Civil Engineering (CE) : Mock Test 8 For GATE


65 Questions MCQ Test Mock Test Series for Civil Engineering (CE) GATE 2020 | Civil Engineering (CE) : Mock Test 8 For GATE


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This mock test of Civil Engineering (CE) : Mock Test 8 For GATE for GATE helps you for every GATE entrance exam. This contains 65 Multiple Choice Questions for GATE Civil Engineering (CE) : Mock Test 8 For GATE (mcq) to study with solutions a complete question bank. The solved questions answers in this Civil Engineering (CE) : Mock Test 8 For GATE quiz give you a good mix of easy questions and tough questions. GATE students definitely take this Civil Engineering (CE) : Mock Test 8 For GATE exercise for a better result in the exam. You can find other Civil Engineering (CE) : Mock Test 8 For GATE extra questions, long questions & short questions for GATE on EduRev as well by searching above.
QUESTION: 1

Select the pair that does not expresses a relationship similar to that expressed in the pair:

Wheel: spokes

Solution:

The spokes are units which radiate of a wheel and make the wheel a complete entity. Similarly, fingers, tentacles and petals are integral units of hand, octopus and flower respectively. On the other hand, roots and leaves do not share this kind of relationship. Thus option 4 does not expresses a relationship similar to that expressed in the given pair.

*Answer can only contain numeric values
QUESTION: 2

If a, b and c are three positive integers such that a and b are in the ratio 3:4 while b and c are in the ratio 2:1, then minimum integer value of a + b + c is _________ 


Solution:

Let a = 3x and b = 4x

Similarly b = 2y and c = y

∴ 4x = 2y ⇒ y = 2x

∴ c = 2x

Now a + b + c = 3x + 4x + 2x = 9x

So, the minimum integer value = 9

QUESTION: 3

Reaching a place of appointment of Friday. I found that I was two days earlier than the scheduled day. If I had reached on the following Wednesday then how many days late would I have been?

Solution:

Friday → 2 days earlier

Therefore, scheduled day = Friday + 2 = Sunday

Sunday + 3 = Wednesday

Therefore, I would have been late by 3 days

QUESTION: 4

Which of the following options is the closest in the meaning to the word given below?

Impeccable

Solution:

The meaning of the given words is:

Impeccable: in accordance with the highest standards

Flawless: without any imperfections or defects

Indelible: (of ink or a pen) making marks that cannot be removed

Intangible: unable to be touched

Collect: bring or gather together (a number of things)

Hence, option 1 is the correct answer.

QUESTION: 5

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

It was hoped at the time that that place would become the centre from which the civilization of Africa would proceed; but this ________ was not fulfilled.

Solution:

The sentence implies that it was hoped that that place would become the point from where the African civilization would proceed, but the belief that it would happen was not fulfilled.

Therefore, the correct word to fill in the blank is expectation as it means a strong belief that something will happen or be the case.

QUESTION: 6

The arithmetic mean of the list of numbers above is 4. If k and m are integers and k ≠ m, what is the median of the list?

Solution:

Mean = 4

k + m = 8

k ≠ m so k = m = 4 is out.

{k, m} = {1, 7} or {2, 6} or {3, 5}

for median of {3, k, 2, 8, m, 3}

{1, 2, 3, 3, 7, 8} or {2, 2, 3, 3, 6, 8} or {2, 3, 3, 3, 5, 8}

Median: (3 + 3) /2 = 3

QUESTION: 7

Consider a random walk on an infinite two-dimensional triangular lattice, a part of which is shown in the figure below.

If the probabilities of moving to any of the nearest neighbour sites are equal. What is the probability that the walker returns to the starting position at the end of exactly three steps?

Solution:

A person can take 1st step in any direction independently

Suppose he moves to A

Now to return to O ( initial point) in 2 steps he can move in 2 directions. Either B or f

Thus probability he will move to either B or F will be =

Suppose he moves to B

Now to return back to O

he has only 1 option

⇒ to move in BO

Probability he will move in BO direction =

Total probability he will return

QUESTION: 8

Twelve straight lines are drawn in a plane such that no two of them are parallel and no three of them are concurrent. A circle is now drawn in the same plane such that all the points of intersection of all the lines lie inside the circle. What is the number of non-overlapping regions into which the circle is divided?

Solution:

The nth line drawn will add n more regions to the circle. 

1st line adds 1 region to the circle for a total of 2 regions. 

2nd line adds 2 more regions to the circle, bringing the total number of regions to 2 + 2 = 4. 

3rd line adds 3 more regions to the circle, bringing the total number of regions to 4 + 3 = 7. 

4th line adds 4 more regions to the circle, bringing the total number of regions to 7 + 4 = 11.

5th line adds 5 more regions to the circle, bringing the total number of regions to 11 + 5 = 16. 

6th line adds 6 more regions to the circle, bringing the total number of regions to 16 + 6 = 22. 

7th line adds 7 more regions to the circle, bringing the total number of regions to 22 + 7 = 29. 

8th line adds 8 more regions to the circle, bringing the total number of regions to 29 + 8 = 37. 

9th line adds 9 more regions to the circle, bringing the total number of regions to 37 + 9 = 46. 

10th line adds 10 more regions to the circle, bringing the total number of regions to 46 + 10 = 56. 

11th line adds 11 more regions to the circle, bringing the total number of regions to 56 + 11 = 67. 

12th line adds 12 more regions to the circle, bringing the total number of regions to 67 + 12 = 79. 

QUESTION: 9

Electromagnetic radiation is an insidious culprit. Once upon a time, the major concern around electromagnetic radiation was due to high tension wires which carry huge amounts of electricity to cities. Now, we even carry sources of this radiation with us as cell phones, laptops, tablets and other wireless devices. While the most acute exposures to harmful levels of electromagnetic radiation are immediately realized as burns, the health effects due to chronic or occupational exposure may not manifest effects for months or years.

Which of the following can be the viable solution for electromagnetic radiation reduction?

Solution:

The correct answer is option 2 i.e. To implement hardware protocols to minimize risks and reduce electromagnetic radiation production significantly. 

The passage states about the electromagnetic radiations released from the devices and how they affect the individuals. Out of the given options, only option 2 states the possible solution which can be implemented to reduce electromagnetic radiation.

QUESTION: 10

In a mock exam, there were 3 sections. Out of all students, 60 students cleared the cut off in section 1, 50 students cleared the cutoff in section 2 and 56 students cleared the cut off in section 3. 20 students cleared the cutoff in section 1 and section 2, 16 cleared cut off in section 2 & section 3, 26 cleared the cut off in section 1 & section 3. The number of students who cleared cutoff of the only one section was equal & was 24 for each section. How many students cleared cut off all the three sections?

Solution:

Let the number of students be X who cleared cut off in all sections.

The number of students who cleared the cutoff in section 1 and section 2, only = 20 – X

The number of students who cleared the cutoff in section 2 and section 3, only = 16 – X

The number of students who cleared the cutoff in section 1 and section 3, only = 26 – X

Now, consider section 1:

24 + 20 – x + x + 26 – x = 60

70 – x = 60

x = 10

QUESTION: 11

The rank of the matrix 

Solution:

R5 → R1 + R2 + R4 + R5

R5 → R5 - R3, R3 → R3 - R1

R4 → 2R4 - R3

R3 → R3 - 4R2

R4 → 5R4 - R3

Now it is in Echelon form.

Rank of matrix = number of non-zero rows = 4.

QUESTION: 12

Let f(z) = z̅ and g(z) = |z|2 for all ϵ C. Then at z = 0.

Solution:

f(z) = z̅ = x - iy

u = x, v = -y

ux = 1, vy = -1

uy = 0, vx = 0

ux ≠ vy, as f(z) is not satisfying CR equations, f(x) is not analytic.

g(z) = |z|2

g(z) = x2 + y2

ux = 2x, vx = 0

uy = 2y, vy = 0

at z = 0, ux = vy and uy = -vx

g(x) is satisfying CR equations at z = 0 but it is not satisfying CR equations at neighborhood of z = 0.

Hence g(z) is not analytic.

*Answer can only contain numeric values
QUESTION: 13

The probability density function of a random variable X is given by


Solution:
QUESTION: 14

  represents the parabola, then the value of m­ will be?

Solution:

The partial differential equation will represent the parabola, if:

B2 – 4AC = 0

Now given equation:

16 – 4 x 1 × m = 0

4m = 16

m = 4

QUESTION: 15

The non-zero of n for which the differential equation (3xy2 + n2 x2y) dx + (nx3 + 3x2y) dy = 0, x ≠ 0 become exact differential equation is:

Solution:

Give equation (3xy2 + n2 x2y) dx + (nx3 + 3x2y) dy = 0, x ≠ 0

Concept:

M dx + N dy = 0, will be exact differential equation if:

Where M and N are functions of x and y.

Here M = 3xy2 + n2 x2y

N = nx3 + 3x2y

Given equation is exact, thus:

6xy + n2 x2 = 3nx2 + 6xy

n2 x2 = 3nx2

n = 3

QUESTION: 16

The normal duration and normal cost of an activity are 25 days and 50,000 rupees respectively. The activity crash duration is 22 days and the indirect cost is 1000 rupees per day. If the cost slope is 1500 rupees per day, then the crash cost of the activity will be

Solution:

The increase in direct cost for 3 days crashing

= 1500×3 

= 4500 rupees

The decrease in indirect cost for 2 days

= 1000×2 

= 2000 rupees

The total cost of the activities after crashing

=  50000 + 4500 - 2000

= 52500 rupees

QUESTION: 17

Which one of the following is incorrect? 

Solution:

Seasoning of timber results in:

1. Increased strength

2. Increased durability

3. Increased resilience

Through the application of preservation of timber,

the life span of the timber can be increased and prevent the growth of fungi in the timber.

The specific gravity of wood is less than 1.

Abel's process treatment is used to make timber fire-resistance.

QUESTION: 18

A doubly reinforced concrete beam with effective cover of 50 mm to centre of both tension and compression reinforcement with effective depth of 550 mm. If the maximum permissible stress in the outermost fibres in both steel and concrete reaches at the same time, what will be the maximum strain in concrete at the level of compression reinforcement satisfying codal provisions of IS 456:2000 according to limit state method.

Take Fe 500 grade of steel for both tension and compression reinforcement and M 20 grade of concrete.

Solution:

As the maximum permissible stress in the outermost fibers in both steel and concrete reaches at the same time, therefore it is the case of balanced section:

So,

Xu = Xulim

For Fe 500

Xulim = 0.46 d

d = effective depth of the beam = 550 mm

Xulim = 0.46 x 550 = 253 mm

d' = effective cover to compression reinforcement = 50 mm

Maximum strain in concrete at outermost compression reinforcement as per IS 456: 2000 = 0.0035

Now from similar triangle,

ϵsc = 0.00281

QUESTION: 19

A close box subjected to the following loading conditions as shown in figure below. The ratio of distribution factor for the member AB to member BC is?

Solution:

Stiffness of member AB at 

Stiffness of member AD at

Distribution factor for AB  = 

Stiffness of member BC at B = 

Stiffness of member BA at B = 

Distribution factor (BC) = 

QUESTION: 20

Clear distance between lateral restrains for continuous reinforced concrete beam of size 250 × 500 mm (effective depth) according to IS 456:2000 shall be limited to:

Solution:

For simply supported and continuous beam: 

clear distance between lateral restrains shall not be greater than:

  whichever is lesser 

Where,

B = width of the beam = 250 mm

d = effective depth of the beam = 500 mm

i) 60 B = 60 × 250 = 15000 mm = 15 m

= 31250mm = 31.25m

lesser will be 15 m.

QUESTION: 21

The number of independent degrees of freedom for the beam shown below is

Solution:

Number of independent degrees of freedom is nothing but kinematic indeterminacy of the beam/structure

Dk = 3j – re + rr

J → No. of Joints

rr = No. of released reactions

re = External support Reactions

J = 3

re = 5

rr = m - 1

m = no. of members meeting at a Joint (hinge)

m = 2

rr = 2 – 1 = 1

Dk = 3 × 3 – 5 + 1

Dk = 5

QUESTION: 22

A state of pure shear is shown in figure below. What will be the radius and centre of moh’r circle at x – x shown in figure below

Solution:

Concept: At x – x

σ′x = τxysin2θ 

σ′y = -τxysin2θ 

τxy′ = τxycos2θ

Calculation:

τxy = 50 MPa

θ = 30°

σ′= 50sin60∘ 

σ′= 43.30MPa 

σ′= −50sin60∘ 

σ′= −43.30MPa 

τxy′ = 50cos60∘ =25MPa 

Centre of moh’r circle = (a, 0)

So, centre = (0, 0)

Radius of moh’r circle

r = 25 MPa

*Answer can only contain numeric values
QUESTION: 23

A transmission tower carries a vertical load of 10000 kg, the vertical stress increment at a depth 6 m directly below the tower as per Boussinesq’s theory is____Pa.


Solution:

Concept:

Vertical stress increment at a any depth (z) at a radial distance (r) as per Boussines’q theory is:

Where,

KB = Boussinesq’s influence factor

Calculation:

At r = 0 m,

Q = 10000 × 9.81 = 98100 N

z = 6 m

*Answer can only contain numeric values
QUESTION: 24

For a ladder shown in figure, the velocity of point A is 0.2 m/min in vertically downward direction the velocity of point B in horizontally rightward direction will be ____ m/s.


Solution:

Velocity at point A, VA = 6 cm × ω

⇒ ω = 0.0556 rad/sec

∴ Velocity of point B, VB

= 0.08 × ω

= 0.08 × 0.0556 rad/sec

= 0.444 m/sec

QUESTION: 25

If the liquid limit and plastic limit of the soil is observed as 60% and 45% respectively and the percentage by weight of particles finer than 2μ is 30%, then identify the correct option:

Solution:

Activity of the soil is defined as the ratio of Plasticity Index of the soil to the % by weight of particles finer than 2μ.

IP = wL - wP

wL = 60%

wP = 45%

IP = 60% - 45% = 15%

As Activity is less than 0.75, Soil is classify as Inactive.

QUESTION: 26

Identify the correct Statement:

Solution:

Toughness index is defined as the ratio of Plasticity index to the flow index.

Toughness Index=

IP = Plasticity Index

If = Flow index, which represent the rate of loss of shear strength in soils.

More will be the toughness index, more will be the soil firm.

QUESTION: 27

If in an old map the line PQ was drawn to a magnetic bearing of 9° 30’ and the magnetic declination at the time being 3° West. If the present magnetic declination is 5° 30’ West, then the present magnetic bearing of the line PQ will be?

Solution:

Magnetic bearing of line PQ = 9° 30’

Magnetic Declination = 3° W

True bearing of line PQ = M.B – Declination

= 9° 30’ – 3°

Present Magnetic Bearing = TB + Present Declination = 6° 30’ + 5° 30’ = 12°

*Answer can only contain numeric values
QUESTION: 28

The ratio of Discharge in prototype (Qp) to the Discharge in model (Qm) for a geometrically. Similar model of spillway is 35000, the scale of the model used is 1 in ______.


Solution:

Concept:

Froude number is used in model analysis of spillways

(F)m = (F)P

Calculation:

= 35000 

QUESTION: 29

If the velocity components in 2 – Dimensional incompressible fluid flow is represented as  Find the rate of shear strain in (x – y) plane.

Solution:

Concept:

Velocity field for a flow is represented as:

Where,

u = velocity component in x – direction

v = velocity component in y – direction

Rate of shear strain in x – y plane is given by:

Calculation:

u = 2xy – 4y3

Check for the flow

   (OK, flow is possible)

For rate of shear stain in x – y plane

 = 

*Answer can only contain numeric values
QUESTION: 30

A wastewater is incubated for 8 days and the BOD8 at 20°C is found to be 250 mg/L. The ratio of oxygen available in the wastewater to the total oxygen required to satisfy its first stage BOD demand expressed as the percentage of total oxygen required will be ____ %.


Solution:

The ratio of oxygen available in the wastewater (as dissolved oxygen) to the total oxygen required to satisfy its first state BOD is called relative stability.

Relative stability,

Where, t20 and t37 represent the time in days for a sewage sample to decolorize a standard volume of methylene blue solution, when incubation is done at 20° or 37°C respectively.

S = 100 × 0.842

S = 84.2%

*Answer can only contain numeric values
QUESTION: 31

A field channel has cultivated command area of 20000 hectares. During the kharif season rice is cultivated in 12000 hectares of area. The Kor depth and Kor period for the rice is given as 25 cm and 28 days. The discharge required from the canal to irrigate the given land for rice crop is _____ × 10-5(hec-m/s).


Solution:

The discharge required from the canal to irrigate 12000 hectares of land = 

Given,

Kor depth = 25 cm

Kor period = 28 days

Area of land to be irrigated = 12000 hec

Discharge through the canal at outlet = 

= 12.4 m3/s

Discharge = 124 × 10-5 hec-m/s

QUESTION: 32

1 TCU is equivalent to the colour produced by

Solution:

TCU represent true colour unit. It is a unit to represent colour impurities in water. 1 TCU is equivalent to the colour produced by 1 mg/L of platinum in form of chloro-platinum ion.

QUESTION: 33

Which kind of plume has minimum downward mixing as its downward motion is prevented by inversion and upward mixing will be quite turbulent and rapid?

Solution:

It is a condition where exists a strong super-adiabatic lapse rate above surface inversion.

Lofting plume has minimum downward mixing due to presence of inversion and is considered as ideal case for dispersion of emission.

QUESTION: 34

Which one of the following test is performed in the laboratory to detect overheated or cracked bitumen and are more sensitive than the other test in terms of detection of cracking

Solution:
  • Softening point denotes the temperature at which the bitumen attains a particular degree of softening under the specifications of the test. The test is conducted by using the Ring and Ball apparatus.
  • Normally the consistency or stiffness of bituminous material can be measured either by penetration test or viscosity test. But for a certain range of consistencies, these tests are not applicable and Float test is used.
  • The bitumen content of a bituminous material is measured by means of its solubility in carbon disulfide and hence can be considered as purity test for bitumen. In the standard test for bitumen content (ASTM D4), a small sample of about 2 g of the asphalt is dissolved in 100 ml of carbon disulfide and the solution is filtered through a filtering mat in a filtering crucible.
  • Spot test is used to find out cracking in bitumen. It is performed in the laboratory to detect overheated or cracked bitumen and are more sensitive than the Solubility test in terms of detection of cracking
*Answer can only contain numeric values
QUESTION: 35

An 8-Lane expressway is to be designed from Delhi to Lucknow with the design speed of 120 kmph. The camber (in %) that can be provided for a horizontal curve of radius 1800 m (such that no super-elevation is required) will be (up to two correct decimal place) ______.


Solution:

The minimum radius of the curve beyond which no superelevation is required

Where, V = speed of the vehicle in kmph

Calculation:

Camber = 0.0256

Camber = 2.56%

QUESTION: 36

Let A be a 3 × 3 matrix such that  and    suppose  Then AQ is

Solution:

These are in the form of AX = λX

Where λ is an Eigen value of A and X is the Eigen vector of A corresponding to the Eigen value λ.

Now, λ1 = 1, λ2 = 1, λ3 = 1

From the properties of matrix, we can write the matrix A as

A = PDP-1

Where P is the matrix formed by Eigen vectors, D is the diagonal matrix and Eigen values as diagonal elements.

⇒ AP = PD

Here,

⇒ AP = PD = 

From the given question, Q = P

⇒ AP = AQ

 

*Answer can only contain numeric values
QUESTION: 37

The area of the region bounded by the parabola y2 = 4(4-x) and the straight line y + x = 4 is______


Solution:

Equation of straight line ⇒ y = 4 – x

Equation of parabola ⇒ y2 = 4 (4 - x)

On solving the above two equations, we get the point of intersection (4, 0) and (0,4)

Area enclosed = ∫∫dxdy

The required area = 

*Answer can only contain numeric values
QUESTION: 38

Maclaurin‘s series expansion of f(x) = ex2+sinx has the form f(x) = A0 + A1x + A2x2 + A3x3 + ….

The coefficient A4 (correct to two decimal place) is equal to _____.


Solution:

So, Maclaurin’s series expansion of

Also, Maclaurin’s series expansion of esinx

g(x) = esinx  (say)

g′(x) = esinx.cosx = f(x).cosx 

g′′(x) = g′(x)cosx − g(x)sinx 

g′′′(x) = g′′(x)cosx − g′sinx − g′sinx − g(x)cosx 

= g′′(x)cosx − 2g′sinx − g(x)cosx

g′′′′(x) = g′′′(x)cosx − 3g′′(x)sinx − 3g′cosx + g(x)sinx 

g(x = 0) = esin 0  = e0  = 1

g'(x = 0) = g(x = 0) × cos (0) = 1

g''(x = 0) = g'(x = 0) cos (0) – g(0) sin (0)

= 1 × 1 – 1 × 0

  = 1

g'''(x = 0) = g''(x = 0) cos (0) – 2g'(x = 0) sin (0) – g(x = 0) × cos (0)

= 1 × 1 – 2 × 1 × 0 – 1 × 1

= 0

g′′′′(x = 0) = g′′′(x = 0) cos0 − 3g′′(x = 0) sin0−3g′(x = 0) cos0 + g(x = 0) sin0 

g''''(x = 0) = -3

Hence,

esinx  = 1 + (1 × x) + 

 

Coefficient of   

QUESTION: 39

For x ϵ R let   then which of the following is FALSE?

Solution:

= 0 × [-1 to 1]

= 0

Since the value of can vary from [-1 to 1]

= 0 × [-1 to 1]

= 0

For maximum and minima

g'(x) = 0

Which can yield infinite value between (0, 1)

Hence,  has infinitely maximum and minima on the interval (0, 1)

Also,

Since,   is continuous in open interval (-∞, 0) U (0, ∞) because it is defined for every x ϵ R, except in

x = 0 while  is continuous all-over real number.

Therefore,    is discontinuous at x = 0 and continuous in (-∞, 0) U (0, ∞)

Similarly, the given function is not differential at x = 0.

Hence, last statement is not correct.

*Answer can only contain numeric values
QUESTION: 40

Use the fourth order Runge-Kutta method to find u(0.2) of the initial value problem u′=−2tu2, u(0) = 1 using h = 0.2. Correct up to three decimal places.


Solution:

Concept:

Runge - Kutta Method of first order:

k1  = h f(x0, y0)

y1  = y0  + k where k = k1  = h f (x0, y0)

y2  = y1  + k where k = h f (x1, y1)

Runge - Kutta Method of Second order:

k1  = h f(x0, y0)

k2  = h f (x0  + h, y0  + k1)

y1  = y0  + k

Runge - Kutta Method of Third order:

k1  = h f (x0, y0)

y1  = y0  + k

Runge - Kutta Method of fourth order:

k1  = h f (x0, y0)

y1  = y0  + k

Calculation:

t0 = 0,u0 = 1

f(u,t) = −2tu2 

k1 = hf(t0,u0) = hf(0,1) = 0.2×[−2×0] = 0

= hf(0.1,1) = 0.2×[−2×0.1×12] = −0.04

= hf(0.1,0.98) = 0.2×[−2×0.1×0.982] = −0.038416

k4 = hf(t0 + h,u0 + k3) = hf(0.2,0.961584) = 0.2×[−2×0.2×0.9615842] = −0.073972

= −0.0384373

u1 = u0 + k = 0.961563

QUESTION: 41

Consider the following statement

P) Crushing strength of first class bricks should not be less than 14.0 N/mm2.

Q) Lime in the brick prevents shrinkage of the raw bricks.

R) Adding 5% and 6% of moisture by weight increases the volume of dry sand from 18% to 38%.

T) Gauged mortar is a mortar in which cement and lime both are mixed.

For the above statement, the correct option is

Solution:

1) The minimum crushing strength of different types of bricks is given below:

Common building bricks - 3.5 N/mm2

Second class bricks - 7 N/mm2

First class bricks - 10.5 N/mm2

2) Lime in Bricks 

For good brick earth, it is desirable not to have lime more than 5%.

Lime prevents shrinkage of raw bricks

Excess of Lime causes the brick to melt and therefore its shape can be lost.

3)The volume of dry sand increases due to absorption of moisture. These volume increase of dry sand is known as bulking of sand. When dry sand comes in contact with moisture, a thin film is formed around the particles, which causes them to get apart from each other. This results in increasing the volume of sand. Addition of 5% and 6% of moisture content by weight increases the volume of dry sand from 18% to 38%.

Graphical representation of bulking of sand is shown below:

4) Gauged mortar is a mortar in which cement and lime both are mixed.

*Answer can only contain numeric values
QUESTION: 42

Two Indian Standard Flat (ISF) 2 m long are to be jointed to make a member of 3 m through lap joint as shown in the figure below. The bolts are of 20 mm diameter and grade 4.6. The two plates to be jointed are 10 mm and 12 mm thick. The efficiency of the lap joint will be _____ %.

(Use steel of grade Fe410)


Solution:

Given:

For Fe410 grade of steel: f­u = 410 MPa

For bolts of grade 4.6: fub = 400 MPa

Diameter of the bolt, d = 20 mm

Bolt-hole diameter (dh or do) = 22 mm

Area of one bolt = 

Ab = 314.16 mm2

The bolts will be under single shearing and bearing.

Strength of the bolt in single shear

= 58 kN

Strength of the joint per gauge length in shear

(since two bolts fall in one-gauge length)

= 2 × 58

= 116 kN

Strength of bolt in bearing

Where, t = thickness of thinner plate

rmb = partial safety factor for

material of bolt = 1.25

Kb is least of 

∴ kb = 0.5

Pb = 89.38 kN

Strength of the joint per gauge length in bearing (two bolts fall lies in each gauge length)

= 2 × 89.38 = 178.76 kN

The net tensile strength of plate per pitch length,

⇒ Pt = 289.296 kN

Hence, the strength of the joint per pitch length = 116 kN

Strength of solid plate per gauge length

= 354.24 kN

Efficiency of the joint

= 32.75%

QUESTION: 43

The development length requirement for concrete of grade M-35 and steel of grade Fe-415 for bars in compression according to limit state method will be:

ϕ = diameter of bar

Solution:

The development length required for bar will be:

Where,

τbd = 1.7 N /mm2

But for bars in compression, the bond stress is increased by 25%.

For HYSD bars, the bond stress is increased by 60%

fy = 415 N/mm2

τbd = 1.7 x 1.6 x 1.25 = 3.4 N /mm2

Ld = 26.55 ϕ

*Answer can only contain numeric values
QUESTION: 44

A 12 mm thick plate is connected through a 20 mm diameter bolt as shown in figure. Assuming the safe stress on the connection to be 120 N/mm2. The strength of the connection (in kN) will be ________.


Solution:

Given

Bolt diameter = 20 mm

Bolt hole diameter, dh = 22 mm

Safe stress on connection = 120 N/mm2

Thickness of plate = 12 mm

Also, Staggered pitch, S = 50 mm

Gauge length, g = 60 mm

Net width of the plate, corresponding to ABCD

= Total width of plate – (number of bolts × dh)

= (40 + 3 × 60 + 40) – 2 × 22

= 216 mm

Net width of the plate corresponding to ABECFG

Where b = width of the plate.

n = total numbers of bolt whole in the path considered

S1, S2, and S3 are the staggered pitch between bolts BE, EC and CF respectively.

g = gauge length

= 203.25 mm

The net width of the plate corresponding ABECFG

= 204.42 mm

The net width of the plate corresponding to ABEFG

= 214.83 mm

Therefore, minimum width for net effective area = 203.25 mm

Strength of the plate = (203.25 × 12) × 120 × 10-3 kN

= 292.68 kN

QUESTION: 45

For the beam shown below, Determine the deflection at point C?

Solution:

Using Conjugate Beam method,

RA + RB = 0

HA = 0

∑ Mz = 0

RA × L – 2M + M = 0

The diagram of the real beam will become loading of the conjugate beam.

QUESTION: 46

Calculate the maximum Bending moment for the beam shown below:

Solution:

Consider B as fixed end

BM at B is zero, so apply – 8.292 kN-m at joint B and half of this is transferred at end A

Final Moment at A  

Or

(Anti clock wise)

∑ Fv = 0

RA + RB = 10 + 2 × 4 = 18 kN

∑ MB = 0

RA × 4 – 8.686 – 2 × 4 × 2 – 10 × 1 = 0

RA = 8.671 kN

RB = 9.329 kN

Maximum Bending moment will occur at A

(BM)max­ = (BM)A = 9.329 × 4 - 2 × 4 × 2 – 10 × 3

(BM)max­ = (BM)A = - 8.684

Or

(BM)max­ = 8.684 kN-m (hogging)

QUESTION: 47

A plane truss is shown below

The members which do not carry any force are:

Solution:

I. Member LK, LA and LB are meeting at joint ‘L’ and AL and LK re co-linear so force in the member LB is zero.

FLB = 0

II. Member FG, FE, and FD are meeting at joint F, So force in the member FD = 0

FFD = 0

III. Removing Member LB, DF, thus member BK, BC and BA meets at point B out of which BA and BC are co-linear and vice versa for member DG, DC and DE.

So, FBK = 0

FDG = 0

IV. Similarly, Member HM, HI and HG are meeting at joint H

FHM = 0

QUESTION: 48

An undisturbed soil sample of thickness 50 mm undergoes 60% consolidation in 28 minutes when double drainage is allowed in the laboratory. If the same clay layer from which the sample was obtained is 3.5 m thick in field conditions, estimate the degree of consolidation occurs when the clay is allowed to consolidate to 600 days with single drainage allowed. Assume the consolidation pressure to be same throughout.

Solution:

U1  = 60%

In lab:

H1  = 50 mm

Double drainage

t = 28 minutes

CV1 = 6.310mm2/min 

In field:

t2  = 600 days = 600 × 24 × 60

t2  = 864000 minutes

d2  = H2  = 3.5 m = 3500 mm [single drainage]

As consolidation pressure is same

CV1 = CV2 = CV = 6.310mm2/min

TV2 = 0.04450 

For U ≤ 60%, TV = 0.2827

So, it is obvious that U < 60%

U= 23.4%

*Answer can only contain numeric values
QUESTION: 49

A vane shear test was performed on saturated clay layer at the site and it is noted that a torque of 45 N-m was required to shear the soil. In order to remould the soil, the vane was rotated rapidly under the certain value of torque say “x”. If the sensitivity of the clay is 8, then calculate the value of “x” in N-m.

Assume diameter and length of the vane to be 70 mm and 130 mm respectively, consider two way shearing.


Solution:

Concept:

Undrained shear strength of clay is given by:

Where,

S =shear strength of soil

H = Length or Height of the vane

D = diameter of the vane

T = Torque applied

Calculation:

At undisturbed state:

T = 45 N-m

H = 130 mm = 0.13 m

D = 70 mm = 0.07 m

c = 38.129 kN/m2

At remoulding state:

Sensitivity of clay soil:

cr = 4.767 kN/m2

Now,

Tr = 5.625 N-m (Ans).

*Answer can only contain numeric values
QUESTION: 50

A wastewater stream in a city with flow rate 200 m3/s, discharges at a point into a river, which is fully saturated with oxygen and flowing at a rate of 2800 m3/s during the lean period with velocity of 25 cm/s. It was estimated that the critical deficit will occur at a location of 30 km downstream of the river from the point where wastewater starts mixing into river water. If the 5 days BOD of the wastewater at 30°C temperature is 350 mg/l and considering the temperature of the mix to be 20°C, then the critical oxygen deficit (in mg/L) will be _______. It is given that kD and kR at 20°C are 0.10 per day and 0.2 per day respectively.


Solution:

Concept:

The critical oxygen deficit, Dc

Where. kD = Deoxygenation constant

KR = Reoxygenation constant

L0 = ultimate BOD

tc = critical time after when critical oxygen deficit occurs.

Solution:

Deoxygenation constant

= 0.10 (1.047)10

= 0.158 per day

Reoxygenation constant

= 0.20 (1.016)10

= 0.234 per day

Critical time = 

= 1.389 days

5-day BOD = 350 mg/l of wastewater

5 – day BOD of the mixture of wastewater stream and river water is given by

= 23.33 mg/l.

BODs at 20°C = L0 (1 – (10)-kD × 5)

23.33 = L0 (1 – (10)-0.10 × 5)

⇒ L0 = 34.12 mg/l

∴ Critical oxygen deficit

DC = 13.89 mg/l

*Answer can only contain numeric values
QUESTION: 51

A retaining wall (acted upon by active thrust) 20m high having smooth vertical back retains a soil having c’ = 25kN/m2 and ϕ’ = 20. To increase the critical depth of unsupported cut by 10%, the ϕ’ required will be _____ (degrees).Take γ of soil = 18 kN/m3


Solution:

Concept:

Critical depth of unsupported cut is given by:

C’ = effective cohesion

γ = unit weight of soil

ka = active earth pressure coefficient 

Calculation:

Case: 1 when c’ = 25 kN/m2, ϕ’ = 20°

Hc2  = 1.1×7.936 = 8.729m

ka2  = 0.405

1−sinϕ′2 = 0.405 + 0.405sinϕ′

0.595 = 1.405sinϕ′2

ϕ′2 = 25.05

*Answer can only contain numeric values
QUESTION: 52

36 Piles are arranged in a square pattern which are embedded in the clayey soil. The shear strength at the base of the pile is 200 kN/m2 and the average shear strength over the depth of the pile is 130 kN/m2. If the Diameter and length of the pile is 0.4 m and 12 m respectively, the safe load that the pile group can carry will be ____kN. Assume, centre to centre spacing between the pile = 0.8 m, Adhesion factor (α) = 0.50 and Factor of safety = 3


Solution:

B = (n – 1)s + d

B = (6 – 1) × 0.8 + 0.4

B = 4.0 + 0.4 = 4.4m

Ultimate load carrying capacity of single pile:

Qup = qb Ab + qs As

Qb = base resistance of pile = 9c

c = cohesion at base of the pile

ϕ = 0° (clay)

Sbase = c = 200kN/m2

qs = average skin friction resistance

Qup = 1206.37 kN

Ultimate load carrying capacity of pile group

= 9 × 200 × (4.4)2 + 0.5 × 130 × 4 × 4.4 × 12

Qug = 48576 kN

Safe load on pile group = 

(Qug)safe = 4825.48kN 

Or

So, safe load on pile group will be lesser of two. Hence safe load = 16192kN

*Answer can only contain numeric values
QUESTION: 53

A rectangular open tank contains water upto a depth of 3m and above it an oil of specific gravity 0.80 for a depth of 1.5m. How much volume of oil is required to be added to the tank such that the pressure intensity at the interface of two liquids is 50% of the pressure intensity at the bottom of the tank. Calculate your answer in _____m3


Solution:

Initial volume of oil in the tank = L × B × H

Vinitial = 3 × 2 × 1.5

Vinitial = 9m3

Let height of oil required in the tank such that pressure at the interface of two liquids is 50% of the pressure at the bottom be “h”

Pressure at the interface of two liquids (P1) = ρoil × g × h

Pressure at the bottom of the tank (P2) = ρoil × g × h + ρwater × g × 3

P1 = 0.50 P2

ρoil × g × h = 0.50 [ρoil × g × h + ρwater × g × 3]

0.50 ρoil × g × h = 1.5 + ρwater × g

ρoil = specific gravity ρw

ρoil = 0.80 × 1000

ρoil = 800 kg/m3

0.50 × 800 × g × h = 1.5 × 1000 × g

h = 3.75m

Final volume of oil in the tank = 3 × 2 × 3.75

Vfinal = 22.5m3

Volume of oil required to be added = Vfinal - Vinitial

ΔV = 22.5 – 9

ΔV = 13.5m3

QUESTION: 54

For a Rectangular, Horizontal and Frictionless channel, the Froude number at the beginning of the jump is 15. It is observed that there is a significant energy loss during the formation of the jump. The energy loss per unit pre-jump depth of the flow and the type of jump is

Solution:

The sequent depth ratio can be obtained through Froude Number:

Calculation:

= 20.72

Hence, Energy loss per unit pre-jump depth of flow is 92.52 m

Since Fr > 9 then the type of jump is classified as strong jump.

*Answer can only contain numeric values
QUESTION: 55

For a hydraulically efficient trapezoidal channel section carrying a discharge of 50 m3/s. If Manning’s coefficient(n) is 0.018 then chezy’s coefficient C for the channel is _______ (Take Bed slope as 1 in 3000)


Solution:

Concept:

For an efficient channel, the discharge should be maximum.

Parameters for the most efficient trapezoidal channel having depth 'y'  are:

1) Depth = y

2) Top width = 

3) Area = 

4) Perimeter = 

5) Hydraulic Radius = y/2

Calculation:

⇒ y = 4.17 m

⇒ C = 62.82

*Answer can only contain numeric values
QUESTION: 56

A jet of water of diameter 50 mm moving with a velocity of 40 m/s, strikes a curved fixed symmetrical plate at the centre. Find the force exerted by the jet (in N) of water in the direction of the jet, if the jet is deflected through an angle of 120° at the outlet of the curved plate.


Solution:

Concept:

Force exerted by the jet in the direction of jet,

Fx = Mass per sec × [V1x - V2x]

Fx = ρaV[V - (-V cos θ)] = ρaV[V + V cos θ]

= ρaV2[1 + cos θ]

Fy = Mass per sec × [V1y - V2y]

Fy = ρaV[0 - V sin θ] = -ρaV2 sin θ

Calculation:

Velocity of jet, V = 40 m/s

Angle of deflection = 120° = 180° - θ or θ = 60°

Force exerted by the jet on the curved plate in the direction of the jet:

Fx = ρaV2 [1 + cos θ]

= 1000 × .001963 × 402 × [1 + cos 60°] = 4711.15 N

*Answer can only contain numeric values
QUESTION: 57

A catchment having 6 rain gauge stations with the precipitation in the month of July is measured as P1 = 600 mm, P2 = 550 mm, P3 = 625 mm, P4 = 475 mm, and P5= 400 mm. The sixth rain gauge station was inoperative during this period. The weight area factor for first four rain gauge station is 0.1, 0.35, 0.16 and 0.15 respectively. If the weight area factor for remaining two rain gauge station are equal, then the Thiessen mean value (in mm, up to two decimal places) of the rainfall is ______. It is given that normal annual precipitation at the different 6 rain gauge station is N1 = 800, N2 = 650 mm, N3 = 500 mm, N4 = 575 mm, N5 = 600 mm, N6 = 700 mm


Solution:

Let x be the weight area factor for the two unknown rain gauge station

Sum of all weight area factor = 1

∴ 0.1 + 0.35 + 0.16 + 0.15 + x + x = 1

⇒ 2x = 0.24

⇒ x = 0.12.

Rainfall data for 6th rain gauge station

P6 = 607.45 mm

∴ Mean rainfall value

= (600 × 0.1) + (550 × 0.35) + (625 × 0.16) + (475 × 0.15) + (400 × 0.12) + (607.45 × 0.12)

= 544.64 mm

Pmean = 544.64 mm

QUESTION: 58

A 3-hr unit hydrograph for a catchment is given in the table below. The maximum ordinate and delay in the peak for a 6-hr unit hydrograph will be

Solution:

When the given 3-hr unit hydrograph is added to the same 3-hr unit hydrograph placed at a lag of 3 hr, then we obtain the ordinates of a 6-hr hydrograph containing 2 cm runoff. Hence, when this hydrograph ordinate is divided by 2, we can obtain the hydrograph of 6 hr duration containing 1 cm runoff, which will be nothing but 6 hr unit hydrograph.

Maximum ordinate of 6 hr UH = 28.5 m3/s

Delay in the peak = 8 hr – 5 hr

= 3 hr.

QUESTION: 59

At a road intersection in New Delhi, during the peak hour about 2500 vehicle ply per hour at an average speed of 30 kmph. Out of these, 50% of the car use leaded petrol. The average fuel consumption is one liter for an average of 10 km of travel. Considering that 60% of the lead present in the fuel is emitted in the form of a particulate aerosol, then the emission rate (in microgram/hours) of lead aerosol in the ambient air is 

(It is given that the concentration of lead in the fuel is 

Solution:

Number of vehicles passing per hour at the intersection of the road = 2500 

Number of lead-emitting vehicles passing per hour on the road =  2500 × 0.50 = 1250 

Average fuel consumption by 1250 vehicles per hour  = 30kmph = 3Liter/hours

Total fuel consumption by 1250 vehicles per hour  = 1250 × 3 Liter/hour = 3750 Liter/hour

Since the total concentration of lead in the fuel =  

The total lead concentration in 3750 Liter/hour of used fuel

=   x 750Liter/hour = 1875μg/hr

 Since 60% of lead present in the fuel is emitted as a particulate aerosol, the lead aerosol released in the air = 1875 × 0.60 =

Hence,  of lead is released in the air as a particulate aerosol.

*Answer can only contain numeric values
QUESTION: 60

A water treatment plant has to treat 0.5 MLD water with 20 filter of equal areas are installed for the operation. Due to maintenance work 5 filters are non-operational and the corresponding hydraulic loading rate is 131.65 m3/m2/day. If the discharge of the treatment plant is 800 L/s, then the area of each filter (in m2) will be ______.


Solution:

Out of 20 filter, 5-are non-operational

∴ Hydraulic loading rate for 15 filters is 131.65 m3/m2/day.

Discharge of water = 800 L/s

= 0.8 m3/s

= 0.8 × 86400 m3/day

= 69120 m3/day

∴ Hydraulic loading rate = 

⇒ A = 525 m2

∴ Area of each filter = 

= 35.

*Answer can only contain numeric values
QUESTION: 61

A 14 m wide two-lane road on a plain terrain is to be constructed along a horizontal curve of radius 400 m. The design speed of the highway is 100 kmph and the design coefficient of lateral friction between the tire and the road is 0.15. The numerical sum of the estimated value of the superelevation required (if full friction is assumed to be developed) and the value of the coefficient of friction needed (if 7% superelevation is provided) will be ________ (up to two correct decimal place)


Solution:

Radius of horizontal curve, R = 400 m

Design speed of the road, V = 100 kmph

Coefficient of lateral friction between the tyre and the road = 0.15

Where, e = super elevation

t = lateral friction

putting f = 0.15

⇒ e = 0.19685 – 0.15

E = 0.04685

Also, putting e = 0.070

⇒ f = 0.19685 – 0.070

⇒ f = 0.12685

∴ Numerical sum of super elevation in first case and lateral friction in second case

= 0.04685 + 0.12685

= 0.1737

*Answer can only contain numeric values
QUESTION: 62

The length of a new runway to be constructed under the standard condition is 1710 m. The elevation of airport site is 750 m above the mean sea level. The airport reference temperature and the effective gradient of the runway are 25.7oC and 1.6% respectively. If the runway is to be constructed as per the International Civil Aviation Organization (ICAO) recommendations, then the corrected length of the runway after elevation and temperature correction (in km, up to two correct decimal place) will be ______.


Solution:

Correction for elevation: 7% increase per 300 m

So, correction =

= 299.25 m

Corrected length = 1710 + 299.25 = 2009.25 m

Correction for temperature: 1% increase per degree rise in temperature

Standard atmospheric temperature = 15 – 0.0065 × 750 = 10.125°C

Rise of temp = 25.7°C – 10.125°C = 15.575°C

15.575 = 311.43375m

Correct length = 2009.25 + 311.43375 = 2320.684 m

Check for total correction for elevation plus temperature

 

According to ICAO, the total correction for elevation plus temperature should not exceed by 35%.

Hence, the total correction for elevation plus temperature correction = 1710 = 598.5m

Therefore, the corrected length of the runway = 1710 + 598.5 = 2308.5 m

*Answer can only contain numeric values
QUESTION: 63

Given the following data: design life n = 15 years, lane distribution factor D = 0.75, annual rate of growth of commercial vehicles r = 6%, vehicle damage factor F = 4 and initial traffic in the year of completion of construction = 3000 Commercial Vehicles Per Day (CVPD). As per IRC:37-2012, the design traffic in terms of cumulative number of standard axles (in million standard axles, up to two decimal places) is ______


Solution:



*Answer can only contain numeric values
QUESTION: 64

Two pegs P and Q were fixed on opposite banks of a river and the following measurements were made during testing a levelling instrument

If the reduced level of station Q is 106.500 m, the Reduced level of station P is _____ m. (up to three decimal places. Neglect earth’s curvature and refraction correction.


Solution:

Let H be the difference of level between P and Q.

Let hp and hq be the staff readings taken from P.

hp  = 3.200 m

hq  = 2.100 m

Let h'and hq′ be the staff readings taken from Q

hp′ = 2.900m 

hq′ = 1.300m

H = 1.35 m

Q is at higher level

R.LP = RLQ – H

R.LP = 106.500 – 1.35

RLP = 105.15 m

QUESTION: 65

The following are the measured values of the angle P with their respective weights

I. 55° 25’ 17’’, weight 4

II. 55° 25’ 22’’, weight 3

III. 55° 25’ 26’’, weight 2

The probable error of the mean will be

Solution:

Concept:

Probable error of the mean (Em)w

v = Measured value – Most probable value

w = weight

n = no. of observations

Calculation:

Most probable value = 

Most probable value = 55° 25’ 20.67’’

v= - 3.67’’

v2 = 1.33’’

v3 = 5.33’’