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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) 2024 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) 2024 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

What is the SI unit of resistivity?

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

CONCEPT:

  • Resistivity (ρ): The property of a conductor that opposes the flow of electric current through them and independent of the shape and size of the materials but depends on the nature and temperature of the materials is called resistivity.
    • The unit for resistivity is the ohm-meter (Ω-m).
    • The resistivity of a material depends on its nature and the temperature of the conductor.
  • Resistance: The property of any conductor that opposes the flow of electric current through it and depends on the shape and size of the materials, temperature, and nature of the materials is called resistance.
    • It is denoted by R and the SI unit is the ohm (Ω).

The resistance is given by:

R = ρL/A

where ρ is resistivity, L is the length and A is the area of the cross-section.

EXPLANATION:

  • The resistance offered by a wire of unit length and unit area of cross-section is called resistivity or specific resistance (r).​

  • The SI unit of length (L) is m and area (A) is m2.

Putting on the respective units, we get:

The SI unit of ρ = Ohm-meter (Ω-m).

  • Thus the SI unit of resistivity is Ω-m.
BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 2

LMTD stands for _______.

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

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.

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

Free float is mainly used to

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

Float:

(i) It is associated with activity times.

(ii) It is analogous to the slack of events in PERT.

(iii) It is the range within which the start or finish time of activity may fluctuate without affecting the project completion time.

(iv) Floats are of the following types:

1. Total float: The time span by which starting or finishing of an activity can be delayed without delaying the completion of the project. Total float of activity affects total float of succeeding as well as preceding activities.

2. Free float: The delay which can be made without delaying succeeding activities. It affects only preceding activities.

3. Independent Float: It is the minimum excess available time which exists without affecting any of succeeding or preceding activities.

4. Interfering Float: It is similar to head event slack.

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

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 5

The angle of inclination of the plane at which the body begins to move down the plane, is called:

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

The main difference between the angle of friction and the angle of repose is that:

The angle of friction is defined as the angle between the normal reaction force and the resultant force of normal reaction force and friction when an object just begins to move,

Whereas the angle of repose is defined as the minimum angle of an inclined plane which causes an object to slide down the plane.

In general, The angle of repose is the maximum angle that a surface can be tilted from the horizontal, such that an object on it is just able to stay on the surface without it sliding down.

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

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 6

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 7
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 7
  • 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 8

The diameter of plunger used in Vicat apparatus is of

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

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 9

Cast iron contains carbon approximately

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

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 10

In a thermodynamic system, a process in which volume remains constant is called ______ process.

Detailed Solution for BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 10
  • An adiabatic process is one in which no heat transfer takes place to or from the system during the process
  • An isenthalpic process or iso-enthalpic process is a process that proceeds without any change in enthalpy, H; or specific enthalpy, h
  • An isentropic process is one in which entropy remains constant (Δs = 0)
  • An isobaric process is a thermodynamic process in which the pressure stays constant: ΔP = 0
  • An isochoric system is one in which volume is held constant (ΔV = 0). Isochoric processes can also be referred to as isometric or isovolumetric
  • An isothermal process is one in which there is no temperature change (ΔT = 0)

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

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 12

In plane surveying, level lines are considered as ________ and plumb lines are considered as __________.

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

Based upon the requirements and magnitude of the survey, the survey has been classified broadly into two main categories :

  1. Plane surveying
  2. Geodetic surveying

Plane Surveying

  • In this type of surveying, the mean surface of the earth is considered as a plane, and the spheroidal shape is neglected.
  • All triangles formed by survey lines are considered plane triangles.
  • The level line is taken as straight, and all plumb lines are considered to be parallel.
  • Plane surveying is done for smaller areas in consideration i.e Areas <195km2

Geodetic surveying

  • This is a type of surveying in which the shape of the earth is taken into account.
  • All lines are taken as curved lines and triangles as spherical triangles.
  • Geodetic survey includes work of larger magnitude and high degree of accuracy.
  • The purpose of geodetic survey is to determine the precise position on the surface of earth, of a system of widely distant points which form control stations to which surveys of less precision may be referred.
  • Geodetic surveys are employed for an area larger than 195 km2

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

An air-standard diesel cycle consists of

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

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 14

The Nusselt number, in the case of natural convection, is a function of:

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

In convection studies, it is common practice to nondimensional the governing equations and combines the variables which group together into dimensionless numbers in order to reduce the number of total variables.

Nusselt number is a parameter to nondimensional the heat transfer coefficient.

Non-dimensional groupings

  • Nusselt No. Nu = hLc/k = (convection heat transfer strength)/ (conduction heat transfer strength)
  • Prandtl No. Pr = n/a = (momentum diffusivity)/ (thermal diffusivity)
  • Reynolds No. Re = Ux/ν = (inertia force)/ (viscous force)

The dimensionless parameter represents the natural convection effects and is called the Grashof number.

Grashof number, Gr, as the ratio between the buoyancy force and the viscous force:

Nusselt number is a function of the Grashof number and the Prandtl number alone. Nu = f (Gr, Pr)

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

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 15

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 16

Which of the following unit works in anaerobic conditions?

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

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 17

The example for continuous flow type equipments is

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

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 18

Find the total current flowing in the given circuit.

Detailed Solution for BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 18

We have given circuit,

Here, both resistance is connected in parallel.

Hence, two parallel combinations of equivalent resistance can be given as,

R = (10 || 10) =

Now, the circuit can be drawn as,

Using Ohm's law,

We have,

V = 50 volts

R = 5 Ω

Hence, I = V/R = 50/5 = 10 A

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

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 19

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 20
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 20

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 21

Carnot cycle consists of the following process -

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

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 22

The possible location of shear centre of the channel section, shown below, is

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

Shear Centre is defined as the point through which the load must be applied to produce zero twisting moment on the section. If load passes through the shear centre, then there will be only bending stress in the cross section and no twisting.

The following points should be remembered for shear center:

1. If a beam has two axes of symmetry, then the shear centre will lie at their point of intersection.

2. For a section having one axis of symmetry, the shear centre lies on that axis of symmetry.

The shear flow in channel section causes clockwise twisting as shown.

To balance that clockwise twisting, external resultant load must cause anticlockwise twisting and it can be possible if load lies an LHS of the channel section.

Further, since channel section has X-X as of symmetry and Y-Y axis is not the symmetric axis so, ‘R’ cannot be shear centre.

‘S’, and ‘θ’ cannot be shear centre because they cause external loading twisting in clockwise rather than anticlockwise.

So, only possible location of shear centre is ‘P’.

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

Which of the following pairs is not correctly matched?

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

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 24
How many units (kWh) of electrical energy have been used in a home if an 8 kW geyser, a 5 kW electric press and four 100 Watt bulbs have been in use for 10 hours?
Detailed Solution for BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 24

Formula Used:

Energy (E) = Power (P) × Time (T)

Application:

We have,

P = 8 kW + 5 kW + 4(0.1) kW = 13.4 kW

T = 10 hours

Hence,

E = PT = 13.4 × 10 = 134 kWh

BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 25

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 25

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 26
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 26
BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 27
A centrifugal pump is running at the speed of 1000 rpm against a head of 40 m. If its speed is changed to 3000 rpm, then it will work against a head of:
Detailed Solution for BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 27

Concept:

Model laws of the pump:

,

,

Calculation:

Given:

N1 = 1000 rpm, H1 = 40 m, N2 = 3000 rpm

⇒ H2 =

⇒ H2 = 9 × 40 = 360 m

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

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

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

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 29

What will be the value of 'y', the distance of center of gravity, for the SOLID CONE given in the figure?

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

1) Centre of gravity of the right circular (SOLID) cone lies at a distance of h/4 from the base at a distance of 3h/4 from the apex and h/4 from the base

2) Centre of gravity for a hollow circular cone is given by

Similarly, the center of gravity for different shapes are:

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

The angle of convergence in venturimeter is

Detailed Solution for BPSC AE Civil Paper 4 (General Engineering) Mock Test - Question 30

Venturimeter

  • A venturimeter is a device used for measuring the rate of flow of a fluid of a liquid flowing through a pipe
  • The venturimeter always have smaller convergent portion and a larger divergent portion.
  • This is done to ensures a rapid converging passage and a gradual diverging passage in the direction of flow to avoid the loss of energy due to separation.
  • In the course of flow through the converging part, the velocity increases in the direction of flow according to the principle of continuity, while the pressure decreases according to Bernoulli’s theorem.
  • The velocity reaches its maximum value and pressure reaches its minimum value at the throat.
  • Subsequently, a decrease in the velocity and an increase in the pressure take place in course of flow through the divergent part.
  • The angle of convergence = 19° - 23°
  • The angle of divergence = 6° - 7° → It should not be greater than 7° to avoid flow separation

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