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RSMSSB JE Civil Mock Test - 4 Free Online Test 2026


Full Mock Test & Solutions: RSMSSB JE Civil Mock Test - 4 (120 Questions)

You can boost your Civil Engineering (CE) 2026 exam preparation with this RSMSSB JE Civil Mock Test - 4 (available with detailed solutions).. This mock test has been designed with the analysis of important topics, recent trends of the exam, and previous year questions of the last 3-years. All the questions have been designed to mirror the official pattern of Civil Engineering (CE) 2026 exam, helping you build speed, accuracy as per the actual exam.

Mock Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 120 minutes
  • - Total Questions: 120
  • - Analysis: Detailed Solutions & Performance Insights
  • - Sections covered: General Knowledge, Engineering Civil

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RSMSSB JE Civil Mock Test - 4 - Question 1

The Rajputana Agency was established in ________ at ________.

Detailed Solution: Question 1

The correct answer is 1832, Ajmer.

Key Points

  • Rajputana Agency was established in 1832 AD in Ajmer.
    • It was the office for Political Activities by British Agent.
    • Its name was A.G.G (Agent to Governor-general).
    • It gives the Report to the British Governor-general.

Additional Information

  • Mr. Abraham's Locket was First A.G.G.
    • Its office is situated in Mount Abu.

RSMSSB JE Civil Mock Test - 4 - Question 2

Which of the following pairs of peasant movement and its starting year is correct?

Detailed Solution: Question 2

The correct answer is Begun Peasant Movement - 1921.

Key Points

  • Begun Peasant Movement (1921)
    • Begun is situated in the Chittorgarh district of Rajasthan.
    • Begun farmer's movement was started in 1921.
    • It was started for fair implementation of the tax system.
    • Farmers decided not to pay taxes and boycott government offices and officials.
    • In reaction to it, the government launched an operation to crush the movement.
    • Mr Trench opened fire on the peasants which led to the killing of peasants Rupaji and Kripaji.

Additional Information

RSMSSB JE Civil Mock Test - 4 - Question 3

When was the state of Rajasthan established?

Detailed Solution: Question 3

The correct answer is March 30, 1949.

Key Points

  • Rajasthan was established on March 30, 1949.
  • It was formed by merging the princely states of Jaipur, Jodhpur, Bikaner, and Udaipur.
  • Jaipur became the capital of the newly formed state.
  • Rajasthan is the largest state by area in India.​

Additional Information

  • It is known as the "Land of Kings."
  • Includes princely states like Jaisalmer, Alwar, Bharatpur, Kota, Bundi, Dholpur.
  • Home to the Thar Desert, covering a large part of its western region.
  • Shares an international border with Pakistan.
  • The official language of the state is Hindi.
  • Major tribal communities include Bhils and Meenas.

Important Points

RSMSSB JE Civil Mock Test - 4 - Question 4

In 1722, the subadari of ______ was given to Sawai Raja Jai Singh.

Detailed Solution: Question 4

The correct answer is Agra.

Key Points

  • Jai Singh II was also referred to as Sawai Jai Singh.
  • He was the 29th Kachwaha Rajput king to lead the Amber Kingdom.
  • He later founded the fortified city of Jaipur and made it his capital.
  • He was born in Amber, the Kachwaha Rajputs capital.
  • When his father, Raja Bishan Singh, passed away on December 31, 1699, he was 11 years old and took over as the monarch of Amber.
  • In 1722, Sawai Jai Singh received the subadari of Agra and made Jaipur the capital of his empire.
  • He was appointed to govern Malwa three times between 1714 and 1737.

Additional Information

  • Sawai Raja Jagat Singh:
    • After Pratap Singh's passing, Jagat Singh II occupied the throne of Jaipur.
    • In the year 1803, Jagat Singh was crowned.
  • Jahandar Shah:
    • He was the ninth Mughal Emperor.
    • He reigned in the years 1712–1713.
    • He was Bahadur Shah's son.
  • Shahjahan:
    • He was the Mughal Empire's fifth emperor.
    • He ruled from January 1628 until July 1658.
    • He was Jahangir's third son.

RSMSSB JE Civil Mock Test - 4 - Question 5

Which in the main centre of salt production in Rajasthan?

Detailed Solution: Question 5

Concept:

  • Salt is a mineral composed primarily of sodium chloride.
  • It is added to food for flavour.
  • Salt is produced naturally when seawater evaporates.
  • Commercial salt is manufactured:
    • from rock salt
    • from seawater
    • from natural and artificial brines
  • ​Brine: it is a solution of salt and water.

Explanation:

  • Rajasthan is the third largest salt-producing state in India.
  • Sambhar Salt Lake is India's largest inland salt lake.
  • The lake is an extensive saline wetland.
  • It produces 196,000 tonnes of clean salt every year.
  • This is around 9% of India's salt production.
  • Salt is produced by the evaporation of brine.
  • Salt production is mostly managed by a government-owned company.

​Thus, Sambhar Salt Lake is the main centre of salt production.

Important Points

  • Sambhar Lake has been designated as a Ramsar site.
  • It is recognized as a wetland of importance.
  • Thousands of pink flamingos and other birds migrate here.
  • Ramsar convention:
    • an international treaty.
    • aims at the conservation and sustainable use of wetlands.
    • the convention was signed in 1971.

Additional Information

  • India is the third-largest salt producing country in the world.
  • Major salt-producing states of India:
    • Gujarat
    • Rajasthan
    • Tamil Nadu

RSMSSB JE Civil Mock Test - 4 - Question 6

Which of the following district is not matched with its division?
A) Ajmer - Bhilwara
B) Bharatpur - Alwar
C) Udaipur - Pratapgarh
D) Jaipur - Jhunjhunu

Detailed Solution: Question 6

The correct answer is Bharatpur - Alwar.

Key Points

  • Rajasthan was established on 1 November 1956. At the time of establishment of Rajasthan, there were total 26 districts in Rajasthan. And at that time the first Chief Minister of Rajasthan was Hiralal Shastri.
  • Divisions of Rajasthan
    • Jaipur Division- Jaipur, Dausa, Sikar, Alwar, Jhunjhunu
    • Jodhpur Division- Jodhpur, Jalore, Pali, Barmer, Sirohi, Jaisalmer
    • Bharatpur Division- Bharatpur, Dhaulpur, Karauli, Sawaimadhopur
    • Ajmer Division- Ajmer, Bhilwara, Tonk, Nagaur
    • Kota Division- Kota, Bundi, Bandra, Jhalawar
    • Bikaner Division- Bikaner, Ganganagar, Hanumangarh, Churu
    • Udaipur Division - Pratapgarh, Udaipur, Rajsamand, Chittorgarh, Dungarpur and Banswara

RSMSSB JE Civil Mock Test - 4 - Question 7

Which of the following pair is incorrect:

Detailed Solution: Question 7

The correct answer is Zinc Industry – Jodhpur.

Key PointsZinc Industry:

  • Zinc is found in large amounts in the southern Rajasthan near Udaipur.
  • Zawar mines are the famous mines situated near Udaipur. It is an undertaking of the Government of India.
  • Hindustan Zinc Ltd. has large zinc smelter plants installed here. It also gets zinc from Zawar mines.

Important Points

Textile Industry:

  • Bhilwara in recent times has emerged as the largest manufacturer of Fabrics.
  • It is famously known as the Textile city of India.
  • It is the traditional industry of Rajasthan.

Salt industry:

  • Rajasthan produces about 12% of the total salt produced in India.
  • The salt-producing regions of Rajasthan are Sambhar, Didwana, Pachpadra, Phalodi, and Kutchaman city.

Marble Industry:

  • Makrana marble is famous all around the world.
  • This stone has various uses from building to statues, toys, artistic goods, decorative items, etc.

RSMSSB JE Civil Mock Test - 4 - Question 8

Which tradition among the following is not of Garsia tribe?

Detailed Solution: Question 8

Garsia tribe does not believe in Gandarv Marriage. It is one of the eight classical types of Hindu marriage. This ancient marriage tradition from the Indian subcontinent was based on mutual attraction between two people, with no rituals, witnesses or family participation.

RSMSSB JE Civil Mock Test - 4 - Question 9

Which of the following statements is/are correct about the Lok Devta in Rajasthan?

1. Ramdev Ji's Horse name was Leelan.

2. Teja Ji's Horse name was Leela.

3. Pabu Ji's Horse name was Leelagar.

Detailed Solution: Question 9

The correct answer is None of The Above.

Key Points

RSMSSB JE Civil Mock Test - 4 - Question 10

Late. Hisamuddin was an accomplished artist of which Handicraft?

Detailed Solution: Question 10

The correct answer is Usta craft.

Key Points

  • Hisam-ud-din Usta was a well-known artist who was born in the former Bikaner State.
  • He was the last officially trained painter in the Bikaner School tradition.
  • He rose to fame as an artist in the Naqqashi and Manoti media. In the Manoti technique, painted flower designs in transparent and opaque oil and watercolors are placed over gold on bronze, wood, gold, silver, plaster of Paris, and camel leather.
  • In 1967, Hisam-ud-din won the National Award for Master Craftsperson, and in 1986, he was given the Padma Shri Award.

Additional Information

  • Jatt Patti craft:
    • Gotta Patti's work, also known as "Lappe ka kaam or gota kinari work," is one of the most famous and among the art craft gems of India.
    • It is a unique embellishment to enhance the fabric look, making it rich and splendid for occasions or wedding wear.
  • Meenakari:
    • It is an Iranian Safavid technique for painting and coloring metal and ceramic surfaces through the use of enamel.
  • Thewa craft:
    • Thewa is a unique method of creating jewelry that requires embossing delicately crafted sheet gold into molten glass.
    • It developed in Rajasthan's Pratapgarh district.

RSMSSB JE Civil Mock Test - 4 - Question 11

The following results are obtained on shear stress (τ) and rate of deformation at constant temperature for a fluid.

The above fluid is classified as

Detailed Solution: Question 11

The mathematical model for describing the behavior of fluids is the Power-Law model, which is given as:

If
n = 1;
fluid is Newtonian Fluid.
n ≠ 1;
fluid is non-Newtonian

n < 1; ⇒ fluid is called Pseudoplastic fluids.
n > 1; ⇒ fluids called dilatants fluids.
The mathematical model for describing the ideal Plastic fluid is given as:

From the given table:
If du/dy = 0 then τ = 10 i.e. non – zero ⇒ Fluid can not be Newtonian and non-Newtonian

Further Shear stress is a linear function of du/dy, so the fluid is ideal plastic.

RSMSSB JE Civil Mock Test - 4 - Question 12

The length of the summit curve having SSD as 150 m and deviation angle as 1 in 30 is

Detailed Solution: Question 12

Concept:

Length of summit curve is given by,

If we assume L > SSD

If we assume L < SSD

Where, N = Deviation angle & S = Stopping distance

Calculation:

Given,

SSD = 150 m and N = 1 in 30 = 0.03333

Case 1: If L > SSD

L ≈ 170 m

RSMSSB JE Civil Mock Test - 4 - Question 13

Calculate the value of Flexural strength of M20 grade concrete.

Detailed Solution: Question 13

Concept:

Modulus of rupture: It is a measure of the tensile strength of concrete beams or slabs.

Flexural strength of concrete/ Bending tensile strength of concrete/Modulus of rupture of concrete (fcr) is given by

fcr = 0.7 × √fck

∴ Flexure strength of concrete is determined as a modulus of rupture.

Calculation:

Given,

Grade of concrete is M20

i.e. fck = 20 MPa

Flexural Strength = 0.7 × √20 = 3.13 MPa

Additional Information

Modulus of Elasticity of Concrete: It can be defined as the ratio of applied stress to corresponding strain within elastic limits.

It may be given as:

i) Short Term Modulus of elasticity = 5000 √fck

ii) Long Term Modulus of elasticity

where θ is the creep coefficient

Modulus of rigidity: It is defined as the ratio of shear stress to shear strain. It is also called shear modulus.

RSMSSB JE Civil Mock Test - 4 - Question 14

Which component of a weir is responsible for diverting water from the river into the canal?

Detailed Solution: Question 14

  • In a weir, the component responsible for diverting water from the river into the canal is typically the crest of the weir.
  • The crest is the highest point of the weir structure, and it determines the level to which water can rise before it starts flowing over the weir.
  • When the water level in the river exceeds the crest level, water is diverted over the weir and into the canal or other downstream channels.
  • The design and dimensions of the weir's crest play a crucial role in controlling the flow of water and diverting it as needed.

Additional InformationAbutment:

  • In a weir, an abutment is a structural support or embankment on one or both sides of the channel where the weir is constructed.
  • Abutments provide stability to the weir structure, resist water pressure, and help control the flow of water over the weir crest, facilitating effective water management and control.

Apron:

  • An apron in a weir is a downstream extension of the structure designed to dissipate energy and prevent erosion.
  • Positioned below the weir crest, the apron helps regulate water flow, minimizes turbulence, and protects the foundation from scouring.

Headwall:

  • A headwall in a weir is a vertical or slightly inclined structure placed at the ends of a spillway or open channel to guide and contain the flow of water over the weir.
  • It helps prevent erosion, ensures uniform flow patterns, and enhances the weir's hydraulic performance.

RSMSSB JE Civil Mock Test - 4 - Question 15

How many positions are placed in prisms with its base on horizontal plane (CAD drawing)?

Detailed Solution: Question 15

When a prism is placed with its base on a horizontal plane in a CAD drawing, it can have four positions or orientations. These are:

  • The base of the prism can be placed flat on the horizontal plane.
  • The prism can be rotated 90 degrees clockwise about an axis perpendicular to the base, so that one of its lateral faces is touching the horizontal plane.
  • The prism can be rotated 180 degrees about an axis perpendicular to the base, so that the opposite face of the base is touching the horizontal plane.
  • The prism can be rotated 270 degrees clockwise about an axis perpendicular to the base, so that another lateral face is touching the horizontal plane.

Each of these positions will have a different appearance in a CAD drawing, and it's important to choose the correct orientation depending on the requirements of the design.

RSMSSB JE Civil Mock Test - 4 - Question 16

Calculate the total charges given to three labourers for their earthwork excavation, with them working from 9 a.m. to 6 p.m. with an hourly working charge of Rs. 50.

Detailed Solution: Question 16

First, determine how many hours each labourer works:
They work from 9 a.m. to 6 p.m. which is 9 hours of work.
Then, since each labourer charges Rs. 50 per hour, find out how much each labourer earns in a day:
9 hours × Rs. 50/hour = Rs. 450
Since there are 3 labourers:
3 labourers × Rs. 450 = Rs. 1,350

So, the total charges given to three labourers for their earthwork excavation is Rs. 1,350.

RSMSSB JE Civil Mock Test - 4 - Question 17

If the circular main scale of a theodolite is divided into degrees and each degree is divided into 1/4 parts, then the main scale division is divided into ______ vernier scale divisions to have a least count of 0.005 degrees.

Detailed Solution: Question 17

Concept:

Least count of Theodolite (LC) = s/n

Where,

s = length of one division in main scale, and n = no of divisions in Vernier scale

The Circular main scale is divided into degrees and each degree is divided into 4 equal parts.

This implies that length of one division in main scale is 1/4 i.e. s = 1/4

S = = =

Given:

LC = 0.005 degrees × 3600 = 18 seconds

Calculation:

LC = 18 =

n = 50

Main scale division is divided into 50 vernier scale divisions.

RSMSSB JE Civil Mock Test - 4 - Question 18

Which of these does not relate to the zone of aeration in the soil profile?

Detailed Solution: Question 18

Zone of Rock Fracture is further divided into two zones:

  1. Zone of Aeration
  2. Zone of Saturation

Zone of Aeration is further divided into three zones:

  1. Root soil water zone
  2. Intermediate zone
  3. Capillary water zone

Hence, the saturation zone does not come under the zone of aeration.

RSMSSB JE Civil Mock Test - 4 - Question 19

A water borne disease is

Detailed Solution: Question 19

Waterborne diseases are caused by a variety of microorganisms, biotoxins, and toxic contaminants.
It leads to devastating diseases such as -

  • Cholera
  • Schistosomiasis
  • Gastrointestinal problems

Additional Information

  • Other Waterborne Diseases
    • Anaemia
    • Arsenicosis
    • Dysentry
    • Ascariasis
    • Campylobacteriosis
    • Cyanobacterial toxins
    • Dengue and Dengue Haemorrhagic fever
    • Diarrhoea

RSMSSB JE Civil Mock Test - 4 - Question 20

A tie bar 5 mm × 8 mm is to carry a load of 80 kN. A specimen of the same quality steel of cross-sectional area 25 mm2 was tested in the laboratory. The maximum load carried by the specimen was 125 kN. Find the factor of safety in the design?

Detailed Solution: Question 20

Concept:

Factor of safety for tension member is given by,

FOS = yield strength/Design strength

Calculation:

Given,

Area of tie member = 5 × 8 = 40 mm2

Design strength of tie member = (80 × 103)/40 = 2000 MPa

Yield strength of tension member = (125 × 103)/25 = 5000 MPa

FOS is given by, FOS = yield strength/Design strength

FOS = 5000/2000 = 2.5

RSMSSB JE Civil Mock Test - 4 - Question 21

In the plastic design of structures if the degree of statical indeterminacy is 'j', the members of hinges 'p' required to convert the structure into a mechanism are given by

Detailed Solution: Question 21

Concept:

Degree of Statical Indeterminacy:

  • The DEGREE OF STATIC INDETERMINACY (DSI) is the number of redundant forces in the structure.
  • Redundant forces are the forces that cannot be found by writing and solving only the equations of equilibrium. They must be independent.
  • Therefore, the degree of static indeterminacy (DSI) represents the difference between the number of static unknowns (reactions and internal forces) and the number of static equations (equilibrium equations).

​Mechanism:

  • If the segment of a beam or any other structure between plastic hinges is able to move without increasing the load, then it is called a mechanism.
  • If a mechanism is formed then it means that the structure becomes unstable.

In plastic analysis to make the statically indeterminate structure a determinate structure, the number of plastic hinges required is "j", where "j" is the degree of static indeterminacy. Further, to make a mechanism, i.e. to make a structure unstable on a statically indeterminate structure, the number of plastic hinges required = j + 1, where j is static indeterminacy.

So the correct answer is Option 3

RSMSSB JE Civil Mock Test - 4 - Question 22

What will be the pressure at point B of this U-tube manometer? Take ρ= density of light liquid and ρ2 = density of heavy liquid.

Detailed Solution: Question 22

Concept:

U-tube manometer:

  • The U-tube manometer is a U-shape bent glass tube, whose one end is open to the atmosphere and other end is connected to a section where the pressure of the liquid is to be measured.
  • The U-tube manometer is filled with a liquid whose specific gravity is more than as compared to the liquid whose pressure is to be measured.

  • The gauge pressure at point B in the above-shown U-tube manometer is calculated by using hydrostatic law and is given as:

The pressure of liquid above section A-A in the right column = ρ2gh2

The pressure of liquid above section A-A in the left column = PB + ρ1gh1  

Equating the above two pressure, we get the gauge pressure at point B as:
PB + ρ1gh= ρ2gh2  

Calculation:

Given:

In this problem, the pressure at point B is:
Pressure of liquid above A-A in right column = 0
Pressure of liquid above A-A in left column = P+ ρ1gh+ ρ2gh2 

Equating both the pressure, we get:
P+ ρ1gh+ ρ2gh= 0
P= −(ρ1gh+ ρ2gh2)
Thus, option (2) is correct answer.

RSMSSB JE Civil Mock Test - 4 - Question 23

Which of the following pairs is correctly matched in case of defects and causes of wood?

Detailed Solution: Question 23

Defects in Timber: The defects occurring in the timber are grouped into the following five divisions:

  • Defects due to conversion
  • Defects due to fungi
  • Defects due to insects
  • Defects due to natural forces
  • Defects due to seasoning.

The various types of defects under each category will now be briefly discussed.

1) Defects due to natural forces: The main natural forces responsible for causing defects in timber are two, namely, abnormal growth and rupture of tissues. The following defects are caused by these forces:

  • Burls
  • Callus
  • Chemical stain
  • Coarse grain
  • Deadwood
  • Twisted fibres
  • Druxiness
  • Foxiness
  • Upsets
  • Knots
  • Rindgalls
  • Shakes
  • Twisted fibres
  • Upsets
  • Water stain
  • Wind cracks
  • Rind galls
  • Wind cracks

Shakes: are longitudinal separations in the wood between the annual rings. These lengthwise separations reduce the allowable shear strength without much effect on compressive and tensile values. The separations make the wood undesirable when appearance is important. Both the shakes and checks if present near the neutral plane of a beam they may materially weaken its resistance to horizontal shear.

Heart hakes: occurs due to shrinkage of heartwood when the tree is overmatured. Cracks start from the pith and run towards sapwood. These are wider at the centre and diminish outwards.

Cup shakes: appears as curved split which partly or wholly separates annual rings from one another. It is caused due to excessive frost action on the sap present in the tree, especially when the tree is young.

Star shakes: are radial splits or cracks wide at the circumference and diminishing towards the centre of the tree. This defect may arise from severe frost and fierce heat of the sun. Star shakes appear as the wood dries below the fibre saturation point. It is a serious fault leading to separate log when sawn.

Knots: are bases of twigs or branches buried by the cambial activity of the mother branch. The root of the branch is embedded in the stem, with the formation of annual rings at right angles to those of the stem. The knots interrupt the basic grain direction of the wood, resulting in a reduction of its strength. In addition, these affect the appearance of the wood. A dead knot can be separated from the body of the wood, whereas a live knot cannot be. Knots reduce the strength of the timber and affect workability and cleavability as fibres get curved. Knots are classified on the basis of size, form, quality and occurrence.

Rindgalls: is characterised by swelling caused by the growth of layers of sapwood over wounds after the branch has been cut off in an irregular manner. The newly developed layers do not unite properly with the old rot, thereby leaving cavities, from where decay starts.

Twisted Fibres: Twisted fibres are caused by wind constantly turning the trunk of a young tree in one direction.

Upset: are caused by the crushing of fibres running transversely during the growth of the tree due to strong winds and unskilled felling consequently resulting in discontinuity of fibres.

Foxiness: is a sign of decay appearing in the form of yellow or red tinge or discolouration of overmatured trees.

2) Defects due to conversion: During the process of converting timber to commercial form, the following defects may occur:

  • Chip mark
  • Torn grain
  • Diagonal grain
  • Wane.

3) Defects due to fungi: The fungi are minute microscopic plant organisms. They attack timber only when the following two conditions are satisfied simultaneously:
i) The moisture content of timber is above 20 percent.
ii) There is a presence of air and warmth for the growth of fungi.

If any of the above conditions is absent, the decay of wood due to fungi would not occur. Hence the dry wood having a moisture content of less than 20% will remain sound for centuries. Similarly, the wood submerged in water will not be attacked by fungi because of the absence of air. The following defects are caused in the timber by the fungi:

  • Blue stain
  • Brown rot
  • Dry rot
  • Sap strain
  • Wet rot
  • White rot
  • Heart rot

4) Defects due to insects: Following are the insects that are usually responsible for the decay of timber:

  • Beetles
  • Marine borers
  • Termites.

5) Defect due to seasoning: Following defects occurs in the seasoning process of wood.

  • Bow
  • Case-hardening
  • Check
  • Collapse
  • Cup
  • Honey-combing
  • Radial shakes
  • Split
  • Twist
  • Warp.

​Hence option (3) is correct.

RSMSSB JE Civil Mock Test - 4 - Question 24

Two stresses exist in a thin-walled pressure vessel. those are:

Detailed Solution: Question 24

Concept:

In a thin wall pressure vessel, two stresses exist:

  • The longitudinal stress
  • The hoop stress

Hoop stress:

Longitudinal stress:

Hence, In thin-walled cylindrical pressure vessel, hoop stress is twice the longitudinal stress.

RSMSSB JE Civil Mock Test - 4 - Question 25

The shear force developed in a cantilever beam subjected to a point load p at the free end is

Detailed Solution: Question 25

Cantilever beam having point load at free end

At the Free end, BM = 0

The shear force developed in the cantilever beam is free to end load P.

Additional Information

Cantilever beam having point load at the middle of its length

At the Free end, BM = 0 but SF = 0.

Cantilever beam having uniformly distributed load over the whole length

At the Free end, BM = 0 but SF = 0.

∴ SF and BM are zero at the free end when the point load is at the middle of its length.

RSMSSB JE Civil Mock Test - 4 - Question 26

The ratio of equivalent length of the column to the maximum radius of gyration is called

Detailed Solution: Question 26

Buckling factor:

  • The ratio of the equivalent length of a column to its maximum radius of gyration is known as the buckling factor.
  • It is an important parameter used in structural analysis to assess the stability of slender columns under compressive loads.
  • The buckling factor provides an indication of the column's susceptibility to buckling, which is a critical failure mode for slender members.

Poisson's ratio (option 1) is a material property that relates lateral strain to axial strain in an elastic material. It is not directly related to the ratio mentioned in the question.

The factor of safety (option 3) is a measure of the capacity of a structure to withstand loads beyond its expected working conditions. It is not related to the ratio of the column length to the maximum radius of gyration.

Therefore, the correct answer is option 2.

RSMSSB JE Civil Mock Test - 4 - Question 27

A simply supported beam is loaded as shown in fig. The maximum shear force in the beam will be

Detailed Solution: Question 27

For Reaction at A and B:
RA + RB = 4W
Taking Moment about B
RA × 4C - W × 3C - 2W × 2C - W × C = 0
RA = 2W
RB = 2W

Shear Force Diagram:

Hence, maximum shear force in the beam will be 2W.

RSMSSB JE Civil Mock Test - 4 - Question 28

Calcium lignosulphonate acts as a

Detailed Solution: Question 28

RSMSSB JE Civil Mock Test - 4 - Question 29

Which of the following can be considered both a 2D and 3D shape under different conditions?

Detailed Solution: Question 29

A cylinder can be considered both 2D and 3D depending on the context.

A 2D representation of a cylinder is a circle or a rectangle in a technical drawing, while its 3D counterpart includes height, allowing it to occupy space.

A square is strictly a 2D shape, while a prism and pyramid are inherently 3D.

RSMSSB JE Civil Mock Test - 4 - Question 30

The optimum depth of a plate girder is given by the following expression:

Detailed Solution: Question 30

Plate girder:

  • A plate girder is a built-up I-beam section, used to carry heavy loads which cannot be carried economically by rolled I-sections.
  • It is made by riveting or welding the steel plates in an I-beam shape.

The optimum thickness for the web (tw)

Depth of web (d)

Where, tw = Optimum thickness of the web, Mz = Moment applied to the flanges, fy = Design strength of the flanges, k = Slenderness ratio of the web = d/tw

d = Depth of web

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