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Practice Test: Civil Engineering (CE)- 13 - Civil Engineering (CE) MCQ


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30 Questions MCQ Test GATE Civil Engineering (CE) 2025 Mock Test Series - Practice Test: Civil Engineering (CE)- 13

Practice Test: Civil Engineering (CE)- 13 for Civil Engineering (CE) 2024 is part of GATE Civil Engineering (CE) 2025 Mock Test Series preparation. The Practice Test: Civil Engineering (CE)- 13 questions and answers have been prepared according to the Civil Engineering (CE) exam syllabus.The Practice Test: Civil Engineering (CE)- 13 MCQs are made for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Practice Test: Civil Engineering (CE)- 13 below.
Solutions of Practice Test: Civil Engineering (CE)- 13 questions in English are available as part of our GATE Civil Engineering (CE) 2025 Mock Test Series for Civil Engineering (CE) & Practice Test: Civil Engineering (CE)- 13 solutions in Hindi for GATE Civil Engineering (CE) 2025 Mock Test Series course. Download more important topics, notes, lectures and mock test series for Civil Engineering (CE) Exam by signing up for free. Attempt Practice Test: Civil Engineering (CE)- 13 | 65 questions in 180 minutes | Mock test for Civil Engineering (CE) preparation | Free important questions MCQ to study GATE Civil Engineering (CE) 2025 Mock Test Series for Civil Engineering (CE) Exam | Download free PDF with solutions
Practice Test: Civil Engineering (CE)- 13 - Question 1
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A monkey is trying to reach the top of a tree. He climbs 3 feet in a minute but slides down by 1.5 feet after each climb. If he reaches the top of the tree in 58 minutes, then the maximum possible height of the tree is ________ feets.

Detailed Solution for Practice Test: Civil Engineering (CE)- 13 - Question 1
You have to carefully observe the point that at the last minute, we DO NOT have to consider the sliding down of the monkey since the question states that the monkey reaches the top of the tree. Out of 58 minutes, the climb by the monkey in the 58th minute is 3 feet whereas for each of his earlier climbs, the monkey covers a net distance of 3 – 1.5 = 1.5 feet.

We can work out the maximum possible height of the tree as 1.5 x 57 + 3 = 85.5 + 3 = 88.5 feet

Practice Test: Civil Engineering (CE)- 13 - Question 2
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In the following question, a sentence has been given in Direct/Indirect Speech. Out of the four alternatives suggested, select the one which best expresses the same sentence in Indirect/Direct Speech.

My husband said to me, “Wait for me outside."

Detailed Solution for Practice Test: Civil Engineering (CE)- 13 - Question 2
This is an imperative sentence as there is instruction being given by the husband to his wife. Since the reported speech is not signifying the mode of behaviour here whether it was a request, a command or was it said in anger, we cannot use verbs related to such emotions. We'll simply convert "said to" into "told". In affirmative imperative sentences, we use "to" for connecting reporting and reported speech. The correct answer is option A as it follows the above rules correctly.
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Practice Test: Civil Engineering (CE)- 13 - Question 3
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What of the following function(s) in an accurate description of the graph for the range(s) indicated?

A. y = 2x + 4 for -3 ≤ x ≤ -1

B. y = |x-1| for -1 ≤ x ≤ 2

C. y = ||x|-1| for -1 ≤ x ≤ 2

D. y = 1 for 2 ≤ x ≤ 3

Detailed Solution for Practice Test: Civil Engineering (CE)- 13 - Question 3
Put value and verify

(i) y = 2x+4 is true in -3 ≤ x ≤ -1

On putting x = -3, y = 2 and x = -2,y = 0 and x = -1, y = 2

(ii) y = |x-1| is also true (x = -1, y = 2), (x = 0 ,y = 1) and (x = 1, y = 0)

(iv) y = 1 in (2 ≤ x ≤ 3) always true

(i), (ii) and (iv) are true.

Practice Test: Civil Engineering (CE)- 13 - Question 4
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A rectangular room having dimensions 3.74 m x 5.78 m is required to be tiled using a minimum number of identical square tiles. The number of such tiles and area of each tile (in cm2) is given by
Detailed Solution for Practice Test: Civil Engineering (CE)- 13 - Question 4
In this question, we need to find the HCF of 374 and 578 which is 34. We can look at 374 = 34 x 11 and 578 = 34 x 17. This gives the number of square tiles required to cover the entire room is 11 x 17 = 187 and the area of each such tile is 342 = 1156 cm2.
Practice Test: Civil Engineering (CE)- 13 - Question 5
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In the following question, out of the four alternatives, select the alternative which best expresses the meaning of the Idiom/Phrase.

To take into account
Detailed Solution for Practice Test: Civil Engineering (CE)- 13 - Question 5
The phrase "take into account" means to consider or remember something when judging a situation. Thus, option A is the correct answer.
Practice Test: Civil Engineering (CE)- 13 - Question 6
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In an examination, a candidate attempts all the questions since there is no negative marking and all the questions carry equal marks. Out of the first 30 questions that he attempts, he has marked correct options for 20 questions whereas out of the remaining questions, he is able to mark only one third question with correct option. His overall score in the examination is 50%.

The total number of questions in the paper are;
Detailed Solution for Practice Test: Civil Engineering (CE)- 13 - Question 6

We can form equations based on the given information. Let ‘x’ be the number of questions attempted by him in the second part.

Or 40+ 2x/3 = 30 + x which leads us to get x/3 = 10 or x = 30

The total number of questions in the examination are x + 30 = 60

Practice Test: Civil Engineering (CE)- 13 - Question 7
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Amol, Brijendra, Chander and Dron are sitting around a roundtable. One of them is an electronics engineer (ECE) who is sitting to the left of the Mechanical Engineer (ME). Amol is sitting opposite a Computer Science Engineer (CSE). It is also known that Dron likes to play computer games and Brijendra is sitting to the right of Civil Engineer (CE). Identify the correct match from the given choices.
Detailed Solution for Practice Test: Civil Engineering (CE)- 13 - Question 7

Based on the given data, we can say that Amol (ME) is sitting opposite to CSE – Brijendra. Civil Engineer has to be to the right of Amol – Mechanical Engineer. However, it is not possible to say whether Chander is an Electronics Engineer or a Civil Engineer AND likewise for Dron.

Practice Test: Civil Engineering (CE)- 13 - Question 8
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In five cities A is more populated than B which is less populated then E. D is more populated than E but not as populated as A. C is less populated than B. Which city has the highest population?
Detailed Solution for Practice Test: Civil Engineering (CE)- 13 - Question 8

From first statement

A > B and B < E

From second statement

D > E and D < A

From third statement

C < B

From all statements together

A > D > E > B > C

Practice Test: Civil Engineering (CE)- 13 - Question 9
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One-third of Ramesh’s marks in Arithmetic is equal to half his marks in English. If he gets 150 marks in the two subjects to gather, how many marks has he got in English?
Detailed Solution for Practice Test: Civil Engineering (CE)- 13 - Question 9

Let marks obtained by Ramesh in Arithmetic be ‘A’ and marks obtained in English be E.

As the information given,

1/3 A = E/2

à (A/3) – (E/2) = 0

à 2A – 3E= 0 ……….(i)

A + E = 150 …………(ii)

From equation (i) and (ii), E = 60

Hence, Ramesh got 60 marks in English.

Practice Test: Civil Engineering (CE)- 13 - Question 10
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In a test given by Prof Virus during the minors, Raju gets 32% marks and fails by 20 marks whereas Chatur gets 30 marks more than passing marks while scoring 42% marks. The marks required to get twice the passing marks are _______.
Detailed Solution for Practice Test: Civil Engineering (CE)- 13 - Question 10

The difference in percentage marks obtained by Raju and Chatur is 42 – 32 = 10% whereas the difference in their marks is 20 + 30 = 50 (Raju gets 20 marks less while Chatur gets 30 marks more than pass marks).

This gives us 1% = 5 marks and the passing marks as 20 + 32% = 180.

The required marks = 360 (twice the passing marks)

Practice Test: Civil Engineering (CE)- 13 - Question 11
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During the subsurface investigations for design of foundations, a standard penetration test was conducted at 4.5m below the ground surface. The record of number of blows is given below.

Q. Assuming the water table at ground level, soil as fine sand and correction factor for overburden as 1.0, the corrected ‘N’ value for the soil would be

Detailed Solution for Practice Test: Civil Engineering (CE)- 13 - Question 11
Standard penetration test – number of penetration for the first 15 cm is not counted and only the number of blows which causes further 30 cm of penetration is taken into account.

So, here N = 6 + 6 + 8 + 7 = 27

Now, here N > 15 and water table at ground level so dilatancy correction is applied.

Ncorrected = 15 + 1/2(N−15)

= 15 + 1/2(27 - 15) = 21

*Answer can only contain numeric values
Practice Test: Civil Engineering (CE)- 13 - Question 12
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A beam-column having overall depth and width is 550 mm and 250 mm. If torsional moment for this beam is 30 kN.m then twisting moment will be-

Detailed Solution for Practice Test: Civil Engineering (CE)- 13 - Question 12
Given, b = 250 mm, D = 550 mm

& T = 30 kN .m

So, the twisting moment

MT = T (1 + D/b) / 1.7

= 30 (1 + 550/250) / 1.7

= 56.47 kN .m

Practice Test: Civil Engineering (CE)- 13 - Question 13
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An isolated ‘T’ beam is used on the walkway. The beam is simply supported with an effective span of 6 m. Effective width of flange for shown figure is :
Detailed Solution for Practice Test: Civil Engineering (CE)- 13 - Question 13
Effective width for isolated T beam is given as

Hence bf = 1000 mm

Practice Test: Civil Engineering (CE)- 13 - Question 14
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What is the maximum possible eccentricity in a prestressed concrete beam of circular cross-section? Diameter of the section is d. Tension is not allowed anywhere and any time in the cross-section. Neglect dead load (self-weight).
Detailed Solution for Practice Test: Civil Engineering (CE)- 13 - Question 14

Practice Test: Civil Engineering (CE)- 13 - Question 15
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Deposit with flocculated structure is formed when
Detailed Solution for Practice Test: Civil Engineering (CE)- 13 - Question 15

  • The process by which individual particles of clay aggregate into coltlike masses or precipitate into small lumps. Flocculation occurs as a result of a chemical reaction between the clay particles and another substance, usually salt water.

  • In the case of flocculated structure, there will be edge to edge and edge to face contact between the particles. This type of formation is due to the net electrical forces between the adjacent particles at the time of deposition being attractive in nature.

  • The concentration of dissolved minerals in water leads to formation of flocculated structures with very high void ratio as in the case of marine deposits.

*Answer can only contain numeric values
Practice Test: Civil Engineering (CE)- 13 - Question 16
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A simply supported prestressed beam of 300 × 500 mm in c/s is subjected to a superimposed load of 20 kN/m over a span of 10m. If a prestressing force of 1000 kN is applied through a straight tendon located along centroidal axis then what is the extreme top fiber stress at the mid section:

Detailed Solution for Practice Test: Civil Engineering (CE)- 13 - Question 16

Practice Test: Civil Engineering (CE)- 13 - Question 17
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What will be the maximum possible uniformly distributed load (inclusive of self-weight) over the entire span of a simply supported beam of span ‘L’ such that the deflection at midspan at service condition is zero? The cross-section is rectangular. The prestressing force ‘P’ is applied with uniform eccentricity ‘e’. Assume no losses.
Detailed Solution for Practice Test: Civil Engineering (CE)- 13 - Question 17

For no deflection at midspan,

[δ due to self (weight + udl)] + [δ due to eccentric force P on cable] = 0

Practice Test: Civil Engineering (CE)- 13 - Question 18
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If Qs4 and Qs6 are the equilibrium discharges of S- curve for the catchment obtained by 4 hour UH summation and 6 hour UH summation, respectively, then Qs4/Qs6 is __________. (Up to 2 decimal places)
Detailed Solution for Practice Test: Civil Engineering (CE)- 13 - Question 18

Qe = A.1D

Qs4/Qs6 = 6/4 = 1.50

*Answer can only contain numeric values
Practice Test: Civil Engineering (CE)- 13 - Question 19
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The beam PQR shown in the given figure is horizontal. The distance to the point of contraflexure from the fixed end P is _____m. (up to two decimal place)

Detailed Solution for Practice Test: Civil Engineering (CE)- 13 - Question 19
Let the point of contraflexure be at a distance ′x′ from the free end.

15x = 26.25(x−1)

x = 2.334m

→ Distance from fixed end = 3 - 2.334 = 00.666m.

Practice Test: Civil Engineering (CE)- 13 - Question 20
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30 KN/m2 load acting on a 4m × 6m 2-way slab. Effective Load acting in a shorter direction is_____ KN/m2.
Detailed Solution for Practice Test: Civil Engineering (CE)- 13 - Question 20

wx = Load acting in shorter direction

wy = Load acting in shorter direction

wx + wy = w

Equating deflection of both strips.

Practice Test: Civil Engineering (CE)- 13 - Question 21
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A stream function is given by:

Ψ = 3xy2 + (x2+2)y

The flow rate across a line joining points A(0,5) and B(1,2) is
Detailed Solution for Practice Test: Civil Engineering (CE)- 13 - Question 21
ΨA = (3 × 0 × 52) + ((02 + 2)×5) = 10

ΨB = (3×1×22) + ((12 + 2) × 2) = 18

Flow rate across a line joining points A and B is

= ΨB − ΨA

= 18 − 10

= 8 units.

Practice Test: Civil Engineering (CE)- 13 - Question 22
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Before passage of a surge, the depth and velocity of flow at a section are 1.8 m and 3.6 m/s. After the surge passage, they are 0.6 m and 8 m/s respectively. The speed of the surge is?
Detailed Solution for Practice Test: Civil Engineering (CE)- 13 - Question 22

Practice Test: Civil Engineering (CE)- 13 - Question 23
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A rectangular plate 1.2 x 3 m is immersed in a liquid of relative density 0.85 with its 1.2 m side horizontal and just at the water surface. If the plane of plate makes an angle of 50o with the horizontal, the pressure force on one side of the plate is
Detailed Solution for Practice Test: Civil Engineering (CE)- 13 - Question 23

Density of liquid = 0.85 x 1000 = 850 kg/m3

h = 1.5 sin 50o = 1.149 m

Pressure force on one side of plate = ρgAh = 850 x 9.81 x (1.2 x 3) x 1.149 = 34.5 kN.

Practice Test: Civil Engineering (CE)- 13 - Question 24
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A beam with both continuous ends has a clear span of 6m and supports 600 mm wide. The effective depth of the beam is 550 mm. As per IS 456-2000, the effective span of the beam (in m) is____________
Detailed Solution for Practice Test: Civil Engineering (CE)- 13 - Question 24

As per IS 456 —2000, the effective span of continuous beam

i) When width of support < 1/12="" of="" clear="" />

ii) when width of support > 1/12 of clear span

a) When one end is fixed & other end continuous or both ends continuous.

leff = lo

b) When one end is continuous & the other end is simply supported.

As per given data, width of support = 600 mm & 1/12 of clear span = 500 mm

Hence leff = lo = 6m [since both ends are continuous]

Practice Test: Civil Engineering (CE)- 13 - Question 25
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A turbine develops 2516 kW at 240 rpm. The torque in the shaft is approximately:
Detailed Solution for Practice Test: Civil Engineering (CE)- 13 - Question 25
As we know P = (2πNT)/60

Where, T = torque and N = speed in rpm

So, T = 60P/(2πN) = (60×2516×1000)/(2×3.14×240) = 100108Nm = 100kNm

Practice Test: Civil Engineering (CE)- 13 - Question 26
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A Kaplan turbine has a runner of diameter 4.0 m. The diameter of the hub is 1.6 m. If the velocity of flow and the swirl velocity at the inlet side of the blade at the hub are 6.0 m/s and 10.0 m/s, respectively, the flow and swirl velocities at the inlet side of the tip are, respectively:
Detailed Solution for Practice Test: Civil Engineering (CE)- 13 - Question 26
Since the area of flow remains the same.

The flow velocity remains the same throughout.

Hence Vf2 = Vf1 = 6 m/s

D1 = 1.6; D2 = 4; u2 = 10m/s

Hence u1 = 4 m/s

Practice Test: Civil Engineering (CE)- 13 - Question 27
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The plastic moment at collapse is
Detailed Solution for Practice Test: Civil Engineering (CE)- 13 - Question 27
Number of plastic hinges needed are

Practice Test: Civil Engineering (CE)- 13 - Question 28
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A pipe of 30 cm diameter conveying 300 l/s of water has a right-angled bend in a horizontal plane. The resultant force exerted on the bend if pressure at inlet and outlet of the bend are 24.525N/cm2 and 23.544N/cm2 is ___ N.
Detailed Solution for Practice Test: Civil Engineering (CE)- 13 - Question 28

D = 30 cm

A1 = A2 = π/4D2 = 0.07068 m2

V = V1 = V2 = Q/A = 0.3/0.07068 = 4.244 m/s

Now, θ = 90°,

P1 = 24.525 N/cm2 = 24.525 × 104 N/m2

P2 = 23.544 × 104 N/m2

Force along x-axis on bend,

LFx = pQ[V1x - V2x]+(P1xA1) + P2xA2

Fx = 1000 × 0.3[4.244-0] + 24.525 × 104 × 0.07068 + 0

Fx = 18607.5N Similarly, force along y-axis on bend,

Fy = pQ [V1y - V2y] + P1yA1 + P2yA2

Fy = 1000 × 0.3 [0- 4.244 ] + 0 - 235440 × 0.07068

Fy = 17914.1 N

FR =

Practice Test: Civil Engineering (CE)- 13 - Question 29
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A sample of saturated cohesionless soil tested in a drained triaxial compression test showed an angle of internal friction of 30°. The deviator stress at failure for the sample at a confining pressure of 200kPa is equal to
Detailed Solution for Practice Test: Civil Engineering (CE)- 13 - Question 29
σ1 = σ3tan2α

Where, α = 45 + ϕ/2

α = 45∘ + 30∘/2 = 60∘, σ1 = 3σ3 = 3 × 200 = 600 kPa

∴ Deviatoric stress = σ1 - σ3

= 600 - 200 = 400 kPα

Practice Test: Civil Engineering (CE)- 13 - Question 30
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The daily cover of MSW landfills consists of which one of the following?
Detailed Solution for Practice Test: Civil Engineering (CE)- 13 - Question 30
The municipal solid waste (MSW) is simply brought and dumped on the ground, spread in layers of about 0.5 m thickness, and compacted Another layer of 0.5 m thickness is then placed on top of the previous layer and also compacted Layering and compacting are repeated until a height of about 1.5 m is reached. At this point and at the end of a working day, a cover of earth of about 0.15 to 0.3 m thickness is compacted on the top and side slopes of the compacted heap, which is called a cell. This cover is called the daily cover.
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