OSSC JE MAINS CIVIL 16 March 2025 PAPER SOLUTION

Q.01. A pitot tube is used to measure Hence velocity of the wave is less than the velocity of
(a) Pressure flow, hence it cannot travel upstream. Here the
(b) Difference in pressure upstream conditions are controlled.
(c) Velocity For supercritical flow upstream is the control section
(d) Discharge
Q.03. Which of the following earth moving machines
Ans. (c) A pitot tube is a pressure measurement has the shorter cycle time?
instrument used to measure fluid flow velocity. This (a) Drag line
works on the principle that when a fluid is brought to (b) Hoe
rest then the velocity head is converted to pressure (c) Clam shell
head. By application of Bernoulli's equation one can (d) Dipper shovel
find the velocity of the flow.
Ans. (b*) Power shovel, backhoe, dragline and
clamshell, all are used as excavation equipment under a
different set of condition and requirements.
The table below shows a comparison between all
different types of excavating equipment.
Item of Power Back Drag Time Clam Shell
Comparison Shovel Hoe
Excavation Good Good Not Good Poor
in hard soil
or rock
Q.02.For subcritical flow in an open channel, the control
section for gradually varied flow profile is: Excavation Poor Poor Moderately Moderately
in wet soil good good
(a) at the downstream end
or mud
(b) at the upstream end
(c) at both ends Distance Small Small Long Long
(d) at any intermediate section between
Ans. (a) In an open channel flow control section is a footing and
section where for a given discharge the depth of flow digging
is known or it can be controlled. Loading Very Good Moderately Precise
The Froude number (FR) is given by Efficiency Good good but slow
𝒗
FR = , √𝒈𝒉 − Celerity
√𝒈𝒉 Cycle time Short Slightly More than More than
V- Velocity of flow more power the other
H - Hydraulic depth than shovel equipment
power
Celerity is the velocity of a wave when the flow is
shovel
disturbed.
Subcritical flow: Power shovel is known as a hoe and Dipper shovel is
In Subcritical flow FR < 1, known as a backhoe.
*As per board answer key correct option is d but correct
√𝒈𝒉 > V
option is b.
Hence velocity of the wave is more than the velocity of
Q.04. Prefabrication is a construction method.
flow, hence it can travel upstream. Here the
downstream conditions are controlled. (a) automotive
(b) modern
For subcritical flow downstream is the control section.
(c) established
Supercritical flow: (d) industrialised
In Supercritical flow FR > 1,
Ans. (b and d*) Prefabricated structures, commonly
√𝒈𝒉 < V known as prefabs, are buildings where key
components, including walls, roofs, and floors, are
fabricated within a controlled factory or

OSSC JE MAINS CIVIL 16/03/2025 1 Civil Ki Goli

manufacturing facility. These components can be fully 3. Drag Lines
or partially assembled at the production site and later 4. Bulldozer
transported to the intended location. 5. Wheel Tractor Scraper
6. Back Hoes
7. Dredgers
8. Ripper
9. Motor Grader
Excavator:- It is the oldest type of machine which
removes earth.
*There is confusion in this question . As per board The general purpose of the excavator is excavation
answer key correct option is d but correct option is b work but other than that it is also used for different
and d both . purposes like heavy lifting, digging of trenches, holes,
Q.05. Which are the characteristics of Materials used for foundations, river dredging, cutting of trees, etc.
construction of PFs (Permanent Formworks) Q.07. In which of the following types, does the
(a) thermal insulation property entrepreneur work as a middle man?
(b) combustibility (a) Joint (b) Private
(c) heavier materials (c) Social (d) Trading
(d) none of the above Ans. (d) In the trading type of entrepreneurship, the
Ans. (a) Permanent formworks (PFs) are designed to entrepreneur acts as a middleman or intermediary.
remain in place after the concrete has set, becoming a Their role is to buy products or goods from
part of the structure. The materials used for manufacturers or suppliers and sell them to customers
permanent formworks typically have the following or retailers. They do not produce the goods themselves
characteristics: but facilitate the movement of goods from producers
Thermal Insulation Property: Materials like expanded to consumers, earning a profit from the margin
polystyrene (EPS) or other insulating materials are between the buying and selling prices.
often used in permanent formworks to provide Q.08. A vertical photograph was taken at an altitude of
thermal insulation to the structure. This helps in 2000 m above MSL. If the focal length of the camera is
improving energy efficiency. 20 m, the scale of photograph for a terrain lying at an
Lightweight: Permanent formwork materials are elevation of 1000 m is:
usually lightweight to ease handling and installation. (a) 1:50 (b) 1:100
Heavier materials (option c) are not typically used. (c) 1:1000 (d) 1:25
Non-combustibility: Materials used for permanent
Ans. (a) Given
formworks are often required to be non-combustible
f = focal length of camera = 20 m
or have fire-resistant properties to ensure safety.
Combustibility (option b) is generally not a desired H = Average height of photograph from MSL = 2000 m
characteristic. h = Average height of terrain from MSL = 1000 m
Durability: The materials must be durable and we know
resistant to environmental factors like moisture, Scale of the photograph
corrosion, and chemical attacks. 𝒇
Scale =
𝑯−𝒉
Q.06. The oldest type of machine which removes earth 𝟐𝟎
is: =
𝟐𝟎𝟎𝟎−𝟏𝟎𝟎𝟎
𝟏
(a) Escalator =
𝟓𝟎
(b) Excavator
Q.09. Which apparatus is generally used to measure the
(c) Elevator soundness of the cement?
(d) Bulldozer (a) Vicat Apparatus
Ans. (b) Following are the different heavy construction (b) Le-Chatelier's apparatus
equipment used for earthwork and other works are as (c) Soundness meter
follows: (d) Abrams apparatus
1. Excavator Ans. (b) Soundness Test [IS : 4031 (Part-III)]
2. Power Shovel

OSSC JE MAINS CIVIL 16/03/2025 2 Civil Ki Goli

• Soundness means the ability to resist volume To get the measurement of materials and work, length
expansion. This test is conducted to detect change in of long wall or short wall, centre line lengths of
volume after setting. individual walls is calculated first. Then the length of
• This test is done with the help of Le - chatelier long wall, (out to out) may be calculated after adding
apparatus and Autoclave test. half breadth at each end to its centre line length.
Le - chatelier's Method (IS : 4031 Part 3-1988) Thus the length of short wall measured into in and may
• Unsoundness of cement is due to free lime only. be found by deducting half breadth from its centre line
length at each end. These lengths are multiplied by
• Weight of cement sample = 100 gm.
breadth and depth to get quantities.
• Quantity of water = 0.78 P.
The length of long wall usually decreases from earth
• Result is given in 'mm'. work to brick work in super structure while the short
• Temperature - 27 ± 20C, humidity 65 ± 5%. wall increases.
Long wall length out to out = centre to centre length +
half breadth on one side + half breadth on other side.
If at both the ends breadth is same
*As per board answer key correct option is b but
correct answer is option a and c both.
Q.11. A cement concrete road is 1000m long, 8m wide
and 15cm thick over the sub base of 10 cm thick gravel.
The box cutting in road crust is:
Autoclave Test (IS : 4031 - Part 3-1988)
(a) 500 m3
• It is used to determine soundness of cement due to
(b) 1000 m3
both free lime and free magnesia.
(c) 1500 m3
• Internal mould dimension = 25 × 25 × 250 mm. (d) 2000 m3
• % expansion of the mould for OPC should not exceed
Ans. (d) Given:
0.8%.
Length of road (L) = 1000 m
Width of road (W) = 8 m
Thickness of concrete (T₁) = 15 cm = 0.15 m
Thickness of gravel (T₂) = 10 cm = 0.10 m
Total thickness of road crust:
T = T1+T2=0.15+0.10=0.25 m
Volume = L × W × T
= =1000 × 8× 0.25 = 2000m3
Q.12. Berms are provided in canals if these are:
(a) Fully in excavation
(b) Partly in excavation and partly in embankment
Q.10. In long wall and short wall method of estimation (c) Fully in embankment
(d) All the above
which one of the following is correct?
(a) Short wall length in-to-in = centre to centre length Ans. (b) Berm is the horizontal distance left at ground
- one breadth level between the toe of the bank and the top edge of
(b) Short wall length in-to-in = centre to centre length cutting. The purpose of providing Berm to help the
+ one breadth channel to attain regime conditions, they protect the
(c) Long wall length out-to-out = centre to centre banks from erosion due to wave action.
length + one breadth Berms are provided in canals that are partially in
(d) Long wall length out-to-out = centre to centre excavation and partially in embankment.Berms are to
length - two breadth be provided in all cuttings when the depth of cutting
is more than 3 meters.
Ans. (a and c*) In long wall and short wall method, the
wall along the length of the room is considered to be Q.13. If the formation level of a highway has a uniform
long wall and the wall perpendicular to the length of gradient for a particular length, and the ground is also
the room is considered to be short wall. having a longitudinal slope, the earthwork may be
calculated by

OSSC JE MAINS CIVIL 16/03/2025 3 Civil Ki Goli

(a) Mid-section formula Ans. (a)
(b) Trapezoidal formula
(c) Prismoidal formula
(d) All the above
Ans. (d) Mid section formula:- In this formula, the
mean depth is to be calculated. First by averaging the
depths of two consecutive sections. From the mean-
depth the area of mid-section is to be worked out and 𝒅 √𝟐
volume of earthwork to be computed by multiplying S0 the ratio = = 1.414
𝒃 𝟏
the area of mid-section by the distance between the Q.17. A uniform beam of span L is rigidly fixed at both
two original sections. supports. It carries a uniformly distributed load W' per
Trapezoidal Formula: unit length over full span. The bending moment at mid
𝒅
Volume(V) = [First area section + Last area section + span is:
𝟐
2(sum of other areas)] (a) 𝐰𝐋𝟐 /𝟖
Prismoidal Formula: (b) 𝐰𝐋𝟐 /𝟏𝟐
𝒅 (c) 𝐰𝐋𝟐 /𝟏𝟔
V= [First area section + Last area section + 4(sum of (d) 𝐰𝐋𝟐 /𝟐𝟒
𝟑
even areas) + 2( sum of odd area]
Ans. (d)
All above can be used for earthwork estimation if
highway has a uniform gradient.
Q.14. A ratio of moment carrying capacity of a circular
beam of diameter D and a square beam of size D is:
(a) 𝛑/𝟒 (b) 𝟑𝛑/𝟖
(c) 𝛑/𝟑 (d) 𝟑𝛑/𝟏𝟔
n

Ans. (d) Given

io
ct

Side of square = D, diameter of circle = D

n
Ju

For a circular beam of diameter D, the section modulus

il
iv

is: Applied Moment equilibrium at X,

C

𝝅𝒅𝟑 Σ Μx = 0
Zcircular =
𝟑𝟐
𝒘𝑳 𝑳 𝒘𝑳𝟐 𝒘𝑳 𝑳
For a square beam of side DDD, the section modulus is: × +𝑴+ = ×
𝟐 𝟒 𝟏𝟐 𝟐 𝟐
𝑫𝟑
Zsquare = 𝒘𝑳𝟐 𝒘𝑳𝟐 𝒘𝑳𝟐
𝟔 𝑴= − −
𝟒 𝟖 𝟏𝟐
The ratio of the moment-carrying capacity is: 𝒘𝑳𝟐
𝝅𝑫𝟑
𝑴=
𝟐𝟒
𝒛𝑪𝒊𝒓𝒄𝒖𝒍𝒂𝒓 𝟑𝟐 𝟑𝝅
= 𝑫𝟑
= Q.18. The degree of static indeterminacy (Ds) of a rigid
𝒛𝒔𝒒𝒖𝒂𝒓𝒆 𝟏𝟔
𝟔
jointed plane frame may be written as (where m = No.
Q.15. What is the ratio of maximum shear stress to of members, j = No. of joints, r = No. of reactions):
average shear stress for a circular section? (a) Ds = (3m + r) - 3j
(a) 2 (b) 2/3 (b) Ds = (3m - r) + 3j
(c) 4/3 (d) 3/4 (c) Ds = (m + r) - 3j
Ans. (c) (d) Ds = (3m - r) + j
We know that Ans. (a) Degree of static determinacy for plane frame
For circular section, (rigid jointed) = 3m + R-3J
𝟒 Where,
𝝉𝒎𝒂𝒙 = 𝝉𝑨𝒗𝒈
𝟑 M = Number of members,
𝝉𝒎𝒂𝒙 𝟒
= R = Number of support reactions,
𝝉𝑨𝒗𝒈 𝟑

Q.16. The ratio of depth to width of a strongest beam J = Number of joints

that can be cut out of a cylindrical log of wood with Q.19. A propped cantilever AB of length L is fixed at A
homogeneous and isotropic properties is: and propped at B, subjected to uniformly distributed
(a) 1.414 (b) 1.25 load ‘w’ per unit length over its full span length. What is
(c) 0.707 (d) 0.504 the reaction at propped end B ?

OSSC JE MAINS CIVIL 16/03/2025 4 Civil Ki Goli

(a) 3wL / 8 forces. Hence in this method it is essential that the
(b) wL / 8 section line should pass through not more than three
(c) 5wL / 8 members in which forces are not known and it should
(d) 7wL / 12 separate the frame into two parts.
Ans. (a) Method of section is preferred over method of joints
when
• There is a large truss and force in only few members
are required.
• In the situation when method of joints fails to start
or proceed with the analysis.
Q.21. What is the value of maximum shear force for a
simply supported beam of length 6.0m, subjected to
uniformly distributed load (w kN/m) through the span
for which the B.M. equation at a section, from left
support, situated at 'X' m is Mx= 30x- 0.5 (wx²) ?
(a) 30 kN
Deflection of end B = 0
(b) 40 kN
(Downward deflection due to uniformly distributed
load) - (upward deflection due to RB) = 0 (c) 50 kN
(d) 60 kN
𝒘𝒍𝟒 𝑹𝑩 𝒍𝟑
− =𝟎 Ans. (a) Given
𝟖𝑬𝑰 𝟑𝑬𝑰
𝟑
𝑹𝑩 = 𝒘𝒍 Length of simply supported beam, L=6.0m
𝟖
𝟑 𝟓 Bending moment equation at a section from the left
𝑹𝑨 = 𝒘𝒍 − 𝒘𝒍 = 𝒘𝒍
𝟖 𝟖 support:
Q.20. Method of joints is applicable to calculate 𝑀𝑥 = 30x – 0.5 wx²
n
io

member forces of a truss, when the number of unknown Shear force (Vx) is the derivative of the bending
n ct

forces at the joint under consideration is not more than moment with respect to x:
Ju

(a) One (b) Two 𝒅𝑴𝒙

il

𝑽𝒙 =
iv
C

𝒅𝒙
(c) Three (d) Four
Differentiate the given B.M. equation:
Ans. (b) Method of Joint:- At each joint the forces in 𝒅𝑴𝒙
𝑽𝒙 = (30x – 0.5 wx²)
the members meeting and the loads acting constitute 𝒅𝒙
a system of concurrent forces. Hence, two 𝑽𝒙 = 30 – wx
independent equations of equilibrium can be formed Maximum shear force occurs at the supports:
at each joint (Horizontal and Vertical Equilibrium). At the left support (x=0 ):
A joint is selected where there are only two unknown 𝑽𝟎 = 30 – w × 0 = 30 kN
forces. First reactions at the support is calculated by
considering the equilibrium of the entire truss frame. Q.22. The hydraulic head that would produce a quick
Then making the use of two equations at the sand condition in sand stratum of thickness 1.5m,
equilibrium at that joint the two unknown forces are specific gravity 2.67 and void ratio 0.67is equal to
found. (a) 1.0m
Method of section:- In method of section, after (b) 1.50m
determining the reactions a section line is drawn (c) 2.0m
passing through not more than three members in (d) 3m
which forces are not known such that that the frame is Ans. (b) Given
cut into parts. Each part should be in equilibrium under Thickness of sand stratum, H=1.5 m
the action of loads, reactions and forces in the Specific gravity, G=2.67
members that are cut by the section line. Equilibrium
Void ratio, e=0.67
of any of these two parts is considered and the
unknown forces in the members cut by the section line hydraulic head causing quicksand condition:
(𝑮−𝟏)
are determined. The system of forces acting on either 𝒉𝒄 = ×𝑯
𝟏+𝒆
part of truss constitutes a non-current force system. (𝟐.𝟔𝟕 − 𝟏)
Since there are only three independent equations of = × 𝟏. 𝟓
𝟏 + 𝟎.𝟔𝟕
equilibrium there should be only three unknown = 1.5 m

OSSC JE MAINS CIVIL 16/03/2025 5 Civil Ki Goli

Q.23. The change that take place during the process of (a) 0.30 (b) 0.40
consolidation of a saturated clay would include: (c) 0.50 (d) 0.60
(a) An increase in pore water pressure and an increase Ans. (b) A lower water cement ratio results in higher
in effective pressure strength and durability but requires more workability.
(b) An increase in pore water pressure and a decrease in
This means that for every 1 part of cement, a minimum
effective pressure
of 0.40 parts of water is required for normal strength
(c) A decrease in pore water pressure and a decrease in concrete.
effective pressure
(d) A decrease in pore water pressure and an increase in Q.27. The modulus of elasticity (E) of concrete is given
effective pressure by
(a) E=1000 fck
Ans. (d) Consolidation:- The decrease in soil volume by
the squeezing out of the pore water on account of the (b) E=5000 √fck
gradual dissipation of excess hydrostatic pressure (c) E=5500 √fck
induced by an imposed total stress is defined as (d) E=1000 √fck
consolidation. It is a time-dependent phenomenon. Ans. (b) As per IS 456: 2000,
The following changes take place during the process of The Short term Static Modulus of elasticity of concrete
consolidation of saturated clay: Ec is given as-
• Volume reduction is due to the expulsion of pore
Ec = 5000 √𝒇𝒄𝒌
water from voids.
Where,
• Decrease in void ratio.
fck = characteristic compressive strength of concrete,
• Decrease in pore-water pressure.
fck and Ec are measured in MPa.
• Increase in effective stress.
Q.28. Dorry's testing machine is used for
Q.24. A soil has bulk density 2.30 g/ cm³ and water
content 15%. The dry density of the sample is: (a) Crushing test of stones
(b) Hardness test of stone
(a) 1.0 g/cm³
n

(c) Impact test of stone

io

(b) 1.5 g/cm³

ct

(d) Water absorption test

n
Ju

(c) 2.0 g/cm³

Ans. (b) Dorry's testing machine is a device used to
il

(d) 2.5 g/cm³

iv
C

determine the hardness of stones. It is widely used in

Ans. (c) Given
the construction industry to determine the suitability
𝜸 = 2.3 g / cm3 of stones for use in various applications.
W = 15% = 0.15
𝜸 Q.29. Enamel paint is made by adding
𝜸𝒅 =
𝟏+𝒘 (a) White paint in varnish
𝟐.𝟑
= (b) Bitumen in varnish
𝟏 + 𝟎.𝟏𝟓
(c) White lead in lacquer
= 2.0 g/cm³
(d) Zinc white in spirit
Q.25. If the time required for 60% consolidation of a
Ans. (a) Enamel Paint:- Enamel paint is obtained by
remolded soil sample of clay with single drainage is "T",
adding a base like white lead, or zinc white, to a vehicle
then what is the time required to consolidate the
which is a varnish.
sample of clay with the same degree of consolidation
but with double drainage? Enamel paints water proof, chemical resistant and also
provide good paint coverage and colour retention.
(a) 4T (b) 2T
(c) T/2 (d) T/4 Q.30.A portion of an embankment having a uniform up-
gradient 1 in 500 is circular of radius 1000m of the
Ans. (d) For same degree of consolidation centre line and subtends 180o at the centre. If the height
𝒕𝟏 𝒅 𝟐
= ( 𝟏) of the bank is 1m at lower end, and side slopes 2:1, the
𝒕𝟐 𝒅𝟐
earth work involved is:
single drainage t₁=t,d₁=H
(a) 5000 m3
𝒉
Double drainage t2=?: d2 =
𝟐
(b) 16, 500 m3
𝑯 𝟐 𝒕
(c) 27000 m3
𝒕𝟐 = 𝒕 × ( ∕ 𝑯) = (d) 40, 500 m3
𝟐 𝟒

Q.26. Minimum required water cement ratio for a Ans. (c) Total length of curve = 1000 × π
workable concrete is: At the lower end, Initial height = 1 m

OSSC JE MAINS CIVIL 16/03/2025 6 Civil Ki Goli

At the 500 m distance, height = 2 m (c) Neutral
At the 1000 m distance, height = 3 m (d) None of the above
Assuming triangular shape of embankment as not data Ans. (b) When
except side slope of 2: 1 (i.e. 2H: 1 V) is given ELR > ALR Super- adiabatic lapse rate Unstable
Environment
ELR < ALR Sub - adiabatic lapse rate Stable
Environment
ELR=ALR Neutral Condition
ELR is negative = Inversion
Area at lower end (A1) = S d2 = 2 × 12 = 2m 2
Adiabatic Lapse Rate (ALR) is - 9.8°C per 1000 m rise
Area at 500 m (A2) = Sd2 = 2 × 22 = 8m2
ELR = Environment lapse rate
Area at 1000 m (A3) = S × d2 = 2 × 32 = 18m2
ALR = Adiabatic lapse rate
As per Simpson's 1/3rd formula
𝒅
V= [(A1 +An )+4(A2 +A4 +A6 +....)+ 2(A3 + A5}) ]
𝟑
𝝅× 𝟏𝟎𝟎𝟎
= [𝟐 + 𝟏𝟖 + 𝟒 × 𝟖]
𝟐×𝟑
= 27227.13 m3
Nearby option is c , hence can take 27000 m3
Q.31. Mobilization advance up to 10% of the cost of
work is given to contractor
(a) For all activities required to start the work at site on
finalization of contract
(b) Shifting electricity poles and cleaning of site only
(c) For procuring materials
Q.34. The chlorine demand of a water sample was found
(d) To construct site office
n

to be 0.2mg / liter. The amount of bleaching powder

io
ct

Ans. (a) Mobilization advance up to 10% of Tendered containing 30% available chlorine to be added to treat
n
Ju

Amount" shall be paid to the contractor on submission one liter of such water sample is:
il
iv

of no revocable Bank Guarantee for an amount of (a) 0.67 mg

C

110% (One hundred ten percent) of an amount of (b) 0.06 mg

mobilization advance demanded, from a nationalized (c) 1.33 mg
/Scheduled Bank. (d) 0.14 mg
(ii) It is provided for all activities required to start the
Ans. (a) Given:
work at site on finalization of the contract documents
Chlorine demand = 0.2 mg/ltr
Q.32. PWD initiates a construction work after Total chlorine demand = 0.2 mg x 1 = 0.2 mg
(a) Technical approval for the work Available chlorine content = 30% = 0.30
(b) Administrative approval for the work
The required amount of bleaching powder
(c) Once Preliminary estimate is made
𝑪𝒉𝒍𝒐𝒓𝒊𝒏𝒆 𝒅𝒆𝒎𝒂𝒏𝒅 𝟎.𝟐
(d) Correctly accessing the probable addition and = = = 0.67 mg
𝑨𝒗𝒂𝒊𝒍𝒂𝒃𝒍𝒆 𝒄𝒉𝒍𝒐𝒓𝒊𝒏𝒆 𝟎.𝟑
alteration of the work
Q.35. Turbidity is measured on
Ans. (b) Before starting any construction work, the (a) Standard silica scale
Public Works Department (PWD) requires (b) Standard cobalt scale
administrative approval from the competent authority. (c) Standard platinum scale
Once administrative approval is obtained, the next (d) Platinum cobalt scale
steps include technical approval and sanctioning of the
Ans. (a) Turbidity can be measured using Turbidity rod
detailed estimate.
method that consists of a platinum needle having
Administrative approval confirms that the project is
Diameter = 1 mm and
necessary and funds are allocated for it.
Length = 25 mm
Q.33. When Environmental Lapse Rate (ELR) is more
at the tip of aluminium rod.
than Adiabatic Lapse Rate (ALR), then the environment
is said to be This rod is inserted in the water sample to be tested
and the height at which this needle just become
(a) Stable
(b) Unstable
OSSC JE MAINS CIVIL 16/03/2025 7 Civil Ki Goli
invisible is noted and represented the turbidity of fck = 25 MPa
water sample. fcr = 0.7√𝒇𝒄𝒌
This representation is done in terms of standard units
= 0.7√𝟐𝟓
which is obtained by the addition of Silica as Silica
Dioxide (SiO2) in milligrams in one litres of pure water. = 3.5MPa
The standard unit is expressed as Silica Turbidity Unit Q.39. Which of the following motivators is the most
(STU) basic need in Maslow’s hierarchy?
Hence, turbidity is measured on Standard silica scale. (a) Safety
Q.36. Minimum clear cover(in mm) to the main steel bar (b) Belonging
in footing, column, beam and slab are respectively: (c) Esteem
(a) 75,40,25,15 (d) Physiological
(b) 40,75,15,25 Ans. (d) According to Maslow's original formulation,
(c) 30,20,25,15 there are five sets of basic needs: physiological, safety,
(d) 50,40,25,20 love, esteem and self-actualization. These needs are
related to each other in a hierarchy of prepotency (or
Ans. (d) Minimum clear cover requirement: strength) beginning with the physiological needs that
1. Slab = 20 mm are the most prepotent of all.
2. Beam = 25 mm Q.40. Which "Pillar of TQM" recognizes that product
3. Column = 40 mm quality is a result of process quality?
4. Footings = 50 mm (a) Customer focus
For Slab: (b) Process Management
For Mild exposure - 20 mm, and For Moderate (c) Employee empowerment
exposure - 30 mm
(d) Continuous improvement
However, if the diameter of the bar does not exceed
Ans. (d) Continuous Improvement is one of the pillars
n

12 mm, or cover may be reduced by 5 mm. Thus for

io
ct

main reinforcement up to 12 mm diameter bar and for of TQM. The output of a production process must not
n
Ju

mild exposure, the nominal cover is 15 mm. only satisfy customer needs but the customer must
il

feel satisfied with the product.

iv

For Column:
C

For a longitudinal reinforcing bar in a column, concrete Q.41. As per IS : 800, the maximum bending moment
cover not less than 40 mm not less than the diameter for design of purlins can be taken as
of such a bar should be provided. In the case of (Where W is total distributed load including the
columns of the minimum dimension of 20 cm or under, wind load on the purlins and L is center distance of
whose reinforcing bars do not exceed 12 mm, the
support) :
concrete cover of 25 mm to be used for reinforcement.
(a) WL/6
Q.37. Minimum percentage of high yield deformed bars
(b) WL/8
in a R.C. slab compared with gross concrete area is:
(c) WL/10
(a) 0.40 (b) 0.15
(d) WL/12
(c) 0.12 (d) 0.10
Ans. (c) Purlins and its design recommendation: Purlins
Ans. (c) The minimum percentage of reinforcement is
are horizontal beam members which run parallel to
0.12% of the gross cross-sectional area if HY SD bars (Fe
the ridge and connect the trusses along the length of
415) are used.
the ridge.
The minimum percentage of reinforcement is 0.15% of
When an iron angle section is used as purlins for steel
the gross cross-sectional area if mild steel bars are
roof trusses then generally following
used.
recommendations are required to be fulfilled. These
Q.38. What is value of flexural tensile strength of M25 recommendations are applicable for roof trusses with
concrete? a pitch angle of less than 30° and for a maximum
(a) 4.0 MPa imposed load of 0.75 kN/m².
(b) 3.5 MPa 1. The depth of the angle purlin in the plane
(c) 3.0 MPa approximate to the incidence of the maximum load
should not be less than (1/45) th of the length of the
(d) 1.75 MPa
purlin.
Ans. (b) Given,
OSSC JE MAINS CIVIL 16/03/2025 8 Civil Ki Goli
2. The width of the angle purlin should not be less than Ans. (b) Given
(1/60) th of the length of the purlin. FC = 30%
3. The maximum bending moment in the purlin in PWP = 10%
taken as (WL/10) where W is the total uniformly
𝜸𝒅 = 1.5g / cc
distributed load on the purlin including wind load in
the normal direction of the roof and L is the center-to- d = 1m
center distance between purlin supports. 𝜸𝒘 = 1g / cc
The depth of water stored (d) in the root zone
Q.42. A butt weld is specified by 𝜸𝒅
(a) Effective throat thickness 𝒅𝒘 = × 𝒅(𝑭𝑪 − 𝑷𝑾𝑷)
𝜸𝒘
(b) Plate thickness =
𝟏.𝟓
× 𝟏 × (𝟎 ⋅ 𝟑𝟎 − 𝟎 ⋅ 𝟏𝟎)
(c) Size of weld 𝟏

(d) Penetration thickness = 0.3 m = 300 mm

Ans. (a) Butt Weld Q.46. In a river, silt excluder and silt ejector are
constructed
• They are better in highly stressed structures where a
smooth flow of stress is a necessity. (a) At a location after the head regulator and at the
head of the canal, respectively
• It is mainly designed for direct compression and
(b) At the head of the canal, and at a location after the
tension and occasionally for shear also.
head regulator, respectively
• It involves no change in the section at the location of (c) At the same location
the joint.
(d) At specific locations depending upon diverse factors
• The butt weld is specified by effective throat and their locations do not follow a set pattern
thickness.
Ans. (b) Silt Regulation works:- The entry of silt into a
Q.43. In a tension member if one or more than one rivet canal, which takes off from a headworks, can be
holes are off the line, the failure of the member depends reduced by constructing certain special works, called
on silt control works. These works may be classified into
n
io

(a) Pitch the following two types:

n ct

(b) Gauge Silt Excluders

Ju

(c) Diameter of the rivet holes

il

Silt Ejectors
iv
C

(d) All of the above Silt Excluders:

Ans. (d) In a tension member, if one or more than one • Silt excluders are those works that are constructed
rivet hole are off the line, the failure of the member on the bed of the river, upstream of the head
depends upon the pitch, gauge, and diameter of the regulator.
rivet holes. • The clearer water enters the head regulator and
silted water enters the silt excluder. In this type of
Q.44. An object weights 289.2N in air and 186.9N in
works, the silt is, therefore,, removed from the water
water. What is relative density of the material of the
before in enters the canal.
object?
Silt Ejector:
(a) 2.83 (b) 2.45
• Silt ejectors, also called silt extractors, are those
(c) 2.15 (d) 3.15
devices that extract the silt from the canal water after
Ans. (a) Given: the silted water has traveled a certain distance in the
Weight of object in air = 289.2 N off-take canal.
Weight of object in water = 186.9 N These works are, therefore, constructed on the bed of
Loss of weight=289.2−186.9=102.3N the canal, and a little distance downstream from the
𝑾𝒆𝒊𝒈𝒉𝒕 𝒊𝒏 𝒂𝒊𝒓 head regulator.
Relative density =
𝑳𝒐𝒔𝒔 𝒐𝒇 𝒘𝒆𝒊𝒈𝒉𝒕 𝒊𝒏 𝒘𝒂𝒕𝒆𝒓
𝟐𝟖𝟗.𝟐
Q.47. The plan of a map was photo copied to a reduced
= = 2.83 size such that a line originally 100 mm measures 90mm.
𝟏𝟎𝟐.𝟑
The original scale of the plan was 1:1000. The revised
Q.45. What is the moisture depth available for
scale is:
evapotranspiration in root zone of 1 m depth soil, if dry
weight of soil is 1.5gm / cc, field capacity is 30% and (a) 1:900
permanent wilting point is 10%? (b) 1:1111
(c) 1:1121
(a) 450 mm (b) 300mm
(d) 1:1221
(c) 200 mm (d) 150 mm

OSSC JE MAINS CIVIL 16/03/2025 9 Civil Ki Goli

 𝟏 Ans. (a) Afflux is an increase in water level that can
Ans. (b) Original sale =
𝟏𝟎𝟎𝟎
occur upstream of a structure, such as a bridge or
Shrunk length = 90 mm
culvert, that creates an obstruction in the flow.
Original length = 100 mm
On the upstream side of the dam, the depth of the
𝑺𝒉𝒓𝒖𝒏𝒌 𝒍𝒆𝒏𝒈𝒕𝒉
Shrunk factor = water will be rising due to the obstruction/barrier. If
𝑶𝒓𝒊𝒈𝒊𝒏𝒂𝒍 𝒍𝒆𝒏𝒈𝒕𝒉
𝟗𝟎 there had not been any obstruction (such as dam) in
S.F. = = 0.9 the path of the flow of water in the channel, the depth
𝟏𝟎𝟎
Shrunk scale or revised scale of water would have been constant parallel to the bed
= S.F. × Original length of the channel. Due to obstruction, the water level
𝟏 𝟏 rises, and it has maximum depth from the bed in some
= 0.9 × =
𝟏𝟎𝟎𝟎 𝟏𝟏𝟏𝟏 sections. This rise of water level on the upstream end
Q.48. The bearings of lines OA and OB are 16° 10' and is termed as afflux.
332° 18', the value of the included angle BOA is : Q.51. Which of the following is NOT TRUE about plane
(a) 316° 10' table surveying?
(b) 158° 28' (a) It is a rapid method of surveying.
(c) 348° 08'
(b) It is a precise method of surveying.
(d) 43° 52'
(c) It is a cost-effective method of surveying.
Ans. (d) Given (d) It is a versatile method of surveying.
Bearing of line OA = 16° 10'
Ans. (b) Plane-table survey:- A plane table surveying is
Bearing of line OB = 332° 18'
a graphical method of surveying. In this method of
Since the bearings are measured clockwise from the surveying. field observation and plotting are done
north, the included angle can be calculated using: simultaneously helping the surveyor to compare the
Included angle = Bearing of OB−Bearing of OA plotted details with actual features of the ground.
= 360∘−(332∘18′−16∘10′) Advantages of Plane Table Survey:
n

= 43° 52' • Plane Table Surveying is suitable for small and

io
ct

Q.49. The highway capacity is expressed in Passenger medium scale mapping (1: 10,000 to 1: 2, 50,000),
n
Ju

Car Unit(PCU). According to IRC, for a passenger car, the where great accuracy is not required As surveying and
il
iv

PCU is: plotting are done simultaneously in the field, chances

C

of getting omission of any detail get less.

(a) 1 (b) 2
• The plotting details can immediately get compared
(c) 3 (d) 4
with the actual objects present in the field. Thus
Ans. (a) Passenger Car Unit:- It is a vehicle unit used for errors, as well as the accuracy of the plot, can be
expressing highway capacity. It is a metric used in ascertained as the work progresses in the field.
Transportation Engineering, to assess traffic-flow rate
• The plane table survey is generally more rapid and
on a highway.
less costly than most other types of survey.
According to IRC
Disadvantages of Plane Table Survey:
Vehicle PCU • The plane table survey is not possible in unfavorable
Car or taxi 1 climates such as rain, fog, etc.
Motor Cycle 0.5 • This method of survey is not very accurate and thus
unsuitable for large scale or precise work.
Bus or truck 3.5
• As no field book is maintained, plotting at different
Cycle 0.2 scale require full exercise.
Bullock Cart 6 • The quality of the final map depends largely on the
Bullock Cart (large) 8 drafting capability of the surveyor
Note:- There is confusion between option b and d but
Q.50. The rise of water level above its normal level most appropriate option is b.
when passing under bridge is called
(a) Afflux Q.52. In an adjustable level, when the bubble is at
(b) Free board centre, the axis of the bubble tube becomes parallel to
(c) Headroom (a) Line of sight
(d) Tailroom (b) Line of collimation
(c) Axis of telescope
(d) None of these
OSSC JE MAINS CIVIL 16/03/2025 10 Civil Ki Goli
Ans. (b*) Line of sight: Ans. (c) A prismatic compass is a navigation and
surveying instrument which is extensively used to find
• The sighting or pointing line of a telescope, defined
out the bearing of the traversing. It is the most
by the optical center of the objective and the
intersection of crosshairs. convenient and portable form of magnetic compass that
can either be used as a hand instrument or fitted on a
Line of collimation:
tripod.
• It is an imaginary line passing through optical centre
The graduated circle of a prismatic compass and
of the objective glass and its continuation.
surveyor compass is shown in the figure below:
Axis of Telescope:
• The axis is an imaginary line passing through the
optical centre of the object glass and the optical centre
of the eye-peace.
Axis of the Bubble Tube:
• It is an imaginary line tangential to the longitudinal
curve of the bubble tube at its middle point.

From observation: The zero of a graduated circle is

marked at the south end for the prismatic compass.
Some other differences between the two compass are
as follows:
Item Prismatic Surveyor’s
Compass Compass
Needle Broad type Edge bar type
Scale Free to float Attached to the
n
io

along with box

n ct

When the bubble of the level tube of a level remains

Ju

broad type
il

central, the line of collimation or line of sight is magnetic

iv
C

horizontal, and the axis of the bubble tube becomes needle

parallel to the Line of sight.
Bearing Whole circle Quadrantal
Note:- According to board answer key correct option is bearing bearing
a but correct answer is option b.
Graduations Inverted (as Direct
Q.53. In a straight line, two lengths A and B are graduation
measured from a point P. The standard errors of have to be
measured lengths A and B are found to be 0.4m and observed
0.3m. The standard error in the length A minus B is: through a
(a) 0.7 m (b) 0.35 m prism)
(c) 0.5 m (d) 0.1 m Sighting & Can be done Sighting is to be
Ans. (c) Given Reading simultaneously done first and
Standard error in length A = 0.4 m then the
surveyor has to
Standard error in length B = 0.3 m
read the
σ A−B = √(𝝈𝑨 )𝟐 + (𝝈𝑩 )𝟐 northern end
= √𝟎. 𝟒𝟐 + 𝟎 ⋅ 𝟑𝟐 of the needle
= 0.5m Tripod Not essential Essential
Q.54. Which one of the following is correct for Prismatic Q.55. Which one of the following methods estimates
Compass? the best area of an irregular and curved boundary?
(a) The graduated ring rotates with line of sight. (a) Trapezoidal method.
(b) Instrument cannot be used without tripod. (b) Simpson's method.
(c) The graduations are engraved inverted. (c) Average ordinate method.
(d) The readings can directly be taken by seeing through (d) Mid-ordinate method
the top of the glass.

OSSC JE MAINS CIVIL 16/03/2025 11 Civil Ki Goli

Ans. (b) Simpson's rule:- In order to apply Simpson's Ans. (b) Free float:- It is the amount of time that the
rule, the area must be divided in even number i.e., the activity completion time can be delayed without
number of offsets must be odd i.e., n term in the last affecting the earliest start time of the immediate
offset 'O,' should be odd. successor activities in the network.
For irregular boundaries, Simpson's rule is preferred Total float: It is the amount of time that the
over the trapezoidal rule to calculate the given area. completion time of an activity can be delayed without
According to this rule the short length of boundaries affecting project completion time.
between the two adjacent ordinates is a parabolic arch Interfering Float: Maximum amount by which an
The area is given by Simpson's rule: activity can be delayed without delaying the project
𝒅 but will cause delay to the Early Start of some
Area = (01+On)+4(02+04+.. where 01, 02, 03, ........ is the
𝟑 following activity
offset
Independent Float: Amount by which an activity can be
Q.56. An image of the top of the hill is 92 mm from the delayed without delaying the project; even if all
principal point of the photograph. The elevation of the predecessors are at Late Finish and all Successors are
top of the hill is 400 m and the flying height is 4000 m at Early Start.
above the datum. The relief displacement will be
Q.59. What is the angle between principal strain axis
(a) 9.2 mm and maximum shear strain axis ?
(b) 12 mm
(a) 0° (b) 30°
(c) 88mm
(c) 45° (d) 90°
(d) 8 mm
Ans. (c) The plane on which the shear stress is zero is
Ans. (a) Given
called the principal plane and the normal stress on the
r = 92 mm.
principal plane is called the principal stress.
H = 4000 m = 4000000 mm.
Planes of maximum shear stress occur at 45° to the
h = 400 m = 400000 mm principal planes.
𝒓𝒉
n

Relief displacement (d) =

io

𝑯 Q.60. A Mohr circle reduces to a point when the body is

ct

𝟗𝟐 𝒙 𝟒𝟎𝟎𝟎𝟎𝟎 subjected to
n

= = 9.2 mm
Ju

𝟒𝟎𝟎𝟎𝟎𝟎𝟎
(a) Pure shear
il
iv

Q.57. To uniquely determine the position of the user (b) Uniaxial stress only
C

using GPS, one needs to receive signals from at least (c) Equal and opposite axial stresses on two mutually
(a) 1 satellite perpendicular planes, the planes being free of shear
(b) 2 satellites (d) Equal axial stresses on two mutually perpendicular
(c) 3 satellites planes, the planes being free of shear
(d) 4 satellites Ans. (d*) When the normal stresses on the two
Ans. (d) In Cartesian coordinate system of space and mutually perpendicular planes are equal and alike
time, we have four variables - x, y, z (three space then the radius of the Mohr circle will be zero.
variables) and t (one time variable). Each satellite Radius of Mohr's Circle,
corresponds to one variable. Therefore, to uniquely 𝟐
determine the position of the user using GPS minimum R = √(𝝈𝒙−𝝈 𝒚 ) + 𝝉𝟐𝒙𝒚
𝟐
4 satellites are required.
However, if we use less than 4 satellite say 2 nos., then Here 𝝈𝒙 = 𝝈𝒚 and 𝝉𝒙𝒚 = 0
out of variable two variables say x and y are known and Therefore, R = 0 means Mohr's circle reduces to a
other two variables z and times are unknown and they point.
can take many values i.e. we have multiple positions *As per board answer key correct option is b but
and hence, we are not able to determine the position correct answer is option d.
of any object uniquely.
Q.61. A steel column is pinned at both ends and has a
Q.58. The time by which activity completion time can be buckling load of 250 kN. If the column is restrained
delayed without affecting the start of succeeding against lateral movement at its mid-height, its buckling
activities is known as: load will be
(a) Total Float (a) 200 kN
(b) Free Float (b) 800 kN
(c) Interfering Float (c) 1000 kN
(d) Zero Float (d) 1200 kN
OSSC JE MAINS CIVIL 16/03/2025 12 Civil Ki Goli
Ans. (c) The buckling load (P₁) of a steel column pinned hours. The cured blocks areimmersed in silicate tanks
at both the ends is (Leff = L) for three days.
𝝅𝟐 𝑬𝑰 a. Victoria stone: These are granite pieces with the
𝑷𝟏 = = 𝟐𝟓𝟎 𝒌𝑵
𝑳𝟐 surfaces hardened by keeping immersed in
When the column is restrained against lateral sodasilicate for about two months.
movement at its mid-height, then Leff = L/2 b. Ransom stone: These are prepared by mixing soda
The buckling load (P2) of the steel column for the above silicate with cement to provide decorative flooring.
case will be These are also known as chemical stones. These have
𝝅𝟐 𝑬𝑰 𝝅𝟐 𝑬𝑰 compressive strenath of about 32 N/mm².
𝑷𝟐 = 𝑳 𝟐
=𝟒×( ) = 4 × 250 = 1000 kN
𝑳𝟐
( )
𝟐 c. Garlic stone: These are produced by moulding a
mixture of iron slag and Portland cement. These
Q.62. The constituent compound in Portland cement
areused as flag stones, surface drains, etc.
which reacts immediately with water, and also sets
earliest, is: d. Bituminous stone: Granite and diorite are
impregnated with prepared or refined tar to
(a) Tricalcium Silicate
formbituminous stone. These are used for providing
(b) Dicalcium Silicate noise, wear and dust resistant stone surfaces.
(c) Tricalcium Aluminate
Q.64. As per IS 456:2000, the maximum admissible
(d) Tetracalcium Alumino ferrite
water-cement ratio for mild environmental exposure
Ans. (c) Tricalcium Aluminate: C3A is formed within 24 for concrete should be
hours of the addition of water in the cement and is (a) 0.55 (b) 0.50
responsible for maximum evolution of heat of (c) 0.45 (d) 0.45
hydration. It is the first compound that is formed after
Ans. (a)
addition of water and sets early.
Tetracalcium aluminoferrite: C4AF is also formed Exposure Minimu Maximu Minimu Maximu Exposure
Condition m m Water- m m Water- Condition
within 24 hours of the addition of water in the cement Cement Cement Cement Cement Description
n

but its individual contribution to the overall strength

io

Content Ratio Content Ratio

ct

of the cement is insignificant. (PCC) (PCC) (RCC) (RCC)

n
Ju

(kg/m³) (kg/m³)
Tricalcium silicate: C3S is formed within a week or so
il
iv

of the addition of water in the cement and is Mild 220 0.6 300 0.55 Concrete
C

responsible for the early development of strength of exposed to

coastal area
the cement.
Dicalcium silicate: C₂S is the last compound that is Moderat 240 0.6 300 0.5 Concrete
e continuousl
formed after the addition of water in the cement
y
which may require a year or so for its formation. It is underwater
responsible for the progressive strength of the cement.
Severe 250 0.5 320 0.45 Concrete
Q.63. Which one of the following stone is produced by immersed
moulding a mixture of iron slag and Portland cement? under
seawater
(a) Imperial stone
(b) Garlic stone Very 260 0.45 340 0.45 Concrete
Severe buried
(c) Ransom stone
under
(d) Victoria stone aggressive
subsoil
Ans. (b) Artificial stone:- Artificial stonesare made with
cement and natural aggregates ofthe crushed stone Extreme 280 0.4 360 0.4 Concrete
and sand with desired surface finish. Where ever under tidal
zone
durable natural stones are not available at reasonable
cost, artificial stones (also known ascast stone) Q.65. The method used for estimation of depreciation
become the choice. of building is known as:
Composition of different artificial stones are given (a) Logistic curve method
below: (b) Rental method
Imperial stone: (c) Constant percentage method
Finely crushed granite is washed carefully and mixed (d) Direct comparison method
with Portland cementand then steam cured for 24 Ans. (c) Method used for estimation of depreciation of
building

OSSC JE MAINS CIVIL 16/03/2025 13 Civil Ki Goli

Straight line method: It is the simplest method of Q.68. Number of bricks (having size of 20 cm x 10 cm x
depreciation. In this method it is assumed that the 10 cm) required for 17 cu.m of brickwork is
book value of an asset will decrease by same amount approximately
every year over the useful life till its salvage value is (a) 6750 (b) 7200
reached. In other words, the book value of the asset (c) 7500 (d) 8500
decreases at a linear rate with the time period.
Ans. (d) Given:
Constant percentage method: In this method, it is
assumed that the property will lose its value by a Volume of brickwork = 17 m³
constant percentage of its value at the beginning of Size of one brick = 20cm×10cm×10cm
every year. Volume of one brick=0.20×0.10×0.10=0.002m3
𝟏𝟕
Sinking Fund Method: In this method, the depreciation Number of bricks = = 8500
𝟎.𝟎𝟎𝟐
of a property is assumed to be equal to the annual
sinking fund plus the interest on the fund for that year, Q.69. The plinth area of a building does not include area
which is supposed to be invested on interest bearing of
investment. (a) the walls at the floor levels.
Q.66. Due to a change in price level, a revised estimate (b) internal shaft for sanitary installations up to 2m²
is prepared if the sanctioned estimate exceeds area.
(a) 2.0% (b) 2.5% (c) lifts
(c) 4.0% (d) 5.0% (d) cantilevered porches

Ans. (d) The revised estimate is a detailed estimate for Ans. (d) The plinth area is the covered built-up area
revised quantities and the rate of items of works measured at the floor level of any storey or at the floor
originally provided in the estimate without material level of the basement.
deviation of a structural nature from the design Following areas are included during measurement of
originally approved for a project. plinth area:
It is required to be prepared for the following reasons: 1) Area of the wall at the floor level, excluding plinth
n
io

i) When a sanctioned estimate is likely to exceed by offsets.

n ct

more than 5% either from the rates being found 2) Areas of the internal shaft for sanitary installations
Ju

and garbage chute, electrical, telecom, and firefighting

il

insufficient or from cause whatsoever except

iv
C

important structural alteration. services

ii) When the expenditure of works exceeds or is likely 3) Vertical duct for air conditioning and lift well
to exceed by more than 10% of the administrative including landing
approval (for work more than 5 lakhs/-) 4) Staircase room or head room other than terrace
iii) When there are material deviations from the level
original proposal but not due to the material deviation 5) Area of barsati at terrace level
of structural nature. Following areas are not included during measurement
iv) When it is found that the sanctioned estimate is of plinth area:
more than the actual requirement. 1) Additional floor for seating in assembly buildings,
Q.67. A wall of 12m in length having a height of 2m is to theatres, auditoriums
be plastered. If the thickness of the wall is 85 cm, the 2) Cantilevered porch, Balcony, Area of loft, Internal
quantity of plastering in (sq. m) required is in the range sanitary shaft, and garbage shaft
(a) 45 to 65 (b) 20 to 40 3) Area of the architectural band, cornice, Open
(c) 65 to 85 (d) 1 to 20 platform, etc.
4) Towers, turrets, domes projecting above the terrace
Ans. (a) Given:
level at the terrace
Length of wall, 𝐿 = 12𝑚
Height of wall, H=2m Q.70. Water charge are usually taken as_____%for rate
analysis of an item
Thickness of wall = 85 cm = 0.85 m
(a) 3 (b) 1
Area to be plastered=2×(L×H)
(c) 1.5 (d) 5
=2×(12×2)
= 48 m2 Ans. (c) The analysis of rates is worked out for the unit
payment of the particular item of work under two
heads: Materials and Labour.

OSSC JE MAINS CIVIL 16/03/2025 14 Civil Ki Goli

The cost of items of work = Material cost + Labour cost • Oven dry method is the most accurate and simplest
• Other costs included to the above cost of items of method for water content determination.
work are: • In this method complete drying of soil sample occurs
• Tools and Plants (T&P) = 2.5 to 3 % of the labour cost and water content in the sample is calculated
• Transportation cost (if conveyance more than 8 km is accurately by a maintained temperature in the oven
considered.) (105°C to 110°C) for 24 hours.
• Water charges = 1.5 to 2% of total cost • For highly organic soils a low temperature of about
60° C is preferable.
• Contractor's profit = 10%
• If Gypsum is present, the temperature should not be
Q.71. A T-beam behaves as a rectangular beam of width more than 80°C but for a long time.
equal to its flange if its neutral axis
Sand bath method:
(a) Coincides with centroid of reinforcement
• It is a quick field method for the determination of
(b) Coincides with centroid of T-section
water content.
(c) Remains within the flange
(d) Remains in the web • This method is used when an electric oven is not
available.
Ans. (c) In general, the portion of concrete which is in • Sand bath method is a rapid method for water
tension region is neglected for design purpose because content determination but not very accurate.
concrete is weak in tension.
• In this method there is no control over heat given to
This is valid for both limit state and working stress soil sample. Therefore, it is not suitable for organic soil
method of design. and soil having higher gypsum content.
In case of T-beam, if depth of the neutral axis lies Pycnometer method:- This is a quick method but is less
within the flange region and there is sagging bending accurate than oven drying method. This method is
moment then only the portion below Neutral Axis will used only when specific gravity of soil solids is known.
be in tension and hence, neglected and T-beam will
behave like a rectangular beam having width equal to Q.73. When the sand in-situ happens to be in its densest
state, then the sand's relative density would be:
n

flange width.
io
ct

:: A T-beam behaves as a rectangular beam of width (a) Greater than 1

n
Ju

equal to its flange if its neutral axis remains within the (b) Between 0 & 1
il

(c) 1
iv

flange is a correct statement.

C

(d) None of these

Q.72. Which of these methods is the most appropriate
method used in site to determine a soil's water content? Ans. (c) In its densest state, the relative density of sand
(a) Pycnometer method is 1. This means that the density of the sand is equal to
(b) Oven drying method the density of water. This occurs when the sand is fully
compacted and the void spaces between the grains are
(c) Calcium carbide method
completely filled.
(d) Sand bath method
Q.74._______ is the measure of loss of strength with
Ans. (c)Calcium carbide method: (Rapid Moisture
remoulding, with water content unchanged.
Meter)
• The calcium carbide method is the quickest and (a) Compressibility
(b) Sensitivity
reasonably accurate method of determining of the
(c) Stability
water content of soil using a portable moisture
(d) Thixotropy
content kit. It usually requires 5 to 7 minutes for
determining the water content. Ans. (b) Sensitivity:- Sensitivity is defined as the ratio
• The method is based on the principle that when the of unconfined compressive strength of the clay in the
water in the soil reacts with calcium carbide, acetylene undisturbed state to the unconfined compressive
gas is produced and the pressure exerted by the strength of the clay in the remoulded state.
acetylene gas on a diaphragm gives a measure of the Sensitivity is the measure of loss of strength with
water content. remoulding, with water content unchanged.
• The water content obtained from the calcium carbide Thixotropy:
method is based on the initial weight of wet soil. It The loss of strength of soil due to remoulding is partly
should be converted to water content based on the dry due to the change in soil structure and partly due to
weight of the soil. disturbance caused to water molecules in the
Oven drying method: absorbing layer.

OSSC JE MAINS CIVIL 16/03/2025 15 Civil Ki Goli

Some of these changes are reversible. If a remoulded (a) 0.531 (b) 0.652
soil is allowed to stand, without loss of water, it may (c) 0.432 (d) 0.622
regain some of its lost strength. In soil engineering, this
Ans. (c) The compression index (Cc) for normally
gain in strength of the true soil with the passage of
consolidated clay can be estimated using the empirical
time after it has been remoulded is called thixotropy.
correlation:
Confusion Point: -The question clearly mentions "loss
Cc=0.009(wL−10)
of strength with remoulding, with water content
unchanged." =0.009(58−10)
Loss of strength occurs immediately after remoulding, =0.009×48
which is related to Sensitivity. =0.432
Thixotropy applies when the strength gradually Q.78. Which one of the following is a secondary
increases over time, which is not directly asked in the pollutant?
question. (a) Carbon Monoxide
Therefore, the correct answer is (b) Sensitivity. (b) Hydrocarbon
Q.75. Triaxial compression test is used to find______of (c) Ozone
soil. (d) Volatile Organic Carbon (VOC)
(a) Compressive strength Ans. (c) Primary pollutant:- Pollutants that are emitted
(b) Permeability directly from identifiable sources, either from natural
(c) Specific gravity hazardous events like dust storms, volcanoes, etc, or
(d) Shear strength from human activities like burning of wood, coal, oil in
homes or industries or automobiles, etc.
Ans. (d) Triaxial test is used to determine the shear
strength parameter of soil. Secondary pollutant:- The primary pollutants often
react with one another or with water vapor, aided and
The triaxial compression test consists of two stages:
abetted by the sunlight, to form entirely a new set of
First stage: In this, a soil sample is set in the triaxial cell, pollutants, called the secondary pollutants.
and confining pressure is then applied.
n
io

These are the chemical substances, which are

ct

Second stage: in this, additional axial stress (also called

n

produced from the chemical reactions of natural or

Ju

deviator stress) is applied which induces shear stresses anthropogenic pollutants or due to their oxidation,
il
iv

in the sample. The axial stress is continuously etc., caused by the energy of the sun.
C

increased until the sample fails.

Q.79. Which of these layers of the atmosphere consists
During both the stages, the applied stresses, axial
of the ozone layer that is responsible for absorbing the
strain, and pore water pressure or change in sample
Ultra-Violet (UV) light?
volume can be measured.
(a) Troposphere
Q.76. A deposit of fine sand has a porosity n and specific (b) Mesosphere
gravity of soil solids is G. The hydraulic gradient of the (c) Stratosphere
deposit to develop boiling condition of sand is given by: (d) None of these
(a) (G-1) (1-n)
Ans. (c) The ozone layer is a natural layer of gas in the
(b) (G-1) (1+n)
upper atmosphere that protects humans and other
(c) (G - 1) / (1 - n)
living things from harmful ultraviolet (UV) radiation
(d) (G - 1) / (1 + n)
from the sun.
Ans. (a) Critical hydraulic gradient (ic): The quicksand / • The ozone layer is typically thicker over the poles
boiling condition occurs at a critical upward hydraulic than over the equator.
gradient typically around 1.0 for many soils, when the
• The ozone layer exists in the stratosphere, a layer 10
seepage force just balances the buoyant weight of an
to 50 km above the Earth's surface.
element of soil.
• The reasons for ozone depletion are a wide range of
At the critical conditions, the effective stress is equal
industrial and consumer applications, mainly
to zero.
𝑮−𝟏
refrigerators,airconditioners(hydrochlorofluorocarbo
𝒊𝒄 = ns(HCFCs), chlorofluorocarbons (CFCs)), and fire
𝟏+𝒆
Or 𝒊𝒄 = (G-1) (1-n) extinguishers.
• Ozone depletion is greatest at the South Pole
Q.77. A normally consolidated undisturbed clay is
(Antarctica)
having a liquid limit of 58%. What is the compression
index (Cc) of this clay? Q.80. Which of these elements is present in the drinking
water that can lead to numerous fatal diseases?
OSSC JE MAINS CIVIL 16/03/2025 16 Civil Ki Goli
(a) Phosphorus Q.83. The purpose of a water storage tank in a
(b) Calcium distribution system is to:
(c) Arsenic (a) Regulate water pressure
(d) None of the above (b) Store water for emergencies
Ans. (c) Arsenic which is found in water is responsible (c) Maintain a constant supply of water
for cancer. (d) All of the above
• The desirable drinking water standard for total Ans. (d) Distribution reservoirs, also called service
dissolved solids is 500 mg/l. reservoirs, are the storage reservoirs, which store the
• If there are no alternate sources of water available treated water for supplying water during emergencies
then the permissible limit is increased to 2000 mg/l. (such as during fires, repairs, etc.) and also to help in
• Dissolved solids are organic/inorganic compounds absorbing the hourly fluctuations in the normal water
such as salts, and heavy metals, etc. demand.
• They cause health hazards like cancer due to the Q.84. Which of the following represents the heavier
presence of heavy metals like arsenic etc. inert matter in wastewater?
• The arsenic problem in India is primarily due to the (a) Screens
overexploitation of groundwater in the affected areas. (b) Grit
• The permitted level of arsenic in drinking water is (c) Debris
0.05 mg/liter, as per WHO. Bureau of Indian Standards (d) Waste
gives the acceptable limit of arsenic in drinking water
to 0.01 mg/litre and in absence of other alternative Ans. (b) The heavier inert matter in wastewater is
sources to 0.05 mg/litre. typically referred to as "grit." Grit can include sand,
gravel, silt, and other heavy materials that settle out
Q.81. One litre of sewage, when allowed to settle for 30 of wastewater due to their weight.
minutes gives a sludge volume of 27 cm³. If the dry
weight of this sludge is 3.0 grams, then its sludge Q.85. In the design of a steel beam using an I-section:
volume index (SVI) in ml/grams will be: (a) Shear capacity of flanges is neglected
n
io

(a) 9 (b) 24 (b) Shear capacity of the web is neglected

nct

(c) Shear capacity of both flange and web is neglected

Ju

(c) 30 (d) 81
(d) None of the above
il
iv

Ans. (a) Given,

C

Ans. (a) A beam having the flange and the web is

Volume of settled sludge = 27 cm³ = 27 ml mainly I beams or H beams in which the horizontal
Drying weight (MLSS) = 3 gm elements are called flanges and vertical elements on
𝑽𝒐𝒍𝒖𝒎𝒆 𝒐𝒇 𝒔𝒆𝒕𝒕𝒍𝒆𝒅 𝒔𝒍𝒖𝒅𝒈𝒆 𝒊𝒏 𝒎𝒍 the web. These are usually made of structural steel and
Sludge volume index =
𝑴𝑳𝑺𝑺 𝒑𝒓𝒆𝒔𝒆𝒏𝒕
are used in construction and civil engineering work.
𝟐𝟕
= = 9 ml/gram Let us consider an I section like the example of a rolled
𝟑
steel section.
Q.82._______Activated sludge process is an example
of Note that more than 85% of shear in the I section is
resisted by the web.
(a) Anaerobic suspended growth process
(b) Anaerobic attached growth process Thus because most of the shear stress is taken by the
(c) Aerobic attached growth process web of a steel section, hence shear capacity of the
(d) Aerobic suspended growth process flange is neglected.

Ans. (d) Activated sludge process: Q.86. Which of the following is the most common type
of failure in a slender steel structure?
• The essential features of the activated sludge process
are an aeration stage, solids-liquid separation (a) Brittle fracture
following aeration, and a sludge recycle system. (b) Fatigue
(c) Buckling
• Wastewater after primary treatment enters an
(d) Yielding
aeration tank where the organic matter is brought into
intimate contact with the sludge from the secondary Ans. (c) The common types of failure in steel beams
clarifier. are:
• It requires less space, does not produce obnoxious Web Crushing: This occurs when the web portion of a
odor, and requires less time for wastewater beam experiences excessive localized compressive
treatment. stress. The web becomes crushed or compacted,
• It requires skilled supervision

OSSC JE MAINS CIVIL 16/03/2025 17 Civil Ki Goli

leading to a loss of structural integrity and potential (d) Elastic recovery of bitumen at 15°C is more than 30%
collapse. Ans. (b) According to IS 73-2013, Number 30 indicates
Shear Buckling: When a beam is subjected to the range of viscosity of bitumen [(100 ±20) × 30] in
significant shear forces, it may buckle or deform out of terms of Poise.
its original shape. This distortion can compromise the For VG 30 Bitumen:
beam's load-carrying ability, leading to further failure.
Maximum Viscosity: (100 + 20 = 120) x 30 = 3600 Poise
Lateral Buckling: This failure occurs when a slender
Minimum Viscosity: (100-20 = 80) x 30 = 2400 Poise
beam experiences buckling out of its plane due to
compressive forces. The buckling may happen Q.89. A car is moving at a speed of 90kmph on a road
laterally, causing the beam to twist or bend, having a 2% upward gradient. The driver applies brakes
weakening its load-bearing capacity. when he sees an obstruction. If his reaction time is 1.5
Flexural Buckling: Caused by excessive bending or seconds, assuming that the coefficient of friction
flexural loads, this failure type results in the beam between pavement and tyre is 0.15, calculate the
buckling in the direction of the applied bending distance traversed before the car finally stops (round to
moments. The beam may become unstable and lose its an integer value).
load-carrying capacity. (a) 200 m
Torsional Buckling: When a beam is subjected to (b) 225 m
unbalanced torsional loads, it may twist, causing an (c) 1056 m
unexpected deformation. The twisting compromises (d) 150 m
the beam's structural performance and can lead to
more significant issues. Ans. (b) Given:
Local Buckling: This happens when a particular section V=90 kmph
of the beam, such as the flanges or web, buckles due Upward gradient=2%
to localized stress. This localized failure can affect the Reaction time = 1.5 sec
overall performance and stability of the entire beam. f=0.15
n

Q.87. As per IRC 96, 1983, the recommended 𝒗𝟐

io

SSD = 𝟎. 𝟐𝟕𝟖𝒗𝒕 +
ct

𝟐𝟓𝟒(𝒇+𝒏%)
carriageway width for a two lane road with kerbs is:
n
Ju

𝟗𝟎𝟐
(a) 7.0 m (b) 7.5 m = 𝟎. 𝟐𝟕𝟖 × 𝟗𝟎 × 𝟏. 𝟓 + 𝟐
il

𝟐𝟓𝟒(𝟎.𝟏𝟓+ )
iv

𝟏𝟎𝟎
(c) 8.0 m (d) 8.5 m
C

= 225.1 m
Ans. (b) Width of Pavement or Carriageway:
• The pavement width depends upon the number of Q.90. Which one of the following is used for drainage
lanes and the width of the single lane required. The lane purpose in pavement design?
width determines on the basis of the width of the (a) Kerb
vehicle and minimum. side clearance. (b) Shoulder
• The maximum permissible width of the vehicle as per (c) Camber
IRC specification is 2.44 m. (d) All of the above
Class of Road Width of Ans. (C) Camber or Cross Slope:
Carriageway Camber is the slope provided to the road surface in the
Single lane 3.75 m transverse direction to drain off the rainwater from
the road surface.
Two lanes, without raised kerbs 7.0 m • Drainage and quick disposal of water from the
Two lanes, with raised kerbs 7.5 m pavement surface by providing camber (cross slope) is
considered important because of the following
Intermediate carriageway (except 5.5 m
reasons:
on important road)
• To prevent the entry of surface water into the
Multi-lane pavement 3.5 m per lane pavement layers and the subgrade soil through the
Q.88. A bitumen sample has been graded as VG30 as per pavement.
IS: 73-2013. The '30' in the grade means that • To prevent the entry of water into the bituminous
(a) Penetration of bitumen at 25°C is between 20 and 40 pavement layers, continued contact with water causes
(b) Viscosity of bitumen at 60°C is between 2400 and the stripping of bitumen from the aggregates and
3600 Poise results in the deterioration of pavement layers.
(c) Ductility of bitumen at 27°C is more than 30 cm
OSSC JE MAINS CIVIL 16/03/2025 18 Civil Ki Goli
• To remove the rainwater from the pavement surface series of interconnected cracks creating small,
as quickly as possible and to allow the pavement to get irregular shapes in pavement.
dry soon after the rain.
Kerb: Kerb indicates the boundary between the
pavement and median or footpath or island or
shoulder. Kerb helps in channeling water away from
the pavement and into the drainage system.
Different types of failure encountered in flexible
Shoulder: Shoulders are provided on both sides of the pavements are as follows:
pavement all along the road in the case of an undivided
1. Alligator cracking or Map Cracking (Fatigue): This is
carriageway. Shoulders are provided along the outer
a common type of failure of flexible pavements. This is
side of the carriageway in the case of a divided
also known as fatigue failure. Following are the
carriageway. Provides additional space for water
preliminary causes
runoff and prevents water accumulation on the main
carriageway. a. Relative movement of pavement layer material
Guard Rail: The guide rail is defined as an additional b. Repeated application of heavy wheel loads
rail placed midway between the two ordinary rails of a c. Swelling or shrinkage of subgrade or other layers due
track that guides the wheels of the train. It is designed to moisture variation
in connection with devices on the engine or car, to 2. Consolidation of pavement layers (Rutting):
keep a train from leaving the track on curves, Formation of rut falls in this type of failure. A rut is
crossings, or steep grades. depression or groove worn into a road by travel of
wheels. This type of failure is caused due to the
Q.91. In highway pavements emulsions are mainly used
following reasons.
in ( JKPSC AE 2021)
a. Repeated application of load along the same wheel
(a) Surface dressing
path resulting in longitudinal ruts.
(b) Patching and maintenance
(c) Bitumen macadam b. Wearing of the surface course along the wheel path
n
io

(d) Asphaltic concrete resulting in shallow ruts.

n ct

3. Shear failure cracking: Shear failure causes

Ju

Ans. (B) Bitumen emulsion:

il

upheaval of pavement material by forming a fracture

iv

• These are used in bituminous road constructions,

C

or cracking. Following are the primary causes of shear

especially in maintenance and patch repair works. cracking
• The main advantages of emulsion is that it can be a. Excessive wheel loading
used in wet weather even when it is raining. b. Low shearing resistance of pavement mixture
• Bitumen emulsion is a liquid product in which 4. Longitudinal Cracking: This type of crack extends
bitumen is suspended in a finely divided condition in to the full thickness of the pavement. The following are
an aqueous medium and stabilized by suitable the primary causes of longitudinal cracking-
material. a. Differential volume changes in subgrade soil
• Normally cationic type emulsions are used in India. b. Settlement of fill materials
• The bitumen content in the emulsion is around 60 % c. Sliding of side slopes.
and the remaining is water. When the emulsion is 5. Frost Heaving: Frost heaving causes upheaval of
applied on the road it breaks down resulting in the the localized portion of the pavement. The extent of
frost heaving depends upon the groundwater table
release of water and the mix starts to set.
and climatic conditions.
Q.92. Alligator or map cracking is the common type of 6. Lack of binding to the lower course: When there
failure in is a lack of binding between surface course and the
(a) Concrete pavements underlying layer, some portion of surface course
(b) Bituminous surfacing looses up materials creating patches and potholes.
(c) Gravel roads Slippage cracking is one form of this type of failure.
(d) WBM construction 7. Reflection cracking: This type of failure occurs
when the bituminous surface course is laid over the
Ans. (B) Alligator cracking are also called as map
existing cement concrete pavement with some cracks.
cracking or fatigue cracking. It is the most common
This crack is reflected in the same pattern on
type of failure in bituminous surfacing consisting of a
bituminous surface/

OSSC JE MAINS CIVIL 16/03/2025 19 Civil Ki Goli

8. Formation of waves and corrugation circular curve and calculation and field
9. Bleeding: Excess bituminous binder occurring on the implementation is very easy.
pavement surface causes bleeding. Bleeding causes a Q.95. Which of the following are related to
shiny, glass-like, reflective surface that may be tacky to maintenance of railway track?
the touch.
1. Jim crow and gauge bar
10. Pumping: Seeping or ejection of water and finds 2. Through packing and boxing
from beneath the pavement through cracks.
3. Buffer stop and sand hump
Q.93. What is curve resistance for a 50 tonnes train on 4. Creep adjustment
a BG on a 4o curve? (RRB JE 29 Aug 2019)
Select the correct answer using the codes given below:
(a) 0.05 tonne
(a) 1, 2 and 3
(b) 0.06 tonne
(b) 1, 3 and 4
(c) 0.08 tonne
(c) 1, 2 and 4
(d) 0.10 tonne
(d) 2, 3 and 4
Ans. (C) Curve Resistance on BG track is given by
Ans. (C)
following formula:
The following are related to maintenance of railway
R = 0.0004 × W × D
track:
Where,
1. Jim crow and gauge bar – Used for adjusting and
W = Weight of train in tonnes and D = Degree of curve maintaining the track gauge and rail curvature.
In this case: D = 4 and W = 50 tonnes 2. Through packing and boxing – Methods used in
We get, track maintenance to ensure proper ballast
Curve resistance, R = 0.0004 × 50 × 4 = 0.08 tonne support and stability.
Q.94. Which one of the following types of transition 3. Buffer stop and sand hump – These are safety
curves is mostly used in Indian Railways? (UKPSC AE 2013) measures at railway terminals to stop trains and
(a) Euler's spiral prevent accidents, not directly related to track
n
io

maintenance.
(b) Cubic Spiral
n ct

4. Creep adjustment – Involves adjusting rail creep

Ju

(c) Lemniscate
(longitudinal movement of rails) to maintain
il

(d) Cubic Parabola

iv
C

proper track alignment.

Ans. (D) A curve of the varying radius is known as a
Different types of tools for railway track maintenance
transition curve. The radius of such a curve varies from
are as follows:
infinity to a certain fixed value. The transition curve is
also called a spiral or easement curve. Name of the Usage
Tool
A transition curve is provided when there is a change
in the horizontal alignment from a straight path to Jim Crow To bend the rails
circular curve gradually. Crow bars To align the track slightly
Rail tongs To lift the rail
Its radius decreases from infinity at the straight end
Claw Bar To take out Dog-spikes from
(tangent point) to the desired radius of the circular
sleeper
curve at the other end (curve point).
Chisel To cut the rails, bolts, etc.
Different types of transition curves are spiral or
Auger To drill holes for spikes
clothoid, cubic parabola, and Lemniscates.
Shovel To handle the ballast
The Railway Board has decided that on Indian Wire Claw To clean the ballast
Railways, transition curves will normally be laid in the
Spirit Level To check cross level
shape of a cubic parabola.
Sleeper tongs To lift sleepers
Note: However, IRC recommends spiral as the
Q.96. Height of the bridge is kept…… above high flood
transition curve for Highways because it fulfills the
level. (JKPSC Lecturer 2022)
requirement of an ideal transition curve, which are:
(a) 1.2 to 1.5 m
(a) Rate of change of centrifugal acceleration is
consistent (smooth) (b) 1.8 to 2.1m
(c) 2.2 to 2.5m
(b) Radius of the transition curve is infinity at the
(d) More than 2.5
straight edge and changes to radius of circular curve.
(c) At the curve point such that length of transition Ans. (A) The height of the bridge is generally 1.2 to 1.5
curve at any point is inversely proportional to radius of meters above the high flood level of the given region.

OSSC JE MAINS CIVIL 16/03/2025 20 Civil Ki Goli

Different dimensions of the bridge: with expansion or construction joint, it is called an
Width of bridges: It is based on the traffic survey. It independent wing wall.
may be a single lane or double lane with a pedestrian
platform on only one side or on both sides.
Length of bridge: It depends upon the waterway.
Span of the bridge: It depends upon the type of
superstructure proposed.
• Masonry arch: 3 to 15 m
• Slab bridges: Up to 9 m
• Girder and beams: 10 to 60 m A bridge structure is divided into two main parts:
• Truss bridges: 30 to 375 m with simply supported Superstructure – The part of the bridge that carries
ends. traffic and spans between supports.
• Suspension bridges: Over 1500 m • Includes girders, slabs, deck, bearings, and parapets.
• Cable-stayed bridges: Over 1000 m Substructure – The supporting structure that transfers
loads to the ground.
Q.97. Sub structure of a bridge does not include
(JKSSB JE 2021) • Includes abutments, piers, wing walls, and return
(a) Abutment walls.
(b) Girder/Slab Q.98. As compared to laminar flow, the boundary layer
(c) Piers in a turbulent flow will be (UKPSC JE 8 May 2022)

(d) Wing and Return (a) Same

(b) Thicker
Ans. (B) Pier: It is a vertical load-bearing member such (c) Thinner
as an intermediate support for adjacent ends of two (d) Cannot say
bridge spans.
Ans. (B)
R.C.C. piers of the following shapes are used:
• Turbulent flow and laminar flow are two types of
• Rectangular
fluid flow.
• Dumb-bell type
• Laminar flow is characterized by smooth, regular
• Trestle bent
movement of fluid, while turbulent flow is
The piers of the following cross-section are used: characterized by chaotic, irregular movement.
• rectangular • Boundary Layer in Turbulent Flow and Laminar Flow
• with triangular edges towards upstream and Boundary layer is a layer of fluid that is in contact
downstream sides with a solid surface.
• with curved faces on upstream and downstream • The characteristics of this layer are affected by the
sides type of fluid flow. In laminar flow, the boundary layer
is thin and smooth.
• In turbulent flow, the boundary layer is thicker and
chaotic.
• Comparison of Boundary Layer in Turbulent Flow
and Laminar Flow In turbulent flow, the boundary
layer is thicker compared to laminar flow.
• This is because the chaotic movement of fluid in
turbulent flow creates more friction between the
Abutment: A masonry or reinforced concrete wall
fluid and the solid surface, resulting in a thicker
that constitutes the end support of bridges or similar
structures by which it joins the bank of the waterway boundary layer.
is called an abutment. • On the other hand, in laminar flow, the smooth
A wing wall :- It is a structural member located at the movement of fluid creates less friction, resulting in a
end of a bridge structure. When it is constructed thinner boundary layer.
integrally with the abutment, it is termed a cantilever
wing wall and when separated from the abutment

OSSC JE MAINS CIVIL 16/03/2025 21 Civil Ki Goli

Q.99. Groynes are adopted for river bank protection force is termed as gravity dam. It may be constructed
works. When it is placed inclined in the downstream in of masonry or concrete.
the direction of flow in the river, it is designated as • Elementary profile is a theoretical profile.
which one of the following?
• Only external forces acting on elementary profile is
(a) Repelling groyne
water pressure.
(b) Attracting groyne
(c) Neither repelling nor attracting groyne • Internal force is self weight.
(d) Fixed groyne • The shape of elementary profile of a gravity dam is
right angled triangle.
Ans. (B) Groynes: These are the embankment type
structures constructed transverse to the river flow. • For full reservoir condition resultant passes through
They extend into the river from the bank and may also outer one third point and for empty (only self weight)
be called as transverse dykes. reservoir condition resultant passes through inner one
• They are constructed to protect the bank by third point.
deflecting the current away from the bank. • By providing a top width for roadway and free board
• As the water is unable to take a sharp embayment, in elementary profile of a gravity dam, for full reservoir
the bank gets protected for certain distance upstream condition the resultant force will shift towards heel of
and downstream of the groyne. the dam.
• However, the nose of the groyne is subjected to a
tremendous action of water, and has to be heavily
protected by pitching, etc. and the action of eddies
reduces from the head towards the bank.
The different type of groynes provided are as follows:
Attracting Groynes – pointing d/s
The Attracting Groyne points D/S of the direction of
normal flow.
Civil Ki Goli 9255624029
• It causes formation of scour holes closer to the banks
than the repelling groynes.
• Therefore, they tend to maintain deep current close
to the bank. ODISHA JE AE CIVIL ENGINEERING PREVIOUS
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• Deflecting Groynes – perpendicular to bank

• Repelling Groynes – pointing u/s
• Sedimenting Groynes – for deposition of sediment
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Q.100. By providing a top width for roadway and WATCH VIDEO SOLUTION
freeboard in the elementary profile of a gravity dam,
the resultant force for full reservoir condition will
(a) shift towards the heel
(b) shift towards the toe
(c) not shift at all
(d) None of the above
Ans. (A) Gravity Dam:- Structure which is designed in
such a way that its own weigh resist all the external

OSSC JE MAINS CIVIL 16/03/2025 22 Civil Ki Goli