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Practice Test: Civil Engineering (CE)- 15 - 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)- 15

Practice Test: Civil Engineering (CE)- 15 for Civil Engineering (CE) 2024 is part of GATE Civil Engineering (CE) 2025 Mock Test Series preparation. The Practice Test: Civil Engineering (CE)- 15 questions and answers have been prepared according to the Civil Engineering (CE) exam syllabus.The Practice Test: Civil Engineering (CE)- 15 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)- 15 below.
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Practice Test: Civil Engineering (CE)- 15 - Question 1

For a > b, if a + b = 5 and ab = 6, then the value of (a2 - b2) is

Detailed Solution for Practice Test: Civil Engineering (CE)- 15 - Question 1
Since, (a + b)2 = a2 + b2 + 2ab and (a – b)2 = a2 + b2 - 2ab

→ (a - b)2 = (a + b)2– 4ab ………………….(equation)

Given, (a + b) = 5 and ab = 6

Substituting the values in equation, we get,

(a-b)2 = 52 – 4 x 6 = 1

→ (a - b) = 1

We know that, (a2 – b2) = (a + b)(a - b)

Substituting the values of (a + b) and (a - b) in above equation, we get,

(a2 – b2) = 5 × 1 = 5

Hence, the required answer is 5.

Another Solutions:

let, a= 3 & b=2 a>b

a+b=5

3+2= 5,

ab =6, 2*3=6,

(a2 - b2)= 32- 22 = 9- 4 = 5.

Practice Test: Civil Engineering (CE)- 15 - Question 2

In the following question, out of the given alternatives, select the word opposite in meaning to the given word.

Inevitable

Detailed Solution for Practice Test: Civil Engineering (CE)- 15 - Question 2
Inevitable (adj.): certain to happen, unavoidable.

Obligatory (adj.): so customary or fashionable as to be expected of everyone or on every occasion.

Prescribed (adj.): state authoritatively or as a rule that (an action or procedure) should be carried out.

Assured (adj.): made (something) certain to happen.

Avoidable (adj.): able to be avoided or prevented.

So, the correct answer is option D.

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Practice Test: Civil Engineering (CE)- 15 - Question 3

What is the difference between the expenditure made by a university for the publication of journals and psychology?

Total Expenditure : Rs. 60 Lacs

Expenditure of funds By University for various purposes

Detailed Solution for Practice Test: Civil Engineering (CE)- 15 - Question 3
Expenditure made by university of publication of journals

= 15% of 60 lacs

= 15 × 60 lacs/ 100

= 9 lacs

Expenditure made by university of psychology

= 10% of 60 lacs

= 10 × 60 lacs/100

= 6 lacs

Difference between the expenditure made by university of publication of journals and psychology

= 9-6

= 3 lacs

Practice Test: Civil Engineering (CE)- 15 - Question 4

In the following questions groups of four words are given. In each group, one word is spelt.

Find the correctly spelled word.

Detailed Solution for Practice Test: Civil Engineering (CE)- 15 - Question 4
The correctly spelled word is ‘Propitiate’ that means proliferate, appropriate, appreciate.
Practice Test: Civil Engineering (CE)- 15 - Question 5

Select the most appropriate antonym of the given word.

INNOCUOUS

Detailed Solution for Practice Test: Civil Engineering (CE)- 15 - Question 5
Let us understand the meaning of the given words :-

Innocuous = not harmful or offensive.

Eg. : The conversation was innocuous and she wasn't identified by name.

Loquacious = tending to talk a great deal; talkative.

Deadly = causing or able to cause death.

Eg. : He was acquitted on charges of assault with a deadly weapon.

Ferocious = frightening and violent.

Harmless = not causing or capable of causing harm.

Hence, option C is the correct answer.

Practice Test: Civil Engineering (CE)- 15 - Question 6

Choose the most appropriate word from the options given below to complete the following sentence.

Communication and interpersonal skills are ......... important in their own ways.

Detailed Solution for Practice Test: Civil Engineering (CE)- 15 - Question 6

The correct word is “both” as communication and interpersonal skills are two skills that are important.

Practice Test: Civil Engineering (CE)- 15 - Question 7

In the following question, some of the sentences have errors and some have none. Find out which part of a sentence has an error. The number of that part is the answer. If there is no error, your answer is (4) i.e., No error.

We requested the watchman (1)/ to clean up the basement (2)/ so that the children had enough space to play (3)/ No error (4)

Detailed Solution for Practice Test: Civil Engineering (CE)- 15 - Question 7

We ordered/asked the watchman should be used as the sense of the given sentence shows order. Hence, there is an error in (1).

Practice Test: Civil Engineering (CE)- 15 - Question 8

The Palghat Gap (or Palakkad Gap), a region about 30 km wide in the southern part of the Western Ghats in India, is lower than the hilly terrain to its north and south. The exact reasons for the formation of this gap are not clear. It results in the neighbouring regions of Tamil Nadu getting more rainfall from the South West monsoon and the neighbouring regions of Kerala having higher summer temperatures.

Q. What can be inferred from this passage?

Detailed Solution for Practice Test: Civil Engineering (CE)- 15 - Question 8

Here the passage is asking, ” what can be inferred ? “, i.e. it’s asking about the hidden conclusion of the passage.

Why not Option A ? – because it is clearly given in the passage : “The exact reasons for the formation of this gap are not clear “.

Why not Option B ? – because the passage nowhere talks about the low-lying regions of Kerala and Tamil Nadu near the Palghat Gap.

Why not option D ? – Because the passage only says : “neighbouring regions of Kerala having higher summer temperatures”, it doesn’t say that this high temperature results in rainfall.

Note: We have to think and conclude only from what is given in the passage. We can’t make our own conclusions.

Why Option C ? – because the passage says : ” It results in the neighbouring regions of Tamil Nadu getting more rainfall from the South West monsoon and the neighbouring regions of Kerala having higher summer temperatures”,

Hence we can conclude here that the weather is getting affected due to the Gap.

Therefore, option C is correct.

Practice Test: Civil Engineering (CE)- 15 - Question 9

It takes 30 minutes to empty a half-full tank by draining it at a constant rate. It is decided to simultaneously pump water into the half-full tank while draining it. What is the rate at which water has to be pumped in so that it gets hilly filled in 10 minutes?

Detailed Solution for Practice Test: Civil Engineering (CE)- 15 - Question 9

Given in the question that; Vhalf = 30(s)

Let us take drawing rate = s

Total volume = 60 S tank

(s1)(10) – (s)10 = 30s

s1(s) – s = 3s

s1 = 4s

s1 = 4 times the drawing rate.

Practice Test: Civil Engineering (CE)- 15 - Question 10

While receiving the award, the scientist said, "I feel vindicated”. Which of the following is closest in meaning to the word ‘vindicated’?

Detailed Solution for Practice Test: Civil Engineering (CE)- 15 - Question 10

Vindicated means freed from any question of guilt, so “substantiated” is the closest.

Practice Test: Civil Engineering (CE)- 15 - Question 11

The area enclosed between the parabola y = x2 and the straight line y = x is

Detailed Solution for Practice Test: Civil Engineering (CE)- 15 - Question 11
Given: y = x2 & y = x.

On solving given equation, we get the intersection points as,

y = x2 puty = x

x = x2

x2−x = 0

x(x−1) = x = 0,1

Then from y = x

For x = 0 y = 0

& x = 1 y = 1

We can see that curve y = x2 and y = x intersects at point (0,0) and (1,1) So, the area bounded by both the curves is

Area is never Negative.

Practice Test: Civil Engineering (CE)- 15 - Question 12

Which of the following statement is INCORRECT?

Detailed Solution for Practice Test: Civil Engineering (CE)- 15 - Question 12
  • Shear parameters ‘c’ and ∅ are dependent of water content of soil

  • Vane shear test is a useful method of measuring the shear strength of clay. It is a cheaper and quicker method. The test can also be conducted in the laboratory.

  • The laboratory vane shear test for the measurement of shear strength of cohesive soils, is useful for soils of low shear strength (less than 0.3kg/{cm}^2) for which triaxial or unconfined tests cannot be performed.

  • The test gives the undrained strength of the soil. The undisturbed and remoulded strength obtained are useful for evaluating the sensitivity of soil.

This test is also useful when the soil is soft and its water content is nearer to the liquid limit.

Practice Test: Civil Engineering (CE)- 15 - Question 13

A manometer is used to measure the pressure in a tank. The fluid used has a specific gravity of 0.86, and the manometer column height is 50cm, as shown in figure If the local atmospheric pressure is 100kPa, calculate the absolute pressure within the tank.

Take the standard density of water to be 1000kg/m3.

Detailed Solution for Practice Test: Civil Engineering (CE)- 15 - Question 13

The specific gravity of the manometer fluid is given to be 0.86

ρ = SG(ρH2O) = (0.86) (1000 kg/m3) =860 kg/m3

Then from Eq. ,

Practice Test: Civil Engineering (CE)- 15 - Question 14

A construction project consists of twelve activities. The estimated duration (in days) required to complete each of the activities along with the corresponding network diagram is shown below.

Total floats (in days) for the activities 5-7 and 11-12 for the project are, respectively,

Detailed Solution for Practice Test: Civil Engineering (CE)- 15 - Question 14

(5)-(7) = Total float = te-tL= 0

(11)-(12) = Total Float = te-tL = 0

Practice Test: Civil Engineering (CE)- 15 - Question 15

A cylindrical shell 1000 mm long, 150 mm internal diameter, having a thickness of 8 mm is filled with a fluid at atmospheric pressure. An additional 25000 mm3 of fluid is pumped into the cylinder, then the hoop stress-induced will be ___________ N/mm2[E=2×105 N/mm2 and μ=0.3]

Detailed Solution for Practice Test: Civil Engineering (CE)- 15 - Question 15
Volume tics train=Change in volume/Original volume=

simplifying and neglecting the products and squares of small quantities,

i.e .δdδL , hence = 2d.δd.L+δL.d2/d2L

= δL/L+ 2.δd/d Bt defintion δL/L

= Longitudinal strain

δdd= hoop strain

Thus Volumetric strain = longitudinal strain + 2 × hoop strain

ϵV = 2ϵh + ϵL

25000/π/4(150)2 × 1000

= pD/4tE(5-4μ) ⇒ 1.4147×10-3

Practice Test: Civil Engineering (CE)- 15 - Question 16

There is a free overfall at the end of a long open channel. For a given flow rate, the critical depth is less than the normal depth. What gradually varied flow profile will occur in the channel for this flow rate?

Detailed Solution for Practice Test: Civil Engineering (CE)- 15 - Question 16
For a given flow rate, if critical depth (Yc) is less than normal depth (Yn)

i.e. if Yc < Yn

Then, the flow profile will be M2.

Practice Test: Civil Engineering (CE)- 15 - Question 17

For a given PERT activity, the standard deviation of the time needed to complete an activity is 3. If the optimistic time and the most likely time required to complete this activity are 8 and 12 minutes respectively. Find out the expected time in order to complete this activity.

Detailed Solution for Practice Test: Civil Engineering (CE)- 15 - Question 17
Reason: σ = tp - t0/6

σ = 3 = (tp-t0)/6

tp-t0 = 6×3, tp-t0 = 18

We know that, t0 = 8

So,

tp- 8 =18 And, therefore, tp = 26

We have, te = t0+4tm+tp/6

Practice Test: Civil Engineering (CE)- 15 - Question 18

The concentration of Sulfur Dioxide (SO2) in the ambient atmosphere was measured as 30 µg/m3. Under the same conditions, the above SO2 concentration expressed in ppm is ___________.

Given: (P/(RT) = 41.6 mol/m3; where, P = Pressure; T = Temperature; R – universal gas constant; Molecular weight of SO2 = 64.

Detailed Solution for Practice Test: Civil Engineering (CE)- 15 - Question 18
1 m3 of air has 30 mg SO2

106 m3 of air has 30g SO2

= 30/64molSO2

Concentration of SO2 in ppm = 0.0113 ppm

Practice Test: Civil Engineering (CE)- 15 - Question 19

A steady, incompressible flow is given by:

u = 2x2 + y2 and v= -4xy , What is the convective acceleration along x-direction at point (1, 2)

Detailed Solution for Practice Test: Civil Engineering (CE)- 15 - Question 19
Convective acceleration along x direction at point (1,2)

Practice Test: Civil Engineering (CE)- 15 - Question 20

A two-faced fair coin has its faces designated as head (H) and tail (T). This coin is tossed three times in succession to record the following outcomes: H, H, H. If the coin is tossed one more time, the probability (up to one decimal place) of obtaining H again, given the previous realizations of H, H and H, would be_____.

Detailed Solution for Practice Test: Civil Engineering (CE)- 15 - Question 20
Given the first three are already heads. If the coin is tossed again, the outcome does not depend on previous outcomes.

Probability getting head = 1/2 = 0.5.

Practice Test: Civil Engineering (CE)- 15 - Question 21

The function f(x) = 2x3 – 3x2 – 36x + 2 has its maxima at

Detailed Solution for Practice Test: Civil Engineering (CE)- 15 - Question 21
For maximum f(x),

⇒ f'(x) = 0

⇒ 6x2−6x−36=0

⇒ x2−x−6=0

⇒ x2−3x+2x−6=0

⇒ x(x-3)+2(x-3)=0

⇒ x = 3, -2

∴ f''(x) = 12x-6

⇒ f''(x)<0 @x=-2

So, f(x) has a maximum value at -2 only.

Practice Test: Civil Engineering (CE)- 15 - Question 22

The ratio of tension and shear force in each bolt of the joint, as shown in figure, is

Detailed Solution for Practice Test: Civil Engineering (CE)- 15 - Question 22

Tension in each bolt

The ratio of tension and shear in each bolt

= Tension in each bolt / Shear in each bolt

= 34.62/14.42 = 2.40

Practice Test: Civil Engineering (CE)- 15 - Question 23

Three values of x and y are to be fitted in a straight line in the form y = a + bx by the method of least squares. Given : N=3, ∑x = 6, ∑y = 21, ∑x2 = 14 and ∑xy = 46, the values of a and b are respectively.

Detailed Solution for Practice Test: Civil Engineering (CE)- 15 - Question 23
Given,

N = 3, ∑x = 6, ∑y = 21, ∑x2 = 14 and ∑xy = 46

There are two normal equations to evaluate a and b of the equation

y = a + bx

∑y = Na + b∑x …(i)

and ∑xy = a∑x + b∑x2 …(ii)

From Eq. (i), 21 = 3a + 6b …(iii)

From Eq. (ii) 46 = 6a + 14b

⇒ 23 = 3a + 7b …(iv)

Now, Eq. (iv) – Eq. (iii),

3a + 7b = 23

3a + 6b = 21

- - -

b = 2

From Eq. (iii), 3a = 21 – 12

3a = 9 ⇒ a = 3

Practice Test: Civil Engineering (CE)- 15 - Question 24

A RC column of square cross-section (400×400mm2) has its column load-moment interaction diagram as shown in figure below.

Detailed Solution for Practice Test: Civil Engineering (CE)- 15 - Question 24

We have from the interaction diagram ∴Mu 0.3 x 20 x 400 x 4002 x 10-3 ⇒ Mu = 384000 kN -mm

e = Mu/Pu = 384000/640 = 600 mm

Practice Test: Civil Engineering (CE)- 15 - Question 25

At two points 1 and 2 in a pipeline the velocities are V and 2V, respectively. Both the points are at the same elevation. Both the points are at the same elevation. The fluid density is ρ. The flow can be assumed to be compressible, inviscid, steady and irrotational. The difference in pressures P1 and P2 at points I and 2 is

Detailed Solution for Practice Test: Civil Engineering (CE)- 15 - Question 25
Applying Bernouli's theorem

Practice Test: Civil Engineering (CE)- 15 - Question 26

In a network, a project activity has normal cost and normal duration of Rs. 400 and 8 days respectively. The same activity has crash cost and crash duration of Rs. 800 and 6 days. The crash cost slope of activity is;

Detailed Solution for Practice Test: Civil Engineering (CE)- 15 - Question 26
Cost Slope: In actual practice, we assume the time-cost curve to be linear between crash point A and Normal point B. The curve(shown below) shows that there is a change in cost for change in time. The term ‘cost-slope’ is defined as the “increase in the cost of the activity per unit decrease in the time”.

Mathematically, we can say,

Cost slope = [Crash cost – Normal cost]/[Normal time – Crash time]

In simple words, we can say that the cost slope of an activity is the increase in cost of an activity by reducing the duration by one unit.

So, Cost Slope = (Crash Cost-Normal Cost)/(Normal time - Crash time) = 800-400/8-6 = 200.

Practice Test: Civil Engineering (CE)- 15 - Question 27

Precast concrete pile of size 50 cm × 50 cm is to be driven into clay strata. The unconfined compressive strength of which is 110 kN/m2. Then the length required to carry a safe working load of 500 kN with FOS = 3 and α = 0.6 is _______ m.

Detailed Solution for Practice Test: Civil Engineering (CE)- 15 - Question 27
Soil is clay ≡ point bearing

Qu = CubNcAb + aCu avAs

Qu = 500 × FOS = 500×3 = 1500

1500 = 110/2 × 9 × 0.52 + 0.6 × 110/2 × 4 × 0.5 × L

L = 20.8522 m.

Practice Test: Civil Engineering (CE)- 15 - Question 28

An RC beam of rectangular cross-section has 250 mm width and 450 mm effective depth. Vertical stirrups 2-legged 10 mm diameter are provided at a spacing of 250 mm centre to centre and assume design shear strength of concrete is 0.6 MPa. The total shear capacity (in kN) of the section is

[Use Fe 415 grade steel for stirrups]

Detailed Solution for Practice Test: Civil Engineering (CE)- 15 - Question 28
Shear resistance of shear reinforcement,

Shear resistance of concrete,

Vus = τcbd = 0.6 × 250 × 450 = 67.5kN

∴ Total shear capacity = Vus + Vus = 169.58kn

Practice Test: Civil Engineering (CE)- 15 - Question 29

The command area of a channel is 4000 hectares. The intensity of irrigation of a crop is 70% and it requires 60 cm of water in 15 days, when the effective rainfall is recorded as 15 cm during that period, then the duty at the head of the channel is _________ ha/cumec.

[Assume total losses as 15%]

Detailed Solution for Practice Test: Civil Engineering (CE)- 15 - Question 29

Depth of water = 60cm

Effective rainfall = 15 cm

Depth of irrigating water = 60 – 15 = 45 cm

⇒ Delta = 45 cm = 0.45 m

B = 15 days

From relation, ∆= 8.64×B/D

⇒ Duty(D) = 8.64×15/0.45 = 288ha/cumec

Due to loss of water, duty at head of channel is reduced

Here losses are 15%

So, duty at head of channel = 288 × 85/100

= 244.80 ha/cumecs

Practice Test: Civil Engineering (CE)- 15 - Question 30

A horizontal water jet with a velocity of 10 m/s and cross-sectional area of 10mm2 strikes a flat plate held normal to the flow direction. The density of water is 1000kg/m3. The total force on the plate due to the jet is

Detailed Solution for Practice Test: Civil Engineering (CE)- 15 - Question 30
Density of water (ρ) = 1000 kg/m3

Velocity of water (υ) =10 m/s

Cross section of jet, A = 10 mm2

Force on Plate due to jet of water F = ρυ2A

= 1000 kg/m3×(10 m/sec)2 × 10 × 10-6 m2

= 1 N

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