UPSSSC JE Civil Mock Test - 9 Free Online Test 2026

Concept:

Stiffness:
A force required to produce unit deformation is called stiffness. Denoted by K.

Where, P = Load; L = Length; A = Area of cross-section; E = Modulus of Elasticity.

Calculation:
Given:
A₁ = 2A₂
As the material is the same, their stiffness must be the same.
∴ K₁ = K₂

∴ L1 = 2L2

Types of shear failure

1) Local shear failure
This type of failure is seen in relatively loose sand and soft clay.

Some characteristics of local shear failure are:
(i) Failure is not sudden and there is no tilting of footing.
(ii) Failure surface does not reach the ground surface and slight bulging of soil around the footing is observed
(iii) Failure surface is not well defined
(iv) Failure is progressive
(v) In the load-settlement curve, there is no well-defined peak
(vi) Failure is characterized by considerable settlement directly beneath the foundation
(vii) Significant compression of soil below the footing and partial development of plastic equilibrium is observed.
(viii) Well-defined wedge and slip surface only beneath the foundation.

(b) General shear failure
Occur in medium to dense soil and stiff clays and has the following properties:
(a) A well – defined failure pattern
(b) Sudden, catastrophic failure accompanied by tilting of foundation
(c) Bulging of the ground surface adjacent to the foundation.

This occurs in soils that show brittle – type stress-strain behavior.

c) Punching shear failure
This occurs in soils possessing stress-strain characteristics of very plastic soil. This type is characterized by
a) Poor defined shear planes
b) Soil zones beyond loaded area being less affected.
c) Significant penetration of a wedge-shaped soil zone beneath the foundation, accompanied by vertical shear beneath the edges of f/x.

Relatively large settlements do occur in this type of failure and ultimate bearing capacity is not well defined.

This type of failure occurs in deep footing in the soil of low compressibility.

Concept:

When two stress are acting at a point, the maximum shear stress(τ_max) is given by

σ₁ = Major principal stress
σ₃ = Minor principal stress

Calculation:

Given:
σ1 = 3 MPa
σ3 = -3 MPa

⇒ 3 MPa

Concept:
Available water in m3 = (Specific Yield) × (Area) × (Change in water level)

Calculation:
Given,
Specific Yield = 10% = 0.1, area = 1 km2 = 1 × 106 m2, and change in water level = 1.0 m
∴ Available water in m3 = (0.1) × (1 × 106 m2) × (1.0 m) = 100000 m3

Concept:
Orientation of plane table by trough compass:
The compass, though less accurate, often proves a valuable adjunct in enabling the rapid approximate orientation to be made prior to the final adjustment. The plane table can be oriented by compass under the following conditions final adjustment.

• When speed is more important than accuracy
• When there is no second point available for orientation
• When the traverse is so long that accumulated errors in carrying the azimuth forward might be greater than the orientation of the compass
• For approximate orientation prior to final adjustment.

Concept:

Plate load test is field test to determine ultimate bearing capacity of soil and settlement under given load.
To conduct a plate load test a pit of size 5 times the size of plate is excavated to a depth equal to depth of foundation.

Limitations of plate load test:

• The test results reflect only the character of the soil located within a depth of less than twice the width of bearing plate. Normally the foundations are larger than the test plates, the settlement and shear resistance of soil against shear failure will depend on the properties of stratum. Thus the results of test could be misleading if the character of the soil changes at shallow depths.

• The plate load test is of short duration. The settlement measured is only the immediate settlement. In granular soil immediate settlement can be taken as total settlement, while in case of cohesive soils it is consolidation settlement. Hence statement B is false.

Hence from above limitation it is clear that statement A is true and statement B is false.

Transverse Stiffeners are provided to increase buckling resistance of the web due to shear, It is provided vertically along the span.

Horizontal Stiffener / Longitudinal Stiffener is designed to prevent web buckling due to bending compression.

End Bearing Stiffeners are provided at the supports & Load Bearing Stiffeners are provided at the points of concentrated loads.

Important Point:

• If d/t_w < 67ϵ ⇒ unstiffened girder can be designed i.e. No girder required.
• If 85 ϵ < d/t_w< 200 ϵ ⇒ Vertical stiffener (C₁ and C₂) may be provided.
• If 200 ϵ < d/t_w < 250 ϵ ⇒ Vertical stiffener along with longitudinal stiffener at 0.2 d may be provided.
• If 250 ϵ < d/t_w < 345 ϵ ⇒ Vertical stiffeners along with two longitudinal stiffeners at 0.2 d and 0.5 d respectively may be provided.

Calculation:

Therefore, intermediate vertical stiffeners are to be provided.

A tension member under factored tensile load may chance to fail by three possible modes as per limit state of strength and limit state of serviceability, which are Gross section yielding or yielding of gross section.

Net section rupture or fracture at a critical section and Block shear of end region. Hence the design strength of a member under axial tension is governed by above three failures as per Limit state Design code of IS-800:2007.

California Bearing Ratio:

CBR % IS Calculated by using standard loads for both 2.5mm and 5mm and the maximum of these is taken as the CBR value
CBR_2.5mm = P_1//1370
CBR _5mm =P₂/2055
The maximum of these two is taken as CBR VALUE

Calculations:
GIVEN P₁ = 54.8 kg, P₂ = 78 kg
Now put the values and calculate
CBR_2.5mm = 54.8/1370 =.04 =4%
CBR_5mm = 78/2055 = 0.037 = 3.7%
Hence, the CBR value of the sample is 4%.

Note:
Generally 2.5mm is more than 5mm CBR but if 5mm CBR comes to more than 2.5mm CBR then it is suggested to repeat the test again and if the same results came then the higher value (5mm) is considered as CBR value
Other Related Points

IRC Recommendations For CBR Test

• Annual rate of increase in the vehicle is taken as 7.5%
• Remoulded soil should be tested in the laboratory after soaking it for 4 days
• At least 3 samples should be tested
• Soil should be compacted at OMC to proctor density

Concept-

A turnout has three distinct portions,

• Switch Assembly
• Lead Assembly
• Crossing Assembly

• Curve lead (CL)- This is the distance from the tangent point (T) to the theoretical nose of crossing (TNC) measured along the length of the main track.
• Switch lead (SL)- This is the distance from the tangent point (T) to the heel of the switch (TL) measured along the length of the main track.
• Lead of crossing (L)- This is the distance measured along the length of the main track as, Lead of crossing (L) = curve lead (CL) – switch lead (SL)
• Heel divergence (D)- This is the distance between the mainline and the turnout side at the heel.
• Correct sequence for a train in order to pass over the turnout from the facing direction is Toe of a switch - Tongue rail - Lead rail - Crossing.

The Indian Standard (IS) classification system for soil categorizes soil based on its grain size distribution and other properties. The key terms here are:

GW: Well-graded gravel
GP: Poorly graded gravel
SW: Well-graded sand
SP: Poorly graded sand

To classify the given soil sample, let's look at the definitions provided by the data:

• 70% of particles passing through a 4.75 mm IS sieve: This indicates that most of the soil is finer than gravel. Gravel is generally defined as particles larger than 4.75 mm, so if 70% pass through, it means the bulk of the soil is sand or finer.
• 40% of particles passing through a 75μ (or 0.075 mm) IS sieve: This indicates the presence of significant fines content (silt and/or clay) or fine sand.
• Uniformity Coefficient (Cu) = 8: The uniformity coefficient is a measure of the range of particle sizes and is used to distinguish between well-graded and poorly graded soils. A higher Cu indicates a wider range of particle sizes. Well-graded sands (SW) and gravels (GW) typically have Cu values greater than 4.
• Coefficient of Curvature (Cc) = 2: This is within the typical range (1 < Cc < 3) for well-graded soils, indicating a good distribution of sizes between the largest and smallest particles.

Given that the majority of particles are finer than 4.75 mm but larger than 75 μm, the soil is primarily sand. The high uniformity coefficient (Cu = 8) suggests a well-graded distribution of particle sizes, and the coefficient of curvature (Cc = 2) falls within the range indicative of well-graded soils.

Therefore, based on the IS classification system and the provided data, this soil is best classified as: SW - Well-graded sand

Aggregate Crushing Value (ACV)

• This test is used to find the strength of the coarse aggregate and indicates the resistance to crushing under a gradually applied crushing load.
• It is measured as the percentage of aggregates passing through 2.36 mm IS sieve after the specimen subjected to a compressive load of 40 tonnes at the rate of 4 tonnes per minute and to the weight of aggregates taken.
• Lower the ACV ⇒ Less aggregates gets crushed during loading ⇒ Higher the resistance against crushing​
• Aggregate crushing value should not be greater than 30 % for the surface course and 45 % for the base course

Given:
Diameter of sewer (d) = 1.2 m
The slope is 1 in 1000
Velocity when sewer runs full (Vfull) = 1 m/s
We know from the manning equation:

where R is the hydraulic radius, n is the manning coefficient

For a sewer running full & half full
we know Hydrualic Radius R = D/4

For a sewer running half full

As n, S is same for the sewer

velocity V ∝ R^2/3

Also for Sewer running full and half full R is same i.e. R = D/4

Vhalf = Vfull = 1 m/s

• Bars are bent near the support at an angle of 45°.
• The purpose of the bend near the support is firstly to resist the negative bending moment which occurs in the region of the support and secondly to resist the shear force which is greater at the support.
• The main reinforcement rods of the beam bend from the ends from L / 4 to L / 5 at 45°. If 'd' is the distance of the centers between those rods, then the extra length for both those bends is taken from 0.9 × d.

Oxygen sag curve of river manifest dissolved oxygen deficit.
The difference between saturated dissolved oxygen content and the actual dissolved oxygen content in the stream at any point during self-purification process is called Oxygen sag

Oxygen sag curve or oxygen deficit curve is obtained by algebraic addition of deoxygenation and reoxygenation curve.

Concept:

The effective span of continuous beam or slab

Case-I: If the width of support <

Then effective span is calculated the same as for the simply supported case

or

whichever is less

where d = effective depth, w = width of support

l_o = clear span, l_eff = effective span

Case-I: If the width of support >

(a)

(i) For one end fixed other continuous

(ii) Both end continuous(Intermediate span)

leff = lo = clear span

(b) One end discontinuous other continuous(simply supported)

whichever is less

Calculation:

Given data

d = 150 mm, w = 300 mm

Span = 5000 mm

300 <

So,

Case-I: If the width of support <

Then effective span is calculated the same as for the simply supported case

= 5150 mm

or

= 5300 mm

whichever is less

Hence, the value of the effective span is 5150 mm.

Alternate bay method

• In this method, if the road is a single lane, it is divided into suitable bays of 6 m to 8 m in length, and the construction work is carried out in alternate bays.
• If the road is a double lane; the construction work is carried out in odd bays of one lane and even bays of the other lane.

Advantages

• The bays which have been cured can serve as additional working space for the preparation of concrete.
• The construction of joints can be easily carried out and its width can be suitably kept as desired.

Drawbacks:

• As the construction activity is spread over the full width of the road, the traffic will have to be completely diverted.
• During the rainy season, the water collects on the surface of bays that are not constructed.
• It requires a large number of transverse joints which may increase the construction cost and reduce the smooth riding quality of the surface.
• It requires more time to complete the work.

Other Related PointsContinuous bay method

• In this method of construction, all the slabs or bays of a strip are constructed continuously without any break from one end to the other. In the continuous bay method construction joints are provided when the day's work is not ended at the specified joint.

Expansion joint method

• An Expansion Joint is a gap in the building structure provided by an Architect or Engineer to allow for the movement of the building due to temperature changes. An Expansion Joint is an assembly designed to safely absorb the heat-induced Expansion and Contraction of various construction materials.

Dowel bar method

• Dowel bars are placed across transverse joints in the concrete pavement to allow movement. They are inserted at the mid-depth of the slab and coated with a bond-breaking material to restrict bonding to the PCC. Thus dowels help to transfer loads allowing the expansion and contraction of adjacent slabs independently.

Concept:
A cross drainage work is a structure constructed for carrying a canal across a natural drain or river intercepting the canal to dispose drainage water without mixing with the continuous canal supplies.

Types of Cross Drainage Works:-
Based on the relative bed levels, water levels of the canal and the drain and their relative discharge the Cross Drainage works are of the following types:-
(1) Cross drainage works carrying the Canal over the Natural Drain:
(i) Aqueduct

• An aqueduct is a hydraulic structure which carries a canal (through a trough) across and above the drainage similar to a bridge in which instead of the road or a railway, a canal is carried over a natural drain.

• In the case of an aqueduct, HFL (highest flood level) of the drainage should remain lower than the level of the underside of the canal trough.
• The drain flows at atmospheric pressure under the work.
• Generally, an inspection road is provided along with the trough

(ii) Syphon aqueduct

• A syphon aqueduct is a cross drainage structure similar to an aqueduct except that the streambed is depressed locally where it passes under the trough of the canal and the barrels discharges the streamflow under pressure.
• A syphon aqueduct is constructed where the water surface level of the drain at high flood is higher than the canal bed.
• The horizontal floor of the barrels is provided with slopes at its ends to join the drain bed on either side.
• The drain water flows under pressure through the barrels which act as inverted syphons and hence this cross drainage work is known as syphon aqueduct.

(2) Cross drainage works carrying the natural drain over the canal-
(i) Super passage
A super passage is also similar to a bridge in which the natural drain is carried over the canal.

• A super passage is reverse of an aqueduct.
• A super passage is constructed where the bed of the drain is well above the canal F.S.L
• The canal flows at atmospheric pressure under the work.

• In this case it is not possible to provide an inspection road along the canal.

(ii) Syphon

• A syphon is similar to a syphon aqueduct with the difference that in the case of a syphon the canal water is carried through the barrels under the drain.
• A syphon is constructed where the full supply level of the canal is higher than the bed of the drain.
• The barrels in this case also act as inverted syphons through which the canal water flows under pressure.

​

3) Cross drainage works admitting the drain water into the canal-
In this type of cross drainage works, the canal water and the drain water are allowed to intermingle with each other. This may be achieved by the following two types of cross drainage works:-
(i) Level crossing
(ii) Inlet and outlet

Concept:
Scissor Cross Over or Double Cross Over

• It enables the trains to change the track from either direction along the main track.
• It consists 4 pairs of points, 6 acute angle crossings, 2 obtuse angle crossing, check rails and straight lengths.
• It is used when space for 2 separate cross over is not available.
• It is the combination of one cross over the other cross over in opposite direction

Turn- Table: Turn table or wheel house is a device for turning railway rolling stock, usually locomotives, so that they can be moved back in the direction from which they came.

Concept:
Lift of material:

• It is the vertical distance calculated from the ground level up to which the laborers have to excavate the soil and remove it for the standard rate.
• It is the average height through which the earth has to be lifted from source to the place of spreading or disposing.

Calculation:

The lift of soil material for the canal = = 1.35 m

Concept:

Prestress concrete is the concrete in which internal stresses are produced due to compression or tension applied before applying external load and these stresses are counter balanced by the applied load to the desired degree.

Calculation:
Given,
Prestressing force = 2500 KN
Effective span length = 10 m
Total external load = 40 kN/m
We know that,

Camber:

• Camber is the slope provided in the transverse direction of the road to drain off the rainwater from the road surface.
• Usually, a camber is provided in the straight roads by raising the center of the carriageway with respect to the edges forming the highest point at the center.
• The straight slope of the camber is easier to be cast and maintained in the case of concrete pavement.
• Parabolic shaped camber is difficult to construct and maintain. It has steeper edges which are inconvenient to use. However, it is kept in bituminous pavements for better drainage of water.

The correct answer is '2748 and 101010111100 respectively'

Explanation

• Converting (ABC)₁₆ to decimal:
  • To convert a hexadecimal number to decimal, we use the formula: Decimal = (An * 16^n) + ... + (A1 * 16^1) + (A0 * 16^0), where A is the hexadecimal digit and n is the position of the digit.
  • For (ABC)₁₆, it translates to: (A * 16^2) + (B * 16^1) + (C * 16^0) = (10 * 256) + (11 * 16) + (12 * 1) = 2560 + 176 + 12 = 2748.
• Converting (ABC)₁₆ to binary:
  • Each hexadecimal digit corresponds to a 4-bit binary number. Thus, A(1010), B(1011), and C(1100) in binary.
  • Combining these gives the binary equivalent of (ABC)16 as 101010111100.

The correct answer is A. Khatouli

The largest sugar mill in Asia is situated in Khatouli, Uttar Pradesh. Specifically, the Triveni Sugar Mill in Khatauli boasts the highest production and storage capacity among sugar mills in the region. This mill has been operational since 1933, making it a significant establishment in the area. Khatouli itself is a rural town that not only features the sugar mill but also offers various tourist attractions, including its location along the upper Ganga canal, which is a key draw for visitors. The town is known for its rich cultural diversity, influenced by Hindu, Muslim, Jain, Christian, and Sikh communities, and is home to nine Jain temples.

Details on Other Options:

• B: Balrampur is known for its sugar production but does not host the largest sugar mill in Asia, making this option incorrect.
• C: Bulandshahar has sugar mills, but it is not the location of the largest one in Asia, thus this option is less appropriate.
• D: Najibabad also has sugar mills but lacks the scale and capacity of the Triveni Sugar Mill in Khatouli, making this option incorrect.

Conclusion:

Option: A, recognized as the most accurate choice, is distinguished by its designation as the largest sugar mill in Asia, setting it apart from the other options. Thus, Khatouli is the correct answer.

The correct answer is Rani Laxmi Bai.

British General Huge Rose referred to Rani Laxmi Bai as ‘The best and bravest military leader of the rebel’. She played a pivotal role in the 1857 revolt, particularly in Jhansi, showcasing her leadership and bravery. Born on November 19, 1828, in Benares (now Varanasi), she is remembered as the fierce queen whose legacy continues to inspire. Her burial site is located at Phoolbagh Maidan in Gwalior, marking her significant contribution to India's fight for independence.

Details on Other Options:

• Begum Hazrat Mahal: This option is incorrect as she led the revolt in Lucknow, Agra, and Awadh and was the wife of the deposed king Wajid Ali Shah. She proclaimed her young son, Birjis Qadar, as the Nawab of Awadh.
• Tantia Tope: This option is incorrect because he was a key leader in the revolt at Kanpur and served as a general under Nana Saheb, not the military leader described by General Rose.
• Birjis Qadar: This option is incorrect as he was merely a figurehead declared Nawab by Begum Hazrat Mahal and did not lead military efforts during the revolt.

Conclusion:

The Rani Laxmi Bai, as the most accurate and relevant option, stands out for her recognized military leadership during the revolt of 1857, clearly differentiating her from the other figures. Thus, Rani Laxmi Bai is the correct choice.

The correct answer is B. Ramkot

Ramkot Temple serves as a significant place of worship located in Ramkot, Ayodhya. This temple is notable for being situated atop a hill, which is historically associated with Lord Rama's fort. Special cultural events are held at the temple, particularly during the festival of Rama Navami, celebrated in the Chaitra month (March-April), honoring Lord Rama.

Details on Other Options:

• A: This option is incorrect because Kushinagar is primarily known for its significance in Buddhism, particularly as the site where Gautam Buddha attained Mahaparinirvana. It is not associated with the Ramkot Temple.
• C: This option is incorrect as Deogarh is renowned for the Dashavatara Temple, built in the 6th century and dedicated to Lord Vishnu, not for the Ramkot Temple.
• D: This option is incorrect since it suggests that none of the listed locations are correct, whereas Ramkot is indeed the location of the temple in question.

Conclusion:

Option: B, as the most accurate and relevant option, stands out for its direct association with the Ramkot Temple, clearly differentiating it from the other options. Thus, Ramkot is the correct choice.

The correct answer is 1773.

Explanation

• The British defeated the Marathas in Rohillakhand in 1773 and expelled them from the Doab region.
• Rohil Sardar Rahmat Khan was defeated by the British in Shahjahanpur in 1774, and Rohillakhand was handed over to the Nawab of Awadh.
• Rahmat Khan ruled Rohillakhand (Rohil Pradesh of Meerut and Doab).
• The Nawabs of Farrukhabad ruled the Central Doab regions.
• The Nawabs of Awadh ruled the regions of Faizabad and Lucknow.
• The Marathas ruled the Bundelkhand region.

The correct answer is C. Vrindavan, Uttar Pradesh

The first all-girls Sainik School in India, known as Samvid Gurukulam Girls Sainik School, was inaugurated in January 2024 by Defence Minister Shri Rajnath Singh in Vrindavan, Uttar Pradesh. This initiative is part of a broader plan to establish 100 new Sainik Schools across the country in collaboration with NGOs and state governments. The establishment of this school represents a landmark advancement in women's empowerment, allowing girls to pursue careers in the Armed Forces and contribute to national defense, thereby promoting gender equality in military opportunities.

Details on Other Options:

• A: This option is incorrect as the first all-girls Sainik School was not inaugurated in Lucknow, Uttar Pradesh, but in Vrindavan.
• B: This option is incorrect because Siliguri, West Bengal, was not the location for the inauguration of the first all-girls Sainik School.
• D: This option is incorrect since Jaipur, Rajasthan, is not the site of the first all-girls Sainik School, which is located in Vrindavan.

Conclusion:

Option: C, as the most accurate and relevant choice, is highlighted by its representation of a significant step towards gender equality in the Armed Forces, clearly distinguishing it from the other options. Thus, Vrindavan, Uttar Pradesh, is the correct choice.

The correct answer is Ghaziabad.
Explanation

• As per Census 2011, Uttar Pradesh has a population of 19.98 Crores, an increase from the figure of 16.62 Crore in the 2001 census.
• Total population of Uttar Pradesh as per 2011 census is 199,812,341 of which male and female are 104,480,510 and 95,331,831 respectively.
• In 2001, total population was 166,197,921 in which males were 87,565,369 while females were 78,632,552.
• The population density of different districts of Uttar Pradesh are:-
  • Ghaziabad - 3971
  • Varanasi - 2395
  • Lucknow - 1816
  • Kanpur Nagar - 1452
• The total population growth in this decade was 20.23% while in the previous decade it was 25.80%.
• The population of Uttar Pradesh forms 16.50% of India in 2011.
• In 2001, the figure was 16.16%.

Other Related Points

• Recently as per Uttar Pradesh census data, 94.70% of houses are owned while 4.11% were rented.
• In all, 64.03% of couples in Uttar Pradesh lived in single-family.
• In 2011, 72.02% of the Uttar Pradesh population had access to Banking and Non-Banking Finance Corporation.
• Only 1.85% of Uttar Pradesh population had an internet facility which is likely to improve in 2021 due to Jio.
• 3.80% of families in Uttar Pradesh owned cars while 19.61% owned two-wheelers.