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
n

in wet soil good good

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(a) at the downstream end

or mud
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(b) at the upstream end

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(c) at both ends Distance Small Small Long Long

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(d) at any intermediate section

C

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%.
n
tio

Length of road (L) = 1000 m

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Width of road (W) = 8 m

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Thickness of concrete (T₁) = 15 cm = 0.15 m

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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. (c*)
(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
volume of earthwork to be computed by multiplying
the area of mid-section by the distance between the
two original sections. *As per board answer key correct option is a but
Trapezoidal Formula: correct option is c
𝒅
Volume(V) = [First area section + Last area section + Q.17. A uniform beam of span L is rigidly fixed at both
𝟐
2(sum of other areas)] supports. It carries a uniformly distributed load W' per
Prismoidal Formula: unit length over full span. The bending moment at mid
𝒅 span is:
V= [First area section + Last area section + 4(sum of
𝟑 (a) 𝐰𝐋𝟐 /𝟖
even areas) + 2( sum of odd area] (b) 𝐰𝐋𝟐 /𝟏𝟐
All above can be used for earthwork estimation if (c) 𝐰𝐋𝟐 /𝟏𝟔
highway has a uniform gradient. (d) 𝐰𝐋𝟐 /𝟐𝟒
Q.14. A ratio of moment carrying capacity of a circular Ans. (d)
beam of diameter D and a square beam of size D is:
(a) 𝛑/𝟒 (b) 𝟑𝛑/𝟖
(c) 𝛑/𝟑 (d) 𝟑𝛑/𝟏𝟔
Ans. (d) Given
Side of square = D, diameter of circle = D
For a circular beam of diameter D, the section modulus
is:
𝝅𝒅𝟑
Zcircular =
𝟑𝟐
For a square beam of side DDD, the section modulus is: Applied Moment equilibrium at X,
𝑫𝟑
Zsquare = Σ Μx = 0
𝟔
The ratio of the moment-carrying capacity is: 𝒘𝑳 𝑳 𝒘𝑳𝟐 𝒘𝑳 𝑳
× +𝑴+ = ×
𝝅𝑫𝟑 𝟐 𝟒 𝟏𝟐 𝟐 𝟐
𝒛𝑪𝒊𝒓𝒄𝒖𝒍𝒂𝒓 𝟑𝟐 𝟑𝝅 𝒘𝑳𝟐 𝒘𝑳𝟐 𝒘𝑳𝟐
= 𝑫𝟑
= 𝑴= − −
𝒛𝒔𝒒𝒖𝒂𝒓𝒆 𝟏𝟔 𝟒 𝟖 𝟏𝟐
𝟔
𝒘𝑳𝟐
Q.15. What is the ratio of maximum shear stress to 𝑴=
𝟐𝟒
average shear stress for a circular section?
Q.18. The degree of static indeterminacy (Ds) of a rigid
(a) 2 (b) 2/3 jointed plane frame may be written as (where m = No.
(c) 4/3 (d) 3/4 of members, j = No. of joints, r = No. of reactions):
Ans. (c) (a) Ds = (3m + r) - 3j
We know that (b) Ds = (3m - r) + 3j
(c) Ds = (m + r) - 3j
For circular section,
𝟒
(d) Ds = (3m - r) + j
𝝉𝒎𝒂𝒙 = 𝝉𝑨𝒗𝒈
𝟑 Ans. (a) Degree of static determinacy for plane frame
𝝉𝒎𝒂𝒙 𝟒
= (rigid jointed) = 3m + R-3J
𝝉𝑨𝒗𝒈 𝟑
Where,
Q.16. The ratio of depth to width of a strongest beam
M = Number of members,
that can be cut out of a cylindrical log of wood with
homogeneous and isotropic properties is: R = Number of support reactions,
(a) 1.414 (b) 1.25 J = Number of joints
(c) 0.707 (d) 0.504
OSSC JE MAINS CIVIL 16/03/2025 4 Civil Ki Goli
Q.19. A propped cantilever AB of length L is fixed at A are determined. The system of forces acting on either
and propped at B, subjected to uniformly distributed part of truss constitutes a non-current force system.
load ‘w’ per unit length over its full span length. What is Since there are only three independent equations of
the reaction at propped end B ? equilibrium there should be only three unknown
(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
(c) 50 kN
load) - (upward deflection due to RB) = 0
(d) 60 kN
𝒘𝒍𝟒 𝑹𝑩 𝒍𝟑
− =𝟎 Ans. (a) Given
𝟖𝑬𝑰 𝟑𝑬𝑰
𝟑 Length of simply supported beam, L=6.0m
n

𝑹𝑩 = 𝒘𝒍
tio

𝟖
Bending moment equation at a section from the left
nc

𝟑 𝟓
𝑹𝑨 = 𝒘𝒍 − 𝒘𝒍 = 𝒘𝒍
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𝟖 𝟖 support:
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iv

Q.20. Method of joints is applicable to calculate 𝑀𝑥 = 30x – 0.5 wx²

C

member forces of a truss, when the number of unknown Shear force (Vx) is the derivative of the bending
forces at the joint under consideration is not more than moment with respect to x:
𝒅𝑴𝒙
(a) One (b) Two 𝑽𝒙 =
𝒅𝒙
(c) Three (d) Four Differentiate the given B.M. equation:
Ans. (b) Method of Joint:- At each joint the forces in 𝑽𝒙 =
𝒅𝑴𝒙
(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
Q.22. The hydraulic head that would produce a quick
considering the equilibrium of the entire truss frame.
sand condition in sand stratum of thickness 1.5m,
Then making the use of two equations at the
specific gravity 2.67 and void ratio 0.67is equal to
equilibrium at that joint the two unknown forces are
found. (a) 1.0m
(b) 1.50m
Method of section:- In method of section, after
(c) 2.0m
determining the reactions a section line is drawn
(d) 3m
passing through not more than three members in
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
hydraulic head causing quicksand condition:
unknown forces in the members cut by the section line

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

 (𝑮−𝟏) 𝒉
𝒉𝒄 = ×𝑯 Double drainage t2=?: d2 =
𝟏+𝒆 𝟐
(𝟐.𝟔𝟕 − 𝟏) 𝑯 𝟐 𝒕
= × 𝟏. 𝟓 𝒕𝟐 = 𝒕 × ( ∕ 𝑯) =
𝟏 + 𝟎.𝟔𝟕 𝟐 𝟒
= 1.5 m
Q.26. Minimum required water cement ratio for a
Q.23. The change that take place during the process of workable concrete is:
consolidation of a saturated clay would include: (a) 0.30 (b) 0.40
(a) An increase in pore water pressure and an increase (c) 0.50 (d) 0.60
in effective pressure
(b) An increase in pore water pressure and a decrease in Ans. (b) A lower water cement ratio results in higher
effective pressure strength and durability but requires more workability.
(c) A decrease 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
(d) A decrease in pore water pressure and an increase in concrete.
effective pressure Q.27. The modulus of elasticity (E) of concrete is given
Ans. (d) Consolidation:- The decrease in soil volume by by
the squeezing out of the pore water on account of the (a) E=1000 fck
gradual dissipation of excess hydrostatic pressure (b) E=5000 √fck
induced by an imposed total stress is defined as (c) E=5500 √fck
consolidation. It is a time-dependent phenomenon. (d) E=1000 √fck
The following changes take place during the process of Ans. (b) As per IS 456: 2000,
consolidation of saturated clay:
The Short term Static Modulus of elasticity of concrete
• Volume reduction is due to the expulsion of pore Ec is given as-
water from voids.
Ec = 5000 √𝒇𝒄𝒌
• Decrease in void ratio.
Where,
• Decrease in pore-water pressure.
fck = characteristic compressive strength of concrete,
• Increase in effective stress.
fck and Ec are measured in MPa.
Q.24. A soil has bulk density 2.30 g/ cm³ and water
content 15%. The dry density of the sample is: Q.28. Dorry's testing machine is used for
(a) 1.0 g/cm³ (a) Crushing test of stones
(b) 1.5 g/cm³ (b) Hardness test of stone
(c) 2.0 g/cm³ (c) Impact test of stone
(d) 2.5 g/cm³ (d) Water absorption test

Ans. (c) Given Ans. (b) Dorry's testing machine is a device used to
determine the hardness of stones. It is widely used in
𝜸 = 2.3 g / cm3
the construction industry to determine the suitability
W = 15% = 0.15 of stones for use in various applications.
𝜸
𝜸𝒅 =
𝟏+𝒘 Q.29. Enamel paint is made by adding
𝟐.𝟑
= (a) White paint in varnish
𝟏 + 𝟎.𝟏𝟓
= 2.0 g/cm³ (b) Bitumen in varnish
(c) White lead in lacquer
Q.25. If the time required for 60% consolidation of a (d) Zinc white in spirit
remolded soil sample of clay with single drainage is "T",
then what is the time required to consolidate the Ans. (a) Enamel Paint:- Enamel paint is obtained by
sample of clay with the same degree of consolidation adding a base like white lead, or zinc white, to a vehicle
but with double drainage? which is a varnish.
(a) 4T (b) 2T Enamel paints water proof, chemical resistant and also
provide good paint coverage and colour retention.
(c) T/2 (d) T/4
Q.30.A portion of an embankment having a uniform up-
Ans. (d) For same degree of consolidation
𝟐
gradient 1 in 500 is circular of radius 1000m of the
𝒕𝟏 𝒅
= ( 𝟏) centre line and subtends 180o at the centre. If the height
𝒕𝟐 𝒅𝟐
of the bank is 1m at lower end, and side slopes 2:1, the
single drainage t₁=t,d₁=H
earth work involved is:
(a) 5000 m3
OSSC JE MAINS CIVIL 16/03/2025 6 Civil Ki Goli
(b) 16, 500 m3 Administrative approval confirms that the project is
(c) 27000 m3 necessary and funds are allocated for it.
(d) 40, 500 m3 Q.33. When Environmental Lapse Rate (ELR) is more
Ans. (c) Total length of curve = 1000 × π than Adiabatic Lapse Rate (ALR), then the environment
At the lower end, Initial height = 1 m is said to be
At the 500 m distance, height = 2 m (a) Stable
At the 1000 m distance, height = 3 m (b) Unstable
(c) Neutral
Assuming triangular shape of embankment as not data
(d) None of the above
except side slope of 2: 1 (i.e. 2H: 1 V) is given
Ans. (b) When
ELR > ALR Super- adiabatic lapse rate Unstable
Environment
ELR < ALR Sub - adiabatic lapse rate Stable
Environment
Area at lower end (A1) = S d2 = 2 × 12 = 2m 2 ELR=ALR Neutral Condition
Area at 500 m (A2) = Sd2 = 2 × 22 = 8m2 ELR is negative = Inversion
Area at 1000 m (A3) = S × d2 = 2 × 32 = 18m2 Adiabatic Lapse Rate (ALR) is - 9.8°C per 1000 m rise
As per Simpson's 1/3rd formula ELR = Environment lapse rate
𝒅
V= [(A1 +An )+4(A2 +A4 +A6 +....)+ 2(A3 + A5}) ] ALR = Adiabatic lapse rate
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
n

work is given to contractor

tio
nc

(a) For all activities required to start the work at site on

Ju

finalization of contract
il
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(b) Shifting electricity poles and cleaning of site only

C

(c) For procuring materials

(d) To construct site office
Ans. (a) Mobilization advance up to 10% of Tendered Q.34. The chlorine demand of a water sample was found
Amount" shall be paid to the contractor on submission to be 0.2mg / liter. The amount of bleaching powder
of no revocable Bank Guarantee for an amount of containing 30% available chlorine to be added to treat
110% (One hundred ten percent) of an amount of one liter of such water sample is:
mobilization advance demanded, from a nationalized (a) 0.67 mg
/Scheduled Bank. (b) 0.06 mg
(ii) It is provided for all activities required to start the (c) 1.33 mg
work at site on finalization of the contract documents (d) 0.14 mg
Q.32. PWD initiates a construction work after Ans. (a) Given:
(a) Technical approval for the work Chlorine demand = 0.2 mg/ltr
(b) Administrative approval for the work Total chlorine demand = 0.2 mg x 1 = 0.2 mg
(c) Once Preliminary estimate is made Available chlorine content = 30% = 0.30
(d) Correctly accessing the probable addition and The required amount of bleaching powder
alteration of the work 𝑪𝒉𝒍𝒐𝒓𝒊𝒏𝒆 𝒅𝒆𝒎𝒂𝒏𝒅 𝟎.𝟐
= = = 0.67 mg
Ans. (b) Before starting any construction work, the 𝑨𝒗𝒂𝒊𝒍𝒂𝒃𝒍𝒆 𝒄𝒉𝒍𝒐𝒓𝒊𝒏𝒆 𝟎.𝟑

Public Works Department (PWD) requires Q.35. Turbidity is measured on

administrative approval from the competent authority. (a) Standard silica scale
Once administrative approval is obtained, the next (b) Standard cobalt scale
steps include technical approval and sanctioning of the (c) Standard platinum scale
detailed estimate. (d) Platinum cobalt scale

OSSC JE MAINS CIVIL 16/03/2025 7 Civil Ki Goli

Ans. (a) Turbidity can be measured using Turbidity rod Q.38. What is value of flexural tensile strength of M25
method that consists of a platinum needle having concrete?
Diameter = 1 mm and (a) 4.0 MPa
Length = 25 mm (b) 3.5 MPa
at the tip of aluminium rod. (c) 3.0 MPa
This rod is inserted in the water sample to be tested (d) 1.75 MPa
and the height at which this needle just become Ans. (b) Given,
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
which is obtained by the addition of Silica as Silica = 0.7√𝟐𝟓
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
Ans. (d) Minimum clear cover requirement: related to each other in a hierarchy of prepotency (or
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
exposure - 30 mm (c) Employee empowerment
However, if the diameter of the bar does not exceed (d) Continuous improvement
12 mm, or cover may be reduced by 5 mm. Thus for Ans. (d) Continuous Improvement is one of the pillars
main reinforcement up to 12 mm diameter bar and for of TQM. The output of a production process must not
mild exposure, the nominal cover is 15 mm. only satisfy customer needs but the customer must
For Column: feel satisfied with the product.
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,
whose reinforcing bars do not exceed 12 mm, the wind load on the purlins and L is center distance of
concrete cover of 25 mm to be used for reinforcement. support) :
Q.37. Minimum percentage of high yield deformed bars (a) WL/6
in a R.C. slab compared with gross concrete area is: (b) WL/8
(a) 0.40 (b) 0.15 (c) WL/10
(c) 0.12 (d) 0.10 (d) WL/12

Ans. (c) The minimum percentage of reinforcement is Ans. (c) Purlins and its design recommendation: Purlins
0.12% of the gross cross-sectional area if HY SD bars (Fe are horizontal beam members which run parallel to
415) are used. the ridge and connect the trusses along the length of
The minimum percentage of reinforcement is 0.15% of the ridge.
the gross cross-sectional area if mild steel bars are When an iron angle section is used as purlins for steel
used. roof trusses then generally following
recommendations are required to be fulfilled. These
OSSC JE MAINS CIVIL 16/03/2025 8 Civil Ki Goli
 𝟐𝟖𝟗.𝟐
recommendations are applicable for roof trusses with = = 2.83
𝟏𝟎𝟐.𝟑
a pitch angle of less than 30° and for a maximum
imposed load of 0.75 kN/m². Q.45. What is the moisture depth available for
1. The depth of the angle purlin in the plane evapotranspiration in root zone of 1 m depth soil, if dry
approximate to the incidence of the maximum load weight of soil is 1.5gm / cc, field capacity is 30% and
should not be less than (1/45) th of the length of the permanent wilting point is 10%?
purlin. (a) 450 mm (b) 300mm
2. The width of the angle purlin should not be less than (c) 200 mm (d) 150 mm
(1/60) th of the length of the purlin. Ans. (b) Given
3. The maximum bending moment in the purlin in FC = 30%
taken as (WL/10) where W is the total uniformly PWP = 10%
distributed load on the purlin including wind load in
𝜸𝒅 = 1.5g / cc
the normal direction of the roof and L is the center-to-
center distance between purlin supports. d = 1m
𝜸𝒘 = 1g / cc
Q.42. A butt weld is specified by
The depth of water stored (d) in the root zone
(a) Effective throat thickness 𝜸𝒅
(b) Plate thickness 𝒅𝒘 = × 𝒅(𝑭𝑪 − 𝑷𝑾𝑷)
𝜸𝒘
(c) Size of weld 𝟏.𝟓
= × 𝟏 × (𝟎 ⋅ 𝟑𝟎 − 𝟎 ⋅ 𝟏𝟎)
(d) Penetration thickness 𝟏
= 0.3 m = 300 mm
Ans. (a) Butt Weld
• They are better in highly stressed structures where a Q.46. In a river, silt excluder and silt ejector are
smooth flow of stress is a necessity. constructed
• It is mainly designed for direct compression and (a) At a location after the head regulator and at the
tension and occasionally for shear also. head of the canal, respectively
(b) At the head of the canal, and at a location after the
• It involves no change in the section at the location of
n

head regulator, respectively

tio

the joint.
nc

(c) At the same location

Ju

• The butt weld is specified by effective throat (d) At specific locations depending upon diverse factors
il

thickness.
iv

and their locations do not follow a set pattern

C

Q.43. In a tension member if one or more than one rivet Ans. (b) Silt Regulation works:- The entry of silt into a
holes are off the line, the failure of the member depends canal, which takes off from a headworks, can be
on reduced by constructing certain special works, called
(a) Pitch silt control works. These works may be classified into
(b) Gauge the following two types:
(c) Diameter of the rivet holes Silt Excluders
(d) All of the above Silt Ejectors
Ans. (d) In a tension member, if one or more than one Silt Excluders:
rivet hole are off the line, the failure of the member • Silt excluders are those works that are constructed
depends upon the pitch, gauge, and diameter of the on the bed of the river, upstream of the head
rivet holes. regulator.
Q.44. An object weights 289.2N in air and 186.9N in • The clearer water enters the head regulator and
water. What is relative density of the material of the silted water enters the silt excluder. In this type of
object? works, the silt is, therefore,, removed from the water
before in enters the canal.
(a) 2.83 (b) 2.45
Silt Ejector:
(c) 2.15 (d) 3.15
• Silt ejectors, also called silt extractors, are those
Ans. (a) Given: devices that extract the silt from the canal water after
Weight of object in air = 289.2 N the silted water has traveled a certain distance in the
Weight of object in water = 186.9 N off-take canal.
Loss of weight=289.2−186.9=102.3N These works are, therefore, constructed on the bed of
𝑾𝒆𝒊𝒈𝒉𝒕 𝒊𝒏 𝒂𝒊𝒓 the canal, and a little distance downstream from the
Relative density =
𝑳𝒐𝒔𝒔 𝒐𝒇 𝒘𝒆𝒊𝒈𝒉𝒕 𝒊𝒏 𝒘𝒂𝒕𝒆𝒓 head regulator.

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

Q.47. The plan of a map was photo copied to a reduced Q.50. The rise of water level above its normal level
size such that a line originally 100 mm measures 90mm. when passing under bridge is called
The original scale of the plan was 1:1000. The revised (a) Afflux
scale is: (b) Free board
(a) 1:900 (c) Headroom
(b) 1:1111 (d) Tailroom
(c) 1:1121 Ans. (a) Afflux is an increase in water level that can
(d) 1:1221
occur upstream of a structure, such as a bridge or
𝟏 culvert, that creates an obstruction in the flow.
Ans. (b) Original sale =
𝟏𝟎𝟎𝟎
Shrunk length = 90 mm On the upstream side of the dam, the depth of the
water will be rising due to the obstruction/barrier. If
Original length = 100 mm
there had not been any obstruction (such as dam) in
𝑺𝒉𝒓𝒖𝒏𝒌 𝒍𝒆𝒏𝒈𝒕𝒉
Shrunk factor = the path of the flow of water in the channel, the depth
𝑶𝒓𝒊𝒈𝒊𝒏𝒂𝒍 𝒍𝒆𝒏𝒈𝒕𝒉
𝟗𝟎 of water would have been constant parallel to the bed
S.F. = = 0.9 of the channel. Due to obstruction, the water level
𝟏𝟎𝟎
Shrunk scale or revised scale rises, and it has maximum depth from the bed in some
= S.F. × Original length sections. This rise of water level on the upstream end
𝟏 𝟏 is termed as afflux.
= 0.9 × =
𝟏𝟎𝟎𝟎 𝟏𝟏𝟏𝟏
Q.51. Which of the following is NOT TRUE about plane
Q.48. The bearings of lines OA and OB are 16° 10' and table surveying?
332° 18', the value of the included angle BOA is :
(a) It is a rapid method of surveying.
(a) 316° 10'
(b) It is a precise method of surveying.
(b) 158° 28'
(c) 348° 08' (c) It is a cost-effective method of surveying.
(d) 43° 52' (d) It is a versatile method of surveying.

Ans. (d) Given Ans. (b) Plane-table survey:- A plane table surveying is
Bearing of line OA = 16° 10' a graphical method of surveying. In this method of
surveying. field observation and plotting are done
Bearing of line OB = 332° 18'
simultaneously helping the surveyor to compare the
Since the bearings are measured clockwise from the plotted details with actual features of the ground.
north, the included angle can be calculated using:
Advantages of Plane Table Survey:
Included angle = Bearing of OB−Bearing of OA
• Plane Table Surveying is suitable for small and
= 360∘−(332∘18′−16∘10′) medium scale mapping (1: 10,000 to 1: 2, 50,000),
= 43° 52' where great accuracy is not required As surveying and
Q.49. The highway capacity is expressed in Passenger plotting are done simultaneously in the field, chances
Car Unit(PCU). According to IRC, for a passenger car, the of getting omission of any detail get less.
PCU is: • The plotting details can immediately get compared
(a) 1 (b) 2 with the actual objects present in the field. Thus
errors, as well as the accuracy of the plot, can be
(c) 3 (d) 4
ascertained as the work progresses in the field.
Ans. (a) Passenger Car Unit:- It is a vehicle unit used for
• The plane table survey is generally more rapid and
expressing highway capacity. It is a metric used in
less costly than most other types of survey.
Transportation Engineering, to assess traffic-flow rate
on a highway. Disadvantages of Plane Table Survey:
According to IRC • The plane table survey is not possible in unfavorable
climates such as rain, fog, etc.
Vehicle PCU
• This method of survey is not very accurate and thus
Car or taxi 1 unsuitable for large scale or precise work.
Motor Cycle 0.5 • As no field book is maintained, plotting at different
scale require full exercise.
Bus or truck 3.5
• The quality of the final map depends largely on the
Cycle 0.2 drafting capability of the surveyor
Bullock Cart 6 Note:- There is confusion between option b and d but
Bullock Cart (large) 8 most appropriate option is b.

OSSC JE MAINS CIVIL 16/03/2025 10 Civil Ki Goli

Q.52. In an adjustable level, when the bubble is at Q.54. Which one of the following is correct for Prismatic
centre, the axis of the bubble tube becomes parallel to Compass?
(a) Line of sight (a) The graduated ring rotates with line of sight.
(b) Line of collimation (b) Instrument cannot be used without tripod.
(c) Axis of telescope (c) The graduations are engraved inverted.
(d) The readings can directly be taken by seeing through
(d) None of these
the top of the glass.
Ans. (b*) Line of sight: Ans. (c) A prismatic compass is a navigation and
• The sighting or pointing line of a telescope, defined surveying instrument which is extensively used to find
by the optical center of the objective and the out the bearing of the traversing. It is the most
intersection of crosshairs. convenient and portable form of magnetic compass that
Line of collimation: can either be used as a hand instrument or fitted on a
• It is an imaginary line passing through optical centre tripod.
of the objective glass and its continuation. The graduated circle of a prismatic compass and
Axis of Telescope: surveyor compass is shown in the figure below:
• 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.
n
tio

Some other differences between the two compass are

nc
Ju

as follows:
il
iv

Item Prismatic Surveyor’s

C

Compass Compass
Needle Broad type Edge bar type
Scale Free to float Attached to the
When the bubble of the level tube of a level remains along with box
central, the line of collimation or line of sight is broad type
horizontal, and the axis of the bubble tube becomes magnetic
parallel to the Line of sight. needle
Note:- According to board answer key correct option is Bearing Whole circle Quadrantal
a but correct answer is option b. bearing bearing
Q.53. In a straight line, two lengths A and B are Graduations Inverted (as Direct
measured from a point P. The standard errors of graduation
measured lengths A and B are found to be 0.4m and have to be
0.3m. The standard error in the length A minus B is: observed
(a) 0.7 m (b) 0.35 m through a
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
Standard error in length B = 0.3 m surveyor has to
σ A−B = √(𝝈𝑨 )𝟐 + (𝝈𝑩 )𝟐 read the
northern end
= √𝟎. 𝟒𝟐 + 𝟎 ⋅ 𝟑𝟐
of the needle
= 0.5m
Tripod Not essential Essential

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

Q.55. Which one of the following methods estimates Q.58. The time by which activity completion time can be
the best area of an irregular and curved boundary? delayed without affecting the start of succeeding
(a) Trapezoidal method. activities is known as:
(b) Simpson's method. (a) Total Float
(c) Average ordinate method. (b) Free Float
(d) Mid-ordinate method (c) Interfering Float
(d) Zero Float
Ans. (b) Simpson's rule:- In order to apply Simpson's
rule, the area must be divided in even number i.e., the Ans. (b) Free float:- It is the amount of time that the
number of offsets must be odd i.e., n term in the last activity completion time can be delayed without
offset 'O,' should be odd. affecting the earliest start time of the immediate
For irregular boundaries, Simpson's rule is preferred successor activities in the network.
over the trapezoidal rule to calculate the given area. Total float: It is the amount of time that the
According to this rule the short length of boundaries completion time of an activity can be delayed without
between the two adjacent ordinates is a parabolic arch affecting project completion time.
The area is given by Simpson's rule: Interfering Float: Maximum amount by which an
𝒅 activity can be delayed without delaying the project
Area = (01+On)+4(02+04+.. where 01, 02, 03, ........ is the
𝟑 but will cause delay to the Early Start of some
offset following activity
Q.56. An image of the top of the hill is 92 mm from the Independent Float: Amount by which an activity can be
principal point of the photograph. The elevation of the delayed without delaying the project; even if all
top of the hill is 400 m and the flying height is 4000 m predecessors are at Late Finish and all Successors are
above the datum. The relief displacement will be at Early Start.
(a) 9.2 mm Q.59. What is the angle between principal strain axis
(b) 12 mm and maximum shear strain axis ?
(c) 88mm
(a) 0° (b) 30°
(d) 8 mm
(c) 45° (d) 90°
Ans. (a) Given
Ans. (c) The plane on which the shear stress is zero is
r = 92 mm.
called the principal plane and the normal stress on the
H = 4000 m = 4000000 mm. principal plane is called the principal stress.
h = 400 m = 400000 mm Planes of maximum shear stress occur at 45° to the
𝒓𝒉
Relief displacement (d) = principal planes.
𝑯
𝟗𝟐 𝒙 𝟒𝟎𝟎𝟎𝟎𝟎 Q.60. A Mohr circle reduces to a point when the body is
= = 9.2 mm
𝟒𝟎𝟎𝟎𝟎𝟎𝟎
subjected to
Q.57. To uniquely determine the position of the user (a) Pure shear
using GPS, one needs to receive signals from at least (b) Uniaxial stress only
(a) 1 satellite (c) Equal and opposite axial stresses on two mutually
(b) 2 satellites perpendicular planes, the planes being free of shear
(c) 3 satellites (d) Equal axial stresses on two mutually perpendicular
(d) 4 satellites planes, the planes being free of shear

Ans. (d) In Cartesian coordinate system of space and Ans. (d*) When the normal stresses on the two
time, we have four variables - x, y, z (three space mutually perpendicular planes are equal and alike
variables) and t (one time variable). Each satellite then the radius of the Mohr circle will be zero.
corresponds to one variable. Therefore, to uniquely Radius of Mohr's Circle,
determine the position of the user using GPS minimum 𝟐
4 satellites are required. R = √(𝝈𝒙−𝝈 𝒚 ) + 𝝉𝟐𝒙𝒚
𝟐
However, if we use less than 4 satellite say 2 nos., then
out of variable two variables say x and y are known and Here 𝝈𝒙 = 𝝈𝒚 and 𝝉𝒙𝒚 = 0
other two variables z and times are unknown and they Therefore, R = 0 means Mohr's circle reduces to a
can take many values i.e. we have multiple positions point.
and hence, we are not able to determine the position *As per board answer key correct option is b but
of any object uniquely. correct answer is option d.

OSSC JE MAINS CIVIL 16/03/2025 12 Civil Ki Goli

Q.61. A steel column is pinned at both ends and has a cost, artificial stones (also known ascast stone)
buckling load of 250 kN. If the column is restrained become the choice.
against lateral movement at its mid-height, its buckling Composition of different artificial stones are given
load will be below:
(a) 200 kN Imperial stone:
(b) 800 kN
Finely crushed granite is washed carefully and mixed
(c) 1000 kN with Portland cementand then steam cured for 24
(d) 1200 kN
hours. The cured blocks areimmersed in silicate tanks
Ans. (c) The buckling load (P₁) of a steel column pinned for three days.
at both the ends is (Leff = L) a. Victoria stone: These are granite pieces with the
𝝅𝟐 𝑬𝑰 surfaces hardened by keeping immersed in
𝑷𝟏 = = 𝟐𝟓𝟎 𝒌𝑵
𝑳𝟐 sodasilicate for about two months.
When the column is restrained against lateral
b. Ransom stone: These are prepared by mixing soda
movement at its mid-height, then Leff = L/2
silicate with cement to provide decorative flooring.
The buckling load (P2) of the steel column for the above These are also known as chemical stones. These have
case will be compressive strenath of about 32 N/mm².
𝝅𝟐 𝑬𝑰 𝝅𝟐 𝑬𝑰
𝑷𝟐 = =𝟒×( ) = 4 × 250 = 1000 kN c. Garlic stone: These are produced by moulding a
𝑳 𝟐 𝑳𝟐
( ) mixture of iron slag and Portland cement. These
𝟐

Q.62. The constituent compound in Portland cement areused as flag stones, surface drains, etc.
which reacts immediately with water, and also sets d. Bituminous stone: Granite and diorite are
earliest, is: impregnated with prepared or refined tar to
(a) Tricalcium Silicate formbituminous stone. These are used for providing
noise, wear and dust resistant stone surfaces.
(b) Dicalcium Silicate
(c) Tricalcium Aluminate Q.64. As per IS 456:2000, the maximum admissible
(d) Tetracalcium Alumino ferrite water-cement ratio for mild environmental exposure
n

for concrete should be

tio

Ans. (c) Tricalcium Aluminate: C3A is formed within 24

nc

(a) 0.55 (b) 0.50

Ju

hours of the addition of water in the cement and is

(c) 0.45 (d) 0.45
il

responsible for maximum evolution of heat of

iv
C

hydration. It is the first compound that is formed after Ans. (a)

addition of water and sets early. Exposure Minimu Maximu Minimu Maximu Exposure
Tetracalcium aluminoferrite: C4AF is also formed Condition m m Water- m m Water- Condition
Cement Cement Cement Cement Description
within 24 hours of the addition of water in the cement Content Ratio Content Ratio
but its individual contribution to the overall strength (PCC) (PCC) (RCC) (RCC)
of the cement is insignificant. (kg/m³) (kg/m³)
Tricalcium silicate: C3S is formed within a week or so Mild 220 0.6 300 0.55 Concrete
of the addition of water in the cement and is exposed to
responsible for the early development of strength of coastal area
the cement. Moderat 240 0.6 300 0.5 Concrete
Dicalcium silicate: C₂S is the last compound that is e continuousl
y
formed after the addition of water in the cement underwater
which may require a year or so for its formation. It is
responsible for the progressive strength of the cement. Severe 250 0.5 320 0.45 Concrete
immersed
Q.63. Which one of the following stone is produced by under
moulding a mixture of iron slag and Portland cement? seawater

(a) Imperial stone Very 260 0.45 340 0.45 Concrete

(b) Garlic stone Severe buried
under
(c) Ransom stone
aggressive
(d) Victoria stone subsoil
Ans. (b) Artificial stone:- Artificial stonesare made with Extreme 280 0.4 360 0.4 Concrete
cement and natural aggregates ofthe crushed stone under tidal
and sand with desired surface finish. Where ever zone
durable natural stones are not available at reasonable

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

Q.65. The method used for estimation of depreciation Height of wall, H=2m
of building is known as: Thickness of wall = 85 cm = 0.85 m
(a) Logistic curve method Area to be plastered=2×(L×H)
(b) Rental method =2×(12×2)
(c) Constant percentage method
= 48 m2
(d) Direct comparison method
Q.68. Number of bricks (having size of 20 cm x 10 cm x
Ans. (c) Method used for estimation of depreciation of
10 cm) required for 17 cu.m of brickwork is
building
approximately
Straight line method: It is the simplest method of
(a) 6750 (b) 7200
depreciation. In this method it is assumed that the
book value of an asset will decrease by same amount (c) 7500 (d) 8500
every year over the useful life till its salvage value is Ans. (d) Given:
reached. In other words, the book value of the asset Volume of brickwork = 17 m³
decreases at a linear rate with the time period.
Size of one brick = 20cm×10cm×10cm
Constant percentage method: In this method, it is
Volume of one brick=0.20×0.10×0.10=0.002m3
assumed that the property will lose its value by a 𝟏𝟕
constant percentage of its value at the beginning of Number of bricks = = 8500
𝟎.𝟎𝟎𝟐
every year.
Q.69. The plinth area of a building does not include area
Sinking Fund Method: In this method, the depreciation of
of a property is assumed to be equal to the annual
(a) the walls at the floor levels.
sinking fund plus the interest on the fund for that year,
which is supposed to be invested on interest bearing (b) internal shaft for sanitary installations up to 2m²
investment. area.
(c) lifts
Q.66. Due to a change in price level, a revised estimate
(d) cantilevered porches
is prepared if the sanctioned estimate exceeds
(a) 2.0% (b) 2.5% Ans. (d) The plinth area is the covered built-up area
(c) 4.0% (d) 5.0% measured at the floor level of any storey or at the floor
level of the basement.
Ans. (d) The revised estimate is a detailed estimate for Following areas are included during measurement of
revised quantities and the rate of items of works plinth area:
originally provided in the estimate without material
1) Area of the wall at the floor level, excluding plinth
deviation of a structural nature from the design
offsets.
originally approved for a project.
2) Areas of the internal shaft for sanitary installations
It is required to be prepared for the following reasons:
and garbage chute, electrical, telecom, and firefighting
i) When a sanctioned estimate is likely to exceed by services
more than 5% either from the rates being found
3) Vertical duct for air conditioning and lift well
insufficient or from cause whatsoever except
including landing
important structural alteration.
4) Staircase room or head room other than terrace
ii) When the expenditure of works exceeds or is likely
level
to exceed by more than 10% of the administrative
approval (for work more than 5 lakhs/-) 5) Area of barsati at terrace level
iii) When there are material deviations from the Following areas are not included during measurement
original proposal but not due to the material deviation of plinth area:
of structural nature. 1) Additional floor for seating in assembly buildings,
iv) When it is found that the sanctioned estimate is theatres, auditoriums
more than the actual requirement. 2) Cantilevered porch, Balcony, Area of loft, Internal
sanitary shaft, and garbage shaft
Q.67. A wall of 12m in length having a height of 2m is to
be plastered. If the thickness of the wall is 85 cm, the 3) Area of the architectural band, cornice, Open
quantity of plastering in (sq. m) required is in the range platform, etc.
(a) 45 to 65 (b) 20 to 40 4) Towers, turrets, domes projecting above the terrace
level at the terrace
(c) 65 to 85 (d) 1 to 20
Q.70. Water charge are usually taken as_____%for rate
Ans. (a) Given:
analysis of an item
Length of wall, 𝐿 = 12𝑚
OSSC JE MAINS CIVIL 16/03/2025 14 Civil Ki Goli
(a) 3 (b) 1 • The water content obtained from the calcium carbide
(c) 1.5 (d) 5 method is based on the initial weight of wet soil. It
should be converted to water content based on the dry
Ans. (c) The analysis of rates is worked out for the unit
weight of the soil.
payment of the particular item of work under two
heads: Materials and Labour. Oven drying method:
The cost of items of work = Material cost + Labour cost • Oven dry method is the most accurate and simplest
method for water content determination.
• Other costs included to the above cost of items of
work are: • In this method complete drying of soil sample occurs
and water content in the sample is calculated
• Tools and Plants (T&P) = 2.5 to 3 % of the labour cost
accurately by a maintained temperature in the oven
• Transportation cost (if conveyance more than 8 km is (105°C to 110°C) for 24 hours.
considered.)
• For highly organic soils a low temperature of about
• Water charges = 1.5 to 2% of total cost 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 • This method is used when an electric oven is not
(d) Remains in the web 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.
n
tio

In case of T-beam, if depth of the neutral axis lies Pycnometer method:- This is a quick method but is less
nc

within the flange region and there is sagging bending accurate than oven drying method. This method is
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moment then only the portion below Neutral Axis will used only when specific gravity of soil solids is known.
il
iv

be in tension and hence, neglected and T-beam will

C

Q.73. When the sand in-situ happens to be in its densest

behave like a rectangular beam having width equal to
state, then the sand's relative density would be:
flange width.
(a) Greater than 1
:: A T-beam behaves as a rectangular beam of width
(b) Between 0 & 1
equal to its flange if its neutral axis remains within the
(c) 1
flange is a correct statement.
(d) None of these
Q.72. Which of these methods is the most appropriate
Ans. (c) In its densest state, the relative density of sand
method used in site to determine a soil's water content?
is 1. This means that the density of the sand is equal to
(a) Pycnometer method the density of water. This occurs when the sand is fully
(b) Oven drying method 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)
(a) Compressibility
• The calcium carbide method is the quickest and (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
of unconfined compressive strength of the clay in the
• The method is based on the principle that when the
undisturbed state to the unconfined compressive
water in the soil reacts with calcium carbide, acetylene
strength of the clay in the remoulded state.
gas is produced and the pressure exerted by the
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.
Thixotropy:

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

The loss of strength of soil due to remoulding is partly Or 𝒊𝒄 = (G-1) (1-n)
due to the change in soil structure and partly due to Q.77. A normally consolidated undisturbed clay is
disturbance caused to water molecules in the having a liquid limit of 58%. What is the compression
absorbing layer. index (Cc) of this clay?
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
gain in strength of the true soil with the passage of Ans. (c) The compression index (Cc) for normally
time after it has been remoulded is called thixotropy. consolidated clay can be estimated using the empirical
Confusion Point: -The question clearly mentions "loss correlation:
of strength with remoulding, with water content Cc=0.009(wL−10)
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
Ans. (d) Triaxial test is used to determine the shear homes or industries or automobiles, etc.
strength parameter of soil. Secondary pollutant:- The primary pollutants often
The triaxial compression test consists of two stages: react with one another or with water vapor, aided and
First stage: In this, a soil sample is set in the triaxial cell, abetted by the sunlight, to form entirely a new set of
and confining pressure is then applied. pollutants, called the secondary pollutants.
Second stage: in this, additional axial stress (also called These are the chemical substances, which are
deviator stress) is applied which induces shear stresses produced from the chemical reactions of natural or
in the sample. The axial stress is continuously anthropogenic pollutants or due to their oxidation,
increased until the sample fails. etc., caused by the energy of the sun.
During both the stages, the applied stresses, axial Q.79. Which of these layers of the atmosphere consists
strain, and pore water pressure or change in sample of the ozone layer that is responsible for absorbing the
volume can be measured. Ultra-Violet (UV) light?
Q.76. A deposit of fine sand has a porosity n and specific (a) Troposphere
gravity of soil solids is G. The hydraulic gradient of the (b) Mesosphere
deposit to develop boiling condition of sand is given by: (c) Stratosphere
(a) (G-1) (1-n) (d) None of these
(b) (G-1) (1+n) Ans. (c) The ozone layer is a natural layer of gas in the
(c) (G - 1) / (1 - n) upper atmosphere that protects humans and other
(d) (G - 1) / (1 + n) living things from harmful ultraviolet (UV) radiation
Ans. (a) Critical hydraulic gradient (ic): The quicksand / from the sun.
boiling condition occurs at a critical upward hydraulic • The ozone layer is typically thicker over the poles
gradient typically around 1.0 for many soils, when the than over the equator.
seepage force just balances the buoyant weight of an • The ozone layer exists in the stratosphere, a layer 10
element of soil. to 50 km above the Earth's surface.
At the critical conditions, the effective stress is equal • The reasons for ozone depletion are a wide range of
to zero. industrial and consumer applications, mainly
𝒊𝒄 =
𝑮−𝟏 refrigerators,airconditioners(hydrochlorofluorocarbo
𝟏+𝒆

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

ns(HCFCs), chlorofluorocarbons (CFCs)), and fire intimate contact with the sludge from the secondary
extinguishers. clarifier.
• Ozone depletion is greatest at the South Pole • It requires less space, does not produce obnoxious
(Antarctica) odor, and requires less time for wastewater
treatment.
Q.80. Which of these elements is present in the drinking
water that can lead to numerous fatal diseases? • It requires skilled supervision
(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
n
tio

to 0.01 mg/litre and in absence of other alternative Ans. (b) The heavier inert matter in wastewater is
nc

typically referred to as "grit." Grit can include sand,

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sources to 0.05 mg/litre.

gravel, silt, and other heavy materials that settle out
il
iv

Q.81. One litre of sewage, when allowed to settle for 30 of wastewater due to their weight.
C

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
(a) 9 (b) 24 (b) Shear capacity of the web is neglected
(c) Shear capacity of both flange and web is neglected
(c) 30 (d) 81
(d) None of the above
Ans. (a) Given,
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
OSSC JE MAINS CIVIL 16/03/2025 17 Civil Ki Goli
Ans. (c) The common types of failure in steel beams Q.88. A bitumen sample has been graded as VG30 as per
are: IS: 73-2013. The '30' in the grade means that
Web Crushing: This occurs when the web portion of a (a) Penetration of bitumen at 25°C is between 20 and 40
beam experiences excessive localized compressive (b) Viscosity of bitumen at 60°C is between 2400 and
stress. The web becomes crushed or compacted, 3600 Poise
leading to a loss of structural integrity and potential (c) Ductility of bitumen at 27°C is more than 30 cm
collapse. (d) Elastic recovery of bitumen at 15°C is more than 30%
Shear Buckling: When a beam is subjected to Ans. (b) According to IS 73-2013, Number 30 indicates
significant shear forces, it may buckle or deform out of the range of viscosity of bitumen [(100 ±20) × 30] in
its original shape. This distortion can compromise the terms of Poise.
beam's load-carrying ability, leading to further failure. For VG 30 Bitumen:
Lateral Buckling: This failure occurs when a slender Maximum Viscosity: (100 + 20 = 120) x 30 = 3600 Poise
beam experiences buckling out of its plane due to Minimum Viscosity: (100-20 = 80) x 30 = 2400 Poise
compressive forces. The buckling may happen
laterally, causing the beam to twist or bend, Q.89. A car is moving at a speed of 90kmph on a road
weakening its load-bearing capacity. having a 2% upward gradient. The driver applies brakes
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
unbalanced torsional loads, it may twist, causing an (b) 225 m
unexpected deformation. The twisting compromises (c) 1056 m
the beam's structural performance and can lead to (d) 150 m
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
to localized stress. This localized failure can affect the Upward gradient=2%
overall performance and stability of the entire beam. Reaction time = 1.5 sec
f=0.15
Q.87. As per IRC 96, 1983, the recommended
𝒗𝟐
carriageway width for a two lane road with kerbs is: SSD = 𝟎. 𝟐𝟕𝟖𝒗𝒕 +
𝟐𝟓𝟒(𝒇+𝒏%)
(a) 7.0 m (b) 7.5 m 𝟗𝟎𝟐
(c) 8.0 m (d) 8.5 m = 𝟎. 𝟐𝟕𝟖 × 𝟗𝟎 × 𝟏. 𝟓 + 𝟐
𝟐𝟓𝟒(𝟎.𝟏𝟓+ )
𝟏𝟎𝟎
Ans. (b) Width of Pavement or Carriageway: = 225.1 m
• 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
vehicle and minimum. side clearance. (a) Kerb
• The maximum permissible width of the vehicle as per (b) Shoulder
IRC specification is 2.44 m. (c) Camber
(d) All of the above
Class of Road Width of
Carriageway Ans. (C) Camber or Cross Slope:
Camber is the slope provided to the road surface in the
Single lane 3.75 m
transverse direction to drain off the rainwater from
Two lanes, without raised kerbs 7.0 m the road surface.
• Drainage and quick disposal of water from the
Two lanes, with raised kerbs 7.5 m
pavement surface by providing camber (cross slope) is
Intermediate carriageway (except 5.5 m considered important because of the following
on important road) reasons:
Multi-lane pavement 3.5 m per lane

OSSC JE MAINS CIVIL 16/03/2025 18 Civil Ki Goli

• To prevent the entry of surface water into the (b) Bituminous surfacing
pavement layers and the subgrade soil through the (c) Gravel roads
pavement. (d) WBM construction
• To prevent the entry of water into the bituminous Ans. (B) Alligator cracking are also called as map
pavement layers, continued contact with water causes cracking or fatigue cracking. It is the most common
the stripping of bitumen from the aggregates and type of failure in bituminous surfacing consisting of a
results in the deterioration of pavement layers. series of interconnected cracks creating small,
irregular shapes in pavement.
• To remove the rainwater from the pavement surface
as quickly as possible and to allow the pavement to get
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 Different types of failure encountered in flexible
the pavement and into the drainage system. pavements are as follows:
Shoulder: Shoulders are provided on both sides of the 1. Alligator cracking or Map Cracking (Fatigue): This is
pavement all along the road in the case of an undivided a common type of failure of flexible pavements. This is
carriageway. Shoulders are provided along the outer also known as fatigue failure. Following are the
side of the carriageway in the case of a divided preliminary causes
carriageway. Provides additional space for water a. Relative movement of pavement layer material
runoff and prevents water accumulation on the main b. Repeated application of heavy wheel loads
carriageway.
c. Swelling or shrinkage of subgrade or other layers due
Guard Rail: The guide rail is defined as an additional to moisture variation
rail placed midway between the two ordinary rails of a
2. Consolidation of pavement layers (Rutting):
track that guides the wheels of the train. It is designed
Formation of rut falls in this type of failure. A rut is
n

in connection with devices on the engine or car, to

tio

depression or groove worn into a road by travel of

nc

keep a train from leaving the track on curves,

wheels. This type of failure is caused due to the
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crossings, or steep grades.

following reasons.
il
iv

Q.91. In highway pavements emulsions are mainly used

C

a. Repeated application of load along the same wheel

in ( JKPSC AE 2021) path resulting in longitudinal ruts.
(a) Surface dressing b. Wearing of the surface course along the wheel path
(b) Patching and maintenance resulting in shallow ruts.
(c) Bitumen macadam
3. Shear failure cracking: Shear failure causes
(d) Asphaltic concrete
upheaval of pavement material by forming a fracture
Ans. (B) Bitumen emulsion: or cracking. Following are the primary causes of shear
• These are used in bituminous road constructions, cracking
especially in maintenance and patch repair works. a. Excessive wheel loading
• The main advantages of emulsion is that it can be b. Low shearing resistance of pavement mixture
used in wet weather even when it is raining. 4. Longitudinal Cracking: This type of crack extends
• Bitumen emulsion is a liquid product in which to the full thickness of the pavement. The following are
the primary causes of longitudinal cracking-
bitumen is suspended in a finely divided condition in
a. Differential volume changes in subgrade soil
an aqueous medium and stabilized by suitable
b. Settlement of fill materials
material.
c. Sliding of side slopes.
• Normally cationic type emulsions are used in India.
5. Frost Heaving: Frost heaving causes upheaval of
• The bitumen content in the emulsion is around 60 %
the localized portion of the pavement. The extent of
and the remaining is water. When the emulsion is frost heaving depends upon the groundwater table
applied on the road it breaks down resulting in the and climatic conditions.
release of water and the mix starts to set. 6. Lack of binding to the lower course: When there
Q.92. Alligator or map cracking is the common type of is a lack of binding between surface course and the
failure in underlying layer, some portion of surface course
(a) Concrete pavements
OSSC JE MAINS CIVIL 16/03/2025 19 Civil Ki Goli
looses up materials creating patches and potholes. Note: However, IRC recommends spiral as the
Slippage cracking is one form of this type of failure. transition curve for Highways because it fulfills the
7. Reflection cracking: This type of failure occurs requirement of an ideal transition curve, which are:
when the bituminous surface course is laid over the (a) Rate of change of centrifugal acceleration is
existing cement concrete pavement with some cracks. consistent (smooth)
This crack is reflected in the same pattern on (b) Radius of the transition curve is infinity at the
bituminous surface/ straight edge and changes to radius of circular curve.
8. Formation of waves and corrugation (c) At the curve point such that length of transition
9. Bleeding: Excess bituminous binder occurring on the curve at any point is inversely proportional to radius of
pavement surface causes bleeding. Bleeding causes a circular curve and calculation and field
shiny, glass-like, reflective surface that may be tacky to implementation is very easy.
the touch. Q.95. Which of the following are related to
10. Pumping: Seeping or ejection of water and finds maintenance of railway track?
from beneath the pavement through cracks. 1. Jim crow and gauge bar
Q.93. What is curve resistance for a 50 tonnes train on 2. Through packing and boxing
a BG on a 4o curve? (RRB JE 29 Aug 2019) 3. Buffer stop and sand hump
(a) 0.05 tonne 4. Creep adjustment
(b) 0.06 tonne Select the correct answer using the codes given below:
(c) 0.08 tonne
(a) 1, 2 and 3
(d) 0.10 tonne
(b) 1, 3 and 4
Ans. (C) Curve Resistance on BG track is given by (c) 1, 2 and 4
following formula: (d) 2, 3 and 4
R = 0.0004 × W × D Ans. (C)
Where, The following are related to maintenance of railway
W = Weight of train in tonnes and D = Degree of curve track:
In this case: D = 4 and W = 50 tonnes 1. Jim crow and gauge bar – Used for adjusting and
We get, maintaining the track gauge and rail curvature.
Curve resistance, R = 0.0004 × 50 × 4 = 0.08 tonne 2. Through packing and boxing – Methods used in
Q.94. Which one of the following types of transition track maintenance to ensure proper ballast
curves is mostly used in Indian Railways? (UKPSC AE 2013) support and stability.
(a) Euler's spiral 3. Buffer stop and sand hump – These are safety
(b) Cubic Spiral measures at railway terminals to stop trains and
(c) Lemniscate prevent accidents, not directly related to track
(d) Cubic Parabola maintenance.
4. Creep adjustment – Involves adjusting rail creep
Ans. (D) A curve of the varying radius is known as a
(longitudinal movement of rails) to maintain
transition curve. The radius of such a curve varies from proper track alignment.
infinity to a certain fixed value. The transition curve is
Different types of tools for railway track maintenance
also called a spiral or easement curve.
are as follows:
A transition curve is provided when there is a change
in the horizontal alignment from a straight path to Name of the Usage
circular curve gradually. Tool
Jim Crow To bend the rails
Its radius decreases from infinity at the straight end
Crow bars To align the track slightly
(tangent point) to the desired radius of the circular
Rail tongs To lift the rail
curve at the other end (curve point).
Claw Bar To take out Dog-spikes from
Different types of transition curves are spiral or
sleeper
clothoid, cubic parabola, and Lemniscates.
Chisel To cut the rails, bolts, etc.
The Railway Board has decided that on Indian Auger To drill holes for spikes
Railways, transition curves will normally be laid in the Shovel To handle the ballast
shape of a cubic parabola.
Wire Claw To clean the ballast
Spirit Level To check cross level
Sleeper tongs To lift sleepers

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

Q.96. Height of the bridge is kept…… above high flood Abutment: A masonry or reinforced concrete wall
level. (JKPSC Lecturer 2022) that constitutes the end support of bridges or similar
(a) 1.2 to 1.5 m structures by which it joins the bank of the waterway
(b) 1.8 to 2.1m is called an abutment.
(c) 2.2 to 2.5m A wing wall :- It is a structural member located at the
(d) More than 2.5 end of a bridge structure. When it is constructed
Ans. (A) The height of the bridge is generally 1.2 to 1.5 integrally with the abutment, it is termed a cantilever
meters above the high flood level of the given region. wing wall and when separated from the abutment
Different dimensions of the bridge: with expansion or construction joint, it is called an
independent wing wall.
Width of bridges: It is based on the traffic survey. It
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
• Includes abutments, piers, wing walls, and return
(JKSSB JE 2021)
n

walls.
tio

(a) Abutment
nc

Q.98. As compared to laminar flow, the boundary layer

Ju

(b) Girder/Slab
in a turbulent flow will be
il

(UKPSC JE 8 May 2022)

iv

(c) Piers
C

(a) Same
(d) Wing and Return
(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.

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

• This is because the chaotic movement of fluid in Q.100. By providing a top width for roadway and
turbulent flow creates more friction between the freeboard in the elementary profile of a gravity dam,
the resultant force for full reservoir condition will
fluid and the solid surface, resulting in a thicker
(a) shift towards the heel
boundary layer.
(b) shift towards the toe
• On the other hand, in laminar flow, the smooth (c) not shift at all
movement of fluid creates less friction, resulting in a (d) None of the above
thinner boundary layer.
Ans. (A) Gravity Dam:- Structure which is designed in
Q.99. Groynes are adopted for river bank protection such a way that its own weigh resist all the external
works. When it is placed inclined in the downstream in force is termed as gravity dam. It may be constructed
the direction of flow in the river, it is designated as of masonry or concrete.
which one of the following? • Elementary profile is a theoretical profile.
(a) Repelling groyne • Only external forces acting on elementary profile is
(b) Attracting groyne
water pressure.
(c) Neither repelling nor attracting groyne
(d) Fixed groyne • Internal force is self weight.
• The shape of elementary profile of a gravity dam is
Ans. (B) Groynes: These are the embankment type
right angled triangle.
structures constructed transverse to the river flow.
They extend into the river from the bank and may also • For full reservoir condition resultant passes through
be called as transverse dykes. outer one third point and for empty (only self weight)
• They are constructed to protect the bank by reservoir condition resultant passes through inner one
deflecting the current away from the bank. third point.
• As the water is unable to take a sharp embayment, • By providing a top width for roadway and free board
the bank gets protected for certain distance upstream in elementary profile of a gravity dam, for full reservoir
and downstream of the groyne. condition the resultant force will shift towards heel of
• However, the nose of the groyne is subjected to a the dam.
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.
• It causes formation of scour holes closer to the banks Civil Ki Goli 9255624029
than the repelling groynes.
• Therefore, they tend to maintain deep current close
to the bank.
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• Deflecting Groynes – perpendicular to bank

• Repelling Groynes – pointing u/s
• Sedimenting Groynes – for deposition of sediment
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OSSC JE MAINS CIVIL 16/03/2025 22 Civil Ki Goli