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BPSC AE Civil Paper 4 (General Engineering) Mock Test - Civil Engineering (CE) MCQ


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30 Questions MCQ Test - BPSC AE Civil Paper 4 (General Engineering) Mock Test

BPSC AE Civil Paper 4 (General Engineering) Mock Test for Civil Engineering (CE) 2025 is part of Civil Engineering (CE) preparation. The BPSC AE Civil Paper 4 (General Engineering) Mock Test questions and answers have been prepared according to the Civil Engineering (CE) exam syllabus.The BPSC AE Civil Paper 4 (General Engineering) Mock Test MCQs are made for Civil Engineering (CE) 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for BPSC AE Civil Paper 4 (General Engineering) Mock Test below.
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BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 1

LMTD stands for _______.

Detailed Solution for BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 1

Concept:

Parallel flow heat exchanger

Counter flow heat exchanger

LMTD CALCULATION FOR PARALLEL FLOW HEAT EXCHANGER-

The formula for calculation of LMTD (Log Mean Temperature Difference) for a parallel flow heat exchanger is given by-

LMTD CALCULATION FOR COUNTER FLOW HEAT EXCHANGER-

  • The LMTD of the counter-flow heat exchanger is always greater than the LMTD of the parallel flow heat exchanger.
  • Now the value of the LMTD of the cross-flow heat exchanger is less than the value of the LMTD of the counter-flow and more than the parallel flow heat exchangers.
  • i.e. LMTDcounter > LMTDcrossflow > LMTDparallel
  • The multi-pass concept is used in heat exchangers is used to keep the length of the heat exchanger small.

Explanation:

LMTD stands for Logarithmic Mean Temperature Difference.

BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 2

If the bulk modulus and shear modulus are 1 and 2 respectively, what is the value of young's modulus of elasticity?

Detailed Solution for BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 2

Concept :

The relation between E, K, and μ

E = 2G (1 + μ)

E = 3K (1 - 2μ)

Where,

E = Young's Modulus of Rigidity = Stress / strain

G = Shear Modulus or Modulus of rigidity = Shear stress / Shear strain = 2

μ = Poisson’s ratio = - lateral strain / longitudinal strain

K = Bulk Modulus of elasticity = Volumetric stress / Volumetric strain = 1

E = 3.6

BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 3

In a simply supported beam of span L subjected to central concentrated load, the central deflection is 24 mm. Then the slope at supports is:

Detailed Solution for BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 3

Concept:

In a simply supported beam of span L subjected to central concentrated load,

then the slope at supports is

Given:

,

So, Deflection

We know slope at supports is

BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 4
As compared to laminar flow, the boundary layers in a turbulent flow will be:
Detailed Solution for BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 4
  • Turbulent flow and laminar flow are two types of fluid flow.
  • Laminar flow is characterized by smooth, regular movement of fluid, while turbulent flow is characterized by chaotic, irregular movement.
  • Boundary Layer in Turbulent Flow and Laminar Flow Boundary layer is a layer of fluid that is in contact with a solid surface.
  • The characteristics of this layer are affected by the type of fluid flow. In laminar flow, the boundary layer is thin and smooth.
  • In turbulent flow, the boundary layer is thicker and chaotic.
  • Comparison of Boundary Layer in Turbulent Flow and Laminar Flow In turbulent flow, the boundary layer is thicker compared to laminar flow.
  • This is because the chaotic movement of fluid in turbulent flow creates more friction between the fluid and the solid surface, resulting in a thicker boundary layer.
  • On the other hand, in laminar flow, the smooth movement of fluid creates less friction, resulting in a thinner boundary layer.
BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 5

The diameter of plunger used in Vicat apparatus is of

Detailed Solution for BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 5

Concept:

Vicat apparatus:

The standard consistency of cement paste test is carried out by the Vicat apparatus having a weight of plunger 300g, diameter and length are 10 mm and 50 mm respectively to penetrate to a depth of 33 to 35 mm from the top of the mold. Mold which is 40 mm deep and 80 mm in diameter.

BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 6

Cast iron contains carbon approximately

Detailed Solution for BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 6

Cast iron

  • Cast iron is an alloy of iron, carbon and silicon and it is hard and brittle.
  • Carbon content in CI may be within 2% to 4% and carbon may be present as free carbon or iron carbide Fe3C.
  • In general, the types of cast iron are (a) grey cast iron and (b) white cast iron (c) malleable cast iron etc

Additional Information

Wrought iro n

  • Wrought iron is a very pure iron where the iron content is of the order of 99.5%.
  • It usually contains less than 0.1% carbon and 1 or 2% slag.
  • It is produced by re-melting pig iron and some small amount of silicon, sulphur, or phosphorus may be present.
  • It is tough, malleable and ductile and can easily be forged or welded.
  • It cannot however take sudden shock. Chains, crane hooks, railway couplings and such other components may be made of this iron.

Pig iron

  • Pig iron is an intermediate product of the iron industry.
  • Pig iron has a very high carbon content, typically 3.8–4.7%, along with silica and other constituents of dross, which makes it very brittle, and not useful directly as a material except for limited applications.

Steel

  • Steel is basically an alloy of iron and carbon in which the carbon content can be less than 1.7% and carbon is present in the form of iron carbide to impart hardness and strength.
  • Two main categories of steel are (a) Plain carbon steel and (b) alloy steel.
  • Thus, wrought iron is an iron alloy with a very low carbon (less than 0.08%) content in contrast to cast iron.
BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 7

The time constant of a thermocouple is the time taken

Detailed Solution for BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 7

Concept:

The thermal time constant indicates a time required for a thermistor to respond to a change in its ambient temperature.
When the ambient temperature is changed from T1 to T2, the relationship between the time elapsed during the temperature change t (sec.) and the thermistor temperature T can be expressed by the following equation.

τ (tau in sec.) in the equation denotes the thermal time constant.

Now, assuming t and τ (tau) are equal (t = τ), the equation can be expressed as follows.

T = (T2 – T1) (1 - e-1) + T1

This shows that the constant τ (sec.) is defined as a time for the thermistor to reach 63.2% of the total difference between its initial and final body temperatures.

Note:

The temperature change rate at n times the constant τ (sec.) is as follows, showing that the thermistor body temperature reaches its ambient temperature approximately within 7 times the constant.
τ = 63.2%, 2τ = 86.5%, 3τ = 95.0%, ・・・・ 7τ ≒ 100%

BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 8

A column with highest equivalent length has-

Detailed Solution for BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 8

A column with maximum equivalent length has one end fixed and the other end is free.

BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 9
Bauxite is used for -
Detailed Solution for BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 9

Concept:

Bayer's process:

In bauxite, in addition to Aluminium Oxide (Al2 O3) impurities like Iron Oxide (Fe2 O3) and Sand (Si O2) are also present.

On refining bauxite by Bayer’s method, pure Aluminum Oxide is obtained which is also called Alumina.

Purification process:

  • First, we dissolve Bauxite in an aqueous Sodium hydroxide (NaOH) by Digestion.
  • The insoluble impurities are separated by Filtration.
  • Purified Bauxite is then dissolved in Cryolite and electrolyzed at 950° C in a carbon lined steel cathode with hard carbon rods as the anode.

Li2 CO3​ is used to

  • Lower the melting point of the electrolyte
  • Permit larger current flow.
  • Reduce Fluorine emission
BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 10

The cracks usually occurs when tree is exposed to sun for seasoning after being fall down are called as

Detailed Solution for BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 10

Concept:

Star shakes: these are cracks which extend from bark towards the sap wood. They are usually confined up to the plane of sap wood. They are wider on the outside ends and narrower on the inside ends. They are usually formed due to extreme heat or severe frost during the growth of the tree. Figure shows the star shakes.

Cup shakes: these are caused by the rupture of tissue a circular direction as shown in figure. It is a curved crack and it separates partly one annual ring from the other. It develops due to non- uniform growth or due to excessive bending of a growing tree during a cyclonic weather. It may not prove to be harmful, if it covers only a portion of ring.

Radial shakes: these are similar to star shakes. But they are fine, irregular and numerous. They usually occur when tree is exposed to sun for seasoning after being felled down. They run for a short distance from bark towards the centre, then follow direction of annual ring and ultimately run towards pith.

Ring shakes: when cup shakes cover the entire ring, they are known as the ring shakes as shown in figure.



BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 11

Select the air pollutant that does NOT belong to the category of secondary air pollutants.

Detailed Solution for BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 11

Secondary air pollutant

1) Sulphuric acid (H2SO4)

2) Formaldehydes

3) Photochemical smog

4) Ozone

5) Peroxy Acetyl – nitrate (PAN)

Primary air pollutant

1) SO2

2) CO & CO2

3) NO, NO2

4) Volatile organic compound

5) Suspended Particulate Matter (SPM)

6) Radioactive compound

7) Halogen compound

Important Points

(i) Secondary air pollutant are those which are formed by the reaction of one or more primary air pollutant either with each other or with water vapour in the presence of sun light.

BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 12

An air-standard diesel cycle consists of

Detailed Solution for BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 12

Diesel cycle:

Processes in compression engine (diesel cycle) are:

  1. Process 1-2: Reversible adiabatic compression
  2. Process 2-3: Constant pressure heat addition
  3. Process 3-4: Reversible adiabatic expansion
  4. Process 4-1: Constant volume of heat rejection

Hence an air-standard diesel cycle consists of one constant pressure, one constant volume and two adiabatic processes.

Additional Information

Otto cycle:

The air-standard-Otto cycle is the idealized cycle for the spark-ignition internal combustion engines.

Otto cycle is the one that has two constant volume heat transfer processes and two adiabatic work transfer processes.

The Otto cycle 1-2-3-4 consists of the following four processes:

  • Process 1-2: Reversible adiabatic compression of air
  • Process 2-3: Heat addition at constant volume
  • Process 3-4: Reversible adiabatic expansion of air
  • Process 4-1: Heat rejection at constant volume

  • During constant volume process heat addition and heat, rejection takes place and no work transfer.
  • During the adiabatic processes [compressions/expansion] only work transfer taken place but no heat transfer occurs.

Dual cycle

The Dual Cycle also called a mixed cycle or limited pressure cycle is a compromise between Otto and Diesel cycles.

The dual cycle is a thermodynamic cycle that combines the Otto cycle and the Diesel cycle. In this cycle, the heat addition occurs partly at constant volume and partly at constant pressure.

Different processes in the dual cycles are given below:

Process 1-2: Reversible adiabatic compression.

Process 2-3: Constant volume heat addition.

Process 3-4: Constant pressure heat addition.

Process 4-5: Reversible adiabatic expansion.

Process 5-1: Constant volume heat rejection.

BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 13

Compression ratio of an internal combustion engine is defined as _______.

[VS = Swept volume, VC = Clearance volume]

Detailed Solution for BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 13

Volume Ratio:

  • The ratio of larger volume to lesser volume during any process is known as the volume ratio of that process.
  • The volume ratio during the compression process is known as the compression ratio and the volume ratio during the expansion process is known as the expansion ratio.
  • Thus compression ratio is given by the ratio of volume before compression to volume after compression.

Important Point

  • The compression ratio of the petrol engine is approximately 6 to 10. As the compression ratio is low the petrol engine is lighter and cheaper.
  • A diesel engine has a compression ratio of approximately 15 to 20. As the compression ratio is high diesel engine is heavier and costlier.
BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 14

Calculate the total impedance for a circuit having the following values:

Resistance = 5 Ω

Inductance reactance = 20 Ω

Capacitance reactance = 8 Ω

Detailed Solution for BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 14

Concept:

In a series RLC circuit, the impedance is given by

R is resistance

L is inductive reactance = ωL

XC is capacitive reactance = -1/ωC

(XL – XC) is net reactance

At the resonant frequency, inductive reactance is equal to capacitive reactance i.e. XL = XC

So, at this condition the impedance is minimum, and it is equivalent to R.

Explanation:

Given that, resistance (R) = 5 Ω

Capacitive reactance (XC) = 20 Ω

Inductive reactance (XL) =8 Ω

Net Impedance Z is calculated as:

BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 15

Which of the following unit works in anaerobic conditions?

Detailed Solution for BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 15

Concept:

Sludge Digestion:

  • Sludge digestion is a biological process in which organic solids are decomposed into stable substances.
  • Digestion reduces the total mass of solids, destroys pathogens, and makes it easier to dewater or dry the sludge.

Sludge Digestion Process:

  • This process is applied for both aerobic and anaerobic conditions.

Confusion Points
The digestion tank can be aerobic also, but the other options ie sedimentation tank and trickling filters are always under aerobic conditions. So, the most appropriate option will be a sludge digestion tank.

Additional Information

The followings are the classification of secondary treatment units:

BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 16

The example for continuous flow type equipments is

Detailed Solution for BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 16

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
BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 17

The frequency distribution of duration of an individual activity takes the shape of ______ as per the PERT analysis.

Detailed Solution for BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 17

PERT is almost identical to the critical path method (CPM) technique except it assumes each activity duration has a range that follows a statistical distribution.

  • In PERT the Normal distribution is used for the completion time of Project.
  • PERT uses three-time estimates for each activity. Basically, this means each activity duration can range from an optimistic time to a pessimistic time, and a weighted avenge can be computed for each activity.
  • Because project activities usually represent work, and because work tends to stay behind once it gets behind, the PERT developers chose an approximation of the beta distribution to represent activity durations.
  • The activity durations can be skewed more toward the high or low end of the data range.

  • Figure A depicts a beta distribution for activity duration that is skewed toward the right and is representative of work that tends to stay late once it is behind.
  • The distribution for the project duration is represented by a normal (symmetrical) distribution shown in Figure B.
  • The project distribution represents the sum of the weighted avenges of the activities on the critical path.

Confusion Points
In PERT individual activities follows Beta distribution whereas whole project duration follows Normal distribution.

BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 18
Determine the value of voltage if a charge of 240 C is delivered to a 200 V source in 1 minute with a resistance of 10 Ω.
Detailed Solution for BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 18

Concept:

Electric charge(q): The property of matter which is responsible for electrostatic force is called an electric charge.

The SI unit of charge is coulomb (C).

The rate of flow of electric charge is called an electric current.

Charge (q) = current (I) × time (t)

Calculation:

Given that;

Q = 240 C

t = 1 min = 60 sec

Hence,

Current = Q/t = 240/60 = 4 A

If current flow through the resistance of 10 Ω, the voltage across it will be = 4 × 10 = 40 V

Given that, the charge is delivered to a 200 V source,

Hence, Total Voltage = 200 + 40 = 240 V

BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 19

Carnot cycle consists of the following process -

Detailed Solution for BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 19

Carnot cycle:

The ideal reversible cycle that has the highest possible efficiency among all heat engines is called the Carnot cycle.

Carnot cycle is one of the best-known reversible cycles. The Carnot cycle is composed of four reversible processes.

  • Reversible Isothermal Expansion (process 1-2)
  • Reversible adiabatic expansion (process 2-3)
  • Reversible isothermal compression (process 3-4)
  • Reversible adiabatic compression (process 4-1)

Fig. P-V and T-S diagrams of Carnot Cycle

The Carnot cycle consists of two isothermal and two isentropic processes.

BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 20

Consider a 390 V, 50 Hz circuit. This inductive circuit has a current of 13 A, lagging 30 degrees. Find the inductive reactance.

Detailed Solution for BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 20

Inductive reactance:

Inductive reactance (XL) is given by:

XL = ωL

With ω = 2πf

XL = ωL = 2πfL

f = frequency in Hz

L = value of inductor in Henry

Observation:

The inductive reactance XL will increase when:

1) L increases

2) f increases

Application:

We have,

V = 390 volts

I = 13 A

Since, the current is lagging 30 degrees, the above is not a purely inductive circuit, but it is an R-L circuit,

Hence, Impedance Z would be

Z = 390/13 = 30 Ω

So, inductive reactance, XL = Z × sin(30°)

XL = 30 sin (30)

XL= 15 Ω

Important Point

Capacitive reactance:

Capacitive reactance is given by:

With ω = 2πf

The capacitive reactance is inversely proportional to the frequency.

BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 21

Which of the following pairs is not correctly matched?

Detailed Solution for BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 21

BOD and COD:

  • Generally COD is always greater than BOD because COD is the amount of inorganic chemicals required to completely oxidize waste material present in the water body, while BOD is the amount of oxygen required to oxidize organic material present in a water body.
  • So,

So, the 4th statement is not correct

Ultimate BOD:

  • The ultimate BOD (Lo) is defined as the maximum BOD exerted by the wastewater. The time required to achieve the ultimate BOD depends upon the characteristics of the wastewater, i.e., the chemical composition of the organic matter present in the wastewater and its biodegradable properties and temperature of incubation.
  • However, the ultimate BOD will be temperature-independent.
  • At higher temperatures for the same concentration and nature of organic matter, BOD will be achieved in a shorter time as compared to lower temperatures, where it will require more time. The ultimate BOD best expresses the concentration of degradable organic matter based on the total oxygen required to oxidize it. However, it does not indicate how rapidly oxygen will be depleted in receiving water.
  • is an indirect measure of the total organic matter widely used for wastewater monitoring, design, and operation of STPs
BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 22

The portions made by cutting standard bricks across their width are known as

Detailed Solution for BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 22

Bat:

  • It is the portion of the brick cut across the width. Thus, a bat is smaller in length than the full brick.
  • If the length of bat is equal to the half the length of original brick, it is known as half bat. A three quarter bat is the one having its length equal to three quarters of the length of a full brick.

Additional Information
Perpends:
The vertical joints between blocks or bricks that have been laid in a horizontal course to form a wall are known as prepend or cross joints. Prepend is normally filled and sealed with mortar.

Hearting: The interior portion of the wall between the face and back.

BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 23

Consider the motion of a point on a circular trajectory. The acceleration in a linear motion (a) and the acceleration in angular motion (α), are related as : (Take r as the radius of circular trajectory)

Detailed Solution for BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 23

CONCEPT:

  • Angular velocity is the velocity associated with objects moving in a curved path.
    • While on a curved path, the displacement is the change in angle about the point of rotation or curved motion.

Angular velocity,

  • The tangent to any point on the curved path gives the direction of the linear velocity of the object at that point.

The angular velocity ω is related to the linear velocity of an object v by the equation

v = rω

Where r is the radius of the curvature.

  • Angular acceleration is the rate of change of angular velocity with time. It is given by

EXPLANATION:

  • Linear acceleration is defined as the rate of change of linear velocity with time.
    • Linear acceleration, (where v is the linear velocity)
  • Angular acceleration, (where ω is the angular velocity)

Ratio =

BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 24

Which of the following classifications are correct with reference to the various types of imperfections in a semi-conductor?

1. Substitutional

2. Vacancies

3. Interstitial

Select the correct answer using the code given below.

Detailed Solution for BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 24

The various imperfections found in semi-conductors are:

1.) Interstitial defect:

  • An interstitial defect occurs when an atom takes the interstitial position of the lattice structure.
  • This interstitial atom may be of the same crystal or of a foreign material.
  • Although the extra atom occupies the empty interstitial space, the size of the atom is usually larger than that of the empty space. Thus the surrounding atoms are compressed and distorted.
  • The presence of a substantial number of interstitial atoms can change the mechanical and thermal properties of the solid.

2.) Substitutional defect:

  • Substitutional Defect occurs when the original atom in the lattice site of a crystalline solid is replaced by a different type of atom.
  • Unlike interstitial defects, foreign atoms should occupy the lattice site only and not the interstitial position.
  • The foreign atom may be of the same size or different (either larger or smaller). Depending on the size of the substituted foreign atom, the neighboring atoms may remain either in tension or in compression.

3.) Vacancy defect:

  • A vacancy is a type of point defect in a crystal where an atom is missing from one of the lattice sites.
  • During the crystallization of a solid, a particle is missing from its regular site in the crystal lattice.
  • The missing particle creates a vacancy in the lattice structure.
BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 25
According to which theory of failure does the ductile material begin to yield, when the maximum principal strain reaches the strain?
Detailed Solution for BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 25
BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 26

Which of the following is NOT a type of centrifugal pump ?

Detailed Solution for BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 26

Centrifugal Pump Classification based on flow direction through impeller:

Radial flow pump:

The liquid enters the impeller axially and the head is developed by centrifugal force. The liquid leaves the impeller radially. Such pumps are called radial flow pumps.

Axial flow pump:

In this, the head is developed by the propelling action of vanes. The liquid enters the impeller axially and also leaves axially. These pumps have very large discharge but a low head.

Mixed flow pump:

Flow through a mixed flow impeller is a combination of axial and radial flow. The head is developed by the combined action of centrifugal force and propelling action of the impeller. The flow enters axially, not necessarily in the center, but leaving at some angle between radially and axially.

BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 27

_________ is the correct formula to determine the elevation difference (dZ) between two points on the ground using a total station.

(VD = Vertical difference, HI = Instrument height and HR = Reflector height)

Detailed Solution for BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 27

Concept:

  • A total station, also known as an electronic tachometer, is an optical instrument.
  • It is a combination of an electronic theodolite for measuring horizontal and vertical angles, an electromagnetic distance measurement (EDM) device for measurement of slope distances, and onboard software to convert the raw observed data to three-dimensional coordinates.
  • Thus with a total station, one may determine the actual positions X, Y, and Z or northing, easting, and elevation of surveyed points or the position of the instrument from known points in absolute terms.
  • Further, the EDM that measures the slope distance can calculate and display horizontal distance and differences in level.
  • This is accomplished with the help of a microprocessor normally working concentric with the telescope eyepiece and generally housed in a casting that forms part of the telescope.
  • The elevation difference can be given as

D = VD + (HI - HR)

where,

VD = vertical difference between two points,

HI =instrument height,

HR = height of centre of reflector.

BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 28

Pelton wheel is which type of turbine?

Detailed Solution for BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 28

Pelton Wheel Turbine:

  • Pelton Wheel Turbine is a tangential flow impulse turbine in which the pressure energy of water is converted into kinetic energy to form a high-speed water jet and this jet strikes the wheel tangentially to make it rotate.

Pelton Wheel Turbine is a high head low discharge turbine.

  • High-head turbine: In this type of turbine, the net head varies from 150m to 2000m or even more, and these turbines require a small quantity of water. Example: Pelton wheel turbine.
  • Medium head turbine: The net head varies from 30m to 150m, and also these turbines require a moderate quantity of water. Example: Francis turbine.
  • Low-head turbine: The net head is less than 30m and also these turbines require a large quantity of water. Example: Kaplan turbine.

The Main parts of the Pelton Wheel Turbine are:

  1. Nozzle and Flow Regulating Arrangement
  2. Runner and Buckets
  3. Casing
  4. Braking Jet
  5. Penstock

BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 29

The quick lime as it comes from kilns is called as-

Detailed Solution for BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 29

Lump Lime:

Calcination or calcining is a thermal treatment process to bring about a thermal decomposition. The process takes place below the melting point of the product. The name calcination is derived from the Latin word ‘Calcinare’ which means to burn lime.

The lime which is obtained by the calcination of comparatively pure limestone is known as quick lime or caustic lime. Its chemical composition is (CaO) oxide of calcium and it has great affinity for moisture.

The quick lime as it comes out from kiln is also known as lump lime.

Additional Information

Fat lime:

(i) This lime slake vigorously due to which it's volume increases by 2 to 2.5 times than it's original volume. Hence it is referred as fat lime.

(ii) It posses perfectly white colour, hence also termed as white lime.

(iii) It is obtained by the calcination of lime stone of approx. 95% purity, hence it is also referred as pure lime, rich lime or highly caustic lime.

(iv) It's properties are:

  • It hardness very slowly with vigorous slaking.
  • It has good solubility in water
  • It possesses a high degree of plasticity.
  • In the presence of air it sets slowly
  • Fat lime is mostly used for plastering and white washing.

Hydraulic lime:

(i) It is also referred as water lime as it is capable of setting in water and in damp locations.

(ii) It is obtained by the calcination of lime stone having the purity in the range of 70 - 92%.

(iii) It is insoluble in water and possess off white colour.

(iv) It hardness comparatively faster than fat lime, hence is used in engineering works where strength is require. Ex: Brick masonry and stone masonry

Poor lime:

(i) It is also referred as impure lime or lean lime as it obtained from the calcination of lime stone having purity less than 70%.

(ii) This lime does not undergoes slaking, hardness very slowly, possess muddy white colour, hence is used in Engineering works of minor importance. Ex: Brick work around foundation.

BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 30
Euler's equation of motion is derived by considering the motion of fluid element along which line ?
Detailed Solution for BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 30

Concept:

Euler's equation of motion:

  • Euler's equation for a steady flow of an ideal fluid along a streamline is a relation between the velocity, pressure, and density of a moving fluid.
  • It is based on Newton's Second Law of Motion which states that if the external force is zero, linear momentum is conserved.
  • The integration of the equation gives Bernoulli's equation in the form of energy per unit weight of the following fluid.

It is based on the following assumptions:

  1. The fluid is non-viscous (i,e., the frictional losses are zero)
  2. The fluid is homogeneous and incompressible (i.e., the mass density of the fluid is constant)
  3. The flow is continuous, steady and along the streamline.
  4. The velocity of the flow is uniform over the section.
  5. No energy or force (except gravity and pressure forces) is involved in the flow.
  6. As there is no external force applied (Non-viscous flow), therefore linear momentum will be conserved.
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