒅 𝝉𝒃𝒅 𝟒𝛟𝝉𝒃𝒅 strain of 0.0035 simultaneously steel yields

Ans : (a) Development Length: A development length is

the amount of rebar length that is needed to be
embedded or projected into concrete to create desired
bond strength between the two materials and also to
develop required stress in steel at that section.
As per IS 456: 2000, clause 26.2.1, The development
length is given by :
𝛟𝛔𝐒
𝐋𝐝 =
𝟒𝛕𝐛𝐝

Where ϕ = nominal diameter of the bar, σs = stress in the

bar at the section considered at design load, τbd = design
bond stress Q.20. For design of concrete structures, failure strain of
concrete under direct compression and flexure
Q.18. What is the minimum nominal cover (mm) to be respectively is ______ and ______.
provided for reinforced concrete member exposed to
(a) 0.02, 0.0035 (b) 0.002, 0.035
severe weather conditions?
(c) 0.002, 0.0035 (d) 0.0035, 0.002
(a) 50 (b) 35 (c) 40 (d) 45
Ans : (d) Cover requirements as per IS 456:2000: Ans : (c) Following assumptions have been made in IS
456: 200 regarding maximum strain in compression
Exposure Minimum Nominal members:
Conditions concrete grade cover
1. As per, Clause 38.1, the maximum compressive strain
Mild M20 20 mm
in concrete in bending compression is taken as 0.0035.
Moderate M25 30 mm
Severe M30 45 mm 2. As per, clause 39.1, the maximum compressive strain
in concrete in axial compression is taken as 0.002.
Very severe M35 50 mm
Extreme M40 75 mm Q.21. For a square reinforced concrete (RC) column
with cross-section of 300 mm × 300 mm having an
• Nominal cover can be decreased by 5 mm for main
effective length of 2500 mm, determine the minimum
reinforcing bars of less than 12 mm diameter.
eccentricity

UPSSSC JE 2018 RE-EXAM 27/05/2023 201 Civil Ki Goli

shall be within the limits given in Table 4 and shall be ii) According to Abram's Law, a higher gel/space ratio
described as fine aggregates, Grading Zones I, II, III and reduces the porosity and increases the strength of
IV: concrete.
As per IS:383-1970:(Table 4) (a) Both statements (i) and (ii) are true
IS sieve (b) Statement (ii) is true and Statement (i) is false
Percentage Passing For
Designation (c) Both statements (i) and (ii) are false
Gradin (d) Statement (i) is true and Statement (ii) is false
Grading Grading Grading
g Zone
Zone I Zone II Zone III Ans : (c) Abram's Law: Abram’s law states that the
IV
10 mm 100 100 100 100 strength of concrete mix entirely depends on the water-
4.75 mm 90-100 90-100 90-100 95-100 cement ratio.
2.36 mm 60-95 75-100 85-100 95-100
• The increase in the water-cement ratio decreases the
1.18 mm 30-70 55-90 75-100 90-100
600 micron 15-34 35-59 60-79 80-100 strength of concrete parabolically i.e. 'the compressive
300 micron 5-20 8-30 12-40 15-50 strength of fully compacted concrete at a given age
150 micron 0-10 0-10 0-10 0-15 and normal temperature is inversely proportional to
Q.29. The following statements ( i and ii ) pertain to the the water-cement ratio.
effect of factors affecting the workability of concrete. • If we decrease the water-cement ratio compressive
i : High ratio of volume of coarse aggregate to fine strength of concrete increases parabolically.
aggregate results in higher workability of the concrete
• The porosity of concrete which governs the strength of
mix (keeping all other parameters of mix constant).
concrete is affected by the gel/space ratio in concrete.
ii : As the placing time of concrete increases, the
workability of the mix decreases. • Power(scientist) has determined the relation between
(a) statement i is False and statement ii is True strength development and gel space ratio. The gel
space ratio is defined to be the ratio of the volume of
(b) Both statement i and statement ii are True
hydrated gel including gel pores to the volume of space
(c) Both statement i and statement ii are False
to accommodate these products. This space will be
(d) statement i is True and statement ii is False
equal to the volume of the cement which has hydrated
Ans : (b) Workability: Workability is used to describe the
plus the volume of mixing water plus any air void. A
ease or difficulty with which the concrete is handled,
transported, and placed between the forms with higher gel/space ratio reduces the porosity and
minimum loss of homogeneity. therefore increases the strength of concrete.
Factors affecting workability:
●Water content ●Mix proportion
●Aggregate size ●Shape of aggregates
●Surface texture ●Grading of aggregates
●Admixtures
Aggregate size: Fine-grained aggregates have more
surface area and therefore require more water to make
them workable. Aggregates with coarser particles have Q.31. Which of the following types of vibrators is
less surface area and hence require less water for suitable for compacting screed concrete layer laid on
workable. existing floors with thickness less than 20 cm?
Placing time: Placing time increases the workability of (a) Form vibrator (b) Internal vibrator
concrete decreases because some losses of water occur (c) Surface vibrator (d) Vibrating table
after making fresh concrete. workability of fresh Ans : (c) If compaction by hand is properly carried out on
concrete continuously decreases with the passing of concrete with sufficient workability, gives good results,
time (placing time). but the strength of hand compacted concrete will be
Hence both statements are correct. necessarily low because of higher water cement ratio
Q.30. Identify whether the following statements required for full compaction. Whenever higher strength
related to concrete are true or false. of concrete is required, it is necessary that stiff concrete,
with low water cement ratio may be used. Hence to
i) According to Abram's Law, the strength of fully
compact such concrete vibrator is used.
compacted concrete at a given age and normal
temperature is directly proportional to the water - Internal vibrator: It is the most commonly used vibrator.
cement ratio. It essentially consists of a power unit, a flexible shaft and
needle. The power may be electrically driven or

UPSSSC JE 2018 RE-EXAM 27/05/2023 204 Civil Ki Goli

Q.35. A beam of same cross section is used in two
different orientations as shown in figure. Bending
moment applied to the beam in both cases are same.
The maximum bending stress induced in case A and B
are related to

Similarly
(a) σa = 2σb (b) σa = σb
(c) σa = σb/2 (d) σa = σb/4
Ans : (a) From the simple bending theory equation:
𝑴 𝝈 𝑹
= =
𝑰 𝒚 𝑹
In the given figure minor principal stress is represented
𝟏 𝒃𝒉𝟑 𝟐 𝒃𝒉𝟐
𝒁= = × = by OB (σ2) and major principal stress is represented by
𝒚 𝟏𝟐 𝒉 𝟔
OA (σ1).
If σb is the maximum bending stresses due to bending
Q.37. The loading, shear force and bending moment
moment M on shaft:
𝑴 𝟔𝒎
diagrams are given in the figure. Determine the
𝝈𝒃 = = position of point of contraflexure y from support B.
𝒁 𝒃𝒉𝟐
𝐛
𝟏 𝛔𝐛𝟏 𝐛𝟐 𝐡𝟐 ( )𝐛𝟐
𝟐 𝟐
𝛔𝐛 ∝ ⇒ = = 𝐛 𝟐
=𝟐
𝐛𝐡𝟐 𝛔𝐛𝟐 𝐛𝟏 𝐡𝟐
𝟏 𝐛( )
𝟐

𝑺𝑶 𝝈𝑨 = 𝟐 𝝈𝑩

Q.36. For the Mohr’s circle shown below, major and

minor principal stresses are represented by :

(a) 0.33 (b) 0.50

(c) 0.25 (d) 0.75
(a) OA & OB (b) EA & EB Ans : (a) For Bending Moment
(c) OQ1 & OQ2 (d) Q1A & Q2B MA = 0
Ans : (a) Mohr's Circle: Mohr's circle is a graphical MB = (2 x 1) x 0. 5 = -1 kN-m
method of finding normal, tangential, and resultant MC = 1 x 2 = + 2 kN-m
stresses on an oblique plane.
MD = 0
Mohr's circle will be drawn for the following cases :
Point of contraflexure
1. A body subjected to two mutually perpendicular
Let P be the point of contraflexure at a distance y from
principal tensile stresses of unequal intensities.
the support B. From the geometry of the figure between
2. A body subjected to two mutually perpendicular B and C. we find that
principal stresses which are unequal and unlike (i.e., 𝒚 𝟏−𝒚
one is tensile and other is compressive ). =
𝟏 𝟐
3. A body subjected to two mutually perpendicular 2y = 1 – y
principal tensile stresses accompanied by a simple 3y=1
shear stress. 𝟏
y = = 0.33
𝟑

UPSSSC JE 2018 RE-EXAM 27/05/2023 206 Civil Ki Goli

Q.38. A cross-sectional bar of area 700 mm2 is Ans : (b) The shear force at 2 m from the fixed end is
subjected to an axial load as shown in the figure below.
What is the value of stress (MPa) in the section PQ?
(a) 30 (b) 40 (c) 50 (d) 90
Ans : (d) Free body diagram

𝟏
= (5 + 15) × 5 = 50 kN
𝟐
Q.41. At a section of a shaft, a bending moment of 8
kN-m and a twisting moment of 6 kN-m act together.
Now for the bar PQ, The equivalent twisting moment in kN-m is given by
𝐅𝐏𝐐 (a) 14 (b) 2 (c) 10 (d) 48
The stress is given as, 𝛔𝐏𝐐 =
𝐀
Ans : (c) Given:
Given:
FPQ = 63 kN, A = 700 mm2 M = 8 kNm, T = 6 kNm
𝐅𝐏𝐐 𝟔𝟑×𝟏𝟎𝟎𝟎
Equivalent Twisting TM:
𝛔𝐏𝐐 = = = 𝟗𝟎𝐌𝐏𝐚
𝐀 𝟕𝟎𝟎
𝐓𝐞𝐪 = √𝐌 𝟐 + 𝐓 𝟐
Q.39. The shape factor for a solid circular section of 𝐓𝐞𝐪 = √𝟖𝟐 + 𝟔𝟐
diameter D is equal to:
(a) D/2π (b) 16/3π (c) πD/8 (d) 15/2π 𝐓𝐞𝐪 = √𝟔𝟒 + 𝟑𝟔 = 10 kN-m

Ans : (b) Q.42. The kernel of square and circular sections will be
respectively
(a) circle and rhombus (b) rhombus and circle
(c) circle and square (d) square and cirlce
Ans : (d) Kernel of a section The kernel of the section is
the area within which the line of action of eccentric load
Solid circular section is shown below. should pass so that the tensile stress is zero
Moment of area A1 and A2 about neutral axis gives Zp. • When the load acts in the kernel of the section, there
𝐀 𝛑𝐃𝟐 is only bending.
𝐀𝐥𝐬𝐨, 𝐀𝟏 = 𝐀𝟐 = =
𝟐 𝟖 • The kernel of the section is also called the core of the
𝛑𝐃𝟐 𝟒 𝐃 𝐃𝟑 section
𝐇𝐞𝐧𝐜𝐞, 𝐙𝐩 = 𝟐 × ( )× × =
𝟖 𝟑𝛑 𝟐 𝟔 (1) Kernel of square section:
Now, Z about natural axis is given by
𝛑𝐃𝟑
𝐙=
𝟑𝟐

𝐙𝐩 𝟏𝟔 𝐃𝟑
So, shape factor = = =
𝐙 𝟑𝛑 𝟔

Q.40. Find shear force at 2 m from the fixed end.

The kernel of the square section is a also a square, while

the shape of kernel for the rectangle is rhombus.
(2) The core of a solid circular section:-

(a) 40 kN (b) 50 kN
(c) 56 kN (d) 80 kN

UPSSSC JE 2018 RE-EXAM 27/05/2023 207 Civil Ki Goli

 Parameter Permissible Cause for The steel used for fishplates should have a minimum
Limit Rejection tensile strength of 5.58 to 6.51 t/cm2 with a minimum
Total suspended 500 2000 elongation of 20 %.
solids (mg/L) Fish plates are designed to have roughly the same
Turbidity (NTU) 1 5 strength as the rail section, and as such the section area
Colour (TCU) 5 15 of two fish plates connecting the rail, ends is kept about
the same as that of the rail section.
Taste & Odour (TON) 1 3
Total dissolved solids 500 2000
(mg/L)
Alkalinity (mg/L as 200 600
CaCO3)
pH 6.5-8.5 No Relaxation
Hardness (mg/L as 200 600 Q.60. The conventional sign shown in the figure below
CaCO3) represents a
Chloride content 250 1000
(mg/L)
Free ammonia 0.15 No Relaxation
(mg/L)
Chloramine 4 No Relaxation
Q.58. In the analysis of unseeded domestic waste water
BOD test, the data used is 6ml of waste in a 300 ml (a) Bridge carrying railway below the road.
bottle, initial DO of 8 mg/l and 5 day DO is 4 mg/l. (b) Bridge carrying road below railway.
Compute 5day BOD of waste water. (c) Bridge carrying road and railway at the same level.
(d) A level crossing.
(a) 220 mg/l (b) 150 mg/l (c) 145 mg/l (d) 200 mg/l
Ans : (a)
Ans : (d) Given:
conventional sign Symbol
DOInitial = 8 mg/L
DOFinal = 4 mg/L Level Crossing
Dilution Factor = 300/6
BOD= (Initial Dissolve Oxygen - Final Dissolve Oxygen) ×
Dilution Factor
BOD = (8 − 4) ×
300
= 200 mg/L Bridge carrying railway
6
over road
❖ RAILWAY
Q.59. Fish bolts are made of:
Bridge carrying railway
(a) Cast iron (b) Low carbon steel
below road
(c) High carbon steel (d) Stainless steel
Ans : (c) Fish bolts are usually made of High carbon steel.
• The name ‘fish plate’ derives from the fish-shaped
section of this fitting. Bridge carrying road &
• The function of a fish plate is to hold two rails together railway
in both the horizontal and vertical planes. Fish plates are
manufactured using a special type of steel
(Indian Railways specification T-1/57) with the Railway over road
composition given below:
• Carbon: 0.30–0.42%
• Manganese: not more than 0.6%
• Silicon: not more than 0.15% Railway below road
• Sulphur and phosphorous: not more than 0.06%
The number of bolts per fish plate is 4

UPSSSC JE 2018 RE-EXAM 27/05/2023 212 Civil Ki Goli

Q.61. Check rails are provided on inner side of inner • It has a life of 12-15 years.
rails if sharpness of a B.G. curve, is more than • Zncl2 solution is used for the increment of sleepers life
(a) 3° (b) 5° (c) 6° (d) 8° (2). Metal sleepers: Cast iron sleepers are more used in
Ans : (d) Check rails are provided parallel to the inner compression to steel sleepers because of corrosion
rail on sharp curves to reduce the lateral wear on the problems.
outer rail. • Life - 35 to 40 years
• They also prevent the outer wheel flange from (3). Concrete sleepers: Life - 40 to 60 years
mounting the outer rail and thus decrease the chances
of derailment of vehicles. ❖ HIGHWAY
• Check rails wear out quite fast but since, normally, Q.64. Traffic density is
these are worn-out rails, further wear is not considered (a) Number of vehicles in a specific direction per lane
objectionable. per day
• According to the stipulations presently laid down by (b) Number of vehicles in a specific direction per hour
Indian Railways, check rails are provided on the gauge (c) Number of vehicles per unit length of a lane at a
face side of the inner rails on curves sharper than 8° on given instant
BG, 10° on MG, and 14° on NG routes. (d) Maximum number of vehicles passing a given point
• The minimum clearance prescribed for check rails is 44 in one hour
mm for BG and MG routes and 41 mm for NG routes. Ans : (c) Traffic Volume: It is defined as the procedure
Q.62. What does the inclined red band at 45° indicates to determine the volume of traffic or no of vehicles
in a shunting signal? moving on the roads at a particular section during a
(a) Stop (b) Proceed particular time period. This time period can be in
‘minutes’, ‘hours’ or ‘days’ etc.
(c) Be ready (d) None of the above
• Unit - Veh/Day
Ans : (b) In railways, the disc signals are provided for
Traffic Density: It is the number of vehicles occupying a
the purpose of shunting.
unit length of the lane of the roadway at a given instant
(1) When the red band of the disc is in a horizontal
• Unit - Veh/Km
position, it indicates “STOP”.
Traffic Capacity: the maximum number of vehicles in a
(2) When the red band of the disc is in an inclined
lane or a road that can pass a given point in unit time,
position, it indicates “PROCEED” slowly for shunting.
usually an hour, i.e., vehicles per hour per lane or
The indications of a disc signal is depicted in following roadway.
diagram:
• Unit - Veh/Hr/lane
Q.65. What is the PCU value of a bullock cart as per
recommendation made by IRC?
(a) 7.0 (b) 8.0 (c) 3.0 (d) 4.0
Ans : (b) Passenger car unit (PCU): A Passenger Car Unit
is a measure of the impact that a mode of transport has
on traffic variables (such as headway, speed, and
density) compared to a single standard passenger car.
PCU is a vehicle unit used for expressing highway
capacity.
Q.63. The standard size of wooden sleepers for broad • For instance, one car is considered as a single unit,
gauge cycle, the motorcycle is considered a half car unit. Bus
(a) 1830 mm × 203 mm × 114 mm and truck cause a lot of inconvenience because of their
(b) 2740 mm × 250 mm × 130 mm large size and are considered equivalent to 3.5 cars or
(c) 1520 mm × 150 mm × 100 mm 3.5 PCU.
(d) 1520 mm × 100 mm × 80 mm The PCU of various vehicles as per IRC is given below:
Ans : (b) Sleepers are used to support or fix the rail. Vehicle PCU
It safely transfers point load into the uniformly Car or taxi 1
distributed load to the ballast. Motor Cycle 0.5
Bus or truck 3.5
Classification of sleepers :
Cycle 0.2
(1). Wooden sleepers: Sal and teak are mainly used in Bullock Cart 6
timber sleepers. Bullock Cart (large) 8

UPSSSC JE 2018 RE-EXAM 27/05/2023 213 Civil Ki Goli

Q.66. What is the requirement of maximum Los 𝐋 𝐋𝟐 𝐋𝟑 𝐋𝟐
(a) (b) (c) (d)
Angeles abrasion value for sub-base course of water 𝟐𝟒𝐑𝟐 𝟐𝟒𝐑 𝟐𝟒𝐑𝟐 𝟐𝟒𝐑𝟐

bound macadam (WBM) road? Ans : (b) Shift: The distance through a curve is shifted is
(a) 15 (b) 60 (c) 40 (d) 50 known as the Shift of curve.
Ans : (d) As per IRC 19:2005 : Physical Requirements of 𝐋𝟐𝐒
mathematically, 𝐒 =
Coarse Aggregates for WBM 𝟐𝟒𝐑

S.
Type of
Test
Where, Ls = Length of the curve, R = Radius of curve
Constructio Test Requirements
No Method Q.69. The road Development Plan for India for the
n
Los period of 1981 to 2001 is also known as:
Angeles IS 2386
Abrasion (Part 4)
Max. 50% (a) Bombay plan (b) Kanpur plan
Value or (c) Nagpur plan (d) Lucknow plan
1. Sub-base
IS 2386 Ans : (d) Road development plans over the different
Aggregate
(Part 4)
Impact Max. 40% plans are as follows:
or IS
Value
5640
Specification 1st 20 Year 2nd 20 Year 3rd 20 Year Plan
Los
Plan Plan
Angeles IS 2386
Max. 40%
Abrasion (Part 4) Name of Plan Nagpur Bombay road Lucknow Road
Value or
Base course road plan plan Plan
IS 2386
with Aggregate
2. (part 4) Duration 1943-1963 1961-1981 1981-2001
bituminous Impact Max. 30%
or IS
surfacing Value Completed -
5640
1961
Flakiness IS 2386
Max. 20% Target Density 16 km/100 32 km/100 82 km/100
Index (Part 1)
km² km² km²
Los
Angeles IS 2386 Total Target 532700 km 10 Lakhs km 27 Lakhs km
Max. 40%
Abrasion (Part 4)
Value or Achieved 709122 km 1502697 km 2702000 km
Surfacing IS 2386 Target NH-66000 km,
3. Aggregate
course (Part 4) SH-145000 km
Impact Max. 30%
or IS
Value
5640 Road Pattern Star and - Square and
Flakiness IS 2386 Grid Block
Max. 15%
Index (Part 1)
Provision of No 1600 km 3200 km
Q.67. The free mean speed on a roadway is found to be Expressway provision
80 kmph. Determine the flow capacity (vehicles/hour)
if the average spacing between vehicles under stopped Road 1. NH NH, SH, MDR, • Primary
conditions is 7 meters. Classification 2. SH ODR, VR and Expressway
3. MDR Expressway NH
(a) 1428 (b) 2857 (c) 5714 (d) 11428
4. ODR • Secondary
Ans : (b) Given
5. VR SH
Free mean speed of a roadway = 80 km/hour
MDR
Stopped condition refers to the jam condition.
• Tertiary
So the spacing between vehicles = 7m ODR
1000
Jam density (Kj) = VR
7
𝐕𝐬𝐟 ×𝐤𝐣
Capacity flow (q) = Development 15% 5% Nil
𝟒
𝟏𝟎𝟎𝟎 allowances
𝟖𝟎× 𝟕
= Q.70. If L is the length of vehicles in meters, C is the
𝟒
= 𝟐𝟖𝟓𝟕. 𝟏𝟒 clear distance between two consecutive vehicles
≈ 𝟐𝟖𝟓𝟕 𝐯𝐞𝐡𝐢𝐜𝐥𝐞𝐬 / 𝐡𝐨𝐮𝐫 (stopping sight distance), V is the speed of vehicles in
Q.68. For a transition curve, the shift S of a circular km per hour, the maximum number N of vehicles/hour
is _________.
curve is given by: 𝟏𝟎𝟎𝟎𝐕 𝐋+𝐂
(a) 𝐍 = (b) 𝐍 =
Where, R is the radius of circular curve and L is the 𝐋+𝐜 𝟏𝟎𝟎𝟎𝐕
length of transition curve. 𝟏𝟎𝟎𝟎𝐋 𝟏𝟎𝟎𝟎𝐂
(c) 𝐍 = (d) 𝐍 =
𝐂+𝐕 𝐋+𝐂

UPSSSC JE 2018 RE-EXAM 27/05/2023 214 Civil Ki Goli

Ans : (a) Traffic Density is given as: Viscosity Penetration
Grade Speed
𝟏𝟎𝟎𝟎 (Centistokes) Residue (mm)
𝐊 (𝐯𝐞𝐡./𝐤𝐦) =
𝐬𝐩𝐚𝐜𝐞 𝐡𝐞𝐚𝐝𝐰𝐚𝐲 𝐢𝐧 𝐦 (𝐒) MC-
Medium 30-60
Where 30
S is space headway means the safe stopping distance or MC-
Medium 70-140
clear distance between two consecutive vehicles + 70
120-150
Length of vehicle MC -
Medium 250-500
S=C+L 250
𝟏𝟎𝟎𝟎 MC-
∴𝐤= Medium 800-1600
𝐂+𝐋 800
Maximum no of vehicle per hour is the traffic volume, N RC-70 Rapid 70-140
=k×V RC-
Rapid 250-500
Where, 250 80-120
V is the traffic speed in km/hr and K is the traffic density RC-
Rapid 800-1600
in veh./km 800
On substituting the values, we get Q.73. What will be the shape and size (mm) for a STOP
𝟏𝟎𝟎𝟎 𝐕 sign if the speed is 120 km/h on an Expressway?
𝐍=
𝐂+𝐋 (a) Octagonal and 900 mm size
Q.71. Match the List – I (Bituminous Material) with List (b) Circular and 1000 mm size
– II (Property/Procedure) and select the most (c) Octagonal and 1200 mm size
appropriate option. (d) Triangle and 1500 mm size
List – I List – II Ans : (c) STOP Sign (As per IRC 067)
P. Asphalt 1. Less viscosity • This is a Mandatory/Regulatory sign.
Q. Bitumen 2. Destructive distillation
• This is for indicating priority for the right of way. The
R. Cut back 3. Fractional distillation
sign is intended for use on roadways where traffic is
S. Tar 4. Inert material required to stop before entering a major road, and
(a) P – 3, Q – 2, R – 4, S – 1 where it is intended that the vehicle shall proceed past
(b) P – 4, Q – 3, R – 1, S – 2 the stop line only after ascertaining that this will not
(c) P – 4, Q – 2, R – 1, S – 3 cause danger to traffic on the main road.
(d) P – 1, Q – 2, R – 4, S – 3 • The sign shall be octagonal in shape and shall have red
background and white border. The word "STOP" written
Ans : (b) Bitumen: It is a petroleum product obtained by
in white (in English or local language) with 150 mm
distillation of petroleum crude. Bitumen is used binding
height letters, centrally positioned. The height of the
material in pavement construction.
octagon and border shall be as per Table 14.1.
Asphalt: It is the bitumen containing inert material like
Table 14.1 Sizes and Dimensions of 'STOP' signs
alumina, silica etc. It is used in pavement construction to
provide void lesser surface. Approach Size Height Border Font
speed on (mm) (mm) Size
Cut back: Cutback bitumen is obtained by blending
minor road (mm)
bitumen binder with suitable volatile diluents in
required proportion to reduce its viscosity. Cutback Up to 50 Small 750 25 125
bitumen is used in surface dressing, soil bitumen kmph
stabilization, etc. 51 - 65 Normal 900 30 150
kmph
Tar: It is the viscous liquid obtained from natural organic
> 65 kmph Large 1200 40 225
material like wood and coal by subjecting them to
destructive distillation in the absence of air. ❖ FLUID MECHANICS
Q.72. The penetration value of residue from distillation Q.74. A streamline is a line:
upto 360°C of rapid curing cutbacks bitumen is - (a) such that the streamlines divide the passage into
(a) 80 to 120 (b) 30 to 120 equal number of parts
(c) 40 to 80 (d) 100 to 160 (b) tangent to which is in the direction of velocity vector
Ans : (a) MC- Medium Curing Cutback Bitumen at every point
RC- Rapid Curing Cutback Bitumen (c) which is along the path of the particle
(d) drawn normal to velocity vector at any point
Ans : (b) Streamlines have the following properties:

UPSSSC JE 2018 RE-EXAM 27/05/2023 215 Civil Ki Goli

 q = 180 KN/m2, γ = 20 KN/m3 and Φ = 30°
𝐪𝐮 𝟏−𝐬𝐢𝐧 𝛗 𝟐
𝐃𝐟 = ( )
𝛄 𝟏+𝐬𝐢𝐧 𝛗
𝟏𝟖𝟎 𝟏−𝐬𝐢𝐧 𝟑𝟎 𝟐
𝐃𝐟 = ( )
𝟐𝟎
𝟏+𝐬𝐢𝐧 𝟑𝟎
𝟏𝟖𝟎 𝟏 𝟐
𝐃𝐟 = ( )
𝟐𝟎 𝟑
Df = 1 m
Q.87. A fully saturated soil sample can be represented
by:
(a) two phase system (b) three phase system
GM > 0 (Metacentre is above Stable (c) one phase system (d) no phase system
centre of Gravity) equilibrium
Ans : (a) In a fully saturated soil, all the void spaces are
GM = 0 (Metacentre coinciding Neutral
filled with water, making the liquid phase dominant.
with centre of Gravity) equilibrium
The presence of these two phases makes the fully
GM < 0 (Metacentre is below Unstable
saturated soil a two-phase system with soil and water.
centre of Gravity) equilibrium
Q.84. The relationship between Cc Cv & Cd is
(a) Cd = Cc/Cv (b) Cv = Cd × Cc
(c) Cd = Cv/Cc (d) Cd = Cv × Cc
Ans : (d) Coefficient of velocity (Cv): The ratio of the
actual velocity of the jet, at vena-contracta, to the
theoretical velocity is known as the coefficient of
velocity.
Q.88. Consider the following statements in relation to
Coefficient of discharge (Cd): The ratio of an actual given sketch :
discharge through an orifice to the theoretical discharge
is known as the coefficient of discharge.
Coefficient of contraction (Cc): The ratio of the area of
the jet, at vena-contracta, to the area of the orifice is
known as the coefficient of contraction (Cc).
Cd = C v × C c
1. Soil is partially saturated at degree of saturation =
Q.85. A stone weight 392.4 N in air and 196.2 N in 60%
water. What is the volume of stone? 2. Void ratio = 40%
(a) 2 × 106 cm3 (b) 4 × 103 m3 3. Water content = 30%
(c) 2 × 104 cm3 (d) 5 × 106 m3 4. Saturated unit weight = 1.3 g/cc.
Ans : (c) Given: Correct answer Of these statements…..
Body weight in air = 392.4 N = (ρbody × g × Vbody)........(1) (a) 1, 2 and 3 are correct (b) 1, 3 and 4 are correct
Body weight in water = 196.2 N = (ρwater × g × Vbody)..(2) (c) 2, 3 and 4 correct (d) 1, 2 and 4 are correct
Buoyancy force, FB = 392.4 - 196.2 = 196.2 Ans : (b) Given, The volume of air Va=0.2,
From (1) and (2) Volume of water Vw =0.3, Volume of solids Vs = 0.5
𝛒𝐛𝐨𝐝𝐲
=𝟐 Weight of water Ww = 0.3, Weight of solids Ws =1
𝛒𝐰𝐚𝐭𝐞𝐫
The degree of saturation is :
⇒ 196.2 N = (1000 × 10 × Vbody) 𝑽𝒘 𝟎.𝟑
𝟏𝟗𝟔. 𝟐 = × 𝟏𝟎𝟎 = × 𝟏𝟎𝟎 = 𝟔𝟎%
𝑽𝒗 𝟎.𝟓
⇒ 𝐕𝐛𝐨𝐝𝐲 = = 𝟎. 𝟏𝟗𝟔𝟐 𝐦𝟑 = 𝟏. 𝟗𝟔𝟐 × 𝟏𝟎𝟒 = 𝟐 × 𝟏𝟎𝟒 𝐜𝐦𝟑
𝟏𝟎𝟒 Void ratio is
𝑽𝒗 𝟎.𝟐+𝟎.𝟑
❖ SOIL 𝒆 =
𝑽𝒔
=
𝟎.𝟓
= 𝟏
Q.86. According to Rankine’s Formula, the minimum Water content is
depth of foundation when q = 180 kN/m2, γ = 20 kN/m3 𝑾𝒘 𝟎.𝟑
and Φ = 30° is = × 𝟏𝟎𝟎 = × 𝟏𝟎𝟎 = 𝟑𝟎%
𝑾𝒔 𝟏
(a) 0.50 m (b) 0.75 m (c) 1.0 m (d)2.0 m Saturated unit weight is
𝑾 𝟏+𝟎.𝟑+𝟎
Ans : (c) Given, = = = 𝟏. 𝟑 𝒈/𝒄𝒄
𝑽 𝟎.𝟐+𝟎.𝟑+𝟎.𝟓

UPSSSC JE 2018 RE-EXAM 27/05/2023 218 Civil Ki Goli

Q.89. In Standard Penetration Test, the term
penetration resistance N, is the number of blows
required to drive the sampler______ beyond the
seating drive.
(a) 22 cm (b) 30 cm (c) 8 cm (d) 2 cm
Ans : (b) S.P.T. Test procedure : -
(1) Suitable for Granular soil.
(2) Split spoon samples are used in the borehole.
(3) The borehole is advanced to a depth at which N-value
is to be calculated.
(4) The split-spoon sampler is allowed to penetrate the
soil by applying an impact load of 65 Kg having a free fall
of 75 cm.
(5) The sampler is allowed to penetrate for 150 mm
depth, but reading is not noted i.e.(No. of blows required
for 150 mm penetration is not Noted).
(6) Then the sampler is allowed to penetrate, further for
300 mm (30 cm) and No. of blows required to penetrate
the sampler to 300 mm (30 cm) is the SPT N-value.
(7) The next test is carried out at level 750 mm below the If it is assumed that the Mohr envelope is a straight line
previous test reference level. passing through the origin (for cohesionless soil or
(8) If borehole depth is large, then the interval of the normally consolidated clays). Therefore the line
next test is taken at a depth of 2.5 to 2 m or the change OP1 must be tangent to the Mohr circle, and the circle
of strata. may be constructed as follows:
• Draw P1C normal to OP1, Point C which is the
intersection point of the normal with the abscissa is
the center of the circle. CP1 is the radius of the circle.
The Mohr circle may now be constructed which gives
the major and minor principal stresses σ1 and σ3
respectively.
• Since the failure is on the horizontal plane, the origin
of planes P0 may be obtained by drawing a horizontal
line through P1 giving P0. P0F and P0E give the
directions of the major and minor principal planes
respectively.
• (Here From the Diagram Given)
Q.90. Which component represents the minor principal
stress in the Mohr’s circle of a direct shear test?

• POE resembles with X (i.e. minor principal stress)

For analytic solution:
(a) W (b) X (c) Y (d) Z tanϕ = P1A/AO, P1O = P1A/sinϕ, P1C = P1O × tanϕ
𝛔𝟏 − 𝛔𝟑
Ans : (a) Mohr Diagram for a Direct Shear Test at 𝐏𝟏 𝐂 =
Failure: 𝟐
In a direct shear test, the sample is sheared along a 𝐬𝐢𝐧 𝛟 =
𝑨𝑪
𝑶𝑪
= 𝝈𝝈𝟏−𝝈𝟑 /𝟐
+𝝈 /𝟐
𝟏 𝟑
horizontal plane. This indicates that the failure plane is
horizontal. Q.91. The results obtained from the grain analysis is
Point P1 on the stress diagram in the Figure represents given below:
the stress condition on the failure plane. The coordinates < 2.0 mm - 90%
of the point are normal stress = σ and shear stress τ = s. < 0.65 mm - 60%

UPSSSC JE 2018 RE-EXAM 27/05/2023 219 Civil Ki Goli

Statement B: The active earth pressure is greater than (a) Plate level Test → Cross Hair Ring Test → Spire Test
the earth pressure at rest. → Vertical Arc Test
(a) Both statements are incorrect. (b) Cross Hair Ring Test → Spire Test → Plate level Test
(b) Both statements are correct. → Vertical Arc Test
(c) Statement B is correct and A is incorrect. (c) Plate level Test → Spire Test → Cross Hair Ring Test
(d) Statement A is correct and B is incorrect. → Vertical Arc Test
(d) Spire Test → Cross Hair Ring Test → Vertical Arc Test
Ans : (d) Active earth pressure arises when the retaining
→ Plate level Test
wall moves away from the soil, allowing the soil to
expand. This results in a decreased earth pressure Ans : (a) The permanent adjustments of a theodolite are
against the wall compared to the at-rest condition. so arranged so that the next adjustment do not creates
• In reality, the active earth pressure is less than the any disturbance in the results obtained from previous
earth pressure at rest. The earth pressure at rest is the adjustments. If the properly arranged sequence is not
pressure exerted by the soil when there is no movement followed it may result in the error in the previous
of the retaining wall, i.e. it's neither moving towards nor adjustment and
away from the backfill. Hence the given order is followed:
(1) Plate level Test: It is done to make the plate bubbles
central to their run when the vertical axis is truly vertical.
(2) Cross Hair Ring Test: It is done to make the vertical
crosshairs lie in a plane perpendicular to the horizontal
axis.
(3) Spire Test: It is done to make the horizontal axis
perpendicular to the vertical axis.
(4) Vertical Arc Test: It is done to make the vertical circle
indicate zero when the line of sight is perpendicular to
Q.104. What is the mass and free drop, respectively, of the vertical axis.
rammer in the standard proctor test, as per IS 2720 - Q.106. Which of the following methods is suitable for
Part 7 - 1980? plane table surveying a township where clearings are
(a) 4.2 kg and 410 mm (b) 2.6 kg and 310 mm rare and distant views are seldom available?
(c) 4.9 kg and 450 mm (d) 3.6 kg and 350 mm (a) Radiation (b) Resection
Ans : (b) (IS: 2720 Part VII - 1974) : The Indian Standard (c) Traversing (d) Intersecting
Equivalent of the Standard Proctor Test is called the light
Ans : (c) Traversing- It is that type of survey in which a
compaction test
number of connected survey lines form the framework.
(IS: 2720 Part VIII - 1983) : The Indian Standard
• It is suitable for a township where clearings are rare
Equivalent of the Modified Proctor Test is called the
and distant views are seldom available.
heavy compaction test
In compaction test: Q.107. Which of the following is not a part of Prismatic
compass?
IS light IS Heavy Modified
Property (a) Lifting pin (b) Objective vane
compaction test compaction test
Weight of (c) Spring brake (d) Rider
2.6 kg 4.9 kg
hammer
Ans : (d)
Number of
3 5
Layers
Number of
25 25
blows
Height of fall 310 mm 450 mm
Volume of
1000 cc 1000 cc
mould

❖ SURVEY
Q.105. The permanent adjustment of theodolite is
made in a specific sequence so as to avoid any
disturbance in the previous adjustment caused due to
next adjustment. The correct sequence is best
represented by

UPSSSC JE 2018 RE-EXAM 27/05/2023 223 Civil Ki Goli

Horizontal Collimation or Line of Sight Error: s = Staff intercept.
• Horizontal collimation or line of sight error is when the • If a tachometer is fitted with an anallatic lens, its
line of sight is not perpendicular to the tilting axis of additive constant is zero and the multiplicative constant
the instrument. This is an axial error. is 100.
• Line of sight error affects the horizontal angle readings Q.113. ___________ is used to carry levelling across a
and increases with steep sightings. The error can be river.
overcome or eliminated by observing two faces. (a) Profile levelling (b) Differential Levelling
Tilting Axis Error or Tilt Error: (c) Precise levelling (d) Reciprocal levelling
• Tilting axis or tilt error is the error when the axis to the
Ans : (d) Reciprocal levelling: Reciprocal levelling is used
total station is not perpendicular to the vertical axis or
to determine the correct difference of elevations of the
plumb line.
two points which are quite far apart and when it is not
• The error affects horizontal readings when the
possible to set up the instrument mid-way between the
instrument is tilted (steep sightings) but has no effect
two points to balance the fore sight and back sight.
on sightings taken when the instrument is horizontal.
• Like horizontal collimation error, the tilting error can be • This reciprocal levelling is usually used to determine
eliminated by two face measurements. Another the difference of elevations of two points on the
method is to apply the measured tilting error at the opposite banks of a river or a deep gorge.
time of the calibration process for all readings. Q.114. The Prismoidal formula for volume
Vertical Collimation Error or Vertical Index Error: If the (Where, V = Volume
horizontal baseline of angle from 0° to 180° in the A = Respective area
vertical circle does not coincide with the vertical axis of D= Distance between the two ends)
the instrument. 𝐃
Compensator Index Error: This error is caused by not (a) 𝐕 = [𝐀𝟏 + 𝟐𝐀𝐦 + 𝐀𝟐 ]
𝟔
𝐃
leveling the total station correctly and carefully. This (b) 𝐕 = [𝐀𝟏 + 𝟒𝐀𝐦 + 𝐀𝟐 ]
𝟑
error can’t be eliminated by taking two faces (face left 𝐃
and face right) readings, unlike the horizontal (c) 𝐕 = [𝐀𝟏 + 𝟒𝐀𝐦 + 𝐀𝟐 ]
𝟔
collimation error. 𝐃
(d) 𝐕 = [𝐀𝟏 + 𝟐𝐀𝐦 + 𝐀𝟐 ]
𝟑
Q.112. If a tachometer is fitted with an anallactic lens,
Ans : (c) The volume (V) of a prismoidal shape is
then,
calculated from the two end-areas (A1 and A2), the area
(a) Additive constant is 100 and multiplying constant is (Am) of a section midway between A1 and A2, and the
zero distance (L) between the two outer sections.
(b) Multiplying constant is 100 and additive constant is (𝐀𝟏 +𝟒𝐀𝐦 +𝐀𝟐 )×𝐋
zero 𝐕= 𝟔
(c) Both additive and multiplying constants are 100 The midpoint area is determined from averaging the
(d) Both multiplying and additive constants are 50 corresponding linear heights and widths of the two end-
areas and not by averaging their areas.
Ans : (b) For tachometer: The stadia method is based on
the principle that the ratio of the perpendicular to the
base is constant in a similar isosceles triangle.

Q.115. In the plane table survey, the accuracy with

which the instrument station can be established in
three-point problem is known as:
As per the principle of stadia method, the horizontal (a) strength of levelling (b) strength of solution
distance method is given by: (c) strength of fix (d) strength of ranging
D = kS + C Ans : (c) Three-Point Problem: The three-point problem
Where, consists in locating the position of a plane table station
multiplying constant(k) = f/i on the drawing sheet by means of observation of three
Additive constant (C) = ( f + d )

UPSSSC JE 2018 RE-EXAM 27/05/2023 225 Civil Ki Goli

vertical crosshairs and the optical center of the object (c) 1-2 percent (d) 1.5-2 percent
glass. Ans : (d) For making any estimate workable, additional
4. Bubble Line (Level tube axis or Altitude level axis): It expenses are to be added.
is a straight line tangential to the longitudinal curve of The followings are the additional expenses that are to
the level tube at its center. It is horizontal when the be considered during estimates:
bubble is in the center.
• Extra Expenses (contingencies) – 3 to 5%
5. Plate level axis: It is perpendicular to the vertical axis
• Contract work establishments – 1.5 to 2% (sometimes
when the bubble is at the center.
2.5%)
Q.119. The whole circle bearing of a line is between 90° • Tools and Machinery – 1 to 1.5%
and 180°. Where does the line lie? • Contractor’s profit – 10%
(a) SW quadrant (b) NE quadrant • Departmental Profit – 10 to 15%
(c) NW quadrant (d) SE quadrant Q.122. Generally, under Brickworks, as per the
Ans : (d) The relation between WCB and RB is given in Indian Standards, the unit of measurement in the MKS
the table below: system for items like Cornices is in
(a) quintal (b) cubic meter
Whole circle Reduced (c) running meter (d) square meter
S.no. Quadrant
bearing (WCB) bearing (RB)
Ans : (c) The different forms of measurement units are
1 0∘−90∘ WCB NE as tabulated below:
2 90∘−180∘ 180°- WCB SE Units of
Particulars of items
3 180∘−270∘ WCB - 180° SW measurements
4 270∘−360∘ 360∘−WCB NW Well sinking, pile driving, Dag-belling,
Grouting of cracks or joints, Supply of Meters
❖ ESTIMATION COSTING pipes, Skirting, cornices
Q.120. Consider the following statements with regard Plastering, pointing, Dado, White
to cubical content method. Washing, distempering, painting,
Square meters
polishing, coal tarring, removing of
I. Length and breadth should be taken as the external paint
dimensions of the buildings at floor level. Tile roof, slate roofing, timber roofing,
II. Height should be taken from floor level to the top of Ceiling, centering and shuttering, Damp
the roof. ( Which of the following options are true ) ? proof course, Turfing or lining of canal,
(a) Only statement II is true surface dressing or leveling, jail work or Square meters
Jafri work, Woodwork indoors and
(b) Both statements I and II are false
shutters, sawing of timber, woodwork in
(c) Both statements I and II are true partition and plywood.
(d) Only statement I is true Boring holes in iron, painting letters,
Ans : (c) Cubical Content Method: and figures, ornamental pillar caps,
cleaning flues, cotton cords in skylight, Numbers
• It is one of the method to prepare approximate
Easing doors and windows, Fixing doors
estimate. and windows
• This method is more accurate than other approximate Earthwork in excavation, puddling,
estimate methods. Quarrying of stones, concreting,
Cubic meters
• It is more suitable to be applied for multistoried brickwork in the foundation, stone
masonry
building.
Rolled steel joists, steel reinforcement
• The cost of a structure is calculated approximately as : bars, binding of steel reinforcement, Quintals
Cost = Volume × cubical rate (rate/unit volume) fabrication and hoisting of steelwork
• The volume of building is obtained by Length Q.123. At a construction site, the reduced level of
× breadth × depth or height. ground is 100 m and reduced level for bottom of footing
• The length and breadth are measured out to out of is 98 m. The quantity of earthwork in excavation for 10
walls excluding the plinth off set. number of square footing of 2 m side is:
• Height should be taken from floor level to the top of (a) 400 cu.m (b) 40 cu.m
the roof. (c) 80 cu.m (d) 800 cu.m
Q.121. In case of abstract of estimated cost what Ans : (c) Given, The square footing of 2m side
percentage of total cost is added to meet the Number of footing = 10, Area of footing = 4 m2
expenditure of a work-charged establishment? The depth of footing
(a) Up to 1 percent (b)2-3 percent

UPSSSC JE 2018 RE-EXAM 27/05/2023 227 Civil Ki Goli

= RL of the ground - RL of the bottom of the footing Area of wall on one side (A) = l × h = 4 × 3 = 12 m2
= 100 - 98 = 2 m Area of plastering on both sides = 12 x 2 = 24 m2
Hence, The quantity of earthwork in excavation is Cost of plastering wall on one side = Rate of plastering ×
= 10 × 4 × 2 = 80 cu.m Area of the wall on one side
Q.124. The covered area of the proposed building is = 12 × 8 = ₹ 96/-
150 m2 and it includes a rear courtyard of 5 m × 4 m. If The total cost of plastering both side of wall = 2 × 96
the plinth area rate for similar buildings is Rs.1,250/m 2, = ₹ 192/-
what is its cost? Q.127. Which of the following statements related to
(a) Rs.1,87,500 (b) Rs. 2,12,500 lead in earthwork estimation is/are true/false?
(c) Rs. 3,75,000 (d) Rs. 1,62,500 1. Distances not exceeding 250 m shall be measured in
Ans : (d) Given units of 50 m.
Covered area = 150 m2 2. Distances exceeding 250 m and not exceeding 500 m
Courtyard area = 5 m × 4 m shall be measured as a separate item.
Covered area of proposed building = 150 m2 (a) Both statements are false.
Courtyard area = 5 x 4 = 20 m2 (b) Statement 2 is true but statement 1 is false.
∴ Plinth area = 150 – 20 = 130 m2 (c) Statement 1 is true but statement 2 is false.
∴ Total cost of the building = 130 x 1250 = 162500 (d) Both statements are true.

Q.125. The quantity of honey-comb wall and partition Ans : (d) Lead : It is the horizontal distance up to which
wall is carried out in the unit of: the excavated earth will be carried or transported on the
ground from the source to the place of spreading.
(a) m (b) m2 (c) m3 (d) Per m
The general lead considered is 50 m for a distance up to
Ans : (b) Quantity of honey-comb wall and partition
500 m and shall be measured separately. It means that it
wall shall be taken in m2. is measured for distance per 50 m.
A brief example is shown for each case: A table for lead distance is given below:
Earthwork in Excavation, Distance Unit of Lead
Earthwork in the filling, 250 m 50 m
Mass
Brickwork with lime or
volume & Cubic Unit 250 - 500 m 50 m
cement mortar, Reinforced
thick work
brickwork, Mass 500 m - 5 km 500 m
Concreting 5 km 1 km
Cutting of large trees,
Q.128. What quantity of stone is required for 1m 3 of
Piece work Painting figures, Manholes
or Job Numbers or Inspection chambers, rubble masonry?
Work Door or Window handles, (a) 0.5 cu.m (b) 0.75 cu.m
Pipe fittings (c) 1.00 cu.m (d) 1.25 cu.m
Thin, Clearing of shrubs,
Shallow, or Square unit or Partition wall, Honeycomb Ans : (d) Volume of Stone Masonry = 1 cum
surface area brickwork, Brick Flat Soling, Considering 25 % Wastage, Volume of stone masonry
works Formwork, DPC required is 1.25 × 1 = 1.25 cum.
Running As thumb rule following wastage are generally
Cornices, Expansion joints,
Linear Metre or considered in various construction material:
Ridge Hip and valley, Eaves
Works linear
Tiles, Handrails Material Standard Wastage in %
measurement
Q.126. Estimate the quantity of plastering (for two Cement 1
faces) required in wall of 4 m long, 3 m high and 30 cm Reinforcing Steel 3
thick. Also calculate the cost of plastering, if the rate of Coarse Aggregate 2.5
plastering is Rs. 8 per sq. m. Fine Aggregate 2.5
(a) 1.2 sq. m & Rs. 9.6/- (b) 2.4 sq. m & Rs. 19.2/- Stone Masonry 25
Structural Steel 5 to 10
(c) 12 sq. m & Rs. 96/- (d) 24 sq. m & Rs. 192/-
Tiles 7 to 10
Ans : (d) Given, Length of wall (l) = 4 m
Q.129. No deduction shall be made for openings like
Height of wall (h) = 3 m ventilators, flues etc. having opening up to __________
The thickness of plastering = 30 cm in section.
Rate of plastering = ₹ 8/m2 (a) 0.1 sqm (b) 1 sqm (c) 0.001 sqm (d) 10 sqm

UPSSSC JE 2018 RE-EXAM 27/05/2023 228 Civil Ki Goli

Ans : (a) Rules for the deduction for openings as per IS meters, and also to have limited space for storage of
- 1200 for brickwork: building materials
No deduction is made for the following: Q.131. Identify the door type shown in the below
(i) Bearing of floor and roof slabs are not deducted from figure.
the masonry in the superstructure.
(ii) Opening up to 0.1 m2
(iii) Ends of the beam, posts, rafters, purlin, etc. chajjas
where thickness does not exceeds 10 cm.
(iv) Bed plates, wall plates, bearing of chajjas where
thickness does not exceed 10 cm.
Deduction in plastering is made in the following
manner: (a) Flush Doors (b) Ledged Doors
(i) No deduction is made for ends of beams, posts, (c) Louvered Doors (d) Glazed Doors
rafters, purlin, etc.
Ans : (c) Louvered doors: Louvered doors are used when
(ii) No deduction is made for opening up to 0.5 m2 and privacy with natural ventilation and quietness for rest is
no addition is made for jambs, soffits, and sills of these desired, as they allow free passage of air even when
openings. closed. You can use louvered doors to help ventilate
(iii) For opening more than 0.5 m2 and up to 3 m2 the certain areas of your home, to add a small amount of
deduction is made for one face only. No addition for privacy to otherwise open space, or as room dividers.
jambs, soffits, and sills of these openings. • They are used for toilets and bathrooms in public
(iv) For opening above 3m2 the deduction is made buildings and residential buildings. These doors are
for both faces of openings and the jambs, soffits, and also used in cabinets, cupboards, utility rooms, etc.
sills shall be added.

❖ BCME
Q.130. The arrangement of supports provided
underneath the existing structure without disturbing
its stability, is known as
(a) scaffolding (b) jacking
(c) shoring (d) underpinning
Ans : (d) Structures provided underneath an existing
foundation to maintain its stability is termed as Flush Doors: A Flush Door has completely smooth
underpinning. surfaces and is made by sandwiching plywood or
• Underpinning is used to repair, strengthen or renewal blackboard over a light timber frame. The hollow part
of the foundation of an existing building. between these doors is filled with cardboard/hardwood.
• During underpinning, the existing structure is required They can be finished with either laminates or veneers.
to be temporarily supported by means of raking shores Flush doors are lighter and cheaper than other types.
• The flush door shutters are manufactured in standard
thicknesses of 25, 30, 35, and 40 mm

Ledged Doors: The door consists of vertical boards i.e.

• Shoring: It is the temporary support provided to the
battens and three or four horizontal ledges. This is the
structure which has become unsafe due to excessive
simplest form of door and the cheapest also.
differential settlement or due to the removal of the
adjacent building, etc. • The vertical boards are tongue and grooved to stop
• Scaffolding: It is a safe working platform provided for draughts and the edges are chamfered to relieve the
plain appearance.
the workers working at an elevation greater than 1.5

UPSSSC JE 2018 RE-EXAM 27/05/2023 229 Civil Ki Goli

 Q.133. Select the INCORRECT statement(s) related to
stone masonry.
(i) The horizontal course provided at suitable levels
between the plinth and the cornice is termed as a string
course.
(ii) A corbel is a projecting stone which is usually
provided to serve as support for roof truss, beam etc.
(iii) The horizontal course provided to strengthen a wall
of irregular small stones is known as coping.
Glazed Doors: This type is used in residential and public
iv) The exposed horizontal surfaces at right angles to
buildings. They supplement the natural lighting provided
the door or window frames are known as jambs.
by windows or make the interior of the room visible from
adjoining rooms. (a) Both (ii) and (iii) (b) Both (i) and (ii)
(c) Both (iii) and (iv) (d) Both (i) and (iv)
• They can be made fully glazed or partly glazed. Fully
glazed doors are recommended where sufficient light Ans : (c) Course : The horizontal course is provided to
is required through the door openings like in shopping strengthen a wall of irregular small stones.
malls, entrance halls, etc. In the case of partly glazed, Jambs: The exposed vertical surfaces at right angles to
the bottom 1/3rd part is usually paneled and the upper the door or window frames.
2/3 part is glazed. The cross-section of substructure and superstructure is
given below

Q.132. Beam which used to support the above wall of

openings is called
(a) Roof (b) Damp proofing course
(c) Floor (d) Lintels
Ans : (d) A lintel is one type of beam which used to
support the above wall when openings like doors,
windows, etc. are necessary to provide a building
structure. The main function of the lintel is to take loads
coming from the above wall and transfer its load to the
side walls. Q.134. Match the following components of stair with
• Damp Proofing course: A layer provided to prevent their definition and select the correct option.
entry of unwanted moisture inside the building either
by seepage or by leakage is known as Damp Proofing Components
Definition
course. of stair
• Floor: The structure which divide a building into stages 1. Scotia A. Underside of staircase
or storeys to give the space to live at different levels 2. Soffit B. The moulding provided under
are termed as floors. the nosing to improve the
elevation of steps,
Different types of floors are Timber floors, cement tile
3. Newel C. Member supporting the
concrete flooring, cement concrete flooring, Mosaic
handrail
floor, brick floor etc.
4. Baluster D. The post usually provided at
• Roof: The horizontal surface used over most part of
the beginning and end of the
the building to keep out rain, sun and wind and to
flights supporting
preserve the interior from exposure is known as Roof.
the handrails.

UPSSSC JE 2018 RE-EXAM 27/05/2023 230 Civil Ki Goli

• Used as ballast in foundation and floors in lime List 1 List 2
concrete. P Iron Oxide (Fe2O3) 1 1% - 2.9%
Q.145. Read the following statements and select the Q Sulphur trioxide (SO3) 2 17% - 25%
correct answer. R Alumina (Al2O3) 3 0.5% - 6%
S Silica (SiO2) 4 3.5% - 9%
Statement A: Conifer trees have distinct annual rings.
(a) P - 4, Q - 3, R - 1, S - 2
Statement B: The presence of the original rounded
(b) P - 4, Q - 1, R - 2, S - 3
surface on the manufactured piece of timber is known
as 'torn grain'. (c) P - 3, Q - 1, R - 4, S - 2
(a) Statement A is true and B is false. (d) P - 2, Q - 4, R - 1, S - 3
(b) Statement A is False and B is True. Ans : (c) Approximate oxide composition limits of
(c) Both statements are correct. ordinary Portland cement are as follows:
(d) Both statements are incorrect. Oxide Percentage
Content
Ans : (a)
CaO 60 - 67
Si02 17 - 25
Al2O3 03 - 08
Fe2O3 0.5 - 6.0
MgO 0.1 - 4.0
Alkalies (K2O, Na2O) 0.4 - 1.3
SO3 1.3 - 3.0
Q.147. Read the following statements and identify the
correct combination of statements.
• The conifers are also known as the evergreen trees and
Statement A: In oil varnishes, soft resins like lac and
the leaves of these trees do not fall till new ones are
shellac are dissolved in linseed oil.
grown, As these trees bear cone-shaped fruits, they
are given the name conifers. These trees yield Statement B: In oil paint, the function of solvent is to
softwoods which are generally light coloured, resinous, provide a binder for the ingredients of paint so that
light in weight, and weak. They show distinct annual they may stick or adhere to the surface.
rings. Statement C: In plywood, three or more veneers in odd
• The deciduous trees are also known as the broad-leaf numbers are placed one above the other with the
trees and leaves of these trees fall in autumn and new direction of grains of successive layers at right angles to
ones appear in spring season. The timber for each other.
engineering purposes is mostly derived from Statement D: The final setting time of low-heat cement
deciduous trees. These trees yield hardwoods which is about 10 hours.
are usually close-grained, strong, heavy, dark coloured, (a) (A)-True, (B)-False, (C)-True, (D)-False.
durable, and nonresinous. They do not show distinct (b) (A)-False, (B)-True, (C)-False, (D)-True.
annual rings. (c) (A)-False, (B)-False, (C)-True, (D)-True.
Defects due to conversion: (d) (A)-True, (B)-True, (C)-False, (D)-False.
Chip mark: This defect is indicated by the marks or signs
Ans : (c) Oil varnishes: The linseed oil is used as a solvent
placed by chips on the finished surface of timber. They
in this type of varnish.
may also be formed by the parts of a planing machine.
• The hard resins such as amber and copal are dissolved
Diagonal grain: This defect is formed due to improper
in linseed oil and if the varnish is not workable, a small
sawing of timber. It is indicated by diagonal mark on
quantity of turpentine is added.
straight grained surface of the timber.
• The oil varnishes dry slowly, but they form hard and
Torn grain: This defect is caused when a small depression durable surface. In fact, these are the hardest and the
is formed on the finished surface of timber by falling off most suitable varnishes.
a tool or so.
• They are specially adopted for exposed works which
Wane: This defect is denoted by the presence of an require frequent cleaning. They are used on coaches
original rounded surface on the manufactured piece of and fittings in houses.
timber.
Ingredient Function of Oil Paint:
Q.146. Match the items under List 1 (oxides) with those (a) Bases: A base is a solid substance in a fine state of
under List 2 (percentage) in an oxide composition of division and it form the bulk of a paint.
ordinary Portland cement.

UPSSSC JE 2018 RE-EXAM 27/05/2023 234 Civil Ki Goli