BPSC AE 2025: Assistant Engineer (Civil/Mechanical/Electrical) (Advt. 29/2025), Paper VI Civil Engineering
Text of page 1
Bihar Public Service Commission
Advt. No.29/2025, Civil Engineering, Paper-VI
Exam Date:19-07-2025, 2nd Shift
Final answer Key (Set-E)
Question No. Ans. Key Question No. Ans. Key
1 D 26 D
2 A 27 D
3 B 28 C
4 A 29 A
5 Deleted 30 B
6 Deleted 31 D
7 B 32 C
8 Deleted 33 C
9 A 34 A
10 Deleted 35 D
11 D 36 B
12 A 37 A
13 A 38 A
14 B 39 C
15 B 40 C
16 A 41 D
17 D 42 A
18 A 43 B
19 Deleted 44 B
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Question No. Ans. Key Question No. Ans. Key
20 D 45 A
21 C 46 A
22 Deleted 47 A
23 D 48 A
24 D 49 C
25 D 50 Deleted
(Set-F)
Question No. Ans. Key Question No. Ans. Key
1 Deleted 26 Deleted
2 B 27 A
3 B 28 Deleted
4 C 29 Deleted
5 B 30 C
6 B 31 D
7 Deleted 32 D
8 B 33 B
9 C 34 A
10 A 35 C
11 D 36 A
12 C 37 B
13 B 38 Deleted
14 C 39 B
15 B 40 D
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Question No. Ans. Key Question No. Ans. Key
16 Deleted 41 A
17 A 42 B
18 A 43 C
19 D 44 C
20 B 45 A
21 C 46 B
22 D 47 C
23 C 48 B
24 C 49 D
25 D 50 B
(Set-G)
Question No. Ans. Key Question No. Ans. Key
1 D 26 A
2 D 27 B
3 Deleted 28 A
4 C 29 C
5 D 30 A
6 Deleted 31 A
7 D 32 A
8 D 33 A
9 C 34 A
10 A 35 Deleted
11 D 36 Deleted
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Question No. Ans. Key Question No. Ans. Key
12 C 37 Deleted
13 C 38 C
14 D 39 A
15 B 40 B
16 B 41 D
17 D 42 B
18 A 43 B
19 D 44 D
20 D 45 C
21 C 46 D
22 A 47 A
23 C 48 D
24 B 49 D
25 Deleted 50 Deleted
(Set-H)
Question No. Ans. Key Question No. Ans. Key
1 D 26 C
2 B 27 D
3 D 28 D
4 C 29 D
5 C 30 C
6 B 31 B
7 A 32 A
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Question No. Ans. Key Question No. Ans. Key
8 Deleted 33 D
9 D 34 D
10 A 35 B
11 Deleted 36 Deleted
12 C 37 Deleted
13 C 38 C
14 Deleted 39 B
15 D 40 D
16 C 41 C
17 Deleted 42 C
18 A 43 B
19 B 44 B
20 Deleted 45 A
21 A 46 C
22 B 47 D
23 D 48 C
24 C 49 A
25 C 50 C
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Reason as per Set-E
Q.No.- 03 - No Change.
Reason :- Concrete pavements need joints. But continuous reinforced pavements
do not require joints as steel is continuously reinforced to control
cracking.
Q.No.- 04 - No Change.
Reason :- The statement for option 'A' is incorrect.
Q.No.- 05 - Deleted.
Reason :- All the four options are correct. Hence, the question is deleted
Q.No.- 06 - Deleted.
Reason :- A transition curve is introduced between a straight alignment and a
circular curve. Hence, the question is deleted
Q.No.- 08 - Deleted.
Reason :- With Usual notations
Q= CAh
For fine sand C = 0.5 m3/hr/m2 per m
A = C/S area of the well
h= depression head
=4m
Q= discharge of the well
= 0.004 m3/s
0.004× 60 ×60= 0.5 × ×4
D=
= 3.028 = say, 3 m
option A is the most appropriate answer.
However in wells, it is assumed that cavity is formed at the
bottom of the well i.e in this case,
0.004× 60×60 = 0.5 × ×4
D=
D= 2.62 m
As the criteria regarding cavity formation is not mentioned.
Therefore, the question is deleted.
Q.No.- 10 - Deleted.
Reason :- Normally Rational method is used for urban areas (≤50 km2) to
determine the peak discharge. However, SCS curve number method is
used for agricultural watersheds (≤10,000 ha) to determine the average
discharge. Therefore, the question is deleted.
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Q.No.- 12 - No Change.
Reason :- Due to blockage of the flow in a river due to contraction of a weir, the
silt charge in the river water reduces. Therefore, the discharge of the
river d/s of the weir has a scouring tendency to compensate the loss of
silt. This process is called retrogression. To control retrogression,
provision is made in the design of d/s protection works.
Q.No.- 15 - No Change.
Reason :- Only the first statement is correct.
Q.No.- 17 - No Change.
Reason :- The main difference between a weir and a barrage is the way the silt is
controlled i.e in a barrage, the silt is controlled in a better way due to
provision of gates. In weirs, the silt control is not proper due to solid
obstruction from the weir wall.
Q.No.- 19 - Deleted.
Reason :- The question is vague i.e There is no fixed number of aspects in the
Hydraulic design of a weir. Hence, the question is deleted
Q.No.- 21 - No Change.
Reason :- According to Newton's second law of motion, the net force F x acting on
a fluid element in the direction of x is equal to mass m of the fluid
element multiplied by the acceleration ax in the x-direction. Thus
mathematically,
Fx = m.ax
In the fluid flow, the following forces are present:
(i) Fg , gravity force.
(ii) Fp , the pressure force.
(iii) F v , force due to viscosity.
(iv) Ft , force due to turbulence.
(v) Fc , force due to compressibility.
Thus in above equation the net force
Fx = (Fg) x +( Fp) x +( F v) x +( Ft) x +( Fc) x.
(i) If the force due to compressibility, Fc is negligible, the resulting
net force
Fx = (Fg) x +( Fp) x +( F v) x +( Ft) x
and equation of motions are called Reynold's equations of
motion.
(ii) For flow, where (Ft) is negligible, the resulting equations of motion
are known as Navier-Stokes Equation.
(iii) If the flow is assumed to be ideal, viscous force (Fv) is zero and
equation of motions are known as Euler's equation of motion.
Q.No.- 22 - Deleted.
Reason :- The amount of electric energy that can be generated by a hydraulic
power plant depends upon more than one factor as given below.
P = η.γ.Qo.H
Where η is the overall efficiency of the power generation, Qo = design
discharge, H= head (m) and γ is the unit weight of water (kN/m3).
Therefore, it depends on all the above four factors. Hence, the question
is deleted.
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Q.No.- 24 - No Change.
Reason :- Coupled joints do not require sockets or spigots.
Q.No.- 25 - No Change.
Reason :- The pathline of a particle in an one dimensional flow is a straight line
along the direction it moves. If the two particles move along the same
direction, their pathlines will be parallel to each other and thus, will
never intersect.
Q.No.- 26 - No Change.
Reason :- The highest CBR number (value) is required for the sub-grade layer of a
pavement. The sub-grade is the foundation, upon which the entire
pavement structure is built, and a higher CBR indicates a stronger, more
load bearing material.
Q.No.- 29 - No Change.
Reason :- Frontal rain occurs when colder air meets warmer air in a weather front .
The warm air is then forced to rise over the cooler air leading to clouds
and rain.
Q.No.- 30 - No Change.
Reason :- Canal capacity must be designed to handle the highest water demand to
prevent the shortages during the critical demand period. Designing a
canal for average demand would lead to insufficient water during peak
usage.
Q.No.- 31 - No Change.
Reason :- For a particle to be transported, the drag force and the lift force
generated by the flowing water must overcome the particle's resistance
forces, mainly gravity and friction.
Q.No.- 32 - No Change.
Reason :- Rigid pavements are stiffer than flexible pavements because cement
concrete has a high modulus of elasticity (E). Therefore, represents high
resistance to elastic deformation.
Q.No.- 37 - The Provisional answer is option 'B' but the correct answer is option 'A'.
Reason :- Hydrological routing methods focus on the continuity equation (mass
balance) and are simpler and faster to compute than hydraulic methods.
The most common method for reservoir routing is modified Puls
method, though it can also be adopted for channel routing. However, the
term modified Puls is commonly used in text books and software to
clearly refer to this specific version of the Puls method.
Q.No.- 38 - No Change.
Reason :- The inverted filter, used in protection works, protects the weir from
piping.
Q.No.- 39 - No Change.
Reason :- Sewer appurtenances includes manholes, inlets, catch basins, pump
houses, etc. Therefore, all options are correct but sewer appurtenances is
the most suitable option.
Text of page 9
Q.No.- 42 - No Change.
Reason :- C/s Area of the channel = 50 m2
Top width of channel = 5 m
mean value of flow =0.1 m/s
absolute viscosity of water = 0.625 N-S/m2
L= = = 10 m
Re =
= = 1600
Since Re >500 to 600 and < 2000
The flow will be transition
Q.No.- 43 - No Change.
Reason :- The region A will receive most precipitation due to the obstruction to the
flow of wind as shown in the figure of the question.
Q.No.- 44 - No Change.
Reason :- Rigid dams transfer high loads. Sand and gravel may not withstand and
show excessive settlement. Also, differential settlement can cause
cracks.
Q.No.- 46 - No Change.
Reason :- Consumptive irrigation requirement (CIR) is the specific amount of
irrigation water needed for a crop's evapotranspiration need during its
growth period, accounting for the water transpired by plants and
evaporated from the soil.
Q.No.- 48 - No Change.
Reason :- Piers and planks are used in the main head regulator or distributary head
regulator to control flow of water entering into the off take channel.
A sluice Gate is a movable barrier placed in the path of flowing
water to control its flow rate and direction. These gates are typically
constructed using sturdy materials, e.g. steel, wood or concrete and
strategically positioned within canals, rivers, dams and other waterways.
Sluice gates can be operated manually, electrically or through hydraulic
systems, depending on their size and intended purpose.
Q.No.- 49 - No Change.
Reason :- It is also called vadose zone and is the subsurface region between the
land surface and the water table where the pores and fractures contain a
mixture of both water and air. Therefore, it is an unsaturated zone.
Text of page 10
Q.No.- 50 - Deleted.
Reason :-
Given:
When Manning's eq.n is used
When Chazy eq.n is used:
Ref. :-HYDRAULICS AND FLUID MECHANICS INCLUDING
HYDRAULIC MACHINES (in Si Units) By Dr. P.N. Modi and Dr.
S.M. Seth. Page No.784 to 787
None of the options are available in choices, therefore the question is
deleted.Page 1 of 11
Source: Bihar Public Service Commission. Question paper published by Bihar Public Service Commission.