UPPSC AE Civil Paper 1 Mock Test - 2 - Civil Engineering (CE) MCQ

शब्‍द "तिरस्‍कार" का सही विशेषण रूप तिरस्कृत है।

मुख्य बिंदु
तिरस्कृत -

• यह शब्द "तिरस्कार" का विशेषण रूप होता है।
• यह शब्द संस्कृत का है और इसका अर्थ होता है "निरादरित" या "उपेक्षित"।
• यह शब्द व्यक्ति या वस्तु को वर्णन करने के लिए उपयोग हो सकता है, जो तिरस्कार की स्थिति में हो।

अन्य संबंधित बिंदु

• विशेषण वह शब्द होता है जो किसी संज्ञा या सर्वनाम की विशेषता बताता है।
• विशेषण के द्वारा हम किसी संज्ञा की गुणवत्ता, रंग, आकार, मात्रा, संख्या, स्थिति, आदि के बारे में जानकारी प्राप्त करते हैं।

• 'वर्त्स्य' शब्द व्याकरणिक और उच्चारण की दृष्टि से शुद्ध है।
• 'वर्त्स्य' शब्द ही सार्थक है।
• 'वर्त्स्य' शब्द का अर्थ:- जिसका उच्चारण जीभ को वर्त्स पर रख कर हो ; मसूढ़े से संबंधित।

अन्य संबंधित बिंदु

• 'र' वर्त्स्य, लुंठ‍ित, सघोष, अल्‍पप्राण व्‍यंजन है।
• वर्त्स्य व्यंजन - ऐसा व्यंजन होता है जिसे उच्चारित करने के लिए जिह्वा को ऊपर के वर्त्स्य कटक से छुआ जाता है या पास लाया जाता है। इनमें 'ट', 'ल', 'ड' और 'स' शामिल हैं।
• अल्पप्राण - जिन व्यंजनों का उच्चारण करते समय प्राण वायु महाप्राण की तुलना में कम निकले या कम प्रयोग हो,अल्पप्राण कहलाते हैं। जैसे- क,ग,ङ,च,ज,ञ,ट,ड,ण,त,द,न,प,ब,म,य,र,ल,व ।
• सघोष - इन व्यंजनों का उच्चारण करते समय स्वर तंत्रियों में अधिक कंपन हो,घोष या सघोष वर्ण कहलाते हैं।जैसे - ख,ग,ङ,ज,झ,ञ,ड,ढ,ण,द,ध,न,ब,भ,म,य,र,ल,व,ह
• लुंठित - यह ऐसा व्यंजन वर्ण होता है जिसमें मुँह के एक सक्रीय उच्चारण स्थान और किसी अन्य स्थिर उच्चारण स्थान के बीच कंपकंपी या थरथराहट कर के व्यंजन की ध्वनि पैदा की जाती है। हिन्दी में इसका सबसे बड़ा उदाहरण "र" की ध्वनि है जिसमें जिह्वा का सबसे आगे का भाग कंपकंपाया या थरथराया जाता है।

सही उत्तर 'जोगन' है।

• 'जोगन' शब्द में महाप्राण व्यंजनों का प्रयोग नहीं हुआ है।
• जोगन' शब्द में अल्पप्राण व्यंजनों का प्रयोग हुआ है।
• अल्पप्राण व्यंजन वह व्यंजन होतें हैं जिन्हें बहुत कम वायु-प्रवाह से बोला जाता है जैसे कि 'क', 'ग', 'ज' और 'प'।
• इनकी संख्या 15 होती है।

व्याख्या
अन्य विकल्पों का विश्लेषण:

• खीझ, झूठ और घाघ शब्दों में महाप्राण व्यंजनों का प्रयोग हुआ है।

• भाषा विज्ञान में महाप्राण व्यंजन वह व्यंजन होतें हैं जिन्हें मुख से वायु-प्रवाह के साथ बोला जाता है, जैसे की 'ख', 'घ', 'झ' और 'फ'।

अन्य संबंधित बिंदु

सही उत्तर "हाथ मलना - हाथ साफ करना" है।

• मुहावरे के अर्थ, 'हाथ साफ करना' और वाक्य प्रयोग 'कड़ाके की सर्दी में वह अपने हाथ मल रहा था' की दृष्टि से, 'हाथ मलना' मुहावरा गलत है।
• हाथ मलना' मुहावरा का सही अर्थ: 'पछताना' है।
• वाक्य प्रयोग-- सुरेन्द्र अपने मित्रों के साथ पुरी नहीं जा सकता, अतः वह हाथ मलने लगा।

व्याख्या

• मुहावरा परिभाषा: मुहावरा का शाब्दिक अर्थ ‘अभ्यास’ है। मुहावरा शब्द अरबी भाषा का शब्द है। हिन्दी में ऐसे वाक्यांशों को मुहावरा कहा जाता है, जो अपने साधारण अर्थ को छोडकर विशेष अर्थ को व्यक्त करते हैं।

प्रमुख मुहावरे:

• काँटा निकलना :- बढ़ा दूर होना।
• चींटी के पर निकलना या जमना :- विनाश के लक्षण प्रकट होना।
• छक्के छूटना :- बुरी तरह पराजित होना।
• जल भुनकर पागल हो जाना :- क्रोध से पागल होना।
• जहर उगलना :- अपमानजनक बातें कहना।
• जीती मक्खी निगलना :- जानबूझकर अशोभन या अभद्र कार्य करना।
• तोते की तरह आँखें फेरना :- बेमुरव्वत होना।
• तीन तेरह होना :- तितर बितर होना।
• बट्टा लगाना :- कलंक लगाना।
• बाँछे खिलना :- अत्यधिक प्रसन्न होना।
• सवा सोलह आने सही :- पूरी तरह सही होना।
• हवा का रुख देखना :- समय की गति पहचान कर काम करना।

सही उत्तर 'प्रव्रजित' है।
'वह (व्यक्ति) जिसने संन्यास ग्रहण किया हो' वाक्यांश के लिए सार्थक शब्द 'प्रव्रजित' है।
व्याख्या  
प्रशमित, प्रवजित और प्रव्राज किसी भी वाक्यांश का सार्थक शब्द नहीं है।
अन्य सार्थक शब्द:

अन्य संबंधित बिंदु
वाक्यांश- भाषा को सुंदर, आकर्षक और प्रभावशाली बनाने के लिए अनेक शब्दों के स्थान पर एक शब्द का प्रयोग किया जाता है तो वह वाक्यांश के लिए एक शब्द कहलाता है।

सही उत्तर 'अधित्यका' है।
'पर्वत के ऊपर की समतल भूमि' वाक्यांश के लिए सार्थक शब्द 'अधित्यका' है।
व्याख्या   अन्य विकल्प:

अन्य संबंधित बिंदु वाक्यांश- भाषा को सुंदर, आकर्षक और प्रभावशाली बनाने के लिए अनेक शब्दों के स्थान पर एक शब्द का प्रयोग किया जाता है तो वह वाक्यांश के लिए एक शब्द कहलाता है।

Concept:

Scaffolding:

Scaffolding is a temporary rigid structure with still, bamboo, or timber platforms raised to increase the building height. It enables the mason to work at different stages of a building and takes up the materials for immediate use at various heights. Different types of scaffolding are used depending upon the types of construction.

Types of Scaffolding:

Different types of scaffolding are described below:

1. Brick layer's scaffolding
2. Mason's scaffolding
3. Steel or tubular scaffolding
4. Needle scaffolding

Brick Layer's Scaffolding:

• This type of scaffolding consists of standards firmly secured on the grounds at 2.4 to 3 m. The standard is connected to each other by ledgers at every rise of 120 to 150 cm. They are provided on the building site of the standards and are secured in position by rope lashing. Putlogs are lashed on the ledgers at one end and into the holes in the wall at the other end.

Mason's Scaffolding:

• Since it is difficult to leave holes in the stone masonry to provide a bearing for the putlogs, in mason's scaffolding two frames of standards, at a distance of 1.5 m from the first one. Thus the mason's scaffolding is entirely independent of the stone wall.

Steel Scaffolding:

• The construction of steel scaffolding is similar to that of timber scaffolding. In this case, wooden members are replaced by 38 mm to 64 mm diameter steel tubes and instead of rope lashings, special types of steel couplets or fittings are used for connecting different members.
• The steel tubes used for scaffolding for normal building construction work are heavy class and of diameter varying from 40 to 60 mm. In this type of scaffolding, the vertical tubes called uprights or standards are spaced 2.5 to 3 m apart. Each standard is welded to a base plate, square or circular in plan. The base plate has holes so that it can be spiked to a timber or concrete base, thus forming a rigid foundation for the scaffolding.

Needle Scaffolding:

• When scaffolding is to provide for a building on the side of a busy street where the construction of ordinary scaffolding will obstruct the traffic on road, a needle scaffold is used.
• The scaffold is erected from window corners of string courses by means of projecting needles. A needle is a timber beam that cantilevers out through the holes cut into the wall. From inside the needles are supported on sole pieces and are prevented from lifting up by vertical struts wedged between the needles and the headpieces. The projected end of the needle is supported by an inclined strut that rests on the window sill.

Concept:
Fineness: It is property of cement that indicates the particle size of cement and specific surface area as it directly affects the heat of hydration.
Fineness of cement is obtained by two methods:
(i) Nurse and Blaine’s method
(ii) Wagner Turbidimeter Test
Fineness of cement is determined by Nurse and Blaine’s method using an apparatus developed by Lea and Nurse.

Apparatus consists of a permeability test cell where cement is placed and air pressure is applied. The flowmeter is installed to determine the quantity of air passing per second through its capillary tube per unit difference of pressure, and manometer is to measure the air pressure.
Thus Nurse and Blaine’s method is based on permeability and not hydration process.
Initial and Final Setting time:
When water is added to cement and mixed the chemical reaction starts and the cement paste remains plastic for a short time. During this time it is possible to mix the paste and this duration is called Initial setting time of cement.
As time passes, the reaction continues and cement begins to harden. The time elapsed between the time of mixing to hardening is called Final setting time of cement.
Determination of initial and final setting time is done through Vicat’s test.
It measures the penetration resistance over time offered due to continued hydration.
Initial setting time test is done using 1 mm² needle and Final setting time is found using 5 mm diameter annular collar. The consistency of cement paste used is 0.85 times the standard consistency.
Hence statement I is false and statement II is true.

Let's analyze the given statements with respect to the characteristics of bending moment diagrams (BMD) for beams subjected to uniformly distributed loads (UDL).
Statement A:
If a beam is subjected to uniformly distributed load (UDL) throughout its span, the shape of BMD will be a cubic parabolic curve.
For a simply supported beam with a uniformly distributed load (UDL), the bending moment diagram (BMD) is a quadratic (parabolic) curve, not a cubic parabolic curve. The equation for the bending moment in such a case is:

where w is the load per unit length, Lis the length of the beam, and x is the position along the beam.
Therefore, Statement A is incorrect.
Statement B:
If a cantilever beam is subjected to a uniformly distributed load (UDL) throughout its span the maximum value of bending moment will be at its mid-span.
For a cantilever beam subjected to a uniformly distributed load (UDL) throughout its span, the maximum bending moment occurs at the fixed end, not at the mid-span. The equation for the bending moment in this case is:

where ww is the load per unit length, L is the length of the beam, and x is the position along the beam.
The maximum bending moment occurs at x = 0 (the fixed end) and is equal to:

Therefore, Statement B is incorrect.
Conclusion:
Both statements A and B are incorrect.
Correct Answer:
4) Both the statements are incorrect.

Explanation:

Sieve Analysis of Soil

• Sieve analysis is a laboratory procedure used to determine the particle size distribution of soil by passing the soil through a set of sieves with progressively smaller openings.
• The effective size (D₁₀) of soil is the diameter of the particle below which 10% of the soil's particles by weight are finer.

Calculation:

Given Data

• Total weight of soil taken: 500 g
• Mass of soil retained over 4.75 mm sieve: 100 g
• Mass of soil retained over 2 mm sieve: 150 g
• Mass of soil retained over 425-micron sieve: 200 g

First, we need to determine the cumulative percentage of soil passing each sieve. The total mass retained and the mass passing through each sieve are calculated as follows:

• Mass retained over 4.75 mm sieve: 100 g
• Mass passing 4.75 mm sieve = 500 g - 100 g = 400 g
• Cumulative percentage passing 4.75 mm sieve = (400 g / 500 g) ×100% = 80%
• Mass retained over 2 mm sieve: 150 g
• Mass passing 2 mm sieve = 400 g - 150 g = 250 g
• Cumulative percentage passing 2 mm sieve = (250 g / 500 g) ×100% = 50%
• Mass retained over 425-micron sieve: 200 g
• Mass passing 425-micron sieve = 250 g - 200 g = 50 g
• Cumulative percentage passing 425-micron sieve = (50 g / 500 g) ×100% = 10%

The effective size (D₁₀) is the diameter of the particle below which 10% of the soil's particles by weight are finer. From the calculations above, we can see that 10% of the soil's particles by weight are finer than the 425-micron sieve.

Types of joints in concrete
Contraction joint

• Contraction joints are provided along the transverse direction to take care of the contraction of the concrete slab due to its natural shrinkage.

Construction joint

• Construction joints are provided whenever the construction work stops temporarily. The joint direction could be either along the transverse or longitudinal direction, their position should be pre-planned before concreting is started.

Expansion joint

• Expansion joints are provided along the transverse direction to allow movement (expansion/ contraction) of the concrete slab due to temperature and subgrade moisture variation. They are provided when there is an abrupt changes in dimensions, shape, etc. occurs.

Warping joint

• Warping joints are provided along the longitudinal direction to prevent warping of the concrete slab due to temperature and subgrade moisture variation.

Classification of Construction Equipment(according to work cycle):
1. Intermittent Type:

• Bulldozers
• Scrappers
• Power Shovels
• Concrete Mixers
• Dragline

2. Continuous Flow Type:

• Air Compressors
• Belt Conveyors

3. Mixed Type:

• Motor Graders

Steel commonly used in the construction industry are:
a) Mild Steel (MS) bars conforming to IS-432, Part-I.
b) High Yield Strength Deformed (HYSD) bars.
Two types of HYSD bars are :

• i) Hot Rolled HYSD bars conforming to IS-1139 and
• ii) Cold Worked HYSD bars conforming to IS-1786.
• These are also known as Cold Twisted Deformed (CTD) bars or Tor steel. The most commonly used HYSD bars for RCC work are the CTD bars.

As per Table Number 22 of IS 456:2000, the minimum characteristic yield strength for hot rolled high yield strength deformed bar is 415 N/mm².

Explanation:
Components of Arch:
Arch Ring: It is a curved ring of masonry forming an arch.

Extrados: Outer curved surface of the arch ring is called Extrados
Intrados: The curved inner surface of the arch is known as Intrados.
Spandrel: It is an irregular triangular space enclosed by the vertical line through the springing point, horizontal line from the crown of the arch, and extrados of the arch.
Skewback: It is the portion where the curved surface meets the abutments or piers.

Important Points Some other important terms are given below:
Rise: It is the clear vertical distance between the springing line and the highest point of intrados.
Springing Line: It is the imaginary horizontal line joining the two springing points.
Voussoir: Wedge-shaped units of plain cement concrete, stone or bricks of which the arch ring is made of, is known as voussoir or arch block.

Explanation:

It refers to the suitable proportioning of length, width, and height of rooms in the building to get the maximum benefit from the minimum dimensions.

• Planning should be such that maximum benefit can be obtained from the minimum required for the functions expected to be available from the space.
• For utility a rectangular room is better than a square room of same floor area.
• Accordingly, the length and breadth ratio should be 1.2: 1 to 1.5:1 because if it is less than 1.2 then it becomes more squarish and if it is more than 1.5 then it creates a tunneling effect.

Hence, In a residential building, the ideal shape of any room is preferably good when its

Length = 1.2 to 1.5 × Breadth

Important Points

The aspect of a room

• It refers to suitable proportioning of length, width and height of rooms in the building to get the maximum benefit from the minimum dimensions.
• Aspect is concerned with the orientation of the building.
• The arrangement of doors, windows in the outside walls of the house should be in such a way that plenty of sunrays, the breeze can enter the house.

Direct Shear Test:

The direct shear test is a common laboratory test used to determine the shear strength parameters of soil, specifically the cohesion (c) and the angle of internal friction (φ). The test involves shearing a soil sample along a predetermined plane. Here’s an overview of how the test is conducted and the role of the proving ring:

• Sample Preparation: A soil sample is placed in a shear box, which is split horizontally into two halves. The sample is typically square or circular in cross-section and can be undisturbed or remolded.
• Application of Normal Load: A normal load is applied to the top of the soil sample to simulate the overburden pressure or the weight of the soil above the test sample. This load is kept constant throughout the test.
• Shearing the Sample: The lower half of the shear box is fixed, while the upper half is moved horizontally to apply shear stress to the soil sample. This movement causes the soil to shear along the plane between the two halves of the shear box.
• Measurement of Shear Load: A proving ring is used to measure the magnitude of the shear load applied to the soil sample. The proving ring is attached to the shear box apparatus and deforms elastically under load. The deformation of the proving ring is measured using a dial gauge or other displacement measuring device, and this deformation is correlated to the applied shear load using a calibration chart specific to the proving ring.
• Data Collection: During the test, both the horizontal displacement (shear displacement) and the corresponding shear load are recorded. Additionally, the vertical displacement (if any) can be measured to observe any volume changes in the soil sample.
• Analysis: The shear stress is calculated by dividing the measured shear load by the area of the shear plane. The test is repeated at different normal loads to generate a series of shear stress versus normal stress data points. These data points are used to plot a Mohr-Coulomb failure envelope, from which the shear strength parameters (cohesion and angle of internal friction) are determined.

The direct shear test provides valuable information about the shear strength of soils, which is crucial for designing foundations, retaining walls, and other geotechnical structures.

Concept:
Centre of gravity of plane figures (T-sections, I-sections, L-sections, etc.)
The plane geometrical figures (such as T-section, I-section, L-section etc.) have only areas but no mass. The centre of gravity of such figures is found out in the same way as that of solid bodies. The centre of area of such figures is known as centroid and coincides with the centre of gravity of the figure. It is a common practice to use the centre of gravity for centroid and vice versa.
Let and be the co-ordinates of the centre of gravity with respect to some axis of reference, then

and
where A1, A2, A3........ etc., are the areas into which the whole figure is divided x1, x2, x3 ..... etc., are the respective coordinates of the areas A1, A2, A3....... on X-X axis with respect to same axis of reference.
y1, y2, y3....... etc., are the respective coordinates of the areas A1, A2, A3....... on the Y-Y axis with respect to the same axis of the reference.
Calculation:
Given







Hence option (1) is correct.

Explanation:
According to IS 875 PART 1: 1987, The unit weight of some common building materials is given below:

Explanation:

Conventional concrete is modified by random dispersal of particles of short discrete fibres. These fibres may be of steel, glass, carbon, asbestos etc. Such concrete is known as fibre reinforced concrete. Asbestos cement fibres so far have proved to be commercially successful.

Other Related Points

Asbestos is natural mineral, which consists of silicate of calcium and magnesium found in the form of very thin fibres.

Characteristics of Asbestos:

• They are light in weight and fire-resistant.
• It has low water permeability hence less porous.
• It is brown, grey, and white in colour.
• It is resistant to acids and alkalis.
• Its specific gravity is 3.10.
• It has excellent heat and electric insulation.
• They do not require any protective paint.
• Asbestos cement is widely used in the manufacturing of roofing slates and tiles.
• These sheets are widely used in industries, factories, cinemas, auditoriums etc.

Explanation:

The fillet welds are subjected to tensile stress. The minimum cross-section of the fillet is at the throat. Therefore the failure due to tensile stress occurs at the throat section. Thus the weakest area of the weld is the throat.
One other approach is that the other parts of the weld are subjected to normal stress and shear stress. But in the throat in addition to both bending is also there which makes it weakest.

Concept:

• Aggregate crushing value gives a relative measure of the resistance of an aggregate to crushing under a gradually applied compressive load. The aggregate crushing value should not exceed 45% for aggregate used for concrete other than for wearing surfaces, and 30 % for concrete for wearing surfaces, such as runways, and roads for pavements.
• Aggregate impact value gives a relative measure of the resistance of an aggregate to sudden shock or impact, As an alternative to crushing value, the aggregate impact value should not exceed 45% by weight for aggregates used for concrete other than for wearing surfaces and 30% by weight for concrete for wearing surfaces, such as runways, roads for pavements.

Concept:
Minimum steel reinforcement prevents sudden failure of beam and gives prior warning before failure.
As per IS 456:2000, clause 26.5.1.1,
The minimum reinforcement is given by:
A_s/bd = 0.85/f_y

Concept:
The different types of sections are as follows:

Explanation:
xu > xu,critical : Over reinforced section
xu = xu,critical : Balanced section
xu < xu,critical : Under-reinforced section
It can be seen that the actual neutral axis is less than the critical neutral axis for the under-reinforced section.
Hence the actual neutral axis fall above the critical neutral axis of the balanced section

Concept:
Velocity Ratio (VR):
The velocity ratio (VR) of a machine is the ratio of the distance moved by the effort to the distance moved by the load.
V.R. = (Distance moved by effort)/(Distance moved by load)Other Related PointsMechanical Advantage (MA):Mechanical advantage is the advantage gained by the use of a machine in transmitting force. Or in other words, it is defined as the ratio of the load to the effort.
M.A. = (Load)/(Effort)

Concept:

Project Cost:

Total project cost is the sum of two separate costs:

(a) The direct cost for accomplishing the work, and

(b) The indirect cost related to the control or direction of that work, financial overhead, lost production, and the hike, etc.

Direct Project Cost:

​​​These include labour cost, material cost, equipment cost etc.

• The direct cost curve, having many segments, thus falls with an increase in duration. However, the total indirect cost curve rises with the increase in duration.
• The project has the highest cost corresponding to the crash duration and has a normal cost corresponding to the normal duration.

Indirect Project Cost:

• Indirect costs on a project are those expenditures that cannot be apportioned or clearly allocated to the individual activities of a project but are assessed as a whole.
• The indirect cost includes the expenditure related to administrative and establishment charges, overhead, supervision, expenditure on a central store organization, loss of revenue, lost profit, penalty, etc.
• The indirect cost rises with increased duration, considering only overhead and supervision. It is represented by a straight line, with a slope equal to daily overhead.​

So the correct answer is Option 4

Explanation:
Crashing: It is a process in which project duration is reduced by deployment of some additional resources to either minimize the project’s cost, duration or to meet the customer’s need.
Reducing project duration by mean of additional resources can prove economical if strategically planned and implemented. To do so, we calculate the cost slope of each activity and start crashing activity with minimum cost slope lying on the critical path.
Critical Path ⇒ It is a path which leads to the maximum duration for completion of the project.
where, Cost Slope = (Crashing Cost - Normal Cost)/(Normal Time - Crashing Time)

Concept:
Vertical Stress:
Vertical stress means the force per unit area imposed on a layer of rock. Vertical stress is the combined stress due to the total weight of rock and interstitial fluids above a specified depth.
The vertical stress just below the entire due to uniformly distributed load over a circular area is given as:-

σ_z = k.q;
where, k = stress intensity factor
q = Load intensity
Here,
k = 1 - cos³ θ

cos θ =

Calculation:
Given that,
Diameter = 6 m, Load intensity = 16 kN/m²
∴ radius r = 3 m, Depth, h = 5 m

cos θ =
k = 1 - (0.857)³ = 0.37
∴ σ_z = 0.37 × 16 = 5.91 kN/m²
So the correct answer is Option D

Concept:
Probability factor (Z) = T_S - T_E/σ
Where σ = Standard deviation
Calculation:
Given:
Scheduled completion time of the project (T_S) = 40 days
Expected completion time of the project (T_E) = 40 days

Probability factor (Z) = 40 - 40/σ = 0
The probability factor (Z) is zero then the probability that the project will be completed in 40 days is 50%.

Concept:
The given information in question can be represented in the given figure:

The water will exert uniform pressure at the bottom plank in the upward direction.
Calculation:
Given:
Since it is given that Plank is floating on water. We get,
Upward force (buoyant force) exerted by water = Downward weight of pressure standing on Plank
(W')(L) = 2W
W = 2W/L
Consider midsection (left) of plank,

Consider Moment at the midpoint to be zero.
∑ Mmid = 0
M_mid + W x L/4) = (W' x L/2) x (L/4)
M_mid + (W x L/4) =  (2 x W/L) x (L/2) x (L/4)
∴ Mmid = 0​​​​​​​