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 CONTENTS
SYLLABUS ......................................................................................................................... 3-4
OPSC AEE (Panchayati Raj) Exam-2021 (Paper-II) ..................................................................5-27
OPSC ASCO 2021 Paper-I ..................................................................................................... 28-37
OPSC ASCO 2021 Paper-II ................................................................................................... 38-48
OPSC AE CIVIL Exam-2020 ................................................................................................ 49-63
OPSC AAO Asst. Agri Engg Paper I (2020) ............................................................................ 64-77
OPSC AAO Asst. Agri Engg Paper II (2020) ......................................................................... 78-92
OPSC AEE Exam-2019 (Paper- I) ......................................................................................... 93-115
OPSC AEE Exam- 2019 (Paper-II) ..................................................................................... 116-135
OSSC JE Exam-2019,Shift -I .............................................................................................. 136-151
OSSC JE Exam-2019,Shift -II ............................................................................................ 152-169
OSSC JE Exam-2019,Shift -III ........................................................................................... 170-188
OPSC Poly. Lect. Exam -2018 (Paper-I) .............................................................................. 189-202
OPSC Poly. Lect. Exam-2018 (Paper-II) .............................................................................. 203-215
OPSC AEE Exam -2016 (Paper-I) ....................................................................................... 227-236
OPSC AEE Exam- 2016 (Paper-II) ..................................................................................... 237-251
Odisha JE (Main) Exam-2014 ............................................................................................ 252-262
OSSC JE Exam -2014 ........................................................................................................ 263-279
Odisha Civil Service Exam-2011 ......................................................................................... 280-304
Odisha Civil Service Exam-2006 .....................................................................................................
Odisha Civil JE & PSC Previous Exam Papers
Analysis Chart

Sl No. Exam Proposed Year Question Paper Total Question

OPSC AEE (Panchayati Raj) Paper-II 24.08.2021 150
OPSC ASCO, Paper-I 2021 100
OPSC ASCO, Paper-II 2021 100
OPSC AE CIVIL 29.11.2020 100
OPSC AAO Asst. Agri Engg, Paper I 2020 100
OPSC AAO Asst. Agri Engg, Paper II 2020 100
OPSC AEE, Paper- I 2019 180
OPSC AEE, Paper- II 2019 180
OSSC JE (Shift-I) 2019 100
OSSC JE (Shift-II) 2019 100
OSSC JE (Shift-III) 2019 100
OPSC Poly. Lect., Paper-I 2018 100
OPSC Poly. Lect., Paper-II 2018 100
OPSC AEE, Paper-I 2016 90
OPSC AEE, Paper-II 2016 90
Odisha JE (Main) 2014 100
OSSC JE 2014 60
Odisha Civil Service 2011 120
Odisha Civil Service 2006 120
Total 2090
 Odisha Staff Selection Commission
Plan and pattern of Examination:-
(a) There shall be two stages of examination.
(i) Preliminary Examination
(ii) Main Written examination
(iii) Certificate verification.
Stage of Type of No of paper & Total Duration Remark
Examination Examination Marks Marks
Stage-I Preliminary One Paper– 150 150 •The question will
Examination • Arithmetic-10th minutes be of MCQ type.
standard. • There shall be
• Data Interpretation negative marking
(Chart, Graph, Table, @ 0.25 marks for
Data, Sufficiency etc.) each wrong answer.
10th standard • Approximately 5
• Logical Reasoning times of number
and Analytical vacancies category
Ability, General wise and post wise
Mental Ability. shall be shortlisted
• Current Events of for the Main
national and written
International Examination.
Importance. • The commission
• Computer/internet at their discretion
Awareness may fix minimum
qualifying mark in
Preliminary
Examination in
different
categories for
different technical
posts/Services.
Stage-II Main Written Technical Paper– 200 3 hours Candidates up to
Examination There shall be (180 2(two) times the
different Technical minutes) vacancies
papers for different advertised in each
posts/services as per category, in each
qualification posts in order of
prescribed for the merit basing on the
post. (Detail Syllabus marks in Written
annexed as Examination shall
Regulation 2 of 2022) be shortlisted for
the verification of
original documents.
Stage-III Certificate The candidate who
Verification fails to attend the
document
verification,
his/her name will
not be considered
for the post.
4
 Syllabus
Civil Engineering Materials Syllabus
Stone, Bricks Clay Products And Refractory Materials Cement Sand Gravel Morrum And Fly
Ash Mortar and concrete Timber Paint Varnish and distemper Iron and steel Bituminous
Materials Plastics Heat Proofing and Acoustic materials.
Construction Technology Syllabus
Introduction to Construction Technology Site Investigation Foundations Walls Damp Proofing
Arches and Lintels Doors and Windows Floors Roofs Stairs Surface Finishes General idea of
Seismic Planning Design of Building Construction Machineries.
Structural Analysis:-
Trusses and frames Slope and deflection Fixed beam Continuous beam Slope deflection Method
Moment Distribution Method Three Hinged Arches.
Transportation Engineering Syllabus
Introduction to transportation Engineering Road Geometric Road Materials Hill Roads
Road Drainage Road maintenance Construction Equipments Traffic Studies Landscaping And
Arboriculture Introduction to Railways Transportation Permanent Way Track materials Geometric for
Broad Gauge Points and Crossings Laying And maintenance to Track Introduction to Bridges
Hydrology and planning Bridge Foundation Bridge Substructure and Approaches Permanent Bridges
Culvert and Causeway Introduction To Docks and harbors Break Waters Docks Introduction to Airport
engineering Components of An Airport Tunnel Engineering
Irrigation Engineering Syllabus
Introduction To Irrigation Engineering Hydrology Water Requirement of Crops Flow Irrigation
Diversion Head Works Regulatory Works Cross Drainage works Dams Water Logging And
Drainage Ground Water Hydrology
Estimating:
Introduction to Estimating Detailed Estimate of Building As per PWD specifications and standards
Analysis of Rates Administrative Setups of Engineering Organizations Detailed Estimate of Culverts and
Bridges Estimate of Irrigation Structures Detailed Estimate of Roads PWD Accounts Works.
Structural Design Syllabus
Introduction to Design And Detailing Working stress Method of Design Limit State Method (LSM) of
design Limit state of collapse of singly Reinforced members In bending Limit state of collapse in shear
Bond Anchorage Development Lengths And Slicing (LSM) Beams (LSM) TWO Way Slabs (LSM)
Axially Loaded short columns (LSM) Ductile detailing of Reinforced concrete structures Design of steel
Design of timber Structures (Limit State) Structural Steel Fasteners and Connections Design of Tension
Members Design of column Bases and foundations Design of Steel Beams Design of Timber structures
Stair case (RCC-LSM) Design of Footings (RCC-ISM)
Public Health Engineering Syllabus
Introduction to water supply Engineering Quantity of water Sources of water Conveyance of Water
Quality of Water and Treatment of Water Distribution system Appurtenance In Distribution System
Water Supply Plumbing In Building Introduction to Sanitary Engineering Quantity of Sewage Sewerage
System Sewer Appurtenance Sewage Characteristics Sewage Disposal Sewage Treatment
Sanitary Plumbing for Building And Rural Water Supply Sanitation
Construction Management Syllabus
Introduction to Construction management Construction Planning Materials management
Site management Construction organization Labor management Equipment management Quality
Control Monitoring Progress in Construction Works Safety Management in Construction Works
Advanced Construction Technology Syllabus
Concrete Mixed Design Handling And Transporting of Concrete Earthquake Resistant Construction
Building Services Construction And Earth Moving Equipment

4
 Odisha Public Service Commission
(AEE, Panchayati Raj)
Exam- 2021 (Paper-II)
1. In solid mechanics the strength of materials cos 2θ + sin 2θ = −1
may be regarded as (cos 2θ + 1) = – sin 2 θ
(a) The statics deformable or elastic bodies 2cos2 θ = –2 sin θ cos θ
(b) The statics of strength and stiffness of bodies tan θ = –1
(c) The statics of rigid bodies θ = 1350
(d) The statics of modulus of resilience θ1 = θ – 90o
Ans. (a) : Strength of material is the study of = 135o – 90o
deformable or elastic bodies and we study different = 45o
properties of material by applying force on it. 3. Modulus of resilience can be defined as
2. A prismatic bar is subjected to axial tension. (a) Strain energy/volume
What is the aspect angle which defines an (b) Kinematic energy/potential energy
oblique section on which the normal and (c) Volume/energy
shearing stresses are equal?
(d) Potential energy/moment of ineria
(a) 900 (b) 450
Ans. (a) : Modulus of resilience– The proof resilience
(c) 1350 (d) 1800
per unit volume of a material is known as modulus of
Ans. (b) resilience.
Strain energy σ y
2

Modulus of resilience =
Volume 2E
4. A prismatic steel rod of length L and cross
sectional area A hangs vertically under its own
weight. What is the strain energy stored in the
bar, if its unit weight per unit volume is y (E is
σx + σy σx – σy Young's Modulus)?
σθ = + cos2θ + τxy sin 2θ
2 2 (a) γ2AL3/6E (b) γAL2/6E3
 σ – σy  (c) γ2AL2/6E (d) γ3AL/6E2
τθ = −  x  sin 2θ + τxy cos 2θ
 2  Ans. (a) : Given, length of rod = L
Cross sectional area = A
Unit weight of bar per unit volume = γ
Young modulus of bar = E
σx = σ
σy = 0
Consider an element at a distance x from the lower free
τxy = 0
end.
σ σ
∴ σθ = + cos 2θ Let the thickness of the element = dx
2 2
The section x-x will be acted upon by the weight of the
σ
τθ = − sin 2θ bar of length x.
2
σθ = τθ (given) wx = weight of the bar of length x
σ σ σ = (volume of the bare of length x) × weight of unit
+ cos 2θ = − sin 2θ volume
2 2 2
σ σ wx = (A × x) × γ = γ Ax.
= − ( sin 2θ + cos 2θ )
2 2 Let due to self weight wx elongation is dδ

OPSC AEE Exam-2021 (Paper-II) 5 YCT

 Elongation in dδ
Strain in portion dx =
Length of dx
dδ
=
dx
Weight action on section x- x
Stress in portion dx =
Area of section
γ.A.x
= = γx 6. A cantilever been of length 5m, subjected to a
A
uniformly distributed load of 20 kN/m. The
Stress γ.x
E= = bending moment of its free end is equal to
Strain  dδ  (a) 20 kNm
 dx 
  (b) 10 kNm
γ.x.dx (c) Zero
dδ = (d) 200 kNm
E
Now the strain energy stored in portion dx is Ans. (c) : L = 5 m, W = 20 kN/m
dU = Average weight × Elongation of dx
 0 + wx 
=   × dδ
 2 
w x γ.x.dx
dU = × x
2 E Mx-x = wx
2
1 2 x 2 dx wx 2
dU = .γ A. Mx-x =
2 E 2
L
x=L=0
U= ∫ dU
0 MB =
wL2
2
L
1 A 2 ∴L=0
∫ 2 .γ
2
U= .x dx
0
E MB = 0
L The bending moment at the free end will be zero.
1 γ 2 A  x3 
U= ×  7. In an element σx = -σy = 30 KPa, if E = 210 KPa
2 E  3  and µ = 0.25, the shearing strain is
0
2 3 (a) 0.0025
1γ A L γ 2 AL3
U= × = (b) 0.0030
2 E 3 6E (c) 0.0035
5. In plane strain protection (d) None of these
(a) The loading is in two directions Ans. (d) :
(b) There is no normal and shear stresses on the Poisson ration (µ) = 0.25
two plane (X any Y) perpendicular to the z σx = – σy = 30 kPa
direction E = 210 kPa
(c) The stress vector is zero across a particular 1− 2µ
plane Shear strain, e =
E
( σ x + σ y + σz )
(d) The loading is axiymmetric and does not very
1 − 0.25 × 2
in axial direction e= ( 30 + 30 )
210
Ans. (d) : Plain strain problem– Involving long bodies
1 − 0.25 × 2 0.5 × 60
whose geometry and loading do not vary significantly in e= ( 60 ) =
the longitudinal direction are referred to as plain strain 210 210
problems. In this case strain component εzz, εyz & exz 3 1
are taken as zero, where z-axis is the longitudinal axis. e= = = 0.142
21 7
Example– Dam and reservoirs are subjected to water
pressure loading. e = 0.142

OPSC AEE Exam-2021 (Paper-II) 6 YCT

8. The ratio of depth of width of strongest beam
that can be cut out of a cylindrical log of wood
with homogenous and isotropic properteis is
(a) 1.414 (b) 1.25
(c) 0.707 (d) 0.504
Ans. (a)
32M
⇒ σb = (Maximum bending stress)
πD3

From bending equation

My max
σ=
I The ratio of the maximum bending stress to maximum
bd 2 σ 32M πD 3
Z= shear stress is · cb max = ×
6 τmax πD 3 16T
b + d 2 = D2
2
 σcb max 2M 
For maximum section modulus  = 
∂  b ( D 2 − b 2 )   τmax T 
=0⇒ 
dz
=0 11. Consider the following statements :
db ∂b The theory of simple bending assumes that :
⇒ D 2 − 3b 2 = 0 (1) The material of the beam is homogenous,
D isotropic and obeys Hook's law
b=
3 (2) The plane section remains plane after bending
(3) Each cross section of the beam is symmetric
2
d= D about the loading plane
3 (4) Young's moduli are the same in tension and
compression of the above statements which
Ratio of d & b are correct?
2 (a) 1 and 2 only (b) 1, 3 and 4 only
D (c) 2, 3 and 4 only (d) 1, 2, 3 and 4 only
d 3
=
b 1 Ans. (d) Assumptions in theory of bending–
D (i) Plane section before bending remains plane after
3
bending.
d
= 2 or 1.414 (ii) Material is homogeneous, isotropic and obey's
b hook's law.
9. Using the Maximum stress theory and (iii)Modulus of elasticity in tension & compression is
maximum shear theory of failure, the ratios of same.
the diameter of circular shaft is as follows (iv) Beam is initially straight and has constant x-section,
(a) 1: 1.90 : 1.26 (b) 1 : 1.52 : 2.15 throughout its length (i.e. prizmatic)
(c) 1 : 1 : 1 (d) 1 : 2.17 : 3.52
(v) The plane of loading must contains a principal axis
Ans. (a) : Using the Maximum stress theory and of cross-section of beam and load must be
maximum shear theory of failure, the ratios of the perpendicularly to the longitudinal axis of the beam.
diameter of circular shaft is 1: 1.90 : 1.26.
12. If the depth of a beam of rectangular section is
10. If a circular shaft is subjected to a torque T reduced to half, strin energy stored inthe beam
and bending moment M, the ratio of maximum
due to bending becomes
bending stress to maximum shear stress is
(a) 8 time (b) 4 time
(a) 2 T/M (b) M/T
(c) 1/4 time (d) 1/8 time
(c) M/2T (d) 2M/T
Ans. (d) : Maximum shear stress developed by the Ans. (a) : Strain energy due to bending.
shaft– d1 = d, d2 = d/2
16T M2l
⇒ τ max = U=
πD 3 2EI

OPSC AEE Exam-2021 (Paper-II) 7 YCT

 1 Steel – 0.27 to 0.3
U1 ∝ Cark – 0
I
16. A beam of uniform strength has constant
12
U1 = (a) Shear force
bd13 (b) Bending moment
12 12 × 8 (c) Cross sectional area
U1 = =
b (d / 2) (d) deflection
3
bd 3
U2 = U1 × 8 Ans. (b) : A beam of uniform strength has constant
bending moment.
U2 = 8 × U1
13. A steel wire of 20 mm diameter is bent into a
circular shape of 10 m radius. If the young's
modulus of elasticity of the wire is 2×106
kg/cm2, then the maximum stress induced in
the wire is Bending moment is the algebraic sum of moment at that
(a) 2 × 103 kg/cm2 (b) 3 × 103 kg/cm2 section but moment at a point is the summation of
(c) 4 × 10 kg/cm
3 2
(d) 5 × 103 kg/cm2 moment due to all loading on the beam produced at that
Ans. (a) : point.
M σ max E dM
= = =V
I y R dx
17. A three hinged parabolic arch of 20 m span is
σmax E
= subjected to 10 kN/m uniformly distributed
y R load. What is the value of BM at 5 m from left
Given, d = 20 mm, R = 10 m = 1000 cm, support?
y = d/2 = 10 mm = 1 cm, E = 2 × 106 kg/cm2 (a) zero (b) 243 kNm
E×y (c) 200 kNm (d) 100 kNm
σ max = Ans. (a) :
R
2 ×106 × 1
=
1000
σ max = 2 ×103
14. If the area under the shear force diagram curve
for a beam between two points C and D is 'X',
then the difference between the moments at the
two points C and D will be equal to
(a) x/4 (b) x/3
(c) x/2 (d) x
Ans. (d)
dM x
= Sf xx 10 × 52
dx ( BM )@5m = R A × 5 − H × y −
C C 2

∫ dM = ∫ Sf
x xx .dx = Area under the shear curve.
= 100 × 5 −
500  4hx ( l − x ) 
  – 125
D D h  l2 
MC − M D = X =0
15. The Poisson's ratio of structural steel is Note : The bending moment throughout the span will be
(a) 0.3 m (b) 1.0 m zero for a three hinged parabolic arch subjected to
(c) 1.2 m (d) None of the above uniformly distributed load.
Ans. (a) : The Poisson's ratio of structural steel is 0.3 18. A fixed beam has how many number of
m. kinematic indeterminacy?
Material Poisson ratio (a) 3 (b) 0
Rubber – 0.5 (c) 2 (d) 1
Gold – 0.42 Ans. (b) : A fixed beam has number of kinematic
Copper – 0.33 indeterminacy is zero. Because every movement is
Cast iron – 0.2 to 0.3 restricted.
OPSC AEE Exam-2021 (Paper-II) 8 YCT
19. Choose the correct and answer Ans. (d) : Muller-Breslau Principle –
(i) The displacement method is more useful • According to this principle ILD for any stress function
when degree of kinematic indeterminacy is can be obtained by removing the restraint offered by
less than the static indeterminacy that stress function and introducing a directly related
(ii) The force method is more useful when degree generalized unit displacement at the location and in the
of static indeterminacy is less then the direction of the function.
kinematic indeterminacy • It is suitable for all type of statically determinate and
(iii) The force method is more useful when degree only linear elastic indeterminate structure.
of kinematic indeterminacy is less tha the • It is not applicable for moving unit point moment and
static indeterminacy for deflection.
(iv) The displacement method is more useful • This principle based virtual work theorem.
when degree of static indeterminacy is less • It determines shape, ordinate and effect of stress
than the kinematic indeterminacy function of the structure.
(a) (i) and (ii) (b) (i) and (iii) 22. When a structure is just on the point of
(c) (ii) and (iii) (d) (iii) and (iv) collapse, the necessary and sufficient conditions
Ans. (a) : In the displacement method, generally all joint attending collapse are
displacements are prevented regardless of the choice of the (i) Equilibrium condition
unknown displacement. A displacement of a joint affects (ii) Yield condition
only the member meeting at the given joint. (iii) Mechanism condition
• When static indeterminacy is the less than the Choose the correct option–
kinematic, the force method is preferred, otherwise the (a) In lower bound theorem (i) and (iii) are
displacement is preferred. considered
• The displacement method is more useful when degree (b) In upper bound theorem (i) and (iii) are
of kinematic indeterminacy is less than the static considered
indeterminacy. (c) In lower bound theorem (ii) and (iii) are
20. A propped cantilever of span L is subjected to considered
concentrated load at mid span. If Mp is plastic (d) In lower bound theorem (ii) and (iii) are
moment capacity of the beam the value of the considered
collapse load will be
(a) 4 Mp/L (b) 8 Mp/L Ans. (b) : When a structure is just on the point of
(c) 16 Mp/L (d) 6 Mp/L collapse, the necessary and sufficient conditions
Ans. (d) : attending collapse are in lower bond theorem
equilibrium condition and mechanism condition are
considered.
23. The carry over factor in a prismatic member
whose far end is hinged is
(a) 0 (b) 1/2
(c) 3/4 (d) 1
Ans. (a) : Carry over factor is defined as the ratio of the
moment at the far end to the moment at the rotating near
end.
By principle of virtual work. • If far end is hinged –
we = wi  0 
COF = M = 0
L  
p   θ = 3 Mpθ
2
PL
3 Mp =
2
6M p • If far end is fixed–
Collapse load, P =
L  M/2 1
COF = M = 2 
21. The Muller-Breslau principle can be used to  
(a) Determinate shape of the influence line
(b) Indicates the parts of the structure to be
loaded to obtain the maximum effect
(c) Calculate the ordinates of influence lines
• If far end is fixed and near is free–
(d) All of the above
OPSC AEE Exam-2021 (Paper-II) 9 YCT
  −M  The maximum free bending moment (M)
COF = M = −1 =
PL
  4
PL PL
The sum of fixed end moment = +
8 8
PL
24. A structure is statically indeterminate to =
second degree. What is the maximum number 4
of plastic hinges required to make this =M
structure a mechanism? 27. In slope deflection equations, the deformation
(a) 1 (b) 4 are considered to be caused by
(c) 3 (d) 2 (a) torsion (b) Axial forces
Ans. (c) : Plastic hinges required to render structure (c) Shear force (d) Bending moment
mechanism is one greater than degree of indeterminacy.
Ans. (d) : It is assumed that deformation are caused due
Plastic hinges (n) = Degree of indeterminacy + 1
n = DS + 1 to bending moment only and axial deformation are
n=2+1 neglected.
n=3 In slope deflection method, established a relationship
• 3 number of plastic hinges required for indeterminacy between degrees of freedom (θ, ∆) member end
of degree 2. moment.
25. If a 100 kN/m, external moment rotates the 28. If in a rigid jointed space frame, (6m+r) > 6j,
near end "A" of a prismatic beam without where 'j' are the number of joints, 'r' are the
translation. What is the value of moment number of unknown reactions and 'm' are the
induced at far fixed end "B". number of structural members, then the frame
(a) 50 kNm in opposite direction of applied is
moment 100 kNm
(a) Stable and statically determinate
(b) 50 kNm in same direction of applied moment
100 kNm (b) Unstable
(c) 100 kNm in opposite direction of applied (c) Stable and statically indeterminate
moment 100 kNm (d) None of the above
(d) 100 kNm in same direction as applied Ans. (c) : In case of space frame–
moment 100 kNm 6m + r > 6 denotes that number of unknown forces are
Ans. (b) : more than the equilibrium equation available. Hence
structure will be indeterminate & stable.
• For stable statically determinate– 6m + r = 6
• For plane frame 3m+r = 3J, stable statically
1
COF = COF = Carryover factor determinate.
2
29. At a joint of a frame four members have joined
1
Moment produced at B = M × and there of the members have distribution
2 factors for moment distribution as 0.21, 0.29
1 and 0.35. What is the value of distribution
= 100 × = 50 kW − m
2 factor for fourth member?
[Same direction as 100 kN-m] (a) 0.75 (b) 0.15
26. Due to some point load anywhere on a fixed (c) 0.02 (d) 0.25
beam ,the maximum free bending moment is Ans. (b) : Distribution factors for fourth member
M. The sum of fixed end moment is
∑ df = 1
(a) M (b) 1.5 M
(c) 2.0 M (d) 3.0 M 0.21 + 0.29 + 0.35 + m4 = 1
m4 = 0.15
Ans. (a)
30. A simply supported beam of length L carries a
load varying uniformly from zero at left end to
maximum at right end. The maximum bending
moment occurs at a distance of
(a) 1/3 from left end
(b) 1/ 3 from left end
(c) 1/ 3 from right end
(d) 1/3 from right end
OPSC AEE Exam-2021 (Paper-II) 10 YCT
Ans. (b) : Ans. (d) : Over-reinforced beam contains steel and steel
is more ductile than concrete.
• But the behaviour of an under reinforced beam is
more ductile than that of over reinforced beam because
in under reinforced beam, steel failure first due to the
maximum permissible stress of steel reading first.
33. Choose the correct one
(a) Modular ratio for M 30 concrete is less than
M 20 concrete
(b) Modular ratio for M 25 concrete is greater
than M 20 concrete
(c) Modular ratio is same for all grade of
concrete
1 (d) As per IS : 456 : 1978, in calculation of
RA + RB = w 0L modular ratio between elastic moduli of steel
2 and concrete the long term effect such as
1 L creep is not taken into consideration
∑ MB = 0 ⇒ R A × L = w 0L ×
2 3 Ans. (a) : Modular ratio –

{ }
w 0L E Es
RA = m= s = without considering
6 E c 5000 f ck creep effect
w 0L
RB = Modular ratio for M30–
3
At the point of maximum bending moment, 2 × 105 2 × 105
m= = = 7.3
Shear force is zero. 5000 30 5000 × 5.478
w x2 Modular ratio for M25–
S.F.xx = R A − 0 = 0
2L 2 × 105 2 × 105
m= = =8
5000 25 5000 × 5
2
w 0L w 0 x
− =0
6 2L Modular ratio for M20–
w 0L w 0 x 2 2 ×105 2 × 105
= m= = = 8.94
6 2L 5000 20 5000 × 4.47
2
L x 34. A doubly reinforced beam is considered less
=
6 2L economical than a singly reinforced beam
1 because
x= From A
3 (a) Concrete is not stressed to full value
31. The deveopment length of bars of diameter φ, (b) Tensile steel required is more than that for a
as per IS : 456 : 1978 is given by (where σx = balanced section
stress in bar τbd = design bond stress (c) Shear reinforcement is more
(a) 4φσ s /τbd (b) φσs/4τbd (d) Compressive steel is under stressed
(c) 2φσs/3τbd (d) φσs/3τbd Ans. (d) : The doubly reinforced beam is less
φ σS economical than singly reinforced beam because
Ans. (b) Development length = Compressive steel remains under stress in doubly
4 τbd
reinforced beam. steel is 10 times costlier than the
φ = nominal diameter of the bar. concrete. The depth reduces but the cost due to steel in
σs=stress in bar at the section considered at design load. compression zone increases.
τbd = design bond stress
35. In limit state design, the maximum limit
32. Which statement is not correct for over
reinforced concrete section imposed by IS : 456-2000 on the redistribution
(a) Steel is not fully stressed of moments in statically indeterminate beam is
(b) Neutral axis lies below the neutral axis for (a) 10% (b) 15%
balanced section (c) 20% (d) 30%
(c) Compressive stress in concrete at extreme Ans. (d) : According to IS code : 456 :2000 (Clause
fiber reaches its maximum permissible stress 37.1.1). The elastic moment at any section in the
value member due to a particular combination of load shall
(d) Steel is stressed to its maximum permissible not be reduced by more than 30% of the elastic
stress maximum moment.
OPSC AEE Exam-2021 (Paper-II) 11 YCT
• The ultimate moment of resistance provided at any 40. The slump recommended for mass concrete is
section of a member is not less than 70% of the moment about
at that section obtained from an elastic maximum (a) 20mm to 50 mm
moment diagram. Covering all appropriate combination (b) 50 mm to 100 mm
of load. (c) 100 mm to 125 mm
36. A reduction factor Cγ to load carrying capacity (d) 125 mm to 150 mm
of a long column is given by Ans. (a) : Recommended slumps of concrete :
(a) Cr = (1.25–Le/24b) No. Type of concrete Slump
(b) Cr = (1.00–Le/48b) 1. Concrete for road construction 20 to 40 mm
(c) Cr = (1.25–Le/48b) 2. Beams and slabs 50 to 100 mm
(d) Cr = (1.5–Le/48b) 3. Mass concrete 25 to 50 mm
Ans. (c) Cr = reduction factor 4. Normal RCC works 80 to 150 mm
 L  5. Impermeable work 75 to 120 mm
Cr = 1.25 – e  or 6. Concrete to be vibrated 10 to 20 mm
 48b 
41. When shear stress exceeds the permissible limit
l
Cr = 1.25 − eff in a slab, then it is reduced by
160i min (a) Decreasing the depth
37. Minimum clear cover (in mm) to the main steel (b) Providing shear reinforcement
bar in footing column, beam and slab are (c) Using high strength steel
respectively
(d) Increasing the depth
(a) 75,40,25,15 (b) 40,75,15,25
(c) 30,20,25,15 (d) 50,40,30,20 Ans. (d) : Shear stress in slabs is controlled by
increasing the depth of slab and not shear reinforcement
Ans. (d) :
while in beams shear stirrups are provided to control
Structural Element Minimum Clear Cover
shear.
Column 40 mm
Slab 20 mm 42. In counter fort retaining walls, the main
Beam 25 mm reinforcement in the stem at support is
Footing 50 mm (a) Note provided
38. In prestressed concrete (b) Provided only on inner face
(a) Forces of tension and compression change but (c) Provided only on front face
lever are remains unchanged (d) Provided both on inner and front faces
(b) Forces of tension and compression remain Ans. (b) : In counter fort retaining walls, the main
unchanged but lever arm changes with the reinforcement in the stem at support is provided only on
moment inner face.
(c) Both forces of tension and compression and
lever arm change
(d) Both forces of tension and compression and
lever arm remain unchanged
Ans : (b) : In pre-stressed concrete forces of tension
and compression remains unchanged but lever arm
changes with moment.
39. In design of two-way slab restrained at all
edges, torsional reinforcement required is
(a) 0.75 time the area of steel provided at
midspan in the same direction
(b) 0.375 time the area of steel provided at
midspan in the same direction
(c) 0.375 time the area of steel provided at 43. Most common method of pre-stressing used for
shorter span factory production is
(d) Not required (a) Long line method
Ans. (a) : Torsional reinforcement should be 0.75 times (b) Freyssinet system
area of steel provided at mid span as per annex D clause (c) Magnet-Blaton system
D-1.8, of IS 456 : 2000.
(d) Lee-MaCall system
• Torsional reinforcement is provided in the form of a
grid or mesh both at the top and bottom of the slab. Ans. (a) : Hoyer's long line method is the system used
• 0.75 Ast if the both the meeting edges are restrained. in pretensioning and the other system like Freyssinet,
• 0.375 Ast if one of the two meeting edges, one in Gifford Udal, Lee-MaCall and Magnet-Blaton are post-
continuous and other discontinuous. tensioning system.
OPSC AEE Exam-2021 (Paper-II) 12 YCT
44. Limit state of serviceability for deflection (c) WL/10 (d) WL/12
including the effects due to creep, shrinkage Ans : (c) Purlins–
and temperature occurring after erection of • It is biaxial bending member
partition and application of finishes as • Maximum spacing between purlin </ 1.4 m
applicable to floors and roofs is restricted to
span
(a) span/150 (b) span/200 • Deflection of purlin =
(c) span/250 (d) span/350 200
wL
Ans. (d) : Limit state of serviceability for deflection • Maximum bending moment in the purlin =
including the effects due to creep, shrinkage and 10
temperature occurring after erection of partition and 49. Minimum spacing of vertical stiffeners for
application of finishes as applicable to floors and roofs plate girder is limited to (where 'd' is the
is restricted to span/350 or 20 mm. distance between flange angles)
45. For bars in tension, a standard hook has an (a) d/4 (b) d/3
anchorage value equivalent to a straight length (c) d/2 (d) d/6
of (where φ is diameter of hook) Ans. (b) : Vertical stiffeners are provided in a spacing
(a) 8 φ (b) 12 φ of 0.33d to 1.5 d, where d is the distance between the
(c) 16 φ (d) 24 φ flanges ignoring the fillets.
Ans. (c) : Anchorage values as per IS code 456 : 2000 50. As per IS : 875, for the purpose of specifying
clause number 26.2.2.1– basic wind velocity, the country has been
Bend and hooks– divided into
• The anchorage value of bend shall be taken as 4 times (a) 4 zones (b) 5 zones
and the diameter of the bar for each 450 bend subjected (c) 6 zones (d) 7 zones
to maximum of 16 times of diameter of bar. Ans. (c) : As per IS 875 (Part 3) for the purposes of
• The anchorage value of standard U-type hook shall be specifying. Basic wind velocity the country has been
equal to 16 times of diameter of bar. divided into 6 zones.
51. As per IS : 800, for compression flanges, the
outstand of flange plates should not exceed, If
''t'' is thickness of thinnest flange plate
(a) 12t (b) 16t
(c) 20t (d) 25t
Ans. (b) : IS : 800, for compression flange, the outstand
46. The channels are angles in the compression of flange plates should not exceed 16t.
chords of the steel truss girder bridges are IS : 800, for tension flange, the outstand of flange plates
turned outward in order to increase should not exceed 20t.
(a) cross-sectional area
52. Intermediate vertical stiffeners in a plate girder
(b) section moduolus need to be provided, if the depth of web exceeds
(c) torsional constant ('t' is thickness of web)
(d) radius of gyration (a) 180t (b) 85t
Ans. (d) : Channels or Angles in the compression (c) 200t (d) 250t
chords of the steel truss girder bridges are turned
outward in order to increase the radius of gyration Ans. (b) : If d < 67 ⇒ unstiffened girder can be
which reduces the slenderness ratio of the member. tw
Hence it increases resistance against buckling. designed i.e. no girder required.
47. Horizontal stiffener in a plate girder is d
provided to safeguard against • If 85 ε < < 200 ε ⇒ Vertical stiffness may be
tw
(a) shear buckling of web plate
provided (C1 & C2)
(b) compression buckling of web plate
d
(c) yielding • If 200 ε < <250 ε ⇒ Vertical stiffener along
(d) all of the above tw
Ans. (b) : Horizontal stiffeners in a plate girder is with longitudinal stiffness at 0.2 d may be provided.
provided to safeguard against compression buckling of d
web plate. Horizontal stiffeners are also called • If 250 ε < t < 345 ε ⇒ Vertical stiffeners along
w
longitudinal stiffeners.
with two longitudinal stiffener at 0.2 d and 0.5 d
48. As per IS : 800, the maximum bending moment
respectively may be provided
for design of purlins can be taken as (where W
is total distributed load including the wind load 53. The number of seismic zones in which the
on the purlins and E is centre distance of country has been divided are
support? (a) 4 (b) 5
(a) WL/6 (b) WL/8 (c) 6 (d) 7

OPSC AEE Exam-2021 (Paper-II) 13 YCT

Ans. (a) : As per IS 1893-2002 [Part-1] – India has 58. The thickness of web for unstiffened plate
been divided in to 4 seismic zones. girder with clear distance 'd' between the
Zone II– Low seismic hazard. flanges shall not be less than
Zone III – Moderate seismic hazard (a) d/200 (b) d/85
Zone IV – Severe seismic hazard. (c) d/100 (d) d/160
Zone V – Very severe seismic hazard. Ans. (b) : As per IS 800 : 1984 clause 67.3.1 the
54. The lacing bars in a steel column should be thickness of the web plate shall be not less than the
designed to resist d τva .cal d 6y
(a) Bending moment due to 2.5% of the column greater of & but not less than
816 1344
load
(b) shear force due to 2.5% of the column load d
for unstiffened web where d is depth of web.
(c) 2.5% of column load only 85
(d) Both 1 and 2 d
≤ 85 then stiffeners are not required.
Ans. (b) : As per IS 800–1984 clause 5.7.2.1. The tw
lacing of compression members shall be proportioned to 59. The effective length of a structural steel
resist a total transverse shear ‘V’ equal to at least 2.5% compression of length 'L' effectively held in
of the axial force in the member. position and restrained against rotation at one
55. Given that the effective area of a tension end but neither held in position nor restrained
member is Ae and the yield stress is σy. In order against rotation at the other end, is member
to obtain the ultimate strength of the tension (a) L
member as per the plastic design concept : Ae (b) 1.2 L
σy is to be multiplied by
(c) 1.5 L
(a) 1.3 (b) 0.95
(d) 2.0 L
(c) 0.85 (d) 0.75
Ans : (d) Effective length of prismatic compression
Ans. (c) : The maximum load capacity of tension member :
member is 0.85 Aeσy . Schematic representation Effective length
• The maximum shear capacity of beam column is 0.55 1. 2.0 L
Aesfy.
The maximum load capacity of compression member is
1.70 Aeσac. 2. 2.0 L
Where,
Ae = Effective area of the member 3. 1.0 L
σy = Yield stress
σac = Allowable compressive stress 4. 1.2 L
Aes = Effective area of the member resisting shear
56. Battens provided for a compression member 5. 0.8 L
shall be designed to carry a transverse shear
equal to
6. 0.65 L
(a) 2.5% of axial force in member
(b) 5% of axial force in member
60. Economical depth of a plate girder is given by
(c) 10% of axial force in member
(where M, σ, and τw are of usual meaning)
(d) 20% of axial force in member
Ans. (a) : As per IS 800 : 1984, battens shall be (a) ( M / σt w ) (b) 1.1 ( M / σt w )
designed to carry the bending moment and shear arising
from transverse shear force ‘V’ of 2.5% of the total
(c) 1.2 ( M / σt w ) (d) 1.3 ( M / σt w )
axial force on the whole compression member.
M
57. Shear buckling of web in a plate girder is Ans : (b) Depth for a plate girder = 1.1
P.t w
prevented by using
(a) Vertical intermediate stiffener Guide lines for selecting depth of plate is given below
(b) Horizontal stiffener along the neutral axis D 1 1
= to for girder in building
(c) Bearing stiffener L 15 25
(d) None of the above 1 1
= to for highway bridges
Ans. (a) : In plate girders, intermediate transverse 12 18
stiffeners are provided to increase buckling resistance of 1 1
= to for railways bridges
web in diagonal or shear buckling. 10 15
OPSC AEE Exam-2021 (Paper-II) 14 YCT
61. Shrinkage cracks in masonry could be (c) split tensile strength
minimized by (d) flexural tensile strength
(a) Avoiding use of rich cement Ans. (d) : Modulus of rupture –
(b) Not delaying plaster work till masonry has • It is a measure of the tensile strength it concrete beam
dried after proper curing or slabs.
(c) By using English bond of bricks • Flexural strength of concrete / bending tensile strength
(d) By providing expansion joints of concrete / modulus of rupture of concrete (fcr)
Ans. (a) : Strong cement mortar are most likely to lead f cr = 0.7 × f ck
to shrinkage cracks. Shrinkage cracks in masonry could
be minimized by avoiding use of rich cement. • The flexural strength of concrete is determined as a
modulus of rupture.
62. Cause of horizontal cracks below RCC slab on
top most storey 66. Compressive strength of brick is
(a) Deflection of slab and lifting up of edge of (a) 4.3 to 6.9 MPa (b) 2 to 3 MPa
the slab (c) 15 to 20 MPa (d) 20 to 25 MPa
(b) Arching of slabs Ans. (a) :The minimum crushing/compressive strengths
(c) Expansion of slab of burnt bricks tested flat-wise prescribed are:
(d) All of the above (i) Common building bricks 3.5 MPa,
(ii) Second class bricks 7 MPa
Ans. (a) : Horizontal cracks below slab level occurs due
(iii) First class bricks 10.5 MPa.
to deflection of a slab and lifting up to the edge of the
bearing slab. 67. Bulking of sand is maximum if moisture
content is about
• At the same time the horizontal movement in slab due
to shrinkage also affect, the horizontal cracks in walls of (a) 2% (b) 3%
the top most story below slab level. (c) 4% (d) 5%
63. Which is not correct for high alumina cement Ans. (d) : Bulking of sand or fine aggregate is the
phenomenon of increase in sand volume due to the
(a) It can withstand high temperature
increase of moisture content. The moisture content in
(b) It resist the action of acid the sand makes thin films around sand particles. Hence,
(c) The initial setting time of this cement is more each particle exerts pressure. Thus they move away
than 3 hours from each other causing increasing in volume.
(d) it can be used in mass concrete
Ans. (d) : High Alumina
Cement (IS : 6452-1989)–
• Bauxite (40%), Limestone (40%), Iron oxide (15%)
• Initial setting time- min. 3 hour 30 minute.
• Final setting time- max. 5 hour.
• It is used for refractory concrete, industries and used
widely in pre-casting and very resistance to
chemical attack.
• Particularly suitable to sea and under water work. The increase in the volume of given mass of fine
• Expansion ≤ 5 mm. aggregate caused by the presence of water is known as
• It can withstand high temperature bulking of sand. The extent of bulking depends upon
• It resist the action of acid the percentage of moisture present in sand and its
64. Pulsed Eddy current (PEC) type no destructive fineness, with ordinary sand bulking varies from 25-40
test is conducted to find percent. It increases with moisture content up to a
(a) Thickness and to detect corrosion on ferrous certain point (4–6%) reaches maximum.
material 68. for a given aggregate content, increasing water-
(b) Compressive strength of concrete used cement ratio in concrete it
(c) Wire bond with concrete (a) Decrease shrinkage
(d) Permeability of concrete (b) Increases shrinkage
Ans. (a) : Pulsed eddy current (PEC) is an advanced (c) Does not change shrinkage
electromagnetic inspection technology used in detecting (d) None of the above
flaws and corrosion in ferrous materials typically Ans. (b) : The water-cement ratio is the ratio of the
hidden under layers of coating fire proofing or weight of water to the weight of cement used in a
insulation.
concrete mix.
65. Modulus of rupture of concrete is a measure of • A lower ratio leads to higher strength and durability,
(a) compressive strength but may make the mix difficult to work with and forms.
(b) direct tensile strength
OPSC AEE Exam-2021 (Paper-II) 15 YCT
• A mix with too much water will experience more 1
Volume of cement = 1.52 × m3
shrinkage as excess water leaves, resulting in internal 7
Volume of cement = 0.22 m3
cracks and visible fractures (particularly around inside
corner), which again will reduce the final strength and 2
durability. Volume of sand = 1.54 × = 0.44m3
7
69. The approximate ratio between the strength of
4
cement concrete at 7 days and 28 days is Volume of aggregate = 1.54 × = 0.86m3  0.85 m3
(a) 3/4 (b) 2/3 7
(c) 1/2 (d) 1/3 73. The role of super plasticizer in a cement paste
Ans. (b) : is to
Age factor for low (a) disperse the particle
Age of cement (b) disperse the particle and to remove the air
strength concrete
7 days 0.65 – 0.7 bubbles
1 month 1.0 (c) Retard setting
3 months 1.1 (d) Disperse the particle and to remove the air
12 months 1.2 bubbles and to retard setting
Strength of concrete at 7 days 2 Ans. (d) : Super-plasticizer– They are admixtures that
= = 0.67
Strength of concrete at 28 days 3 work on surfactant properties, in which they disperse
70. Sum of tread and rise (in mm) for a staircase and deflocculate cement particles thus making concrete
must lie between flowing pourable and easily placed.
(a) 300 to 350 (b) 400 to 450 74. Choose the most correct statement with regard
(c) 500 to 550 (d) 600 to 650 to Queen closer
Ans. (b) : Sum of tread and rise must lie between 400 (a) Brick laid with its breadth parallel to the face
mm to 450 mm. or direction of wall
Some thumb rule of sum of tread and rise (b) Brick having the same length and depth as the
(i) 2R + T = 60 cm other bricks but half the breadth
(ii) R + T = 40 to 45 cm (c) Brick with half the width at one end and full
(iii) R × T = (400 to 450) cm width at the other
Where R = Rise in cm (d) To break the continuity of vertical joints and
T = Tread in cm to provide proper bond in brick masonry work
71. In c concrete mix the fineness modulus of Ans. (b) : When the bricks is cut along the length it is
coarse aggregate is 7.6, the fineness modulus of
called queen closer.
the aggregate is 2.8 and economical value of the
• When the bricks is cut at one end by half header and
fineness modulus of combined aggregate is 6.4,
then the proportion of fine aggregate is half stretcher, it is known as king closer.
(a) 66.67% (b) 25% 75. The type of bond provided in brick masonry
(c) 50% (d) 33.33% for carrying load is
Ans. (d) : Given, (a) English bond
Fineness modulus of coarse aggregate =7.6 (b) Single flemish bond
Fineness modulus of fine aggregate =2.8 (c) Double flemish bond
Fineness modus of combined aggregate =6.4 (d) Zigzag bond
Fcoure − Fcomb Ans. (a) For the load bearing brick wall, the English
× 100
Fcomb − Ffine bond is provided.
7.6 − 6.4
× 100 =
1.2
× 100
• English bond in brick masonry has one course of
6.4 − 2.8 3.6 stretcher only and a course of header above it. Header
33.33% are laid centered on the stretchers in course below and
72. To make one cube meter of 1 : 2 : 4 by volume each alternate row is vertically aligned.
concrete, the volume of coarse aggregate
required is
(a) 0.85 m3 (b) 0.95 m3
3
(c) 0.90 m (d) 0.75 m3
Ans. (a) : 1 m of dry concrete = 1.52 m3 of wet
3

concrete
M15 = 1:2:4 = 1 + 2 + 4 = 7
Dry volume of concrete
= 1.52
Wet volume of concrete
OPSC AEE Exam-2021 (Paper-II) 16 YCT
76. Which of the following is a weakness of bar (c) longed duration
chart (d) highest cost slope
(a) Interdependencies of activities Ans. (b) : In the time-cost optimization, using CPM
(b) Project progress method for network analysis, the crashing of the
(c) Time Uncertainties activities along the critical path is done starting with the
(d) All of the above activity having least cost slope .
Ans. (d) : Limitation of bar chart– • The objective of time cost optimisation is to
1. Lack of degree of details determine optimum project duration corresponding to
2. Does not show project progress the minimum total cost.
3. Does not show activity inter-relationship 81. There are three parallel paths in a part of a
4. Time uncertainties network between a bursting node and the next
5. It does not indicate the critical activities of the merging node with only one activity in each
project. path. The minimum number of dummy arrows
6. No cost optimization. needed will be
(a) 3 (b) 2
77. The earthwork quantities are calculated
(c) 1 (d) 0
(a) By mid-sectional method
(b) By mean sectional method Ans. (b) : Certain activities that neither consume time
nor resources, but are used simply to represent a
(c) By prismoidal method
connection or a link between events are known as
(d) All of the above methods dummies.
Ans. (d) : The earthwork quantities can be calculated by:
• Mid-section method.
• Mean-sectional method.
• Prismordial method.
78. In 1.0 cubic meter of 1 : 2 : 4 cement concrete,
how many bags of cement (approximately is
required)? 2 dummy arrows to
(a) 6.6 (b) 16.6 correctly represent the networks
(c) 26.6 (d) 36.6 82. In long wall and short wall method of
Ans. (a) : Given 1 : 2 : 4 estimation which one of the following is correct
The sum of ratio = 7 (a) short wall length in to in = centre to centre
One cu.m. volume of wet cement concrete = 1.54 m3 length - one breadth
1) Quantity of cement in one cu.m. concrete = (b) short wall length in to in = centre to centre
1.54 × 1 length + one breadth
= 0.22 m3
7 (c) Long wall length out to out = centre to centre
kg length + one breadth
2) In kg = 0.22 × 1440 cum × = 316.8 kg
cum (d) Long wall length out to out = centre to centre
316.8 length - two breadth
3) In bags = = 6.3 bags ≈ 6.6 bags Ans. (a & c) : Long wall – Short wall method–In this
50
method, the wall along the length of room is considered
79. The detailed estimate of the cost of the project to be long wall while the wall perpendicular to long
is done by
wall is said to be short wall.
(a) Unit quantity method To get the length of long wall or short wall, calculate
(b) Total quantity method first the centre line lengths of individual walls.
(c) BOQ method
83. The direct cost of a project with respect to
(d) By first two methods
normal time is
Ans. (b) : Detailed estimate is done by total quantity (a) minimum (b) maximum
method.
(c) zero (d) infinite
• Detailed estimates are prepared by carefully and
separately calculating in detail the costs of various items Ans. (a) : Direct cost:-
of the work that continue the whole project. Direct cost include the cost of materials, labour
80. In the time-cost optimization, using CPM equipment etc. direct cost of a project reduces with
method for network analysis, the crashing of time
the activities along the critical path is done 84. if the optimistic time, most likely time and
starting with the activity having pessimistic time for activity x are 10, 18 and 20
(a) shortest duration respectively and for activity b are 12, 18 and 30
(b) least cost slope respectively, then
OPSC AEE Exam-2021 (Paper-II) 17 YCT
 (a) expected time of activity x is greatest than the (c) character's profit
expected time of activity y (d) all of the above
(b) expected time of activity y is greatest than the
Ans. (d) : Rate analysis include :
expected time of activity x 1. Cost of labour wages.
(c) expected time of activity x is same as that the
2. Cost of material.
expected time of activity y 3. Overhead charges.
(d) none of the above is correct 4. Location of site.
t 0 + 4t m + t p 5. Contractor profit.
Ans. (b) : t e = 89. Cost slope
6
10 + 4 × 18 + 20 (a) (Crash cost – normal cost ) / crash time
For X → te = = 17 (b) Crash cost / (normal cost – crash time)
6
(c) (Crash cost – normal cost ) / normal time
12 + 4 × 18 + 30
For Y → te = = 19 (d) (Crash cost – normal cost / normal time –
6 crash time)
Expected time of activity Y is greater than the expected
time of activity X. Ans. (d) : Cost slope :
It is given by difference between crash cost and normal
85. Slack time refers to cost divided by difference between crash time and
(a) an activity normal time.
(b) an event
 Crash cost - Normal cost 
(c) both event and activity Cost Slope =  
(d) critical event only  Normal time - Crash time 
Ans. (b) : Slack–The difference between the two times C − Cn ∆c
Cost Slope = c =
of an activity indicates the range between which the tn − tc ∆t
occurrence time of an event can vary.
• Slack may be simply defined as the difference
between the latest allowable time and the earliest 90. Free float is mainly used to
expected time of an event. (a) identify the activities which can be delayed
without affecting the total float of preceding
86. The probability of competition of any activity
within its expected time is activity
(a) 50% (b) 81.1% (b) identify the activities which can be delayed
(c) 67% (d) 100% without affecting the total float of succeeding
activity
Ans : (a) In PERT analysis the probability of
(c) Establish priorities
completion of any activity within its expected time is
50% whereas in CPM it is 100% (d) identify the activities which can be delayed
without affecting the total float of either of
preceding or succeeding activities
Ans. (b) : Free float–Free float is that portion of
positive total float that can be used by an activity
without delaying any succeeding activity (or without
affecting) the total float of the succeeding activity.
87. The PERT calculations yield a project length of • The concept of free float is based on the possibility,
75 weeks, with a variance of 9. Within how that all the events occur at their earliest time.
many weeks would you expect the project to be 91. The line of action of the buoyancy force acts
completed with probabilility of 95%, Take through the
probability factor Z equal to 1.65 for 95% (a) Centre of gravity of the submerged body
probability 54.95 56.6 60 79.95
(b) Centroid of the volume of any floating body
(a) 54.95 (b) 56.6
(c) 60 (d) 79.95 (c) Centroid of the displaced volume of fluid
(d) Centroid of the volume of fluid vertically
Ans. (d) : Given σ2 = 9, so σ = 3, TE = 75 weeks
above the body
T – TE
Z= S ⇒ TS = TE + Zσ Ans. (c) : Buoyancy – When a body is immersed is a
σ
= 75 + 1.65 × 3 fluid either wholly as partially, it is buoyed as lifted up
T95% = 79.95 weeks. by a force, which is equal to the weight of fluid
displaced by the body.
88. In analysis of rates which is/are included from
the following • The point of application of the force of buoyancy on
(a) cost of quantities of materials the body is known as the centre of buoyancy. It is
(b) cost of labour and other miscellaneous always the centre of gravity of the volume of fluid
expenditures displaced.
OPSC AEE Exam-2021 (Paper-II) 18 YCT
92. Choose the correct statement called attention. Further, the peak of the outflow occurs
(a) Standard project flood (SPF) is always greater after the peak of inflow; the time differences between
than probable maximum flood (PMF) the two peak known as lag.
(b) PMF > SPF • The storage capacity of the reservoir and the
(c) The ctachement characteristics decides characteristic of spillways and other outlet control the
whether PMF is greater than SPF log and attenuation of an inflow hydrograph.
(d) PMF = SPF 95. In sequent peak method for calculating
Ans. (b) : Standard project flood (SPF)– reservoir capacity, which one of the following is
the correct statement.
• The flood that would result from a sever combination
(a) The difference between the first peak and the
of meteorological and hydrological factors that are
through following it is the reservoir storage
reasonably applicable to the region.
required under normal condition
Extremely rare combinations of factors are excluded. (b) Cumulative inflow volume is plotted in Y-
Probable maximum flood (DMF)– axis against time in X-axis
• The extreme flood that is physically possible in a (c) The cumulative difference of inflow and
region as a result of severemost combination, included demand is plotted in Y-axis against
rare combination of meteorological and hydrological cumulative inflow in X-axis
factors. (d) The difference in summation of trough gives
PMF > SPF storage required under normal inflows
93. What is the limitation of rational formula for Ans. (c) : Sequent peak method– The mass curve
flood peak estimation? method is widely used for the analysis of reservoirs
(a) Duration of rainfall intensity should be less capacity demand problems.
than the time of concentration • Sequent peak algorithm is particularly suitable for the
(b) Rainfall intensity must be constant over the analysis of large data with help of computer.
entire watershed during the 90% time of • Sequent peak algorithm is particularly suitable for the
rainfall duration analysis of large data with help of computer.
(c) It gives base of hydrograph but not the peak • The surplus or deficit of storage in that period is the
of hydrogen net flow volume given by
(d) Formula is application to watershed area up to Net flow volume = Inflow volume – Outflow volume
50 square kilometers • In the sequent peak algorithm a mass curve of
Ans. (d) : Rational method– cumulative net flow volume acquaint chronological
The peak value of runoff by rational formula is given time is used this curve known as residual mass curve.
by– 96. In Newton formulation the law of fluid friction
1 (a) Shear stress is proportional to shear stress
QP = CiA (b) Shear stress is inversely proportional to shear
3.6
C = Coefficient of runoff stress
(c) Shear stress is proportional to shear strain
QP = Peak discharge (m3/sec)
(d) Shear stress is proportional to rate of shear
i = Mean intensity of precipitation (mm/h) strain
A = Drainage area in km2.
Ans. (d) : Newton's law of viscosity –
• The rotional formula is found to be suitable for peak
For Newtonian Fluid shear stress (τ)
flow prediction in small catchments up to 50 km2 in
area. dv
τ∝ (velocity gradient)
• It finds considerable application in urban drainage dy
design and in the designs of small culvert and bridges. dv
τ=µ µ = dynamic viscosity
94. Why in flood routing the peak of outflow dy
hydrograph is less than the peak of inflow • For Newtonian fluid, coefficient of viscosity remain
hydrograph
constant.
(a) As the velocity of flood wave increases with Example of Newtonian fluid water, alcohol, petrol, air
time etc.
(b) due to the effect of storage and channel
friction 97. An object weight 289.9 N air and 186.9 N in
water. What is the relative density of the
(c) As the outflow hydrograph contains more material of the object?
volume of water than inflow hydrograph
(a) 2.83 (b) 2.45
(d) As the time base of the outflow hydrograph
(c) 2.15 (d) 2.00
reduces.
Ans. (b) : Owing to the storage effect, the peak of the Ans. (a) : Weight in water = 186.9 N
outflow hydrograph will be smaller than that of the Weight in air = 289.9 N
inflow hydrograph. This reduction in the peak valve is W–FB = 186.9

OPSC AEE Exam-2021 (Paper-II) 19 YCT

FB = 289.9 – 186.9 = 103 Ans. (*) : Velocity of river is given by–
V × ρ w × g = 103 v = 2gh
103 s 
Volume of body (V) = h = y  m − 1
ρw g  s 
Q W = 289.9  13.6 
Vρg = 289.9 h = 76  − 1 =957.6 mm = 0.9576 m
 1 
103 v = 2 × 9.81× 0.9576 = 4.33 m / s
× ρg = 289.9
ρw g
102. The main function of a divide wall is to
ρ (a) Control the silt entry in the canal
Specific gravity = = 2.81
ρw (b) Prevent river floods from entering the canal
98. The pressure 44.1 kPa is equivalent to (c) Separate the under sluices from weir proper
(a) 5.94 m of water (d) Provide smooth flow at sufficiently low
(b) 0.33 m of mercury velocity
(c) 154.84 kN/m2 absolute Ans. (c) : Divide wall– A divide wall is masonry or
(d) 15.84 m of water absolute concrete wall constructed at right angles to the axis of
the weir or barrage.
Ans. (b) : Given,
• It separates the weir bays the under-sluices portion.
pressure (P) = 44.1 kPa
• The top width of the divide wall is kept 1.5 to 2.5 m.
P = ρgh
103. The hydraulic mean depth laid at an
44.1×103
h= = 4.4 m of water longitudinal slope of 0.004 is 0.837. What is the
9.81× 1000 minimum size of stone that will remain at rest?
Taking mercury, then (a) 3.70 cm (b) 4.50 cm
4.49 (c) 5.30 cm (d) 6.45 cm
h= = 0.33 m of mercury
13.6 Ans. (a) : Hydraulic mean depth (R) = 0.837 m
99. Choose the correct value of friction factor (f) of Longitudinal slope (S0) = 0.004
the circular pipe for the laminar flow eight Minimum size of stone = 11×R× S0
Reynold's number 640 = 11× 0.837×0.004
(a) 0.1 (b) 0.15 = 0.0368 m
(c) 0.20 (d) 0.25 = 3.68 cm
Ans. (a) : For laminar flow– 104. The rainfall in four successive 12 hours period
64 on a catchment are 40, 80, 90 and 30 mm. If the
Friction factor (f) =
Re infiltration index for the soil is 5 mm/hr, then
the total surface run off will be
64
f= = 0.1 (a) 0 (b) 50 mm
640 (c) 120 mm (d) 180 mm
100. In supercritical open channel flow P−R
(a) The critical depth is always above normal Ans. (b) : Q =
t
depth
Only value above 5 mm/hr consider only calculation of
(b) The critical depth and normal depth merges
40 30
(c) Critical depth is always below the normal runoff (mean mm/hr and mm/hr excluded)
depth 12 12
(d) Insufficient information for any comment Q=
(80 + 90 ) − R
Ans. (a) : Super critical flow– Depth of flow less than 12 + 12
critical depth resulting from relatively steep slopes.
5=
( 80 + 90 ) − R
• Fraud number is greater than one. 24
• Flow of this type is most common is steep streams. 120 = 170 – R
101. The differential gauge attached to pitot tube R = 170 –120
shows 76 mm deflection of mercury, when the R = 50 mm
placed against the flow direction of water in the 105. A confined aquifer 2.0 km wide discharges
river. What is the value of velocity of river 0.06m3/day/km to a dry river in the month of
water? April. What is the value of transmissivity of
(a) 3.444 m/s (b) 4.17 m/s aquifer, if the slope of the piezometric surface
(c) 2.87 m/s (d) 4.19 m/s is 0.375 m/km
OPSC AEE Exam-2021 (Paper-II) 20 YCT
 (a) 0.08 m2/day (b) 0.16 m2/day 110. Select the correct statement
2
(c) 0.32 m /day (d) 0.04 m2/day (a) 5 day BOD is the ultimate BOD
Ans. (b) : Q = kiA (b) 5 day BOD is greater than 4 day BOD
Q = ki (B × L) keeping other conditions same
(c) BOD does not depend on time
Q
  = kiL (d) 5 day BOD is less than 4 day BOD keeping
B other condition same
0.06 = k × 0.375 × 2 Ans. (b) : BOD (Biochemical oxygen demand)–
k = 0.08 m/day • It is the amount of oxygen required for the
Transmissivity decomposition if biodegradable organic matter present
T= kB in the system.
T = 0.08 × 2 • BOD during 5 days at 200C is taken as standard BOD
T = 0.16 m2/day and is approximately 0.8% of the ultimate BOD.
106. The non-scouring limiting velocity (in m/s) for • 5 Day BOD is greater than 4 day BOD is keeping
cement concrete sewers is other conditions same.
(a) 4.5 to 5.5 (b) 3.5 to 4.5 111. The working condition of imhoff tanks are
(c) 3.0 to 4.0 (d) 2.5 to 3.0 (a) Aerobic only
Ans. (d) : Non-scouring velocity– The maximum (b) Anaerobic only
permissible velocity at which no such scouring action (c) Aerobic in lower compartment and aerobic in
will occur is known as non-scouring velocity and it will lower
mainly depend on the material used in the constructed (d) Anaerobic in lower compartment and aerobic
of sewers. in upper compartment
• The non scouring velocity for cement concrete sewer Ans : (d) An Imhoff tank is an improvement over septic
lies between 2.5 to 3.0 m/sec. tank, in which the incoming sewage is not allowed to
107. The dissolved oxygen level in natural get mixed up with the sludge produced, and the
unpolluted waters at normal temperature is outgoing effluent is to allowed to carry with it large
found to be of the order of amount of organic load as in the case of a septic tank.
(a) 1 mg/litre (b) 10 mg/litre An Imhoff tank is, in fact, at two storeyed tank. They
(c) 100 mg/litre (d) 1000 mg/litre are sometimes also known as Two-storey Digestion
Ans. (b) : It is the maximum quantity of DO that can tanks.
remain in water at a particular temperature. Hence at 112. Sludge volume index is defined as the ratio of
normal temperature, DO content is nearly 10 mg/l d (a) Percentage of sludge by volume to percentage
108. For a given discharge, the efficiency of of suspended solids by weight
sedimentation tank can be increased by (b) Percentage of sludge by volume of percentage
(a) Decreasing surface area of the tank total solids by weights
(b) Increasing the depth of the tank (c) Percentage of suspended solids by weight to
(c) Decreasing the design of the tank percentage of sludge by volume
(d) Increasing surface area of the tank (d) Percentage of total solids by weight to
Ans. (d) : The efficiency of the sedimentation tank percentage of sludge by volume
increase is the verflow rate reduces (more time available Ans. (a) : Sludge volume index 'SVI' =
Q Percentage of sludge by volume
to particles for settle) overflow rate (V3) =
B× L percentage of suspended solids by weight
• For the equation, it is clear that if the surface area (B× • Sludge volume index is the volume occupied in mL
L) of the tank increase the overflow rate reduces, and by one gm of solids in mixed liquor after setting for 30
efficiency increases for a given discharge. minuts.
109. The process in which the chlorination is done 113. In the two pipe system of house plumbing, the
beyond the break point is known as pipes required are
(a) Pre chlorination (a) One soil pipe, one waste pipe and one vent
(b) Post chlorination pipe
(c) Break point chlorination (b) One soil pipe, two waste pipe and one vent
(d) Super chlorination pipe
Ans. (d) : Super chlorination– When excess chlorine (c) One soil pipe, one waste pipe and two vent
(5 to 15 mg/l) is added during an epidemic such that it pipe
gives a residual at 1 to 2 mg/l beyond break point is (d) Two soil pipe, one waste pipe and one vent
called super chlorination. pipe
• It is most commonly added at the end of filtration.
OPSC AEE Exam-2021 (Paper-II) 21 YCT
Ans. (c) : Two pipe system– This is the most common (a) 3% (b) 5%
system used in India. (c) 75% (d) 97.5%
• This method provided an ideal solution, where it is not Ans. (c) : V1 (100 − P ) = V (100 − P )
possible to fix the fixtures closely.
V1 100 − 99 1
• One pipe collect the foul soil and water closet wastes = = = 0.25
and the second pipe collects the water from kitchen, V 100 − 96 4
bathrooms, house washings etc. V − V1
Volume reduction (in %) = × 100
• The soil pipes are directly connected to the V
manhole/drain, where as the waste pipes are connected  V
through fully ventilated gully trap. = 1 − 1  × 100 ⇒ (1 − 0.25 ) ×100
 V
114. Select the primary air pollutants among the
= 75%
following :
(a) Sulpher dioxide and nitrogen oxides 118. Areosol is
(b) Ozone and carbon monoxide (a) Carbon particles of microscopic size
(c) Sulpher dioxide and ozone (b) Dispersion of small solids or liquid particles
(d) Nitrigen and ozone in gaseous media
(c) Finely divided particles of ash
Ans. (a) : Primary pollutant– A primary pollutant is
an air pollutant emitted directly from a source. (d) Diffused liquid particles
• Ex Carbon Monoxide (CO), Sulpher dioxide and Ans. (b) : Aerosols– Aerosols refer to the depression of
nitrogen oxides etc. solid or liquid particles of microscopic size in gaseous
media, such as dust, smoke, of mist.
115. When Environmental Lapse Rate (ELR) is
more than Adiabatic Lapse Rate (ALR), then 119. A city supply of 15000 cubic meter of water per
the environment is said to be day is treated with a chlorine dosages of 0.5
(a) Stable (b) Unstable ppm. For this purpose, the requirement of 25%
bleaching powder per day would be
(c) Neutral (d) None of the above
(a) 300 kg (b) 75 kg
Ans. (b) : When Environmental Lapse Rate (ELR) is (c) 30 kg (d) 7.5 kg
more than adiabatic Lapse Rate (ALR), then the
environment is said to be unstable- Ans. (c) :
E.L.R > A.L.R = Unstable Chlorine required for the city = Does × Discharge
E.L.R < A.L.R = Stable 15000 × 103
= 0.5 × = 7.5 kg / day
E.L.R = A.L.R = Neutral 106
Lapse rate– In the troposphere, the temperature of the 7.5
ambient (surrounding) air normally decreases with Bleaching powder required =
0.25
increases in the altitude (height). This rate of change of = 30 kg/day
temperature is called lapse rate.
120. The detention period for oxidation ponds are
116. Two samples of water X and Y have pH values usually kept as
of 4.4 and 6.4 respectively. How many times (a) 4 to 8 hrs (b) 24 hrs
more acidic sample X is than sample Y? (c) 10 to 15 days (d) 3 months
(a) 0 (b) 50 Ans. (c) : Oxidation ponds–
(c) 100 (d) 1000 (i) Depth = 1 m to 1.8 m
Ans. (c) : pH = –log [H+] Detention period = 10 to 15 day.
[H+] = Concentration in moles/litre Organic loading = 150 to 300 kg/hour/day
For sample A 121. The type of footing which is used to transmit
4.4 = –log [H+] heavy loads through steel column is
[H+] = 3.98 × 10–5 moles/litre (a) Raft foundation
For sample 'B'– (b) Grillage foundation
6.4 = –log [H+] (c) Well foundation
[H+] = 3.98 × 10–7 moles/litre (d) Isolated foundation
 H +  of sample A 3.98 × 10−5 Ans. (b) : Grillage foundation – This type of
= −7
=100 foundation is foundation is found suitable when load
 H  of sample B 3.98 × 10
+
transmitted by a column of a wall is exceptionally
100 times more acidic sample × is than sample y heavy and bearing capacity of the soil is very low.
137. Fresh sludge has moisture content 99% and Raft foundation– It is used in those place where the
after thickening, its moisture content reduces heavy concentrated loads of the structures are to be
to 96%. The reduction in volume of sludge is distributed over the whole floor area.
OPSC AEE Exam-2021 (Paper-II) 22 YCT
122. Under a given load, a clay layer attains 30% Ans. (d) : Coefficient of compression/compression
degree of consolidation in 1000 days. The taken index (Cc)–
by the same clay layer to attain 60% degree of
consolidation will be (in days)
(a) 1600 (b) 400
(c) 800 (d) 200
π
Ans. (b) : Time factor, 'Tv' = × u 2 ( IS 4 ≤ 60% )
4
Where, u = degree of consolidation, ∆e
Cc =
Cv = Coefficient of consolidation σ
d = length of drainage log 2
σ1
t = time taken for consolidation
Equal (i) and (ii) e1 − e 2 e1 − e 2
Cc = Cc =
π 2 Cv t log σ
10 2 − log σ
10 1 σ 
4 = 2 log10  2 
4 d  σ1 
4 Cv t
4=
π d2 125. The soils most susceptible to liquefaction are
For a given Cv and d (a) Saturated gavels and cobbles
u∝ t (b) Saturated clays of uniform size
u1 t 30 100 (c) Saturated dense sands
= 1 ⇒ = (d) Saturated fine and medium sands of uniform
u2 t2 60 t2
particle size
t2 = 400 day Ans. (d) : Liquefaction is a phenomenon in which the
123. At a site having a deposit of dry sandy soil, an strength and stiffness of a soil is reduced by earthquake
average soil of standard penetration resistance shaking or other rapid loading. The phenomenon is
N equal to 6 was recorded. The compactness of most often observed is saturated, loose (low density or
the soil deposit can be described as un-compacted), sandy soils. This is because a loose
(a) Loose (b) Dense sand has a tendency to compress when a load is applied.
(c) Medium (d) Very loose
126. Contact pressure beneath a rigid footing
Ans. (a) : Standard Penetration Resistance – resting on cohesive soil is
(IS-2131)
(a) More at edge compared to middle
• Used for determining relative density/ density index
(b) Uniform throughout
• Angle of shearing resistance
(c) Less at edges compared to middle
• Unconfined compressive strength
(d) Zero at edges and maximum at middle
• Pile load capacity
Ans. (a) : Rigid footing on clayey soil has constant
• Ultimate bearing capacity on the basis of shear centre
settlement and variable pressure. Its pressure is more at
• Allowable bearing pressures on the basis of settlement edge and lower at center.
criteria
• Test suitable for medium and dense sand
SPT Value Relative Density of φ( in º)
consistency
0-4 Very loose <30
5-10 Loose 30º-35º
11-30 Medium dense 35-38
31-50 Dense 38-41º 127. For a base failure, the depth factor Df is
>50 Very Dense 41-44º (a) Df = 1.0
* Above table confirms to IS : 2911-1-2 (b) Df > 1.0
124. The slope of the e-log p curve for a soil mass (c) Df < 1.0
gives (d) 0
(a) Coefficient of consolidation, Cv Ans. (b) : Base failure– In this case the failure surface
(b) Coefficient of permeability, k passes below the toe. This generally occurs when the
(c) Coefficient of volume compressibility, mv soil below the toe is relatively soft and weak in
(d) Compressive index, Cc comparison to soil mass above the toe.
OPSC AEE Exam-2021 (Paper-II) 23 YCT
 4 N.C & clay 0-5 to 0.6
5 O.C clay 1.0-4.0
129. Select the incorrect statement
(a) Unconfined compression test can be carried
out on all types of soils
(b) Stress distribution on the failure plane in the
case of triaxial compression test is uniform
(Depth factor >1) (c) In a direct shear box test, the plane of shear
Toe failure– In this case the failure surface passes failure is predetermined
through the toe. This occurs when the slope is steep and (d) Better control is achieved on the drainage of
soil mass is homogenous above and below the toe. the soil in a triaxial compression test
Ans. (a) : Unconfined compression test can be carried
out on cohesive soils only
• It is an special case of triaxial test with σ3=0
•No confining cell pressure is used
• Load is rapidally applied, Hence it is an untrained test
• Stress distribution on the failure plane in the case of
tri-axial compression test is uniform.
(Depth factor =1) • In direct shear box test the plane of shear failure is
predetermined
Face failure– This type of failure occurs when the
slope angle is large and when the soil at the toe position • Better control is achieved on the drainage of the soil
is strong. in a tri-axial compression test

(Depth factor <1)

128. Cohesive soils are
(a) Good for backfill because of large lateral
pressure
(b) Good for backfill because of high lateral
pressure
(c) Poor for granular in nature and drains water
quickly
(d) Poor for backfill because of large lateral
pressure
Ans. (d) : Cohesive soils are- poor for backfill due to
large lateral pressure on wall and their swelling
characteristics.

130. The hydraulic head that would produce a quick

condition in a sand stratum of thickness 1.5 m,
specific gravity 2.67 and void ratio 0.67 is equal to
(a) 1.5 m (b) 1.0 m
(c) 2.3 m (d) 1.75 m
 G −1 
Ans. (a) : Hydraulic head 'H' = t  
 1+ e 
As per Rankine's theory value of ko for various soil
 2.67 − 1 
are as tabulated below H = 1.5  
 1 + 0.62 
Soil type Ko
1 For dense sand 0.9-0.5  1.67 
H = 1.5  
2 Loose sand 0.45-0.5  1.67 
3 Mechanically compacted sand 0.8-1.0 H = 1.5 m
OPSC AEE Exam-2021 (Paper-II) 24 YCT
131. In a deposit of normally consolidated clay Ans. (b) : In the under reamed pile, the ratio of double
(a) Effective stress and undrained strength under reamed pile to the single UR pile is 1.5.
increase with depth but water content It means if the number of bulbs is increased from one
decreases with depth to two, the carrying capacity increases by about 50%.
(b) Effective stress, water content and underained 135. If the proportion of soil passing 75 micron sieve
strength decrease with depth is 50% and the liquid limit and plastic limit are
(c) Effective stress and water content increases 40% and 20% respectively, then the group
with depth but underained strength decrease index of the soil is
with depth (a) 6.5 (b) 65
(d) Effective stress and undrained strength (c) 38 (d) 3.8
decrease with depth but water content
Ans. (a) : Group index, GI = 0.2 a + 0.005 ac + 0.01 bd
increases with depth
a = % passing 75 mm sieve greater than 35 but not
Ans. (a) : In a deposit of normally consolidated clay exceeding 75 (between 0 to 40).
Effective stress and undrained strength increase with
b = % passing 75 mm sieve greater than 15 but not
depth but water content decreases with depth. exceeding 55 (between 0 to 40).
132. Coefficient of consolidation for clays normally c = liquid limit greater than 40 not exceeding 60
(a) First increases and then decreases with (between 0 to 20)
increase in liquid limit d = Ip greater than 10 and not exceeding 30 (between 0
(b) Increases with increase in liquid limit to 20)
(c) Remains constant at all liquid limit a = 50-35 = 15 < 40
(d) Decrease with increase in liquid limit b = 50-15 = 35 < 40
Ans. (d) : We know that – c = 40-40 = 0
Cc = 0.007 (wL-10) d = 20-10 = 10 < 20
Cc ∝ wL G.I. = 0.2 × 15 + 0.005 × 40 × 0 + 0.01 × 35 × 10
Now, = 3 + 0 + 3.5 = 6.5
∆H
∝ Cc 136. The minimum design speed for hairpin bends
H in hills roads is taken as
∆H ∆e (a) 10 kmph (b) 20 kmph
= ⇒ C c ∝ ∆e
H 1 + e0 (c) 30 kmph (d) 40 kmph
Now Ans : (b)
∆e (i) The minimum design speed for hair pin bends in hill
av = ⇒ a v ∝ Cc roads is taken as 20 kmph.
1 + e0
(ii) Minimum radius of the inner curve = 14 m.
av (iii) Minimum length of transition = 15 m
mv = ⇒ m v ∝ Cc
1 + e0 (iv) Super elevation in circulation portion of the curve =
Now, we know 1 in 10.
K = Cv mv γw (v) Minimum width of carriageway
(a) At the apex of the curve are 11.5 and 9.0 m for two
1
Cv ∝ lane and single lane pavement of National highway &
mv State highway.
1 1 137. Expansion joints in cement concrete pavements
Cv ∝ ⇒ Cv ∝ are provided at an interval of
Cc wL
(a) 18 m to 21 m (b) 25 m to 30 m
It means by increasing the liquid limit the coefficient of (c) 10 m to 15 m (d) 30 m to 40 m
consolidation decreases.
Ans. (b) : Cement concrete pavements, expansion
133. For a damped vibrating system with single joints should be at an interval of 25m to 30m
degree of freedom resonance occurs at a As per IS 6509 : 1485 clause 5.3.1.2
frequency ratio of
Type Spacing for Spacing for
(a) 0 (b) 1 Width of slab
of expansion contraction
(c) Less than 1 (d) Greater than 1 slab (m)
joint joint
Ans. (b) : For a damped vibrating system with single 0.25 51 m 17 m
degree of freedom resonance occurs at a frequency ratio
of 1. 0.20 45 m 14 m
RCC
0.15 36 m 13 m
134. The ratio of bearing capacity of double under
reamed pile to that of single under reamed pile 0.15 30 m 7.5 m
is nearly 0.20 & above 36 m 4.5 m
(a) 2 (b) 1.5 PCC 0.15 27 m 4.5 m
(c) 1.2 (d) 1.7 0.10 27 m 4.5 m
OPSC AEE Exam-2021 (Paper-II) 25 YCT
 As per IRC the maximum specify speeding of 0.20 45 m 14 m
contraction joints in rigid joints in rigid pavements is 0.15 36 m 13 m
4.5 m. 0.15 30 m 7.5 m
138. For sandy soils the most common method of 0.20 & above 36 m 4.5 m
stabilization is
(a) Soil lime stabilization PCC 0.15 27 m 4.5 m
(b) Soil bitumen stabilization 0.10 27 m 4.5 m
(c) Soil cement stabilization As per IRC the maximum specify speeding of
(d) Mechanical stabilization contraction joints in rigid joints in rigid pavements is
4.5 m.
Ans. (c) : For Sandy soil stabilization method is most
common method is soil cement stablization. 142. Select the correct statement
• Quantity of cement depend on quality and quantity of (a) Minimum and maximum values of group
fines in sandy soils and final compacted density. index can be zero and 20 respectively
• It ranges between 5 to 12% of cement by weight. (b) More the value of CBR, greater thickness of
pavement will be required
• Quantity of cement in stabilization increases as soil
plasticity increases. (c) More the value of group index, less thickness
of pavement will be required
139. For the construction of water bound macadam
(d) All of the above
roads, the correct sequence of operations after
spreading coarse aggregate is ODISHA PSC 24.08.2021
(a) Dry rolling, wet rolling application of Ans. (a) : Design of pavement thickness–
screening and application of filter Group index method– This method is based on index
(b) Dry rolling, application of filter, wet rolling properly of soil index properties are those which one
and application of screening used only for classification of soil such as liquid limit,
(c) Dry rolling, application of screening, wet plastic limits plasticity index. etc.
rolling and application of filter • The GI method of pavement design is essential an
(d) Dry rolling, application of screening and empirical method based on physical properties of the
application of filter and wet rolling subgrade soil.
Ans. (c) : For construction of water bound macadam • This method does not consider the strength
road we follow– characteristics of the subgrade soil and therefore is open
dry rolling to question regarding the reliability of the design based
↓ on the index properties of the soil only.
application of screening Note :
↓ GI value lies between 0 to 20.
wet rolling Higher the GI value, poor the soil, hence higher
↓ thickness of pavement required.
application of filler
Total thickness of pavement depends upon GI Value
140. Traffic flow is calculated by only whereas thickness of base and surface course
(a) Multiplying measured depends upon GI value and traffic volume.
(b) Multiplying measured traffic flow rate with
143. Bitumen of grade 80/100 means
road density
(c) Multiplying road density with measured (a) Its penetration value is 8 cm to 10 cm
travel speed (b) Its penetration value is 8 mm to 10 mm
(d) Multiplying measured travel speed with road (c) Its penetration value is 8 cm
density (d) Its penetration value is 0 mm
Ans. (a) : Traffic flow is the product of measured Ans. (b) : Penetration test is the most commonly
density and travel speed. adopted to determine the grade of the bitumen in terms
Traffic flow = measured density × travel speed of its hardness because of its simplicity. Softer the
141. The maximum spacing of contraction joints in bitumen, the greater will be the penetration values.
rigid joints in rigid pavements is • 80/100 bitumen denotes that the penetration values of
(a) 5.5 m (b) 4.5 m the binder ranges between 8 to 10 mm.
(c) 3.5 m (d) 2.5 m 144. The maximum design gradient for vertical
Ans. (b) : As per IS 6509 : 1485 clause 5.3.1.2 profile of a road is
Type Spacing for Spacing for (a) Rulling gradient
Width of slab (b) Limiting gradient
of expansion contraction
(m)
slab joint joint (c) Minimum gradient
RCC 0.25 51 m 17 m (d) Exceptional gradient
OPSC AEE Exam-2021 (Paper-II) 26 YCT
Ans. (a): Ruling gradient: It is the maximum gradient Ans. (b) :
within which the designer attempts to design the vertical
profile of a road. It is also known as design gradient.
Limiting gradient– The gradient steeper than the ruling
gradient, which may be used for a limited road length, is
called limiting gradient or maximum gradient.
Minimum gradient– The minimum desirable slope
essential for effective drainage of rainwater from the
purpose is 0.5% if the side drains are lined and 1% if
the side drains are unlined.
Exceptional gradient- The gradient steeper than the Interior angle N = 60030' + 1800–1220
limiting gradient which may be used in short length of = 118030'
road, only in an extraordinary situation is called 149. A 3000 m long line lying at an elevation of 450
exceptional gradient. m measures 10 cm on a vertical photograph.
145. The ruling design speed on a National highway The focal length of the camera is 21 cm. The
in plain terrain as per IRC recommendations is scale of the photograph for the area having an
(a) 60 kmph (b) 45 kmph elevation of 1000 m will be
(c) 120 kmph (d) 100 kmph (a) 1 : 25008
Ans. (d) : Design speed for various classes of road in (b) 1 : 27381
plan terrain is as follows - (c) 1 : 37231
(i) NH & SH → 100 kmph
(d) 1 : 22222
(ii) MDR → 80 kmph
(iii) ODR → 65 kmph f
Ans. (b) : Scale S =
(iv) VR → 50 kmph H−n
(v) Expressways → 120 kmph For Ist case –
146. In triangulation, the best shape of the triangle 10 1
Scale = =
would be 3000 × 100 30000
(a) Equilateral 1 0.21
Now =
(b) Right angled isocreles 30000 H − 450
(c) Isoscreles with two base angles of 65014' Flying height, H = 6740 m
(d) Isosceles with two base angles of 56014' For 2nd case–
Ans. (d) : The shape of the triangle in which any error f
in angular measurements has a minimum effect upto the Scale =
lengths of the computed sides is know of well H−n
conditioned triangle. Hence the best shape of a triangle 0.25
S= = 1 in 27381
is an isosceles triangle whose base angles are 56º14'. 6740 − 1000
147. The length of transition curve for a circular 150. Which of the following is not a part of a total
curve of radius 300 m and for a design speed of station
15 m/s, when the rate of change of centrifugal (a) Electronic transit theodolite
acceleration is 0.3 m/s2, is (b) Electronic distance bar
(a) 60.53 m (b) 45.25 m
(c) Microprocessor
(c) 30.75 m (d) 37.5 m
(d) Subtensebar
Ans. (d) : Radius of curve (R) = 300 m
Design speed (V) = 15 m/s Ans. (d) : Total station– The total station is a
Rate of change acceleration (c) = 0.3 m/s2 combination of an electronic theodolite and an
electronic distance meter (EDM)
V3
Length of transition curve (LS) = Parts of total stations–
CR
• Electronic transit theodolite
153 • Electronic distance bar
= =37.5 m
0.3 × 300 • Microprocessor
148. If the bearing of a line MN is 60030' and that of
• Data input/output connector
NO is 1220 of a closed traverse MNOPQ, then
the measures of interior angle N is • Instrument height mark
(a) 154 0 0
(b) 118 30' • Display
0
(c) 122 00' 0
(d) 240 30' • Operation panel.
OPSC AEE Exam-2021 (Paper-II) 27 YCT
 Odisha Public Service Commission
Assistant Soil Conservation Officer (ASCO)
Exam- 2021 (Paper-I)
1. What is the contribution of agriculture in the Ans. (c):
national GDP of India? Masanobu Fukuoka–Japanese former- Natural farming
(a) 10% (b) 15% Albert Howard–Founder of organic farming movement
(c) 30% (d) 50% Lady Eva Balfour–Farmer-Organic farming 1st time
Ans. (*) : According to economic survey 2021-22 the Charles Walton–Patent holder for RFID device.
contribution of agriculture in the national GDP of India 7. The Kisan Credit Card (KCC) Scheme was
is about to 20.2% in 2020-21 and 18.8% in 2021-2022. launched in the year :
2. The Major emission of CO2 to the atmosphere (a) 1971 (b) 1956
is from : (c) 1998 (d) 1988
(a) Biosphere (b) Hydrosphere Ans. (c) : The Kisan Credit Card scheme was launched
(c) Lithosphere (d) Stratosphere in 1998 for helping farmers avail short term loan and a
Ans. (a) : The major emission of CO2 to the atmosphere credit limit to purchase equipments for agriculture.
is from biosphere at which lithosphere, hydrosphere and These are of two types–
atmosphere meets. (i) Cash credit
3. The total Geographical area of India is: (ii) Term credit
(a) 144 mha (b) 197 mha KCC is launched on the recommendation of R.V. Gupta
(c) 276 mha (d) 329 mha committee.
Ans. (d) : The total geographical area of India is 3.28 8. Which among the following gas destroys the
million sq. kilometer chlorophyll content in plants ?
Q 1 hac = 104 m2 (a) CO2 (b) CO
∴ 1 km2 = 100 hac (c) SO2 (d) O3
6
= 3.28 × 100 × 10 hac Ans. (c) : Sulphur dioxide (SO2) is a major air pollutant
= 328 mha and plants are mainly sensitive to SO2 because the
4. The country having largest irrigation area in process of photosynthesis is mainly affected by the
the world : increased concentration of sulphur dioxide and leaves
become yellow.
(a) China (b) India
(c) Egypt (d) USA 9. What is green GDP?
(a) Net value of GOP after discounting the cost
Ans. (a) : China having the largest irrigation area
incurred due to man-made destruction
country in the world.
(b) Net value of GDP after discounting the
1. China (≈ 70 Mha)
money earned by NRl
2. India (≈ 67 Mha)
(c) Net value of GDP after discounting the cost-
3. USA (≈ 22 Mha) incurred due to natural calamities
5. Total horticulture crop production of India in (d) Net value of GDP after discounting the cost
2020-21 : incurred due to environmental degradation
(a) 98 mt (b) 185 mt
Ans. (d) : Green GDP–Net value of gross domestic
(c) 292 mt (d) 313 mt product after discounting the cost incurred due to
Ans. (d) : The horticulture crops (i.e. fruits, vegetable, environmental degradation.
medical aromatic plants etc.) production of India in Green GDP = GDP – Environment degradation cost
2020-21 is about to 310 million tonne and 296 million
tonne in 2019-2020. 10. Flow of energy in the ecosystem is :
6. Which among the following input can be used (a) Unidirectional
in organic farming ? (b) Bidirectional
(a) Masanobu Fukuoka (b) Albert Howard (c) Circular
(c) Lady Eva Balfour (d) Charles Walton (d) Both uni-and bidirectional

OPSC ASCO Exam. 2021 (Paper-I) 28 YCT

Ans. (a): The flow of energy in an ecosystem is always Ans. (b) : The ozone layer exists in the lower regions of
unidirectional because some energy is lost in form of the stratosphere. The ozone has a protective role in the
heat when moving from one trophic level to the next for upper atmosphere to prevent the UV rays from the
the maintenance of the homeostasis of an organism. reaching the earth surface.
11. Pradhan Mantri Krishi Sinchayee Yojana 17. Which state is having the highest cropping
(PMKSY) launched in the year : intensity in India?
(a) Launched in lst July. 2015 (a) Punjab (b) Haryana
(b) Launched in 2nd February, 2006 (c) UP (d) WB
(c) Launched in 21st January 2004 Ans. (a) : Punjab have the highest cropping intensity in
(d) Launched in 2nd October, 1953 India.
Ans. (a) : Pradhan Mantri Krishi Sinchayee Yojana To complete the agricultural demand of the country can
(PMKSY) launched on 1st July 2015 with the motto of be done by expanding the cultivation area or
"Har Khet Ko Pani". It is being implemented to expand intensifying cropping over existing area.
cultivated area with assured irrigation reduce wastage of Gross cropped area
water and improve water use efficiency. Cropping Intensity = × 100
Net sown area
12. Who is known as father Of Indian Ecology ? Thus, the cropping intensity means the portion of the
(a) R. Mishra (b) B. Sahani net area which is being cropped.
(c) R. Carson (d) Eamst Haeckel
18. Which gas is likely to increase by afforestation?
Ans. (a) : Prof. Ramdeo Mishra is known as father of (a) N2 (b) O2
Indian Ecology. He laid the foundation of ecology and (c) CO2 (d) N2O
environmental science in the country.
Ans. (b) : Afforestation is the process of creating new
Birbal Sahani–Paleobotanist
forests which play an important role in maintaining
Rachel Carson–American marine biologist proper balanced environment. Afforestation increases
Earnest Haeckel–German zoologist the % of oxygen and reduces the % of carbon dioxides.
13. What is the Percentage share of export earning Note–Deforestation increases the % of CO2.
from Agricultural Export ? 19. Seed rate of Papaya for one hectare is:
(a) 10% (b) 25% (a) 150 – 200 gm (b) 250 – 450 gm
(c) 40% (d) 55% (c) 450 – 700 gm (d) 1000 – 1200 gm
Ans. (b) : The share of Agriculture Export in total Ans. (b) : Papaya is a tropical fruit grows well in the
earning from export in 2020-21 is about to 25%. mild sub-tropical regions of the country. Sandy loam
14. Which is not true for ecosystem? soil is ideal for cultivation of papaya. It requires 250-
(a) Energy flow 450 gm seed for every hectare of cultivation area for
(b) Nutrient cycling proper growth.
(c) Interacting components Seed rate = (250-450 g)/ha
(d) Static 20. What is the contribution of agriculture in
Ans. (d) : The ecosystem consists of energy flow, employment generation of India?
nutrient cycling interacting components etc. It is a (a) 15% (b) 35%
dynamic in nature so that static is not a nature of (c) 45% (d) 65%
ecosystem. Ans. (c) : According to the Economics times of India
15. Highest nutry-cereal producing country in the the contribution of agriculture in employment
world is : generation in India is about to 45% in 2019-20 and
(a) USA (b) China approximately 35% in 2017-18.
(c) Ethiopia (d) India 21. Which one among the following is the biotic
Ans. (b) : Highest nutry-cereal producing country in the component of an ecosystem?
world. (a) Producer (b) Consumer
(i) China (c) decomposer (d) All of these
(ii) USA Ans. (d) : Biotic components are made up of organisms,
(iii) India which are living and dead both.
16. The protective layer of Ozone is located in Components–
which of the following atmospheric layer? (i) Autotrophs/producers
(a) Troposphere (b) Stratosphere (ii) Heterotrops/consumers
(c) Mesosphere (d) Thermosphere (iii) Decomposers

OPSC ASCO Exam. 2021 (Paper-I) 29 YCT

22. What would happen to the temperature of the 27. The Chipko Movement started by Sunderlan
Earth if there were no atmosphere Bahuguna in 1973 to conserve :
surrounded? (a) Soil (b) Water
(a) No effect (c) Forest (d) Endangered animals
(b) Go on increasing Ans. (c) : The Chipko movement started by Sunderlal
(c) Go on decreasing Bahuguna in 1973 to conserve the forest. It is a
(d) Increasing in day and decreasing in night time nonviolent social and ecological movement by rural
Ans. (d) : If there were no atmosphere around the earth, villagers particularly women in India.
then the temperature of the earth will increase during 28. Agriculture price support policy of the
day and decreasing during night (like moon) because air Government mainly aims to benefit:
is a ban conductor of heat and it help in temperature (a) Farmers and consumers
control of surrounding. (b) Middleman
23. Which among the following is a Non-renewable (c) Industrialist
Natural Resource ? (d) Government
(a) Soil (b) Water Ans. (a) : Agriculture price support policy of the
(c) Coal (d) Wildlife government mainly aims to benefit farmers and
Ans. (c) : (A) Non-renewable Natural Resource. consumers.
(i) Coal Objectives:
(ii) Petroleums • Protecting farmers interests
(iii) Bio gases/fuel gases etc. • Maintaining a reasonable price for agricultural
(B) Renewable Resources products
(i) Solar Energy • Increasing agricultural production
(ii) Wind 29. The amount of energy that transfer from
(iii) Water etc. producer to primary consumer in food chain is:
24. Total food grain production of India in 2020- (a) 10% (b) 30%
21: (c) 50% (d) 90%
(a) 230 mt (b) 281 mt Ans. (c) : The amount of energy that transfer from
(c) 292 mt (d) 313 mt producer to primary consumer in food chain is about to
Ans. (d) : The total food grain production in India in 50%. The energy flow in ecosystem is one of the major
2020-21 is about to (30 g) million metric tonnes and it factors that support the survival of many organisms.
is assumed that the growth in further years it will (2- 30. The new Agriculture Policy was established in :
3%) or more than previous. (a) 2010 (b) 2005
25. Which among the following are the components (c) 2000 (d) 2008
of Liquified Petroleum Gas? Ans. (c) : The GOI announced a new agriculture policy
(a) Methane and Ethane on July 28, 2000 to promote economically viable,
(b) Methane and propane environmentally non-degrading and acceptable use of
(c) Propane and Ethane natural resources for sustainable development of
(d) Propane and Butane agriculture.
Ans. (d) : The main components of liquefied petroleum 31. Ten Ecosystem was coined by :
gas are propane, butane, propylene, bulylene and (a) AG Tansley (b) EP Odom
isobutane. It is a highly flammable in nature which is (c) Charles Elton (d) Lindeman
used as a fuel in household cooking applications. Ans. (a) : Ecosystem–The community or group of
26. "More crop per drop" is the Slogan for which living organisms that live in and interact with each in a
among the following scheme ? specific environment. The term ecosystem was coined
(a) Soil Health card (b) PMKSY by A.G. tansley in 1935.
(c) PKW (d) PMFBY 32. Amritsagar is important cultivar of :
Ans. (b) : "Per drop more crop" It is a centrally (a) Beal (b) Mango
sponsored micro irrigation scheme administered by the (c) Banana (d) Guava
department of agriculture, cooperation and farmers Ans. (c) : Amritsagar Banana–They are known for
welfare on 1st July 2015 under the Pradhan Mantri their medium size and bright yellow colour. It is a good
Krishi Sinchayee Yojana (PMKSY). source of potassium, vitamin C; and fiber and also
Soil health card– 19th Feb 2015 contain small amount of protein, magnesium and
Pradhan Mantri Fasal Bima Yojana– 18th Feb 2016 vitamin B6. It is widely available in Bangladesh.

OPSC ASCO Exam. 2021 (Paper-I) 30 YCT

33. Out of total 35 global biodiversity hotspots, (i) Zero/No/Minimum tillage
how many are located in India ? (ii) Mulch tillage
(a) 4 (b) 6 (iii) Strip or zonal tillage
(c) 7 (d) 10 (iv) Ridge tillage etc.
Ans. (a) : Biodiversity–It refer as the variation of plant 38. Which among the following is an in-situ
and animal species in a particular habitat. There are 4 Agricultural Waste Management Process ?
major biodiversity hotspots in India. (a) Green Manuring
1. The western ghats (b) Mukhing
2. The Himalayas (c) Green Leaf Manuring
3. Sundaland (d) All of these
4. Indo-Burma Region Ans. (d) : In-situ Agricultural waste management is the
34. Which among the following sector is the Major process of decomposition of wastage at their origin or
Non Point source of pollution ? existence.
(a) Agriculture (b) Industry These are of following types of process–
(c) Manufacture (d) Infrastructure (i) Mulching
(ii) Green Manuring
Ans. (a) : Types of water pollution on the basis of
(iii) Green Leaf manuring
source.
(i) Point source–The source of pollution which are 39. In India Cyclonic storms generally observed
close to the water source are called as point source of during :
pollution. This is more harmful then non point source. (a) April-May (b) June-July
Ex. Sewage, power plant wastes, Oil wells, coal mines (c) October-November (d) Both (a) and (c)
etc. Ans. (c) : Cyclones–These are caused by atmospheric
(ii) Non-point source–The source of pollution are disturbance around a low-pressure area distinguished by
scattered and not specified their location. This is less swift and often destructive air circulation.
harmful. Ex. runoffs from gardens, roads, constructions, These are two types–
agricultural herbicides etc. (i) Cyclones
35. Which among the following Radiation is (ii) Anti-Cyclones
responsible for Ozone layer formation ? In India cyclonic storms are generally observed in May-
(a) Cosmic rays (b) U.V. rays June or October-November.
(c) Visible rays (d) Infra rays 40. How many soil parameters been taken to
Ans. (b) : Ultra violet rays are responsible for ozone develop soil health card ?
layer formation. Ultra violet rays splits O2 molecules (a) 4 (b) 8
into oxygen atoms these single atoms then react with (c) 12 (d) 16
other O2 molecules to produced O3 (ozone). Ans. (c) : Soil Health Card–It is a report of soil which
36. What is the cropping intensity of India ? contain the status of soil with respect to 12 parameter
(a) 111% (b) 126% i.e. Nitrogen, phosphorous, potassium (known as micro-
nutrients), sulphure, zinc, iron, copper, manganese,
(c) 136% (d) 146%
boron and physical parameters i.e. pH, Ec, OC.
Ans. (c) : Cropping Intensity–It is the way to increase Soil health card will also indicate fertilizer
crop production from the same area of land and it recommendations and soil amendment if required.
defined as the no. of crops grows on the field in given
agricultural year. 41. The best soil structure for favourable Physical
Properties is :
Gross total cropped area
C.I. = × 100 (a) Crumby and Granular
Net sown area (b) Platy and Laminar
The current cropping intensity in India is about to (c) Columnar and prismatic
136%. (d) Blocky
37. Which among the following is not a Ans. (a) : Types of soil structure–
conservation tillage option ? (i) Platy (ii) Prismatic
(a) Zero tillage (b) Minimum tillage (iii) Columnar (iv) Blocky
(c) Clean tillage (d) Both (A) and (B) (v) Crumby and granular
Ans. (c) : Conservation tillage–Any method of soil Crumby and granular–These are the most favorable
cultivation that leaves the previous year's crop residue structure of soil for the growth of plant. It is a
on the field before and after planting the next crop in combination of sand, silt and clay in small and spherical
order to decrease soil erosion. These are of following grains due to which circulation of water easily takes
options/tillage. place.

OPSC ASCO Exam. 2021 (Paper-I) 31 YCT

42. Which one of the following is the aim Ans.(d):Phalaris minor–It is a noxious weed
Integrated Pest Management ? associated with the wheat crop which causes severe
(a) Increasing natural enemies of the pest yield losses if these are not controlled by herbicides.
(b) Strengthening the host Phenotypic mimicry with wheat crop and resistance
(c) Keeping pest populations below injurious development has created a major hurdle in controlling
levels them.
(d) Billing the pest 47. National Bureaus of Soil Survey and Land Use
Planning (NBBS and LUP) was established in
Ans. (c) : Integrated Pest Management–The main aim the year :
of integrated pest management is not to eliminate all (a) 1934 (b) 1955
types of pests. There are many types of pests available (c) 1976 (d) 1985
which are essential for the crops. Rather the aim is to
Ans. (c) : The research aspects of soil survey,
keep pests population below injurious levels.
classification and further in 1976 an independent
43. The degree to which a soil resists deformation institute of ICAR named as National Bureau of soil
when a force applied is termed as : survey and land use planning (NBSS & LUP) was
(a) Field capacity (b) Capillary capacity established.
(c) Consistency (d) Friability 48. Rill erosion usually begins in the :
Ans. (c) : Consistency is the degree to which the soil (a) Lower part of land slope
resists its deformation under an action of applied force. (b) Upper part of land slope
Degree of consistency is defined as the range of (c) Middle of land slope
consistency index of soil. (d) Entire length of land slope
Field capacity = Max. water holding capacity Ans. (b) : Rill erosion–It is oftenly called as micro
Capillary capacity = Max. height of capillary rise channel irrigation. It is the removal of soil by running
Friability = Softness of soil. water through cutoff and water course.
When, water from precipitation does not infiltrate into
44. Which of the following is not the selective
the soil and run off over the surface of the soil at the
herbicides ?
same place continuously then rill erosion takes place.
(a) 2, 4D (b) Simazine
49. Consider the following statements :
(c) Paraquat (d) Butachlor
Assertion (A) : Phosphorus, availability is poor
Ans. (c) : Selective herbicides–The herbicide which is in acid soils.
applied tot he crop will control and suppresses the Reason (R) : Phosphorus, is leached from
largeted weed species without affecting the growth of acidic soils codes.
existing crop in the field is termed as selective (a) (A) is true but (R) is wrong
herbicides. (b) (R) is true but (A) is wrong
Example– (c) Both (A) and (R) are true
(i) Simazine (d) Both (A) and (R) are false
(ii) Butachlor Ans. (c) : The phosphorus content of soil is influenced
(iii) 2, 4 Di amine etc. by the pH of soil. When soil is very acidic, phosphorus
45. At which pH better growth of plants is found ? reacts with iron (Fe) and aluminium (Al) which results
(a) 4-5 (b) 5-6 in lack of phosphorus for the plants. But when soil is
too alkaline then phosphorus react with calcium (Ca)
(c) 6.5-7.5 (d) 8-9 and also become unavailable for plants.
Ans. (c) : Better pH range of soil for the optimum Water is the primary driver of phosphorus loss in the
growth of plants is 5.5-6.5, if pH decreases below 5.5 form of dissolved phosphorus.
then soil become very acidic which is not suitable for 50. Nutrient deficiency caused chlorosis in older
many crops. So according the option 6.5-7.5 will be leaves of plants :
more closed to optimum condition. (a) Sodium (b) Calcium
46. Which of the following is an example of (c) Magnesium (d) Nitrogen
Mimicry Weed ? Ans. (c) : Chlorosis is the less of the green coloration of
(a) Green gram RMG-344 in green gram RMG- leaves of plants due to deficiency of nutrient which
492 crop results in yellowing of leaves chlorosis caused by the
(b) Black gram in green gram crop deficiency of iron, nitrogen, potassium, magnesium etc.
(c) Chenopodium murale in berseem crop Magnesium is the main constituent of chlorophyll due
(d) Phalaris minor in wheat crop to which leaves become green.
OPSC ASCO Exam. 2021 (Paper-I) 32 YCT
51. Vertical section of soil is called as : 56. The deficiency symptoms of calcium and Boron
(a) Soil profile (b) Soil Horizon are generally observed on:
(c) Soil column (d) Regolith (a) Young leaves in terminal bud
Ans. (a) : Soil Profile–The vertical section of soil mass (b) Older leaves
is termed as soil profile which allow to examine the (c) Lower leaves
structure of the soil particles. (d) Older and new leaves
A soil profile is divided into 4 layers- Ans. (a) : The deficiency symptoms of calcium and
(i) A-horizon Boron are generally observed on young leaves in
(ii) B-horizon terminal bud. The growth is inhibited and plants have a
Most important for plant growth (Top 2 layers) pushy appearance.
(iii) C-horizon–Consists of weathering rock 57. The decomposition of organic matter in soil
(iv) D-horizon–Consists of bed rock. reduces:
52. The essential constituent of chlorophyll is : (a) Acidity (b) Alkalinity
(a) Calcium (b) Sulphur (c) Fertility (d) Salinity
(c) Nitrogen (d) Iron Ans. (b) : Decaying organic matter produces H+ which
Ans. (c) : The essential constituent of the chlorophyll is is responsible for acidity so that acidity will increases
magnesium which is located at the centre of pigment in and it reacts with the alkane presents into the soil
chlorophyll. Nitrogen is a mineral element required by therefore alkalinity of soil reduces and salinity also
plants in large amount. It is also a constituent of the increases.
pigment of chlorophyll.
58. The propagating, material of Napier grass,
53. Plays a Major role in energy storage and Guinea grass, Para grass etc called :
transfer of ADP into ATP molecules : (a) Setts (b) Rotted slips
(a) Mg (b) Fe (c) Corms (d) Tubers
(c) P (d) Mo
Ans. (a) : The stem cutting or section of the stalks
Ans. (c) : Phosphorus is a major constituent of energy which is used for propagating Napier grass, Guinea
rich nucleotids like (ATP, ADP etc). Therefore it is grass is called as 'setts'.
indispensable for reactions involving in energy transfer
in living organisms. 59. The disease resistance in plants is imported by :
(a) Potassium (b) Molybdenum
54. Kresek Phase Symptom found in plant disease :
(c) Nitrogen (d) Manganese
(a) Black arm of cotton
(b) Bacterial leaf blight of rice Ans. (c) : The disease resistance in plants is imported
(c) Leaf curl of chilli by Nitrogen and iron mainly because are the one of the
(d) Late blight of potato most important micronutrients having a significant
impact on the pathogens.
Ans. (b) : Kresek Phase Symptom–Wilting and
yellowing of leaves or wilting of seedlings is called as 60. Which is an illustration of relay cropping ?
Kresek phase symptom. It is found in rice. The bacterial (a) Paddy-Wheat
leaf blight of rice is the symptoms of kresek phase in (b) Soyabean-Maize
rice. Infected leaves turn grayish green and roll up and (c) Maize-Toria-Wheat
further leaves turn yellow to straw coloured and wilting (d) Maize - Potato - Wheat -Moong
takes place. Ans. (b) : Relay cropping–The cropping system in
55. The following soil water is held due to which succeeding crop is sowing before the harvesting
adsorption forces : of preceding crop.
(a) Gravitation water Soyabean-Maize
(b) Capillary water 61. Dioceous variety of papaya is :
(c) Hygroscopic water (a) Pusa Majesty (b) Pusa Nanha
(d) Hygroscopic and Capillary water
(c) Pusa Delicious (d) Sunrise solo
Ans. (c) : Hygroscopic water–Hygroscopic water is
Ans. (b) :
defined as the water that is held by the soil particles at
the suction of more than 31 bars due to force of Papaya variety - Characteristics
adsorption. Pusa Dwarf/Nanha - Dioecious
Hygroscopic water is held so tenaciously that plants are Pusa Giant - Dioecious
not able to absorb it and therefore it is unavailable to Pusa Majesty - Gynodioecious
plants. Pusa delicious - Gynodioecious

OPSC ASCO Exam. 2021 (Paper-I) 33 YCT

62. The natural aggregates of soil are known as : 68. Arrangements of Primary and Secondary soil
(a) Peds (b) Clods particles in a certain pattern called :
(c) Fragments (d) Grid (a) Soil concentration (b) Soil structure
Ans. (a) : The natural aggregates of soil are known as (c) Soil texture (d) Soil stabilization
soil structure or peds. Ans. (b) : Soil structure–An arrangement of primary
Types of soil peds– and secondary soil particles in a certain pattern is called
(i) Granular and crumb structure as soil structure.
(ii) Blocky and subangular blocky Soil structure according to shape of individual
(iii) Prismatic and columnar aggregate
(i) Granular and crumb
(iv) Platy structure
(ii) Blocky
63. Edible part of coconut is :
(iii) Prismatic and columnar
(a) Mesocarp (b) Endocarp
(iv) Platy
(c) Endosperm (d) Perianth
69. Mridula is a variety of :
Ans. (c) : The edible part of coconut is endosperm
(a) Guava (b) Ber
(coconut meat and water). The hard cover over the
(c) Pomegranate (d) Aonla
endosperm is termed as endocarp and fiber husk of
coconut is called as mesocarp. Ans. (a) : Mridula–
64. Which one of the following parent material is • It is a variety of guava.
transported by gravity? • Plants are semi-tall and spreading.
(a) Glacial (b) Colluvial • Fruits are round in shape.
(c) Eolicum (d) Alluvial • It is a open pollinated seedling of Allahabad safeda
Ans. (b) : • White flesh and yellow skin.
Material - Transporting agency 70. Flame Photometer is used in the determination
Colluvial - Gravity of :
Glacial - Glacier (a) Nitrogen (b) Phosphorous
Alluvial - Running water (c) Potassium (d) Boron
Aeoline - Wind Ans. (c) : Flame photometer–It is an instrument which
is used to determine the concentration of certain metal
65. ETLro. Of aphids for successful seed Plot
ions like sodium potassium, lithium and calcium in
Technique (SPT) in Potato is :
terms of emission.
(a) 10 (b) 20
The intensity of the emission is directly proportional to
(c) 30 (d) 40 the number of atoms return to the ground state.
Ans. (b) : Seed plot technique (SPT) is described as Element Flame colour
raising the crop during a period when aphid population Sodium Yellow
is very low after taking precaution such as use of Potassium Violet
insecticides against aphids.
Barium Green
Economical threshold level of aphids for successful
Lithium Red
seed plot technique in potato is 20.
Calcium Orange
66. Passive factors of soil formation includes :
71. Bio 212 is a somaclonal variety of :
(a) Parent Material (b) Climate
(a) Brassica junoea
(c) Organisms (d) None of these
(b) Lathyrus sativus
Ans. (a) : Topography, time duration and parent (c) Lycopersican esculentum
material are noted as passive factors because their (d) Citronella java
effects are not observed immediately.
Ans. (b) : Somaclonal variation–It is defined as
Active factors–Rainfall, heat temperature, wind
genetic variation observed among progeny plants
climate, microorganism etc.
obtained after somatic tissue culture in vitro.
67. Chemical uses for control for Pre-harvest drop Bio 212 is a somaclonal variety of lathyrus (Lathyrus
of citrus is : sativas).
(a) 2-4D@20ppm (b) 2-4D@200ppm 72. Which of the following clay mineral is
(c) NAA@50ppm (d) NAA@500ppm dominant in red soil ?
Ans. (a) : 20ppm@2-4D chemical is used for control of (a) Smectite (b) Mica
pre harvest drop of citrus. (c) Kaolinite (d) IIlite
OPSC ASCO Exam. 2021 (Paper-I) 34 YCT
Ans. (c): Kaolinite is a dominant clay mineral of red Ans. (d): Golden rice–It is a genetically modified
soil with small quantity of illite mineral. Kaolinite is a (GM) crop which can mitigate the deficiency of vitamin
two layer unit that is formed by stacking a gibbsite sheet 'A'. The deficiency of vitamin 'A' causes serious health
on a silica sheet bonded with strong hydrogen bond. problem like blindness and premature death.
73. Consider the following statements : 78. The suitable Nitrogen fertilizer for sodic
ln Paddy crop, fertilizer application should be (Alkali) soil is :
directed to the : (a) Sodium Nitrate
(I) Oxidised zone (b) Ammonium sulphate
(II) Reduced zone (c) Urea
(III) Surface (d) Ammonium chloride
(a) Only (I) is correct Ans. (b) : The suitable nitrogen fertilizer for sodic
(b) Only (II) is correct (alkali) soil is ammonium sulphate or Ammonium
nitrate. The pH of sodic soil is more than 8.5.
(c) (Il) and (III) is correct
(d) (I) and (III) is correct 79. In grafting lower part of plant is know as :
(a) Stock (b) Apical
Ans. (a) : In paddy crop, fertilizer application should be
(c) Bud (d) Scion
directed in the oxidised zone of crop.
Ans. (a) : Grafting is a technique whereby tissues of
plants are jointed so that they continue their growth
together.
The upper part of the combined plant (graft) is called as
scion while the lower part is called as rootstock.
74. Which one of the soil component has highest 80. Which of the following instrument is used for
CEC ? measuring soil strength ?
(a) Kaolinite (b) Montmorillonite (a) Dynamometer (b) Penetrometer
(c) Vermiculite (d) Organic Matter (c) Hydrometer (d) Thermometer
Ans. (c) : The sum of exchangeable cutions that a soil Ans. (b) : Penetrometer–Penetrometer, drop-cone
can absorb is called as cation exchange capacity penetrometer and pocket penetrometer are used to
vermiculite has highest cation exchange capacity of measure the strength of soil at different depth up to 150
100-150 milli equivalent/100 g. mm in cultivated or uncultivated seed beds in a loam
and sandy clay loam.
Soil component CEC
Kaolinte 3-15 Meg/100g Instrument Used
Montmorillonite 80-100 Meg/100g Dynamometer – Mechanical force
Thermometer – Temperature
Halloysite 40-50 Meg/100g
Hydrometer – Density
Illite 10-40 Meg/100g
Smectite 80-120 Meg/100g 81. Hermaphrodite variety of papaya is :
(a) Sunrise solo
75. Grassy shoot disease of sugarcane is caused by:
(b) Coorge Honey Dew
(a) Fungi (b) Bacteria
(c) Surya
(c) Virus (d) Mycoplasma (d) CO-3
Ans. (b) : Grassy shoot disease–It is charactrised by Ans. (b) : Coorge Honey Dew–It is a gynodiocious,
proliferation of vegetative buds from the base of the semi dwarf hermaphrodite variety of papaya. The fruits
cane giving rise to crowded bunch of tillers bearing are big weighing 1.75 to 2 kg dark green in colour with
narrow leaves. It is caused by a bacteria named as phyto slight ridging skin surface.
plasma. Other varieties of papaya–
76. The rate of decomposition of lignins and (1) Mexican Red Papaya
phenolic compound is : (2) Hawaiian Sunrise Papaya
(a) Rapid (b) Very slow (3) Hawaiian Sunset Papaya
(c) Slow (d) Medium (4) Bettina Papaya
Ans. (b) : The rate of decomposition of lignins and (5) Guinea Gold Papaya
phenolic compound is very slow when it is heated up by (6) Hortus Gold etc.
10oC/min. It losing only 40% of its initial mass below 82. Baby com is harvested at the stage of:
700oC. (a) 2-3 days after silking
77. Golden rice can mitigate the deficiency of: (b) 2-3 days after tasseling
(a) Vitamin D (b) Vitamin C (c) Milk stage
(c) Calcium (d) Vitamin A (d) R4 stage
OPSC ASCO Exam. 2021 (Paper-I) 35 YCT
Ans. (a): The baby corn is harvested at the stage of 2-3 Ans. (c): Concentric ring or target board symptoms
days after silking. It is ready for harvest after 50-60 found in disease is caused by alternaria solani.
days after sowing. Symptoms appears first on the oldest foliage. Affected
83. use of crop rotation fame reduction of pest leaves develop circular to angular dark brown lesions 3-
population pest attack is a method : 4 mm in diameter.
(a) Cultural (b) Physical 88. Maximum number of fruit plants can be
(c) Chemical (d) Mechanical planted in the orchard by the system :
Ans. (b) : Crop rotation replaces a crop that is (a) Square (b) Triangular
susceptible to a serious pest with another crop that is not (c) Hexagonal (d) Rectangular
susceptible is a physical method of reduction of pest Ans. (c) : The maximum number of fruit plants can be
population of crops. planted in orchard by the system of hexagon.
• Cultural methods–by tillage works
• Chemical methods–by pesticides
84. Which one of the following is a double cross
hybrid variety of Mango?
(a) Sindhu
(b) Alphanso
(c) Neelam
(d) Arka suprabhat (H-14)
Ans. (d) : Arka suprabhat (H-14)–It is a double cross
hybrid variety of mango between Amrapal (Deshehari ×
Neelum) × Arka anmal (Alphanso × Janardhan).
Note–Sindhu is almost seedless variety of mango. 89. Which of the fertilizer contains citric acid
85. Application of Muriate of potash is generally soluble phosphoric acid?
not recommended for : (a) Basic slag
(a) Wheat, Batty, Mustard (b) Rock phosphate
(b) Bajara, Sorghum, Maize (c) Single super phosphate
(c) Sugarcane, Sugarbeet, Tobacco (d) Triple super phosphate
(d) Groundnut, Sesamum Ans. (b) : Types of phosphorous fertilizer–
Ans. (c) : Muriate of potash–It is also known as Phosphorous fertilizers are classified into three main
potassium chloride contains 60% potash which is categories.
essential for plant growth and quality by producing (1) Water soluble–Single super sulphate or DAP
proteins and sugars. (2) Citric soluble–Guano bold and rock phosphates
It is not recommended for sugarcane tobacco and (3) Organic soluble–Manures and compost
sugarbeet crops. 90. Gulabi is important cultivar of:
86. Consider the following statements : (a) Grape fruit (b) Pomegranate
(I) Maize is C4 plant (c) Strawberry (d) Litchi
(II) Kranz anatomy present tin C4 plant Ans. (a) : The Gulabi and Kali champa are the
(III) Photo-respiration does not occur in C3 plant important varieties of grape fruit. Gulabi or Muscut
(a) (I) only (b) (I) and (III) only hamburg grapes harvest in the month of January-March
(c) (II) and (III) only (d) (I) and (II) only and June-December.
Ans. (d) : Coplants–In C4 plants photo-respiration Other varieties–
does not occur because they have a mechanism that Sultana - Light green oval shape
increases the concentration of CO2 at enzyme site. Kal Champa - For juice and wine purpose
Kranz anatomy is a specialized structure in C4 plants 91. Which of the following element is required by
where mesophyll cells are clustered around the cells in plants for uptake and utilization of calcium?
ring like structure. Ex. : Maize, sugarcane, sorphum etc. (a) Manganese (b) Chlorine
87. Concentric ring or target board Symptoms (c) Boron (d) Selenium
found in disease is caused by : Ans. (c) : Boron (B)–It is a essential for the uptake and
(a) Phythium (b) Cercospora utilization of calcium by plants. It facilitates the
(c) Alternaria (d) Phytopthora movement of calcium into the root zone of plants.
OPSC ASCO Exam. 2021 (Paper-I) 36 YCT
92. Best harvesting or plucking stage of tea is : 97. Growing only one crop year after year on a
(a) Two leaves, two bud stage piece of land is called :
(b) Two leaves, single bud stage (a) Sole cropping
(c) Single leaves, single bed stage (b) Cropping pattern
(d) All of these (c) Companion cropping
Ans. (d) : Harvesting or plucking stage of tea–It is (d) Mono-cropping
the time interval in days between the successive Ans. (d) : Mono cropping–Growing only one crop year
plucking operations. It varies from 4-14 days. after year on a piece of land is called as mono cropping
Sole cropping–Crop composed of individual plants of
the same variety of one species.
Other cropping system–
(i) Alley cropping
93. Which of the weed is controlled by the use of (ii) Inter cropping
Zygogramma Bicolorata as a bio-agent ? (iii) Mixed cropping
(a) Prickly pear opuntia (iv) Raton cropping
(b) Xanthinm Strumarium (v) Sequential cropping
(c) Lantana camera 98. Sindhu seedless cultivar of Mango has which ?
(d) Parthenium Hysteraphous (a) vegetative (b) Stimulative
Ans. (d) : Parthenium Hysteraphous is a weed which is (c) Stenospermocarpy (d) None of these
controlled by the use of zygograma bicolorata as a bio- Ans. (c) : Sindhu is seedless mango variety from
agent. hybrids of mango varieties has a rich, sweet and
94. Inflorescence of Banana is : distinctive flavour and less fibre than other varieties.
(a) Spadix (b) Comby The production of abortive development of seed less
(c) Solitary (d) Panicle mango is termed as steno spermocarpy.
Ans. (a) : Inflorescence of Banana–Cymose groups of 99. Which of the following is not a Kharif season
flowers are arranged acropetally on the fleshy axis. weed ?
Each cymose group is subtended by he spathe, an older (a) Chenopodium album
spathe subtending the next younger. The mixed spadix (b) Amaranthus viridis
is the cymose group of flowers acropetally on fleshy (c) Echynochto colonum
axis of banana which is termed as inflorescence of (d) Commelina Benghalensis
banana. Ans. (a) : Kharif seasoned weed–
95. Herbicide used to control (i) Amarnathus virdis
Orobancheaegyptiaca parasite in Mustard crop (ii) Echynochto colonum
is : (iii) Commelina Benghalensis
(a) Pendemethalin and fluchloalin Chenopodium album is a rabi season weed which is
(b) Alachlor and bentazon commonly known as 'Bathua' which is a very rich
(c) Nitrofen and trifluratlin source of vitamin 'A'.
(d) MSMAand DSVA 100. Which cereal crop is having least amount of
Ans. (a) : Herbicide used to control lysine and tryptophan?
Orobancheaegyptiaca parasite in Mustard crop is (a) Maize
pendemethalin and fluchloalin (b) Rice
96. Weight of sword sucker in Banana : (c) Wheat
(a) 250 gm (b) 450 gm (d) Sorghum
(c) 750 gm (d) 800-900 gm Ans. (b) : Rice is a cereal which is having least amount
Ans. (c) : Sucker in Banana–There are two types of of lysine and tryplophan.
sucker in banana plant– Ex. Maize, rice, wheat, barley, sorghum, millet etc.
1. Sword sucker – Broad Rhizome Lysine–Alpho amino acid
2. Water Sucker – Small Rhizone, broad leaves weight Tryptophan–α-amino acid which is used in the
of sword sucker lies from 500-750 gm. biosynthesis of protein.

OPSC ASCO Exam. 2021 (Paper-I) 37 YCT

 Odisha Public Service Commission
Assistant Soil Conservation Officer (ASCO)
Exam- 2021 (Paper-II)
1. Aman rice is sown in the month of: Ans. (b): The area required for nursery
(a) April-may (b) June-July A = 1000 sq.m/hectare
(c) November-December (d) May-June 1000 sq.m
=
Ans. (b) : Aman, boro and aus is the variety of rice 10000sq.m
which are sown in the month of June-July. In India the
percentage production of Aman is 48.79%. 1
= or total area
Climatic condition– 10
(i) Avg. Temperature – (21-37ºC) Note–Generally 800-1000 sq. m. is required for nursery
of paddy for 1 hectare
(ii) Blooming temperature – (26-29ºC)
(iii) Ripening temperature – (20º-25ºC) 5. Ammonium sulphate Fertilizer is preferred for
nitrogen application in ground nut because it
(iv) suitable soil – Clay loam
also provides :
(v) pH of soil – 5.5-6.5
(a) Calcium (b) Magnesium
Note–Boro rice is sown is winter and harvested in
summer so that it is known as spring rice. (c) Sulphur (d) Boron
2. A short duration crop is sown in between the Ans. (c) : Ammonium sulphate fertilizer is preferred for
two rows of main crops is termed as : ground nut because of both nitrogen and sulphur
(a) Cash crop (b) Catch crop content.
(c) Companion crop (d) Ephemeral 6. Nitrogen is taken up by the plants in the form
of:
Ans. (b) : Catch crop–The crop which is sown in
between the two rows of main crops is termed as catch (a) Chloride (b) Oxide
crops. These are also used for increasing the fertility of (c) Nitrate (d) Sulphate
soil. Ans. (c) : Nitrogen is taken up by the plants in the
Cash crop–The crops which are used to make profit are forms of nitrate for their growth.
known as cash. i.e. Sugarcane, cotton etc.
7. The deficiency symptoms on lower leaves are
Compaction crops–When two crops are grown near seen due to :
each other for the benefit of one those crops or both is
termed as companion crops. (a) Nitrogen (b) Phosphorus
(c) Potassium (d) Zinc
3. Crop production and, Animal Husbandry
collectively is known as : Ans. (a) :
(a) Mixed cropping (b) Relay cropping Deficiency symptoms Nutrients Location
(c) Mixed Farming (d) None of these
Whole leaves turn Nitrogen Lower leaves
Ans. (c) : Mixed farming–The crop production and yellow
animal husbandry/rearing simultaneously known as
mixed farming. It is different from mixed cropping Purple of bronze Phosphorus Lower leaves
which is the cultivation of two or more crops discolouration in upper
simultaneously in a field. and lower side of leave
Relay cropping = Crop about to harvest+Crops seeding Browning or yellowing Potassium Lower leaves
Mixed farming = Crop production + Animal rearing on leaves edges of newly
Mixed cropping = Crops + crops at same time. leaves
4. How much area of nursery is needed for sowing 8. Which one of the following nutrient is more
of paddy crop in one hectare ? required for berseem crop ?
(a) 1/5 of total area (b) 1/10 of total area (a) Nitrogen (b) Phosphorus
(c) 1/8 of total area (d) 1/16 of total area (c) Potash (d) Boron
OPSC ASCO Exam. 2021 (Paper-II) 38 YCT
Ans. (b): Phosphorous is more required for barseem 14. How much kgN2/ha/year can be fixed by
crop. Barseem is a leguminous crop so it does not Azolla?
required large amount of nitrogen. Leguminous crops (a) 80 to 100 (b) 30 to 40
are those which are used to regain the nitrogen content (c) 10 to 15 (d) 100 to 120
in field by natural means.
Ans. (d) : The rate of N2 fixation by Azolla is 100-170
9. Acidic soil can be reclaimed by the application kg/ha/year.
of :
15. What is crop Rotation?
(a) CaCO3 (b) H2SO4
(a) Growing more than one crop at a time
(c) CaSO4.2H2O (d) HNO3
(b) Growing of crops one after another to
Ans. (a) : Reclamation of soil –
maintain soil fertility
(i) Acidic soil – Limestone – CaCO3
(c) Growing of an associate crop in between the
(ii) Alkaline soil – Gypsum – CaSO4·2H2O rows of two main crop
(iii) Sodic soil – Gypsum – CaSO4·2H2O (d) Growing of crops together in strips
10. Addition of following material makes it Ans. (b) : Crop rotation–The process of growing crops
possible to take good crop in sodic soils; one after another to maintain the fertility of soil is called
(a) FYM (b) Green Manuring as crop rotation.
(c) Gypsum (d) Vermicompost • Growing more than one crop at same time – Mixed
Ans. (c) : Sodic soil–The soil containing high sodium cropping
ion concentration as compared to others then the soil is • Growing of an associate crop in between the rows of
termed as sodic soil. The reclamation of sodic soil can two main crops – Catch crops
be done with the help of gypsum.
• Growing of crops together in strips – Strips cropping
11. Ammonia is lost through volatilization in
significant amount from : 16. Which is the most important source of
Irrigation in India ?
(a) Alkaline soils (b) Acidic soils
(a) Canal (b) Pond
(c) Saline soils (d) Sodic soils
(c) Tubewell (d) Charsa
Ans. (b) : Volatilization–It is the loss of nitrogen
through the conversion of ammonium to ammonia gas. Ans. (c) : The most important source of irrigation in
Which is released to the atmosphere. As pH of soil India is tubewells. It is about to 46% of total.
increases volatilization losses are increased 17. When only two irrigations are available, the
significantly. wheat crop should be irrigated at ?
Acidic soil – high H+ ion (a) CRI and Tillering stage
Alkaline soil – high OH– ions (b) CRI and Flowering stage
Sodic soil – high Na+ ions (c) CRI and Milking stage
Saline soil – high salt concentration (d) CRI and late joining stage
12. An aerobic environmental condition of paddy Ans. (b) : When only one irrigations is available then
soil is responsible for gaseous tosses of fertilizer the wheat crop should be irrigated at crown root
nitrogen : initiation (CRI) and when two irrigations is available
(a) Ammonification (b) Nitrification then it is provided at crown root initiation and flowering
(c) Denitrification (d) Volatilization stage.
Ans. (c) : In an aerobic environmental condition of 18. Which crop is Considered as King of the fodder
paddy soil which is responsible for losses of fertilizer Crop ?
nitrogen in gaseous form is termed as denitrification. (a) Lucem (b) Berseem
13. Application of nitrogen in pulses at the time of (c) Oat (d) Sudan grass
sowing is known as : Ans. (b) : Berseem (Trifolium alexandrium) is a major
(a) Additional dose (b) Starter dose fodder crop and it is considered as king of fodder crop.
(c) Synergistic dose (d) Basic dose It is a leguminous crop.
Ans. (b) : The application of nitrogen in pulses at the 19. Which of the following operation is not a
time of sowing is known as starter dose because pulses primary practice ?
with the help of rhizobium bacteria and no further (a) Ploughing (b) Planking
nitrogen does required. (c) Harrowing (d) Weeding
OPSC ASCO Exam. 2021 (Paper-II) 39 YCT
Ans.(c): Primary tillage–These operations which 24. Capillary movement of water in the soil is
consume more power per unit area in order to prepare complemented by :
the soil for seeding and planting is termed as primary (a) Stem Elongation (b) Root Extension
tillage. (c) Leaf Orientation (d) Fruit formation
Ex.: Ploughing, Planking, Weeding etc. Ans. (b) : The water which is useful for plant growth is
Seconary tillage–These operations which consume less called as capillary water. The movement of water and
power per unit area is termed as secondary tillage. minerals from soil to the plant is possible through the
Ex.: Cultivators, Harrow, Rollers. root extension of plants due to capillary action.
20. Phalaris minor weed is associated with: 25. Topping in tobacco crop means a process of :
(a) Gram crop (b) Wheat crop (a) Removal of buds in the axil of Leaves
(c) Paddy crop (d) Soybean crop (b) Removal of leaves
Ans. (b) : Phalaris minor weed is a major weed of (c) Removal of terminal buds
wheat crop. The intensity of this weed is so high that it (d) Burning of leaves
has become rather impossible to a grow wheat generally Ans. (a) : The removal of buds in the axil of leaves of
in certain localities it is called as Gehunsa. tobacco crop is called as topping and the process of
Control–To control phalaris minor tribunil or dosonex drying the leaves is called as curing.
or isoproturon is spryed over wheat crops after 32-35 Tobacco is obtained from the leaves of the 'Nicotiana
days of sowing. speicies.'
21. Basal application of fertilizers means : 26. Diameter of Breast Height (DBH) Of tree is
(a) Application of fertilizers at the time of measured at :
sowing (a) 4½ feet above ground level
(b) Application of fertilizers in two split doses (b) Merchantable height of the tee
(c) Application of fertilizers in sending Crop (c) 1.37 meter from the ground level
(d) Application of fertilizers many times (d) Just below the canopy of the tree
Ans. (a) : Application of fertilizers at the time of Ans. (a) : The Diameter of Breast Height (DBH) of the
sowing is termed as basal application which is done to tree is measured at 4.5 feet (1.37 m) above the ground
obtain maximum crop yield. level.
22. Bio fertilizer is a:
(a) Mixture of organic matter and micro
organisms
(b) Mixture of inorganic fertilizers and micro
organisms
(c) Culture having the desired strain of micro
organisms
(d) Decomposed organic matter enriched with
bacteria
Ans. (a) : Biofertilizer–These are biological 27. Tree species which give fuel, fodder, food. fruit
preparation of efficient microorganisms that promote and fiber are called :
plant growth by improving nutrient acquisition. (a) Fodder Trees
23. Biological Nitrogen fixation is : (b) Multipurpose Trees
(a) Aerobic and anaerobic (c) Agroforestry Trees
(b) Denitrification (d) Leguminous Trees
(c) Leaching Ans. (b) :
(d) Removal Trees Uses
Ans. (a) : Biological Nitrogen fixation–These bacteria • Multipurpose Trees – Fuel, fodder, food, fruit,
that convert atmospheric nitrogen into utilizable fiber, fertilizers
compound of nitrogen are known as nitrogen fixing
• Fodder Trees – Food for animals
bacteria.
Azotobacter is aerobic and clostridium is anaerobic • Agroforestry Trees – Agricultural + forest plant
nitrogen fixing bacteria. • Leguminous Trees – Nitrogen fixation

OPSC ASCO Exam. 2021 (Paper-II) 40 YCT

28. A tree which completely becomes leafless in • Nitrogen fixing tree
rainy season is : → Acacia Nilotica (Babul)
(a) Faiderbia albida → Dalbergia sissoo (Sheesham)
(b) Prosopis julifrora
→ Leucanea leucacephala
(c) Leucaena leuoooephala
34. Sal tree bears flower:
(d) Acacia nilotica
(a) Every year (b) Once in two years
Ans. (*) :
(c) Once in five years (d) None of these
29. Which is the scented portion in Sandal wood ?
Ans. (a) : The sal flowers are whitish and yellowish in
(a) Flower (b) Seed colour which appear in every summer season.
(c) Soft wood (d) Heart Wood
35. The criteria of essentiality of nutrients is given
Ans. (d) : Heart wood is the scented portion of the by :
sandal wood. It is the most precious part of sandal wood (a) Amon (b) Tandon
tree. The unique smell comes from a naturally occurring
(c) Rattan (d) Dargan
compound α-santalol (C15 H24 O)
Ans. (a) : The channels that are termed necessary
30. Tree age can be measured by:
essential for plants without which the processes and life
(a) Altimeter (b) Increment borer cycle of plant is incomplete, are catigorised in to
(c) Hygrometer (d) Relascope various groups.
Ans. (b) : The age of tree can be measured by The criteria for essentiality was given by Amon and
increment borer foresters use tree borers to extract core stout in 1939. There are 17 essential element have been
samples from trees to determine age and growth rate. found for growth and metabolism of plants.
• Altimeter – Height of aeroplanes 36. In Alley Cropping, the row to row spacing
• Hydrometer – Humidity varies from:
• Relascope – Height of tree and basal area (a) 1-4m (b) 4-8m
31. Head Quarter Of ICRAF is located at: (c) 2-4m (d) 4-6m
(a) Dehredun (b) Jhansi Ans. (d) : Alloy cropping–It i defined as the planting
(c) Nairobi (d) Europe of rows of tree and shrubs to create alley. The row to
row spacing varies from (4-6)m.
Ans. (c) : ICRAF – International Centre for Research
in Agro forestry (1978) 37. Home garden is found extensively in:
It is an international institute headquartered in Nairobi. (a) Low rainfall areas
(Kenya) (b) Medium rainfall areas
32. Global warming is due to: (c) High rainfall areas
(a) Deforestation (b) Mixed farming (d) Arid areas
(c) Crop diversity (d) Agroforestry Ans. (c) : Home garden agriculture–A farming system
Ans. (a) : Global warming–The global annual that combines different physical, social and economical
temperature has increased due to release of green house function in the land area around the family home.
gases into atmosphere. Deforestation is the main cause Tomatoes, peppers, garlic, cucumber, peas, etc. are the
of global warming because it causes more increase in % plants widely used in home gardens. Home gardens are
of CO2 into atmosphere. CO2 absorbs the sunlight and found in highly rainfall areas.
create a thermal blanket and average temperature 38. The mature trees removed in one operation is
increases. called :
33. Which of the following tree species is not (a) Shelter wood system
nitrogen fixing? (b) Clear Felling system
(a) Acacia nilotica (c) Improvement Felling
(b) Dalbergia Sissoo (d) Coppice with Reserve System
(c) Azadirachta Indica
Ans. (b) : Clear felling system–It is defined as a
(d) Leucaena Leucocephala silviculture system in which equal or equi-productive
Ans. (c) : Azadirachta Indica which is commonly area of a mature crops are successively clear felled in
known as Neem is not nitrogen fixing tree species as one operation to be regenerated, most frequently
compare to other given options. artificial but naturally also.

OPSC ASCO Exam. 2021 (Paper-II) 41 YCT

Shelter wood system–The shelterwood system is the 44. World Forestry Day is celebrated on date :
system in which the mature crop is removed in a series (a) 5th June (b) 15th June
of operations. The first operation is called seeding (c) 22nd March (d) 25th September
felling and last is the final felling.
Ans. (*) :
39. Katha is extracted from which part of Khair
Days Importance
trees ? st
(a) Fruit (b) Heart Wood 21 March World forestry day
th
(c) Seed (d) Roots 5 June World environment day
15th June World wind day
Ans. (b) : Katha is obtained by boiling the heartwood of th
acacia catechu which is generally called as "Khair tree". 25 September Antyoday Diwas
The process of kattha making is a long and ardous 45. Growing of trees with the crop is called :
process which takes up to 45 days time period. (a) Agri-silviculture-System
40. For cooking 1 kg of food, how much quantity of (b) Agri-horticulture system
fuel wood is required ? (c) Agro-silvopastoral System
(a) 1 kg (b) 1.2 kg (d) Agro-horti-silviculture System
(c) 1.5 kg (d) 1.7 kg Ans. (a) :
Ans. (b) : For cooking 1 kg of food the amount of the Agri - silviculture – Crop + forestry
fuel wood of 1.2 kg is required. Fuel wood is also
Agro - horticulture – Crop + Grass or ornamental plant
known as fire wood which is commonly used for known
as fire house hold cooking operations. Agro - silvopostral – Crop + tree + animals
Agro - horti - silviculture – Crop + fruits + ornamental
41. An operation is carried out for the benefit of a
forest crop at any stage between seedling to trees
maturity, is called : 46. Silvipastoral system means.
(a) Cultural operation (b) Plant protection (a) Growing of trees with pasture
(c) Regeneration (d) Tending Operations (b) Growing of trees with crop
Ans. (d) : Tending–An operation is carried out for the (c) Growing of grasses with fruit tree
benefit of forest crop at any stage between seeding to (d) Growing of pasture + crop
maturity is termed as tending operations.
Ans. (a) : Silvipastoral system–The combined
Cultural operations–These are carried out to assist the production of woody plants with pasture is termed as
crop to complete regeneration. If some operation done silvipastoral system. The trees and shrubs used
before planting or sowing is done it will also be cultural primarily to produce fodder for live stock.
operation. There is no income generated from cultural
Silvipastoral – Tree + Pasture/animals
operation.
Regeneration–The natural process of replacing or 47. The ideal patting mixer used in filling polybags
restoring of missing cells, tissues organs, and entire for raising seedlings contains ratio of Soil :
plant body is termed as regeneration. Sand : FYM :
(a) 1:3:1 (b) 1:2:1
42. Which is the most important physical
characteristics of wood ? (c) 1:1:1 (d) 2:1:1
(a) Strength (b) Grain Ans. (d) : The ideal patting mixer used in filling
(c) Specific gravity (d) Elasticity polybags for raising seedling contains two part of fine
earth/soil, one part of sand and one part of FVM to
Ans. (c) : The fundamental physical property of wood is
improve aeration and fertility of the soil.
specific gravity. The hardwoods have a specific gravity
ranging from (1-1.54). FYM–Organic fertilizer which stands for farm yard
manure.
Strength–Maximum strength in the direction of fibers
tensile strength 0.1 N/mm2. Shear strength 0.15 N/mm2. 48. Which of the following is an example of
Silvipastoral System ?
43. First Inspector General of Forest of India was :
(a) Hardwickkia binate + Cenchrus Ciliaris +
(a) K.F.S. King (b) H.G. Champion
Goat
(c) R.S. Troop (d) Dietrich Brandis
(b) Eurblica officinalis + Cowpea
Ans. (d) : Dietrich Brandis was the first Inspector
(c) Gmelina arborea + Dicanthium annufatum
General of forests in India. He helped the Britishes to
formulate the Indian forest services act 1865. (d) Acaica nilotica + Paddy

OPSC ASCO Exam. 2021 (Paper-II) 42 YCT

Ans.(a):Silvipastoral system–A agroforestry (ii) Broadcasting–The most common and oldest
arrangement that combines fodder plant (i.e. grasses method of sowing in which seeds are spread over the
leguminous herbs) with shrubs (plants) and trees for soil. It may covered or in covered with soil.
animal nutrition and complementary uses. (iii) Drilling–In this method seeds are dropped into the
49. Raising of trees on bunds or farm boundaries is holes, then covered and compacted by soil with the help
known as : of seed drill.
(a) Social Forestry (iv) Planting
(b) Community Forestry (v) Transplanting
(c) Farm Forestry (vi) Sowing
(d) Extension Forestry 53. For the planting of turmeric crop, which of the
Ans. (c) :Types of social forester– following vegetative material is used ?
(1) Farm forestry–Raising of tree on bunds or farm (a) Tubers (b) Bunches
boundaries is known as farm forestry. (c) Setts (d) Rhizomes
(2) Community forestry–The planning, managing and Ans. (d) : For the planting of turmeric crop, rhizomes is
harvesting of forest crops by the local population in used which is a modified subterranean plant stem that
order to encourage the involvement of locals is termed sends out roots and shoots from its nodes. It is also
as community forestry. known as creeping root stalks.
(3) Agro-forestry
(4) Extension forestry
(5) Scientific or Silviculture
50. Central Agroforestry Research Institute
(CAFRI) is located at :
(a) Hyderabad (b) New Delhi
54. Maximum rainfall occurs on the :
(c) Jhansi (d) Solan
(a) Leeward side
Ans. (c) : Central Agroforestry Research Institute
(CAFRI) was established on 8th, May 1988 at Jhansi (b) East-West Direction
(Uttar pradesh) in order to strengthen and coordinate the (c) North-South Direction
agroforestry research. (d) Windward side
Now it is upgraded on 2014 as Central Agroforestry Ans. (d) : The windward slope of the mountain receives
Research Institute (CARI) maximum rainfall and leeward slope remains drier so
51. The putting of plant propagules in the field to that it is also known as rain shadow zone.
grow as crop plants is called :
(a) Sowing (b) Gap filling
(c) Planting (d) Transplanting
Ans. (d) : Planting–The putting of seeds or progagules
in the field to grow as crop plants is called planting.
Transplanting–The method in which planting of
seedlings in main field after pulling out from the 55. The optimum range of temperature required
nursery to grow as crop. for the grain formation in Wheat crop is :
Sowing–Seeds are placed into the furrows continuously (a) 8 to 10ºC (b) 10 to 12ºC
or at specific distance. (c) 12 to 26ºC (d) 5 to 10ºC
52. Which of the following method of sowing gives Ans. (c) : The optimum range of temperature means the
rapid and uniform germination with good temperature at which grain formation of crop is
seedling vigour? maximum. For wheat crop the optimum range of
(a) Broadcasting (b) Dibbling temperature lies between 12ºC-26ºC.
(c) Drilling (d) Line sowing 56. Which of the statement is not related to deep
ploughing ?
Ans. (b) : Methods of sowing–
(a) Favours break up of clots
(i) Dibbling–In this method seeds are placed in holes or
pits at equal predetermined distances and depth which (b) Kills the weeds
gives rapid and uniform germination with good seedling (c) Increases soil erosion
vigour. (d) Incorporates organic residues
OPSC ASCO Exam. 2021 (Paper-II) 43 YCT
Ans. (a): Deep ploughing–The ploughing more then 50 (iv) Nitrus oxide N2O
cm is termed as deep ploughing which is done to change These are major green house gases which are mainly
the soil water retention and to kill the weeds. Deep responsible for global warming and ozone layer
ploughing causes more soil erosion and incorporates depletion.
organic residues that present over the surface of field.
61. Absorption of which of the following gas
Break up of clots is done by leveller after ploughing.
reduces the level Of chlorophyll pigments in
57. Which of the following instrument is not used cell and affects photosynthesis ?
for ploughing ?
(a) CO2 (b) SO2
(a) Country Plough
(c) NO2 (d) NH3
(b) Mould Board Plough
Ans. (b) : Sulphur dioxide SO2 is a major air pollutant
(c) Ridge Plough
and plants are mainly sensitive to SO2 because the
(d) Disc Plough process of photosynthesis is mainly affected by the
Ans. (a) : Instruments used for ploughing. increased concentration of sulphur dioxide and leaves
• Mould Board plough became yellow due to reduced level of chlorophyll
• Ridge plough pigments in cells.
• Disc plough 62. Quantity, Quality, Intensity and Duration are
the important parameters of :
58. Find the odd one out-In Sigmoid Growth
Curve, there are three well marked regions : (a) Soil Temperature (b) Rain Fall
(a) Lag phase (b) Economic phase (c) Solar Radiation (d) Soil Moisture
(c) Log phase (d) Steady state phase Ans. (b) : Quantity, quality, intensity and duration are
Ans. (b) : Sigmoid Growth Curve– the important parameter of rain fall which is mainly
affects the growth of crops.
Rainfall–
Quantity– Amount of water required for crops
Quality– Acidic, Alkali, neutral water
Intensity–Rate of rainfall
Duration–Time interval
63. The science of identification and classification
of earth surface features using electromagnetic
59. Which one of the following operation is not radiation as a medium of interaction refers to :
helpful in controlling floods ? (a) Weather Forecasting
(a) Construction of dams (b) Remote Sensing
(b) Afforestation (c) Geographic Information System
(c) Provision of adequate drainage (d) Global Positioning System
(d) Deforestation Ans. (b) : Remote sensing–The science of
Ans. (d) : Loss of vegetation, which is known as identification and classification of earth surface feature
deforestation causes more chances of flooding. So that using electro-magnetic radiation as a remote sensing.
deforestation is not a flood controlling operation.
Weather forecasting–Prediction of conditions of
Flood controlling operation– atmosphere.
(i) Construction of dams GIS–Computer system for geographic information.
(ii) Afforstation GPS–Navigation system (satellite based radio
(iii) Provision of adequate drainage navigation)
(iv) Development of canals
64. The process of destruction of soil aggregates by
60. Which of the following are not related to mechanical force in soils with a moisture
greenhouse gases ? content exceeding the moisture equivalent :
(a) Methane (b) Carbondioxide (a) Puddling (b) Ploughing
(c) Chlorofluoro carbons (d) NO3 (c) Mulching (d) lntereultivating
Ans. (d) : Green house gases– Ans. (a) : Puddling–The process of destruction of soil
(i) Methane CH4 aggregates by mechanical force in soils with a moisture
(ii) Carbon dioxide CO2 content exceeding the moisture equivalent is termed as
(iii) Chlorofluoro carbons (CFC) puddling.

OPSC ASCO Exam. 2021 (Paper-II) 44 YCT

Indercultivating–A process operations in the rows of If wind velocity increases this wings are being removed
standing crop. Weedings, tilting and cultivating are the from seeds. Generally wings are white and seeds are
examples of intercultivation. black or dark brown in colour.
Ploughing–Ploughing is the process of turning over the 69. The removal of seeds in case of flashy fruits is
uppermost soil, bringing fresh nutrients to the surface called :
while burying weeds and crop remains for decaying.
(a) Depulping (b) Extraction
Mulching–The process of covering the open surface of
ground by some materials in order to prevent the growth (c) Soaking (d) Winnowing
of weeds near the plants. Ans. (b) : The flashy part of a fruits is called the
65. In urea, the nitrogen is available in : mesocrop. It is a eatable part of fruit and the removal of
(a) Nitrate form (b) Sulphate form seeds in case of flashy fruits is called extraction.
(c) Amide form (d) Ammonium form Depulping–Removal of pulp
Ans. (c) : Urea in a organic matter which is produced in Winnowing–Removal of chaff from grain
lab. It is a source fo nitrogen for the plants. Nitrogen Soaking–Deep into water
present in urea in the form of amide. 70. The decomposition 6f litter (Leafy matter) is
Urea → NH2 CO NH2 faster in case of :
66. The maximum Anion Exchange Capacity is (a) Narrow leaves (b) Pointed leaves
found in : (c) Broad leaves (d) Wax-coated leaves
(a) Kaolinite mineral
Ans. (c) : Leaf litter decomposition occurs at faster rate
(b) Montmorillonite mineral in case of tropical rain forests, which have broad leaves.
(c) Granite mineral
71. Which of the statement is not related to deep
(d) Basalt mineral
ploughing ?
Ans. (a) : Anion exchange capacity–It is found (a) Favors break-up of clots
maximum in kaolinte mineral (43 me/100g) and (5
me/100g) montmorillonite minerals. Anion exchange (b) Kills the weeds
property is useful for crops in order to extract (c) Increases soil erosion
phosphose for their growth (d) Incorporates organic residues
Ans. (a) : Repeat Q. 56
72. The natural geo-hydrological unit whereby all
streams are draining into a common point is
called :
(a) Water conservation
(b) Water shed management
(c) Soil conservation
(d) Dug wells
Ans. (b) : Water shed management–The natural geo
hydrological unit whereby all streams are draining into
67. Castor belongs to the family:
a common point is called as water shed.
(a) Leguminoceae (b) Euphorbiaceae
Water conservation–Protection of water losses.
(c) Cruciferae (d) Compositae
Soil conservation–Prevention of soil erosion.
Ans. (b) : Castor oil plant is belongs to euphorbiaceae
which is also called caster bean. It is useful for 73. The science which deals with water, concerning
pharmaceutical and industrial use. with distribution, physical and chemical
reaction and in relation to the life of the earth :
68. The removal of wings from the seeds is called :
(a) Deheading (b) Dewinging (a) Precipitation (b) Flooding
(c) Washing (d) Cleaning (c) Hydrology (d) Agrostology
Ans. (b) : The removal of wings of seeds is called as Ans. (c) : Hydrology–The science which deals with
dewinging. These seeds are transported by wind but water concerning with distribution, physical and
chemical reaction and in relation to the life of the earth
is known as hydrology.
Precipitation–Rainfall/snow fall
Flooding–Uncontrolled discharge of water
Agrostology–Study of grasses

OPSC ASCO Exam. 2021 (Paper-II) 45 YCT

74. The removal of thin uniform layer of soil from 78. The direct or indirect harmful effect by one
the land surface by the action of runoff water is plant on another through the production of
called: inhibitory substances is called :
(a) Splash erosion (b) Sheet erosion (a) Allelopathy (b) Competition
(c) Rill erosion (d) Gully erosion (c) Stimulation (d) Interaction
Ans. (b) : Sheet erosion–The removal of thin uniform
Ans. (a) : The direct or indirect harmful effect by one
layer of soil from the land surface by the action of
plant on another through the production of inhibitory
runoff water is called sheet erosion of the soil.
substances is called as allelopathy.
Splash erosion–The removal of soil particles due to
rain drops is called splash erosion 79. The living organisms (bio-agent) used to limit
Rill erosion–The lest of soil from small channels or the infestation of Parthenium Hysterophorus
rills by runoff water is called rill erosion. weed is :
Gully erosion–The formation of gullies due to (a) Cactoblastic Cactorum
excessive rill erosion is called as gully erosion. (b) Crylophagous salvinia
75. The making of small depressions of about 10-15 (c) Zygograma bicolorata
cm depth around the vegetation before sowing (d) Delias hypareta
of crop is called :
(a) Bunding (b) Bench terracing Ans. (c) : Zygograma bicolorata–A living organism
(bio-agent) used to limit the infestation of parthenium
(c) Basin listing (d) Graded buns
hysterophorus weed is called as zygograma biocolorata.
Ans. (c) : The making of small depression of about 10-
15 cm depth around the vegetation before sowing of 80. The process by which a herbicide passes from
crop is called as basins. one system to another system is called:
Bunding–The process of water retention and prevention (a) Adsorption (b) Formulation
of soil erosion through a spur or bunds of small height (c) Incorporation (d) Absorption
is called as bunding.
Ans. (d) : The process by which a herbicide passes
76. The weeds which normally start and complete from one system from another system is called
their life-cycle on the land is called: absorption.
(a) Alien weeds (b) Obligate weeds
Adsorption–The process of film formation on the
(c) Parasitic weeds (d) Terrestrial weeds surface of particle by herbicides or any other substance
Ans. (d) : The weed which normally start and complete is termed as adsorption.
their life cycle on the land is called as terrestrial weeds.
Other types of weed–
1. Perennial weeds
2. Broad leaf weeds
3. Herbaceous weeds
4. Parasitic weeds 81. The Lucknow-49 is a variety of:
5. Crop-associated weed (a) Mango (b) Guava
6. Alien weed (c) Amia (d) Ber
7. Facultative or obligate weeds
Ans. (b) : Lucknow-49 is a variety of guava. These
8. Noxious weed etc. fruits are spherical and meaty, seeds are soft and in
77. A weeds that has become an integral part of a plenty, pulp is while and contains 130 mg vitamin per
crop-ecosystem is called: 100 gm pulp.
(a) Facultative weeds (b) Satellite weeds Pulp–The fleshy part of fruit which is between outer
(c) Associated weeds (d) Noxious weds layer and seeds.
Ans. (b) : The weeds that has become an integral part of 82. Growing of two or more crops on the same
a crop-ecosystem is called satellite weeds. field per year, where the succeeding crop is
Facultative weeds–Those weed species that grow planted after the preceding crop has been
primarily in wild communities but often escape to harvested is called:
cultivated field, associated themselves closely with
(a) Multiple cropping
men's affairs. These are also known as apopytes.
Noxious weeds–It is a plant arbitrarily defined as being (b) Relay cropping
especially undesirable, troublesome and difficult to (c) Sequential cropping
control. (d) Ratoon cropping
OPSC ASCO Exam. 2021 (Paper-II) 46 YCT
Ans.(c): Sequential cropping–Growing of two or more Foliar application–Application of fertilizer in liquid
crops on the field where the succeeding crop is planted for though spray on standing crops for quick recovery
after the preceding crop has been harvested is called as from deficiency.
sequential cropping. Soil application–Direct application of liquid fertilizer
to the soil need special injecting equipment.
Relay cropping–Growing of two or more crop on the
same field before the harvesting of succeeding crop is Fertigation–Application of fertilizer with irrigation
called as relay cropping. water in either open or closed system.
Multiple cropping–Growing of many crop at a same 86. The moisture of the soil at which plants can no
field in a year is called multiple cropping. longer obtain enough moisture to meet the
transpiration requirement and water is held by
83. The quantity of water required by a crop in a soil so tightly as thin film around soil particles
given period of time of their normal growth is called:
under field condition is called: (a) Field capacity
(a) Water requirement (b) Available moisture
(b) Irrigation requirement (c) Permanent Wilting Point
(c) Consumptive use of water (d) Water-holding capacity
(d) Irrigation frequency Ans. (c) : Permanent wilting point–The moisture of
the soil which plants can be longer obtain enough
Ans. (c) : Consumptive use of water–The amount of
moisture to meet the transpiration requirement and
water required by a crop in a crop period for their
water is held by soil so tightly as thin film around soil
normal growth under field condition is called as particles is called permanent wilting point.
consumptive use of water. It does not contains
evaporation and infiltration of irrigation water from the
field.
84. When fertilizers are applied close to the seed or
plant which is adopted when relatively small
quantity of fertilizer has to be applied for
widely spaced crop is called:
(a) Deep placement
87. The downward movement of water through
(b) Localized placement
saturated soil when water is under pressure
(c) Drill placement and tension is less than 12 atmosphere is called:
(d) Band placement
(a) Percolation (b) Water intake
Ans. (b) : Localized placement–When fertilizers are (c) Permeability (d) Seepage
applied close to the seed or plant which is adopted when
Ans. (a) : Percolation–The downward movement of
relatively small quantity of fertilizer has to be applied
water through saturated soil when water is under
for widely spaced crop is called as localized placement.
pressure and tension is less than 12 atmosphere is called
Deep placement–Application of fertilizer in the
reduced zone to avoid nitrogen losses in lowland rice. as percolation.
Seepage–The horizontal movement of water through
Drill placement–When drilling seed and fertilizer
saturated soil under pressure.
simultaneously at the time of sowing is known as drill
placement or contact placement. Permeability–The movement of water through inter-
connected voids is called as permeability.
85. When the fertilizer solutions of low
88. The scientist who has given the concept of Law
Concentrations prepared for soaking seeds or of Minimum :
dipping roots of seedlings for early
(a) Mitscherlich (1909)
establishment is called:
(b) Justusvon Liebig (1840)
(a) Foliar application (b) Soil application
(c) Blackman (1005)
(c) Starter solutions (d) Fertigation
(d) Wilkrox(1942)
Ans. (c) : Application of fertilizers in liquid form.
Ans. (b) : Law of minimum–It is a law which
Starter solution–When the fertilizer solution of low described how to plant growth relied on the scarcest
concentration prepared for soaking seeds or dipping nutrient resource rather than the total amount of
roots of seedlings for early establishment is called resource available. This is given by Justusvon Liebig in
starter solution. 1840.
OPSC ASCO Exam. 2021 (Paper-II) 47 YCT
89. Plant that grows on extremely dry soil are 95. The reasons of enormous increase in the
classified under : livestock population of the country is :
(a) Thalophytes (b) Hydrophytes (a) Availability of green fodder
(c) Xerophytes (d) Hyroponics (b) Social attitude of people
(c) Multiple uses of animals
Ans. (c) :
(d) Large number of family members
Xerophytes – Extremely dry soil
Ans. (c) : The reasons of enormous increase in the
Thalophytes – Dry and moist soil
livestock population of the country is the multiple uses
Hydrophytes – Water of animals i.e. wool, milk, meat, eggs, etc.
Hydroponics – Technique of growing plant using water 96. The characteristics of tree species for Shelter
Halophytes – Saline soil belt is :
90. The C:N ratio, of humus is: (a) Nitrogen fixing (b) Fast growing
(c) Profused branching (d) Deep rooted
(a) 20 : 1 (b) 100 : 1
(c) 10 : 1 (d) 400 : 1 Ans. (c) : The tree species for shelter belt should have
profused branching which means repeated branching of
Ans. (c) : C:N Ratio–Carbon to Nitrogen ratio for trees. So that vertical growth will prevent. The presence
humus C : N ratio is 10 : 1 generally it is varies from of numerous branches is desirable agronomic feature of
(8–15) : 1 fodder crops.
91. Cutting of green branches and leaves of a tree Shelter belts–This is a row of trees along fance line.
for feeding the cattle is known as : they are planted to protect animals or crops from cold
winds and give shade in hot weather.
(a) Lopping (b) Pruning
97. Mulching is useful for:
(c) Pollarding (d) Thinning
(a) Conserving moisture
Ans. (a) : Lopping–Cutting of green branches and (b) Reducing crop growth
leaves of a tree for feeding the cattle is known as
(c) Nutrient depletion
lopping.
(d) High evaporation
Pruning–Removal of live or dead branches or multiple
leaders standing trees for the improvement of the tree or Ans. (a) : Mulching–The process of covering the soil
surface in order to conserve the moisture and prevent
its timber.
the weed growth near the vegetation or plants.
92. In the pellet method of sowing, the 98. The growth of seedling destroyed by animals is
homogeneous paste is prepared in the ratio : due to :
(a) 3 : 1 : 1 : 1 (b) 1 : 2 : 1 : 3 (a) Browsing (b) Grazing
(c) 2 : 1 : 1 : 2 (d) 4 : 1 : 2 : 1 (c) Trampling (d) Up-rooting
Ans. (*) : Ans. (c) : Trampling–The growth of seeding destroyed
by animals is trampling and then by grazing when crop
93. The optimum depth of sowing of grasses in the
grow few centimeters above the ground.
rangeland should be :
99. Which is not the component of agro forestry?
(a) 3 to 5 cm (b) 5 to 7 cm
(a) Land (b) Animal
(c) 0.5 to 1 cm (d) 2 to 4 cm
(c) Tree (d) Water
Ans. (c) : The optimum depth of sowing of grasses in
Ans.(d): There are three main components of agro
the range land should be 0.5 cm-1 cm. The thumb rule forestry
for optimum depth is 5 times the diameter of seeds.
(1) Animal
94. Which method is used for the entire plant (2) Crops/land
removal of scattered shrubs or tree seedlings of (3) Trees
small diameter?
• Water is not a main component of agroforestry.
(a) Grubbing (b) Girding
100. The tree species suitable for bio drainage in the
(c) Chopping (d) Mowing water logged areas :
Ans. (a) : Grubbing–The method which is used for (a) Eicane (b) Allanthus exceisa
removal of entire plant, scattered shrubs or tree seedling (c) Albizia procera (d) Gmelina arborea
of small diameter is called grubbing. Ans. (*) :
OPSC ASCO Exam. 2021 (Paper-II) 48 YCT
 Odisha Public Service Commission
Assistant Engineer (AE) Exam-2020
1. The radius of Mohr's circle for two equal 4. If the depth of simply supported beam carrying
unlike principal stresses of magnitude P is : an isolated load at its centre, is doubled the
(a) p (b) p/2 deflection of the beam at its centre will be
(c) Zero (d) None of these changed by a factor of :
Ans. (a) : Radius of Mohr's circle (a) 2 (b) 1/2
2 (c) 8 (d) 1/8
 σx − σ y 
 + τ xy [Given, σx = p, σy = –p, τxy = 0]
2
r=  Ans. (d) : Deflection at centre of a s/s beam is given as
 2 

 p − ( −p ) 
2

Then, r =   +0
 2 
WL3 WL3 ×12 WL3
( 2p ) δ= = =
2
4p 2 3
r= = =p 48EI 48Ebd 4 E bd 3
4 4
WL3
2. Rate of change in Bending Moment is equal to : δ1 =
(a) Shear force (b) Deflection 4bEd 3
(c) Slope (d) Rate of loading For d2 = 2d1
Ans. (a) : Rate of change of bending moment along the WL3 WL3 δ
length of beam is equal to shear force. δ2 = = = 1
4bEd 2 4bE ( 2d1 )3 8
3

dm
= Sx 5. For a given material Young's modulus is 200
dx
Rate of change of shear force is equal to the load GN/m2 and modulus of rigidity is 80 GN/m2.
The value of Poisson's ratio is :
ds
=w, w = load per unit length (a) 0.15 (b) 0.20
dx (c) 0.25 (d) 0.40
3. The maximum compressive stress at the top of
Ans. (c) : Young modulus E = 200 GN/m2
a beam is 1,600 kg/cm2 and the corresponding
tensile stress at the bottom of the beam is 400 Modulus of rigidity (G) = 30 GN/m2
kg/cm2. If the depth of the beam is 10 cm, the E = 2G (1+µ)
neutral axis from the top is at : E = 2G + 2Gµ
(a) 2 cm (b) 4 cm E − 2G 200 − 2 × 80
(c) 6 cm (d) 8 cm µ= =
2G 2 × 80
Ans. (d) : From stress relation (from stress diagram)
40
= = 0.25
160
6. If the dynamic viscosity of a fluid is 0.5 poise
and specific gravity is 0.5, then the kinematic
viscosity of that fluid in stokes is :
(a) 0.25 (b) 0.5
Ca t a / m f (c) 0.75 (d) 1.0
= = cbt Ans : (c) µ = 0.5 poise
xa d − xa D − xa
Ca = 1600 kg/cm2 NS
µ = 0.5 × 10–1
D = 10 cm m2
fcbt = 400 kg/cm2 specific gravity = 0.5
1600 400 ρ
= specific gravity = l
xa 10 − x a ρw
16000 – 1600 xa = 400 xa ρl =0.5× 1000 = 500 kg/m3
xa =
16000
= 8cm µ
kinematic viscosity, ν =
2000 ρ
OPSC AE Exam-2020 49 YCT
 0.5 ×10−1 τ0  y
ν= = τ0  1 − 
500 3  R
m2 2
ν = 1× 10−4 or 1 stokes y= R
s 3
7. Centre of buoyancy always : 10. In series-pipe problems :
(a) Coincide with the Centre of Gravity (a) The head loss is same through
(b) Coincide with the centroid of the volume of (b) The discharge is same through each pipe
liquid displaced (c) A trial solution is not necessary
(c) Remains above the Centre of Gravity (d) The discharge through each pipe is added to
(d) Remains below the Centre of Gravity obtain discharge
Ans. (b) : Centre of Buoyancy is centroid of the volume Ans. (b) : For pipes in series discharge through all pipes
of liquid displaced. will be same. For pipe joined in parallel, the head loss
Centre of gravity is the point where total mass of the due to friction will be same for pipes.
body is assumed to act. 11. The best hydraulic channel cross section is the
(It is Geometrical centre for body having uniform mass one which has a :
distribution) (a) Minimum roughness co-efficient
Note–Centre of buoyancy coincides with centre of (b) Least cost
gravity in case when body of uniform mass distribution (c) Maximum area for a given flow
is completely immersed in liquid. (d) Minimum wetted perimeter
8. A rectangular block 2 meters long, 1 meter Ans. (d) : Most economical or most efficient or best
wide and 1 meter deep floats in water, the section of the channel.
depth of immersion being 0.5 meter. If the • Discharge is maximum when hydraulic radius (R) is
water weighs 10 kN/M3, then the weight of the
maximum and wetted perimeter is minimum (ρw) .
block is :
A
(a) 5 kN (b) 10 kN Hydraulic radius 'R' =
(c) 15 kN (d) 20 kN ρw
Ans. (b) : Given data, 12. For maximum discharge in a circular channel
L=2m section, the ratio of depth of flow to that of the
B=1m diameter of the channel is :
D=1m (a) 0.95 (b) 0.81
Depth of immersion (h) = 0.5 m (c) 0.50 (d) 0.30
Weight density of water = 10 kN/m3 Ans. (a) : Most economical circular channel section.
We know that, • For maximum discharge through a circular channel,
Weight of water displaced = weight of block the depth of flow is equal too 0.95 time its diameter.
γw (0.5 × 2 × 1) = ρb (2 × 1 × 1) • Maximum velocity occurs when the depth of flow is
10 ×1×1 0.81 time the diameter of the circular channel.
ρb = = 5 kN / m3
2 13. If the conjugate depths before and after the
Hence weight of block (wb) = ρb × Vb jump are 0.5 m and 2.5 m respectively, then the
loss of energy in the hydraulic jump will be :
=5×2×1×1
(a) 0.8 m (b) 1.6 m
= 10 kN/m3
(c) 3.2 m (d) 6.4 m
9. The distance from pipe boundary, at which the
'turbulent shear stress' is one-third the 'wall Ans. (b) : The energy less due to hydraulic jump is
shear stress', is : given s.
( y 2 − y1 ) ( 2.5 − 0.5 )
3 3
(a) 1/3 r (b) ½ r 8
E= = = = 1.6 m
(c) 2/3 r (d) 3/4 r 4y1 , y 2 2 × 2.5 × 0.5 5
Where r is radius of the pipe.
14. Hydraulic pressure on a dam depends upon its:
Ans. (c) : The distance from pipe boundary at which the
(a) Length (b) Depth
turbulent shear stress is one-third the wall shear stress is
(c) Shape (d) SHape and depth
 y
τ = τ0  1 −  Ans. (b) : Hydraulic pressure on a dam depends upon
 R  its depth
τ0 P = wH (w = ρg)
τ=
3 • The horizontal water pressure acts at a height of H/3.
OPSC AE Exam-2020 50 YCT
15. Manometers are used to measure : (c) High flood drainage discharge is large and
(a) Pressure in water channels, pipes etc continues for a long time
(b) Difference in pressure at two points (d) None of these
(c) Atmospheric pressure Ans. (a) : Aqueduct–
(d) Very low pressure When the HFL of the drain is sufficiently below the
Ans. (b) : Manometer are used to measure difference of bottom of the canal such that the drainage water flows
pressure between two points. freely under gravity, the structure is known as
• Pressure in water channels/pipes can be measured Aqueduct.
using pressure gauge. H.F.L.

• Atmospheric pressure is measured using barometer. Bank of canal

stream
16. The normal annual precipitation of stations A,
B, C and D are 700 mm, 1,000 mm, 900 mm
and 800 mm respectively. If the storm
precipitation at three stations B, C and D
where 100 mm, 90 mm and 80 mm respectively,
then the storm precipitation for station A will F.S.L.
piers canal
be :
(a) 70 mm (b) 80 mm
(c) 90 mm (d) 105 mm
Ans. (a) : Normal precipitation (N) at station A, B, C, D 20. Hydrograph is the graphical representation of :
are 700 mm, 1000 mm, 900 mm and 800 mm (a) Rainfall and time
respectively. (b) Surface run off and time
Stream precipitation (P) at station B, C, D are 100mm, (c) Ground water flow and time
90 mm and 80 mm respectively. (d) Run off and time
As per normal ratio method– Ans. (a) : Hydrograph : Hydrograph is a graph that
PA 1  PB PC P  shows plot of discharge in a stream against time (in
=  + + D  hours or days)
NA n  NB NC ND 
Hydrograph is the graphical representation of runoff
700  100 90 80  and time. Or
PA =  + + 
3  1000 900 800  Hydrograph is a plot of discharge in a stream plotted
700 3 against time chronology.
PA = × = 70 mm
3 10
17. S-hydrograph is used to obtain hydrograph of :
(a) Shorter duration from longer duration
(b) Longer duration from shorter duration
(c) Both (A) and (B)
(d) None of these
Ans. (c) : A 'S' hydrograph is nothing but a hydrograph
generated by continuous effective rainfall occurring at • [Runoff = Direct runoff + base flow]
an uniform rate for an indefinite period, it is called 'S'
hydrograph comes out like alphabet 'S' though slightly 21. Infiltration rate is always :
deformed. (a) More than the infiltration capacity
18. A major resistive force in a dam is : (b) Equal to or less than the infiltration capacity
(a) Water pressure (b) Self weight of dam (c) Less than the infiltration capacity
(c) Wave pressure (d) Uplift pressure (d) Equal to or more than the infiltration capacity
Ans. (b) : Self weight of the dam– The weight of the Ans : (c) Infiltration rate is always less or equal to
dam is major resisting force for gravity dam. infiltration capacity.
Uplift force– It is the pressure exerted by the seepage This infiltration rate is the velocity or speed at which water
water on the base of the dam. enters into the soil. It is usually measured by the depth (in
• It depends on head of water upstream and mm) of the water layer that can enter the soil in one hour.
downstream side. 22. Seepage through embankment in an earthen
19. Aqueduct or super passage type of works are dam is controlled by :
used when : (a) Drain trenches
(a) High flood drainage discharge is small (b) Drainage filters
(b) High flood drainage discharge is large and (c) Relief Wells
short lived (d) Provision of down streams beams
OPSC AE Exam-2020 51 YCT
Ans. (b) : Seepage through embankment of earthen dam 26. For water bound macadam roads in localities
can be controlled by providing drainage trenches. of heavy rainfall, the recommended value of
• Seepage through foundation is controlled by camber is :
impervious cut-off whereas seepage through (a) 1 in 30 (b) 1 in 36
embankment is controlled by rock toe, horizontal (c) 1 in 48 (d) 1 in 60
blanket and chimney drain. Ans. (a) : IRC values for camber–
23. Which of the following is least suited for an Surface type Heavy rain Light rain (%)
earthen dam ? (%)
(a) Ogee spillway 2% 1.7%
Concrete/
(b) Chute spillway
bitumenous
(c) Side channel spillway
Gravel/ WBM 3% 2.5%
(d) Shaft spillway
Earthen 4% 3.0%
Ans. (a) : Ogee spillway– Ogee spillway is an
improvement upon the free overfall spillway and is widely • 3% Means 1 in 33.
used with concrete, masonry arch and buttress dams. Hence most appropriate answer is 1 in 30.
• Such spillway is made in accordance with the shape 27. In the stopping distance is 60 meters, then the
of the lower nappe of a free falling Jet, over a minimum stopping sight distance for two lane,
ventilated sharp crested weir. two way traffic is :
• The least suited spillway for earthen dams is the ogee (a) 180 m (b) 120 m
spillway as these are overflown spillways that can (c) 30 m (d) 60 m
not be constructed on the earthen dam as it will wash
away soil and pose other problem also. Ans. (d) : Stopping sight distance (SSD) = 60 m
24. The main function of a divide wall is to : • SSD for two lane two way traffic = SSD = 60 m
(a) Control the silt entry into the canal • SSD for single lane two way traffic = 2 × 60
(b) Prevent river floods from entering into the canal = 120 m
(c) Separate the under sluices from weir proper 28. If the average centre to centre spacing of
(d) Provide smooth flow at sufficiently low velocity vehicles is 20 meters, then the basic capacity of
Ans. (c) : The divide wall– The divided wall is a traffic lane at a speed of 50 kmph is :
masonry or a concrete wall constructed at right angle to (a) 2,500 vehicles per day
the axis of the weir, and separates the ''weir proper'' (b) 2,000 vehicles per hour
from the 'under-sluices'. (c) 2,500 vehicles per hour
• The main function of a divide wall is to separate the (d) 1,000 vehicles per hour
under sluices from weir proper. Ans : (c) Basic capacity
• The divide wall extends on the upstream side beyond 1000V
the beginning of the canal head regulator, and on the C=
downstream side, it extends up to the end of loose S
protection of the under-sluices. V= angular speed (in kmph)
• The top width of divide wall is about 1.5 to 2.5 meters. V=50 kmph
25. A divide wall is provided : S = centre to cetnre spacing of vehicle (in m) = 20 m
(a) Parallel to the axis of weir and up stream of it 1000 × 50
C= = 2500 veh./ hr
(b) At right angles to the axis of weir 20
(c) Parallel to the axis of weir and downstream of it 29. The background colour of inforatory sign
(d) At an inclination to the axis of weir board is :
Ans. (a) : Divide wall– It is built at right angles to the (a) Red (b) Yellow
axis of the weir separating the weir and the under sluices. (c) Green (d) White
Ans. (c) : Prohibitory sign– White background, red
border, prohibition black in colour.
Warning sign– White background, red colour border,
black symbol.
Informatory sign– Green colour background, white
border, white symbol.
30. The ductility value of bitumen for suitability in
road construction should not be less than :
(a) 50 cm (b) 60 cm
(c) 40 cm (d) 30 cm
OPSC AE Exam-2020 52 YCT
Ans. (a) : Ductility test of bitumen– Bitumen binder 35. The suitable gradient within which an engineer
should be sufficiently ductile, i.e. it should be capable must endeavour to design the road is called :
of being stretched without breaking. Ductility is the (a) Limiting gradient (b) Ruling Gradient
opposite of brittleness. (c) Average gradient (d) Exceptional gradient
• Ductility is measured by stretching a standard Ans. (b) : Ruling gradient– It is the maximum gradient
briquette of bitumen having a cross-sectional area of with which the designer attempts to design the vertical
1 cm2 at a temperature of 270C the rate of pull profile of the road.
being 5 cm/min. • Ruling gradient is adopted by the designer by
• It value varies from 5 to 100. considering a particular speed and vehicle with
• A minimum values of 50 cm is commonly specified. standard dimensions.
• IFC has recommended a minimum ductility value of Limiting gradient – It is adopted when ruling gradient
75 cm for grade of 45 and above. results in enormous cost of construction.
• It is provided frequently in hilly terrain.
31. Bitumen grade 80/100 means : Exceptional gradient– These gradients are provided in
(a) Its penetration value is 8 mm very short stretches of road in unavoidable situation.
(b) Its penetration value is 10 mm • Length of stretch >/ 100 m.
(c) Its penetration value is 8 to 10 mm
36. The ranging of the line between two stations
(d) Its penetration value is 8 to 10 cm across the raised ground is called :
Ans. (c) : Penetration test of bitumen– (a) Direct ranging (b) Indirect ranging
• A measure of the hardness of bitumen is indirectly (c) Random line ranging (d) None of these
obtained by the penetration test.
Ans. (b) • Indirect ranging is adopted when two ends
1 of survey line are not visible from either of the station
• The unit of penetration is mm.
10 due to hill or high ground between them.
80 • Direct ranging minimum number of ranging rods
• Thus penetration means a penetration of 8-10 mm.
100 required 3 and in indirect ranging minimum number of
32. If V is the speed of a moving vehicle, r is the ranging rods required 4.
radius of the curve, g is the acceleration due to • Direct ranging is adopted when two ends of survey
gravity, W is the width of the carriage way, the line are visible from either of the station.
super elevation is : 37. The correct sequencing of setting up a plane
(a) WV/gr (b) W2V/gr table at a working station is :
(c) WV/gr2 (d) WV2/gr (a) Levelling, Centering, Orienting
(b) Centering, Orienting, Levelling
V2
Ans. (*) : Superelevation (e) = −f (c) Orienting, Levelling, Centering
gR
(d) Levelling, Orienting, Centering
WV 2 Ans. (a) : Sequencing of sitting up a plane table at a
Centrifugal force (CF) =
gR working station is levelling, centering, orienting.
Hence none of the above options is correct. Levelling– The table is levelled by placing the level on
33. Reinforcement in cement concrete slab of road the board in two positions at right angles and getting the
pavements is placed : bubble control in both directions.
(a) In the form of welded mesh Centering– Table should be so placed over the station
on the ground that the point plotted on the sheet
(b) Longitudinally
corresponding to the station occupied should be exactly
(c) Transversally over the station on the ground.
(d) Longitudinally and transversally Orientation– Orientation is the process of putting the
Ans. (a) : Reinforcement in cement concrete slab of plane table into some fixed direction.
road pavement is provided in the form of welded wire
38. Which of the following scale is the largest one ?
mesh for avoiding the slipping failure of the
(a) 1 cm = 50 m (b) 1 : 42000
reinforcement or shear failure.
(c) RF = 1/300000 (d) 1 cm = 50 cm
34. Minimum thickness of a layer of fine sand
required to cut off the capillary rise of water Ans. (a) : 1
= scale ↑= R.F. ↑
completely should be : small value
(a) 40 cm (b) 52 cm 1 cm = 50 m > 1 : 42000 > RF = 1/300000>1 cm = 50 km
(c) 64 cm (d) 76 cm 39. The length of a chain is measured from :
Ans. (d) : Minimum thickness of a layer of the sand (a) Centre of one handle to centre of other handle
required to cut off the capillary rise of water completely (b) Outside of one handle to outside of other
should be 76 cm. handle
OPSC AE Exam-2020 53 YCT
 (c) Outside of one handle to inside of other 43. If the fore bearing of a line is 36º 15' its back
handle bearing will be :
(d) Inside of one handle to inside of other handle (a) 36º 15' (b) 126º 15'
(c) 143º 15' (d) 216º 15'
Ans. (b) Chains are formed straight links of galvanised o
mild steel wire bent into rings at the end and joined Ans. (d) : FB = 36 15'
each other by three small circular or oval wire rings. BB = FB ± 180o
The length of a link is the distance between the centers = 36o15' + 180o
of two consecutive middle rings while the length of the = 216o15'
chain is measured from the outside of one handle to the 44. The theodolite is an instrument used for
outside of other handle. measuring very accurately :
(a) Horizontal angles only
40. The horizontal angle between the true meridian
(b) Vertical angles only
and magnetic meridian is called :
(c) Horizontal and vertical angles
(a) Azimuth (b) Declination
(d) Linear measurement
(c) Local attraction (d) Magnetic bearing
Ans. (c) : Theodolite - Theodolite is an instrument used
Ans. (b) : Bearing Angle-Direction of a line wrt of for -
fixed meridian is called bearing • Measurement of horizontal angle
Magnetic bearing- Line joining magnetic north pole, • Measurement of vertical angle
magnetic south pole and point of reference is called • Measurement of horizontal distance
magnetic Meridian. Bearing taken art magnetic • Measurement of vertical distance
meridian is called Magnetic bearing • Ranging etc.
Magnetic declination-At any place horizontal angle The theodolite has been the most-used instrument
between true median and magnetic meridian is called for measuring horizontal and vertical angles since, at
Magnetic declination. least, the sixteenth century when it was first described.
TB = MB ± declination It consists of a telescope, mounted on a tripod that
For eastern declination θ is positive rotates 360° around a vertical axis and vertically around
a horizontal axis, which are perpendicular.
For western declination θ is negative
45. The horizontal distance between any two
41. A series of closely spaced contour lines consecutive contours is called :
represent a : (a) Vertical equivalent
(a) Steep slope (b) Gentle slope (b) Horizontal equivalent
(c) Uniform slope (d) Plane surface (c) Contour interval
Ans. (a) : • A series of closely spaced contour lines (d) Contour gradient
represent a steep slope. Ans. (b) : ● The horizontal distance between two points
• Flat ground is indicated where the contours are widely on two consecutive contours is known as the horizontal
equivalent and depend upon the slope of the ground.
spaced.
● The vertical distance between any two consecutive
• A uniform slope is indicate when the contour lines are contour is called contour interval. The contour interval
uniformly spaced. is kept constant for a contour map.
• They indicated a gentle slope if they are for apart. If 46. The maximum frictional force which comes
they are equally spaced uniform slope is indicated. into play when a body just beginning to slide
42. If the intercept on a vertical staff is observed as over the surface of another body is known as :
0.75 m from a tacheometer, the horizontal (a) Static friction
distance between the tacheometer and staff (b) Dynamic friction
station is : (c) Limiting friction
(a) 7.5 m (b) 25 m (d) Coefficient of friction
(c) 50 m (d) 75 m Ans. (c) : The maximum friction that can be generated
Ans. (d) : Given that, between two static surfaces in contact with each other.
S = Staff intercept = 0.75 m Once a force applied to the two surfaces exceeds the
limiting friction motion will occur.
D = Distance between the tacheometer
The law of limiting friction are
and the staff station = ?
• The direction of limiting friction force is always
D = KS + C opposite the direction of motion.
K = Multiply constant = 100 • It always acts tangential to the two surfaces.
C = additive constant = 0 (zero) • It is dependent on the material and the nature of the
D = 0.75 × 100 + 0 surfaces in contact.
D = 75 m • It is independent of the shape and area.
OPSC AE Exam-2020 54 YCT
47. Two balls of equal mass and of perfectly elastic • This is also equal to the length of the chain pulled
material are lying on the floor. One of the balls over the larger pulley. Since the smaller pulley also
with velocity is made to strike the second ball. turns with larger one, there fore length of the chain
Both the balls after impact will move with a released by the smaller pulley = πd
velocity : • Net shortening of the chain = πD– πd = π(D-d)
(a) v (b) v/2 This shortening of chain will be equally divided
(c) v/4 (d) v/8 between the two portions of the chain,
Ans. (b) : • The two perfectly elastic spherical bodies π(D − d)
=
are A and B. 2
• Mass of both the bodies is same that is m, the initial Distance moved by the effort
velocity of object A is VA, the initial velocity of body B Velocity ratio, V.R. =
Distance moved by the load
is VB, V0 is common velocity after impact.
• Moment of conservation πD 2D
= =
MVA + MVB = ( m + m ) V0
π
(D − d) ( D − d)
2
m ( V ) = 2m × V0 50. A rubber ball is dropped from a height of 2
V meters. If there is no loss of velocity after
V0 = rebounding, the ball will rise to a height of :
2
(a) 1 meter (b) 2 meter
48. The angular velocity (in radians/second) of a
(c) 3 meter (d) 4 meter
body rotating at N RPM is :
Ans. (b) : After rebounding the height of ball
(a) πN/60 (b) πN/180
x = h × e2
(c) 2πN/60 (d) 2πN/180 = 2×1 (body is fully elastic no loss of
Ans. (c) : Angular velocity is defined as the rate of energy)
change of angular displacement with respect to time. =2m
If body is rotating at the rate of N rpm (revolutions per 51. The law of motion involved in recoil of a gun is:
minute), then it angular velocity – (a) Newton's First Law of Motion
2πN (b) Newton's Second Law of Motion
w= rad / s
60 (c) Newton's Third Law of Motion
49. The velocity ratio of a differential pulley block (d) None of these
with D and d as diameter of larger and smaller Ans. (c) : Newton's Third Law of Motion states that to
pulley is : every action there is an equal and opposite reaction. A
(a) D/(D – d) (b) D/(D + d) gun recoils due to reaction force of bullet fired by it.
(c) 2D/(D – d) (d) 2D/(D + d) 52. The moment of inertia of circular section about
Ans. (c) : Differential pulley block– it's diameter (d) is :
(a) πd3/16 (b) πd3/32
(c) πd4/32 (d) πd4/64
Ans. (d) :

53. Two forces are acting at angle of 120º. The

Let, bigger force is 40 Newton and the resultant is
D = Diameter of the pulley P1 perpendicular to the smaller force. The smaller
d = Diameter of the pulley P2 force is :
w = Weight lifted, and (a) 20 N
P = Effort applied to lift the weight. (b) 40 N
We know that displacement of the effort in the (c) 80 N
revolution of the upper pulley black, = πD (d) None of these
OPSC AE Exam-2020 55 YCT
Ans. (a) : 56. The water content of soil is defined as the ratio
of :
(a) Volume of water to volume of given soil
(b) Volume of water to volume of voids in soil
(c) Weight of water to water to weight of air in
voids
(d) Weight of water to weight of solids of given
By Lami's equation mass of soil
P 40 Ans. (d) : Water content of soil– It is the ratio of weight
=
sin1500 sin 900 of water and weight of solids of given mass of soil.
P = 20 N W
54. The acceleration of a particle moving with w= w
Ws
simple harmonic motion, at any instant is given
by : 57. The minimum size of grains of silt is about :
(a) ω . y (b) ω . y
2
(a) 0.0002 mm (b) 0.002 mm
(c) ω2/ y (d) ω3. y (c) 0.02 mm (d) 0.2 mm
Where ω is the angular velocity of the particle Ans. (b) : As per IS particle size classification–
in rad/sec and y is the displacement of the
d < 2µ – Clay
particle from mean position.
2µ ≤ d < 75µ – Silt
Ans. (b) : Distance traveled by the object is
75µ ≤ d < 4.75 mm – Sand
y = a.sin (ω.t) ......... (i)
To obtained velocity differentiate equation (i) with 4.75 mm ≤ d < 80 mm – Gravel
respect to 't'. 80 mm ≤ d < 300 mm – Cobbles
dy d > 300 mm – Boulders
dv = = a.cos ( ω.t ) × ω ......... (ii)
dt 58. Gravel and sand are :
To obtained acceleration differentiate equation (ii) with (a) Cohesive coarse grained soil
respect to 't' (b) Cohesive fine grained soil
dv d (c) Non-cohesive coarse grained soil
= {a cos ( ω.t ) × ω}
dt dt (d) Non-cohesive fine grained soil
a = −a sin ( ω.t ) × ω 2
.......... (iii) Ans. (c) : Cohesive soils are the silt and clays or fine
grained soil. A cohesion less (soil) (non-cohesive) soil
a = –y × ω = y ω
2 2
are soil that do not adhere to each other and rely on
55. The moment of inertia of a rectangular section, friction.
3 cm wide and 4 cm deep, about the X-X axis is These soil are sand and gravels.
(a) 9 cm4 (b) 12 cm4
4 59. The ratio of settlement at any time 't' to the
(c) 16 cm (d) 20 cm4
final settlement, is known as :
Ans. (c) : Given,
(a) Coefficient of consolidation
width (b) = 3 cm
(b) Degree of consolidation
depth (d) = 4 cm
(c) Consolidation index
bd 3 (d) Consolidation of undisturbed soil
I xx =
12 Ans. (b) : The ratio of settlement at any time ‘t’, to the
final settlement is called as degree of consolidation
∆h
(%u) = × 100
∆H
Where,
∆H = Final total settlement at the end of completion of
primary consolidation i.e., at t = ∞
3cm × ( 4cm )
3

Ixx = ∆h = settlement occured at any time ‘t’.

12 60. According to Terzaghi, the net ultimate
3× 4× 4× 4 4 bearing capacity of Clay is given by :
= cm
12 (a) c Nq (b) c Nγ
= 16 cm4 (c) c Nc (d) 1.3 c Nc
OPSC AE Exam-2020 56 YCT
Ans. (c) : Ultimate bearing capacity of clay – Ans. (c) : Given

{ }
N =0
q u = CN C + γDf N q + 0.5BγN γ N γ = 1
q
bulk density (γb) = 2.3 g/cm3
Water content (w) = 15% = 0.15
q u = CN C + γD f γ
Dry density, γd = b
Net ultimate bearing capacity (qnu) 1+ w
2.3
qnu = qu – γDf γd =
= CNC + γDf – γ Df 1 + 0.15
γd = 2.0 g/cm3
q nu = CN C
64. The plasticity index is the numerical difference
61. If w is the water content and γ is the unit between :
weight of soil mass, then the unit weight of dry (a) Liquid limit and plastic limit
soil (γd) is equal to : (b) Plastic limit and shrinkage limit
(a) (w/γ) + 1 (b) (γ/w) + 1 (c) Liquid limit and shrinkage limit
(c) γ/(1 + w) (d) (1 + w)/γ (d) None of these
Gγ w (1 + w ) Ans. (a) : Plasticity index is the numerical difference
Ans. (c) : Bulk density (γ) = .......(i)
(1 + e ) between liquid limit and plastic limit. Ip=WL−WP
65. Mechanical stabilization of soil is done with the
Gγ w
Dry density (γd) = .......(ii) help of :
1+ e (a) Cement (b) Lime
Relation b/w γb/γd – (c) Bitumen (d) Proper Grading
γ Gγ w (1 + w ) (1 + e ) Ans. (d) : Mechanical stabilization–
∴ = ×
γd (1 + e ) ( Gγ w ) (i) Mechanical stabilization consist of – Mechanical
stabilization is the process of improving the properties
γ of the soil by changing its gradation.
∴ γd =
1+ w • The gravity of the soil particles (by changing the
62. The relation between void ratio(e) and porosity composition of the soil mixture. By adding or removing
ratio(n) is : the different soil particles)
(a) n = (1 + e)/(1 – e) • Compacting the soil to improve the stability and
(b) e = (1 + n)/(1 – e) strength.
(c) n = e/(1 – e) (ii) No chemicals are added to the soil in the mechanical
(d) e = n/(1 – e) stabilization.
(iii) It is used in preparing base course of roads.
V
Ans. (b) : Void ratio (e) = v 66. A load 'W' is moving from left to right
Vs supported on a simply supported beam of span
Vv 'L'. The maximum bending moment at 0.4 L
Porosity (n) = × 100 from the left support is :
V
(a) 0.16 WL (b) 0.20 WL
(c) 0.24 WL (d) 0.25 WL
Ans. (c) :

∴ (0 < n < 100)

∴Relation between e and n–
e n 0.4L × 0.6wL
n= , e= So Mmax = = 0.24wL
1+ e 1− n L
Where n = Porosity
e = void ratio
63. A soil has bulk density of 2.30 g/cm3 and water
content 15 percent, the dry density of the
sample is :
(a) 1.0 g/cm2 (b) 1.5 g/cm3
3
(c) 2.0 g/cm (d) 2.5 g/cm3
OPSC AE Exam-2020 57 YCT
67. In Moment Distribution Method, the sum of Ans. (a) :
distribution factors of all the members meeting
at any joint is always :
(a) Zero (b) Less than 1
(c) 1 (d) Greater than 1
Ans. (c) : The moment distribution method is a Given —
structural analysis method for statically indeterminate l1= l, l2 = l, A1 = A, A2 = A, F1 = F, F2 = F
beams and frames developed by Hardy cross. The
Fl
method only accounts for flexural effects and ignores ∆l1 =
axial and shear effects. AE1
The sum of distribution factors of all members meeting Fl
at any joint is always one '1'. ∆l 2 =
AE 2
68. When a uniformly distributed load, longer than
the span of the girder moves from lef to right, ∆l1 Fl AE 2
= ×
then maximum bending moment a mid section ∆l 2 AE1 Fl
occurs when the uniformly distributed load ∆l1 E 2 5
occupies : = =
∆l 2 E1 2
(a) Less than the left half span
(b) Whole of the left half span E1 2
=
(c) More than the left half span E2 5
(d) Whole span 71. The shear force diagram for a cantilever beam
Ans. (d) : When a uniformly distributed load, longer of length l and carrying a gradually varying
than the span of the girder moves from left to right, then load from zero at the free end and W per unit
maximum bending moment at mid section occurs when length at the fixed end is a :
the uniformly distributed load occupies whole span. (a) Horizontal straight line
From the ILD, it is clear that maximum bending (b) Vertical straight line
moment at C will occur when the UDL covers the
(c) Inclined line
whole span.
(d) Parabolic curve
Maximum bending moment at C
= Area of IlD for MC × w Ans. (d) :
wa ( L − a )
Mmax = unit
2
69. Degree of static indeterminacy of a rigid-
jointed plane frame having 15 members, 3
reaction components and 14 joints is :
(a) 2 (b) 3
(c) 6 (d) 8 (S.F.)XX = Area of loading diagram between X-X and
A.
Ans. (c) : Given data,
1
Number of members (m) = 15 (S.F. )XX = ( w x ) ( x )
Reaction (R) = 3 2
Number of joints J = 14 1  wx 
=  x
Degree of static indeterminacy 2 L 
Ds = (3m – 3J + R) wx 2
= 3 × 15 – 3 × 14 + 3 (S.F. )XX = .......... (1)
2L
= 45 – 42 + 3= 6 Hence equation (1) shows parabolic variation of S.F.
70. Two bars of different material and same size
72. A simply supported beam carries a varying
are subjected to the same tensile force. If the
load from zero at one end and ω at the other
bars have unit elongation in the ratio of 5 : 2,
then the ratio of the modulus of elasticity of the end. If the length of beam is α, the maximum
two materials will be : bending moment is :
(a) 2 : 5 (b) 5 : 2 (a) ωα/27 (b) ωα2/27
(c) 4 : 3 (d) 3 : 4 (c) ω2α/ 27 (d) ωα2/ 9 3
OPSC AE Exam-2020 58 YCT
Ans. (d) : ∑ M B = 0 73. The equivalent length of a column of length L,
having one end fixed and the other end free, is :
ωα
RA + RB = (a) 2L (b) L
2
(c) L/2 (d) L/ 2
1 α
× ω× α × Ans. (a) :
RA = 2 3
Type of support Effective
α
Length
ωα 2 ωα both end hinged Leff = L
RA = =
6α 6 both end are fixed Leff = L/2
ωα one end fixed other end free Leff = 2L
RB =
3 One end fixed other end hinged Leff = L/ 2
74. The single rolling load of 8 kN rolls along a
girder of 15 m span. The absolute maximum
bending moment will be :
(a) 8 kN.m (b) 15 kN.m
(c) 30 kN.m (d) 60 kN.m
Ans. (c) :

When single load is moving then maximum bending

moment will occur at mid span. where the ordinate will
7.5 × 7.5
be = 3.75
For the maximum bending moment. 15
ωα 1  ω  So, the maximum moment = Load × ordinate of the ILD
Vx = 0 ⇒ −  x × x = 0 for the moment.
6 2α 
= 3.75 × 8 = 30 kN-m
ωx 2 ωα
= 75. For a single point load W moving on a
2α 6 symmetrical three hinged parabolic arch of
α2 span L, the maximum sagging moment occurs
x2 = at a distance x from the ends. The value of x is :
3
(a) 0.211 L (b) 0.25 L
α (c) 0.234 L (d) 0.5 L
x=
3 Ans. (a) :
→ Maximum bending moment
 ωα  1ω  x
=  ×x− x × x ×
 6   2 α  3
α
→ But, x =
3
ωα  ω
2
α3 
MBmax = − × 
6 3  6α 3 3 
ILD for bending moment
ωα 2
BM max = Bending moment at any section = Beam moment by
9 3 From similar triangle
OPSC AE Exam-2020 59 YCT
 A 'B' x ( l − x ) Ld
= (iii) Compression – lap length = max 
x L/2 24 φ
2x 2
A 'B' = 2 ( l − x ) Ld = Development length
l φ = Smaller dia of bar.
x ( l − x ) 2x 2
= − 2 (l − x ) 78. If H is the overall height of a retaining wall
x l retaining a surcharge, the width of the base
2x 3 + xl 2 − 3x 2 l slab usually provided, is :
⇒ (a) 0.3 H (b) 0.4 H
l2
(c) 0.6 H (d) 0.7 H
∂B'm'
taking =0 Ans. (d) : If H is the overall height of a retaining wall
∂x retaining a surcharge, the width of the base slab usually
6x 2 + l 2 − 6xl = 0 provided 0.70 H.
x = 0.211 (from both ends) 79. Workability of concrete is inversely
76. If the length of a wall on either side of a linter proportional to :
opening is at least half of its effective span L, (a) The time of transit
the load W carried by the lintel is equivalent to (b) The water-cement ratio
the weight of brickwork contained in an (c) The air in the mix
equilateral triangle, producing a maximum (d) The size of aggregate
bending moment: Ans. (a) : Workability of concrete is inversely proportional
(a) WL/2 (b) WL/4 to the time of transit and aggregate cement ratio.
(c) WL/6 (d) WL/8 The workability of concrete is defined as the case with
Ans. (c) : which concrete can be prepared, transported, and stored
on site without losing its homogeneity.
80. If diameter of a reinforcement bar is d, the
anchorage value of the hook is :
(a) 4d (b) 8d
(c) 12d (d) 16d
Ans. (d) : Anchorage values as per IS code 456 : 2000
clause number 26.2.2.1–
Bend and hooks–
• The anchorage value of bend shall be taken as 4 times
and the diameter of the bar for each 450 bend subjected
to maximum of 16 times of diameter of bar.
• The anchorage value of standard U-type hook shall be
equal to 16 times of diameter of bar.

wL2 wL
Maximum bending moment = =
6 6
77. The length of the lap in a compression member
is kept greater than bar diameter x 81. According to Indian Standards, the pozzolana
(Permissible stress in bar/Five times the bond content in Portland Portland Pozzolana
stress) or : Cement is :
(a) 12 bar diameters (b) 18 bar diameters (a) 10% to 25% (b) 25% to 35%
(c) 24 bar diameters (d) 30 bar diameters (c) 35% to 35% (d) More than 50%
Ans. (c) : Following lap length must be ensured for Ans. (a) : Portland pozzolana cement– It is
splicing. manufacturingd by grinding OPC and pozzolana
(usually flyash 10 to 25% by mass)
L d
(i) Bending tension – lap length = max  • The hydration of PPC is a slower process than
30φ hydration of OPC pozzolanic cement increase
 2 L d strength compared to OPC.
(ii) Direct tension – lap length = max  • Mass concrete (dam), retaining walls, large
30φ foundation etc.
OPSC AE Exam-2020 60 YCT
82. For longitudinal reinforcing bars in a column, 86. For rivet diameter up to 24 mm, the diameter
the cover should not be lees than : of the rivet hole is larger than the diameter of
(a) 10 mm (b) 20 mm the rivet by :
(c) 30 mm (d) 40 mm (a) 1.0 mm (b) 1.5 mm
Ans. (d) : According to IS 456 : 2000 : Longitudinal (c) 2.0 mm (d) 2.5 mm
reinforcing bar in a column nominal cover shall in any Ans. (b) : In case of rivets,
case not be less than 40 mm, or less than the diameter of Diameter of hole
such bars. = Dia of rivet (d) + 1.5 mm, d ≤ 25 mm
• In case of columns of minimum dimension of 200 mm = Dia of rivet (d) + 2 mm, d > 25 mm
or under, whose reinforcing bars not exceed 12 mm, a ⇒ 24+1.5 =25.5 mm
nominal cover of 25 mm may be used. 87. The effective length of a fillet weld is taken as :
Note : For footing minimum cover shall be 50 mm. (a) The actual length plus twice the size of weld
Nominal cover : The design depth of concrete cover (b) The actual length minus twice the size of
provided at all type of steel reinforcement including weld
links is called "nominal cover." (c) The actual length plus thrice he size of weld
83. For the design of retaining walls, the minimum (d) The actual length minus thrice the size of
factor of safety against overturning is taken as : weld
(a) 1.5 (b) 2.0 Ans. (b) : : Effective length of fillet weld,
(c) 2.5 (d) 3.0 l = over all length − 2s
Ans. (a) : For the design of retaining walls, the s = size of weld
minimum factor of safety against overturning is taken as • Effective length of fillet weld is the length for which
F.S > 1.5 for granular soil. the specified size and throat thickness of weld exist.
F.S > 2.0 for cohesive soil. • Effective thought thickness = k × S
84. For deflection of a simply supported beam to
be within permissible limits, the ratio of span to
effective depth as per IS 456-1978 should not
exceed :
(a) 7 (b) 20
(c) 26 (d) 35

Ans. (b) : For deflection of a simply supported beam to

88. The average shear stress for rolled beams is
be within permissible limits, the ratio of span to calculated by dividing the shear force at the
effective depth as per IS 456-1978 should not exceed cross section by the :
20. (a) Gross section of the web
• To control deflection limit state, the basic values of (b) Depth of the beam
span to effective depth ratio is given– (c) Web thickness
Cantilever beam 7 (d) Width of flange
simply supported beam 20 Ans. (a) : The average shear stress for rolled beams is
continuous beam 26 calculated by dividing the shear force at the cross
simply supported slab 35 section by the gross section of the web.
continuous slab 40 The gross-cross section of the web is defined as the
85. The live load to be considered for an depth of the beam or channel multiplied by its web
inaccessible roof, is : thickness.
(a) Nil (b) 75 kg/m2 Average shear stress for rectangular beam is given by
(c) 150 kg/m 2
(d) 200 kg/m2 F
τv =
Ans. (b) : b×d
Floor uses LL(kN/m2) Average shear stress for I-beam is given by
F
Inaccessible roof 0.75 τv =
h × tw
Lift motor rooms 7.5
Service areas 4 The most shear stress is taken by the web of a steel
section, hence shear capacity of the flange is neglected.
Exhibition hall roof 2
• Average shear stress is calculated by the dividing the
Exhibition hall floors 10 shear force at the section with the gross sectional area of
Toilets and bathrooms 2 the web only and not that of the complete beam.
OPSC AE Exam-2020 61 YCT
89. The diameter of cold driven rivets range from: Ans. (c) : Allowable working stress for rolled steel
(a) 6 to 12 mm (b) 12 to 22 mm beam section compression members may be assumed as
(c) 22 to 32 mm (d) 32 to 42 mm 100 N/mm2.
Ans. (b) : The small size rivets ranging from 12 mm to 93. The net cross sectional area of a tension
22 mm in diameter may be cold driven rivets. member is equal to :
• Minimum pitch for rivet joint </ 2.5 d. (a) Gross sectional area
(b) Gross sectional area minus the maximum
• Rivet diameter, dmin = 6.05 t mm deduction for rivet holes
90. The longitudinal space between the effective (c) Gross cross sectional area plus the maximum
length of intermittent butt welds is taken not deduction for rivet holes
more than : (d) Two times the gross sectional area
(a) Four times the thickness of the thicker part Ans : (d) : Net cross-sectional area = Gross sectional
joined area -Area of rivet holes.
(b) Four times the thickness of the thinner part • In tension members.
joined
(c) Sixteen times the thickness of the thicker part
joined
(d) Sixteen times the thickness of the thinner part
joined
Ans. (d) : Intermittent butt welds are used to shear only.
• The effective length of intermittent weld is taken not For plated P1 :
less than four times the thickness of thinner part joined. • Diameter of rivet hole = dh, pitch of rivet = P
• The longitudinal space between the effective length of • Thickness of plate = t1, gauge of rivet = g
welds is taken not more than 16 times the thickness of • Net cross sectional area along section
thinner part joined (Cl 10.5.5.3). 1-2-4-5 = (b-2× dh) × t1
91. Maximum permissible slenderness ratio of a • Net cross sectional area along section.
member carrying loads resulting from wind is : 6-7- = (B- dh) × t1
(a) 180 (b) 250
• Net cross sectional area along
(c) 300 (d) 350
P2
Ans. (b) : As per IS : 800-1964 (Cl.3.7) 1-2-3-7 = (B - 2 dh) ×t1 + × t1
4g
Maximum slenderness Ratio (λ max)
• Net cross sectional area along
Member λ max
P2 P2
A member carrying compressive load 180 1-2-3-4-5 = (B-3 dh) × t1 + t1 + × t1
resulting from dead load and imposed 4g 4g
load • The Failure (rupture) section will be one with least net
A Tension member in which reversal 180 cross-sectional area.
of direct stress due to load other than 94. The stress in the wall of thin cylinder subjected
wind and seismic force to internal pressure is :
A member subjected to compressive 250 (a) Hoop tension (b) Shear
forces resulting from wind EQ force (c) Hoop compression (d) Torsional shear
provide deformation of such member Ans. (a) : Internal water pressure causes hoop tension
does not affect stress in any part of in pipe shell.
the structure • Thin cylinder subjected to internal pressure–
Compression flange of beam 300
A member normally act as a tie in 350
Roof Truss
Tension member other than Pre- 400
tension
92. Allowable working stress for rolled steel beam
sections compression members may be
assumed as :
Pt
(a) 60 N/mm2 (b) 80 N/mm2 Hoop stress σ h =
(c) 100 N/mm 2
(d) 120 N/mm 2 2t

OPSC AE Exam-2020 62 YCT

95. Stiffeners are used in plate girders to : Ans. (d) :
(a) Reduce the compressive stress
(b) Reduce shear stress
(c) Take bearing stress
(d) Avoid buckling of web plate
Ans. (d) : Stiffeners are used in plate girders to avoid
the buckling of web plates by providing resistance to
longitudinal and transverse shear.
96. When a body is subjected to two equal and
opposite forces, acting tangentially across the
resisting section, as a result of which, the body
tends to shear off across the section, the stress
and strain induced is : According to bending moment diagram point of
(a) Tensile stress, tensile strain contraflexure does not exist.
(b) Compressive stress, compressive strain 100. A rectangular beam A has length l, width b and
(c) Shear stress, tensile strain depth d. Another beam B has the same length
(d) Shear stress, shear strain and width but depth is 2d. The elastic strength
Ans. (d) : When a body is subjected to two equal and of beam B will be :
opposite forces acting tangentially across the resting (a) Same (b) Double
section, as a result of which, the body tends to shear off (c) Four times (d) Six times
across the section, the stress and strain induced is shear Ans. (c) : Given,
stress and shear strain. l1 = l 2 = l
b1 = b 2 = b
d1 = d
d2 = d
97. Hook's law holds good up to : Strength of beam depend on section modulus.
(a) Yield point (b) Elastic limit
Sectional modulus
(c) Plastic limit (d) Breaking point
I
Ans. (b) A law, which states that within elastic limits, Z=
y max
strain produced material is proportional to stress
producing it, is known as Hooke's law. I1
Z1 =
98. The Poisson's ratio of steel varies from : ( y max3 )1
(a) 0.23 to 0.27 (b) 0.25 to 0.33 b1d1
b d2
(c) 0.31 to 0.34 (d) 0.32 to 0.42 = 12 = 1 1
d1 6
Ans. (b) :
2
Material Poission ratio
bd 2
Cork 0 Z1 =
6
Steel 0.25 to 0.33
I2
Cast iron 0.23 to 0.27 Z2 =
( y max3 )2
Elastic material 0.85 to 0.80 b2d 2
Rubber 0.5 b d2
= 12 = 2 2
Wrought iron 0.30 d2 6
99. In a simply supported beam, carrying a 2
b. ( 2d )
2
uniformly distributed load w per unit length, 4bd 2
the point of contraflexure : Z2 = =
6 6
(a) Lies in the centre of the beam
 bd 2 
(b) Lies in the end of the beam Z2 = 4.  = 4Z1
(c) Depends on the length of the beam  6 
(d) Does not exist Z2 = 4 Z1
OPSC AE Exam-2020 63 YCT
 Odisha Public Service Commission
Assistant Agriculture Officer (AAO)
Exam- 2020 (Paper-I)
1. Which of the following is not considered a data Strings are useful for holding data that can be
type in computer programming? represented in text form.
(a) Symbolic Data (b) Numeric Data Char–The char type is used to store single characters
(c) Alphanumeric Data (d) Alphabetic Data (Letters, digits, symbols etc.)
Ans. (a) : Data types are of three basic types– 5. The lathe bed is usually manufactured with the
(i) Numeric–Numeric data consists of only numbers following material:
(ii) Alphabetic–Alphabetic data consists of only letters (a) Cast iron (b) Wrought iron
(iii) Alphanumeric–A blank character and Alpha (c) Stainless steel (d) High carbon steel
numeric data consists of symbols. Ans. (a) : Lathe–It is a basic machine tool used to
2. What does GUI stand for? perform a number of operations. The engine lathe is its
(a) Ground User Interface oldest version which was driven by a stream engine.
(b) Graphical User Instruction The important parts of a lathe include head stock,
(c) General User Instruction tailstock, carriage, tool post, cross-slide and bed.
(d) Graphical User Interface Generally, the bed is cast by using high grade cast iron
Ans. (d) : GUI–It stands for "Graphical user Interface" because it exhibits good range of mechanical properties.
and enables users to interact and navigate with To produce castings of high quality the meehanite
information on their computer display using a mouse or process is usually used.
other pointing device to point, click and drag data and 6. The endpoint of the drill bit is ______ is shape.
icons around on the screen, instead of typing in words (a) circular (b) semi-circular
and phrases as commands. (c) flat (d) conical
3. Which of the statement is most suitable for Ans. (d) : Parts of Drill bit–
taking a decision based on multiple choices? 1. Body–The body between the shank and the point of
(a) if (b) if-else drill is called the body.
(c) if-else-if (d) All of these 2. Point–The end point of the drill is called point and it
Ans. (c) : • If-else-if types of statements are known as is cone shaped. It is generally found at 180o.
conditional statements. 3. Shank–The driving end of the drill which fits into the
• These statements are followed by an optional else drilling machine is called a shank.
if...else statement. 4. Tang–Tang is a part of the shank that fits into the
• It helps in testing various conditions using single if... machine spindle.
else if statement. (v) Flutes, (vi) Land, (vii) Body clearance (viii) Web
• These statements are used for handling decisions. (ix) Web, (x) Neck
4. Which of the following is a collection of
7. The _____ machine is superior to other
different types of data?
machines as mentioned below in terms of
(a) String (b) Structure
accuracy and surface finish.
(c) Char (d) All of these
(a) Lathe machine (b) Milling machine
Ans. (b) : Data structure–A data structure is a storage
(c) Slotting machine (d) Shaping machine
that is used to store and organize data, it is way of
arranging data on a computer so that it can be accessed Ans. (b) : Milling machine–It is a process of producing
and updated efficiently. flat and complex shapes with the use of a multipoint
• A data structure is not only used for organizing the cutting tool. Milling machines are superior to other
data, it is also used for processing, retrieving and machines in terms of accuracy and surface finish. There
storing data. are two types of milling operations–
String–The string object is used to represent and (i) Down milling/climb milling
manipulate a sequence of characters. (ii) Up milling/conventional milling

OPSC AAO Exam. 2020 (Paper-I) 64 YCT

8. The length of stroke for a shaper machine is
210 mm, the number of double strokes per
minute is 28 and the ratio of return time to
cutting time is 2 : 3, then the approximate
cutting speed is:
(a) 650 m/min (b) 9.80 m/min
(c) 11.25 m/min (d) None of these
Ans. (b) : Given that,
Number of double stroke per minute (N) = 28
Length of stroke (L) = 210 mm
ratio of return (m) = 2/3 Shockley diode (A) Basic construction arrangement
NL (1 + m ) (B) Schematic symbol
We know, cutting of shaper (V) = m / min
1000 10. LVDT is a common type of electromechanical
28 × 210 × (1 + 0.667) ) transducer. What does LVDT stand for?
V= (a) Linear Virtual Differential Transducer
1000
V = 9.80196 m/min (b) Linear Virtual Double Transformer
(c) Linear Variable Differential Transformer
9. Which of the following diodes is a four-layer
(P-N-P-N) with two terminals (namely anode (d) Linear Variable Differential Transducer
and cathode)? Ans. (c) : LVDT (linear variable differential
(a) Tunnel diode transformer)–It is an electromechanical sensor used to
(b) Light emitting diode convert mechanical motion or vibrations, specially
(c) Zener diode rectilinear motion, into a variable electrical current,
(d) Shockley diode voltage or electric signals and the reverse actuating
Ans. (d) : Shockley Diode–It is a two-terminal four- mechanism used primarily for automatic control
layer device. The basic construction consisting of four systems. The classification of electromechanical
semi conductor layers forming the p–n–p–n structures transducer includes conversion principal of types of
and the schematic or circuit symbol is shown the p–n– output signals.
p–n structure can easily be seen as a combination of two 11. What is the value of absolute zero
BJTs, one a p–n–p transistor (Q1) and the other in p–n– temperature?
p transistor (Q2) (a) 0ºC (b) 100ºC
• The first member of the thyristor family of devices is (c) –273ºC (d) None of these
the P-N-P-N (four layer) diode, also known as the Ans. (c) : Absolute zero, temperature at which a thermo
Shockley diode. dynamic system has the lowest energy. It corresponds to
• It consists of a four layer P-N-P-N structure the diode –273.15oC on the Celsius temperature scale and to –
has an anode (A) and a cathode (K). 459.67o F on the Fahrenheit temperature scale. All the
thermal energy is released at absolute 0 temperature.
This temperature range can be achieved theoretically
not practically.
12. Which of the following can be used as
mechanical pressure sensing element?
(a) Diaphragm (b) Bellows
(c) Bourdon tube (d) U-tube
Ans. (c) : Bourdon tube–Bourdon tubes operate on the
same principle as the aneroid barometer but instead of
an evacuated capsule or bellow arrangement, ac-shaped
or helical tube is used. The tube are closed at one end
connected to the pressure at other end, which is fixed in
position. Bourdon tubes are usually used by gauge
pressure sensing application but different sensing is
possible by connecting two tubes to one pointer.

OPSC AAO Exam. 2020 (Paper-I) 65 YCT

 The institute is located at phulnakhara, cuttak the side of
NH-5. It is almost midway between the state capital
Bhubanesar and cuttak. The institute was inaugurated
on 16th June 1992 by Sri Kanhu Charana Lenka, then
hon'eble minister of state for Agriculture and Animal
Husbandry, Government of India, New Delhi.
16. Which one of the below is announced by the
Government of India to support a crop?
(a) Minimum support price
(b) Maximum support price
(c) Moderate support price
(d) Influential support price
13. The Golden Revolution is related to the
production of: Ans. (a) : Minimum support price was announced by
the government of India for the first time in 1966-67 for
(a) Milk
wheat. The MSP is the price at which government
(b) Cotton and jute purchases crop from the formers. Its announced on the
(c) Grain and pulses basis of the recommendation of the commission for
(d) Horticulture and honey agriculture costs of the recommendations of the
commission for agriculture costs on prices (ACP).
Ans. (d) : Golden Revolution is related to–Fruits
production or horticulture and honey 17. Which of the following is a Kharif Crop?
Some import revolution production– (a) Paddy (b) Maize
(c) Jowar (d) All of these
Yellow Revolution – Oil seeds production
Ans. (d) : There are three types crop seasons in India
Red Revolution – Tomato and meat production
which are given below:
Pink Revolution – Jhinga fish production (i) Ravi crop
White Revolution – Milk production (ii) Kharif crop
Silver Revolution – Egg production (iii) Zaid
Almond Revolution – Spices production Kharif crop–Kharif crops are sown in the monsoon in
Round Revolution – Potato production June and July harvested in September and October- rice,
Black Revolution – Petrolium production jowar, cotton, maize are kharif crops.
Ravi crop example–Wheat, mustard, gram, pea, linseed
14. How much percent of the GDP in Odisha
etc.
comes from the Agriculture Sector in the year
18. How many joules contain one watt-hour?
2021-22?
(a) 3.6 × 108 J (b) 3.6 × 102 J
(a) 17% (b) 21% 3
(c) 3.6 × 10 J (d) 10–3 J
(c) 28% (d) 35%
Ans. (d) :
Ans. (b) : The Agriculture & Alied sector exhibited 1 watt hour = 1 watt × 1 hour
contraction of 3.37 percent in 2021-22, due to multiple = 1 watt × 3600 sec
natural calamities like heavy rain, unseasonal cyclonic = 3600 watt-sec
rain and severe cyclonic storm 'yaas' followed by flood
= 3.6 × 103 Joule [Q1watt = 1 joule ]
water inundation, cyclonic and jawaad in December
2021. In 2021-22, Agriculture and Alied Activities = 3.6 × 103J
contributed 20.61 percent, industry contributed 39.51 19. What is the term used for an imaginary line on
percent and services contributed the remaining 39.88 the ground joining points of equal elevation?
percent to GSVA. (a) Level line (b) Contour
(c) Line of sight (d) Datum
15. Animal Disease Research Institute (ADRI) is
Ans. (b) : Contour is an imaginary line on the ground
situated in which place in Odisha?
surface joining points of equal elevation or a line m
(a) Brahmagiri (b) Phulnakhara which every point is at the same level above or below a
(c) Ranpur (d) Chandrapur chosen reference surface is sea level. This line on map
Ans. (b) : The Animal Disease Research Institute, represents a contour and is called contour line.
Phulnakhara, cuttak was conceptualized by the direction 20. Which of the physical quantity has the same
of Animal Husbandry and veterinary services. Odisha dimensional formula as that of energy?
and established by an order of the Government of (a) Work done (b) Pressure
Odisha vide letter no. 2492/FARD dt. 01.02.1992. (c) Force (d) Power
OPSC AAO Exam. 2020 (Paper-I) 66 YCT
Ans. (a): Dimensions and Dimensional formula Ans. (c) : Revenue chain–The revenue chain is 33 ft.
All the physical quantities of interest can be divided long and consists of 16 links, each link being 2 1 ft
from the base quantities. The power of base quantity is 16
called the dimension of the quantity in that base. long. The chain is mainly used for measuring fields in
cadastral survey
Dimension formula of work done,
Engineer's chain– length = 100 ft
= force × displacement
link = 100
= [MLT–2] × [L]
24. What is the length of the survey chain?
= [ML2T–2] (a) 15 m (b) 20 m
Energy (E) = [ML2T–2] (c) 25 m (d) 30 m
Power (P) =
W
, Pressure ( P ) =
F Ans. (b) : • It is also called surveyor's chain
t A • A Gunter's chain or surveyor's chain is 66 ft long and
2 −2
 ML T  consists of 100 links, each link being 0.66 ft or 7.92
P=  inches long.
[T ] • The length of 66 ft or [66 × 0.305 = 20.13 m] was
Force = Mass × Acceleration originally adopted for convenience in land measurement
since 10 square chains are equal to 1 acre.
 V  LT  
−1
25. The greenhouse built against the side of an
Q Acceleration = = 
F = [M] × [LT–2]  t [T ]  existing building structure is known as:
  (a) Even span type greenhouse
 a =  LT −2   (b) Uneven span type greenhouse
(c) Lean to type greenhouse
F = [MLT–2]
(d) Ground to ground greenhouse
21. Which surveying instrument is used for Ans. (c) : Green house type based on shape
marking the position of stations, and for The commonly followed types of green houses based on
sightings of those stations, as well as for shape are–
ranging straight lines? • Lean to type of green house
(a) Arrow • Even span type green house
(b) Peg • Uneven span type green house
(c) Ranging rod • Ridge and furrow type
(d) All of these • Saw tooth type
Ans. (c) : Ranging Rods– • Quonset green house
• Interlocking ridges and furrow type Quonset green
Ranging rods have a length of either 2 m or 3 m, the 2
house
more length being more common. They are shed at the
bottom with a heavily iron point and are pointed in • Ground to ground green house.
alternative bonds of either black and white. Ranging Lean to type green house–It is used when a green
rods are used to range some intermediate points in the house is placed against the side of an existing building.
survey line. It is also used for marking the position at It is built against a building, using the existing structure
stations and for sighting of those stations, as well as for for one or more of its side. It is usually attached to a
ranging straight line. house.
22. What is the shrinkage factor if the area of a 26. Which of the following pairs of gases are the
plan of an old survey plotted on a sheet shrunk main contributors to the greenhouse effect?
from 10 cm to 9.8 cm in length? (a) Ozone and Ammonia
(a) 0.098 (b) 0.98 (b) Oxygen and Nitrogen
(c) Carbon dioxide and Methane
(c) 9.8 (d) None of these
(d) Nitrogen and Sulphur dioxide
Ans. (b) : Given that,
Ans. (c) : Green house effect
Original length = 10 cm
• It is a phenomenon that happens when heat from a
Shrunk length = 9.8 cm planet's host star passes through its atmosphere.
shrunk length 9.8 • Carbon dioxide, methane, nitrous oxide, hydro chloro
shrinkage factor = = = 0.98
original length 10 fluoro carbon is (HCFCs), hydro-fluoro carbons HFCs)
and ozone in the lower atmosphere are the principal
23. How many links are in a Revenue chain?
green house gases whose concentrations are rising.
(a) 10 (b) 12
• The main green house effect components are carbon
(c) 16 (d) 19 dioxide and methane.
OPSC AAO Exam. 2020 (Paper-I) 67 YCT
27. The process of adding water to calcium oxide to Ans. (c): Seed metering mechanism–
produce calcium hydroxide is referred to as: The mechanism picks up seeds from the seed box and
(a) Watering (b) Slaking delivers them into the seed tube is called seed metering
(c) Baking (d) Soaking mechanism may be of several type (A) Fluted type (B)
Ans. (b) : Calcium hydroxide is manufactured by Internal double run type (C) Cup feed type (D) Cell feed
adding water to the oxide (line or quick lime), a process type (E) Brush feed type (F) Augar feed type (G) Picker
known as slaking so that the oxide turns to a fine wheel type.
powder, the product is slaked lime. If more water is
Usually seed metering mechanism is provided at the
added, a thick suspension manufacture of mortar,
whitewash and bleaching powder and for the softening bottom of the box.
of temporary hard water. 32. Quality of work of performance evaluation of
28. The greenhouse effect is related to: tillage implement does not depend on:
(a) Global warming (a) Depth of cut (b) Width of cut
(b) Increased growth of green algae (c) Soil inversion (d) Soil pulverization
(c) Cultivation of vegetables in the house Ans. (d) : Tillage in agriculture, the preparation of soil
(d) Development of terrace gardens for planting and the cutlivation of soil after planting.
Ans. (a) : Green house effect refers to the absorption of Tillage is the manipulation of the soil into a desired
the sun's radiations by the atmospheric gases like condition by mechanical.
carbon dioxide and trapping of these radiations within 33. The centre of resistance lies at a distance equal
the atmosphere. These are subsequently reflected onto
to ______ from the share wing.
the earth's surface and cause and increase in the global
temperature. So it is called global warming, therefore (a) 3/4th size of tractor (b) 3/4th size lf plow
green house effect is related to the global warming. (c) 3/4th size of share (d) 3/4th size of frog
29. The anthropogenic CO2 emission is related to Ans. (b) : Centre of resistance–It is the point, on
_______ emissions. mould board plough, at which the resultants of all the
(a) Industrial CO2 (b) Natural CO2 horizontal and vertical forces act it lies at a distance
(c) Human-made CO2 (d) None of these equal to 3/4th of size of plough from the wing of share.
Ans. (c) : The anthropogenic CO2 emission is related to It is also known as centre of load.
Human-made CO2 emission, while the issue of CO2
emission deserves attention, also in relation to the
choice of materials for hydraulic gates.
CO2 emissions from energy use are which provides CO2
associated with human activities.
30. Which one is the most abundant greenhouse
gas in the Earth's atmosphere?
(a) Nitrous oxide (b) Carbon dioxide
(c) Water vapour (d) Methane 34. Vertical Conveyor Reaper (VCR) is most
popular for harvesting of:
Ans. (c) : Most abundant green house gases in earth's
(a) Grass (b) Paddy
atmosphere are
(c) Vegetable crops (d) All of these
• Water vapours
• CO2 Ans. (b) : Most of the crops are harvested manually
with sickles. However, there is good scope to introduce
• CH4
improved and efficient designs of sickles to improve the
• N2O, O3
output and remove the labours drudgery. The harvesting
So, Water vapour accounts for the largest percentage of
equipment namely, combine harvesters. Propelled
the green house effect between 36% and 66% for clear
reapers and tractor drawn reapers developed in the
sky conditions and between 66% and 85% when
including clouds, therefore water vapour is the most country are commercially available for used. The small
important and most abundant green house gas, other self propelled vertical conveyor reaper windrower of
gases are CO2, CH4, N2O and O3. one meter width has been found very useful for
harvesting paddy, wheat, soyabean and rope seed
31. The seed rate of the seed metering mechanism
mustard crops.
does not depend upon:
(a) Speed of operation 35. The disk plow is forced into the ground by the
(b) Types of seeds effect of:
(c) Types of the metering mechanisms (a) Gravity (b) Suction
(d) Number of furrow openers (c) Direction of travel (d) Both (a) and (b)

OPSC AAO Exam. 2020 (Paper-I) 68 YCT

Ans. (d): Disc plow–It is the form of a disc. Disc plows 40. The power available at the engine crankshaft
are designed to reduce friction by making a rolling plow which is measured by a suitable dynamo meter
bottom instead of sliding plow bottom. is called:
Disc tools are forced into ground by effect of gravity (a) Brake power (b) Indicated power
upon its mass rather than suction. Total mass of disc (c) Friction power (d) Drawbar power
plow varies from 200 to 600 kg. Ans. (a) : Brake Power–The net power available at the
36. The specific gravity of electrolyte can be crankshaft is called the brake power. It is called 'brake'
checked with: power because some type of brake is used to measure it.
(a) Hygrometer (b) Hydrometer The brake places a load on the engine crankshaft.
(c) Voltmeter (d) Multimeter The brake power of the engine
Ans. (b) : The specific gravity of the electrolyte 2 πNT
B.P = watts
provides the definitive means of checking the charged 60
condition of a lead-acid cell, this must be checked with
Where T = Torque in N-m
a hydrometer on a periodic basis. The electrolyte must
always cover the plates. It can be topped up with N = Engine rpm
distilled water. • Friction power (FP) = IP – BP
37. The sticky belt method is associated with the: 41. In a tractor engine, the complete path of power
(a) Testing of seed uniformity from the engine to the wheels is called:
(b) Testing of seed rate (a) Transmission path (b) Power wheel
(c) Power consumption of seed drill (c) Power train (d) Hitch system
(d) Both (a) and (b) Ans. (c) : Power transmission system–Transmission is
Ans. (a) : Sticky belt method was used in IARI, New a speed reducing mechanism, equipped with several
gears. It may be called at sequence of gears and shafts,
Delhi for determining of seed spacing at different
through which the engine power is transmitted to the
rotational speed of seed metering unit. Speeds of 30, 40,
tractor wheels. The system consists of various devices
50 and 60 rpm of seed metering unit were used for
that cause forward and backward movement of tractor
determining the seed spacing. The sticky belt method is
to suit different field conditions. The complete path of
associated with the testing of seed uniformity. the power from the engine to the wheels is called power
38. The tank capacity of the knapsack sprayer is train.
about: 42. A turbocharger is a centrifugal compressor
(a) 5-10 litres (b) 5-15 litres driven by ______ and employed in engines to
(c) 10-23 litres (d) 10-33 litres boost the charge air pressure.
Ans. (b) : Knapsack sprayers loaded on the back of (a) Engine (b) Exhaust gas
worker during operations. Tanks be of plastic or metal. (c) Dynamo (d) PTO
Common knapsack sprayers are hydraulic, manual Ans. (b) : A turbocharger is a centrifugal compressor
pneumatic or motorized pneumatic. driven by exhaust gas and employed in engine to boost
Hydraulic knapsack sprayers–Manually-operated the charge air pressure.
tank capacity is 15 litres, mechanical agitation works Methods of turbocharging–
with a hand lever to maintain constant pressure, (i) Constant pressure turbocharging
particularly used for spot treatment. Equipped with a (ii) Pulse turbo charging
boom. It holding, farmers and hand treatment. Equipped (iii) Pulse converter
with a boom. It is good for blanket application.
(iv) Miller turbo charging
39. What is the main objective of puddling? (v) Two-stage turbocharger
(a) Destroy insects and pests (vi) Hyperbar turbo charging
(b) Mix the fertilizer in the soil
43. The average speed of the engine connected to
(c) Reduce the soil erosion the governor is 1000 rpm. If the fluctuation in
(d) Decrease water loss by percolation speed is ±100 rpm, then % governor regulation
Ans. (d) : A majority of the formers growing rice in is about:
India use this method. Under paddy rice cultivation, the (a) 15% (b) 17%
rice field is permanently flooded with water varying (c) 20% (d) None of these
from 5 to 10 cm depth during the period after
Ans. (c) : Given that,
transplanting until 2 weeks before harvest. The main
objective of puddling is to reduce the percolation losses Average speed of the engine = 1000 rpm
and central weed growth. The recommended plant Fluctuation speed = ±100 rpm
density is 33 hills/m2 during kharif (rainy season) and Nmin = 1000–100 = 900 rpm
44 hills/m2 during rabi (winter) season. Nmax = 1100 rpm

OPSC AAO Exam. 2020 (Paper-I) 69 YCT

 1100 − 900 t = 0.045D + 1.6
Governor regulation = t = 0.045 × 150 + 1.6
1000
t = 6.750 + 1.6
200
= t = 8.35 mm
1000
48. Water circulation is a radiator of a water
= 0.2
cooling system takes place from:
= 20%
(a) Upper tank to lower tank
44. The Cetane rating, also known as Cetane (b) Lower tank to upper tank
number is a measurement of: (c) Engine to upper tank
(a) Ignition quality of fuel (d) Engine to water pump
(b) Calorific value of fuel Ans. (a) : Water cooling system–In water cooling
(c) Viscosity of fuel system, water is used as a cooling medium. The water
(d) Specific gravity of fuel jackets are provided around the each engine cylinder,
Ans. (a) : Cetane Number (CN)–A measure of the cylinder head, combustion space and the valve
ignition quality of fuel, also known as cetane rating or openings. The water cooling systems are of the
cetane number. Higher the cetane rating of the fuel following two types:
lesser is the propensity for diesel knock. 1. Thermosyphon system
• Cetane number is the most important delay single fuel 2. Pump circulation system
propensity which affects period the exhaust emissions, Thermosyphon system–The thermosyphon system is a
noise and startability of a diesel engine. very simple system, which was used in earlier
automobiles and is now absolute.
In this system, there is no pump to circulate the water in
the system. The radiator is connected to the engine
through flexible hoses. The radiator header tank is kept
above the level of water jackets. When the water in the
water jackets become hot due to the engine combustion,
it expands and become lighter.
The radiator consists of an upper tank, a lower tank and
45. Which are the main constituents of fuel? in between the upper and lower tank radiator cores are
(a) Oxygen and Hydrogen provided. The upper tank is connected to water outlet of
(b) Oxygen and Nitrogen the engine through a rubber hose. The lower tank is
(c) Carbon and Nitrogen connected to the water pump through hoses. This water
(d) Carbon and Hydrogen circulation takes place from the upper tank to lower
tank.
Ans. (d) : Carbon and hydrogen are the main
constituents of a fuel. Fuel also contains sulphur 49. The basic purpose of the ignition coil in the
oxygen and hydrogen. Fossils fuels are found in ignition systems of a petrol engine is used to:
sedimentary rocks. (a) Step-up current (b) Step-down current
46. When the crankshaft rotates, which part does (c) Step-up voltage (d) Step-down voltage
the job of splashing oil from the oil sump to the Ans. (c) : Ignition coil–
cylinder walls? • It is used to step up low voltage to high voltage from
(a) Spoon (b) Dipper 12V to 22,000 V.
(c) Hooper (d) Scooper • It consists of two windings, one wound over the soft
Ans. (b) : The lubricating oil is stored in a sump. A iron core.
dipper is made at the lowest part of the connecting rod. • The secondary winding is wound over the core and it
When the crankshaft rotates the dipper dips in the oil consists of about 21,000 turns.
once in every revolution of the crank shaft and splashes 50. What is the firing order of 4 cylinders 4-stroke
oil on the cylinder walls. vertical engine?
47. What is the thickness of the cylinder wall if the (a) 1–2–3–4 (b) 1–3–4–2
diameter of the cylinder bore is 150 mm? (c) 3–4–1–2 (d) 4–3–2–1
(a) 7.15 mm (b) 7.62 mm Ans. (b) : Firing order is an essential part of engine
(c) 8.35 mm (d) 8.80 mm design manufacture carefully decide firing orders to
Ans. (c) : Given that, tame vibrations and improve heat dissipation.
diameter of the cylinder bore (D) = 150 mm Most 4 cylinder engines have a firing order of 1-3-4-2
thickness of the cylinder = t although other firing orders such as 1-3-2-4, 1-4-3-2, 1-
we know, 2-4-3 are possible.

OPSC AAO Exam. 2020 (Paper-I) 70 YCT

51. The towed force (TF) of a pneumatic tire is 55. The horizontal component of pull
given by (symbols have their usual meanings): perpendicular to the direction of travel is:
(a) TF = 1.2/Cn + 0.046 (a) Pull (b) Draft
(b) TF = 1.2/Cn + 0.04 (c) Unit draft (d) Side draft
(c) TF/W = 1.2/Cn + 0.04
Ans. (d) : Side draft–It is the horizontal component of
(d) TF = 1.2/W + 0.04 + Cn
the pull perpendicular to the direction of motion. This is
Ans. (c) : Towed Force (TF)–The towed force or
developed if the centre of resistance is not directly
motion resistance of a pneumatic tire is dependent on
load, size and inflation pressure as well as soil strength. behind the centre of pull.
For soils that are not very soft and tires that are operated Pull–It is the total force required to pull an implement
at nominal tire inflation pressures the towed force are Unit draft–It is the draft per unit cross sectional area of
given below the burrow.
TF 1.2 56. A tractor has a rear wheel of diameter 1.5
= + 0.04
W Cn meters, the final drive gear ratio is 5 : 1, the
differential gear ratio is 3 : 1 and the gearbox
CIbd reduction is 2 : 1. Find the travel speed of the
where, Cn = wheel number =
W tractor when the speed of the engine is 1250
CI = Cone index rpm and transmission efficiency is 86%:
52. Moving the center of gravity of a tractor (a) 10.13 kmph (b) 11.77 kmph
towards its front wheel creates the problem of: (c) 13.68 kmph (d) None of these
(a) Instability (b) Steering
(c) Overturning (d) All of these Ans. (a) : Given,
Ans. (b) : Moving the center of gravity of a tractor Rear wheel diameter = 1.5 m
towards its front wheel creates the problem of steering. Final drive gear ratio = 5 : 1
53. What is the purpose of ballasting in the front Differential gear ratio = 3 : 1
tires of a tractor? Gear box reduction = 2 : 1
(a) Increase traction Speed of the engine = 1250 rpm
(b) Decrease front wheel slippage Efficiency = 86%
(c) Increase stability Rear wheel speed = Engine speed × Gear box reduction
(d) Decrease tractor vibration × differential gear ratio × Final drive gear ratio
Ans. (b) : Filling of liquid in the tractor tyres is called 1 1 1
ballasting. = 1250 × × ×
2 3 5
• To prevent slippage weight are added to the tyre of
tractor. Weights can be added in two ways, on is dry = 41.66 rpm
weight which are attached to the rim of the tyre and  πDN × 60 × η 
Speed of travil =  
other method is by adding water to the tyre, this is  1000 
called ballasting of tyres.
 86 
54. Drawbar power can be also expressed as  3.14 × 1.5 × 41.66 × 60 × 100 
(symbols have their usual meanings): =  
GTR  1000 
(a) × (1 + s ) × AHP  
NTR
GTR = 10.13 km/h
(b) × (1 − s ) × AHP 57. The relationship between Theoretical Field
NTR
NTR Capacity (TFC), Effective Field Capacity
(c) × (1 − s ) × AHP (EFC), and Field Efficiency (FE) of a machine
GTR is given by:
NTR
(d) × (1 + s ) × AHP (a) FE (%) = (EFC/TFC) × 100
GTR (b) FE (%) = (TFC/EFC) × 100
Ans. (b) : Drawbar power–Drawbar power is power (c) FE (%) = (EFC × TFC) × 100
measured at the point implements are attached to the (d) FE (%) = (EFC – TFC) × 100
tractor, drawbar or 3 point hitch, during tractor testing a
load cell is placed between the tractor and a load. Ans. (a) : The relationship between Theoretical field
capacity (TFC), Effective field capacity (EFC) and
FV
Db ( HP ) = Field efficiency (FE) of a machine is given by
375
EFC
Where, F = Force FE ( % ) = × 100
V = Speed TFC

OPSC AAO Exam. 2020 (Paper-I) 71 YCT

Field efficiency–The ratio of effective field capacity to
theoretical field capacity is called the machines field
efficiency.
Field efficiency accounts for failure to utilize the full
operating width of the machine.
58. The basic difference between seed planter and
seed drill is in respect to:
(a) Power transmission
(b) Metering mechanism 61. The difference between the actual sales and
(c) Furrow opener break-even sales is known as:
(d) Calibration process (a) Margin of safety (b) Price-cost margin
(c) Fixed cost (d) Profit
Ans. (b) : Seed drill–It is a machine used for placing
the seeds in a continuous stream in furrows at uniform Ans. (a) : Margin of safety–It is an important concept
in Marginal costing approach total sales minus the sales
rate and at controlled depth with an arrangement of
at break-even point is known as the margin of safety.
cavening the seeds with soil. That is margin of safety is the excess of normal or
Functions of a seed drill– actual sales over sales at break-even point.
• To carry the seeds The margin of safety refers to the amount by which
• To open furrow at uniform depths sales revenue can fall before a loss is incurred. It is the
difference between the actual sales and sales at the
• To meter the seeds break-even point.
• To cover the seeds and compact the soil around the 62. Which of these accounts for more than half of
seed. tractor-related deaths?
Planter–Planter is a sowing equipment used for sowing (a) Overturns (b) Run-overs
those seeds which are larger in size and can not be (c) Highway collisions (d) Field collisions
handled by seed drills row to row and plant spacing is Ans. (a) : Tractor rollover occurs when a tractor tips
maintained in a planter. sideways or backward and overturns, potentially
Functions of a planter– crushing the operator. A review of the relevant literature
• To open the furrow reveals that more than 800 people are killed each year in
tractor accidents and for every person killed, at least 40
• To meter the seed others are injured. This paper focuses on tractor
• To deposit the seed in the furrow overturns because they account for more than half of all
• To cover the seed and compact the soil over the seed. the tractor related deaths.
59. Convert 37ºC to Kelving scale: 63. What is the number of splines for a 1000 rpm
PTO shaft?
(a) 097 K (b) 274 K
(a) 17 (b) 21
(c) 298 K (d) 310 K
(c) 23 (d) 25
Ans. (d) : Given that, Ans. (b) : Check the coupling of the power shafts to
The temperature on the Celsius scale is given as 37ºC make sure it has the same number of splines or grooves
K = ºC + 273.15º as the stub shaft on the tractor.
K = 37 + 273.15º • The 540 rpm stub shaft has 6 splines
K = 310.15 K • The 1000 rpm stub shaft has 21 splines
60. The break-even point is the point at which: • If the tractor has a high speed (1000 rpm) and low
speed (540 rpm) PTO, use the correct one to suit the
(a) Total expenses = Total revenue
implement to be operated, otherwise refer to the tractor
(b) Fixed cost = Variable cost operator's manual for information on how to convert or
(c) Total expenses < Total revenue adjust the tractor PTO speed.
(d) Total expenses > Total revenue • If the implement requires a peak torque in excess of
Ans. (a) : The break-even point in economics, business 2000 Nm you must use 1000 rpm.
and specifically cost accounting is the point at which 64. What des ROPS stand for?
total cost and total revenue are equal i.e. 'even'. There is (a) Roof Protective Structure
nahet loss and one has "broken even" though (b) Rangeland Occupation Protection System
opportunity costs have been paid and capital has (c) Recovery Operation Program Status
received the risk-adjusted expected return. (d) Rollover Protection Structure

OPSC AAO Exam. 2020 (Paper-I) 72 YCT

Ans.(d): ROPS stand for–Roll-over protection structure
P 
or Roll-over protective structure. dB = 10 log10  2 
The Roll-over protection structure for tractors used is  P1 
agriculture operations. Where,
65. Which one is not included in the safety P1 = Reference standard quantity of sound pressure or
program for the prevention of accidents? sound power
(a) Development of safe working conditions P2 = Measured quantity of sound pressure or sound
(b) Promotion of employees participation in power
safety 69. The most frequently used controls in the
(c) Corrective measures for the prevention of dashboard are arranged in:
accidents (a) Left side (b) Right side
(d) Compensation and medical payment (c) Central location (d) All of these
Ans. (d) : Accidents are an unplanned and unexpected Ans. (c) : The most frequently used controls in the
occurrences that upsets the planned sequence of events. dashboard are arranged in central location.
Safety program generally does not talk anything about
The Dash boards toolkit contains the controls and
after-effects of accidents (compensation and medical
services that are used by coaches in a human service,
payment). It always focuses on the preventation of
for example an instance UI a dashboard such as the
accidents and how to develop a safe working procedure
and environment. process performance dashboard or a task completion UI.
66. What is the process in which the rate of heat 70. What are common ergonomic risk factors in
transfer is maximum? manually operated agricultural tools?
(a) Conduction (b) Convection (a) Awkward working postures
(c) Radiation (d) Evaporation (b) Excessive repetition
(c) Excessive force
Ans. (c) : There are three modes of transmission of
heat. They are conduction, convection and radiation. (d) All of these
Radiation–It transfers heat in the form of Ans. (d) : Ergonomic is the study of work and of the
electromagnetic waves. It can heat any form of material. relationship between humans and their working and
It does not need any medium for the transfer of heat. So physical environment. Ergonomics is the science of
radiation is the process for the rate of heat transfer is fitting the task or the job to the worker. As a field of
maximum. study, ergonomic deals with job design, work
67. Which principle is used to calculate the energy performance, health and safety, stress, posture, body
requirements based on the heart rate and mechanics bio mechanics, anthropometry, manual
oxygen consumption of a person? material handling, equipment design, quality control,
(a) Biomechanical principle environment, workers, education and training
(b) Physiological principle ergonomics reduces risk factors known to contribute to
occupational ergonomic related injuries.
(c) Anthropometric principle
(d) Psychological principle 71. What does ergonomics mean?
(a) The laws of work
Ans. (b) : Physiological principle is used to calculate
the energy requirements based on the heart rate and (b) Keep your back straight
oxygen consumptions of a person. (c) Posture at work
For each person, heart rate and oxygen uptake relate (d) All of these
linearly throughout a broad range of aerobic exercise Ans. (d) : Ergonomics–It is derived from Greek terms
intensities, By knowing this precise relationship, ergon, meaning work and names meaning law. So it is
excessive heart rate provides an estimate of oxygen the study of work and of the relationship between
uptake (and this energy expenditure) during physical humans and their working and physical environment.
activity. 72. Calculate the Body Mass Index (BMI) for a
68. Noise level is measured in: person whose height is 1.70 m and whose
(a) Decibel (b) Hertz weight is 70 kg:
(c) ppm (d) Pascal (a) 20.5 (b) 24.2
Ans. (a) : Noise level is measured in decibel. The (c) 41.2 (d) None of these
decibel is a logarithmic unit used to measure sound Ans. (b) : Body mass index (BMI)–It is the ratio of
level decibel is also used in electronics, signals and mass to square height of the person.
communication, Decibel is the ratio between two power mass
levels expressed in logarithmic terms with relation to BMI =
height 2
reference to reference level.

OPSC AAO Exam. 2020 (Paper-I) 73 YCT

Where, mass in kilograms and height in meters Ans.(b): Variable rate technology in precision
Given that, Weight = 70 kg agriculture focuses on many areas of crop production,
Height = 1.70 m including applying herbicides and pesticides, lime,
gypsum and other common crop nutrients, seeding and
70
BMI = detecting weeds and diseased crops. VRT works by
1.702 using GPS and GIS technology to locate precise
BMI = 24.2 locations in the field for material application.
73. What is the daily exposure action value for 77. Which one is not directly related to precision
hand-arm vibration? farming?
(a) 1.5 m/s2A(8) (b) 2.5 m/s2A(8) (a) Prevents forest degradation
2
(c) 3.5 m/s A(8) (d) 3.8 m/s2A(8) (b) Increase agriculture productivity
Ans. (b) : The exposure action value is a daily amount (c) Efficient use of water resources
of vibration exposure above which employers are (d) Prevents soil degradation
required to take action to control exposure. The greater Ans. (a) : • Precision agriculture is a method of
the exposure level, the greater the risk and the more increasing average yields by using precise amounts of
action employers will need to take to reduce the risk. inputs.
For hand-arm vibration the EAV is a daily exposure of Benefits of precision agriculture–
2.5 m/s2 A(8). It represents a high risk above which • Increases agriculture productivity
employees should not be exposed. • Assists in the prevention of soil degradation
74. What does HMI stand for? • Reduces the use of chemicals in crop cultivation
(a) Human Machine Interaction 78. Drone in agriculture is most widely used for:
(b) Human Machine Interface (a) Planting (b) Spraying
(c) Human Machine Implementation (c) Irrigation (d) Harvesting
(d) Human Machine Involvement Ans. (b) : In China, India and emerging Asia drone
Ans. (b) : HMI stand for Human Machine Interface. It technology for precision agriculture is mostly used for
is defined as a feature or component of a certain device plantation management and for industrial crops
or software application that enables humans to engage including cereal based feeds.
and interact with machines. Human machine interface One of the most critical uses of drones in Agriculture is
devices that we encounter in our daily lives include its flexibility to move around in swift motions and
touch screens and keyboards. maneuver to destined locations. This ability of drones
75. Grey color is associated with the revolution in: helps spray fertilizers and insecticides to nurture crops
(a) Milk production and provide them with needed nutrients. The drone
operator are free to monitor the drone spraying fertilizer
(b) Grain production
that keep insects, pests and worms away and increase
(c) Fertilizer production crop life longevity.
(d) None of these
79. Which power is the primary source for running
Ans. (c) : Grey colour is associated with the revolution small equipment and tools at the farm?
in fertilizer production. (a) Human power (b) Animal power
Some important revolution– (c) Mechanical power (d) Electrical power
Black revolution – Petroleum production Ans. (a) : There are different sources of farm power,
Silver revolution – Egg production which is given below
Red revolution – Meat and Tomato (i) Human power (ii) Animal power
White revolution – Milk (iii) Electrical power (iv) Mechanical power
Yellow revolution – Oil seeds (v) Renewable energy
Green revolution – High yielding varieties of crops Human power–It is the primary source for running
Golden revolution – Horticulture small equipment and tools at the farm. Static activity
Brown revolution – Coffee such as chaff slicing, threshing, lifting, water and so
Blue revolution – Aquaculture forth also are carried out employing labour. An average
person can produce a maximum strength of about 0.1 hp
76. What is VRT in precision agriculture? for doing agriculture activities.
(a) Variable Rate Transfer
80. Technician 'X' says that engine oil is used do
(b) Variable Rate Technology clean, cool and lubricate the engine. Technician
(c) Virtual Reality Technology 'Y' says that engine oil helps seal some internal
(d) Voltage Reduction Technology engine parts. The correct statement is said by:
OPSC AAO Exam. 2020 (Paper-I) 74 YCT
 (a) X only (b) Y only 84. The main composition of biogas is:
(c) Neither X nor Y (d) Both X and Y (a) Methane (b) Carbon dioxide
Ans. (d) : The functions of a lubricant in an engine– (c) Nitrogen (d) Hydrogen
• To reduce the friction between the components and Ans. (a) : The main components of biogas are carbon
metal to metal contact. dioxide and methane. In biogas, methane constitutes 50-
• To reduce the overheating of the components that's 75%, carbon dioxide constitutes 25-0% and nitrogen
mean to keep the parts cool. constitutes 2 to 8%. It also includes trace gases
• To prevent wear and corrosion of the components due including nitrous oxide and hydrogen sulphide.
to rubbing. 85. Which one is correct for feeding biomass
• The dirt particle present if any deteriorates the life of material and gasification agent into an updraft
the engine, the lubricant also helps in removing it. gasifier?
81. A differential system of tractor is fitted in (a) Biomass from top, gasifying agent from top
between: (b) Biomass from top, gasifying agent from
(a) Engine and gearbox bottom
(b) Gearbox and final drive (c) Biomass from bottom, gasifying agent from
(c) Clutch and gearbox top
(d) Engine and final drive (d) Biomass from bottom, gasifying agent from
bottom
Ans. (b) : Parts of differential–Following are the parts
of the differential system– Ans. (b) : In an updraft gasifier, the biomass material is
fed from the top of the reactor and the gasification agent
• Differential side gear
is fed from the bottom of the reactor. As the gasifying
• Pinion shaft or cross pin
agent flows through the biomass, gas is generated and
• Axle shafts or half shaft exhausted from the top. Updraft gasifiers are one of the
• Ring gear or crown wheel simplest and most common types of gasifier for
• Drive pinion or bevel pinion biomass.
• Differential pinions or planet gears 86. Which feature is not true for fixed dometype
• Differential case or housing biogas plants?
A differential system of tractor is fitted in between (a) Constant gas pressure
gearbox and final drive. Differential is a basically a pair (b) Less maintenance
of bevel gears working together to transmit torque to (c) No corrosion problem
rear wheel when vehicle is turning either left or right, (d) Better heat insulation
rear wheels will be at different radius fro the center
Ans. (a) : Fixed dome type biogas plants–In fixed
point of virtual circle.
dome (gas collector) are combined and enclosed in the
82. Which one is the most appropriate fuel to blend same camber. These types of plants are best suited for
with bio-ethanol for transport fuel? batch type gas plants. These types of plants are more
(a) Diesel (b) Petrol economical compared to floating dome type since only
(c) Kerosene (d) All of these the masonry work is needed for their construction.
Ans. (b) : Bio-ethanol can be used in petrol engines as Advantages of fixed dome type–
an alternative for gasoline. It can be mixed with • Cast of plant is less compared to floating drum type
gasoline to virtually any percentage. Most of the plant.
existing petrol operate on blends of up to 15% bio- • Loss of heat is negligible since these are constructed
ethanol with petroleum. underground.
83. What is the name of a process that converts • No corrosion problems
fats and oils into biodiesel and glycerin?
• It is maintenance free
(a) Fermentation (b) Transesterification
Disadvantage–
(c) Distillation (d) All of these
• Needs skilled labour to operate.
Ans. (b) : Transesterification–Transesterification
stands out as a good alternative due to its ease of use • Gas production of digestive is at variable pressure.
and the premise that the batty acid easter produced by 87. Which of the following is a major drawback for
this procedure is like that of diesel. Transesterification most renewable energy sources?
converting fats and oils into biodiesel and glycerin is (a) High pollution
used to make a gasoline (a by product). (b) High running cost
Transesterification is the chemical process that turns (c) Available only in a new places
vegetable oil of animal fat into biodiesel. (d) Unreliable supply
OPSC AAO Exam. 2020 (Paper-I) 75 YCT
Ans. (d): Unreliable supply is a major drawback of Ans. (b) : The wind speed are measured using an
most renewable energy sources. The renewable energy instrument called anemometer and wind direction is
also depends on the weather condition of its power measured using a wind vane or cock. A typical
source. anemograph consisting of wind speeds recorded at there
88. Determine wind power if the wind speed is 10 heights during strong wind.
m/s and has a blade length of 20 m. (air density Wind rose–The wind speed is measured and recorded
ρ = 1.23 kg/m3): at an effective height of 10 m from the ground. Wind
rose is drawn showing mean wind speeds based on
(a) 625 kW (b) 773 kW
averaging period which may vary from 10 min to 1
(c) 851 kW (d) 855 kW hour.
Ans. (b) : Given that, 92. The angle between the beam from the Sun and
Wind speed (V) = 10 m/s the Vertical is known as:
Blade length (l) = 20 m (a) Solar Azimuth angle (b) Altitude angle
Air density (P) = 1.23 kg/m3 (c) Zenith angle (d) Hour angle
The wind energy (power) formula Ans. (c) : Zenith Angle–It is the angle between the
1 vertical plane and the line to the sun, that is the angle of
P= ρAV 2 incidence of the beam radiation on a horizontal surface.
2
Hour angle–It is the angular displacement of the sun
Area (A) = πr2
east or west of the local meridian, due to the rotation of
= π × 202 the earth on its axis at 150 h–1, morning is negative and
= 400π m2 afternoon is positive.
1 93. Which of the following is the phase of Project
Wind power (P) = ρAV 2
2 Management?
1 (a) Project planning
P= × 1.23 × 400π × 102
2 (b) Project scheduling
P = 772440 W (c) Project controlling
p = 772.44 kW  773 kW (d) All of these
89. A photovoltaic cell or solar cell converts: Ans. (d) : Five phases of project management–The 5
(a) Thermal energy into electricity basic phases in the project management process are:
1. Project initiation
(b) Solar radiation into thermal energy
2. Project planning
(c) Electromagnetic radiation into electricity
3. Project execution
(d) Solar radiation into kinetic energy
4. Project monitoring and controlling
Ans. (c) : Solar or photovoltaic cell–
5. Project closing
• A photovoltaic cell converts solar radiation into
94. Who introduced the bar charts in Project
electrical energy.
Management?
• The process of conversion of solar energy into electric (a) Henry Gantt (b) Williams Henry
energy is called a photovoltaic effect. It primarily (c) Jane Gantt (d) Joseph Henry
consists of methane and carbon dioxide.
Ans. (a) : Bar chart–Bar chart is a very good tool for a
• Solar radiation to thermal energy is done by the solar preliminary understanding of the project and helps in
electricity generation system. the preparation of the feasibility report.
90. The theoretical maximum efficiency of a wind The bar chart was originally conceived and develop by
turbine is given by the Betz Limit and is about: an American, Henry Gantt, prior to world war-I for use
(a) 30% (b) 48% in ship building projects and therefore it is also known
(c) 59% (d) 65% as the Gantt chart. It is perhaps the oldest tool used for
Ans. (c) : The theoretical maximum efficiency of a scheduling the project. The project manager can carry
wind turbine is given by the Betz limit and is around the bar chart with him for ready reference and
59%. Practically wind turbines operate below the Betz monitoring on a daily basis.
limit, this is called the "Power coefficient". 95. What does PERT stand for?
91. The wind speed is measured using an (a) Program Evaluation and Review Technique
instrument called: (b) Program Evaluation and Robot Technology
(a) Pyrometer (b) Anemometer (c) Program Evaluation and Robot Technique
(c) Manometer (d) Wind vane (d) Program Evaluation and Rate Technology
OPSC AAO Exam. 2020 (Paper-I) 76 YCT
Ans. (a): PERT stands for the method of analyzing 1. North-west corner rule
project schedules called "Program Evaluation and 2. Lowest cost entry method
Review Technique" PERT was developed in the late 3. Vogel's Approximation method
1950s to account for the impact of project schedule risk • It does not take into account the cost of transpiration.
on the planned completion of major projects. So it is one drawback of the North-west corner rule to
• Most of the Activities of PERT network, follows Beta finding the initial solution for the transportation
(β)-probability distribution curve. problem.
96. The activity which can be achieved under ideal 99. The transportation problem deals with the
circumstances within the shortest possible time transportation of:
is known as:
(a) Single product from a source to several
(a) Optimistic time estimate
destinations
(b) Most likely time estimate
(b) Single product from several sources to a
(c) Expected time estimate
destination
(d) Pessimistic time estimate
(c) Several products from a source to a
Ans. (a) : Optimistic time estimate–The estimate of destination
shortest possible time, when the works can be
(d) Several products from several sources to
completed under ideal condition, is known as optimum
several destinations
time estimate or optimistic time estimate. In this
estimate no provisions of any delay or conditions Ans. (d) : Transportation Problem–The transportation
beyond control are taken into account. problem deals with the transportation of goods at
Most likely time estimate–While preparing this minimum cost, from several sources to a number of
estimate of time provisions of few set backs, delays and different destinations. In this type of problem, the goods
breakdowns. Which have occurred while doing such are only allowed to be directly transported between
works in the past are taken into account. This estimate sources or between destinations. The capacities of the
is more nearer to the actual completion time of the demands of the destinations and the unit transportation
acitivites. costs from each source to each and every destination are
97. The method for solving LPP without using known.
artificial variables is called: 100. The finding of initial solution in transportation
(a) Simplex method problem which method is used?
(b) Big-M method (a) Least cost method
(c) Dual simplex method (b) Hungarian
(d) Graphical method (c) Big-M
Ans. (c) : The method for solving LPP without using (d) Simplex
artificial variables is called dual simplex method.
Ans. (a) : Lowest cost entry method or least cost
Big-M method–The method of solving a linear method–
programming problem in which a high penalty cost has
• It is also called matrix minima method
been assigned to the artificial variable is known as the
chame's method of penalties or Big-M method. • In this method, allocations are made on the basis of
98. What is one drawback of the North-West economic desirability. The steps involved in
Corner rule to finding the initial solution for determining an initial solution using least-cost method
the transportation problem? are as follows:
(a) It leads to a degenerate initial solution. Step 1–Write the given transportation problem in
(b) It does not take into account the cost of tabular form (if not given)
transportation Step 3–Choose the cell with minimum cost it is not
(c) It is complicated to implement and use unique, anyone can be chosen suppose it is the (i, j)th
(d) All of these cell.
Ans. (b) : Transportation Problem–Transportation Step 3–Allocate min (ai, bj) to this cell. If the min (ai,
problems are one of the sub classes of linear bj) = ai then the availability of the ith origin is exhausted
programming problems. and demand at the jth destination remains as bj - ai and ith
Methods for solving transportation problem–There row is deleted from the table.
are three methods used for finding initial solution to Step 4–Repeat steps 2, 3, until all origins are exhausted
transportation problems, which are– and all demands are fulfilled. The process must end as.
OPSC AAO Exam. 2020 (Paper-I) 77 YCT
 Odisha Public Service Commission
Assistant Agriculture Officer (AAO)
Exam- 2020 (Paper-II)
1. Bernoulli's equation is applicable for: Ans. (b): Viscosity of tomato juice increases rapidly
(a) Viscous and compressible fluid flow with increase in concentration of solid, therefore, the
(b) Inviscid and compressible fluid flow juice must be concentrated in a recirculating batch type
(c) Viscous and incompressible fluid flow falling film evaporator.
(d) Inviscid and incompressible fluid flow Falling film evaporator are more commonly used than
Ans. (d) : Bernoulli's equation–The Bernoulli's rising film types.
equation was derived on the assumption that fluid is 4. The peak wavelength of radiation emitted by a
inviscid (non-viscous). black body at a temperature of 2000 K is 1.45
(i) The fluid is ideal, i.e. viscosity is zero. µm. If the peak wavelength of emitted
(ii) The flow is steady radiation changes to 2.90 µm, then the
(iii) The flow is incompressible temperature (in K) of the black body is:
(iv) The flow is irrotational (a) 500 (b) 1000
2. Which one of the following statements is (c) 4000 (d) 8000
correct for a superheated vapour? Ans. (b) : Given,
(a) Its pressure is less than the saturation pressure
at a given temperature. Black body (λ)peak = 1.45 µm at 2000 K
(b) Its temperature is less than the saturation (λ')peak = 2.90 µm
temperature at a given pressure. From wein's displacement low
(c) Its volume is less than the volume of the (λ)max T = 2898 µm – K (constant)
saturated vapour at a given temperature.
(λ)peak T = (λ')peak T'
(d) Its enthalpy is less than the enthalpy of the
saturated vapour at a given pressure. 1.45 × 2000 = 2.90 × T'
Ans. (a) : Superheated vapour– T' = 1000 K
5. During a non-flow thermodynamic process (1-
2) executed by a perfect gas, the heat
interaction is equal to the work interaction (Q1-
2 = W1-2) when the process is:
(a) Isentropic (b) Polytropic
(c) Isothermal (d) Adiabatic
Ans. (c) : During a non-flow thermodynamic process
(1-2) executed by perfect gas, the heat interaction is
equal to the work interaction (Q1-2 = W1-2) when the
Psat@T1 – Saturation pressure at T1 temperature process is isothermal.
P1 = Pressure of superheated vapour at states 6. A slender rod of length L, diameter d (L>>d)
So, P1 < Psat@T1, i.e. pressure less than the saturation and thermal conductivity k1 is joined with
pressure at a given temperature. another rod of identical dimensions, but of
3. Viscosity of tomato juice increases rapidly with thermal conductivity k2, to form a composite
increase in concentration of solid, therefore, the cylindrical rod of length 2L. The heat transfer
juice must be concentrated in a: in radial direction and contact resistance are
(a) Recirculating batch type rising film negligible. The effective thermal conductivity of
evaporator the composite rod is:
(b) Recirculating batch type falling film (a) k1 + k2 (b) k1 k 2
evaporator
(c) Backward feed multiple effect evaporator k1 k 2 2k1k 2
(c) (d)
(d) Forward feed multiple effect evaporator k1 + k 2 k1 + k 2

OPSC AAO Exam. 2020 (Paper-II) 78 YCT

Ans. (d): • Application of natural convection are is many
chemical industries and food industries.
Natural convection is associated with grashof number
only.
9. Thermal Death Time (TDT) curve is between:
According to question, (a) Time Vs Temperature
heat transfer in radial direction (Rth)eq = R1 + R2 (b) Temperature Vs log (time)
2L L L (c) Time Vs log (number of survivors)
= +
A k eq Ak1 Ak 2 (d) None of these
2 L L k1 + k 2 Ans. (c) : • Thermal Death Time (TDT) curve is
= + = between time vs number of survivors.
k eq k1 k 2 k1 k 2
• The thermal death time curve provides information
2k1k 2 about the time required to kill a particular
k eq =
k1 + k 2 microorganism in a particular food at variety of
temperatures.
7. Convective heat transfer coefficient of a body is 10. Larger decimal reduction time implies:
dependent partly upon its: (a) Microorganisms (m.o.) more heat resistant
(a) Temperature (b) Microorganisms less heat resistant
(b) Composition (type of material) (c) Microbial growth redness with time
(c) Shape and size (d) No effect on Microorganisms
(d) None of these Ans. (b) : Larger decimal reduction time implies
Ans. (b) : • Conventional heat transfer occurs when microorganisms less heat resistant.
there is bulk fluid motion, which carries away the heat 11. Psychrometric relations developed are valid at
from the heated surface. only:
Convention heat transfer is given by (Q) = h×A×(Ts-T∞) (a) Any pressure
Where, h = Heat transfer coefficient (b) One atmosphere
A = Area along the heat transfer (c) Below one atmosphere
Ts = It is the surface temperature (d) Above one atmosphere
T∞ = T∞ is the temperature of the surrounding fluids. Ans. (d) : Psychrometric relations developed are valid
at only above one atmosphere.
• The convection heat transfer coefficient is not the
property of the fluid rather it depends upon the thermo- 12. Consider an ideal vapour compression
physical properties. refrigeration cycle. If the throttling process is
replaced by an isentropic expansion process,
Convective heat transfer coefficient depends on keeping all the other processes unchanged,
several factor like- which one of the following statements is true
1. Surface area for the modified cycle?
2. Temperature difference (a) Coefficient of performance is higher than that
3. Motion of fluid of the original cycle
4. Thermal and physical properties of fluids (b) Coefficient of performance is lower than that
of the original cycle
5. Geometry and orientation in space of the surface.
(c) Coefficient of performance is same as that of
8. Natural convection is associated with: the original cycle
(a) Prandtl number only (d) Refrigerating effect is lower than that of the
(b) Reynolds number only original cycle
(c) Grashof number only Ans. (a) : In case isentropic expansion process,
coefficient of performance is higher than that of the
(d) Nusselt number only
original cycle
Ans. (c) : Natural convection, also called as free
13. The "degrees of freedom" for a pure substance
convection, occurs as a result of density difference at its triple point is:
between the heated fluid and cold fluid, which bring
(a) 3 (b) 2
about, convection currents so as to transfer the heat
from one point to another. (c) 1 (d) 0

OPSC AAO Exam. 2020 (Paper-II) 79 YCT

Ans. (d): Triple point–The temperature and pressure at 17. Ratio of Schmidt number to Lewis number is:
which, the solid liquid and vapour phase of a pure (a) Prandtl number (b) Reynolds number
substance coexists in equilibrium are called the triple (c) Nusselt number (d) Sherwood number
points.
Ans. (a) : Ratio of Schmidt number to Lewis number is
prandtl number.
18. 'Red dog' is one of the byproducts during
milling of:
(a) Corn (b) Rice
(c) Ragi (d) Wheat
Ans. (d) : 'Red dog' is one of the byproducts during
milling of wheat.
19. Potato slices have been dehydrated from an
According to Gibb's phase rule is given by initial solid content of 12% to a final solid
P + F = C + 2 ......(i) content of 94%. If the peeling and other losses
are to the tune of 10%, final yield (percent) of
P = No. of phase the dried chips per ton of fresh potato taken is
F = Degree of Freedom approximately:
C = No. of components (a) 11.5 (b) 10.0
Given, (c) 8.5 (d) 5.0
For pure substance (C) = 1 Ans. (*) : j
At triple point (P) = 3 20. Which of the following radiation methods is
According to equation (i) used in preserving food from spoilage by
3+F=1+2=3 microorganisms?
(a) Radio waves (b) Microwaves
F=0
(c) Non-ionizing (d) Ionizing
14. Colloidal stability of milk casein is because of
the highly hydrated carbohydrate residues in: Ans. (d) : • Food irradiation the application of ionizing
radiation to food) is a technology that improves the
(a) αs1 casein (b) αs2 casein safety and extends the shelf life of foods by reducing or
(c) β casein (d) k casein eliminating micro organisms and insects. Like
Ans. (d) : • Colloidal stability of milk casein is because pasteurizing milk and canning fruits and vegetables,
of the highly hydrated corbohydrate residues in k- irradiation can make food safer for the consumer.
casein. 21. Which of the following is oil soluble pigment
present in fruits and vegetables?
15. Rice bran is stabilized prior to oil extraction to
protect it from the activity of: (a) Flavonoids (b) Carotenoids
(a) Polyphenol oxidase (b) Peroxidase (c) Anthocyanins (d) Tannins
(c) Lipase (d) Lipoxygenase Ans. (b) : Carotenoids–
Ans. (d) : Rice bran is stabilized prior to oil extraction • Carotenoids also called tetraterpenoids are yellow,
to protect it from the activity of lipoxygenase. orange and red organic pigments that are produced by
plant and algae, as well as several bacteria and fungi.
16. Sticking of powder to wall of the chamber
• Carotenoids give the characteristic colour of
during spray drying of fruit juice is due to: pumpkins, carrots, parsnips, corn, tomatoes, canaries,
(a) Low glass transition temperature of the etc.
compounds in juice • Carotonoids can be produced from fats, and other
(b) High glass transition temperature of the basic organic metabolic building blocks by all these
compounds in juice organisms.
(c) Improper processing parameters of spray 22. Which of the following represent the group of
dryer saturated fatty acids?
(d) Presence of gums in feed material (a) Lauric, Myristic, Arachidic
Ans. (b) : Sticking of powder to wall of the chamber (b) Palmitic, Linoleic, Linolenic
during spray drying of fruit juice is due to high glass (c) Capric, Stearic and Oleic
transition temperature of the compounds in juice (d) Behenic, Caprylic, Arachidonic
OPSC AAO Exam. 2020 (Paper-II) 80 YCT
Ans. (a): Fatty acids are two main types– (c) Plate heat exchanger
(i) Saturated fatty acid– (d) Evaporator tubes
• Animal fats are mostly saturated fats. Ans. (c) :
1. Arachidic Acid– • In heat exchanger analysis, the log mean temperature
• It is also known as icosonic acid. difference is widely used for heat exchanger selection in
• It is a saturated fatty acid with a 20-carbon chain. which the temperature are known and the size of the
• It is minor constituent of cupuacu butter, perilla oil, heat exchanger is required.
peaut oil, corn oil, and cocoa butter. • Log mean temperature difference (LMTD) correction
• The salt and esters of arachidic acid are known as factor is applicable to plate heat exchanger.
arachidates. 26. In refining of the rice bran oil for edible
2. Lauric Acid– purpose, neutralization process is carried out
• Breast milk, coconut oil, palm kernel oil. mainly to remove:
3. Myristic Acid– (a) Wax and organic impurities
• Dairy products and cow milk. (b) Gums and mucilage
ii. Unsaturated fatty acid (c) Saturated glycerides
23. Irradiation carried out to reduce viable non- (d) Free fatty acids
spore forming pathogenic bacteria using a dose Ans. (d) : • In refining of the rice bran oil for edible
between 3 to 10 kGy is called: purpose, neutralization process is carried out mainly to
(a) Radurization (b) Thermoradiation remove free fatty acid.
(c) Radappertization (d) Radicidation
27. One hundred kilogram spice is extracted for
Ans. (d) : Radicidation– essential oil using twice the amount of a pure
• Radicidation is a specific case of food irradiation organic solvent. The extracted solid mass
where the dose of ionizing radiation applied to the food contains 5% residual oil (oil-free solid mass
is sufficient to reduce the number of viable specific non basis). The liquid extracted mass contains 20%
spore forming pathogenic bacteria to such a level that oil. Assume no solvent retained by the
none are detectable when the treated food is examined extracted solid mass. Initial mass of the oil in
by any recognized method. the spice in kg is:
• The term radiciation is derived from radiation and (a) 52.5 (b) 55.0
caldera.
(c) 57.5 (d) 60.0
24. Identify an example of a classical diffusional
mass transfer process without involving heat, Ans. (*) : j
among the following: 28. A fat globule of 1.5 µm diameter is rising up in
(a) Drying of food grains a stagnant skim milk medium of 1005 kg m–3
(b) Carbonation of beverages density and 1.5 cP viscosity. If the density of
(c) Distillation of alcohol the fat globule is 915 kg m–3, the steady rising
(d) Concentration of fruit juice velocity of the globule in µms–1 is:
(a) 0.074 (b) 0.082
Ans. (b) : Diffusional mass transfer–
• Diffusion is a mass transfer phenomenon that causes (c) 0.095 (d) 0.125
the distribution of a chemical species to become more Ans. (*) : j
uniform in space as time passes. 29. Milk and fruit juice are deaerated before they
• It this case species is a chemical dissolved in a solvent are allowed to flow through pasteurizer. This is
or a component in a gas mixture, such as the oxygen in done in order to:
air.
(a) Reduce fouling of pasteurizer
Ex.:
(b) Increase rate of heat transfer
(i) Carbonation of beverages
(c) Reduce oxidative deterioration
(ii) Purification of blood in the kidneys and liver.
(d) Avoid discoloration
25. Log Mean Temperature Difference (LMTD)
correction factor (F) is applicable to: Ans. (a) : Milk and fruit juice are deaerated before they
(a) Spiral tube heat exchanger are allowed to flow through pasteurizer. This is done in
order to reduce fouling of pasteurizer
(b) Steam jacketed kettle
OPSC AAO Exam. 2020 (Paper-II) 81 YCT
30. When heat flows in steady state through two Ans. (c): Vertical issues–
metal plates separated by a negligibly small air • Sweets, confectionary, sugar and honey, milk and
gap between them, the rate of heat flow will be milk products.
governed by:
• Water flavoured, water and beverage.
(a) Thermal diffusivity of air
• Oil and fats
(b) Thermal diffusivity of metal plate
• Fruits and vegetable and their products.
(c) Thermal conductivity of air
• Cereals, pulses, legumes, and their products.
(d) Thermal conductivity of metal plates
• Fish and fisheries product.
Ans. (a) : When heat flows in steady state through two
metal plates separated by a negligibly small air gap • Meat and meat products including poultary.
between them, the rate of heat flow will be governed by Horizontal issues–
thermal diffusivity of air • Food additives, flavourings processing aids and
31. The temperature at which the water in a fruit materials in contact with food.
will freeze is dependent on the amount of the • Pesticides and Antibiotics residues.
following present in the fruit. • Genetically modified organisms and foods.
(a) Sugar (b) Fibre
35. Temperature yielding maximum storage life of
(c) Oil (d) Water food materials decreases in the order:
Ans. (a) : • Freezing preservation is one of the most (a) Potato < Fish < Milk < Orange
beneficial preservation methods. (b) Fish < Milk < Orange < Potato
• It involves, conversion of liquid content of food into (c) Fish < Potato < Orange < Milk
ice crystals, which lowers down water activity and (d) Milk < Fish < Orange < Potato
microbial growth is arrested due to cold shock.
Ans. (d) : Temperature yielding maximum storage life
• Pure water is frozen at 0ºC but since fruits and of food materials decreases in the order
vegetables contain number of dissolved solids like,
Milk < Fish < Orange < Potato
sugar, acid they freeze at below 0ºC.
36. The falling drop of water become sphere due
32. Thermal vapour compression is used in a/an:
to:
(a) Evaporator (b) Homogenizer
(a) Surface tension of water
(c) Pasteurizer (d) None of these
(b) Compressibility of water
Ans. (d) : Thermal vapour compression is used in a (c) Viscosity of water
distillation.
(d) Capillarity of water
33. The dimension of mass transfer coefficient is:
Ans. (a) : • Because of a phenomenon known as surface
(a) M0L–2θ–1 (b) M0L–1θ–1 tension water drops or any other liquid drops, are
(c) M0L2θ–1 (d) M0Lθ–1 spherical in shape.
Ans. (d) : The mass transfer coefficient is a diffusion • This acts on the surface of a freely falling drop in a
rate constant that relates the mass transfer rate, mass liquid to reduce its area.
transfer area, and concentration change as driving force. • In comparison to other things, such as an ellipsoid, the
η&A ratio of surface area to volume in the case of a sphere is
KC = the smallest.
A ∆C A
• Spherical shapes haves less potential because they can
where, KC = mass transfer coefficient withstand any external force in the atmosphere.
η
& A = mass transfer rate (mol/sec) • Water forms a spherical shape because of its surface
tension.
A = transfer area (m2)
37. In a U-tube manometer, one end is open to the
∆C A = Force concentration (md/m3)
atmosphere, the other end attached to a
The dimension of mass transfer coefficient is M0Lθ–1 pressurized gas of gauge pressure 40 kPa. The
34. The vertical issues of FSS Act includes: height of the fluid column on the atmospheric
side is 60 cm, and that on the gas side is 30 cm.
(a) Functional foods
The Manometric fluid used is : (Take g = 9.8
(b) Food Additives m/s2)
(c) Water, flavoured water and beverages (a) Liquid ammonia (b) Water
(d) Genetically modified organisms and foods (c) Mercury (d) Oil
OPSC AAO Exam. 2020 (Paper-II) 82 YCT
Ans. (c): Given, 41. For compressible fluid flow in a pipe, having a
Gauge pressure of Gas (∆P) = 40 kPa = 40,000 Pa decrease in specific gravity, what will be the
effect of a reduction in diameter?
Height difference (∆h) = (60–30) = 30 cm (a) It will cause a decrease in velocity
we know that, (b) it will cause an increase in velocity
∆ = ρg ( ∆h ) (c) It remains constant
( )
4000 = ρ × 9.8 30 × 10−2 (d) None of these
Ans. (b) : According to continuous equation.
ρ = 13606 kg/m3 ρAV = Constant
38. If the tank is moving vertically, which of its • Density will be decrease because specific gravity
component is subjected to maximum total decrease.
pressure?
• Reduction of diameter of pipe, so, area will be
(a) The higher part of vertical walls decrease
(b) The lower part of vertical walls So, For find out the constant value, velocity will be
(c) Base increase.
(d) Centre of the tank 42. Which equation must be perfunctorily satisfied
Ans. (c) : • If the tank is moving vertically, which of its while dealing with fluid flow problems?
components is subjected to maximum to pressure on the (a) Newton's Third law
base. (b) Newton's Second law
39. In a stationary fluid, how does the local (c) Continuity equation
pressure of the liquid vary? (d) Law of Conservation or momentum
(a) In the horizontal direction, only Ans. (c) : While dealing with fluid flow problems
(b) With depth only continuity equation must be perfunctorily satisfied.
(c) Neither with depth nor along the horizontal 43. If there is a change in angle contained by two
direction sides, the average of the curve is:
(d) Both with depth and along the horizontal (a) Linear Deformation
direction (b) Linear Translation
Ans. (b) : According to Pascal's law, the local pressure (c) Rotation
of a fluid is same in all directions. Hence, the pressure (d) Angular Deformation
won't vary along the x and y direction. The local Ans. (d) : If there is an change in angle contained by
pressure will increase with an increase in depth due to two sides, the average of the curve is angular
the extra weight of water column above that point. deformation.
40. A uniform body of size 4 m long × 2.5 m wide × 44. The flow of fluid along a curvilinear or curved
1.5 m deep floats in water. What is the weight path is known as:
of the body if the depth of immersion is 1 m? (a) Vortex Flow (b) Sink Flow
(a) 73.5 kN (b) 147.1 kN (c) Circular Flow (d) Curvilinear Flow
(c) 294.3 kN (d) 588.6 kN Ans. (a) : Vortex flow is defined as the flow of a fluid
Ans. (b) : Length of body = 4m along a curved path or the flow of a rotating mass of
Width of body = 2.5 m fluid is known a vortex flow.
Depth of body = 1.5 m The vortex flow is of two types namely.
1. Forced vortex flow
2. Free vortex flow
45. The loss of head due to friction in a pipe of
uniform diameter in which a viscous flow is
taking place, is (where RN = Reynold number)
(a) 1/RN (b) 4/RN
We know that, (c) 16/RN (d) 64/RN
Weight of the body = weight of water displaced Ans. (c) : The loss of head (or energy) in a pipes due to
= ρ × g × volume of displaced friction is calculated from Darcy-weisbach equation.
= 9.81 × 1000 × 4 × 2.5 × 1.5 4fLV 2
hf =
= 147.1 kN 2gd

OPSC AAO Exam. 2020 (Paper-II) 83 YCT

Where, hf = head loss due to friction 48. The total number of independent equations
f = co-efficient of friction which is function of Reynolds that form the Lacey regime theory is:
number (a) 6 (b) 10
16 (c) 3 (d) 4
= (For RN < 2000) → Viscous flow
RN Ans. (c) : The total number of independent equation
0.079 6 that form the Lacey regime theory is three.
= (For Re varying from 4000 to 10 )
(RN )
1/ 4
2
Vo = fR
RN = Reynolds Number 5
L = Length of pipe
V = Mean velocity of flow Af 2 = 140 Vc5
d = diameter of pipe
f = 1.76 d
46. Pick up the incorrect statement from the
following:
Vc = 10.8 R 2 / 3S1/ 3
(a) In free flooding irrigation, water is admitted
at one corner of a field and is allowed to Where, Vc = Channel relating regime velocity
spread over the entire area. f = Silt factor
(b) In check method of irrigation, the field is
R = Hydraulic radius
divided into smaller compartments and water
is admitted to each in turn. A = Area
(c) In furrow irrigation, water is admitted d = Sediment size (mm)
between the rows of plants in the field. S = Bed slope
(d) None of these
49. During land levelling of agricultural land for
Ans. (d) : • In free flooding method water if admitted at irrigation and drainage purposes, the
one corner of a field and is allowed to spread over the acceptable deviation in elevation from the
entire area.
design value in metre is:
• The check method of irrigation, the field is divided in
(a) 0.015 (b) 0.025
to smaller compartments and water is admitted to each
in turn. (c) 0.055 (d) 0.150
• In furrow irrigation, water is admitted between the Ans. (a) : • The acceptable deviation in elevation from
rows of plants in the field. the design value in meter is = 0.015
47. The infertility of the soil in waterlogged area is 50. A crop has effective root zone depth of 1200
due to: mm and monthly (30 days) crop
(a) Death of bacteria causing nitrification evapotranspiration of 260 mm. The effective
(b) Increase in salinity rainfall during 30 days period is 20 mm. The
(c) Growth of weeds field capacity and permissible soil moisture
(d) All of these depletion (volume basis) are 16% and 8%
Ans. (d) : Water logging–An agricultural land is said respectively. The irrigation interval in days for
to be water logged when its productivity or fertility is the crop will be:
affected by high water table. (a) 30 (b) 18
Effects of water logging– (c) 12 (d) 8
(i) Inhibiting activity of soil bacteria–The liberation
Ans. (c) : Effective root zone depth = 1200 mm
of plant food is dependent upon the activity of soil
bacteria, which requires adequate amount of oxygen in 260
evapotranspiration = = 8.67 mm / day
the air for proper functioning. 30
(ii) Rise of salt/Increase in salinity–The rise of water 20
table also causes accumulation of alkali salts in the Effective rain fall/day = = 0.67 mm / day
surface soil by the upward flow of water which is 30
established in waterlogged lands. Peak moisture use rate of crop
(iii) Growth of wild flora–In water logged soils, = 8.67 – 0.67 = 8 mm/day
natural flora such as water hyacinth growth profusely. 8 × 1 × 100
This reduced the crop yield. A cultivator has to waste Irrigation interval = 16 − = 12 day
money and time both for clearing it out. 200

OPSC AAO Exam. 2020 (Paper-II) 84 YCT

51. The consumptive use of water for a crop: Ans.(d):
(a) Is measured as the volume of water per unit
area
(b) Is measured as depth of water on irrigated
area
(c) May be supplied partly by precipitation and
partly by irrigation
(d) All of these
Ans. (d) : Consumptive use– • The bottom edge of the notch or the top of a weir over
• Consumptive use of water by a crop is the depth of which water flow is known as sill or crest.
water consumed by the plant in the process of • The height above the bottom of the tank or channel is
transpiration and evaporation during crop growth. known as crest height.
• Consumptive use of water by a crop is expressed as 55. The ratio of total volume of water delivered to
the depth of water per unit area for specified periods a crop to the area on which it has been spread,
such as days months or seasons. is called:
(a) Critical depth (b) Duty
• The value of consumptive use of water is needed to
(c) Delta (d) Base
calculate the irrigation requirement of the crop.
Ans. (b) : • The ratio of total volume of water delivered
52. Irrigation for cereal crop is generally done by: to a crop to the area on which it has been spread is
(a) Check flooding called duty.
(b) Basic flooding • Duty is expressed in hectare per cumec.
(c) Furrow 56. Tile drains increase crop yield by:
(d) Subway surface irrigation (a) Removing the free gravity water that is not
directly available to the plants
Ans. (a) : Check flooding–
(b) Increasing air circulation
• Check flooding is similar to free flooding except that
(c) Reducing and removing toxic substances such
the water is controlled by surrounding the check area
as sodium and other soluble salts
with low, flat levels are provided and the strips are
(d) All of these
divided by field channels.
Ans. (d) : • Tile drainage is a form of agricultural
• Check flooding or basin method are suitable for crops
drainage system that removes excess sub-surface water
that are unaffected by standing water present over a from fields to allow sufficient air space with in the soil,
long period of time like-mangoes, cereals, fruits. proper cultivation and access by heavy machinery to
53. Effective precipitation for a crop may be tend and harvest crops.
defined as: Advantage of tile drainage–
(a) Total precipitation during the crop period • Increases the volume of soil from which the roots can
(b) Available water stored in soil within root obtain food.
zone of the crop • Removes the free gravity that is not directly available
to the plants.
(c) Total precipitation minus the loss due to
infiltration • Increasing air circulation.
(d) Total precipitation minus the loss due to • Reducing and removing toxic substances as sodium
evaporation and other soluble salts.
57. The grid iron pipe drainage system is more
Ans. (b) : Effective precipitation is the amount of
economical than the Herring-bone pipe
precipitation that is actually added and stored in the soil.
drainage system because:
During drier period less than 5 mm of daily rainfall
would not be considered effective, as this amount of (a) It is adopted in the fields which do not require
precipitation would likely evaporate from the surface complete drainage
before soaking in to the ground. (b) The number of main or sub-main lines is
reduced
54. Top of the weir is called:
(c) The number of junctions and the double-
(a) Ridge (b) Peak drainage area are reduced
(c) Head (d) Crest (d) It has only main or sub-main lines
OPSC AAO Exam. 2020 (Paper-II) 85 YCT
Ans. (c): • Herringbone drainage and grid drainage 60. Which type of alternate layouts system for tile
system are the most common pipe drainage systems. drainage has two mains?
Herringbone drainage system– (a) Double Main System
• The hearing bone system is the design for the main (b) Natural System
central pipe. (c) Grid-iron System
• The pipe run down a slope and at acute angles with (d) Herring Bone System
numberous lateral pipes connecting to it. Ans. (a) : • This system has two mains with separate
• This type of design is popular with large general lateral for each main.
amenity playing fields and by rectangular lawns. • This type of system is used when the bottom of
• It is also suitable for area that are irregular in shape. depression is wide.
• Its design is more complicated than the grid iron • It also helps in reducing the length of the laterals and
drainage system. eliminates the break in slope at the edge of the
depression of the lateral.
Grid iron pipe drainage–In this system pipe or the
main pipe, is situated at the perimeter of an area, with 61. How many cubic metres of water will be
lateral joining the central pipe at acute angles or right removed for this particular period of a system
angles. This is a form of drainage design, that is suitable designed to use a D.C. of 1.5 cm draining 20
hectares for a capacity of 5 days?
for bowling greens, football pitches and other areas that
boast a regular shape. (a) 10000 m3 (b) 15000 m3
Note–It is easier to install than a herringbone system, (c) 20000 m3 (d) 25000 m3
and this is often because it has fewer junctions. Ans. (b) : • Drainage coefficient of 1.5 cm means that
58. If the drainable porosity of a command area is 1.5 cm of water depth will be removed from the
5% and the design rate of drop of the water drainage in 24 hours.
table is 0.25 m day–1, the drainage coefficient of • Volume of water entering the drain per day
the command area in mm day–1 will be: = (1.5 × 100)20 × 104
(a) 250 (b) 12.5 = 3000 m3/day
(c) 1.25 (d) 0.0125 • Volume of water passing the drain within 5 days of
Ans. (b) : Drainage coefficient of the command area, flow = 5 × 3000 = 15000 m3
0.25 × 1000 × 5 62. What is the cross sectional shape of shallow
= surface drains?
100
(a) Triangular Shape (b) Circular Shape
= 12.5
(c) Rectangular Shape (d) Trapezoidal Shape
59. What is the size of the tile at an outlet of a 12
Ans. (d) • Since the shallow drains are designed to
hectare drainage system, if the D.C. is 2 cm and carry normal storm water and excess irrigation water.
the tile grade is 0.5%. Assume the rugosity
coefficient as 0.015 for the drain material? • These are designed for normal runoff from the storm
water, but we have to take excess irrigation in to the
(a) 10 cm (b) 15 cm account. So,
(c) 20 cm (d) 25 cm • For maximum discharge the shallow drains are given
Ans. (c) : Volume of water passing the drain in one day a trapezoidal cross sectional shape.
 1  63. In surface inlet, what is provided to deal with
=  × 12 × 104  = 1200 m3 / day trash?
 100 
(a) Cunnette (b) Beehive Grate
Volume of water passing the drain in one second
(c) Blind Inlet (d) Surface Inlet
 1200 
=  × 3600 Ans. (b) : The bottom of the surface inlet a concrete
 24  collar is provided around the intake on to the prevent
growth of vegetation or weeds. And to prevent the trash
1 m3
= from entering the tile a beehive grate is provided on the
72 sec top of the riser.
1
× A × ( R ) × (S) 64. Which type of drains is used for small quantity
2/3 1/ 2
Q=
n of waters removal?
1  1   πD 2   0.5  (a) Blind Inlet
= × ×
72  0.015   4 
  100 
(b) Shallow Surface Drains
(c) Deep Surface Drains
D = 17.4 cm = 20 cm (d) Open Drains
OPSC AAO Exam. 2020 (Paper-II) 86 YCT
Ans. (a): • When the water quantity to be removed 68. If the axial length of a drainage basin is 35 km
from the pits or depression is small, then a blind inlet or and its form factor is 0.2, the total area of the
depression is small, then a blind inlet or also named as basin is:
French drain is constructed over the tile drain. (a) 205 sq. km. (b) 215 sq. km.
• These drains are constructed by back filling with (c) 225 sq. km. (d) 245 sq. km.
graded material in the trench of the tile drain. Ans. (d) :
65. An aquifer can hold water _______ and the Area of the basin
state of water is _______. Form factor =
Square of the axial length
(a) Permanently–State of flow
(b) Temporarily–State of flow Area of the basin
0.2 =
( 35)
2
(c) Permanently–State of stagnancy
(d) Temporarily–State of stagnancy Area of the basin = 0.2 × (35)2 = 245 km2
Ans. (b) : • It must be remembered than an aquifer can 69. A well of diameter 25 cm fully penetrates a
hold water only temporarily, as the water is always in confined aquifer. After a long period of
state of flow in it. However, pumping at a rate of 55 l/sec, the observations
• The rat of flow of water through an aquifer is very of drawdown taken at 12 m and 120 m
flow compared to surface waters. distances from the centre of the well are found
to be 3.5 m and 0.75 m respectively. The
66. If P and A are the perimeter and area of a transmissivity of the aquifer is _____ m2/day.
drainage basin, its compactness coefficient, is:
(a) 633.5 (b) 623.5
(a) P2/2πA (b) P/2πA (c) 600 (d) None of these
(c) P/2√πA (d) P3/π3A
Ans. (a) : Transmissivity (T) = k.D
Ans. (c) : Shape of a drainage basin is generally Where, k = coefficient of permeability
expressed by form factor and compactness coefficient.
D = Depth of aquifer
Average width of the basin r 
Form Factor = 2.303q·D
Axial length of the basin T= × log10  2 
2 πD ( h 2 − h 1 )  r1 
Perimeter of the basin Given, q = 55 l/sec = 55 × 10–3 × 24 × 60 × 60 m3/day
Compactness coefficient =
Circumference of a circle q = 4752 m3/day, h1 = 0.75 m, h2 = 3.5 m
whose area is equal to the
2.303 × 4752  120 
T= × log10  
area of the basin 2π ( 2.75)  12 

=
P
=
P T = 633.37 m2/day
4 πA 2 πA 70. Pick up the correct statement from the
67. The yield of an open well depends upon: following:
(a) Permeability of a soil (a) A confined bed of impervious material laid
over an aquifer, is known as an aquiclude.
(b) Area of aquifer opening into the wells
(b) The topmost water bearing strata having no
(c) Actual flow velocity
aquifer is known as non-artesian aquifer.
(d) All of these
(c) The ordinary gravity wells which supply
Ans. (d) : Following factor depends on yield of an water from the topmost water bearing strata,
open well– are called water table wells.
(1) Permeability of soil– (d) All of these
Permeability ↑ → yield ↑ Ans. (d) : Aquiclude–
• A confined bed of impervious material laid over on
(2) Area of aquifer opening in to the wells–
aquifer is known as an aquiclude.
Area of aquifer opening ↑ → yield↑
• It contains a large amount of water in it does not
(3) Actual flow velocity– permit water through it and also does not yield water.
Flow velocity↑ = yield↑ Ex.: Aquiclude
OPSC AAO Exam. 2020 (Paper-II) 87 YCT
Non-Artesian aquifer– From equation (i)
• The top most water bearing strata having no confining H ∝N
impermeable over burden is known as unconfined From equation (ii)
aquifer. Q∝N
• This aquifer is also know non-artesian aquifer or Combining both equation (i) and (ii)
water table aquifer.
H ∝Q
Water table wells–The ordinary gravity wells which
supply water form the top most water bearing strata are Q H ∝ Q2
called water table wells.
71. A commonly use of hand pump is the:
(a) Centrifugal Pump
(b) Reciprocating Pump
(c) Rotary Pump
(d) Axial flow Pump
Ans. (b) : Reciprocating pump–
• It is a hydraulic machine which converts mechanical
energy into hydraulic energy.
73. A centrifugal pump having an overall efficiency
• These are used to lift water against high head at low of 75% requires 6 kW power at 1450 rpm to
discharge. deliver water against a suction head of 5 m and
• Volume of water discharge per sec a delivery head of 12 m. If the pump runs at
1650 rpm and frictional head losses are
ALN 3 negligible, the total head developed by the
Q= m / sec
60 pump will be:
where, A = Area of cylinder (m2) (a) 22.01 m (b) 25.05 m
N = Crank speed (in rpm) (c) 29.35 m (d) 31.72 m
L = Length of cylinder (in m) Ans. (*) : j
• To increase discharge and to maintain it more 74. A piston pump is driven by a 5 m diameter
uniform, double acting reciprocating pumps are used. horizontal axis wind turbine for supplying
water from a borehole with a total pump head
2ALN of 10 m. The mean velocity of air is 18 km h–1
Q=
60 and the density of air is 1.29 kg m–3. The actual
power coefficient of the wind turbine is 0.30
Hence, Power also get double.
and the overall pump efficiency in 60%.
Example of reciprocating pump– Neglecting the transmission losses, the expected
• Wind mill pump pump discharge in L S–1 will be:
• Hand pump (a) 2.90 (b) 5.80
• Axial piston pump (c) 28.50 (d) 32.27
• Radial piston pump Ans. (*) : j
72. For the operating point of the pump, a system 75. Negative slip occurs in reciprocating pumps,
characteristic between the head required 'H' when delivery pipe is:
and the discharge to be maintained 'Q' is (a) Long and suction pipe is short and pump is
generally expressed as: running at low speed
(a) Linear equation (b) Parabolic equation (b) Long and suction pipe is short and pump is
(c) Exponential equation (d) Cubic equation running at high speed
(c) Short and suction pipe is long and pump is
Ans. (b) : We know that
running at low speed
 H  H (d) Short and suction pipe is long and pump is
  =   .......(i)
 DN  m  DN  p running at high speed
Ans. (d) : • Negative slip occurs in reciprocating
 Q   Q 
 D3 N  =  D 3 N  .......(ii) pumps, when delivery pipe is short and suction pipe is
 m  p long and pump is running at high speed.
OPSC AAO Exam. 2020 (Paper-II) 88 YCT
• Slip in percentage is given by 79. For the upstream face of an earthen dam, the
most adverse condition for stability of slope is:
Q th − Qact  Q 
%slip = × 100 =  1 − act  × 100 (a) Sudden drawdown
Q th  Q th  (b) Steady seepage
%slip = (1 – Cd) × 100 (c) During construction
Where, Cd = coefficient of discharge
(d) Sloughing of slope
76. A centrifugal pump acts as a reverse of:
Ans. (a) : • The most critical of the slide of the
(a) Outward radial flow reaction turbine
upstream slope is the sudden draw-down of water-level
(b) Pelton turbine in the reservoir and the downstream slope is most likely
(c) Inward radial flow reaction turbine to slide, when the reservoir is full.
(d) Reciprocating pump Note–End of construction, critical condition upstream
Ans. (c) : • A centrifugal pump acts as a reverse of side.
inward radial flow reaction turbine.
80. Precipitation caused due to striking of air
• It work on principle of forced vortex motion. masses with a topographical feature, is called:
It has high discharging capacity and can be used for (a) Orographic precipitation
lifting highly viscous liquid e.g = sewage water,
chemical etc. (b) Convective precipitation
♦ Priming in centrifugal pump can be done by the any (c) Cyclonic precipitation
one of the following method. (d) None of these
(i) Automatic or self priming Ans. (a) : Orographic precipitation–Due to mountain
(ii) Priming with suction or vacuum pump. barrier.
(iii) Priming with jet pump Cyclonic precipitation–It occurs due to pressure
(iv) Priming with separation. difference.
77. Hydrograph is a graphical representation of: Convective precipitation–It occurs due to heating of
(a) Surface run off air.
(b) Ground water flow 81. Pick up the correct statement from the
(c) Rainfall following:
(d) Discharge flowing in the river (a) The unit hydrograph of a specified unit
Ans. (d) : Hydrograph–A hydrograph is a showing the duration obtained from the past data can be
rate of flow versus time past of specific point in a river, used to obtain the hydrograph of future
channel, or conduit carrying flow. storms of like duration.
78. If a gauge is installed perpendicular to the (b) To obtain the ordinates of storm hydrograph,
slope, its measurement is reduced by the ordinates of unit hydrograph are
multiplying: multiplied by the multiplying factor.
(a) sine of the angle of inclination with vertical (c) The multiplying factor for storm hydrograph
(b) cosine of the angle of inclination with vertical may be obtained by dividing the run off in
(c) Tangent of the angle of inclination with mm by 25 mm.
vertical (d) All of these
(d) Calibration coefficient of the gauge
Ans. (d) : Unit hydrograph–A hydrograph of direct
Ans. (b) : If a gauge is installed perpendicular to the runoff resulting from 1 cm of effective rainfall applied
slope, its measurement is reduced by multiplying cosine during a specified period of time.
of the angle of inclination with vertical.
• It is given by sherman.
• To obtain the ordinates of stern hydrograph may be
obtained by dividing the run-off in mm by 25 mm.
82. For determination of average annual
precipitation in a catchment basin, the best
method is:
(a) Arithmetical method
(b) Thiessen's mean method
y (c) Isohyetal method
cos θ = ⇒ y = x cos θ
x (d) None of these

OPSC AAO Exam. 2020 (Paper-II) 89 YCT

Ans. (c): Average annual precipitation–It is average We know that,
value of annual rainfall value for last 35 years. Discharge (Q)
Normal precipitation–It is average value of rainfall 30 Volume Area × height
yrs data of particular date or month. = =
time time
• The isohyetal method is superior to other methods,
when stations are large in number. 120 × 106 × 0.01
=
• Thiessen-polygon method of calculating the average 6
precipitation is superior to the arithmetic average Q = 0.2 × 106m3/hours
method as some weighted is given to the various station
on a rational basis. 86. The flow in a chute spillway is generally:
• In calculating average rainfall in a catchment area (a) Supercritical (b) Critical
arithmetic mean method is a very crude method and it is (c) Sub-critical (d) Uniform
rarely used. Ans. (a) : Chute or Trough Spillway–
83. For efficient working of a control meter, its • A chute spillway is the one which passes the surplus
throat length is approximately kept: discharge through a steep sloped open channel, called a
(a) Equal to the critical depth chute or trough, placed either along a dam abutment or
(b) Twice the critical depth through a saddle.
(c) Three times the critical depth • This type of spillway is provided on earth or rock fill
(d) Four times the critical depth dam.
Ans. (c) : For efficient working of a control meter its • The flow in a chute spillway is generally supercritical.
throat length is approximately kept three times of the
87. What is the volume of groundwater which can
critical depth.
be extracted by gravity drainage from a soil
84. For the estimate of high floods in fan-shaped stratum when expressed as percentage fraction
catchment, the formula used is: of the volume of the soil stratum?
(a) Dicken's formula (b) Inglis formula (a) Pellicular water (b) Specified yield
(c) Ryve's formula (d) None of these
(c) Available water (d) Field capacity
Ans. (b) : Emperical formula for flood peak–
Ans. (b) : Yield is the volume of ground water
A. Dicken formula used in central & Northern parts of extracted by gravity drainage from a saturated water
the country. bearing material.
Qp = CD A3/4
It is known as the specific yield when it is expressed as
where Qp = m3/sec ratio of the volume of the total material drained.
Cd = 6 to 30 Specific yield =
A = km2
Volume of the water obtained by gravity drainage
B. Inglish formula used in western ghat of Maharashtra × 100
Total volume of drained material
124A
Qp = 88. The permeability of an aquifer (m/day) will:
A + 10.4
(a) Increase with an increase in temperature of
• Inglish formula is used for estimate high floods in fan water flowing through the aquifer
shaped catchment area. (b) Decrease with an increase in temperature of
C. Ryves formula used in tamilnadu, parts of Andhra water flowing through the aquifer
Pradesh & Karnataka.
(c) Not get affected by the change in temperature
Qp = CRA2/3 of water flowing through the aquifer
where CR = Ryves coefficient (d) Increase upto 20º C and then decreases with
85. For a catchment area of 120 km2, the the increase in the temperature of water
equilibrium discharge in m3/hour of a S-curve flowing through the aquifer
obtained by the summation of 6 hour unit
hydrograph is: Ans. (a): As per Kozney karman equation,
6 6
(a) 0.2 × 10 (b) 0.6 × 10 1 γ e3
6 6 Permeability ( K ) = × w × D10
2
(c) 2.4 × 10 (d) 7.2 × 10 K µ 1+ e
Ans. (a) : Given,
1
Area, A = 120 km2 = 120 × 106 m2 So, K∝
µ
Time, t = 6 hr,
Height, h = 1 cm = 0.01 m (for unit hydrograph) • Temperature↑→dynamic viscosity↓→permeability↑

OPSC AAO Exam. 2020 (Paper-II) 90 YCT

89. Normal annual rainfall at metrological stations 92. Muskingum method of flood routing is used
P, Q and R 1606 mm, 1803 mm and 1653 mm for:
respectively. In 1999 station Q was inoperative (a) Hydraulic channel routing
and stations P and R recorded annual rainfall (b) Hydraulic reservoir routing
of 1530 mm and 1451 mm respectively.
Estimated value of rainfall at Q in 1999 is: (c) Hydrologic channel routing
(a) 1490.5 mm (b) 1650.2 mm (d) Solving Saint-Venant equations
(c) 1687.3 mm (d) 1803.0 mm Ans. (c) : Muskingum method–
Ans. (b) : Normal ratio method–It is used when • It is most widely used hydrological channel routing
normal precipitation of the selected station is beyond method.
10% of that of station with missing data. • It involves the concept of wedge and prism storage.
PQ 1  Pp P  S = K  xI m + (1 − x ) Q m 
=
 + R 
N Q n  N p N R  Where, S - storage of channel
Where, P = Rain fall or precipitation K = Storage time constant, has dimension of time
N = Normal precipitation or Rainfall x = weighing factor, depending on shape of wedge
n=2 I = Inflow
Q = Outflow
N Q  Pp P 
PQ =  + R  m= (1) 0.6 for artificial channel
n  N p N R 
(2) 1, for natural channel
Given, NQ = 1803, Pp = 1530, Np = 1606, PR = 1451, 93. In a rectangular channel, the flow is in critical
NR = 1653 state with a specific energy of 3.0 m. The
1803  1530 1451  discharge per unit channel width is:
PQ = + = 1650.2 mm (a) 2.00 m3s–1m–1 (b) 3.13 m3s–1m–1
2  1606 1653  3 –1 –1
(c) 5.75 m s m (d) 8.86 m3s–1m–1
90. A flow duration curve indicates:
3
(a) The stream flow available for different Ans. (d) : Critical specific energy (EC) = y C
percent of time 2
(b) The duration of floods or droughts where, yC = Critical depth
(c) The effect of storage  q2 
1/ 3

(d) The power available for different percent of y C =  

time  g 
Ans. (a) : • A flow duration curve is a plot of discharge
v/s percent of time that a particular discharge was
equalled or exceeded.
• The area under the flow duration curve gives the The specific energy corresponds to the critical depth of
average daily flow, and the medium daily flow is the the flow is known as critical specific energy.
50% value. Where, q = discharge per unit width (m3/sec/m)
• It is useful to graph the data on probability paper. g = acceleration due to gravity (m/sec2)
91. A S-curve hydrograph derived from a 4-hr UG Given, EC = 3.0 m
(time base T = 24 hr) for a 640 km2 basin: 3
(a) Is the result of six 4-hr unit storms occurring EC = y C
2
in succession over the basin
3
(b) Produces a constant outflow 445 cumec after 3= yC
24 hours 2
(c) The shape of the curve in the first 24 hours is yC = 2 m
in the form of letter S 1/ 3
 q2  q2
yC =   ⇒ ( 2 ) =
3
(d) All of these
Ans. (d) : A S-curve hydrograph derived from a 4-hr  g  g
3
UG (time base T = 24 hr) for a 640 km2 basin– q = 8.86 m /sec/m
• Is the result of six 4-hr unit storms occurring in 94. Bench terraces are constructed on a 10% hill
succession over the basin slope. If the vertical interval is 1.5 m, the
• Produces a constant outflow 445 cumec after 24 hours earthwork per hectare is:
• The shape of the curve in the first 24 hours is in the (a) 1875 m3 (b) 3750 m3
form of letter S (c) 7500 m3 (d) 18750 m3
OPSC AAO Exam. 2020 (Paper-II) 91 YCT
Ans. (a): (a) 0.002 (b) 0.017
100ws 100 × 1.5 × 10 (c) 0.033 (d) 0.050
Earth work/hectare = = = 1875m 3
8 8 Ans. (b) : Given,
95. A flownet below a dam consists of 24 Width of channel = 5 m
equipotential drops and 7 flow channels. The Depth of channel = 2 m
difference between the upstream and Slope = 0.001
downstream water levels is 6m. The length of Flow velocity = 2 m/sec
the flow line adjacent to the toe of the dam at
exit is 1 m. The specific gravity and void ratio We know that
of the soil below the dam are 2.70 and 0.70, According to Manning's equation
respectively. The factor of safety against piping
1
velocity ( v ) = ( R ) (S )
2/3 1/ 2
is:
(a) 1.67 (b) 2.5 n
(c) 3.4 (d) 4 Area 5×2 10
R= = = = 1.11
Ans. (d) : Given, Perimeter ( 5 + 2 + 2 ) 9
Equipotential drop (Nd) = 24 1
2 = (1.11) ( 0.001)
2/3 1/ 2
Flow channel (Nf) = 7 n
Difference in water level (∆H) = 6 m 1.072 × 0.03162
Specific gravity (G) of soil = 2.7 n= = 0.0169  0.017
2
Void ratio of soil (e) = 0.70 n = 0.017
ic 98. The most suitable hydraulic structure for
Factor of safety against piping (FOS) =
ie conveying water from higher elevation to lower
elevation across the earthen bund is:
G − 1 2.7 − 1 1.7
ic = = = =1 (a) Drop structure (b) Pipe drop structure
1 + e 1 + 0.7 1.7
(c) Chute spillway (d) Gabion structure
∆h ∆h 6
ie = = = = 1/ 4 Ans. (b) : Pipe drop structure–
L N d × 1 24 × 1
• A pipe drop structure or flume is temporary pipe
1 structure or constructed flumed placed from the top of a
FOS = =4
1/ 4 slope to the bottom of the slope.
FOS = 4 99. If the width of bench terrace is W, drop D and
existing land slopes S; then for 150% batter
96. In a rectangular channel, the ratio of the slope, the drop D will be:
velocity head to the flow depth for critical flow
condition, is: WS WS
(a) (b)
(a) 1/2 (b) 2/3 100 (100 − S)
(c) 3/2 (d) 2 2WS 3WS
(c) (d)
Ans. (a) : We know that, (200 − S) (300 − S)

q2 3WS
Velocity head for a critical depth = (Vn) = .... (i) Ans. (d) : Drop (D) =
2gyC2 (300 − S)
1/ 3
Where, W = Width of bench terrace
 q2  S = Existing land slope.
Critical depth (yC) =  
 g  100. In an arid region:
From equation (i), we get (a) Rainfall is mostly during summer
(b) Each fall of rain is considered as a separate
( yC ) =  yC 
3
 q2  1 unit
Vn =   × =  2 
 g  2yC 2 ( yC )
2
  (c) 'Drip irrigation' is preferred
(d) Major loss is by transpiration
Vn 1
= Ans. (d) : Arid regions by definition receive little
yC 2 precipitation and major loss is by transpiration.
97. A rectangular open channel has a width of 5m Region Precipitation/year
and a bed slope of 0.001. For a uniform flow of Arid regions less than 25 cm
depth 2m, the velocity is 2 m/s. The Manning's
roughness coefficient for the channel is: Semi arid regions 25 to 50 cm

OPSC AAO Exam. 2020 (Paper-II) 92 YCT

 Odisha Public Service Commission
(AEE) Exam-2019 (Paper-I)
1. Strain energy per unit volume that a material 4. A prismatic bar is subjected to axial tension.
can absorb without exceeding its proportional What is the aspect angle (φ) which defines an
limit is called : oblique section on which normal and shearing
(a) Strain hardening stresses are equal?
(b) Shear modulus of material (a) 300 (b) 450
0
(c) Bulk modulus of material (c) 60 (d) 900
(d) Modulus of resilience Ans. (b)
Ans. (d) : The modulus of resilience is maximum
strain energy that can be absorbed per unit volume
without creating a permanent distortion (within the
elastic limit)
Maximum strain energy which a body stores upto the
elastic limit is called proof resilience.
Proof resilience
Modulus of resilience =
Unit volume of the body σx + σy σx – σy
i.e. the strain energy stored below the elastic limit σθ = + cos2θ + τxy sin 2θ
2 2
called resilience. σx – σy
2. The Lueders' Lines in a material indicate that : τθ = sin 2θ – τxy cos 2θ
2
(a) The material is failing in flexure
(b) The material is failing due to fatigue
(c) The material is failing due to its crushing
(d) The material is failing in shear σx = σ
Ans. (d) : Lueder's lines are elongated surface σy = 0
markings or depressions, often visible to the unaided τxy = 0
eye, that form along the length of a tension specimen σ σ
at an angle of approximately 450 to the loading axis ∴ σθ = + cos 2θ
2 2
(i.e. the plane of maximum shear) it is caused by
σ
localized plastic deformation and result from τθ = sin 2θ
discontinuous (in homogeneous) yielding it is also 2
known as luedert band or stretcher strain. σθ = τθ (given)
3. For structural steel, experiment indicates, the σ σ σ
+ cos 2θ = sin 2θ
value of Poison's ratio (µ) is 2 2 2
(a) 1.3 (b) 0.01 σ σ σ
= sin 2θ – cos 2θ
(c) 0.75 (d) 0.3 2 2 2
Ans. (d) σ σ
1 = (sin 2θ – cos 2θ)
– (Lateral Strain) 2 2
Poisson's Ratio (µ) =
Longitudinal Strain sin 2θ = cos 2θ
– Lateral Strain tan 2θ = 1 tan 450
(µ) = 2θ = 450
Longitudinal Strain
∴ Aspect Angle = 450
µ ≤ µ ≤ 0.5
5. What is the total elongation of a prismatic bar
µ = 0 for cork
of length (L) and cross-sectional area (A) and
µ = 0.5 for perfectly plastic body (rubber) Young's Modulus of its material (E) hangs
µ = 0.25 to 0.42 for elastic metals vertically under its own weight (W)?
µ = 0.1 to 0.2 for concrete WL WL
(a) (b)
µ = 0.286 mild steel 2AE 8AE
⇒ µ is greater for ductile metals than for brittle WL WL
metals. (c) (d)
6AE 4AE
OPSC (AEE) Exam-2019 (Paper-I) 93 YCT
 Ans. (a) Ans. (b)
d = 20 mm
d
ymax = = 10 mm
2
R = 10 m = 10 × 103 mm
E = 2 × 106 kg/cm2
Where, Wx = weight of the portion below the section = E
from bending equation, Fmax = y
γAX R max
The extension of the entire bar can be obtained by 2 × 106
= ×10
integrating the equation, 10 ×103
L L
Wx d x fmax = 2 × 103 kg/cm2
∫ δdx = δL ∫
O O
AE 8. The of width to depth of a strongest beam that
can be out of a cylindrical log of wood with
L
γ AX dx
δL = ∫ homogeneous and isotropic properties is
AE
O 1 1
L
(a) (b)
γ 5 4
E O∫
δL = x.dx
1 1
(c) (d)
L 3 2
γ  x2 
δL =
E  2  O Ans. (d)

γL2
δL =
2E
if W is the total weight of the bar, then
W
γ= d
AL for strongest rectangle b =
3
WL
δL = 2
2AL h= d
3
The deformation of the bar under its own weight is
equal to the Half of the deformation, if the body is d
subjected to the direct load equal to the weight of the b 3 1
= =
body. h 2 2
d
6. Select in which case of the following biaxial 3
stress, pure shear condition prevails : 9. The maximum shear stress caused due to a
(a) σx = 2 σy (b) 2 σx = σy shear force in a beam of rectangular cross-
section is how much more is percentage than its
(c) σx = σy (d) σx = – σy
average value?
Ans. (d) : When normal stresses acting on (a) 200% (b) 150%
perpendicular planes are equal in magnitude and (c) 100% (d) 50%
opposite in direction is known as pure shear condition
Ans. (d) : for rectangular cross-section,
3
τmax = τavg
2
τmax 3
= = 1.5
τavg 2
τmax = ↑ 50% τavg
σx = – σy
10. The diameter of the core of solid circular
7. A steel wire of 20 mm diameter is bent into a column of diameter "D", where stress induced
circular shape of 10 m radius, then the due to a normal concentrated load of any
maximum stress induced in the wire is eccentricity with respect to the centre of the
[Take E = 2 × 106 kg/cm2] column is :
(a) 1 × 103 kg/cm2 (b) 2 × 103 kg/cm2 (a) 0.20D (b) 0.25D
(c) 4 × 103 kg/cm2 (d) 6 × 103 kg/cm2 (c) 0.33D (d) 0.5D
OPSC (AEE) Exam-2019 (Paper-I) 94 YCT
 Ans. (b) : Condition for no tension is σmin = 0 M R square Z N.A square a3 6
= =
P M
– =0 M R circle Z N.A circle π 3
d
A Z 32
P M a3 6
= = = 1.18
A Z 3
π  2a 
32  π 
P Pe
=
π 2 π 3
D D
4 32 14. A simply supported beam with rectangular
D cross-section is subjected to a central
∴e = concentrated load. If the width and depth of
8
the beam are doubled, then the deflection at the
D D
∴ Diameter of core = 2× = = 0.250 centre of the beam will be reduced to :
8 4 (a) 50% (b) 25%
11. A beam of rectangular cross-section is 100 mm (c) 112.5% (d) 6.25%
wide and 200 mm deep. If the section is Ans. (d)
subjected to a shear force of 20 kN, then the
maximum shear stress in the section is :
(a) 1.25 N/mm2 (b) 1.5 N/mm2
2
(c) 1.6 N/mm (d) 1.75 N/mm2
Ans. (b)
S = 20 kN = 20 × 103 N
3
τmax = (τavg)
2
τmax = 3/2  
S
bd 3
 
bd y I 1
∴ 2
= 1
= 12 = × 100 = 6.25%
3  20 × 103  y1 I2 (2b)(2d)3 16
= 
2  100 × 200  12
τmax = 1.5 N/mm2 15. If the deflection at the free end of a uniformly
loaded cantilever beam is 15 mm and the slope
12. For no torsion, the plane of bending should
of the deflection curve at the free end is 0.02
(a) Be parallel to one of the principal axes
radian, then the length of the beam is :
(b) Pass through shear centre of section
(a) 0.8 m (b) 1.0 m
(c) Pass through neutral axis of the section
(d) Pass through centre of gravity of the section (c) 1.2 m (d) 1.5 m
Ans. (b) : Shear center of a section is the centroid of Ans. (b)
all the shear forces in the section that's why if load is
applied at shear center it produce zero twisting
moment.
13. Two beams, one of circular cross-section and
other of square of cross-section, have equal
wl 3
areas of cross-section. If subjected to bending θB = = 0.02
(a) Circular section is more economical 6EI
(b) Square section is more economical wl 3
= 0.02 × 6 -------------- (1)
(c) Both sections are equally strong EI
(d) Both sections are equally stiff wl 4
yB = = 15 mm
Ans. (b) 8EI
wl 3 l
yB = × = 15
EI 8
l
0.02 × 6 × = 15 (∴ from (1) )
8
Acircle = Asquare l = 1000 mm = 1.0 m
π 2 2 16. Two ratio of maximum shear stress developed
d =a
4 in a solid shaft of diameter D and a hollow
2a shaft of external diameter D and internal
d=
π diameter d for the same torque is given by

OPSC (AEE) Exam-2019 (Paper-I) 95 YCT

 (a)
(D 2
+ d2 ) (b)
(D 2
– d2 ) 19. If the depth of a rectangular section is reduced
to half, strain energy stored in the beam
D2 D2 becomes :

(c)
(D 4
+ d4 ) (d)
(D 4
– d4 ) (a)
1
4
times (b)
1
8
times
4 4
D D (c) 4 times (d) 8 times
Ans. (d) 1
Ans. (d) : Strain energy due to bending U ∝
T 16T I
(Zmax)Solid = =
ZP πD3  
3

16.TD U 2 I1  d 
= = =8
(τmax)Hollow = U1 I 2  d 
(
π D4 – d2 )  
2
Q THollow = TSolid Q b = constant
16T 20. The phenomenon of decreased resistance of a
( τmax )solid πD3 D4 – d 4 material to reversal of stress is called
= =
( τmax )Hollow 16TD D4 (a) Creep (b) Fatigue
π D –d
4 4
( ) (c) Resilience (d) Plasticity
Ans. (b) : The phenomenon of decreased resistance of
17. Strain energy stored in a member is given by : a material to reversal of stress is called fatigue.
(a) 0.5 × stress × strain Some sort of microscope cracks will begin to appear
(b) 0.5 × strain × volume on the material when the loads on the material cross
(c) 0.5 × stress × volume the certain limit which leads to the fatigue in that
(d) 0.5 × stress × strain × volume material.
Ans. (d) 21. The property of metal which allows it to
deform continuously at slow rate without any
Strain energy further increase in stress is known as :
● Work has to be done to stretch a material. (a) Fatigue (b) Creep
● Before the elastic limit all this work done is stored (c) Plasticity (d) Resilience
as potential energy. Ans. (b) : The phenomenon of metal which allows it to
● This stored energy is called elastic strain energy. deform continuously at slow rate without any further
1 increase in stress is known as creep.
S.E. = U = × stress × strain × volume
2 22. If a circular shaft is subjected to a torque T
18. In plane stress problem there are normal and bending moment M, the ratio of maximum
tensile stresses σx and σy accompanied by shear bending stress to maximum shear stress is :
stress τxy at a point along orthogonal Cartesian 2M M
(a) (b)
coordinates X and Y respectively, If it observed T 2T
that the minimum principle stress on a certain M 2T
(c) (d)
plane is zero then : T M
(a) τ xy = ( σx + σ y ) (b) τ xy = ( σx – σ y ) Ans. (a)
32M
( σx × σ y ) ( σx / σ y ) Fmax
= πd =
3
2M
(c) τ xy = (d) τ xy =
τmax 16T T
Ans. (c) πd 3
2 23. The identical bars, one simply supported and
σx + σy  σx – σy  other fixed at ends, are acted upon by equal
σ2 = 0 = –   + τxy
2
2  2  loads applied at the midpoints. The ratio of
strain energy stored in the simply supported
2
σx + σy  σx + σy  beam and the fixed ended beam is :
 + τxy
2
=  (a) 1 (b) 2
2  2  (c) 3 (d) 4
2 2
 σ x + σ y   σx – σ y  Ans. (d)
⇒  2  –  2  = τxy
2

   
τ xy = σ x .σ y

OPSC (AEE) Exam-2019 (Paper-I) 96 YCT

 We know, work done = Strain energy 26. In terms of bulk modulus (K) and modulus of
1 1 wl 3
w l2 3 rigidity (G), the Poisson's ratio can be
USSB = W.δ max = .W. = expressed as :
2 2 48EI 96EI
(3K – 4G) (3K + 4G)
1 1 Wl3 W 2 l3 (a) (b)
Ufixed = .W. y max = W. = (6K + 4G) (6K – 4G)
2 2 192EI 384EI
2 3 (3K – 2G) (3K + 2G)
w l (c) (d)
USSB (6K + 2G) (6K – 2G)
= 96EI2 3
=4
U fixed w l Ans. (c)
384EI (3K – 2G)
ymax = maximum deflection in fixed beam due to (6K + 2G)
point load (w) act at the mid of span.
27. The deflection at the free end of a cantilever
δmax = maximum deflection in simply supported subjected to a couple of M at its free end
beam due to point load (w) act at mid of span. having a uniform flexural rigidity EI
24. For ductile materials, the most appropriate throughout its length "L" is equal to
failure theory is :
ML2 ML2
(a) Maximum shear stress theory (a) (b)
2EI 3EI
(b) Maximum principal stress theory
2
(c) Maximum principal strain theory ML ML2
(c) (d)
(d) Shear strain energy theory 6EI 8EI
Ans. (d) : Maximum shear stress theory (Guest and Ans. (a)
Tresca’s theory) :
According to this theory, failure of a specimen
subjected to any combination of loads when the
maximum shearing stress at any point reaches the
failure value equal to that developed at the yielding is
an axial tensile or compress –ve test of the same
material. YB–YA = Ax (YA=0)
Graphical representation :
YB =   l ×
M l
σy
⇒ τ max ≤ for no failure  EI  2
2
Ml 2
 σy  YB =
⇒ σ1 – σ2 ≤   for design 2EI
 fos  28. The shear centre of a section is defined as that
point :
(a) Through which load must be applied to
produce zero twisting moment on the section
(b) At which shear force is zero
(c) At which shear force is maximum
(d) At which shear force is minimum
σ1 and σ2 are maximum and minimum principal Ans. (a) : Shear Centre– Shear centre of a section is
stresses respectively. defined as that point through which the load must be
Here, applied to produce zero twisting moment on the
τmax = maximum shear stress section. Generally, if a beam has two axes of
σy = permissible stress symmetry, then shear centre coincides with the
This theory is well justified for ductile materials. centroid.
25. The stress below which a material has a high 29. If a three hinged parabolic arch carries a
probability of not failing under reversal of uniformly distributed load over the entire
stress is known as : span, then any section of the arch is subjected
(a) Tolerance limit (b) Elastic limit to :
(c) Proportional limit (d) Endurance limit (a) Normal thrust only
Ans. (d) : Endurance limit is the stress level below (b) Normal thrust and shear force
which a specimen can with stand cyclic stress (c) Normal thrust and bending moment
indefinitely without exhibiting fatigue failure. Rigid, (d) Normal thrust, bending moment and shear force
elastic, low damping materials such as thermosetting Ans. (a) : In this case the arch will be subjected to
plastics and some crystalline thermoplastics do not normal thrust only. No bending moment and shear
exhibit an endurance limit. force at any section.
OPSC (AEE) Exam-2019 (Paper-I) 97 YCT
30. If the area under the shear force diagram for a Ans. (a)
beam between the two points C and D is 'K", E = 2G (1+µ)
then the difference between the moments at the E = 2G (given)
two points C and D will be equal to : 2G = 2G (1+µ)
(a) K (b) 2K 1 = 1+µ
K ⇒ µ=0
(c) (d) K2
2 34. If a composite bar of steel and copper is heated,
Ans. (a) then the copper bar will be under :
dm (a) Tension (b) Compression
=F (c) Shear (d) Torsion
dx
dm = f.dx Ans. (b) : Copper having higher coefficient of linear
expansion (α) will have a tendency to expand more but
f.dx = k (given) it’s moment will be resisted by steel having smaller
⇒ dm = f.dx = k coefficient of linear expansion (α). So, copper having
31. Given that for an element in a body of large coefficient of linear expansion (α) will be in
homogeneous isotropic material subjected to compression.
plane stress; εx, εy and εz are normal strains in 35. Shear stress on principal planes is :
x, y and z directions respectively and µ is the (a) Zero (b) Maximum
Poisson's ratio, the magnitude of unit volume (c) Minimum
change of the element is given by : (d) Depends on axial forces
(a) εx + εy + εz (b) εx – µ (εy + εz) Ans. (a) : Principal plane may be defined as, “the
(c) µ (εx + εy + εz) (d) µ (εy + εz) – εx plane on which normal stress attains its maximum and
minimum value.” So there planes are also called as
Ans. (a) major principal plane and minor principal plane.
δv The shear stress on principal plane is zero.
EV = = εx + εy + εz
v 36. A simply supported beam of span L carries
δv over its full span of load varying linearly from
⇒ = εx + εy + εz zero at either end to w/unit length at midspan.
1
The maximum bending moment occurs at :
32. If a material has identical properties in all
directions, it is said to be : WL2
(a) Quarter points and is equal to
(a) Homogeneous (b) Orthotropic 8
(c) Elastic (d) Isotropic WL2
(b) Quarter points and is equal to
Ans. (d) 12
⇒ Isotropic : Identical properties in all direction. WL2
(c) Midspan and is equal to
⇒ Homogenous Material : 8
A material is said to be homogenous if the WL2
(d) Midspan and is equal to
materials elastic properties (E, µ) are same at all 12
the points or throughout the body Ans. (d)
e.g→ A uniform elastic field
⇒ Orthotropic Material
Orthotropic materials have different properties in
different direction but normal strain does not
depend on shear strain. While in anisotropic
material normal strain depends on shear strain. Mmax = Mmidpoint
⇒ Elastic Material
= R A   –  .w 
l 1 l l 1
Elastic material are material which show high  . 
   2 z  2 3 
2
elasticity i.e. which can be stretched several times
its original length and upon releasing of the stress wl  l  wl 2 wl 2
= =
4  2  24
–
(stretching force), it returns back to its original 12
shape and dimensions. 37. Consider the following statements about
33. If the Young's modulus of elasticity of a flitched beams :
material is twice its modulus of rigidity, then 1. A flitched beam has a composite section
the Poisson's ratio of the material is : made of two more materials joined together
in such a manner that they behave as a unit
(a) Zero (b) 0.5
piece and each material bends to the same
(c) – 0.5 (d) – 11 radius of curvature.
OPSC (AEE) Exam-2019 (Paper-I) 98 YCT
 2. The total moment of resistance of a flitched (a) 150 mm (b) 350 mm
beam is equal to the sum of the moments of (c) 500 mm (d) 550 mm
resistance of individual sections. Ans. (d)
3. Flitched beams are used when a beam of one
material, if used alone, would require quite a
large cross-sectional area.
(a) 1, 2 and 3 are correct (b) 1 and 3 are correct
(c) 2 and 3 are correct (d) 1 and 2 are correct
Ans. (a)
(1) A flitched beam has a composite section made of E
M = S = 20
two more materials joined together in such a EW
manner that they behave as a unit piece and each bev = b + 2mt
material bends to the same radius of curvature.
= 150 + 2 × 20 × 10
(2) The total moment of resistance of a flitched beam
is equal to the sum of the moments of resistance = 550 mm
of individual sections. 41. Find out the wrong statement from the
(3) Flitched beams are used when a beam of one followings :
material, if used alone, would require quite a (a) The elastic section modulus of a section does
large cross-sectional area. not affect Shape Factor.
(b) The Shape Factor is a function of the cross-
38. Consider the following statements :
sectional shape.
The theory of simple bending assumes that : (c) The Shape Factor represents the increase of
1. The material of the beam is homogeneous, strength due to plasticization.
isotropic and obeys Hooke's law. (d) The Shape Factor of a section is a measure of
2. The plane section remains plane after reserve strength available in the section after
bending. initial yielding.
3. Each cross-section of the beam is symmetric Ans. (a) : It is a function of the cross-section form or
about the loading plane. shape and is represented by S.
4. Young's moduli are the same in tension and M Z
compression. S= P = P
MY Z
Of the above statements which are correct?
MP = Plastic moment
(a) 1 and 2 only (b) 1, 3 and 4 only
MY = Yield moment
(c) 2, 3 and 4 only (d) 1, 2, 3 and 4
ZP = Plastic modulus
Ans. (d) : In case of the theory of simple bending there Z = Elastic modulus
are the following assumptions It is a geometrical property of a session. It depends
(1) Only pure bending can occur-there’s no shear wholly on the shape of cross-section.
force, torsion nor axial load. 42. Find the correct statement with regards to
(ii) The material of the beam is homogeneous, Plastic Hinge :
isotropic and obeys Hooke's law. (a) Plastic Hinges are reached first at sections
(iii) The plane section remains plane after bending. subjected to least curvature.
(iv) Each cross-section of the beam is symmetric (b) The plastic hinge will not form at the point of
about the loading plane. zero shear in a span under distributed load.
39. If the diameter of a shaft, subjected to a torque (c) Where three structural members meets,
alone, is doubled, then the horse power "P" plastic hinge will form in all members
can be increased to : irrespective of their capacities of taking the
(a) 16P (b) 8P moment.
(c) 4P (d) 2P (d) A plastic hinge is a zone of yielding due to
shear in a structural member.
Ans. (b) : P α T α d3
Ans. (c)
3 3
P2 T2  d 2   2d  ⇒ The plastic hinges are formed first at the sections
= =  =  =8
P1 T1  d1   d  subjected to the greatest deformation (curvature).
⇒ The plastic hinge is defined as a yielded zone due
P2 = 8P
to flexure in a structure.
40. A flitched beam consists of a wooden joist 150 ⇒ The plastic hinge will form at point of zero shear
mm wide and 300 mm deep strengthened by
in a span under distributed load.
steel plates 10 mm thick and 300 mm deep one
on either side of the joist. If modulus of ⇒ When three or four members meet at a point,
elasticity of steel is 20 times that of wood, then plastic hinges are formed in all the members
the width of equivalent wooden section will be : irrespective of their plastic modulus.

OPSC (AEE) Exam-2019 (Paper-I) 99 YCT

43. Choose the correct statement : where,
(a) Through equilibrium condition will always be Dsc = External indeterminacy
satisfied, a solution arrived at on the basis of Dsc = r–s, for a plane frame
an assumed mechanism will give a loading where,
that is either correct or too high. s = Number of equilibrium equations = 3
(b) The load obtained using the assumed moment r = Reaction components = 4
diagram that does not violate the plastic Dsi = Internal Indeterminacy
moment condition will be either correct or too
high. Dsi = 3C for rigid jointed plane frame
(c) The strain at the onset of strain hardening is C = Number of cuts required for obtaining an open
about 30 to 40 times the elastic strain in configuration (or) number of closed boxes
structural steel. D s = D sc + dsi = 1
(d) The load factor of a rectangular steel section Number of independent mechanism (I) = N – Ds
is 3.75, if the Factor of safety is 1.65. number of possible plastic hinges (N) = 3
Ans. (a) : I=3–1=3
⇒ A solution arrived at on the basis of assumed 46. What is the value of kinematic indeterminacy
mechanism will yield a load carrying capacity that and statical indeterminacy of a fixed beam
is either correct or too high. respectively?
⇒ A solution arrived at on the basis of a statical (a) 3, 1 (b) 1, 2
moment condition, will either be correct or too low. (c) 2, 0 (d) 0, 3
⇒ The equilibrium condition however has to be Ans. (d)
satisfied in both the cases.
Load factor = FOS × shape factor
= 1.65 × 1.5 = 2.475
For ordinary steel the elastic strain is about At fixed support, DOF = 0
1 1 Kinematic indeterminacy = 0
to of strain at the beginning of the strain Ds = Dsc = r – s = 6 – 3 = 3
12 15
hardening. 47. What is the value of Bending Moment in kN-m
at 3m from the left support of a three-hinged
44. A fixed beam of 8 meters length is subjected to parabolic arch of span 10m and rise 4m which
a single central concentrated load. If the plastic carries a uniformly distributed load of 5 kN/m
moment value is MP, the ultimate load by over the whole span?
mechanism method is : (a) 27.5 (b) 120
(a) MP (b) 2MP (c) 0 (d) 225
(c) 1.5 MP (d) 4 MP
Ans. (c)
Ans. (a)

8M P
WC = = MP
8 wl 2
45. How many independent mechanism can be H =
8h
formed for a symmetrical portal frame of In this case the arch will be subjected to normal thrust.
single bay and single storied, statically
48. Which method of structural analysis is a Force
indeterminate to first degree and subjected to a
Method of analysis?
concentrated load on the beam portion and one
(a) Moment distribution method
at the beam-column junction?
(b) Slope deflection method
(a) 1 (b) 4
(c) None of these
(c) 3 (d) 2
(d) Both of these
Ans. (d)
Ans. (c) : Force method of analysis are :
● Method of consistent deformation.
● Three moment theorem : clay per ron.
● Column Analogy Method : Hardy cross.
● Elastic Centre Method.
● Maxwell – Mohr equations.
Static indeterminacy (DS) ● Castigliano’s theorem of minimum strain energy.
Ds = Dsc + Dsi ● Flexibility Matrix method.
OPSC (AEE) Exam-2019 (Paper-I) 100 YCT
49. How many simultaneous equations, you have to 51. What is the value of the bending moment at
solve to find the support moments for a built-in ends of a prismatic beam of length L,
continuous two spanned beams ABC of two carrying single concentrated load P at the
consecutive spans AB and BC of equal length middle?
carrying uniformly distributed loads of 5 kN/m PL PL
with end supports are simple supported? (a) (b)
4 8
(a) 4 (b) 6
(c) 1 (d) 3 3PL PL
(c) (d)
8 2
Ans. (c)
Ans. (b)

The number of simultaneous equations to be solved in

the slope deflection method is equal to kinematic
indeterminacy FEB = fixed end moment
DOF = 3 (rotation at ‘A’, ‘B’ and ‘C’) 52. The deflection at the free end of a cantilever of
50. The stiffness of a member of length (L), Length (L) and moment of inertia of its cross-
moment of inertia of the section of the member section (I) and Young's modulus of the
is (I) and Young's Modulus of the material of material is (E), subjected to uniformly
the member is (E) is : distributed load of "w" is
(a)
3EI
(b)
4EI wL4 wL4
(a) (b)
L2 L 30EI 8EI
4
(c)
2EI
(d)
4EI wL wL4
L 2 (c) (d)
L 2EI 48EI
Ans. (a) : Ans. (b)
(A) Far end fixed :

M 4EI
K= =
θ L
(b) Far end hinged :

53. The slope at the end of a simple supported beam

of length (L), subjected to concentrated load (P),
M 3EI
K= = mement of inertia of its cross-section (I) and
θ L Young's modulus of the material is (E) is
(c) If far end is a guided rollar
PL2 PL2
(a) (b)
16EI 24EI
PL2 PL2
(c) (d)
EI 48EI 8EI
K=
L Ans. (a)
(d) If far end is a roller

3EI
K=
L
OPSC (AEE) Exam-2019 (Paper-I) 101 YCT
54. A propped cantilever beam, AB, fixed at B is Ans. (d)
subjected to the action of a couple of moment DS = (6m + r) – 6j, for rigid joined space frame
MA at the end A. What is the value of the m = Number of member forces
reactive moment MB will be induced at the r = Reaction components
built-in end B?
59. Castigliano's first theorem is applicable
M (a) For statically determinate structures only
(a) MB = A (b) MB = MA
4 (b) When the system behaves elastically
M M (c) Only when principle of superposition is valid
(c) MB = A (d) MB = A
8 2 (d) For statically indeterminate structures only
Ans. (d) Ans. (c) : Castigliano's first theorem is applies to both
elastic and inelastic material behaviour.
Castigliano’s first theorem use full in analyzing
statically indeterminate structure. Castigliano’s second
theorem :
55. Choose the correct statement : ● useful for finding the defections of statically
(a) At the point of contra flexure, the shear force determinate structures.
changes sign. ● Applicable only when principle of super position
(b) At the point of contra flexure, the bending is valid.
moment at not zero. 60. The carry over factor in a prismatic member
(c) At the point of contra flexure, the bending whose far end is hinged is :
moment is maximum. 1
(d) At the point of contra flexure, the gradient of (a) 0 (b)
2
elastic line changes sign. 3
Ans. (d) : In a bending beam, a point is known as a (c) (d) 1
4
point of contra flexure if it is a location where bending Ans. (a) : Carry over factor (COF) it is defined as the
moment is zero (changes its sign). In a bending ratio of the moment at the far end to the moment at the
moment diagram, it is the point at which the bending rotating near end.
moment curve intersects with the zero line.
Carry over moment at far end
So that, the point of contra flexure, the gradient of COF =
applied moment at near end
elastic changes sign.
Case 1 : if far end is fixed
56. The basic perfect frame is a :
M/2 1
(a) Triangle (b) Rectangle COF = =
(c) Square (d) Hexagon M 2
Case 2 : if far end is hinged
Ans. (a)
0
COF = =0
M
Case 3 : if far end is guided roller
–M
COF = = –1
M
61. The moment required to rotate the near end of
The basic perfect frame is a triangle. a prismatic beam through a unit angle without
57. Method of joints is applicable only when the translation, the far end being simply
number of unknown forces at the joint under supported, is given by (where EI is the flexural
consideration is not more than : rigidity and L is span of beam) :
(a) One (b) Two 3EI 4EI
(a) (b)
(c) Three (d) Four L L
Ans. (b) : At every joint two equations of equilibrium 2EI EI
(c) (d)
ΣV = 0, ΣH = 0 are available and so at one joint L L
unknowns can be evaluated method of joints is suitable Ans. (a)
only when two unknown forces at a joint.
58. The degree of static indeterminacy of a rigid-
jointed space frame is where m, r and j are
unknown as member forces, r is unknown as
reaction components and j is number of joints :
M 3EI
(a) M + r – 2j (b) M + r – 3j K= =
(c) 3m + r – 3j (d) 6 m + r – 6j θ L
OPSC (AEE) Exam-2019 (Paper-I) 102 YCT
62. If the sinking of a support of a fixed beam 66. Due to some point load anywhere on a fixed
causes the beam to rotate in the clockwise beam, the maximum free bending moment is
direction, then the moments induced at both M. The sum of fixed end moment is
the ends of the beam will be (a) M (b) 1.5 M
(a) In anticlockwise direction and of equal magnitude (c) 2.0 M (d) 3.0 M
(b) In clockwise direction and of different
magnitude Ans. (c)
(c) In opposite directions and of equal magnitude
(d) In opposite directions and of different magnitude
Ans. (a)

The maximum free bending moment (M)

63. The Muller-Breatau principle can be used to PL
=
(a) Determine the shape of the influence line 4
(b) Indicates the parts of the structure to be PL PL
loaded to obtain the maximum effect The sum of fixed end moment = +
8 8
(c) Calculate the ordinates of the influence lines PL
(d) All these are correct = =M
4
Ans. (d) : The use of Muller-Breatau principle for
67. In the slope deflection equations, the
drawing the influence line for the reaction, shear, and
moment at a point in a statically determinate beam. deformations are considered to be caused by
(a) Shear force
64. To generate the jth column of the flexibility
matrix : (b) Bending moment
(a) A unit force is applied at coordinate j and the (c) Axial force
displacements are calculated at all coordinates. (d) Bending moment and shear force both
(b) A unit displacement is applied to coordinate j Ans. (b) : In this method, it is assumed that
and the forces are calculated at all coordinates. deformation are caused due to bending moment only
(c) A unit force is applied at coordinate j and the and axial deformation are neglected.
forces are calculated at all coordinates.
68. An ordinate in a funicular polygon represents :
(d) A unit displacement is applied at coordinate j
and the displacements are calculated at all co- (a) Shear force (b) Resultant force
ordinates. (c) Bending moment (d) Equilibrium
Ans. (a) : To generate the jth column of the flexibility Ans. (c) : The shape of the loaded chord is same as
matrix, we apply a unit force at co-ordinate ‘j’ only that of BMD or funicular polygon.
and compute displacement at all the co-ordinates. 69. The number of independent equations to be
65. For stable structures, one of the important satisfied for static equilibrium of a space
properties of flexibility and stiffness matrices is structure is :
that the element of main diagonal : (a) 1 (b) 2
(a) Of a stiffness and flexibility matrix must be (c) 3 (d) 6
negative.
(b) Of a stiffness matrix must be negative and Ans. (d) : For space frames number of equation of
flexibility matrix must be positive. equilibrium is 6.
(c) Of a stiffness matrix must be positive and Equations of equilibrium for space frames
flexibility matrix must be negative. Σ fx = 0 , Σ Mx = 0
(d) Of a stiffness and flexibility matrix must be Σ fy = 0 , Σ My = 0
positive. Σ fz = 0 , Σ Mz = 0
Ans. (d)
70. Independent displacement at each joint of a
● While the other elements of the flexibility matrix
rigid-jointed plane frame are :
may be positive or negative, the elements lying on
the leading diagonal are always positive. (a) Three linear movements
● While the other elements may be positive or (b) Two linear movements and one rotation
negative, the elements lying on the leading (c) One linear movement and two rotations
diagonal are always positive. (d) Three rotations
OPSC (AEE) Exam-2019 (Paper-I) 103 YCT
 Ans. (b) : Independent displacement components or 75. A simple supported beam AB of length "L = 3
degree of freedom for different types of joints : m". If the displacement at one-third length
Type of Joint Degree of freedom from the right hand support, at point D, due to
(i) Pin joint of plane 2 (Translations) load "W" at 1.0 m from left hand support, at
point C is 5 mm, then the displacement at C
(ii) Fram 3 (Translations)
due to a load of 0.2 W at D will be
pin joint of space
(a) 1 mm (b) 2 mm
(iii) Fram 3 (One rotation and two (c) 5 mm (d) 25 mm
Rigid joint of a translations)
plane frame Ans. (a)
(iv) Rigid joint of a 6 (3 rotations and 3
space frame translations)
71. Degree of kinematic indeterminacy of a pin-
jointed plane frame is given by (where j is
number of joints and r is unknown reactions) :
(a) 2j – r (b) j – 2r
(c) 3j – r (d) 2j + r
Ans. (a) : Dk = 2j–r, for pin jointed plane frame.
Maxwell Betties Theorem :
72. A beam of length "L", fixed at A and B of
uniform flexural rigidity carries two loads "P" Virtual work done by the 1st loading due to
at one-third interval of span length C and D displacement caused by IInd loading is equal to virtual
nd
respectively. The maximum bending moment work done st by the II loading due to displacement
will occur : caused by I loading
(a) At C and D and will be equal to 2PL/9 W × δC = 0.2 W × 5
(b) Between C and d and will be equal to PL/0 δC = = 1mm
(c) At A and at B will be equal to 2PL/9 76. A propped cantilever AB is fixed at A and
(d) Between A and C and also between B and D simply supported at B. If a 1 kN concentrated
will be equal PL/9 load is acting at B, what is the vertical reaction
Ans. (b) : The maximum bending moment will occur at support A?
between C and D and will be equal to PL/9 (a) 0 kN (b) 1 kN
73. The strain energy stored in a simply supported (c) 0.5 kN (d) 2 kN
beam of length "L" and flexural rigidity EI Ans. (a)
due to a central concentrated load "W" is :
W 2 L3 W 3 L2
(a) (b)
48EI 48EI
W 2 L3 W 3 L2 77. The principle of virtual work can be applied to
(c) (d)
96EI 96EI elastic system by considering the virtual work
Ans. (c) of :
W (a) Internal force only
L/2 L/2 (b) External force only
(c) Internal as well as external forces
W 2 L3 (d) None of these
U=
96EI Ans. (c) : The principle of virtual work can be applied
74. The strain energy of a structure due to bending to elastic system by considering the virtual work of
is given by : Internal as well as external forces.
 M 2 dx   M 2 dx  78. If one end of a prismatic beam AB of length
(a)  ∫
 EI 
 ∫
(b) 0.5 
 EI 
 "L" with flexural rigidity "EI" with fixed ends
is given a transverse displacement ∆ without
 M 2 dx   M 2 dx  any rotation, then the transverse reactions at A
∫
(c) 0.5 
 EI 

∫
(d) 0.33 
 EI 

and B due to displacement is :
6EI∆ 6EI∆
Ans. (b and c) (a) 2
(b)
L L L3
M2
Strain energy (U) = ∫ dx 12EI∆ 121EI∆
2EI (c) 2
(d)
O
L L3
OPSC (AEE) Exam-2019 (Paper-I) 104 YCT
 Ans. (*) (c) Smallest concrete section and minimum area
of reinforcement.
(d) Largest concrete section and minimum area
of reinforcement.
Ans. (a) : In limit state method, balanced design of a
reinforced concrete beam gives, smallest concrete
section and maximum area of reinforcement.
83. The development length of bars of diameter φ,
as per IS : 456-2000 is given by
79. A simply supported beam of length L over 4φ σs φ σs
(a) (b)
hanged to the right by
L
at support B. The τ bd τbd
2
maximum reaction at support B will be when 2φ σs φ σs
(c) (d)
the uniformly distributed load is loaded : 3 τ bd 3 τbd
(a) Only in AB (b) Only in BC Ans. (*)
(c) Entire span (d) None of these φ σs
Ans. (c) No option is correct. Answer should be
4τ bd
φ σS
Development length =
4 τbd
φ = nominal diameter of the bar.
σs = stress in bar at the section considered at design
load.
τbd = disign bond stress given in 26.2.1.1
Volue of τbd is obtained from IS 456–2000 clause
To get the max reaction at B, place the UDL over 26.2.1.1
entire span. 84. In prestressed concrete :
Max. reaction at B = Intensity of UDL × Area of ILD (a) Forces of tension and compression change but
under the UDL lever arm remains unchanged.
(b) Forces of tension and compression remain
= W  × L × 2
1
unchanged but lever arm changes with the
2 
moment.
= W L kN (c) Both forces of tension and compression and
80. The element of flexibility matrix of a structure : lever arm change.
(a) Are dependent on the choice of coordinates (d) Both forces of tension and compression and
(b) Are independent on the choice of coordinates lever arm remain unchanged
(c) Are always dimensionally homogeneous Ans. (b) : In prestressed concrete
(d) Both (A) and (C) are correct The principle behind prestressed concrete is that
Ans. (d) : The elements of a flexibility matrix are not compressive stresses induced by high strength steel
necessarily dimensionally Homogeneous they tendons in a concrete member before loads are applied
represent either a translation or a rotation due to a unit will balance the tensile stresses imposed in the member
load or so a couple. during service.
81. In limit state design of concrete for flexure, the In prestressed concrete forces of tension and
area of stress block diagram is taken as 9 compression remain unchanged but lever arm changes
where fck is characteristic compressive strength with the moment.
of concrete and Xu is depth of neutral axis from 85. Modulus of rupture of concrete is a measure of :
top of the section : (a) Flexural tensile strength
(a) 0.36 fck Xu (b) 0.42 fck Xu (b) Direct tensile strength
(c) 0.446 fck Xu (d) 0.56 fck Xu (c) Compressive strength
Ans. (a) : 0.36 fck Xu (d) Split tensile strength
82. In limit state method, balanced design of a Ans. (a) : Modulus of rapture of concrete is measure
reinforced concrete beam gives : of flexural tensile strength.
(a) Smallest concrete section and maximum area 86. According to IS : 456-2000, the maximum
of reinforcement. strain in concrete at the outermost compression
(b) Largest concrete section and maximum area fibre in the limit state design of flexural
of reinforcement. member is :
OPSC (AEE) Exam-2019 (Paper-I) 105 YCT
 (a) 0.0020 (b) 0.0035 Ans. (c) : The yield line analysis is an upper bound
(c) 0.0050 (d) 0.0065 method in which the predicted failure load of a slab for
Ans. (b) : According to IS : 456-2000, the maximum given moment of resistance (capacity) may be higher
strain in concrete at the outermost compression fibre in than the true value. Thus, the solution of the upper
the limit state design of flexural member is 0.0035 bound method (yield line analysis) may result into
unsafe design if the lowest mechanism could not be
chosen.
92. A reduction factor Cr to load carrying capacity
of a long column is given by :
 L   L 
(a) Cr = 1.25 – e  (b) Cr = 1.00 – e 
 24b   48b 
 L   L 
Stress-strain curve for concrete (c) Cr = 1.25 – e  (d) Cr = 1.5 – e 
 48b   48b 
87. Concordant profile represents for a certain set
of external loads to some scale, the : Ans. (c)
(a) Bending moment diagram  L 
(b) Williot-Mohr diagram Cr = 1.25 – e 
 48b 
(c) Shear force diagram
93. The effect of creep on modular ratio is :
(d) Influence line diagram
(a) To decrease it
Ans. (a) : The Bending moment diagram represents (b) To increase it
concordant profile for a certain set of external loads to
(c) Either to decrease or to increase it
some scale.
(d) To keep it unchanged
88. For prestressed structural elements, high
Ans. (d, a) : Modular ratio is WSM
strength concrete is used primarily because :
(a) Both shrinkage and creep are more 280
M=
(b) Shrinkage is less but creep is more 3σ cbc
(c) Modulus of elasticity and creep values are Long term modular ratio;
higher E E
(d) Of high modulus of elasticity and low creep M= s = s , θ = creep coefficient
Ec Ec
Ans. (d) : For high strength concrete modulus of 1+ θ
elasticity is high but low creep.
94. For walls, columns and vertical faces of all
89. For bars in tension, a standard hook has an structural members, the form work is generally
anchorage value equivalent to a straight length removed after :
of (where φ is diameter of hook) : (a) 24 to 48 hours (b) 3 days
(a) 8 φ (b) 12 φ (c) 7 days (d) 14 days
(c) 16 φ (d) 24 φ Ans. (a) : Form work– form work is the term given to
Ans. (c) either temporary or permanent molds into which
concrete or similar materials are poured.
⇒ For walls, columns and vertical faces of all
structural members, the form work is generally
removed after 24 to 48 hours.
Anchorage values for standard U type hook is 16 φ
95. The maximum compressive stress in concrete
90. The relation between modulus of rupture fcr,
for design purposes is based on a partial safety
splitting strength fcs and direct tensile strength
factor of :
fcs and direct tensile strength fct is given by :
(a) 1.15 (b) 1.50
(a) fcr = fcs = fct (b) fcr > fcs > fct
(c) 1.85 (d) 2.20
(c) fcr < fcs < fct (d) fcs > fcr > fct
Ans. (b) : For design the partial safety factor for
Ans. (b)
concrete is 1.5
2
fS = fcr 96. In the design of prestressed concrete
3 structures, which of the following limit state
fcr > fcs > fct will come under the limit state of
91. Yield line theory results in : serviceability?
(a) Elastic solution 1. Flexure 2. Shear
(b) Lower bound solution 3. Deflection 4. Cracking
(c) Upper bound solution (a) 1 and 4 (b) 3 and 4
(d) Unique solution (c) 2, 3 and 4 (d) 2 and 3
OPSC (AEE) Exam-2019 (Paper-I) 106 YCT
 Ans. (b) : Limit state of serviceability combination of principal stresses equals or exceeds the
● Deflection value obtained for the shear stress at yielding in the
● Crack width uniaxial tensile test
● Vibrations ⇒ Maximum strain energy Theory (Beltrami’s theory)
97. For the deflection of simply supported beam to failure would occur when the total strain energy
be within permissible limits the ratio of its span absorbed at a point per unit volume exceeds the
to effective depth as per IS : 456 : 1978 should strain energy absorbed per unit volume at the
not exceed : tensile yield point.
(a) 7 (b) 20 101. The design yield stress of steel according to IS :
(c) 26 (d) 35 456 - 978 is :
Ans. (b) : Span to effective depth ratio for simply (a) 0.37 f y (b) 0.57 f y
supported beam to satisfy vertical deflection limit is
L (c) 0.67 f y (d) 0.87 f y
= 20
d Ans. (*) : Design yield stress = 0.87 fy
98. Most common method of prestressing used for 102. Pre-stressing losses in post-tensioned and pre-
factory production is : tensioned beams are respectively :
(a) Long line method (a) 15% and 20% (b) 20% and 15%
(b) Freyssinet system (c) 5% and 15% (d) 20% and 20%
(c) Magnel-Blaton system Ans. (b) : Pre-stressing losses 20% and 15% in post-
(d) Lee-Macall system tensioned and pre-tensioned beams are respectably.
Ans. (a) : Pretensioning method is also called long line 103. In limit state of design, the maximum limit
method. It is suitable for short span and mass imposed by IS:456-1978 on the redistribution
preduction. of moments in statically indeterminate beams is
Pretensioning (a) 10% (b) 15%
In this method the concrete is prestressed with (c) 20% (d) 30%
tendons before it is placing in position. This method is
Ans. (d) : The maximum limit imposed on the
developed due to bonding b/w the concrete and steel
redistribution of moment in statically indeterminate
tendons. Pre-tensioning is preferred when the
beams is 30%.
structural element is small and easy to transported.
104. At limit state of collapse in shear, in case of
99. Shrinkage deflection in case of rectangular
web shear cracks, it is assumed that the
beams and slabs can be eliminated by putting :
concrete cracks when the maximum principal
(a) Compression steel equal to tensile steel tensile stress exceeds a value of ft equal to :
(b) Compression steel more than tensile steel
(c) Compression steel less than tensile steel (a) 0.24 fck (b) 0.20 fck
(d) Compression steel 25% more than tensile (c) 0.16 fck (d) 0.30 fck
steel
Ans. (a) : The maximum principal tensile stress should
Ans. (a) : Shrinkage deflection can be eliminated by
putting tension steel is equal to compression steel. not exceed 0.24 fck
100. In the limit state method of design, the failure 105. Minimum clear cover (in mm) to the main steel
criterion for reinforced concrete beams and bars in slab beam column and footing
columns is : respectively are :
(a) Maximum principal theory (a) 0, 15, 20, 25 (b) 15, 25, 40, 75
(b) Maximum principal strain theory (c) 20, 25, 30, 40 (d) 20, 35, 40, 75
(c) Maximum shear theory Ans. (b) : As per IS : 456–2000 minimum cover
(d) Maximum strain energy theory As per old code
Ans. (b) Slab : 20 mm : 15 mm
⇒ Maximum Principal Strain Theory Beam : 25 mm : 25 mm
Or Column : 40 mm : 40 mm
Saint Venant’s Theory Footing : 50 mm : 75 mm
According to this theory, the maximum principal strain 106. Which one of the following statements is
in the complex stress system must be less than the correct?
elastic limit in simple tension if there is to be no (a) Maximum longitudinal reinforcement in an
failure. axially loaded short column is 6% of gross
⇒ Maximum Shear Stress Theory sectional area.
The maximum shear stress theory states that failure (b) Columns with circular section are provided
occurs when the maximum shear stress from a transverse reinforcement helical type only.
OPSC (AEE) Exam-2019 (Paper-I) 107 YCT
 (c) Spacing of lateral ties can not be more than 112. As per IS : 800, the maximum bending moment
16 times the diameter of the bar. for design of purlins can be taken as (where W
(d) Longitudinal reinforcement bar need not be is total distributed load on the purlins and L is
contact with lateral ties. centre distance of support) :
Ans. (a) : Maximum longitudinal reinforcement to an WL WL
axially loaded short column is 6% of gross cross (a) (b)
6 8
section area.
WL WL
107. Limit state of serviceability for deflection (c) (d)
10 12
including the effects due to creep, shrinkage
Ans. (c) : As per IS 800–1984 clause 6.9.2. The
and temperature occurring after erection of
partitions and application of finishes as maximum bending moment in a purlin may be taken as
applicable to floors and roofs is restricted to : WL
when W is the total distributed load on the purlin
Span Span 10
(a) (b) including wind load.
150 200
Span Span 113. The number of seismic zones in which the
(c) (d) country has been divided are :
250 350
Ans. (d) : The deflection including the effects of (a) 3 (b) 5
temperature, creep and shrinkage occurring after (c) 6 (d) 7
erection partitions and application of fine shear should Ans. (b) : As per IS 1893 (Part)–1970. Based on the
not exceed span/350 (or) 20 mm which ever is smaller. levels of intensities sustained during damaging past
108. Flexural collapse in over reinforced beams is earthquake, the zone map subdivided India into 5 zone.
due to : 114. As per IS:875, for the purposes of specifying
(a) Primary compression failure basic wind velocity, the country has been
(b) Secondary compression failure divided into :
(c) Primary tension failure (a) 4 zones (b) 5 zones
(d) Bond failure (c) 6 zones (d) 7 zones
Ans. (a) : Primary compression failure is due to flexure. Ans. (c) : As per IS 815 (Part 3) for the purposes of
109. Approximate value of shrinkage strain in specifying. Basic wind velocity the country has been
concrete is : divided into 6 zones.
(a) 0.003 (b) 0.0003
115. The thickness of web for unstiffened plate
(c) 0.00003 (d) 0.03
girder with clear distance "d" between the
Ans. (b) : As per IS 456 : 2000 (clause 6.2.4.1) in the flanges shall not be less than :
absence of data, the approximate value of total
d d
shrinkage strain for design may be taken as 0.0003. (a) (b)
200 85
110. According to IS : 456-1978, the modulus of
elasticity of concrete Ec (in N/mm2) can be d d
(c) (d)
taken as : 100 160
(a) Ec = 5000 fck (b) Ec = 570 f ck Ans. (b) : As per IS 800 : 1984 clause 67.3.1 the
thickness of the web plate shall be not less than the
(c) Ec = 5700 fck (d) Ec = 700 f ck d τ va .cal d 6y
greater of & but not less than
Ans. (a) : As per IS : 456–1978, the modulus of 816 1344
elasticity of concrete is Ec = 5700 fck d
for unstiffened webs where d is depth of web.
85
111. The lacing bars in a steel column should be
designed to resist : 116. At a section along the span of a welded plate
(a) Bending moment due to 2.5% of the column girder, where the web is sliced, the bending
load. moment is M. If the girder has top flange, web
(b) Shear force due to 2.5% of the column load and bottom flange plates of equal area, then
(c) 2.5% of the column load only the share of the bending moment which would
(d) Both (a) and (b) be taken by the splice plates would be
Ans. (b) : As per IS 800–1984 clause 5.7.2.1. The M
(a) M (b)
having of compression members shall be proportioned 3
to resist a total transerse shear ‘V’ equal to atleast M M
2.5% of the axial force in the member. (c) (d)
7 13
OPSC (AEE) Exam-2019 (Paper-I) 108 YCT
 Ans. (c) : Moment taken by splice plate = moment 119. Battens provided for a compression member
taken by web plate shall be designed to carry a transverse shear
equal to
(a) 2.5% of axial force is member
(b) 5% of axial force in member
(c) 10% of axial force in member
(d) 20% of axial force in member
Ans. (a) : As per IS 800 : 1984, battens shall be
designed to carry the bending moment and shears
arising from transverse shear force ‘V’ of 2.5% of the
total axial force on the whole compression member.
MW M
= 120. The channels or angles in the compression
IW I chords of the steel truss girder bridges are
t .B3 turned outward in order to increase :
∴ IW = (a) Cross-section area (b) Section modulus
12
(c) Torsional constant (d) Radius of gyration
I = M . I of entire plate
Ans. (d) : Channel or Angles in the compression
 Bt 3 B + 2t  
2
+ Bt × 
chords of the steel truss girder bridges are turned
I=   + outward in order to increase the radius of gyration
 12  2  
which reduces the slenderness ratio of the member.
 t × B3  Hence it increase resistance against buckling.
 12  121. The effective length of a structural steel
 
compression of length "L" effective held in
7tB3 position and restrained against rotation at one
I=
12 end but neither held in position nor restrained
I against rotation at the other end is member :
Mw = w × M (a) L (b) 1.2L
I (c) 1.5L (d) 2.0L
tB2 Ans. (d) : As per IS 800 table 5.2
M
M w = 12 3 × M = For compression member with one end fixed and other
7tB 7 free.
12 Effective length = 2 × unsupported length.
M
Mw =
7
117. Given that the effective area of a tension
member is Ae and the yield stress is σy. In order
to obtain the ultimate strength of the tension 122. A column, base is subjected to mement. If the
member, as per the plastic design concept Ae σy intensity of bearing pressure due to axial load
is to be multiplied by is equal to stress due to the moment, then the
(a) 1.1 (b) 0.95 bearing pressure between the base and
concrete is
(c) 0.85 (d) 0.75
(a) Uniform compression throughout.
Ans. (c) : The maximum load capacity of tension (b) Zero at one end and compression at the other end.
member is 0.85 AcFy for compression member the load (c) Tension at one end and compression at the
capacity is 1.7 Aeσac other end.
118. For a compression member with double angle (d) Uniform tension throughout
section, which of the following section will give Ans. (b)
larger value of minimum radius of gyration?
(a) Equal angles back to back.
(b) Unequal legged angles with long legs back to
back.
(c) Unequal legged angles with short legs back to
back.
(d) Both (b) and (c)
Ans. (c) : For a double angle section back to back it is
desirable to use unequal angles with the long legs back
to back to achieve a balance b/w radius of gyration
values about zz and yy axises.
OPSC (AEE) Exam-2019 (Paper-I) 109 YCT
123. Economical depth of a plate girder is given by : Ans. (b) : Bearing stiffness are provided at the points of
 M   M  concentrated loads and at support to resist reactions
(a)  σt  (b) 1.1   bearing stiffeners are provided to avoid local bending
 w  σt w  failure of the flange, web crippling and buckling of web.
 M   M  129. As per IS:800, for compression flange, the
(c) 1.2   (d) 1.3   outstand of flange plates should not exceed, if
 σt w   σt w  "t" is thickness of thinnest flange plate :
Ans. (b) : Economical depth of plate girder (d) (the (a) 12t (b) 16t
depth at which weight of girder is minimum) (c) 20t (d) 25t
M Ans. (b) : As per IS 800 – 1984 clause 3.5-2.1 for
for riveted plate girder, D = 1.1 compression flange, the outstand of flange plates
σ.t w
should not exceed 16t, where it is the thickness of the
124. Shear buckling of web in a plate girder is flange of a section or of a plate.
prevented by using : 130. Horizontal stiffener in a plate girder is
(a) Vertical intermediate stiffener provided to safequard against :
(b) Horizontal stiffener along the neutral axis (a) Shear buckling of web plate
(c) Bearing stiffener (b) Compression buckling of web plate
(d) None of these (c) Yielding
Ans. (a) : In plate girders, intermediate transverse (d) All of these
stiffeners are provided to increase buckling resistance Ans. (b) : Horizontal stiffeners in a plate girder is
of web in diagonal or shear buckling. provided to safe guard against compression buckling
125. Intermediate vertical stiffeners in a plate of web plate.
girder need to be provided, if the depth of web 131. The most common admixture which is used to
exceeds ("t" is thickness of web) : retarder in cement is :
(a) 50t (b) 85t (a) Gypsum (b) Calcium chloride
(c) 200t (d) 250t (c) Calcium carbonate (d) None of these
Ans. (b) : Intermediate vertical stiffeners in a plate girder Ans. (a) : Gypsum and common sugar are the
are needed when the depth of web exceeds 85 kw. admixtures generally used as retarders.
126. The range of economical spacing of trusses Calcium chloride is an accelerator which increase the
varies from (where L is span) : rate of hydration (accelerate the initial set)
L L L 2L 132. Which of the following test is used to determine
(a) to (b) to the rate of wear of stones?
3 5 4 5
L L 2L 3L (a) Crushing test (b) Abrasion test
(c) to (d) to (c) Attrition test (d) Impact test
3 2 5 5
Ans. (a) : The spacing of the truss depends on the type Ans. (c) : To determine the rate of wear of stones
truss used, the truss span, the function of building, the Attrition Test is done using Deval’s testing machine.
subsoil conditions etc. ⇒ Crushing Test :
1 1 Aggregate crushing value test on coarse aggregates
A good thumb rule for the truss spacing is to of gives a relative measure of the resistance of an
5 3
aggregate crushing under gradually applied
their span.
compressive load. Aggregate crushing value is a
127. In case of plastic design, the calculated numerical index of the strength of the aggregate and it
maximum shear capacity of a beam as per is used in construction of roads and pavements.
IS:800 shall be (where Aw effective cross- ⇒ Abrasion Test :
sectional area resisting shear and fy is the yield Abrasion Test is the measure of aggregate
stress of the steel) : toughness and abrasion resistance such as crushing,
(a) 0.55 Awfy (b) 0.65 Awfy degradation and disintegration.
(c) 0.75 Awfy (d) 0.85 Awfy
⇒ Impact Test :
Ans. (a) : The maximum shear capacity of beam The aggregate impact test value is a measure of
column is 0.55 Awfy where Aw is the effective cross- resistance to sudden impact or shock, which way
sectional area resisting shear. vary from its resistance to gradually applied
128. Bearing stiffener in a plate girder is used to : compressive load.
(a) Transfer the load from the top flange to the 133. According the IS specification, the compressive
bottom one. strength of ordinary Portland cement after
(b) Prevent buckling of web three days should not be less than :
(c) Decrease the effective depth of web (a) 7 MPa (b) 1.5 MPa
(d) Prevent excessive deflection (c) 16 MPa (d) 21 MPa
OPSC (AEE) Exam-2019 (Paper-I) 110 YCT
 Ans. (c,b) : For 33 grade OPC, the 3-day strength Ans. (b) : Low heat cement contains high C2S and low
should not be less than 16 MPa and for 43 grade OPC, C3S content compared to OPC
the 3-day strength should not be loss than 21 MPa. (1) Low Heat Cement :
134. Which of the following cements is suitable for use It is a Portland Cement with low C3S and C3A and high
in massive concrete structures such as large dams? C2S content.
(a) Ordinary Portland Cement ● It is suitable for mass concreting works such as dams
(b) Low-Heat Cement due to low heat of hydration.
(c) Rapid Hardening Cement ● Rate of strength development is slow but the
ultimate strength is same as the OPC.
(d) Sulphate Resisting Cement
(2) Rapid Hardening Cement :
Ans. (b) : For the construction of mass concrete ● Also called as high early strength Portland Cement
structure like dams, cements with low heat of ● Contains high C3S content and less C2S content
hydration are preferred Ex : Low heat cement, Portland
● Suitable for road repairs work and is structures
Pozzolana cement, Blast furnace slag cement etc. where load is applied in a short period of time.
Low Heat Portland Cement : ● Unsuitable for mass concreting because excessive
• It is a Portland Cement with low C3S and C3A and heat evolution can result in serious cracking.
high C2S content. (3) Sulphate Resisting Cement is a type of Portland
• It is suitable for mass concreting works such as Cement in which the amount of tricalcium aluminate
dams due to low heat of hydration. (C3A) is restricted to lower than 5% and (C3A + C4AF)
• Rate of strength development is slow but the lower than 25%, which reduces the formation of
ultimate strength is same as the opc. sulphate salts. The reduction of sulphate salts lowers
Rapid Hardening Portland Cement : the possibility of sulphate attack on the concrete.
• Also called as high early strength Portland Cement. Ordinary Portland Cement
It reacts with silica and is helpful in development of
• Contains high C3S content and less C2S content strength.
• Suitable for road repair work and in structures
138. Number of bricks required for one cubic meter
where load is applied in a short period of time. of brick meter of brick masonry is :
• Unsuitable for mass concreting because excessive (a) 400 (b) 450
heat evaluation can result in serious cracking. (c) 500 (d) 550
Sulphate Resisting Cement :
Ans. (c) : Nominal size of a modular brick = 20 × 10 ×
Sulphate Resisting Cement is a type of Portland 10 cm Number of bricks required for 1 cum of
Cement in which the amount of tricalcium Aluminate
1
(C3A) is restricted to lower than 5% and (C3A + C4AF) masonry = × 0.1× 0.1 = 500
lower than 25%, which reduces the formation of 0.2
sulphate salts. The reduction of sulphate salts lowers 139. Distemper is used to coat :
the possibility of sulphate attack on the concrete. (a) Compound wall
Ordinary Portland Cement : (b) Wood work
It reacts with silica and is helpful in development of strength. (c) External concrete surface
135. The slump recommended for mass concrete is (d) Interior surfaces not exposed to weather
about : Ans. (d) : Distemper is a powder made point so it is
(a) 25 mm to 50 mm (b) 50 mm to 100 mm not used for the external purpose.
(c) 100 mm to 25 mm (d) 125 mm to 50 mm 140. Neoprene is suitable for use in :
Ans. (a) : Concrete for road construction 20 to 40 mm. (a) Bearings of bridge (b) Floors of dance hall
Concrete for tops of curbs, parapets, piers, slabs and (c) Joinery work (d) Coating of floor
wall 40 to 50 mm. concrete for canal lining 70 to 80 Ans. (a) : Neoprene is generally used for making
mm. Normal RCC work 80 to 150 mm. bearings of bridges.
Mass concrete 20 to 50 mm. 141. Minimum width of landing should be :
136. The most common admixture which is used to (a) Equal to width of stairs
accelerate the initial set of concrete : (b) Half the width of stairs
(a) Gypsum (b) Calcium Chloride (c) Twice the width of stairs
(c) Calcium Carbonate (d) None of these (d) One-fourth the width of stairs
Ans. (b) : Calcium chloride is the most commonly used Ans. (a) : The minimum width of landing should be
admixture to accelerate initial strength of concrete. equal to the width of the stairs
137. Which of the following cements contains 142. Doglegged stairs are :
maximum percentage of dicalcium silicate? (a) Half turn stairs
(a) Ordinary Portland Cement (b) Quarter turn stairs
(b) Low Heat Cement (c) Straight stairs
(c) Rapid Hardening Cement (d) Three quarter turn stairs
(d) Sulphate Resisting Cement Ans. (a) : Doglegged stairs are half turn stairs.
OPSC (AEE) Exam-2019 (Paper-I) 111 YCT
143. The type of windows provided on the sloping (c) Compact the soil
side of a pitched roof is called : (d) Replace the poor soil
(a) Dormer window (b) Gable window Ans. (d) : Generally, before the construction of
(c) Lantern (d) None of these foundation on black cotton soil sand bed is laid to act
Ans. (a) : Windows provided on the sloping roof are as a drainage layer. Drainage of water decreases the
called Dormer window. swelling nature of this soil.
144. Sum of tread and rise must lie between : 150. A queen closer is a
(a) 300 mm to 350 mm (b) 400 mm to 450 mm (a) Brick laid with its length parallel to the face
(c) 500 mm to 550 mm (d) 600 mm to 650 mm or direction of wall.
Ans. (b) : Sum of tread and rise must lie between 400 (b) Brick laid with its breadth parallel to the face
or direction of wall.
mm to 450 mm
(c) Brick having the same length and depth as the
145. The function of king post in a king post roof other bricks but half the breadth.
truss is (d) Brick with half the width at one end and full
(a) To support the framework of the roof width at the other.
(b) To receive the ends of principal rafter Ans. (c) : Queen closer is a brick having the same length
(c) To prevent the walls from spreading outward and depth as the other bricks but half the breadth.
(d) To prevent the tie beam from sagging at its
151. Le Chatelier's device is used for determining
centre
the :
Ans. (d) : A king post (or king-post or king post) is a (a) Setting time of cement
central vertical post used in architectural or bridge designs, (b) Soundness of system
working in tension to support a beam below from a truss (c) Tensile strength of cement
apex above (whereas a crown post, through visually
similar, supports items above from the beam below). (d) Compressive strength of cement
The function of king post in a king post roof truss is to Ans. (b) : Le Chatelier's device is used for determining
prevent the tie beam from sagging at its centre. the soundness of system.
Soundness test
146. The pitched and sloping roofs are suitable for :
● This test is done by Le-Chatelier method
(a) Coastal regions (b) Pain regions
● For OPC, soundness value should not exceed 10 mm.
(c) Covering large areas (d) All of these
The various tests and their uses are tabulated below
Ans. (a) : Pitched Roofs are those which have the Type of Test Use
decks or surfaces with the considerable slope for
(1) (i) Air permeability test To measure degree of
covering the building structure. These roofs are fineness of cement
generally lighter than a flat roof. Sloping roof is
constructed of wood or steel. Sloping roof is most (ii) Sieve Method
suitable for the area with the heavy rainfall and (iii) Sedimentation test
snowfall. (2) Vicat apparatus test To measure
The pitched and sloping roofs are suitable for coastal consistency of cement
regions. and setting time of
cement.
147. The lintels are preferred to arches because : (3) (i) Le-Chatelier method To measure soundness
(a) Arches require more head-room to span the of test.
opening like doors and windows. (ii) Autoclave test
(b) Arches require strong abutments to withstand (4) Briquette method To determine tensile
arch thrust. strength of cement.
(c) Arches are difficult to construct. (5) Crushing test To Determine
(d) All of these. Compressive strength
Ans. (d) : The lintels are prefferred to arches because of cement.
arches require more headroom to span the openings 152. The ultimate tensile strength of structural mild
like doors, windows etc. steel is about :
148. The type of footing which is used to transmit (a) 30 N/mm2 (b) 260 N/mm2
2
heavy loads through steel columns is : (c) 420 N/mm (d) 840 N/mm2
(a) Raft foundation (b) Grillage foundation Ans. (c) : The yield strength of mild steel is around
(c) Well foundation (d) Isolated footing 250 N/mm2 and its ultimate strength is around 420
Ans. (b) : The type of footing which is sued to transmit N/mm2.
heavy loads through steel column is isolated feeting. 153. The type of bond provided in brick masonry
149. In case of foundation on black cotton soils, the for carrying heavy load is
most suitable method to increase the bearing (a) English bond
capacity of the soil is : (b) Single Flemish bond
(a) Increase the depth of foundation (c) Double Flemish bond
(b) Drain the soil (d) Zigzag bond
OPSC (AEE) Exam-2019 (Paper-I) 112 YCT
 Ans. (a) : English bond is the strongest bond. Hence it 156. For a given environment, the most significant
is generally used for load bearing constructions. factor that influences the total shrinkage of
English bond consists of alternate courses of headers concrete is
and stretchers (a) Cement content of mix
(b) Total amount of water added at the time of
mixing
(c) Size of the member concreted
(d) Maximum size of the coarse aggregate used
Ans. (b) : Factors affecting shrinkage of concrete
(1) Humidity (Drying condition)
⇒ Single Flemish Bond is a combination of English (2) Water Cement ratio
Bond and Flemish bond in this type of construction, (3) Hardness of aggregates
the front exposed surface of wall consists of (4) Moisture movement in concrete
Flemish bond and the back surface of the wall (5) Type of coarse aggregate
consists of English bond in each course minimum (6) Shape of aggregate.
thickness required for single Flemish bond is one So, for the given environment, the most significant
and a half brick thickness.
factor that influences the total shrinkage of concrete is
total amount of water added at the time of mixing.
157. A good brick, when immersed in water for 24
hours, should not absorb more than
(a) 20% of its dry weight
⇒ Double Flemish bond : (b) 30% of its saturated weight
Every course consist of headers and structure (c) 0% of its dry weight
placed alternatively, the facing and backing of the (d) 20% of its saturated weight
wall in each course have the same appearance.
Ans. (a) : A good brick, when immersed in water for
⇒ Zigzag Bond : This Bond is similar to herring-bone 24 hours, should not absorb more than 20% of its dry
bond, except that the bricks are laid in zigzag
weight.
fashion. This Bond is commonly used for making
ornamental panels in the brick flooring. 158. The approximate ratio between the strength of
154. Paints with white lead base are suitable for cement concrete at 7 days and 28 days is
painting of : 3 2
(a) (b)
(a) Wood work 4 3
(b) Iron work 1 1
(c) Both wood work and iron work (c) (d)
2 3
(d) None of these Ans. (b) : The approximate ratio between the strength
Ans. (a) White lead : 2
This is a carbonate of lead and forms the base of lead of cement concrete at 7 days and 28 days in
3
paints it is dense, permanent and waterproof. It is not
suitable for dedicated works as lead becomes discolored 159. If X is the standard consistency of cement, the
when exposed to sulphur vapours it is most suitable for amount of water used in conducting the initial
wood surfaces; since it does not afford protection against setting time test on cement is :
rusting, it is not suitable for Iron surfaces. (a) 0.65X (b) 0.85X
155. The role of super plasticizer in a cement paste (c) 0.6X (d) 0.8X
is to : Ans. (b)
(a) Disperse the particle Test Performed Amount of Water
(b) Disperse the particle and to remove the air Required
bubbles ● Initial setting time test 0.85 X
(c) Retard setting
● Final setting time test 0.85 X
(d) Disperse the particle and to remove the air
bubbles and to retard setting ● Soundness test 0.78 X
Ans. (d) : Plasticizers and superplasticizers retard the ● Compressive strength test x/4 + 3.5
curing of concrete. Their addition to concrete of mortar ● Tensile strength test x/5 + 2.5
allows the reduction of the water to cement ratio 160. For complete hydration of cement the
without negativity affecting the workability of the
water/cement ratio needed is :
mixture, and enables the production of self
consolidating concrete and high performance concrete. (a) Less than 0.25
So, the role of super plasticizers in a cement paste is to (b) More than 0.25 but less than 0.35
disperse the particle and to remove the air bubble and (c) More than 0.35 but less than 0.45
to retard setting. (d) More than 0.45 but less than 0.55
OPSC (AEE) Exam-2019 (Paper-I) 113 YCT
 Ans. (a) : Water required for complete chemical 165. The Schedule of Rates is prepared based on :
reaction with cement particles is around 23% (a) Analysis of rates
Water required for filling the gel pores of cement is (b) Experience
around 15% hence, for complete hydration of cement (c) General abstract of cost
the water percentage should be around 38%. (d) Building cost index
161. Approximate cost of building commonly Ans. (a) : Analysis of rates is done to prepare schedule
estimated by of rates.
(a) Service unit method 166. In analysis of rates which is/are included from
(b) Comparison of costs at relative dates the following?
(c) Rough grouped quantities or elemental bill method (a) Cost of quantities of materials.
(d) All of these methods (b) Cost of labour and other miscellaneous
Ans. (d) : Approximate method of estimation expenditures.
(1) Service unit method. (c) Contractor's profit
(2) Plinth area method. (d) All of these
(3) Cubical rate method. Ans. (d) : Rate analysis include :
(4) Elemental bill method. 1. Cost of labour wages.
(5) Bay method. 2. Cost of material.
(6) Cost comparison method. 3. Overhead charges.
4. Location of silt.
162. In estimation of earthwork the measurements shall
be taken separately for what lead and height? 5. Constructor’s profit.
(a) 30 m, .5 m (b) 25 m, 2.0 m 167. In .0 cubic meter of 1:2:4 cement concrete,
(c) 20 m, 1.0 m (d) 40 m, 3.0 m how many bags of cement (approximately) is
required?
Ans. (a) (a) 6.6 (b) 16.6
Lead → 30 m (c) 26.6 (d) 36.6
Lift → 1.5 m Ans. (a) : Given
163. In long wall and short wall method of estimation 1:2:4 cement = 1/7
which one of the following is correct? Volume of wet cement concrete = 1.54 m3
(a) Short wall length in-to-in = centre to centre Volume of 50 kg cement bag = 0.035 m3
length – one breadth
1
(b) Short wall length in-to-in = centre to centre Volume of cement share in concrete = × 1.54
length + one breadth 7
= 0.22 m3
(c) Long wall length out-to-out = centre to centre
length + one breadth 0.22
Number of cement bags =
(d) Long wall length out-to-out = centre to centre 0.035
length – two breadth = 6.33 bags.
Ans. (a) : Long wall – Short wall method–In this 168. In case of approximate estimation, which one is
method, the wall along the length of room is correct?
considered to be long wall while the wall (a) It gives a rough idea of the probable
perpendicular to long wall is said to be short wall. expenditure for project financing.
To get the length of long wall or short wall, calculate (b) It is used for administrative approval for the
first the centre line lengths of individual walls. project.
164. Revised estimate is a detailed estimate and is (c) It is used for valuation and rent fixation.
required to be prepared under the condition when (d) All of these
(a) The original estimate exceeds or likely to exceed Ans. (d) : uses of approximate estimate
by more than specified in OPWD manual. 1. It gives rough idea of cost of project.
(b) The expenditure on a work exceeds or likely 2. To get administrating approval.
to exceed the amount of adminis-trative 3. To decide between completing project.
sanction as specified in OPWD manual. 4. For valuation and rent fixation.
(c) There are material deviation from the original 169. The earthwork quantities are calculated :
proposal even through the cost may be met (a) By mid-sectional method
from the sanctioned strength. (b) By mean-sectional method
(d) Any or all such case prevails.
(c) By Primordial method
Ans. (b) : Revised Estimate (d) All of these
(1) In calculating quarterly GDP, a third estimate
published approximately three months after the Ans. (d) : The earthwork quantities are calculated by :
end of a quarter. It includes information not available 1. Mid-section method.
at the time of the advance estimate or preliminary 2. Mean-sectional method.
estimate, as well as any necessary data revisions. 3. Prismordial method.
OPSC (AEE) Exam-2019 (Paper-I) 114 YCT
170. The detailed estimated of the cost of the project 176. Slack time refers to
is done by (a) An activity
(a) Unit-quantity method (b) An event
(b) Total-quantity method (c) Both event and activity
(c) BOOQ method (d) Critical event only
(d) By first two methods Ans. (d) : Slack is the time with which an event can be
Ans. (a) : Detailed estimate is done by total quality delayed without delaying the project scheduled time.
method 177. If the optimistic time, most likely time and
171. Which of the following is a weakness of bar pessimistic time for activity A are 4, 6 and 8
chart? respectively and for activity B are 5, 5.5 and 9
(a) Interdependencies of activities respectively, then :
(b) Project progress (a) Expected time for activity A is greater than
(c) Uncertainties the expected time of activity B
(d) All of these (b) Expected time of activity B is greater than the
expected time of activity A
Ans. (d) : Bar chart short comings.
(c) Expected time of activity A is same as that of
1. Lack of inter-dependencies of activities. the expected time of activity B
2. Critical path cannot be found. (d) None of these is correct
3. project progress cannot be monitored.
Ans. (c)
4. Uncertainties in activity time cannot be Analyzed
by bar chart. 4 + 6 × 4 + 8 36
TE for activity A = = =6
172. Cost slope : 6 6
5 + 5.5 × 4 + 9 36
(a)
(Crash cost – Normal cost) TE for activity B = = =6
Crash time 6 6
(Crash cost) 178. Free float is mainly used to :
(b) (a) Identify the activities which can be delayed
(Normal time – Crash time) without affecting the total float of preceding
(Crash cost – Normal cost) activity.
(c)
Normal time (b) Identify the activities which can be delayed
(Crash cost – Normal cost) without affecting the total float of succeeding
(d) activity.
(Normal time – Crash time)
(c) Establish priorities.
Ans. (d) (d) Identify the activities which can be delayed
Crash cost – Normal cost without affecting the total float of either of
Cost slope =
Normal time – Crash time preceding or succeeding activities.
173. The PERT calculations yield a project length of 50 Ans. (b) : Free float is time with which activity can be
weeks, with a variance of 9. Within how many delayed without affecting completion of succeeding
weeks would you expect the project to be completed activities.
with probability of 95%, take probability factor Z 179. There are three parallel paths in a part of a
equal to .65 for 95% probability : network between a bursting node and the next
(a) 54.95 (b) 56.6 merging node with only one activity in each
(c) 60 (d) 70 path. The minimum number of dummy arrows
Ans. (a) needed will be
(a) 0 (b) 1
T1 – Ts T − 50
= z, = 1.65 (c) 2 (d) 3
σ 3
Ans. (c)
T90 = 54.95 weeks.
174. The probability of competition of any activity
within its expected time is :
(a) 50% (b) 84.1%
(c) 99.9% (d) 100%
Ans. (a) : Probability of completion of any activity in 180. In the time-cost optimization, using CPM
its expected time is 50%. method for network analysis, the crashing of
175. The direct cost of a project with respect to the activities along the critical path is done
normal time is starting with the activity having :
(a) Minimum (b) Maximum (a) Shortest duration (b) Least cost slope
(c) Zero (d) Infinite (c) Longest duration (d) Highest cost slope
Ans. (a) : Minimum cost is calculated using normal Ans. (b) : Least cost slope is used to ensure minimum
time. increase in direct cost.
OPSC (AEE) Exam-2019 (Paper-I) 115 YCT
 Odisha Public Service Commission
(AEE) Exam-2019 (Paper-II)
1. F.S.L. of a canal at its head with respect to then the water will flow freely and structure called
parent channel is kept : aqueduct, but if the HFL of drainage is above the
(a) At the same level (b) 15 cm lower trough of canal then water will flow due to syphonic
(c) 15 cm higher (d) None of these action (pipe flow) and structure provided is called
Ans. (b) : F.S.L. of a canal at its head with respect to syphon aqueduct.
parent channel is kept 15 cm lower. Super Passage– Canal bed level below the drainage
F.S.L (Full supply level) is canal cross-section and drainage bed level above the full supply levels.
which can holding (carrying) the amount of water 6. Inundation of habitation is worse close to
headed for irrigation/supply. (a) Reservoirs (b) Diversion Weirs
Parent channel means that type of canal which can (c) Deltaic plains (d) None of these
share the water towards the sub canal. Ans. (a) : Inundation of habitation is worse close to
2. The measure to remove water logging of land, is Reservoirs
(a) To reduce percolation from canals and water 7. The measure to remove water logging of land, is
courses (a) To reduce percolation from canals and water
(b) To increase out low from the ground water courses
reservoir (b) To increase outflow from the ground water
(c) Both (a) and (b) reservoir
(d) Neither (a) nor (b) (c) Both (a) and (b)
Ans. (c) : However we do the measure to remove (d) Neither (a) nor (b)
water legging of land to increase outflow from the Ans. (c) : To measure to remove water logging of land is
ground water reservoir. (i) To reduce percolation from canals and water
3. For smooth entry of water in a canal, the angle courses.
between head regulator and water is generally (ii) To increase outflow from the ground water.
kept : 8. Groynes are generally built :
(a) 800 (b) 900 (a) Perpendicular to the bank
0
(c) 110 (d) 1200 (b) Inclined up stream up to 300
Ans. (c) : Angle between 90 to 1100 is provided for
0
(c) Inclined down stream up to 300
smooth entry of water into regulator slightly more than (d) All these
900 angle is better hence provide 1100.
Ans. (b) : As type of Groynes is net mentioned answer
4. If L is total length of a canal in kilometers, P is should be (d).
total perimeter of its lining in metres and C is ⇒ Arracting Groynes – Pointing downstream (300
the cost of lining per square metre, the to 600 inclination)
additional expenditure involved on lining, is
⇒ Repelling Groynes – Pointing upstream [with
PLC respect to bank 600 to 800 or with alone
(a) 1000 PLC (b)
1000 perpendicular to the bank varies from 100 to
(c)
PL
(d)
PC 300]
1000C 100L ⇒ Dylecting Groynes – either perpendicular to
Ans. (a) : the bank or pointing slightly upstream.
Length L = 000m, Perimeter = PM 9. The top of the capillary zone
Total area = 000 PL m2 (a) Lies below the water table at every point
Total cost = 1000 PLC (b) Lies above the water table at every point
5. When a canal is carried over a natural (c) Coincides the water table at every point
drainage, the structure provided, is known as (d) None of these
(a) Siphon (b) Aqueduct Ans. (b) : The top of the capillary zone lies above the
(c) Super passage (d) Siphon-aqueduct water table at every point.
Ans. (b and d) : In both aqueduct and syphon 10. For the design of major hydraulic structures on
aqueduct, canal is passed over the drainage, but if the the canals, the method generally preferred to,
HFL of drainage is much below the though of canal is based on
OPSC (AEE) Exam-2019 (Paper-II) 116 YCT
 (a) Bligh's theory Ans. (b)
(b) Electrical analogy method i – intensity of rainfall
(c) The relaxation method fc – infiltration capacity
(d) Khosla's method of independent variables f – infiltration rate
Ans. (d) : For the design of major hydraulic structures i – fc, then f = fc
on the canals, the method generally preferred to, is 16. For a given discharge in a channel, Blench
based on Khosla's method of independent variables curves given the relationship between the loss
11. The depth of rice root zone is of head (HL) and
(a) 50 cm (b) 70 cm (a) Specific energy up stream
(c) 80 cm (d) 90 cm (b) Specific energy down stream
Ans. (d) : The depth of rice root zone is 90 cm (c) Critical depth of water downstream
(d) Depth of water downstream
12. The saturation line is the line up to which
banks get saturated after the canal runs for Ans. (b) : For a given discharge in a channel, blench
some time. The saturation gradient in ordinary curves give the relationship b/w the loss of head (HL)
loam soil is generally and specific energy down stream.
(a) 2 : 1 (b) 3 : 1 17. The velocity of drainage water in the barrels of
(c) 4 : 1 (d) 5 : 1 a siphon-aqueduct, is normally limited to
(a) 1 to 2 m per second (b) 2 to 3 m per second
Ans. (c) : For ordinary loamy soil the saturation
gradient is 4 : 1 (c) 3 to 4 m per second (d) 4 to 5 m per second
Saturation line is the line up to which banks get Ans. (b) : Velocity of flow through the barrels is
saturated after the canal runs for some time. normally limited to 2 to 3 m/s.
13. The depth of the crest of a scouring sluice below 18. According to Khosla, the exit gradient of
the crest of a head regulator, is generally kept surface flow
(a) 0.2 m (b) 1.2 m (a) Depends upon the b/d ratio
(c) 2.2 m (d) 3.2 m (b) Is independent of the depths of d/s cut off
walls
Ans. (b) : To obtain control on entry of silt into the (c) Is independent of the b/d ratio
canal, the crest of the head regulator should be higher
(d) None of these
than the crest of the under sluices by a minimum of 1.2
m if silt excluder is not provided and by a minimum of Ans. (a)
1.8m of silt excluder is provided. H
GE =
14. If q is the discharge per unit width of a channel πd λ
and D1, D2 are the depths of water before and 1 + 1 + α2
after hydraulic jump, the following λ=
2
relationship is true
d
2q 2 α=
(a) D1D2(D2 – D1) = d
g 19. If the irrigation efficiency is 80%, conveyance
D2 2q 2 losses are 20% and the actual depth of
(b) (D2 – D1) = watering is 6cm, the depth of water required at
D1 g
the canal outlet, is
2q 2 (a) 10 cm (b) 15 cm
(c) D1D2(D2 + D1) =
g (c) 20 cm (d) 25 cm
Ans. (d) : Depth of water required in canal
D1 2q 2
(d) (D2 + D1) = 16
D2 g = = 25cm
0.8 × 0.8
Ans. (c) 20. The scour depth D of a river during flood, may
2q 2 be calculated from the Lacey's equation
= D1D2(D1+D)
g 1/ 2
Q Q
(a) D = 0.47   (b) D = 0.47  
15. If the intensity of rainfall is more than the f  f 
infiltration capacity of soil, then the infiltration 1/ 3 2/3
rate is Q Q
(c) D = 0.47   (d) D = 0.47  
(a) Equal to rate of rainfall f  f 
(b) Equal to infiltration capacity 13
(c) More than rate of rainfall Q
Ans. (c) : Lacey's scour depth D = 0.47  
(d) More than infiltration capacity f 
OPSC (AEE) Exam-2019 (Paper-II) 117 YCT
21. If V0 is the critical velocity of a channel, its silt Ans. (d) : Rigid modular outlet
transporting power, according to Kennedy, is These modules allow constant discharge within
proportional to : reasonable working limits of head.
(a) V01 2 (b) V03 2 27. Canals constructed for draining off water from
water logged areas, are known as
(c) V05 2 (d) V07 2 (a) Drains (b) Inundation canals
Ans. (c) : Qt = aB V05 2 (c) Valley canals (d) Contour canals
Ans. (a) : Canals constructed for draining off water
22. The main cause of silting up a channel
from water logged areas, are known as Drains.
(a) Defective outlets
Inundation canal are long canals taken off from large
(b) Inadequate slope rivers. They receive water when the river is high
(c) Defective head regulator enough and especially when in flood while perennial
(d) All of these canals are lined to dams and barrages to provide water
Ans. (d) : The main cause of silting up a channel are throughout the year and they irrigate a vast area.
(i) Defective outlets A contour canal is an artificially dug navigable canal
(ii) Inadequate slope which closely follows the contour line of the land it
(iii) Defective head regulator traverses in order to avoid costly engineering marks
23. The most suitable location of a canal head such as boring a funnel through higher ground,
work, is building an embankment over lower ground or
constructing a canal lock (or series of locks) to change
(a) Boulders stage of the river
the level of the canal. Because of this, these canals are
(b) Delta stage of the river characterised by their meandering course.
(c) Rock stage of the river
28. A minimum of 90 cm free board is provided if
(d) Trough stage of the river
the discharge in the canal is between
Ans. (a and d) : Delta and rocky stages are net (a) 30 to 33 cumecs (b) 35 to 60 cumecs
suitable canal head marks may be located either in the (c) Over 60 cumecs (d) Over 00 cumecs
boulder stage or in the trough stage of the river.
Ans. (c) : A minimum of 90cm free board is provided if
24. Bligh's theory of seepage assumes the discharge in the canal is between over 00 cumecs.
(a) Equal weightage to the horizontal and vertical
creep 29. If water table is comparatively high, the
irrigation canal becomes useless, due to
(b) More weightage to horizontal creep than
(a) Large amount of seepage
vertical creep
(c) Less weightage to horizontal creep than (b) Water logging of the cultivated areas
vertical creep (c) Spread of malaria
(d) Loss of head follows the sine curve (d) All of these
Ans. (a) : Bligh's theory of seepage assumes equal Ans. (d) : If water table is comparatively high, the
weightage to the horizontal and vertical creep. irrigation canal becomes useless, due to water legging
of the cultivated areas.
25. Regime conditions in a channel may occur if
(a) Discharge is constant 30. If DC is the critical depth, D is the depth of
water downstream and HL is the total head loss
(b) Channel flows uniformly in incoherent
at the cistern of a vertical fall, the depth x of
alluvium as that transported in suspension
the cistern below the bed level downstream
(c) Silt grade and silt charge are constant suggested by Blench, is
(d) All of these
(a) DC – D +  H L – DC 
1 3
Ans. (d) : Regime conditions in a channel may occur if
4 8 
(i) Discharge is constant
(b) 2DC – D +  H L – DC 
(ii) Channel flows uniformly in incoherent arruvium 1 3
as that transported in suspension 4 8 
(iii) Silt grade and silt change are constant.
(c) 2DC – D +  H L – DC 
3 1
26. An outlet which maintains a constant discharge 8 4 
irrespective of fluctuation in the water levels of
(d) 3DC – D +  H L – DC 
the supplying channel or water course, is 3 1
known as 8 4 
(a) Non-modular outlet Ans. (b) : Blench formula
(b) Semi-modular outlet
X = 20c – D +  H L – D C 
1 3
(c) Flexible modular outlet
4 8 
(d) Right modular outlet
OPSC (AEE) Exam-2019 (Paper-II) 118 YCT
 X = depth of cistern below d/s bed in m Perennial canals – These canals flow all the year
HL = Height of drop in m round and thus they are known as perennial canals they
D = Depth of water in the d/s channel in m draw water from perennial rivers or from artificial
DC = Critical depth. lakes which maintain a high level of water on the
upstream side.
31. The ratio of the rate of change of discharge of
an outlet and parent channel, is known as 35. V and R are the regime mean velocity and
(a) Efficiency (b) Sensitivity hydraulic mean depth respectively in metres.
(c) Flexibility (d) Modular limit Lacey's silt factor f is
Ans. (c) : Flexibility – It provides a criteria for 2V 2 3V 2
(a) (b)
judging the behaviour of modules and semi-modules. It 3R 4R
is generally denoted by letter 'F'. It is the ratio of the
2
rate of change of outlet discharge to the rate of change 5V 2V 2
(c) (d)
of discharge of the parent channel. 2R 5R
32. If average particle size of the silt in millimetres 5V 2

in m, the Lacey's silt factor f is proportional to Ans. (c) : Lacey's silt factor 'f' =
3 2R
(a) m (b) m
(c) m 1/2
(d) m 1/3 36. Which is not an estimate for consumptive use?
(a) Blaney-Criddle equation
Ans. (c) : Silt factor f = .76 m (b) Manning's equation
33. If A is the area of the surface, x is the depth of (c) Hargreaves class A pan evaporation method
its C, G from the surface of the water and ω is (d) Penman's equation
the density of water, then
(a) Total pressure on the surface is equal to ωx. Ans. (b) : Manning's equation is net an estimate for
(b) Depth of the point at which total pressure acts is consumptive use.
equal to its moment of inertia divided by Ax. 37. Which strata does not hold water?
(c) Depth of the centre of pressure is 2/3 H (a) Confined aquifer (b) Unconfined aquifer
vertically below the surface. (c) Perched aquifer (d) Aquiclude
(d) All of these Ans. (*) : Aquifer : is a saturated formation of earth
Ans. (d) material which not only stores water but yields it in
Given area of the surface = A significant quantity relatively easily due to its high
Depth of C.G. from the surface of water = x permeability.
Density of water = ω eg : sand, gravel.
∴ Total pressure on the surface = ωx Aquiclude is a formation like clay which is highly
∴ option (a) right pores but not permeable due to very small size of
Depth of point at which total pressure acts = centre of pores.
pressure A confined aquifer is an aquifer below the land surface
1 that is saturated with water layers of impermeable
=x+
Ax material are both above and below the aquifer, causing
∴ option (b) is wrong it to be under pressure so that when the aquifer is
1 penetrated by a will, the water will rise above the top
Depth of centre of pressure = x +
Ax of the aquifer.
2 38. The water level in unconfined aquifer is
it may or may not be H
3 (a) At the water table level
∴ option (c) is wrong (b) Below water table level
34. The diversion of flood water of rivers, the type (c) Above water table level
of canal constructed, is (d) Flowing well
(a) Ridge canal (b) Perennial canal Ans. (a) : Unconfined aquifer are these into which water
(c) Inundation canal (d) Canal seeps from the ground surface directly above the aquifer.
Ans. (c) : The diversion of flood water of rivers, the
type of canal constructed, is Inundation canal. 39. Volume of water released from the aquifer of
unit cross-sectional area and of the full height
Ridge Canal : The dividing ridge line between the
catchment area of two streams (drains) is called the of aquifer is called
watershed or ridge canal. (a) Specific Yield
It is suitable for plain areas, where slopes are relatively (b) Specific Retention
flat and uniform this type alignment ensures gravity (c) Transmissibility
irrigation on both sides of the canal. (d) Storage coefficient
OPSC (AEE) Exam-2019 (Paper-II) 119 YCT
 Ans. (a) : Volume of water released from the aquifer Ans. (b)
of unit cross-sectional area and of the full height of 1/ 6 1/ 6
 Qf 2   So × 1.12 
aquifer is called Specific Yield. V=  =  = 0.87
Specific Retention– The ratio of the volume of water  40   140 
that a given body of rack or soil will hold against the Q = AV
pull of gravity to the volume of the body itself. Q 50
A= = = 57.50m 2
40. Potential evapotranspiration is V 0.87
(a) Evaporation where there is sufficient 46. Garrett's diagrams are used for graphical
moisture available to a fully vegetated area solution of design equations of a canal by
(b) The evapotranspiration of a forest area (a) Lacey's theory (b) Kennedy's theory
(c) Actual evapotranspiration of a crop before (c) Gibbs theory (d) Lindlay's theory
application of irrigation water Ans. (b) : Garrett's diagrams are used for graphical
(d) Amount of water needed to bring the solution of design equations of a canal by kennedy's
moisture content of a soil to its field capacity theory.
Ans. (a) : Potential evapotranspiration is evaporation 47. For a discharge of 64 cumecs and silt factor of
where there is sufficient moisture available to a furry 0.6, the normal regime scour depth is
vegetated area. (a) 2.24 m (b) 2.00 m
41. A line on a map connecting points having the (c) 2.36 m (d) 2.90 m
same amount of rainfall in a given period 3
Ans. (a) : Q = 64 m /s f = 0.6
(a) Isobars (b) Isovels 1/ 3
Q
(c) Isohyets (d) Isochrone RL = 0.47   = 2.23m
Ans. (c) : Isohyets : A line on a map connecting points f 
having the same amount of rainfall in a given period.48. A measure of safety in driving on the service
Isobar : A line on a weather map that joins places that road of canal is called
have the same air pressure at a particular time. (a) Spoil banks (b) Counter beams
Isochrone : A line on a diagram or map connecting (c) Dowlas (d) Berms
points relating to the same time or equal times. Ans. (d) : A measure of safety in driving on the
42. The consumptive use of water for a crop service road of canal is called Dowlas.
49. Value of free board in canal for discharge
(a) Is measured as the volume of water per unit area
(b) Is measured as depth of water on irrigated area between 30 to 50 cumecs
(a) 0.5 m (b) 0.6 m
(c) May be supplied partly by precipitation and
partly by irrigation (c) 0.75 m (d) 0.90 m
(d) All of these Ans. ((c) or (d))
Value of free board in canal for discharge between 30
Ans. (d) : The consumptive use of water for a crop is
to 50 cumecs is 0.75m or 0/90 m
quantity of water used by plants for evapotranspiration.
50. If the total accumulated precipitation of the
43. The Lacey's regime velocity is proportional to storm is plotted against time, the curve is
(a) R1/2S3/4 (b) Q3/4S1/3 knows as
3/4 1/3
(c) R S (d) R2/4S1/2 (a) Rain hydrograph
Ans. (*) (b) Mass curve
V = 0.8 R2/3 S1/3 (c) Depth area duration curve
1 3 / 4 1/ 2 (d) Intensity duration frequency curves
V = R S
Na Ans. (b) : If the total accumulated precipitation of the
storm is plotted against time, the curve is knows as
44. If the average particle size is 1.44 mm, the silt Mass curve.
factor is Rain hydrograph : The hydrograph deals with the
(a) 1.425 (b) 1.76 accurrence and distribution of water over and under the
(c) 1.584 (d) 1.00 earth surface is called hydrograph.
Ans. (b) Depth Area Duration (DAD) : Analysis of a storm is
done to determine the maximum amounts of rainfall
f = 1.76 d = 1.76 1.44 = 2.112m
within various durations over areas of various sizes.
45. What is the cross-sectional area of regime The insensity duration frequency curves are used in
channel for 50 cumecs discharge and silt factor hydrology to express in synthetic way, fixed a return
1.1? period T and a duration d of a rainfall event, and for a
(a) 58.4 m2 (b) 58.3 m2 given location, the information on the maximum
(c) 694.2 m 2
(d) 763.63 m2 rainfall height h and the maximum rainfall intensity.

OPSC (AEE) Exam-2019 (Paper-II) 120 YCT

51. What is the per head water requirement for Ans. (b) : For wells placed closely, the minimum
Domestic purposes in litres per day? distance between them is twice the radius of circle of
(a) 45 (b) 50 influence.
(c) 75 (d) 35 58. Manholes on sewer lines are provided for
Ans. (d) : The IS code lays down a limit on domestic (a) Periodic cleaning
water consumption b/w 35 to 225 lpcd under ordinary (b) Provide additional water for easy disposal
conditions (as per IS 1172 : 1993) the minimum (c) Providing air for oxidation
domestic water demand for a town with full flushing (d) Removal of part of sewage
system should be taken at 200 l/h/d although it can be
Ans. (a) : Manholes on sewer lines are provided for
reduced to 35 l/h/d for economically weaker sections
periodic cleaning.
and LIG colonies
Low income group, depending upon prevailing condition. 59. For public water supply, the threshold number
for taste and odour should not be more than
52. What is the fire demand in litres per head for a
(a) 1 (b) 2
city of 25 lakh population for moderate fire to
last for 3 hours requiring three fire streams (c) 3 (d) 4
with discharge of 1100 pm and the provision Ans. (c) : For public water supply, the threshold
for four fires at a time? number for taste and odour should not be more than 3.
(a) 0.95 (b) 0.24 60. The measurement of colour of water is carried
(c) 0.32 (d) None of these out by means of
Ans. (a) (a) Osmoscope
Time = 3 hours, Q = 1100 lpm (b) Tintometre
Population = 25 lakhs (c) Nephelometric turbidimetre
No. of fires = 4 (d) Thermometer
fire streams = 3 Ans. (b) : The measurement of colour of water is
fire demand = 1100 × 3 × 60 × 3 × 4 = 2376000L carried out by means of tintometre.
2376000 61. The hardness is expressed in degrees as per
fire demand (L/head) = = 0.95 L/head (a) Clark's scale
25, 00, 000
(b) Carbonate hardness
53. Aeration of water is done to remove
(c) Total hardness
(a) Odour (b) Colour
(d) Non-carbonate hardness
(c) Bacteria (d) Turbidity
Ans. (a) : The hardness is expressed in degrees as per
Ans. (a) : Aeration of water is done to remove gases clarks scale.
that import odour and taste.
62. The alkalinity is caused by which positively
54. In sewar treatment plants, the oil and grease is charges ions of
removed by
(a) Ca++ (b) Na+
(a) Oxidation (b) Filteration ++
(c) Mg (d) Sr++
(c) Skimming (d) Screening
Ans. (b) : The alkalinity is caused by which positively
Ans. (c) : In sewar treatment plants, the oil and grease
charges ions of Ca++
is removed by skimming.
63. The permissible dissolved solid for drinking
55. Type of joint used in metallic sewer is
water as per BIS is
(a) Collar joint (b) Flexible joint
(a) 200 mg/l (b) 330 mg/l
(c) Mechanical joint (d) Simplex joint
(c) 40 mg/l (d) 500 mg/l
Ans. (c) : Mechanical joint is used in metallic sewer.
Ans. (d) : The permissible dissolved solid for drinking
56. Which is the surface source of water supply
water as per BIS is 500 mg/l
scheme?
(a) Infiltration well (b) Springs 64. Activated carbon removes from water
(c) Wells (d) Streams (a) Iron (b) Organic matter
Ans. (d) : Streams is the surface source of water (c) Manganese (d) All of these
supply scheme. Ans. (d) : Activated carbon is used in water filter
57. For wells placed closely, the minimum distance purifiers because activated carbon removed from the
between them is water most toxic organic compounds in water like
(a) Twice the diameter of the wells pesticides and heavy metal organic compounds.
(b) Twice the radius of circle of influence Activated carbon also removes smalls in water and
(c) Twice the depth of well makes cloudy water clear by removing color causing
(d) Twice the depth of water in the well compounds in the water.

OPSC (AEE) Exam-2019 (Paper-II) 121 YCT

65. In Kanpur the major source of pollution is Vermicompost is the product of the decomposition
(a) Aircraft factory (b) Cotton mills process using various species of worms, usually red
(c) Tanneries (d) All of these wigglers, white warms, and other easth worms, to
create a mixture of decomposing vegetable of food
Ans. (c) : Among the major causes of air pollution in
waste, bedding materials, and vermicast, vermicast is
Kanpur are industrial sector, vehicles, road dust and
the end-product of the break down of organic matter
domestic cooking. The industrial sector is the biggest
by earthworms.
cause of air pollution in Kanpur.
73. The simplest and most common method used in
66. Coagulation is to be adopted when the the cities is to collect and dump the waste in a
Turbidity of water exceeds
(a) Landfill (b) River
(a) 20 ppm (b) 30 ppm
(c) Road side (d) Any of these
(c) 40 ppm (d) 50 ppm
Ans. (a) : The simplest and most common method
Ans. (c) : Coagulation is to be adopted when the used in the cities is to collect and dump the waste in a
Turbidity of water exceeds 50 ppm. Landfill.
67. Which coagulant is also called Alum? 74. What is the order of waste management
(a) Sodium Aluminate hierarchy from most to least favoured?
(b) Aluminium Sulphate (a) Prevention-Recycle-Reuse-Disposal
(c) Magnesium Carbonate (b) Prevention-Reuse-Disposal-Recycle
(d) Chlorinated Copperas (c) Prevention-Disposl-Reuse-Recycle
Ans. (b) : Aluminium Sulphate (d) Prevention-Reuse-Recycle-Disposal
68. Jar test is carried out to decide the dosage of Ans. (d) : order of waste management hierarchy from
(a) Coagulant (b) Ozone most to least favoured is prevention-Reuse-Recycle-
(c) Chloride (d) Lime in hard water Disposal.
Ans. (a) : Jar test is carried out to decide the dosage of 75. Organic matter, which can be decomposed by
Coagulant. bacteria is known as
(a) Biodegradable organic matter
69. What is the method of removal of permanent
hardness in water? (b) Degradation
(a) Zeolite process (b) Lime-soda process (c) Eutrophication
(c) Reverse osmosis (d) All of these (d) Decomposers
Ans. (a) : Organic matter, which can be decomposed
Ans. (d) : Permanent hardness of water can be
by bacteria is known as Biodegradable organic matter.
removed using zeolite process, Lime–soda process and
reverse osmosis. Eutrophication–The process of too many plants
growing on the surface of a river, lake etc. often
70. The equivalent of seven grams per litre in parts because chemicals that are used to help crops grow
per million would be have been carried there by rain.
(a) 7 (b) 70 Decomposer– Decomposer are organisms that break
(c) 700 (d) 7000 down dead or decaying organisms, and in doing so,
Ans. (d) : The equivalent of seven grams per litre in they carry out a process possible by only certain
parts per million would be 7000. kingdoms/including fungi for example decomposition.
71. The most frequent method of becterial 76. With passage of time, filled up solid wastes will
reproduction is by get stabilized by
(a) Cell division (b) Sexual process (a) Eutrophication (b) Decomposition
(c) Binding (d) Colony formation (c) Hydrolysis (d) Urbanization
Ans. (a) : The most frequent method of becterial Ans. (b) : With passage of time, filled up solid wastes
reproduction is by cell division. will get stabilized by Decomposition.
72. What is called for the process of burning 77. Allowed noise level in residential areas during
municipal solid waste in a properly designed day time in India
furnace under suitable temperature and (a) 25 dB (b) 55 dB
operating conditions? (c) 75 dB (d) 00 dB
(a) Landfill (b) Recycling Ans. (b) : Allowed noise level in residential areas
(c) Cermicomposting (d) Incineration during day time in India 55 dB.
Ans. (d) : The process of burning municipal solid 78. The major contributor of Carbon monoxide in
waste in a properly designed furnace under suitable atmosphere is
temperature and operating conditions is called (a) Motor vehicle
incineration. (b) Industrial processes
OPSC (AEE) Exam-2019 (Paper-II) 122 YCT
 (c) Stationary fuel combustion ⇒ Loess is a silt deposit made by wind. These
(d) None of these deposits have low density and high
Ans. (a) : The major contributor of Carbon monoxide compressibility. The permeability in the vertical
in atmosphere is stationary fuel combustion. direction is large.
79. What is the permissible limit of 24 hourly ⇒ Deposits directly made of melting of glaciers are
called till.
particulate matter (size less than 0 µm) or PM
10 concentration in residential area as per 84. Which of the statement are correct for Black
Indian Standard? Cotton Soils?
(a) 60 µg/m3 (b) 80 µg/m3 (a) High swelling and low shrinkage
(b) Low bearing capacity and high shearing strength
(c) 00 µg/m 3
(d) None of these
(c) Highly compressible and low swelling
Ans. (a) : 100 mg/m3 is the permissible limit of 24 characteristics
hourly particulate matter (size less than 0 µm) or PM (d) High plasticity and low shearing strength
10 concentration in residential area as per Indian
Ans. (d) : Black cotton soils are clays of high
Standard
plasticity, high shrinkage and swelling characteristics.
80. The pH value of potable water is Shear strength of soils is extremely low.
(a) 4.5 (b) 5.6 Highly compressible.
(c) 6.0 (d) 6.8 Very low bearing capacity.
Ans. (d) : The pH value of potable water is 6.8. 85. Degree of saturation is
81. Aeolian is the soil deposited by (a) Ratio of volume of voids to the volume of
(a) Wind transported soil solids
(b) Water transported soils (b) Ratio of volume of water to the volume of
(c) Glacier deposited soils solids
(d) Gravity deposited soil (c) Ratio of volume of water to the volume of
Ans. (a) : Soil deposited by wind are known as voids
Aeolian deposits. (d) Ratio of volume of water to the total volume
Alluvial deposit : deposited by river water. VW
Ans. (c) : Degree of saturation, S = ×100
Lacustrine deposit : deposited by still water like lakes. VV
Marine deposit : deposited by sea water. VW = volume of water
Aeolian deposit : Transported by wind. VV = volume of voids.
Glaciar deposit : Transported by ice. 86. Water content is defined as
Celluvial soil : Deposited by gravity (Example : Ta----) (a) Ratio of volume of water to the volume of
82. Loess is silt-deposited by wind and deposits solids.
have (b) Ratio of mass of water to the mass of solids.
(a) High density and low compressibility (c) Ratio of weight of water to the weight of
(b) Low density and high compressibility solids.
(c) High density and low permeability (d) Ratio of weight of water to the volume of
(d) High bearing capacity and low solids.
compressibility Ans. (b or c)
Ans. (b) : Loess is a silt deposit made by wind they W
have low density and high compressibility. Water content W = W × 100
WS
83. Soil transported by Gravity is called
MW
(a) Talus (b) Loess or W= × 100
MS
(c) Drift (d) Aeolian
Ans. (a) : Colluvial soils, such as talus, have been Ratio of mass of water to the mass of solids.
deposited by the gravity. Ratio of weight of water to the weight of solids.
⇒ Soil carried and deposited by river water are 87. Unit weight of solid is
known as alluvial deposits. (a) Ratio of weight of solids to the volume of
⇒ Deposits made in lake are called lacustrine solids
deposits. (b) Ratio of mass of solids to the volume of
⇒ Marine deposits are formed when the focusing solids
water carries soil to ocean or sea. (c) Ratio of mass of solids to the mass of solids
⇒ Soil deposited by wind are known as aeolian (d) Ratio of weight of solids to the weight of
deposits solids
OPSC (AEE) Exam-2019 (Paper-II) 123 YCT
 Ans. (a) : Unit weight of solid is ratio of weight of (c) Moisture content of soil
solids to the volume of solids. (d) Ratio of mass density and moisture content
88. Specific gravity is defined as Ans. (b) Sand bath method–It is used to determine
(a) Ratio of mass of solid to the mass of equal the water content of soil. This method is not be used
0 for organic soils, or for soils having higher percentage
volume of water at –4 C.
(b) Ratio of weight of solid to the weight of equal of gypsum.
volume of water at 00C. The methods commonly used for the determination of
(c) Ratio of mass of solid to the mass of equal in-situ unit weight of a natural soil deposit or
volume of water at 00C. compacted earth fill are–
(d) Ratio of mass of solid to the mass of equal • Core-cutter method
volume of water at 40C. • Sand-replacement method
Ans. (d) : Specific gravity is defined as Ratio of mass • Water displacement method
of solid to the mass of equal volume of water at 40C. 92. A soil mass is 20 kg and volume 0.011 m3. After
89. Porosity is defined as oven drying the mass reduces to 6.5 kg. G = 2.7.
(a) Ratio of volume of void to the total volume The void ratio is
(b) Ratio of volume of void to the volume of (a) 0.2121 (b) 0.444
solids (c) 0.8 (d) 0.7158
(c) Ratio of volume of air to the volume of solids Ans. (c)
(d) Ratio of volume of water to the volume of M = 20 kg, M = 16.5 kg, V = 0.011m3 G = 2.7
d
solids
Md 16.5
V δd = = = 1500 kg/m3
Ans. (a) : Porosity, n = V v 0.011
V
GPW
90. Soil water content and specific gravity can be δd = 1 + e
determined by
(a) Oven drying method 2.7 ×1000
1500 =
(b) Density bottle method 1+ e
(c) Pycnometer method e = 0.8
(d) Alcohol method 93. A moist soil mass weights 3.52 N and after oven
Ans. (c) : Soil water content and specific gravity can drying the weight is 2.9 N. Specific gravity of
be determined by Pycnometer method. solids and mass specific gravity are 2.65 and
Water content and specific gravity is determined with 1.85 respectively. The degree of saturation is
the help of Pychometer method. (a) 0.4253 (b) 0.2138
⇒ Other method for water content determination : (c) 0.74 (d) 0.7656
● Oven drying method. Ans. (d)
● Torsion balance method. W = 3.52 N
● Sand bath method. Wd = 2.9 N = WS
● Alcohol method. W 3.52 – 2.9
● Calcium coreside method. W = W × 100 = × 100
Wd 2.9
● Radiation method.
W = 21.38%
⇒ Other method for specific gravity determination
WS WS 2.9
● Density bottle method. Gs = VS = =
● Measuring flash method. V X
S W G X
S W 2.65 × 9810
–4
● Gas jar method. GS = 1.1155 × 10 MS
● Shrinkage limit method. W W 3.52
Gm = V= =
⇒ The following methods are generally used for V × XW G m × X W 1.85 × 9810
determining of bulk unit weight Gm = 1.94 × 10–4 m3
● Core cutting method.
VV = V–VS = 8.241 × 10–5 m3
● Water displacement method.
V 8.241×10 –5
● Sand replacement method. e= V = = 0.739
● Water balloon method. VS 1.1155 × 10–4
● Radiation method. ωG
e=
91. Sand replacement method is used to determine S
(a) Specific gravity of soil ωG 0.2138 × 2.65
s= = = 0.7667
(b) Mass density of soil c 0.739
OPSC (AEE) Exam-2019 (Paper-II) 124 YCT
94. The sedimentation analysis for soil particles Ans. (d)
finer than 75µ is based on D60
(a) Boyle's law (b) Stoke's law Cu =
D10
(c) Charle's law (d) Darcy's law
D60 = Particle size such that 60% of the soil is finer
Ans. (b) : Stoke's Law– It is applicable for spheres of than this size
diameters between 0.2 mm and 0.0002 mm. D10 = Particle size such that 10% of the soil is finer
Spheres of diameter larger than 0.2 mm falling through than this size.
water cause, turbulence, whereas, for spheres with ● The larger the numerical value Cu, the more is the
diameter less than 0.0002 mm, Brownian motion takes range of particles
place and the velocity of settlement is too small for ● For soil; Cu = 1 uniformly graded
accurate measurement.
Cu ≥ 1 always
According to stokes law, terminal velocity is given as
for other value, poorly graded or gap graded.
 γ − γ w  2 g ( G − 1) γ w d
2

Vs = g  s d = 99. Plasticity index is the numerical difference

 18µ  18µ between
Where, (a) Plastic limit and shrinkage limit
µ = Dynamic viscosity of water (b) Plastic limit and liquid limit
d = particle size (c) Liquid limit and plastic limit
95. Hydrometre is used for determination of (d) Liquid limit and shrinkage limit
(a) Specific gravity of liquids Ans. (c) : Plasticity Index : Range of consistency
(b) Density of liquid (water convent) within which soil behave as a plastic
material is called plasticity index
(c) Particle size distribution of soil
IP = WL–WP
(d) Turbidity of water
WL = Liquid Limit
Ans. (a and c) : Hydrometre is used for determination
WP = Plastic Limit
of specific gravity of liquids and particle size
IP = Plasticity Index
distribution of soil particles finer than 75µ.
100. What is Toughness Index?
96. Coefficient of uniformity is defined as
(a) Ratio of plasticity index to the consistency
D60 D60 index.
(a) (b)
D10 D30 (b) Ratio of plasticity index to the liquidity index.
D10 D10 (c) Ratio of plasticity index to the flow index.
(c) (d) (d) Ratio of consistency index to the flow index.
D60 D30
Ans. (c) : Toughness Index (It) of a soil is defined the
Ans. (a)
ratio of the plasticity Index (IP) to the flow Index (If)
D60
Cu = I Plasticity Index
D10 It = P =
If Flow Index
D60 = Particle size such that 60% of the soil is giver
than this size 101. In unified soil classification system, the group
D10 = Particle size such that 10% of the soil is giver system SM stands for
than this size. (a) Silty sand (b) Clayer sand
● The larger the numerical value Cu, the move is the (c) Well graded sand (d) Silty graveis
range of particles. for soil ; Cu = 1 uniformily Ans. (a) : In unified soil classification system, the
graded, for will graded sand Cu > 6 and for well group system SM stands for Silty Sand.
graded gravel Cu > 4 SM = silty Sand
97. According to Terzaghi, the net ultimate Soil Symbol
bearing capacity of clay is given by Sand S
(a) CNQ (b) CN1 Gravel G
(c) CNC (d) 1.3 CNC
Silt M
Ans. (c) : qnu = CNC for stripfooting
Clay C
98. Uniformity coefficient of soil is
(a) Always less than 1 102. Liquid limit test is performed on soil samples
(b) Always equal to 1 passing through IS sieve of size
(c) Equal to less than 1 (a) 25 µ (b) 2 mm
(d) Equal to greater than 1 (c) 425 µ (d) 250 µ
OPSC (AEE) Exam-2019 (Paper-II) 125 YCT
 Ans. (c) : Liquid limit is found our using 104. The minimum water content at which the soil
(a) cassagrande's feel just begins to crumble when rolled into threads
(b) cone penetration 3 mm in diameter, is known as
(c) cassagrande's feel (a) Liquid limit
Procedure : (b) Plastic limit
● Soil about 120 gm of an air dried sample passing (c) Shrinkage limit
through 425µ is sieve is taken in a disn and mixed (d) Permeability limit
with distilled water to form a uniform paste Ans. (b) : The minimum water content at which the
● The soil is put casagrande's apparaties and a groove soil just begins to crumble when rolled into threads
of 2mm size is cut 3mm in diameter is known as plastic limit.
● Number of blows which is required to close 2mm
groove over a rubber pad is noted.
● Water content at which 25 blows close the groove
is called liquid limit

Shrinkage limit–Shrinkage limit is the water consent

at which the soil stop shrinkage further and attains a
constant volume.
Shrinkage limit– Shrinkage limit dyined as that
● Slope of the above curve is called flow index If and maximum water content at which further reduction in
the curve is called flow curve. water content does not cause reduction in the volume
W1 – W2 of soil sample is called shrinkage.
If =
log10N2 – log10N1 Shrinkage limit may also be dyined as the lowest water
content at which the soil is just saturated.
W – W2 Below the shrinkage limit, the soil does not remain
If = 1
N saturated. Air enters the voids of the soil.
log10 2
N1 Note :
1 Liquid Limit– Minimum water content at which soil
Flow Index α has a tendency to flow is called liquid limit water
Shear Strength
content.
(b) Cane Penetration method
All soils at liquid limit will have similar shear strength
● The cup is placed below the cone and the cone which is negligible i.e. 2.7 kN/M2
is gradually lowered so as to just touch the
surface of the soil in the cup. Plastic limit– Minimum water content at which soil is
in plastic stage is called plastic limit water content.
● The water content at which penetration is
25mm is reported as the liquid limit. At plastic limit water content, a soil rolled into a thread
of 3 mm starts to crumble.
103. The active earth pressure of a soil is
105. The water held by electro-chemical forces
proportional to (where φ is the angle of friction
existing on the soil surface is called
of the soil).
(a) Absorbed water (b) Bonded water
2  0 φ
(a) tan (45 – φ)
0
(b) tan  45 +  (c) Molecular water (d) Adsorbed water
 2
Ans. (d) : Absorbed water – Water which is held on
2  0 φ the surface of a material by electro-chemical forces; its
(c) tan  45 –  (d) tan (45 + φ)
0
physical properties are substantially different from
 2
those of absorbed water of chemically combined water
Ans. (c) : Pa = Ka σa at the same temperature and pressure.
1 106. The quantity of seepage of water through soils
Ka =
2 φ is proportional to
tan  45 + 
 2  (a) Coefficient of permeability of soil
(b) Total head loss through the soil
 φ
Ka = tan2  45 –  (c) Neither (a) nor (b)
 2 (d) Both (a) and (b)
OPSC (AEE) Exam-2019 (Paper-II) 126 YCT
 Ans. (d) ∆e change in void ratio
av = – =
Nf ∆σ change in effective stress
Q=K×H×
Nd Strain
=
Q= Discharge passing through flow channel under Stress
total head of H
K= Coefficient of permeability
H= Hydraulic head
Ng = Number of flow channels
= (Number of flow lines–1)
ND = Number of equipotential drop
= (Number of equipotential line–1)
Nf
= Shape factor
ND ● With each increment of effective stress soil become
107. When drainage is permitted under initially more densified, hence resistance to further
applied normal stress only and full primarily compression with same effective stress increment
consolidation is allowed to take place, the test is increase.
known as 111. If the failure of a finite slope occurs in which
(a) Drainged test the failure surface passes below the toe to
(b) Consolidated undrained test known as
(c) unconsolidated undrained test (a) Slope failure (b) Face failure
(d) Quick test (c) Base failure (d) Toe failure
Ans. (b) : When drainage is permitted under initially Ans. (c) : In case of finite slope, three types of failure
applied normal stress only and full primarily of surfaces occur, which has failure surface as circular
consolidation is allowed to take place, the test is arc.
known as Consolidated undrained test. (i) Base failure–In base failure, rapture surface
108. The minimum water content at which the soil passes below the toe and it occurs when soil mass
retains its liquid state and also possesses a below the toe is relatively weak of soft.
small shearing strength against following, is (ii) Face failure or slope failure–It is when failure
known as surface intersects above the toe of slope. This
(a) Liquid limit (b) Plastic limit occurs when soil close to the toe is quite strong
(c) Shrinkage limit (d) Permeability limit but in upper part soil is weak, and slope angle B is
very high.
Ans. (a) : The minimum water content at which the (iii) Toe failure–Failure surface posses through the toe
soil retains its liquid state and also possesses a small and it occurs in steep slope. Soil mass above the
shearing strength against following, is known as liquid base and below the bare is homogeneous. This is
limit. the most common mode of failure.
109. Minimum size of the particle of silt soil in 112. Rankine's theory of active earth pressure
International Classification Systems is assumes
(a) 0.02 mm (b) 0.0425 mm (a) Soil mass is homogeneous, dry and
(c) 0.002 mm (d) 0.04 mm cohesionless.
Ans. (c) : Minimum size of the particle of silt soil in (b) Ground surface is a plane which may be
International Classification Systems is 0.002 mm. horizontal or inclined.
110. The coefficient compressibility of soil, is the (c) Back of the wall is vertical and smooth.
ratio of (d) All of these
(a) Stress to strain Ans. (d) : Rankine's theory–Rankine's theory
(b) Decrease in void ratio to increase in stress considers stress in soil mass when it attains plastic
(c) Strain to stress equilibrium.
(d) Stress to settlement By plastic equilibrium we infer that every point in the
Ans. (b) : Coefficient of compressibility soil mass experience shear failure, under the effects of
∆e shear stress developed.
av = Assumption in Rankine theory.
∆σ '
● The slope of void ratio u/s effective stress curve for 1. Soil is semi-infinite, homogeneous, isotropic dry
normally consolidated soil is called coefficient of and cohesion less.
compressibility which reduces with increase in 2. Soil is in state of plastic condition at the time of
effective stress active and passive pressure generation.
OPSC (AEE) Exam-2019 (Paper-II) 127 YCT
 3. The backfill soil is horizontal. 117. For determining the moisture content of a soil
4. Back of wall is vertical and smooth. smple, the available data are : Weight of
5. Rapture surface is a planar surface which is container = 260 g, Weight of soil sample = 320
obtained by considering the plastic equilibrium of g with container and weight of soil sample
soil. (dried) = 30 g with container. The moisture
content of the soil sample is
113. If the plasticity index of a soil mass is zero, the
(a) 15% (b) 18%
soil is
(c) 20% (d) 25%
(a) Sand (b) Silt
(c) Clay (d) Clayey silt WW W – Wd
Ans. (c) : W = × 100 = × 100
Ans. (a) : Plasticity Index–Range of consistency Wd Wd
(water content) within which soil behaves as a plastic W = 320–260 = 60
material is called plasticity index Wd = 310–260 = 50
IP = W L – W P 60 – 50
W = × 100 = 20%
if plasticity index comes out hagsited their it shall be 50
reported as zero. 118. A partially saturated soil is classified as
IP Consistency (a) One phase soil (b) Two phase soil
O Non plastic (c) Three phase soil (d) Four phase soil
<f Low Plastic Ans. (c)
7 – 17 Medium plastic
< 17 Highly plastic
114. If S, L and R are the arc length, long chord and
radius of the sliding circle then the
perpendicular distance of the line of the
resultant cohesive force, is given by
S.R S.L
(a) a = (b) a =
L R
L.R S
(c) a = (d) a =
S R.L
Ans. (a)
S.R
a=
L
S = A length
L = Long chord where,
R = Radius of the sliding circle. Va = Volume of air
Vw = Volume of water
115. Settlement occurs in short time for the
Vs = Volume of soil solid
foundation resting on
V = Total volume of soil
(a) Coarse grained soil
Vv = Volume of voids
(b) Fine grained soil
(c) Mixed soil 119. The Westergaard Analysis is used for
(d) Not related to type of soil (a) Sandy soils (b) Cohesive soils
(c) Stratified soils (d) Clayey soils
Ans. (a) : Settlement occurs in short time for the
foundation resting on coarse grained soil. Ans. (c) : Westergaard theory of stress distribution in
soil deals with layered soil while beussines q equation
116. The factors leading to the failure of slopes are
can be applied to actual field problems.
classified as
(a) Increase in the shear stress 120. If L and B are the length and breadth of a
footing, e the eccentricity along the length and
(b) Decrease in the shear strength of the soil
P and Q are the axial force and bearing
(c) No shear stress in the failure plain
capacity of the soil, then, to avoid tension
(d) Both (a) and (b)
(a) BL =  1 +  (b) BL =  1 + 
P 6e Q 6e
Ans. (d) : The factors leading to the failure of slopes
Q L P L
are classified as :
(i) Increase in the shear stress P  6e  P  3e 
(c) BL = (d) BL =
Q 
1– 
Q 
1– 
(ii) Decrease in the shear strength of the soil. L L
OPSC (AEE) Exam-2019 (Paper-II) 128 YCT
 Ans. (c) 123. A phreatic line is defined as the line within a
L = Length dam section below which there are
B = Breadth (a) Positive equipotential lines
P = Axial force (b) Atmospheric pressure
Q = Bearing capacity (c) Positive hydrostatic pressure
Q  6e  (d) Negative hydrostatic pressure
amin = 1–
BXL  B  Ans. (c) : A phreatic line is the top flow line at which
water pressure is zero (only atmospheric pressure).
121. The direct shear test suffers from the following It is perpendicular to upstream face AB which is
disadvantage : 100% equipotential line its shape is approximately a
(a) Drain condition cannot be controlled parabola (correction at u/s face, rest a parabola) given
(b) Pore water pressure cannot be measured by kozney as kozney's basic parabola.
(c) Shear stress on the failure plane is not uniform
(d) The area under the shear and vertical loads
does not remain constant throughout the test
Ans. (c) : (All options are correct)
Disadvantages of direct shear test
● Drainage condition cannot be controlled and pore
water pressure can not be measured.
● Failure plane is always horizontal and pre- 124. What is the maximum settlement allowed for
determined which may not be the weakest plane. isolated foundations for RCC structures on
● Non-uniform stress distribution on shear plane sand?
failure starts at edge and progresses towards centre. (a) 40 mm (b) 50 mm
● Area of specimen under normal and shear does not (c) 75 mm (d) 00 mm
remain constant during the test. Hence calculation Ans. (b) : 50 mm is the maximum settlement allowed
of normal and shear stresses are done on the basis for isolated foundation for RCC structures on sand.
of nominal area (original area) which is not correct. 125. The maximum gross pressure the soil can carry
● Direction of principal plane are not known at every safely without shear failure is called
stage of the test it is only when ....... failure (a) Net safe bearing capacity
envelope is known that direction of principal stress (b) Safe bearing capacity
will be known.
(c) Gross safe bearing capacity
122. The equation τ = C + σ tan φ is given by (d) Net ultimate bearing capacity
(a) Rankine (b) Terzaghi Ans. (b or c) : The maximum gross pressure the soil
(c) Mohr (d) Coulomb can carry safely without shear failure is called safe
Ans. (d) bearing capacity and gross safe bearing capacity.
τf = C + σ tan φ 126. For determining the ultimate bearing capacity
τf = Shear strength of soil of soil, the recommended size of a square
σ = Normal stress on the plane of rapture bearing plate to be used in load plate test
φ = Angle of internal friction should be 30 to 75 cm square with a minimum
initially coulomb believed that σ = total stress. Later thickness of
after the knowledge of effective stress it was realised (a) 10 mm (b) 16 mm
that 'σ' is actually the effective stress, hence new (c) 25 mm (d) 32 mm
definition of shear stress come out. Ans. (c) : Circular or square bearing plates of weld
τf = c' + σ tan φ' steels are used of thickness not less than 25 mm and
varying in size from 300 mm to 750 mm (300, 450,
σ = (σ–u) = effective stress
600, 750) are used smaller size plates are used in dense
σ = Total stress or stiff soils where as larger plates are being used in
u = Pore water pressure loose or soft soil.
c' and φ' are effective stress shear parameters 127. A cohesionless soil attains quick condition due
Note : to
c and φ are known as shear strength parameters (a) Head causing upward flow is equal to the
of soil. stress from the top
c and φ are not the inherent properties of soil. (b) When the vertical load increases
These are related to the type of test and the condition (c) When water volume is reduced
under ........ they are measured. (d) None of these
OPSC (AEE) Exam-2019 (Paper-II) 129 YCT
 Ans. (a) Number of 25 25 25 25
blows per
● Quick sand is not a type of sand, it is hydraulic layer (R)
condition which exist in cohesion less soil mass, when
Number of 3 5 3 5
vertical effective pressure in it reduced to zero. Layer (S)
● If seepage pressure is such that it equals the Volume of 944 944 1000 1000
submerged weight of soil mass, effective vertical mould
pressure reduces to zero. In such condition (CC)
coherioncess soil looses all its shear strength and In India, the Indian Standard specification are followed in
soil particles have the tendency to lift up along with laboratory testing.
the flowing water. 131. What is the Ruling Design speed in plain
● It is also referred as quick sand condition, quick terrain for State Highways?
condition, piping and soil boiling condition. (a) 120 KMPH (b) 100 KMPH
Note : (c) 80 KMPH (d) 65 KMPH
Quick sand condition is found only in fine sand or Ans. (b) : 100 KMPH is the ruling design speed in
coarse silt and not in clay, gravel, coarse sand. plain terrain for state high.
Effective stresses throughout the soil become zero. 132. The basic equation relating to speed of vehicle
x G –1 (V), the radius of curve (R), the super-elevation
icr = sub = s
xw 1+ e (e) and the coefficient friction (µ) is
if i ≥ icr quick sand condition will occur. V2 R2
To avoid quick sand condition, hydraulic gradient must (a) e + µ = (b) e + µ =
127R 127R
be less than critical hydraulic gradient 2
V R.V 2
i < icr (c) e – µ = (d) e – µ =
127R 127
i
fos = c Ans. (a)
i
(factor of safety provided) e + µ = V2/127 R
V = speed of vehicle
128. The length/diameter ratio of cylindrical
specimens used in triaxial test, is generally R = Radius of curve
(a) 1.0 (b) 1.5 e' = super elevation
(c) 2.0 (d) 2.5 µ = coefficient of friction
Ans. (c and d) : length/diameter ratio of cylindrical 133. What is the ruling minimum radius for
specimens used in triaxial test, is generally 1.5 and 2 National Highway in plain terrain?
129. In the Standard Penetration Test what is the (a) 155 m (b) 230 m
driving depth of the sample for which the (c) 360 m (d) 400 m
number of hammer blows are counted? Ans. (c) : 230 m is the ruling minimum radius for
(a) 100 mm (b) 150 mm National Highway in plain terrain.
(c) 200 mm (d) 300 mm 134. What is the radius of the curve at 80 KMPH
Ans. (b) : In the Standard Penetration Test 300 mm beyond which no super-elevation is needed for
depth of the sample is driving for which the number of the value of 2.5% camber?
hammer blows are counted. (a) 950 m (b) 1100 m
130. What is the mass of the hammer in modified (c) 1400 m (d) 1800 m
proctor test? Ans. (b) : 1100 m is the radius of the curve at 80
(a) 2.5 kg (b) 3.93 kg KMPH beyond which no super-elevation is needed for
(c) 4.89 kg (d) 6.1 kg the value of 2.5% camber
Ans. (c) : 4.89 kg is the mass of the hammer in 135. What is the type of curve adopted for the
modified proctor test. trnsition curve in India?
Indian Standard
Equivalent of Proctor
(a) Clothoid (b) Cubic parabola
Test (c) Cubic spiral (d) Lemniscate
Standard Modified Light Heavy Ans. (a) : Clothoid is the type of curve adopted for the
Proctor Proctor Compact Compaction trnsition curve in India
Test Test -ion Test Test
Weight of 2.495 4.54 2.6 4.9 136. The wall constructed for the stability of an
hammer excavated portion of a road on the hill side, is
P(kg) known as
Fall of 304.8 457.2 310 450
hammer
(a) Retaining wall (b) Breast wall
Q(mm) (c) Parapet wall (d) All of these
OPSC (AEE) Exam-2019 (Paper-II) 130 YCT
 Ans. (b) : The wall constructed for the stability of an 143. What are the intersections known as where
excavated portion of a road on the hill side, is known traffic is controlled on the minor road by
as Breast wall. STOP of GIVEWAY signs and markings?
Retaining wall– Retaining walls are relatively rigid (a) Rotary intersection
walls used for supporting soil laterally so that it can be (b) T junction
retained at different levels on the sides. (c) Priority intersections
Parapet–A parapet is a barrier which is an extension (d) Staggered junction
of the wall at the edge of a roof, terrace, balcony, Ans. (c) : Priority intersections known as where traffic
workway or other structure. is controlled on the minor road by STOP of
137. What is the extra width of pavement in hairpin GIVEWAY signs and markings.
bend for two lane carriageways? 144. What is the type of interchange when two high
(a) 2.0 m (b) 1.5 m volume and high speed roads intersect each
(c) 1.2 m (d) 0.9 m other where through traffic on both roads are
Ans. (b) : 1.5 m is the extra width of pavement in unimpeded?
hairpin bend for two lane carriageways. (a) Diamond interchange
138. What is the reason for providing adequately (b) Clover leaf interchange
long transitional curve between two reverse (c) Rotary interchange
curves? (d) Directional interchange
(a) Intervisibility of the curve Ans. (b) : Clover leaf interchange is the type of
(b) Accommodating sight distance between two interchange when two high volume and high speed
curves roads intersect each other where through traffic on
(c) For effective drainage both roads are unimpeded.
(d) For accommodating super-elevation run off Diamond interchange– A diverging diamond
interchange (DDI), also called a double crossever
Ans. (d) : For accommodating super elevation runoff.
diamond interchange (DCD), is a type of diamond
139. What is the exceptional gradient in plain interchange in which the two directions of traffic on
terrain? the non-freeway road cross to the opposite side on both
(a) 1 in 20 (b) 1 in 15 sides of the bridge at the freeway.
(c) 1 in 12 (d) 1 in 16.7 Directional interchange is mainly characterized by
Ans. (b) : 1 in 16.7 is the exceptional gradient in plain providing free flow movement in all directions with a
tervain. small duration from the direction of travel (direct
140. For the designing of valley curves, the height of connections), as opposed to loop ramps, which require
headlight above the road surface is assumed at large deviation from the original trajectory.
(a) 0.15 m above the road surface 145. What is the Vehicle Damaged Factor for
(b) 1.2 m above the road surface vehicle having single axle with dual wheel type
(c) 0.75 m above the road surface weighs 118 kN when the weight of the standard
(d) 1.5 m above the road surface axle of same configuration of wheels is 80 kN?
(a) 3.535 (b) 3.20
Ans. (c) : For the designing of valley curves, the
(c) 4.733 (d) 6.515
height of headlight above the road surface is assumed
at 0.75 m above the road surface. Ans. (c) : 4.733 is the Vehicle Damaged Factor for
vehicle having single axle with dual wheel type weighs
141. What is the perception and break reaction time
118 kN when the weight of the standard axle of same
taken together in India?
configuration of wheels is 80 kN.
(a) 1.5 seconds (b) 2.0 seconds
146. Calculate cumulative traffic in million
(c) 2.5 seconds (d) 3.5 seconds
standard axle for per day commercial vehicles
Ans. (c) : 2.5 seconds is the perception and break is 000 in the year of completion for a two lane
reaction time taken together in India two way road, VDF of 3.9, growth rate of 5%
142. What is the level of driver's eye and the object and design period of 5 years.
height between two points for the Safe (a) 15.36 msa (b) 23.53 msa
Stopping Distance? (c) 23.03 msa (d) 30.73 msa
(a) 1.2 m and 0.15 m (b) 0.9 m and 0.15 m Ans. (d) : 30.73 msa
(c) 0.75 m and 0.3 m (d) 0.90 m and 0.6 m
147. Which layer in flexible pavement crust
Ans. (a) : 1.2 m and 0.15 m is the level of driver's eye functions as a drainage cum strength layer?
and the object height between two points for the Safe (a) Wet Mix Macadam
Stopping Distance (b) Water Bound Macadam
OPSC (AEE) Exam-2019 (Paper-II) 131 YCT
 (c) Sub-grade concrete slab due to temperature and subgrade
(d) Granular Sub-base moisture variation.
Ans. (d) : Granular Sub-base layer in flexible The maximum spacing between expansion joint is
pavement crust functions as a drainage cum strength 140m.
Water Bound Macadam (WBM) is a dense and Warping joint–Warping joints are provided along the
compact course of a road pavement composed of stone longitudinal direction to prevent warping of the
aggregates held together by a film consisting of gravel concrete slab due to temperature and subgrade
or screenings with a minimum amount of water. moisture variation.
148. What test in undertaken to evaluate the 151. Dowel bars in the plain jointed rigid pavement
toughness of stones used in road projects? are provided
(a) Los Angeles Abrasion Test (a) Providing continuity to slab
(b) Aggregate impact Test (b) Preventing differential settlement
(c) Aggregate Crushing Test (c) For transferring of load from one panel to the
(d) Soundness Test next
Ans. (b) : Los Angeles Abrasion Test– It is the (d) Increasing flexural strength of concrete
resistance of an aggregate to wear it measures hardness Ans. (b) : Dowel bars in the plain jointed rigid
of the aggregate. pavement are provided preventing differential
Impact Value–It measure the resistance of aggregate settlement.
under impact loading. It measures toughness of 152. What is the shape of the cautionary signs?
aggregate. (a) Octagonal (b) Circular
Shear resistance– It is the resistance of an aggregate (c) Rectangular (d) Triangle
to shear loading. Ans. (d) : The shape of the cautionary signs triangle.
Gushing resistance– It is the resistance of an
153. Which of the following represents hardest
aggregate under static loading. It measures strength of
grade of bitumen?
the aggregate. It is the measure of strength of particle
under static loading. 30 40
(a) (b)
Soundness–The soundness test on aggregate is carried 40 60
out to learn the resistance of aggregate to weathering 60 80
(c) (d)
action like thawing, freezing, alternate wetting and 70 100
drying in normal conditions or the effect of hydration Ans. (a)
of cement and water.
30
149. At a road junction, 16 cross conflict points are represents hardest grade of bitumen.
40
severe, if
154. In highway construction on super-elevated
(a) Both are one-way roads
curves, the rolling shall proceed from
(b) One is two-way road and other is one-way
(a) Sides towards the centre
road
(c) Both are two-way roads (b) Centre towards the sides
(d) Both are four lane roads (c) Lower edge towards the upper edge
(d) Upper edge towards the lower edge
Ans. (c) : At a road junction, 16 cross conflict points
are severe, if both are two-way roads. Ans. (c) : In highway construction on super-elevated
150. Tie bars in cement concrete pavements are at curves, the rolling shall proceed from lower edge
(a) Expansion joints (b) Contraction joints towards the upper edge.
(c) Warping joints (d) Longitudinal joints 155. What is the value of rigidity factor for an
Ans. (d) : Longitudinal joints is the interface between average type pressure greater than 7 kg/cm2?
two adjacent and ............ HMA mats improperly (a) 1.0 (b) > 1
constructed longitudinal joints can cause premature (c) < 1 (d) 0
deterioration of multilane HMA pavement in the form Ans. (c)
of cracking. < 1 the value of rigidity factor for an average type
Contraction joints– Contraction joints are provided pressure greater than 7 kg/cm2
along the transverse direction to take care of the
contraction of concrete slab due to its natural shrinkage 156. When the path travelled along the road surface
is more than the circumferential movement of
When reinforcement is not provided, the maximum
spacing between contraction joint is taken as 4.5 m. the wheels due to their rotation is called
Expansion joint– Expansion joints are provided along (a) Slip (b) Skid
the transverse direction to allow movement of the (c) Rolling (d) Traction
OPSC (AEE) Exam-2019 (Paper-II) 132 YCT
 Ans. (b) : When the path travelled along the road (a) Only (b) Both 1 and 3
surface is more than the circumferential movement of (c) Only 3 (d) Both 2 and 3
the wheels due to their rotation is called skid. Ans. (d) : Principle of surveying :
Slip– slip occurs when the wheel revolves more than the (1) Working from whole to part to prevent the
corresponding longitudinal movement along the road accumulation of errors.
157. Softening point of bitumen to be used for read (2) Location of a point w.r.t. to atleast two well
construction at a place where maximum defined point for better accuracy of fix.
temperature is 400C should be 162. What is the use of two vertical cross hairs?
(a) Less than 400C (b) Greater than 400C (a) To check plumpness of vertical staffs
(c) Equal to 400C (d) None of these (b) Required for Tachometry
Ans. (b) : Softening point of bitumen to be used for (c) Required for levelling operations
read construction at a place where maximum (d) None of these
temperature is 400C should be greater than 400C. Ans. (c) : Horizontal cross hairs : for vertical distance
158. In CBR test the value of CBR is calculated at Vertical cross hairs : for accurate bisection of vertical
(a) 2.5 mm penetration only ranging rods.
(b) 5.0 mm penetration only 163. Type of lens used for aberration free images
(c) 7.5 mm penetration only (a) Convex (b) Plano convex
(d) Both 2.5 mm and 5.0 mm penetrations (c) Plano concave (d) Compound lens
Ans. (d) : Generally, the CBR value at 2.5mm Ans. (d) : Aberrations are optical defects of telescopes
penetration is higher than that at 5.0 mm penetration it can be eliminated by using compound lens.
and reported as CBR value of material.
164. The maximum tolerance in a 20m chain is
However, if the CBR value obtained from test at 5.0
mm penetration is higher than that at 2.5 mm, then the (a) ± 1 mm (b) ± 2 mm
test is to be repeated for checking if the check test (c) ± 3 mm (d) ± 5 mm
again gives similar test results, the higher value Ans. (d) : A chain tested under 80N pull at 200C
obtained at 5.0 mm penetration is reported as CBR should be within the following units.
value. The average CBR value of three test specimens 20m → ± 5 mm
is reported to first decimal as the CBR value of 30m → ± 8 mm
material.
165. Dimensions of the embankment was measured
159. If the radii of a compound curve and a reverse with 20 m chain and the volume was calculated
curve are respectively the same, the length of as 400 cum. It was then found that the chain
common tangent? was 8 cm too long. The true volume of the
(a) Of compound curve will be more embankment is
(b) Of reverse curve will be more (a) 395.24 (b) 403.2
(c) Of both curves will be equal (c) 404.82 (d) 40.6
(d) None of these Ans. (c)
Ans. (c) : If the radii of a compound curve and a L = 20 m
reverse curve are respectively the same, the length of 8 cm too long
common tangent of both curves will be equal. L' = 20 m + 0.08 m = 20.08 m
160. To compensate the loss of tractive force of r' = 400 cumec
vehicles along curves of radius R, the 3 3

v = v '   = 400 × 
L' 20.08 
percentage reduction of gradient, is 
 
L  20 
50 75
(a)
R
(b)
R ≈ 400 (1 + 3e )
100 125  0.08  
(c)
R
(d)
R ≈ 400 1 + 3 ×  
  20  
Ans. (b) : To compensate the loss of tractive force of
400 × 3 × 8
vehicles along curves of radius R, the percentage ≈ 400 +
reduction of gradient, is 75/R. 20 × 100
4 × 3× 8
161. Fundamental principle of surveying : ≈ 400 +
1. Working from part to whole. 20
2. Working from whole to part. 24
≈ 400 +  400 + 408
3. Establishing any point by at least two 5
independent measurements. ≈ 404.8 cum

OPSC (AEE) Exam-2019 (Paper-II) 133 YCT

166. The correction of sag is always Note :
(a) Additive (1) Level Tube/spirit level
(b) Substractive ● level tube/spirit level is used for levelling the
(c) Zero plane table.
(d) Depends on the atmospheric temperature ● it is placed on the board in two positions mutually
at right angles and the bubbles is centered in arch
Ans. (b) : error due to sag is cumulative and always position to make the bard horizontal.
(+ve) (2) Trough compass
∴ correction is always –ve Generally it is 15 cm long and provided to plot the
167. The permissible error in changing for magnetic meridian (N–S direction) to facilitate
measurement with chain on rough or Hilly orientation of the plane table in the magnetic
ground is meridian.
(a) 1 in 100 (b) 1 in 250 (3) Alidade
(c) 1 in 500 (d) 1 in 1000 An alidade is straight–edge, ruler having some
sighting device, it is used for sighting the object
Ans. (b) : The permissible error in changing for
and drawing the line.
measurement with chain on rough or Hilly ground is 1
Now a days telescopic alidade are also much is
in 250.
use in place of plane alidade when the point is too
168. The angle of intersection of plane mirrors of an high or low are to be sighted, the accuracy and the
optical square is range are considerably increased by providing a
(a) 300 (b) 450 telescopic alidade.
(c) 600 (d) 900 171. Which line is tangential to the level line at a
Ans. (b) point?
(a) Datum line (b) Vertical line
(c) Horizontal line (d) Plumb line
Ans. (c) : Radiation is adopted when points to be
surveyed are intervisible and accessible
(1) Radiation
● In this method the instrument is setup at station
and rays are drawn to various stations which are to
be rotted
This method is suitable only when the area to be
surveyed is small and all the required stations to
169. The process of rotating the telescope about the be plotted are clearly visible and accessible from
vertical axis in horizontal plane is known as the instrument station.
(a) Transiting (b) Reversing (2) Traversing :
(c) Swinging (d) Centring ● This method is similar to compass or theodolite
Ans. (c) : The process of rotating the telescope about traversing.
the vertical axis in horizontal plane is know as ● This method is most suited when a narrow strip of
swinging. terrain is to be surveyed e.g. survey of roads,
railways etc.
170. The instrument used for accurate centring of
(3) Intersection :
plane table surveying is
● In this method two stations are selected such that
(a) Spirit level (b) Alidade
all the other stations to be plotted are visible from
(c) Plumbing fork (d) Trough compass these.
Ans. (c) : Plumbing fork–It is used for the centering ● A line joining these two stations is called base
of the table over the station line. The length of this base line is measured very
accurately.
(4) Resection :
This method of orientation is employed when the
plane table occupies a position not yet plotted on
the drawing sheet.
Resection can be dyined as the process of locating
the instrument station occupied by the plane table
by drawing rays from the stations whose positions
are already plotted on the drawing sheet.
OPSC (AEE) Exam-2019 (Paper-II) 134 YCT
172. Which line is tangential to the level line at a 176. A surface whose elevation is known or assumed
point? is known as
(a) Datum line (a) Level surface
(b) Vertical line (b) Reduced level
(c) Horizontal line (c) Datum surface
(d) Plumb line (d) Horizontal plane
Ans. (c) Ans. (c) : A surface whose elevation is known or
assumed is known as Datum surface.
177. The degree of precision required for
establishment of bench marks (K is the
distance in kilometres) :
(a) ± 100√K (b) ± 24√K
(c) ± 12√K (d) ± 4√K
Ans. (d) : The degree of precision required for
establishment of bench marks is ± 4√K.
173. Which term is used for the surface to which 178. In the method of repetition for measuring
elevations are referred? horizontal angles, to rotate the instrument
without changing the reading.
(a) Level surface
(a) Lower clamp screw is tightened and upper
(b) Level line
clamp is loosened.
(c) Horizontal plane (b) Lower clamp screw is loosened and upper
(d) Datum clamp is tightened.
Ans. (d) : Datum is an imaginary surface w.r.t. to (c) Any one of the clamp screw is loosened.
which elevation are expressed. (d) Both the clamp screw are loosened.
174. If the RL of the BM is 100.00m, the back sight Ans. (b) : In the method of repetition for measuring
is 1.435 m and the fore sight is 1.620 m, the RL horizontal angles, to rotate the instrument without
of the forward station is changing the reading lower clamp screw is loosened
(a) 98.380 (b) 99.815 and upper clamp is tightened.
(c) 98.565 (d) 100.85 179. The adjustment of horizontal cross hair is
Ans. (b) required particularly when the instrument is
used for
(a) Levelling
(b) Prolonging a straight line
(c) Measurement of horizontal angle
(d) All of these
Ans. (a) : In levelling the level of a point on ground is
calculated using staff. The reading of staff is read by
using horizontal cross-hairs. So the adjustment of
RLP = 100 + 1.435 – 1.620 horizontal cross hair is required in levelling.
= 99.815 m 180. In electronic theodolite, the horizontal and
175. For a close travers, sum of latitudes is –0.46 m vertical circles made of
and sum of departures is –0.22 m. What is the (a) Glass having specially coded graduations
closing error? read by photo diodes
(a) 0.475 (b) 0.226 (b) Special metal read electronically
(c) – 0.680 (d) – 0.240 (c) Steel plates with specially coded graduations
Ans. (a) (d) Plastic with coded graduations
ΣL = – 0.46M Ans. (b) : Angles are measured by means of electro-
ΣD = – 0.22M optical scanning of extremely precise digital bar codes
imprinted on rotating glass cylinders or dises within
e = ΣL2 + ΣD 2 the instruments in most of the modern total stations of
= 0.475M digital theodolites.
OPSC (AEE) Exam-2019 (Paper-II) 135 YCT
 Odisha Staff Selection Commission
Junior Engineer (Civil)
Exam- 2019 (Shift-I)
1. Select the CORRECT sequence of given steps Flemish bond : When in a course header sand stretcher
in the manufacturing process of Bricks. is alternate laying then it is called Flemish bond.
1. Moulding of Bricks English bond : The type of bond consists of alternate
2. Preparation of Brick Earth course of headers and stretchers bond are called as
3. Burning of Bricks English bond.
4. Drying of Bricks 4. During a motion of fluid, the path traced by the
(a) 2 - 1 - 4 - 3 (b) 1 - 2 - 4 - 3 fluid can be stream lines, streak lines or path
(c) 4 - 1 - 2 - 3 (d) 3 - 4 - 1 - 2 lines. These three may coincide in the case of
(a) turbulent flow only (b) non-uniform flow
Ans. (a):Correct sequence of manufacturing of brick–
(c) steady flow (d) unsteady flow
Preparation of brick earth
↓ Ans. (c) : Stream line – A stream line is an imaginary
line drawn in a flow field such that a tangent drawn at
Moulding of brick
any point on this line represents the direction of velocity
↓ vector at that point.
Drying of brick Streak line- When a dye is injected in a liquid or smoke
↓ in a gas so as to trace the subsequent motion of fluid
Burning of brick particles passing a fixed point the path followed by the
2. Read the following statements and choose the dye or smoke is called the streak line.
Correct option. Path line– A path line may be defined as the line traced
(i) A freshly cut stone carries some natural by a single fluid particle as it moves over a period of
moisture known as Quarry Sap, which makes time.
it soft and workable. These three may coincide in the case of steady flow.
(ii) To make the stone weather resistant and 5. The general form of few empirical formulae for
useful for construction, a process called estimating flood discharge is Q = CAn. If in the
dressing is done that removes the Quarry Sap expression Q = Flood discharge and A =
from the stone. Catchment Area, then C and n respectively
(a) (i) is false and (ii) is false denotes
(b) (i) is true and (ii) is false
(a) Flood Index and Flood Coefficient
(c) (i) is true and (ii) is true
(b) Water Coefficient and Water Index
(d) (i) is false and (ii) is true
(c) Flood Coefficient and Flood Index
Ans. (b) : Quarry sap– A freshly cut stone carries
some natural moisture known as Quarry Sap, which (d) Flood Coefficient and Flood Index
makes it soft and workable. Ans. (c) : Several empirical formula have been
Dressing– Giving the required shapes to the stone is developed for estimating the maximum or peak value of
called dressing. the flood discharge.
Seasoning– To removal of quarry sap from stone. Q = CA n
3. Generally in a brickwork, the type of bond in Where, Q = Flood discharge
which all bricks are laid as headers on the faces
of the wall is known as A = Catchment area
(a) English Bond (b) Stretcher Bond C = Flood coefficient
(c) Header Bond (d) Flemish Bond n = Flood index
Ans. (c) : Header bond–In this type of bond, all the 6. The maximum daily demand at a water
bricks are laid as headers towards the face of the wall. purification plant has been estimated as 12
This is used for one brick thick walls and also useful for million litres per day. Find the length of the
curved wall construction. sedimentation tank required if a detention
Stretcher bond–In this bond, all the bricks are laid with period of 5 hours and the velocity of flow of 20
their lengths parallel to the longitudinal direction of the cm/minute is assumed.
wall. This bond is only useful for half brick (10 cm) (a) 60 m (b) 50 m
thick partition wall. (c) 80 m (d) 70 m
OSSC JE Exam-2019 (Shift-I) 136 YCT
Ans. (a) : Given, Ans. (a) : The contour lines are perpendicular to the
Detention time (td) = 5 hrs. ridge line as well as valley line.
Flow velocity (Vf) = 20 cm/min • The highest contours along ridges and the lowest
= 0.20 × 60 m/hr contours is valley always go in pairs.
=12 m/hr • Contours do not pass through permanent structure
Length of tank (L) such as buildings.
Detention time (td) = • Contours deflect uphill at valley lines and downhill at
Flow velocity (Vf )
ridge lines.
L
td = • Contour lines cross a watershed or ridge line at right
Vf angles. They from curves of U-shape round it with
L=t d × Vf = 5 ×12 = 60 m the concave side of the curve towards the higher
ground.
7. For determining the strength of a structure,
under the limit state of collapse as per IS 456: • Contour lines cross a valley line at right angles. They
2000, the partial safety factors for Concrete from sharp curves of V-shape across it with convex
and Steel respectively are side of the curve towards the higher ground.
(a) 2.5 and 2.15 (b) 1.5 and 1.15 10. If to, tp and tm are optimistic, pessimistic and
(c) 1.15 and 1.5 (d) 2.15 and 2.5 most likely time estimates of an activity
respectively, the expected time (te) of the
Ans. (b) : For determining the strength of a structure,
activity will be one sixth of
under the limit state of collapse as per IS 456: 2000, the
partial safety factors for Concrete and Steel respectively (a) (3to+ tm + tp) (b) (tm+ 3to + tp)
are 1.5 and 1.15. (c) (tm+ 4to + tp) (d) (to+ 4tm + tp)
8. In a RCC column, as per IS 456: 2000, the  T + 4Tm + TP 
Ans. (d) : Expected time, Te =  o 
maximum pitch of the lateral ties (transverse  6 
reinforcement) should be equal to what among
Te = Expected time
the following options?
(i) Least lateral dimension of the compression Tm = Mean time
members. TP = Pessimistic time
(ii) sixteen times the smallest diameter of the T0 = Optimistic time
longitudinal reinforcement bar to be tied. 11. The minimum time in which activity can be
(iii) 300 mm. finished but leads to the maximum cost of
(a) Always (i) project is known as
(b) Always (iii) (a) Normal Time (b) Standard Time
(c) Maximum of (i), (ii) and (iii) (c) Crash Time (d) Slow Time
(d) Least of (i), (ii) and (iii) Ans. (c) : Crash time (tc)–Crash time is the minimum
Ans. (d) : Design criteria for column– possible time in which an activity can be completed by
Minimum diameter of bar in column = 12 mm employing extra resources crash time is that time
beyond which the activity cannot be shortened by any
Maximum distance between longitudinal bar
amount of increase in resources.
= 300 mm
Normal time (tn)–Normal time is the standard time that
Traverse reinforcement (Ties)
an estimator would usually allow for an activity.
 1
φ = max  4 φmain
12. According to ASTM standards, Coarse
aggregates are identified when they are
6mm
retained on
φmain = dia of main longitudinal bar (a) 10 mm sieve (b) 8.75 mm sieve
φ = dia of tie bar (c) 4.75 mm sieve (d) 6 mm sieve
Pitch for ties (P) Ans. (c) : Sieve analysis (for coarse grained soils)–
least lateral dimension • The fraction retained on 4.75 mm sieve is called the
φ = Min. 16φ main min gravel fraction which is subjected to coarse sieve
300 mm analysis.
φ mm = min dia of main longitudinal bar. • The material passing through 4.75 mm sieve is
9. A line on a map joining points of equal further subjected to fine sieve analysis if it is sand or
elevation is a contour line. These lines intersect to a combined sieve and sedimentation analysis if silt
the ridge lines usually at what angle? and clay sizes are also present.
(a) 90º (b) 0º • Sieve analysis test of sand is carried out to find its
(c) 30º (d) 45º fineness modulus.

OSSC JE Exam-2019 (Shift-I) 137 YCT

13. Viscosity is a measure of the reluctance of the 16. As per IS: 456 - 2000, the lap length including
fluid to yield to shear when the fluid is in anchorage value of hooks for direct tension in
motion. The dimensions of dynamic viscosity bars shall be (Given: Ld = Development Length
and kinematic viscosity respectively are of bars and φ is the nominal diameter of the
(symbols/notations carry their usual meaning) bar.)
(a) FL-1 T-1, L2 T-1 (b) L2 T-1, FL2 T (a) Ld or 18 φ whichever is greater
-2
(c) FL T, L T 2 -1
(d) FL-2 T, FL-1 (b) 2Ld or 30 φ whichever is greater
(c) 0.50Ld or 16 φ whichever is greater
Ans. (c) : Unit of dynamic viscosity (µ) –
(d) Ld or 16 φ whichever is greater
N −S
= [In S.I. unit] Ans. (b) : According to IS 456-2000 and clause No.
m2 26.2.1 ⇒
Dimension =
[ F] × [ T ] = F1L−2T1  • Lap length in compression shall not be less than 24φ
 L2   
• Lap length in flexural tension shall not be less than
30φ and Ld
Units of kinematic viscosity (ν)
• Lap length in direct tension shall not be less than
m2 30φ and 2Ld
= [In S.I. unit]
s Where Ld = development length, φ = diameter of bars
 L2  17. The process of cutting off the excess wet
Dimension = =  L2 T −1 
[T]  concrete to bring the top surface of a slab to the
proper grade and smoothness is known as
14. If base period for a particular crop is 50 days (a) trowelling (b) floating
and the duty of the canal is 500 hectares per (c) screeding (d) finishing
cumec, then the depth of water will be Ans. (c) :
(a) 86.4 cm (b) 8.64 cm • Screeding as the process of cutting off the excess
(c) 0.864 cm (d) 864 cm wet concrete to bring the top surface of slab to the
proper grade and smoothness or to removing humps
Ans. (a) : Given,
and hollows of uniform concrete surface.
Base period (B) = 50 days • Floating consists in removing the irregularities on
Duty of the canal (D) = 500 hect./cumec the surface of concrete which are left after screeding.
8.64.B This is done by wooden float.
∆= • Troweling produces a hand, smooth, dense, surface
D
and should be done immediately after floating.
864 × 50
Delta(∆) = = 86.40 cm 18. Which of the following options is a project
500 planning technique that considers the
15. Which of the following options is a statically uncertainty factor of various activities?
indeterminate structure? (a) Both CPM and PERT
(a) Fixed Beam (b) PERT
(b) Simply Supported Beam (c) CPM
(c) Three-Hinged Arch (d) Neither CPM nor PERT
(d) Single Overhanging Beam Ans. (b) : PERT (Program Evaluation and Review
Ans. (a) : Technique).
(i) Three time estimates are made.
(ii) It is based on probabilistic approach.
(iii) Useful for research and development oriented
problem.
(iv) Each activity follow β distribution.
(v) Uncertainties in the completion time.
Reaction = 4 = Re (vi) Event oriented uncertainty and certainty of any
activity as any project is determine by its variance.
Equilibrium Equation = 2
⇒ More is the variance, more is the uncertainty.
Ds =Re – E
=4–2 19. Identify the test to check the presence of poorly
weathering calcium carbonate in sandstone as a
=2
building material among the given options.
• A fixed beam loaded transversely is statically (i) Crushing Test (ii) Smith's Test
indeterminate by 2 degree (iii) Acid Test (iv) Slump Test
OSSC JE Exam-2019 (Shift-I) 138 YCT
 (a) only (iii) (b) only (iv) Ans. (b) : The members of a pin-jointed truss are
(c) Both (i) and (ii) (d) Both (iii) and (iv) usually subjected to bending when the loads are applied
Ans. (a) : Acid test–This test is carried out to at intermediate point on the member.
understand the presence of calcium carbonate in 24. When an inclined load is acting on a beam,
building stone. then the inclined load is resolved into two
• A sample of stone weighing about 50 to 100 g is components namely vertical and horizontal.
taken. Read the following statements and choose the
• It is placed in a solution of hydrophobic acid having CORRECT option with respect to effects of
strength of one percent and is kept there for seven these components.
days. (i) The Vertical Component will cause axial
• A good building stone maintains its sharp edges and thrust in the beam.
keeps its surface free from powder at the end of this (ii) The Horizontal Component will cause Shear
period. Force and Bending Moment in the beam.
• This test is usually carried out on sandstones. (a) (i) is FALSE and (ii) is FALSE
20. For a steel construction as per CPWD (b) (i) is FALSE and (ii) is TRUE
specifications, Rivets shall be made from rivet (c) (i) is TRUE and (ii) is TRUE
bars of mild steel according to which of the
(d) (i) is TRUE and (ii) is FALSE
following IS Codes?
(a) IS 814 (b) IS 228 Ans. (a) : When an inclined load is acting on a beam,
(c) IS 1367 (d) IS 1148 then the inclined load is resolved into two components
namely vertical and horizontal.
Ans. (d) : For a steel construction as per CPWD
specifications, Rivets shall be made from rivet bars of Effect of these components on a beam–
mild steel according to IS 1148. • The Vertical Component will cause Shear Force and
21. The percentage of estimated cost of building Bending Moment in the beam.
works for sanitary and water works are usually • The Horizontal Component will cause axial thrust in
provided as the beam.
(a) 4% to 6% (b) 50% to 60% 25. Which of the following instruments for
(c) 8% to 10% (d) 22% to 42% chaining in surveying, is used as a centering aid
Ans. (c) : Work % of Estimate in plane table, compass and theodolites?
Contingencies charges 3-5% (a) Plumb Bob (b) Ranging Rod
Work charge establishment 1.5-2% (c) Tape (d) Offset Rod
Sanitary and water supply charge 8-10% Ans. (a) : For chaining in surveying plumb bob is used
Electrification 8% as a centering aid in plane table, compass and
Labour charge 25% theodolites.
Contractor charge 10% Plum bob– It is suspended from the vertical axis of the
22. In weirs and barrages, for the oscillating jump level, to be centered over the station from where
of flows, the Froude number usually lies in the measurement are made.
range of 26. Match the following different Gauges adopted
(a) 10 to 11 (b) 4.6 to 9.0 in India in List I with their corresponding
(c) 1 to 2.0 (d) 2.5 to 4.5 Widths in List II.
Ans. (d) : When the Froude number lies in the range of List I List II
2.5 to 4.5, the jump is troublesome and oscillating as in
(A) Broad Gauge (i) 1000 mm
case of weirs and barrages there is on oscillating jet
entering the jump bottom to the surface and back again (B) Meter Gauge (ii) 1676 mm
which produces a large wave of irregular period doing (C) Narrow Gauge (iii) 310 mm
ultimate damage. (iv) 762 mm
23. The members of a pin-jointed truss are usually (a) (A) - (iv), (B) - (ii), (C) - (iii)
subjected to bending when
(b) (A) - (i), (B) - (ii), (C) - (iii)
(a) the material of the truss is brittle
(c) (A) - (iii), (B) - (ii), (C) - (i)
(b) the loads are applied at intermediate point on
the member (d) (A) - (ii), (B) - (i), (C) - (iv)
(c) the truss is of span less than 3m and loads are Ans. (d) : Clear horizontal distance between inner face
applied at the joints of two rails forming the track at the top is Gauge
(d) the supports of the truss are hinged distance.

OSSC JE Exam-2019 (Shift-I) 139 YCT

Different Gauge– 30. The sewer pipes usually have to be designed
and checked for
Distance
Gauge (a) only maximum flow
between rails
(b) only minimum flow
Broad Gauge 1.676 m (c) both maximum and minimum flow
Metre Gauge 1.0 m (d) only average flow
Narrow Gauge 0.762 m Ans. (c) : Sewers are designed to carry maximum
Light Gauge (feather track) 0.610 m hourly discharge and are checked at minimum hourly
discharge for development of self cleaning velocity.
Standard Gauge 1.435 m
(used in Delhi metro) 31. Which of the following general guidelines is
INCORRECT while planning a Staircase for a
27. The versatile equipment used for clearing the building?
construction sites of debris and rubbish, and (a) The minimum vertical headroom above any
very efficient for short haul application up to step should be 2 m
100 m is (b) Generally, the number of risers in a flight
(a) Crane (b) Dumper Truck should be restricted to twelve
(c) Bulldozer (d) Tipper (c) The minimum width of stair should be 250 mm
Ans. (c) : Bulldozer– The versatile equipment used for (d) The respective dimensions of tread and riser
clearing the construction sites of debris and rubbish, and for all the parallel steps should be the same in
very efficient for short haul application up to 100 m is consecutive floor of a building
bulldozer.
Ans. (c) : Planning a staircase for a building–
Dump truck–A dump truck know also as a dumper truck
• The minimum vertical headroom above any step
or tipper truck, is used for taking dumps (such as sand,
should be 2 m.
gravel or demolition waste) for construction as well as coal.
• Generally, the number of risers in a flight should be
28. The Roof which slopes in two directions so that restricted to twelve.
the end formed by the intersection of slopes is a • The respective dimensions of tread and riser for all
vertical triangle, is the parallel steps should be the same in consecutive
(a) Gambrel roof (b) Hip roof floor of a building.
(c) Gable roof (d) Shed roof • The minimum width of stair in residential building is
Ans. (c) : Gable roof– This is the common type of 85 cm and in commercial building is 1.0 m, in
sloping roof which slopes in two directions so that the domestic building 90 cm and in public building 1.5 to
end formed by the intersection of slopes is a vertical 1.8 m.
triangle. 32. Extra widening refers to the additional width
Gambrel roof– The type of roof which slopes in two of carriageway that is required on a curved
directions with a break in the slope on each side is section of a road over and above that required
called gambrel roof. on a straight alignment. In this regard, the
Hip roof– This roof is formed by four sloping surface additional width required for a vehicle taking a
in four directions at the end faces, sloped triangles are horizontal curve is known as
formed. (a) Psychological Widening
29. The total force on a curved surface submerged (b) Motor Widening
in liquid is the resultant of horizontal and (c) Physical Widening
vertical forces. In this regard, which force will (d) Mechanical Widening
be equal to the total pressure force on the Ans. (d) : Mechanical Widening– The widening
projected area of the curved surface on the required to account for the off-tracking due to the
vertical plane? rigidity of wheel base is called mechanical widening
(a) Horizontal force on curved surface (Wm).
(b) Vertical force on curved surface  nl2 
W =
 m 2R 
(c) Both Horizontal and Vertical forces on the
 
curved surface
R is the mean radius of the curve and n is number of
(d) Neither Horizontal nor Vertical force on the
traffic lanes.
curved surface
Psychological Widening– Extra width of pavement is
Ans. (a) : The total force on a curved surface also provided for psychological reasons such as, to
submerged in liquid is the resultant of horizontal and provide for higher speeds to allow for the extra space
vertical forces. In this regard, horizontal force on curved requirement for the overhangs of vehicles and to
surface will be equal to the total pressure force on the provide greater clearance for crossing and overtaking
projected area of the curved surface on the vertical plane. vehicles on curves.
OSSC JE Exam-2019 (Shift-I) 140 YCT
  V  Ds = 3C – rr
 Wpsy =  C = 1, rr = 3
 9.5 R
=3–3=0
V = Design speed in Kmph.
Structure is statically determinate
R = Radius of horizontal curve, m.
36. Which of the following construction
l = Length of wheel base of longest vehicle, m.
equipments is suitable for improving the
33. Normally, shortening of a project duration strength, by decreasing the porosity and
results in which of the following options? increasing the density of Soil?
(a) Direct cost decreases and there is no effect on
(a) Hoisting Equipment
indirect cost
(b) Hauling Equipment
(b) Increase in the direct cost and decrease in the
indirect cost (c) Excavating Equipment
(d) Compaction Equipment
(c) Increase in the direct cost and also an increase
in the indirect cost Ans. (d) : Compaction Equipment is suitable for
(d) Direct and indirect cost both decreases
improving the strength, by decreasing the porosity and
increasing the density of soil.
Ans. (b) : Direct cost– It is the cost that is directly
dependent on the number of resources involved for the
Hauling Equipment– Elevator and conveyors are the
completion of activities when delay of project it is increase.
hauling equipments.
Indirect cost– It rises with increased duration,• These are used to haul aggregate, soil and RMC
considering only overhead and supervision. It is mixing material.
represented by straight line.
• These are the equipment to shift material with
34. In long wall and short wall method of a minimal human effort by mechanical action.
building estimation, if ''l" is the center to
37. Granite stone is a widely used building stone,
center length of wall "d" is the breadth of the
wall, then length of long wall (out to out) is mainly composed of quartz, feldspar and mica.
given by Which of the following options is INCORRECT
(a) l–2d (b) l+2d about this stone in general?
(c) l+d (d) l–d (a) Used in the preparation of tiles for floors and
Ans. (c) : • Long wall length out to out = centre to countertops of kitchens
centre length + Half breadth on one side + Half breadth (b) Used as a sound reflecting material
on the other side = centre to centre length + one (c) Used in the building walkways, driveways
breadth. and pavements
• Short wall length in to in = centre to centre length –(d) Used in the preparation of various types of
one breadth. wall putties
Ans. (d) :
• Granite is an igneous rock and is hard and durable.
• It most of these rocks possess excellent building
properties, like high strength, very low abrasion
value, good fire resistant.
Uses–
• Used in the preparation of tiles for floors and
countertops of kitchens
35. Which of the following types of arches is a
• Used as a sound reflecting material
statically determinate structure?
(a) Three Hinged Arch • Used in the building walkways, driveways and
(b) Single Hinged Arch pavements.
(c) Two Hinged Arch 38. As a general rule in India, the relation between
(d) Fixed Arch height and width of a Door is given as:
Ans. (a) : (a) Height = (Width - 1.2 ) meters
(b) Height = (Width x 1.2 ) meters
(c) Height = (Width + 1.2 ) meters
(d) Height = (Width /1.2 ) meters
Ans. (c) : The height and the width of the door are
related in a way such that the height of the door is 1.2 m
more than the width of the door.
OSSC JE Exam-2019 (Shift-I) 141 YCT
39. Read the following statements and choose the Ans. (c) : (Ac+mAs), expression is called the equivalent
CORRECT option with respect to the area of the section in terms of concrete. It means that
precautions taken against cavitation. the area of steel Ast, can be replaced by an equivalent
(i) The pressure of the flowing liquid in any part area of concrete equal to m.Ast.
of the hydraulic system should be above its 43. Surveying is done to take measurements of
vapour pressure. objects in a horizontal plane for locating
(ii) The cavitation resistant materials such as objects on the surface of the earth, before any
aluminium-bronze and stainless steel should construction of a structure is started. During
be used for Centrifugal Pumps or components this process the fundamental principle that is
followed is to work from the
of the hydraulic system.
(a) Lower level to higher level
(a) (i) is FALSE and (ii) is TRUE (b) Higher level to lower level
(b) (i) is TRUE and (ii) is FALSE (c) Whole to the part
(c) (i) is TRUE and (ii) is TRUE (d) Part to the whole
(d) (i) is FALSE and (ii) is FALSE Ans. (c) : j Work from whole to part– The survey area
Ans. (a) : Precautions taken against cavitation – is always totally covered with the simplest possible
(i) The pressure of the following liquid in any part of frame-work of high quality measurements. It the rest of
hydraulic system should not be allowed to fall the survey work is carried out within this control, the
below its vapour pressure. possible accumulation of error can be contained and
(ii) The cavitation resistant materials such as compensate error in a ways.
aluminium-bronze and stainless steel should be used 44. Standard gauge is the most widely used gauge
for Centrifugal Pumps or components of the in the World, for which the width is equal to
hydraulic system. (a) 1435 mm (b) 1525 mm
40. For a crop, if the optimum depth of kor (c) 1676 mm (d) 1365 mm
watering is 19 cms, then the outlet factor Ans. (a) : Clear horizontal distance between inner face of
(outlet duty) for the crop for a four week two rails forming the track at the top is Gauge distance.
period in hectares/cumec, is Different Gauge–
(a) 1472.26 hectares/cumec Distance
Gauge
(b) 1372.26 hectares/cumec between rails
(c) 1720.63 hectares/cumec Broad Gauge 1.676 m
(d) 1273.26 hectares/cumec Metre Gauge 1.0 m
Ans. (d) : Given, Narrow Gauge 0.762 m
Base period = 4 weeks = 28 days Light Gauge (feather track) 0.610 m
Duty = 19 cm = 0.19 m Standard Gauge 1.435 m
8.64B (used in Delhi metro)
delta =
D 45. Read the following statements and choose the
8.64 × 28 CORRECT option with respect to the rules for
= = 1273.26 hect/cumec drawing a network for project management.
0.19
(i) Each activity is represented by one and only
41. The surge tank in a water supply system is one arrow in the network.
generally provided to Score (ii) All the arrows must run from right to left.
(a) increase velocity of flow in a pipe (iii) For coding alphabets are used for all activities
(b) remove impurities from water including the dummy activity and numbers
(c) supply of chlorine alone required for for events.
disinfection process (a) (i) and (iii) are TRUE and (ii) is FALSE
(d) relieve excess pressure caused by water hammer (b) All (i), (ii) and (iii) are TRUE
Ans. (d) : The surge tank is provided to decreased the (c) All (i), (ii) and (iii) are FALSE
water hammer effect. (d) (i) and (iii) are FALSE and (ii) is TRUE
• It is a water storage device used as a pressure Ans. (a) : Rules to draw network–
neutralizer in hydropower water conveyance system to • A network will have only one initial event and one
resist excess pressure rise and pressure drop. finish event.
42. If in a reinforced concrete column the cross • An event can not occur twice.
sectional area of a steel bar is As and that of • No activity can start until its tail event occurs.
concrete is Ac, then the equivalent area of the • One arrow should represent only one activity and
section in terms of concrete is expressed as number of activity should be equal to number of
(Given m= modular ratio) arrow (excluding dummy arrow).
(a) As+mAc (b) Ac - mAs • Length and direction of arrow has no significance but
(c) Ac+mAs (d) As-mAc generally the direction of arrow is from left to right.
OSSC JE Exam-2019 (Shift-I) 142 YCT
46. Given below are statements with respect to the 48. A valve that is used to completely shut off fluid
components of a Staircase in general. flow or, in the fully open position, provide full
A. A vertical member is placed at the ends of flow in a pipeline and generally NOT suitable
flights in a staircase to connect the ends of the for throttling applications, is known as
Strings and Handrails. (a) Washout valve (b) Sluice valve
B. A vertical member of wood or metal usually (c) Gate valve (d) Check valve
supports the Handrail in a Staircase. Ans. (c) : A valve that is used to completely shut off
Select the CORRECT option among the fluid flow or, in the fully open position, provide full
following options. flow in a pipeline and generally NOT suitable for
(a) Statement (a) refers to -->> Nosing and throttling applications, is known as gate valve.
Statement (b) refers to -->> Soffit
• Gate valve is also known as sluice valve or shut off
(b) Statement (a) refers to -->> Newel Post and
valve.
Statement (b) refers to -->> Baluster
(c) Statement (a) refers to -->> Soffit and 49. Read the following statements and choose the
Statement (b) refers to -->> Nosing CORRECT option with regards to Splicing of
(d) Statement (a) refers to -->> Baluster and reinforcement bars.
Statement (b) refers to -->> Newel Post (i) For the constructional requirements at site,
Splicing of reinforcing bars is done that
Ans. (b) : Newel post – A post at the head or foot of a
flight of stairs, supporting a handrail. transfers the axial force from the terminating
bar to the connecting bar.
• The central supporting pillar of spiral or winding
stair case. (ii) Lap splices should not be used for bars larger
than 12 mm and End-bearing splices shall be
Baluster – It is a short column used in a group to
used only for bars in tension.
support a nail as commonly found on the side of a
stairway. (a) (i) is TRUE and (ii) is TRUE
(b) (i) is TRUE and (ii) is FALSE
47. Gauge pressure is usually defined as a pressure
which is measured using Atmospheric Pressure (c) (i) is FALSE and (ii) is TRUE
as Datum. Gauge pressure at a point in a fluid (d) (i) is FALSE and (ii) is FALSE
is equal to Ans. (b) :
(a) Absolute pressure + Atmospheric pressure • For the constructional requirements at site, Splicing
(b) Absolute pressure + Vacuum pressure of reinforcing bars is done that transfers the axial
(c) Absolute pressure - Atmospheric pressure force from the terminating bar to the connecting bar.
(d) Absolute pressure - Vacuum pressure • Lap splices shall not be used for bars larger than 36
Ans. (c) : Atmospheric pressure– It is the pressure mm and for larger diameter bars may be welded.
exerted by atmosphere • by amendment No. 3 IS 456: 2000 lap splice shall not
Vacuum pressure = Atmospheric pressure – Absolute be used for bars larger than 32 mm
pressure • End-bearing splices shall be used only for bars in
compression.
Pv = Patm − Pabs
50. As per IRC recommendations, on a National
Vacuum pressure is defined as the pressure below the
Highway in rolling terrain, the ruling design
atmospheric pressure.
speed should be
atm = 1.013 bar at measure level
(a) 60 kmph (b) 80 kmph
Absolute pressure– Pressure measured were to
absolute zero or complete vacuum is called absolute (c) 70 kmph (d) 150 kmph
pressure. Ans. (b) :
It is measured using aneroid barometer. Design speed (kmph)
Plain Rolling Mountainous Steep
Gauge pressure– It is pressure measured w.r.t. local
atmospheric pressure as datum. It is measure using Road Ruling Min. Ruling Min. Ruling Min. Ruling Min.
classification
manometer or bourdon gauge.
Expressways 120 100 100 80 80 60 80 60
• Pabs = Patm – Pvacuum NH & SH 100 80 80 65 50 40 40 30
• Pgauge = Pabs – Patm MDR 80 65 65 50 40 30 30 20
∴ Pabs − Patm − Pgauge = 0 ODR 65 50 50 40 30 25 25 20
VR 50 40 40 35 25 20 25 20
51. Under which of the following conditions needle
scaffolding is usually adopted for construction?
(a) When construction work is carried out at very
high level in case of high-rise buildings
(b) When minor repairs or painting works are
required inside the rooms
OSSC JE Exam-2019 (Shift-I) 143 YCT
 (c) Construction of the lower part of the wall is to (c) The foundations of residential building take
be carried out inside a room the load of the Slab alone from the
(d) Enough space is available to provide bracings superstructure above it
and ledgers (d) The bearing pressure at the base of foundation
Ans. (a) : Needle scaffolding proves highly effective in can exceed the allowable soil pressure
situations such as. Ans. (a) :
(i) When the ground surface does not allow for a 56. While submitting the tender form, the
conventional scaffold set up. contractor has to deposit a certain amount
(ii) When the project is on the side of a busy road or called Earnest Money. Usually this deposit can
street that cannot be blocked. be approximately taken as
(iii) When construction or maintenance is required at a (a) 0.1% to 0.2% of the estimated cost of work
great height.
(b) 0.1% to 0.2% of the estimated cost of work
52. Which of the following options is a type of (c) 1% to 2% of the estimated cost of work
Reinforced Concrete Slab directly supported
(d) 20% to 30% of the estimated cost of work
on columns without the use of beams?
(a) Flat slab (b) Tiled slab Ans. (c) : The amount of deposit paid by the
(c) Ribbed slab (d) Block slab tenderer/bidder at the time of submitting the tender is
called earnest money deposit.
Ans. (a) : Flat slab are also known as beamless slab is a
type of slab in which the flooring slab is directly • While submitting a tender the contractor is to deposit
supported on columns without the agency of beam or a certain amount about 1-2% of estimated cost, with the
girders. The diameter of the column head supporting a department as earnest money as guarantee of the tender.
flat slab 0.25 times the span. Security deposit– On acceptance of the tender the
53. The irrigation capacity of a unit of water is contractor has to deposit 10% of the tendered amount as
usually known as Duty. In this regards, Duty security money with the department which is exclusive
on capacity is also called as of the earnest money already deposit.
(a) capacity coefficient 57. The depth of the centre of pressure on a
(b) quality duty vertical rectangular gate (4 m wide, 3 m deep)
(c) outlet duty with water up to top surface, is
(d) full supply coefficient (a) 1.5 m (b) 2.5 m
Ans. (d) : Full Supply Coefficient–The number of (c) 1.0 m (d) 2.0 m
hectares irrigable per cumec of the canal capacity at its Ans. (d) : Given, b = 4 m and h = 3 m
head is known as full supply coefficient. Depth of the center of pressure on a vertical rectangular
Time factor–No. of days canal actually run to the total gate.
no. of days the watering period. 2h 2 × 3
Capacity factor–Ratio of mean supply discharge of h= = = 2m
3 3
canal to maximum discharge capacity.
58. The water is filtered through the beds of fine
Duty on capacity–Duty at the head of canal.
granular material, such as sand for removing
54. Find the hypotenusal allowance per chain of 30 or reducing impurities in water through the
m length, if the slope is 1 in 4. (Given: value of
process of Filtration. The Rate of filtration
1/cosθ = 1.03077)
(Litres/hour/m2) for Rapid Sand Filter is
(a) 0.16 m (b) 0.34 m generally in the range of
(c) 0.92 m (d) 0.30 m (a) 150 to 2000 (b) 8000 to 10000
Ans. (c) : Given, (c) 10 to 100 (d) 3000 to 6000
Chain length (l) = 30 m
Ans. (d) : Rapid sand filter–
Slope, tanθ = 1/4
• Cu = 1.2 to 1.6
Hypotenusal allowance = l (sec θ–1)
• D10 = (0.35 to 0.55) mm
= 30 × (1.03077 –1)
• Rate of filtration (3000 to 6000) l/m2/hr almost 30
= 0.9231 m
time of SSF
55. Which of the following statements is
• Depth of tank 1-2 m
CORRECT for foundations in general?
(a) It prevents the differential settlement of a • Period of cleaning 1-3 days.
structure • Removal of turbidity up to 40 ppm
(b) It transfers the load to the column directly • Washing period 24-48 hrs.
OSSC JE Exam-2019 (Shift-I) 144 YCT
59. The process of ‘lagooning’ serves which of the (a) 0.03% (b) 0.02%
following purposes in a wastewater treatment (c) 0.05% (d) 0.04%
plant? Ans. (d) : Grade compensation for on curves in Indian
(a) Reduces the excessive flow in sewer railways :
(b) Helps in the disposal of sludge 1. 0.04% per degree of curve → B.G track.
(c) Increases the capacity of storage reservoir
2. 0.03% per degree of curve → M.G track.
(d) Increases flow of a sewage through imhoff
tanks 3. 0.02% per degree of curve → N.G track.
Ans. (b) : The process of lagooning is primarily a 63. In a project network analysis, Forward Pass
means of disposal of sludge. method is used to determine which aspect of
Use of lagoons for disposal of raw sewage– This various activities?
method is sometimes used at smaller places for (a) Only the total float
disposing of raw sludge without digestion. In this (b) Earliest Finish Time
method, the raw sludge is kept at rest in a large shallow (c) Latest Finish Time
open ponds called a lagoon. (d) Only the slack time
• The detention periods is 1 to 2 months and may Ans. (b) : The forward pass gives an estimate of the
extended upto 6 month. During its detention in the earliest time when an activity can be started.
lagoon, the sludge undergoes anaerobic digestion 64. Which of the following options is a raw
thereby getting established. material used in the manufacture of cement,
60. In case of a Cantilever beam, as per IS:456- that is responsible for the grey color of OPC
2000, what is the basic value of span to effective (Ordinary Portland Cement)?
depth ratio for spans up to 10 m? (a) Sulphur trioxide
(a) 7 (b) 20 (b) Aluminium oxide
(c) 26 (d) 30 (c) Iron oxide
Ans. (a) : Span to effective depth ratio for beam span (d) Silica
upto 10 m to satisfy vertical deflection limit –
Ans. (c) : The four most important materials that go to
Beam Span to effective depth make Portland cement are Lime, Silica, Alumina and
ratio Iron Oxide
Simply supported 20
Iron oxide – 0.5 to 6%. It imparts grey colour and helps
Continuous beam 26 in fusion of different ingredients.
Cantilever beam 7
Alumina – 3 to 8% – quick setting the cement.
For slab – Span/overall depth
Lime – 60-67% – Act as a binding material and main
One way slab 30
part of the cement. It controls strength and soundness.
Two way slab 35
Continuous slab 40 65. Which of the following options is NOT a type of
dam classified on the basis of function?
Cantilever slab 12
(a) Diversion Dam (b) Flood Control Dam
61. With regards to the shape test of aggregates for
(c) Storage Dam (d) Buttress Dam
pavement construction, the elongation index of
an aggregate is defined as the percentage by Ans. (d) : Classification of dams–
weight of particles whose greatest dimension 1. Based on material of construction–
(length) is (a) Earth fill dam (b) Rock fill dam
(a) 0.8 times their mean dimension (c) Concrete dam (d) Masonry dam
(b) 1.8 times their mean dimension (e) steel dam (f) Timber dam
(c) 2.8 times their mean dimension 2. Based on flow over its top–
(d) 3.8 times their mean dimension (a) Overflow dams (b) Non overflow dams
Ans. (b) : Elongation Index Test–Percentage by 3. Based on the use of the dam–
weight of particles whose greater dimension is more (a) Storage dam
than 1.8 times their mean dimension. Applicable to size (b) Diversion dam
larger than 6.3 mm.
(c) Detention dam
Note–Flakiness Index test followed by elongation index
test on the non-flaky aggregates. 66. The length of a survey line measured with a 20
62. The reduction in gradient at the horizontal m chain was found to be 520 m. When the
curve because of the additional tractive force chain was compared with a standard chain it
required due to curve resistance is known as was found to be 10 cm too long. The true length
Grade Compensation. As per Indian Railway of the line in 'm' is
provision, the grade compensation per degree (a) 517.4 m (b) 532 m
of curve for Broad Gauge track is (c) 524.2 m (d) 522.6 m
OSSC JE Exam-2019 (Shift-I) 145 YCT
Ans. (d) : Wrong length of chain (l') = 20 m 71. A rectangular plane surface that lies in a
Wrong measured length (L') = 520 m vertical plane in water is 2 m wide and 3 m
Correct length of chain (l) = (20.00 – 0.10) m deep. Find the Total pressure if the upper edge
Correct measured length (L) = ? is horizontal and 2 m below the free water
We know – l '× L ' = l × L surface.
(a) 206.01 kN (b) 106.01 kN
20 × 520
=L (c) 406.01 kN (d) 306.01 kN
19.90
522.613 m = L Ans. (a) :
67. In the figure given below, the network of a
small part of a project represents what among
the following options?

(a) Excavation and sand filling & ramming may

be start together
(b) Event 3 will occur after 2 days after
occurrence of event 1
(c) Excavation of a footing takes 2 days to complete
(d) Event 3 will occur after 2 days after
occurrence of event 2
Ans. (d) : Event 3 will occur after 2 days after
occurrence of event 2
68. Which of the following types of lime usually Centre of pressure ( h ) – 2m + 1.5m = 3.5 m
slakes with difficulty and is mostly used in the Total pressure (P) = ρg × h × A
constructions near damp places?
(i) Feebly Hydraulic Lime = 1000 × 9.81 × 3.5 × 2 × 3
(ii) Moderately Hydraulic Lime = 206010 N
(iii) Eminently Hydraulic Lime = 206.01 kN
(a) All (i), (ii), (iii) (b) only (ii) 72. In an open channel flow of fluids, Hydraulic
(c) Only (i) (d) Only (iii) Jump usually occurs when the flow changes
Ans. (d) : Eminently hydraulic lime has clay impurities from
20-30% and slakes with difficulty. Its initial setting time (a) sub-critical to super-critical state
is 2 hrs and final setting time is 48 hrs. (b) laminar to turbulent state
• It is used in dam places and for all structural purposes. (c) super-critical to sub-critical state
69. A horizontal structural member provided (d) critical to turbulent state
above the doors and windows of a building to Ans. (c) : Hydraulic Jump in a rectangular channel is a
support load of the wall above it is known as natural phenomenon that occurs when the flow changes
(a) Stair (b) Slab from super critical to subcritical flow. The jump can
(c) Lintel (d) Masonry occur in the rectangular as well as non-rectangular
Ans. (c) : Lintel : A lintel is a beam placed across the channels such as triangular, Parabolic, trapezoidal
opening like doors, window etc. in building to support gradually expanding and abruptly expanding channels.
the load from the structure above.
73. Read the following statements and choose the
70. The revised estimates are prepared when the CORRECT option with respect to objectives of
expenditure on work exceeds or is likely to Sewage collection and disposal.
exceed the amount of administrative sanction
by more than (i) An efficient sewerage scheme can provide a
(a) 3.5% (b) 1.5% good sanitary environmental condition of city
protecting public health.
(c) 5% (d) 0.5%
(ii) An efficient sewerage scheme can also
Ans. (c) : Revised estimate is a detailed estimate, which dispose of all liquid waste generated from
is prepared when– community to a proper place to prevent a
(i) When the original sanctioned estimate is likely to favorable condition for mosquito breeding,
exceed by 5% fly developing or bacteria growing.
(ii) When the expenditure exceeds the administrative
(a) (i) is TRUE and (ii) is TRUE
sanction by 10%
(iii) There are major deviations from the original (b) (i) is FALSE and (ii) is FALSE
proposal. (c) (i) is FALSE and (ii) is TRUE
(d) (i) is TRUE and (ii) is FALSE
OSSC JE Exam-2019 (Shift-I) 146 YCT
Ans. (a) : objectives of Sewage collection and disposal– Tie lines – It is the line which runs between the tie
• An efficient sewerage scheme can provide a good stations. It is helpful for locating interior details.
sanitary environmental condition of city protecting Check or proof line – It is provided to check the
public health. accuracy of work.
• An efficient sewerage scheme can also dispose of all 77. Read the following statements and choose the
liquid waste generated from community to a proper CORRECT option with respect to the
place to prevent a favorable condition for mosquito components of a railway track.
breeding, fly developing or bacteria growing. (i) The Granular bed resting on formation and
74. For the conveyance of water various types of supporting sleepers over it is known as
valves are used for different purposes in a Ballast.
water distribution system. In this regard, the (ii) The Steel sections laid along two parallel
valve, which allows the flow of water only in lines over sleepers are known as Rails.
one direction, is known as (a) (i) is FALSE and (ii) is FALSE
(a) Butterfly valve (b) Check valve (b) (i) is TRUE and (ii) is FALSE
(c) Sluice valve (d) Gate valve (c) (i) is TRUE and (ii) is TRUE
Ans. (b) : Check valve or reflux valve or non- (d) (i) is FALSE and (ii) is TRUE
returning valve: In this process some automatic device Ans. (c) : Both statements are correct-
which allows the water to flow in one direction only. • The Granular bed resting on formation and supporting
• This valve is provided in the pipe line which draws sleepers over it is known as Ballast.
water from the pump. • The Steel sections laid along two parallel lines over
• When the pump is operated, the valve is opened but sleepers are known as Rails.
when the pump is suddenly stopped the valve is 78. The radius of the horizontal curve (R), for a
automatically closed and the water is prevented from design speed of (V) 100 kmph with the
returning to the pump. maximum values of (superelevation) e = 0.07
Butterfly valve– The valve is used in water tank and (coefficient of friction) f = 0.15, can be
and cisterns to maintain constant water level and to calculated using which of the following
prevent outflow. It is activated by means of a lever and equations?
float (ball) the rise and fall of ball controls the flow of V3 V2
water. (a) e + f = (b) e + f =
127 R 127 R
Sluice valve– These valves are provided to regulate
V3 V2
the flow of water through the pipe where it is essential (c) e + f = (d) e + f =
to divide the main line into several section. 137 R 12 R
• These valves are usually placed at the summits of Ans. (b) : Superelevation– Super-elevation as cant is
the pressure conduit, where pressure is low. the transverse slope provided at horizontal curve to
75. The size of the fillet weld used as connection in counteract the centrifugal force, by raising the outs edge
steel structures is specified by of the pavement with respect to the inner edge,
(a) total length of the weld throughout the length of the horizontal curve.
(b) thickness of the weld • Superelevation with coefficient of friction –
(c) minimum leg length of weld V2
(d) maximum leg length of weld e+f=
127R
Ans. (c) : According to IS 800-2007 , the size of • For mixed traffic condition as per IRC 37 highway is
normal fillets shall be taken as the minimum weld leg design for 75%. Design speed and neglating lertral
size. friction,
• For deep penetration beyond the root run is a
( 0.75V )
2
minimum of 2.4 mm, the size of the fillet should be e+f =
taken as the minimum leg size plus 2.4 mm. 127R
2
76. Which of the following options is generally V
e=
considered as the longest of the main survey 225R
lines in Chain Surveying? V = kmph, R = radius (m)
(a) Base Line (b) Tie Line 79. In CPM network analysis of a project, the
(c) Oblique Offset (d) Check Line disconnection of an activity before the
Ans. (a) : Base line– Longest line in chain survey completion of all activities in a network
which divided the area two part is called base line. It diagram is known as (symbols and notations
measured by invar tape (generally). carry their usual meaning)

OSSC JE Exam-2019 (Shift-I) 147 YCT

 (a) Dangling error (b) Redunancy error 81. The ground for the formation level of a
(c) Mathematical error (d) Looping error Highway construction is of uniform gradient
Ans. (a) : Dangling error– To distance an activity with longitudinal slope. Here the quantity of
before the completion of all activities in a network earthwork can be calculated using which of the
diagram is known as dangling. As shown in the figure following options?
activities (5-10) and (6-7) are not the last activities in A. Trapezoidal formula
the network. So the diagram is wrong and indicates the B. Prismoidal formula
error of dangling. C. Mid-section formula
(a) Only B (b) Only C
(c) All A, B and C (d) Only A and C
d
Ans. (c) : Trapezoidal formula– Volume, V = [First
2
area section + Last area section + 2 (sum of other
Looping or cycling error– It is also known as cycling areas)]
error in a network diagram. Drawing an endless loop in Mid section formula– In this formula, the mean depth
a network is known as error of looping as shown in the is to be calculated. First by averaging the depths of two
following figure. consecutive sections. From the mean-depth the area of
mid-section is to be worked out and volume of
earthwork to be computed by multiplying the area of
mid-section by the distance between the two original
section.
d
Prismoidal formula– V = [First area section + Last
3
area section +4 (sum of even areas) +2 (sum of odd
Redunancy error– Unnecessarily inserting the dummy area]
activity in network logic is known as the error of All above can be use for earthwork estimation if
redundancy as shown in the following diagram. highway has a uniform gradient.
82. With reference to CPWD specifications for
formwork in general read the following
statements and choose the CORRECT option.
(i) Centering, and shuttering where exceeding
1.5 metre height in one floor shall be
80. The average mass of a standard brick is measured and paid for separately.
approximately in the range of (ii) No deductions from the shuttering due to the
(a) 1 to 1.5 kg (b) 3 to 3.5 kg openings/ obstructions shall be made if the
(c) 5 to 5.5 kg (d) 7 to 7.5 kg area of each openings/ obstructions does not
exceed 0.1 square metre.
Ans. (b) : Size and weight of bricks–
(a) (i) is TRUE and (ii) is FALSE
For India,
(b) (i) is TRUE and (ii) is TRUE
• Standard size of bricks is 19 cm × 9 cm × 9 cm (c) (i) is FALSE and (ii) is FALSE
• Nominal size (with mortar) is 20 cm × 10 cm × 10 cm (d) (i) is FALSE and (ii) is TRUE
• The commonly adopted nominal size of traditional Ans. (c) :
bricks is 23 cm × 11.4 cm × 7.6 cm
83. The ratio of the area irrigated in the Rabi
• It is found that the weight of 1 m3 of bricks earth is season to the area irrigated in the Kharif
about 1800 kg hence the average weight of a brick will
season is known as
be about 3 to 3.50 Kg.
(a) Overlap Ratio
(b) Season Ratio
(c) Crop Ratio
(d) Irrigated Ratio
Ans. (c) : Crop Ratio– The ratio of the area irrigated in
the rabi season to the area irrigated in the Kharif season.
It is approximately 1 : 2

OSSC JE Exam-2019 (Shift-I) 148 YCT

84. Read the following statements and choose the 86. Which of the following options is NOT an
CORRECT option with respect to the forces in example of Instrumental errors in Plane Table
a member of a Truss in general. Surveying?
i. The force in a member will be Compressive if (a) The fiducial edge of the alidade might not be
the member pushes the joint to which it is straight
connected. (b) The needle of the trough compass may not be
ii. The force in a member will be Tensile if the perfectly balanced
member pulls the joint to which it is (c) The alidade may not be correctly centered on
connected. the station point
(a) (i) is TRUE and (ii) is FALSE (d) The table may be loosely joined with the
(b) (i) is FALSE and (ii) is TRUE tripod stand
(c) (i) is FALSE and (ii) is FALSE Ans. (c) : Instrumental errors are the primary source of
(d) (i) is TRUE and (ii) is TRUE errors in plane table surveying which can occur to the
following ways–
Ans. (d) : Correct statements are with respect to the
forces in member of a truss in generally– • Errors will occur if the top surface of the plane table
is not flat or contains.
→ The force in a member will be Compressive if the
• The fittings of the tripod and plane table should be
member pushes the joint to which it is connected.
tightly fastened. Loose fitting can make the plane
→ The force in a member will be Tensile if the member while drawing.
pulls the joint to which it is connected.
• The magnetic compass used to plane table surveying
85. Which of the following options is the should represent accurate direction otherwise an error
CORRECT statement, according to Lami's may occur due to the wrong orientation of plane
Theorem? table.
(a) The three forces acting at a point kept in • When the edge of fiducial edge of the alidade is not
equilibrium must be equal to each other straight or curved, an error occurs in the drawing.
(b) If three forces acting at a point are in • Both sight vane and object vane of alidade should be
equilibrium, then each force is proportional to perpendicular to the base of an alidade. If not, there
the tangent of the angle between the other two will be an error in sighting.
(c) If three forces acting at a point are in • Faulty spirit level or level tube may not keep the
equilibrium, then each force is proportional to plane table in horizontal position, therefore as error
the cosine of the angle between the other two may occurs.
(d) If three forces acting at a point are in 87. The process of mixing some mortar in the
equilibrium, then each force is proportional to mixer at the beginning of the first batch
the sine of the angle between the other two concrete mixing is called as
Ans. (d) : Lami's theorem–It states that if three forces (a) buttering (b) initiating
coplanar and concurrent forces acting at a point be in (c) accelerating (d) borrowing
equilibrium, then each force is proportional to the sine Ans. (a) : Buttering–In machine mixing a certain
of the angle between the other two forces. amount of cement mortar adheres to the inside of drum
Mathematically. walls and remains adhered to after discharging the first
P Q R batch of concrete mix. Thus, it is recommended that
= = before mixing the first batch of concrete mortar (i.e.
sin α sin β sin γ
cement, sand and water) is fed into the mixer and mixer
is rotated . A quantity of mortar should be just enough
to adhere to the walls. This procedure of mixing cement
mortar prior to the mixing of first batch of concrete is
called as buttering the mixer.
88. An aggregate is termed as elongated when its
greatest dimension (length) is greater than
Score
(a) nine-fifth of its mean dimension
(b) one-third of its mean dimension
where, P, Q and R are the three forces and α, β and γ (c) four times its mean dimension
are the angles as shown in fig. (d) thrice its mean dimension
OSSC JE Exam-2019 (Shift-I) 149 YCT
Ans. (a) : Elongation index - It is the percentage by 92.
weight of particle whose greatest dimension (length) is
9
greater than 1.8 time or times/(180%) their mean
5
dimension.
• Flaky aggregate is said to be elongated if its length is
twice the mean size. For the pin-jointed truss shown in the figure
89. For Stone Masonry or structural concrete, the above, identify whether the truss is
categories for measurement of items, should be (i) Statically determinate
measured separately and the heights should be (ii) Statically indeterminate
described first from foundation to plinth level (iii) Unstable
and then from (a) Only (i) (b) Both (ii) and (iii)
(a) plinth level to third floor (c) Only (iii) (d) Only (ii)
(b) plinth level to second floor Ans. (d) :
(c) plinth level to first floor • Redundant frame
(d) second floor to fourth floor
Ans. (c) : In case of stone or brick masonry or structural
concrete, the categories shall be measured separately
and heights shall be described as––
• From foundation to plinth level
Ds = m + r – 2j
• From plinth to first floor level
=6+3–2×4
• From first floor to second floor level and so on.
=9–8=1
90. As per the General Building Requirements in Ds ≠ 0, so redundant frame will be indeterminate
India, the horizontal distance of a building site structure
from High voltage lines upto and including 93. Read the following statements and choose the
11,000 volts should be atleast CORRECT option with respect to Moment
(a) 0.5 m (b) 1.2 m Distribution Method of analysis of structures.
(c) 0.2 m (d) 0.9 m i. Moment Distribution Method of analysis is a
convenient way to analyze only the
Ans. (b) : As per the General Building Requirements in determinate beams and frames.
India, the horizontal distance of a building site from ii. In this method, the distributed moments in the
High voltage lines upto and including 11,000 volts ends of members meeting at a joint cause
should be atleast 1.2 m. moments in the other ends, which are
91. For analysing two-dimensional irrotational assumed to be fixed.
flow problems, a flownet is required that is These induced moments at the other ends are
called carry-over moments.
usually drawn using a series of lines. In this
(a) (i) is TRUE and (ii) is FALSE
regard, a line along which the velocity potential
(b) (i) is FALSE and (ii) is FALSE
is constant is known as
(c) (i) is TRUE and (ii) is TRUE
(a) Equipotential Line (b) Streak Line (d) (i) is FALSE and (ii) is TRUE
(c) Stream Line (d) Di-potential Line
Ans. (d) : Moment distribution method is a
Ans. (a) : Equipotential Line– For analysing two- displacement method. This method offers a convenient
dimensional irrotational flow problems, a flownet is way to analyse statically indeterminate beams and rigid
required that is usually drawn using a series of lines. In frames.
this regard, a line along which the velocity potential is • In this method, the distributed moments in the ends of
constant. members meeting at a joint cause moments in the other
Stream line – A stream line is an imaginary line drawn ends, which are assumed to be fixed.
in a flow field such that a tangent drawn at any point on 94. Generally, in R.C.C sections, the depth of the
this line represents the direction of velocity vector at neutral axis usually determines the type of
that point. section. For an under reinforced section, the
actual neutral axis lies:
Streak line- When a dye is injected in a liquid or smoke
(a) in the same line along the critical neutral axis
in a gas so as to trace the subsequent motion of fluid
(b) above the critical neutral axis
particles passing a fixed point the path followed by the (c) below the critical neutral axis
dye or smoke is called the streak line. (d) both above or below the critical neutral axis
OSSC JE Exam-2019 (Shift-I) 150 YCT
Ans. (b) : Under reinforced section– Strain in steel Ans. (c) : Gap graded aggregate is recommended for
reaches its ultimate value earlier than strain in concrete. mixes of low workability since they show greater
• The failure is in steel, causes a ductile failure and tendency to segregation in high workability mixes.
gives clear warning before failure.
98. In a certain type of scaffolding the working
Xa < Xc (under reinforced beam)
Xa = Xc (balanced reinforced beam) platform is supported on movable tripods or
Xa > Xc (over reinforced beam) ladders. This is generally used for work inside
the room, such as paintings, and repairs up to a
height of 5m. Identify this type of scaffolding
among the following options.
(a) Trestle Scaffolding
(b) Cantilever Scaffolding
(c) Patented Scaffolding
(d) Double Scaffolding
Ans. (a) : Trestle Scaffolding – In this type of
95. A track structure which permits movement of scaffolding the working platform is supported on
train from one track to another refers to which movable tripods or ladders. This is generally used for
of the following options? work inside the room, such as paintings, and repairs up
(a) Bearing Plate to a height of 5 m.
(b) Turnout
99. Which of the following options is an example of
(c) Stuart Key Sub-surface raw water source?
(d) Anchor (a) Lakes
Ans. (b) : Turnout– A turnout enabling railway trains (b) Streams
to be guided from one track to another, such as at a (c) Infiltration Wells
railway junction or where a spur or siding branches off. (d) Rivers
• A complete set of points and crossings, along with
Ans. (c) : Infiltration wells– It is the shallow wells
lead rails.
constructed in series along the banks of a river, in order
to collect the river water seeping through their bottoms.
These well are generally constructed of brick masonry
with open joints and covered at the top and kept open at
the bottom. For inspections purposes, manholes are
provided in the top slab cover.

96. The vertical plank which remains in contact

with side of the trench and directly resist
pressure from the side of a trench during
excavation and trenching, is called as
(a) Ledgers
(b) Sheathing
(c) Wale
100. The revenue chain is 33 feet long and divided
(d) Bracings
into 16 links. It is used for
Ans. (b) : Sheathing – The vertical plank which
(a) Determining the magnetic bearing of a line in
remains in contact with side of the trench and directly
a cadastral survey
resist pressure from the side of a trench during
(b) Distance measurement in a cadastral survey
excavation and trenching.
(c) Horizontal angle measurement in a cadastral
97. Gap graded aggregate is recommended for survey
mixes of low workability since
(d) Vertical angle measurement in a cadastral
(a) they have less percentage of coarse survey
aggregates and specific surface is more
Ans. (b) : Revenue chain–It is 33 ft. long and consists
(b) they have high tensile strength
(c) they show greater tendency to segregation in of 16 links, being 2 161 ft. long.
high workability mixes ⇒ The chain mainly used for measuring field in
(d) they high compressive strength cadastral survey.
OSSC JE Exam-2019 (Shift-I) 151 YCT
 Odisha Staff Selection Commission
Junior Engineer (Civil)
Exam - 2019 (Shift-II)
1. Read the following statements and choose the (i) PERT system of analysis uses event oriented
CORRECT option with respect to the shape of network diagrams.
aggregates and their workability. (ii) This system of analysis is usually preferred
(A) Rounded Aggregates usually show a high for projects which are repetitive.
workability as compared to angular, flaky or (a) (i) is TRUE and (ii) is TRUE
elongated aggregates. (b) (i) is TRUE and (ii) is FALSE
(B) Irregular or partly rounded aggregates may (c) (i) is FALSE and (ii) is TRUE
result in 65- 70% of voids and give higher
(d) (i) is FALSE and (ii) is FALSE
workability as compared to rounded
aggregates. Ans. (b) : PERT–PERT was developed by US Navy in
(a) (A) is TRUE and (B) is FALSE late 1950s to accelerate the development of polaris fleet
ballistic missile.
(b) (A) is FALSE and (B) is FALSE
(c) (A) is TRUE and (B) is TRUE • PERT is used in R & D type projects.
(d) (A) is FALSE and (B) is TRUE • Most of the activities of PERT Network follows Beta
(β) - Probability distribution curve.
Ans. (a) : • Angular & flaky aggregate have large
Note–CPM, system of analysis is usually preferred for
surface area, so they are less workable.
projects which are repetitive.
• Rounded cubical shape aggregates have less surface
area. Therefore, less amount of paste is required for 4. During a project planning, which type of float
lubrication, Hence they are more workable. determines the flexibility available or the
amount of time that an activity can be delayed
Workability–Property of concrete which determines
without the activity interfering/delaying with
the amount of useful internal work necessary to produce
any of its successor activities?
full compaction.
(a) Free float (b) Interfering float
• Slump test used for in-situ determination of
workability. (c) Total float (d) Independent float

2. Polyvinyl chloride (PVC) is a type of Ans. (a) : Float–It indicates the time by which, starting
or finishing of an activity be delayed without affecting
(a) Thermoplastic material the project completion time.
(b) Rigid-plastic material
(i) Total float–It is the difference between maximum
(c) Thermosetting material time available & actual required for the completion of
(d) Elastoplastic material the activity.
Ans. (a) : Classification of the plastics– (ii) Free float–It is the amount of time by which an
A. Thermoplastic–These are the types of polymers, activity can delayed without affecting the EST of the
which become soft upon heating and become hard on succeeding activity.
cooling, such plastic can undergo infinite cycles of (iii) Independent float–It is the excess of minimum
heating and cooling provides the temperature is not so available time over the required activity duration.
high to cause any chemical decomposition. (iv) Interfering float–It is the difference of total float
B. Thermosetting plastic–These are the plastics which and free float of an activity. It is also equal to the slack
of head event of an activity.
set after application of heat and pressure and then do not
change their form an further application of heat 5. A possible fluid flow is represented by which of
pressure. the following velocity fields that satisfies the
3. Read the following statements and choose the continuity equation?
CORRECT option with respect to PERT (a) u = xy, v = x2y2 (b) u = x, v = –y
network analysis of construction management. (c) u = x, v = y2 (d) u = x2, v = y2

OSSC JE Exam-2019 (Shift-II) 152 YCT

Ans. (b): Given, u = x, v = –y Ans. (d) : Reclamation of alkaline soil–
du dv (i) Reclamation of alkaline soils can be done by
= 1, = −1 chemical method in which some chemical like gypsum
dx du
is added to the soil in order to brine the alkalinity
For continuity equation-
desired load.
du dv
+ (ii) Also this reclamation can be done by mechanical
dx du practices such as improving drainage and leaching,
1–1=0 mechanical shattering of clay pans and scrapping.
6. To check for the occurrence of Cavitation in a 9. As per IS 456:2000, in the absence of test data,
Centrifugal pump or a turbine, the Thoma's the approximate value of total shrinkage strain
Cavitation Factor (σ) and Critical Cavitation for design of concrete should be taken as
Factor (σc) for that pump is usually compared. (a) 0.0005 (b) 0.0002
The cavitation will NOT occur in a particular
(c) 0.0004 (d) 0.0003
turbine or pump when
(a) σ > σc (b) σ = σc/2 Ans. (d) : As per IS 456:2000, in the absence of test
data, the approximate value of total shrinkage strain for
(c) σ < σc (d) σ = σc design of concrete should be taken as 0.0003.
Ans. (a) : The cavitation will not occurs in a particular • The volumetric change of concrete structures due to
turbine or pump when σ > σc, the loss of moisture by evaporation is known as
Where, σ = Thoma's cavitation factor concrete shrinkage.
σc = Critical cavitation factor • It is time dependent deformation which reduces the
7. The foundation of a weir consists of a volume of concrete without the impact of external
horizontal floor of 30 m length, an upstream forces.
pile of depth 8 m, and a downstream pile of 10. Given below are the two main types of weld
depth 12 m. The creep length according to used for steel structures.
Bligh's Creep Theory is (i) Butt Weld
(a) 70 m (b) 90 m (ii) Fillet Weld
(c) 110 m (d) 50 m Which of the following options is a type of
Ans. (a) : weld that is used when members overlap with
each other?
(a) Neither (i) nor (ii) (b) Only (i)
(c) Only (ii) (d) Both (i) and (ii)
Ans. (c) : The two main types of weld used for steel
structure–
Creep length = l + 2 × d1 + 2 × d2 (i) Butt weld (ii) Fillet weld
= 30 + 2 × 8 + 2 × 12 Butt weld–Butt weld or groove weld is used when the
= 70 m plate to be jointed are in the same plane or when a 'T'
joint is desired.
8. Read the following statements and choose the
• A butt weld is designated to the shape of groove made
CORRECT option with respect to reclamation
during the preparation of ends of the pieces to be
of alkaline soils.
jointed.
(i) Reclamation of alkaline soils can be done by,
• A butt weld is specified by the size of the weld. The
chemical method in which some chemicals
size of the butt weld is defined by the effective throat
like Gypsum is added to the soil in order to
thickness.
brine the alkalinity to desired level.
Fillet weld–When the lapped plates are to be jointed,
(ii) Also this reclamation can be done by
fillet weld are used. These are generally of right angle
mechanical practices such as improving
triangle shape. The outer surface is generally made
drainage and leaching, mechanical shattering
convex. It is used when member overlap with each
of clay pans, and scrapping. other.
(a) (i) is TRUE and (ii) is FALSE
• A fillet weld is specified by the–
(b) (i) is FALSE and (ii) is TRUE
1. Size of weld
(c) (i) is FALSE and (ii) is FALSE
2. Throat thickness
(d) (i) is TRUE and (ii) is TRUE 3. Length of weld
OSSC JE Exam-2019 (Shift-II) 153 YCT
11. The required slope correction for a measured 14. Which of the following materials is NOT
length of 80 m along a gradient of 1 in 20 is generally used as a binder in oil paints?
approximately equal to (a) Linseed oil (b) Turpentine
(a) 10 m (b) 0.75 cm (c) Resins (d) Bitumen
(c) 10 cm (d) 5.75 cm Ans. (b) : Binder–
Ans. (c) : Given, • Also known as vehicle. Vehicle is an oil to which the
L = 80 m base is mixed.
Slope = 1 in 20 • It hold the constituents of paint in suspension and
80 helps spread it over the surface to be painted imparts
then, h= = 4m durability, toughness and water proofness to the paint
20 film and resistance to weathering and glass to the
h2 4×4 1 painted and forms the body of the paints.
Slope correction (cslope) = = = = 100 cm
2l 2 × 80 10 Ex. linseed oil, resins, bitumen, etc.
Cslope = 10 cm Note–Turpentine oil is used as a 'salvent' and salvent
also known as thinner.
12. If the magnetic bearing of a line AB, is 123° 15. What is the full form of PERT, a network
35', find its true bearing, if the magnetic based scheduling system for managing a
declination is 10°10' W. project in construction?
(a) 113° 25' (b) 124° 35' (a) Project Execution and Review Technique
(c) 134° 35' (d) 10° 10' (b) Program Execution and Review Technique
Ans. (a) : (c) Process Evaluation and Review Technique
(d) Program Evaluation and Review Technique
Ans. (d) : PERT–Program Evaluation and Review
Technique.
• PERT is event oriented.
• CPM is activity oriented.
16. For the Flow through pipes or tubes, the type
of flow will be turbulent if the Reynolds
Number is
(a) greater than or equal to 4000
(b) equal to 1000
(c) in between 1000 to 2000
(d) equal to 2000
Ans. (a) :
True bearing (T.B) = M.B ± Declination
T.B. = 123º35' – 10º10' Laminar Transition Turbulent
T.B. = 113º25' Flow in pipe Re < 2000 < Re Re ≥ 4000
2000 < 4000
13. The method of repair and strengthening of
building foundation in which a additional Flow between Re < 1000 < Re Re > 2000
foundation is installed so that additional depth parallel plate 1000 < 2000
and bearing capacity is achieved is known as Flow in open Re < 500 500 < Re < Re > 200
(a) Grouting (b) Jacketing channel 2000
(c) Shorting (d) Underpinning Flow through Re < 1 1 < Re < 2 Re > 2
Ans. (d) : Underpinning– soil
• It is a method of construction that sees the depth of 17. Rainfall resulting from lifting of warm
the foundation to a building being increased. moisture laden air masses due to topographic
• The soil beneath the existing foundation is excavated barriers, is known as
and is replaced with foundation material normally (a) Converging precipitation
concrete in phases. (b) Orographic precipitation
• Underpinning requires close attention to design, (c) Cyclonic precipitation
methodology and safety procedures. (d) Convective precipitation
OSSC JE Exam-2019 (Shift-II) 154 YCT
Ans. (b): Orographic Precipitation–Rainfall resulting Ans. (c) : Pozzolanic materials–The terms puzzolana
from lifting of warm moisture laden air masses due to is derived from puzzouli, a town in Italy on the Bay of
topographic barriers. Naples near mount reservoirs.
Cyclonic precipitation–It occurs due to pressure It is two types–
difference. (i) Natural Puzzolanas–All puzzolanas are rich in
Convective precipitation–It occurs due to heating of silica and alumina and contain only a small quantity of
air. alkalis.
18. As per IS:456-2000, a continuous beam shall be • Clays and shales which must be calcined to become
considered as a deep beam if the ratio of its active.
effective span to overall depth is less than • Rhenish and Bavarian truss.
(a) 5 (b) 3.5 (ii) Artificial Puzzolanas–
(c) 2.5 (d) 7 • Fly ash
Ans. (c) : A beam is said to be deep beam, if the • Ground blast furnace slag
l • Silica fume
(i) <2 (for simply supported beam)
D • Surkhi
l 22. Which of the following options is an example of
(ii) < 2.5 (for continuous beam)
D an event in the network analysis for managing
Where, l = Effective span a project in construction?
D = Over all depth (a) Excavate Foundation
19. Generally in Chain Surveying, the area to be (b) Mix Concrete
surveyed is divided into number of small (c) Excavation Completed
triangles which should be well conditioned. (d) Assemble Parts
Certain lines control the accuracy of this
Ans. (c) : Event–It is a stage that refers to a particular
triangulation work. In this regard, the line
instant of time that indicates the starting or completion
joining the apex point of a triangle to some
of an activity.
fixed point on its base is known as
Ex. Excavation completed an event in the network
(a) Offset Line (b) Main Line
analysis for managing a project in construction.
(c) Tie Line (d) Check Line
23. Which of the following options is NOT one of
Ans. (d) : Check line– the main objectives of a Site Investigation for a
• It is provided to check the accuracy of the field work. particular construction with respect to soil
• Check line joining the apex point of a triangle to some exploration?
fixed point on its base. (a) To access the soil condition at the site of
• It is also known as proof line. construction from a geological, geotechnical
and aerial perspective
20. As per Indian Standards, the Masonry units for
(b) To access the ground water conditions at the
earthquake resistant buildings, should be
site of construction
constructed with solid concrete blocks having
crushing strength NOT less than (c) Measurement of mechanical properties such
as strength of different soil or rock strata at
(a) 4.5 MPa (b) 5.5 MPa
the site of construction
(c) 2.5 MPa (d) 3.5 MPa
(d) The measurement of vertical distance
Ans. (d) : During the construction of earthquake between two points on an undulated ground at
resistant building, it should be kept in mind that those the site of construction
solid, concrete blocks should be used which have
Ans. (d) : Soil exploration–The object of site
crushing strength not lesser than 35 mPa. However,
exploration is to provide reliable, specific and detailed
depending on the thickness of wall and the number of
information about the soil and ground water conditions
strings, measuring units of higher strength may be
of the site which may be required for a safe excavation
required.
and foundation works, and also an sites liable to be
21. Pozzolanic materials are those that do not have water logged.
cementitious properties themselves, but form Method of soil exploration–
cementitious compounds by reacting with lime
in presence of water. Identify the Artificial 1. Open excavations
Pozzolanic material among the given options. 2. Borings
(a) Clays (b) Volcanic tuffs 3. Sub-surface sounding
(c) Fly Ash (d) Shale 4. Geophysical method

OSSC JE Exam-2019 (Shift-II) 155 YCT

24. Read the following statements and choose the (a) Q = 2500 (P/5 +10)
CORRECT option with respect to the stream (b) Q = 1136.5 (P/5 +10)
function in Kinematics of flow.
(c) Q = 5663 √P
(i) The stream function is defined for only three-
(d) Q = 3182 √P
dimensional flow.
(ii) If the stream function exists, the flow has to Ans. (d) : Fire demand - 1
be rotational. (A) Kuichling's formula–
(a) (i) is FALSE and (ii) is TRUE Q = 3182 P l / min
(b) (i) is TRUE and (ii) is TRUE
Where P = population in thousands
(c) (i) is TRUE and (ii) is FALSE
(B) National Board of fire under writer's formula–
(d) (i) is FALSE and (ii) is FALSE
• For population ≤ 2 lakhs,
Ans. (d) : Stream friction–
• It is defined as the scalar function of space and time, ( )
Q = 4637 P 1 − 0.01 P l / min
such that its partial derivative with respect to any
direction given the velocity component at right angles to • For population > 2 lakhs
that direction. 546 l/min with an extra additional provision of 9100 to
nd
• It is denoted by ψ (Psi) and defined only for two- 36,400 l/m for 2 fire.
dimension flow. 27. In a R.C.C beam section, if the area of steel
Mathematically reinforcement is of such magnitude that the
For steady flow, ψ = f(x, y) permissible stresses in concrete and steel are
developed simultaneously, then the section is
∂ψ known as
Such that =v
∂x (a) Under-reinforced section
∂ψ (b) Over-reinforced section
= −u
∂x (c) Critical section
(d) Prestressed section
• Existence of φ means a possible case of fluid flow.
The flow may be rotational or irrotational. Ans. (c) : Critical section–If the area of steel
reinforcement is of such magnitude that the permissible
25. The moisture content or a point below which a
stresses in concrete and steel are developed
crop or a plant can no longer extract moisture
from the soil for its growth is known as simultaneously, then the section is known as critical
section.
(a) Capillary Point
(b) Hydrated Water x u = ( x u ) l im
(c) Field Capacity Under reinforced section–An under reinforced section
(d) Permanent Wilting Point means, sections, where the yield strain of the steel
Ans. (d) : Permanent welting point (PWP)–It is the occurs at loads lower than the load at which the failure
moisture content at which the films of water around the strain of the concrete occurs.
soil particles are held so tightly that plant roots can not x u < ( x u ) l im
extract enough moisture at suitable rapid rate of
transpiration requirements. Over reinforced section–When the stress in concrete
Value of permanent welting %may be as low as 2% for reaches its permissible value before steel, then the
light sandy soils & may be as high as 30% for heavy section is failed with the sudden collapse of the
clay soils. structure. The resulting beam is known as an over-
26. Fire demand is one of the factors for estimating reinforced beam.
the quantity of water required in a certain x u > ( x u ) l im
community. The Kuchling's formula for
estimating the fire demand (Q) in litres per 28. In case of a R.C.C. rectangular beam, what is
minute is given by the shape of shear stress distribution curve for
(Here P= Population in thousands) the complete section?
(Symbols and notations carry their usual (a) Parabolic (b) Sigmoid
meaning) (c) Zigzag (d) Triangular

OSSC JE Exam-2019 (Shift-II) 156 YCT

Ans. (a): In a homogenous rectangular, section of
concrete, shear stress distribution is parabolic through
the depth. But R.C.C. section is non homogenous
sections and analysis purpose, we assume that concrete
section below neutral axis is under tension and would
not be able to resist any load we ignore this part.

Parallel Activities–Parallel activities in project

management can be defined as a situation where two
activities take place simultaneously without affecting
the performance of each other. It allows to finish
multiple steps at once and helps in finishing the given
task in a shorter frame of time than the frame.
29. Raveling is usually the disintegration of an
asphalt road surface. Which of the following is 31. Which of the following refers to the level
NOT a cause of raveling in bituminous between the building’s window base and floor
pavements? level above ground level?
(a) Inadequate Compaction of the surface (a) Lintel Level
(b) Construction done during rainfall (b) Beam Level
(c) Sill Level
(c) Construction done at high temperatures
(d) Plinth Level
(d) Low bitumen content in the mixture
Ans. (c) : Sill level–It is the base where the windows of
Ans. (c) : Raveling–
the building rest. The level from the floor level to where
• Raveling is the slow disintegration of your asphalt the window opening starts is known as the sill level
from the top down resulting from aggregate loss. • The sill level helps to hold the water out of the
• Essentially the top layer of aggregate breaks free from window and keep the inside dry.
the asphalt binder, which causes raveling issues.
• Issues with raveling begin when your asphalt's
aggregate particles separate from the asphalt's binder.
There are several causes for asphalt raveling including.
(i) Weather issues
(ii) Poor installation
(iii) Separation of aggregated particles
(iv) Mechanical dislodging 32. The geological formation, which does NOT
(v) Low bitumen content in the mixture contain any amount of ground water is an
(vi) Inadequate compaction of the surface. (a) aquitard (b) aquifer
(c) aquiclude (d) aquifuge
30. Which of the following options is an example of
parallel activity during the construction of a Ans. (d) : Aquifuge–The geological formation which
structure in general? are very dense and certain no water in voids and neither
porous nor permeable are termed as aquifuge.
(a) Casting of roof and construction of parapet
wall • Granite rock is an example of aquifuge.
(b) Construction of walls and casting of roofs 33. The type of layout of water supply distribution
(c) Construction of walls and carpentry work of system that has only one main from which sub-
doors and windows mains and laterals branch off, and also most
suitable for an irregularly grown city is
(d) Digging of a septic tank pit and the
(a) ring system
concreting of a septic tank
(b) grid iron system
Ans. (c) : Activity–A project can be broken in to
(c) dead-end system
various jobs in the form of operation & each operation
consuming time or resource is called an activity. (d) radial system

OSSC JE Exam-2019 (Shift-II) 157 YCT

Ans. (c): Dead end system– • The lane width determines on the basis of the width of
the vehicle and minimum side clearance and number of
lanes.
Class of road Width of carriage way
Single lane 3.75 m
Two lanes, with raised kerbs 7.5m
Two lanes without raised kerbs 7.0 m
Multi lane pavement 3.5 m per lane
36. Which of the following options is considered as
• It is followed for old town. general indirect overhead expense in
• It is also called tree system. construction industry?
• Water can reach at a particular at a particular point (a) Travelling expenses
only through one route. Hence it is unidirectional only. (b) Amenities like fitness club to labour
• Many dead end which prevent free circulation of (c) Interest on investment
water. (d) Losses on advance payment
• Stagnant water has to removed periodically by Ans. (a) : Overhead expenses–The ongoing costs
providing scour valves. associated with running a business are different for
construction companies than for a firm operating in any
34. One of the Tacheometric constants in other industry.
Surveying is called Additive, and the other • Overhead expenses for a construction firm can be
constant is termed as: placed in to two broad categories. Direct & indirect.
(a) Dividing Constant The estimate for any construction job includes projected
(b) Indicative Constant overhead coast.
(c) Subtractive Constant A. Indirect or general–Overhead expenses are those
(d) Multiplying Constant that are not specific to any particular job but are fees the
contractor pays on a regular basis.
f Ex.–Travelling expenses.
Ans. (d) : K = , C =f +d
i B. Direct of job–Overhead costs are unique to a
Where, C =Additive constant should be zero specific project and change from job to job.
K = Multiplying constant should be 100 37. For a residential building the livable
area(carpet area) should be 50% to 65% of the
i = Internal between the stadia hairs of the diaphragm
(a) Plinth Area (b) Area of Balcony
d = Horizontal distance between centre × vertical axis
(c) Area of Porch (d) Staircase Area
of tacheometer.
Ans. (a) : Carpet area–
35. Read the following statements and choose the
CORRECT option with respect to Width of • Carpet area of a building is the useful area or livable
area.
Pavement or Carriageway in Highway
Engineering. • Carpet area is calculated by total floor area minus the
circulation area, verandahs, corridors, passages,
(i) Width of the carriage way or the width of the staircase lift, entrance hall, etc.
pavement depends on the width of the traffic
lane and number of lanes. • For a residential building the livable area or carpet
area should be 50% to 60% of the plinth area.
(ii) Width of a traffic lane usually depends on the
height of the vehicle and whether it is geared 38. Which of the following statements is
INCORRECT with respect to the factors
or non-geared type.
considered for selecting the materials for a
(a) (i) is TRUE and (ii) is TRUE Sewer Pipe in general?
(b) (i) is TRUE and (ii) is TRUE (a) The material selected should be highly
(c) (i) is FALSE and (ii) is TRUE pervious
(d) (i) is FALSE and (ii) is FALSE (b) The material should be corrossion resistance
for long life of the pipe
Ans. (b) : Carriage way–
(c) The sewer should have smooth interior
• It is also called width of pavement. The pavement surface
width depends upon the number of lanes and the width
(d) The material selected for sewer should have
of single lane required.
less specific weight
OSSC JE Exam-2019 (Shift-II) 158 YCT
Ans. (a): Important points to be considered before S.N Types of road surface Range of camber in
selecting sewer material– areas of rainfall
(i) Impervious– range
• This property is important in selecting a sewer Heavy Light
material. To eliminate chances of sewage exfiltration
1 Cement concrete and 1 in 50 1 in 60
and infiltration. the material selected for pipe should be
high type bituminous
impervious.
surface
• The material should be corrosion resistance for long
life of the pipe. 2 Thin bituminous 1 in 40 1 in 50
surface
• The sewer should have smooth interior surface.
• The material selected for sewer should have less 3 Water bound macadam 1 in 33 1 in 33
specific weight. and gravel pavement
39. A fixed beam AB of span L carries a uniformly 4 Earth 1 in 25 1 in 33
distributed load of w/unit length over the entire 41. Extra widening refers to the additional width
length, with constant flexural rigidity EI. The of carriageway that is required on a curved
deflection at the centre of the beam will be: section of a road over and above that required
(symbols/notations carry their usual meaning) on a straight alignment. In this regard, the
(a) wL2/248EI (b) wL4/384EI additional width required, due to the tendency
3
(c) wL /48EI (d) wL4/48EI of the drivers to ply away from the edge of the
carriageway as they drive on a curve is known
Ans. (b) : as
(a) Mechanical widening
(b) Psychological widening
(c) Physical widening
(d) Motor widening
wL4
Deflection ( δ ) = Ans. (b) : Extra widening–On horizontal curves,
384 EI especially when they are not of very large radii, it is
Note–The deflection at the centre of a simply supported common to widen the pavement slightly more than the
beam carrying a uniformly distributed load over the normal width.
5wL4 The extra widening of pavement on horizontal curves is
entire length is . This means that the central divided in two parts–
384 EI
deflection for the fixed beam is one-fifth of the central (i) Mechanical (ii) Psychological
deflection of the simply supported beam. Psychological widening–Extrawidth of the pavement is
also provided for psychological reason such as, to
40. Which of the following shapes of the camber is
usually preferred for cement concrete provide for greater maneuverability of steering at high
pavements? speeds, to allow for the extra space requirements for the
overhangs of vehicles and to provide greater clearance
(a) Combination of straight and parabolic
for crossing and overtaking vehicles on the curves.
(b) Parabola
Psychological widening is therefore important in
(c) Straight line pavements with more than one lane.
(d) Semi-Circle
V
Ans. (c) : Camber– (
Psychological widening Wps = )
9.5 R
• Camber is the slope provided to the road surface in the
transverse direction to drain off the rainwater from the Where, V = Design speed (kmph)
road surface. R = Radius of horizontal curve
• It also known as cross slope of the road. 42. Match the following IS-Codes in List I with
Types of camber in road– their corresponding specifications for different
(i) Sloped or straight camber building materials in List II.
(ii) Parabolic or Barrel camber A. IS 3495-1976 i. Specifications for ordinary
(iii) Composite camber portland cement
Straight line camber–This type of camber is usually B. IS 432-1982 ii. Specifications for building
preferred for cement concrete pavements. bricks
OSSC JE Exam-2019 (Shift-II) 159 YCT
 C. IS 269-1989 iii. Specifications for coarse Types of strip foundation–
and fine aggregates (i) Deep strip foundation–The most common type of
foundation is deep strip foundation, which is also the
D. IS 383-1970 iv. Specifications for mild
steel cheapest provided the soil conditions are suitable.
• A reinforced concrete strip support the walls.
v. Specification for timber
(ii) Wide strip foundation–Wide strip foundation is
(a) A - i, B - ii, C - iv, D - iii seen in areas having soils with low load bearing
(b) A - iii, B - iv, C - i, D - v capacity.
(c) A - i, B - iv, C - ii, D - v 45. According to Terzaghi what is the condition for
(d) A - ii, B - iv, C - i, D - iii a shallow foundation?
Ans. (d) : (a) The depth should be always five times the
width
IS 3495-1976 Specifications for building bricks
(b) The depth should be three times the width
(c) The depth should be equal to or less than the
IS 432-1982 ii. Specifications for mild steel width
IS 269-1989 Specifications for ordinary Portland (d) The depth should be greater than the width
cement Ans. (c) : According to Terzaghi–
IS 383-1970 Specifications for coarse and fine • For a shallow foundation, the depth should be equal or
aggregates less than the width.
IS 3629-1986 Specification for timber 46. Vortex flow is defined as the flow of a fluid
43. Read the following assertion and reason along a curved path. Which of the following
statements and choose the CORRECT option. options is NOT an example of 'Free Vortex
Flow'?
Assertion (A): Lime mortars are usually
considered suitable for the construction of (a) Flow of water through the runner of a turbine
chimneys. (b) Flow of water around a circular bend in a pipe
Reason (R): Lime mortars have a property of (c) Flow of fluid through the hole at the bottom
continued strength development over a long of a container
period. (d) Flow of fluid in a centrifugal pump casing
(a) A is true but R is false Ans. (a) : Vortex flow–The motion of a fluid in a
(b) Both A and R are false curved path is known as vortex flow. The vortex flow is
(c) Both A and R are true and R is the correct of two types namely.
explanation of A Free vortex flow–A free vortex motion is that in which
(d) Both A and R are true but R is NOT the the fluid may rotate without any external force applied
correct explanation of A on it.
Ans. (c) : Lime mortars–Lime mortars are usually Example–
considered suitable for the construction of chimneys 1. Flow of liquid through a hole provided at bottom of a
because lime mortars have a property of continued container.
strength development over a long period.
2. Flow of liquid around a circular bend in pipe.
44. Which of the following options is a type of strip 3. A whirlpool in a river.
footing?
4. Flow of fluid in a centrifugal pump casing.
(a) Bearing pile footing
Forced vortex flow–Forced vortex flow is defined as
(b) Friction pile footing
that type of vortex flow, in which some external torque
(c) Well footing is required to rotate the fluid mass.
(d) Shallow foundation Example–
Ans. (d) : Strip foundation–Strip foundation are a type (i) A vertical cylinder containing liquid which is rotated
of shallow foundation that are used to provide a about its central axis with a constant angular velocity ω.
continuous level strip of support for linear structures
(ii) Flow of liquid inside the impeller of a centrifugal
such as walls or closely spaced rows of columns that are
pump.
built on top of the foundation, placed centrally along
their length. (iii) Flow of water through the runner of a turbine.

OSSC JE Exam-2019 (Shift-II) 160 YCT

47. The fluid flows can be defined as Compressible 50. Which of the following options is NOT an
or Incompressible on the basis of Mach advantage of a fixed beam over a simply
Number. The flow will be Incompressible if the supported beam?
Mach Number is (a) For the same loading, the maximum
deflection of a fixed beam is less than that of
(a) more than 0.3 (b) less than 0.3 a simply supported beam
(c) equal to 0.5 (d) more than 0.5 (b) For the same loading, the fixed beam is
Ans. (b) : • If mach number is very high then the fluid subjected to a lesser maximum bending
moment
is considered compressible, and if mach number is low
then for change in pressure the volume remains constant (c) The slope at both ends of a fixed beam is
maximum
making the flow incompressible.
(d) The fixed beam is more stable and stronger
• If mach number is less than 0.3 then it is
Ans. (c) : Advantage of fixed beam over a simply
incompressible flow.
supported beam–
Mach Number Regins • For the same loading, the maximum deflection of a
<1 Subsonic fixed beam is less than that of a simply supported beam.
=1 Sonic • For the same load the fixed beam is subjected to a
lesser maximum bending moment.
0.8 -1.2 Transonic
• The slope at the both ends of a fixed beam is
1.2 - 5 Super sonic maximum.
5 -10 Hypersonic • The fixed beam is more stable and stronger.
> 10 High - hypersonic 51. Read the following statements and choose the
CORRECT option.
48. During the treatment of water, the process (i) The effective depth of a T-Beam is the
which involves addition of chlorine beyond distance between the top of the flange and the
break point chlorination, is known as bottom of the web.
(a) dechlorination (b) post chlorination (ii) For designing purposes, the overall depth of a
simply supported T-Beam is usually assumed
(c) prechlorination (d) super chlorination as 1/12 to 1/15 of the span.
Ans. (d) : Super chlorination– (a) (i) is FALSE and (ii) is FALSE
• Super chlorination is the application of chlorine (b) (i) is FALSE and (ii) is TRUE
beyond the stage of breakpoint. (c) (i) is TRUE and (ii) is TRUE
• The addition of chlorine sufficient to give a residual (d) (i) is TRUE and (ii) is FALSE
chlorine content of 1 to 3 ppm has proved useful to Ans. (b) : Effective Depth of T-beam–Effective Depth
destroy odours and tastes resulting from chloro-products of T-Beam is the distance, between the top compression
formed between the decomposition product from edge to the centre of the tensile reinforcement.
vegetable matter and algae.
• Super chlorine is followed by a retention or contact
period of 30 to 60 minutes, when the residual is
discharged by means of dechlorination agents.
• Super chlorination is adopted when there is an
epidemic in the locality.
49. With respect to wastewater treatment, MPN
index is the measure of
(symbols/notations carry their usual meaning) Over all Depth (D)–
(a) BOD (b) COD For design–Over all depth of a simply supported T-
(c) coliform bacteria (d) dissolved oxygen beam is usually–
1 1
Ans. (c) : MPN Index– • For light loads = to of the span
15 20
• MPN index means the most probable number of
coliform organisms in a given volume of wastewater • For medium loads = 1 to 1 of the span
which, in accordance with statistical theory, would yield 12 15
the observed test result with the greatest frequency. 1 1
• For heavy loads = to of the span
• MPN index is the measure of coliform bacteria. 10 12

OSSC JE Exam-2019 (Shift-II) 161 YCT

52. PIEV theory is used to interpret the reaction 55. What should be the minimum height of a door
time of a driver for calculating sight distance/s opening as a general rule in India?
in highway geometric designs. In this theory 'P' (a) 2.8 to 3 meters (b) 1.8 to 2 meters
stands for (c) 4.8 to 5 meters (d) 3.8 to 4 meters
(a) Performance Time (b) Perception Time
Ans. (b) : • Minimum height of a door opening as
(c) Panic Time (d) Penetration Time general rule in India is 1.8 to 2 meters.
Ans. (b) : According to this theory the total reaction 56. A neutral solution with pH = 7, will have
time of the driver is split in to four parts. hydroxyl ion concentration, equal to
P = Perception (a) 10–9 (b) 109
I = Intellection (c) 10–7 (d) 105
E = Emotion
Ans. (c) : pH = 7
V = Volition
pH = –log[H+]
53. As per I.R.C : 66 -1976, for a design speed of 80
7 = –log[H+] ⇒ [H+] = 10–7
kmph the absolute minimum stopping distance
of a moving vehicle should be approximately We know that
equal to [H+] × [OH–] = 10–14
(a) 80 m (b) 120 m 10–7 × [OH–] = 10–14
(c) 160 m (d) 200 m [OH–] = 10–14 + 7
Ans. (b) : [OH–] = 10–7
Design speed 20 25 30 40 50 60 65 80 100 57. A prismatic compass is a navigation and
kmph surveying instrument which is extensively used
Safestoppingsight to find out the bearing of the traversing and
20 25 30 45 60 80 90 120 180 included angles between them, waypoints (an
dis tan cefor design ' m'
endpoint of the lcourse) and direction. What is
54. Prefabrication concept in construction industry the least count of a prismatic compass?
involves, combination of good design with (a) 20 minutes (b) 30 minutes
modern high performance components and
(c) 1 minute (d) 18 minutes
quality controlled manufacturing procedures.
Which of the following options is an advantage Ans. (b) :
of prefabrication method used for • Least count of prismatic compass is = 30 minute (30')
construction? • Least count of surveyor compass is = 15 minute (15')
(a) This method leads to non‐monolithic • Least count of levelling staff is = 5 mm
construction
58. Using the PERT formula find the approximate
(b) Shuttering and scaffolding is usually not expected time (Te) to paint a room in a
necessary in this method building with the data given below.
(c) Handling and transportation may cause Optimistic Time (O) to paint = 2 Hours
breakages of members during the transit
Most Likely Time (M) to paint = 3 Hours
(d) Skilled labour and supervision is required for
Pessimistic time to paint = 5 Hours
this method
(a) Te = 1.2 Hours (b) Te = 2.2 Hours
Ans. (b) : Prefabrication–Prefabrication construction
(c) Te = 3.2 Hours (d) Te = 4.2 Hours
or prefab is a method of construction that uses
components made off site in a factory, which are then Ans. (c) : Given that
transported put together on-site to create a structure. Optimistic time (tO) = 2 hours
Advantages of prefabrication– Most likely time (tM) = 3 hours
• Building time is thereby decreased, resulting in lower Pessimistic time (tP) = 5 hours
labour costs. We know that
• Reduce the quantity of waste materials relative to t P + 4t M + t O
building an site. Approximate expected time (tE) =
6
• Reduction in construction tie to allow an earlier return
5 + ( 4 × 3) + 2 19
of the invested principal. tE = == 3.2 hours
• Shuttering and scaffolding is usually not necessary in 6 6
this method. tE = 3.2 hours

OSSC JE Exam-2019 (Shift-II) 162 YCT

59. If center to center length of long wall L is 5 m,
and center to center length of short wall S is 3
m and wall thickness is 0.30 m, then the length
of short wall will be
(a) 3.6 m (b) 2.4 m
(c) 2.7 m (d) 3.3 m
Ans. (c) : Length of short wall = (3 – 0.30) = 2.7 m

62. Reciprocal ranging is generally adopted when,

which of the following options is encountered?
(a) A hillock (b) A river
(c) A plain land (d) A pond
Ans. (a) : Reciprocal Ranging–
60. The quality of irrigation water may be affected • When end station is not visible due to there being high
by Toxicity of specific ions, turbidity of water ground between them intermediate ranging rod fixed on
and Sodium-Absorption ratio. This Sodium- the line in cm indirect way this method is known as
Absorption ratio (SAR) is defined as indirect ranging or reciprocal leveling.
(Symbols and notations carry their usual • Reciprocal ranging is generally adopted when a
meaning)
hillock is between two end points.
(a) SAR = Ca+ / { √(Na++ + Mg++)/2}
63. Piles used to transfer load through water or
(b) SAR = Na+ / {√(Ca++ + Mg++)/2}
soft soil to a suitable bearing stratum to
(c) SAR = Mg++/ {√(Ca++ + Na++)/2} minimize settlement is termed as:
(d) SAR = Na+ / {√(Ca++ - Mg++)/2} (a) Fender Piles
Ans. (b) : The quality of irrigation water may be (b) End Bearing Pile
affected by Toxicity of specific ions, turbidity of water
(c) Friction Pile
and Sodium-Absorption ratio.
(d) Compaction Pile
Na +
Sodium-absorption ratio (SAR) = Ans. (b) : Pile–Small diameter shaft which is driven or
Ca ++ + mg ++ bored in to ground.
2
Classification of piles based on various factor–
61. A simply supported beam is subjected to a (i) Function/Action pile–Fender, sheet, batter, tension,
moment M at the centre. The shear force load bearing etc.
diagram will be
(ii) Installation method–Driven, jack, screw & bored
(a) circular in shape piles.
(b) parabolic in shape (iii) Material–Steel, timber, concrete & composite
(c) rectangular in shape piles.
(d) triangular in shape (iv) Displacement of soil–Displacement and non-
Ans. (c) : According to figure– displacement piles.
Using equilibrium equation ∑fy = 0 (v) Mode of load transfer–End bearing, friction and
RA + RB = 0 combined piles.
Taking moments about point A– End bearing pile–
RB × L – M = 0 • These are used in stiff clay and dense sand.
M • End bearing piles, which is also known as paint
RB =
L bearing piles are used to transfer superstructure load
M through low bearing capacity soil to a strong stratum
Therefore, RA = – such as rock or very dense sand and gravel.
L
OSSC JE Exam-2019 (Shift-II) 163 YCT
 67. Box like structures–circular or rectangular
built above ground level and sunk to required
depth from the surface of either land or water,
for major and heavy foundation work such as
under water as in lakes, rivers, seas, and
oceans, are known as
(a) Locks (b) Cofferdam
(c) Caisson (d) Scaffolding
Ans. (c) : Caisson–Box lime staircase-circular or
64. Mild steel is widely used as a construction rectangular built above ground level and sunk to
material and proven to be highly durable. This required depth from the surface of either land or water,
mild steel is usually which type of steel, among for major and heavy foundation work such as under
the given options? water as in lakes, rivers, seas and oceans.
(i) Low Carbon Steel 68. Match the following types of plastics in List I
(ii) Medium Carbon Steel with their corresponding uses in List II.
(iii) High Carbon Steel List I List II
(a) Only (iii)
Types of Uses
(b) Both (i) and (iii) Plastics
(c) Both (ii) and (iii)
A. Polyvinyl i. Paints, Varnishes and Wood
(d) Only (i) Chloride (PVC) adhesive for plywood
Ans. (d) : • Mild steel is a type of carbon steel that B. Phenol ii. Manufacture of Roofing
contains a low level of carbon, otherwise known as low Formaldehyde Sheets
carbon steel.
iii. Manufacture of Rainwater
• Mild steel contains roughly between 0.05% and
Pipes
0.25% of carbon by weight.
65. As per the National Building Code of India (a) (A - i, iii), (B - ii) (b) (A - i), (B - ii, iii)
2005, for a Habitable room, the minimum (c) (A - ii, iii), (B - i) (d) (A - iii), (B - i, ii)
height from the surface of the floor to the Ans. (c) : Polyvinyl chloride (PVC)–
ceiling or bottom of slab should NOT be less Use– Manufacture of roofing sheets
than
Manufacture of Rainwater pipes
(a) 0.75 m (b) 1.5 m
Phenol Formaldehyde–
(c) 2.75 m (d) 1.75 m
Use–Paints, Varnishes and wood adhesive for plywood.
Ans. (c) : • The height of habitable shall not be less
than 2.75m. 69. How many number of storeys can be
constructed for a residential plot of 600 m2 if
• In residential building the height of bathroom should floor space index is 2 and covered area at
not be less than 2.4m. ground floor is 150m2 ?
• Height of store room shall not be less than 2.2 m. (a) 8 (b) 10
66. A frame is usually known as Deficient Frame if (c) 12 (d) 6
the number of members are less than (Given : j
Ans. (a) : Given,
= number of joints)
Plot area = 600 m2
(symbols/notations carry their usual meaning)
Covered area at ground floor = 150 m2
(a) 3j + 2 (b) 2j - 3
Floor spacing index = 2
(c) 3j - 2 (d) 2j + 3
1200
Ans. (b) : No. of floor (Storey) = =8
150
• For deficient frame M < ( 2J − 3) 70. In which type of flow the fluid particles will
flow along stream lines and also rotate about
• For imperfect frame M = ( 2J − 3) their own axis?
(a) Vertical flow (b) Irrotational flow
• For redundant frame M > ( 2J − 3)
(c) Rotational flow (d) Horizontal flow
OSSC JE Exam-2019 (Shift-II) 164 YCT
Ans. (c): Rotational flow–Rotational flow is that type • It is possible, therefore to set out angle of either 45o or
of flow in which the fluid particles while flowing along 90o with this instrument.
stream lines, also rotate about their own axis.
Irrotational flow–If the fluid particles while flowing
along stream lines, do not rotate about their own axis
then that type of flow is called irrotational flow.
71. With respect to Khosla's Theory on diversion
head-works the permissible exit gradient for
fine sand is
(a) between 1/4 to 1/5
(b) 0
(c) 1
(d) between 1/6 to 1/7
Ans. (d) : Permissible Exit Gradien–
Type of soil GE 74. Activity and event are the two basic elements of
a project network. In this regard, the term
1 1
Finesand to Total Float in an activity is equal to
6 7
(a) Maximum time available to complete an
1 1 activity / Activity duration
Coarse sand to
5 6 (b) Activity duration - Maximum time available
1 1 to complete an activity
Shingle to
4 7 (c) Maximum time available to complete an
72. Tension/flexure cracks in a R.C.C beam usually activity - Activity duration
develop (d) Maximum time available to complete an
(a) in a horizontal straight line for the total length activity + Activity duration
of the beam Ans. (c) : Total Float = Maximum time available to
(b) in circular patterns compete activity – Activity duration
(c) in a vertical direction 75. Generally, if the density of a fluid changes from
(d) initially in a horizontal straight line and then point to point in a fluid flow, this type of flow is
incline towards the top of the beam known as
(a) Rotational Flow
Ans. (c) : Tension crack–
(b) Incompressible Flow
• Tension cracks in a R.C.C. beam usually develop in a
vertical direction. (c) Compressible Flow
• These type of crack occurs usually due to shrinkage (d) Irrotational Flow
and temperature variation. Ans. (c) : Compressible flow–If the density of fluid
• Tension crack usually appears those member where changes from point to point in a fluid flow. The type of
restrain is provided in the longitudinal direction and flow is known as compressible flow.
they propagate to the full length of the beam. ρ ≠ constant
73. Cross-staff is usually an instrument used for
Incompressible flow–Incompressible flow is that type
setting out right angles. Which of the following
of flow in which the density is constant for the fluid
angles can be set with the help of French cross-
flow.
staff?
Liquid are generally incompressible while gases are
(a) Both 45° and 90°
compressible.
(b) Any angle between 95° and 120°
ρ = constant
(c) 45° only
(d) 90° only 76. According to IS:456-2000, in a R.C.C column
Ans. (a) : French cross staff–It consists of a hollow the spacing of longitudinal bars measured
octagonal box. along the periphery of the column should NOT
exceed
• Vertical sighting silts are cut in the middle of each
face, such that the lines between the centres of opposite (a) 300 mm (b) 200 mm
silts make angles of 45o with each other. (c) 100 mm (d) 50 mm
OSSC JE Exam-2019 (Shift-II) 165 YCT
Ans. (a): 80. A truss containing 'j' joints and 'm' members,
• Maximum C/C spacing of reinforcement = 300 mm will be a perfect frame if
• Minimum diameter of longitudinal bar = 12 mm (a) j = 2m-3
• Minimum number of bars (b) j = 3m-2
For rectangular column = 4 (c) m = 2j-3
For circular column = 6 (d) m = 3j-2
77. Quantities for the brick masonry work are Ans. (c) : For perfect frame, M = 2J − 3
generally computed in
(a) m (b) cu m For deficient frame M < 2J − 3
(c) kg (d) sq m
Redundant frame M > 2J − 3
Ans. (b) : • Earthwork, stone work, Brick work, wood
work sunshade work = m3 where, M = Member, J = Joints
• Pointing work, distemper, colour washing, mesh work 81. Data given for a concrete mixer is as follows:
= m2 (i) the effective working = 10 hours
• Cornice = unit running meter (ii) the time per batch of concrete = 3 minutes
• Cleaning, Fixing glass panel = No (iii) efficiency = 90%.
78. The unit of measurement in MKS system for The output of concrete mixer of 150 litres
providing skirting in internal walls is generally capacity will be
in (a) 25000 litres (b) 27000 litres
(a) Meters (b) Cubic Meters
(c) 15000 litres (d) 20000 litres
(c) Quintals (d) Number of Pieces
Ans. (b) : The effective working = 10 hours
Ans. (a) : Measurement of striking in internal walls =
meter. The time for batch of concrete = 3 minute
Striking–Striking is a basically a board that runs along efficiency = 90%
the border between interior wall and the floor. 150 × 90
output of concrete mixture = = 135 litres
• It covers the improper edges, avoids scratches from 100
furniture and gives a room a good finish. 10 × 60
79. If a fixed beam is carrying a concentrated load Total no. of batch = = 200
3
W at the centre, with uniform rigidity EI
throughout its span, then its maximum Capacity = 135 × 200 = 27000 litres
deflection would be 82. A road roller is used to compact soil, concrete,
(a) WL3 / 48EI (b) WL3 / 96EI gravel, or asphalt in the construction of
(c) WL3 / 192EI (d) 5WL3 / 384EI foundations and roads. The type of roller that
is large in size, ride-on roller with several rows
Ans. (c) :
of rubber tyres on the front or rear ends which
provide uniform pressure on pavement is:
(a) Tandem Vibratory Roller
(b) Sheepsfoot Roller
(c) Pneumatic Roller
(d) Single Drum Roller
Ans. (c) :
Types of Suitable of soil Nature of
equipment type project
Rammers or All soils In confined area
tempers as files behind
retaining behind
retaining wall,
• There will be two points of contra flexure at a distance basement walls,
of L/4 from the ends. Trench fills
Frog hammer Cohesionless soil For small
WL3
Maximum deflection (y) max = restricted &
192 EI confined areas

OSSC JE Exam-2019 (Shift-II) 166 YCT

Pneumatic tyred Gravel sills, sands Bas, sub base & 85. For a stream of vehicles, if q = flow, k = density
rollers clayey soils, not embankment for and v=space mean speed, then the fundamental
suitable for highways, equation of the Traffic Flow is given by
uniformly graded airfields, earth (symbols and notation carry their usual
soils dam meaning)
Sheep foot Clayey soil pure core of earth (a) q = k - v (b) q = k × v
rollers clay dams (c) q = k / v (d) q = k + v
Vibratory rollers sands Embankments Ans. (b) :
for oil storage Traffic flow (q) = Traffic density (k) × space mean
tank speed (v)
83. Member designed or constructed to resist Traffic density (k)–It is the number of vehicles
inclined thrust in an arch is known as occupying a unit length of a lane or roadway at a given
instant expressed in vehicle/kilometer.
(a) Soffit (b) Voussoirs
Space mean speed (v)–Average speed of vehicles over
(c) Abutment (d) Crown
a certain road length at any time.
Ans. (c) : Arch–An arch is a structure constructed of
• It is the harmonic mean of the speed.
wedge shaped units (bricks or stone), jointed together
with mortar, and spanning an opening to support the • It applicable for study of traffic studies.
weight of the wall above it along with other super 86. In a network diagram of project planning, the
imposed loads. Critical Path is a path that moves along
Abutment–This is the end support of an arch. activities having Total Float equal to
Pier–This is an intermediate support of an arch. (a) Zero (b) Infinity
Soffit–It is the inner surface arch, sometimes intrados (c) Unity (d) Negative
and soffit are used synonymously. Ans. (a) : On a critical path, the total float is zero as the
Voussoirs–These are wedged shaped units of masonry critical path already gives the longest sequence of
forming an arch. activities in a project plan which must be completed on
Crown–It is the highest part of extrados. time for the project to complete on the due date.
84. Read the following statements and choose the 87. Which of the following pile foundations is
CORRECT option with respect to hardness of classified based on the function?
water. (a) Driven Piles (b) Bored Piles
i. The hardness in water is mainly caused by (c) Composite Piles (d) Friction Piles
calcium and magnesium salts that can cause Ans. (d) : Classification of pile based on function–
deposits in the water distribution pipes
1. End bearing pile
leading to damage or reduced efficiency of
flows. 2. Friction pile
ii. When the soluble salts of magnesium and 3. Compaction pile
calcium are present in the form of chlorides 4. Tension pile or uplift pile
and sulphides in water, we call it permanent 5. Anchor pile
hardness because this hardness cannot be 6. Fender pile and dolphins
removed by boiling.
7. Batter pile
(a) (i) is TRUE and (ii) is TRUE
8. Sheet pile
(b) (i) is FALSE and (ii) is TRUE
• Friction piles are used to transfer load through water
(c) (i) is TRUE and (ii) is FALSE
or soft soil to a suitable bearing stratum.
(d) (i) is FALSE and (ii) is FALSE
Ans. (a) : Hardness of water–
• The hardness in water is mainly caused by calcium
and magnesium salts that can cause deposits in the
water distribution pipes loading to damage or reduced
efficiency of flows.
• When the soluble salts of magnesium and calcium are
present in the form of chlorides and sulphides in waters
we call it permanent hardness because this hardness
cannot be removed by boiling.
OSSC JE Exam-2019 (Shift-II) 167 YCT
88. The quantity of door and frame work is 92. The chlorine demand of a water sample was
measured in found to be 0.2 mg/litre. The amount of
(a) meters bleaching powder containing 20% available
(b) cubic meters chlorine to be added to treat one litre of such a
water sample is
(c) square meters
(d) kg/m³ (a) 0.14 mg
(b) 0.06 mg
Ans. (b) : • The quantity of door and frame work is
measured cubic meters. (c) 1.0 mg
89. Floor area ratio (FAR) denotes the maximum (d) 2.33 mg
floor space that can be constructed on a piece Ans. (c) : Given,
of land. FAR is the ratio of Chlorine demand for 1 litre of water = 0.2 mg
(a) Total built-up area of the building to the plot Amount of bleaching powder containing 20% available
area chlorines
(b) Area of verandah to plot area Amount of available chlorine = 0.20
(c) Plot area to carpet area The required amount of bleaching powder–
(d) Plot area to the sum of covered area of all
0.20
floors of the building = = 1 mg
0.20
Ans. (a) : Floor Area Ratio (FAR)–
93. A pipe installed in a drainage system, that
Total builtup area of the building
= conveys the foul air upwards out of a building
Plot area is generally known as
• Floor space index (F.S.I)/Plan efficiency/FAR (a) waste pipe
• FAR is used in classifying type of construction. (b) vent pipe
• The covered area is governed by FAR or FSI (c) soil pipe
• FAR values are specific in National Building code for (d) anti-siphonage pipe
different occupancies and types of construction. Ans. (b) : Vent pipe–A pipe which is provided for the
90. If the maximum built up area is 800 m² and its ventilation purpose to facilitate the exit of foul gases in
covered area at ground floor is 200 m², then the to atmosphere.
number of floors that can be constructed is Soil pipe–A pipe which carries human excreta from
(a) 4 (b) 1 water closet to septic tank is called soil pipe.
(c) 2 (d) 3 • It is not connected to any other pipe except vent pipe.
Ans. (a) : Given, 94. For the pin-jointed truss shown in the figure
Built up area = 800 m2 above, identify whether the truss is
Covered area at ground floor = 200 m2
Maximum builtup area
Number of floor =
Covered area at ground floor
800
= =4
200
91. An irrigation outlet is said to be proportional
when its setting is equal to
(a) channel index + outlet index (i) Statically determinate
(b) channel index - outlet index (ii) Statically indeterminate
(c) outlet index / channel index (iii) Unstable
(d) channel index × outlet index (a) Only (i) (b) Only (ii)
Ans. (c) : Setting of a canal outlet–The ratio of the (c) Only (iii) (d) Both (ii) and (iii)
depth of the water level of the outlet to the full supply Ans. (b) : Given, m = 6, r = 3, j = 4
depth of the distributary is known as setting. Ds = m + r – 2j = 6 + 3 – 2 × 4
Outlet index =9–8=1
Setting of a proportional canal outlet =
Channel index So, member will be statically indeterminate.

OSSC JE Exam-2019 (Shift-II) 168 YCT

95. A three-hinged parabolic arch of span l and 98. Various fixtures like Fish Plate, Chairs,
rise h carries a uniformly distributed load of w Bearing Plates and Anchors are required to
per unit run over the whole span. The keep the rails in position. In this regard, which
Horizontal Thrust at each support will be of the following options is NOT a function of
(a) wl3 / 4h (b) wl2 / 4h Fish Plate in general?
3
(c) wl / 8h (d) wl2 / 8h
(a) It provides full expansion and contraction
Ans. (d) : (b) It holds only the Double Headed rail in
position
(c) It provides resistance against wear
(d) It bears vertical and lateral stresses without
distortion
Ans. (b) : Fish plate–
• These are used for connecting one rail to the next rail.
• Also use to resist heavy transverse shear.
Parabolic arch carries a UDL of w per unit run an entire
span. If the span of the arch is L & its rise is h. • Minimum 4 fish bolt are required to connect 2 fish
plate.
wl 2
HA = HB = • The buckling occurs if fish plates are bolted so tightly
8h
that rails are not allowed to slip/expansion.
96. In the design of R.C.C. footings by Limit State 99. Pensky-Martens apparatus is used to test the
Method, there are two types of shears
sustainability of bitumen and bituminous
considered namely one-way shear and two way
shear. The critical section for the two-way binders used in highways. This test is called as
shear lies at a distance equal to (a) Water content test
(Given d= effective depth of footing) (b) Specific gravity test
(a) d/8 from the face of the column (c) Softening point test
(b) d/12 from the face of the column (d) Fire and flash point test
(c) d from the face of the column
Ans. (d) :
(d) d/2 from the face of the column
• Two types of test apparatus may be used for
Ans. (d) : In LSM
conducting flash point test on bitumen, namely the
The critical section for the two way shear lies at a
pensky mortens closed cup tester and open cup tester.
distance equal to = d/2 fro the face of the column.
For one way shear = d from the face of the column Fire paint–It is the lowest temperature at which
application of a test flame causes the binder material to
where, d = Effective depth of footing.
ignite and burn at least for five second under specified
97. To continue a survey line AB past on obstacle,
test condition.
a line BC 300 m long was set out perpendicular
to AB, and from 'C' angle BCD and was set out 100. Excessive surface trowelling of concrete should
at 45°. Determine the obstructed length BD in be avoided because it may cause a phenomenon
'm'. known as:
(a) 400 m (b) 300 m (a) Flaking
(c) 100 m (d) 150 m (b) Laitance
Ans. (b) : (c) Peeling
(d) Finishing
Ans. (b) : Excessive surface trowelling of concrete
should be avoided because it may cause a phenomenon
known as laitance/
• Laitance is the weak, milky or powdery layer of
300 cement dust, lime and sand fines that appear on the
tan 45o = ⇒ BD = 300 m surface of concrete.
BD
OSSC JE Exam-2019 (Shift-II) 169 YCT
 Odisha Staff Selection Commission
Junior Engineer (Civil)
Exam - 2019 (Shift-III)
1. The Roof which slopes in two directions so that 4. The members of a pin-jointed truss are usually
the end formed by the intersection of slopes is a subjected to bending when
vertical triangle, is (a) the loads are applied at intermediate point on
(a) Hip roof (b) Gambrel roof the member
(c) Gable roof (d) Shed roof
(b) the supports of the truss are hinged
Ans. (c) : Gable roof–The roof slopes in two directions
(c) the material of the truss is brittle
so that the end formed by the intersection of the slopes
is a vertical triangle these are not ideal for the area with (d) the truss is of span less than 3m and loads are
high wind. applied at the joints
Shed roof–This type of roof slopes in one direction only Ans. (a) : Pin-Jointed Trusses–The members of a pin-
and it is used for smaller spans. jointed truss are usually subjected to bending when the
Gambrel roof–This type of roof slopes in two loads are applied at intermediate point on the member.
directions but there is a break in the slope on each side. The bending moments is the member of a truss are after
2. For a steel construction as per CPWD secondary effects. Since the members, the strain energy
specifications, Rivets shall be made from rivet of the system is determined by the displacements of the
bars of mild steel according to which of the joints. A pin-joint truss is essentially a system with
following IS Codes?
finite degree of freedom.
(a) IS 1148 (b) IS 228
(c) IS 814 (d) IS 1367 5. Extra widening refers to the additional width of
Ans. (a) : As per CPWD specification– carriageway that is required on a curved
IS 1148–Hot rolled steel rivet bars (up to 40 mm section of a road over and above that required
diameters) for structural purposes. on a straight alignment. In this regard, the
IS 228–Structural steel (Standard quality) additional width required for a vehicle taking a
IS 814–Covered electrodes for manual metal are horizontal curve is known as
welding of carbon and carbon manganese steel. (a) Mechanical Widening
IS 1367–Technical supply conditions for threaded steel (b) Motor Widening
fasteners. (c) Psychological Widening
3. The sewer pipes usually have to be designed (d) Physical Widening
and checked for
Ans. (a) : Analysis of extra widening on curves–The
(a) only maximum flow
(b) only minimum flow extra widening of pavement on horizontal curve is
(c) both maximum and minimum flow divided into two parts (i) Mechanical widening (ii)
(d) only average flow Psychological widening.
Ans. (c) : The sewer pipes usually have to be designed Mechanical widening–The widening required to
and checked for both maximum and minimum flow of account for the off-tracking due to the rigidity of wheel
velocity. base is called mechanical widening and may be
Minimum velocity–It is also known as self-cleaning calculated as given below formula.
velocity it is the velocity at which there is non- nl 2
settlement of solid particle in a sewer. wm =
The minimum velocity generally followed are– 2R
• Sanitary sewer = .6 m/sec V
Psychological widening (wp) =
• Partially combined sewer = 0.7 m/sec 9.5 R
• Storm sewer = 1 m/sec Where, n = number of traffic lanes
Maximum velocity–It is also known as limiting l = length of wheel base (m)
velocity. It is the velocity that a sewer should not
exceed. The exceeding limit is 2.4 m/sec. If a sewer pipe R = Radius of the curve (m)
exceeds this limit. V = velocity (kmph)

OSSC JE Exam. 2019 (Shift-III) 170 YCT

6. Match the following different Gauges adopted
in India in List I with their corresponding F = Fx 2 + Fy2
Widths in List II.
List I List II Fx = ρg ∫ hdA sin θ
A. Broad Gauge i. 1000 mm
Fy = ρg ∫ hdA cos θ
B. Meter Gauge ii. 1676 mm
C. Narrow Gauge iii. 310 mm So the total pressure force on the projected area of the
iv. 762 mm curve surface on the vertical plane. Thus Fx = Total
pressure force on the projected area of the curved
(a) A - iii, B - ii, C - i
surface on vertical plane.
(b) A - i, B - ii, C - iii
(c) A - iv, B - ii, C - iii 9. The irrigation capacity of a unit of water is
usually known as Duty. In this regards, Duty on
(d) A - ii, B - i, C - iv
capacity is also called as
Ans. (d) : Gauges in Railway track–It is the clear (a) outlet duty
distance between inner faces/running faces of two track
(b) capacity coefficient
rails.
(c) quality duty
Broad gauge = 1.676 m = 1676 mm
(d) full supply coefficient
Narrow gauge = 0.762 m = 762 mm
Meter gauge = 1.0 m = 1000 m Ans. (d) : Duty on capacity–Duty at the head of canal.
It is also known as full supply coefficient. The number
Light gauge = 0.610 m = 610 mm
of hectares irrigable per cumec of the canal capacity at
7. Which of the following options is a type of it's head is called full supply coefficient. The irrigable
Reinforced Concrete Slab directly supported on capacity of a unit of water is usually known as duty.
columns without the use of beams? Duty is expressed in hectare per cumec.
(a) Flat slab (b) Block slab 10. Which of the following general guidelines is
(c) Tiled slab (d) Ribbed slab INCORRECT while planning a Staircase for a
Ans. (a) : Flat slab–It means a reinforced concrete slab building?
with or without drop, supported generally without (a) The respective dimensions of tread and riser
beams, by column with or without flared column heads. for all the parallel steps should be the same in
A flat slab may be a solid slab or may have recesses consecutive floor of a building
formed on the soffit so that the comprises a series of ribs (b) The minimum vertical headroom above any
in two directions. The minimum thickness of the flat step should be 2 m
slab should be 125 mm. (c) Generally, the number of risers in a flight
8. The total force on a curved surface submerged should be restricted to twelve
in liquid is the resultant of horizontal and (d) The minimum width of stair should be 250
vertical forces. In this regard, which force will mm
be equal to the total pressure force on the
Ans. (d) : General guide lines to be considered while
projected area of the curved surface on the
planning a staircase:
vertical plane?
• The respective dimensions of tread and riser for all the
(a) Vertical force on curved surface
parallel steps should be the same on the consecutive
(b) Horizontal force on curved surface floor of a building.
(c) Neither Horizontal nor Vertical force on the
• The minimum width of stairs should be 850 mm.
curved surface
(d) Both Horizontal and Vertical forces on the • The minimum vertical head room above any step
curved surface should be 2m.
• Generally, the number of risers in a flight should be
Ans. (b) : Total pressure on the curved surface–
restricted to 12.
F = ∫ ρghdA • The minimum width of the landing shall be equal to
staircase width.
11. Find the hypotenusal allowance per chain of 30
m length, if the slope is 1 in 4. (Given: value of
1/cosθ = 1.03077)
(a) 0.34 m (b) 0.30 m
(c) 0.16 m (d) 0.92 m
Ans. (d) : Given that,
The total force in x and y directions i.e. Fx and Fy are Chain length (l) = 30m
obtained by integrating dFx and dFy, then total fire on Slope = 1 in 4
the curved surface is
OSSC JE Exam. 2019 (Shift-III) 171 YCT
 1 Soda and These are residues 0.5-1
= 1.03077 potash and if in excess cause
cos θ
sec θ = 1.03077 (Na2O + efflorescence and
K2O) cracking
Hypotenusal allowance = l(sec θ – 1)
= 30(1.03077 – 1) Sulphur Excess of it makes 1-2
= 0.9231m trioxide cement unsound
SO3
0.92m
14. Which of the following options is NOT a type of
12. Normally, shortening of a project duration
dam classified on the basis of function?
results in which of the following options?
(a) Diversion Dam (b) Buttress Dam
(a) Direct and indirect cost both decreases
(c) Storage Dam (d) Flood Control Dam
(b) Direct cost decreases and there is no effect on
indirect cost Ans. (b) : Classification of the DAM
(c) Increase in the direct cost and decrease in the (A) Based on the function–
indirect cost • Storage dam
(d) Increase in the direct cost and also an increase • Diversion dam
in the indirect cost • Detention dam
Ans. (c) : Total project cost in the sum of two separate • Debris dam
cost • Coffer dam
A. The direct cost for accomplishing the work. (B) Based on structure–
B. The indirect cost related to the control or direction of • Gravity dam
the financial overhead, lost production and the hike. • Rock-fill dam
• If indirect cost rises with increased project duration. • Earth dam
• If we increase amount of labour, equipment and time • Arch dam
saving material that too at extra charges which simply • Buttress dam
means increase in direct cost.
• Steel dam
13. Which of the following options is a raw
15. The length of a survey line measured with a 20
material used in the manufacture of cement, m chain was found to be 520 m. When the chain
that is responsible for the grey color of OPC was compared with a standard chain it was
(Ordinary Portland Cement)? found to be 10 cm too long. The true length of
(a) Silica (b) Iron oxide the line in 'm' is
(c) Aluminium oxide (d) Sulphur trioxide (a) 524.2 m (b) 522.6 m
Ans. (b) : Chemical comparison of raw materials (c) 532 m (d) 517.4 m
used in the manufacture of cement Ans. (b) : Given that,
Oxide Function Composition Actual length of chain (l) = 2 m
(%) Measure length of line (L') = 520 m
Lime, Cao Control strength and 60-65 Corrected length of chain (l') = 20.10 m
soundness its True length of line (L) = ?
deficiency reduce therefore, we know
strength and setting L × l = l' × L'
time l'
L = ×L'
Silica, SiO2 Gives strength, 17-25 l
excess of it causes 20.1
slow setting L= × 520
20
Alumina, Responsible for quick 3-8 L = 522.6 m
Al2O3 setting, if in excess, it
lower strength 16. In the figure given below, the network of a
small part of a project represents what among
Iron oxide, Gives colour and 0.5-6 the following options?
Fe2O3 helps in fusion of
different ingredients.
It is responsible for
the grey colour of (a) Excavation and sand filling & ramming may
OPC be start together
Magnesia Imparts colour and 0.5-4 (b) Event 3 will occur after 2 days after
hardness occurrence of event 2

OSSC JE Exam. 2019 (Shift-III) 172 YCT

 (c) Excavation of a footing takes 2 days to 19. Read the following statements and choose the
complete CORRECT option with respect to the
(d) Event 3 will occur after 2 days after components of a railway track.
occurrence of event 1 (i) The Granular bed resting on formation and
Ans. (b) : supporting sleepers over it is known as
Ballast.
(ii) The Steel sections laid along two parallel lines
over sleepers are known as Rails.
As per given diagram, event 3 will occur after 2 days (a) (i) is TRUE and (ii) is TRUE
after occurrence of event 2 (b) (i) is TRUE and (ii) is FALSE
17. A horizontal structural member provided (c) (i) is FALSE and (ii) is TRUE
above the doors and windows of a building to (d) (i) is FALSE and (ii) is FALSE
support load of the wall above it is known as Ans. (a) : Track or permanent way is the single costliest
(a) Slab (b) Stair of Indian Railways. It consist of rails, sleeper, fitting
(c) Lintel (d) Masonry and fastenings, ballast and formation
Ans. (c) : Lintels–Lintels is horizontal structural • The ballast is a layer of broken stones, gravel,
member which is fixed over the openings viz. doors, maximum or only other granular material placed and
windows etc. to support the structure or opening lintels packed below and around sleepers to the formation
behave just like beam. It provides a bearing for the • The steel sections laid along two parallel lines over
masonry above the openings and transfers all the loads sleepers are known as rails or rails are the members of
acting over the opening to the supporting walls. the track laid in two parallel lines to provide on
unchanging, continuous level surface for the movements
Or
of trains.
A horizontal structural member provided above the
doors and windows of a building to support load of the
wall above it is known as lintel.
18. Read the following statements and choose the
CORRECT option with respect to objectives of
Sewage collection and disposal.
(i) An efficient sewerage scheme can provide a
good sanitary environmental condition of city
protecting public health. 20. While submitting the tender form, the
(ii) An efficient sewerage scheme can also contractor has to deposit a certain amount
dispose of all liquid waste generated from called Earnest Money. Usually this deposit can
community to a proper place to prevent a be approximately taken as
favorable condition for mosquito breeding, fly (a) 20% to 30% of the estimated cost of work
developing or bacteria growing. (b) 0.1% to 0.2% of the estimated cost of work
(a) (i) is FALSE and (ii) is FALSE (c) 1% to 2% of the estimated cost of work
(b) (i) is TRUE and (ii) is FALSE (d) 50% to 60% of the estimated cost of work
(c) (i) is TRUE and (ii) is TRUE Ans. (c) : Earnest money Deposit–
(d) (i) is FALSE and (ii) is TRUE While submitting any tender, the contractor is required
to deposit certain amount with tender document to the
Ans. (c) : Objectives of sewage collection and department as a guarantee that is his tender is accepted.
disposal– This amount of deposit is called 'Earnest Money
(i) An efficient sewerage system can provide a good Deposit'. The amount of earnest money is 1 to 2 percent
sanitary environmental condition of city protecting of estimated cost of work.
public health. 21. The process of ‘lagooning’ serves which of the
(ii) An efficient sewage scheme can also disposal of all following purposes in a wastewater treatment
liquid waste generated from community to a proper plant?
place to prevent a favorable condition for mosquito (a) Helps in the disposal of sludge
breeding, fly developing or bacteria growing. (b) Increases the capacity of storage reservoir
(iii) Sewage is collected by sewer pipes and It is (c) Increases flow of a sewage through imhoff
transferred for treatment or is discharged. Lack of tanks
sewage system is taken care by collecting waste from (d) Reduces the excessive flow in sewer
house holds through pipes into septic tanks and it is then Ans. (a) : Use of lagoons for Disposal of Raw sludge–
treated or collected for disposal. This method is used at smaller places for disposing of
(iv) Sewage is disposal of using on-site sanitation raw sludge without digestion In this method the raw
procedures or removed by water borne sewer system. sludge is kept at rest in a large shallow open pond called
OSSC JE Exam. 2019 (Shift-III) 173 YCT
a lagoon. The detention period is 1 to 2 months and may Long wall length (out-to-out) = centre to centre length +
extend up to 6 months. During its detention in the one breadth
lagoon, the sludge undergoes anaerobic digestion, Long wall length = l + d
thereby getting stablised. Short wall length (in-to-in) = centre to centre length –
22. The reduction in gradient at the horizontal one breadth
curve because of the additional tractive force Short wall length = l – d
required due to curve resistance is known as 25. For analysing two-dimensional irrotational flow
Grade Compensation. As per Indian Railway problems, a flownet is required that is usually
provision, the grade compensation per degree drawn using a series of lines. In this regard, a
of curve for Broad Gauge track is line along which the velocity potential is
(a) 0.03% (b) 0.05% constant is known as
(c) 0.04% (d) 0.02% (a) Equipotential Line (b) Stream Line
Ans. (c) : Grade compensation on curves (c) Streak Line (d) Di-potential Line
70 Ans. (a) : Velocity potential–When the flow is
• On BG tracks, 0.04% per degree of the curve or ,
R P

whichever is minimum irrotational the value of the integral ∫

A
q s ds (where qs
• On MG tracks, 0.03% per degree of the curve or represents the component of velocity along an element
52.5 ds of the curve AP) is independent of the path between
, whichever is minimum A and P
R
P
• On NG tracks, 0.02% per degree of the curve or
35
, − φ = ∫ q s ds
A
R
whichever is minimum
Where R = radius of curve in meters.
• The gradient of a curved portion of the section should
be flatter than the ruling gradient because of the extra
resistance offered by the curve.
23. Which of the following types of lime usually ∂φ
We have δφ = –qs δs or qs = − (The reason for the
slakes with difficulty and is mostly used in the ∂s
constructions near damp places? minus sighn will appear in a moment). The function φ is
(i) Feebly Hydraulic Lime termed the velocity potential if the line element of
(ii) Moderately Hydraulic Lime length δs, is perpendicular to a streamline, then qs = 0
(iii) Eminently Hydraulic Lime and so ∂φ = 0. Thus the velocity potential is constant
(a) only (ii) (b) only (i) along lines perpendicular to streamlines. These lines of
(c) All (i), (ii), (iii) (d) only (iii) constant velocity potential are known as equipotential
Ans. (d) : Hydraulic lime (water lime)–This lime lines.
contains clay and some amount of ferrous oxide. It sets The velocity potential provides an alternative means of
under water. I.S.I has divided hydraulic lime in three expressing the velocity components parallel to the
classes namely– coordinates axis in irrotational flow
A. Class A – Eminently hydraulic −∂φ −∂φ
B. Class B – Semi-hydraulic u= , v=
∂x ∂y
C. Class C – Non-hydraulic
Class A (Eminently hydraulic–This lime contains 26. In an open channel flow of fluids, Hydraulic
about 25% clay content and set readily under water Jump usually occurs when the flow changes
within a day or so. This lime slakes with difficulty. The from
mortar and lime concrete prepared from this lime a very (a) sub-critical to super-critical state
useful for construction under water or in damp places. (b) super-critical to sub-critical state
24. In long wall and short wall method of a (c) laminar to turbulent state
building estimation, if 'l" is the center to center (d) critical to turbulent state
length of wall "d" is the breadth of the wall, Ans. (b) : Hydraulic jump–The hydraulic jump is
then length of long wall (out to out) is given by based on the froude number hence the critical for an
(a) l + d (b) l – d open channel flow is another important physical issue.
(c) l + 2d (d) l – 2d Hydraulic jump is basically issue. Hydraulic jump is
Ans. (a) : Given that, basically a physical situation involving both
Centre to centre length of wall = l supercritical flow and subcritical flow. In an open
breadth of wall = d channel flow of fluids, hydraulic jump occurs whereever
For long wall and short wall method a supercritical flow changes to a subcritical flow.

OSSC JE Exam. 2019 (Shift-III) 174 YCT

Classification of hydraulic jump If the depth of actual neutral axis (xa) is lesser than the
Froude number depth of critical neutral axis (xc) the the section is called
(i) Undular jump – 1.0-1.7 under reinforced section and in the case the steel in the
(ii) Weak jump – 1.7-2.5 tensile zone attains its maximum stress earlier than the
(iii) Oscillating jump – 4.5-9.0 concrete in compressive zone. It is failure is ductile. In
this section rate of increase of stress in steel is more than
27. Which of the following types of arches is a
concrete.
statically determinate structure?
Note–
(a) Single Hinged Arch (b) Two Hinged Arch
• When xa = xc for balanced section
(c) Three Hinged Arch (d) Fixed Arch
• If xa > xc then it is called over reinforced section.
Ans. (c) : A three Hinged Arch–This type of arch has
two hinges one at each end supports and one additional 29. Which of the following options is generally
intermediate hinge generally at the crown. considered as the longest of the main survey
A three hinged arch is a statically determinate structure lines in Chain Surveying?
since only four independent reaction components, with (a) Base Line (b) Oblique Offset
two reactions at each end support, develop in such (c) Tie Line (d) Check Line
arches. Ans. (a) : Base line–In triangular survey a single line is
• Three hinged parabolic arc of span L and rise 'h' measured very accurately and the sides of the other
carrying a UDL over the whole span triangles of the connected network are computed from
the measured length by solving the triangles
trigonometrically such line is called the 'base line'. In
chain survey the term 'Base line' is also used as the
longest line of the main survey lines in chain surveying

DS = 0
BMR = 0
wl 2
H=
8h
wx 2
mA = VPx – − Hy
2
where H = horizontal thrust
Where AB, CD – Base line
wl
VP = Vertical reaction at P = AB, BC, CD, DE, EA, AC – Main survey lines
2 The line – Dr, Ef, Eh, sh
Hy = H-moment Check line – mn, pf
DS = Degree of static indeterminacy
30. Which of the following construction equipments
BmR = Bending moment at C
is suitable for improving the strength, by
28. Generally, in R.C.C sections, the depth of the decreasing the porosity and increasing the
neutral axis usually determines the type of density of Soil?
section. For an under reinforced section, the (a) Compaction Equipment
actual neutral axis lies:
(b) Hauling Equipment
(a) below the critical neutral axis
(c) Hoisting Equipment
(b) in the same line along the critical neutral axis
(d) Excavating Equipment
(c) above the critical neutral axis
(d) both above or below the critical neutral axis Ans. (a) : Compaction of soil is defined as the process
by which soil particles are rearranged and packed
Ans. (c) : For under reinforced section together to decrease its porosity and increase its dry out
xa < xc weight, density of soil. The process of compaction
involves the expulsion of pore air only. Compaction of
soil mass is done to improve its engineering properties.
Compaction generally increases the shear strength of the
soil and hence the stability and bearing capacity. It is
also useful in reducing the compressibility and
permeability of the soil, therefore compaction
equipment is suitable for improving the strength, by
decreasing the porosity and increasing the density of
soil.
OSSC JE Exam. 2019 (Shift-III) 175 YCT
31. Gap graded aggregate is recommended for Duty (D) = ? Ha/cumec
mixes of low workability since 8.64 B
(a) they have high tensile strength ∆=
D
(b) they show greater tendency to segregation in 8.64 B
high workability mixes D=
∆
(c) they have less percentage of coarse aggregates
and specific surface is more 8.64 × 28
D=
(d) they high compressive strength 19 ×10−2
Ans. (b) : Gap-graded aggregate– D = 12.7326 × 102
Aggregate particles of a given size pack so as to form D = 1273.26 hectares/cumec
voids that can be penetrated only if the next smaller size 34. Surveying is done to take measurements of
of particles is sufficiently small. This means that there objects in a horizontal plane for locating objects
must be a minimum difference between the sizes of any on the surface of the earth, before any
two adjacent particle fractions. construction of a structure is started. During
Gap graded aggregate can be used in any concrete but this process the fundamental principle that is
there are particular uses, pre-placed aggregate concrete followed is to work from the
and exposed aggregate concrete where a pleasing finish (a) Part to the whole
is obtained. However, to avoid segregation gap-grading (b) Lower level to higher level
is recommended mainly for mixes of relatively low (c) Higher level to lower level
workability that are to be compacted by vibration, good (d) Whole to the part
control and care in handling are essential. Ans. (d) : Surveying–Surveying may be defined as an
32. Granite stone is a widely used building stone, art to determine the relative positions of points on,
mainly composed of quartz, feldspar and mica. above or beneath the surface of the earth, with respect to
Which of the following options is INCORRECT each other, by measurements of horizontal and vertical
about this stone in general? distances, angles and directions.
(a) Used as a sound reflecting material Principle of surveying–There two basic principles of
surveying–
(b) Used in the preparation of various types of
wall putties (i) To work from whole to part.
(ii) To locate a point by at least two measurements.
(c) Used in the building walkways, driveways
and pavements (i) It is the main principle of surveying and a method
violating the principle of working from whole to part
(d) Used in the preparation of tiles for floors and should not be adopted until and unless there is no
countertops of kitchens alternative. The main idea of working from whole to
Ans. (b) : part is to localize the errors and prevent their
accumulation.
35. Identify the test to check the presence of poorly
weathering calcium carbonate in sandstone as a
building material among the given options.
Igneous Rock–Igneous rocks are those that have (i) Crushing Test (ii) Smith's Test
crystallized from molten rock (magma). The most (iii) Acid Test (iv) Slump Test
common igneous rock used for building in the UK is (a) Both (i) and (ii) (b) Both (iii) and (iv)
granite. Granites are coarse grained acidic rocks and (c) only (iii) (d) only (iv)
consist mainly of quartz, mica and feldspars. Their Ans. (c) : Testing of stones–There are many type of
porosity is usually very low, granite is commonly used testing–
in buildings and elsewhere including for facade cladding (i) Durability Test, (ii) Smith Test, (iii) Brad's Test
panels, paving, internal floors and accessories such as (iv) Acid Test, (v) Crystallization Test (vi) Crushing
kitchen worktops. Test
33. For a crop, if the optimum depth of kor Acid Test–It is to check weather resistance. It confines
watering is 19 cms, then the outlet factor (outlet the power of stones to withstand the atmospheric
duty) for the crop for a four week period in conditions. 100g of stone chips are kept in 5 percent
hectares/cumec, is solution of H2SO4 or HCl for 3 days. Then the chips are
(a) 1720.63 hectares/cumec taken out and dried. The sharp and firm corners and
edges are indication of sound stone. This test is used to
(b) 1472.26 hectares/cumec
test the cementing material of sand stones.
(c) 1273.26 hectares/cumec
36. For Stone Masonry or structural concrete, the
(d) 1372.26 hectares/cumec
categories for measurement of items, should be
Ans. (c) : Given that measured separately and the heights should be
Delta (∆) = 19 cms = 19 × 10–2 m described first from foundation to plinth level
Base period (B) = Four week = 4 × 7 = 28 day and then from
OSSC JE Exam. 2019 (Shift-III) 176 YCT
 (a) plinth level to second floor So, the given truss is statically indeterminate.
(b) plinth level to third floor 40. When an inclined load is acting on a beam, then
(c) second floor to fourth floor the inclined load is resolved into two
(d) plinth level to first floor components namely vertical and horizontal.
Ans. (d) : In case of masonry (stone or brick) or Read the following statements and choose the
structural concrete, the categories shall be measured CORRECT option with respect to effects of
separately and the height shall be described as: these components.
• From foundation to plinth level i. The Vertical Component will cause axial
thrust in the beam.
• From plinth to first floor level
ii. The Horizontal Component will cause Shear
• From first floor to second floor level and so on. Force and Bending Moment in the beam.
37. The percentage of estimated cost of building (a) (i) is TRUE and (ii) is TRUE
works for sanitary and water works are usually (b) (i) is FALSE and (ii) is FALSE
provided as
(c) (i) is FALSE and (ii) is TRUE
(a) 50% to 60% (b) 22% to 42%
(d) (i) is TRUE and (ii) is FALSE
(c) 8% to 10% (d) 4% to 6%
Ans. (b) : Beam with inclined loading–When a beam
Ans. (c) : The percentage of estimated cost of building
subjected to load P which is inclined at an angle θ with
works for sanitary and water works are usually provided
axis of beam so P will resolved in two components:
as 8% to 10%.
1. Vertical or transverse component
• Tools and plants = 1-1.5% of project cost
2. Horizontal or axial component
• Sanitary and water supply = 8% of project cost
• Electrification = 8% of project cost
• Departmental cost = 10-15% of project cost
• Contingency = 3-5% of project cost
38. In weirs and barrages, for the oscillating jump
of flows, the Froude number usually lies in the
range of
(a) 10 to 11 (b) 4.6 to 9.0
(c) 1 to 2.0 (d) 2.5 to 4.5
Ans. (d) : In weirs and barrages, for the oscillating jump
of flows, the froude number lies in the range at 2.5 to • Vertical component – Shear force and bending
4.5. This gives rise to the heavy moment
Classification of hydraulic jump– • Horizontal component – Direct force (push or pull)
Jump Froude number 41. Which of the following instruments for
Undular jump – 1–1.7 chaining in surveying, is used as a centering aid
Weak jump – 1.7–2.5 in plane table, compass and theodolites?
Oscillating jump – 2.5-4.5 (a) Offset Rod (b) Ranging Rod
Steady jump – 4.5-9 (c) Plumb Bob (d) Tape
Strong jump – >9 Ans. (c) : Plumb bob–While chaining along sloping
39. For the pin-jointed truss shown in the figure ground, a plumb bob is required to transfer the points to
above, identify whether the truss is the ground. It is also used to make ranging poles vertical
(i) Statically determinate and to transfer points from a line ranger to the ground.
(ii) Statically indeterminate In addition, it is used as centering aid in theodolites,
(iii) Unstable compass, plane table and a variety of other surveying
instruments.

(a) Only (i) (b) Only (iii)

(c) Only (ii) (d) Both (i) and (iii)
Ans. (c) : Degree of static indeterminacy,
Ds = m + r – 2j
=6+3–8
=1

OSSC JE Exam. 2019 (Shift-III) 177 YCT

42. A track structure which permits movement of • It is found that the weight of 1 m3 of bricks earth is
train from one track to another refers to which about 1800 kg. Hence the average weight of a brick will
of the following options? be about 3 to 3.50 kg.
(a) Bearing Plate (b) Anchor 46. Read the following statements and choose the
(c) Stuart Key (d) Turnout CORRECT option with respect to the
Ans. (d) : Turnout–It is an arrangement of points and precautions taken against cavitation.
crossings with lead rails by means of which the rolling (i) The pressure of the flowing liquid in any part
stock may be diverted from one track to another. of the hydraulic system should be above its
Or vapour pressure.
A complete set of points and crossings, along with lead (ii) The cavitation resistant materials such as
rails is called a turnout. aluminium-bronze and stainless steel should
43. The versatile equipment used for clearing the be used for Centrifugal Pumps or components
construction sites of debris and rubbish, and of the hydraulic system.
very efficient for short haul application up to (a) (i) is FALSE and (ii) is TRUE
100 m is (b) (i) is TRUE and (ii) is TRUE
(a) Dumper Truck (b) Tipper
(c) (i) is TRUE and (ii) is FALSE
(c) Crane (d) Bulldozer
(d) (i) is FALSE and (ii) is FALSE
Ans. (d) : Bull dozers–A heavy steel blade mounted on
the front of a tractor. This blade pushes the material Ans. (a) : Precaution Against Cavitation–The
from one place to another. These are efficient following precautions should be taken against cavitation
excavating tools for short hauling application upto (i) The pressure of the flowing liquid in any part of the
100m. hydraulic system should not be allowed to fall below its
44. Read the following statements and choose the vapour pressure. If the flowing liquid is water, then the
CORRECT option. absolute pressure head should not be below 2.5 m of
(i) A freshly cut stone carries some natural water.
moisture known as Quarry Sap, which makes (ii) The certain materials are more resistant to cavitation
it soft and workable. or pitting action, such as stainless steel, aluminium and
(ii) To make the stone weather resistant and bronze. The coating of these materials or the materials
useful for constructions, a process called as such are used in the manufacturing of the pumps or
dressing is done that removes the Quarry Sap components of the hydraulic system.
from the stone. In centrifugal pumps, the cavitation can occur at inlet of
(a) (i) is TRUE and (ii) is FALSE the impeller of the pump.
(b) (i) is TRUE and (ii) is TRUE 47. The general form of few empirical formulae for
(c) (i) is FALSE and (ii) is FALSE estimating flood discharge is Q = CAn. If in this
(d) (i) is FALSE and (ii) is TRUE expression Q = Flood discharge and A =
Ans. (a) : Seasoning of Stone–A freshly cut stone Catchment Area, then C and n respectively
carries some natural moisture known as quarry sap denotes
making it soft and workable. The quarry sap is mineral (a) Flood Index and Flood Coefficient
solution and reacts chemically with the mineral (b) Flood Coefficient and Flood Index
constituents when the stone is exposed to atmosphere (c) Water Coefficient and Water Index
after quarrying the process takes about 6 to 12 months (d) Flood Factor and Water Index
for complete seasoning when the quarry sap evaporates.
It leaves a crystalline film on the faces of the stone and Ans. (b) : Empirical method–Some of the empirical
makes them weather resistant. The dressing before formulae for estimating flood are given below, most of
seasoning improves the weather resistance as such the these are in the form:
dressing carving and moulding etc. should be done as Q = CA n
early after quarrying.
Where, Q = Flood distance, A = Catchment Area
45. The average mass of a standard brick is
approximately in the range of C = Flood coefficient, n = Flood index
(a) 1 to 1.5 kg (b) 7 to 7.5 kg Dicken formula (1865)
(c) 5 to 5.5 kg (d) 3 to 3.5 kg Q p = C D A 3/ 4
Ans. (d) : Size and weight of bricks–
Where, QP = Maximum flood discharge (m3/s)
Standard size (modular brick) of brick = 19 cm × 19 cm
× 9 cm A = Catchment area (km2)
Nominal size of brick = 20 cm × 10 cm × 10 cm CD = Dicken constant with value between 6 to 30

OSSC JE Exam. 2019 (Shift-III) 178 YCT

 Region Value of CD in above equation (iii), the expression (AC + mAS) is
called the equivalent area of the section in terms of
North − Indian plains 6 concrete.
North − Indian hilly regions 11 − 14 A e = A C + mAS = ( A − AS + mAS )
Central India 14 − 28 Ae = A + (m – 1)AS
Coastal Andhra and Orissa 22 − 28 Where, A = total cross-sectional area of the original
48. The ratio of the area irrigated in the Kharif column section.
season to the area irrigated in the Rabi season 50. In a RCC column, as per IS 456: 2000, the
is known as maximum pitch of the lateral ties (transverse
(a) Crop Ratio (b) Irrigated Ratio reinforcement) should be equal to what among
(c) Season Ratio (d) Overlap Ratio the following options?
(i) Least lateral dimension of the compression
Ans. (a) : Crop ratio–It is the ratio of the area irrigated members.
in the kharif season to the area irrigated in the rabi (ii) sixteen times the smallest diameter of the
season is known as crops ratio. longitudinal reinforcement bar to be tied.
Kharif season (iii) 300 mm.
Crop ratio =
Rabi season (a) Least of (i), (ii) and (iii)
(b) Always (i)
49. If in a reinforced concrete column the cross (c) Maximum of (i), (ii) and (iii)
sectional area of a steel bar is As and that of (d) Always (iii)
concrete is Ac, then the equivalent area of the
Ans. (a) : Lateral ties–The arrangement of lateral ties
section in terms of concrete is expressed as
should be effective in fulfilling the above requirements
(Given m = modular ratio) in RCC column as per IS 456 : 2000, the maximum
(a) AS - mAC (b) AS + mAC pitch of the lateral ties are given below
(c) AC - mAS (d) AC+mAS
long max /4
Ans. (d) : Equivalent areas of composite section–A tie diameter φt ≥ 
reinforced concrete column area to cross-section AS of 6 mm
steel bar and AC of that of concrete such that D
A = AS + AC 
tie Spacing (pitch) s t ≥ 16 φ long min
If pC is the stress in concrete and pS is stress in steel 300mm
under an external compressive load P 
P = PS + PC Where, D = least lateral dimension of compression
members.
P = ps.As + pC.AC ------ (i)
Again if we assume a perfect bend between steel and 51. Which of the following options is NOT an
concrete then compressive strain in both be equal. example of Instrumental errors in Plane Table
Surveying?
pS p C
Hence, = (a) The needle of the trough compass may not be
ES E C perfectly balanced
E (b) The table may be loosely joined with the
pS = S .pC = mpC ------- (ii) tripod stand
EC
(c) The alidade may not be correctly centered on
the station point
(d) The fiducial edge of the alidade might not be
straight
Ans. (c) : Errors in plane Table Surveying–The errors
in plane table surveying are of three types–
1. Instrumental errors 2. Errors in plotting
3. Errors due to manipulation and sighting
1. Instrumental errors–Instrumental errors are the
primary source of errors in plane table surveying which
can occur in the following ways:
Substituting the value of pS in (i), we get • Errors will occur if the top surface of the plane table is
AS.m.pC + AC.pC = P not flat or contains undulations.
P P • The fittings of the tripod and plane table should be
pC = = ------ (iii) tightly fastened loose fittings can make the plane table
A C + m.AS A e unstable and cause errors while drawing.

OSSC JE Exam. 2019 (Shift-III) 179 YCT

• The magnetic compass used in plane table surveying (a) (i) is TRUE and (ii) is FALSE
should represent accurate direction. (b) (i) is TRUE and (ii) is TRUE
• When the edge or fiducial edge of the alidade is not (c) (i) is FALSE and (ii) is FALSE
straight or curved an errors occurs in the drawing. (d) (i) is FALSE and (ii) is TRUE
52. Read the following statements and choose the Ans. (c) : Form work (Centering & Shuttering)
CORRECT option with respect to the rules for Centering–In case of structure with two or more floors,
drawing a network for project management. the weight of concrete, centering and shuttering of any
(i) Each activity is represented by one and only upper floor being cost shall be suitably supported on
one arrow in the network. one-floor below the top most floor already cost.
(ii) All the arrows must run from right to left. • Formwork and concreting of upper floor shall not be
(iii) For coding alphabets are used for all activities done until concrete of lower floor has set at least for 14
including the dummy activity and numbers for days.
events. Shuttering–Shuttering used shall be of sufficient
(a) (i) and (iii) are FALSE and (ii) is TRUE stiffness to avoid excessive deflection and joints shall be
(b) (i) and (iii) are TRUE and (ii) is FALSE tightly butted to avoid leakage of slurry. In case the
(c) All (i), (ii) and (iii) are TRUE height of centering exceeds 3.5 meters, the pro may be
(d) All (i), (ii) and (iii) are FALSE provided in multi stages.
Ans. (b) : Rules for drawing a network– • No deductions from the shuttering due to the
openings/obstructions shall be made if the area of each
• There is only one starting event. Arrows come out of
this event and none of the arrows enters it. openings/obstructions does not exceed 0.4 square metre.
Nothing extra shall be paid for forming such openings.
• Each activity is represented by an arrow.
56. For the conveyance of water various types of
• An event can not occur more than once, i.e. the
valves are used for different purposes in a
network cannot loop back.
water distribution system. In this regard, the
• An end event occur only after completion of all valve, which allows the flow of water only in
activities. one direction, is known as
• The length of the arrow does not have any relation (a) Sluice valve (b) Butterfly valve
with the duration of the activity. (c) Check valve (d) Gate valve
53. The minimum time in which activity can be Ans. (c) : Check valve–These posses some automatic
finished but leads to the maximum cost of device which allow the water to flow in one direction
project is known as only
(a) Standard Time (b) Slow Time
• It is also called reflux valve.
(c) Normal Time (d) Crash Time
• This valve is provided in the pipe line which draws
Ans. (d) : Crash Time–Crash time is the minimum water from the pump.
possible time in which an activity can be finished or
Sluice valves–
completed, by employing extra resources, but leads to
the maximum estimator would usually allow for an • It is also known as gate valve
activity. • It is provided to regulate the flow of water through the
54. The revised estimates are prepared when the pipe and are essential to divide the main line into several
expenditure on work exceeds or is likely to sections.
exceed the amount of administrative sanction 57. Which of the following options is the
by more than CORRECT statement, according to Lami's
(a) 5% (b) 1.5% Theorem?
(c) 3.5% (d) 0.5% (a) If three forces acting at a point are in
equilibrium, then each force is proportional to
Ans. (a) : Revise estimate–Estimate is revised if–
the sine of the angle between the other two
(A) When cost of work exceeds 5% of original
sanctioned cost (b) If three forces acting at a point are in
equilibrium, then each force is proportional to
(B) When cost of work exceeds 10% of administrative the tangent of the angle between the other two
sanction.
(c) The three forces acting at a point kept in
55. With reference to CPWD specifications for equilibrium must be equal to each other
formwork in general read the following
(d) If three forces acting at a point are in
statements and choose the CORRECT option.
equilibrium, then each force is proportional to
(i) Centering, and shuttering where exceeding 1.5 the cosine of the angle between the other two
metre height in one floor shall be measured
and paid for separately. Ans. (a) : Lami's theorem–Lami's theorem states that,
if three forces acting on a particle keep it in equilibrium,
(ii) No deductions from the shuttering due to the
then each force is proportional to the sine of the angle
openings/ obstructions shall be made if the
between the other two and the constant of
area of each openings/ obstructions does not
proportionality is the same
exceed 0.1 square metre.
OSSC JE Exam. 2019 (Shift-III) 180 YCT
 Ans. (d) : Gauge pressure–When pressure is measured
by taking atmospheric pressure as datum, it is called
gauge pressure. All pressure gauges show zero value
when open to the atmosphere it indicates only the
difference between the fluid pressure and the
atmospheric pressure.
Absolute pressure = Atmospheric pressure + Gauge
pressure
Gauge pressure = Absolute pressure – Atmospheric
P Q R pressure
= = =K
sin α sin β sin γ
Where, α, β and γ are the angles between Q, R, P, R and
P, Q respectively. The triangle of forces, in this case,
should close to provide the resultant force zero for
equilibrium.
58. The radius of the horizontal curve (R), for a
design speed of (V) 100 kmph with the The scale of pressure
maximum values of (superelevation) e = 0.07 61. The depth of the centre of pressure on a vertical
and (coefficient of friction) f = 0.15, can be rectangular gate (4 m wide, 3 m deep) with
calculated using which of the following water up to top surface, is
equations? (a) 1.5 m (b) 1.0 m
V3 V3 (c) 2.5 m (d) 2.0 m
(a) e + f = (b) e + f =
127 R 137 R Ans. (d) : Given that
breadth (b) = 4m
V2 V2
(c) e + f = (d) e + f = height (h) = 3m
12 R 127 R
Ans. (d) : Given that,
design speed (V) = 100 kmph
Maximum super elevation (e) = 0.07
Coefficient of friction (f) = 0.15
V2
We know, e+f =
127 R
h 3
So, using this equation for calculation of horizontal x= = = 1.5 m
2 2
curve (R)
bh 3
59. An aggregate is termed as elongated when its IG =
greatest dimension (length) is greater than 12
(a) thrice its mean dimension 4 × 33
IG =
(b) four times its mean dimension 12
(c) one-third of its mean dimension IG = 9m4
(d) nine-fifth of its mean dimension Centre of pressure for rectangular plate
Ans. (d) : Flakiness index–The flakiness index of an I sin 2 θ
aggregate is the percentage of weight of particles in it h=x+ G
Ax
whose least dimension is less than 0.6 of their mean
dimension. I G sin 2 90
h=x+  θ = 90o 
Elongation index–It is the percentage by weight of Ax
particles whose greatest dimension (length) is greater I
than 1.8 times their mean dimension. h=x+ G
Ax
60. Gauge pressure is usually defined as a pressure 9
which is measured using Atmospheric Pressure h = 1.5 +
as Datum. Gauge pressure at a point in a fluid 1.5 × 3 × 4
is equal to h = 1.5 + 0.5
(a) Absolute pressure + Vacuum pressure h = 2m
(b) Absolute pressure + Atmospheric pressure 62. Read the following statements and choose the
(c) Absolute pressure - Vacuum pressure CORRECT option with regards to Splicing of
(d) Absolute pressure - Atmospheric pressure reinforcement bars.
OSSC JE Exam. 2019 (Shift-III) 181 YCT
 (i) For the constructional requirements at site, Ans. (a) : Needle or cantilever scaffolding–This type
Splicing of reinforcing bars is done that of scaffolding is used under following situations–
transfers the axial force from the terminating • Hard firm ground is not available for standards to rest.
bar to the connecting bar. • When construction is to be carried on the side of a
(ii) Lap splices should not be used for bars larger very busy road where construction of single or double
than 12 mm and End-bearing splices shall be scaffolding is not possible.
used only for bars in tension.
• When construction is to be carried out a very large
(a) (i) is TRUE and (ii) is TRUE height of a multistoried building.
(b) (i) is TRUE and (ii) is FALSE
65. In a project network analysis, Forward Pass
(c) (i) is FALSE and (ii) is TRUE
method is used to determine which aspect of
(d) (i) is FALSE and (ii) is FALSE various activities?
Ans. (b) : Splicing of reinforcement–Splices are (a) Only the slack time
required when bars places short of their required length (b) Latest Finish Time
(due to non-availability of longer bars) need to be (c) Earliest Finish Time
extended. Splices are also required when the bar
(d) Only the total float
diameter has to be changed along the length. The
purpose of splicing is to transfer effectively the axial Ans. (c) : Network analysis–It is a process by which
force from the terminating bar to connecting bar with activity and project start and completion dates are
the same line of action at the junction. identified. Network analysis is performed all activities.
• Lap splices are usually not permitted for every large The critical path method is used as a means for network
analysis. Network analysis using CPM involves forward
diameter bars (φ > 36mm) For which welded splices are
and backward pass computation
recommended.
The forward pass computation identifies the early start
• The straight length of the lap should not be less than and finish times for each activity comprising a project in
15φ or 200m. time unit of calendar dates. The backward pass
• End bearing splices are permitted by the code for bars computation identifies the late start and finish times for
subject to compression. each activity comprising a project in time units.
63. In case of a Cantilever beam, as per IS : 456- 66. Select the CORRECT sequence of given steps in
2000, what is the basic value of span to effective the manufacturing process of Bricks.
depth ratio for spans up to 10 m? 1. Moulding of Bricks
(a) 7 (b) 20 2. Preparation of Brick Earth
(c) 26 (d) 30 3. Burning of Bricks
Ans. (a) : Code recommendations for span/Effective 4. Drying of Bricks
Depth Ratios–For prismatic beams of rectangular (a) 3 - 4 - 1- 2 (b) 1 - 2 - 4- 3
sections and slabs of uniform thickness and spans up to (c) 4 - 1 - 2- 3 (d) 2 - 1 - 4- 3
10m the limiting l/d ratios are specified by the IS code :
456-2000. Ans. (d) : Manufacture of bricks
Correct sequence in the manufacturing process of
l l bricks–
 d  =  d  × kt × kc
  max   basic • Preparation of brick earth
• Moulding of brick
l
where   • Drying of moulded bricks
 d  basic
• Burning of dried bricks
= 7 for cantilever spans
Preparation of brick earth–
= 20 for simply supported spans
1. Unsoiling 2. Digging 3. Clearing
= 26 for continuous spans
4. Weathering 5. Blending
kt, kc = modification factor 6. Tempering or pugging
64. Under which of the following conditions needle 67. In a certain type of scaffolding the working
scaffolding is usually adopted for construction? platform is supported on movable tripods or
(a) When construction work is carried out at very ladders. This is generally used for work inside
high level in case of high-rise buildings the room, such as paintings, and repairs up to a
(b) Construction of the lower part of the wall is to height of 5m. Identify this type of scaffolding
be carried out inside a room among the following options.
(c) When minor repairs or painting works are (a) Trestle Scaffolding
required inside the rooms (b) Patented Scaffolding
(d) Enough space is available to provide bracings (c) Cantilever Scaffolding
and ledgers (d) Double Scaffolding
OSSC JE Exam. 2019 (Shift-III) 182 YCT
Ans. (a): Trestle scaffolding–Such type of scaffolding since the velocity in our Eulerian system does not
is used for painting and repair works inside the room, change with time hence the pathlines and streak lines
upto a height of 5m. The working platform is supported coincide the velocity vector of a particle at a given point
on the top of movable contrivances such as tripods, will be tangent to the line that the particle is moving
ladders etc. mounted on wheels. along, thus the line is also a streamline.
Patented scaffolding–Many patented scaffolding made 70. Which of the following options is an example of
of steel are available in the market. These scaffoldings
Sub-surface raw water source?
are equipped with special couplings, frames etc. The
working platform is supported on brackets which can be (a) Streams (b) Rivers
adjusted at any suitable height. (c) Infiltration Wells (d) Lakes
68. Viscosity is a measure of the reluctance of the Ans. (c) : Various sources of water for supply
fluid to yield to shear when the fluid is in (A) Surface water source–Streams, rivers, natural
motion. The dimensions of dynamic viscosity ponds and lakes, impounding reservoirs.
and kinematic viscosity respectively are (B) Ground water source or sub-surface water
(symbols/notations carry their usual meaning) source–In filtration galleries, infiltration wells including
(a) FL–2 T, L2 T–1 (b) FL–2 T, FL–1 tubewells
–1 –1 2 –1
(c) FL T , L T (d) L2 T–1, FL–2 T • It is also called under ground water source
Ans. (a) : Dimension of dynamic and kinematic Infiltration wells–Infiltration wells are the bank of the
viscosity– rivers, in order to collect the river water seeping through
Units of viscosity– their bottoms
du
τ = µ.
dy
τ Shear stress Force / Area
µ= = =
du Change of velocity  Length  1
  ×
dy Change of dis tan ce  Time  Length
Force / length 2 Force × Time
µ= =
1 (length)2
Time
Dimension of dynamic viscosity– 71. Which of the following options is a statically
Force × Time indeterminate structure?
µ=
(length) 2 (a) Three-Hinged Arch
F×T (b) Simply Supported Beam
µ= 2 (c) Single Overhanging Beam
L
(d) Fixed Beam
=  FTL−2 
Ans. (d) : Statically indeterminate structure–When
Viscosity the number of unknown reaction or stress (force)
Kinematic viscosity (υ) =
Density components exceeds the number of conditions of
Force × Time equilibrium, the system is said to be statically
length indeterminate or redundant structure.
( length )
2 Mass × × Time
Force × Time Time
υ= = = Statically indeterminate structures are very sensitive to
Mass Mass  Mass  such factors as the settlement of their support
 length 
( length )
3
length   temperature variation, lack of fitness of members which
( length ) given rise to additional stresses. A continuous beam is a
2

υ= typical example of externally indeterminate structure,

Time fixed beam and propped cantilever are statically
L2 inderminate structure.
Dimension of kinematic viscosity (υ) = =  L2T −1 
T  72. For determining the strength of a structure,
69. During a motion of fluid, the path traced by the under the limit state of collapse as per IS 456:
fluid can be stream lines, streak lines or path 2000, the partial safety factors for Concrete and
lines. These three may coincide in the case of Steel respectively are
(a) steady flow (b) turbulent flow only (a) 1.15 and 1.5 (b) 2.5 and 2.15
(c) non-uniform flow (d) unsteady flow (c) 2.15 and 2.5 (d) 1.5 and 1.15
Ans. (a) : In a steady flow, path lines, streak lines and Ans. (d) : Limit state of collapse–It induces flexure,
streamlines are all coincident. All particles passing a compression, shear, torsion, over turning, sliding,
given point will continue to trace out the same path buckling and fracture due to fatigue.

OSSC JE Exam. 2019 (Shift-III) 183 YCT

Under the limit state of collapse as per IS 456 2000, the 76. As a general rule in India, the relation between
partial safety factors for concrete and steel are 1.5 and height and width of a Door is given as:
1.15. (a) Height = (Width - 1.2 ) meters
Limit state of serviceability–In includes vibration, (b) Height = (Width × 1.2 ) meters
durability deflection and deformation and repairable (c) Height = (Width + 1.2 ) meters
damage due to fatigue cracking etc. (d) Height = (Width /1.2 ) meters
73. The revenue chain is 33 feet long and divided Ans. (c) : Size of Doors–
into 16 links. It is used for Height–Height of door is fixed on the basis that longest
(a) Vertical angle measurement in a cadastral man may enter or come out through the door unbent. In
survey India general height of the people is generally less than
(b) Determining the magnetic bearing of a line in 2m, hence door height of 2m is considered most
a cadastral survey suitable. Height of door is sometime, approximated by
(c) Horizontal angle measurement in a cadastral the following thumb rule
survey Height of door = (width of the door + 1.2 meters)
(d) Distance measurement in a cadastral survey Height of window is generally kept nearly 1.10 m to
Ans. (d) : Chaining and taping methods–In actual 1.20 m. Bottom of still of the window is kept 750 mm to
surveying, distance is measured by a chain or a tape of 900 mm above the floor level and top of the window is
different lengths and accordingly it is called chaining or always kept at level with the top of the doors fixed in
taping method formerly engineer's chain or Gunter chain the room. The minimum height of doors is not be less
was used to measure the distance. The engineer's chain than 180 mm.
was 100 feet long and 100 links each 1 foot long. At • Width of the door varies from 800 mm to 1.2m.
every 10 feet or links brass tags were fastened. The 77. A rectangular plane surface that lies in a
Gunter's chain or Surveyor's chain was 66 feet long and vertical plane in water is 2 m wide and 3 m
divided into 100 links each link equal to 0.6 foot and deep. Find the Total pressure if the upper edge
revenue chain of 33 feet long consisting 16 links had is horizontal and 2 m below the free water
been also used in cadastral survey. surface.
74. The process of cutting off the excess wet (a) 106.01 kN (b) 206.01 kN
concrete to bring the top surface of a slab to the (c) 306.01 kN (d) 406.01 kN
proper grade and smoothness is known as Ans. (b) : Given that
(a) screeding (b) trowelling
(c) floating (d) finishing
Ans. (a) : Screeding–The process of striking off the
excess concrete to bring the top surface upto proper
grade.
Floating–It consist of removing the irregularities on the
surface of concrete and it is generally done by wooden
float.
Trowelling–It is final operation of finishing and it gives F = ρghA
a very smooth finish. 3
75. If to, tp and tm are optimistic, pessimistic and h = 2+ = 3.5
2
most likely time estimates of an activity σ = 1000 × 9.8 × 3.5 × 6
respectively, the expected time te of the activity
will be one sixth of F = 205.8 kN ≈ 206 kN
(a) (tm+3to+tp) (b) (3to+ tm + tp) 78. The surge tank in a water supply system is
(c) (to+4tm+tp) (d) (tm+4to+tp) generally provided to
(a) remove impurities from water
Ans. (c) : Expected of an activity (te)–The average or
mean time taken for the completion of an activity. It is (b) increase velocity of flow in a pipe
calculated by weighted average method and the (c) supply of chlorine alone required for
probability of completion of work on expected time is disinfection process
50%. (d) relieve excess pressure caused by water
hammer
t 0 + 4t m + t p
te = Ans. (d) : Surge Tanks–In hydro-electric installations
6 the turbines are often with water through a long pipeline
Where, t0 = optimistic time estimate or tunnel. High and destructive pressures can be
tp = pessimistic time estimate developed if there is a change in the discharge, which
tm = most likely time estimate can occur quite suddenly if the electric load is taken of
OSSC JE Exam. 2019 (Shift-III) 184 YCT
the generator: the turbines start to race and the
governors react, closing the guide blade openings,
reducing the flow as quickly as possible without causing
water hammer.
Surge tank are usually provided in high or medium head 2. Looping error–Looping error is also known as cyclic
plants when there is a considerable distance between the error in the network. Drawing an endless loop in a
water source and power unit. The main functions of the network diagram is known as error of looping.
surge tank are when the load decreases, the water moves
backwards and gets stored in it and when the load
increases an additional supply of water will be provided
by the surge tank.
79. The size of the fillet weld used as connection in
steel structures is specified by
(a) thickness of the weld
(b) maximum leg length of weld 82. Which of the following statements is
(c) total length of the weld CORRECT for foundations in general?
(d) minimum leg length of weld (a) The bearing pressure at the base of foundation
Ans. (d) : Types of welds–The most common types of can exceed the allowable soil pressure
welds used in steel structures are fillet and butt welds. (b) It prevents the differential settlement of a
Fillet welds are used to connect structural components structure
meeting at an angle (60º and 120º), while butt welds are (c) It transfers the load to the column directly
used to connect horizontal members. (d) The foundations of residential building take
Fillet welds–The size of a fillet weld is generally taken the load of the Slab alone from the
as the minimum leg length. The leg length is the superstructure above it
distance from the root to the toe of the fillet weld it is Ans. (b) : General requirements for foundation–
measured by the largest right angle triangle. • Shear failure criteria or bearing capacity i.e.
80. Standard gauge is the most widely used gauge foundation must be safe against shear failure.
in the World, for which the width is equal to • Settlement criteria i.e. settlement of foundation esp.
(a) 1435 mm (b) 1676 mm differential settlement must be within the permissible
(c) 1365 mm (d) 1525 mm limit
Ans. (a) : Standard gauge is the most widely used gauge • Location depth criteria–Foundation must be located
in the world, for which the width is equal to 1435 mm. at such a depth that is performance is not affected by
seasonal volume changes of soil duel to swelling &
Gauge Distance between rails shrinkage and also by the presence of adjoining
Broad guage 1.67m structure.
Meter gauge 1.0m 83. Given below are statements with respect to the
Narrow gauge 0.762m components of a Staircase in general.
(A) A vertical member is placed at the ends of
Light gauge 0.610m
flights in a staircase to connect the ends of the
Standard gauge Strings and Handrails.
1.435m
(used in delhi metro) (B) A vertical member of wood or metal usually
supports the Handrail in a Staircase.
81. In CPM network analysis of a project, the
Select the CORRECT option among the
disconnection of an activity before the
following options.
completion of all activities in a network
(a) Statement (A) refers to -->> Nosing and
diagram is known as Statement (B) refers to -->> Soffit
(symbols and notations carry their usual (b) Statement (A) refers to -->> Soffit and
meaning) Statement (B) refers to -->> Nosing
(a) Mathematical error (b) Redundancy error (c) Statement (A) refers to -->> Baluster and
(c) Dangling error (d) Looping error Statement (B) refers to -->> Newel Post
Ans. (c) : Error in Drawing Network–There are three (d) Statement (A) refers to -->> Newel Post and
types of errors which are common in network diagrams. Statement (B) refers to -->> Baluster
1. Dangling error–To disconnect an activity before the Ans. (d) : Components in stair case in general–
completion of all activities in a network diagram is (i) Newel post–This is the vertical post placed at the end
known as dangling of flights to connect the ends of strings and hand rails.

OSSC JE Exam. 2019 (Shift-III) 185 YCT

(ii) Baluster–This is the vertical member which (a) 10 to 100 (b) 8000 to 10000
supports the hand rail. (c) 3000 to 6000 (d) 150 to 2000
(iii) Winder–This is a tapering step used for changing Ans. (c) : Rapid sand filter–
the direction of stair.
• Particle more than and less than 1 µm dia are
84. The ground for the formation level of a efficiently removed.
Highway construction is of uniform gradient
• Removes suspended and colloidal matter
with longitudinal slope. Here the quantity of
earthwork can be calculated using which of the • Coefficient of uniformity (Cu) = 1.2-1.6
following options? • Effective particle size D10 = (0.35-.55) mm
(A) Trapezoidal formula • Rate of filtration = 3000-6000 l/m2/hr
(B) Prismoidal formula • Removal of turbidity up to 40 ppm.
(C) Mid-section formula • Washing period is 24-48 hrs.
(a) Only A and C (b) Only C
• It is construction is complicated and skilled labour
(c) Only B (d) All A, B and C
required.
Ans. (d) : The ground for the formation level of a
highway construction is of uniform gradient with the 87. A valve that is used to completely shut off fluid
longitudinal slope. The quantity of earthwork can be flow or, in the fully open position, provide full
calculated using some formula who is given below– flow in a pipeline and generally NOT suitable
(i) Trapezoidal formula–This method is based on the for throttling applications, is known as
assumption that the mid area is the mean of the end (a) Check valve (b) Sluice valve
areas. (c) Gate valve (d) Washout valve
 A + An  Ans. (c) : Gate valve–A gate is a linear motion valve
v =d 1 + A 2 + A 3 + .....A n −1 
 2  that uses a typically flat closure element perpendicular
to the process flow which slides into the flow stream to
d ( A1 + A n ) + 4 ( A 2 + A 4 + A n −1 ) + 2 ( A 3 + A5 ) provide shut-off with a gate value, the direction of fluid
v=  
2  + ....A n − 2  passes is essentially equal to that of pipe. Gate valves
This is also known as Simpson's rule for volume. Here are designed to minimize pressure drop across the valve
also it is necessary to have an odd number of cross- in the fully opened position and stop the flow of fluid
sections. completely, gate valves are not used to regulate fluid
85. As per the General Building Requirements in flow. Typically these are used on pipelines, on gas walls
India, the horizontal distance of a building site and plant block valves.
from High voltage lines upto and including • The butterfly valve is used for throttling high pressure
11,000 volts should be atleast applications.
(a) 0.9 m (b) 0.5 m 88. Read the following statements and choose the
(c) 1.2 m (d) 0.2 m CORRECT option with respect to the forces in
Ans. (c) : As per the general building requirements in a member of a Truss in general.
India. i. The force in a member will be Compressive if
Distance from electric live the member pushes the joint to which it is
Vertically Horizontally connected.
A Low and medium 2.5 m 1.2 m ii. The force in a member will be Tensile if the
voltage lines and member pulls the joint to which it is
service lines. connected.
B High voltage lines 3.70 m 1.20 m (a) (i) is FALSE and (ii) is TRUE
upto and including (b) (i) is TRUE and (ii) is TRUE
11,000 votls (c) (i) is TRUE and (ii) is FALSE
C High voltage lines 3.70 m 2.00 m (d) (i) is FALSE and (ii) is FALSE
above 11,000 volts Ans. (b) : Forces in a member of a truss in general
and upto and requirements–
including 33,000
volts • The force in a member will be compressive if the
member pushes the joint to which it is connected.
86. The water is filtered through the beds of fine
granular material, such as sand for removing • The force in a member will be tensile if the member
or reducing impurities in water through the pulls the joint to which it is connected.
process of Filtration. The Rate of filtration • If there are two inclined members connected to a joint
(Litres/hour/m2) for Rapid Sand Filter is that has no external force, then both member are zero
generally in the range of force members.
OSSC JE Exam. 2019 (Shift-III) 186 YCT
89. Read the following statements and choose the Ans. (d) : The ruling design speeds suggested for the
CORRECT option with respect to Moment National and State highways of our country passing
Distribution Method of analysis of structures. through plain terrain is 100 kmph and through rolling
i. Moment Distribution Method of analysis is a terrain is 80 kmph.
convenient way to analyze only the Design speeds on rural highways
determinate beams and frames. Road Design speed in kmph for various terrains
ii. In this method, the distributed moments in the classification
ends of members meeting at a joint cause Plain Rolling Mountainous Steep
Ruling Min Ruling Min Ruling Min Ruling Min
moments in the other ends, which are assumed
National & 100 80 80 65 50 40 40 30
to be fixed. State
These induced moments at the other ends are highways
called carry-over moments. Major District 80 65 65 50 40 30 30 20
roads
(a) (i) is TRUE and (ii) is TRUE
Other District 65 50 50 40 300 25 25 20
(b) (i) is TRUE and (ii) is FALSE roads
(c) (i) is FALSE and (ii) is TRUE Village roads 50 40 40 35 25 20 25 20
(d) (i) is FALSE and (ii) is FALSE 92. The process of mixing some mortar in the
Ans. (c) : Moment distribution method– mixer at the beginning of the first batch
• The method was first introduced by Prof. Hardy cross concrete mixing is called as
in 1930. (a) accelerating (b) borrowing
(c) buttering (d) initiating
• The moment distribution method could be used for the
analysis of all types of statically indeterminate beams or Ans. (c) : Buttering–The process of mixing some
rigid beams. mortar in the mixer at the beginning of the first batch
concrete mixing is called buttering.
• The method essentially consists of first locking all the
93. The verticle plank which remains in contact
joints (which are not fixed) so that each span of the
with side of the trench and directly resist
structure behaves like a fixed beam. It should be
pressure from the side of a trench during
remembered that a fixed joint does not need any excavation and trenching, is called as
balancing and it absorbs all the moments carried over to (a) Bracings (b) Wale
it from the other end. (c) Sheathing (d) Ledgers
• When the moments are carried over from the near ends Ans. (c) : Sheathing–The vertical plank which remains
to the far ends there is no carry moment from joint are in contact with side of the trench and directly resist
originally fixed. pressure from the side of a trench during excavation and
90. With regards to the shape test of aggregates for trenching is called sheathing.
pavement construction, the elongation index of Wall–It is a piece of timber that is inserted between the
an aggregate is defined as the percentage by sheeting and strut. It runs in directions normal to
weight of particles whose greatest dimension sheeting.
(length) is Bracings–Bracings are the diagonal members of frame
(a) 1.8 times their mean dimension supporting sheetings.
(b) 0.8 times their mean dimension 94. According to ASTM standards, Coarse
(c) 3.8 times their mean dimension aggregates are identified when they are
(d) 2.8 times their mean dimension retained on
(a) 4.75 mm sieve (b) 6 mm sieve
Ans. (a) : Elongation index–The elongation index of (c) 10 mm sieve (d) 8.75 mm sieve
aggregate is the percentage by weight of the particles
whose longest dimension (i.e. length) is greater than 1.8 Ans. (a) : According to American Association of State
Highway and Transportation Officials (AASHTO) and
times their mean dimension. The elongation test, two is
American Society for Testing and Materials (ASTM)
not applicable to aggregate of size less than 6.3 mm.
standards, coarse aggregate is retained on the 4.75 mm
Flakiness index–The flakiness index of aggregate of sieve (Also known as the No. 4 sieve) for aggregate
that mass of the aggregate whose least dimension is less passes through the 4.75 mm sieve and is retained on the
than 0.6 time their mean dimension. The flakiness test is 0.075 mm sieve (or No. 200 sieve) for fines, all particles
applicable to the aggregate of size larger than 6.3 mm. pass through the 0.75 mm sieve.
91. As per IRC recommendations, on a National 95. Generally in a brickwork, the type of bond in
Highway in rolling terrain, the ruling design which all bricks are laid as headers on the faces
speed should be of the wall is known as
(a) 150 kmph (b) 70 kmph (a) Stretcher Bond (b) Header Bond
(c) 60 kmph (d) 80 kmph (c) Flemish Bond (d) English Bond
OSSC JE Exam. 2019 (Shift-III) 187 YCT
Ans. (b) : Header Bond–In header bond, all the bricks 98. As per IS:456-2000, the lap length including
are laid as headers on faces of walls. In header bond, the anchorage value of hooks for direct tension in
bond is formed by three quarter bats at the quion. The bars shall be (Given: Ld = Development Length
width as brick is thus along the direction of wall. It is of bars and φ is the nominal diameter of the
suitable for partition wall is equal to one brick. This type bar.)
of bond is useful for the construction of one brick thick
(a) Ld or 18φ whichever is greater
walls.
(b) 2Ld or 30φ whichever is greater
(c) 0.50Ld or 16φ whichever is greater
(d) Ld or 16φ whichever is greater
Ans. (b) : Lap length in tension–
For direct tension–2Ld or 30φ (whichever is larger)
For flexural tension–Ld or 30φ (whichever is larger)
Lap length in compression–It is equal to the
96. If base period for a particular crop is 50 days development length calculated in compression but
and the duty of the canal is 500 hectares per should not be less than 24φ.
cumec, then the depth of water will be 99. A line on a map joining points of equal
(a) 8.64 cm (b) 0.864 cm elevation is a contour line. These lines intersect
(c) 864 cm (d) 86.4 cm the ridge lines usually at what angle?
Ans. (d) : Given that, (a) 30º (b) 45º
Base period (B) = 50 days (c) 90º (d) 0º
Duty (D) = 500 hectares/cumec
Ans. (c) : Characteristics of contour lines–
We know that,
• All the points of a contour line have the same
8.64 B
∆= elevation. The elevations of the contours are shown
D either by inserting the figure in a break in the respective
8.64 × 50 contour or printed close to the contour.
∆=
500 • A contour line is a closed curve. They may close either
∆ = 0.864 m on the map or outside the map it depends on the
∆ = 0.864 × 100 cm topography
∆ = 86.400 • A watershed or ridge line (line joining the highest
∆ = 86.4 cm point of a series) and the that weg or valley (line joining
97. The maximum daily demand at a water the lowest points of a valley) cross the contours at right
purification plant has been estimated as 12 angles.
million litres per day. Find the length of the • Irregular contours represent an uneven ground surface.
sedimentation tank required if a detention • Equally spaced contours represent a uniform slope and
period of 5 hours and the velocity of flow of 20 contours that are well about represents slope.
cm/minute is assumed.
100. Which of the following options is a project
(a) 50 m (b) 70 mm
planning technique that considers the
(c) 60 m (d) 80 m uncertainty factor of various activities?
Ans. (c) : Given that, (a) CPM
Discharge (Q) = 12 million = 12 × 106 × 10–3 m3/day (b) Neither CPM nor PERT
12000 (c) Both CPM and PERT
Q= = 8.33 cum / min
24 × 60 (d) PERT
Velocity of flow (vf) = 20 cm/min
Ans. (d) : PERT stands for programme evaluation and
Detention period (td) = 5 h
review technique.
L
Detention time td = • Three time estimates are made.
vf
• It is based on probabilistic approach.
Length of sedimentation tank L = td × vf • Time is directly proportional to time hence minimum
0.2 cost will be achieved corresponding to minimum
=5×
1 completion time.
60 • Suitable for new type of projects.
= 5 × 0.2 × 60 • Each activity follows & distribution.
L = 60 m • Uncertainty of various activities occur.
OSSC JE Exam. 2019 (Shift-III) 188 YCT
 Odisha Public Service Commission
(Polytechnic Lecturer)
Exam- 2018 (Paper-I)
1. A metal cube of volume 10000 cc is subjected to Ans. (d) :
all round stress of 130 MPa. The bulk modulus • On the plane of maximum and minimum principle
of the material is 1.3 × 105 MPa. The material is stress, shear stress is zero.
1.3 × 105 MPa. The volumetric changes is :
σ − σ2
(a) 0.8 cc (b) 1 cc • Maximum shear stress (τmax) = 1
(c) 8 cc (d) 10 cc 2
Ans. (d) : V = 10.000 cc 4. The Euler's buckling load for a column with
Stress (P) = 130 MPa one end fixed and other end pin is :
Bulk modulus (K) = 1.3 × 105 MPa (a) π 2 EI / L2 (b) 4π 2 EI / L2
P (c) 2π 2 EI / L2 (d) π 2 EI / 2L2
K=
∆V / V Ans : (c)
PV Euler's
∆V = Effective
Types of column formula for
K length
buckling load
130 × 10, 000
= Both ends hinged L=l π2 EI
1.3 × 105
l2
130
= = 10 One ends fixed L=2l π2 EI
13 other free
4l 2
∆V = 10 cc
One ends fixed l 2π2 EI
2. The relation between the Young's modulus of other hinged L=
2 l2
elasticity (E), modulus of rigidity (G) and
Both ends fixed l 4π2 EI
Poisson's ratio (µ) is : L=
(a) G = E/2 (1+µ) (b) G = E/2 (1–µ) 2 l2
(c) G = E (1+µ) (d) G = E (1–µ) 5. Euler's buckling stress of the mild steel column
Ans. (a) : For a isotropic material, the relationship of 250 MPa and Young's modulus of elasticity
between the Young's modulus (E), shear modulus (G) is 2 × 105 MPa. Euler's formula is applicable if
and Poisson's ratio (µ) is slenderness is nearly equal to or greater than :
E (a) 50 (b) 70
G= (c) 79 (d) 89
2 (1 + µ )
Ans. (d) :
3. Consider the following statements :
π2 EI
(1) On planes having maximum and minimum P = 2 , I = Ar2
principal stresses, there will be tangential L
2
stress π 2 E.Ar 2 P r
P= ⇒ = π 2
E.  
(2) On planes having maximum and minimum L2 A L
principal stresses, there will be no tangential
stress πE2
E
σ= 2 ⇒λ=π
(3) Maximum shear stress is half of the sum of λ σ
the maximum and minimum principal stresses 2 × 10 5

(4) Maximum shear stress is half of the ∴ λ = π× = 88.8 ≃ 89

250
difference of the maximum and minimum
principal stresses 6. The ratio of section modulus of a circular beam
Which of the above statement (s) is / are true? of diameter d and square beam of size d is
(a) 1 only (b) 2 only (a) π/4 (b) π/8
(c) 1 and 3 (d) 2 and 4 (c) 3π/8 (d) 3π/16

OPSC Polytechnic Lecturer Exam-2018 (Paper-I) 189 YCT

 I 9. Which point on the stress-strain curve of mild
Ans. (d): Section modulus (Z) = steel occurs after lower fluid point?
y
(a) Yield plateau
I = Moment of inertia
(b) Upper yield point
y = Depth from neutral axis
(c) Ultimate point
(d) None of the above
Ans. (a) : The stress-strain curve of mild steel occurs
after lower yield point– yield plateau.
10. For a material, Young's modulus is given as 2.4
πd 4 × 105 MPa and Poisson's ratio 0.3. Calculate
Moment of inertia for circular section (I) =
64 the bulk modulus.
y = d/2 (a) 0.8 × 105 MPa (b) 1.2 × 105 MPa
Section modulus for circular section (c) 2.0 × 105 MPa (d) 2.4 × 105 MPa
5
πd 4 πd3 Ans. (c) : E = 2.4 × 10 MPa
Zc = =
64 32 µ = 0.3
d/2 E = 3 K (1 – 2µ)
πd 3 2.4 × 105 = 3K (1 – 2 × 0.3)
Zc = 2.4 × 105 = 3K × 0.4
32
2.4 × 105 = 1.2K
Moment of inertia for square section
d4 K = 2 × 105 MPa
I=
12 11. A circular shaft is subjected to a twisting
y = d/2 moment T and bending moment M. The ratio
of maximum bending stress to maximum shear
stress is given by
(a) 2 M/T (b) M/T
(c) 2 T/M (d) M/2T
Section modulus for square section Ans. (a) : Maximum shear stress developed by the
shaft–
d4
ZS = d / 2 = d3 / 6 16T
12 ⇒ τ max =
πD 3
ZC πd 3 6π 3π
= d3 / 6 = =
ZS 32 32 16
7. The property of a material by which it can be
beaten or rolled into thin plates, is called :
(a) Malleability (b) Plasticity
(c) Ductility (d) Elasticity
32M
Ans. (a) : Malleability : It is that property of a metal ⇒ σb = (Maximum bending stress)
due to which a piece of metal can be converted into a πD3
thin sheet.
Ductility : Ductility of a metal is that property due to
which a metal can be drawn into wire of thin section.
Elasticity : It is that property of material by which the
original dimension can be recovered after unloading.
8. Where in the stress-strain curves, the Hooke's
law is valid?
(a) Strain hardening region The ratio of the maximum bending stress to maximum
(b) Necking region σ 32M πD3
shear stress is · cb max = ×
(c) Elastic range τmax πD3 16T
(d) Valid everywhere
 σcb max 2M 
Ans. (c) : The stress-strain curve hooks law is valid to  = 
elastic range.  τmax T 

OPSC Polytechnic Lecturer Exam-2018 (Paper-I) 190 YCT

12. The cement deflection of a fixed beam carrying Ans. (a) : According to the maximum shear stress
a central point is X times of the central failure criterion, yielding in material occurs when τmax =
deflection of simply supported beam carrying σy/2.
same point load. The value of X is
Theory Given Suitable Graphical
(a) 0.25 (b) 0.50 by for Representation
(c) 1.0 (d) 2.0 Material
Ans. (a) : Maximum Rankin Suitable
Principal Stress e for
Brittle

(Square)
Pl 3 Maximum St. Can be
Simply supported beam, δc =
48EI Principal Strain Venant Applied
s for
Pl 3 1
δc = δ max = = × δ max in SSbeam Ductile
192EI 4
13. A beam of channel cross-section with vertical
web is subjected to a point load at the mid-span Maximum Guest Suitable
in a plane perpendicular to the plane of Shear Stress & for
symmetry passing through the centroid of the Trasca' Ductile
section. The beam is subjected to : s
1. Bending moment
2. Twisting moment
3. Axial thrust
Maximum Haigh Ductile
4. Shear-force Strain Energy &
(a) 1, 2 and 3 Beltra
(b) 2, 3 and 4 mi
(c) 1, 3 and 4
(d) 1, 2 and 4 Maximum Von- Ductile
Ans. (d) : Shear Strain mises
Energy &
Hencky

15. A block of size 100 mm × 100 mm × 100 mm is

subjected to shear stress 100 MPa. If the
modulus rigidity of the material is 1 × 106 MPa,
the strain energy stored will be
(a) 1000 N-mm (b) 2000 N-mm
(c) 3000 N-mm (d) 5000 N-mm
τ2 V
The load does not pass through shear centre of the Ans. (d) : Strain energy stored =
channel section, so the cross section is subjected to 2G
torsion in addition to bending and shear. V = 100 × 100 × 100
(i) Torsion or twisting moment τ = 100 MPa
(ii) Bending moment G = 1 × 106 MPa
(100 )
2
(iii) Shear force. × 100 × 100 × 100
Strain energy stored =
14. According to maximum shear stress failure 2 × 1× 106
criterion, the yielding in material occurs when : = 5000 N-mm
(a) Maximum shear stress = yield stress / 2 16. An element is subjected to σx = 120 MPa, σy=
(b) Maximum shear stress = 1.414 × yield stress 60 MPa, σy = 60 MPa and τxy = τyx = 40 MPa.
The radius of Mohr's circle is :
(c) Maximum shear stress = 1.723 × yield stress
(a) 50 MPa (b) 80 MPa
(d) Maximum shear stress = 2.0 × yield stress (c) 100 MPa (d) 120 MPa
OPSC Polytechnic Lecturer Exam-2018 (Paper-I) 191 YCT
Ans. (a) : Ans. (a) :
2
 σx − σ y 
 + τ xy
2
Radius of circle = 
 2 
2
 120 − 60 
=   + 40 = 50 MPa
2

 2 
17. An element is subjected to σx = 100 MPa σx + σ y  σx − σy 
2

(tension), σy = 40 MPa (compression). The σ1,2 = ±   +τ

2

normal stress acting on the plane of maximum 2  2 

shear stress is : Given, σ x = 100, σ y = 0
(a) 30 MPa compression 100
(b) 30 MPa tension σ2 = − 502 + τ2xy = 0
2
(c) 70 MPa compression
τxy = 0
(d) 70 MPa tension
Ans. (b) : Normal stress on the plane of maximum 21.
For the validity of principle of superposition,
σ x + σ y 100 + ( −40 ) material should behave in which manner?
shear = = = 30 MPa (Tension) (a) Linear elastic
2 2 (b) Non-linear elastic
18. The shear centre of a i-section is located : (c) Non-linear inelastic
(a) At the centroid of the section (d) Linear-inelastic
(b) At the interaction of the web and top flange
Ans. (a) : For the validity of principle of superposition,
material should behave in linear elastic manner.
(c) At the interaction of web and bottom flange
(d) Outside the section Principle of super position– For linear elastic
structure, the load effects caused by multiple loads are
Ans. (a) : For a section having two axes of symmetry,
the sum of load effects caused by each load.
shear centre lies at intersection of them i.e. at its
centroid. 22. How many equilibrium equations do we need
to solve generally on each joint of a truss?
(a) 1 (b) 2
(c) 3 (d) 4
Ans. (b) : Due to hinge joint in a truss no BM comes on
a joint hence only two equilibrium equation are needed
19. If the area of cross-section of bar is 100 mm2, for analysis of truss.
then the gauge length for the measurement of 23. What is the degree of indeterminacy of a fixed
ductility will be supported portal frames?
(a) 56.5 mm (b) 5.65 mm (a) 1 (b) 2
(c) 65.6 mm (d) 6.56 mm (c) 3 (d) 4
Ans. (a) : Given, Ans. (c) :
Area of cross-section of bar,Ao = 100 mm2
Gauge length= ?
L o = 5.65 Ao
L o = 5.65 100 = 56.5 mm
20. In a plane stress problem, there are normal DS = 3C– R
tensile stresses σx = 100 MPa accompanied by C=1
shear stress τxy at a point along orthogonal R=0
cartesian co-ordinates x and y, respectively. If D S=3

it is observed that the minimum principal 24. In deflection diagram, which of the following
stress on a certain plane is zero then : can have zero angular deflection?
(a) τxy = 0 MPa (a) Pin support (b) Roller support
(b) τxy = 1 MPa (c) Fixed support (d) Hinge
(c) τxy = 14.14 MPa Ans. (c) : In deflection diagram, fixed support can have
zero angular deflection
(d) τxy = 100 MPa
OPSC Polytechnic Lecturer Exam-2018 (Paper-I) 192 YCT
 28. A fixed end beam of uniform cross-section is
loaded with uniformly distributed load
throughout the span. The proportion of
bending moment at the centre to the end
moment :
(a) 1 : 1 (b) 1 : 2
At A and B angular deflection is zero.
(c) 1 : 4 (d) 2 : 3
25. The maximum deflection of a beam occurs at :
Ans. (b) : Bending moment diagram for UDL
(a) Zero slope location
(b) Zero shear force location
(c) Zero bending moment location
(d) None of the above
Ans. (a) : The maximum deflection of a beam occurs at
zero slope location.

26. A beam carries a uniformly distributed load wL2

throughout its length. In which of the following Bending moment at centre ( M C )
configuration will have maximum strain = 242
energy? Bending moment at fixed end ( M F ) wL
(a) Cantilever beam 12
(b) Simply supported beam 12 1
= =
(c) Propped cantilever beam 24 2
(d) Fixed beam MC 1
=
Ans. (a) : Strain energy will be maximum when the MF 2
maximum moment will be high.
Moment of different beam–
wL w 2 L5
For cantilever beam, M = , Strain energy =
2 40EI
wL 2 (i)
FB =
12
wL
SSB =
8 (ii)
2
wL
Propped cantilever beam =
8
27. A fixed beam AB of span 4 m has a hinge C at
mid span. A concentrated load of 100 kN is (iii)
applied at C. What is the fixed end moment at
A?
(a) 50 kN-m (b) 100 kN-m
(c) 200 kN-m (d) 400 kN-m
Ans. (b) : (iv)
29. The slope and deflection of beams of varying
flexural rigidity may be easily computed by :
(a) Macauly method
(b) Mohr method
(c) Conjugate beam method
(d) Moment distribution method
Ans. (c) : The slope and deflection of beams of varying
flexural rigidity may be easily computed by Conjugate
MA = 50 × 2 = 100 kN-m beam method.

OPSC Polytechnic Lecturer Exam-2018 (Paper-I) 193 YCT

 A conjugate beam is a fictious beam of the φ
same length as the corresponding real beam, but it is φ = max  main or 6mm {where φmain = dia of main bar
externally supported and internally connected such that,  4
if the conjugate beam is loaded with the M/EI diagram least lateral dim ension
of the real beam, the shear and bending moment at any Pitch (φ) = Min. 16 φmin
point on the conjugate beam are equal respectively to 300 mm
the slope and deflection at the corresponding point on
the real beam. {φwhere= Minimum dia of main longitudinal bar
min
The conjugate beam method essentially
34. The maximum compressive strain permitted in
involved determining the slope and deflection of beam RCC columns subjected to axial load and
by computing the shear and bending moment in the bending is :
corresponding conjugate beam. (a) 0.0020 (b) 0.0030
30. Sway analysis of a portal frame becomes (c) 0.0035 (d) 0.0040
essential when : Ans. (c) : The maximum compressive strain permitted
(a) Loading is unsymmetric in RCC columns subjected to axial load and bending is
(b) Frame is unsymmetric 0.0035
(c) Both frame and loading are unsymmetric 35. The ratio of the strength of helically reinforced
(d) all of the above concrete short columns to that of tied column
Ans. (d) : Sway analysis of a portal frame becomes is:
essential when– (a) 0.85 (b) 1.00
• Loading is unsymmetric (c) 1.06 (d) 1.15
• Frame is unsymmetric Ans. (c) : As per IS 456 : 2000 clause 39.4 – The
strength of compression members with helical
• Both frame and loading are unsymmetric.
reinforcement shall be taken as 1.05 times the strength
A frame, in general, will undergo sideway if its joints
of similar member with lateral ties.
are not restrained against translation, unless it is a
Helical Reinforcement– The diameter of helical bars
symmetric frame subjected to symmetric loading.
should not be less than 1/4th the diameter of largest
31. Limit state method is based on : longitudinal bar and not less than 6 mm.
(a) Calculation on service load condition alone • The pitch should not exceed (if helical reinforcement
(b) Calculation on ultimate load conditions alone is allowed)
(c) Calculation on working loads and ultimate ⇒ 75 mm
loads ⇒ 1/6 time of core diameter of the column.
(d) Calculation on earthquake loads 36. The primary compression failure in RCC beam
Ans. (c) : Limit state method is based on calculation on is caused in :
working loads and ultimate loads. (a) Under-reinforced beams
32. The minimum clear cover (in mm) to the main (b) Over- reinforced beams
reinforcement bars in column, beam and slab (c) Balanced beams
respectively are : (d) All beams
(a) 20, 15, 10 (b) 40, 25, 15 Ans. (b) : The failure of an over reinforced concrete
(c) 30, 25, 20 (d) 40, 35, 20 beam is due to primary compression failure.
Ans. (b) : Minimum clear cover (in mm) to the main • The concrete attain its maximum stress earlier its
steel bars in means that tension steel does not yield up to ultimate
slab = 15 mm strength and the section will be over reinforced (Brittle
beam = 25 mm failure).
column = 40 mm xa > xc
footing = 75 mm 37. Steel beam theory is used for :
33. The maximum longitudinal reinforcement in (a) Design of simple steel beams
an axially loaded short column is ______ of (b) Steel beams encased in concrete
gross sectional area. (c) Doubly reinforced beams ignoring
(a) 4% (b) 5% compressive street in concrete
(c) 6% (d) 8% (d) Beams if the shear exceeds 2 times allowable
Ans. (c) : Longitudinal reinforcement of column – shear stress
Min area of steel = 0.8% of the gross area of column Ans. (c) : Steel beam theory is used for the design of
Max area of steel = 6% of the gross area of column doubly reinforced beams by ignoring compressive stress
Transverse reinforcement (Ties) – in concrete.

OPSC Polytechnic Lecturer Exam-2018 (Paper-I) 194 YCT

• We have seen that total moment of resistance of 41. Limit state of serviceability for deflection
doubly reinforced beam consists of two terms- including the effects due to creep, shrinkage
The moment of resistance M1 of the compression and temperature occurring after erection of
concrete and the corresponding tensile steel (Ast1) to partitions and application of finishes as
balance if and the moment of resistance M2 of the applicable to floors and roofs is restricted to :
compression steel (Asc) and the remaining tensile steel (a) Span/ 150 (b) Span/250
(Ast2) to balance it. (c) Span/300 (d) Span/350
38. The shear reinforcement in RCC beam is Ans. (d) : Permissible limit of deflection–
provided to resist : span
(1) Final deflection >/
(a) Flexural tension 250
(b) Flexural compression • Due to all loads = DL & LL
(c) Diagonal compression • Including all effect of temperature, creep, shrinkage
(d) Diagonal tension • measurement from cost level of support.
Ans. (d) : Shear reinforcement in RCC beam is span
provided to resist diagonal tension. Stirrups is provided (2) The deflection >/ or 20 mm
350
in the beam for shear reinforced method. Whichever is less
• Bent up bars and vertical stirrups are provided to • Only due dead load + finishing + partition load.
resist diagonal tension in member. • Occuring after erection of partition and applications of
• Diagonal tension occurs in the tensile zone of the finishing.
beam, generally tensile zone in a beam lies below the • This is an intermediate stage deflection.
neutral axis and compression zone is above the neutral • Inducing the effect of temperature, creep & shrinkage.
axis.
• Diagonal tension in beam increase below the neutral 42. In case of two-way, the limiting deflection of
axis and decreases above the neutral axis. the slab is :
(a) Primarily a function of the long span
39. The unsupported length between end restraints
(b) Primarily a function of the short span
shall not exceed _______ times the least lateral
dimension of a column. (c) Independent of long or short span
(a) 50 (b) 60 (d) Dependent of both long and short spans
(c) 70 (d) 80 Ans. (b) : The strip of a two way slab may be checked
against shorter-span to over all depth ratio.
Ans. (b) : According to IS 456 : 2000 (Cl.25.3.1) Span to overall depth ratio –
The unsupported length between end restraints shall not
Type of reinforcement
exceed 60 times the least lateral dimension of a column. Type of slab
for Mild steel for HYSD
40. In limit state design, permissible bond stress in S.S.B. 35 28
case of deformed bars is more than that in Continuous 40 32
plain bars by
43. Side face reinforcement shall be provided in
(a) 60% (b) 50%
the beam subjected to bending moment, shear
(c) 40% (d) 25%
force and torsion when the depth of the beam
V exceeds:
Ans. (a) : τ bd =
∑ P.j (a) 450 mm (b) 500 mm
Where, (c) 600 mm (d) 750 mm
ΣP.j = sum of perimeter of diameter Ans. (a) : As per IS 456: 2000, side face reinforcement
of reinforcement ( nπφ ) is provided when–
j = coefficient of lever arm. • Depth of beam (web) is more than 750 mm (when
Maximum Permissible values (bond) – beam is not subjected to torsion)
Steel in WSM (N/mm2) LSM (N/mm2) • Depth of beam (web) is more than 450 mm (when
Tension beam is subjected to bending moment, shear force
M 15 0.6 ≈ 1.0 and torsion)
M 20 0.8 1.2 • Side reinforcement is provided 0.1% of web area
M 25 0.9 1.4 equally distributed on both face.
M 30 1.0 1.5 • Maximum spacing of side face reinforcement = 300
M 35 1.1 1.7 mm or width of beam or width of web which ever
M 40 1.2 1.9 less.
• For high yield strength deformed bar (HYSD) and 44. The deflection can be controlled by using the
CTD bars, above value in tension is increased by 60% appropriate :
Note : If bars are in compression, the value should be (a) Aspect ratio (b) Modular ratio
further increased bond strength by 25%. (c) Water/cement ratio (d) Span/depth ratio
OPSC Polytechnic Lecturer Exam-2018 (Paper-I) 195 YCT
Ans. (d): According to IS 456 : 2000, Clause–23.2 • Torsional reinforcement is provided in the form of a
• The vertical deflection limits may generally be grid or mesh both at the top and bottom of the slab.
assumed to be satisfied provided that the span to depth • 0.75 Ast if the both the meeting edges are restrained.
ratios are not greater than the value obtained as below. • 0.375 Ast if one of the two meeting edges, one in
• Thus, the basic value of span of effective depth ratios, continuous and other discontinuous.
for span upto 10 m are as follows : 49. In brick masonry, arch action is possible only
Cantilever – 7 when the minimum height of wall above the top
Simply supported beam – 20 of lintel is ______ times the height of triangular
Continuous – 26 distribution.
45. In limit state approach, spacing of main (a) 1.00 (b) 1.25
reinforcement controls primarily : (c) 1.50 (d) 1.75
(a) Collapse (b) Cracking Ans. (b) : For the transfer of load to the sides by arch
(c) Deflection (d) Durability action, it is necessary that the height of the wall above
Ans. (b) : Cracking– In limit state approach spacing of the opening is 1.25 times the height of triangle of
the main, reinforcement primarily controls, cracking masonry load transferred on the lintel and the length of
width with in the permissible limit under normal the wall on each side of the opening equal to at least
exposure condition. half of the effective span of the lintel.
The code specifies minimum and maximum limit for the
spacing between parallel reinforcing bars in a layer. The
minimum limits are necessary to ensure that the concrete
can be placed easily in between and around the bars during
the placement of fresh concrete. The maximum limits are
specified for bar in tension for the purpose of controlling
crack-widths and improving bond. 50. In the design of RCC beams, which will come
under the limit states of serviceability?
46. The maximum area of compression steel in a
(a) Flexure (b) Shear
beam of width = b, over all depth = D and
effective depth = d is : (c) Torsion (d) Cracking
(a) 0.12 bd (b) 0.12 bD Ans. (d) : Lime state of serviceably include–
(c) 0.04 bd (d) 0.04 bD (i) Deformation cracking and deflection adversely
affection appearance or effective use of structure.
Ans. (d) : Maximum area of tension reinforcement (ii) Vibrations in structure or any part of its compound
Ast max = 0.04 BD, i.e 4% of total C/S
limiting its functional effectiveness.
• Maximum area of compression reinforcement (iii) Repairable repair or crack due to fatigue.
Asc max = 0.04 bD i.e. 4% of total C/S (iv) Corrosion
Total = 4 + 4 = 8% of total C/S
(v) Fire.
47. A reinforced concrete slab is 90 mm thick. The
51. The drawback of Lee-McCall system cannot
maximum size of reinforcement bar that can be
used is : use:
(a) 12 mm diameter (b) 10 mm diameter (a) Straight tendons
(c) 8 mm diameter (d) 6 mm diameter (b) Elliptical tendons
(c) Curved tendons
Ans. (b) : The diameter of the bars shall not exceed one
eight of the total thickness of the slab. (d) Trapezoidal tendons
90 Ans. (c) : In this system, the forces are transmitted by
Maximum size = = 11.25mm the bearing at the end blocks while the system
8
So, the diameter will be 10 mm. eliminates the loss of stress due to anchorage slip by
screwing a nut and washer against end blocks and the
48. In the design of two-way slab restrained at all drawback of this system is that it cannot use curved
edges, torsional reinforcement required is :
tendons.
(a) 0.75 times the area of steel provided at
midspan in the same direction 52. The method of tensioning Magnel-Balton
(b) 0.375 times the area of steel provided at system involves :
midspan in the same direction (a) Hydraulic jack and wires
(c) 0.375 times the area of steel provided in the (b) Hydraulic jack and bars
shorter span (c) Multi strand hydraulic jack
(d) 0.375 times the area of steel provided in the (d) Jack inserted at centre of beam
longer span Ans. (a) : Magnel Blaton system– In this system, the
Ans. (a) : Torsional reinforcement should be 0.75 times anchorage device consists of sandwhich plate having
area of steel provided at mid span as per annex D clause grooves to hold the wires and wedges which are also
D-1.8, of IS 456 : 2000. grooved. Each plate carries eight weirs.

OPSC Polytechnic Lecturer Exam-2018 (Paper-I) 196 YCT

• Between the two ends the spacing of the wires is (c) 4 (d) 6
maintained by spacers, wires of 5 mm to 7 mm are Ans. (d) : Loss of pre-stress in post tensioned members-
adopted. (i) Losses due to elastic determination of concrete.
(ii) Losses due to relaxation of stress in steel.
(iii) Losses due to to creep in concrete
(iv) Losses due to to shrinkage of concrete
(v) Losses due to slip of anchorages.
(vi) Losses due to friction.
57. The frictional and anchorage slip losses are
observed in :
(a) Post-tensioned members
(b) Pre-tensioned members
53. If the beam supports uniformly distributed (c) Ruptured members
load the tendon follows : (d) Tensile members
(a) Straight (b) Ellipse profile Ans. (a) : The frictional losses and losses due to
(c) Random profile (d) Parabolic profile anchorage slip are observed in post-tensioned members
Ans. (d) : The cable for a prestressed concrete simply only because pre-tensioned members do not require
supported beam subjected to uniformly distributed load anchorages for pre-stressing.
over the entire span should ideally be parabolic with 58. Which of the following is a loss of pre-stress in
zero eccentricity at the ends and maximum eccentricity post-tensioned members?
at the centre of span. (a) Loss due to slip of anchorage
(b) Loss due to deformations
(c) Loss due to tensioning
(d) Loss due to pumped concrete
Ans. (a) : Loss of pre-stress in post tensioned members-
54. The concept of load balancing is useful in (i) Losses due to elastic determination of concrete.
selecting the :
(ii) Losses due to relaxation of stress in steel.
(a) Anchorage profile
(iii) Losses due to to creep in concrete
(b) Bending profile
(iv) Losses due to to shrinkage of concrete
(c) Tendon profile
(v) Losses due to slip of anchorages.
(d) Jack profile
(vi) Losses due to friction.
Ans. (c) : The concept of loading balancing is useful in
selecting the tendon profile which can supply the most 59. The loss of stress in steel due to elastic
desirable system of forces in concrete, straight portion shortening or deformation is :
of cable profile does not produce any reactions at the (a) α e f c (b) α c f c
end, while the curve and sharp angles of cable develop (c) α e / f c (d) α e / f c
uniformly distributed and concentrated load
Where αe = Es/Ec = modular ratio, fc = pre-
respectively.
stress in concrete at the level of steel, ES =
55. How many types of losses in pre-stress are modulus of elasticity of steel, Ec = modulus
observed in pre-tensioned member? of elasticity of concrete
(a) 7 (b) 8
Ans. (a) : In pre-tensioning, loss of prestress due to
(c) 4 (d) 2
elastic shortening of concrete = = α e f c
Ans. (c) : Loss of pre-stressing in pre-tensioned
αe = modular ratio
members -
fc = Stress in concrete at the level of prestressing steel.
(i) Losses due to elastic deformation of concrete.
αe In post-tensioning, there is no loss of prestress due to
(ii) Losses due to relaxation of stress in steel. elastic shortening of concrete if all the wires are
(iii) Losses due to creep in concrete. tensioned simultaneously. When the wires are tensioned
(iv) Losses due to shrinkage of concrete. successively, the stretched wire is shortened by the
56. How many types of loss in pre-stress are subsequent stretching of all the other tendons, and the
observed in post-tensioned members? last tendon is not shortened by any subsequent
(a) 8 (b) 10 stretching.

OPSC Polytechnic Lecturer Exam-2018 (Paper-I) 197 YCT

60. A pre-tensioned concrete beam 100 mm wide equal to the maximum frictional loss and jacking the
and 250 mm deep in pre-stressed by straight tendons from both ends of the beam adopted generally,
wires and modulus of elasticity of steel and when the tendons are long or when the angles of bearing
concrete are 2.1 × 105 MPa and 35000 N/m2. are large.
What will be the modular ratio? 65. The term anchorage slip means :
(a) 4 (b) 5 (a) Distance moved by friction wedges
(c) 6 (d) 10 (b) Radius by friction wedges
Ans. (c) : Given, (c) Rotation by friction wedges
Modulus of elasticity of steel 'ES' 12.1 ×105 N/mm2 (d) Twisting movement by friction wedges
= 210 kN /m2 Ans. (a) : Anchorage slip is the distance moved by the
Modulus of elasticity of concrete = 35000 N/m2 friction wedges (in post tensional the members) after
Ec = 35 kN/m2 releasing the jacks at the ends of the member and before
the wire get fixed perfectly in wedges, the loss during
Es 210
Modular ratio 'm' = = =6 anchoring which occurs with wedge type grips is
Ec 35 normally, allowed for on the site by over-extending the
61. Which curing is adopted in case of pre- tendon in the pre-stressing operation by the amount of
tensioned members to prevent shrinkage? the draw in before anchoring.
(a) Surface curing (b) Edge curing 66. Elastic Modulus and Poisson's ratio of steel are
(c) Moist curing (d) Total curing _______ and______ respectively.
Ans. (c) : In the case of pretensioned members, (a) 1.5 × 109 N/mm2 and 0.1
generally moist curing is resolved to in order to prevent (b) 2.0 × 105 N/mm2 and 0.3
shrinkage until the time of transfer, the magnitude of (c) 1.5 × 109 N/mm2 and 0.2
relative strain and the stresses induced depend on the (d) 2.5 × 105 N/mm2 and 0.3
concrete composition and surrounding environment to
Ans. (d) : Elastic Modulus and Poisson's ratio of steel
which the composite member is exposed.
are 2.0 × 105 N/mm2 and 0.3 respectively.
62. The loss of stress due to creep of concrete can
67. Which of the following is added to steel to
be estimated by :
increase resistance to corrosion?
(a) Ultimate creep strain
(a) Carbon (b) Manganese
(b) Ultimate load
(c) Sulphur (d) Copper
(c) Ultimate creep stress
Ans. (d) : Addition of small quantity of copper
(d) Ultimate creep tension
increases resistance to corrosion.
Ans. (a) : Loss of prestress due to creep of concrete can
• Even chrome and Nickel are added to impart
be determined from the following.
corrosion resistance property to steel.
(1) Ultimate creep strain method.
(2) Creep coefficient method. 68. Which of the following factors is included in
the limit state of serviceability?
63. The phenomena of reduction of stress in steel at
(a) Brittle fracture
a constant strain are known as :
(b) Fracture due to fatigue
(a) Relaxation of stress
(c) Failure by excessive deformation
(b) Shrinkage of concrete
(d) Deformation and deflection adversely
(c) Creep of concrete
affecting appearance or effective use of
(d) Anchorage slip structure
Ans. (a) : The phenomena of reduction of stress in steel Ans. (d) : Lime state of serviceably include–
at a constant strain are known as relaxation of stress. (i) Deformation cracking and deflection adversely
64. The total loss of pre-stress due to friction is of: affection appearance or effective use of structure.
(a) 2 types (b) 4 types (ii) Vibrations in structure or any part of its compound
(c) 6 types (d) 8 types limiting its functional effectiveness.
Ans. (a) : The total loss due to friction is divided into (iii) Repairable repair or crack due to fatigue.
two types. (iv) Corrosion
(v) Fire.
• Loss of prestress due to effect of curvature, loss pre
prestress due to wobble effect and frictional losses can 69. The partial factor of safety for resistance
be reduced by over tensioning the tendons by an amount governed by ultimate strength is :

OPSC Polytechnic Lecturer Exam-2018 (Paper-I) 198 YCT

 (a) 1.10 (b) 1.5 (c) 1.5 × nominal diameter of fastener
(c) 2.0 (d) 1.25 (d) 2.5 × nominal diameter of fastener
Ans. (d) : In limit state of design: Ans. (d) : Minimum pitch– Rivet should be placed at a
Partial safety factors. sufficient distance and the minimum pitch is ensured for
Partial safety the following reasons :
Definitions
factors • To prevent bearing failure of members between the
rivets.
Resistant governed by yielding 1.10
• To prevent efficient installation of rivets
Resistant of member to bulking 1.10 As per IS 800 : 1984,
Resistant governed by ultimate 1.25 Minimum pitch = 2.5 d
stress d = nominal dia of rivet
70. For roofs of slope greater than 100, the imposed Maximum pitch –
load is reduced by _____ for every degree rise • For tension member → P >/ (16t, 200 mm)min
in slope. • For compression member → P >/ (12t, 200 mm)min
(a) 0.02 kN/m2 (b) 0.05 kN/m2
2 • For rows near the edge → P >/ (100 mm + 4t, 200
(c) 0.75 kN/m (d) 0.5 kN/m2 mm)min
Ans. (a) : As per IS 875 for roof of shape greats than 75. Nominal bearing strength of bolt is 2.5 kb.d.t.fu
100, the imposed load is reduced by 0.02 kN/m2 for where kb depends on :
every degree rise in slope. 1. End distance
Normally no access is provided for sloping roofs with
2. Pitch distance
sheet. In such cases IS 875 part II make the following
provision : 3. Ultimate tensile stress of bolt
(a) upto 10o slope : 0.75 kN/m2 4. Shank area of bolt
(b) For more than 10o slope : 0.75 – 0.2 kN/m2 5. Yield stress of bolt
Where, θ is slope of sheeting. 6. Diameter of hole
71. In simple connections, the connection between (a) 1, 2, 4, 5 (b) 1, 2, 3, 6
members at their junction will : (c) 2, 3, 4, 5 (d) 3, 4, 5, 6
(a) Resist moment only Ans. (b) : Nominal bearing strength of bolt
(b) Resist force and moment Vnpb = 2.5 k b tdf u
(c) Not resist force  e p f 
(d) Not resist moment k b = min  , − 0.25, ub ,1
 3d 0 3d 0 fu 
Ans. (d) : In simple construction, connection between
members at their junction will not resist any appreciable e = edge distance
moment and shall be assumed to be hinged. p = pitch, d0 = hole diameter
fub = ultimate tensile stress of bolt
72. In first order elastic analysis, equilibrium is fu = ultimate tensile stress of plate
expressed in terms of :
76. Depth of intermediate batten = _______ depth
(a) Geometry of deformed structure
of end batten.
(b) Geometry of underformed structure
(a) 1/2 (b) 3/4
(c) Geometry of both deformed and underformed
(c) 1 (d) 2
structure
(d) Geometry of any structure Ans. (b) : The depth of the intermediate batten is
taken as ¾ of the effective depth of the end batten and
Ans. (b) : In first order elastic analysis, equilibrium is
should be more than twice the width of one component
expressed in terms of geometry of unreformed structure.
members.
• This assumption is valid when plastic displacement
d1' > 2b
are small compared to dimension of structure.
73. What is the yield strength of bolt of class 4.6? 3
d1' = d'
(a) 400 N/mm2 (b) 240 N/mm2 d
2
(c) 250 N/mm (d) 500 N/mm2 Overall depth of intermediate batten d1 = d1' + 2 × edge
Ans. (b) : For bolt of grade property 4.6 represents the distance.
ultimate tensile strength is 400 N/mm2 and yield The effective depth of the end batten should not be
strength 0.6 times 400 which is 240 N/mm2. less than distance between the centre of gravity of the
74. What is the minimum pitch distance? component members and this should be more than twice
(a) 2.0 × nominal diameter of fastener the width of one component members
(b) 3.0 × nominal diameter of fastener Effective depth of end batten–

OPSC Polytechnic Lecturer Exam-2018 (Paper-I) 199 YCT

d' = S'+2Cyy 81. Which chemical composition has highest
d' > 2b content in OPC?
Over all depth of end batten d = d' + 2 × edge distance (a) Alumina (b) Silica
(c) Lime (d) Iron Oxide
Ans. (c) : The properties of various ingredient of
Portland cement are as follows :
Ingredient Percentage range
Lime (CaO) 62-67
Silica (SiO2) 17-25
Alumina (Al2O3) 3-8
CuSO4 (Gypsum) 2-3
Ferric oxide (Fe2O3) 3-4
MgO 0.1-3
Sulphur 1-3
Alkalies 0.2-1
77. Which of the following is not a compression 82. Which cement contains high percentage of C3S
member? and less percentage of C2S?
(a) Strut (b) Boom (a) Rapid hardening cement
(c) Tie (d) Rafter (b) Ordinary portland cement
Ans. (c) : Tie is a tension member, whereas strut, rafter (c) Quick setting cement
and boom are compression members. (d) Low heat cement
78. If imperfection factor α = 0.49, then what is the Ans. (a) : Rapid hardening cement – Also known as
buckling class? early gain in strength of cement, this cement contain
(a) a (b) b more percentage of C3S and less percentage of C2S high
(c) c (d) d proportion of C3S and lime grounded finer than normal
Ans. (c) : value of imperfection factor– cement will impart quicker hydration.
Buckling class A → 0.21 Rapid hardening Cement (8041) –
Buckling class B → 0.34 • Larger proportion of lime grounded finer than in
ordinary cement.
Buckling class C → 0.49
• It is similar to ordinary portland cement but with
Buckling class D → 0.76
higher C3S content and finer grinding.
79. Members used in bridges parallel to traffic are
• High heat of hydration characteristics.
called?
• Initial setting time - 30 min.
(a) Spandrel (b) Stringers
(c) Purline (d) Joist • Final setting time - 600 min
• The high strength at early stage is due to finer
Ans. (b) : Stringers– Stringers run longitudinally and
grinding.
pallel with roadway.
• They are spaced across the length of the floor beams • This cement attains strength at the age of 3 day,
and either rest upon the floor beam or are connected by equipment to the strength of O.P.C. at 7 day.
angles to the web of the floor beam. 83. Which cement is used in sewage and water
80. Local buckling can be prevented by : treatment plants?
(a) Limiting width-thickness ratio (a) Rapid hardening cement
(b) Increasing width-thickness ratio (b) Low heat cement
(c) Changing material (c) Sulphate resisting cement
(d) Changing load on member (d) Quick setting cement
Ans. (a) : The individual elements of a column i.e. Ans. (b) : Sulphate resisting cement is used for canal
flange or web may buckle locally forming wrinkles. lining.
This type of buckling causing a column failure is called Cement type Suitability
local buckling and can be prevented by providing • Rapid hardening Situation where a rapid
suitable width– to thickness ratio of the element, development of strength is
• Buckling are three types : desired e.g. repair of
1. Flexural buckling roads, bridges and in water
2. Torsional buckling front structures.
3. Flexural–torsional buckling. • Quick setting cement For underwater concreting.

OPSC Polytechnic Lecturer Exam-2018 (Paper-I) 200 YCT

• Pozzolana cement Used in marine work, Ans. (b) : Initial setting time and final setting time–
mass concreting operation It is measured from the instant water is added into the
such as dams, large cement upto time, it start loosing its plasticity and final
foundation etc. setting time is referred as the time, which is measured
84. Which cement is used to create bond with old from the instant water is added in the cement upto the
concrete surface? extent it completely losses it plasticity and attain
(a) Rapid Hardening Cement sufficient firmness to resist definite loading.
(b) Expansive Cement
(c) Sulphate Resisting Cement • For OPC initial setting time is ≮ 30 min and final
(d) Low heat Cement setting time is ≯ 600 min (10 hours).
Ans. (b) : It is mostly used in structural elements, such 89. What is the size of fine aggregates?
as slabs, pavement, beams and roofs. Expansive
(a) 4.75 mm
cement is used to create bond with old concrete surface
When expanding or expansive cement is used to fill (b) < 4.75 mm
concrete cracks or opened up joints, it expands while (c) > 4.75 mm
hardening and thus effectively packs the crack or joint. (d) 12 mm
85. How many times in each layer of concrete Ans. (b) :
rodded in a slump cone? Fine aggregate < 4.75 mm
(a) 75 (b) 25 Coarse aggregate > 4.75 mm
(c) 12 to 15 (d) 35 to 65 Boulder > 80 mm
Ans. (b) : Slump test– It is the most popular method Corbel > 300 mm
available to find the workability of concrete both in
field and in laboratory due to ease of its performance. 90. Crushed stone, gravel and ordinary sand are
examples of :
• For proper compaction, It should be subjecting to 25
numbers of blows with the help of tamping rod. (a) Light-weight aggregate
86. Which apparatus is generally used to measured (b) Normal-weight aggregate
the soundness of the cement? (c) Heavy-weight aggregate
(a) Vicat apparatus (d) Both normal-weight aggregate and Heavy-
(b) Le-Chatelier apparatus weight aggregate
(c) Soundness meter Ans. (b) : Normal weight aggregate– Crushed stone,
(d) Duff Abrams apparatus gravel and ordinary sand are examples of normal weight
Ans. (b) : Soundness test : the purpose of this test is to aggregates. They are commonly used in the
detect the presence of uncombined lime in cement after manufacture of normal weight concrete, asphalt
setting. It is performed with the help of :
concrete, and road way sub base.
1
• The average values of specific gravity for sand and
87. Before testing setting time of cement one
should test for : granite are 2.6 and 2.65 respectively.
(a) Soundness (b) Strength 91. The bulking of coarse aggregate is :
(c) Fineness (d) Consistency (a) More than fine aggregate
Ans : (d) Consistency Test– This is a test to estimate (b) Less than the aggregate but appreciable
the quantity of mixing water to from a paste of normal (c) Equal than fine aggregate
consistency defined as that percentage water
(d) Negligible
requirement of the cement paste, the viscosity of which
will be such that the vicat's plunger penetrates up to a Ans. (d) : • The bulking of coarse aggregate is
point 5 to 7 mm from bottom of the vicat's mould. negligible.
Importance– The water requirement for varies tests of Bulking of fine aggregate– The increase in the volume
cement depends on the normal consistency of the of a given mass of fine. Aggregate caused by the
cement, which itself depends upon the compound
presence of water is known as bulking. The extent of
composition and fineness of the cement.
bulking depends upon the percentage of moisture
88. Initial and final setting time of OPC should not
present in the sand and its fineness.
be less than _______ and not be more than
_____ respectively. 92. Which machine is preferred for abrasion for?
(a) 15 minutes and 1 hour (a) Vicat's mould (b) Los angles
(b) 30 minutes and 10 hour (c) Flakiness Gauge (d) Elongation Gauge
(c) 30 minutes and 1 hour Ans. (b) : Abrasion test– Abrasion test is carried out to
(d) 1 hour and 10 hour test the hardness of stone.
OPSC Polytechnic Lecturer Exam-2018 (Paper-I) 201 YCT
• The abrasion value of stone may be determined by (c) Minimum void method
either Deval machines or by Los Angeles machine. (d) Talbot Richard test
w − wB Ans. (b) : Slump test:- It is a test done to check the
Abrasion value = A × 100 workability (consistency) of concrete It can be done on
wA
site or in laboratory It shows the uniformity of concrete
93. Which of the following option doesn't come in in different batches.
chemical admixtures? • Slump test specifies the procedure to be adopted,
(a) Plasticizers either in the laboratory during the progress of work in
(b) Pozzolanic the field, for determining the consistency of concrete,
(c) Super plasticizer where the nominal maximum size of the aggregate does
(d) Accelerators not exceed 38 mm.
Ans. (b) :

94. Admixtures which cause early setting and 98. A lower water cement ratio leads to :
hardening of concrete are called : (a) Higher strength and durability
(a) Plasticizers (b) Retarders
(b) Higher strength but low durability
(c) Super plasticizer (d) Accelerators
(c) Lower strength but high durability
Ans. (d) : Accelerators– Accelerators normally reduce
(d) Lower strength and durability
the setting time, accelerate the rate of hydration of
cement and consequently the rate of gain strength. Ans. (a) : At lower water content ratio strength of
Examples of accelerators– Silicate, Cacl2 concrete is increased.
95. What is plasticizer? Durability– Durability of concrete may be defined as
(a) Which adds water for workability the ability of concrete to resist weathering action.
(b) Which reduces water for workability • Durability of concrete is inversely proportional to
(c) Which decreases workability at the same water-cement ratio.
water content 99. The modulus of elasticity of M 25 concrete as
(d) Which oxidizes water for workability per IS 456 : 2000 is :
Ans. (b) : Plasticizer– It reduce water content C (a) 20000 MPa (b) 22000 MPa
improve workability for a given water content 3. (c) 25000 MPa (d) 28000 MPa
Doses, 0.1–0.4% of weight of cement. Ans. (c) : Given,
Ex – • Ligno sulphonic acid M 25
• Hydroxylated carboxylic acid fck = 25 N/mm2
• Polyglycol esters. Ec = 5000 f ck N/mm2
96. Setting time of cement increase by adding :
Ec = 5000 × 25 = 25000 N/mm2
(a) Gypsum (b) CaCl2
(c) NaOH (d) Hydrogen peroxide 100. An ultrasonic pulse velocity test is an :
(a) ex-situ, non-destructive test
Ans. (a) : Gypsum is used to increase the setting time of
cement. (b) in-situ, non-destructive test
• Gypsum is used in cement during final grinding which (c) ex-situ, destructive test
slows down the chemical reactions/chemical process of (d) in-situ, destructive test
hydration so that flash set not occurs and setting time Ans. (b) : An ultrasonic pulse velocity test is based on
increases. the pulse velocity method to provide information on the
97. Workability of concrete is measured by : uniformity of concrete, cavities, cracks and defects.
(a) Vicat apparatus test • This test is an in situ-nondistructive test to check the
(b) Slump test quality of concrete and natural rocks.

OPSC Polytechnic Lecturer Exam-2018 (Paper-I) 202 YCT

 Odisha Public Service Commission
(Polytechnic Lecturer)
Exam- 2018 (Paper-II)
1. The consumptive use of water for a crop : Ans. (c) : Super-Passage –
(a) Is measured as the volume of water per unit • When full supply level (F.S.L.) of the canal is much
area below the bed level of the drainage trough, so that
(b) Is measured as depth of water on irrigated the canal water flows freely under gravity, the
area structure provided is known as super passage.
(c) May be supplied partly by precipitation and Syphon–
partly by irrigation • When the full supply level (F.S.L.) of the canal is
(d) All of the above much above the bed level of the drainage trough, so
Ans. (d) : The consumptive use of water for a crop– that the canal water flows under the trough, the
structure provided is known as canal syphon.
• Is measured as the volume of water per unit area.
• Is measured as depth of water on irrigated area.
• May be supplied partly by precipitation and partly by
irrigation
Factors affecting consumptive use–
• Mean monthly temperature
• Wind velocity in the locality.
• Soil and topography
• Evaporation
2. The field capacity of a soil is 25%, its
permanent wilting point is 15% and specific
dry unit weight is 1.5. If the depth of root zone 4. The measure to remove water logging of land,
of a crop is 80 cm, the storage capacity of the is:
soil is : (a) To reduce percolation from canals and water
(a) 8 cm (b) 10 cm courses
(b) To increase outflow from the ground water
(c) 12 cm (d) 14 cm
reservoir
Ans. (c) : Given, (c) Both (A) and (B)
Fc= 25% (d) Neither (A) nor (B)
φ = 15% Ans. (c) : However we do the measure to remove water
ϒd = 1.5 logging of land to increase outflow from the ground
d = 80 cm water reservoir.
5. A hydraulic structure is designed to withstand:
γ (a) Seepage forces
The storage capacity = ( Fc − φ ) × d × d (b) Hydraulic jump
γw
(c) Hydraulic pressure
( 25 − 15) 1.5 (d) All of the above
Storage Capacity = × × 80
100 1 Ans. (d) : A hydraulic structure is designed to
10 withstand–
= × 1.5 × 80 • Seepage forces
100
• Hydraulic jump
= 12 cm
• Hydraulic pressure
3. When a canal flowing under pressure is carried
below a natural drainage such that its F.S.L. 6. According to Lacey, depth of scour in a river
does not touch the underside of the supporting depends upon the straightness of the reach. If
D is the depth of scour in regime flow in a right
structure, the structure so provided, is called :
angled bend, it is :
(a) Syphon (b) Aqueduct (a) 1.25 D (b) 1.50 D
(c) Super passage (d) Syphon-aqueduct (c) 1.75 D (d) 2.00 D
OPSC Polytechnic Lecturer Exam-2018 (Paper-II) 203 YCT
Ans. (d): 10. For a standing crop, the consumptive use of
Nature of river Depth of scour water is equal to the depth of water :
Greatly constricted section 1.0 R (a) Transpired by the crop
At severe bend 1.75 R (b) Evaporated by the crop
At a right angle bend 2.00 R (c) Transpired and evaporated by the crop
or at nose of piers (d) Used by the crop in transpiration, evaporation
At the moderate bend condition 1.5 R and also the quantity of water evaporated
from adjacent soil
e.g. along the apron of guide bond
In a straight reach 1.27 R Ans. (d) : Consumptive use – Consumptive use for
particular crop may be defined as the total amount of
7. The most suitable section of a lined canal, is : water used by the plant in transpiration and evaporation
(a) Triangular section with circular bottom for from adjacent soil or from plant leaves in any specified
small canals time.
(b) Trapezoidal section with rounded corners for 11. Irrigation canals are generally aligned along :
large canals
(a) Ridge line (b) Contour line
(c) Both (A) and (B)
(c) Valley line (d) Straight line
(d) None of the above
Ans. (a) : Cannels are normally aligned along ridge
Ans. (c) : In general, we adopt two types of lining canal lines and as a general rule all the ridge in the command
section- area from the alignment of distributaries and minors.
• Triangular shaped channel with circular bottom (for The distributions system is so aligned that maximum
small discharge). area is served with the least length of channels.
• Trapezoidal shaped channel with rounded comes (for 12. A river is said to be of :
large discharge). (a) Aggrading type if it builds up its bed to a
8. If the straight sides of a triangular section of a certain slope
lined canal with circular bottom of radius D, (b) Degrading type if it builds up its bed to a
make the horizontal, the hydraulic mean depth certain slope
is : (c) Meandering type if it flows in sinous curve
(a) D (b) D/2 (d) All of the above
(c) D/3 (d) D/5 Ans. (d) : Aggrading or accreting type– An aggrading
Ans. (b) : If the straight sides of a triangular section of river is a silting river, such a river increases its bed
a lined canal with circular bottom of radius D, make the slope, which is called building up of slope.
horizontal, the hydraulic mean depth is D/2.
9. According to Bligh's creep theory, percolation
water flows along :
(a) Straight path under the foundation of the dam
(b) Circular path under the foundation of the dam
(c) The outline of the base of the foundation of
the dam
(d) None of the above
Ans. (d) : Bligh's creep theory, Degrading Type– If the river bed is constantly getting
• Bligh developed a theory for the sub surface flow scoured (eroded) to reduce and dissipate available
which come to known as Bligh's creep theory. excess land scope then river is known as degrading
• Bligh assumed that the percolating water follows river.
the outline of the base of the structure which is in
contact with the subsoil.
• The length of the path traversed by the percolating
water is called the length of creep or creep length.
• Bligh further assumed that the head loss per unit
length of creep which is called the hydraulic
gradient is constant throughout the percolating
passage i.e. the loss of the head is proportional to 13. A deficit of sediments in flowing water may
the length of creep. causess a river :
The creep length = L = b + 2d1 + 2d2 + 2d3 (a) Meandering type
hydraulic gradient = H/L = H/b + 2d1 + 2d2 + 2d3 (b) Aggrading type
Note– The reciprocal of hydraulic gradient i.e. (L/H) is (c) Degrading type
known as Bligh coefficient of creep C hence (L = CH). (d) Sub-critical type
OPSC Polytechnic Lecturer Exam-2018 (Paper-II) 204 YCT
Ans. (c): Degrading river– It causes the channel banks
Ans. (d) : Diversion head works– The work which are
to get shifted and lower the river bed depth. This is
constructed at the of the canal, in order to divert the
caused by a deficit of sediments in flowing water. It
river water towards the canal, so as to ensure a
takes in area when the supply of sediment is less than
regulated continuous supply of silt-free water with a
the amount of material which can be transported by the
certain minimum heat into the canal, are known as
system. diversion head works.
14. The ratio of the rate of change of discharge on Crest level is depends–
• FSL of the canal
an outlet to the rate of change in level of water
surface in a distributary of its normal depth, is
• Discharge perimeters
known as : • Pond level
(a) Efficiency (b) Sensitivity
18. A land is said to be water-logged if its soil pores
(c) Flexibility (d) Modulur limit
within :
Ans. (b) : Sensitivity – The ratio of the rate of change (a) A depth of 40 cm are saturated
of discharge of an outlet to the rate of change in level of (b) A depth of 50 cm are saturated
water surface in a distributary at its normal depth.
(c) Root zone of the crops are saturated
Sensitivity of rigid module is zero.
(d) All of the above
15. When a canal and a drainage approach each
Ans. (c) : Water logging also leads to salinity, if the
other at the same level, the structure so
water table has risen up, or if the plant roots happen to
provided, is :
come within the capillary fringe, water is continuously
(a) An aqueduct (b) A syphon
evaporated by capillarity thus, a continuous upward
(c) A level crossing (d) Inlet and outlet
flow of water from the water table to the land-surface,
Ans. (c) : A level crossing– It is a cross drainage work
gets established.
in which the drainage and the canal meet each other at
approximately the same level. Level crossing is 19. Canals constructed for draining off water from
generally adopted when canal and the drainage are water logged areas, are known as :
practically at the same level and for high flood drainage (a) Drains
discharge but short lived. Aqueduct for super passage is (b) Inundation canals
adopted when high flood drainage discharge is large. (c) Valley canals
16. The difference in level between the top of a (d) Contour canals
bank and supply level in a canal, is called :Ans. (a) : Canals constructed for draining off water
(a) Berm from water logged areas, are known as Drains.
(b) Free board Inundation canal are long canals taken off from large
(c) Height of bank rivers. They receive water when the river is high enough
(d) None of the above and especially when in flood while perennial canals are
lined to dams and barrages to provide water throughout
Ans. (b) : The difference in level between the top of a
bank and full supply level in a canal is called Freethe year and they irrigate a vast area.
board. A contour canal is an artificially dug navigable canal
♦ The margin b/w FSL and bank level is known as freewhich closely follows the contour line of the land it
board. The amount of freeboard depends upon the sizetraverses in order to avoid costly engineering marks
of the channel. such as boring a funnel through higher ground, building
an embankment over lower ground or constructing a
canal lock (or series of locks) to change the level of the
canal. Because of this, these canals are characterized by
their meandering course.
20. A fall in a canal bed is generally provided, if :
Discharge (m3/s) Free board (m) (a) Ground slope exceeds the designed bed slope
1 to 5 0.50 (b) Designed bed slope exceeds the ground slope
5 to 10 0.60
(c) Ground slope is practically the same as the
10 to 30 0.75
designed bed slope
30 to 150 0.90
(d) None of the above
17. The crest level of a canal diversion head work,
depends upon : Ans. (a) : Canal fall– A fall is a structure constructed
(a) F.S.L. of the canal across a channel to permit lowering down of water level
(b) Discharge perimeters in order to permit the surface energy processed by the
(c) Pond level falling water which may otherwise scour the bed and
(d) All of the above bank of the channel.

OPSC Polytechnic Lecturer Exam-2018 (Paper-II) 205 YCT

• When the available natural ground slope is steeper Ans. (d) : • The unit hydrograph of a specified unit
than the designed bed slope of the channels the duration obtained from the past data can be used to
difference is adjusted by constructing vertical falls or obtain the hydrograph of future storms of like duration.
drops in the canal bed at suitable intervals. • To obtain the ordinates of storm hydrograph, the
ordinates of unit hydrograph are multiplied by the
multiplying factors.
• The multiplying factor for storm hydrograph may be
obtained by dividing the run off in mm by 25 mm.
25. The deficiency in rain catch due to vertical
21. The time required by rain water to reach the acceleration of air forced upward over the
outlet of drainage basin, is generally called :
gauge, is :
(a) Time of concentration
(a) Greater for heavy rain
(b) Time of overland flow
(c) Concentration time of overland flow (b) Greater for lighter rain
(d) Duration of the rainfall (c) Greater for large drops
Ans. (a) : Time of concentration is a concept used in (d) Lesser for small rain drops
hydrology to measure the response of a watershed to a Ans. (b) : j
rain event. It is defined as the time needed for water to 26. The minimum recommended diameter of
flow from most remote point to the outlet. sewers is :
22. A river training work is generally required (a) 5 cm (b) 10 cm
when the river is : (c) 15 cm (d) 20 cm
(a) Aggrading type
(b) Degradinet type Ans. (c) :
(c) Meandering type • The minimum recommended diameter of sewers is
(d) Both (A) and (B) 15 cm.
Ans. (c) : • A river training work is generally required • The maximum recommended diameter of sewers is
when the river is meandering type. 300 cm.
• River training work done in stable channel to guide • Design of sewers is based on peak flow discharge.
the axis of flow at ordinary and low stages and safe • The flow of waste water in sewer is steady and
passage of floods without overtopping the banks. uniform.
23. Pick up the correct statement from the 27. The rate of accumulation of sludge in septic
following: tanks is recommended as :
(a) The zone below water table is called zone of
(a) 30 litres / person / year
the saturation
(b) The zone above water table is called zone of (b) 25 litres / person / year
aeration (c) 30 litres / person / month
(c) The water which exists in the zone of (d) 25 litres / person / month
saturation is called ground water Ans. (a) : The rate of accumulation of sludge in septic
(d) All of the above tank is recommended as 30 lit/person/year.
Ans. (d) : The zone below water table is called zone of 28. If 2% solution of a sewage sample is incubated
the saturation. for 5 days at 200C and depletion of oxygen was
The zone above water table is called zone of aeration. found to be 5 ppm, B.O.D. of the sewage is :
The water which exists in the zone of saturation is (a) 200 ppm
called ground water. (b) 225 ppm
24. Pick up the correct statement from the (c) 250 ppm
following: (d) None of the ahove
(a) The unit hydrograph of a specified unit Ans. (c) : Given,
duration obtained from the past data can be Dilution factor (DF) = 2% = 0.02
used to obtain the hydrograph of future Dipletion of oxygen (DO) = 4 ppm
storms of like duration.
Consumed DO 5
(b) To obtain the ordinates of storm hydrograph, BOD = = = 250 ppm
the ordinates of unit hydrograph are DF 0.02
multiplied by the multiplying factors. 29. If D is the diameter of upper circular portion,
(c) The multiplying factor for storm hydrograph the overall depth of a standard egg shaped
may be obtained by dividing the run off in section is
mm by 25 mm (a) D (b) 1.25 D
(d) All of the above (c) 1.5 D (d) None of the above

OPSC Polytechnic Lecturer Exam-2018 (Paper-II) 206 YCT

 Ans. (c) : • Hydraulic design of sewer, if D is the • If the temperature of sewage is more the D.O.
diameter of upper circular portion, the overall depth content will be less.
of a standard egg 1.5 D. • The solubility of oxygen in sewage is 95% of that in
• Egg shape sewer maintain self cleaning velocity of distilled water.
flow of sewage in dry and rainy seasons. 33. Self-cleansing velocity is :
• This sewer circular side walls resists internal and (a) Velocity at dry weather flow
external pressure. (b) Velocity of water at flushing
30. The non-clog pump which permits solid matter (c) Velocity at which no accumulation remains in
to pass out with the liquid sewage is : the drain
(a) centrifugal pump (d) Velocity of water in a pressure filter
(b) reciprocating pump Ans. (c) : Self cleansing velocity is that which do not
(c) pneumatic ejector permit the silting of solid in sewer more over also carryout
(d) none of these removed of solid which have already settled in it.
Ans. (a) : Centrifugal pump– • Maximum velocity of flow in sewer is also
• Centrifugal pumps are most commonly used for constructed as higher velocity leads to wear and tear
pumping sewage, because these pumps can be easily (scouring) of sewer material thereby reduces their life
installed in pits and sumps, and can easily transport the span and discharge carrying capacity.
suspended matter present in the sewage. • This limiting or non-scouring velocity mainly
• A centrifugal pump consists of a revolving wheel depends upon the material of the sewer.
called impeller which is enclosed in an air tight casing 34. An inverted siphon is designed generally for :
to which suction pipe and delivery pipe or rising main (a) One pipe (b) Two pipes
are connected. (c) Three pipes (d) Four pipes
• The clearance between the vanes of the impeller is Ans. (c) : An inverted siphon is designed generally for
kept large enough to allow any solid matter entering the three pipes.
pump to pass out with liquid so that the pump does not 35. For treating the sewage of a large city, you will
get clogged. recommended.
31. Assertion (A) : Discharge the effluents from the (a) A sedimentation tank and an activated sludge
oxidation ponds just up stream of lakes or treatment plant
reservoirs is undesirable. (b) A plant consisting of Imhoff tanks with low
Reason (R) : The discharge algae get settled in rate trickling filters
the reservoirs and cause anaerobic (c) Sedimentation tanks with high rate trickling
decomposition and other water qualities. fliters
(a) Both A and R are true and R is the correct (d) None of the above
explanation of A
Ans. (a) : For treating the sewage of a large city, for
(b) Both A and R are true but R is not a correct
treating the sewage of large city, sedimentation tank
explanation of A
with an activated sludge treatment plant is
(c) A is true but R is false recommended.
(d) A is false but R is true
• Activated sludge treatment process is secondary
Ans. (a) : • Discharge the effluents from the oxidation treatment and aerobic treatment.
ponds just up stream of lakes or reservoirs is 36. A technique used to determine the
undesirable. concentration of odour compounds in a sample
• Reason (R) : The discharge algae get settled in the is known as :
reservoirs and cause anaerobic decomposition and other (a) Stripping (b) Settling
water qualities. (c) Flushing (d) Chlorination
32. Pick up the correct statement from the
Ans. (a) : • A technique used to determine the
following:
concentration of odour compounds in a sample is
(a) In treated sewage, 4 ppm of D.O. is essential known as stripping.
(b) Only very fresh sewage contains some
• Chlorination is the process of adding the element
dissolved oxygen
chlorine to water as a method of water purification to
(c) The solubility of oxygen in sewage is 95% make it fit for human consumption as drinking water.
that is in distilled water
37. The total dissolved solids (TDS) can be reduced
(d) All of the above
by which one of the following methods?
Ans. (d) : Dissolved oxygen (D.O.)– The (a) Distillation
determination of dissolved oxygen present in sewage is
(b) Reverse osmosis
very-very important because while discharge the
treated sewage into some river stream it is necessary to (c) Ion exchange
ensure at least 4 ppm of D.O. in it. (d) All of the above
OPSC Polytechnic Lecturer Exam-2018 (Paper-II) 207 YCT
Ans. (d) : There are a few treatment options to reduce • It is due to addition of artificial and natural substance
total dissolved solids in water. such as nitrogen and phosphate, through fertilizers as
Such– Reverse osmosis system sewage to an aquatic life.
Water filters 42. Which of the following is not a water borne
Distillation disease?
Ion-exchange (a) Typhoid (b) Scabies
Softeners (c) Cholera (d) Hepatitis
38. Which one of the following cause alkalinity as Ans. (b) : Water borne disease–
well as hardness in natural water? (i) Typhoid
(a) Calcium carbonate (ii) Hepatitis
(b) Calcium bicarbonate (iii) Cholera
(c) Magnesium carbonate (iv) Amoebiasis
(d) All of the above (v) Dysentery
Ans. (d) : Calcium carbonate, bicarbonate, magnesium 43. Which one of the following is an organic gas?
carbonate, all the above salts cause the hardness of (a) Hydrocarbons (b) Aldehydes
water. (c) Ketones (d) Ammonia
39. Fluorides can be removed by : Ans. (d) : Aldehydes, Ketone and hydrocarbon is an
organic compound but ammonia is an inorganic gas.
(a) Reverse osmosis
(b) Lime softening 44. The major contributor of Carbon monoxide is
(c) Ion exchange (a) Motor vehicle
(d) All of the above (b) Industrial processes
(c) Stationary fuel combustion
Ans. (d) : The common methods used for the removal
(d) None of the above
of fluoride from drinking water are divided in the
following categories– Ans. (a) : The major contributor of Carbon monoxide in
atmosphere is stationary fuel combustion.
• Precipitation
• Carbon monoxide is a colorless, odorless and toxic
• Adsorption and ion-exchange has, produced when organic material like natural gas,
• Membrane filtration coal or wood are incompletely burn off. Vehicles
• Distillation exhausts are the single largest source of carbon
• Lime softening monoxide.
40. The source of Arsenic in water is : 45. Fugitive emissions consist of :
(a) Industrial waste (a) Street dust
(b) Fertilizers (b) Dust from construction activities
(c) Phosphate rocks (c) Dust from farm cultivation
(d) All of the above (d) All of the above
Ans. (d) : The source of arsenic in water is – Ans. (d) : Fugitive emission consist of–
• Street dust
• Industrial waste
• Dust from farm cultivation
• Fertilizers
• Dust from construction activities
• Phosphate rocks
46. Ozone is found in :
→ Arsenic is used for the preparation of fertilizers, it is (a) Mesosphere (b) Ionosphere
also present in phosphate rocks and it as also released as
(c) Stratosphere (d) Exosphere
an industrial waste.
Ans. (c) : Ozone is found in stratosphere at an altitude
→ When these things are dumped into the water bodies extending from 11 km to 16 km.
they release arsenic in water.
47. The principal source of volatile organics
41. The process of nutrient enrichment is termed (Hydrocarbon) is :
as: (a) Transportation
(a) Eutrophication (b) Industrial processes
(b) Limiting nutrients (c) Stationary fuel combustion
(c) Enrichment (d) Volcanoss
(d) Schistosomiasis Ans. (b) : The principal source of volatile organics
Ans. (a) : Eutrophication : Eutrophication is an (hydrocarbons) is industrial processes.
enrichment of water by nutrient salts that covers Industrial processes– In the industrial processes, when
structural changes to the ecosystem such as; increased fractional distillation takes place, it produces petrol,
production of algae, and aquatic plant, depletion of fish diesel and many more compound which consists mainly
species etc. of hydrocarbons (volatile organics).

OPSC Polytechnic Lecturer Exam-2018 (Paper-II) 208 YCT

48. The function of automobile catalytic converter 52. In the _______ zone of purification in filter
is to control emissions of : media, bacteria oxidizes the organic matter
(a) Carbon dioxide and hydrogen completely.
(b) Carbon monoxide and hydro carbons (a) Autotrophic
(c) Carbon monoxide and carbon dioxide (b) Heterotrophic
(d) Carbon monoxide and nitrogen dioxide (c) Schmutzdecke
(d) Electroytic
Ans. (b) : Catalytic converters help in oxidizing CO and
HC into CO2 they also reduce NO into nitrogen. These Ans. (b) : In the heterotrophic zone bacteria multiplies
converters are generally made of enable metals like in large number, breaks down not only the organic
platinum, palladium etc. Automobile emissions also matter but also destroys each other to maintain a
balance of life in the filter.
possess SO2 and load compounds.
53. Which water treatment process is done after
49. The list of industrial sources of air pollution
filtration of water?
and their emissions are given. Match the
(a) Primary sedimentation
following :
(b) Disinfection
(a) Ammonia (1) Carbon monoxide
(c) Secondary sedimentation
(b) Plating (2) Particulates
(d) Flocculation
(c) Fertilizers (3) Metal fumes
Ans. (b) : Disinfection– After the water has been
The correct order is : filtered water treatment plants may add one or more
(a) a -1, b -2, c-3 chemical disinfectants (Such as, chlorine, chloramine or
(b) a -3, b -2, c-1 chlorine dioxide) to kill any remaining or chlorine
(c) a -1, b -3, c-2 dioxide) to kill any remaining parasites, bacteria or
(d) a -3, b -1, c-2 viruses.
Ans. (c) : Ammonia– Carbon monoxide 54. The pathogenic bacteria do not last long at a
Plating– Metal fumes pH_______.
Fertilizers– Particulates (a) > 11
(b) < 11
50. Which of the following is used as antiknock
compound in gasoline? (c) < 8
(d) > 8
(a) Tetraethyl lead
(b) Tetraethyl lead Ans. (a) : The pathogenic get killed at pH > 11 which is
very alkaline in nature or at pH = 3, which is highly
(c) Trimethyl lead
acidic.
(d) Triethyl lead
55. In which of the following distribution system,
Ans. (b) : Tetraethyl lead– Tetraethyl lead (TEL) is an the clean water flows entirely under gravity?
organolead compound. (Chemical compounds (a) Gravity system
containing a chemical bond between carbon and lead) (b) Pressure system
with the formula (CH3CH2)4 Pb. (c) Combined gravity and pumping system
TEL was mixed with gasoline (petrol) beginning in the (d) Pumping system
1920 as a patented octane rating. booster that allowed
engine compression to be raised substantially, which in Ans. (a) : In gravity distribution system the clean water
turn increased vehicle performance or fuel economy. flows entirely under gravity.
• In gravitational system, the pumping is normally not
51. Which is the first zone of purification in sand
required at any stage. However in case where the
bed?
source, such as a lake is situated at a hill and the
(a) Autotrphic zone
treatment plant is also situated almost at the same level
(b) Heterotrophic zone on the hill itself then their the some mechanical system
(c) Schmutzdecke zone are setup to intake water from lake
(d) Electrolytic zone 56. The pressure in the distribution mains does not
Ans. (c) : A slow sand filter works by a combination of depend on :
both straining as well as microbiological action. (a) Altitude to supply water
• Three zones of purification in the filter media. (b) Fire fighting requirements
(i) First zone– The surface coating, called (c) Availability of funds
schmutzdecke. (d) Quality of water
(ii) Second zone– The auto-trophic zone existing a few Ans. (d) : The pressure in the distribution mains
millimeters below the schmutzdecke. depends on the height to which water is required to be
(iii) Third zone– The meterotrophic zone which may supplied, fire fighting requirements, whether the supply
extend some 30 cm in the bed. is metered or not and availability of funds.
OPSC Polytechnic Lecturer Exam-2018 (Paper-II) 209 YCT
57. Pick up correct statement from the following : Ans. (b) : length of the long chord (L)–
(a) A eyepiece plays no part in defining the line L = T1T2
of sight ∆
(b) The diaphragm plays no part in defining the = 2OT1 sin
2
line of sight
1
(c) The optical centre of the objective plays no = 2R sin ∆
part in defining the line of sight 2
(d) None of the above ∆
L = 2R sin
Ans. (a) : Line of sight–It is a line which passes 2
through the optical centre of the objective and the 62. In reciprocal levelling, the error which is not
intersection of cross hairs. completely eliminated, is due to :
• A eyepiece plays no part in defining the line of sight. (a) Earth's curvature
58. The intercept of a staff : (b) Non-adjustment of line of collimation
(a) Is maximum if the staff is held truly normal to (c) Refraction
the line of sight (d) Non-adjustment of the bubble tube
(b) Is minimum if the staff is held truly normal to Ans. (c) : Reciprocal levelling eliminates following
the line of sight errors–
(c) Decreases if the staff is tilted away from (i) Error in instrument adjustment i.e. error due to
normal
collimation.
(d) increases if the staff is tilted towards normal
(ii) The combined effect of Earth's curvature and the
Ans. (b) : The intercept of a staff is minimum if the refraction of the atmosphere.
staff is held truly normal to the line of sight. (iii) Variation in average refraction.
• If staff is not vertical, the reading will be higher and if • Reciprocal levelling elements the error due to
staff is normal to the line of sight the will be lesser and curvature of earth completely but as the refraction
accurate
depends upon the atmosphere which may change every
59. If the angular measurements of a traverse are minute.
more precise than its linear measurements,
balancing of the traverse, is done by : 63. Back bearing of a line is equal to :
(a) Bowditch's rule (b) Transit rule (a) For bearing ± 900
(c) Empirical rules (d) All of the above (b) For bearing ± 1800
Ans : (b) In case of angular measurements being more (c) For bearing ± 3600
precise than the linear measurements, the transverse can (d) For bearing ± 2700
be balanced by Transit rule. Ans. (b) : If the bearing of a line AB is measured from
In case of angular measurements and linear A towards B it is known as forebearing.
measurements are equally precise then transverse can be If the bearing of a line AB is measured from B towards
balanced by Bowditch rule. A it is known as back bearing.
60. The branch of surveying in which both
horizontal and vertical positions of a point, are
determined by making instrumental
observations, is known as :
(a) Tacheometry (b) Tachemetry
(c) Telemetry (d) All of the above
Ans. (d) : Tacheometry (or tachemetry or telemetry) is
a branch of surveying in which horizontal and vertical
distance are determined by taking angular observation
with and instrument known as tacheometer.
• Tacheometry is mainly used preparation of contoured 64. The longitudinal section of the surface of
maps or plans reversing both the horizontal as well as bubble tube is :
vertical control. (a) Straight (b) Circular
61. If D is the degree of the curve of radius R, the (c) Parabolic (d) Elliptic
exact length of its specified chord, is : Ans. (b) : Bubble tube–A level tube or Bubble tube is
(a) Radius of the curve × sine of half the degree an essential part of most of the surveying instrument. It
(b) Diameter of the curve × sine of half the consist of glass tube partially filled with a liquid. The
degree space above the liquid surface is occupied by vapour
called bubble.
(c) Diameter of the curve × cosine of half the
degree • The longitudinal section of the tube cut by a vertical
(d) Diameter of the curve × tangent of half the plane through the axis of tube is a part of the circular
degree arc.
OPSC Polytechnic Lecturer Exam-2018 (Paper-II) 210 YCT
65. Imaginary line passing through points having Ans. (d) : In Levelling–
equal magnetic declination is termed as : • The line of sight while observing back sight and fore
(a) Isogonic line sight lie in the same horizontal plane.
(b) Agonic line • The staff readings are measurement made vertically
(c) Isoclinit line downwards from a horizontal plane.
(d) None of the above • The horizontal plane with reference to which staff
Ans. (a) : Isogonic Line– Isogonic line is the line readings are taken, coincides with the level surface
drawn through the points of same declination. through the telescope axis
• The distribution of earth magnetism is not regular and
69. Surveys which are carried out to depict
consequently, the isogonic lines do not form complete
great circle, but radiating from the north and south mountains, rivers, water bodies, wooded areas
magnetic regians they follow irregular paths. and other cultural details, are known as :
66. In case of reduction of levels by the height of (a) Cadastral surveys
instrument method : (b) City surveys
(a) ∑ B.S. − ∑ F.S. = difference in R.L.S. of the (c) Topographical surveys
first station and last station (d) Guide map surveys
(b) ∑ ( R.L. + I + F.S.) – first R.L. = Ans. (c) : Topographical surveys–Conducted to obtain
data and to make a map indicating inequalities of land
∑ ( H.I. + No.of R.L.s.) surface (mountains, rivers, water bodies, wooded area).
(c) Both (A) and (B) Cadastral surveys–Done to produce plans of property
(d) All of the above boundaries for legal purpose.
Ans. (d) : Height of Instrument method City survey–The survey made in connection with the
• This method is generally adopted if there are 10 + construction of streets, water supply and sewage lines
many numbers of intermediate stations and also if the fall under this category.
number of shifting of instrument is less. 70. The total change in level along the line is equal
• It does not provide a check on intermediate sights. to total back sights :
• Height of instrument is calculated at each set up of the (a) Minus total fore sights
instruments. (b) The total rises minus total falls
• HI - Back sight + elevation of bench mark. (c) The reduced level of last point minus reduced
• Generally used in contour survey; in small area. level of the first point
• Check Σ B S – Σ F. S = Last R.L. = fist R.L.) (d) All of the above
67. A theodolite is said to be in perfect adjustment Ans. (d) : The total change in level along the line is
if: equal to total back sights–
(a) Rotation axis is vertical to the transit axis • Minus total fore sight.
(b) Transit axis is perpendicular to line of • The rises minus total falls
collimation • The reduced level of last point minus reduced level of
(c) Line of collimation sweeps out a vertical the first point.
plane while the telescope is elevated or 71. ABCD is a rectangular plot of land. If the
depressed bearing of the side AB is 750, the bearing of DC
(d) All of the above is :
Ans. (d) : A theodolite is said to be in perfect (a) 750
adjustment if– (b) 2550
• Rotation axis is vertical to the transit axis. (c) 1050
• Transit axis is perpendicular to line of collimation. (d) 2850
• Line of collimation sweeps out a vertical plane while Ans. (a) : ABCD is a rectangular plot
the telescope is elevated or depressed.
68. Pick up the correct statement from the
following:
(a) The lines of sight while observing back sight
and fore slight lie in the same horizontal
plane
(b) The staff readings are measurement made
vertically downwards from a horizontal plane
(c) The horizontal plane with reference to which
staff readings are taken, coincides with the Bearing of DC = 180° – (90° + 15°)
level surface through the telescope axis = 180° – 105°
(d) All of the above = 75°

OPSC Polytechnic Lecturer Exam-2018 (Paper-II) 211 YCT

72. Total latitude of a point is positive if it is lies :
(a) North of the reference parallel
(b) South of the reference parallel
(c) East of the reference parallel
(d) West of the reference parallel
Ans. (a) : Total latitude of a point is positive if it is lies
north of the reference parallel.
73. Contours of different elevations may cross each
other only in the case of : Where, x = error due to inclined line of sight, and
(a) An over handing cliff e = error due to curvature and refraction
(b) A vertical cliff
(c) A saddle
(d) An inclined plane
Ans. (a) : Contours of different elevations cannot cross
each other. If contour lines cross-each other, it show
existence of over hanging cliffs or a cave.

h A − h B = h 'A − h 'B If instrument is correct.

( h B − h A ) + ( h 'B − h 'A )
H=
2
Where 'H' is the true difference of R.L. between A.B.
76. Maximum number of vehicles that can pass a
given point on a lane during one hour without
creating unreasonable delay, is known as :
Contour lines generally do not interest each other. If (a) Traffic density of lane
countour lines are meeting in same portion, it shows (b) Basic capacity of lane
existence of vertical cliff. (c) Probable capacity of lane
(d) Practical capacity of lane
• If they did, the point of unit section would have two
different elevation which is absurd. Ans. (d) : Practical capacity– It is the maximum
number of vehicle that can pass a given point on a lane
74. Systematic errors are those errors : or roadway during one hour, without traffic density
(a) Which cannot be recognized being so high as to cause unreasonable delay, hazard or
(b) Whose character is understood restriction to the drivers freedom to manoeuvre under
(c) Whose effects are cumulative and can be the prevailing roadway and traffic condition.
Basic capacity– It is the maximum number of
eliminated
passenger cars that can pass a given point on a lane or
(d) None of the above roadway during one hour under nearly ideal roadway
Ans. (c) : Systematic error:– A systematic error or and traffic condition which can possibly be attained.
cumulative error is an error that, under the same Possible capacity– It is the maximum number of
conditions, will always be of the same size and sign. A vehicles that can pass a given point on a lane or
systematic error always follows some definite roadway during one hour under prevailing roadway and
traffic condition.
mathematical or physical law, and a correction can be
determined and applied. 77. How many types of methods are there to design
a flexible pavement?
75. The method of finding out the difference in (a) One (b) Two
elevation between two points for eliminating (c) Three (d) Four
the effect of curvature and refraction is : Ans. (c) : There are three types of methods to design a
(a) Reciprocal levelling flexible pavement–
(b) Precise levelling • Empirical method
(c) Differential levelling • semi-empirical method
(d) Flying levelling • Theoretical method
Ans. (a) : Reciprocal levelling– This method is 78. CBR is a :
adopted to accurately determine the difference of level (a) Measure of soil strength
between two point which are far apart. It is also used (b) Flexible pavement design method
when it is not possible to set up level in mid way (c) Rigid pavement design method
between two point. (d) Measure of soil characteristics
OPSC Polytechnic Lecturer Exam-2018 (Paper-II) 212 YCT
Ans. (b): The method of design of flexible pavement as Ans. (a): The cement concrete roads are designed with
recommended by IRC is the CBR method. plain concrete with dowel bar longitudinally which is
• The CBR method was developed originally by the not considered is reinforced cement concrete.
california state highway department. 83. The dowel bars are provided :
• The advantage of the CBR method is that is can be (a) Longitudinally
used to find the pavement and that of individual courses (b) Laterally
in addition thickness of the CBR values of the materials (c) Any direction required
of the courses are also known. (d) In the base of pavement
Ans. (a) : : Dowel bars in concrete pavement are
placed along direction of traffic.

84. The surface of highway pavement should be

designed to allow :
79. The top 500 mm of soil-sub grade should be (a) High rolling resistance
compacted at : (b) Low rolling resistance
(a) OMC (b) MDD (c) No rolling resistance
(c) Dry density (d) Saturated density (d) Very high rolling resistance
Ans. (a) : The top 500 mm layer in the soil subgrade be Ans. (b) : The surface of highway pavement should be
compacted at OMC (optimum moisture content) to designed to allow low rolling resistance.
obtained maximum dry density (MDD), only at MDD
the soil gets its maximum strength. 85. Which of the following pavement has greater
life?
80. The Westergaard equation was modified by : (a) Bituminous pavements
(a) Bradbury (b) Cement concrete pavements
(b) Burnister (c) Gravel roads
(c) Teller and sutherland (d) Earth roads
(d) Telford Ans. (b) : The cement concrete roads have a greater life
Ans. (c) : Modified equation for wheel load stress– than remaining all pavements which may last even upto
Westergaard's edge load stress formula was modified by 100 years.
Teller and Sutherland. Similarly Westergaard's corner 86. The branch of engineering that deals with
load stress formula was modified by Kelley. improvement of traffic performance, traffic,
Edge load stress equation by Teller and Sutherland– studies and traffic network is called :
P
Se = 0.529 (1 + 0.54µ ) × 4log10 ( l / b ) + log10 b − 0.4048 (a) Highway engineering
h2 (b) Railway engineering
Corner load stress equation by Kelley– (c) Traffic engineering
3P   a 2  
1.2 (d) Traffic Management
Sc = 2 1 −    Ans. (c) : The branch of engineering that deals with
h   l  
  improvement of traffic performance, traffic, studies and
81. The pavement thickness is usually assumed in traffic network is called traffic engineering.
rigid pavement as : 87. The ''3-Es'' of traffic engineering stand for :
(a) 20 cm (b) 25 cm (a) Enforcement, empowerment and eradication
(c) 30 cm (d) 35 cm (b) Engineering, education and expulsion
Ans. (b) : The pavement thickness is usually assumed (c) Engineering, education and enforcement
in rigid pavement as 25 cm. (d) Engineering, education and enthusiasm
82. The cement concrete roads are designed with : Ans. (c) : • The various measure to decrease the
(a) Plain concrete accident rates may be divided into three groups–
(b) RCC (i) Engineering
(c) ICPB (ii) Enforcement
(d) Bitumen (iii) Education
OPSC Polytechnic Lecturer Exam-2018 (Paper-II) 213 YCT
88. Pedestrian data is used for planning : Ans. (b): Grade separated intersection :
(a) Highway • Grade separated intersection design in the highest
(b) Sidewalks and cross-walks from of intersection treatment. This type of intersection
(c) kerb causes least delay and hazard to the crossing traffic and
(d) camber in general is much superior to intersection at grade from
Ans. (b) : Pedestrian data is use for indicate safety the point of view of traffic safety and efficient
analysis workability analysis such as sidewalks design operation.
cross walks and other pedestrian facilities. • A highway grade separation is achieved by means of
89. Which of the following relationship is correct? vertical level. Separation of interchanged roads by
(a) Travel speed = running speed means of a bridge thus eliminating all crossing confilict
(b) Travel speed < Running speed at the intersection.
(c) Travel speed > running speed 93. If the angle of merging is low, then the relative
(d) Travel speed = 1.5 times of running speed speed will be :
Ans. (b) : Running speed : It is the average speed (a) Low
maintained by a vehicle over a particular stretch of road, (b) High
while the vehicle is in motion, this is obtained by
dividing the distance covered by the time during which (c) Medium
the vehicle is actually in motion. (d) Depends on width of pavement
Overall speed or travel speed : It is the effective speed Ans. (a) : Relative speed is an important factors in
with which a vehicle travel a particular route between traffic flow at grade. Relative speed is the vector
two terminals; this is obtained by dividing the total different in the velocities of two vehicles in the same
distance travelled by the total time taken including all
flow and is the sum of the speeds of approaching
delay and stoppage.
vehicles from opposite direction.
Running speed > Travel speed
• Relative speed depend on the absolute speeds of
90. If the distance of a vehicle moved is 25 m and intersecting vehicles and the angles between them.
the observed travel time is 15 sec then the space
mean speed is : • When the angle of merging is small the relative
speed will be also low.
(a) 4 m/s (b) 5 m/s
(c) 6 m/s (d) 7 m/s 94. If an additional pavement is provided for lane
change, then that intersection is called :
Ans. (*) : Distance of vehicle moved = 25m
(a) Tee intersection
Observed travel time = 15 sec.
(b) Rotary intersection
nL n
Space mean speed = = (c) Flared intersection
∑ ti ∑ 1 (d) Skewed intersection
ui
Ans. (c) : In un-channelized intersections, the entire
1 25 intersection area is paved and there is absolutely no
= = = 1.67 m/s
 
15 15 restriction to the vehicles to use any part of intersection
  area.
 25 
91. The geometric design in India are designed for: • When no additional pavement width for turning
(a) 85th percentile speed movement is provided, it is called 'palm intersection'.
(b) 15th percentile speed But when the pavement is widened at the intersection
(c) 98th percentile speed area, by a traffic lane or more, it is known as 'flared
(d) 100th percentile speed intersection.
Ans. (c) : The road geometries in India are designed 95. The various regulations imposed through the
for– traffic control devices do not include :
• Design speed for checking design element of highway (a) Clear visibility
= 98th percentile speed. (b) Easy recognition
• The lower speed limit for regulation is mixed traffic (c) Sufficient time for driver
flow = 15th percentile speed.
(d) Traffic population
• The upper - speed limit for regulation of mixed traffic
flow. 85th percentile speed. Ans. (d) : The various aids and devices used to control
regulate and guide traffic may be called traffic control
92. An intersection that is provided for different
levels of road is called : device. The general requirement of traffic control
(a) Intersection at grade device are attention, meaning time of response and
(b) Grades separated intersections respect and road users.
(c) Channelized intersection • Traffic population on is not include and under
regulations imposed through the traffic control devices.
(d) Rotary intersection
OPSC Polytechnic Lecturer Exam-2018 (Paper-II) 214 YCT
96. Which of the following is disadvantage in one (c) Equal to a fill of height Z equal to q/r, where r
way traffic? is the density of the backfill
(a) Increase in average travel speed (d) None of the above
(b) More effective coordinate of signal system Ans. (c) : Back fill with uniform surcharge– If the
(c) More stream lined movement of vehicles backfill is horizontal and carries, a surcharge of uniform
(d) More chances of overtaking intensity q per unit area, the vertical pressure increment,
at any depth h, will increase by q. The increase in the
Ans. (d) : Advantage and disadvantage of one way lateral pressure due to this will be Ka × q. Hence the
traffic– lateral pressure at any depth h is given by
Advantage– Possible increase roadway capacity Pa = K a γh + K a .q
In expensive to implement
May reduce traffic volume
disadvantage–
Possible increase in speed
Chance of over taking.
97. In a liquid limit test, the moisture content at 10
blows was 70% and that at 100 blows was 20%.
The liquid limit of the soil, is :
(a) 35%
(b) 50%
(c) 65%
(d) None of the above
0.121
Figure (a) and (b) show two alternative methods of
N plotting the lateral pressure diagram for this case. The
Ans. (c) : Liquid limit = w ×  
 25  laterial pressure increment Ka × q due to the surcharge is
the same at every point of the back of the wall and does
Where,
not vary with height h. The height of fill he equivalent
W = water content at N blows to uniform surcharge intensity is given by the relation.
N = number of blows K a γh e = K a q
0.121
 100  q
Liquid limit = 70 ×   = 82.78% he =
 25  γ
0.121
 10  100. Pick up the correct statement from the
Liquid limit = 20 ×   = 17.90%
 25  following:
(a) Isotropic consolidation of clay can be
Required liquid limit = 82.78–17.90
obtained in the tri axial apparatus under equal
= 64.87% all round pressure.
 65% (b) If the present effective stress is the maximum
98. Buoyant unit weight equals the saturated to which the clay has ever been subjected, it
density : is called normally consolidated clay.
(a) Multiplied by unit weight of water (c) If the present effective stress in the past was
(b) Divided by unit weight of water more than present effective stress, it is called
(c) Plus unit weight of water over-consolidated clay
(d) Minus unit weight of water (d) All of the above
Ans. (d) : Buoyant unit weight equals the saturated Ans. (d) : Isotropic consolidation– The ories of
density minus unit weight of water. isotropic consolidation for geometrical with
compressible phases require two loading tests, a full
• Buoyant unit weight is also known as the submerged drainage test at constant pore-pressure and a mixture
unit weight. test in which pure-pressure varies.
• Buoyant unit weight is the ratio of submerged weight Consolidation clay– Normally consolidated clays are
to the volume of soil mass. these that are currently experiencing the maximum
99. If the back fill is having a uniform surcharge of vertical overburden effective pressure they have ever
Intensity q per unit area, the lateral pressure experienced is their history.
will be : Over consolidated clay– Clay soil deposit that has
(a) q times the lateral pressure within the surface been fully consolidated under pressure 'PC' in the past,
larger than the present overburden pressure 'P0' i.e.
(b) 1/q times the lateral pressure within the
P c > P o.
surface
OPSC Polytechnic Lecturer Exam-2018 (Paper-II) 215 YCT
 Odisha Public Service Commission
(AEE), Exam- 2016 (Paper-I)
1. If a material has identical elastic properties in 4. Every material obeys the Hooke's law within
all directions, it is said to be- its-
(a) Homogenous (b) Isotropic (a) Elastic limit
(c) Elastic (d) Orthotropic (b) Plastic limit
Ans. (b) : (c) Limit of proportionality
• Isotropic– Elastic properties are same in each and (d) None of these
every direction. ex- Glass Ans. (c) : The limit of proportionality is slightly less
• Homogenous – Materials having properties are than the limit of elasticity in stress-strain graph
always constant they would not change any other. σ∝ε
• Anisotropic – Elastic properties are not same in any Elastic Limit : It can be defined as the point in the
diretion, Ex- wood. graph upto which material comes back to original shape
• Orthotropic – Elastic properties are same in all when loads are removed.
direction other than that in perpendicular direction eg- 5. In a rectangular element subjected to like
cooled Rolled steel. principal tensile stresses P1 and P2 in two
2. If a composite bar of steel and copper is heated, naturally perpendicular directions x and y, the
maximum shear stress would occur along the-
the copper bar will be under-
(a) Plane normal to y-axis
(a) Tension (b) Compression
(b) Plane normal to x-axis
(c) Shear (d) Torsion
(c) Planes at 45o and 135o to the y-direction
Ans. (b) : Compression – Coefficient of thermal (d) Plane at 45o to the y-direction
expansion of copper is greater than steel thereby
resulting stress in copper will be compression and in Ans. (c) :
steel will be tensile.

 σx − σy 
τθ =   sin 2θ + τ xy cos 2θ
Temp↑→ more value of α → compression  2 
Temp ↓→ more value of α → tension τ xy = 0
Thermal coefficient (α) @ 200C τθ max at θ = 450
αsteel = 13 ×10-6/0C
P1 − P2
αcopper = 17 ×10-6/0C =
2
3. Maximum bending moment in a beam occurs
for maximum shear
where-
(a) Deflection is zero  σ − σy 
tan 2θ =  x 
(b) Shear force is maximum  2τxy 
(c) Shear force is minimum ∵ θ = 45o or 135o
(d) Shear force changes sign
6. Principle of superposition is applicable when-
Ans. (d) : The maximum bending moment occurs in
(a) Deflections are linear functions of applied
beam, when shear force at the section is zero or change
forces
the sign because at the point of contra flexure the
(b) The action of applied forces will be affected
Bending Moment is zero.
by small deformations of the structure
dM (c) Material obeys Hook's law
V= or V = 0.
dx (d) None of these
OPSC AEE Exam-2016 (Paper-I) 216 YCT
Ans. (c): The Principle of superposition says that when a 10. The shear stress on a beam section is
number of load are acting on a body, the resulting strain, maximum-
according to the principle of superposition, will be the (a) On the extreme bottom surface fibres
algebraic sum of strain caused by individual loads. (b) On the extreme top surface fibres
7. Principal plane is defined as a plane on which (c) At the free edges
the shear stress is- (d) At the neutral axis of the section
(a) Maximum (b) Half of the normal
Ans. (d) : Shearing stress in a beam is maximum at the
(c) Zero (d) None of these
neutral axis. Shearing stress is defined as the resistance
Ans. (c) : According of Mohr's circle the principle offered by the internal stress to the shear force.
stress is plotted on x-axis and shear stress is plotted on
y-axis so principal plane will be on x-axis and if we 11. The variation of the bending moment in the
taken a point on a x-axis then y will be zero (y=0). So segment of a beam where the load is uniformly
the shear stress will be zero on principal plane. distribute is-
8. Poisson's ratio is an involving- (a) Zero (b) Linear
(a) Elastic Modular (b) Stresses (c) Parabolic (d) Cubic
(c) Strains (d) None of these Ans. (c) :
Τransverse strain
Ans. (c) : Poisson's ratio, µ =
AxialStrain
9. The shear force and bending moment is zero at
the free end of a cantilever beam, if it carries-
(a) Point load at the free end
(b) Point load at the middle of its length
(c) Uniformly distributed load over the whole
length
(d) None of these
Ans. (b, c) : UDL over entire length

12. The maximum bending moment (M) caused by

a concentrated load (W) at the mid span of a
simply supported beam is-
WL  WL 
(a) M =   (b) M =  
 2   8 
WL  WL 
(c) M =   (d) M =  
 4   12 
Point load at the middle
Ans. (c) :

Total load
Ra = × Dist.of opposite side
Total Distance
W
RA = RB =
Point load of free end 2
W L WL
M max = M C = × =
2 2 4
13. A simply supported beam with rectangular
cross-section is subjected to a central
concentrated load. If the width and depth of
beam is doubled, the deflection at centre of the
beam will be reduced to-
(a) 50% (b) 25%
(c) 12.5% (d) 6.25%

OPSC AEE Exam-2016 (Paper-I) 217 YCT

Ans. (d) : 17. The most appropriate failure theory for ductile
materials is-
(a) Maximum principal stress theory
(b) Maximum shear stress theory
2b × ( 2d )
3
PL3 bd 3 (c) Maximum shear stain energy theory
δ1 = , I1 = , I2 = = 16I1 (d) Maximum principal stain theory
48 EI1 12 12
Ans. (c) :
1 PL3 • Maximum principal stress theory is not applicable
δ2 = ×
16 48EI1 for ductile materials.
1 • Maximum principal strain theory over estimates the
= × δ1 = 0.0625 of δ1 elastic strength of ductile material.
16
14. The maximum tensile stress in a cantilever • Most appropriate failure theory for ductile material
beam with concentrated load acting is maximum shear strain energy theory.
downwards on the span is caused at- 18. A cantilever beam having length 'L' is
(a) Top fibre at mid span subjected to a moment 'M' at its free end. If
(b) Bottom fibre at mid span flexural rigidity of beam is EI, the deflection at
(c) Bottom fibre at support free end will be-
(d) Top fibre at support ML ML
(a) (b)
Ans. (d) : Maximum moment will be at support. Hence EI 2 EI
maximum tensile stress will occur at support. Since it is ML2 ML2
hogging moment, tensile stress will occur on top. (c) (d)
EI 2 EI
Ans. (d) :

15. A fixed beam of uniform section is carrying a

point load at its mid span. If the moment of
inertia of the middle half length is reduced to
half its previous value, then the fixed end
moments will-
(a) Decrease (b) increase
(c) Change their direction M
(d) Remain constant Conjugate beam as loading
EI
Ans. (d) : Moment do not depend upon the geometry of Now, deflection in real beam = BM in conjugate beam
the Cross-section in fixed End.
16. At a point 'P', the state of stress is px = 6 MPa, δc = M c = × L ×
M L ML2
=
py = -2 MPa and qxy = 3 MPa, the magnitude of EI 2 2EI
principal stresses for this state of stress will be- 19. The relationship between Young's Modulus of
(a) 9 MPa and – 1 MPa (b) 7 MPa and – 3 MPa Elasticity E, Bulk Modulus K and Poisson's
(c) 7 MPa and – 1 MPa (d) 8 MPa and – 3 MPa ratio µ is given by-
Ans. (b) : (a) E = 2K(1 – 2µ) (b) E = 3K(1 + µ)
(c) E = 3K(1 – 2µ) (d) E = 2K(1 + µ)
Ans. (c) : E = 3K(1 – 2µ)
and
E = 2G(1 + µ)
20. A beam simply supported at both the ends of
length 'L' carries two unequal unlike couples M
2 at the both ends. If the Flexural Rigidity EI is
 σx + σ y   σx − σ y 
 + ( τ xy )
2 constant, the central deflection of beam will be-
σmax/ min =  ± 
 2   2 
ML2 ΜL2
(a) (b)
2
 6 + ( −2 )   6 − ( −2 )  2
4 EI 16 EI
= ±   +3 2
 2   2  ML ML2
(c) (d)
= 2 ± 5 = 7 , −3 64 EI 8EI

OPSC AEE Exam-2016 (Paper-I) 218 YCT

Ans. (d) : 24. For approximate analysis of building frames
under vertical loads, the point of inflection is
assumed at-
(a) Centre of each beam
(b) One-tenth of the span length from each end of
the beam
(c) Centre of each column
(d) Both (a) and (c)
Ans. (b) :

BM
Conjugate beam with as loading.
EI
Now deflection at C in real beam = BM at C in
conjugate
ML L M L L ML2
δC = × − × × = 25. A fixed beam of uniform section is carrying a
2 EI 2 EI 2 4 8EI
point load at its mid span. If the moment of
21. Free body diagram is an- inertia of the middle half length is reduced to
(a) Isolated joint with only body forces acting on half of its previous value, then the fixed end
it moments will-
(b) Isolated joint with all the forces internal as (a) Decrease
well as external, acting on it (b) Increase
(c) Isolated joint with internal acting on it (c) Change their direction
(d) None of these (d) Remain constant
Ans. (b) : Ans. (b) : Increase
26. The deformation of a spring produced by a unit
load method is called-
(a) Flexibility (b) Stiffness
(c) Unit strain (d) None of these
Ans. (a) :
Free body diagram is a graphical illustration used to
Deflection
visualise the applied forces, moments and resulting Stiffness is inverse of flexibility =
reactions on a body in a given condition. Force
22. Independent displacement components at each 27. In the displacement method of structural
analysis, the basic unknowns are-
joint of a rigid jointed plane frame are-
(a) Forces
(a) Three linear movements
(b) Displacements
(b) One linear movement and two rotations
(c) Displacements and forces
(c) Two linear movements and one rotation
(d) None of these
(d) Three rotations
Ans. (b) : In displacement method of analysis, unknown
Ans. (c) : For a rigid jointed plane frame, at each joint, joint displacement (∆, θ) are taken as basic unknowns.
3 displacements are possible –
28. If the displacement at coordinate i due to unit
(i) linear movement in ∆x in x direction
force at coordinate j is δij and displacement at
(ii) Linear movement in ∆y in y direction coordinate j due to unit force at coordinate i is
(iii) Rotation about Z axis, θz. δji then according to Maxwell's Reciprocal
23. Which of the following methods of structural Theorem-
analysis is a force method? (a) δij = δji (b) δij > δji
(a) Column analogy method (c) δij < δji (d) δij ≠ δji
(b) Slope deflection method Ans. (a) : According to Maxwell's reciprocal theorem
(c) Moment distribution method
δij = δ ji
(d) None of these
th
Ans. (a) : 29. To generate the j column of flexibility matrix-
• Column analogy method (a) A unit force is applied at coordinate j and the
displacement are calculated at all coordinates
• Theorem of least works
(b) A unit force is applied at coordinate j and the
• Method of consistent deformation forces are calculated at all coordinates
OPSC AEE Exam-2016 (Paper-I) 219 YCT
 (c) A unit displacement is applied at coordinate j Ans. (b): Influence lines are drawn for a particular
and the forces are calculated at all coordinates section.
(d) A unit displacement is applied at coordinate j 34. The moment required to rotate the near end of
and the forces are calculated at all coordinates prismatic beam of length 'L' and flexural
Ans. (a) : To generate flexibility matrix, rigidity EI through unit angle, without
 f11 f12 f13  translation (the far end being fixed), is given by-
  2EI EI
f =  f 21 f 22 f 23  (a) (b)
L L
 f31 f32 f33 
3EI 4EI
∆ Deflection (c) (d)
Flexibility = = L L
F Force Ans. (d) :
30. For a statically indeterminate pin jointed plane
frame, the relation between number of
members 'm' and number of joints 'j' is
expressed as- Given, θ = 1
(a) m = 3j – 6 (b) m = 2j – 3
ML
(c) m > 2j – 3 (d) m > 3j – 6 θ= =1
4 EI
Ans. (c) : if m = 2J – r, Determinate Truss
4EI
if m > 2J – r, Indeterminate Truss ∴ M =
if m < 2J – r, Unstable Truss L
31. Ratio of strain energy stored by solid shaft of 35. Castigliano's first theorem is applicable-
diameter 'D' and strain energy stored by (a) For statically determinate structure only
hollow shaft (external diameter 'D' and (b) When the system behaves elastically
internal diameter 'd') is given by- (c) Only when principle of superposition is valid
D2 D2 (d) None of these
(a) (b)
(D2 − d 2 ) (D2 + d 2 ) Ans. (c) : Castigliano's first theorem is applicable only
when principle of superposition is valid.
D4 D4 Castigliano's first theorem : The first partial
(c) (d) derivative of the total internal energy (strain energy) in
(D4 + d 4 ) (D4 − d 4 ) a structure with respect to any particular deflection
T 2L T 2L component at a point is equal to the force applied at that
Ans. (*) : SE1 = , SE2 = point and in the direction corresponding to that
2GJ1 2GJ 2 deflection component.
SE1 J 2 • This first theorem is applicable to linearly or non-
= linearly elastic structures in which the temperature
SE2 J1
is constant and the supports are unyielding.

= ×
(
32 π D − d
4 4
=
)
D4 − d 4 Castigliano's second theorem : The first partial
derivative of the total internal energy (strain energy)
πD 4 32 D4 in a structure with respect to the force applied at any
32. Muller Breslau principle in structural analysis point is equal to the deflection at the point of
is used- application of that force in the direction of its line of
(a) To obtain virtual work equation action.
(b) To draw influence line diagram for any force • The second theorem of Castigliano's is applicable to
function linearly elastic (Hookean material) structures with
constant temperature and unyielding supports.
(c) For superposition of load effect
(d) None of these 36. The deflection at any point of a perfect frame
can be obtained by applying a unit load at the
Ans. (b) : If an internal stress component or reaction is joint in-
considered to act through some small distance and (a) Vertical direction
thereby to deflect or displace the structure the curve of (b) Horizontal direction
the deflected structure will be to some scale of influence (c) Inclined direction
line for the stress or reaction component.
(d) The direction in which the deflection is
33. Influence line for forcing function gives its required
variation at-
∂U
(a) Mid span (b) A given section Ans. (d) : ∆ =
(c) A support ∂P
(d) Everywhere in the beam Where, P and ∆ are in the same direction.
OPSC AEE Exam-2016 (Paper-I) 220 YCT
37. The strain energy stored in a simply supported Ans. (d) :
beam of span 'L' and flexural rigidity EI due to
central concentrated load W is given by-
W 2 L2 W 2 L3
(a) (b)
48EI 48 EI
2 2 2 3
Wl 2
W L W L M BA 3
(c)
96 EI
(d)
96 EI = 202 =
M AB Wl 2
Ans. (d) : 30
40. In moment distribution method, the sum of
distribution factors of all members meeting at
any joint is always-
(a) Less than one
The deflection under the load, (b) Zero
(c) One
WL3
δ= (d) Greater than one
48EI
 Stiffness i 
Strain Energy due to load, Ans. (c) : DFi =  
1 WL3 W 2 L3  Total Stiffness of jo int 
u== × = k + k + .......kn
2 48EI 96EI ∑ DF = 1 2
38. A beam AB is fixed at both ends and carries a ∑k
uniformly distributed load of intensity W per 41. In a slab, the minimum reinforcement for Fe
unit length run over its centre length. Due to 250 provided, is-
some construction defects, the end B is now (a) 0.10% of its gross sectional area
reduced to a simple support. The percentage (b) 0.12% of its gross sectional area
increase in bending moment at A is- (c) 0.15% of its gross sectional area
(a) 25 (b) 50
(d) None of these
(c) 75 (d) 100
Ans. (c) : Minimum R/F in slab for
Ans. (b) :
Fe 250 = 0.15% mild steel
Fe 415 = 0.12% HYSD
42. Whenever the earthquake or wind loading is
considered in design of a member, the
permissible stresses may be increased by-
(a) 25% (b) 30%
(c) 35% (d) 33.33%
Ans. (d) :
• When the effect of wind or seismic load is taken in to
account, the permissible stress in steel are increased by
1
33 %
3
% age of increase in Bending Moment of A :
• For rivets, bolts and tension rods, the permissible
3  Wl 2  Wl 2 stress are increased 25%, when the effect of wind or
. −
= 2  12 2 12 × 100 = 50%
seismic load is taken in to account.
Wl 43. The purpose of lateral ties in short RC columns
12 is to-
39. A fixed beam AB is subjected to a triangular (a) Avoid buckling of longitudinal bars
load varying from zero at end A to W per unit (b) Facilitate construction
length at end B. The ratio of fixed end moment (c) Facilitate compaction of concrete
at B to A will be- (d) Increase the load carrying capacity of the
1 1 columns
(a) (b) Ans. (a) : The main function of lateral ties in RC
2 3
column is to :
2 3
(c) (d) • Resist the shear force and hence contributes avoiding
3 2 shear failure
OPSC AEE Exam-2016 (Paper-I) 221 YCT
• Restrains the spliced bars and hence prevents their (b) Elastic shortening of concrete
slip. (c) Loss due to friction
• Prevent longitudinal reinforcement bars from (d) Creep of concrete
buckling. Ans. (c) :
• Confines the concrete core to provide sufficient
Pre-tensioned PSC beam :
ductility or deformability. • Loss due to shrinkage.
44. For vertical stirrups, the maximum spacing of• Loss due to creep.
shear reinforcement measured along the axis of
• Loss due to relaxation of steel.
the members shall not exceed-
• Loss due to elastic shortening.
(a) 0.70 d (b) 0.75 d
Post-tensioned PSC beam :
(c) 0.80 d (d) 0.90 d
• Loss due to creep.
Ans. (b) : According to IS code says maximum shear
• Loss due to friction.
reinforcement for vertical reinforcement should be
0.75d or 300 mm, whichever is less. • Loss due to anchorage slip.
• Loss duet to relaxation of steel.
45. In the limit state design of concrete section of
• Loss due to shrinkage.
limiting value of the depth of the neutral axis
Xu(max) 48. Minimum reinforcement in a circular column
for steel grade Fe415 is- as per IS 456 is-
d
(a) 0.53 (b) 0.48 (a) 4 bars of 12 mm
(c) 0.46 (d) 0.42 (b) Greater of 0.8% of cross sectional area and 4
bars of 12 mm
Ans. (b) : (c) 6 bars of 12 mm
(d) Greater of 0.8/% of cross sectional area and 6
bars of 12 mm
Ans. (d) : According to IS : 456-2000, of clause
26.5.3.1, The cross-section area of longitudinal
reinforcement shall not be less than 0.8% of cross
sectional area of concrete.
The minimum number of bars of 4 in
rectangular/square column and 6 in circular column
with bar diameter not being less than 12 mm in the
xu lim d − xu lim same clause (26.5.3.1).
=
0.0035 0.87 f y 49. The vertical retaining wall of the RCC
0.002 +
Es Counterfort is designed as a _____.
Where, fy = 415 MPa and Es = 2.1 × 105 Pa (a) Cantilever
xu lim (b) Simply supported slab
= 0.48 (c) Continuous slab
d
(d) None of these
46. Most common method of pre-stressing used for
factory production is- Ans. (a) :
(a) Freyssinet system
(b) Long line method
(c) Lee-Macall system
(d) Magnel-Blaton system
Ans. (b) : This method of prestressing also known as
Hoyer system. In a retaining wall, vertical wall, toe and base slab all
are designed as cantilever.
50. When width b, effective depth d, overall depth
D, the maximum area of reinforcement in RCC
beam shall not exceed-
In this the steel cables are prestressed by using (a) 0.04 bd (b) 0.04 bD
anchorage blocks and then concrete blocks are casted (c) 0.05 bd (d) 0.05 bD
around it. After the concrete has hardened, the cables Ans. (b) : According to IS code the maximum
are snapped. reinforcement in an RCC beam is 4% of gross cross
47. Which of the following losses occurs only in sectional area.
post-tensioning? 51. In T-shaped RCC retaining walls, the main
(a) Shrinkage of concrete reinforcement in the stem is provided on-
OPSC AEE Exam-2016 (Paper-I) 222 YCT
 (a) The front face in one direction For grade Fe415, the values given above should
(b) The front face in both direction be multiplied by 0.8
(c) The inner face in one direction Span to effective depth ratio for a continuous
(d) The inner face in both direction slab = 40 × 0.8 = 32
Ans. (c) : 56. The gross diameter of a rivet is the diameter of-
(a) Cold rivet measured before driving
(b) Rivet measured after driving
(c) Rivet hole
(d) None of these
Ans. (c) Gross diameter : Diameter of the hole,
assuming that rivet fills the holes completely.
52. If nominal shear stress τv exceeds the design
for d ≤ 25 mm, dg = d + 1.5 mm
shear strength of concrete τc' the nominal shear
reinforcement as per IS : 456-2000 shall be for d < 25 mm, dg = d + 2 mm
provided for a shear stress equal to- Where, d = Nominal diameter of rivet.
(a) τv (b) τc 57. The maximum permissible slenderness ratio of
(c) τv - τc (d) τv + τc tension members liable to reversal of stress due
to action of wind and earthquake is-
Ans. (c) : Nominal shear reinforcement is provided for (a) 300 (b) 350
nominal shear stress minus the design shear strength of
concrete. (c) 400 (d) 425
Ans. (b) :
53. As per IS 456, the minimum grade of concrete
for the design of reinforced concrete structure Maximum
in moderate exposure condition is- Type of member slenderness
(a) M 20 (b) M 25 ratio
(c) M 15 (d) M 30 1 Tension member. 400
Ans. (b) : According to IS Code 456 : 2000 2 Member subjected to 250
compressive loads resulting from
Minimum grade wind/earthquake force.
Exposure Condition
of RCC required
3 Member normally carrying 350
Mild M20 tension but subjected to reversal
Moderate M25 of stress due to wind or
Severe M30 earthquake force.
Very Severe M35 4 Member carrying compressive 180
Extreme M40 loads resulting from dead and
54. In reinforced concrete footing on soils, the superimposed loads.
minimum thickness at the edge should not be 58. The maximum deflection for a steel beam as
less than- per IS code should not exceed-
(a) 150 mm (b) 250 mm 1 1
(c) 100 mm (d) 200 mm (a) of span (b) of span
150 250
Ans. (a) : The thickness at edge shall be not less than 1 1
150 mm for footing on soils nor less than 300 mm (c) of span (d) of span
above the tops of piles for footing on Piles. 325 350
55. Span to effective depth ratio for a two way OPSC AE -2016 (I)
continuous slab (upto span 3.5m) with steel of Ans. (c) : Maximum deflection shall not exceed 1/325
grade Fe 415 should not be more than- of the span.
(a) 7 (b) 40 59. The average shear stress in a member
(c) 26 (d) 32 calculated on the cross section of unstiffened
Ans. (d) : According to IS : 456 : 2000 web shall not exceed-
(a) 0.45 fy (b) 0.40 fy
• For slabs spanning in two directions, the shorter of
(c) 0.65 fy (d) 0.66 fy
the two spans should be used for calculating the
span to effective depth ratio. Ans. (b) : 0.40 fy
• For two way slabs of shorter spans (up to 3.5 m) Cases Permissible stress
with mild steel reinforcement, the span to overall 1. Axial tension
depth ratio given below may generally be assumed end 0.6 fy
to satisfy vertical deflection limits for loading class compression
upto 3 kN/m2 3. In bearing (eg-
0.75 fy
Continuous slab → 40 at support)

OPSC AEE Exam-2016 (Paper-I) 223 YCT

2. In bending 0.66 fy Ans. (b) : The components of typical plate girder are as
4. In shear Maximum permissible avg= 0.40 fy follows :
Maximum permissible = 0.45 fy 1. Web
60. Generally the purlins are placed at the panel 2. flanges
points so as to avoid- 3.stiffeners
(a) Axial force in rafter In a plate girder, bending is primarily resisted by flange
(b) Shear force in rafter plates and shear by web.
(c) Deflection of rafter 65. The effective length of a steel compression
(d) Bending moment in rafter member which is effectively held in position at
both ends but restrained in direction at one end
Ans. (d) : only-
(a) L (b) 0.8 L
(c) 1.2 L (d) 1.5 L
Ans. (b) : 0.8 L effective length of a steel compression
member which is effectively held in position at both
ends but restrained in direction at one end only.

In trusses load application is desired at panel points so

as to avoid Bending moment.
61. The yield stress of mild steel of normally rolled
structural steel is about (in N/mm2)-
(a) 240 to 260 (b) 330 to 360
(c) 420 (d) 550
Ans. (a) : Yield stress ≈ 250 MPa
Ultimate stress ≈ 410 MPa
62. The effective length of fillet weld should not be
less than-
(a) Two times weld size 66. The addition of pozzolana to Portland cement
may cause-
(b) Four times weld size
(a) Decrease in early strength
(c) Six times weld size
(d) Weld size (b) Increase in early strength
(c) Decrease in curing time
Ans. (b) :
(d) Increase in permeability
Ans. (a) : Addition of pozzolana reduces the initial rate
of increase in strength.
67. The two main compounds imparting strength
for Ordinary Portland Cement are-
(a) Tricalcium silicate and dicalcium silicate
(b) Dicalcium silicate and aluminates
Minimum effective length of fillet weld should not be (c) Tricalcium aluminates and silicate
less than 4 times the weld size. (d) Tricalcium silicate and tricalcium aluminates
63. The rolled 'I' section for a given depth having Ans. (a) : C3S and C2S together constitute about 70% of
largest moment of inertia, IXX is designated as- cement and are responsible for providing most of the
(a) ISMB (b) ISHB strength to it C3A is responsible for flash set and C4AF
(c) ISLB (d) ISWB generates very less strength.
Ans. (b) : ISHB – Indian standard heavy beam 68. For a satisfactory workable concrete with a
ISMB – Indian standard medium beam constant water cement ratio, increase in
ISLB – Indian standard light beam aggregate cement ratio-
ISWB – Indian standard wide beam (a) Decrease the strength of concrete
64. In a plate girder, bending is primarily resisted (b) Does not change the strength of concrete
by- (c) Increase the strength of concrete
(a) Web plate (d) None of these
(b) Flange plate only Ans. (a) : The water to cement ratio is calculated by
(c) Flange angle only dividing the water in one cubic yard of the mix by the
(d) Flange plate and flange angle cement in the mix.

OPSC AEE Exam-2016 (Paper-I) 224 YCT

69. The strength of concrete is directly Ans. (c) : Frogs created an extra recess for the mortar,
proportional- resulting in a stronger bond between bricks to reduce
(a) Water cement ratio the weight of the bricks, so that the bricks can be laid
(b) Cement water ratio with convenience.
(c) Sand cement ratio 75. A type of bond in a brick masonry consisting of
(d) Water aggregate ratio alternate course of headers and stretchers, is
Ans. (b) : As the strength of concrete is inversely called-
proportional to w/c ratio, it is directly proportional to (a) English bond (b) Flemish bond
cement to water ratio. (c) Stretching bond (d) Heading bond
70. The most commonly used admixture which Ans. (a) :
prolongs the setting time is-
(a) Calcium chloride
(b) Gypsum
(c) Sodium silicate
(d) All of these
Ans. (b) :
• Gypsum– Gypsum is a retarder and delays the
setting of concrete.
• Calcium chloride– CaCl2 is an accelerator and
76. Preventative maintenance for a building work
increases the setting rate of concrete.
means-
71. In ordinary residential buildings, D.P.C. may (a) Taking action before break-down
be provided- (b) Breakdown maintenance
(a) At ground level (c) Taking action after break-down
(b) Between ground level and water table level (d) None of these
(c) At water table level
Ans. (a) : Preventive maintenance is maintenance that
(d) At plinth level
is regularly performed to decrease the likelihood of
Ans. (d) : D.P.C. is short form of damp proof course. It failing. It is performed so that the building does not
is always provided at plinth level. break down unexpectedly.
72. The tolerance in the width of mould of a class I 77. The slenderness ratio for masonry walls should
brick is about- not be more than-
(a) ± 3 mm (b) ± 6 mm (a) 10 (b) 20
(c) ± 10 mm (d) ± 12 mm (c) 30 (d) 40
Ans. (a) : Tolerance is width = ± 3 mm Ans. (b) : The slenderness ratio of masonary walls
Tolerance is length = ± 6 mm should not be more than 20.
73. The slump recommended for mass concrete is 78. Expansion joint in masonry wall is provided
about- when length of wall is greater than-
(a) 50 mm to 100 mm (a) 20 m (b) 10 m
(b) 25 mm to 75 mm (c) 40 m (d)30 m
(c) 100 mm to 125 mm Ans. (c) : Expansion joint in masonry wall is provided
(d) None of these when length of wall is greater than 40 m.
Ans. (b) : Recommended slumps of concrete : 79. The height between two floors is 3.00 m and
No. Type of concrete Slump risers are of 150 mm. Assuming two flights
1. Concrete for road construction 20 to 40 mm between the floors, the number of treads will
2. Beams and slabs 50 to 100 mm be-
3. Mass concrete 25 to 50 mm (a) 18 (b) 19
(c) 20 (d) 21
4. Normal RCC works 80 to 150 mm
Ans. (b) :
5. Impermeable work 75 to 120 mm
3000
6. Concrete to be vibrated 10 to 20 mm No. of treads = = − ( 2 − 1) = 19
150
74. The most important purpose of frog in a brick
is to- 3000
No. of risers = = 20
(a) Reduce the weight of brick 150
(b) Emboss manufacture's name 80. The type of flooring suitable for the use in
(c) Form keyed joint between brick and mortar theatres and public libraries and other places
(d) Improve insulation by providing 'hollows' where noiseless floor covering is desire-
OPSC AEE Exam-2016 (Paper-I) 225 YCT
 (a) Wooden flooring 86. A contract is an agreement between-
(b) Linoleum flooring (a) Two parties valid in law
(c) Cork flooring (b) Several agencies
(d) None of these (c) Three agencies
Ans. (c) : Cork flooring will be suitable for theatres and (d) Two parties without legal binding
public libraries since it will dampen noise level. Ans. (a) : A contract is an agreement between two
81. One of the main disadvantages of the bar chart parties enforced by law. It has some terms and
for construction management is- conditions.
(a) The time schedule is not shown properly 87. Measurement of 50 mm thick concrete flooring
(b) Progress of the work cannot be monitored will be done in-
(c) The financial aspect is not shown (a) Cubic m (b) % sq m
(c) Meter (d) Sq. m
(d) Does not show the interdependencies of the
activity Ans. (d) : Since the thickness has been fixed, we are
left will measurement in terms of area only.
Ans. (d) : A bar chart shows progress of activities with
time but fails to depict interdependencies between 88. The reduction in project time normally results
various activities. in-
(a) Increasing the direct cost and decreasing the
82. Which of the following does not represent an
indirect cost
activity?
(b) Decreasing the direct cost and increasing the
(a) Foundation is being dug
indirect cost
(b) Site located (c) Increasing the direct cost and the indirect cost
(c) The office area is being cleaned both
(d) None of these (d) Decreasing the direct cost and the indirect
Ans. (b) : Activity is defined as the condition in which cost both
things are happening or being done. Ans. (a) : Increasing the direct cost and decreasing the
Hence option (a) and (c) can be classified as activities indirect cost.
but not option (b).
83. Critical Path Method (CPM) network is-
(a) Activity oriented
(b) Event oriented
(c) Both activity as well as event oriented
(d) None of these
Ans. (a) : CPM networks is activity oriented whereas
PERT network is event oriented.
89. A document containing detailed description of
84. The security deposit deducted from all the items of work together with their
contractor's bill is- current rates is called-
(a) Refunded as soon as the construction is over (a) Analysis of rates
(b) Not refunded (b) Abstract of estimate
(c) Refunded in the middle of the contract (c) Schedule of rates
(d) Refunded after maintenance period (d) None of these
Ans. (d) : The security deposit shall be collected by Ans. (a) : A documents containing detailed description
deduction from each running bill of the contractor 5% of all the items of works together with their current
of the Gross amount of the Bill. rates is called Analysis of rates.
85. Earliest finish of an activity is always- 90. Work Breakdown structure for a construction
(a) Less than earliest event of the following node project will help in-
(b) Greater than earliest event of the following (a) Breaking the project into several elements
node (b) Identifying the activities
(c) Less than or equal to earliest event of the (c) Identifying the functional elements of a
following node project and their interrelationship
(d) Greater than or equal to earliest event of the (d) None of these
following node Ans. (a) : Work Breakdown structure for a construction
Ans. (c) : Less than or equal to earliest event of the project will help in Breaking the project into several
following node. elements.
OPSC AEE Exam-2016 (Paper-I) 226 YCT
 Odisha Public Service Commission
(AEE), Exam- 2016 (Paper-II)
1. The pressure of a liquid measured with the A
help of a piezometer tube is- HMD =
(a) Vacuum pressure P
(b) Gauge pressure πd 2 1 d
= × =
(c) Absolute pressure 4 πd 4
(d) Atmospheric pressure 4. In case of flow through parallel pipes-
Ans. (b) : Piezometer is a type of simple manometer (a) The head loss for all the pipes is same
which measures the gauge pressure at any point. (b) The head loss is different in different pipes
(c) The head loss is the sum head losses in the
various pipes
(d) None of the above
Ans. (a) :
• For pipes in parallel, head loss through each pipe
remains same whereas for pipes in series, discharge
PA = γH through each pipe remains same.
Where, PA = gauge pressure at A. • Series connection
2. When a fluid is flowing through a pipe, the
velocity of the liquid is-
(a) Maximum at the center and minimum near
the walls
(b) Minimum at the center and maximum near
the walls
(c) Zero at the center and maximum near the • Parallel connection
walls
(d) Maximum at the center and zero near the
walls
Ans. (d) : j
• Velocity variation in pipe flow.

Q = av
Q = Q1 + Q2 + Q3
hLAB = hL1 = hL2 = hL3
• Due to no slip condition, velocity of fluid will be 5. The critical depth for a channel is given by-
zero near the walls. 1/ 2 1/ 3
q  q 2 
3. The hydraulic mean depth for a circular pipe (a)   (b)  
 g 
of diameter d is-  g 
d d  q 3 
1/ 4
 q 4 
1/ 5
(a) (b)   
6 4 (c)    (d)
 g   g 
d
(c) (d) d Ans. (b) : For critical depth Fr = 1
2
Ans. (b) : V
fr = =1
• Hydraulic mean depth is defined as area of flow gyc
section divided by the wetted perimeter.
Q
∴ Q = AV, q =
B
Q q
⇒ V= =
By y

OPSC AEE Exam-2016 (Paper-II) 227 YCT

 10.
The graphical representation of average
q2 q2
⇒ = 1 ⇒ yc3 = rainfall and rainfall excess (i.e., rainfall minus
gyc3 g infiltration) rates over specified areas during
1/ 3 successive unit time intervals during a storm is
q 
2
known as-
yc =  
 
g (a) Hydrograph
6. When the Mach number is less than unity, the (b) Unit hydrograph
flow is called- (c) Hyetograph
(a) Sub-sonic flow (d) None of the above
(b) Sonic flow Ans. (c) : A hyetograph is graphical representation of
(c) Super-sonic flow the distribution of rainfall intensity over time.
(d) Hyper-sonic flow
Ans. (a) : Mach number is given as
V
M=
C
Where C = velocity of sound in the medium
If M > 1 supersonic flow
If M = 1 sonic flow
If M < 1 subsonic flow
7. The power developed by a tubine is-
(a) Directly proportional to H1/2
(b) Inversely proportional to H1/2 11. According to Lacey's equation, the scour depth
(c) Directly proportional to H 3/2 is equal to-
3/2 1/2 1/3
(d) Inversely proportional to H Q Q
(a) 0.47   (b) 0.47  
Ans. (c) : The power devloped by the turbine:-  f   f 
V = 2gH Q
1/4
Q
1/5

and (c) 0.47   (d) 0.47  

 f   f 
P = γ QH
Ans. (b) : The source depth according to lacey's
= γ(a 2gH )H 1/ 3
P = kH3/2  q2  Q
1/ 3
equation is given as 1.35   or 0.47  
8. Which of the following pumps is suitable for  f   f 
small discharge and high head? 12. The phenomenon occurring in an open channel
(a) Centrifugal pump when a rapidly flowing steam abruptly changes
(b) Axial flow pump to a slowly flowing stream causing a distinct
(c) Mixed flow pump rise of liquid surface, is-
(d) Reciprocating pump (a) Water hammer
Ans. (d) : (b) Hydraulic jump
• Reciprocating pump-small discharge and high head. (c) Critical discharge
• Centrifugal and axal flow pump-large discharge and (d) None of the above
lower head.
Ans. (b) :
9. The ratio between the area of crop irrigated
and quantity of water required during its
entire period of the growth, is known as-
(a) Delta (b) Duty
(c) Base period (d) Crop period
Ans. (b) :
• Duty of a water simply expresses the number of Hydraulic jump is the jump or standing wave formed
hectare of land that can be irrigated for the full
when the depth of flow of water changes from super
growth of the given crop by supplying 1 cumec
critical to subcrtical state. It is accompanied by loss in
water continuously during the entire base period of
that crop. energy due to turbulences.
A  Hectance  13. Dimensions of the dynamic viscosity (µ) are-
D=   (a) MLT-2 (b) M-1L-1T-1
Q  Cumec  -1 -1
(c) ML T (d) None of the above
OPSC AEE Exam-2016 (Paper-II) 228 YCT
 du Inertal force
Ans. (c) : τ = µ Ans. (a) : Reynold's number =
dy Viscous force

⇒
kgm
= [µ ]
kg ρL2V 2 ρVL
= =
2
S m 2 S×m µVL µ
kg
[µ ] = = ML-1T -1 18. Differential manometers are used to measure-
ms (a) Pressure in water channels, pipes, etc
14. When a canal is carried over a natural (b) Difference in pressure at two points
drainage, the structure provided, is known as- (c) Atmoshpheric pressure
(a) Syphon
(d) Very low pressure
(b) Aqueduct
(c) Super passage Ans. (b) : Differential manometer is used to measure
(d) Syphon-aqueduct pressure deference between two point in a pipe or
between two different pipes.
Ans. (b, d) : There are 3 types of cross drainage works

15. The maximum vacuum at the summit of a

syphon is- P1+G1γw(a+h)-G2γwh-G1γwa = P2
(a) 2.7 m of water (b) 7.4 m of water P1-P2 = (G2-G1)γwh
(c) 74 mm of water (d) 74 m of water 19. Highest dam in India, is-
Ans. (b) : (a) Hirakud dam
• If the pressure at summit becomes less than 2.7 m of (b) Nagarjuna Sagar dam
water absolute, the dissolved air and other gases (c) Idukki dam
would come out from water and collect at the (d) Bhakra dam
summit.
• The pressure at summit point may be reduced to Ans. (d) : The biggest dams in India–
10.3 m of water theoretically but in actual practice (i) Nagarjuna Sagar Dam – Telangana
this pressure is only reduced to about 7.4 m or 10.3 (ii) Sardar Sarovar dam – Gujrat
– 7.4 = 2.9 m or water absolute. (iii) Hirakud dam – Odisha
16. An ideal flow of a liquid obeys- (iv) Tehri dam – Uttarakhand
(a) Continuity equation 20. If Cv, Cc, Cd and Cr are the hydraulic
(b) Newton's law of viscosity coefficients of an orifice, then-
(c) Newton's second law of motion
Cv2
(d) Dynamic viscosity law (a) Cd = Cc. Cv (b) Cr = 1 +
Cd
Ans. (a) : An ideal flow of fluid will always satisfy
continuity equation. Cv
(c) Cv = Cc + Cd (d) Cc =
• Continuity equation states that mass entering a Cd
control volume = mass exiting from that control
volume Ans. (a) : Cd = CC × CV
Q1 = Q2 Where, Cd = Coefficient of discharge
ρ1A1V1 = ρ2A2V2 CC = Coefficient of contraction
CV = Coefficient of velocity
21. For the stability of a structure against seepage
pressure according to Khosla's Creep Theory
the critical gradient is-
(a) 0.0 (b) 0.5
It is based on law of conservation of mass. (c) 1.0 (d) 0.75
17. Reynold's number is the ratio of inertial force Ans. (c) : If at the exit point at the downstream side, the
and-
net gradient is such that the upward force at exit is just
(a) Viscosity (b) Elasticity equal to the submerged weight of the soil particle, then
(c) Gravitational force (d) Surface tension that gradient is called critical gradient.
OPSC AEE Exam-2016 (Paper-II) 229 YCT
22. The main function of a diversion head works of Ans. (a) : The geometric progression method gives
a canal from a river is- highest value of forecasted population.
(a) To remove silt For large cities, the suitable method for forecasting
(b) To control floods population is "Arithmetic increase method".
(c) To raise water level dP
=C
(d) To store water dt
Ans. (c) : 27. The maximum permissible total solid content in
• Weirs and barrages help in storing water and in water for domestic purposes should not exceed-
controlling floods. (a) 300 ppm (b) 400 ppm
• A diversion head works is a structure constructed (c) 500 ppm (d) 1000 ppm
across a river so that it can be diverted into the off Ans. (c) : Acceptable limit for TDS = 500 mg/l and
taking canals. cause for rejection = 2000 mg/l.
• To remove silt, silt excluders or silt ejectors are 28. The maximum permissible chlorine content for
used. public supplies should be between-
23. The standard height of a standard rain gaugs (a) 0.1 to 0.2 ppm (b) 0.3 to 0.4 ppm
is- (c) 1.2 to 4.0 ppm (d) 6.5 to 8.5 ppm
(a) 10 cm (b) 20 cm Ans. (b) : Permissible chlorine content is between 0.3
(c) 30 cm (d) 40 cm to 0.4 ppm.
Ans. (c) : 29. The bacteria which require oxygen for their
survival is known as-
(a) Anaerobic bacteria
(b) Pathogenic bacteria
(c) Non-pathogenic bacteria
(d) Aerobic bacteria
Ans. (d) : Aerobic bacteria require oxygen to perform
cellular respiration and derive energy to survie whereas.
Anaerobic bacteria boes not require oxygen to survive.
30. The self cleaning velocity, recommended for
Diameter = 20 cm Indian conditions, in order to prevent setting
Height = 30 cm above ground level down sewage at the bottom or on the sides of a
large sewer is-
24. For determination of average annual
(a) 0.25 m/s (b) 0.50 m/s
precipitation in a catchment basin, the best
(c) 0.75 m/s (d) 1.5 m/s
method is-
(a) Arithmetical Method Ans. (c) : Minimum velocity of 0.8 m/s at design peak
(b) Thiessen's mean Method flow and 0.6 m/s at current peak flow is recommended
in sanitary sewer.
(c) Isohyetal Method
(d) None of the above 31. Sludge treatment is mainly done in order to-
(a) Stabilize the organic matter
Ans. (c) : Isohyetal method is the best method for
(b) Destroy the pathogenic bacteria
determination of average annual precipitation as it
includes effect of both weighted area and elevation (c) Reduce the water content
difference effect of rainfall. (d) All of the above
1 N Ans. (d) :
P = ∑ Ai Pi • Sludge treatment is focused on reducing sludge
A i =1 weight and volume, to reduce disposal costs and on
25. Isohytes are the imaginary lines joining the reducing potential health risks of disposal options.
points of equal- • Sewage sludge treatment describes the processes
(a) Pressure (b) Height used to manage and dispose of sewage sludge
(c) Humidity (d) Rainfall produced during sewage treatment.
Ans. (d) : Isohyet is a line on a map connecting points • Water removal is the primary means of weight and
having the same amount of rainfall in a given period. volume reduction, while pathogen destruction is to
26. For large cities, the suitable method for reduce potential health risks of disposal options.
forecasting population is- 32. Aerobic bacterias-
(a) Arithmetical increase method (a) Flourish in the presence of free oxygen
(b) Graphical method (b) Consume organic matter as their food
(c) Geometrical increase method (c) Oxidise organic matter in sewage
(d) Comparative method (d) All of the above
OPSC AEE Exam-2016 (Paper-II) 230 YCT
Ans. (d) : Aerobic bacteria require oxygen to perform Ans. (a) : Generally in water treatment plant, the
cellular respiration and derive energy to survive Some sequence of treatment followed is :
examples of aerobic bacteria are : Coagulation → flocculation → sedimentation →
(i) Nocardia filtration → chlorination.
(ii) Bacillus 38. An area is declared drought affected if its mean
(iii)Lactobacillus rainfall is less than-
(iv)Mycobacterium tuberculosis (a) 50% (b) 60%
33. Most commonly used pump for lifting water in (c) 75% (d) 85%
water supply mains, is-
Ans. (c) : A meteorological sub-division is considered
(a) Reciprocating pump to be affected by drought if it receives a total seasonal
(b) Axial flow pump rainfall less than of 75% of the normal value.
(c) Rotary type pump
39. The permissible pH value for public water
(d) Centrifugal pumps supplies may range between-
Ans. (b) : Axial flow pumps are usually designed for (a) 4.5-5.5 (b) 5.5-6.5
very large discharges and comparatively lower heads. (c) 6.5-8.5 (d) 7.0-8.5
34. By boiling water, hardness can be removed if it Ans. (c) : Permissible pH vale for public water supplies
is due to-
ranges between 6.5 – 8.5.
(a) Calcium sulphate
(b) Calcium bicarbonate 40. Water losses in water supply is assumed as-
(a) 5% (b) 7.5%
(c) Magnesium sulphate
(d) Calcium nitrate (c) 10% (d) 15%
Ans. (d) : Water losses in water supply are assumed to
Ans. (b) :
be nearly 15.
• However this method is not commonly used since it
is very expensive. 41. The soil transported by wind is called-
• Temporary hardness can be removed by simply (a) Aeolian soil
boiling the water (b) Marine soil
Heated (c) Alluvial soil
Ca(HCO3 )2 → CaCO3 ↓+ H 2 O + CO 2 ↑
ppt.
(d) Lacustrine soil
35. Aeration of water is done to remove- Ans. (a) : According to the transporting agency, sol are
classified as
(a) Odour (b) Colour
(c) Bacterias (d) Turbidity Aeolian deposit → transported by wind.
Marine deposit → deposited by sea water.
Ans. (a) : Aeration brings water and Air in close
contact in order to remove dissolved gaves (Such as Lacustrine deposit → deposited by still water like
carbon dioxiede and oxidizes dissolved metals such as lakes.
iron, hydrogen sulfide, and volatile organic chemicals Alluvial deposit → deposited by river water
(VOCs). Glacial deposit → transported by ice.
36. Increase in population of a rapidly growing 42. A soil sample is having a specific gravity of 2.6
city, may be estimated by- void ratio of 0.78. The water content in
(a) Arithmetical mean method percentage required to fully saturate the soil at
(b) Geometric growth method that void ratio is-
(c) Incremental increase method (a) 10% (b) 30%
(d) Graphical comparison method (c) 50% (d) 70%
Ans. (b) : es
• Geometric growth method, gives highest values of Ans. (b) : We know that, W = G
forecasted population.
• Incremental method is considered to be the best for 0.78 × 1
= , S = 1 for fully saturated soil
any city whether old or new. 2.6
• For old cities, arithmetic method may be better. w = 30%.
• This method is suitable for new younger cities 43. The ratio of the unconfined compressive
expanding at faster rate. strength of undisturbed soil to the unconfined
37. During treatment of water, sedimentation ism compressive strength of soil in a remolded
done- state, is called-
(a) Before filtration (a) Sensitivity
(b) After filtration (b) Thixotropy
(c) Simultaneously with filtration (c) Relative strength
(d) Along with chlorination (d) None of the above
OPSC AEE Exam-2016 (Paper-II) 231 YCT
Ans. (a) : Thixotropy– It is that property of soil due to 47. A vertical cut in a clayey soil with unit weight
which loss of strength (Shear strength) or remolding can of 19 kN/m3 failed when the depth of cut 4.0 m.
be regained if left undisturbed for some time. The cohesive strength of clay is-
q u (undisturbed) (a) 76.0 kN/m2 (b) 8.5 kN/m2
St = (c) 38 kN/m 2
(d) 19.0 kN/m2
q u (remoulded)
Ans. (d) : Critical depth for vertical cut in clayey soil is
44. The ratio of settlement at any time, to the final
settlement is known as-  4C  4C
given as   i.e. = 4m
(a) Compression ratio  γ ka  γ ka
(b) Coefficient of consolidation
For clayey soil φ = 0, ka = 1
(c) Compression index
4×C
(d) Degree of consolidation =4
Ans. (d) : Degree of consolidation represents the stage 19 × 1
of consolidation ata certain location in the consolidating 1 × 19 × 4
layer. C=
4
∆h
U= × 100 ∴ C = 19kN/m2
∆H
48. According to IS classification, the range of silt
If ∆h = settlement at any stage (at any time t) size particles is-
(a) 4.75 mm to 2.00 mm
∆H = final consolidation settlement then degree of
(b) 2.00 mm to 0.425 mm
consolidation (U).
(c) 0.425 mm to 0.075 mm
45. The suitable method of finding the shear
(d) 0.075 mm to 0.002 mm
strength of very plastic cohesive soils is by
means of- Ans. (d) : According to IS classification
Boulder Cobble Gravel Sand Silt Clay
(a) Cone test (b) Peretration test > 300 mm 300 – 80 Coarse Fine Coarse Medium Fine sand
mm Gravel gravel sand sand
(c) Vane shear test (d) Torsional shear test 80 – 20 20 – 4.75 4.75 – 2.0 2.0 – 0.425 0.425 – 0.075 0.075–0.002 < 0.002
mm mm mm mm mm mm mm
Ans. (c) : Coarse grained Fine grained
• Shear strength of such soil may be significantly 49. Void ratio of soil mass can-
affected during sampling and handling.
(a) Take any value greater than zero
• For such soil, vane shear test can be done in the
field. (b) Be zero
(c) Never be greater than unity
• Plastic cohesive soils are very sensitive and hence it
is very difficult to obtain undisturbed specimen. (d) Take values between 0 and 1 only
46. The coefficient of earth pressure at rest is given Ans. (a) : Void ratio is defined as the ratio of volume of
in- voids to the volume of solids
µ 1 + µ  V
(a) (b)   e= V
(1−µ )  µ  VS

µ (1−µ ) • It can take any value greater than zero.

(c) (d) 50. Relative density of a compacted dense sand is
(1 + µ ) µ
approximately equal to-
Ans. (a) : For plain strain condition, (a) 0.4 (b) 0.6
εx = 0 (c) 0.95 (d) 1.2
Ans. (c) : Relative density is given
e −e
I D = max
emax − emin
For a compacted dense sand
e = emin
but will be slightly greater than emin.
Here σx = σy = σh and σz = σv
Hence ID ≈ 0.95.
 −σ   −σ y   −σ z  51. According to Terzaghi's theory, the ultimate
ex =  x  − µ   − µ =0
 E   E   E  bearing capacity at the ground surface for a
σh(1 - µ) = µσv purely cohesive soil with cohesion c and for a
σh smooth base of a strip footing is-
 µ 
= k0 =   (a) 2.57 c (b) 5.14 c
σv  1− µ  (c) 6.2 c (d) 5.7 c
OPSC AEE Exam-2016 (Paper-II) 232 YCT
Ans. (d): According to Terzaghi's, for clayey soil (a) Capillary compression
NC = 5.7, Nq = 1 and Nγ = 0 (b) Capillary tension
Hence, qu = (5.7 c + q) (c) Capillary pore pressure
and qnu = 5.7 c (d) None of the above
52. Select the correct statement- Ans. (d) :
(a) Both negative skin friction and skin frictional
resistance are caused by relative settlement of
soil
(b) Both negative skin friction and skin frictional
resistance are caused by relative settlement of
pile
(c) Negative skin friction is caused by relative
settlement is caused by relative settlement of
pile
• The stress responsible for retaining water in
(d) Negative skin friction is caused by relative capillary tube above the free water surface is called
settlement of pile and skin frictional as surface tension.
resistance is caused by relative settlement of
soil 4σ cos θ
capillary rise hc =
Ans. (c) : γ×d
• Negative skin friction or 'down drag' phenomenon, 56. The effective size of particles of soil is denoted
which occurs when a portion of soil layer by-
surrounding a pile settles more than the pile. (a) D10 (b) D20
• When a compressive load (P) is applied at the top of (c) D30 (d) D60
pile, the pile will tend to move vertically downward Ans. (a) :
relative to the soil. Due to this, shear or friction • Effective size of particles of soil was found close to
develops between soil and surface of shaft. D10 as a result of experiments conducted by Allen
Hazen.
• The effective size of particles refers to effective
optical size of a particle and is defined as the
average diameter of the sphere.
57. The ultimate settlement of a soil is directly
proportional to-
(a) Depth of the compressible soil strata
(b) Compressive index
(c) Void ratio
53. Taylor's stability number is given by- (d) Both (a) and (b)
Fc γH Ans. (d) : Ultimate settlement of a soil is given as
(a) (b)
CγH cFc  σ + ∆σ 
C H
∆H = C 0 log  0 
C H 1 + e0  σ0 
(c) (d)
Fc γH cFc γ
58. A coarse-grained soil has a voids ratio 0.75 and
Ans. (c) : Taylor's stability number is a dimensionless specific gravity as 2.75. The critical gradient at
parameter given as : which quick and condition occurs, is-
 C  (a) 0.25 (b) 0.50
Sn =   (c) 0.75 (d) 1.00
 FC γH 
 G −1 
54. Sand stone is- Ans. (d) : Critical gradient icr =  S 
(a) Sedimetary rock  1+ e 
(b) Metamorphic rock  2.75 − 1 
=  =1
(c) Igneous rock  1 + 0.75 
(d) Volcanic rock 59. Bouyant unit weight equals the saturated
Ans. (a) : Sandstone is a type of sedimentary rock. density-
55. The stress which is responsible for retaining (a) Multiplied by unit weight of water
water in capillary tube above the free water (b) Divided by unit weight of water
surface of the water body in which the capillary (c) Plus unit weight of water
tube is inserted, is- (d) Minus unit weight of water
OPSC AEE Exam-2016 (Paper-II) 233 YCT
Ans. (d): Buoyant unit weight = saturated unit weight –
unit weight of water
60. A phreatic line is defined as the line within a
dam section below which there are-
(a) Positive equipotential lines
(b) Positive hydrostatic pressure
(c) Negative hydrostatic pressure
(d) Negative equipotential lines
Ans. (b) : The phreatic line is the line within a dam
section in which atmospheric pressure exists. It is the 66. The most efficient traffic signal is-
top most equipotential line and hence below it we have (a) Simultaneous system
positive hydrostatic pressure. (b) Flexible progressive system
61. The minimum centre to centre distance of (c) Simple progressive system
friction piles of 1 m diameter, is- (d) Alternate system
(a) 1.0 m (b) 2.0 m Ans. (b) :
(c) 3.0 m (d) 4.0 m
• This is the most efficient system of all the four types
Ans. (c) : Minimum spacing between friction piles of traffic signal system.
according to IS code : • In flexible progressive system, it is possible to
2 × diameter → for loose sand or fill deposit automatically vary the length of signal cycle and
2.5 × diameter → for point bearing piles signal phase at each signalized intersection with the
3 × diameter → for friction piles help of sensors to detect vehicle arrival and
62. Stoke's law does not hold good if the size of connecting to a master computer.
particle is smaller than- 67. The rate of rise of fall of the road surface along
(a) 0.0002 mm (b) 0.002 mm its length is called-
(c) 0.02 mm (d) 0.2 mm (a) Cant
Ans. (a) : The lower limit of particle size for validity of (b) Super-elevation
Stoke's law is 0.0002 mm. However the upper limit for (c) Gradient
the same is 0.2 mm. For particle size less than 0.0002 (d) Banking
mm, Brownian movement affects their settlement and in Ans. (c) : Rate of rise or fall of the road surface along
the case of particles large than 0.2 mm, turbulence its length is called gradient whereas across its length is
affects the settlement. called cant or super-elevation.
63. The ultimate bearing capacity of a soil, is- 68. In hill roads, the minimum sight distance
(a) Load at which soil consolidates required is-
(b) Load at which soil fails (a) Stopping sight distance
(c) Total load on the bearing area (b) Passing sight distance
(d) Safe load on the bearing area (c) Braking distance
Ans. (b) : Ultimate bearing capacity of a soil is the load (d) None of the above
at which the soil fails. Ans. (a) : Stopping sight distance.
64. Raft foundation are generally preferred to 69. Which of the following methods is
when the area required for individual footing, recommended by IRC for the design of flexible
is more than- pavement?
(a) 25% to total area (a) Group index method
(b) 30% of total area (b) CBR method
(c) 40% of total area (c) Westegard method
(d) 50% of total area (d) None of the above
Ans. (d) : Raft foundation are generally preferred when Ans. (b) :
area required for individual footing is more than 50% of
• CBR- California bearing ratio method.
total area.
• Flexible pavement design by CBR method is used to
65. The arrangement of supporting an existing
determine the total thickness of pavement.
structure by providing supports underneath, is
known as- 70. The Indian Railway has been divided into-
(a) Shoring (b) Underpinning (a) Six zones
(c) Jacking (d) Piling (b) Eight zones
Ans. (b) : Underpinning – a solid foundation laid below (c) Twelve zones
ground level to support or strengthen a building. (d) Sixteen zones

OPSC AEE Exam-2016 (Paper-II) 234 YCT

Ans. (d) : There are 16 zones in Indian railways as 75. Coning of wheels is provided-
given below : (a) To check lateral movement of wheels
(i) Northern (ii) North Eastern (b) To avoid damage to inner faces of rails
(iii) North Frontier (iv) Eastern (c) To avoid discomfort to passangers
(v) South Eastern (vi) South Central (d) All of the above
(vii) Southern (viii) Central Ans. (d) : The surface of wheels are made in cone shape
(ix) Western (x) South Western at an inclination of 1 in 20, and the same slope is
(xi) North Western (xii) West Central provided in the rails.
(xiii) North Central (xiv) South East Central
(xv) East Coast (xvi) East Central
** In present time there are 18 Zones of Railway
these two are –
Kolkata Metro Railway is the 17th Zone in
place of Konkan Railway and 18th South Coast
Zone – Visakhapatnam. Purposes :
71. Which of the following sleepers provide the (i) To avoid damage to inner face of rails.
best elasticity of track? (ii) To keep the train in its central position of the rails,
(a) Wooden sleeper coning does not allow any sidewise movement.
(b) Cast iron sleeper 76. For a vehicle moving with a speed of 80 km per
(c) Steel sleeper hour, the brake reaction time, in ordinary
(d) RCC sleeper cases, is-
(a) 1 sec (b) 1.5 sec
Ans. (a) : Wooden sleeper provide best elasticity of
tracks. (c) 2.0 sec (d) 2.5 sec
72. Maximum super-elevation on hill roads should Ans. (d) : The brake reaction time in ordinary cases for
vehicle moving at 80 kmph is taken as 2.5 sec.
not exceed-
(a) 5% (b) 7% 77. Bull headed rails are generally provided on-
(c) 8% (d) 10% (a) Points and crossing
(b) Straight tangents
Ans. (d) : As per IRC, maximum super-elevation in
(c) Curved tracks
areas :
(d) Meter gauge tracks
(i) Which are not bound by snowfall – 10%
Ans. (a) :
(ii) Which are bound by snowfall – 7%
• The are especially used for making points and
73. The type of transition curves generally, crossings.
provided on hill roads, is-
• The rail section whose head dimensions are more
(a) Circular (b) Cubic parabola than that of their foot is called bull headed rails.
(c) Spiral (d) Lemniscate's • In this type of rail, the head is made little thickener
Ans. (c) : The type of transition curves generally, and stronger than the lower part by adding more
provided on hill roads, is Spiral. metal to it.
74. The road foundation for modern highways 78. Width of vehicles affects the width of-
construction, was developed by- (a) Lanes
(a) Tresgue (b) Shoulders
(b) Telford (c) Parking paces
(c) Telford and Macadam simultaneously (d) All of the above
(d) Macadam Ans. (d) : Width of lanes, shoulders and parking spaces
Ans. (c) : Development of highway : all depend of width of the vehicle. If the width of
(i) Macadam Construction vehicle changes, then all of these widths will be
• Suggested no requirement of heavy stones in affected.
foundation. 79. The maximum safe speed on roads, depends on
(ii) Roman Roads the-
• No gradient, thickness about 0.75 m to 1.2 m. (a) Type of the highway
(iii) Tresguet construction (b) Type of the road surface
• Gradient provided, thickness reduced to 0.3 m. (c) Type of curves
(iv) Telford Construction (d) All of the above
• Subgrade width 9 m, binding layer of wearing Ans. (d) : Maximum safe speed on roads depends upon
course 4 cm. all of the given factors.

OPSC AEE Exam-2016 (Paper-II) 235 YCT

80. The head of Public Works Department of any • Reciprocal leveling eliminates errors due to
Indian state is- refraction, curvature and collimation.
(a) Transport Minister 85. The contour lines can; cross one another on
(b) Chief Engineer map only in the case of-
(c) Superintending Engineer (a) An overhanging cliff (b) A valley
(d) Executive Engineer (c) A ridge (d) A vertical cliff
Ans.(c): Superintending Engineer– He is the technical Ans. (a) :
and administrative head of a Circle within a Wing or
Zone of the Roads and Highways Department.
He has technical and administrative control over the
personnel in his Circle. He is authorised to make direct
contact with Development Partners and other foreign
organisations, which are working directly with his Circle.
81. The optical square is used to measure angles
by-
(a) Refraction (b) Reflection
(c) Double refraction (d) Double reflection
86. A curve of varying radius is known as-
Ans. (d) : The optical square uses a pentaprism to
reflect and refract a beam or sighting 90o. It is used in (a) Varying curve (b) Compound curve
pairs in surveying and in a singular block in metrology. (c) Transition curve (d) Reverse curve
Ans. (c) : Transition curve – a curve of constantly
changing radius, used to connect a circular arc to a
straight line or to an arc of different curvature.
87. The method of finding out the difference in
elevation between two points for eliminating
the effect of curvature and refraction, is-
(a) Precise leveling (b) Differential leveling
82. When the magnetic declination is 5o20' east, the (c) Reciprocal leveling (d) Fly leveling
magnetic bearing of the sun is at noon will be-
Ans. (c) : Reciprocal leveling removes errors due to
(a) 95o20' (b) 174o40'
o collimation, curvature and refraction.
(c) 185 20' (d) 354o40'
o 88. The horizontal angle between true meridian
Ans. (d) : 354 4' and magnetic meridian, is known as-
(a) Convergence (b)Magnetic declination
(c) Bearing (d) Dip
Ans. (b) : The horizontal angle which a magnetic
meridian makes with the true or astronomic meridian is
called the magnetic declination.
89. A relatively fixed point of known elevation
83. A line joining the point of intersection of the above datum, is called-
cross-hairs of the diaphragm and the optical (a) Bench mark (b) Datum point
center of the object glass, is known as- (c) Reduced level (d) Reference point
(a) Fundamental line (b) Axis of telescope Ans. (a) : A bench mark is a point of reference by
(c) Axis of level tube (d) Line of collimation which something can be measured. In surveying, a
Ans. (d) : Line of collimation-imaginary line passing bench mark is a post or other permanent mark
through the intersection of the cross hairs of the established at a known elevation that is used as the basis
diaphragm and the optical centre of the objective lens. for measuring the elevation of other topographical
Axis of telescope-line joining the optical centre of the points.
objective to the centre of the eyepiece. 90. The curvature of the Earth's surface, is taken
84. The error which is not completely eliminated in into account only if the extent of survey is more
reciprocal leveling, is- than-
(a) Error due to curvature (a) 100 sq km (b) 160 sq km
(b) Error due to refraction (c) 200 sq km (d) 260 sq km
(c) Error due to non-adjustment of line of collimation Ans. (d) : Geometric survey – In this survey the
(d) Error due to non-adjustment of bubble tube curvature of surface is taken into the account. If the
Ans. (d) : If the bubble is disturbed, the error cannot be extent of survey is more than 250 sq km. That type of
eliminated. survey is used.

OPSC AEE Exam-2016 (Paper-II) 236 YCT

 Odisha Staff Selection Commission
JE (Main), Exam-2014
1. For a cantilever beam of span L and flexural Ans. (c) : Two span continuous fixed support beam.
rigidity EI applied with a point load W at free Therefore two point of contra flexure is generated for
end, the magnitude of maximum deflection each span hence total 4 points of contra flexure.
occurring is : 3. The maximum negative bending moment in
(a) WL2/2EI (b) WL3/3EI case of a fixed beam carrying uniformly
4
(c) WL /4EI (d) WL4/5EI distributed load is at :
(e) None of these (a) Mid span (b) 1/3 of the span
Ans. (b) : (c) 1/4 of the span (d) Supports
(e) 1/5 of span
Ans. (d) :

Mxx = –Px
As we know,
d2y
EI 2 = M xx
dx
d2y In fixed beam carrying uniformly distributed load,
So, EI 2 = − Px maximum negative bending moment is at support.
dx
wL2
dy Px 2 (BM)max. at support = = MA = MB
EI =− + C1 .....(1) 12
dx 2
wL2
Px 3 (BM)at center =
EI × y = − + C1x + C 2 .....(2) 24
6
4. The moment required to rotate the near end of
Applying boundary condition – a prismatic beam through unit angle without
dy translation (the far end is fixed, El = flexural
x = L, =0
dx rigidity and L is the span of the beam), is given
PL2 PL2 by :
So, 0=− + C1 ⇒ C1 = (a) El/L (b) 2 El/L
2 2
(c) 3 El/L (d) 4 El/L
& x = L, y = 0
(e) 5 El/L
PL3 PL2
0=− + × L + C2 Ans. (d) : It is the moment that must be applied at one
6 2 end of a constant section member (which is unyielding
PL3 PL3 PL3 supports at both ends to produce a unit rotation of that
C2 = − + =− end when the other end is fixed).
2 6 3
Then, deflection at free end, x = 0
PL3
EI × y = 0 + 0 + C 2 = −
3
PL3 4EI
∴ y= (In downward↓) i.e. k=
3EI L
3EI
2. A two span continuous beam with both ends Stiffness (k) = (far end is hinged)
fixed have : l
(a) 6 points of contraflexure Where, k → Stiffness of BA at joint B when the farther
(b) 5 points of contraflexure end is fixed.
(c) 4 points of contraflexure EI → Flexural rigidity
(d) 3 points of contraflexure L → Length of the beam
(e) 2 points of contraflexure M → Moment at B
OSSC JE (Mains) Exam-2014 237 YCT
5. If a 3 hinged semi-circular arch of span L and 8. The carry over factor in a prismatic member,
radius R is loaded uniformly with w per unit whose far end is fixed, is :
run throughout the span, the horizontal thrust (a) 0 (b) 1/2
is : (c) 3/4 (d) 1
(a) wL (b) wR (e) None of these
(c) wR/2 (d) wL/2 Carry over moment
(e) None of these Ans. (b) : Carry over factor =
Applied moment
Ans. (c) : (i) When far end is fixed
M/2 1
COF = =
M 2
0
(ii) When far end is hinged = COF = =0
M
9. In a riveted joint of rivet dia, d, the minimum
In three hinged arch uniformly distributed w/m load, the pitch of the rivets shall not be less than :
2 (a) 1.5 d (b) 2.0 d
wl
H is = ………….. (i) (c) 2.5 d (d) 3.0 d
8h
(e) 3.5 d
When arch is semicircular then L = 2R and h = R is
Ans. (c) : Minimum pitch of rivet, should be 2.5 times
taken.
of nominal dia i.e, = 2.5 d
Putting the value in (i) Pitch : It is the distance between the centre of two
H = 4wR2 / 8R = wR/2 consecutive rivets measured in the direction of force
6. The slope deflection method falls in the applied.
category of :
(a) Matrix analysis
(b) flexibility analysis
(c) Indeterminate analysis
(d) Determinate analysis
(e) Force analysis Maximum pitch–The tacking rivets are provided at a
pitch in line should not more than 32 times the thickness
Ans. (c) : There are various displacement methods used
of outside plate or 300 mm (whichever is less)
to find moments.
For tension – minimum (16 t, 200 mm) whichever is less
(i) Slope deflection method For compression – Minimum (12 t, 200 mm) whichever
(ii) kani's method is less
(iii) Moment distribution method 10. The heaviest I section for same depth is :
(iv) Stiffness method (a) ISMB (b) ISLB
(v) Matrix method (c) ISHB (d) ISWB
• These methods are used for indeterminate structural (e) None of these
analysis. Ans. (c) : ISWB– Indian standard wide flange beam has
7. The bending moment diagram of a cantilever high moment of inertia about y-axis as compared to
beam subjected to a couple at the free end other beam section.
would be : ISLB– Indian standard light beam are generally used as
(a) Triangle (b) Parabola purlins in roof or as secondary beams where loading is
(c) Cubic parabola (d) Rectangle less.
(e) Trapezium ISMB– Indian standard medium weight beam can be
Ans. (d) The bending moment diagram of cantilever used as structural element in frames, floor beam etc.
beam subjected to moment at the free end will be with high moment of inertia about x-axis.
rectangular. ISHB– Indian standard heavy weight beam have thicker
webs and flanges and are used as column section.
11. The accuracy of measurement in chain
surveying does not depend on :
(a) Importance of features
(b) General layout of the chain lines
(c) Length of the offset
(d) Scale of the plotting
(e) Topography of the uite
OSSC JE (Mains) Exam-2014 238 YCT
Ans. (b) : Chain surveying is suitable only for areas of triangle and particularly when it is at the orthocentre
small extent on open ground. The accuracy of and the middle station is much leaner than the other. On
measurement in chain surveying does not depends upon the other hand, the fix is bad or poor when the
general layout of the chain lines because any layout are instruments station T is near the circumference of the
measured by chain but length of the offset, scale of the circumscribing circle. So the accuracy of fix depend on
plotting and importance of the features are depend on the relative positions of the plotted points and that of
accuracy of measurement in chain surveying. location of the plane table station. Thus the choice of
12. If R is the radius of the main curve, θ is the plotted objects and location of table should be made to a
angle of deflection, S is the shift and L is the strong fix.
length of the transition curve, then total 15. ABCD is a rectangular plot of land. If the
tangent length of the curve is : bearing of the side AB is 75º, the bearing of DC
(a) (R + S) tan θ/2 – L/2 is :
(b) (R + S) tan θ/2 + L/2 (a) 345º (b) 75º
(c) (R – S) tan θ/2 – L/2 (c) 105º (d) 180º
(d) (R – S) tan θ/2 + L/2 (e) 255º
(e) None of these Ans. (b) :
∆ L
Ans. (b) : Ts = ( R + S ) tan +
2 2
Where,
Ts = total tangent length of the curve
R = Radius of the curvature
∆ = Deflection angle
L = Length of the transition curve bearing of side AB = 105o
L2 bearing of CD = 270o – 15o – 1800
S = Shift =
24R = 75o
13. The angle of intersection of a contour and ridge 16. For true difference in elevations between two
line is : points A and B, the level must be set up :
(a) 120º (b) 90º (a) Near the point B
(c) 60º (d) 45º (b) Near the point A
(e) 30º (c) At any point between A and B
Ans. (b) : • Angle of intersection of contour line and (d) Either at A o at B
ridge line be 90º. (e) Exactly at the midpoint between A and B
• Pointing towards lower elevation Ans. (e) : For true difference in elevation between two
• U shape contour. points A and B the level must be set up at the exact
mid-point of A and B in case of points are not properly
visible.
Note : The levelling instrument for the elimination of
collimation error must be placed exactly mid-way
between the BS and FS.
17. The slope correction for a length of 30 m along
a gradient of 1 in 20 is :
(a) 0.375 cm (b) 37.5 cm
14. The 'fix' of a plane table station with three
(c) 3.75 cm (d) 1.75 cm
known points, is bad if the plane table station
lies : (e) 2.75 cm
(a) In the great triangle Ans. (c) :
(b) Outside the great triangle
(c) On the circumference of the circumscribing
circle
(d) Inside the circumference of the
circumscribing circle h2
(e) None of these Slope correction =
2L
Ans. (c) : Fix of a plane table from three known points. Given,
The accuracy with which a plane table station can be chain length = 30 m
located through three point problem is known as its fix. slope = 1 in 20
The fix is good when instrument station T is within the As gradient is 1 in 20
OSSC JE (Mains) Exam-2014 239 YCT
 for 20 m = 1 m (a) For thin structures subjected to wetting and
So, for 30 m drying, the water-cement ratio should be 0.45
h = 1.5 m (b) For mass concrete structures subjected to
(1.5 ) wetting and drying, the water-cement ratio
2

Slope correction = should be 0.55

2 × 30 m (c) For thin structures which remain continuously
(1.5 )
2
under water, the water-cement ratio by weight
= 0.0375 m should be 0.55
2 × 30
(d) For massive concrete structures which remain
Slope correction = 3.75 cm continuously under water, the water-cement
18. For the construction of highway (or railway) : ratio by weight should be 0.65
(a) Longitudinal sections are required (e) All of these
(b) Both longitudinal and cross sections ae Ans. (e) :
required
Exposure Min grade of Max free w/c
(c) Cross sections are required
concrete ratio
(d) Either (a) or (c)
Mild M20 0.55
(e) None of these
Moderate M25 0.50
Ans. (b) : We need both longitudinal and cross-section
because at every interval there might difference in land Severe M30 0.45
profile so that we will get clear idea of landfilling or Very severe M35 0.45
cutting to maintain level surface. Extreme M40 0.40
19. Transition curves are introduced at either end 21. Portland Pozzolana Cement possesses :
of a circular curve, to obtain : (a) Higher resistance to chemical attack
(a) Gradual increase of super elevation from zero (b) Lower heat of hydration
at the tangent point to the specified amount at (c) Lower shrinkage on diying
the junction of the transition curve with main (d) Water tightness
curve
(e) All of these
(b) Gradually decrease of curvature from zero at
the tangent point to the specified quantity at Ans. (e) : Portland pozzolana cement contain 20-30%
the junction of the transition curve with main of pozzolanic materials.
curve Pozzolana cement is produced either by grinding
(c) Gradual change of gradient from zero at the together Portland cement clinker and pozzolana or by
tangent point to the specified amount at the intimately or uniformly bleeding Portland cement with
junction of the transition curve with main pozzolana.
curve Advantages :
(d) All of these • Higher resistance to chemical attack.
(e) None of these • Water tightness
• Lower shrinkage of an drying
Ans. (a) : Transition curves are introduced at either end
of a circular curve to obtain– • Lower heat of hydration.
• Gradual increase of superelevation from zero at the 22. The timber having maximum resistance against
tangent point to the specified amount at the junction of white ants, is obtained from :
the transition curve with main curve. (a) Teak (b) Sal
• To introduce gradually the centrifugal force between (c) Shisam (d) Chir
the tangent point and the beginning of the circular (e) All of these
curve. Ans. (a) : Teak :
• It is moderately hard, durable and fire resistance.
• It is not attacked by white ants and dry rot.
• It can be easily seasoned and worked.
• It is used for house construction, railway, carriage,
flooring, structural works, ship building etc.
• It's use is limited to superior work only as it is
• It is used on both highway and railway between comparatively very costly.
tangent and a circular curve in order to have a smooth 23. For one cubic meter of brick masonry, number
transition from tangent to the curve and from curve to of bricks of IS size required is about :
the tangent. (a) 600 (b) 525
20. Pick up the correct statement from the (c) 550 (d) 500
following : (e) 650
OSSC JE (Mains) Exam-2014 240 YCT
Ans. (d) : Standard brick size = 19 cm × 9 cm × 9 cm Compacted stone has following properties -
with mortar .. size of brick = 20 cm × 10 cm × 10 cm • It is a sedimentary rock
= 0.20 m × 0.10 m × 0.10 m • It is fire resistant
volume of brick in masonry = 0.002 m3 • It is specific gravity 2.5.
∴ No. brick in 1 cum =
1
= 500 • It's crushing strength various 35-40 MN/m2
0.002 • Its colour depends upon that of the feldspar.
24. Cement paints usually : 28. Which of the following bricks are used for
(a) Contain 5% sodium chloride lining of furnaces ?
(b) Are prepared with white cement (a) Over burnt bricks (b) Under burnt bricks
(c) Contain 5% to 10% colour pigments (c) Refractory bricks (d) First class bricks
(d) Contain hydrated lime (e) None of these
(e) All of these Ans. (c) : Refractory brick – A refractory brick is built
Ans. (e) : Cement paints (IS 5410) : primarily to withstand high temperature but will also
• White or coloured Portland cement with (OPC usually have a low thermal conductivity for greater
minimum 65%) forms the base with a small addition of energy efficiency.
water proofer and accelerator. • It is a block of ceramic material used in lining
• They are thinned with water during application. furnace, kilns, fire boxes and fireplace.
• Proper curing is necessary for strength and durability. 29. A piezometer opening in pipes measures :
• Cement paints are durable, strong and display better (a) Negative static pressure
water-proofing qualities and are used on exterior (b) Velocity head
surface of building. (c) Static pressure
• Cement paints contain 5% sodium chloride. (d) Total pressure
• 5% to 10% colour pigments. (e) None of these
• Prepared with white cement. Ans. (c) : A piezometer opening in pipes measures
static pressure.
25. To give a brilliant finish, the type of vamish • A piezometer is either a device used to measure liquid
used, is : pressure in a system by measuring the height to which a
(a) Turpentine varnish (b) Oil varnish column of the liquid rises against gravity or a device which
(c) Water varnish (d) Spirit varnish measures the pressure (more precisely the piezometric
(e) None of these head) of ground water at a specific point.
Ans. (d) : Spirit varnish is resin of soft variety such as 30. Maximum efficiency of transmission of power
lac or shallac dissolved in spirit. Spirit varnish gives through a pipe is :
brilliant finish. It is not more durable. This varnish used (a) 76.67% (b) 25%
in high quality furniture. (c) 50% (d) 33.33%
26. Sea sand used in structures, causes : (e) 66.67%
(a) Dampness (b) Efflorescence Ans. (e) : Maximum efficiency of transmission of
(c) Disintegration (d) All of these power → efficiency of power transmission through pipe
(e) None of these is given by–
Ans. (d) : Sea sand – As it obtained from sea it contain H − hf
η=
salt, which is used in attracting moisture from H
atmosphere. for maximum power through pipe the condition is given
by
• Such absorption causes efflorescence, dampness and
H
disintegration of work. hf =
3
• It is generally not used for engineering purpose due substituting the value of hf in efficiency, we get
to its retard setting action of cement. maximum,
27. Which of the following has more fire resisting H − H/3 1 2
ηmax = = 1 − = or 66.70%
characteristics : H 3 3
(a) Marble H = total head available at the inlet of pipe
(b) Lime stone hf = loss of head due to friction,
(c) Compact sand stone ηmax = maximum efficiency
(d) Granite 31. The momentum correction factor (β) for the
(e) Wood viscous flow through a circular pipe is :
Ans. (c) : Compacted stone - It is a type of rock which (a) 1.33 (b) 1.5
has quartz as the sand bound together with the (c) 1.5 (d) 1.67
cementing minerals like mica, felspar etc. (e) 0.67
OSSC JE (Mains) Exam-2014 241 YCT
Ans. (a) : Momentum correction factor– It is define Ans. (d) : A triangular channel is hydraulically most
as the ratio of momentum/second based on average economical when each of its sloping side is inclined to
velocity. the vertical at an angle of 450.
1 y
AV 2 ∫
β= u 2 .dA R=
2 2
Momentum correction factor (β)
35. Mach number us the ratio of intertie force to :
Momentum per second across a section
(a) Viscosity
calculated on the basis of actual velocity
= (b) Elasticity
Momentum per second across the section (c) Gravitational force
calculated on the basis of mean velocity (d) Surface tension
for a circular tube β = 1.33 (e) Shear force
Ans. (b) : Mach number is the ratio of inertia force to
32. The maximum vacuum created at the summit elastic force.
of a syphon is : inertia force V
(a) 2.6 m of water (b) 3.14 m of water Mach number = =
elastic force K/ρ
(c) 7.4 m of water (d) 9.81 m of water
(e) None of these Where, K / ρ is the velocity in the medium and
Ans. (c) : The maximum vacuum created at the summit represented by C.
of a syphon is 7.4 m of water. 36. To avoid the force of surface tension in an
Then highest point of the syphon is called the incline manometer, the minimum angle of
summit. An air vessel is provided at the summit in order inclination is :
to avoid interruption in the flow. (a) 2º (b) 3º
The pressure is only –7.6 m of water or 10.3– (c) 4º (d) 5º
7.6 = 2.7 m of water absolute (e) 6º
Pvaccum = Patm – Pabsolute
= 10.3 – 2.7 = 7.6 m Ans. (c) : To avoid the force of surface tension in an
incline manometer, the minimum angle of inclination is
33. A pitot tube is used to measure :
4 0.
(a) Pressure
37. The velocity of the fluid particle at the center of
(b) Difference in pressure
the pipe section is :
(c) Velocity of flow
(a) Minimum (b) Maximum
(d) Discharge
(c) Equal throughout (d) Zero
(e) Thrust
(e) None of these
Ans. (c) : To measure static pressure in a pipe, pressure
gauge connected to a pitot tube. It is a device used for Ans. (b) : Flow through circular pipe– The velocity of
measuring the velocity of flow at any point in a pipe or the fluid particle at the centre of the pipe section is
a channel. The velocity is determined by measuring the maximum.
rise of liquid in the tube.

38. If the average daily demand of a city of 50,000

population, is 20 mld, the maximum daily
34. Most economical section of a triangular demand is :
channel, is :
(a) Right angled triangle (a) 36 mld (b) 30 mld
(b) Equilateral triangle (c) 24 mld (d) 18 mld
(c) Isosceles triangle with 45º vertex angle (e) 12 mld
(d) Right angled triangle with equal sides
Ans. (a) : Given,
(e) Any isosceles triangle

OSSC JE (Mains) Exam-2014 242 YCT

Average daily demand = 20 mld (b) Degradation
Maximum daily demand = 1.8 × 20 mld (c) Recovery
(d) Cleaner water
= 36 mld
(e) None of these
39. The dilution ratio at which the odour is hardly
detectable is generally called threshold odour Ans. (b) : Zone of degredation or zone of pollution–
number and for public supplies it should not This zone is found for a certain length just below the
point where sewage is discharged into the river -
exceed :
stream.
(a) 10 (b) 7
(c) 5 (d) 3 • This zone is characterized by water becoming dark
and turbid with formation of sludge deposits at the
(e) 1
bottom DO is reduced to about 40% of the saturation
Ans. (d) : The dilution ratio at which the odour is value.
hardly detectable is generally called threshold odour
• These conditions are unfavorable to the development
number and for public supplies it should not exceed 3. of aquatic life, and as such, algae dies out, but fish life
40. The chloride content of treated water for may be present feeding on fresh organic matter.
public supplies should not exceed : 43. Self-cleansing velocity is :
(a) 100 ppm (b) 200 ppm (a) Velocity of water at flushing
(c) 250 ppm (d) 300 ppm
(b) Velocity at dry weather flow
(e) 150 ppm
(c) Velocity at which no accumulation remains in
Ans. (c) : For potable water, the amount of chloride the drains
should not exceed 250 ppm. (d) Velocity of water in a pressure filter
Water quantity Permissible Cause for (e) None of these
parameter units Rejection Ans. (c) : Self-cleansing velocity is velocity at which
value no accumulation remains in the drains.
1. Total dissolved 500 mg/l 2000 mg/l
Self-cleansing velocity → (Vsc)
solids
• Sewers are designed to carry maximum hourly
2. Alkalinity 200 mg/l 600 mg/l
discharge and are checked at minimum hourly
3. pH 7 – 8.5 < 6.5 discharge for development of self cleansing
4. Hardness 200 mg/l 600 mg/l velocity.
5. Fluoride 1 mg/l 1.5 mg/l • Self cleansing velocity is that which do not permit
41. The yield of a rapid gravity filter as compared the silting of the solids in the sewer moreover also
to that of slow sand filter is : carried out the removal of the solids. Which are
(a) 30 times (b) 25 times already settled in it.
(c) 20 times (d) 15 times • Self cleansing velocity in sewer is given by :
(e) 10 times Here,
8k
Ans. (a) : The yield of a rapid gravity filter as VSC = (G − 1)g.d f → Friction factor
compared to that of show sand filter is 30 times SSF. f
G → Specific gravity
Slow sand filter Rapid sand filter
k→ Constant which depends upon type of solids in
Cu = 3-5 Cu = 1.2 to 1.6
the sewage
D10 = 0.2 to 0.3 mm D10 = (0.35 to 0.55) mm
G→ Acceleration due to gravity.
Rate of filtration Rate of filtration
d→ Size of the particles.
(100 to 200) l/m2/hr (3000 to 6000) l/m2/hr
almost 30 time of SSF 44. If 2% solution of a sewage sample is incubated
Depth of tank 2.5 to 3.5 m Depth of tank 1-2 m for 5 days at 20ºC and depletion of oxygen was
found to be 5 ppm, B.O.D. of the sewage is :
Period of cleaning 1-3 Period of cleaning 1-3
months days. (a) 150 ppm
(b) 200 ppm
Removal of turbidity up to Removal of turbidity up to
50 ppm 40 ppm (c) 250 ppm
• Washing period 24-48 (d) 300 ppm
hrs. (e) 350 ppm
42. The algae dies out in the zone of : Ans. (c) : Depletion of oxygen = 5 ppm
(a) Active decomposition
OSSC JE (Mains) Exam-2014 243 YCT
 100
Dilution factor = = 50
2
B.O.D. = 5 × 50
= 250 ppm
45. Ventilating shafts are provided to a sewer line
at every :
(a) 100 m (b) 150 m
(c) 200 m (d) 250 m
(e) 300 m
Ans. (e) : Ventilating shafts are provided to a sewer
line at every 300 m.
• The decomposition and putrefaction of sewage
inside the sewers may result in the production of 48. When a canal and a drainage approach each
various sewer gases. Such as CO2, CO, methane, other at the same level, the structure so
hydrogen sulphide, ammonia etc. provided is :
• These gases are disposed of into the atmosphere by (a) A siphon aqueduct
exposing the sewage to the outside atmosphere by (b) A siphon
suitable method of ventilation. (c) A level crossing
• In order to achieve proper ventilation, ventilating (d) At the head of the distributaries
shafts are generally placed at intervals of 300 m (e) Same at all places
along the sewer line. Ans. (c) : A level crossing– It is a cross drainage work
in which the drainage and the canal meet each other at
• They are also provided at the upper end of every approximately the same level. Level crossing is
branch sewer and also at every change in the size of generally adopted when canal and the drainage are
the sewers. practically at the same level and for high flood drainage
46. The digested sludge from septic tanks, is discharge but short lived. Aqueduct or super passage is
removed after a maximum period of : adopted when high flood drainage discharge is large.
(a) 2 years (b) 3 years
(c) 4 years (d) 5 years
(e) 6 years
Ans. (b) : Design parameters of septic tank –
• Flow of sewage 40-70 lpcd
• Rate of accumulation = 30 lpc per year
• Detention time 12-36 hours
• Length/ width ratio = 2 to 3 49. In a barrage, the crest level is kept :
• Depth 1.2 to 1.8 m (a) High with no gates
• Cleaning period = 6 month to 1 year (b) High with large gates
(c) Low with large gates
• Maximum period of removal of digested sludge is 3
(d) Low with no gates
years.
(e) None of these
47. The duty is largest :
Ans. (c) : In a barrage, the crest level is kept low with
(a) At the head of water course large gates.
(b) At the head of a main canal • The crest level is the barrage (top of solid obstruction)
(c) On the field is kept of low level. During flood gates are raised to
(d) At the head of the distributaries clear of the high flood level. As a result there is less
silting and provide better regulation and control than the
(e) Same at all places
weir.
Ans. (c) : Duty is the area that can be irrigated by the 50. Canals constructed for draining off water from
discharge of 1 cumec of water. water logged areas are known as :
• Duty of water changes from place to place, it will be (a) Drains (b) Inundation canals
maximum at the field and minimum at the head of the canal. (c) Contour canals (d) Valley canals
(e) Ridge canals

OSSC JE (Mains) Exam-2014 244 YCT

Ans. (a) : Canal constructed known as drains– 54. If the grain size of soil increases :
• A properly designed drainage system is an effective (a) Water supply in well increases
means to prevent land from getting water logged. (b) Specific retention decreases
• Before undertaking a drainage project few things (c) Specific yield increases
should kept into mind. (d) All of (1), (2) and (3)
(i) Investigations such as topographical, geological and (e) None of these
soil surveys should be carried out. Ans. (d) : If the grain size of soil increases the void
(ii) The nature of soil from the point of view of space increases the water supply is more.
permeability should be studied. • When size ↑ than specific retention decrease, and
(iii) A knowledge of water table and its fluctuations and specific yield is increased
the quality of ground in the area proposed for irrigation. S y + Sr = 1
Two type of drains can be provided–
(i) Surface drains (ii) Sub surface drains. 55. Isohytes are the imaginary lines joining the
points of equal :
51. Consumptive use of a crop during growth is the
(a) Rainfall (b) Pressure
amount of :
(c) Height (d) Humidity
(a) Interception (b) Evaporation
(e) Reduced level
(c) Transpiration (d) Both (a) and (b)
(e) All of (a), (b) and (c) Ans. (a) : Isohytes lines– • These lines are the lines
joining the points of equal rainfall.
Ans. (e) : Consumptive use of water (wcu)– It is the
normal consumptive use of water which is easily • These lines are also termed as isopluvial lines.
available for evaportational and transpirational need of 56. The foundations are placed below ground level
crop. to increase :
w (a) Structural stability
ηcu = cu × 100 (b) Workability
wd
(c) Strength
wd – it is the net amount of water depleted from the root
zone. (d) Stiffness of structures
(e) All of these
52. The best unit period of a unit hydrograph, is
equal to the basin lag divided by : Ans. (c) : Foundation is the element of a structure
which connect it to the ground and transfer load from
(a) 2 (b) 3
the structure to the ground. Foundation provide the
(c) 4 (d) 5 structure's stability from the ground, foundation are
(e) 6 generally consider either shallow or deep.
Ans. (c) : Best duration of a unit hydrograph is equal to 57. If the height of the first storey of a building is
1 3.2 m and riser is 13 cm, the number of treads
the of basin lag.
4 required is :
1 1 (a) 12 (b) 18
The duration of rainfall should be to of the basin
5 3 (c) 24 (d) 30
lag, and storm should be isolated storm occurring (e) 36
individually. 3.20
Ans. (c) : Height of a building / rise = = 25
53. The earthen embankments constructed parallel 0.13
to the river banks at some suitable distance for Number of riser = 25.
flood control, are known as : Number of tread = Riser – 1
(a) Levees i.e. 25 –1 = 24
(b) Dykes
58. The piece of a brick cut with its one corner
(c) River walls equivalent of half the length and half the width
(d) Both dykes and levees of a full brick is known as :
(e) None of these (a) King closer
Ans. (d) : During floods, when the river are enclosed by (b) Queen closer
dikes. The water spills over the natural banks of the (c) Beveled closer
river and will spread in the area confined between the (d) Half king closer
river and the dikes. Due to this the velocity of flow
(e) Half queen closer
reduces and silt gets deposited. But infact, this reduction
in velocity in a diked river is much less than that in an Ans. (a) : King closer – The portion of a standard brick
undiked river because of the spread area is less in a made by cutting off the triangular piece between the
diked river. Hence it can be conducted the silt centre of one header and one stretcher face.
deposition will be less in a diked river compared to that • A king closer is used near door and window opening
in an undiked river. to get a satisfactory arrangement of the mortar joint.
OSSC JE (Mains) Exam-2014 245 YCT
 62. Open test pit is only suitable up to a depth of :
(a) 3 m (b) 4 m
(c) 5 m (d) 6 m
(e) 1 m
Queen Closer : The portion of a standard brick made Ans. (a) : Open test pit is only suitable up to a depth of
by cutting in along the length into two equal parts is 3 m.
called Queen Closer. 63. The projections of head or sill of a door or
59. In horizontal D.P.C. thickness of cement window frame are known as :
concrete (1 : 2 : 4) is : (a) Chocks (b) Stops
(a) 3 cm (b) 4 cm (c) Horns (d) Transoms
(c) 5 cm (d) 6 cm (e) Lintels
(e) 8 cm Ans. (c) : Horn–These are the horizontal projections of
Ans. (b) : In horizontal DPC thickness of cement frames.
concrete (1 : 2 : 4) is 4 cm. Damp proof course (DPC) Jamb–These are the vertical portion of the door frame
shall be applied at the plinth level in a horizontal layer onto which the door is secured.
of 2.5 cm thickness. Damp proof course shall consist of Rebate–It is the depression or recess made inside the
cement, coarse sand and stone aggregate of door frame to receive the door shutter.
1
1:1 : 3 proportion with 2% of impermo or cem-seal, or 64. For effective drainage, the finished surface of
2 flat roof should have a minimum slope of :
According proof by weight of cement or other standard (a) 1 in 5 (b) 1 in 10
water proofing compound (1 kg per bag of cement).
(c) 1 in 15 (d) 1 in 20
60. The voussoir placed at crown of an arch, is (e) 1 in 40
known as a :
(a) Soffit Ans. (d) : For effective drainage the finished surface of
the flat roof should have a minimum stop of 1 in 20.
(b) Key
(c) Haunch 65. Which of the following earth moving
(d) Gauged brick arch machineries has the shortest cycle time ?
(e) Springer (a) Dipper shovel (b) Drag line
(c) Hoe (d) Clam shell
Ans. (b) : Key stones : A species that plays a critical
role in maintain the structure of an ecological (e) Bulldozer
community and whose impact on the community is Ans. (a) : Dipper shovel : A dipper shovel is oldest and
greater than would be expected. most important excavating machine.
Through/bond stone : Thoughts are stones, which • Consist of heavy relatively short boom and a dipper
extend the full thickness of the wall. stick (A beam that pivots on the boom) with digging
Note : Through should not be used in the external walls. bucket attached.
• As moisture may be conducted through then and • It move shortest distance and the efficiency is more,
cause dampness on the internal face. and it moving cycle is fast.
61. The construction joints in buildings are 66. The first stage of a construction is :
provided after : (a) Preparation of estimate
(a) 10 m (b) 20 m
(b) Survey of the site
(c) 30 m (d) 40 m
(c) Initiation of proposal
(e) 50 m
(d) Preparation of tender
Ans. (d) : Joints in buildings : (e) Allotment of funds
• The construction joints in buildings are provided
Ans. (c) : A brief description specifically about stage
after 40 m.
are explained below.
• All building materials expand or contract with
• Initiation phase of construction project.
change in temperature and variation in moisture
contents. Thus, major dimensional changes are • Planning phase of construction project.
caused in structure due to expansion or contraction of • Execution phase of construction project.
materials used in their construction. • Performance and monitoring phase of construction
• The magnitude of these changes varies with the type project.
of material used. • Closure phase of construction project.
OSSC JE (Mains) Exam-2014 246 YCT
67. The critical activity has : 71. Ordinary concrete may not be used for the
(a) Normal float following grade of concrete :
(b) Maximum float (a) M 10 (b) M 15
(c) Minimum float (c) M 20 (d) M 25
(d) Zero float (e) M 40
(e) None of these Ans. (e) :
Specified
Ans. (d) : Critical activity is the activity which has zero characteristic
float or zero slack. Grade compressive
68. Bar charts are suitable for : Group
designation strength of 15
(a) Minor works mm cube at
(b) Major works 28 day.
(c) Large projects (i) Ordinary M 10 10 N/mm2
(d) All of (a), (b) and (c) concrete M 15 15 N/mm2
(e) None of these M 20 20 N/mm2
(ii) Standard M 25 25 N/mm2
Ans. (a) : Bar charts : concrete M 30 30 N/mm2
• Bar charts are suitable for minor projects only. M 35 35 N/mm2
• It is a graphical representation of activity v/s time. M 40 40 N/mm2
• It is also being referred as Gantt chart. M 45 45 N/mm2
69. Separation of water or water sand cement from M 50 50 N/mm2
M 55 55 N/mm2
a freshly mixed concrete, is known as :
M 60 60 N/mm2
(a) Flooding (b) Bleeding
(iii) High M 65 65 N/mm2
(c) Segregation (d) Creeping strength M 70 70 N/mm2
(e) None of these concrete M 75 75 N/mm2
Ans. (b) : Bleeding : Bleeding is one form of M 80 80 N/mm2
segregation, where water or water cement comes out to M 85 85 N/mm2
the surface of the concrete. Being lowest specific M 90 90 N/mm2
gravity among all the ingredients of concrete. M 95 95 N/mm2
M 100 100 N/mm2
72. For normal RCC works, i.e. slabs, columns,
beams, walls etc, the grade of concrete mix
used is generally :
(a) 1 : 3 : 6 (b) 1 : 2 : 4
(c) 1 : 1/2 : 3 (d) 1 : 1 : 2
(e) 1 : 4 : 8
Ans. (b) :
To reduce bleeding in concrete- Type of Proportional
(i) Reduced the water content Nature of work
concrete mix
(ii) Reduce w/c ratio For general RCC works
M 15 1:2:4
(iii) Reduce slump (slab, beams, columns, etc.
(iv) Use chemical admixture Water retaining
(v) Entrained concrete M 20 1 : 1.5 : 3 structures, piles and
(vi) Increase the amount of fineness in the sand. general RCC structure
Heavily loaded RCC
70. Percentage of pozzolanic material containing M 25 1:1:2 structure long span slab,
clay up to 80%, used for manufacture of beam etc.
Pozzolana cement is :
73. Workability of concrete is increased due to an
(a) 70% (b) 60%
excess of :
(c) 50% (d) 40% (a) Cement
(e) 30% (b) Water
Ans. (e) : Percentage of pozzolanic material containing (c) Rounded aggregates
clay up to 80%, used for manufactures of pozzolana (d) All of (a), (b) and (c)
cement is 30%. (e) None of these
OSSC JE (Mains) Exam-2014 247 YCT
Ans. (d) : Due to excess of cement, water and rounded Ans. (b) :
aggregate, Internal friction of concrete reduces, Internal Multiplying factor for
Type of door
friction of concrete reduces which is cause of increase is one side face
workability of concrete. Collapsible 1.50
74. Which of the following is a method for door/collapsible steel
estimating the building works? shutter
Sliding/Rolling shutters 1.10
(a) Long and short wall method
Framed, ledged and doors 1.30
(b) Out-to-out and in-to-in method
Flush doors 1.20
(c) Center line method Wire gauged doors 1.00
(d) Crossing method
78. For one cubic meter of cement concrete prop
(e) All of these with stone chips 1 : 2 : 4, the required number
Ans. (e) : Method for estimating the building works– of cement bags is :
• Long and short wall method (a) 4.54 (b) 5.34
• Out-to-out and in-to-in method (c) 6.34 (d) 6.5
• Centre line method (e) 7.0
• Crossing method Ans. (c) : Quantity of cement concrete = 1 m3
75. For RCC works, no deduction is made in Proportion of cement concrete = 1 : 2 : 4
estimation for openings up to : Dry concrete increase 1.50 extra.
(a) 0.1 m2 (b) 0.2 m2 Note : Where compaction is not done. There are 50%
extra dry concrete taken.
(c) 0.3 m2 (d) 0.4 m2
2 1×1.5 ×1
(e) 0.5 m Required number of cement bag = = 0.21 m3
1+ 2 + 4
Ans. (a) : In case of rules the deduction for brickwork,
1 cubic cement = 30 bags
no deduction or addition shall be made for following–
0.21 m3 cement = 0.21 × 30 = 6.3
• Opening upto 0.1 sqm (1sq.ft) in section.
79. For 2.5 cubic meter thick cement concrete (1 : 2
• Ends of joints, beams, lintels posts, rafters, purlins,
: 4) damp-proof-course; the quantity of stone
corbels etc.
chips required will be :
• Wall plates and bed plates, bearing of slabs, chajjas
(a) 2 m3 (b) 2.2 m3
and the like where the thickness does not exceed 10 3
(c) 2.5 m (d) 2.7 m3
cm and the bearing does not extend over the full 3
(e) 3 m
width (thickness) of wall.
For plastering work– Ans. (b) : Quantity of cement concrete = 2.5 m3
• Ends of beams post, rafters, etc. up to 500 sq. cm. or Proportion of cement concrete = 1 : 2 : 4
0.05 sq. m in section. Dry concrete 1.52 extra taken.
• Bed plate, wall plate, bearing of balcony (Chajja)
2.5 ×1.52 × 4
Quantity of stone chips = = 2.17  2.2 m3
and the like up to 10 cm. depth, bearing of floor and 7
roof slabs are not deducted from masonry. 80. The quantity of brick work in foundation and
76. For paving or floor finishes, dado and skirting plinth per day per mason should be :
deduction for ends of dissimilar materials or (a) 0.5 m3 (b) 0.75 m3
3
other articles embedded shall note be made for (c) 1 m (d) 1.25 m3
3
areas exceeding : (e) 1.5 m
(a) 0.01 m2 (b) 0.1 m2 Ans. (d) : The quantity of brick work in foundation and
(c) 0.2 m 2
(d) 0.3 m2 plinth per day per mason should be 1.25 m3.
(e) 0.4 m2 Brick work in lime or cement mortar in super structure
Ans. (b) : For paving or floor finishes, dado and per day per mason should be 1.00 m3.
skirting deduction for ends of dissimilar materials or 81. As per IRC recommendations, the maximum
other articles embedded shall note be made for areas limit of super elevation for mixed traffic in
exceeding 0.1 m2. plain terrain is :
77. The multiplying factor for painting collapsible (a) 1 in 15
gate measured that (size of opening) all over is : (b) 1 in 12.5
(a) 1.3 (b) 1.5 (c) 1 in 10
(c) 2 (d) 2.5 (d) 1 in 7.5
(e) 3.0 (e) Equal to camber
OSSC JE (Mains) Exam-2014 248 YCT
Ans. (a) : As per IRC recommendations, the maximum (a) N = VT
limit of super-elevation for mixed traffic in plain terrain (b) V = NT
is 1 in 15. (c) T = VN
• Mixed traffic facilities are most appropriate on roads (d) N = square root of VT
with low volumes of traffic operating at low speeds. (e) V = square root of NT
 1 Ans. (a) : The fundamental relationship between traffic
0.07 = 15 for plain & rolling terrain volume, density and speed may be given by the general
 1 equation of traffic flow–
e max = 0.1 = for hilly terrain N = VT
 10
1 N = Traffic volume (vehicle per hour)
0.04 = for urban road
 25 V = Traffic speed of vehicle (kmph)
T = Traffic density (vehicle/km)
82. The bottom most layer of a flexible pavement is
85. The best material for sleepers is :
:
(a) Reinforced concrete
(a) Base
(b) Pre stressed concrete
(b) Sub-base (c) Mild steel
(c) Subgrade (d) Cast iron
(d) Base course (e) HYSD steel
(e) None of these Ans. (b) : • The best material for sleeper is pre-stressed
Ans. (c) : Bottom most layer of pavement is known as concrete.
sub-grade. • In pre- stressed concrete sleepers, the concrete is put
• Pavements are generally classified into two categories under a very high initial compression. The design is
on the basis of structural behavior. based on–
(1) Flexible pavement (i) The maximum permissible compressive strength of
• Flexible pavement is described in IRC : 37 – 2012 211 kg/cm2
(ii) The maximum cube crushing strength of concrete in
the sleeper is 422 kg/cm2 at 28 days.
(iii) All the disadvantage of reinforced concrete sleepers
have been eliminated by pre-stressing technique for
sleepers.
86. The overall length of a turnout is the distance
between the end of the stock rail and :
(2) Rigid pavement (a) Heel of crossing (b) Toe of crossing
(c) Throat of crossing (d) Nose of crossing
(e) None of these
Ans. (a) : Overall length of turnout– It is the distance
between the end of stock rails and heel of crossing.
87. Points are a group of :
83. As per IRC, the width of the pavement of a (a) Stock rails (b) Tongue rails
single lane should be a minimum of : (c) Switches (d) Stretcher bars
(a) 3.25 m (b) 3.50 m (e) Bearing plate
(c) 3.75 m (d) 4.0 m Ans. (c) : Points– It is the set of switches.
(e) 4.5 m Turnout– It is combination of points & crossing.
Ans. (c) : Width of Carriage way–It is decided on the Switch– It is combination of stock rail & tongue rail.
basis of capacity which depend on traffic lane and 88. The place used for servicing and repairing of
number of lanes. the aircraft is called :
No. of lane Carriage way (a) Apron
width (m) (b) Hanger
Single lane 3.75 (c) Terminal building
Two lane, no kerb 7.0 (d) Holding apron
Two lane, raised kerbs 7.5 (e) Stock yard
Intermediate carriage 5.5
Ans. (b) : Hanger it is large shed erected at the airport
Multi-lane 3.5 m per lane for the purpose of housing and repairing aircrafts.
84. The traffic volume N, traffic density T and Hangars are usually constructed by using steel frames
traffic speed V are interrelated as : and are covered with sheets of galvanised iron.
OSSC JE (Mains) Exam-2014 249 YCT
89. The most popular method of constructionof Ans. (c) : Truss carry only axial load, no bending
wall break water is : moment or transverse load.
(a) Barge method • Beam carry transverse load mainly.
(b) Staging method • Column only resist compressive loads.
(c) Low level method 92. For a bar in tension, a standard hook has an
(d) Mooring system method anchorage equivalent to a straight length of :
(e) None of these (a) 8 φ (b) 12 φ
Ans. (b) : The most popular method of construction of (c) 16 φ (d) 20 φ
wall breakwater is staging method. (e) 24 φ
• The caisson is placed on relatively thin stone bedding. Ans. (c) : Anchorage values as per IS code 456 : 2000
Advantage of this type is the minimum use of natural clause number 26.2.2.1–
rock. Bend and hooks–
Wall Breakwater • The anchorage value of bend shall be taken as 4 times
and the diameter of the bar for each 450 bend subjected
to maximum of 16 times of diameter of bar.
• The anchorage value of standard U-type hook shall be
equal to 16 times of diameter of bar.
93. The purpose of provision of lateral ties in
columns is to :
(a) Avoid buckling of longitudinal bars
(b) Facilitate construction
• Wave walls are generally placed on shore connected (c) Facilitate compaction of concrete
(reduce overtopping). (d) Increase the load carrying capacity of column
90. Drift method of tunneling is used to construct (e) None of these
tunnels in : Ans. (a) : The main functions of lateral
(a) Rock reinforcement in RC columns are–
(b) Soft ground • It prevent longitudinal reinforcement bars from
buckling.
(c) Self supporting ground
• It resist the shear force and hence contributes
(d) Broken ground avoiding shear failure.
(e) Granite leyer • It confines the concrete core to provide sufficient
Ans. (a) : Drift method of tunneling is used to construct ductility or deformability.
tunnels in rock. • It restrains the spliced bars and hence prevents their
slip.

Drift method–
• It consists of driving small sized healing, centrally at
94. In case of two way slab, the limiting deflection
top or bottom of face, which is later enlarged by
of the slab is :
widening and benching.
(a) Primarily a function of the long span
• It is suitable for large size tunnels in difficult or hard
(b) Primarily a function of the short span
rock.
(c) Independent of long or short span
91. A member carrying an axial load can be best (d) Dependent on both long and short span
observed in case of a : (e) None of these
(a) Beam Ans. (b) As per clause 24.1 of IS 456 : 2000
(b) Column In case of two way slab the limiting deflection of the
(c) Truss slab is primarily a function of the short span.
(d) Truss or column • The strip of a two way slab may be checked against
(e) Truss and column shorter span to effective depth ratios.

OSSC JE (Mains) Exam-2014 250 YCT

 Type of reinforcement • In 'I' section the web resists shear forces, while the
Type of slab Fe 415 grade flanges resist more than 80% of the bending moment.
Mild steel
steel Beam theory shows that the I-shaped section is a very
Simply supported 35 28 efficient form for carrying both bending and shear loads
Continuous 40 32 in the plane of the web.
• For same area it gives much higher moment of inertia
95. For a beam, if d = effective depth, b = width
which results in stiff section and hence deflection is
and D = overall depth, then the maximum area very less.
of compression reinforcement in a beam is :
(a) 0.04bd (b) 0.04bD
(c) 0.12bd (d) 0.12bD
Ans. (b) : • Maximum area of tension reinforcement
Ast max = 0.04 BD, i.e 4% of total C/S
• Maximum area of compression reinforcement 99. The slenderness ratio of a tension member
Asc max = 0.04BD i.e. 4% of total C/S subjected to reversal of stresses due to wind
• Total = 4 + 4 = 8% of total C/S shall not exceed:
(a) 60 (b) 120
96. Half of the main steel in a simply supported
(c) 150 (d) 180
slab of span L is bent up near the support at a
(e) 350
distance of X from the centre of the bearing,
where X is equal to : Ans. (e) : Maximum slenderness ratio for tension
member As per IS 800 : 2007.
(a) L/3 (b) L/5
Maximum Effective
(c) L/7 (d) L/9 Type of Member
Slenderness Ratio
(e) L/10 A tension member in which a
Ans. (c) : As per IS 456-2000, in design of shear reversal of direct stress occurs
reinforcement for single bar or single of parallel bars, 180
due to loads other than wind or
all bent-up at the same cross section the strength of seismic forces.
shear reinforcement (Vus) = 0.87 fy× a s v × sinα A members normally acting as
fy = Characteristic strength a tie in a roof truss or a bracing
system. But subject to possible
a s v = Area of shear reinforcement. 350
reversal of stress resulting from
α = angle between the inclined stirrups. the action of wind or
• Half of the main steel in a simply supported slab of earthquake forces.
span L is bent up near the support at a distance of L/7 A tension member permanently
from the centre of the bearing. in tension except pretensioned 400
member.
97. The minimum and maximum values of the 100. As per IS : 800, the outstand of flange plates
longitudinal reinforcement in a column as for a compression flange should not exceed (t =
percent of the sectional area of the column are : thickness of the thinnest plate) :
(a) 0.6 and 6 (b) 0.6 and 8 (a) 12 t (b) 16 t
(c) 0.8 and 6 (d) 0.8 and 8 (c) 20 t (d) 25 t
(e) None of these (e) 30 t
Ans. (c) : The minimum and maximum value of cross- Ans. (b) : The load is transferred from flange plate to
sectional area of longitudinal reinforcement, shall be not web plate through flange angle only.
less than 0.8 percent nor more than 6 percent of the • Width of outstand in compression flanges.
gross-sectional area of the column. b >/ 16t f (tf is thickness of flange)
• The use of 6 percent reinforcement may involve • The width of outstand in tension zone.
practical difficulties in placing and compacting of b >/ 20t f
concrete hence lower percentage is recommended.
98. The most efficient steel section for a beam is :
(a) Angle section (b) Rectangular section
(c) Channel section (d) T section
(e) I section
Ans. (e) :
• The most efficient and economical section used as a
steel beam is I section.

OSSC JE (Mains) Exam-2014 251 YCT

 Odisha Staff Selection Commission
(JE), Exam-2014
1. A rod of length L and diameter D is subjected wL4 wL3
to a tensile load P. Which of the following is Deflection (δ) = , Slope (θ) =
sufficient to calculate the resulting change in 8EI 6EI
diameter? 8δ 4δ 4 × 7.5
∴ θ= = = Error! Not a valid
(a) Young's modulus 6L 3L 3 × 1000
(b) Shear modulus link. 0.01 radian
(c) Poisson's ratio 4. Resilience is
(d) Both Young's modulus and shear modulus (a) maximum strain energy
(b) recoverable strain energy
Ans. (d) : From the relation E = 2G (1 + µ), (c) total potential energy
• We can calculate Poisson's ratio if we known the (d) shear strain energy
value of both shear modulus (G) and Young's modulus Ans. (b) : Resilience is the total strain energy stored
(E). in a given volume of material within the elastic limit.
2. Match List I with List II and select the correct • On removal of load, this energy is released, hence if
answer using the codes given below the Lists: is recoverable strain energy.
List I List II 1
= P×δ
A. Ultimate strength 1. Internal structure 2
B. Natural strain 2. Change of length σ 2y
per unit Instantaneous Modulus of resilience =
2E
length σy = yield stress
C. Conventional strain 3. Change of length E = young modulus
per unit gauge length
5. All the theories of failure gives nearly the same
D. Stress 4. Load per unit area
result
Codes:
(a) when one of the principal stresses at a point is
A B C D large in comparison to other
(a) 1 2 3 4 (b) when shear stresses act
(b) 4 3 2 1 (c) when both the principal stresses are
(c) 1 3 2 4 numerically equal
(d) 4 2 3 1 (d) for all situations of stress

Ans. (a) Ans. (a) : When beam subjected to pure bending –

A. Ultimate strength 1. Internal structure All the theories of the failure will give the same result
B. Natural strain 2. Change of length per unit when one of the principle stresses is very large as
Instantaneous length compared to the other principle stresses.
C. Conventional 3. Change of length per unit For pure shear state of stress, all the theories of failure
strain gauge length
will give the different result.
D. Stress 4. Load per unit area
3. The deflection at the free end of a uniformly 6. In a cantilever of span ‘L’, subjected to a
loaded cantilever of length 1m is 7.5mm. What concentrated load of ‘W’ acting at a distance of
is the slope at free end? L/3 from the free end, the deflection under load
(a) 0.01 radian will be
(b) 0.015 radian (a) WL3 / 3 EI
(c) 0.02 radian (b) WL3 / 81 EI
(d) 0.025 radian (c) 14 WL3 / 81 EI
(d) 8WL3 / 81 EI
Ans. (a) For uniformly loaded cantilever :
 2L 
Ans. (d) : Deflection under the load at distance  ,
 3 
from support (A).

OSSC JE Exam-2014 252 YCT

 • If structure has achieved limit state, it cannot be used
satisfactory in future.
• Limit state can be achieved due to either collapse or
serviceability failure.
W × (2L / 3)3 10. Which of the following pairs is not matched
=
3EI correctly?
8 WL3 (Cement test) (Apparatus)
= (a) Fineness Nurse and Blains
81EI
(b) Consistency Vicat
7. As per IS : 456, side face reinforcement, not (c) Soundness Le-Chatelier
less than 0.05% of web area, is provided on
(d) Sp. gravity Le-chatellier's flask
each side when the depth of web is not less than
(a) 300 mm (b) 400 mm Ans. (d) :
(c) 500 mm (d) 750 mm (Cement test) (Apparatus)
(a) Fineness Nurse and Blains
Ans. (d) : As per IS 456 Cl. 26.5.1.3–
(b) Consistency Vicat apparatus
• The depth of the web in a beam exceeds 750 mm,
side face reinforcement shall be provided along the (c) Soundness Le-Chatelier apparatus
two faces. (d) Sp. gravity Le-chatellier's flask
• The total area of such reinforcement shall be not 11. Consider the following oxides:
less than 0.1 percent of the web area and shall be 1. Al2O3 2. CaO 3. SiO2
distributed equally on two faces at a spacing not The correct sequence in increasing order of
exceeding 300 mm or web thickness which ever is their percentage in an ordinary Portland
less. cement is
8. What is the number of plastic hinges which will (a) 2, 1, 3 (b) 1, 3, 2
cause the overall total collapse of a structure? (c) 3, 1, 2 (d) 1, 2, 3
(a) One more than the order of static Ans (b) Increasing order of percentage in an ordinary
indeterminacy Portland cement is
(b) Equal to order of static indeterminacy Alkalies < MgO < Fe2O3 < Al2O3 < SiO2 < CaO
(c) One less than the order of static In cement various ingredients are in the following
indeterminacy order:-
(d) Not determinable Ingredients %age
Ans. (a) : The number of plastic hinges required for Lime (CaO) 60-65
total collapse of a structure is one more than the order Silica (SiO2) 17-25
of statical indeterminacy.
Alumina (Al2O3) 3-8
• In order to make an indeterminate beam determinate
Ferric oxide (Fe2O3) 3-4
the number of plastic hinges, required is equal to the
Calcium sulphate 2-3
static degree of indeterminacy of the beam.
MgO 0.1-3
9. Consider the following statements
Sulphur 1-3
1. The limit state of collapse is defined as the
Alkalies 0.2-1
acceptable limit for the stresses in the
materials. 12. The compacting factor test of cement concrete
2. Limit state method is one that ensures determines its
adequate safety of structure against collapse. (a) strength
3. In the limit state design method, actual (b) porosity
stresses developed at collapse differ (c) degree of compaction under loads
considerably from the theoretical values. (d) workability
Which of the above statements is/are correct? Ans. (d) : The compacting factor test of cement
(a) 1 and 2 (b) 1 and 3 concrete determines its workability.
(c) 2 and 3 (d) None Note : CF = 0.85 low workability
Ans. (a) : Limit state– It is the condition of a structure = 0.92 Medium workability
once achieved, the structure cannot perform its intended = 0.95 High workability
function satisfactory in future interms of safety or Note : Workability of concrete mix with low water
service ability. cement ratio is determined by compaction factor test.

OSSC JE Exam-2014 253 YCT

13. In a forced vortex 16. The flow rate between stream lines with values
(a) the fluid velocity is inversely proportional to ψ1 and ψ2 is given by
the radius (a) ψ1 + ψ2 (b) ψ1 + Cψ2
(b) the fluid rotates without any relative velocity (c) ψ2 – ψ1 (d) Cψ1 + ψ2.
(c) the rise depends on the specific weight Ans. (c) : Stream line function (ψ)
(d) the rise is proportional to the cube of angular ∂ψ −∂ψ
velocity v= , u=
∂x ∂y
Ans. (b) : Forced vortex flow – When a fluid is rotate If stream function satisfies Laplace equation then it is
about a vertical axis at constant speed, such that every case of irrotational flow. The flow rate b/w stream lines
particle has the same angular velocity motion is known with values ψ1 & ψ2 is
as the forced vortex.
dQ = ψ 2 − ψ1
ω2 r 2
v = rω or h =
2g 17. A path line describes
Forced vortex require constant supply of external (a) the velocity direction at all points on the line
energy / torque. (b) the path followed by particles in a flow
• In a forced vortex the fluid velocity is directly (c) the path over a period of times of a single
particle that has passed out at a point
proportional to the radius.
(d) the instantaneous position of all particles that
have passed a point.
Ans. (c) : Path lines– the path over a period of times of
a single particle that has passed out at a point. It is
based on Lagrangian concept.
• Two path line can intersect each other.
• In general, this is the curve in 3D space.

Exp. Rotating cylinder. Flow inside centrifugal pump.

14. In a differential manometer a head of 0.6 m of Path line
fluid A in limb 1 is found to balance a head of 18. Reynolds number signifies the ratio of
0.3 m of fluid B in limb 2. The ratio of specific (a) gravity forces to viscous forces
gravities of A to B is (b) inertial forces to viscous forces
(a) 2 (c) inertia forces to gravity forces
(b) 0.5 (d) buoyant forces to inertia forces.
(c) 0.18 Ans. (b) : Reynolds number is the ratio of inertia force
(d) cannot be determined to the viscous force.
Ans. (b) : s1h1 = s2h2 inertia force
Reynolds Number =
s1 h 2 0.3 viscous force
= = = 0.5
s 2 h1 0.6 ρvd
Re =
15. The centre of pressure of a rectangular plane µ
with height of liquid h m from base 19. The shear stress at the wall of a 16 cm dia pipe
(a) is h/2 m from bottom in laminar flow is 36 N/m2.
(b) is h/3 m from top The shear stress at a radius of 4 cm in N/m2 is
(c) is h/3 m from bottom (a) 9 (b) 18
(d) can be determined only if liquid specific (c) 6 (d) 72
weight is known. ∂p R
Ans. (b) : τ0 = .
∂x 2
Ans. (c) : The centre of pressure of a vertical plane τ R /2
= 1
01

immersed in water is at one third height from bottom. τ 02 R 2 / 2

τ01 R1
=
τ 02 R2
36 8 / 2
= ⇒ τ02 = 18N / m 2
τ02 4 / 2

OSSC JE Exam-2014 254 YCT

20. The Muller-Bresalu principal in structural 22. As per the Indian Standard soil classification
analysis is used for system, a sample of silty clay with liquid limit
(a) drawing influence line diagram for any force of 40% and plasticity index of 28% is classified
function as
(b) superimposition of load effects (a) CH (b) CI
(c) to write virtual work equation (c) CL (d) CL-ML
(d) None of these Ans. (b) : I P of soil I P = wL–wP
= 40 – 28 = 12%
IP A line = IP = 0.73 (wL–20)
Ans. (a) : Muller Breslau principle – = 0.73 (40–20)
Assumption– = 14.6%
(a) Material should obey Hooke's law. Note– IPsoil < IPAline, then soil lies C.I.
(b) Loads should be within elastic limit. 23. Two series of compaction tests were performed
Limitation– Muller Breslau principle gives qualitative in the laboratory on an inorganic clayey soil
ILD for internal stress components such as BM, shear employing two different levels of compaction
force only. . energy per unit volume of soil. With regard to
21. Road roughness is measured using the above tests, the following two statements
(a) Benkelman beam are made.
I. The optimum moisture content is expected to
(b) bump integrator
be more for the tests with higher energy.
(c) dynamic cone penetrometer
II. The maximum dry density is expected to be
(d) falling weight deflectometer more for the tests with higher energy.
The CORRECT option evaluating the above
Ans. (b) : Bump integrator– statements is
• The roughness of road is measurement by bump (a) Only I is TRUE
integrator. (b) Only II is TRUE
• Uneveness is a cumulative measure of vertical (c) Both I and II are TRUE
undulations of pavement per unit length of the road. (d) Neither I nor II is TRUE
• Uneveness is measure by using bump integrator Ans. (b) : (i) With higher energy of compaction
developed by CRRI. optimum moisture content decreases.
• For good pavement, cumulative undulation should (ii) The higher compactive energy results maximum dry
not be more than density.
150 cm/km length of road 24. Group I contains parameters and Group II
{B.I. = 630 [IRT]1.12 } methods/instruments.
IRT = International roughness Group I Group II
Unevenness (mm/km) Index P. Streamflow velocity 1. Anemometer
0-1500 Good Q. Evapo-transpiration rate 2. Penman’s method
1500-2500 Satisfactory R. Infiltration rate 3. Horton’s method
2500-3200 Bad-repairing work
S. Wind velocity 4. Current meter
>3200 uncomfortable
The CORRECT match of Group I with Group II is
Dynamic cone pentrometer– The dynamic cone
(a) P – 1, Q – 2, R – 3, S – 4
penetrometer (DCP) is used to determine underlying
soil strength by measuring the penetration of the device (b) P – 4, Q – 3, R – 2, S – 1
into the soil after each hammer below. (c) P – 4, Q – 2, R – 3, S – 1
Falling weight deflect meter– A falling weight (d) P – 1, Q – 3, R – 2, S – 4
deflectometer is a testing device used by civil engineers
the evaluate the physical properties of pavement in Ans. (c) : Stream flow velocity is measured by current
highways, local roads, airport pavements, harbor areas, meter.
railing tracks and elsewhere. • Evapo-transpiration rate is calculated with the help of
Benkelman beam– The Benkelman beam enables Penman's method.
precise and non-destructive measurements of the load • Infiltration rate is calculated with help of Horton's
bearing capacity of road surface layers mode of asphalt equation.
or pavement. • Anemometer is used for measuring wind velocity.

OSSC JE Exam-2014 255 YCT

25. In its natural condition, a soil sample has a Ans. (a) : Moment curvature relation–
mass of 1.980 kg and a volume of 0.001 m3. M f E
After being completely dried in an oven, the = =
I y R
mass of the sample is 1.800 kg. Specific gravity
is 2.7. Unit weight of water is 10 KN/m3. The 1 f
Curvature =
degree of saturation of the soil is: R E y
(a) 0.65 (b) 0.70 At fully plastic state, y = 0
(c) 0.54 (d) 0.61
Ans. (c) : Given,
M = 1.980 kg, Md = 1.800 kg
V = 0.001 m3
G = 2.70
γw = 10 kN/m3
S=?
1 f
Mass of water (Mw) = M – Md then, = =∞
= 1.980 – 1.800 = 0.18 kg R E×0
Mw 0.18 28. A steel plate is 300 mm wide and 10 mm thick.
Water content (w) = ×100 = × 100 = 10%
Md 1.800 A rivet of nominal diameter of 16 mm is driven
M 1.980kg into it, what is the net sectional area of the
γb = = plate
V 0.001m 3
(a) 2600 mm2
= 1980 kg / m = 19.8 kN / m
3 3

(b) 2760 mm2

γb Gγ w
γd = = (c) 2830 mm2
1+ w 1+ e
(d) 2840 mm2
Gγ w (1 + w )
(1 + e ) = Ans. (c) : Given,
γb
B = 300 mm
2.70 ×10 × (1 + 0.1) t = 10 mm
e = − 1 = 0.5
19.8 Gross dia, d = 16 mm = 16 + 1.5 = 17.5
Degree of saturation of soil– n=1
wG 0.1 × 2.7 So, net sectional area of the plate is given by
S= = = 0.54
e 0.5 A net = [ B − nd ] × t
26. In a diamond riveting for a plate of width ‘b’
= [300 − 1× 17.5] × 10
and rivet dia ‘d’ the efficiency of the joint is
(a) (b-d)/b (b) (b-2d)/b = 2825 2830 mm 2
(c) (b-d)/2b (d) (b-d)/d 29. Lateral ties in RC columns are provided to
Ans. (a) : resist
strength of joint
Efficiency of joint η = (a) bending moment
strenth of solid plate (b) shear

η=
( b - d ) t × permissible stress (c) buckling of longitudinal steel bars
bt × permissible stress (d) both bending moment and shear
b−d
η=
b Ans. (c) : The main functions of lateral reinforcement in
b = width of plate RC columns are–
d = gross dia of rivet • It prevent longitudinal reinforcement bars from
27. At a location of plastic hinge of a deformed buckling.
structure • It resist the shear force and hence contributes
(a) curvature is infinite avoiding shear failure.
(b) radius of curvature is infinite • It confines the concrete core to provide sufficient
(c) moment is infinite ductility or deformability.
(d) flexural stress is infinite • It restrains the spliced bars and hence prevent their slip

OSSC JE Exam-2014 256 YCT

 6
= 4 × 10−5 × 6 × = 8 × 10–5 m3/min
18
= 8 × 24 × 60 × 10–5 = 11.52 × 10–2 m3/day
33. Match List I(in situ test) with List II
(Measurement) and select the correct answer :
List I List II
A. SPT test 1. Penetration resistance
(N value)
B. Plate load test 2. Load settlement data
C. Field vaneshear test 3. Point resistance and
skin friction
30. Partial safety for concrete and steel are 1.5 and
D. CPT test 4. In situ shear strength
1.15 respectively, because
A B C D
(a) concrete is heterogeneous while steel is
homogeneous (a) 1 2 4 3
(b) 1 2 3 4
(b) the control on the quality of concrete is not as
good as that of steel (c) 2 1 3 4
(c) concrete is weak in tension (d) 2 1 4 3
(d) voids in concrete are 0.5% while those in
steel are 0.15% Ans. (a) :
In-situ test Measurement
SPT test Penetration resistance
Ans. (b) : The partially safety factor for concrete (1.5)
Plate load test Bearing capacity and settlement
is higher than that of steel (1.15) because of higher
Vane shear test In-situ shear strength of cohesive
variability associated with concrete i.e. quality control
soil
of concrete is not as good as that of steel.
CPT test Point resistance and skin friction
31. Concordant cable profile is
34. The method of orienting a plane table with two
(a) a cable profile that produces no support inaccessible points is known as
reactions due to pre-stressing
(a) Intersection
(b) a cable profile which is parabolic in nature (b) Resection
(c) a cable profile which produces no bending (c) Back sighting
moment at the supports of a beam
(d) two-point problem
(d) a cable profile laid corresponding to axial
stress diagram
Ans. (d) : Two point problem– Consist locating the
position of a plane table station on drawing sheet by
Ans. (a) : A cable profile that produces no support observation of two well defined point whose position
reactions, due to prestressing. The concordant cable have already been platted on the plane.
profile does not produce secondary moments and there
35. Reciprocating pumps are suitable for
will be no reactions at the support due to prestressing
(a) low discharge and high head
action.
(b) high discharge and low head
32. A flownet of a Coffer dam foundation has 6
(c) low discharge and low head
flow channels and 18 equipotential drops. The
(d) high discharge and high head
head of water lost during seepage is 6 m. If the
coefficient of permeability of foundation is
4×10−5m/min., then the seepage loss per m Ans. (a) Reciprocating pump are used mainly in village
length of dam will be areas and are suitable for high head with low volume of
(a) 2.16×10−2m3/ day discharge.
(b) 6.48 ×10−2 m3/ day • It is also used in oil gas industries petrochemical and
(c) 11.52× 10−2 m3/ day refinery industry.
(d) 34.56×10−2 m3/ day • The reciprocating pump is a positive displacement
pump as its sucks and raises the liquid by actually
Nf
Ans. (c) : q = kH. displacing it with a piston / plunger that executes a
Nd reciprocating motion in a closely fitting cylinder .

OSSC JE Exam-2014 257 YCT

1187. A 2% solution of a sewage sample is kept at an 81 − 39 − 7 − 3
incubation temperature of 20oC, if initial DO So, φ-index = = 8 mm / hr.
6−2
(dissolved oxygen) and final DO values after 5
39. The unit-hydrograph theory is based on the
days’ incubation period are 8.5 mg/l and 5.5
assumption of
mg /l respectively, then the BOD will be
(a) nonlinear response and time invariance
(a) 100 mg/l (b) 250 mg/l
(b) linear response and nonlinear time variance
(c) 150 mg/l (d) 350 mg/l (c) time invariance and linear response
Odisha SSC JE 2014 (d) nonlinear response and nonlinear time
Ans. (c) : Given, Dilution factor given in 2% variance
solution. Ans. (c) : Theory of unit hydrograph is based on–
DOi = 8.5 mg/l (i) Time invariance– As per this DRH for a given
DOf = 5.5 mg/l effective rainfall is always same in a catchment
BOD = (DOi–DOf) × dilution factor irrespective of the time when the rainfall or storm
100 takes place.
= (8.5–5.5) × (ii) Linear response– According to this any change in
2
input proportionately reflected in the output.
100
= 3× mg / l 40. For which one of the following purposes is the
2
double mass curve used?
BOD = 150 mg/l (a) Checking on the consistency of precipitation
37. The correct statement of comparison of records
ultimate BOD, COD. Theoretical oxygen (b) Prediction of annual precipitation
demand (ThOD) and 5-day BOD (BOD5) is (c) Defining which periods of storm should be
(a) BODu > COD > ThOD > BOD5 analysed to obtain the maximum useful
(b) COD > ThOD > BODu > BOD5 information from storm rainfall records
(c) ThOD > COD > BODu > BOD5 (d) For estimating the capacity of a reservoir
(d) COD > BODu > BOD5 > ThOD Ans. (a) : Inconsistency of record is corrected by using
Ans. (c) : Theoretical oxygen demand– double mass curve technique. Thus on correction the
• Theoretical method of oxygen demand. previous record becomes consistent with the present day
• If chemical formula of organic matter is known environmental and land use condition.
T.O.D. can be easily computed. M
Pcor = P × c
BOD– It is used for measurement of the quantity of Ma
oxygen required for oxidation of bio-degradable
Mc = Corrected slope of double mass curve.
organic matter present in water sample by aerobic
Ma = Original slope of double mass curve.
biochemical action.
M C
COD– It measure the content of organic matter of Correction factor - c =
waste water, both bio degradable and non-bio Ma S
degradable. 41. A 6 hour storm has 6 cm of rainfall and the
Generally – ThOD ≥ COD > BOD u > BOD5 > TOC resulting runoff was 3 cm. If o-index remains at
38. A 6-hour rainstorm with hourly intensities of 7, the same value, which one of the following is
the runoff due to 12 cm of rainfall in 9 hours in
18, 25, 17, 11 and 3 mm/hour produced a
the catchment?
runoff of 39 mm. Then, the φ-index is
(a) 4.5 cm (b) 6.0 cm
(a) 3 mm/hour (b) 7 mm/hour
(c) 7.5 cm (d) 9.0 cm
(c) 8 mm/hour (d) 10 mm/hour
Ans. (c) : For 6 hour storm–
Ans. (c) : Total rainfall = 7 + 18 + 25 + 17 + 11 + 3 =
Rainfall (P) = 6 cm
81 mm
Run off (R) = 3 cm
Total runoff = 39 mm
P −R 6−3 1
duration = 6 hour φ-index = = = cm / hr
t 6 2
P − R 81 − 39
φ-index = = = 7mm / hr. For 9-hour storm
t 6
Rainfall (P) = 12 cm
but rainfall intensity of 7 mm/hr. and 3 mm/hr will
1
totally infiltrate. φ-index = cm / hr
2
OSSC JE Exam-2014 258 YCT
 P−R 4. Limit of withdrawal from well without
φ − index = depletion of the aquifer
t
5. Water-bearing capacity of aquifer
1 12 − R
= ⇒ R = 7.5 cm / hr A B C D
2 9
(a) 4 3 2 5
42. Match List I (Hydrological Terms) with List II (b) 3 4 1 2
(Relationship/Nature of Curve), and select the
(c) 4 3 1 2
correct answer:
(d) 3 4 2 5
List I
Ans. (b) : Specific yield– Volume of water that can be
A. Theissen Polygon
removed by pumping which was extracted by force of
B. Mass Curve
gravity.
C. Hyetograph Safe yield– Max water which can be extracted from
D. DAD curve well without damaging the aquifer.
List II Specific capacity– Discharge per unit draw down at
1. Average depth of rainfall over an area well.
2. Relationship of rainfall intensity and time Field capacity and specific retention is same thing.
3. Relationship of accumulated rainfall and 44. The total observed runoff volume during a 4
time hour storm with a uniform intensity of 2.8
4. Relationship of river run-off and time cm/hr is 25.2×106 m3 from a basin of 280 km2
5. Always a falling curve area. What is the average infiltration rate for
A B C D the basin?
(a) 1 3 2 5 (a) 3.6 mm/hr (b) 4.8 mm/hr
(b) 1 5 3 2 (c) 5.2 mm/hr (d) 5.5 mm/hr
(c) 4 3 2 5 Ans. (d) : Total storm precipitation depth–
(d) 4 5 3 2 P = 2.8 × 10 × 4 = 112 mm
Total storm runoff depth =
25.2 ×106 ×103
Ans. (a) : Theissen polygon method is used to R= = 90 mm
calculated average depth of rainfall over an area. 280 ×106
• Mass curve is plot between accumulated rainfall with According to question –
time. t c = 4hr, l a = 0
• Hyetograph is a graph plot between rainfall intensity P − R − la
w index =
with time. t
• DAD (Depth area duration) curve express graphically 112 − 90 − 0
the relation between progressively decreasing = = 5.5 mm / hr.
4
average depth of rainfall over a progressively
45. For a certain loading condition, a saturated
increasing area from centre of storm outward to its
clay layer undergoes 40% consolidation ma
edges for a given duration of rainfall. period of 178 days. What would be the
43. Match List I (Well Hydraulics Parameters) additional time required for further 20%
with List II (Definition) and select the correct consolidation to occur?
answer (a) 89 days (b) 222.5 days
List I (c) 329.5 days (d) 400.5 days
A. Specific yield Ans. (b) : Give : t1 = 178 days
B. Safe yield Consolidation = 40% = U%
C. Specific capacity fore soil, under certain range of loading.
D. Field capacity Tv .H 2
List II C v =
t
1. Discharge per unit drawdown of well π
2. Same as specific retention ∴ Tv = (U%)2 {for U% ≤ 60%
4
3. Measure of water that can be removed by
pumping

OSSC JE Exam-2014 259 YCT

 π (0.4) 2 × H 2 π (0.6) 2 × H 2 48. The whole circle bearings of lines AB and BC
= are 30° 15' and 120° 30'. What is the included
4 178 4 t2
angle ABC between the lines AB and BC?
t2 = 400.5 days (a) 229° 45' (b) 89° 45'
∴ Additional time required t = t2 – t1 = 400.5 – 178
(c) 269° 45' (d) 90° 15'
t = 222.5 days
Ans. (b) :
46. Match List — I with List — II and select the
correct answer using the code given below the
Lists:
List—I (Equipment) List—II (Parameter)
A. Tintometer 1. Temperature
B. Nephelometer 2. Colour
C. Imhoff cone 3. Turbidity
D. Muffle furnace 4. Settleable solids F.B. of line AB = 30o15'
5. Volatile solids F.B. of line BC = 120o30'
B.B. of line AB = (180o + 30o15')
A B C D
= 210o15'
(a) 4 3 1 5
∠ABC = 30 15' + (1800–120030')
0
(b) 2 5 4 3
=30015' + 59030'
(c) 4 5 1 3
= 89o45'
(d) 2 3 4 5
49. What is the value of “off tracking” while a
Ans. (d) Tintometer is colour measuring instrument
vehicle is negotiating a curve of radius 40m
which compares colour to Nesler Tubes which contains
with a wheel base of 7.0 m?
solution of platinum cobalt dissolved in water.
(a) 0.75 m
• Nephelometer is based on the scattering principle
(b) 0.69 m
for measurement of turbidity. It uses formazin, a
chemical compound and is mostly used now a days. (c) 0.61 m
In Imhoff cone, settleable solids are measured. (d) 0.52 m
• Waste water is allowed to stand in it for two hours
and the quality of solids settled down is directly Ans. (c) : Given–
read out. Radius (R) = 40 m
47. What is the minimum length of a transition Wheel base length (l) = 7 m
curve for a design speed of 80 kmph in a
l2 72
horizontal curve of 240 m radius? Off tracking = = = 0.61m
(a) 32 m (b) 42 m 2R 2 × 40
(c) 52 m (d) 72 m 50. What is the super elevation for a horizontal
highway curve designed for a speed of 100
Ans. (d) : Rate of change of centrifugal acceleration
kmph and radius 500 m in mixed traffic
for driver comfort criteria–
conditions?
C=
80
75 + V
{0.5 ≤ C ≤ 0.80 m / sec3 } (a) 8.9% (b) 6.2%
(c) 0 (d) 7%
80
C= = 0.516 m / s3
75 + 80
Length of transition curve– Ans. (d) : Given –
0.0215V 3 0.0215 × 803 Design speed (v) = 100 kmph
Ls = = Radius (R) = 500 m
CR 0.516 × 240
Ls = 88.9 m ≈ 89 m For mixed traffic conditions,
(100 ) = 8.9%
2
For plain and rolling train v2
e= =
2.7V 2 2.7 × 802 225R 225 × 500
Ls = = = 72 m
R 240 So,
For hilly train e < 0.07 → provide
V 2 802 Hence, super-elevation for a horizontal highway curve
Ls = = = 26.66 m
R 240 → 7%

OSSC JE Exam-2014 260 YCT

51. Fresh sludge has moisture content of 99% and, Ans. (d) : Losses in pre-stress in pre-tensioning–
after thickening its moisture content is reduced (i) Loss due to elastic shortening of concrete.
to 96%. The reduction in volume of the sludge (ii) Loss due to creep of concrete.
is
(iii) Loss due to shrinkage of concrete.
(a) 3% (b) 5%
(iv) Loss due to relaxation of steel.
(c) 75% (d) 97.5%
54. Consider the following statements:
Ans. (c) : V1 (100 − P1 ) = V2 (100 − P2 )
Low percentage of C3S and high percentage of
V1 (100 − 99 ) = V2 (100 − 96 ) C2S in cement will result in
V1 = 4V2 1. Higher ultimate strength with less heat
V − V2 generation
% Reduction in volume = 1 ×100 2. Rapid - hardening
V1
3. Better resistance to chemical attack
4V2 − V2
= ×100 Which of the statements given above are
4V2 correct?
3V2 (a) 1 and 2 (b) 2 and 3
= ×100
4V2 (c) 1 and 3 (d) 1, 2 and 3
= 75 % Ans. (c) Low percentage of C3S and high percentage of
52. Self cleansing velocity is: C 2S in cement will result in

(a) the minimum velocity of flow required to • Higher ultimate strength with less heat generation.
maintain a certain amount of solids in the • Better resistance to chemical attack.
flow
55. Consider the following statements about the
(b) the maximum velocity of flow required to
characteristics of contours:
maintain a certain amount of solids in the
flow 1. Closed contour lines with higher values inside
(c) such flow velocity as would be sufficient to show a lake.
flush out any deposited solids in the sewer 2. Contour is an imaginary line joining points of
(d) such flow velocity as would be sufficient to equal elevations.
ensure that sewage does not remain in the 3. Closely spaced contours indicate steep slope.
sewer 4. Contour lines can cross each other in case an
Ans. (c) : Self cleansing velocity– Self cleansing overhanging cliff.
velocity is the minimum velocity at which no solid
Which of the statements given above are
gets deposited at the bottom of sewer.
correct?
• It is generally 0.45 – 0.6 m/sec.
(a) 2, 3 and 4 (b) 1 and 2 only
• The silting of sewers can be avoided by generating
success high velocity that wound not permit the (c) 1 and 4 (d) 1,2 and 3
solid to settle down i.e., the velocity should be such Ans. (a) : Characteristics of contours– The following
as to cause automatic self-cleaning effect known as characteristics features may be used while plotting or
self - cleaning velocity. reading a contour plan.
• The generation of such a minimum self - cleaning (i) Two contour lines of different elevations cannot
velocity in the sewer, at least one a day, is important cross each other.
because if contain deposition takes place and is not (ii) Contour line of different elevations can unite to
removed it will obstruct free flow, causing further
form one line only in the case of a vertical cliff.
deposition and finally leading to the complete
blocking of the sewer. (iii) Contour lines close together indicate steep slope.
They indicate a gentle slope if they are far apart. It they are
53. The losses in prestress in pre-tensioning system
are due to equally spaced uniform slope is indicated. A series of
1. elastic deformation of concrete when wires straight, parallel and equally spaced contours represent a
are tensioned successively plane surface.
2. friction (iv) A contour passing through any point is perpendicular to
3. shrinkage and creep of concrete the line of steepest slope at the point.
(v) A closed contour line with one or more higher ones
Select the correct answer using the codes
inside is represents a hill. Similarly, a closed contour
given below:
line with one or more lower ones inside it indicates a
(a) 1, 2 and 3 (b) 2 and 3
depression with but an outlet.
(c) 1 alone (d) 3 alone
OSSC JE Exam-2014 261 YCT
56. Match List — I with List — II and select the 58. Match List I with List II and select the correct
correct answer using the code given below the answer:
Lists: List I
List-I
A. Flow development
A. Adjustment of surveying instruments
B. Pipe network
B. Bowditch rule
C. Triangulation C. Water hammer
D. Bessel’s method D. Friction loss
List-II List II
1. Bringing the various fixed parts of the 1. Surge tank
instruments into proper relations with one 2. Entrance length
another
3. Darcy-Weisbach equation
2. Solution of three point problem
3. Measuring all the angles and base line 4. Hardy-Cross method
4. Balancing the latitudes and departures A B C D
A B C D (a) 2 4 1 3
(a) 1 2 3 4 (b) 4 2 3 1
(b) 3 4 1 2 (c) 2 4 3 1
(c) 1 4 3 2 (d) 4 2 1 3
(d) 3 2 1 4
Ans. (a) :
List I List II
Ans. (c) :
Adjustment of Bringing the various fixed ports Flow development Entrance length
surveying of the instrument into proper Pipe network Hardy-Cross method
instrument– relation with one another. Water hammer Surge tank
Bowdith rule Balancing the latitude and Friction loss Darcy-Weisbach equation
departure.
Triangulation Measuring all the angles and 59. Sewage sickness signifies:
baseline. (a) diseases caused by sewage
Bessels method Solution of three point problem. (b) soil pores getting dogged and preventing free
57. Match List I (Treatment units) with List II circulation of air when sewage is
(Types of processes) and select the correct continuously applied on land
answer: (c) raw sewage is applied and used for irrigating
List I vegetables which are eaten raw
A. Trickling filter (d) disposal of septic sewage on land
B. Activated sludge process
Ans. (b) : Sewage sickness– When untreated sewage is
C. Oxidation ditch
continuously applied on a piece of land, during course
D. Oxidation pond
of time the soil void gets clogged thereby free
List II
circulation of air is prevented and anaerobic condition
1. Symbiotic develop.
2. Extended aeration
• Then the land can not take any further sewage load
3. Suspended growth
and foul gases will evolve. This phenomenon is known
4. Attached growth
as 'sewage sickness'.
A B C D
(a) 3 4 2 1 60. Leaching is a process
(b) 4 3 1 2 (a) by which alkali salts present in the soil are
(c) 3 4 1 2 dissolved and drained away
(d) 4 3 2 1 (b) by which alkali salts in soil come up with
Ans. (d) water
List - I List - II (c) of draining excess water of irrigation
(Treatment units) (Types of processes) (d) which controls water logging
Trickling filter - Attached growth Ans. (b) : Leaching is the process in which land is
Activated sludge process - Suspended growth flooded with adequate depth of water. The alkali salts
Oxidation ditch - Extended aeration present in soil dissolved in this water. It will percolate
Oxidation pond - Symbiotic down or drained away by sub-surface drain.

OSSC JE Exam-2014 262 YCT

 Odisha Public Service Commission
(Civil Services), Exam- 2011
1. The dimension of linear momentum in MLT (c) A – 4, B – 5, C – 2, D – 1
system is: (d) A – 3, B – 4, C – 2, D – 5
(a) MLT (b) MLT–1 Ans. (d) : A. Coplanar forces – Lines of actins of all
(c) MT (d) LT–1 forces lies in the same plane.
Ans. (b) : Linear momentum = mass × velocity B. Concurrent forces – Lines of actions of all forces
=m×v pass through a common point.
–1
= [M] [LT ] C. Concurrent coplanar forces – Lines of actions of
–1 all forces lie in the same plane and pass through a
dimension of linear momentum = [MLT ]
common point.
2. The angles between two vectors to make their
Collinear forces – Lines of actions of all forces lie
resultant a minimum and maximum
along same line.
respectively, are
4. When a body is in equilibrium undergoes an
(a) 0º and 90º
infinitely small displacement, work imagined to
(b) 180º and 90º
be done, is knwon as:
(c) 90º and 180º
(a) Imaginary work
(d) 180º and 0º (b) Negative work
Ans. (d) : The resultant of two vector is minimum (c) Virtual work
where both vectors are equal and in opposite direction, (d) None of the above
i.e. the angle between the vector is 1800. Resultant is
maximum when vectors are parallel to each other i.e. Ans. (c) : Explanation virtual work – Virtual work is
angle between vectors is zero. the total work done by the applied forces and the
internal forces of a mechanical system as its waves
3. Match the list-I with List-II and selct the through a set of virtual displacement.
correct answer using the codes given below the
Negative work– Negative work is done when an object
lists:
moves in opposite direction of the direction of
List- I List-II application of forces.
A. Coplanar 1. Lines of action of all
5. On a ladder resisting on a smooth ground and
force forces lie in the same
leaning against a rough vertical wall, the force
plane, but do not pass
of friction acts:
through a common point
(a) Towards the wall at its upper end
B. Concurent 2. Lines of actions of all
(b) Away from the wall at its upper end
forces forces lie in the same
plane and pass through a (c) Upwards at its upper end
common point (d) Downwards at its lower end
C. Current 3. Lines of actions of all Ans. (c) : On a ladder resisting on a smooth ground and
Coplanar forces lie in the same leaning against a rough vertical wall the fore of friction
forces plane. acts upwards at its upper end.
D. Collinear 4. Lines of actions of all 6. The moment of inertia of a circular lamina of a
forces forces pass through a diameter, d, about an axis perpendicular to the
common point plane of the lamina and passing through its
5. Lines of actions of all centre, is:
forces lie along same
πd 4 πd 4
line. (a) (b)
12 16
Codes:
(a) A – 5, B – 4, C – 1, D – 3 πd 4 πd 4
(c) (d)
(b) A – 4, B – 3, C – 2, D – 1 32 64
OPSC (Civil Services) Exam-2011 263 YCT
Ans. (c) : The moment of inertia of a circular lamina of ev α = eu cos θ
a diameter d about an axis perpendicular to the plane of time taken by the ball of collide using wall
πd 4 t (u cos θ) = d ............... (1)
lamina and passing through its where is .
32 time taken by ball to come back to point
7. A particle is dropped from the top of a tower of 2t eu cos θ = d ............... (2)
60 m high and another is projected upwards from equation (1) and (2)
from the foot of the tower to meet the first t u cos θ = 2t eu cos θ
particle at a height of 15.9m. The velocity of the
1
second particle is: e = = 0.50
2
(a) 16 m/sec (b) 18 m/sec
coefficient of restitution = e
(c) 20 m/sec (d) 22 m/sec
9. The property of a material by which it can be
Ans. (c) : Distance traveled by 1st particle when they
drawn to a smaller section due to tension is
meet = 60 – 15.9
called:
= 44.1 m
(a) Plasticity
1
By equation S = ut + gt 2 (b) Ductility
2 (c) Elasticity
for 1st particle
(d) Malleability
1
44.1 = 0 × t + × 9.8 × t 2 Ans. (b) Ductility : The property of a material by
2 which it ca be drawn to a smaller section due to tension
t = 2.99 ≃ 3 sec. is called plasticity.
For section particle Plasticity : Plasticity is the ability of a material to
1 undergo permanent deformation under stress without
S = ut + gt 2
2 cracking.
Elasticity : The ability of an object or material to
15.9 = U × 3 + ( −9.8 × 32 )
1
2 resume its normal shape after being stretched or
U = 20 m/sec compressed.
Malleability : The state of being malleable or capable
8. A glass ball is shot to hit a wall from a point on
of being shaped as by hammering or pressing.
a smooth floor. If the ball returns back to the
point of the projection in twice the time taken 10. A solid cube is subjected to equal normal forces
in reaching the wall, the coefficient of on all its faces. The volumetric strain will be x-
restitution between the glass and the wall is: times the linear strain in any of the three axes
when:
(a) 0.25 (b) 0.33
(a) x = 1
(c) 0.40 (d) 0.50
(b) x = 2
Ans. (d) : (c) x = 3
(d) x = 4
dv σ  2 
= 3  1 − 
Ans. (c) :
v  E  m 
It is 3 times because cube is subjected to 3 actually
perpendicular stresses.
11. If a composite bar of steel and copper bar is
heated, then the copper bar is subjected to :
(a) Compression
(b) Tension
(c) Shear
Velocity along horizontal does not change in case of (d) Torsion
projectile motion. Ans. (a) : When composite bar of steel and copper is
µ cos θ = V cos α heated then copper bar is in compression and steel bar is
for collision with wall in tension.

OPSC (Civil Services) Exam-2011 264 YCT

12. Match the List-I with List-II and select the 13. Consider the following statement with
correct answer by using codes given below the reference to a continuous beam supported at
lists : A, C and E for which shear force diagram is
List-I List-II shown in fig. below
(Simply supported Bending moment There is :
beams) diagrams 1. A concentrated load acting at point B
A 1 2. A concentrated load acting at point B
3. A uniformly distributed load acting in the
portion AC
4. A uniformly distributed load acting in the
B 2 portion CE

C 3
(a) 1 and 2 are correct
(b) 2 and 3 are correct
(c) 3 and 4 are correct
D 4 (d) 1 and 4 are correct
Ans. (d) :

A B C D
(a) 4 1 2 3
(b) 3 4 1 2
(c) 1 2 3 4
(d) 2 3 4 1 14. Simple bendiung equation is :
Ans. (b) M R σ M E σ
(a) = = (b) = =
(Simply supported Bending moment diagrams I E y I R y
beams) I E y M R y
(c) = = (d) = =
M R σ I E σ
M E σ
Ans. (b) : Bending equation = = =
I R y
15. Match the List-I with List-II and select the
correct answer using the codes given below the
lists :
List-I List-II
(Members) (Strain
energy)
A. Axial loaded members l V
2
K∫ dx
0 2GA

B. Members under bending l T

2

∫0 2GJ dx
C. Members under shearing l M
2

∫0 2EI dx
D. Circular members under l P
2

section ∫0 2EA dx
OPSC (Civil Services) Exam-2011 265 YCT
 Codes ; C. Interior 3. Fixed support
A B C D support
(a) 1 4 2 3 D. Hinge 4. Free end
(b) 2 1 3 4
Codes :
(c) 3 2 4 1
A B C D
(d) 4 3 1 2
(a) 1 2 3 4
Ans. (d) :
(b) 2 3 4 1
(Members) (Strain
(c) 3 4 1 2
energy)
(d) 4 3 2 1
A. Axial loaded members P2
l
∫0 2EA
dx Ans. (d) :
Real Beam Conjugate beam
B. Members under bending M2
l
∫0 2EI dx (support conditions) (support conditions)
A Fixed support Free End
C. Members under shearing V2 l
K∫ dx B Free end Fixed support
0 2GA
C Interior support Hinge
D. Circular members under T2
l
D Hinge Interior support
section ∫
0 2GJ
dx
18. The ratio of the torsional resistance of a solid
16. A cantilever carries a uniformly distributed circular shaft of diameter D and a hollow
downward load W over its whole length and a circular shaft of external diameter d with same
concentrated upward load W at its free end. material is :
The net deflection of the free end will be :
(a) Zero D4 D3
(a) (b)
D4 − d 4 D3 − d3
5 WL3
(b) upward D4 − d 4 D3 − d3
24 EI (c) (d)
4
5 WL3 D D3
(c) downward Ans. (a) : The polar modulus for a solid circular shaft
24 EI
(d) None of the above πD3
z1 =
Ans. (b) : 16
The vertical deflection at end B due to point load W is Polar modulus for a hollow shaft

∆1 =
WL3
( Upwords ) π ( D4 − d 4 )
3EI z2 =
16D
The vertical deflection at end B due to the total load W
over its whole length is– z1 πD 3
16D D4
= × =
WL3 z2 16 π ( D 4 − d 4 ) D 4 − d 4
∆2 = (Downwards)
8EI 19. If p is internal pressure in a thin cylinder of
Net deflection = ∆1 − ∆ 2 diameter d and thickness t, the developed loop
WL3 WL3 stress is :
= − pd pd
3EI 8EI (a) (b)
5 WL3 t 2t
= (Upwards) pd pd
24 EI (c) (d)
17. Match the List-I with List-II and select the 3t 4t
correct answer using the codes given below the Ans. (b)
lists :
Real Beam Corresponding conjugate
(support beam (support
conditions) conditions)
A. Fixed 1. Interior support
support
B. Free end 2. Hinge
OPSC (Civil Services) Exam-2011 266 YCT
Hoop stress : Hoop stress is the stress induced in the 21. iF 2j > m + r in a plane truss with number of
hallow cylinder or hollow sphere which acts points 'j' number of members 'm' and number
tangentially to the circumference of the cylinder and of reactions 'r' the truss is :
sphere in direction due to the action internal fluid (a) Redundant
pressure. It is tensile in nature. (b) Determinate
P.d
• Hoop or circumferential stress ( σ h ) = (c) Stable
2t
(d) Unstable
P.d
• Longitudinal stress ( σ L ) = Ans. (d) : Static indeterminacy DS = M + γe – 2j
4t
For plane truss–
20. Match List-I with List-II and select the correct
answer using the codes given the lists : If, m + r = 2J → Perfect stable truss
List -I List- II m + r < 2J → Deficient or unstable truss
(Loaded members) (Maximum m + r > 2J → Redundant or indeterminate truss
deflection) Where,
A 1. WL3 m = total number of member
8EI r = support reaction
J = total joint
B 2. WL3 22. A moment M is applied at the propped end of a
48EI propped cantilever beam of span l and flexural
rigidity EI. The moment at fixed end is
(a) 2 M
C 3. WL3 (b) M
3EI (c) M/2
(d) M/3
D 4. 5WL3 Ans. (c) :
384EI

Codes : IF a moment is applied at propped end then M/2

A B C D moment is transferred at fixed end.
(a) 3 1 2 4
23. A portal frame is shown in fig. below. The
(b) 1 2 3 4
deflection shape of the frame for constant EI is
(c) 2 3 4 1
as per :
(d) 4 2 3 1
Ans. (a) :
Loaded members Maximum deflection
WL3
3EI

WL3 (a) (c)

8EI

WL3
48EI

(b) (d)
5WL3
384EI

OPSC (Civil Services) Exam-2011 267 YCT

Ans. (a) : Ans. (b) : In a two hinged stiffening girder of span 'l'
the maximum positive or negative moment due to a
Wl
moving concentrated load W is
8
26. In Kani's method the iteration will converge if
there is some mistake in calculation of
moments
(a) True in some cases
24. The three hinged arch shown in fig. below will (b) False
have the value of H as : (c) True
(d) True for beams only
Ans. (c) Kani's Method : This methjod was introduced
by Gasper Kani's in 1940. It involves distributing the
unknown fixed end moments of structural members to
adjacent joints, in order to satisfy the conditions of
continuity of slops and displacement. Convergence is
generally fast.
(a) 20 kN 27. The structure coordinates are as indicated on
a-simply supported beam shown in fig. below.
(b) 30 kN
The flexibility matrix is :
(c) 40 kN
(d) 50 kN
Ans. (c) :

 L3 L2 
 
(a) 12EI
2
2EI 
 L L 
 2EI 12EI 

Taking moment about point A  L3 L2 

 
VB × 16 – 80 × 4 = 0 (b) 12EI
2
2EI 
VB = 20 kN  L L 
 2EI 12EI 
VA + VB = 80
VA = 80–20 = 60 kN  L3 L2 
 
Taking moment about point C (c) 12EI
2
2EI 
 L L 
VA × 8 – H × 4 – 80 × 4 = 0
 2EI 12EI 
60 × 8 – H × 4 – 80 × 4 = 0
H = 40 KN  L3 L2 
 
25. In a two hinged stiffening girder of span 'l' the (d) 12EI
2
2EI 
maximum positive or negative moment due to a  L L 
moving concentrated load W is :  2EI 12EI 
Wl  L3 L2 
(a)  
4 Ans. (d) : Flexibility matrix = 12EI
L2
2EI 
L
(b)
Wl  
8  2EI 12EI 
Wl
(c)
16
Wl
(d)
24

OPSC (Civil Services) Exam-2011 268 YCT

28. The influence line for structural function is End condition Effective Buckling
used for obtaining the maximum value due to : length (Le) load (Pb)
(a) Single point load only Both ends Le = L π2 EI
(b) Uniformaly distributed load only hinged Pb =
L2
(c) Several point loads Both ends fixed
Le =
L 4π2 EI
(d) All of the above Pb =
2 L2
Ans. (d) The influence line diagram is drawn for One end fixed Le = L π2 EI
maximum value due to structural element is – and another end Pb =
4L2
• Single point load is free
• Uniformly distributed load One end fixed L 2π2 EI
Le = P =
• Several points load and another is 2 b
L2
hinged
29. In the truss shown fig. below the force in
Hence, The maximum buckling load will be in both
member CD is :
end fixed conditions.
4π2 EI
Pb =
L2
31. If the volume of solids is equal to the volume of
voids in a soil mass, then the values of porosity
and void ratio are respectively :
(a) 0 and 0.5 (b) 0 and 1.0
(c) 0.5 and 1.0 (d) 1 and 0.5
(a) 100 kN compressive Ans. (d) : Volume of solids = Volume of voids in soil
(b) 100 kN tensile mass VS = VV
(c) Zero V V
Void Ratio (e) = V = V = 1
(d) Intermediate VS VV
Ans. (c) VV VV V
Porosity (η) = = = v = 0.5
V VV + Vs 2Vv
32. The soil transported by wind is called :
(a) Aeolian soil
(b) Talus
(c) Alluvial soil
(d) Lacustrine soil
Ans. (a) :
In the truss shown fig. below the force in member CD is • Soil transported by wind – Aeoline soil
Zero.
• Soil formed by gravity forces – Talus
30. Column of given length, cross-section and • Soil deposited by surface water,
material property has different values of in floodplains and deltas – Alluvial soil
buckling loads for different and condition. The • Soil formed in lakes – Lacustrine soil
strongest column is one whose :
33. The critical hydraulic gradient ic of a soil mass
(a) One end if fixed and the other end is hinged of specific gravity G and void ratio e is given
(b) Both the ends are hinged by:
(c) One end is fixed and the other end is entirely (a) ic = G+1
free G −1
(b) ic =
(d) Both the ends are fixed 1+ e
Ans. (d) : For same length, cross-section and G +1
(c) ic =
material property both ends are fixed is strongest 1+ e
column because buckling load according to Euler's G −1
formula is maximum for both end fixed condition. (d) ic =
1− e
OPSC (Civil Services) Exam-2011 269 YCT
 G −1 1 − sin φ 1 + sin φ
Ans. (b) : Critical hydraulic gradient iC = Ka = , Kp = , Ka = 1 – sin φ
1+ e 1 + sin φ 1 − sin φ
G −1 Where,
iC = = (G–1) (1+η)
1+ e Ka = Active earth pressure coefficient
G = specific gravity Kp = Passive earth pressure coefficient
e = void ratio K0 = At rest earth pressure coefficient
η = porosity 37. The slope of the e-log p curve for a soil sample
ic = critical hydraulic gradient gives :
(a) Coefficient of permeability, k
34. The ratio of unconfined compressive strength
(b) Compression index, cc
of undisturbed soil to the unconfined
(c) Coefficient of consolidation, cv
compressive strength of the soil in remoulded
state, is called : (d) Coefficient of volume compressibility, mv
(a) Relative strength Ans. (b) :
(b) Sensitivity
(c) Thixotrophy
(d) Strength reduction factor
Ans. (b) :
unconfined compressive
strength of undisturbed soil
Sensitivity =
uncompressive compressive
strength of remoulded soil
Thixotropy – If a soil remoulded and left undisturbed
at a same water content for some time. It may regain
∆e
part of its lost strength. This gain of strength in the soil Compression Index CC =
with the passage of time after remoulding is called ∆ log p
thixotropy. 38. Coarse grained soils are best compacted by a :
• Strength reduction factor : It is defined as the ratio (a) Drum roller
of elastic strength to yield strength. (b) Rubber tyred roller
35. Cohesive soil are : (c) Sheep's foot roller
(a) Good for backfill because of low lateral (d) Vibratory roller
pressure Ans. (d) : Vibratory roller – Coarse grained soil
(b) Poor for backfill because of largte lateral Sheep foot roller – Cohesive soil
pressure Rubber tyred roller –Cohesive + cohesionless both
(c) Good for backfill because of high shear 39. If the time required for 50% consolidation of a
strength remoulded sample of clay with single drainage
(d) Ideal for backfill soil is t, then the time required to consolidate the
Ans. (b) : Cohesive soils are poor for back-fill because same sample of clay with same degree of
of large lateral pressure. consolidation but with double drainage
(a) t/4
36. Coefficient of earth pressure at rest is :
(b) t/2
(a) Less than the active earth pressure but greater
than the passive earth pressure (c) 2t
(b) Greater than the active earth pressure but less (d) 4t
than the passive earth pressure Ans. (a) : 50% consolidation U = 50%
(c) Greater than both the active and passive earth time for consolidation single drainage = t
pressure π 2 Cv t
(d) Less than both the active and passive earth Times factor TV = 4 v = d 2
pressure 2
π 1 π
Ans. (b) : Coefficient of earth pressure at rest greater = ×   = =
than active earth pressure but less than the passive earth 4  2  16
pressure. For double drainage

OPSC (Civil Services) Exam-2011 270 YCT

 CvT Ans. (c) In compaction test with increases in
TV = 2 compactive effort maximum dry density increases but
d
  optimum moisture content decrease.
2
43. The useful method of finding the shear strength
Cv t CvT
= of soft sensitive clayey soils is by means of :
(d / 2)
2 2
d (a) Cone penetration test
t (b) SPT test
T= (c) Vane shear test
4
(d) Tensional shear test
40. The ultimate bearing capacity of a purely
cohesive soil at ground surface for a based Ans. (c) : Vane shear test is used to determine the
footing as given by Terzaghi, is (cu = undrained undrained shear strength of soils especially soft
strength of the soil, Df = depth of foundation clays.
and γ = unit weight of soil). Field test Penetration to determine
(a) 5.7 cu Plate load test Bearing capacity of soil
(b) 5.14 cu Settlement of footing
(c) 5.14 cu + γDf Modulus of subgrade
(d) 5.7 cu + γDf Cone penetration Penetration resistance of soil
Ans. (a) : test strata
1 Standard Penetration resistance of soil
Ultimate bearing capacity = 5.7 cu + γ DFNq + NγBγ
2 penetration test strata
At ground DF = 0 Vane shear test Undrained shear strength of soft
purely cohesive Nγ = 0 sensitive clays
qu = 5.7 cu Proctor test Maximum dry density of soil
41. Taylor's stability number is given by : along with the optimum moisture
Fc content and amount of
(a)
cγH compaction.
γH
(b) 44. Rise of water table up to the ground surface for
cFc
a cohesionless soil reduces the net ultimate
c bearing capacity approximately by :
(c)
Fc γH (a) 25%
H (b) 50%
(d) (c) 33.3%
cFc γ
(d) 66.67%
Ans. (c) : Taylor's stability number
Ans. (b) : Rise of water table upto ground surface for
C cohesionless soil reduces the net ultimate bearing
Sη =
f c γH capacity approximately by : 50%
C = cohesion 45. The value of bearing capacity factor Nc for
fc = factor of safety with respect to cohesion. piles as per Meyerhof is taken as :
H = vertical height of the slope. (a) 5.14
• It is the method used to evaluate slope stability for (b) 5.70
homogenous soils having cohesion. (c) 9.0
42. In a compaction test, with increase in (d) 6.2
compactive effort : Ans. (a) : Value of bearing capacity factor NC for piles
(a) Both maximum dry density and optimum as per meyorhof is 5.14.
moisture content increases
46. A dry sand specimen was failed at a deviator
(b) Both maximum dry density and optimum
stress of 100 kPa in a tri-axial test when the all
moisture content decreases
round pressure was 50 kPa. The angle of
(c) Maximum dry density increases but optimum
internal friction for the sand specimen is :
moisture content decreases
(a) 150 (b) 350
(d) Maximum dry density decreases but optimum 0
moisture content increases (c) 30 (d) 330
OPSC (Civil Services) Exam-2011 271 YCT
Ans. (c): (c) abs.pr.= atm. pr. + gauge pr.
Deviator stress σd = 100 kPa (d) atm. pr.=abs. pr.+ gauge pr.
Cell pressure σ0 = 50 kPa Ans. (c) : abs.pr.= atm. pr. + gauge pr.
Normal stress = σ1 = 100 + 50 = 150 kPa
 φ
σ1 = σ3 tan2  45 + 
 2
 φ
σ1 = 50 tan2  45 + 
 2
φ = 300
47. Contain liquid has a specific mass 1260 kg/m3
at 40C. Its specific weight is :
(a) 1.236 × 104 N/m3
(b) 1.236 × 103 N/m3
(c) 1.236 × 103 N/m3 51. The centre of pressure on a plane surface
(d) 1.236 × 104 N/m3 which is immersed in a liquid is :
m (a) Above centre of gravity
Ans. (a) : Specific mass = = 1260 kg/m3
v (b) At the same point
m (c) Can be below or above the centre of gravity
Specific weight = ×g
v (d) Always below the centre of gravity
= ρg Ans. (d) : When a plane surface immersed in a liquid
= 1260 × 9.81 then confer of pressure is always below the centre of
gravity.
= 1.236 × 104 N/m3
52. The metacentric height of an object is related
48. The relation between dynamic viscosity (µ)
to period of oscillation by the relation :
mass density (ρ) and kinematic viscosity (υ) is
expressed as : K2 2π.K 2
(a) T = 2π (b) T =
ρ g.GM g.GM
(a) υ = (b) ρ = µ.υ
µ K 2 .g 2πK 2 .g
(c) T = 2π (d) T =
µ υ GM GM
(c) υ = (d) υ =
ρ ρ Ans. (a) : Periods of oscillation in metacentric height.
Ans. (c)
K2
dynamic vis cosity ( µ ) T = 2π
kinematic viscosity (v) = gGM
density ( ρ )
K = least radius of gyration
49. Surface tension on a liquid surface is caused GM = metacentric height
by: 53. Stability of a completely submerged body is
(a) Adhesion normally determined by considering :
(b) Cohesion (a) Centre of gravity and metacentre
(c) Adhesion and Cohesion (b) Centre of buoyancy and centre of gravity
(d) Capillarity (c) Centre of buoyancy and metacentre
Ans. (b) : Surface tension on a liquid surface is caused (d) Centre of gravity, metacentre and centre of
by cohesion. buoyancy
• Cohesion refers to clinging of like molecules and Ans. (b) : When a body fully submerged in water then
Adhesion refers to clinging of unlike molecules. buoyancy and gravity force is worked at that body.
50. The relationship between atmospheric Stability of body determined by center of buoyancy and
pressure, absolute pressure and gauge pressure centre of gravity.
is given as : 54. The acceleration in x direction of a flow field is
(a) abs, pr = atm pr. + gauge pr. expressed as ( u = vel. in x direction, v = vel. in
(b) abs. pr. = atm.pr.–gauge pr. y direction, w= vel. in z direction) :
OPSC (Civil Services) Exam-2011 272 YCT
  ∂u ∂u ∂u  ∂u 58. Condition of irrotational flow is :
(a) a x =  u +v +w +
 ∂x ∂x ∂x  ∂t (a) When φ (vel. potential) exists
 ∂u ∂u ∂u  ∂u (b) When φ (velocity potential) and ψ (stream
(b) a x =  u +v −w − function) exists
 ∂x ∂x ∂x  ∂t
(c) When φ (vel. potential) and ψ (stream
 ∂u ∂u ∂u  ∂u
(c) a x =  u +v +w − function) donot exist
 ∂x ∂x ∂x  ∂t (d) None of the above
 ∂u ∂u ∂u  ∂u Ans. (a) : Condition of irrigational flow
(d) a x =  u −v −w +
 ∂x ∂x ∂x  ∂t • When φ (vel. potential) exists.
Ans. (a) : Acceleration in x direction of a flow field • Angular velocity along x, y, z is zero and velocity of
 ∂u ∂u ∂u  ∂u flow is also zero is value.
ax =  u + v + w  +
 ∂x ∂x ∂x  ∂t 59. HELE SHAW model is used for drawing :
u = velocity in x direction (a) Streak lines
v = velocity in y direction (b) Stream lines
w = velocity in z direction (c) Equipotential lines
(d) Stream lines and equipotential lines
55. The continuity equation in three dimensional
flow field is expressed as : Ans. (d) : Hele shaw model is used to drawing stream
lines and equipotential lines.
∂u ∂v ∂w
(a) − − =0 60. An example of steady non-uniform flow is :
∂x ∂y ∂z
(a) Flow through a tapering of increasing or
∂u ∂v ∂w decreasing diameter at a constant rate
(b) ± ± =0
∂x ∂y ∂z (b) Flow through a tapering pipe of increasing or
∂u ∂v ∂w decreasing diameter at a varying rate
(c) ∓ ∓ =0
∂x ∂y ∂z (c) Flow through a constant diameter pipe at a
constant rate
∂u ∂v ∂w
(d) + + =0 (d) Flow through a constant diameter pipe at a
∂x ∂y ∂z varying rate
Ans. (d) : The continuity equation in three dimensional Ans. (a) : Example of steady non-uniform flow through
∂u ∂v ∂w a tapering pipe of increasing or decreasing diameter at a
flow field is + + =0 constant rate.
∂x ∂y ∂z
Steady non uniform flow– Conditions change from
56. The difference between values of two stream - point to point in the stream but do not change with time.
lines or stream - functions will give : 61. Integration of EULER's equation gives :
(a) Shear stress (a) Continuity equation
(b) Pressure (b) Bernoulli's equation
(c) Velocity (c) Continuity equation and Bernoulli's equation
(d) Volumetric flow rate (d) Momentum equation
Ans. (d) : The difference between the stream function Ans. (b) : The Euler's equation for steady flow of an
values at any two points gives the volumetric flow rate ideal fluid along a stream line is a relation between the
or volumetric flux through a line connecting the two velocity, pressure and density of a moving fluid. It is
points. based on the Newton second law of motion. The
57. Flow net represents a group of : integration of equation gives bernoullis equation in the
(a) Stream lines and streak lines form of energy per unit weight.
(b) Stream lines and equipotential lines 62. Practical equation of Bernoulli's equation :
(c) Equipotential lines and path lines (a) Venturimeter (b) Orificemeter
(c) Pitot tube (d) All of the above
(d) Stream lines and path lines
Ans. (d) : Application of Bernoulli's equation–
Ans. (b) : A flow net is a graphical representation of
(1) Venturimeter
two dimensional steady state ground water flow through
(2) Orificemeter
aquifers. It is a curvilinear net formed by the
combination of stream lines and equipment lines. (3) Pitot tube

OPSC (Civil Services) Exam-2011 273 YCT

63. The kinetic energy correction factor is defined 67. The flow is said to be laminar if Reynold's
as : number (Re) in a pipe is :
K.E per second based on actual velocity (a) Re < 2000 (b) Re > 2000
(a)
K.E per second based on average velocity (c) Re < 4000 (d) Re > 4000
Ans. (a) :
K.E per second based on actual velocity
(b) Reynold number in pipe flow Flow
K.E per second based on average velocity
Re < 2000 Laminar flow
K.E per second based on actual velocity 2000 < Re < 4000 transition flow
(c)
K.E per second based on average velocity Re > 4000 turbulent flow
K.E per second based on actual velocity 68. The phenomenon of water hammer is
(d) associated with :
K.E per second based on average velocity
(a) Pipes
Ans. (a) : Kinetic energy correction factor = (b) Pipes and channels
K.E per second based on actual velocity (c) Channels
K.E per second based on average velocity (d) None of the above
Its value for fully developed laminar flow is around 2 Ans. (a) : Water hammer is induced in pipe flow with
where as for a turbulent pipe flow it is between 1.04 to the closure of gate valve.
1.11 It is usually to tape it is 1 for a turbulent flow. 69. Moving hydraulic jump with a wave front
64. Dimensions of Torque are : moving upstream or downstream has :
(a) ML2T2 (b) ML2T–3 (a) Negative surge
2 –2 2 3
(c) ML T (d) ML T (b) Positive surge
Ans. (c) (c) No surge
Torque (T) = Moment of Inertia x angular acceleration (d) Negative and positives surge
0 0 –2 Ans. (b) : A positive surge results from a sudden
dimension of angular acceleration = [M L T ]
dimension of moment of Inertia = [ML2T0] change in flow that increases the depth. It is an abrupt
dimension of torque = [ML T ] 2 –2 wave front. when surge is of tidal origin it is usually
fermed a tidal bare. Moving hydraulic jump with a
65. If velocity distribution in a boundary layer is wave front moving upstream or downstream has
u y
given by = , where v is velocity at a positive surge.
v δ 70. For a most economical triangular shaped
distance y from plate and u = v at y = δ. The channel, the hydraulic radius is :
displacement thickness δ* is : y y
−δ δ (a) (b)
(a) (b) 2 2 2
2 2
(c) 2y (d) 2 2y
δ δ
(c) (d) Ans. (a) : Most economical triangular shaped channel
4 4
y
Ans. (b) : Velocity distribution in a boundary layer is the hydraulic radius is
2 2
u y
= 71. If discharge per unit width in a rectangular
v δ
displacement thickness channel is 1.8 m3/sec/m and depth of flow is 0.8
m, calculate critical depth which is equal to :
66. MAGNUS effect which is produced by a
(a) 0.80 m (b) 0.69 m
spinning cylinder is associated with :
(c) 0.90 m (d) 1.0 m
(a) Rotation
Ans. (b) : discharge per unit width (q) = 1.8 m3/s/m
(b) Drag
1/ 3
(c) Lift  q2 
Critical depth Hc =  
(d) Drag and Lift  g 
Ans. (c) : Magnus effect : The magnus effect is the lift 1/ 3
 1.82 
produced by a rotating cylinder in a uniform stream. It =  
can be predicted using the appropriate potential flow  9.81 
solution for a rafting cylinder in a uniform stream. Hc = 0.69 m

OPSC (Civil Services) Exam-2011 274 YCT

72. Loss of energy in a hydraulic jump is given by Ans. (c) : In suppressed weir crest length is equal to
(y1 = prejump depth, y2= post-jump depth) : width of channel.
3
(a) (y2+y1) /c1y1y2 Equation for standard suppressed rectangular weir with
(b) (y2+y1)3/y1y2 full bottom contraction.
(c) (y2+y1)3/c1y1y2 Q = 3.33 LH13 / 2
(d) (y2+y1)3/y1y2 77. Coincident draft in relation to water demand is
Ans. (c) : Loss of energy in hydraulic jump is given based on:
( y − y1 ) (a) Peak hourly demand
3

∆E L = 2 (b) Maximum daily demand

C1 y1 y 2
(c) Maximum daily + free demand
Where C1 is usually taken 4. head loss due to violent (d) Greater of (a) and (c)
turbulent mixing and description that occur within the
Ans. (d) : Coincident draft is the demand of water
jump itself.
maximum of peak hourly demand or minimum daily
73. In critical flow condition, the specific energy is: demand and fire demand.
(a) Minimum 78. If the specific capacity of a well is 1.166
(b) Maximum litres/sec, then the discharge from this well
(c) Zero under a depression head of 3 m head of 3 m
(d) Maximum or Minimum will be :
Ans. (a) : Critical flow condition is the state of flow at (a) 1.166 l/s (b) 3.5 l/s
which the specific energy is a minimum for a given (c) 10.5 l/s (d) None of the above
discharge. When the depth of flow is greater is smaller Ans. (b) : Specific capacity of well = 1.166 l/s/m
than the critical velocity for given discharge. Depression head = 3 m
74. If the slope of free water surface in a gradually So,
varied flow is zero, it means that the : discharge, Q = specific capacity × depression head
(a) Free water surface is parallel to bed of Q = 1.166 × 3
channel Q = 3.498 l/s = 3.5 l/s
(b) Free water surface is not parallel to bed of 79. The measure of the amount to which light is
channel absorbed or scattered by the suspended
(c) Free water surface is inclined to bed to material in water is called :
channel (a) Opacity (b) Turbidity
(d) Free water surface is parallel and inclined to (c) Diffraction (d) None of the above
bed of channel Ans. (b) : The measure of the amount to which light is
Ans. (a) When slope of the surface water in a gradually absorbed or scattered by the suspended material in
varied flow is zero, it means free water surface is water is called turbidity.
parallel to bed of channel. Turbidity is the cloudiness or haziness of a fluid caused
75. In Manning's equation, the Manning's constant by large numbers of individual particles that are
depends on : generally invisible to the naked eye, similar to smoke in
(a) Width of channel air.
(b) Length of channel 80. A blow off valve is provided in water
(c) Slope of channel distribution system at :
(d) Type of surface of channel (a) Low points
(b) High points
Ans. (d) : The mannings coefficient after denoted as n,
is an evpirically derived coefficient which is deponent (c) Junction points
on many factors, including surface roughness and (d) All of the above
sinousity. Ans. (a) : Blow off valve or scour valve or drain
76. Suppressed weir has : valve–
(a) Crest length less than width of channel • These valves are also known as wash out valves.
(b) Crest length more than width of channel • To remove the entire water from pipe after closing the
supply, small gate valves are provided of low point.
(c) Crest length equal to width of channel
• These valves are necessary at low level points for
(d) Crest length can be less or more than width of
completelly emptying the pipe for inspection, repair etc.
channel
OPSC (Civil Services) Exam-2011 275 YCT
81. The maximum allowable concentration of iron Ans. (d) : In a city the roads radiating from centre then
in water is : most suitable layout will be radial system.
(a) 1.0 ppm (b) 0.05 ppm • Dead end system – Suitable for old turns and cities
(c) 0.3 ppm (d) 0.03 ppm having no definite pattern of road.
Ans. (c) : Maximum allowable concentration of iron in • Grid system – Suitable for cities with rectangular
water is 0.3 ppm. layout.
82. The efficiency of sediment removal in a Ring system – The supply main is laid all among the
continuous flow sedimentation tank does not peripheral roads and sub main branch out from the
depend upon the mains.
(a) Discharge of the tank
(b) Width of the tank 87. The flow velocity in a sewer does not depend
upon :
(c) Length of the tank
(d) Depth of the tank (a) Its grade
Ans. (d) : Efficiency of sediment removal in a (b) Its length
continuous flow sediment tank depends upont– (c) Its hydraulic mean depth
• discharge of the tank (d) Its roughness
• width of tank Ans. (b) : Flow velocity of a sewer depends upon–
• length of tank • Its grade
83. A clari-flocculator is a : • Its roughness
(a) Plain sedimentation unit • Its hydraulic mean depth
(b) Aeration unit
88. A drop manhole may be provided along the
(c) Coagulation cum sedimentation unit sewer line :
(d) None of the above (a) To provide inspection channels in the sewer
Ans. (c) : Clarifloccular is a combination of coagulation line
sedimentaion unit in this tank flocculation and (b) When a branch sewer outfalls into it from a
clarification is done in a single tank. height of more than 0.6 or so
84. Process which involves chlorination beyond the (c) To provide self cleaning velocity in the sewer
break point is known as : line
(a) Pre-chlorination (b) Post-chlorination (d) None of the above
(c) Super- chlorination (d) Hype-chlorination
Ans. (b) : A drop manhole may be provided along the
Ans. (c) Superchlarination – superchlorination also sewer line when a branch sewer outfalls into it from a
known as hyperchlorination temporarity increases the height of more than 0.6 or so.
free chlorine residual in a water beyond break point.
High free chlorine resudual (i.e. above 5 mg/l) is 89. An intercepting trap is provided at the junction
effective against most bacteria. of :
Prechlorination – During pre chlorination process (a) A house and a municipal sewer
chlorine is applies to raw water that may contain high (b) Any two houses drains
concentrations of natural organic matter. (c) An un-foul room or roof drain and a foul bath
85. The main disadvantages of lime soda process of or a kitchen drain
water softening is that : (d) None of the above
(a) It is unsuitable for turbid and acidic waters Ans. (a) Intercepting trap – An intercepting trap is
(b) Zero hardness effluent can not be obtained often provided at the junction of a house sewer and a
(c) Excessive hard water can not be softened municipal sewer, so as to prevent entry of foul gases
(d) Huge amount of precipitate is formed posing from municipal sewer into houses drainage system.
the problem of disposal • An intercepting trap is provided to disconnect the
Ans. (d) All above are the disadvantages of time soda house drain from the street sewer.
process of water softening. But main disadvantage is
90. Minimum dissolve oxygen (D.O.) prescribed
huge amount of precipitate of formed posing the
for a river stream to avoid fish kills is :
problem of disposal.
(a) 2 ppm (b) 4 ppm
86. If a city is having the roads radiating from
centre then the most suitable layout will be : (c) 8 ppm (d) 10 ppm
(a) Dead and system (b) Grid iron system Ans. (b) Minimum 4 ppm dissolved oxygen is required
(c) Ring system (d) Radial system for survival of aquatic life (like fishes etc).
OPSC (Civil Services) Exam-2011 276 YCT
91. if a 2% solution of sewage sample is incubated 96. In an oxidation pond the sewage is treated by :
for a 5 days at 200C and the dissolved oxygen (a) Aerobic bacteria
depletion is 10 mg /l then BOD of the sewage (b) Anaerobic bacteria
would be :
(c) Dual action of aerobic bacteria and algae
(a) 50 mg/l (b) 200 mg/l
(d) Sedimentation
(c) 500 mg/l (d) 5000 mg/l
Ans. (c) Oxidation pond : Also called lagoons or
Ans. (c) BOD = (initial D.O. – final D.O.) × dilution stabilization ponds are large, shallow ponds designed to
factors depletion of oxygen is given = 10 mg/l. treat waste water through the interaction of sunlight
100 bacteria and Algae.
dilution factor = = 50
2% 97. The difference in elevation between inner and
BOD = 10 × 50 = 500 mg/l outer edges of carriageway of 7.0 m width at a
92. The neutral process, under width the flowing horizontal curve of a highway 0.42 m. What is
river water gets cleaned, is known as : the rate of super elevation provided?
(a) Oxidation (a) 6.0 %
(b) Self purification (b) 6.5%
(c) Photosynthesis (c) 5.9%
(d) Reduction (d) 6.1%
Ans. (b) Self purification – It is natural process of Ans. (a) : Super elevation E = tane
rivers, lakes or canals to recover the rate of D.O. values 42
of the height concentration of oxygen which is one of = = 0.06 or 6%
700
the best indicates of water quality.
98. What is the recommended shape of camber?
93. If the sewage contains greases and fatty oils,
they can be removed in : (a) Straight
(a) Grit chambers (b) Parabolic
(b) Sedimentation tanks (c) Straight at edges and parabolic in middle
(c) Skimming tanks (d) Parabolic at edges and straight at middle
(d) Aeration tanks Ans. (c) Recommended shape of camber = straight at
Ans. (c) Skimming tanks – A skimming tank is a edges and parabolic in middle.
chamber so arranged that the floating matter like oil 99. The traffic volumes during six consecutive 15
grease etc. rise and remain on the surface of water minutes interval are 60, 90, 75, 78, 70. 30. What
coater (sewage) untill removed, while the liquid flows is the peak hour factor?
out continuously under partition or baffles. (a) 0.842
94. Which of the following unit work on the (b) 0.869
principle of anaerobic decomposition? (c) 0.811
(a) Oxidation ditch (d) 0.769
(b) Tickling fitter
Ans. (b) : Peak hour volume is just the sum of the
(c) Activated sludge process volumes of four 15 minute interval within the peak
(d) Sludge digestion tank hour.
Ans. (d) Sludge digestion tank – Sludge digestion is a 4 peak hour volume = 90, 75, 78, 70
process in which organic solids are decomposed into Sum of 4 peak hour volume = 901 +75 +78 + 70
stable substances. The sludge then flows into a second
= 313
tank, where the dissolved matter is converted by other
bacteria into biogas, a mixture of carbon di oxide and Peak hour volume = 90
methane under anaerobic process. 313
Peak hour factor = = 869
95. Lower food to micro-organism ratio in a 4 × 90
conventional activated sludge plant will mean : 100. What is the layer above the bottom most layer
(a) Lower BOD removal in flexible pavement?
(b) Higher BOD removal (a) Base
(c) Lower bacteria removal (b) Sub-base
(d) None of the above (c) Sub-grade
Ans. (b) Lower food to micro : organism ratio in a (d) None of the above
conventional activated sludge plant will mean higher Ans. (b) Layer above bottom most layer in a flexible
BOD removal. pavement is sub-base.
OPSC (Civil Services) Exam-2011 277 YCT
 Ans. (b) Water bound macadam road– The roads
having its wearing surface consisting of clean, crushed
aggregates, mechanically interlocked by rolling and
bound together with a filler material and water. Stone
dust is used in water bound macadam roads.
105. How many rails are required for 1 km. length
of B.G. Track?
101. If the thickness of cement concrete pavement is (a) 154 (b) 167
22 cm and the radius of contact area is 15 cm, (c) 150 (d) 160
determine the radius of resisting section :
(a) 14.07 cm (b) 15.17 cm Ans. (a) : Length of a rail = 12.8 ≈ 13 m
(c) 14.22 cm (d) 13.92 cm number of rail in 1 km in both side =
Ans. (c) Radius of resisting section 1000
2× = 153.84 ≃ 154 rails.
b = (1.6 a2 + n2)1/2 – 0.675 h 13
When a is less than 1.724 h = 1.724 × 22 = 37.92 > a 106. On a single track, loaded goods trains run from
h = 22 cm X to Y and these goods trains run empty in the
a = 15 cm reverse direction. The amount of creep will be:
b = (1.6 × 15 + 22 ) – 0.675 ×22
2 2 1/2
(a) Zero
= 14.22 cm (b) More in direction from Y to X
102. What is the limiting gradient recommended by (c) More in direction from X toY
Indian Roads congress for roads in plain (d) None of the above
terrain?
Ans. (c) : When loaded goods train run from x to y then
(a) 5.0 % (b) 4.0 %
reverse pressure is more applied at track so creep is
(c) 6.0 % (d) 4.5 % more compare to empty train y to x.
Ans. (a) :
107. The misalignment of rails due to temperature
limiting changes is known as :
Tension
gradient
(a) Buckling (b) Creeping
(1) Plain or rolling 5% (c) Hogging (d) Bulging
(2) Mountaneous and steep terrain 6%
Ans. (a) Buckling of track- Track buckling in
more than 3000 m above MSL
formation of large lateral misalignments in continuous
(3) Steep terrain upto 3000 m height 7% welded rail track, often resulting in catastrophic
above MSL derailments.
103. What is the stopping sight distance, as per IRC
108. What is the safe speed on a high speed BG
recommendations, to be provided on a level
Track with radius, 1750 m and cant 80 mm?
road with design speed of 100 km/hour?
Assume the coefficient of friction as 0.35 (a) 151 kmph
(a) 210 m (b) 200 m (b) 141 kmph
(c) 190 m (d) 180 m (c) 161 kmph
v2 (d) 156 kmph
Ans. (c) : Stopping sight distance (SSO) = vt + Ans. (a) : Given, Radius of BG track (R) = 1750 m
2gF
v = 100 kmph Cant = 80 mm
For broad gauge track–
5
= 100 × = 27.77 m / s (i) By Martin's formula –
18
= 4.35 R − 67
27.77 2
SSD = 27.77 × 2.5 + = 181.726 m = 4.35 1750 − 67 = 178.45 kmph
2 × 9.81× 0.35
10 4. What is the binding material in water bound (ii) By Indian railway formula –
macadam roads? Vmax = 0.27 ( ea + ed ) × R
(a) Brick powder
(b) Stone dust = 0.27 (100 + 80 )1750 = 151.537 kmph
(c) Construction waste So, safe speed of track is 151 kmph (Lesser of the above
(d) Lime powder two values)

OPSC (Civil Services) Exam-2011 278 YCT

 109. What is the runway number marked at SW end (a) Head activity (b) Head event
of a runway oriented in SW-NE direction? (c) Dual role event (d) Tail event
(a) 27 (b) 4 Ans. (b) Head event – Head event slack of an activity
(c) 22 (d) 9 in a network is the slack at the head (or terminal point)
Ans. (c) : 22 is the runway number marked at SW end of an activity.
of a runway oriented in SW-NE direction. 117. The cost of labour in a construction activity is :
110. If the design aircraft requires 1800 m to come (a) Fixed cost (b) Marginal cost
to stop during landing, what is the length of (c) Variable cost (d) Sunk cost
runway required? Ans. (c) Fixed cost – Insurance rent CEO salary
(a) 2250 m (b) 2750 m materials.
(c) 3600 m (d) 3000 m Variable costs – Construction labour, fuel cost,
Ans. (d) : The length of landing = 0.6 times of the supplies.
runway length 118. The first year cost of maintenance of a building
1000 is Rs. 100 and it increases at a uniform rate of
= runway
0.6 10% per year. If the interest rat is 8%, what is
Runway length = 3000 m the present worth of cost of the first 5 years of
maintenance?
111. Critical path connects the events that have :
(a) 490.42 (b) 480.42
(a) Negative slack (b) Positive slack
(c) 500.42 (d) 470.42
(c) Zero slack (d) High slack
Ans. (c) : Critical path connects the events which has   1 + g n 
A 1 −   
zero slack or zero float.   1 + i  
112. How do you represent the various activities of a Ans. (b) : P = g =/ i
i−g
project on bar chart?
Where,
(a) Dots (b) Crosses
P = Present of all cash flow between periods 1 and n
(c) Vertical lines (d) Horizontal lines
A = Cash flow in period 1
Ans. (d) : Activity of a project on bar chart represented
g = rate of change per period
by horizontal lines. An activity bar chart also known as
Gantt chart clearly indicates now different tasks are i = effective interest rate per period
dependent on each other in a project. So,
113. An activity that shows dependency consumes   1.10 5 
1 −   
no time or resources is called?   1.08  

(a) Critical (b) Dummy P = 100 = 480.02 Rs.
−0.02
(c) Event (d) Float
119. The amount of examination of the difference
Ans. (b) : Dummy activity – A dummy activity is a between mutually exclusive alternatives is
simulated activity of sorts, one that is of a zero duration known as :
and consumes no resources called dummy activity.
(a) Life cycle analysis
114. The waiting time required between activities is (b) Benefit cost analysis
known as : (c) Incremental analysis
(a) Slack (b) Float (d) Present worth analysis
(c) Lag (d) Free float
Ans. (c) Incremental Analysis – It is a decision making
Ans. (c) : Lag is the delay of a successor activity and technique used in business to determine the trace cost
represents time that must pass before the second activity difference between alternatives also called the relevant
can begin. There are no resources associated with a lag. cost approach, marginal analysis or differential analysis.
115. The time of completion of an activity which can Incremental analysis disregards any sunk cost or past
not be reduced further irrespective of cost cost.
considerations is : 120. The money that is already spent as a result of a
(a) Crash time (b) Normal time past decision is called.
(c) Slack time (d) Float time (a) Sunk cost (b) Fixed cost
Ans. (a) : The time of completion of an activity which (c) Variable cost (d) Marginal cost
can not be reduced further irrespective of cost Ans. (a) Sunk cost : A sunk cost is a cost that has
considerations is crash time. already been incurred an cannot be recovered. Sunk
116. The completion of an activity in PERT costs are contrasted with prospective costs, which are
techniques is called : future costs that may be avoided if action is taken.
OPSC (Civil Services) Exam-2011 279 YCT
 Odisha Public Service Commission
(Civil Services)
Exam- 2006
1. A steep flow curve indicates soil of Ans. (d) :
(a) low shear strength (b) high shear strength
Property Dry of Wet of
(c) low compressibility (d) low permeability Optimum Optimum
Ans. (a) : Structure after Flocculated Dispersed
compaction (random) (oriented)
Water More Less
deficiency
Permeability More, isotropic Less,
anisotropic
Curve 1–Large shear strength Compressibility
Curve 2– Low shear strength 1. At low stress Low Higher
2. A flat grain size distribution curve shows a 2. At high stress High Lower
(a) narrow range of grain sizes.
Swellability High Low
(b) wide range of grain sizes.
Shrinkage Low High
(c) uniform grain sizes.
(d) certain range of missing grain sizes. Stress-strain Brittle, high Ductile, no
behaviour peak, higher peak, lower
Ans. (d) : Gap graded soil–This is the case when
elastic modulus elastic modulus
intermediate sizes are absent. This could be the case
when two separate soil are mixed. Pore water Low High
Well graded soil–The curve is smooth and covers a pressure
wide range of sizes, this indicates that the soil is non- Strength High Much lower
uniform.
Sensitivity more low
4. The value of compression index (Cc) for a
remoulded sample whose liquid limit is 40% is
(a) 0.21 (b) 0.28
(c) 0.021 (d) 0.028
Ans. (a) : Given,
wL = 40%
Cc = ?
3. Which of the following describes a soil
compacted dry of optimum relative to the same Compression index (Cc) for remoulded soil–
soil compacted wet of optimum? Cc = 0.007 [wL – 10]
(a) Dispersed structure, higher strength, higher Cc = 0.007 [40 – 10]
swelling.
Cc = 0.007 × 30
(b) Flocculated structure, higher strength, lower
Cc = 0.21
swelling.
(c) Dispersed structure, lower strength, less 5. The only stress of Boussinesq suitable for
permeability. computation of stress in soils is
(d) Higher permeability, higher strength, higher (a) vertical stress (b) horizontal stress
swelling. (c) tangential stress (d) shear stress
OPSC Civil Services Exam. 2006 280 YCT
Ans.(a): Boussinesq's equation–The application of a (ii) CU Test–
concentrated vertical load to the horizontal surface of (a) To check stability under sudden drawdown &
any solid load to the horizontal surface of any solid unloading
body produces a set of vertical stresses on every (b) Analysis of post construction phase stability
horizontal plane with in the body. (iii) UU Test–
5/ 2
(a) Sudden loading such as rapid construction
3Q  1 
∆σ v =   (b) Short term (construction phase) stability
2 πz  1 + ( r / z ) 
2 2
  8. Negative skin friction on a pile
Where, Q = Concentrated vertical load (a) acts downward and decreases the load
z = Vertical distance between N and the surface of the carrying capacity of the pile.
mass. (b) acts downward and increases the load
r = Horizontal distance from N to the line of action of carrying capacity of the pile.
the load. (c) acts upward and decreases the load carrying
capacity of the pile.
(d) acts upward and increases the load carrying
capacity of the pile.
Ans. (a) : Negative skin friction [Down Drag]–It is a
phenomenon which occurs when a portion of a soil
layer surrounding a pile more than the pile.

6. For piping phenomenon to occur in soils the

most important condition to be satisfied is that
the
(a) specific gravity of soil solids is more than 2.8.
(b) void ratio is more than 2.0.
(c) hydraulic gradient is nearly unity.
(d) soil is fine grained. • It reduce the load carrying capacity of pile.
• It is develop due to lowering of ground water table,
Ans. (c) : Critical hydraulic gradient–
sudden compression loading etc.
G −1
I cr = • It occurs m soft/loose unconsolidated layers of soil.
1+ e
9. A normally consolidated clay settled 10 mm
for most of the soils
when effective stress was increased from 50
G = 2.6-2.7 kN/m2. If the effective stress is further
& e = 0.6-0.7 increased from 100 kN/m2, then further
hence I cr ≅ 1 settlement of the clay shall be
(a) 10 mm (b) 20 mm
7. For stability analysis of an earth dam for
(c) 30 mm (d) 40 mm
steady seepage case, the most appropriate test
would be the Ans. (b) : ∆σ1 = 50 kPa
(a) unconsolidated undrained test ∆σ2 = 100 kPa
(b) consolidated undrained test
we know that,
(c) unconsolidated drained test
∆e ∆H
(d) consolidated drained test av = =
∆σ ∆σ
Ans. (d) : Uses of triaxial tests– ∆H1 ∆σ1
(i) CD test– =
∆H 2 ∆σ 2
(a) Analysis of gradual loading condition
10 × 100
(b) To check long term stability of embankment under ∆H 2 = = 20 mm
steady state seepage. 50

OPSC Civil Services Exam. 2006 281 YCT

10. The correct increasing order of specific surface du du
(d) τ = ( µ )
3/ 2
i.e. surface area per mass of the given soils is (c) τ = µ
dy dy
(a) silt, sand, colloids, clay
(b) sand, silt, colloids, clay Ans. (c) : Newton's law of viscosity–Newton's law of
viscosity states that shear stress is directly proportional
(c) sand, silt, clay, colloids
to velocity gradient or rate of angular deformation.
(d) clay, silt, sand, colloids
du dθ
Ans. (c) : The specific surface area is inversely τ∝ ∝
proportional to grain size. dy dt
1. Sand – 0.075 mm to 4.75 mm 15. Surface tension is expressed in
2. Silt – 0.002 mm to 0.075 mm (a) N/m (b) N/m2
3. Clay – < 0.002 mm 2
(c) N /m (d) N/m3
• Colloids are basically finer clay particles whose Ans. (a) : Surface tension–Surface tension occurs at
surface area is very high.
the interface of liquid and gas or out the interface of two
• Increasing order ⇒ Sand < Silt < Clay < Colloids liquid.
11. For stability analysis of and earth dam for • Surface tension is inversely proportional to
steady seepage case, the most appropriate test temperature and it also act when fluid is at rest.
(a) unconsolidated undrained test
• Surface tension is caused by force of cohesion
(b) consolidated undrained test between liquid molecules.
(c) unconsolidated drained test • Surface tension measured as force/length.
(d) consolidated drained test
Unit = N/m
Ans. (d) : Repeat question no. 7
16. Gauge pressure is defined as
12. While designing the abutments of a bridge, the
(a) pressure measured with respect to
lateral earth pressure to be considered is
atmospheric pressure.
(a) active earth pressure.
(b) pressure measured from zero pressure.
(b) earth pressure at rest with value of K0 for
loose back fill. (c) absolute pressure plus atmospheric pressure.
(c) earth pressure at rest with value K0 for dense (d) pressure given by a gauge.
back fill. Ans. (a) : Gauge pressure–It is the pressure with
(d) none of them. respect to atmospheric pressure at datum.
Ans. (c) : Earth pressure at rest–The lateral Earth • It is measured using manometer or bourdon gauge.
pressure is called at rest pressure when the soil mass is • It can be +ve, –ve, zero
not subjected to any lateral yielding or movement.
• An abutments of bridge is at rest earth pressures
because it restrained at its top by the bridge deck.
13. The rise of water table below the foundation
influences the bearing capacity of soil mainly
by reducing
(a) cohesion and effective unit weight of soil.
(b) cohesion and effective unit weight of soil.
(c) effective unit weight of soil and effective 17. The pressure intensity at a point in water is
angle of shearing resistance. given by 49.05 kN/m2. The depth to that point
(d) effective angle of shearing resistance. below surface is
Ans. (c) : Rise of water table below the foundation level (a) 0.198 m (b) 481.2 m
reduces the effective unit weigh below the foundation (c) 5 m (d) 10 m
level hence bearing capacity reduces.
Ans. (c) : Given,
• We know that from triaxial test results
Pressure P = 49.05 kN/m2 = 49.05 × 103 N/m2
φdrained > φundrained
ρwater = 1000 kg/m3
hence rise of water table also reduces the angle of
shearing resistance (φ). g = 9.81 m/s2
14. Newton's law of viscosity is given by the h = ?
relation So, ρgh = 49.05 × 103
du du 1000 × 9.81 × h = 49.05 × 103
(a) τ = µ 2 (b) τ = µ
dy dy h=5m
OPSC Civil Services Exam. 2006 282 YCT
18. The total pressure an a vertical plate 2 m wide (c) neutral equilibrium
and 4 m depth when held normal to the free (d) none of the above
surface in water with its width on the free
surface is Ans. (b) : When Meta centre of a floating body is
(a) 16 kN (b) 16,000 kN below the centre of gravity of the body, then body is
said to be in unstable equilibrium condition.
(c) 78.48 kN (d) 32 kN
Ans. (b) : Condition Submerged Floating body
body
Stable equilibrium G below B M above G
BM > BG
Unstable equilibrium G above B M below G
BM < BG
Neutral equilibrium G and B M and G
Given, coincide coincide
b=2m GM = 0
h=4m
21. The type of flow in which the velocity at any
h =2m given time does not change with space is called
ρ = 1000 kg/m3 (a) steady flow (b) unsteady flow
g = 10 m/s2 (c) rotational flow (d) uniform flow
Total pressure = ρgAh Ans. (d) : Uniform flow–A flow is said to be uniform
= 1000 × 10 × 2 × 4 × 2 flow in which velocity and flow both in magnitude and
= 160,000 N direction do not change along the direction of flow for
given instance of time.
= 160 kN
19. A submerged body is supposed to be in stable dv
=0
equilibrium when ds
(a) centre of buoyancy coincides width concrete Steady flow–At any given location the flow and fluid
of gravity. properties do not change with time is known as steady
(b) center of gravity lies above centre of flow.
buoyancy.
dv dp dρ
(c) center of buoyancy lies above center of = 0, = 0, =0
gravity. ds dt dt
(d) when buoyant forces is less than the 22. The path followed by a fluid particle in motion
gravitational force. is called a
Ans. (c) : A submerged body is supposed to be in (a) stream line (b) path line
stable equilibrium when centre of buoyancy lies above (c) streak line (d) equipotential line
centre of gravity.
Ans. (b) : Path line–Actual path traveled by an
Condition Submerged Floating body individual fluid particle over individual fluid particle
body over some time period is called path line.
Stable equilibrium G below B M above G • These are actual lines not imaginary ones
BM > BG • These line may cut one another
Unstable equilibrium G above B M below G Streak line–It is the locus of all fluid particles at an
BM < BG instant, which have crossed through the same point.
• Introduction of dye or smoke from a point in a flow
Neutral equilibrium G and B M and G
will from streak line.
coincide coincide
Stream line–A stream line is an imaginary curve drawn
GM = 0
through a flowing fluid in such a way that the tangent to
20. If the position of metacentre M lies below C.G. it at any point gives the direction of the instantaneous
of the floating body G, the body will remain in velocity (V) of flow at that point.
a state of 23. In fluid mechanics, the continuity equation is a
(a) stable equilibrium mathematical statement embodying the
(b) unstable equilibrium principle of
OPSC Civil Services Exam. 2006 283 YCT
 (a) conservation of momentum Ans. (d) : Stagnation point–A point in the flow, where
(b) conservation of mass the velocity of fluid is zero, is called a stagnation point.
(c) conservation of energy Stagnation points exist at the surface of objects in the
(d) none of the above flow field, where the fluid is brought to rest by the
object. The Bernoulli equation shows that the static
Ans. (b) : Continuity equation–It is based on the pressure is highest when the velocity is zero and hence
principle of conservation of mass. According to this static pressure is at its maximum value at stagnation
principle mass inflow in a fixed region should be equal
point.
to mass out flow that fixed region in a particular time.
• For steady and incompressible flow– 27. A stream function is given by ψ = x2 + y2. The
a1v1 = a2v2 velocity component in the x-direction at
appoint (1, 3) will be
• for flow to be possible continuity equation must be
satisfied. (a) 6 (b) 2
24. _____ is defined as a scalar function of space (c) –6 (d) 40
and time such that its negative derivative with
respect to any direction gives the fluid velocity Ans. (c) : Given,
in that direction. ψ = x2 + y2
(a) velocity potential function −∂ψ −∂ 2
(b) stream function u=
∂y
=
∂y
(
x + y2 )
(c) circulation
u = –2y ....... (i)
(d) vorticity
from equation (i) at point (1, 3) value put
Ans. (a) : Velocity potential or Potential function
(φ)–It is the scalar function of space and time in such a u = –2 × 3
way that negative derivative with respect to any u = –6
direction gives velocity of flow in that direction 28. Euler's equation (in differential form) is
• It must satisfy Laplace equation. written as
−∂φ −∂φ −∂φ dp
=u =v =z (a) + v 2 .dv + g.dz = 0
∂x ∂y ∂z ρ
25. The streamlines and equiptotential lines are dp
(b) + v.dv + g.dz = 0
(a) normal to each other ρ
(b) parallel to each other dp
(c) always intersecting at acute angles (c) + v.dv + g 2 .dz = 0
ρ
(d) lie one over the other
dp
Ans. (a) : Lines of constant ψ are stream line, that is (d) + v 2 .dv + g.dz = 0
ρ2
∂y u
= .....(i) Ans. (b) : Euler's equation of motion in the differential
∂x v form given by
Lines of constant φ are equipotential line, that is
dp
+ gdz + vdv = 0
∂y − u ρ
= .....(ii)
∂x v
It is based on the assumption that the flow is ideal and
A comparison of equation (i) and (ii) shows that lines of viscous forces are zero.
constant φ are orthogonal to lines be drawn that consist
of a family of stream lines of constant ψ at all points 29. The piezometric head is the summation of
where they intersect. The equipotential lines are drawn (a) velocity head and pressure head.
everywhere normal to the stream lines. The spacing of (b) pressure head and elevation head.
the equipotential line are selected in such a way that the
change in velocity potential from one equipontential (c) velocity head and elevation head.
lines to the next is constant. (d) total head.
26. A stagnation point in a fluid flow is a point at Ans. (b) : Piezometric head–
which
• It is also called the hydraulic head without considering
(a) pressure is zero the velocity head is then written as
(b) total energy is zero
p p
(c) pressure and velocity are zero h= + z or + z
(d) velocity is zero pg γ

OPSC Civil Services Exam. 2006 284 YCT

 p Where, Fr = Force ratio
Where, = pressure head or static head (Fi)p = Inertia force at a point in prototype
pg
z = datum head or elevation head (Fv)p = Viscous force at a point in prototype
(Fg)p = Gravity force at a point in prototype
p
• Summation of and z is called piezometric head at 32. The boundary layer separation occurs when
pg
every point in a fluid at rest is constant. dp  ∂u 
(a) <0 (b)   = 0
30. A Venturimeter is used for measuring dx  ∂y  y = 0
(a) pressure (b) piezometric head
 ∂u   ∂u 
(c) total energy (d) flow rate. (c)   > 0 (d)   < 0
∂y
 y=0  ∂y  y = 0
Ans. (d) : Venturimeter–It is used for measuring the
discharge (i.e. rate of fluid flow) through a pipe. Ans. (b) : Location of separation point–The
A venturimeter is based on the principle of Bernoulli's separation point(s) is determined from the condition
equation. When the velocity head increases in an  ∂u 
accelerated flow, there is a corresponding reduction in   =0
pressure head.  ∂y  y = o
Main parts of venturimeter–The following are three For a given velocity profile. It can be determined
main parts of a venturimeter. whether the boundary layer has separated.
• A short converging part  ∂u 
• If   is negative – the flow has separated
• Diverging pat  ∂y  y = o
• Throat
 ∂u 
Length of converging part = 2.5 D = for Angle = 19º to • If   = 0 the flow is on the verge of separation
23º.  ∂y  y = o
Length of diverging part = 7.5 D for angle 5º to 7º
 ∂u 
• If   = 0 is positive–the flow will not separate
d
= 0.25 to 0.75  ∂y  y = o
D
33. Loss of head due to sudden enlargement is
Length of throat = d given as
( V1 − V2 ) ( V1 − V2 )
3 2

(a) (b)
2g 2g

V12 − V22 V1 − V2
(c) (d)
2g 2g
Where V1 and V2 are respectively velocities
before and after the enlargement and g is the
acceleration due to gravity.
31. Dynamic similarity between the model and Ans. (b) : Minor losses–Minor losses occur whenever
prototype is the there is a sudden change in the area of flow or direction
(a) similarity of motion of flow. In almost all cases minor losses are determined
(b) similarity of lengths by experiment. Minor losses are given below–
(c) similarity of forces (i) Loss of head due to sudden enlargement

( v1 − v 2 )
2
(d) similarity of flow
Ans. (c) : Dynamic similarity–Dynamic similarity he =
2g
means the similarly of forces between the model and
prototype. This dynamic similarity is said to exist (ii) Loss of head due to sudden contraction
between the model and the prototype if the ratios of the
( v − v2 )
2
corresponding forces acting at the corresponding points he = 1
are equal. Also the directions of the corresponding 2g
forces at the corresponding points should be same.
(iii) Loss of head due to entrance in pipe
For dynamic similarity–
( Fi ) p = ( Fv ) p = ( Fg ) p = Fr
2
v 22  1 
he =  
( Fi ) m ( Fv ) m ( Fg ) m 2g  Cc 

OPSC Civil Services Exam. 2006 285 YCT

 36. When the depth of flow changes abruptly over
0.5v 2 a short distance in a free surface flow, the flow
he =
2g is known as
(iv) Loss of head at the bends (a) uniform flow
(b) spatially varied flow
kv 2 (c) gradually varied flow
hb =
2g (d) rapidly varied flow
(v) Loss of head at the exit of pipe Ans. (d) : Rapidly varied flow–The depth changes
2
abruptly over a comparatively short distance. Rapidly
v varied flow is known as local phenomenon e.g.
ho =
2g hydraulic jump and hydraulic drop.
(vi) Loss of head in various pipe fitting Gradually varied flow–If the depth of flow varies
gradually over a long reach of the channel, the flow is
kv 2 called gradually varied flow.
hb =
2g 37. For flow in open channels, uniform flow is
characterized by
34. In a laminar flow, Reynolds number is (a) a constant slope of channel bottom.
(a) less than 4000 (b) a constant depth of flow.
(b) more than 2000 (c) a changing depth of flow.
(c) more than 2000 but less than 4000 (d) none of the above.
(d) less than 2000
Ans. (b) : Uniform flow channel–A flow in the
Ans. (d) : Reynold's number (Re)–It is the ratio of channel is said to be uniform if the depth of the flow
inertia force and viscous force. remains constant over a given length of the channel. In
Inertia force uniform flow, bed slope, water surface slope and energy
Reynold number = slope are paralleled to each other.
Viscous force
38. The hydraulic mean depth is given by
Mathematically Reynold's number is given by
P P2
ρVD (a) (b)
Re = A A
µ
A A
Where, ρ = Fluid density (c) (d)
P P
D = Diameter of the pipe
Ans. (c) : Hydraulic mean depth (m)
µ = Viscosity of the fluid
V = Velocity A
m=
• For laminar flow, Re < 2000 P
• For transition flow, 2000 < Re < 4000 Where, m = hydraulic mean depth
• For turbulent flow, Re > 4000 A = Area of cross-section of flow
35. The main function of a surge tank is P = Wetted perimeter of the cross section
(a) to regulate the flow in penstock. 39. For the best rectangular section
(b) to increase the storage capacity of the (a) y = b/3 (b) y = b
reservoir. (c) y = b/4 (d) y = b/2
(c) to absorb water hammer pressures. Ans. (d) : The best hydraulic section is the channel
(d) to create water hammer action. proportions that yield a minimum wetted perimeter for a
Ans. (c) : Surge tank–A surge tank is a small reservoir given flow area. It can be shown that the best hydraulic
or tank in which the water level or balls to reduce the section has a depth of flow equal to one-half of the
pressure swings so that they are not transmitted in full channel width (the shape of a half square) for
to a closed circuit. rectangular channels.
• To reduce the distance between the free water surface • The ideal shape of the best hydraulic section is the
and turbine thereby reducing the water hammer effect semi circle.
on penstock and also protect upstream tunnel from high
pressure rises.
• to serve as supply tank to the turbine when water in
the pipe is accelerating during increased load conditions
and storage tank when the water is decelerating during
reduced load conditions.
OPSC Civil Services Exam. 2006 286 YCT
A = area = B.y
P = Perimeter = B + 2y
dp
For an economical channel section, =0
dy
A = By
A  A By 
B=
y Q R = P = B + 2y 
 
B
B.P A B.P  2 8 θ
B=  =  Q= c d ( L − 0.2H ) 2H 3/ 2 + cd 2g tan H 5/ 2
( + 2y )
B  y ( B + 2y )  3 15 2
B + 2y = P 42. The surface profile lying in zone 1 of a channel
with mild slope is called
A
+ 2y = P (a) C1 profile (b) draw down
y
(c) M2 profile (d) M1 profile
d A  dP Ans. (d) : Zone I : y > yo, yc
 + 2y  = =0
dy  y  dy yc y
Zone II : <y< o
A = 2y2 = By yo yc
B Zone III : The space above channel bed & below
B = 2y y=
2 NDL/CDL 0 ≤ y < y o
40. The hydraulic jump occurs when Qualitative analysis of GVF Profile
(a) the bed slope is steep.
(b) the flow changes from super critical to sub
critical.
(c) the flow changes from sub-critical to super
critical.
(d) the flow is critical.
Ans. (b) : The hydraulic jump occurs when the flow
changes from super critical flow meets the subcritical
flow, hydraulic jump is used for: 1. M1
(i) Dissipation of energy downstream of hydraulic y > yo > yc
structures such as dams, sluice gates y > yo so .....f
(ii) Mixing of chemicals y > yc Fr < 1
(iii) Desalination of sea water dy + ve
= = + ve
(iv) Efficient operation of flow measuring flames etc. dx + ve
41. Cipolletti weir is a trapezoidal weir having side 43. A surge with an advancing front moving in the
upstream direction with increased depth is
slope of
called a
(a) 1 horizontal to 2 vertical (a) hydraulic bore (b) negative surge
(b) 4 horizontal to 1 vertical (c) positive surge (d) hydraulic jump
(c) 1 horizontal to 4 vertical Ans. (c) : A surge with an advancing front moving in
(d) 1 horizontal to 3 vertical the upstream direction with increased depth is called as
negative surge.
Ans. (c) : Cipolletti weir or Notch–The Cippolette
weir is trapezoidal weir having side slopes of 1
horizontal to 4 vertical. The advantage of this weir is
that the factor of end contraction is not required (while
using Francis formula)
Total discharge (Q) = Q1 + Q2
= Rectangular + Triangular Where, vw = Velocity of wave or surge

OPSC Civil Services Exam. 2006 287 YCT

44. The present population of a community is 1
28000 with an average water demand of 150 Mn + 2OH − O 2 → MnO2 ↓ ( brown ) + H 2 O
2
lpcd. The existing water treatment plant has a
When O2 is absent
design capacity of 6000 m3/d. It is expected that
1
the population will increase to 44000 during the Mn + 2OH − O 2 → MnO2 ↓ ( white )
next 20 years. The number of years from now 2
when the plant will reach its design capacity, Therefore sample S1 does not contain DO, this
assuming an arithmetic rate of population producing white precipitate and sample S2 contain DO
growth, will be and develop brown precipitate.
(a) 5.5 years (b) 8.6 years 46. The turbidity of surface water is measured
(c) 15.0 years (d) 16.5 years with the help of a turbidity meter – an
Ans. (c) : Given that, instrument measuring value so obtained is
Po = 28000 expressed in the units of
for, n = 20, P20 = 44000 (a) CFU (b) FTU
Population after n years– (c) JTU (d) NTU
Ans. (d) : The turbidity of surface water is measured
from Pn = Po + nx
with the help of a turbidity meter an instrument
P20 = Po + 20x measuring the intensity of light scattered by suspended
44000 − 28000 16000 material present in water. So, this works on
x= = = 800 per year Nephelometer. It measures the turbidity in terms of
20 20
NTU (Nephelometric turbidity unit).
The population when design capacity will be required
47. Hardness of water is measured by titration
6000 × 1000
Pn = = 40000 with ethylene-di-amine-tetra-acetic acid
150 (EDTA) method using
Number of years to reach the plant at design capacity (a) Eriochrome black T indicator
40000 − 28000 (b) Ferroin indicator
n=
800 (c) Methyl orange indicator
12000 (d) Phenolphthalein indicator
n=
800 Ans. (a) : Hardness measurement–The hardness can
n = 15 years be measured by spectro photo metric or chemical
45. Water samples (S1 and S2) from two different titration techniques used to determine the amounts of
sources were collected for the measurement of calcium and magnesium ions in a given sample. The
dissolved oxygen (DO) using modified Winkler hardness can be measured directly by titration with
method. Samples were transferred to 300 mL Ethylene Diamine Tetra Acetic Acid (EDTA) with the
BOD bottles. 2 mL of MnSO4 solution and 2 Black reagent Eriochrome Black T indicator. EBT
mL of alkali-iodide-azide reagents were added reacts with divalent metal cations to form a red
to each bottle containing the sample and mixed. complex, EDTA substitutes EBT in the complex and
Sample S1 developed a white precipitate, when the substitution is complete the soluble color
whereas sample S2 developed a brown changes from red to blue.
precipitate. In reference to these observations, 48. The design parameter of a flocculation unit is
the correct statement is given by a dimensionless number Gt, where G
(a) Both the samples were devoid of DO is the velocity gradient and t is the detention
(b) Sample S1 was devoid of DO while samples time. Values of Gt ranging from 104 to 105 are
S2 contained DO commonly used, with t ranging from 10 to 30
(c) Sample S1 contained DO while sample S2 was minutes. The most preferred combination of G
devoid of DO and t to produce smaller and denser flocs is
(d) Both the samples contained DO (a) large G values with short t
Ans. (b) : Modified Winkler method–The test is based (b) large G values with long t
on the addition of divalent manganese solution, (c) small G values with short t
followed by strong alkali, to the water sample in a (d) small G values with long t
glass-stoppered bottle. Any DO present in the sample
rapidly oxidizes an equivalent amount of the dispersed Ans. (a) : • If larger G and smaller t will make small
divalent manganous hydroxide precipitate to hydroxides and dense floc
of higher oxidation states. • If smaller G and larger t will make large and light
When O2 is present floc.

OPSC Civil Services Exam. 2006 288 YCT

Therefore large flocs are easily removed in the tank it is VSA 1
advantageous to vary the G value over the length of the = = 0.25
VSB 4
flocculation tank.
Where, G = Velocity gradient VSA
= 0.25
t = Detention time VSB
49. Particles whose surface properties are such
that they aggregate, or coalesce, with other 51. Assertion (A) : A discrete particle (having
particles upon contact, thus changing size, diameter d0) settling in a circular sedimentation
shape and perhaps specific gravity with each tank follows a parabolic path.
contact, are called Reason (R) : The downward settling velocity (v0)
(a) colloidal particles of the discrete particle (having diameter d0) in the
circular sedimentation tank do not change with
(b) discrete particles time.
(c) flocculating particles Codes:
(d) suspended particles (a) Both (A) and (R) are true and (R) is the
Ans. (c) : Flocculating particles–Particles whose correct explanation of (A).
surface properties are such that they aggregate upon (b) Both (A) and (R) are true but (R) is NOT a
contact thus, changing in size, shape and perhaps correct explanation of (A).
specific gravity with each contact. (c) (A) is true, but (R) is false.
Discrete particles–Particles whose size, shape and (d) (A) is false, but (R) is true.
specific gravity do not change with time.
Ans. (b) : When the flow in circular sedimentation tank
50. Two particles are released in water at the same enters at the centre and is baffled to flow radially
time. Particle A has a diameter dA of 0.45 mm. towards the perimeter, the horizontal velocity of the
Particle B has a diameter dB of 0.90 mm. water is continually decreasing as the distance from the
Assuming equal densities for both the particles centre increases. Thus, a discrete particle with a settling
and laminar flow conditions, the ratio of the velocity (V) is continually undergoing velocity. Thus,
terminal settling velocity of particle A to that of the particle path in a circular basin is a parabola as
particle B will be compared to the straight particle path line in the long
(a) 4.00 (b) 2.00 rectangular tank.
(c) 0.50 (d) 0.25 52. Assertion (A) : A small quantity of ammonia is
Ans. (d) : Given that, added to water before carrying out disinfection
using chlorine.
diameter of particle A (dA) = 0.45 mm
Reason (R) : Chloramines are persistent
diameter of particle B (dB) = 0.90 mm disinfectant, which provides continued protection
density of A & B = ρA = ρB = P against regrowth of microorganisms in water
From stoke's law distribution system.
Codes:
g d2
Terminal setting velocity Vs = ( G − 1) (a) Both (A) and (R) are true and (R) is the
18 υ correct explanation of (A).
Where, Vs = Velocity of settlement of particle in m/s (b) Both (A) and (R) are true but (R) is NOT a
d = Diameter of the particle in meter correct explanation of (A).
G = Sp. gravity of the particle (c) (A) is true, but (R) is false.
γ (d) (A) is false, but (R) is true.
G= s Ans. (a) : A small quantity of ammonia is added to
γw
water before carrying out disinfection using chlorine.
υ = Kinematic viscosity (m2/sec) The treatment of drinking water with a chloramine
g d A2 disinfectant is termed as chloramination. Both chlorine
VSA = ( G − 1) ..... (i) and small amounts of ammonia are added to the water
18 υ
one at a time which react together to form chloramine
g d B2 (Also called combined chlorine), a long lasting
VSB = ( G − 1) ..... (ii)
18 υ disinfectant. As such chloramine disinfection is
From equation (i) and (ii) sometimes used in large distribution system. So
chloramines are persistent which provides continued
VSA d A2 protection against regrowth of microorganisms in the
=
VS d 2 water distribution system.
B B

OPSC Civil Services Exam. 2006 289 YCT

53. Match list-I (Equation/method) with list-II after the obstruction is passed. It is necessary to
(Application) and select the correct answer ascertain the minimum flows and the peak flows for
using the codes given below the lists: design. To ensure self-cleansing velocities for the wide
List-I List-II variation in flows, generally, two or more pipes not less
(Equation/Method) (Application) than 200 mm diameter are provided in parallel so that
up to the average flows, the first pipe is used and when
A. Manning's 1 Frictional head loss the flow exceeds the average the balance flow is taken
Equation estimation in pipe by the second and subsequent pipes. It is necessary to
flow
have a self-cleansing velocity of 1 m/s for the minimum
B. Darcy-Weisbach 2 Domestic sewer flow to avoid deposition in the line.
Equation design
55. A combined sewerage system is designed to
C. Hardy Cross 3 Storm water sewer carry
method design (a) domestic sewage and industrial wastewaters
D. Rational Method 4 Water distribution together.
system design (b) storm water and domestic sewage together.
Codes: (c) industrial wastewaters and storm water
A B C D together.
(a) 2 1 4 3 (d) domestic sewage only.
(b) 1 4 3 2 Ans. (b) : A combined sewerage system is designed to
(c) 4 3 2 1 carry storm water and domestic sewage together. It has
(d) 3 4 1 2 high initial cost but less maintenance cost and due to its
Ans. (a) : larger size, less chances of choking. It is not suitable for
region where heavy and concentrated rainfall occurs for
Manning's Equation– Domestic sewer design very short period of time because the remaining period
Darcy-Weisbach Equation–Frictional head loss where rainfall occurs very less or not at all.
estimation in pipe flow
56. In a circular sewer of diameter D, if the wetted
Hardy Cross method–Water distribution system design
πD
Rational method–Storm water sewer design perimeter is , the depth of flow will be
3
Rational method–In this method it is assumed that the
maximum flood flow is produced by a certain rainfall equal to
intensity which lasts for a time equal to or greater than (a) 0.25 D (b) 0.50 D
the period of concentration time. (c) 0.75 D (d) 1.00 D
The concentration time is the time required for the Ans. (a) : Given that,
surface runoff part of the catchment area to reach the
basin outlet. πD
Wetted perimeter (P) =
54. Assertion (A) : Inverted siphons normally include 3
multiple pipes and an entrance structure designed
to divide the sewage flow among them so that the
velocity (at least 0.9 m/s) in those pipes in use will
be adequate to prevent deposition of solids.
Reason (R) : A single pipe of smaller diameter
may be enough to maintain the required velocity
at the minimum flow, but the velocity at peak flow Arc = Angle × Radius
will produce very high head losses leading to πD D
damage of the pipe itself. =θ×
Codes: 3 2
(a) Both (A) and (R) are true and (R) is the π D θπ D
= ×
correct explanation of (A). 3 180 2
(b) Both (A) and (R) are true but (R) is NOT a θ = 120º
correct explanation of (A). D D θ
(c) (A) is true, but (R) is false. Depth of flow (d) = − cos
2 2 2
(d) (A) is false, but (R) is true. D D 120º
Ans. (a) : Inverted siphon–When a sewer line dips d= − cos
2 2 2
below the hydraulic grade line, it is called an inverted
D D1
siphon. The purpose is to carry the sewer under the d= − ·
obstruction and regain as much elevation as possible 2 2 2

OPSC Civil Services Exam. 2006 290 YCT

 D D 63.145 − 62.485
d= − Inorganic solid in sample = × 106
2 4 50
2D − D Organic fraction = Total solid – Inorganic solids
d= Organic fraction = 50700 – 13200
4
D Organic fraction = 37500 mg/L
d= 59. Chemical Oxygen Demand (COD) of a
4
wastewater containing organic matters is
d = 0.25D estimated by dichromate method titrating with
57. 3 mL of wastewater containing no dissolved ferrous ammonium sulphate solution using
oxygen is mixed with 297 mL of dilution water (a) Eriochrome black T indicator
containing 8.6 mg/L of dissolved oxygen in a (b) Ferroin indicator
300 mL BOD bottle. After a 5-day incubation (c) Methyl orange indicator
at 20º C, the dissolved oxygen content of the
(d) Phenolphthalein indicator
mixture is 5.4 mg/L. The BOD5 of the
wastewater is Ans. (b) : Chemical oxygen demand (COD)–It is used
(a) 317 mg/L (b) 320 mg/L to measure the content of organic matter of waste water
both biodegradable and non-biodegradable.
(c) 950 mg/L (d) 960 mg/L
• COD of a wastewater containing organic matters is
Ans. (b) : Ist method– estimated by dichromate method titrating with ferrous
( Do ) w =0 ammonium sulphate solution using ferrion indicator.
• COD test is also called dichromate oxygen demand
Qw = 3 ml test.
( Do ) Dw = 297 60. It is important to control and maintain the
horizontal velocity at approximately 0.3 m/s in
0 × 3 + 8.6 × 297
( Do )mix = a channel type horizontal-flow grit chamber
300 irrespective of variations in flow rate through
( Do )mix = 8.514 mg / l the chamber. The controlled horizontal velocity
is achieved by providing a velocity control
( BOD )5day = ( Doi − Dof ) × dilution factor section at the end of the channel. The shape of
velocity control section is dependent on the
300 section of the channel. If the channel is made of
= (8.515 – 5.4) × a rectangular section, the velocity control
3
section should be
= 311.4 mg/l
(a) a sharp-crested weir (b) an inboard weir
58. An analysis for determination of solids in the
(c) a proportioning weir (d) an ogee weir
domestic sewage was carried out as follows:
1. A crucible was dried to a constant mass of Ans. (c) : Velocity control devices for grit
62.485 g. chamber–
2. 50 mL of a well-mixed sample was taken in Velocity control device Cross section shape of
the crucible. grit chamber
3. The crucible with the sample was dried to a Proportional weir Rectangular channel
constant mass of 65.020 g in drying oven at
Parshall flume Parabolic channel
104ºC.
4. The crucible with the dried sample was 61. Activated sludge process is designed for
placed in a muffle furnace at 600ºC for an operation in
hour. After cooling, the mass of the crucible (a) endogenous growth phase of microorganism.
with residues was 63.145 g. The (b) lag growth phase of microorganism.
concentration of organic fraction of solids (c) log growth phase of microorganism.
present in the sewage sample is (d) stationary growth phase of microorganism.
(a) 13200 mg/L (b) 33800 mg/L
Ans. (d) : The relationship between food concentration
(c) 37500 mg/L (d) 50700 mg/L and sludge bio-mass (microorganism) concentration is a
Wfinal − Winitial fundamental one in activated one in activated sludge
Ans. (c) : Total solids in sample = operation. In the activated sludge plant, the mass of
Vol. of sample test
microorganisms multiply rapidly in the presence of
65.020 − 62.485 oxygen, food and nutrients. Activated sludge process
= × 10 6

50 operate towards the end of the log phase and in the

Inorganic fraction = 50700 mg/L declining/stationary growth phase.

OPSC Civil Services Exam. 2006 291 YCT

 63. Assertion (A) : A sewage lagoon is a suspended-
culture biological system consisting of a large,
shallow earthen basin in which sewage is retained
long enough for natural purification processes to
provide the necessary degree of treatment.
Reason (R) : The natural purification occurs in
the presence of oxygen, which is largely provided
by artificial aeration.
62. Match List-I (Terms) with List-II (Definitions)
and select the correct answer using the codes Codes:
given below the lists (a) Both (A) and (R) are true and (R) is the
List-I List-II correct explanation of (A).
(Terms) (Definitions) (b) Both (A) and (R) are true but (R) is NOT a
correct explanation of (A).
A. Concentrated 1 Whose size, shape
Suspension and specific gravity (c) (A) is true, but (R) is false.
do not change with (d) (A) is false, but (R) is true.
time. Ans. (a) : A waste water pond, also called stabilization
B. Flocculating 2 Concentration of pond, oxidation pond and sewage lagoon comprises a
Particles particles is not large shallow earthen basin in which waste water is
sufficient to cause retained long enough for natural purification processes
significant to provide the necessary degree of treatment. Al least a
displacement of part of the system must be aerobic to produce an
water as they settle. acceptable effluent. Even though some oxygen is
C. Dilute 3 Whose surface provided by the diffusion from the air, the bulk of the
Suspension properties are such oxygen in ponds is provided by photosynthesis.
that they aggregate Lagoons are distinguished from ponds, as oxygen for
or coalesce with lagoons is provided by artificial aeration.
other particles upon
contact. 64. The highest power-to-weight ratio amongst the
following vehicles:
D. Discrete Particles 4 Particles are close
enough together so (a) Car (b) Truck
that their velocity (c) Motorcycle (d) Bicycle
field overlaps with
Ans. (a) : Power to weight ratio–Power-to-weight
those of
ratio is the engine's power output to the weight of the
neighbouring
particles. vehicle, it evaluates engine performance, regardless of
vehicle size. Generally, a power-to-weight ratio value is
Codes: a compromise between comfort, fuel economy and
A B C D emissions. The power-to-weight ratio varies based on
(a) 2 1 4 3 the speed range of the vehicle. The ratio reaches its
(b) 4 3 2 1 peak value as speed increases and begins declining as
the speed continues to rise. Power-to-weight ratio has a
(c) 2 3 4 1 direct relation to the maximum acceleration of a
(d) 4 1 2 3 vehicle. Diesel vehicles generally have a smaller power-
Ans. (b) : Concentrated suspension–Particles are to-weight ratio in comparison to gasoline vehicles.
close enough together so that their velocity field Brake power developed
overlaps with those of neighbouring particles. Power-to-weight ratio =
Laden weight of vehicle
Flocculating particles–Whose surface properties are
such that they aggregate or coalesce with other particles 65. The standard lane width in multi-lane roads as
upon contact. per IRC in highways
Dilute suspension–Construction of particles is not (a) 3.6 m (b) 5.75 m
sufficient to cause significant displacement of water as (c) 7.0 m (d) 3.5 m
they settle. Ans. (d) : Width of carriageway for various classes
Discrete Particles–Whose size, shape and specific of roads
gravity do not change with time.
OPSC Civil Services Exam. 2006 292 YCT
 According to IRC Width of carriageway 35 7
n2 = 3.5 = =
(meters) 10 × 100 200
Class of Singe Two Two Multi lane 7 1
n = n2 – n1 = –
Roads lane lanes lanes pavements 200 200
without with per lane 6 3
raised raised n= =
200 100
kerbs kerbs
Ns 2
National and 3.75 7.0 7.5 3.5 L=
1.5 + 0.0355
state
3
× ( 270 )
2
highways
L= 100
Major district 3.75 1.5 + 0.0355 × 270
roads L = 201.20 m
Other district 3.75 68. In a flexible pavement the layer below the top
roads surface is
Village roads 3.75 (a) Sub-grade (b) Sub-base
(c) Base (d) None of these
66. Extra widening on the horizontal curves is Ans. (c) : Flexible pavement–It is a pavement structure
provided for that maintains intimate contact with and distributes
(a) Accommodating off-tracking. loads to subgrade and depends on aggregate interlock,
(b) To counter various forces at sharp turns. particle friction and cohesion for stability. It consists of
(c) For extra safety. a series of layers with highest quality materials at or
near the surface.
(d) None of these.
Ans. (a) : Extra widening of pavement on horizontal
curve–It is the extra width provided at horizontal
curves. it is usually provided for curves of radius less
than 300 m.
Purpose of Extra widening of roads–
• When a vehicle takes a turn on horizontal curve, rear
wheels do not follow front wheels which is called as off
tracking.
• Drivers tend to take outer edge for better visibility.
• Psychological tendency to maintain larger clearance
between vehicles.
Extra widening (we) = wm + wps 69. Bitumen is obtained from
(a) Destructive distillation of coke
nl 2 V
we = + (b) Distillation of petroleum
2R 9.5 R (c) Fractional distillation of petroleum
where, n = number of traffic lanes (d) Mixing solvents in tar
l = length of wheel base Ans. (c) : Bitumen is obtained from the fractional
distillation of petroleum. The bitumen is the binding
R = Radius of horizontal curve (m) material which is present is asphalt. It is also sometimes
V = Design speed (kmph) called as the mineral tax. It is obtained by partial
67. The stopping sight distance (270 m) in a sag distillation of crude petroleum. It is chemically hydro-
vertical curve (with tangent slopes + 0.5% and carbon. Bitumen is insoluble in water but it completely
dissolves in carbon-bi-sulphide, chloroform, benzol,
+ 3.5%) is greater than the curve length. The coal tar, naptha, alkaline, carbonates petroleum spirit
curve length when the rest other factors are and oil of turpentine. It is found on analysis to compose
taken as per IRC standard, is of 87 percent carbon, 11 percent hydrogen and 2 percent
(a) 266 m (b) 200 m oxygen by weight.
(c) 175 m (d) None of these 70. In a flexible pavement, the following four
Ans. (b) : Given, materials with the CBR values given are
available : 80%, 60%, 15% and 4%. Indicate
Stopping sight distance (s) = 270
the order (top to bottom) in which the
5 1 materials are to be placed for making a good
Slope n1 = +0.5 = =
10 × 100 200 pavement

OPSC Civil Services Exam. 2006 293 YCT

 (a) 4%, 15%, 60%, 80% Ans. (b) : Gauges in Railway Track–It is the clear
(b) 4%, 80%, 15%, 60% distance between inner faces of two track rails
(c) 80%, 60%, 15%, 4% Broad gauge = 1.676 m
(d) 4%, 60%, 80%, 15% Narrow gauge = 0.762 m
Ans. (c) : Increasing order of CBR value of various Meter gauge = 1.0 m
road layers is Standard gauge = 1.435 m (Delhi metro, Mumbai,
subgrade < sub base < base < surface Ahemdabad bullet train)
• At top material of highest CBR value should be used 74. The number of crossing indicates the
as it is directly subjected to abrasion of wheels of (a) number of rails that criss-cross a joint.
vehicles. (b) angle between main rail and crossing rail.
71. Ideally a permanent way comprises of (c) number of sleepers all through the crossing.
(a) rails, sleepers, ballast cushion and sub ballast (d) number of sleepers all through the crossing
moorum. plus a number depending on the connecting
(b) rails, sleepers and ballast cushion. rail length.
(c) rails, sleepers, ballast cushion and sub grade. Ans. (a) : A crossing or frog is a device introduced at
(d) rails, sleepers, sub ballast murum and sub the point where two gauge faces cross each other to
grade. permit the flanges of a railway vehicle to pass from one
track to another.
Ans. (b) : Requirements of an ideal permanent way–
The following are the requirements of an ideal 75. In a broad gauge track with 1 in 8 turnout the
permanent way. curve lead (in m) is
• A uniform and correct gauge are necessary. (a) 17.00 (b) 34.00
• The level of both the rails should be the same on a (c) 14.25 (d) 28.49
horizontal track. Ans. (d) : Given that,
• At the curved locations, superelevation in a permanent N=8
way must be provided properly by raising the outer rail Gauge length for broad gauge (G) = 1.676 m
of the track. Curve lead = 2 GN
• The design of the permanent way should be done in Curve lead = 2 × 1.676 × 8
such a way that the load is uniformly distributed
between two rails. Curve lead = 16 × 1.676
Component of permanent way Curve lead = 26.816
Curve lead = 26.81
• Rails
• Sleepers 76. The component in the airport system where the
aircrafts land/take off
• Ballast cushion
(a) Apron (b) Taxiway
• Fixtures and fastening
(c) Runway (d) None of these
• Cross section of permanent
Ans. (c) : The component in the airport system where
72. The gauge widths (in m) for broad, standard the aircrafts land/take off is called runway. Runway is
and narrow gauges respectively are oriented such that the direction of wind in opposite to
(a) 1.767, 1.650, 0.760 the direction of landing and take off.
(b) 1.676, 1.500, 0.676 77. The external aid for en route overwater
(c) 1.676, 1.435, 0.762 navigation of aircraft
(d) 1.876, 1.656, 0.800 (a) Doppler navigation
Ans. (c) : Gauges in Railway Track–It is the clear (b) Inertial navigation
distance between inner faces of two track rails (c) Celestial navigation
Broad gauge = 1.676 m (d) Long range navigation
Narrow gauge = 0.762 m Ans. (d) :
Meter gauge = 1.0 m 78. A circuit/loop in a construction network
Standard gauge = 1.435 m (Delhi metro, Mumbai, represents
Ahemdabad bullet train) (a) cyclic pattern of activities
73. The gauge width (in m) for a metre gauge (b) illogical interpretation of activities
(a) 0.760 (b) 1.00 (c) Routine maintenance activities
(c) 1.656 (d) None of these (d) None of these
OPSC Civil Services Exam. 2006 294 YCT
Ans. (b): In a network diagram there should be unique to reflect uncertainty in the times of completion of
start event & unique end event but in cyclic loop both activity. Due to involvement of uncertainty, three times
start & end event will be same which is illogical are assigned to each activity of the network.
sequence of activities. 83. For building a culvert across a road, the
79. A dummy activity is introduced to optimistic, pessimistic and most likely times for
(a) for getting logical dependencies completion were given as 4, 7, 5, days
(b) make the network diagram look elegant respectively. The mean or expected value and
variance time (in the same order) of completion
(c) make network efficient of this activity in days are
(d) None of these (a) 5, 4 (b) 5.17, 0.25
Ans. (a) : Dummy activity–It is an artificial activity (c) 5.33, 4 (d) 5.33, 0.50
represented by dashed arrows which neither consumes
any time nor resources. Activity G is dummy activity. Ans. (b) : Given that,
Dummies maintain the logic of network diagram & Optimistic time (to) = 4
keeps the numbering system of the network unique. Pessimistic time (tp) = 7
Dummy activities can also be used to represent the Most likely (tm) = 5
precedence logic in a network. Expected completion time of value–
t o + 4t m + t p
tE =
6
4 + 4×5+ 7
tE =
6
b
4 + 20 + 7
80. A critical path is essentially used to find ..... of a tE =
6
project
31
(a) total float (b) interfering float tE =
6
(c) free float (d) completion time
Ans. (d) : Critical path–Critical path is the shortest t E = 5.17
path in terms of time in which the project can be 2
completed as soon as possible or as early as possible. t −t 
σ2 =  p o 
Events on critical path, will have zero or minimum  6 
slack.
2
81. Critical Activities are those that have  7−4
= 
(a) zero float time  6 
(b) infinite float time  3
2

(c) small but negligible float time = 

6
(d) none of these
1
Ans. (a) : If maximum available time for an activity is Variance σ2 = = 0.25
equal to the activity time, then total float will be zero. 4
Such activities are called critical activities.
1
Critical activities are those activities whose float is zero σ=
and they lie on the critical path. 4
82. Program evaluation and review technique 1
assumes that activity completion times are σ=
2
(a) deterministic
σ = 0.5
(b) time dependent
(c) probabilistic 84. Economy of scale exists when
(d) fuzzy (a) Fixed and variable costs are nearly equal.
Ans. (c) : PERT was developed by US Navy in late (b) Fixed costs are high compared to variable
1950's to accelerate the development of Polaris Fleet costs.
Ballistic Missile. PERT is used in R & D type projects. (c) Fixed costs are low compared to variable
PERT stands for Program Evaluation and Review costs.
Technique. PERT also belongs to the family of Network (d) No specific relation between fixed and
Analysis Technique. PERT assumes probabilistic model variable costs.
OPSC Civil Services Exam. 2006 295 YCT
Ans.(c):Economy of scale–It refers to the cost (c) metre, kilogram and second
advantage experienced by a firm when it increases the (d) millimeter, Newton and second
level of output. It is inversely relationship between per
Ans. (c) : The international system of units is based on
unit fixed cost and the quantity produced. The greater
seven units. It has been observed that although the three
the quantity of output produced, the lower the per-unit
basic units of length, mass and time in International
fixed cost. system of units are metre, kilogram and second.
85. The worth of an infrastructural asset changes SI Fundamental units
with time due to
(a) inflation/deflation Physical Quantity SI Unit Abbreviation
(b) utility it provides 1. Length Metre m
(c) Both (a) and (b) 2. Mass Kilogram Kg
(d) None of these
3. Time Second S
Ans. (c) : Worth of an asset changes due to–
4. Electric current Ampere A
A. Change in its utility
B. Technological advancements 5. Thermodynamical Kelvin K
temperature
C. Inflation/deflation
D. Depreciation 6. Luminous intensity Candela Cd
E Wear & tear 7. Amount of substance Mole mol
F. Reduction in efficiency 89. Varignon's theorem of moments states
86. The present worth of Rs. 100 cr asset valued 2 (a) arithmetical sum of the moments of two
years ahead, with an annual interest rate of forces about any point, is equal to the
10% moments of their resultant about that point.
(a) Rs. 121.00 cr (b) Rs. 82.64 cr (b) arithmetical sum of the moments of the forces
(c) Rs. 100 cr (d) Rs. 110 cr about any point in their plane, is equal to the
Ans. (b) : Given that, moment of their resultant about the point.
Cash flow at a period (C) = 100 cr (c) algebraic sum of the moments of the forces
number of period (n) = 2 yrs. about any point in their plane, is equal to the
moment of their resultant about that point
rate of interest (r) = 10% = 0.1
(d) algebraic sum of the moments of two forces
C about any point, is equal to the moment of
Present value or worth (PV) =
(1 + r )
n
their resultant about that point.
100 Ans. (c) : Varignon's Theorem (or Principle of
PV = moments)–"The moment of a force about any point is
(1 + 0.1)
2

equal to the algebraic sum of the moments its

100 components about that point." As stated by Varignon's
PV =
(1.1) theorem.
2

Principle of moments states that the moment of the

PV = 82.64 cr
resultant of a number of forces about any point is equal
87. Variable costs refer to to the algebraic sum of the moments of all the forces of
(a) cost incurred for use/operation. the system about the same point.
(b) fixed expenses irrespective of use/operation. 90. A force of 800 N acts on a bracket as shown in
(c) costs that vary with time. Figure. The moment of the force about B is
(d) none of these equal to
Ans. (c) : Variable costs are expenses that do not
remain constant. Instead they vary with time and
production levels. Variable expenses are directly
proportional to production quantity. It is the opposite of
fixed costs which remain constant irrespective of
production levels.
88. The base unit of length, mass and time in
International system of units are
(a) metre, kilonewton and second (a) 203 N-m (b) 230 N-m
(b) metre, kilogram and minute (c) 302 N-m (d) 320 N-m
OPSC Civil Services Exam. 2006 296 YCT
Ans. (a): The moment of the force about 92.
A 20-Mg rail road car moving at a speed of
0.50 m/s to the right collides with a 35-Mg car
which is at rest. If after collision the 35-Mg car
is observed to move to the right at a speed of
0.30 m/s, then the co-efficient of restitution
between the two car is
(a) 0.56 (b) 0.65
(c) 0.55 (d) 0.66
Ans. (b) : Given that,
M1 = 20 Mg
MB = [800 sin 60º × 200 + 800 cos 60º × 160] × 10–3 V1 = 0.50 m/s
3 1 M2 = 35 Mg
MB = 800 × × 0.2 + 800 × × 0.16 V2 = 0.3 m/s
2 2
We know V2' = 0 (at rest)
MB = 138.56 + 64
M1V1 = M2V2' + M1 V1'
MB = 202.56 N-m
20 × 0.50 = 35 × 0.3 + 20 × V1'
B  203 N-m 10 = 105 + 20 V1'
91. Three loads are applied to the beam as shown 20 V1' = – 0.5
in Figure. Neglecting the weight of the beam, 0.5
the vertical reactions at A and B are V1' = = − 0.025 m / s
20
V1' = – 0.025 m/s
V − V1'
Coefficient of restitution (e) = 2 '
V1 − V2
0.3 − ( − 0.025)
e=
(a) 30 kN and 105 kN 0.5 − 0
(b) 35 kN and 100 kN 0.3 + 0.025
e=
(c) 105 kN and 30 kN 0.5
(d) 100 kN and 30 kN 0.325
e=
0.5
Ans. (a) : From Equilibrium conditions–
e = 0.65
∑ H = 0, ∑ V = 0, ∑ M = 0
93. The mass moment of inertia of a slender roof
∑V = 0 length 'l' and mass 'm' with respect to axis
RA + RB = 75 + 30 + 30 perpendicular to the rod and passing through
RA + RB = 135 kN ..... (i) one end of the rod is
(a) ml2/4 (b) m2l/3
2
(c) ml /3 (d) m2l2/3
Ans. (c) : Given that,
Mass of rod = m
length of rod = l
If the rod is supported at an end, either l, or l2 is zero
∑M = 0
depending upon at which end it is supported. For
Moment about A support at its left end.
R A × 0 + 75 × 1.5 − R B × 4.5 + 30 × 5.5 + 30 × 6.5 = 0 l1 = 0, l2 = l
RB × 4.5 = 472.5 1 1
Io = pal 3 = ml 2
472.5 3 3
RB =
4.5 If the rod is supported at its mid point
RB = 105 kN l
l1 = l2 =
From equation (i) 2
RA + 105 = 135 1 1
Io = pal 3 = ml 2
RA = 30 kN 12 2

OPSC Civil Services Exam. 2006 297 YCT

94. In portal frame with columns of same height + PL
and section having fixed base, the sum of the M BA =
8
displacement factors of column in Kani's
method is −Pab 2
(a) –2.0 (b) –1.5 M AB =
L2
(c) –1.0 (d) –0.5 2.
+ Pa 2 b
Ans. (b) : Displacement Factor in Kani's method– M BA =
L2
Kani's method is a similar method as moment
distribution method which is often used in Analysis of − wL2
continuous beam or indeterminate frames. M AB =
12
In Kani's method displacement 3.
+ wL2
Factor for column is given by M BA =
12
k ij
Dij = −1.5 − wL2
∑ k ij M AB =
30
4.
I + wL2
where, k ij = i M BA =
hi 20
−5 2
M AB = wL
96
5.
+5 2
M BA = wL
96
Where lij and hij is the moment of inertia and height of + Mo
the column connecting ith and jth node of the frame. M AB =
4
I I 6.
k AB = 1 , k CB = 2 + Mo
h1 h2 M BA =
4
k AB k CD 96. Match List-I with List-II and select the correct
D AB = −1.5 DCD = −1.5 answer using given codes:
k AB + k CD k AB + k CD
95. Match List-I with List-II and select the correct
answer

Codes:
Codes: A B C D
A B C D (a) 4 3 1 2
(a) 4 3 1 2 (b) 4 1 2 3
(b) 4 1 2 3 (c) 3 4 1 2
(c) 3 4 1 2 (d) 3 4 2 1
(d) 3 4 2 1 Ans. (d) : j
Ans. (c) : Fixed end moment Cross section Shear stress
Sr. 1. Rectangle
Loading fixed and moments
No.
3  v 
τ max = τavg  τ avg = bd 
−PL 2  
M AB =
1. 8 3v
τ max =
2bd

OPSC Civil Services Exam. 2006 298 YCT

 2. Triangle Equating equation (i) and (ii) we get
3v Rl 3 5wl 3
τ max = =
bd 3EI 48EI
5w
R=
3. Circular 16
4
τ NA = τavg ( at NA ) R=
10w
3 32
4v 4v × 4
τ max = = 98. The movable bracket shown in Figure may be
3A 3πd 2 placed at any height on the 60 mm diameter
 4v  pipe. If the coefficient of static friction between
 τavg = 2
πd  the pipe and bracket is 0.25, minimum distance
 x at which the load W can be supported
16v (neglecting the weight of bracket) is
τ max =
3πd 2
4. Diamond
9  2v 
τ max = τavg  τavg = d 2 
8  
9 2v 9v
τ max = × 2 = 2
8 d 4d
97. The cantilever beam AB with loading is shown
in Fig. 10. The deflection at the free end B, will
be zero, if the value of R is (a) 200 mm (b) 240 mm
(c) 250 mm (d) 220 mm
Ans. (b) : Given that,
Coefficient of static friction between the pipe and
bracket = 0.25
When w is placed at the minimum distance x from the
axis of the pipe, the bracket is just about to slip and the
12 w 8w force of friction at A and B have reached their minimum
(a) (b) values.
32 32
15 w 10 w FA = µsRA = 0 0.25 RA, FB = µsRB = 0.25 RB
(c) (d) From Equilibrium conditions
32 32
Ans. (d) : Let R = Reaction at the rigid propped →∑Fx = 0 ↑∑Fy = 0
RA – RB= 0 FA + FB – w = 0 RA×120–FA×60
RA = RB 0.25RA+0.25RB = w (x–30)
0.25RA+0.25RA= w 120 RA–60FA –
0.50 RA = w wx + 30w = 0
Now we know that down ward deflection at point B due 2w = RA 120 × 2w – 60 ×
to load w 0.25 × 2w – wx +
3 2 30w = 0
l l 240w – 30w – wx
w  w 
l
=   +   × 
2 2 + 30w = 0
3EI 2EI 2 240w = wx
x = 240 mm
wl 3 wl 3
= +
24EI 16EI 99. A pin-jointed simple truss is shown in figure. In
which the zero force members are
5wl 3
= ..... (i)
48EI
The upward deflection at the point B due to prop
Rl 3
reaction R alone = = ..... (ii)
3EI

OPSC Civil Services Exam. 2006 299 YCT

 (a) AB, BC, CD and DE
(b) BC, CD, DE and BE
(c) AB, BC, CD and AE
(d) AB, CD, BD and DE
Ans. (c) : Zero force members–
• If three members join at a point and out of them, two
are collinear & also no external load acts at joint, then
the third member is a zero force member.
If only two non-linear members exist at a truss joint and
no external force or support reaction is applied to the
joint then the members must be zero force member
Codes:
100. The flexibility matrix of the cantilever shown in
Figure is (a) 1 and 2 (b) 2 and 3
(c) 1 and 4 (d) 2 and 4
Ans. (b) : Condition for sway–
(i) If the supports are not in the same position, sway
occurs.
(ii) If the applied load is not acting at the centre of the
l  3l 2l 2  l  2l 2 3l  structure (eccentric loading) sway will occur.
(a)   (b)   (iii) When the loading is asymmetrical
6EI  2l 2 6  6EI  3l 6
(iv) Different end conditions of the columns of the
l 3l 2 6  l  6 3l  frame.
(c)   (d)
6EI  3l 2l 2  6EI 3l 2l 2  (v) Non-uniform sections of the members
Ans. (b) : (vi) Horizontal loading on column of the frame,
settlement of the supports of the frame.
102. A motorist in travelling on a curved section of
highway of radius 750 m at a speed of 100
km/h. The motorist suddenly applies the
brakes. If after 8 second, the speed has been
reduced to 75 km/h, then the acceleration of the
Pl 3 automobile immediately after the brakes have
F11 =
3EI been applied is
l3 (a) 1.351 m/s2 (b) 1.135 m/s2
P = 1 ⇒ F11 =
3EI (c) 1.531 m/s2 (d) 1.315 m/s2
Pl 2 l2 Ans. (a) : Given,
F21 = ⇒ R = 750 m
2EI 2EI
Ml l u = 100 km/h = 27.78 m/sec
F22 = ⇒ v = 75 km/h = 20.83 m/sec
EI 2EI
t = 8 sec
l2
F12 = Tangential component of acceleration
2EI
∆v ( v − u )
 l3 l2  at = =
  ∆t t
3 EI 2 EI 
[ ]  2
F = ( 20.83 − 27.78 ) = −6.95 = −0.86875 m / sec2
l l  =
  8 8
 EI
2 EI 
Normal component of acceleration
 2l 2 3l  v 2 ( 27.78 )
2
l
=  an = = = 1.0289 m / sec2
6EI  3l 6 R 750
101. Which of the following structures will
(1.0289 ) + ( −0.8675)
2 2
a 2 = a 2n + a 2t =
experience sway? Select the correct answer
from the given codes? a = 1.3458 m/sec2

OPSC Civil Services Exam. 2006 300 YCT

103. A three hinged parabolic arch with hinged Codes:
supports at the same level and third hinge at A B C D
the central rise 'h', is subjected to uniformly
distributed load of w/m throughout its (a) 2 4 3 1
horizontal span, 'l'. The moment and (b) 3 4 1 2
horizontal thrust at quarter span are (c) 4 2 3 1
respectively
(d) 2 3 4 1
wl 2 wl 2 wl 2 wl 2
(a) and (b) and Ans. (d) :
8 8h 8h 8
Type of beam with type B.M diagram
wl 2 wl 2 wl 2
(c) and (d) 0 and of loading
8h 8h 8h
Ans. (d) : j

We know that, In case of 3 hinged arch subjected to

UDL
BM at every section = 0.
wl 2
and H =
8h
106. The ratio of maximum deflection of beam 'P' to
Where, w = loading intensity
maximum deflection of beam 'Q' as shown in
l = span Figure is
h = central rise
104. In a simply supported beam subjected to the
action of a train of moving loads, the bending
moment at a given section maximizes when
(a) the total load acting on the portion of the
beam left to the section is equal to that acting
on the portion of the right to the section.
(b) the average load acting on the portion of the (a) 2 (b) 3
beam left to the section is equal to that acting
on the portion of the beam left to the section (c) 5 (d) 7
is equal to that acting on the portion of the Ans. (c) : In a simple supported beam carrying UDL on
beam right to the section. whole span, the maximum deflection of beam
(c) the section and the C.G. of the load system 5wl 4
are equidistant from the center of the span. ( δmax )s.s =
384EI
(d) the section and the center of the span are
equidistant from the C.G. of the load system. In fixed beam carrying UDL on whole span the
maximum deflection of beam.
Ans. (c) : In a simple supported beam subjected to the
action of a train of moving loads, the bending moment wl 4
at a given section maximizes when the section and the ( δmax )fixed =
384
C.G of the load system are equidistant from the centre
of the span. 5wl 4
105. Match List-I with List-II and select the correct Ratio = 384EI =5
answer using codes as given below: wl 4
384EI
107. The bending moment of the conjugate beam
has the same value of the function of
corresponding real beam known as
(a) slope (b) shear force
(c) bending moment (d) deflection
OPSC Civil Services Exam. 2006 301 YCT
Ans. (d): • The slope and deflection at any section in
the real beam are given by the shear and bending
moment at that section in the conjugate beam.
• The elastic curve of the real beam is given by the
bending moment diagram of the conjugate beam.
• The end slope and end deflection of the real beam are Hence option (d) is correct
given by the end reaction and end moment of the 110. The ratio of shear stress and shear strain of an
conjugate beam. elastic material is termed as
108. The conjugate beam for the real beam in (a) Modulus of Rigidity (b) Young's Modulus
Figure is (c) Modulus of Elasticity (d) None of the above
Ans. (a) :
Longitudinal stress
Modulus of Elasticity, E =
Longitudinal strain
Normal stress
Bulk modulus, K =
Ans. (b) : Volumetricstrain

Real Beam Conjugate Beam Shear stress

Modulus of Rigidity, G =
Shear strain
111. Match List-I with the List-II and select the
correct answer using given codes:
List-I List-II
(Euler's Formula) (End conditions of
long column)
2π2 EI One end fixed
A. P= 1.
l2 other end free

π 2 EI
B. P= 2. Both ends fixed
4l 2
π 2 EI
C. P= 3. Both ends hinged
l2
4π2 EI One end fixed,
D. P= 4.
other end hinged
l2
Hence
Codes:
A B C D
(a) 4 1 3 2
Real beam Conjugate beam
(b) 3 1 2 4
109. The ordinates of the influence line for support
(c) 4 2 3 1
moment of a cantilever beam of length 'l' at
free end and support are (d) 3 4 2 1
(a) l and –l respectively Ans. (a) :
(b) zero and zero respectively End condition Leff
(c) zero and l respectively
Both end fixed l/2
(d) none of the above
Both end hinged l
Ans. (d) :
One end fixed & other is 2l
free
One end is fixed & other l
Using Muller Breslau principle- is hinged 2
Provide unit rotation at B

OPSC Civil Services Exam. 2006 302 YCT

112. The magnitude of the concentrated load to be tension/compression. failure.
applied at the free end B of a cantilever beam
shown in figure to produce deflection '∆' at the B. Maximum/minimum 2. St. Venant's
point (B) is principal strain equal theory of
to the maximum / elastic failure.
minimum strain at
the elastic limit in
4EI ∆ 3EI ∆ simple tension /
(a) (b) compression
l3 l3
2EI ∆ 3EI ∆ C. Maximum shear 3. Rankine's
(c) (d) stress must equal the theory of
l2 l3
maximum shear elastic failure.
Ans. (b) : stress at elastic limit
in simple tension.
D. Strain energy per 4. Haigh's theory
unit volume in it of elastic
equals the maximum failure
In cantilever beam, the load is applied at free end the
magnitude of deflection on free end. strain energy per
unit in the material
wl 3 at elastic limit in
∆=
3EI simple tension.
The magnitude of concentrated load. Codes:
3EI ∆ A B C D
w=
l3 (a) 2 3 4 1
113. A portal frame is shown in figure. The (b) 4 3 2 1
deflected shape of the frame for constant EI (c) 1 2 4 3
will be as per
(d) 3 2 1 4
Ans. (d) : Maximum principal stress theory–This
theory way proposed by Rankine.
• It states that failure will occur when the
maximum/minimum principal stress in the complex
system, reaches the value of maximum stress at the
elastic limit simple tension compression.
• This theory is valid for brittle metals.
Maximum principal strain theory–
• This theory was proposed by Saint Vennant.
• It states that the failure of a material occurs when the
major principal tensile strain reaches the strain at the
elastic limit in simple tension or when the minor
principal strain reaches the strain at elastic limit in
simple compression.
Maximum shear stress theory–This theory is also
Ans. (c) : When a structural element is subjected to called coulomb. Guest's or Treasca's theory.
bending, it curves causing one side to stretch and the • It states that the material will fail when the maximum
other to contract. shear stress in the complex. System reaches the value of
• If the flexural rigidity of the frame EI is constant, then maximum shear stress in simple tension at the elastic
the curvature is directly proportional to the bending limit.
moment as a result of the applied loading. Maximum strain energy theory–This theory was
114. Match List-I with List-II and choose the proposed by Beltrani Haigh.
correct answer from given codes: • It states that the failure of a material occurs when the
List-I List-II total strain energy in the material reaches the value of
maximum shear stress in simple tension at elastic limit.
A. Maximum/Minimum 1. Coulomb
principal stress guest's theory 115. For the given S.F. and B.M. diagrams as shown
reaches the elastic of elastic in figure the appropriate beam with loading is
limit stress in simple as in
OPSC Civil Services Exam. 2006 303 YCT
 Ans. (b): Inertial is the inharent property of body which
opposed change in its state from rest to motion or vice
versa.
117. Knowing M, I, R, E, F and Y are the bending
moment, moment of inertia, radius of
curvature modulus of elasticity, stress and
depth of neutral axis at a section in flexure, the
M E F M R F
(a) = = (b) = =
I R Y I E Y
I R F M EE F
(c) = = (d) = =
M E Y I R Y
Ans. (a) : Bending equation is
M E F
= =
I R Y
Ans. (c) : 118. A grading curve is
(a) the results of a sieve analysis.
(b) a plot of mass retained in each sieve against
particle size.
(c) the plot of cumulative fraction smaller than a
given size against the logarithm of that size.
(d) none of the above.
l l w×l l Ans. (c) : The particle size distribution curve also
RA + RB = w × × + × known as a gradation or grading curve, it is the plot of
2 2 2 2 cumulative fraction (The percentage finer) smaller than
wl wl a given size against the logarithm of that size.
= +
4 4 119. Aeolian soil is a
wl (a) soil occurring in flood plain
=
2 (b) glacial clayey soil
wl (c) soil deposited in lake
RA = RB = (d) wind-borne soil
4
wl Ans. (d) : Aeolian soil–This type of soil particles are
SF at support = transported by winds.
4
• The particle size of the soil depends upon the velocity
Max B.M equation at center of wind.
l wl l  l  • The finer particles are carried for away from the place
RB × × +
2 2 3  2  of the formation.
• A dust storm gives a visual evidence of the soil
l l wl l
RB × − × × particles carried by wind
2 2 2 6
120. A soil sample has a shrinkage limit of 10% and
wl l l wl 2 specific gravity of soil solids as 2.7. The
× − × porosity of the soil at shrinkage limit is
4 2 2 12
(a) 21.2% (b) 27.0%
wl 2 wl 2
× (c) 73.0% (d) 78.8%
8 24
Ans. (a) : ws = 10%, G = 2.7, n=?
3wl 2 − wl 2 2wl 2 wl 2 at shrinkage limit s = 1
= =
24 24 12 s.e = w.h
116. The inherent property of a body which offers 1 × e = 0.10 × 2.7 = 0.27
reluctance to change its state of rest or uniform e 0.27
motion, is n= =
1 + e 1.27
(a) momentum (b) inertia n = 0.2125
(c) mass (d) weight n = 21.2%
OPSC Civil Services Exam. 2006 304 YCT