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


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30 Questions MCQ Test - UPPSC AE Civil Paper 2 Mock Test - 2

UPPSC AE Civil Paper 2 Mock Test - 2 for Civil Engineering (CE) 2025 is part of Civil Engineering (CE) preparation. The UPPSC AE Civil Paper 2 Mock Test - 2 questions and answers have been prepared according to the Civil Engineering (CE) exam syllabus.The UPPSC AE Civil Paper 2 Mock Test - 2 MCQs are made for Civil Engineering (CE) 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for UPPSC AE Civil Paper 2 Mock Test - 2 below.
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UPPSC AE Civil Paper 2 Mock Test - 2 - Question 1

The main feature of the Kshatrapal form of governance in the Shaka Empire was _______.

Detailed Solution for UPPSC AE Civil Paper 2 Mock Test - 2 - Question 1
The correct answer is "Joint rule of a Mahakshatrapa and a Kshatrapa".

Explanation

  • The Indo-Greek rule lasted from about 180 BC to about 55 BC.
  • The Sakas (also written Shakas), alternatively known as Indo-Scythians, invaded northwest India in the first century BC onwards.

Satrap system of government:

  • The Ancient Sakas in India had introduced Satrap system of government, along with the Parthians.
  • Which was quite similar to the Iranian Achaemenid and Seleucid.
  • Under this system, the kingdom was divided into provinces, each under military governor Mahakshatrapa (great satrap).
  • The governors of lower status were called kshatrapas (satraps). These governors had the power to issue their own inscriptions and mint their own coins.

Other Related PointsShakas-Rulers:

  • Maues (Reign 98/50 BC – 60/57 BC)
    • Maues, also known as Moga was the earliest Indo-Scythian king.
    • He ruled over Gandhara (present Pakistan and Afghanistan).
    • He invaded the Indo-Greek territories but unsuccessfully.
    • His capital was at Sirkap (Punjab, Pakistan).
    • Many coins issued by Maues have been found. They contain Buddhist and also Hindu symbols.
    • The languages used in these coins were Greek and Kharoshti.
  • Chashtana (Reign 78 AD – 130 AD)
    • He was a Saka ruler of the Western Kshatrapas (Satraps) dynasty who ruled over Ujjain.
    • The Saka Era is believed to have started at his ascension to power in 78 AD.
    • Ptolemy mentions him as “Tiasthenes” or “Testenes”.
    • He was the founder of one of the two major Saka Kshatrapa dynasties in northwest India, the Bhadramukhas.
      • The other dynasty was called Kshaharatas and included the king Nahapana (who was defeated by Satavahana king Gautamiputra Satakarni).
  • Rudradaman I (Reign 130 AD – 150 AD)
    • He is considered the greatest of the Saka rulers.
    • He is from the Western Kshatrapa dynasty.
    • He was the grandson of Chastana.
    • His kingdom included Konkan, Narmada valley, Kathiawar, other parts of Gujarat and Malwa.
    • He conducted the repair work of the Sudarshana Lake at Kathiawar.
    • He married a Hindu woman and had converted to Hinduism.
    • He also issued the first long inscription in chaste Sanskrit.
    • He took up the title of Makakshatrapa after becoming king.
    • He maintained marital relationships with the Satavahanas. Vashishtiputra Satakarni was his son-in-law. But he also fought numerous wars with them.
    • He regained through conquests most of the territories previously under Nahapana.
    • He supported Sanskrit literature and cultural arts
UPPSC AE Civil Paper 2 Mock Test - 2 - Question 2

Which among the following is a homogenous solution?

Detailed Solution for UPPSC AE Civil Paper 2 Mock Test - 2 - Question 2

The correct answer is Sugar in water.
Explanation

  • A Homogeneous mixture has the same proportion of its compound in a sample.
    • Sugar in water is a homogeneous mixture.
  • A Heterogeneous mixture has a different proportion of its compound in a sample.
  • A Mixture in which a solute particle does not dissolve, but gets suspended is a suspension.
  • A solution in which the substances are regularly suspended in a fluid is a Colloidal Solution.
  • A true solution is a homogeneous mixture of two or more substances in which the substance dissolved in the solvent has a particle size of less than 10-9 m or 1 nm.

Mistake Points

  • Ice cubes in cola: Ice can be physically separated from cola when it is in a mixture.
  • Once the ice melts in the cola, it can no longer be separated, thus it is no longer a mixture.
  • So ice in cola is a heterogeneous solution.​

Confusion Points

  • Sugar and water can be separated by removing the water from the solution using the distillation process.

Important Points

  • Mixture: Two or three compounds that are not fused chemically, and have no physical interactions.
  • Solution: Two substances that are chemically mixed to form a new compound.

Other Related Points

UPPSC AE Civil Paper 2 Mock Test - 2 - Question 3

The provision for Anti Defection Act is mentioned in which of the following Schedules of the Constitution of India?

Detailed Solution for UPPSC AE Civil Paper 2 Mock Test - 2 - Question 3

The correct answer is 10th Schedule.

Explanation

  • 10th Schedule of the Constitution of India deals with the provision for Anti Defection Act.
  • The Tenth Schedule was inserted into the Constitution in 1985 by the 52nd Amendment Act.
  • It deals with the Anti defection law i.e, provisions as to disqualification on the ground of defection.
  • Decision on questions as to disqualification on the ground of detection:-
    • If any question arises as to whether a member of a House has become subject to disqualification under this Schedule, the question shall be referred for the decision of the Chairman or, as the case may be, the
      Speaker of such House and his decision shall be final.
    • Provided that where the question which has arisen is as to whether the Chairman or the Speaker of a House has become subject to such disqualification, the question shall be referred for the decision of a such member of the House as the House may elect on this behalf and his decision shall be final.

Other Related Points

  • 9th Schedule:
    • It became a part of the Constitution in the First Constitutional Amendment Act, of 1951.
    • It contains a list of central and state laws which cannot be challenged in courts.
    • Currently, 284 such laws are shielded from judicial review.
    • It was created by the new Article 31B which shields specific laws or enactments.
    • Article 31A was brought in to protect laws related to agrarian reform and for abolishing the Zamindari system.
  • Eleventh Schedule-
    • Provisions that specify the powers,
      authority, and responsibilities of
      Panchayats.
    • It has 29 subjects.
  • Twelfth Schedule-
    • Provisions that specify the powers,
      authority, and responsibilities of
      Municipalities.
    • It has 18 subjects.
UPPSC AE Civil Paper 2 Mock Test - 2 - Question 4
Which one of the following is NOT correctly matched?
Detailed Solution for UPPSC AE Civil Paper 2 Mock Test - 2 - Question 4

The correct answer is M.A Ansari - Kanpur Conspiracy Case.

Explanation

  • M.A. Ansari was NOT associated with the Kanpur Conspiracy Case of 1924.
  • Kanpur Bolshevik conspiracy case 1924:
    • In this case, newly emerged communists of India were trailed by the British for conspiring against the Government.
    • Some newly turned communists named M N Roy, Muzaffar Ahamed, SA Dange, Shaukat Usmani, Nalini Gupta, Singaravelu Chettiar, and Ghulam Hussain were caught by the Government.
    • They were trialed for conspiring against the Government.
    • In this case, MN Roy was charged in absentia, so he was not arrested.
    • Ghulam Hussain turned into a British informer and was pardoned.
    • Rest all people were arrested and sent to jail for 4 years. Hence statement 1 is wrong.

Other Related Points

  • The Central Assembly Bomb Case (1929):
    • It was filed against Batukeshwar Dutt.
    • Bhagat Singh and Batukeshwar Dutt on 8th April 1929, threw bombs at the Central Legislative Assembly, Delhi.
    • They started raising slogans of 'Inquilab Zindabad' from the visitor's gallery at the central assembly hall.
    • Their main motive was to make the deaf hear, rather than to injure people.
    • Both of them were members of the Hindustan Socialist Republican Association. Hence option 2 is correct.
  • Kakori Train Conspiracy:
    • It was a political robbery and the incident that took place in the small town of Karori which was only 16 km away from Lucknow on 9 August
      1925.
    • Hindustan Republican Association (HRA) under the leadership of Ram Prasad Bismil and supported by Ashfaqulla Khan, Rajendra Lahiri, Chandrashekhar Azad, Sachindra Bakshi, Keshab Chakravarty, Manmathnath Gupta, Murari Lal Gupta (Murari Lal Khanna), Mukundi Lal (Mukundi Lal Gupta) and Banwari Lal.
    • Ram Prasad Bismil, Thakur Roshan Singh, Rajendra Nath Lahiri, and Ashfaqullah Khan were given a death sentence and the rest were given Kala Pani (Port Blair Cellular Jail). Hence option 3 is correct.
  • ​Chittagong Armoury Raid:
    • The Chittagong armory raid took place on 18 April 1930 and was an attempt at raiding the armory of police and auxiliary forces, led by freedom fighter Surya Sen. Hence option 4 is correct.
    • The volunteers he gathered were part of a revolutionary group of nationalists who believed in using force to fight the British.
    • The original plan was to capture two armories in Chittagong, subsequently destroying the telegraph and telephone office.
    • The group would then take hostages from the European Club, as most of its members were involved in the colonial government in some form.
    • Amongst these volunteers were two incredible women: Pritilata Waddedar and Kalpana Datta.
UPPSC AE Civil Paper 2 Mock Test - 2 - Question 5

As per ICAO, for airports serving big aircrafts, the crosswind component should not exceed

Detailed Solution for UPPSC AE Civil Paper 2 Mock Test - 2 - Question 5

Crosswind component:

  • It interrupts safe landing and take-off operations.
  • The crosswind component is computed by multiplying the wind speed by the sine of the angle between the wind and the direction of travel while the headwind component is computed in the same manner, using cosine instead of sine.

According to ICAO, the permissible limit of crosswind are as follows:

UPPSC AE Civil Paper 2 Mock Test - 2 - Question 6

Number of fish bolts per fish plate is

Detailed Solution for UPPSC AE Civil Paper 2 Mock Test - 2 - Question 6

The name ‘fish plate’ derives from the fish-shaped section of this fitting.
The function of a fish plate is to hold two rails together in both the horizontal and vertical planes. Fish plates are manufactured using a special type of steel (Indian Railways specification T-1/57) with the composition given below:
a) Carbon: 0.30–0.42%
b) Manganese: not more than 0.6%
c) Silicon: not more than 0.15%
d) Sulphur and phosphorous: not more than 0.06%
The number of bolts per fish plate is 4
The steel used for fish plates should have a minimum tensile strength of 5.58 to 6.51 t/cm2 with a minimum elongation of 20%.
Fish plates are designed to have roughly the same strength as the rail section, and as such the section area of two fish plates connecting the rail ends is kept about the same as that of the rail section.

UPPSC AE Civil Paper 2 Mock Test - 2 - Question 7
The classification of survey on the basis of nature of area is called-
Detailed Solution for UPPSC AE Civil Paper 2 Mock Test - 2 - Question 7

Explanation:

Classification of survey-based upon the nature of filed survey:

1) Topographical survey:

  • Used to determine the natural features of a country such as lakes, streams, rivers, hills, etc.

2) Cadastral survey:

  • Used to fix the property lines, calculate the area or transfer land property from one owner to another.

3) Marine/ Hydrographical survey:

  • Deals with water bodies for navigation, water supply, harbour works, or for determining the mean sea level

4) Astronomical survey:

  • Used to determine the absolute location of any point or direction of any line on the surface of the earth.
  • Consists of a heavy body such as the sun or any fixed star.

Classification of surveying based on the object of the survey

1) Engineering survey:

  • Used to determine the quantities or to afford sufficient data for the design of engineering work

2) Military survey

  • Used for determining points of strategic importance

3) Mine survey:

  • Used for exploring the mineral wealth

4) Geological survey:

  • Used to determine the various strata in the earth's crust and their properties
UPPSC AE Civil Paper 2 Mock Test - 2 - Question 8
Which of the following is provided to prevent the river flood from entering the canal and regulating the supply of water entering the canal?
Detailed Solution for UPPSC AE Civil Paper 2 Mock Test - 2 - Question 8

Concept:

Head Sluices (or Canal Head Regulator): A head sluices or canal head regulator (CHR) is provided at the head of the off-taking canal, and serves the following functions:

  • It controls the entry of the silt into the canal.
  • It regulates the supply of water entering the canal.
  • It prevents river floods from entering the canal.

Silt Ejectors (or Silt extractors): These are the devices that extract the silt from the canal water after the silted water has traveled a certain distance in the off-take canal. These works are, therefore, constructed on the bed of the canal, and a little distance downstream from the head regulator.

Under-Sluices (or Scouring Sluices): The under sluices are the openings that are located on the same side as the off-taking canal and are fully controlled by gates, provided in the weir wall with their crest at a low level. Their main functions are:

  • It helps in controlling the silt entry into the canal.
  • It scours the silt deposited on the river bed above the approach channel.
  • It passes the low floods without dropping the shutter of the main weir.
  • It preserves a clear and defined river channel approaching the regulator.

Divide Walls: Divide wall is a masonry or concrete wall constructed at the right angle to the axis of the weir. Its main objective is to form a still and comparatively less turbulent water pocket in front of the canal head regulator so that the suspended silt can be settled down which then later be cleaned through the scouring sluices from time to time.

UPPSC AE Civil Paper 2 Mock Test - 2 - Question 9

Equation of continuity is based on the principle of conservation of

Detailed Solution for UPPSC AE Civil Paper 2 Mock Test - 2 - Question 9

Continuity equation:

Continuity equation:
The continuity equation
is the equation of conservation of mass in a fluid flow.
According to the law of conservation of mass:

Rate of flow in section 1 - 1 = Rate of flow at section 2 - 2

For Incompressible fluids

UPPSC AE Civil Paper 2 Mock Test - 2 - Question 10

For a submerged body, if the centre of buoyancy coincides with the centre of gravity, the equilibrium is called

Detailed Solution for UPPSC AE Civil Paper 2 Mock Test - 2 - Question 10

Centre of buoyancy:

  • When a solid body is either wholly or partially immersed in a fluid, the hydrostatic lift due to the net vertical component of the hydrostatic pressure forces experienced by the body is called the buoyant force.
  • The buoyant force on a submerged or floating body is equal to the weight of the liquid displaced by the body and acts vertically upward through the centroid of a displaced volume known as the centre of buoyancy.

Condition of equilibrium for a submerged body in terms of centre of buoyancy and the centre of gravity:

  • Stable equilibrium: If its centre of gravity is directly below the centre of buoyancy
  • Neutral equilibrium: If its centre of gravity is coincident with the centre of buoyancy
  • Unstable equilibrium: If its centre of gravity is directly above the center of buoyancy

Important Points

  • Condition of stable equilibrium for a floating body in terms of metacentric height (GM) as follows:
    • Stable equilibrium : GM > 0 (M is above G)
    • Neutral equilibrium : GM = 0 (M coinciding with G)
    • Unstable equilibrium : GM < 0 (M is below G)

  • A submerged body is in equilibrium when the centre of gravity is located below the centre of buoyancy.
  • If the body is tilted slightly in any direction, the buoyant force and the weight always produce a restoring couple trying to return the body to its original position.
UPPSC AE Civil Paper 2 Mock Test - 2 - Question 11

Major energy losses occur due to:

Detailed Solution for UPPSC AE Civil Paper 2 Mock Test - 2 - Question 11

There are generally two types of losses that occur in a pipe flow problem:

  1. Major loss: Major head loss occurs due to friction, which is given by:
    • Where f = friction factor = 64/Re
  2. Minor head loss: Minor head loss occurs due to:
    1. Sudden enlargement
    2. Sudden contraction
    3. Due to pipe bends
    4. head loss at the entrance and exit of pipe etc.

Minor loss:
Different type of minor losses in pipe


UPPSC AE Civil Paper 2 Mock Test - 2 - Question 12

The second twenty year road plan is also called _______.

Detailed Solution for UPPSC AE Civil Paper 2 Mock Test - 2 - Question 12

Three road plans developed for the planning and construction of the road network are:

UPPSC AE Civil Paper 2 Mock Test - 2 - Question 13

Which one of the following expresses the height of rising or fall of a liquid in a capillary tube?
Where w = Specific weight of the liquid, θ = Angle of contact of the liquid surface, σ = Surface tension, d = diameter of the capillary tube.

Detailed Solution for UPPSC AE Civil Paper 2 Mock Test - 2 - Question 13

Capillarity:
When a small diameter tube of less than 10 mm is inserted into the liquid, there may be a rise and fall of liquid in the tube with respect to the outer surface is known as capillarity.
In case of rising or fall of the capillary, the weight of the liquid is balanced by the surface tension on the surface of the liquid.
Weight of liquid, W = mg = Vρg = Vw
where, V = volume of the tube, w = ρg = specific weight
Surface tension, Fs = σcosθ × πd
where, σ = surface tension, θ = angle of contact of fluid and tube
On equating W = Fs
Vw = σcosθ × πd
(π/4)
d2 hw = σcosθ × πd
h =

for Pure water θ = 0
h =

UPPSC AE Civil Paper 2 Mock Test - 2 - Question 14

Gradients are compensated to the following extent on curves, on NG (now gauge) tracks, _____ per degree of the ​​curve or _____ whichever is minimum

Detailed Solution for UPPSC AE Civil Paper 2 Mock Test - 2 - Question 14

Concept:
Grade Compensation on Curves:
Curves provide extra resistance to the movement of trains. As a result, gradients are compensated to the following extent on curves
(a) On BG tracks, 0.04% per degree of the curve or 70/R, whichever is minimum
(b) On MG tracks, 0.03% per degree of curve or 52.5/R, whichever is minimum
(c) On NG tracks, 0.02% per degree of curve or 35/R, whichever is minimum
where R is the radius of the curve in meters.
Note:- The gradient of a curved portion of the section should be flatter than the ruling gradient because of the extra resistance offered by the curve.

UPPSC AE Civil Paper 2 Mock Test - 2 - Question 15

When the surface of the spillway is made to coincide with the shape of the lower nappe of free falling water jet, then it is known as ________

Detailed Solution for UPPSC AE Civil Paper 2 Mock Test - 2 - Question 15

Ogee spillway

  • An ogee spillway is a type of spillway design where the flow of water is controlled by a curved structure that matches the shape of the lower nappe of the water jet.
  • The spillway is designed in such a way that the water smoothly flows over the curved surface, reducing the impact and erosion caused by the water flow.
  • This design helps to prevent excessive turbulence and maintain the stability of the spillway structure.

Chute spillway:

  • A chute spillway is a type of spillway that consists of a steep, straight, and lined channel or chute. It is designed to quickly discharge water from a dam or reservoir.
  • The flow of water down the chute is controlled by its steep slope, which allows for high velocity and efficient water discharge.

Shaft spillway:

  • A shaft spillway, also known as a drop inlet spillway or bellmouth spillway, is a type of spillway structure that consists of a vertical shaft or tunnel-like structure with a large opening at the top. Water flows into the shaft and is then conveyed through an underground passage or tunnel to a lower elevation or downstream.

Drop spillway:

  • A drop spillway is a type of spillway that utilizes a vertical drop to control and release water. Water flows over a vertical structure, such as a weir or crest, and drops vertically to a lower level, often into a stilling basin or plunge pool. The purpose of the drop is to dissipate the energy of the flowing water and prevent erosion downstream.
UPPSC AE Civil Paper 2 Mock Test - 2 - Question 16
Which of the following statements regarding the requirements of a good ballast is INCORRECT?
Detailed Solution for UPPSC AE Civil Paper 2 Mock Test - 2 - Question 16

Explanation:

A good ballast should be non-porous and water-resistant because:

  • If ballast absorbs water, it loses strength and deteriorates over time.

  • Water retention can lead to vegetation growth, which reduces the effectiveness of ballast in providing stability.

  • Excess moisture can weaken the track foundation, affecting train safety.

Other Related Points

"It should be durable and should not get pulverized or disintegrated under adverse weather conditions." – Correct. A good ballast must withstand harsh weather.
"It should be tough and wear-resistant." – Correct. Toughness ensures that the ballast remains effective under repeated train loads.
"It should resist both attrition and abrasion." – Correct. Resistance to wear and tear increases the lifespan of the ballast.

UPPSC AE Civil Paper 2 Mock Test - 2 - Question 17
Rail is designated by
Detailed Solution for UPPSC AE Civil Paper 2 Mock Test - 2 - Question 17

Concept:

Rail Section:

A rail is designated by its weight per unit length.

The various important factors to be considered in deciding the weight of rail are:

1. Speed of Train

2. The axle load

3. The Gauge of Track

4. Type of rails

5. Spacing of sleepers.

UPPSC AE Civil Paper 2 Mock Test - 2 - Question 18

Which of the following methods of designation of crossing is mostly used in India?

Detailed Solution for UPPSC AE Civil Paper 2 Mock Test - 2 - Question 18

Explanation:
Crossing angle:
The angle which is formed between the gauge faces of the vee is called gauge angle.
Method of representing crossing angle:
1) Right angle method or Cole's method:

This method is generally used in Indian railway
Tanα = 1/N
α = Cot-1N
Where α = crossing angle and N = Number of crossing

2) Centerline method:
In this the measurements are taken along the line bisecting the crossing angle.
Tan (a/2) = 1/2N
α = 2Cot-1 (2N)

3) Isosceles triangle method:
In this, the measurements are done along one side of an isosceles triangle.
Sin (a/2) = 1/2N
a/2= Sin-1(1/2N)
α = 2Cosec-1(2N)

UPPSC AE Civil Paper 2 Mock Test - 2 - Question 19
Zero hardness of water is achieved by
Detailed Solution for UPPSC AE Civil Paper 2 Mock Test - 2 - Question 19

Concept:

  • Water softening: It is the process of hardness removal from the water. It is caused by multivalent cation and affects water quality.
  • Lime soda method: It is a water softening method in which lime and soda ash are added to the water, which causes the precipitation of multivalent cation as CaCO3.
  • Precipitation of CaCO3 occurs only when the pH of water is greater than 9, so in case of less pH alkalinity is added to the water. In this process small amount of Ca2+ and Mg2+ precipitates very late, which will create incrustation in the pipe, so to avoid this recarbonation is done to dissolve back this small amount of cation.
  • Due to this, the method does not give zero hardness.
  • Ion Exchange Process: Ion-exchange resin, (zeolite) exchanges one ion from the water being treated for another ion that is in the resin (sodium is one component of softening salt, with chlorine being the other). Zeolite resin exchanges sodium for calcium and magnesium. It can produce water with zero hardness.
UPPSC AE Civil Paper 2 Mock Test - 2 - Question 20
The camber provided on a sloping road is 1 in 48. Which one of the following is the ruling gradient?
Detailed Solution for UPPSC AE Civil Paper 2 Mock Test - 2 - Question 20

Explanation:

Camber:

It’s mostly found in highways the median part of the road surface is elevated with regards to the edges. This slope from the diagonal direction is known as the Camber.

  • The purpose of the camber is to drain out the rainwater from the road surface around the edges.
  • The rate of camber is normally indicated by 1: n ( 1 in n ) (1 vertical to n horizontal) or in proportions as n %

gradient = 2 × Camber

Calculation

Camber = 1 in 48

So gradient = 2 × (1 in 48) = 1 in 24

UPPSC AE Civil Paper 2 Mock Test - 2 - Question 21

The product of traffic density and traffic speed is termed as

Detailed Solution for UPPSC AE Civil Paper 2 Mock Test - 2 - Question 21

Explanation:
The relationship between the fundamental variables of traffic flow, namely speed, volume, and density is called the fundamental relations of traffic flow.
The relation is given as:
q = K × V
Where,
q is traffic flow (Volume)
K is traffic density
V is space mean speed
The product of traffic density and traffic speed is traffic volume.
The flow and density vary with time and location. A typical relation between flow- density is shown below in the graphical form:

UPPSC AE Civil Paper 2 Mock Test - 2 - Question 22

​If the base period of a 6 hr. unit hydrograph of a basin is 84 hr. then, the base period of a 12 hr. unit hydrograph of the same basin will be

Detailed Solution for UPPSC AE Civil Paper 2 Mock Test - 2 - Question 22

Explanation:
Unit hydrograph:
A unit hydrograph is a direct runoff hydrograph resulting from one unit (one cm) of constant intensity uniform rainfall occurring over the entire watershed.

Base period of 12 hr unit hydrograph = Base period of 6 hr unit hydrograph + 6
Base period of 12 hr unit hydrograph = 84 + 6 = 90h

UPPSC AE Civil Paper 2 Mock Test - 2 - Question 23
While testing for COD of sewage, organic matter is oxidised by K2Cr2O7 in the presence of
Detailed Solution for UPPSC AE Civil Paper 2 Mock Test - 2 - Question 23

Explanation

Chemical Oxygen Demand (COD)

Chemical Oxygen Demand (COD) is a test that measures the amount of oxygen required to chemically oxidize the organic material and inorganic nutrients, such as Ammonia or Nitrate, present in water.

The COD test uses potassium dichromate (K2Cr2O7) in presence of concentrated sulfuric acid (H2SO4) solution that oxidizes both organic (predominate) and inorganic substances in a waste water sample.

The regents of COD test are

  • Standard potassium dichromate(K2Cr2O7) solution (.25N)
  • Concentrated sulfuric acid (H2SO4)
  • Standard Ferrous ammonium sulphate titrant
  • Ferroin indicator solution

UPPSC AE Civil Paper 2 Mock Test - 2 - Question 24
The 'track modulus' is an index of measure of which of the following?
Detailed Solution for UPPSC AE Civil Paper 2 Mock Test - 2 - Question 24

Concept:

Track Modulus:

  • Track foundation modulus, commonly referred to as track modulus, is defined as the supporting force per unit length of rail per unit deflection.
  • Track modulus is thus a measure of the vertical stiffness of the track foundation and is related to the vertical deflection of the rail under a specified or known vertical wheel load.
  • A related parameter, track stiffness, is a measure of the vertical stiffness of the whole track structure (effects of rail included).
  • Track modulus is closely related to track performance as it is a measure of the structural state of the track.
  • The effects of the fasteners, ties, ballast, sub-ballast, and subgrade are included in the track modulus.
UPPSC AE Civil Paper 2 Mock Test - 2 - Question 25
The Bernoulli's equation is applicable only for
Detailed Solution for UPPSC AE Civil Paper 2 Mock Test - 2 - Question 25

Explanation:

Bernoulli’s equation:

  • It can be derived from the principle of conservation of energy.
  • It states that, in a steady flow, the sum of all forms of energy in a fluid along a streamline is the same at all points on that streamline.
  • It represented in head form (the total energy per unit weight).

Bernoulli’s equation assumptions:

  1. The fluid is ideal, i.e. fluid has zero viscosity
  2. Flow is steady
  3. Flow is continuous
  4. Flow is incompressible
  5. Flow is irrotational
  6. Flow is along a streamline.

Hence, the given answer is most appropriate.

Note:

An inviscid flow is a type of flow in which viscous forces are very small in comparison with inertial ones.

UPPSC AE Civil Paper 2 Mock Test - 2 - Question 26

Creep is the

Detailed Solution for UPPSC AE Civil Paper 2 Mock Test - 2 - Question 26

Concept:
Creep of rail:
Creep is defined as the longitudinal movement of the rail with respect to the sleepers. Rails tend to gradually move in the direction of dominant traffic.
Theories for the development of creep
1) Wave motion theory:

According to this theory the rails have a wavy formation, as the wheels move forward, the depression also moves with them and the previously depressed portion springs back to the original one.

2) Percussion theory:
According to this theory, creep is developed due to the impact of wheels at the rail end ahead of the joint. The continuous movement of several wheels passing over the joint pushes the facing or landing rail forward, thereby causing creep

Major causes for creep in rail are as follows:
a) Forces developed when the train is starting or stopping by application of brakes.
While starting, the wheels push the rail backward and hence the direction of creep is in the backward direction.
When brakes are applied then the wheels of the vehicles push the rails in the forward direction and hence the creep is in the forward direction.
b) Due to wave motions developed when the wheels of the vehicle hit the crest.
c) Unequal expansion/contraction due to the different temperature gradients along the track.

UPPSC AE Civil Paper 2 Mock Test - 2 - Question 27

If the length of a chord/arc is 20 m in a curve, then the relationship between R and D in the curve will be

Detailed Solution for UPPSC AE Civil Paper 2 Mock Test - 2 - Question 27

Concept:

  • Arc definition of degree of curve
    • According to this, the degree of curve is the central angle subtended by an arc of length 20 m or 30 m depending on the case.
  • ​​Chord definition of degree of curve
    • According to this, the degree of curve is the central angle subtended by a chord of length 20 m or 30 m depending on the case.

Calculation:

R = Radius of the curve
D = degree of the curve as per arc definition for 20 m arc length i.e 20 m arc length subtends an angle of Dº at the center.
Dº = D(
π/180) Radians
θ (radians = Arc/Radius = 20/R
RDπ/180 = 20
R = 1146/D
Similarly, for 30 m arc
R = 1719/D

UPPSC AE Civil Paper 2 Mock Test - 2 - Question 28

The unit power Pu of a turbine developing a power P under a head H is equal to

Detailed Solution for UPPSC AE Civil Paper 2 Mock Test - 2 - Question 28

Concept:
Unit Power (Pu)

  • The Unit Power of a given turbine is the power output of the turbine when operating under a head of 1 meter assuming no change in efficiency.
  • Used for single Turbine to find power at various Heads.

Unit Power
= P/H3/2Other Related PointsUnit speed (Nu)
= N /√ H
Unit discharge (Qu)
= Q /√ H

UPPSC AE Civil Paper 2 Mock Test - 2 - Question 29
Following errors are eliminated during reciprocal levelling
Detailed Solution for UPPSC AE Civil Paper 2 Mock Test - 2 - Question 29

Explanation:

Reciprocal levelling is adopted to accurately determine the level difference between two points which are separated by obstacles like a river, ponds, lakes, etc.

It eliminates the following errors:

  • error in instrument adjustments i.e error due to collimation
  • the combined effect of Earth's curvature and the refraction of the atmosphere
  • variation in the average refraction

Other Related Points

  • Reciprocal levelling eliminates the error due to collimation and error due to curvature of Earth completely, but as the refraction depends upon the atmosphere which may change every minute;
UPPSC AE Civil Paper 2 Mock Test - 2 - Question 30
If 'f' is the focal length of camera and 't' is the tilt angle, distance of the photo nadir from the principal point will be
Detailed Solution for UPPSC AE Civil Paper 2 Mock Test - 2 - Question 30

Concept:

Distance of the photo nadir from the principal point

= f tan t

f = focal length of the camera lens and t = tilt angle

Other Related Points

Isocenter:

  • The point on the photograph at which the bisector of the angle of tilt meets the photographic plane.

Principal point:

  • Point of the intersection of the optical axis of the camera with the photographic plane.

Plumb points:

  • The points at which the vertical line through the optical center meets the photographic plane and the ground surface are the plumb points
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