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# Gate Mock Test Civil Engineering (CE)- 7

## 65 Questions MCQ Test GATE Civil Engineering (CE) 2022 Mock Test Series | Gate Mock Test Civil Engineering (CE)- 7

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This mock test of Gate Mock Test Civil Engineering (CE)- 7 for GATE helps you for every GATE entrance exam. This contains 65 Multiple Choice Questions for GATE Gate Mock Test Civil Engineering (CE)- 7 (mcq) to study with solutions a complete question bank. The solved questions answers in this Gate Mock Test Civil Engineering (CE)- 7 quiz give you a good mix of easy questions and tough questions. GATE students definitely take this Gate Mock Test Civil Engineering (CE)- 7 exercise for a better result in the exam. You can find other Gate Mock Test Civil Engineering (CE)- 7 extra questions, long questions & short questions for GATE on EduRev as well by searching above.
QUESTION: 1

### Choose the most approximate word from the options given below to complete the following sentence. If I had known that you were coming, I _______ you at the airport

Solution:

The sentence talks about a past incident and a condition has been mentioned. Thus option 2 fits here correctly. 'Would have' is the correct tense to be used here. Option 1 is illogical. 'Would' must be used here and not 'will' as it is in the past tense. Past perfect tense is incorrect here.

QUESTION: 2

### Choose the most approximate word from the options given below to complete the following sentence. I believe in the _____ of positive thinking thus always recommend these books to my clients.

Solution:

The word that fits here is a noun thus options 3 and 4 are eliminated. The form should be singular as for abstract nouns we do not use the plural form. Option 1 is thus the correct answer.

The word powerful is an adjective and 'empowering' is a verb.

QUESTION: 3

### Consider a circle of radius r. Fit the largest possible square inside it and the largest possible circle inside the square. What is the radius of the innermost circle?

Solution:

The radius of outer circle = r

∴ Diagonal of square = 2r

Side of square (a)=√2r

Now, the radius of the inner circle = r

QUESTION: 4

A bird files along the three sides of a field in the shape of an equilateral triangle at speeds of 3, 6, 8 km/hr respectively. The average speed of the bird is

Solution:

Average speed = Total distance/Total time

Total distance = a + a + a = 3a

Total time =

Average speed =

QUESTION: 5

Choose the most appropriate words from the options given below to fill in the blanks.

She was ______ to travel abroad and _____ in the field of commerce as per her wishes.

Solution:

The sentence mentions a person doing something as per her wishes thus the first word must be a positive reaction. 'Restive' which means 'restless' and 'jinxed' which means 'cursed' are incorrect here. Between options 1 and 3,the word 'elated' which means 'too happy' and the word 'venture' which means 'undertake a risky or daring journey or course of action' fit here correctly. The word 'cease' means 'stop' and does not convey a proper meaning. Thus option 3 is the correct answer.

QUESTION: 6

In a 1600 m race around a circular track of length 400 m, the faster runner and the slowest runner meet at the end of the sixth minute, for the first time after the start of the race. All the runners maintain uniform speed throughout the race. If the faster runner runs at twice the speed of the slowest runner. Find the time taken by the faster runner to finish the race.

Solution:

As, the faster runner is twice as fast as the slowest runner, the faster runner would have complete two rounds by the time the slowest runner completes one round.

Their first meeting takes place after the fastest runner takes 6 min to complete two rounds and the slowest runner completes one round.

Time taken to complete one round by fastest runner = 6/2 = 3 minutes

Rounds needed to complete the race = 1600/400 = 4

Time taken to complete the race = 4 × 3 = 12 min

QUESTION: 7

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

Horse: Foal

Solution:

The young one of a horse is called a foal. Similarly, the young one of a goose is known as gosling. Thus, option 2 contains the pair that best expresses a relationship similar to that expressed in the given pair. In all the other pairs, the first word expresses the male form of the latter.

QUESTION: 8

Read the following passage and find out the inference stated through passage.

Juvenile delinquency is also termed as Teenage Crime. Basically, juvenile delinquency refers to the crimes committed by minors. These crimes are committed by teenagers without any prior knowledge of how it affects the society. These kind of crimes are committed when children do not know much about outside world.

Which of the following Inferences is correct with respect to above passage?

Solution:

Since teenagers here have tender age and commit crimes unknown to its consequences, clearly they should not be treated as any criminal. They need to be taught so that they can make difference between right and wrong and do not commit these crimes in future. Hence inference in option 1 is correct.

QUESTION: 9

If n and y are positive integers and 450 y = n³, which of the following must be an integer?

Solution:

450 y = n³ implies that 450 y is the cube of an integer.

When we prime-factorize the cube of an integer, we get 3 (or a multiple of 3) of every prime factor.

8 is the cube of an integer because 8 = 2³ = 2*2*2.

Thus, when we prime-factorize 450 y, we need to get at least 3 of every prime factor.

Here's the prime factorization of 450 y:

450 y = 2 * 3² * 5² * y

Since 450 provides only one 2, two 3's, and two 5's, and we need at least 3 of every prime factor, the missing prime factors must be provided by y. Thus, y must provide at at least two more 2's, one more 3, and one more 5.

Smallest possible case:

y = 2² * 3 * 5

QUESTION: 10

The graph shows cumulative frequency % of research scholars and the number of papers published by them. Which of the following statements is true?

Solution:

In this problem, cumulative frequency is given, so converting into frequency

From the table, it is clear that option b, i.e.

60% of the scholars published at least 2 papers is correct.

QUESTION: 11

For which value of x, the following matrix A is a singular matrix.

Solution:

A matrix is singular if and only if its determinant is zero.

x(3+1)−2(x2+1)=0

−2x2+4x−2 = 0

x2−2x+1 = 0

(x−1)2 = 0

x = 1

QUESTION: 12

Find the laplace transform of t2sin (2t).

Solution:

According to formula, L(tf(t)) =

Here, f(t) = sin (2t) ⇒ F(s) =

∴ L(t2sin (2t)) =

QUESTION: 13

Consider the following function:

f(x) at x=2 is

Solution:

ii) (4 - x)x=2 = 2

Hence at x = 2, f(x) is continuous

∴ f(x) is not differentiable.

QUESTION: 14

value of b = ?

Solution:

The given limit is in 0/0 form, So

Applying L hospitals rule

⇒ log b - 1 = -1 - log b

⇒ 2 log b = 0

⇒ b = 1

QUESTION: 15

The integrating factor of equation y log y dx + (x – log y) dy = 0 is

Solution:

y log y dx + (x – log y) dy = 0

It is a Leibnitz’s equation in x.

Integrating factor =

QUESTION: 16

For the Real Beam shown in the figure below, identify the correct conjugate beam

Solution:

In context of the Real and conjugate Beam:

(i) The fixed end of the real beam will become the free end of the conjugate beam

(ii) The diagram of the Real beam will be loading diagram for the conjugate beam

(iii) Hogging moment diagram (-ve) in the real beam will give downward loading in the conjugate beam

As end A is fixed in the real beam, so it will be free in conjugate beam. Similarly for end B in real beam will become fixed end in conjugate beam. The  diagram of the real beam will become loading diagram for the conjugate beam.

*Answer can only contain numeric values
QUESTION: 17

A 25 cm long rod of uniform rectangular section of size 10 mm (wide) and 1.4 mm thickness is bend to form an arc of a circle. If the longitudinal strain at the surface of the rod is 0.00084, the central displacement of the rod will be _____ mm. Neglect second order quantities.

Solution:

Concept:

Using the relations/ formula

At surface y = ymax , ε = εmax

εsurface

Calculation:

εsurface = 0.00084

L = 250 mm

ymax  = = 0.7mm

εsurface =

εsurface =

εsurface = 0.00084

0.00084 =

δ = 9.375 mm

QUESTION: 18

Consider the following statements and Identify the correct option with regard to provisions of IS 456: 2000:

P. The minimum stripping time for props to slab spanning over 4.5 m is 7 days

Q. The minimum stripping time for props to arches spanning over 6 m is 14 days.

R. To ensure strength of Designed concrete mix, Random sampling procedure is adopted

S. When bar of two different diameter are to be spliced, the lap length shall be calculated on the basis of bar of the larger diameter.

Solution:

As per IS 456 : 2000,

(i) The minimum stripping time for props to slab spanning up to 4.5 m and spanning over 4.5 m is 7 days and 14 days respectively.

(ii) The minimum stripping time for props to aches spanning up to 6 m and spanning over 6 m is 14 days and 21 days respectively.

(iii) When bar of two different diameter are to be spliced the lap length shall be calculated on the basis of bar of larger diameter.

(iv) A random sampling procedure shall be adopted to ensure that each concrete batch shall have a reasonable chance of being tested that is, the sampling should be spread over the entire period of concreting and cover all mixing units.

QUESTION: 19

For a truss shown in figure below. Find the force in the member GC ______ kN.

Solution:

Member AB and AG are co-linear,

So FAB = FAG = 0 kN

Now removing the member AB and AG. Now member BC and BG are also collinear. So, FBC = FBG = 0 kN

Consider Joint G

∑Fx = 0

FGC Sin 63.43° = 0

FGC = 0 kN

∑FV = 0

20 + FGF + FGC Cos 63.43° = 0

FGF = - 20 kN

*Answer can only contain numeric values
QUESTION: 20

A tension member 1SLB 250 @ 273.7 N/m is connected with two 200 wide and 12 mm thick plate with two lines of 16 mm diameter bolt in each flange. If the net section area is 2800 mm2 , then the tensile strength (in kN) on the basis of net area cracking of ISLB section using Fe 410 grade of steel will be _________. (Take Partial Safety Factor as 1.25)

Solution:

Concept:

Tensile strength of a section, (Tdn)

where An = net section area, mm2

fu = ultimate strength of the material

γm = pantial safety factor.

Calculation:

For Fe 410 grade of steel

fu = 410 N/mm2, fy = 250 N/mm2

γm = 1.25

Tdn = 826.56 kN.

*Answer can only contain numeric values
QUESTION: 21

A circular footing of diameter 4m is loaded with 200 kN/m2 of load intensity. The vertical stress increment (kN/m2) at 6m depth below the footing will be?

Solution:

Concept:

Vertical stress increment is given by:

r = radius of circular area

z = depth of soil stata

q = uniformly distributed circular load

Calculation:

z = 6 m

q = 200 kN/m2

σz = 29.2 kN/m2

*Answer can only contain numeric values
QUESTION: 22

A square footing of size 3m × 3m is placed at 2m depth below the Ground surface. If the permissible settlement of the foundation rests on sand strata is 40 mm and the final corrected S.P.T Number is 28. The net allowable bearing pressure from settlement consideration according to Peck, Hanson equation will be _____kN/m2

Assume water table is at 1m below the base of the footing

Solution:

Concept:

Using Peck, Hanson equation:

The net allowable bearing pressure (kN/m2) is given by

qa(net) = 0.41 N S CW

N = Final corrected SPT Number

Cw = Water table correction factor

S = Permissible Settlement of foundation (mm)

Calculation:

N = 28

S = 40 mm

Cw = 0.8

qa(net) = 0.41 × 28 × 40 × 0.8

qa(net) = 367.36 kN/m2

QUESTION: 23

The liquid limit and plastic limit of the soil is 55% and 25% respectively. If the natural water content of the soil is 30%. The correct statement is

Solution:

Consistency Index (IC) of the soil is:

IP = WL – WP

Wn = 30%

WL = 55%

WP = 25%

Liquidity Index (IL) will be:

Wn = 30%

WP = 25%

WL = 55%

As the soil is in Plastic stage of consistency because both liquidity index and consistency index lies in between 0 to 1.

*Answer can only contain numeric values
QUESTION: 24

A capillary rise in a glass tube when immersed in oil having surface tension 0.0825 N/m at 20° C and specific gravity of 0.96 is 6.25 mm. For 30% increase in height of capillarity, the diameter of glass tube should be decreased by ______ (percentage)

Take angle of contact between glass tube and oil will be zero

Solution:

Concept:

Height of capillary rise is given by:

Where,

σ = Surface tension of fluid at 20° C

ρ = Density of fluid (kg/m3)

D = Diameter of the glass tube

h = height of capillary rise/fall

Calculation:

σ = 0.0825 N/m

θ = 0°

ρ = 1000 × 9.6 = 960 kg/m3

Diameter (D) = 5.60 mm

For 30% increase in “h”

hI = 8.125 mm

DI = 4.31 mm

% Decrease

QUESTION: 25

As per IS 10500:2012, the total concentration of manganese (as Mn) and iron (as Fe) for drinking water purpose shall not exceed

Solution:

As per IS10500:2012, the total concentration of manganese (as Mn) and iron (as Fe) for drinking water purpose shall not exceed 0.3 mg/L.

QUESTION: 26

A water sample analysis produces alkalinity and hardness of 200 mg/L and 250 mg/L as Caco3 respectively. The carbonate and non-carbonate hardness respectively will be

Solution:

Concept:

Carbonate hardness = minimum of {Alkalinity, Total Hardness}

Total hardness = Carbonate Hardness + Non-carbonate Hardness.

Calculation:

Carbonate Hardness= minimum of {200 mg/L , 250 mg/L}

Carbonate Hardness = 200 mg/L

Non – Carbonate Hardness = Total Hardness – Carbonate Hardness

= 250 – 200

= 50 mg/L

QUESTION: 27

The ultimate BOD of waste water is 300 mg/L and the reaction rate constant (to the base ‘e’) at 20°C is 0.3585/day, then the 5 day BOD will be.

Solution:

Concept:

Ultimate BOD = Lo

BOD5­ = 5day BOD, of water/waste water

BOD at any time ‘t’ (Lt)

Lt = Lo(1−e−kDt

Where Lo = ultimate BOD

KD = Rate constant

Calculation:

Lt = 300 × (1 – e-0.3585 × t)

L5 = BOD5 = 300 × (1 – e-0.3585 × 5)

= 250 mg/L

QUESTION: 28

As per IRC 37: 1984, the practical capacity values suggested for the purpose of design of two lane roads with 7.0 m wide carriageway and earthen shoulders in both direction is

Solution:

Capacity of different types of roads in rural areas.

QUESTION: 29

In order to allow smooth movement of Locomotive with a maximum permissible speed of 120 kmph on a Broad Gauge track in transition curve, the radius of curvature by Martin’s formula will be

Solution:

Concept:

We know that by Martin’s formula for a Broad Gauge track in a transition curve. For speed greater than 100 kmph.

Speed.

Where V is speed in kmph

R is the radius of curvature in m.

Calculation:

QUESTION: 30

In an open traverse PQR. If the Bearing of the line PQ and RQ is 60° and N50° W respectively. The included angle ∠PQR will be

Solution:

Bearing of PQ = 60°

Bearing of RQ = N50° W (WCB)

Bearing of RQ = 360° - 50° = 310° (QBS)

Bearing of line QR = Bearing (RQ) – 180°

Bearing of line QR = 310° - 180° = 130°

Bearing of line QP = 180° + Bearing of PQ = 180° + 60° = 240°

∠PQR = (Bearing)QP = - (Bearing)QR

∠PQR = 240° - 130°

∠PQR = 110°

QUESTION: 31

The closing error in a closed loop traverse of 500 m total length is 0.289 m, then the relative precision of this traverse will be

Solution:

Relative precision =

where p = perimeter of traverse

e = closing error

Relative precision =

Relative precision =

QUESTION: 32

For specification of split spoon sampler one should follow:

Solution:

IS 6403 – 1981: Indian standard code of Practice for determination for Bearing Capacity of shallow foundation

IS 4640 – 1980: Indian standard specification for split spoon sampler

IS 1892 – 1979: Code of Practice for site investigations for foundations

IS 2132 – 1972: Code of Practice for Thin walled tube sampling for soils

*Answer can only contain numeric values
QUESTION: 33

A rectangular block 6m long, 4m wide and 2m high is immersed in a oil of specific gravity 0.90. if the metacentric height for the stable equilibrium is 1.4m, then the depth of immersion of the rectangular block in a oil will be _____m.

(up to two decimal places)

Solution:

Concept:

Metacentric height of the rectangular block is given by:

Imin = Moment of inertia of the plan about X – X or Y – Y (minimum value should be taken)

VDis = Volume of the displaced fluid

B G = Distance between centre of Buoyancy and centre of gravity of the block

Calculation:

Let the Depth of immersion of the block will be ‘x’

GM = 1.4 m

Imin = Iyy = 32 m4

BG = PG – PB

PG = 1 m

VDis = x × 4 × 6 = 24x

3x2 – 14.4x + 8 = 0

X1 = 4.16 m

X2 = 0.641 m

X1 = 4.16 m is not possible

So, X2 = x = 0.641 m

So, depth of immersion will be 0.641 m.

*Answer can only contain numeric values
QUESTION: 34

The average surface area of a reservoir for a period of two months is 50 hectare-m. The difference of saturated vapour pressure and actual vapour pressure is 15 mm of Hg. If the wind velocity at a height of 9.0 m above the ground is 10m/s, then the daily average evaporation in (mm/day) from the reservoir will be (use km = 0.36)

Solution:

Concept:

The empirical method to evaluate evaporation based on Dalton’s law in developed by Meyer’s and the equation is as follow.

E = Rate of evaporation in mm/day

ew = saturated vapour pressure in mm of Hg

ea = vapour pressure of air in mm in Hg

V9 = Mean wind velocity in km/hr at a height of 9 m from the ground.

km = constant, 0.36 for large deep water and 0.50 for small shallow water.

Calculation:

(es - ea) = 15 mm of Hg

V= 10 m/s

= 36 km/hr

E = 17.55 mm/day

*Answer can only contain numeric values
QUESTION: 35

The ratio of average normal flow of traffic to the saturation flow in a two intersecting road P and Q are 0.3 and 0.25 respectively. If the total lost time per cycle is 11 seconds, then the optimum cycle length as per Webster’s approach in seconds will be ________.

Solution:

Concept:

The optimum signal cycle is given by as per Webster’s Method

Where L = total lost time per cycle, in sec

= 2n + R (n = number of lanes and sum of R is all red-time)

Y = Sum of Ratio of normal flow to the saturation flow.

= y1 + y2

y1, y2 are the ratio of normal flow to saturation flow in direction 1 and 2.

Where q1 = Normal flow on each approach

q2 = Normal flow on each approach

S = saturation flow per unit time.

Calculation:

Given:

L = 11 sec

yp = 0.3 sec

yQ = 0.25 sec

∴ Y = yp + yQ = 0.3 + 0.25

Y = 0.55

The optimum cycle length

Co = 47.77 sec

QUESTION: 36

The column vector is a simultaneous Eigen vector of    and

Solution:

Given that,    is an Eigen vector of A.

⇒ AX = λ1X

For all values of a and b, is an Eigen vector of A.

Given that,  is an Eigen vector of B

⇒ BX = λ2X

being an eigen vector of matrix B

a + b = λ2 a      ----(1)

2a = λ2 b      ----(2)

From (1) and (2)

⇒ ab + b2 = 2a2

⇒ 2a2 - ab - b2 = 0

⇒ 2a2 - 2ab + ab - b2 = 0

⇒ 2a (a - b) + b (a - b) = 0

⇒ b = a, b = -2a

For b = a, (or) b = -2a, X is an eigen vector of matrix B.

QUESTION: 37

Satisfy the assumption of Rolle’s theorem in the interval [-1, 1], then ordered pair (p, q) is

Solution:

Since the above function f(x) satisfy the assumption of Rolle’s theorem in interval [-1, 1]

∴ f(x) is continuous in [-1, 1]

F(x) is differentiable in (-1, 1)

F(-1) = f(1)

∴ for continuity

LHL = RHL = Functional value at 0 i.e f(0)

= 1

= q

∴ LHL = RHL

⇒ q = 1

Also for differentiability

LHD = RHD

at x = 0

at x = 0

p – q = 1

⇒ p = 1 + q

⇒ p = 1 + 1

⇒ p = 2

∴ ordered pair (p, q) ≡ (2, 1)

*Answer can only contain numeric values
QUESTION: 38

Let y(x) be the solution of the initial value problem     = x and y(0) = 0. Find the value of     (Correct up to 2 decimal places).

Solution:

Integrating on both sides:

Now given y (0) = 0

Thus,

C = 0

ln (1 – 2y) = -x2

1−2y = e − x2

2y =1 − e - x2

QUESTION: 39

2000 cashew nuts are mixed thoroughly in flour. The entire mixture is divided into 1000 equal parts. And each part is used to make 1 biscuit. Assume that no cashews are broken in the process. A biscuit is picked at random. The probability it contents no cashew is _____.

Solution:

Assuming poisson distribution

Alternate:

• Probability a particular cashew is in particular biscuit is
• Probability a particular cashew is not in a particular biscuit =
• Probability that no cashew in =
QUESTION: 40

If the following formula is used for the numerical integration of a function f(x), then what should be the values of the constants a, b, c so that for f(x) = 1 or f(x) =  x or f(x) =  x2, there is no error in the formula

Solution:

For f(x) = 1, x, x2 there is no error in the formula

f(x) = 1

= a(1) + b(1) + c(1)

2h = a + b + c

f(x) = x

= a(−h) + b(0) + c(h)

a – c = 0

f(x) = x2

= a(h2) + b(0) + c(h2)

⇒ a = c =

*Answer can only contain numeric values
QUESTION: 41

An I-section bean is fabricated with plates having an elastic section modulus of 200 × 10-5m3. If the laterally unsupported steel beam is classified as a semi-compact section, then the design plastic moment capacity of the beam considering Fe 410 grade of steel and the partial safety factor of material as 1.1 will be ____ kNm. (Use Bending stress reduction factor (XLT= 1.0)

Solution:

Concept:

The bending strength or the plastic moment capacity of a laterally unsupported beam is given by:

Md = βb Zp fbd

Where

βb = 1 → for plastic and concept section

for semi - compact section.

Ze = Elastic section modulus

Zp = Plastic section modulus

fbd = design bending compressive strength

ymo = Partial safety factor for material = 1.1

XLT = bending stress reduction factor to account for lateral torsional buckling

XLT ≤ 1.0

Calculation:

For a semi-compact section of a laterally unsupported steel bean plastic moment capacity

XLT = 1.0

γmo = 1.1

fy = 250 Mpa for grade fe 410

Md = 200 =  454.54 kNm

QUESTION: 42

The specific gravity of the material of the dam is 2.65 and it is designed as an elementary profile with uplift pressure coefficient as 0.5. If the height of the dam to be built is 45m, then the minimum width of the dam for no tension condition will be approximately

Solution:

For No tension failure

Where

Bmin = Minimum width of the dam

H = Height of the dam

G = specific gravity of the dam

C = coefficient of uplift pressure.

For critical condition

c = 0

Calculation:

Minimum width

= 30.69

= 31m

*Answer can only contain numeric values
QUESTION: 43

A 10 m wide rectangular channel carries a distance of 22 m3/s at a depth of 1.2 m. The minimum height of the hump (in cm) required to be placed at a section to cause critical flow over the hump, neglecting energy loss will be ______.

Solution:

Concept:

Minimum height of the hump

Where y1 = depth of flow

F1 = Froude numbers at section depth y1.

Calculation:

ΔZm = 0.186 m

ΔZm = 18.6 cm

Alternate solution:

= 1.371 m

yc = 0.79

Ec = 1.5 × 0.79 = 1.185

Δzm = E1 – Ec = 1.371 – 1.185 = 0.186 m = 18.6 cm

Δzm = 0.186, Δzm = 18.6 cmZ

*Answer can only contain numeric values
QUESTION: 44

A layer of normally consolidated clay of thickness 1.5 m is undergoing one dimensional consolidation under a pressure increment of 30 kPa. If the final primary consolidation settlement (mm) by using Terzaghi’s theory is 20 mm, the initial effective stress within the soil layer will be ______ kN/m2. Assume the following soil properties:

initial void ratio (e0) = 1.345.

Compression index (Cc) = 0.40.

Solution:

Concept:

Primary consolidation settlement is given by

CC = compression index

e0 = initial void ratio

H0 = thickness of clay layer.

initial effective stress within the clay layer.

increase in effective stress

Δ H = primary consolidation settlement

Calculation:

CC = 0.40

e0 = 1.345

Δ H = 20 mm

H0 = 1.5 m

*Answer can only contain numeric values
QUESTION: 45

As per IRC 37:2012, the design traffic in terms of the cumulative number of standard axis (in million standard axes, up to two decimal places) based on below data will be_____.

Initial traffic in the year of completion of construction = 4000 commercial vehicle per day (CPVD)

Vehicular damage factor = 5

Lane distribution factor = 0.75

Annual rate of growth of commercial vehicle = 9%

Design life = 12 years.

Solution:

Concept:

To find the cumulative standard axle load repetition using Modified CBR method

Where A = Number of commercial vehicle per day when construction is completed.

r = Rate of growth of traffic

n = design life of pavement.

LDF = Lane distribution factor

VDF = Vehicle damage factor.

Calculation:

Ns = 110.27 million standard axles.

*Answer can only contain numeric values
QUESTION: 46

A traffic survey is conducted on an intersection to compute practical capacity yields the following results:

The width of the weaving section = 12m

Length of the weaving section = 54 m

The proportion of weaving traffic = 60%

The average width of entry and exit = 8m

The practical capacity (in vehicle/hr) of the rotary on the basis of the above data will be

Solution:

Concept:

Practical capacity of Rotary

Where W = width of weaving section.

e = Average width at entry and exit =

P = proportion of weaving traffic

L = length of the weaving section.

The above expression will be valid only when these four conditions will be satisfied

A. 6m ≤ w ≤ 18m

D. 0.4 ≤ P ≤ 1

Calculation:

Given:

W = 15m

L = 54m

e = 8m

P = 60%

Practical capacity of Rotary

Qp = 3665.5 vehicle/hr

*Answer can only contain numeric values
QUESTION: 47

The measured co-ordinates on the vertical photographs are

Xa = +15.50 mm

Ya = -71.20 mm

Xb = +109.20 mm

Yb = -25.20 mm

The elevations of a and b are 502 m and 525 m respectively. If the flying height above the datum is 1670 m, the ground distance of a line AB will be ______ m.

Assume focal length to be 158 mm.

Solution:

Concept:

The distance AB on ground is given by:

Calculation:

ha = 502 m, hb = 525 m

H = 1670 m

xa = +15.50 mm

xb = = +109.20 mm

ya = -71.20 mm

yb = -25.20 mm

Xb = 791.35 m

Xa = 114.58 m

Yb = -182.62 m

Ya = -526.34 m

AB = 759.05 m

*Answer can only contain numeric values
QUESTION: 48

A vertical cut is made in a pure saturated clay having undrained shear strength of 150 kN/m2 (consider smooth vertical back). How much actual height (m) of this vertical cut be made such that there is a 400% factor of safety against shear failure.

Take γsat of clay = 19 kN/m3.

Solution:

Concept:

Maximum height of vertical cut is given by:

γ = γsat

For pure clay ɸ = 0°

cu = undrained cohesion in clay

Where,

τu = undrained shear strength of clay

Cu = undrained cohesion in clay

ɸu = undrained friction angle

Calculation:

As, ɸu = 0°

τu = Cu + 0

150 = Cu

Hc = 31.58 m

Ho = Actual height of vertical cut

Ho = 7.895 m

*Answer can only contain numeric values
QUESTION: 49

A simply supported rectangular prestressed concrete beam of span 9 m has to be prestressed with a force of 2000 kN. The profile of the cable is parabolic with zero eccentricity at ends. If the maximum dip (allowable) of the tendon at the mid-span to balance the external load is 0.18 m, the maximum uniformly distributed load (throughout the span) the beam can carry will be _____kN/m. (Neglect the self weight of the beam).

Solution:

Equating the moments

P = Prestressing force = 2000 kN

l = span of the beam = 9 m

w = load on the beam

e = eccentricity of the tendon = 0.18 m

w = 35.55 kN/m

*Answer can only contain numeric values
QUESTION: 50

In order to prepare 50m3 Ready Mix concrete for a rigid pavement for a mix 1 : 3 : 6 by weight, with water/cement ratio of 0.5 and unit weight of concrete as 2250 kg/m3, the amount of coarse aggregate (in tonnes) will be ______ .

Solution:

Calculation:

Let us assume the weight of cement will be w.

Therefore, For 1m3 of concrete

(1 × w + 3 × w + 6 × w) + 0.5w = 2250 kg

10.5 w = 2250

w = 214.29 kg

Therefore, the amount of course aggregate for 1mof concrete will be

= 3 × 214.29

= 642.87 kg

For 50m3 of concrete

= 32.14 tonnes

*Answer can only contain numeric values
QUESTION: 51

A hollow circular shaft of length 4 m is to transmit 500 kW of power at 350 rpm. If the maximum permissible shear stress developed in the shaft is limited to 70 Mpa, the maximum twist in the shaft will be _____ radian. (up to two decimal places.)The ratio of internal diameter to external diameter of the shaft will be 0.75 and shear modulus of the shaft maternal will be 90 Gpa.

Solution:

Concept: Power developed in the shaft will be

Where, P = power developed / transmitted

T = Applied Torsion

N = Speed of rotation of shaft (rpm)

Torsion Equation:

IP = Polar section modulus of the shaft

L = Length of the shaft

θ = Angle of rotation of the shaft

G = shear modulus of the shaft material

τ = shear stress developed in the shaft material

r = radial distance of the shaft

Calculation:

P = 500 kW

T = 13.642 kN-m

Dex = 112.33 mm

Dint = 0.75 × 112.33

Dint = 84.25 mm

*Answer can only contain numeric values
QUESTION: 52

For the state of stress (MPa) shown in the figure below. The ratio of Major principal stress to minor principal stress will be _____ (up to two decimal places)

Solution:

Concept:

Major (σ1) and Minor (σ2) principal stresses are given by:

Calculation:

σx = 15 MPa

σy = 10 MPa

τxy = 5 MPa

σ1 = 18.10 MPa

σ2 = 12.5 – 5.59

σ2 = 6.91 MPa

*Answer can only contain numeric values
QUESTION: 53

A flocculation chamber 50m long, 15 m wide and 5m deep is to treat 250 MLD of water. It is equipped with 18m long and 0.45m wide paddles. The tank contains 2500m3 of water which is mixed using paddles at an average velocity gradient of 106/second. If the kinematic viscosity of water at 28° C temperature is 8.9 × 10-7m2/s, then the power required (in kW, up to two decimal place) to achieve adequate mixing will be ________.

Solution:

Concept:

μ = dynamic viscosity of water,

V = Volume of the raw water to be treated, m3

P = Power dissipated, watts

Calculation:

Given:

G’ = 106/second

V = 2500 m3

νν = 8.9 × 10-7 m2/s

∴ dynamic viscosity (μ) = ρν

μ = 1000 kg/m3 × 8.9 × 10-7 m2/s

⇒ P = μVG’2

⇒ P = 8.9 × 10-4 × 2500 × 106 × 10-3

⇒ P = 0.23585 kW

*Answer can only contain numeric values
QUESTION: 54

A continuous beam ABC of two equal span AB and BC carries load of P at a distance 5 m from end C as shown in given figure. If the length of the beam and the moment carrying capacity of beam is 20 m and MP then the collapse load P of the beam will be ______ MP.

Solution:

Possible mechanism for the beam is

By Principle of Virtual work

P × 5 × Q = MP × θ + QMP × 2θ

⇒ P = 0.6 MP

*Answer can only contain numeric values
QUESTION: 55

A block of weight 50kN is supported by a rope that is wrapped one-and-one-half times around the circular shaft. If the coefficient of static friction between the rope and the shaft material is 0.15, then the largest possible value of P according to the figure (given below) will be ____x 104 N.

Solution:

Concept:

∴ T2 = P = T1 eμθ ​

If the tension in the one side of the belt is known, then the tension in the other side of the belt can be calculated as

T2 = T1 eμθ​

Where, T1 and T2 are tension in the rope

μ = coefficient of friction between rope and shaft material.

q = contact angle in radian.

Calculation:

T2 = P and T1 = 50 kN

q = 1.5 × 2π

q = 3π (q must be in radian)

∴ T2 = P = T1 eμθ

P = 50 × (e)0.15×3π

P = 205.6 kN

P = 20.56 x 104 N

*Answer can only contain numeric values
QUESTION: 56

For a construction project, PERT calculation yield a project duration of 50 weeks with a variance of 25 weeks. The probability of completion of the project in 63 weeks will be________%, using the below table.

Solution:

Concept:

Probability factor

Where

TS = Scheduled time required to complete the project

TE = expected completion time of the project

σ = standard deviation

Calculation:

Expected completion time of the project, TE = 50 weeks

variance = 25 weeks

Scheduled Completion time of project = 63weeks

∴  standard deviation

Probability factor

z = 2.6

Probability = 97.72 + (2.6 - 2)

Probability = 99.01%

*Answer can only contain numeric values
QUESTION: 57

A completely saturated clay sample was tested in a UU test under a cell pressure of 250 kN/m2 and the principal stresses at failure was noted as 350 kN/m2 and 200 kN/m2 respectively. If the effective shear strength parameters of this clay layer are cI= 25 kN/m2 and ɸ= 23°, the pore water pressure at failure will be _____ kN/m2

Solution:

Concept:

From Mohr Coulomb failure criterion:

ɸI = effective angle of friction

cI = effective cohesion

= major effective principal stress at failure

= Effective cell pressure

Calculation:

CI = 25 kN/m2

ɸI = 23°

Because in unconsolidated undrained test, total stress parameters are used

350 - uf = (200 - uf) × 2.283 + 75.54

350 - uf = 456.6 - 2.283 uf + 75.54

(2.283-1) uf = 456.6 + 75.54 – 350

uf = 141.96 kN/m2

*Answer can only contain numeric values
QUESTION: 58

Pervious sand having air content 15% is used in foundation of a masonry dam. If the maximum permissible upward gradient for a factor of safety of 3 against boiling is 0.315. The percentage air voids in the pervious sand will be ______ %.

Solution:

Concept:

F.O.S against boiling is giving by

η = percentage air voids

ac = air content

η = porosity

calculation:

icr = iper x FOS

icr = 0.315 × 3

icr = 0.945

e = 0.746

η = 0.427 = 42.7%

ac = 15%

ηa = 0.427 × 0.15

ηa = 0.06405

ηa = 6.405%

QUESTION: 59

A hydraulically efficient trapezoidal section conveying a discharge of 210 m3/s. If the Froude number at a section is 0.489, the bed slope required will be?

Assume the flow to be uniform and manning’s roughness coefficient to be 0.017.

Solution:

Concept:

Discharge through the trapezoidal section is given by

n → manning’s roughness coefficient

R → Hydraulic mean depth

For efficient cross-section,

y0 = depth of flow

A = Area of cross-section

For efficient cross-section,

S0 = Bed slope of the channel

Calculation:

Q = 210 m3/sec

Fr = 0.48

D = Hydraulic depth

For efficient cross-section,

y0 = 6.132 m [depth of flow]

QUESTION: 60

The velocity components in a two-dimensional flow is given as

V = 4x4 – y2

Identify the correct statement:

Solution:

Concept:

To check the possibility of flow:

Stream function (Ψ)

Velocity potential function (ϕ):

Calculation:

Hence flow is possible

Angular velocity (WZ)

WZ ≠ 0

Flow is rotational

Also, For rotational floes velocity potential (ϕ) function does not exist.

Proof:

But WZ ≠ 0

So, ∇ 2 ϕ ≠ 0

ϕ not exist

check for stream function (Ψ)

For stream function, if there is a flow, stream function (Ψ) always exist.

Integrating

From equation (i) and (ii)

*Answer can only contain numeric values
QUESTION: 61

A plate load test is carried out on a 40 cm × 40 cm square plate which is placed at a depth of 3m below the ground surface. Consider the soil strata below the ground surface which is extended up to great depth. Further the data obtained from the above test is

The settlement of a foundation of size 2.5m × 2.5m carrying a load of 1600 kN located at a depth of 4m below ground surface is _____mm.Assume water table is at great depth below Ground surface and depth correction factor to be 0.88.

Solution:

Concept:

Settlement of foundation on sands is given by:

Sf = settlement of foundation (mm)

Sp = settlement of plate (mm)

Bf = width of foundation (m)

Bp = width of plate (m)

Calculation:

Load intensity on the foundation =

From the chart,

Sp = 13.5 mm

Bf = 2.5m

Bp = 0.40 m

Settlement of foundation at 3m depth is:

Sf = 32.96 mm

Now correcting it for the depth

So, final settlement of foundation at 4m below the ground surface is

Sf = (Depth correction factor) × Sf

Sf = 0.88 × 32.96

Sf = 29.00 mm

*Answer can only contain numeric values
QUESTION: 62

In a catchment there are five rain gauge stations A, B, C, D and E are located on a circular shape basin of diameter 24 km as shown in figure below. If the rainfall at station A, B, C, D and E in the month of August are 250 cm, 200 cm, 225 cm, 240 cm and 210 cm respectively, then the average rainfall (in m, up to two decimal place) over the catchment in month of August will be __________.

Solution:

Thiessen Polygon Method:

This method is also known as weighted area method. The rainfall data at every station is given a weightage on the basis of an area close to the station

Where Pavg = Average Precipitation

P1, P2, ........Pn are precipitation at station 1, 2, ....., n respectively

A1, A2, ........An is the area for station 1, 2, ....., n respectively

This method of fining average rainfall is suitable

→ when the area is large

→ rainfall varies from place to place

1) In order to find the representative area, all the stations are joined, so as to form a network, of the non-intersection triangle.

2) Perpendicular bisector is then drawer from the sides of the triangle in such a way that a representative area is obtained for each of the rain gauge stations.​

Calculation:

Catchment diameter = 24 km

Catchment area for station A

= (12 × 12) km2

= 144 km2

∴ Total area of catchment

= 452.4 km2

∴ Catchment area for B, C, D, and E will be equal and

= 77.1 km2

∴ Mean Rainfall

Mean Rainfall =228.7 cm = 0.2287 m

*Answer can only contain numeric values
QUESTION: 63

A rapid sand filter with area of filter 200m2 is used to process the water for a town with a population of 10 million. The rate of filtration of the filter is 25 m3/m2/hour. The porosity of the bed and specific gravity of the material used in the bed are 0.42 and 2.72 respectively. If the bed is expanded to 0.75 m from its original undisturbed depth of 0.675 m, then the head loss in (mm) due to expanded bed will be________.

Solution:

Concept:

Hydraulic Head Loss and Expansion of the Filter bed during Backwash:

Backwashing is accomplished by reversing the flow in filter media to force clean water to move upward through the filter media. In order to clean the filter bed, it is required to expand the bed, so that granular media filters are no longer in contact with each other.

To hydraulically expand a porous bed, the head loss (hLe) must be at least equal to the buoyant weight of the filter bed. Even when the bed gets expanded to depth De, the head loss through the expanded bed essentially remains unchanged, because the total buoyant weight of the bed is constant.

Head loss, hLe = D (1 – n) (G – 1)        …..(i)

where D = undisturbed depth of filter bed in m.

n = porosity

G = specific gravity

Also, hLe = De (1 - n­e)(G – 1)        ..…(ii)

where De = the depth of expanded bed in m.

ne = porosity of expanded bed

∴ from equation (i) and (ii) we get:

Calculation:

⇒ ne = 0.478

Therefore,

hLe = De (1 – ne) (G – 1)

hLe = 0.75 (1 – 0.478) (2.72 – 1)

hLe = 0.67338m

∴ hLe = 673.4 mm

Alternate Solution:

Head loss in both the condition will be same i.e.

hLe = D (1 – n) (G – 1) = De (1 – ne)(G – 1)

hLe = 0.675 (1 – 0.42) (2.72 – 1)

hLe = 0.67338 m

hLe = 673.38 mm

QUESTION: 64

Consider the following statement about the municipal solid waste:

P. If it is assumed that air contains 23.15 percent oxygen by mass, then the amount of air required for the complete oxidation of 1 kg of carbon would be equal to 11.25kg.

Q. The anaerobic digestion will require an optimum C/N ratio of about 30 – 50.

R. Pulverization refers to the action of cutting and tearing, whereas shredding refers to the action of crushing and grinding.

S. Electrostatic precipitator is also installed in the incineration plant to reduce air pollution, particularly due to the emission of dioxin.

T. Pyrolysis is highly exothermic in nature and hence also known as destructive distillation.

Which one of the following options is correct?

Solution:

1) Amount of oxygen required for the complete combustion of solid waste, it is necessary to compute the oxygen requirement for the oxidation of carbon, hydrogen, and sulfur contained in the waste.

The basic reactions involved are:

For carbon

C + O2 → CO2

C - (12), O2 - (32) and CO2 - (44)

For Hydrogen

2H2 + O2 → 2H2O

H - (4), O2 - (32) and 2H2O(64)

The amount of air required for the complete oxidation of 1 kg of carbon would be

= 11.52 kg.

2) Pyrolysis is a highly endothermic process and known as destructive distillation whereas combustion process is a highly exothermic process.

3) For proper development of anaerobic digestion, C/N ratio of the digestive material should be between 30 to 50 for optimum digestion.

4) Pulverization refers to the action of crushing and grinding, whereas shredding refers to the action of cutting and tearing.

5)  ​An electrostatic precipitator is also installed in the incineration plant to reduce air pollution, particularly due to the emission of dioxin.

*Answer can only contain numeric values
QUESTION: 65

While recording sound level two reading have been taken at a site within an hour. 80 dB noise lasting for 40 minutes is followed by 100 dB noise lasting for 20 minutes. The Lequivalent of the above noise levels (in dB) will be ______.

Solution:

Concept:

LequivalentLequivalent is that statistical value of sound pressure level that can be equated to any fluctuating noise. Thus, Leq is defined as the constant noise level, which over a given time, expand the same amount of energy, as it expanded by the fluctuating levels over the same time.

The value of Leq can be expressed as

Where n = Total number of sound samples

Li = The noise level of any ith sample

ti = Time duration of ith sample, expressed as fraction of total sample time.

Calculation:

Total time for which observation has been done

= 40 min + 20 min

= 60 min

Leq= 10 × 9.531 dB

Leq = 95.31 dB