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Practice Test: Civil Engineering (CE)- 1 - Civil Engineering (CE) MCQ


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30 Questions MCQ Test - Practice Test: Civil Engineering (CE)- 1

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Practice Test: Civil Engineering (CE)- 1 - Question 1

In each question below are given two statements followed by two conclusions numbered I and II. You have to take the given two statements to be true even if they seem to be at variance from commonly known facts. Read the conclusion and then decide which of the given conclusions logically follows from the two given statements, disregarding commonly known facts.

Give answer:

  • (A) If only conclusion I follows
  • (B) If only conclusion II follows
  • (C) If either I or II follows
  • (D) If neither I nor II follows and
  • (E) If both I and II follow.

Statements: All bags are cakes. All lamps are cakes.

Conclusions:

  1. Some lamps are bags.
  2. No lamp is bag.

Detailed Solution for Practice Test: Civil Engineering (CE)- 1 - Question 1

Since the middle term 'cakes' is not distributed even once in the premises, no definite conclusion follows. However, I and II involve only the extreme terms and form a complementary pair. So, either I or II follows.

Practice Test: Civil Engineering (CE)- 1 - Question 2

Man does not live by __________ alone.

Detailed Solution for Practice Test: Civil Engineering (CE)- 1 - Question 2

Man does not live by bread alone.

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Practice Test: Civil Engineering (CE)- 1 - Question 3

Extreme focus on syllabus and studying for tests has become such a dominant concern of Indian students that they close their minds to anything ___________ to the requirements of the exam.

Detailed Solution for Practice Test: Civil Engineering (CE)- 1 - Question 3

Extraneous – irrelevant or unrelated to the subject being dealt with

Practice Test: Civil Engineering (CE)- 1 - Question 4

Select the pair that best expresses a relationship similar to that expressed in the pair:
LIGHT : BLIND

Detailed Solution for Practice Test: Civil Engineering (CE)- 1 - Question 4

Light cannot be seen by blind.
On the similar logic, Speech cannot be spoken by dumb. Sound cannot be heard by deaf. 

Practice Test: Civil Engineering (CE)- 1 - Question 5

If = log(a+b), then:

Detailed Solution for Practice Test: Civil Engineering (CE)- 1 - Question 5

Practice Test: Civil Engineering (CE)- 1 - Question 6

In an exam, the average was found to be 50 marks. After deducting computational errors the marks of the 100 candidates had to be changed from 90 to 60 each and average came down  to 45 marks. Total No of candidates who took the exam were.

Detailed Solution for Practice Test: Civil Engineering (CE)- 1 - Question 6

Let the total number of candidates who took the exam be NN.

Step 1: Compute the total marks before the corrections

The average marks before corrections were 5050. Hence, the total marks before corrections can be written as:

Total Marks Before Correction=50N\text{Total Marks Before Correction} = 50N


Step 2: Compute the adjustment due to the corrections

100 candidates originally scored 9090 marks each, contributing:

100×90=9000marks.100 \times 90 = 9000 \, \text{marks}.

After the correction, these candidates scored 6060 marks each, contributing:

100×60=6000marks.100 \times 60 = 6000 \, \text{marks}.

Thus, the total reduction in marks due to the correction is:

90006000=3000marks.9000 - 6000 = 3000 \, \text{marks}.


Step 3: Compute the total marks after the corrections

The new total marks after corrections are:

Total Marks After Correction=50N3000\text{Total Marks After Correction} = 50N - 3000

The new average marks are 4545, so:

Total Marks After Correction=45N\text{Total Marks After Correction} = 45N


Step 4: Set up the equation and solve for NN

Equating the two expressions for total marks after corrections:

50N3000=45N50N - 3000 = 45N

Simplify:

50N45N=300050N - 45N = 3000 5N=30005N = 3000 N=600N = 600


Final Answer:

The total number of candidates who took the exam was 600.

Practice Test: Civil Engineering (CE)- 1 - Question 7

A solid 4cm cube of wood is coated with red paint on all the six sides. Then the cube is cut into smaller 1cm cubes. How many of these 1cm cubes have no colour on any side?

Detailed Solution for Practice Test: Civil Engineering (CE)- 1 - Question 7

Number of cubes with 3 faces red = 8
Number of cubes with 2 faces red = 24
Number of cubes with 1 face red = 24
Number of cubes with no face red = 64-(24+24+8) = 8

Practice Test: Civil Engineering (CE)- 1 - Question 8

In deriving the equation for the hydraulic jump in a rectangular channel in terms of the conjugate depths and the initial Froude number.

Detailed Solution for Practice Test: Civil Engineering (CE)- 1 - Question 8

Continuity equation and momentum equation are used for the hydraulic jump in terms of the conjugate depth and Froude number.
Energy equation is further used to determine the loss of energy in the jump.

Practice Test: Civil Engineering (CE)- 1 - Question 9

This paragraph states that warm weather affects the consumer’s inclination to spend the age of father is 4 times more than the age of his son Amit. After 8 years, he would be 3 times older than Amit. After further 8 years, how many times will he be older than Amit?

Detailed Solution for Practice Test: Civil Engineering (CE)- 1 - Question 9

Let Amit’s age be n years
Age of his father = (4n+n) = 5n years
After 8 years:
Amit’s age = n+8
Father’s age =
3(n+8) = 5n+8
3n+24 = 5n+8
16 = 2n
n = 8 years
After further 8 years:
Amit’s age = 24 years
Father’s age = 5n+16 = 56 years
So father is 2.33 times older than Amit

Practice Test: Civil Engineering (CE)- 1 - Question 10

Pipe A, B and C are kept open and together fill a tank in t minutes. Pipe A is kept open throughout, pipe B is kept open for the first 10 minutes and then closed. Two minutes after pipe B is closed, pipe C is opened and is kept open till the tank is full. Each pipe fills an equal share of the tank. Furthermore, it is known that if pipe A and B are kept open continuously, the tank would be filled completely in t minutes. Find t?

Detailed Solution for Practice Test: Civil Engineering (CE)- 1 - Question 10

Let pipe A, B, C fills the tank in a, b, c minutes respectively.
Pipe A is kept open for t minutes
Pipe B is kept open for 10 minutes
Pipe C is kept open for t-10-2 minutes

t/a+10/b+(t-12)/c=1
t/a=10/b=(t-12)/c since each pipe has equal share of tank
Also, 
t/a+t/b=1
Both pipes fill the tank in time t.
Solving 1st two equations we get 30/b =1  b= 30 minutes
Solving last 2 equations we get 10+t = b   t =20 minutes
Therefore, time taken by C to fill it will be  24 minutes from equation 2.   

Practice Test: Civil Engineering (CE)- 1 - Question 11

The plastic modulus of a section is 4.8 x 10-4 m3. The shape factor is 1.2. The plastic moment capacity of the section is 120 kN.m. The yield stress of the material is:

Detailed Solution for Practice Test: Civil Engineering (CE)- 1 - Question 11

Moment Capacity of the Section = Plastic Modulus of Section x Yield Stress of Material
120 x 106 Nmm = 4.8 x 10-4+9 mm3 x Yield stress of Material
Yield stress of Material = 250 N/mm2

Practice Test: Civil Engineering (CE)- 1 - Question 12

MPN index is a measure of one of the following:

Detailed Solution for Practice Test: Civil Engineering (CE)- 1 - Question 12

MPN- Most Probable Number used to measure Coliform Bacterial
Wrinkler’s Method is used to measure dissolved oxygen and further, which is used to determine BOD5
Hardness is measured in mg/L as CaCO3 via various methods like lime treatment, boiling.

Practice Test: Civil Engineering (CE)- 1 - Question 13

Consider the following statements:
1) Strength of concrete cube is inversely proportional to water-cement ratio.
2) A rich concrete mix gives higher strength than a lean concrete mix since it has more cement content.
3) Shrinkage cracks on concrete surface are due to excess water in mix.
Which of the following statements is correct?

Detailed Solution for Practice Test: Civil Engineering (CE)- 1 - Question 13

Lower the water cement ratio higher the concrete strength.
A rich concrete due to greater cement content will possess higher strength in comparison to lean cement.
Higher the water content in mix higher will be the shrinkage in concrete.

Practice Test: Civil Engineering (CE)- 1 - Question 14

Standard 5-day BOD of a waste water sample is nearly x% of the ultimate BOD, where x is:

Detailed Solution for Practice Test: Civil Engineering (CE)- 1 - Question 14

BOD5 = BODu [ 1-10-k x 5] and the usual value of deoxygenation constant is 0.1
Substituting, the value of k BOD5 = BODu [0.6837] 
Therefore, value of x will be 68.37%

Practice Test: Civil Engineering (CE)- 1 - Question 15

The dimensions for the flexural rigidity of a beam element in mass (M), length (L) and time (T) is given by:

Detailed Solution for Practice Test: Civil Engineering (CE)- 1 - Question 15

In a beam, the flexural rigidity (EI) varies along the length as a function of x as shown in equation:

Where E is the young's modulus (in Pasual, Pa), I is the second moment of area (in m4). Y is the traverse displacement of the beam x and M(x) is the bending moment at x. The SI unit of flexural rigidity is thus Pa. m4 or Nm4
∴ Dimension is ML3T–2

Practice Test: Civil Engineering (CE)- 1 - Question 16

The superelevation needed for a vehicle travelling at a speed of 60 kmph on a curve of radius 128 m on a surface with a coefficient of friction of 0.15 is:

Detailed Solution for Practice Test: Civil Engineering (CE)- 1 - Question 16

A Correct the options by inserting a zero after the decimal in each option.

e- superelevation; f- coefficient of friction = 0.15; v= speed of vehicle in m/s = 60 kmph = 60x0.278 m/s = 16.68 m/s; R = radius of curvature = 128m
ⅇ+f=v2/gR  ⅇ+0.15=〖16.68〗2/(9.81 x 128)  e= 0.0716 

Practice Test: Civil Engineering (CE)- 1 - Question 17

Which of the following raingauge gives a plot of the accumulated rainfall against the elapsed time?

Detailed Solution for Practice Test: Civil Engineering (CE)- 1 - Question 17

Weighing bucket type

Practice Test: Civil Engineering (CE)- 1 - Question 18

If the time period between centroid of the rainfall diagram and peak of the hydrograph is 5 hour, using Snyder’s equation the value of the base width of unit hydrograph in hours is?

Detailed Solution for Practice Test: Civil Engineering (CE)- 1 - Question 18

Practice Test: Civil Engineering (CE)- 1 - Question 19

A river 5 m deep consists of a sand bed with saturated unit weight of 20 kN/m3 . vw = 9.81 kN/m3. The effective vertical stress at 5 m from the top of sand bed is:

Detailed Solution for Practice Test: Civil Engineering (CE)- 1 - Question 19

Effective Vertical Stress = Total Vertical Stress – Pore Water Pressure
= γ_(sat ) z - γ_(w ) z = 20x 5 – 9.81 x 5 = 50.95 kN/m2

Practice Test: Civil Engineering (CE)- 1 - Question 20

A hydraulic model of a spillway is constructed with a scale 1:16. If the prototype discharge is 2048 cumecs, then the corresponding discharge for which the model should be tested is:

Detailed Solution for Practice Test: Civil Engineering (CE)- 1 - Question 20

Q = A×V
Spillway model obeys Froude’s Law

 

Practice Test: Civil Engineering (CE)- 1 - Question 21

A droplet of water at 200C (σ = 0.0728 N/m) has internal pressure 1 kPa greater than that outside it, its diameter is nearly:

Detailed Solution for Practice Test: Civil Engineering (CE)- 1 - Question 21

For a droplet;

Practice Test: Civil Engineering (CE)- 1 - Question 22

A reinforced concrete structure has to be constructed along a sea coast. The minimum grade of concrete to be used as per IS : 456-2000

Detailed Solution for Practice Test: Civil Engineering (CE)- 1 - Question 22

Minimum Grade of Concrete on Sea Cost:
Plain Cement Concrete – M20
Reinforced Cement Concrete – M30

Practice Test: Civil Engineering (CE)- 1 - Question 23

The order and degree of the differential equation given, is  

Detailed Solution for Practice Test: Civil Engineering (CE)- 1 - Question 23


Order = 2
Degree = 3

Practice Test: Civil Engineering (CE)- 1 - Question 24

The probability distribution taken to represent the completion time in PERT analysis is:

Detailed Solution for Practice Test: Civil Engineering (CE)- 1 - Question 24

The distribution curve for the time taken in to complete each activity of a project resembles a beta-distribution curve and the distribution curve for the time taken to complete entire project ( consisting of several activities) in general resembles a normal distribution curve.

Practice Test: Civil Engineering (CE)- 1 - Question 25

Consider the following statements
1) Cambium layer is between sapwood and heartwood
2) Heartwood is otherwise termed as deadwood
3) Timber used for construction is obtained from heartwood
Which of the following is/are correct statements?

Detailed Solution for Practice Test: Civil Engineering (CE)- 1 - Question 25

Cambium layer is between inner bark and sapwood. Heartwood is dead wood which does not take part in the growth of tree and is used for construction.

Practice Test: Civil Engineering (CE)- 1 - Question 26

The true bearing of a tower T as observed from station A was 3560 and the magnetic bearing of the same was 40.The back bearing of the line AB when measured with prismatic compass was found to be 2960.Then the true fore bearing of line AB will be _____ degrees.

Detailed Solution for Practice Test: Civil Engineering (CE)- 1 - Question 26

Practice Test: Civil Engineering (CE)- 1 - Question 27

The degree of static indeterminacy of the rigid frame having 2 internal hinges is as shown in the figure below

Detailed Solution for Practice Test: Civil Engineering (CE)- 1 - Question 27

Number of restraints required = 6
Number of cuts required = 3
= 3×3-6 = 3

Practice Test: Civil Engineering (CE)- 1 - Question 28

If the deformations of the truss-members are as shown in parentheses, the rotation of the member bd is:

Practice Test: Civil Engineering (CE)- 1 - Question 29

The problem of lateral buckling can arise only in those steel beams which have:

Detailed Solution for Practice Test: Civil Engineering (CE)- 1 - Question 29

Lateral buckling occurs when an applied load causes lateral displacement of a member which generally occurs when moment of inertia about the bending axis smaller than the other.

Practice Test: Civil Engineering (CE)- 1 - Question 30

The available moisture holding capacity of soil is 15cm per meter depth of soil. If a crop with a root zone of 0.8m and consumptive use of 6 mm/day is to be grown, the frequency of irrigation for restricting the moisture depletion to 60% of available moisture is?

Detailed Solution for Practice Test: Civil Engineering (CE)- 1 - Question 30

Available moisture holding capacity = 15×0.8 = 12cm
Readily available moisture content = 0.4×12 = 4.8cm
Frequency of irrigation = 48/6 = 8 days

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