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30 Questions MCQ Test UPSSSC JE Civil Mock Test Series 2026 - UPSSSC JE Civil Mock Test - 1

UPSSSC JE Civil Mock Test - 1 for Civil Engineering (CE) 2026 is part of UPSSSC JE Civil Mock Test Series 2026 preparation. The UPSSSC JE Civil Mock Test - 1 questions and answers have been prepared according to the Civil Engineering (CE) exam syllabus.The UPSSSC JE Civil Mock Test - 1 MCQs are made for Civil Engineering (CE) 2026 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for UPSSSC JE Civil Mock Test - 1 below.
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UPSSSC JE Civil Mock Test - 1 - Question 1

Which of the following types of shapes is INCORRECT for plain sedimentation tanks?

Detailed Solution for UPSSSC JE Civil Mock Test - 1 - Question 1

Concept:

A sedimentation tank is structure in which wastewater is filled and stored for some time to remove the suspended particles present in the water by the action of gravitational forces. These particles may settle at the bottom of the tank and are removed by using scrapers.

Based on shape, a sedimentation tank has three types:

1. Circular tank: In this case influent is sent through central pipe of the tank and radial flow takes place. Mechanical sludge scrappers are provided to collect the sludge and collected sludge is carried through sludge pipe provided at the bottom.

2. Rectangular tank: Rectangular sedimentation tanks are mostly preferred sedimentation tanks and are used widely. The flow takes place in horizontal direction that is length wise in rectangular tanks

3. Hopper bottom tank: In case of hopper bottom tank, a deflector box is located at the top which deflects the influent coming from central pipe to downwards. Sludge is collected at the bottom and it is disposed through sludge pump.

The triangular tank with radial flow is not used for plain sedimentation tank.

Additional InformationA. Types of Sedimentation Tanks based on Location:

1. Primary Sedimentation Tank

Primary sedimentation tank is a normal sedimentation tank in which water is stored at rest for some time and sludge collected at bottom and oily matter collected at top are removed.

2. Secondary Sedimentation Tank

After activated sludge process the wastewater enters secondary sedimentation tank in which suspended particles contains microbes are removed .

B. Types of Sedimentation tanks based on Methods of Operation

1. Fill and Draw Type Sedimentation Tank

In case of fill and draw type sedimentation tank, water from inlet is stored for some time. The time may be 24 hours. In that time, the suspended particles are settled at the bottom of the tank. After 24 hours, the water is discharged through outlet. Then settled particle are removed.

2. Continuous Flow Type Sedimentation Tank

In this case, water is not allowed to rest. Flow always takes place but with a very small velocity. During this flow, suspended particles are settled at the bottom of the tank. The flow may be either in horizontal direction or vertical direction.

UPSSSC JE Civil Mock Test - 1 - Question 2

In a non-tilting type drum mixer,

  1. Large size aggregate up to 20 – 25 cm can be handled
  2. Mixing time is less than 2 minutes
  3. Discharge is through buckets onto the platform
  4. For large-size mixers, the mixing time should be slightly increased if handling more than 800 litres of the mix

Which of the above statements are correct?

Detailed Solution for UPSSSC JE Civil Mock Test - 1 - Question 2

The characteristics of non-tilting type drum mixer:

  1. The drum is cylindrical and always rotates about a horizontal axis .
  2. They do not tilt.
  3. The concrete is obtained by either inserting a chute/bucket into the drum or by reversing the dire­ction of rotation of the drum.
  4. Due to the slow rate of discharge of concrete, it may segregate.
  5. Large size aggregates up to 80 mm can be used.

Mixing time: As per IS 456:2000, Cl 10.3.1, mixing time should be at least 2 minutes. (Statement II –False)
As per IS 4925, mixing time directly proportional to the capacity of concrete mixer i.e. on increasing its capacity, mixing time will be increased. (Statement IV may be true).

UPSSSC JE Civil Mock Test - 1 - Question 3

For a saturated clay soil Skempton's pore pressure parameter 'B' is

Detailed Solution for UPSSSC JE Civil Mock Test - 1 - Question 3

Pore Pressure Parameters:

(i) Sometimes it is not possible to determine pore pressure practically, then the theoretical approach given by skempton can be adopted.

(ii) Pore pressure parameters A and B are empirical coefficients that are used to express the response of pore pressure to changes in vertical pressure and lateral pressure under "undrained condition".

Parameter B:

(i) This parameter is defined under cell pressure stage and represents the ratio of change in pore pressure to the change in cell pressure.

B = ΔUc/Δσc

(ii) B varies from o to 1 depending on the degree of saturation. "B" is zero for dry soil and is equals unity for fully saturated soil.

Parameter A:

(i) This parameter is valid in deviator stage and is defined in terms of another parameter A̅, such that

A̅ = A.B

(ii) The parameter A̅ represents the ratio of change in pore pressure to change in deviator stress during shear stage.

(iii) The parameter depends upon strain in soil, degree of saturation, over consolidated ratio, stratification of soil, sample disturbance, etc. Its value may be as low as -0.5 for over consolidated soil with high O.C.R. to as high as 3 or loose saturated sand.

UPSSSC JE Civil Mock Test - 1 - Question 4

Match List-I with List-II and select the most appropriate answer using the codes given below the list:

Detailed Solution for UPSSSC JE Civil Mock Test - 1 - Question 4

Concept:
(i) Average Daily Demand (q)
Average daily demand = (per capita average consumption in litre/person/day) × population
(ii) Maximum Daily Demand
Maximum daily demand = 1.8 × Avg. daily demand = 1.8 × q
(iii) Maximum Hourly Demand:
Maximum hourly demand of the maximum day i.e. Peak demand
= 1.5 × Avg hourly demand of the maximum day = 2.7 × annual avg hourly demand
(iv) Monthly peak demand = 1.28 × monthly average demand
(v) Yearly peak = 1.0 × yearly average demandOther Related Points The percentage ratio of maximum demand to the average demand can be computed using Good Rich equation;
p = 180 × t-0.10

where
t = time(days)

Fluctuation in hourly demand is also termed as Peak factor which depends upon the population as follows;

UPSSSC JE Civil Mock Test - 1 - Question 5
During a sampling operation, the drive sampler is advanced 600 mm and the length of the sample recovered is 525 mm. What is the recovery ratio of the sample?
Detailed Solution for UPSSSC JE Civil Mock Test - 1 - Question 5

Concept:

Recovery Ratio

The sample disturbance depends on the design features of a sampler (cutting edge, inside wall friction, and non-return valve) and the method of sampling.

Disturbance can be measured in terms of recovery ratio, which is

If, Lr = 1, Good recovery

Lr > 1, Sample of swelled

Lr < 1, Sample of shrunk or compressed

Calculation:

Given data,

the length of the sample recovered = 525 mm

Penetration length of sample = 600 mm

Hence, the recovery ratio is

Lr = 87.5 % OR 0.875

UPSSSC JE Civil Mock Test - 1 - Question 6

A rectangular beam 60 mm wide and 150mm deep is simply supported over a span of 4 m. If the beam is subjected to a central point load of 10 kN, find the maximum bending stress induced in the beam section.

Detailed Solution for UPSSSC JE Civil Mock Test - 1 - Question 6

Concept:
The Bending equation:

Where M= bending moment, I= moment of inertia, E= modulus of elasticity, Ris radius of curvature, = bending stress, y is the distance of the neutral axis.
Calculation:
Given: b = 60 mm, d = 150mm, l = 4 m, y = d/2 = 75 mm
Bending moment is
M = wl/4

M = 10 kNm
The moment of inertia is
I = bd3/12 = = 16875000 mm4
The maximum bending stress induced in the beam section


UPSSSC JE Civil Mock Test - 1 - Question 7

Consider the following statements:

  1. Rich mixes are less prone to bleeding than lean ones
  2. Bleeding can be reduced by increasing the fineness of cement

Which of the above statements is /are correct?

Detailed Solution for UPSSSC JE Civil Mock Test - 1 - Question 7
  • Lean concrete contains less cement and more water as compared to rich mix concrete.
  • In lean concrete, Water is pushed upwards because it has low specific gravity compared to other materials in concrete. Hence lean mix concrete is more prone to bleeding than rich mix concrete.
  • Rich mix has sufficient cement aggregates, and less water as compared to lean mix hence it will have a proper lubrication, thus increasing workability.

UPSSSC JE Civil Mock Test - 1 - Question 8

In laboratory compaction tests, the optimum moisture content of soil decreases

Detailed Solution for UPSSSC JE Civil Mock Test - 1 - Question 8

Proctor test :

  • A definite relationship is established between the degree of dry density and soil moisture content
  • The competitive effort is a moisture content of mechanical energy applied to the soil mass.
  • OMC is the moisture content at which a particular soil attains maximum dry density (MDD)
  • Maximum dry unit weight obtained is a function of compactive effort and methods of compaction for a particular type of soil
  • On increasing compactive effort, the curve shifts backward and upwards and OMC decreases, and MDD increases as shown by the graph below

UPSSSC JE Civil Mock Test - 1 - Question 9
A soil sample has a porosity of 40 percent. The specific gravity of solids is 2.70. What is voids ratio ?
Detailed Solution for UPSSSC JE Civil Mock Test - 1 - Question 9

Concept:

Void ratio (e):

It is defined as the ratio of the volume of voids to the volume of solids.

Porosity(n):

It is defined as the ratio of the volume of voids to the total volume of the soil.

Calculation:

Relationship between void ratio (e)and porosity(n) is given by :

e = 0.667

UPSSSC JE Civil Mock Test - 1 - Question 10

The omitting error of a line lies in the South-West quadrant having a length ‘l’ and reduced bearing θ. The latitude ‘L’ and departure ‘D’ are computed by:

Detailed Solution for UPSSSC JE Civil Mock Test - 1 - Question 10

Concept:
For any segment AB of length L and θ with North,


Latitude = L cos θ & Departure = L sin θ
In whole circle bearing, Bearings are taken w.r.t North Direction
Calculation: For example let us assume an angle of 210° which will be in southwest direction as the question is saying

Let length of the line PQ is ‘L’
Latitude of line PQ = Lcos 210° = -0.867l and for any angle θ in the southwest direction it will be, - L × Cosθ
Department of line PQ = Lsin 210° = -0.5l and for any angle θ in the southwest direction it will be, - L × Sinθ

UPSSSC JE Civil Mock Test - 1 - Question 11

Which of the following is an example of a hypabyssal rock?

Detailed Solution for UPSSSC JE Civil Mock Test - 1 - Question 11

The correct answer is D. Dolerite

Hypabyssal rocks are formed from the rapid cooling of magma at relatively shallow depths beneath the Earth's surface, resulting in a fine-grained crystalline structure. Dolerite is a prime example of such a rock, distinguishing it from plutonic and volcanic rocks, which cool at greater depths or on the surface, respectively.

Details on Other Options:

  • A: This option is incorrect because granite is a plutonic rock formed from the slow cooling of magma at significant depths, resulting in a coarse-grained structure.
  • B: This option is incorrect as gabbro is also a plutonic rock, similar to granite, with a coarse texture due to its formation at deep levels in the Earth.
  • C: This option is incorrect since syenite is classified as a plutonic rock, characterized by its formation from slow-cooling magma at depth, similar to granite.

Conclusion:

The D, as the most accurate and relevant option, stands out for its representation of hypabyssal rock formation, clearly differentiating it from the other options. Thus, Dolerite is the correct choice.

UPSSSC JE Civil Mock Test - 1 - Question 12

What is the correct sequence of preparing an estimate?

  1. Detailed estimate
  2. Approximate estimate
  3. Supplementary estimate
  4. Revised estimate
Detailed Solution for UPSSSC JE Civil Mock Test - 1 - Question 12



The correct sequence of preparing an estimate is Approximate estimate, Detailed estimate, Supplementary estimate, and Revised estimate.

UPSSSC JE Civil Mock Test - 1 - Question 13

If the odour intensity or pO value is 4, then what is the interpretation?

Detailed Solution for UPSSSC JE Civil Mock Test - 1 - Question 13

Concept:
PO value:

  • The Odour in water changes with temperature. the water Odour is tested at 20oC to 25oC
  • The apparatus used to measure Odour in water is known as osmoscope.
  • A commercial osmoscope is graduated with a PO value from 0 - 5.

 PO value is 4 then the interpretation is a very distinct Odour.

UPSSSC JE Civil Mock Test - 1 - Question 14

Consider the following statements:

A. Stores and Balcony are included in the calculation of the carpet area.

B. Staircases within the Property unit are included in the carpet area

C. Roof Terrace are excluded from the carpet area.

The correct statements are:

Detailed Solution for UPSSSC JE Civil Mock Test - 1 - Question 14

Explanation:

It is the distance between the inner walls. The carpet area would include the areas of the bedroom, living room, kitchen, bathrooms, balconies & staircases within the house/flat. It does not include the external and internal walls, terraces, common areas, lifts, corridors, utility ducts, etc.

The carpet area shall be the area shall be excluding the area of the following portion:

Verandah, Corridor and passage, Entrance hall and porch, Staircase (outside the house) and stair-cover (Mumty), Shaft and machine room for lift, Bathroom and lavatory, Kitchen and pantry, Store, and Canteen.

UPSSSC JE Civil Mock Test - 1 - Question 15

What is the principle behind the two theodolite methods used to establish a simple circular curve?

Detailed Solution for UPSSSC JE Civil Mock Test - 1 - Question 15

Explanation:
The various methods used for setting curves are as follows:

UPSSSC JE Civil Mock Test - 1 - Question 16

Propped cantilever ABCD is loaded as shown in figure. If it is of uniform cross-section, the collapse load of the beam will be nearly

Detailed Solution for UPSSSC JE Civil Mock Test - 1 - Question 16

There are two possibilities of hinge Formation.
Case 1: Plastic hinge at C alone
Case 2: Plastic hinges at A & V
Case 1 : Plastic hinge at C alone

(w/8) × (ℓ/3 × θ) = Mp × θ ⇒ W = (24 Mp)/ℓ
Case 2 : Plastic hinges at A & V


w × (1/2 × θ) - w/8 × (ℓ/3 × θ) = Mp × θ + Mp × 2θ
w = 6.5Mp/ℓ
∴ w = minimum { Case 1 Case 2 } = (6.5 Mp)/ℓ

UPSSSC JE Civil Mock Test - 1 - Question 17

The total station is the one which is the combination of:

Detailed Solution for UPSSSC JE Civil Mock Test - 1 - Question 17

The correct answer is D. Electronic Theodolite and EDM

A total station integrates an electronic theodolite with an electronic distance meter (EDM), enabling precise determination of a reflector's coordinates by aligning the instrument's crosshairs on it. The device utilizes a microprocessor to manage recordings, readings, and calculations, enhancing accuracy and efficiency in surveying tasks.

Details on Other Options:

  • A: This option is incorrect as a combination of a plane table and a dumpy level does not provide the electronic capabilities necessary for precise distance measurement and angular calculations.
  • B: This option is incorrect because a plane table and theodolite are separate instruments and do not integrate the electronic distance measurement capabilities found in a total station.
  • C: This option is incorrect since a combination of EDM and a dumpy level lacks the angular measurement functionality provided by an electronic theodolite, which is essential for a total station.

Conclusion:

Option: D, as the most accurate and relevant option, stands out for its comprehensive integration of electronic measurement tools, clearly differentiating it from the other options. Thus, Electronic Theodolite and EDM is the correct choice.

UPSSSC JE Civil Mock Test - 1 - Question 18
As per IS: 800-1984 specification, how to calculate the gross diameter of a rivet? (If diameter of the rivet is less than or equal to 25 mm.)
Detailed Solution for UPSSSC JE Civil Mock Test - 1 - Question 18

As per IS 800-1984 clause 3.6.1.1,

In making a deduction for rivets less than or equal to 25 mm in diameter, the diameter of the hole shall be assumed to be 1.5 mm in excess of the nominal diameter of the rivet unless specified otherwise. If the diameter of the rivet is greater than 25 mm, the diameter of the hole shall be assumed to be 2.0 mm in excess of the nominal diameter of the rivet unless specified otherwise.

Important Points

In making a deduction for bolts, the diameter of the hole shall be assumed to be 1.5 mm in excess of the nominal diameter of the bolt, unless otherwise specified.

For countersunk rivets or bolts, the appropriate addition shall be made to the diameter of the hole.

UPSSSC JE Civil Mock Test - 1 - Question 19

What is the name of following equipment?

Detailed Solution for UPSSSC JE Civil Mock Test - 1 - Question 19

Explanation:
Wheelbarrow, truck mixer, dumper truck, belt conveyor, pipeline etc. are the various ways concrete is transported to the construction site.
A wheelbarrow is employed for hauling concrete in the longer distance in case of concrete road construction. If the distance is long or ground is rough it is likely that the concrete gets segregated due to vibration. To avoid this, wheelbarrows are provided with the pneumatic wheel, which helps in absorbing these vibrations.|

UPSSSC JE Civil Mock Test - 1 - Question 20

The hydraulic efficiency of an impulse turbine is maximum, when the ratio of the velocity at wheel to jet velocity is

Detailed Solution for UPSSSC JE Civil Mock Test - 1 - Question 20

Concept:
Blade speed ratio is the ratio of blade velocity to the inlet jet velocity. It is denoted by ρ = u/v
For Pelton turbine (Impulse turbine) maximum blade efficiency is ηb = cos2a  and Speed ratio (ρ) = cos a/2 
For maximum efficiency,
ηb = 1
1 =  cos2a ⇒ a = 0
The Speed ratio
u/v = (cos α/2) = (cos 0/2) = 1/2
Therefore, u = v/2
where, u = blade velocity, v = inlet absolute fluid flow velocity

Important Points

UPSSSC JE Civil Mock Test - 1 - Question 21

Which statements are true for Quartz (a silicate-based mineral)?

I. Its chemical formula is SiO2.

II. Its hardness on the Mohs scale is 7.

III. Insoluble in hydrofluoric acid.

IV. Specific gravity is 2.6 - 2.65.

Detailed Solution for UPSSSC JE Civil Mock Test - 1 - Question 21

Explanation:

Quartz: It is one of the rock-forming siliceous minerals.

Note: Quartz is soluble in hydrofluoric acid.
So, statement III is incorrect.
Therefore, correct statements are I, II ad IV.

UPSSSC JE Civil Mock Test - 1 - Question 22

Due to each end contraction, the discharge of rectangular sharp-crested weir is reduced by

Detailed Solution for UPSSSC JE Civil Mock Test - 1 - Question 22

Concept:
Classification of weir based on the shape of crest:
1. Sharp crested weir:
The crest shape of the weir is very sharp. It is the most widely used weir.

2. Broad-crested weir:
These are constructed only in a rectangular shape and are suitable for larger flows.

3. Narrow-crested weir:
It is similar to a rectangular weir with a narrow-shaped crest at the top. The discharge over the narrow crested weir is similar to the discharge over the rectangular weir.
If 2b is less than H, the weir is called a narrow-crested weir.
Where b = Width of the crest of the weir and H = Height of water above the weir crest
4. Ogee-shaped weir:
Generally, ogee shaped weirs are provided for the spillway of a storage dam. The crest of the ogee weir is slightly risen and falls into parabolic form.

Explanation:
Weir is a physical structure of masonry constructed across the channel width to calculate the discharge of the channel section.
Weir can be rectangular or triangular.
For rectangular weir:

The discharge through the weir is given by:

When water flows over a rectangular sharp-crested weir, the velocity of the fluid increases as it approaches the weir's sharp crest. This results in a contraction of the flow at the crest, leading to a reduction in discharge.
The amount of discharge reduction due to end contraction can be determined using the following equation:
Qc = Cd x L x (Hc - K)^1.5 Qc = Cd x L x (Hc - K)3/2 
Where,
Qc = discharge coefficient with end contraction
Cd = discharge coefficient without end contraction
L = length of the weir
Hc = height of water above the weir crest (measured at the downstream side)
K = end contraction coefficient (typically ranges from 0.1 to 0.2)
The discharge coefficient without end contraction (Cd) for a rectangular sharp-crested weir is 0.62. If we assume an end contraction coefficient (K) of 0.2, then the discharge coefficient with end contraction (Qc) can be calculated as follows:
Qc = 0.62 × L × (Hc - 0.2Hc)3/2
Qc = 0.62 × L × (0.8Hc)3/2
Qc = 0.62 × L × 0.794 × Hc3/2
The discharge reduction due to end contraction can be calculated as follows:
Discharge reduction = (Cd - Qc) / Cd x 100%
Discharge reduction = (0.62 - 0.62 x 0.794 x Hc3/2 ) / 0.62 x 100%
Discharge reduction = 0.1 x Hc3/2 x 100%
Therefore, the discharge reduction due to end contraction is approximately 10% for a rectangular sharp-crested weir.

UPSSSC JE Civil Mock Test - 1 - Question 23
Maximum value of 'throw of switch' for Broad gauge track is:
Detailed Solution for UPSSSC JE Civil Mock Test - 1 - Question 23

Explanation

Throw of switch:

(i) It is the distance between the running face of the stock rail and the toe of the tongue rail.

(ii) Its limiting values are 95-115 mm for BG routes and 89-100 mm for MG routes.

Other Related Points

Some other important terms are as follows:

Switch angle: It is the angle between the gauge faces of the stock rail and the tongue angle.

Heel clearance: It is the distance between the running edge of the stock rail and the switch rail at the switch heel.

Its recommended value on BG is 133 mm, 121 mm to 117 mm for MG and 98 mm for NG track respectively.

Check rails: These are the rails provided to guide the wheel flanges, while the opposite wheel is jumping the gap.

Flange way clearance: It is the distance between the adjacent faces of the stock rail and the check rail. Its minimum value is 60 mm.

UPSSSC JE Civil Mock Test - 1 - Question 24
Empirical relationship between tensile strength and compressive strength of concrete is given by:
Detailed Solution for UPSSSC JE Civil Mock Test - 1 - Question 24

Concept-

The theoretical compressive strength of concrete is eight times larger than its tensile strength. This implies a fixed relation between the compressive and tensile strength of concrete. In fact, there is a close relation but not a direct proportionality. The ratio of tensile to compressive strength is lower for higher compressive strengths.

One of the most common relations is given by the following relation:

Tensile strength = K (compressive strength)n.

The value of K may be taken as 6.2 for gravel and 10.4 for crushed aggregate. The average value for both may be taken as 8.3 and the value of n may vary from 0.5 to 0.75.

The I.S. 456-2000 has suggested the following relation between the compressive strength and flexural strength of concrete.

Flexural strength = 0.7 x √fck

Where fck is the compressive strength cylinder of concrete in MPa (N/mm2).

UPSSSC JE Civil Mock Test - 1 - Question 25

1 Kilobyte is equal to ______ bytes

Detailed Solution for UPSSSC JE Civil Mock Test - 1 - Question 25

The correct answer is 1024.Explanation

  • The memory of a computer is usually measured in Bytes.
    • The memory of a computer is similar to the human brain.
    • It is used to store data and instructions.
    • One kilobyte is equal to 1024 bytes.
    • The half byte is known as a nibble.
    • The smallest unit of memory is called a bit.
    • Bit stands for binary digit.
    • The storage capacity of a hard disk is measured in Megabytes, Gigabytes, and Terabytes.

Other Related Points Memory Measurements

UPSSSC JE Civil Mock Test - 1 - Question 26
In Microsoft Excel, if a column is too narrow for a long number then Excel will automatically convert the number to a series of ________ signs.
Detailed Solution for UPSSSC JE Civil Mock Test - 1 - Question 26

The correct answer is #.

Explanation

  • In Microsoft Excel, if a column is too narrow to display the full width of a long number, Excel will display a series of "#" signs (hash symbols) instead of the actual number.
  • This is known as the "number truncation" or "number display overflow" feature.

Other Related Points

  • By default, Excel adjusts the column width to fit the contents of the cells.
  • However, if the column width is narrower than the number being displayed, Excel replaces the visible digits with "#" signs to indicate that the full number cannot be shown within the available space.
  • To resolve this,
    • Adjust column width: Hover your cursor between the column headers, and when it changes to a double-headed arrow, double-click. Excel will automatically adjust the column width to fit the widest number in that column.
    • Manually adjust column width: Click and drag the column header boundary to the right to increase the column width manually until the full number is visible.
UPSSSC JE Civil Mock Test - 1 - Question 27

Which of the following is NOT a network topology?

Detailed Solution for UPSSSC JE Civil Mock Test - 1 - Question 27

The correct answer is DiskExplanation Network topology is the schematic description of a network arrangement, connecting various nodes (sender and receiver) through lines of connection. Various types of Network Topologies are described below:

  • Ring: In a ring topology, each node is connected to exactly two other nodes, forming a circular pathway for signals. This topology ensures that each piece of data can reach any node, but a break in the ring can disrupt the entire network.
  • Mesh: In a mesh network topology, every device (node) is connected to every other device in the network. This results in robustness (if one link fails, other links can continue functioning) but requires a large number of connections, making it expensive and difficult to manage.
  • Star: In a star topology, all nodes are connected to a central node – often a hub, switch, or controller. The central node typically manages the network, and each peripheral node communicates through this central one. This topology simplifies adding or removing nodes but a failure in the central node affects the entire network.
  • Bus: In a bus topology, all nodes are connected to a common backbone or bus. The bus transmits the signal from one end to the other. All nodes along the bus can receive the signal, but only the intended recipient (identified by an address in the signal) processes it. A problem in the bus can disrupt the entire network.
UPSSSC JE Civil Mock Test - 1 - Question 28

In Microsoft word, the word document orientation is set ________ by default.

Detailed Solution for UPSSSC JE Civil Mock Test - 1 - Question 28

The correct answer is Portrait.Explanation

  • Portrait and Landscape mode are two types of Page Orientation.
  • Page orientation is the direction in which a document is displayed or printed.
  • The two basic types of page orientation are portrait (vertical) and landscape (horizontal).
  • Most monitors have a landscape display, while most documents are printed in portrait mode.
  • Before printing a document, you may be able to change the page orientation by selecting "Page Setup..." from the program's File menu.
    • The default orientation is typically portrait, but you can change it to landscape if you want the width to be longer than the height.
    • This may be useful for printing signs, cards, or other documents that require a wide display.

Other Related Points

  • Page size
    • By default, the page size of a new document is 8.5 inches by 11 inches.
    • Depending on your project, you may need to adjust your document's page size.
    • It's important to note that before modifying the default page size, you should check to see which page sizes your printer can accommodate.
  • Page margins
    • A margin is a space between the text and the edge of your document.
    • By default, a new document's margins are set to Normal, which means it has a one-inch space between the text and each edge. Depending on your needs, Word allows you to change your document's margin size.
  • ​​
UPSSSC JE Civil Mock Test - 1 - Question 29
Slow left-arm spinner Kuldeep Yadav belongs to which district of Uttar Pradesh?
Detailed Solution for UPSSSC JE Civil Mock Test - 1 - Question 29

The correct answer is A. Kanpur

Kuldeep Yadav, born on December 14, 1994, is an Indian international cricketer hailing from the Kanpur district of Uttar Pradesh. He represents India in international cricket and plays for Uttar Pradesh at the domestic level, as well as for the Kolkata Knight Riders in the Indian Premier League (IPL). Notably, he is the second Indian, after Bhuvneshwar Kumar, and the third spinner globally, following Imran Tahir and Ajantha Mendis, to achieve five-wicket hauls across all three formats of the game. On September 21, 2017, he made history by becoming the third Indian bowler to take a hat-trick in One Day Internationals (ODIs), joining the ranks of Chetan Sharma and Kapil Dev.

Details on Other Options:

  • B: This option is incorrect as Agra is not the district from which Kuldeep Yadav originates; he is specifically from Kanpur.
  • C: This option is incorrect because Ghaziabad is not the correct district associated with Kuldeep Yadav's background.
  • D: This option is incorrect since Meerut is not the district linked to Kuldeep Yadav's roots, which are firmly in Kanpur.

Conclusion:

Option: A, as the most accurate and relevant option, stands out for its representation of Kuldeep Yadav's origin, clearly differentiating it from the other options. Thus, Kanpur is the correct choice.

UPSSSC JE Civil Mock Test - 1 - Question 30
In which district of Uttar Pradesh did the Chauri Chaura incident take place?
Detailed Solution for UPSSSC JE Civil Mock Test - 1 - Question 30

The correct answer is D. Gorakhpur

The Chauri Chaura incident took place on February 4, 1922, in the Gorakhpur district of Uttar Pradesh, during British rule. This event involved a large group of protesters from the Non-Cooperation Movement who clashed with police. In retaliation, the demonstrators attacked and set fire to a police station, resulting in the deaths of all officers inside. The violent confrontation led to the deaths of three civilians and 22 policemen. Following this incident, Mahatma Gandhi, who opposed violence, suspended the Non-Cooperation Movement on February 12, 1922, at a national level.

Details on Other Options:

  • A: This option is incorrect because the Chauri Chaura incident did not occur in Lucknow, which is a different district known for its own historical significance.
  • B: This option is incorrect as Agra, while historically important, is not the site of the Chauri Chaura incident.
  • C: This option is incorrect because Moradabad is not associated with the Chauri Chaura incident, which specifically took place in Gorakhpur.

Conclusion:

Option: D, as the most accurate and relevant option, stands out for its direct association with the Chauri Chaura incident, clearly differentiating it from the other options. Thus, Gorakhpur is the correct choice.

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