SSC JE Civil Previous Year Papers 2010–2015 (Model Practice Paper with Detailed Solutions)

Updated for SSC JE Aspirants

Preparing for SSC Junior Engineer (Civil) requires more than reading theory. The best way to improve accuracy is to solve concept-based questions that reflect the SSC JE exam level.

This model practice paper is designed using the important concepts repeatedly asked in SSC JE examinations between 2010 and 2015. Every question is original and includes a detailed solution, formula, and exam tip to help you understand the concept instead of memorizing answers.

Whether you are preparing for SSC JE 2026 or any upcoming Civil Engineering exam, this practice set will help you identify important topics and strengthen your technical fundamentals.

SSC JE Civil Model Paper Highlights

Particular	Details
Exam Level	SSC JE Civil
Practice Set	Model Paper 1
Questions	15
Difficulty	SSC JE Standard
Solution	Detailed
Suitable For	SSC JE, State JE, RRB JE

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Question 1

Subject: Strength of Materials

A steel bar having a cross-sectional area of 600 mm² is subjected to a tensile load of 150 kN. Determine the tensile stress developed.

Options

A. 200 MPa	B. 225 MPa
C. 250 MPa	D. 275 MPa

Correct Answer = ✅ C. 250 MPa

Solution

Stress = Load / Area

Given Load = 150 kN = 150000 N

Area = 600 mm²

Stress =150000/600

=250 N/mm² =250 MPa

Hence, Correct Answer = 250 MPa

Formula σ=P/A

Concept Revision: Stress is the internal resisting force acting per unit area.

Unit = MPa

Exam Tip: Always convert kN into Newton before calculation.

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Question 2

Subject: RCC

Which of the following grades represents the minimum concrete grade recommended for reinforced cement concrete work under normal conditions?

A. M10	B. M15
C. M20	D. M25

Correct Answer = ✅ C. M20

Solution

According to IS 456,

Minimum grade of reinforced concrete = M20

Lower grades are generally used only for plain cement concrete.

Concept Revision

PCC → M10, M15

RCC → Minimum M20

Exam Tip: M20 is among the most frequently asked values in SSC JE.

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Question 3

The primary purpose of a benchmark in surveying is to:

A. Measure horizontal distancen	B. Establish reference elevation
C. Measure bearing	D. Fix property boundary

Correct Answer = ✅ B. Establish reference elevation

Solution

A benchmark is a permanent reference point having a known Reduced Level (RL). All levelling observations are connected to this point.

Concept Revision

Benchmark → Known RL

Levelling → Unknown RL

Exam Tip: Benchmark questions are asked almost every year.

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Question 4

The pressure exerted by water at a depth depends mainly upon:

A. Shape of tank	B. Volume of water
C. Depth below free surface	D. Diameter of tank

Correct Answer = ✅ C. Depth below free surface

Solution

Hydrostatic pressure P=ρgh

Pressure depends upon Density, Gravity, Depth It is independent of container shape.

Formula P=ρgh

Remember: Pressure depends only on depth.

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Question 5

Which test is commonly performed to determine the compressive strength of bricks?

A. Water Absorption Test	B. Hardness Test
C. Crushing Strength Test	D. Soundness Test

Correct Answer = ✅ C. Crushing Strength Test

Solution

The compressive strength of bricks is determined by conducting the crushing strength test in a compression testing machine (CTM).

Concept Revision

Water absorption → Durability

Crushing test → Strength

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Question 6

The water content of soil is expressed as:

A. Weight of water / Weight of dry soil ×100	B. Weight of water / Total weight ×100
C. Volume of water / Volume of soil ×100	D. Weight of solids / Water ×100

Correct Answer= ✅ A

Formula

w=(Weight of Water / Weight of Dry Soil)×100

Exam Tip: Never confuse moisture content with degree of saturation.

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Question 7

The camber provided on roads mainly helps in:

A. Reducing traffic	B. Increasing speed
C. Draining rainwater	D. Increasing pavement thickness

Correct Answer = ✅ C

Concept Revision

Camber

↓

Quick drainage

↓

Longer pavement life

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Question 8

Which process removes suspended impurities from drinking water?

A. Aeration	B. Filtration
C. Sedimentation	D. Chlorination

Correct Answer = ✅ C

Solution

Heavy suspended particles settle under gravity during sedimentation.

Treatment Sequence = Aeration → Coagulation → Sedimentation → Filtration → Disinfection

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Question 9

The modulus of elasticity of structural steel is approximately:

A. 100 GPa	B. 150 GPa
C. 200 GPa	D. 250 GPa

Correct Answer = ✅ C

Formula: E≈200 GPa

Exam Tip

Steel =200 GPa

Concrete ≈25 GPa

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Question 10

Which item is generally not included in the plinth area?

A. Staircase	B. Internal walls
C. Open courtyard	D. Covered balcony

Correct Answer = ✅ C

Concept Revision

Open areas are excluded from plinth area calculations.

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Question 11

The continuity equation is based on the law of:

A. Conservation of Energy	B. Conservation of Mass
C. Newton’s Law	D. Bernoulli’s Principle

Correct Answer = ✅ B

Formula Q=AV

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Question 12

The duty of water is generally expressed in:

A. Hectares per cumec	B. Cubic metre
C. Litres	D. Hectares

Correct Answer = ✅ A

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Question 13

A Gantt Chart is mainly used for:

A. Soil Testing	B. Project Scheduling
C. Concrete Mix Design	D. Levelling

Correct Answer = ✅ B

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Question 14

The resultant of two equal forces acting at 90° is:

A. F	B. √2F
C. 2F	D. Zero

Correct Answer = ✅ B

Formula

R=√(F²+F²)

=√2F

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Question 15

Which reinforcement primarily resists tensile stress in a simply supported RCC beam?

A. Compression steel	B. Stirrups
C. Bottom reinforcement	D. Top reinforcement

Correct Answer = ✅ C

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Question 16

A steel rod of 2 m length elongates by 1 mm under an axial load. Find the strain developed.

Options

A. 0.00025	B. 0.00050
C. 0.00100	D. 0.00200

Correct Answer = ✅ B. 0.00050

Solution

Strain = Change in Length / Original Length

Original Length = 2 m = 2000 mm

Strain = 1 / 2000

= 0.00050

Formula

ε = ΔL / L

Exam Tip: Always convert the length into the same unit before calculating strain.

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Question 17

The main purpose of providing development length in reinforced concrete members is to:

A. Reduce dead load	B. Improve workability
C. Transfer stress safely between steel and concrete	D. Increase cover

Correct Answer = ✅ C

Solution

Development length ensures sufficient bond between reinforcement and concrete so that steel can develop its full tensile strength.

Formula

Ld = (φ × σs) / (4 × τbd)

Concept Revision

Development Length → Bond Strength → Safe Load Transfer

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Question 18

If the Back Sight is 2.150 m and the Fore Sight is 1.420 m, the rise between two stations is:

A. 0.530 m	B. 0.730 m
C. 0.930 m	D. 1.150 m

Correct Answer = ✅ B

Solution

Rise = BS − FS

= 2.150 − 1.420

= 0.730 m

Formula

Rise = BS − FS

Exam Tip

If BS > FS → Rise

If FS > BS → Fall

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Question 19

Which soil property mainly governs the bearing capacity of shallow foundations?

A. Specific Gravity	B. Shear Strength
C. Water Content	D. Porosity

Correct Answer = ✅ B

Solution

The bearing capacity depends primarily on the soil’s shear strength parameters, namely cohesion and angle of internal friction.

Concept Revision

Bearing Capacity = f(Cohesion, Friction Angle)

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Question 20

Water flows through a pipe of diameter 0.25 m at a velocity of 2.5 m/s. The discharge is approximately:

A. 0.061 m³/s	B. 0.092 m³/s
C. 0.123 m³/s	D. 0.150 m³/s

Correct Answer = ✅ C

Solution

Area = πD²/4

= 3.1416 × (0.25)² / 4

= 0.0491 m²

Discharge = Area × Velocity

= 0.0491 × 2.5

≈ 0.123 m³/s

Formula

Q = AV

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Question 21

Which of the following properties is most desirable for good building stone?

A. High Water Absorption	B. High Porosity
C. High Crushing Strength	D. Low Density

Correct Answer = ✅ C

Solution

A good building stone should have high compressive strength, durability, and low water absorption.

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Question 22

The primary objective of road super-elevation is to:

A. Reduce pavement thickness	B. Counteract centrifugal force
C. Increase traffic speed limit	D. Improve visibility

Correct Answer = ✅ B

Solution

Super-elevation balances centrifugal force on vehicles negotiating horizontal curves.

Formula

e = V² / (225R)

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Question 23

Which unit process is mainly responsible for removing dissolved organic matter from water?

A. Screening	B. Filtration
C. Sedimentation	D. Biological Treatment

Correct Answer = ✅ D

Concept Revision

Suspended Matter → Sedimentation

Dissolved Organics → Biological Treatment

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Question 24

A steel member primarily subjected to compression is known as:

A. Tie	B. Strut
C. Girder	D. Beam

Correct Answer = ✅ B

Solution

Compression Member

↓

Strut

Tension Member

↓

Tie

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Question 25

The centre line method is most suitable for:

A. Buildings having symmetrical walls	B. Bridges
C. Dams	D. Highways

Correct Answer = ✅ A

Concept Revision

Symmetrical wall layout

↓

Centre Line Method

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Question 26

The hydraulic radius of a full circular pipe is:

A. D/2	B. D/3
C. D/4	D. D/8

Correct Answer = ✅ C

Solution

Hydraulic Radius

= Area / Wetted Perimeter

For a full circular pipe,

R = D/4

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Question 27

The canal outlet supplying water to agricultural fields is generally called:

A. Aqueduct	B. Sluice
C. Watercourse Outlet	D. Syphon

Correct Answer = ✅ C

Concept Revision

Canal → Outlet → Watercourse → Field

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Question 28

Which network technique mainly identifies the critical path in project planning?

A. Bar Chart	B. CPM
C. Histogram	D. Flow Diagram

Correct Answer = ✅ B

Solution

Critical Path Method (CPM) identifies the sequence of activities that determines the minimum project completion time.

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Question 29

Two equal forces of 15 kN act in opposite directions along the same line. The resultant force is:

A. 15 kN	B. 30 kN
C. Zero	D. 7.5 kN

Correct Answer = ✅ C

Solution

Equal and opposite collinear forces cancel each other completely.

Resultant = 0

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Question 30

Which reinforcement is primarily provided to resist diagonal tension in RCC beams?

A. Main Bars	B. Distribution Bars
C. Stirrups	D. Compression Bars

Correct Answer = ✅ C

Solution

Diagonal tension caused by shear forces is resisted by stirrups (shear reinforcement).

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Question 31

Topic: Bending Stress

A simply supported beam is subjected to a bending moment of 24 kN-m. If the section modulus of the beam is 600 × 10³ mm³, determine the maximum bending stress.

Options

A. 30 N/mm²	B. 35 N/mm²
C. 40 N/mm²	D. 45 N/mm²

Correct Answer = ✅ C. 40 N/mm²

Detailed Solution

Bending stress is calculated by dividing the bending moment by the section modulus of the beam.

Formula: Bending Stress (σ) = Bending Moment (M) / Section Modulus (Z)

Given
Bending Moment (M) = 24 kN-m
Section Modulus (Z) = 600 × 10³ mm³

Step 1: Convert Bending Moment into N-mm
1 kN-m = 10⁶ N-mm
Therefore,
M = 24 × 10⁶ N-mm

Step 2: Calculate the Bending Stress

Bending Stress (σ) = (24 × 10⁶) / (600 × 10³)
= 40 N/mm²

Therefore, the maximum bending stress in the beam is 40 N/mm².

Concept Revision

Bending stress is directly proportional to the bending moment and inversely proportional to the section modulus.

Exam Tip: Always convert bending moment from kN-m to N-mm before using the bending equation.

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Question 32

Topic: Effective Depth of Beam

An RCC beam has an overall depth of 500 mm, clear cover of 25 mm, stirrup diameter of 8 mm, and main reinforcement diameter of 20 mm. Determine the approximate effective depth.

Options

A. 447 mm	B. 457 mm
C. 467 mm	D. 477 mm

Correct Answer = ✅ B. 457 mm

Detailed Solution

The effective depth of an RCC beam is the distance measured from the compression face to the centre of the tensile reinforcement.

Formula: Effective Depth (d) = Overall Depth (D) − Effective Cover

Where
Effective Cover = Clear Cover + Stirrup Diameter + (Main Bar Diameter / 2)

Given
Overall Depth (D) = 500 mm
Clear Cover = 25 mm
Stirrup Diameter = 8 mm
Main Bar Diameter = 20 mm

Step 1: Calculate the Effective Cover

Effective Cover
= 25 + 8 + (20 / 2)
= 25 + 8 + 10
= 43 mm

Step 2: Calculate the Effective Depth

Effective Depth (d)
= 500 − 43
= 457 mm

Therefore, the effective depth of the RCC beam is 457 mm.

Concept Revision

Overall depth is the total depth of the beam, while effective depth is measured up to the centre of tensile reinforcement.

Exam Tip

Do not subtract only the clear cover. Stirrup diameter and half the main bar diameter must also be considered.

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Question 33

Topic: Dry Density

A soil sample has a bulk density of 1.92 g/cm³ and water content of 20%. Calculate its dry density.

Options

A. 1.40 g/cm³	B. 1.50 g/cm³
C. 1.60 g/cm³	D. 1.70 g/cm³

Correct Answer = ✅ C. 1.60 g/cm³

Detailed Solution

Dry density is calculated from the bulk density and the water content of the soil.

Formula
Dry Density (ρd) = Bulk Density (ρ) / (1 + Water Content)

Where
ρd = Dry Density (g/cm³)
ρ = Bulk Density (g/cm³)
w = Water Content (in decimal)

Given
Bulk Density (ρ) = 1.92 g/cm³
Water Content (w) = 20% = 0.20

Calculation
Dry Density (ρd)
= 1.92 / (1 + 0.20)
= 1.92 / 1.20
= 1.60 g/cm³

Therefore, the dry density of the soil is 1.60 g/cm³.

Concept Revision

Bulk density includes both soil solids and water, whereas dry density considers only the dry soil mass.

Exam Tip: Convert percentage water content into decimal before using the formula.

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Question 34

Topic: Height of Instrument Method

The reduced level of a benchmark is 102.500 m. A backsight reading of 1.365 m is taken on the benchmark. If the foresight reading on the next point is 2.140 m, determine the reduced level of that point.

Options

A. 101.725 m	B. 102.275 m
C. 103.275 m	D. 104.005 m

Correct Answer = ✅ A. 101.725 m

Detailed Solution

The Height of Instrument (HI) method is used in levelling to determine the Reduced Level (RL) of unknown points.

Formula

Height of Instrument (HI) = Reduced Level (RL) + Backsight (BS)
Reduced Level (RL) = Height of Instrument (HI) − Foresight (FS)

Given

Reduced Level of Benchmark (RL) = 102.500 m
Backsight (BS) = 1.365 m
Foresight (FS) = 2.140 m

Step 1: Calculate the Height of Instrument

HI

= 102.500 + 1.365
= 103.865 m

Step 2: Calculate the Reduced Level of the Next Point

RL = 103.865 − 2.140
= 101.725 m

Therefore, the Reduced Level (RL) of the next point is 101.725 m.

Concept Revision

Backsight is added to a known reduced level to calculate the height of instrument. Foresight is subtracted from the height of instrument.

Exam Tip

Remember:

• BS is added
• FS is subtracted

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Question 35

Topic: Velocity Head

Water flows through a pipe with a velocity of 6 m/s. Calculate the velocity head. Take (g = 9.81 m/s²).

Options

A. 1.24 m	B. 1.47 m
C. 1.83 m	D. 2.20 m

Correct Answer = ✅ C. 1.83 m

Detailed Solution

Velocity head represents the kinetic energy of flowing water expressed as an equivalent height of water.

Formula

Velocity Head (hv) = V² / (2 × g)

Where

V = Velocity of Flow (m/s)
g = Acceleration Due to Gravity (9.81 m/s²)

Given

Velocity (V) = 6 m/s
Acceleration Due to Gravity (g) = 9.81 m/s²

Calculation

Velocity Head (hv)

= 6² / (2 × 9.81)
= 36 / 19.62
= 1.83 m (Approx.)

Therefore, the velocity head of the flowing water is approximately 1.83 m.

Concept Revision

Velocity head represents the kinetic energy of flowing water per unit weight.

Exam Tip: Velocity head increases with the square of velocity.

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Question 36

Topic: Super-Elevation

A vehicle travels at a speed of 54 km/h on a horizontal curve of radius 180 m. Determine the required super-elevation, neglecting lateral friction.

Options

A. 0.050	B. 0.060
C. 0.072	D. 0.090

Correct Answer = ✅ C. 0.072

Detailed Solution

Super-elevation is the transverse slope provided on a horizontal curve to counteract the effect of centrifugal force and improve vehicle stability.

Formula

Super-elevation (e) = V² / (225 × R)

Where

V = Speed of the Vehicle (km/h)
R = Radius of the Curve (m)

Given

Speed (V) = 54 km/h
Radius (R) = 180 m

Calculation

Super-elevation (e)

= 54² / (225 × 180)
= 2916 / 40500
= 0.072

Percentage of Super-elevation

= 0.072 × 100
= 7.2%

Therefore, the required super-elevation is 0.072 or 7.2%.

Concept Revision

Super-elevation is provided on horizontal curves to counteract the effect of centrifugal force.

Exam Tip: When speed is given in km/h, use the denominator (225R).

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Question 37

Topic: Water Demand

A town has a population of 40,000 and the average water demand is 135 litres per capita per day. Calculate the total daily water demand.

Options

A. 4.40 MLD	B. 5.40 MLD
C. 6.40 MLD	D. 7.40 MLD

Correct Answer = ✅ B. 5.40 MLD

Detailed SolutionThe total daily water demand is calculated by multiplying the population by the per capita daily water demand.

Formula

Total Water Demand = Population × Per Capita Water Demand

Given

Population = 40,000
Per Capita Water Demand = 135 litres/person/day

Step 1: Calculate the Total Water Demand

Total Water Demand
= 40,000 × 135
= 5,400,000 litres/day

Step 2: Convert Litres per Day into MLD

1 MLD = 1,000,000 litres/day

Therefore,

Total Water Demand

= 5,400,000 ÷ 1,000,000
= 5.40 MLD

Therefore, the total daily water demand of the town is 5.40 MLD.

Concept Revision

MLD means million litres per day and is commonly used for municipal water supply calculations.

Exam Tip: Divide litres per day by (10^6) to convert it into MLD.

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Question 38

Topic: Slenderness Ratio

A steel column has an effective length of 3.2 m and a minimum radius of gyration of 40 mm. Calculate its slenderness ratio.

Options

A. 60	B. 70
C. 80	D. 90

Correct Answer = ✅ C. 80Detailed Solution

The slenderness ratio is the ratio of the effective length of a column to its least radius of gyration. It indicates the tendency of a column to buckle under compressive load.

Formula

Slenderness Ratio (λ) = Effective Length (Le) / Least Radius of Gyration (r)

Given

Effective Length (Le) = 3.2 m
Least Radius of Gyration (r) = 40 mm

Step 1: Convert Effective Length into Millimetres

Le = 3.2 × 1000
= 3200 mm

Step 2: Calculate the Slenderness Ratio

Slenderness Ratio (λ)
= 3200 / 40
= 80

Therefore, the slenderness ratio of the column is 80.

Concept Revision

A higher slenderness ratio indicates a greater tendency of the compression member to buckle.

Exam Tip: Always use effective length and minimum radius of gyration.

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Question 39

Topic: Brickwork Quantity

A wall is 5 m long, 3 m high, and 230 mm thick. Calculate the volume of brickwork.

Options

A. 2.45 m³	B. 3.15 m³
C. 3.45 m³	D. 4.15 m³

Correct Answer = ✅ C. 3.45 m³

Detailed Solution

Delta is the total depth of water required by a crop during its base period. It is calculated using the relationship between duty and base period.

Formula: Delta (Δ) = (8.64 × Base Period) / Duty

Where

Δ = Delta (m)
B = Base Period (days)
D = Duty (hectares per cumec)

Given

Base Period (B) = 120 days
Duty (D) = 1800 hectares per cumec

Calculation

Delta (Δ)
= (8.64 × 120) / 1800
= 1036.8 / 1800
= 0.576 m

Therefore, the delta (depth of irrigation water) required for the crop is 0.576 m.

Concept Revision

Brickwork is generally measured in cubic metres when wall thickness is more than one brick.

Exam Tip: Convert all dimensions into metres before calculating volume.

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Question 40

Topic: Duty and Delta

The duty of water is 1800 hectares per cumec and the base period is 120 days. Calculate the delta of the crop.

Options

A. 0.432 m	B. 0.576 m
C. 0.648 m	D. 0.720 m

Correct Answer = ✅ B. 0.576 m

Detailed Solution

Delta is the total depth of water required by a crop during its base period. It is calculated using the relationship between duty, base period and delta.

Formula: Delta (Δ) = (8.64 × Base Period) / Duty

Where

Δ = Delta (m)
B = Base Period (days)
D = Duty (hectares per cumec)

Given

Base Period (B) = 120 days
Duty (D) = 1800 hectares/cumec

Calculation

Delta (Δ) = (8.64 × 120) / 1800
= 1036.8 / 1800
= 0.576 m

Therefore, the delta of the crop is 0.576 m.

Concept Revision

Duty and delta are inversely proportional. A higher duty means a lower depth of irrigation water.

Exam Tip: Use base period in days and duty in hectares per cumec.

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Question 41

Topic: Critical Path Method

A project has three activities on its critical path with durations of 4 days, 6 days, and 5 days. What is the minimum project completion time?

Options

A. 10 days	B. 11 days
C. 15 days	D. 20 days

Correct Answer = ✅ C. 15 days

Detailed Solution

The critical path is the longest sequence of activities in a project network. The total duration of all activities on the critical path gives the minimum time required to complete the project.

Calculation

Project Completion Time

= 4 + 6 + 5

= 15 days

Therefore, the minimum project completion time for the project is 15 days.

Concept Revision

Activities on the critical path normally have zero total float.

Exam Tip: Any delay in a critical activity may delay the entire project.

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Question 42

Topic: Water Absorption of Brick

A dry brick weighs 3.20 kg. After immersion in water, its weight becomes 3.68 kg. Calculate the percentage water absorption.

Options

A. 10%	B. 12%
C. 15%	D. 18%

Correct Answer = ✅ C. 15%

Detailed Solution

Water absorption is the percentage increase in the weight of a brick after it is immersed in water.

Formula: Water Absorption (%) = [(Wet Weight − Dry Weight) / Dry Weight] × 100

Given

Wet Weight (Ww) = 3.68 kg
Dry Weight (Wd) = 3.20 kg

Step 1: Calculate the Increase in Weight

Increase in Weight
= 3.68 − 3.20
= 0.48 kg

Step 2: Calculate Water Absorption

Water Absorption
= (0.48 / 3.20) × 100
= 0.15 × 100
= 15%

Therefore, the water absorption of the brick is 15%.

Concept Revision

Water absorption indicates the porosity and durability of bricks.

Exam Tip: Always divide the increase in weight by the original dry weight.

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Question 43

Topic: Moment of Force

A force of 25 kN acts perpendicular to a lever arm at a distance of 1.2 m from the point of rotation. Calculate the moment of the force.

Options

A. 20 kN-m	B. 25 kN-m
C. 30 kN-m	D. 35 kN-m

Correct Answer = ✅ C. 30 kN-m

Detailed Solution

The moment of a force is the turning effect produced when a force acts at a perpendicular distance from a fixed point or axis.

Formula: Moment of Force (M) = Force (F) × Perpendicular Distance (d)

Given

Force (F) = 25 kN
Perpendicular Distance (d) = 1.2 m

Calculation

Moment of Force
= 25 × 1.2
= 30 kN-m

Therefore, the moment of the force is 30 kN-m.

Concept Revision

The perpendicular distance between the line of action of force and the point of rotation is called the moment arm.

Exam Tip: Use perpendicular distance, not the inclined distance.

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Question 44

Topic: Hydraulic Radius

A rectangular channel is 3 m wide and carries water to a depth of 1.5 m. Calculate the hydraulic radius.

Options

A. 0.50 m	B. 0.60 m
C. 0.75 m	D. 1.00 m

Correct Answer = ✅ C. 0.75 m

Detailed Solution

The hydraulic radius is the ratio of the cross-sectional flow area to the wetted perimeter of the channel.

Formula

Hydraulic Radius (R) = Flow Area (A) / Wetted Perimeter (P)

Given

Width of Channel (b) = 3 m
Depth of Water (y) = 1.5 m

Step 1: Calculate the Flow Area

A = b × y
= 3 × 1.5
= 4.5 m²

Step 2: Calculate the Wetted Perimeter

P = b + 2y
= 3 + (2 × 1.5)
= 3 + 3
= 6 m

Step 3: Calculate the Hydraulic Radius

R = 4.5 / 6
= 0.75 m

Therefore, the hydraulic radius of the rectangular channel is 0.75 m.

Concept Revision

Hydraulic radius is the ratio of the flow area to the wetted perimeter.

Exam Tip: Do not include the top water surface in the wetted perimeter.

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Question 45

Topic: Percentage of Reinforcement

An RCC beam has a width of 300 mm, effective depth of 450 mm, and tensile steel area of 1350 mm². Calculate the percentage of tensile reinforcement.

Options

A. 0.75%	B. 1.00%
C. 1.25%	D. 1.50%

Correct Answer = ✅ B. 1.00%

Detailed Solution

Formula: Percentage of Tensile Reinforcement = (Ast / (b × d)) × 100

Given
Ast = 1350 mm²
b = 300 mm
d = 450 mm

Calculation

Percentage of Tensile Reinforcement

= (1350 / (300 × 450)) × 100
= (1350 / 135000) × 100
= 0.01 × 100
= 1.00%

Therefore, the percentage of tensile reinforcement is 1.00%.

Concept Revision

The percentage of reinforcement represents the ratio of steel area to the effective concrete cross-sectional area.

Exam Tip: Use effective depth, not overall depth, in the reinforcement percentage formula.

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Question 46

Topic: Compressive Strength of Concrete

A concrete cube of size 150 mm × 150 mm × 150 mm fails under a compressive load of 675 kN. Determine its compressive strength.

Options

A. 20 N/mm²	B. 25 N/mm²
C. 30 N/mm²	D. 35 N/mm²

Correct Answer = ✅ C. 30 N/mm²

Detailed Solution

Compressive strength is calculated by dividing the failure load by the loaded area of the concrete specimen.

Formula: Compressive Strength = Failure Load / Loaded Area

Given

Failure Load (P) = 675 kN
Loaded Area (A) = 150 mm × 150 mm

Step 1: Convert Load into Newton

P = 675 × 1000 = 675000 N

Step 2: Calculate Loaded Area

A = 150 × 150 = 22500 mm²

Step 3: Calculate Compressive Strength

Compressive Strength = 675000 / 22500 = 30 N/mm²
Therefore, the compressive strength of the concrete cube is 30 N/mm².

Concept Revision

Concrete cube strength is generally determined by testing a standard cube under axial compression.

Exam Tip: Use the loaded face area, not the total surface area of the cube.

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Question 47

At a particular section of a beam, the shear force is zero. What generally occurs at that section?

Options

A. Bending moment is always zero	B. Bending moment may be maximum or minimum
C. Loading becomes zero throughout the beam	D. Support reaction becomes zero

Correct Answer = ✅ B. Bending moment may be maximum or minimum

Detailed Solution

The relationship between shear force and bending moment is:
Shear Force = Rate of Change of Bending Moment
Formula: V = dM / dx
Where:
V = Shear Force
M = Bending Moment
x = Distance along the beam

Condition Given

Shear Force (V) = 0
Therefore,
dM / dx = 0

Conclusion

When the shear force becomes zero at a section of a beam, the bending moment reaches a stationary value. At this point, the bending moment may be maximum or minimum, depending on the loading and support conditions. Therefore, the correct answer is: Bending moment may be maximum or minimum.

Concept Revision

• Rate of change of bending moment = Shear force
• Rate of change of shear force = Load intensity

Exam Tip: A change in the sign of shear force generally indicates a maximum or minimum bending moment.

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Question 48

Topic: Consolidation Settlement

A saturated clay layer undergoes a primary consolidation settlement of 80 mm. If 75% consolidation has taken place, determine the settlement that has already occurred.

Options

A. 50 mm	B. 60 mm
C. 70 mm	D. 75 mm

Correct Answer = ✅ B. 60 mm

Detailed Solution

The degree of consolidation is the ratio of settlement that has occurred at a particular time to the final consolidation settlement.

Formula: Degree of Consolidation (U) = Settlement at a Particular Time / Final Settlement or Settlement at a Particular Time (St) = U × Sf

Given

Degree of Consolidation (U) = 75% = 0.75
Final Consolidation Settlement (Sf) = 80 mm

Calculation

Settlement at a Particular Time

= 0.75 × 80 = 60 mm

Therefore, the settlement already completed is 60 mm.

Concept Revision

Consolidation is the gradual reduction in soil volume caused by the expulsion of pore water under sustained loading.

Exam Tip: Do not confuse degree of consolidation with percentage reduction in void ratio.

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Question 49

Topic: Continuity Equation

Water flows through a pipe whose diameter reduces from 300 mm to 150 mm. If the velocity in the larger section is 2 m/s, determine the velocity in the smaller section.

Options

A. 4 m/s	B. 6 m/s
C. 8 m/s	D. 10 m/s

Correct Answer = ✅ C. 8 m/s

Detailed Solution

For steady incompressible flow: According to the Continuity Equation, the discharge remains constant throughout the pipe.

Formula: A₁ × V₁ = A₂ × V₂

Since the cross-sectional area of a circular pipe is proportional to the square of its diameter,

A₁ / A₂ = D₁² / D₂²

Given

D₁ = 300 mm
D₂ = 150 mm
V₁ = 2 m/s

Calculation

V₂ = (D₁² / D₂²) × V₁
= (300² / 150²) × 2
= (90000 / 22500) × 2
= 4 × 2
= 8 m/s

Therefore, the velocity of water in the smaller pipe is 8 m/s.

Concept Revision

When pipe area decreases, flow velocity increases for the same discharge.

Exam Tip: If the diameter becomes half, the area becomes one-fourth and the velocity becomes four times.

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Question 50

Topic: Stopping Sight Distance

A vehicle travels at 54 km/h. The driver’s reaction time is 2.5 seconds. Calculate the approximate lag distance.

Options

A. 27.5 m	B. 32.5 m
C. 37.5 m	D. 42.5 m

Correct Answer = ✅ C. 37.5 m

Detailed Solution

Lag distance is the distance travelled by a vehicle during the driver’s perception and reaction time before the brakes are applied.

Formula: Lag Distance = 0.278 × V × t

Where

V = Speed of the vehicle (km/h)
t = Driver’s reaction time (seconds)

Given

Speed (V) = 54 km/h
Reaction Time (t) = 2.5 seconds

Calculation

Lag Distance
= 0.278 × 54 × 2.5
= 15.012 × 2.5
= 37.53 m
≈ 37.5 m

Therefore, the lag distance travelled by the vehicle is approximately 37.5 m.

Concept Revision

Stopping sight distance consists of: Formula
SSD = Lag Distance + Braking Distance

Exam Tip: Use the factor 0.278 when speed is given in km/h.

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Question 51

Topic: Biochemical Oxygen Demand

The dissolved oxygen of a wastewater sample before incubation is 8.5 mg/L, and after five days it is 3.0 mg/L. Assuming no dilution, calculate the five-day BOD.

Options

A. 3.5 mg/L	B. 4.5 mg/L
C. 5.5 mg/L	D. 6.5 mg/L

Correct Answer = ✅ C. 5.5 mg/L

Detailed Solution

For an undiluted water sample, the Biochemical Oxygen Demand (BOD₅) is calculated by subtracting the dissolved oxygen after five days from the initial dissolved oxygen.

Formula

BOD₅ = Initial Dissolved Oxygen (DO₁) − Final Dissolved Oxygen (DO₂)

Given

Initial Dissolved Oxygen (DO₁) = 8.5 mg/L
Final Dissolved Oxygen (DO₂) = 3.0 mg/L

Calculation

BOD₅ = 8.5 − 3.0 = 5.5 mg/L

Therefore, the five-day Biochemical Oxygen Demand (BOD₅) of the sample is 5.5 mg/L.

Concept Revision

BOD represents the oxygen required by microorganisms to biologically decompose biodegradable organic matter.

Exam Tip: Higher BOD generally indicates greater organic pollution.

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Question 52

Topic: Gradient

A road rises by 3 m over a horizontal distance of 150 m. Determine the gradient.

Options

A. 1 in 30	B. 1 in 40
C. 1 in 50	D. 1 in 60

Correct Answer = ✅ C. 1 in 50

Detailed Solution

Gradient is the ratio of the vertical rise (or fall) to the horizontal distance.

Formula: Gradient = Vertical Rise / Horizontal Distance

Given

Vertical Rise = 3 m
Horizontal Distance = 150 m

Calculation

Gradient = 3 / 150 = 1 / 50
Therefore, the gradient of the road is 1 in 50.

Concept Revision

A gradient of 1 in 50 means a vertical rise or fall of 1 unit for every 50 horizontal units.

Exam Tip: First simplify the ratio before selecting the answer.

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Question 53

Topic: Cost of Cement Concrete

The quantity of cement concrete in a foundation is 12 m³. If the rate is ₹6,500 per m³, calculate the total cost.

Options

A. ₹68,000	B. ₹72,000
C. ₹78,000	D. ₹84,000

Correct Answer = ✅ C. ₹78,000

Detailed Solution

The total cost of an item is calculated by multiplying its quantity by the unit rate.

Formula: Total Cost = Quantity × Rate

Given

Quantity = 12 m³
Rate = ₹6,500 per m³

Calculation

Total Cost
= 12 × 6,500
= ₹78,000

Therefore, the total cost of the cement concrete work is ₹78,000.

Concept Revision

An abstract of cost is prepared by multiplying the measured quantity of each item by its approved unit rate.

Exam Tip: Check whether the rate includes labour, materials and carriage before using it in practical estimates.

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Question 54

Topic: Effective Length of Column

A steel column is hinged at both ends and has an actual length of 4 m. Determine its effective length.

Options

A. 2 m	B. 2.8 m
C. 4 m	D. 8 m

Correct Answer = ✅ C. 4 m

Detailed Solution

The effective length of a column depends on its end conditions. For a column hinged (pinned) at both ends, the effective length is equal to its actual length.

Formula: Effective Length (Le) = K × L

Where

K = Effective Length Factor
L = Actual Length of the Column

Given

Actual Length (L) = 4 m
For a column hinged at both ends,
K = 1.0

Calculation

Effective Length (Le) = 1.0 × 4
= 4 m

Therefore, the effective length of the column is 4 m.

Concept Revision

Common ideal effective length factors are:

• Hinged–hinged: (K=1.0)
• Fixed–free: (K=2.0)
• Fixed–fixed: (K=0.5)
• Fixed–hinged: (K\approx0.7)

Exam Tip: SSC JE frequently asks direct questions based on end conditions of columns.

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Question 55

Topic: Intensity of Irrigation

A canal system has a culturable command area of 20,000 hectares. If the area irrigated during a season is 8,000 hectares, calculate the intensity of irrigation.

Options

A. 30%	B. 40%
C. 50%	D. 60%

Correct Answer = ✅ B. 40%

Detailed Solution

The intensity of irrigation is the percentage of the culturable command area that is actually irrigated during a particular crop season.

Formula: Intensity of Irrigation = (Area Irrigated / Culturable Command Area) × 100

Given

Area Irrigated = 8,000 hectares
Culturable Command Area (CCA) = 20,000 hectares

Calculation

Intensity of Irrigation = (8,000 / 20,000) × 100
= 0.40 × 100
= 40%

Therefore, the intensity of irrigation is 40%.

Concept Revision

Intensity of irrigation indicates the percentage of culturable command area proposed or actually irrigated during a crop season.

Exam Tip: Use culturable command area in the denominator, not the gross command area.

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Question 56

Topic: Water-Cement Ratio

A concrete mix contains 180 litres of water and 360 kg of cement. Determine the water-cement ratio.

Options

A. 0.40	B. 0.45
C. 0.50	D. 0.60

Correct Answer = ✅ C. 0.50

Detailed Solution

The water-cement ratio is the ratio of the weight of water to the weight of cement used in the concrete mix.

Formula: Water-Cement Ratio (w/c) = Weight of Water / Weight of Cement

Given

Water = 180 litres
Cement = 360 kg

Since, 1 litre of water ≈ 1 kg
Therefore,
Weight of Water = 180 kg

Calculation

Water-Cement Ratio = 180 / 360
= 0.50

Therefore, the water-cement ratio of the concrete mix is 0.50.

Concept Revision

The water-cement ratio significantly affects concrete strength, permeability, durability and workability.

Exam Tip: The ratio is based on weight, not volume.

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Question 57

Topic: Bearing Pressure

A square footing of size 2 m × 2 m carries a vertical load of 800 kN. Neglecting the footing weight, calculate the uniform bearing pressure on the soil.

Options

A. 100 kN/m²	B. 150 kN/m²
C. 200 kN/m²	D. 250 kN/m²

Correct Answer = ✅ C. 200 kN/m²

Detailed Solution

Bearing pressure is the load transmitted to the soil per unit area of the footing.

Formula

Bearing Pressure (q) = Total Load / Footing Area

Given

Total Load (P) = 800 kN
Footing Size = 2 m × 2 m

Step 1: Calculate the Footing Area

Area = 2 × 2
= 4 m²

Step 2: Calculate the Bearing Pressure

Bearing Pressure

= 800 / 4
= 200 kN/m²

Therefore, the bearing pressure on the soil is 200 kN/m².

Concept Revision

Uniform bearing pressure is assumed when the resultant load acts through the centroid of the footing without eccentricity.

Exam Tip: For eccentric loading, the soil pressure will not remain uniform.

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Question 58

Topic: Manning’s Equation

In Manning’s equation, the velocity of flow is directly proportional to which of the following?

Options

A. Manning’s roughness coefficient	B. Square of wetted perimeter
C. Two-third power of hydraulic radius	D. Inverse of channel slope

Correct Answer = ✅ C. Two-third power of hydraulic radius

Detailed Solution

Manning’s equation is used to calculate the average velocity of flow in an open channel.

Formula

V = (1 / n) × R^(2/3) × S^(1/2)

Where

V = Mean Velocity of Flow
n = Manning’s Roughness Coefficient
R = Hydraulic Radius
S = Bed Slope

Concept

According to Manning’s equation:

• Velocity is directly proportional to the two-third power of the hydraulic radius (R^(2/3)).
• Velocity is directly proportional to the square root of the bed slope (S^(1/2)).
• Velocity is inversely proportional to the Manning’s roughness coefficient (n).

Therefore, the correct answer is: Two-third power of the hydraulic radius.

Concept Revision

A smoother channel, larger hydraulic radius and steeper slope generally produce greater flow velocity.

Exam Tip Remember the powers:

• Hydraulic radius: (2/3)
• Bed slope: (1/2)

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Question 59

Topic: Self-Cleansing Velocity

The main purpose of maintaining self-cleansing velocity in a sewer is to:

Options

A. Increase sewage temperature	B. Prevent the deposition of solids
C. Reduce sewer diameter to zero	D. Stop all turbulence

Correct Answer = ✅ B. Prevent the deposition of solids

Detailed Solution

Sewage contains suspended matter that may settle at the bottom of a sewer when the velocity is too low.

Self-cleansing velocity is the minimum velocity required to prevent solids from settling and accumulating inside the sewer.

Therefore, its main purpose is:

Therefore, the primary purpose of maintaining self-cleansing velocity is to prevent the deposition of solids in the sewer.

Concept Revision

• Velocity below the required level may cause silting.
• Excessively high velocity may damage the sewer surface through scouring.
• Sewer design must therefore maintain a suitable velocity range.

Exam Tip: Self-cleansing velocity prevents silting, while limiting velocity prevents scouring.

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Question 60

Topic: Centroid of a Triangle

Determine the distance of the centroid of a triangular area from its base if the height of the triangle is 900 mm.

Options

A. 200 mm	B. 300 mm
C. 450 mm	D. 600 mm

Correct Answer = ✅ B. 300 mm

Detailed Solution

The centroid of a triangular area is located at one-third of its height measured from the base.

Formula: Distance of Centroid from Base = Height / 3

Given

Height of Triangle (h) = 900 mm

Calculation

Distance of Centroid from Base

= 900 / 3
= 300 mm

Therefore, the centroid of the triangular area is located 300 mm above the base.

Concept Revision

For a triangular area:

• Centroid lies at (h/3) from the base.
• It lies at (2h/3) from the opposite vertex.

Exam Tip: Read carefully whether the distance is being asked from the base or from the vertex.

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SSC JE Civil Model Paper Questions 1–60

The complete model paper covers a broad range of Civil Engineering subjects commonly included in the SSC JE syllabus. The questions progress from basic factual concepts to calculation-based applications.

Subject	Approximate Questions	Level
RCC and Concrete Technology	8	Basic to Moderate
Strength of Materials and Structural Analysis	6	Moderate
Surveying	5	Basic to Moderate
Soil and Foundation Engineering	6	Moderate
Fluid Mechanics and Hydraulics	7	Moderate
Transportation Engineering	5	Basic to Moderate
Environmental and Wastewater Engineering	6	Basic to Moderate
Building Materials	4	Basic
Steel Structures	4	Basic to Moderate
Irrigation Engineering	4	Basic to Moderate
Estimation and Costing	4	Basic
Construction Management	3	Basic to Moderate
Engineering Mechanics	4	Basic to Moderate

Some questions involve concepts that overlap more than one Civil Engineering subject.

Most Important Topics Covered

Candidates should revise the following topics carefully after completing this model paper:

• Stress, strain and bending stress
• Shear force and bending moment relationship
• Effective depth and reinforcement percentage
• Concrete grade and water-cement ratio
• Concrete cube compressive strength
• Dry density, bearing pressure and consolidation
• Levelling, gradient and reduced level calculations
• Continuity equation and velocity head
• Hydraulic radius and Manning’s equation
• Super-elevation and stopping sight distance
• BOD, sedimentation and sewer velocity
• Duty, delta and intensity of irrigation
• Slenderness ratio and effective column length
• Centre line method and quantity estimation
• CPM and critical path
• Moment, resultant force and centroid

How to Use This Model Paper

First, attempt all 60 questions without checking the solutions. Mark the questions in which you are unsure about the formula or concept.

After completing the paper:

1. Check every answer carefully.
2. Create a separate list of incorrect questions.
3. Revise the related formula or theory.
4. Solve the same questions again after two or three days.
5. Attempt another model paper under a fixed time limit.

Regular revision of mistakes is more useful than repeatedly solving only easy questions.

Continue Your SSC JE Preparation

Completed this model paper? Strengthen your preparation with the SSC JE Civil syllabus, exam pattern, formula notes, subject-wise mock tests and previous-year concept practice sets available on JobApplyIndia. Use the related study resources to revise weak topics before attempting the next paper.

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Frequently Asked Questions

Is this an official SSC JE Civil previous year question paper?

No. This is an original model practice paper prepared according to the SSC JE Civil Engineering syllabus and commonly tested concepts. It is not an official Staff Selection Commission question paper.

Are these questions copied from SSC JE papers conducted between 2010 and 2015?

No. The questions have been independently created for educational practice. Their concepts and difficulty level are aligned with topics commonly relevant to SSC JE preparation, but the paper does not reproduce an official examination paper.

Is this model paper useful for the current SSC JE examination?

Yes. Fundamental Civil Engineering subjects such as RCC, Strength of Materials, Surveying, Soil Mechanics, Hydraulics, Transportation Engineering and Environmental Engineering remain important for SSC JE preparation. Candidates must also check the latest official syllabus and notification.

How many questions should I attempt daily?

Beginners may solve 15–20 questions daily with complete analysis. Candidates who have completed the syllabus may attempt all 60 questions under timed conditions.

Should I memorise the solutions?

No. Focus on understanding the formula, units and solution process. SSC may test the same concept using different values or wording.

Can this model paper help in other Junior Engineer examinations?

Yes. The questions may also support preparation for RRB JE, state-level JE examinations, diploma-level technical exams and other Civil Engineering recruitment tests. However, candidates should separately verify the syllabus of each examination.

Where can I find official SSC JE papers and answer keys?

Candidates should refer to the official Staff Selection Commission website and its official notices, answer keys and examination-related resources.

What should I study after completing this paper?

Revise the questions answered incorrectly and then move to subject-wise practice sets, formula revision, full-length mock tests and the next SSC JE Civil model paper.

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Disclaimer

This SSC JE Civil Model Practice Paper has been created by JobApplyIndia solely for educational and self-practice purposes.

All questions, numerical values, options, explanations and solutions presented in this article are independently prepared original practice material based on general Civil Engineering concepts and the SSC JE syllabus. This article is not an official SSC question paper, official answer key or reproduction of any examination conducted by the Staff Selection Commission.

The use of the term “2010–2015” indicates the syllabus period and concept-oriented practice category of this model paper. It does not mean that every question was asked in an official SSC JE examination during those years.

Although reasonable care has been taken while preparing the questions and solutions, candidates should verify technical standards, code provisions, examination patterns, eligibility conditions and current information through the latest official SSC notification and recognised engineering references.

JobApplyIndia is an independent educational and career-information platform and is not affiliated with, authorised by or endorsed by the Staff Selection Commission or any government department.