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QUESTION: 1

They have come a long way in _______ trust among the users.

Solution:

They have come a long way in "creating" trust among the users.

QUESTION: 2

The CEO’s decision to quit was as shocking to the Board as it was to __________.

Solution:

“me”

QUESTION: 3

The lecture was attended by quite _________ students, so the hall was not very___________.

Solution:

a few, quiet

QUESTION: 4

If E = 10; J = 20; O = 30; and T = 40, what will be P + E + S + T?

Solution:

P = 16 × 2 = 32

E = 5 × 2 = 10

S = 19 × 2 = 38

T = 20 × 2 = 40

P + E + S + T = 120

*Answer can only contain numeric values

QUESTION: 5

On a horizontal ground, the base of straight ladder is 6m away from the base of a vertical pole. The ladder makes an angle of 45° to the horizontal. If the ladder is resting at a point located at one-fifth of the height of the pole from the bottom, the height of the pole is _____meter.

Solution:

QUESTION: 6

P, Q, R, S and T are related and belong to the same family. P is the brother of S. Q is the wife of P. R and T are the children of the siblings P and S respectively. Which one of the following statements is necessarily FALSE?

Solution:

‘T’ is child of ‘S’. So option (b) is right.

QUESTION: 7

A square has sides 5cm smaller than the sides of a second square. The area of the larger square is four times the area of the smaller square. The side of the larger square is ________cm.

Solution:

Given,

(Area)B = 4 × (Area)A

⇒ x^{2} = 4(x – 5)^{2}

⇒ x^{2} = 4[x^{2} + 25 – 10x]

⇒ x^{2} = 4x^{2} + 100 – 40x

⇒ 3x^{2} – 40x + 100 = 0

⇒ 3x^{2} – 30x – 10x + 100 = 0

⇒ 3x(x – 10) – 10(x – 10) = 0

⇒ x = 10

QUESTION: 8

In a sports academy of 300 people, 105 play only cricket, 70 play only hockey, 50 play only football, 25 play both cricket and hockey, 15 play both hockey and football and 30 play both cricket and football. The rest of them play all three sports. What is percentage of people who play at least two sports?

Solution:

Total = 300 C = 105

H = 70

F = 50

C + H = 25

H + F = 15

C + F = 30

C + H + F = 300 – (295) = 5

% of people playing at least 25 sports

QUESTION: 9

The increasing interest in tribal characters might be a mere coincidence, but the timing is of interest. None of this, though, is to say that the tribal hero has arrived in Hindi cinema, or that the new crop of characters represents the acceptance of the tribal character in the industry. The films and characters are too few to be described as a pattern.

Q. Who does the word ‘arrived’ mean in the paragraph above?

Solution:

The word ‘arrived’ mean in the above para is to “Attained a status”

QUESTION: 10

The new cotton technology, Bollgard-II, with herbicide tolerant traits has developed into a thriving business in India. However, the commercial use of this technology is not legal in India. Notwithstanding that, reports indicate that the herbicide tolerant Bt cotton had been purchased by farmers at an average of Rs 200 more than the control price of ordinary cotton, and planted in 15% of the cotton growing area in the 2017 Kharif season.

Q. Which one of the following statements can be inferred from the given passage?

Solution:

Farmers want to access the new technology even if it is not legal

QUESTION: 11

In a soil specimen, the total stress, effective stress, hydraulic gradient and critical hydraulic gradient are σ, σ’, i and ic, respectively. For initiation of quicksand condition, which one of the following statement is TRUE?

Solution:

During quick sand condition, the effective stress is reduced to zero [i.e, σ ‘ = 0]

QUESTION: 12

Assuming that there is no possibility of shear buckling in the web, the maximum reduction permitted by IS 800-2007 in the (low-shear) design bending strength of a semi-compact steel section due to high shear is

Solution:

Maximum reduction permitted by IS 800-2007 is zero.

QUESTION: 13

The coefficient of average rolling friction of a road is fr and its grade is +G%. If the grade of this road is doubled, what will be the percentage change in the braking distance (for the design vehicle to come to stop) measured along the horizontal (assume all other parameters are kept unchanged)?

Solution:

Case I: Braking distance

Case II: Braking distance

Percentage change

*Answer can only contain numeric values

QUESTION: 14

A circular duct carrying water gradually contracts from a diameter of 30cm to 15cm. The figure (not drawn to scale) shows the arrangement of differential manometer attached to the duct.

When the water flows, the differential manometer shows a deflection of 8cm of mercury (Hg). The values of specific gravity of mercury and water are 13.6 and 1.0, respectively. Consider the acceleration due to gravity, g = 9.81 m/s^{2}. Assuming frictionless flow, the flow rate (in m^{3}/s, round off to 3 decimal places) through the duct is __________.

Solution:

h = 100.8 cm = 1.008 m

Flow rate

A1 = 4A_{2}, g 9.81m/sec^{2}

= 0.01767 m^{2}

*Answer can only contain numeric values

QUESTION: 15

A concentrated load of 500 kN is applied on an elastic half space. The ratio of the increase in vertical normal stress at depths of 2m and 4m along the point of the loading, as per Boussinesq’s theory, would be _______.

Solution:

Boussinesq’s theory

QUESTION: 16

A retaining wall of height H with smooth vertical backface supports a backfill inclined at an angle β with the horizontal. The backfill consists of cohesionless soil having angle of internal friction ϕ. If the active lateral thrust acting on the wall is Pa which one of the following statements is TRUE?

Solution:

active thurst act at a height H/3 from the base of the wall and at an angle equal to backfill inclination.

QUESTION: 17

In a rectangular channel, the ratio of the velocity head to the flow depth for critical flow condition, is

Solution:

Velocity head for a critical flow

So, ratio of velocity head to critical flow depth = 1/2

*Answer can only contain numeric values

QUESTION: 18

The probability that the annual maximum flood discharge will exceed 25000 m3/s, at least once in next 5 years is found to be 0.The return period of this flood event (in years, round off to 1 decimal place) is _____________.

Solution:

Probability exceed maximum discharge at least once in next 5 years is given by = 1 (1– p)^{n}

⇒ 0.25 = 1 - (1 - P)^{n} [n = 5 year]

⇒ P = 0.559

⇒ 1/T = 0.559

⇒ T = 17.9 year

QUESTION: 19

The interior angles of four triangles are given below:

Q. Which of the triangles are ill-conditioned and should be avoided in Triangulation surveys?

Solution:

A triangle is said to be ill condition when angle is less than 30° and more than 120°.

So, triangle S is ill conditioned.

For well conditioned of triangulation two angle should not be almost equal.

So, only triangle Q or triangle R is ill-conditioned

So, ill-condition S and Q or S and R. option S and Q is given. So option (b) correct.

QUESTION: 20

A catchment may be idealised as a rectangle. There are three rain gauges located inside the catchment at arbitrary locations. The average precipitation over the catchment is estimated by two methods: (i) Arithmetic mean (P_{A}) and (ii) Thiessen polygon (P_{T}).

Q. Which of the following statements is correct?

Solution:

There is no definite relationship between arithmetic mean and Thiessen polygon method.

Only it can be says that in Thiessen polygon method average value is more uniformly distributed as compared to arithmetic mean.

QUESTION: 21

An isolated concrete pavement slab of length L is resting on a frictionless base. The temperature of the top and bottom fibre of the slab are T_{t} and T_{b}, respecitvely. Given: the coefficient of thermal expansion = α and the elastic modulus = E. Assuming T_{t} > T_{b} and the unit weight of concrete as zero, the maximum thermal stress is calculated as

Solution:

Here thermal stress is zero.

*Answer can only contain numeric values

QUESTION: 22

For a given loading on a rectangular plain concrete beam with an overall depth of 500 mm, the compressive strain and tensile strain developed at the extreme fibers are of the same magnitude of 2.5 × 10^{-4}. The curvature in the beam cross-section (in m^{-1}, round off to 3 decimal places), is __________.

Solution:

Given, ∈ = ∈_{t }= ∈_{c} = 2.5x10^{-4}

y = 250 mm = 0.25 m

As per flexure formula:

⇒ Curvature of beam cross-section

⇒

= 0.001 m^{–1}

QUESTION: 23

For a small value of h, the Taylor series expansion of f(x + h) is

Solution:

For the small value of h, the Taylor’s series expansion of

So option (b) is correct

QUESTION: 24

If the path of an irrigation canal is below the level of a natural stream, the type of cross drainage structure provided is

Solution:

If the path of an irrigation canal is below the bed level of a natural stream, the type of crossdrainage work provided is super passage.

QUESTION: 25

In the reinforced beam section shown in the figure (not drawn to scale), the nominal cover provided at the bottom of the beam as per IS 456–2000, is

All dimensions are in mm

Solution:

QUESTION: 26

Consider a two-dimensional flow through isotropic soil along x-direction and z-direction. If h is the hydraulic head, the Laplace’s equation of continuity is expressed as

Solution:

*Answer can only contain numeric values

QUESTION: 27

A soil has specific gravity of its solids equal to 2.The mass density of water is 1000 kg/m^{3}. Considering zero air voids and 10% moisture content of the soil sample, the dry density (in kg/m^{3}, round off to 1 decimal place) would be _______.

Solution:

G_{s} = 2.65

ρ_{w }= 1000 kg/m^{3}

ηa = 0

w = 10%

= 0.10

= 2094.862 kg/m^{3}

QUESTION: 28

Which one of the following is correct?

Solution:

So, option (a) is correct.

QUESTION: 29

A plane truss is shown in the figure (not drawn to scale).

Q. Which one of the options contains ONLY zero force members in the truss?

Solution:

Only 4 member having zero force member GF, FI and SR, PR

QUESTION: 30

A simple mass-spring oscillatory system consists of a mass m, suspended from a spring of stiffness k. Considering z as the displacement of the system at any time t, the equation of motion for the free vibration of the system is The natural frequency of the system is

Solution:

For simple harmonic motion

Standard equation is

⇒

⇒

*Answer can only contain numeric values

QUESTION: 31

A completely mixed dilute suspension of sand particles having diameters 0.25, 0.35, 0.40, 0.45 and 0.50mm are filled in a transparent glass column of diameter 10 cm and height 2.50 m. The suspension is allowed to settle without any disturbance. It is observed that all particles of diameter 0.35 mm settle to the bottom of the column in 30 s. For the same period of 30s, the percentage removal (round off to integer value) of particles of diameters 0.45 and 0.50 mm from the suspension is _____________.

Solution:

Since sand particle of size 0.35 mm settles to the bottom of the column in 30 secparticles having size greater than 0.35 mm i.e. 0.45 and 0.50 mm will also settle insuspension at the bottom of column by 100% in 30 sec, infact these bigger sized particlewill settle by 100% in less than 30 sec. So answer is 100%.

*Answer can only contain numeric values

QUESTION: 32

The maximum number of vehicles observed in any five minute period during the peak hour, If the total flow in the peak hour is 1000 vehicles, the five minute peak hour factor (round off to 2 decimal places) is ___________.

Solution:

Five minute peak hour factor

QUESTION: 33

Which one of the following is secondary pollutant?

Solution:

Ozone is a secondary pollutant

QUESTION: 34

An element is subjected to biaxial normal tensile strains of 0.0030 and 0.0020. The normal strain in the plane of maximum shear strain is

Solution:

(∈)_{at max shear strain}

QUESTION: 35

Consider the pin-jointed plane truss shown in the figure (not drawn to scale). Let R_{P}, R_{Q}, and RR denote the vertical reactions (upward positive) applied by the supports at P, Q, and R, respectively, on the truss. The correct combination of (R_{P}, R_{Q}, R_{R}) is represented by

Solution:

F_{2} = F_{3} …(iii)

F_{2} × 3 + 30 × 3 – F_{3} × 1 = 0

F_{2} × 3 – F_{2} × 1 + 90 = 0

2F_{2} = – 90

F_{2} = – 45

⇒ F_{3} = –45

Q + R = 0 …(iv)

F3 × 2 + R × 3 = 0

– 45 × 2 + R × 3 = 0

QUESTION: 36

For the following statements:

P – The lateral stress in the soil while being tested in an oedometer is always at-rest.

Q – For a perfectly rigid strip footing at deeper depths in a sand deposit, the vertical normal contact stress at the footing edge is greater than that at its centre.

R – The corrections for overburden pressure and dilatancy are not applied to measured SPTN values in case of clay deposits.

The correct combination of the statements is

Solution:

P – TRUE; Q – TRUE; R – TRUE

*Answer can only contain numeric values

QUESTION: 37

Tie bars of 12 mm diameter are to be provided in a concrete pavement slab. The working tensile stress of the tie bars is 230 MPa, the average bond strength between a tie bar and concrete is 2 MPa, and the joint gap between the slab is 10mm. Ignoring the loss of bond and the tolerance factor, the design length of the tie bars (in mm, round off to the nearest integer) is ___________.

Solution:

Given:

d = 12 mm

σ_{st} = 230 MPa = 230 N/mm^{2}

S_{b} = 2 MPa = 2 N/mm^{2}

t = 10 mm

Length of tie bar

= 10 + 690

= 700 mm

*Answer can only contain numeric values

QUESTION: 38

Average free flow speed and the jam density observed on a road stretch are 60 km/h and 120 vehicles/km, respectively. For a linear speed-density relationship, the maximum flow on the road stretch (in vehicles/h) is _________.

Solution:

Vf = 60km/h

kJ = 120 Veh/km

*Answer can only contain numeric values

QUESTION: 39

The network of a small construction project awarded to a contractor is shown in the following figure. The normal duration, crash duration, normal cost, and crash cost of all the activities are shown in the table. The indirect cost incurred by the contractor is INR 5000 per day.

If the project is targeted for completion in 16 days, the total cost (in INR) to be incurred by the contractor would be _____

Solution:

Path PRTV is critical path and corres-ponding normal duration is 18 days.

For 18 days:

Direct cost = 66000

Indirect cost = 18 × 5000 = 90000

Total project cost = 156000

1st stage crashing :

Crash activity R by 1 day.

New project duration = 17 days.

T.P.C = 156000 + 1 × 750 – 1 × 5000 = 151750

2nd stage crashing :

Crash activity Q & R simultaneously by 1 day. New project duration = 16 days.

T.P.C = 151750 + 1 × (750 + 2000) – 5000 = 149500

*Answer can only contain numeric values

QUESTION: 40

A sample of air analyzed at 0°C and 1 atm pressure is reported to contain 0.02 ppm (parts per million) of NO_{2}. Assume the gram molecular mass of NO_{2} as 46 and its volume at 0°C and 1 atm pressure as 22.4 litres per mole. The equivalent NO_{2} concentration (in microgram per cubic meter, round off to 2 decimal places) would be _______

Solution:

0.02 ppm of NO_{2} means

*Answer can only contain numeric values

QUESTION: 41

Traffic on a highway is moving at a rate 360 vehicles per hour at a location. If the number of vehicles arriving on this highway follows Poisson distribution, the probability (round off to 2 decimal places) that the headway between successive vehicles lies between 6 and 10 seconds is _________

Solution:

λ = 360 veh/hr

*Answer can only contain numeric values

QUESTION: 42

Consider the ordinary differential equation Given the values of y(1) = 0 and y(2) = 2, the value of y(3) (round off to 1 decimal place), is ______

Solution:

y(1) = 0 & y(2) = 2

Putting this into ordinary differential equation

D(D – 1)y – 2Dy + 2y = 0

[D^{2} – 3D + 2] = 0

Auxillary equation is

m^{2} – 3m + 2 = 0

(m – 1) (m – 2) = 0

i.e. m = 1 & m = 2

∴ solution of equation

y = c_{1}e^{t} + c_{2}e^{2t}

⇒ Putting et as x

⇒ y = c_{1}x + c_{2}X^{2}

⇒ y(1)= 0

⇒ c_{1} + c_{2} = 0 ...(i)

& y(2) = 2

⇒ 2c_{1} + 4c_{2} = 2

⇒ c_{1} + 2c_{2} = 1 …(ii) (ii) – (i)

⇒ c_{2} = 1

c_{1 }= –1

⇒ y = –x + x^{2}

Then y(3) = –3 + 3^{2} = 6

*Answer can only contain numeric values

QUESTION: 43

A 3 m × 3 m square precast reinforced concrete segments to be installed by pushing them through an existing railway embankment for making an underpass as shown in the figure. A reaction arrangement using precast PCC blocks placed on the ground is to be made for the jacks.

At each stage, the jacks are required to apply a force of 1875 kN to push the segment. The jacks will react against the rigid steel plate placed against the reaction arrangement. The footprint area of reaction arrangement on natural ground are: c = 17 kPa; ϕ = 25^{o} and γ = 18 kN/m^{3}. Assuming that the reaction arrangement has rough interface and has the same properties that of soil, the factor of safety (round off to 1 decimal place) against shear failure is __________.

Solution:

FOS against shear failure = strength/Appliedload

= (c + σtanϕ)A/P

σ = N/A

= (24kN/m^{3} × 37.5m^{2} × 7.5m)/37.5m^{2}

= 24 × 7.5 kN/m^{2}

⇒ FOS = (c + σtanϕ)A/P

= [(17 + 24 × 7.5 × tan25∘)×37.5]/1875

FOS = 2.0187

*Answer can only contain numeric values

QUESTION: 44

A parabolic vertical curve is being designed to join a road of grade + 5% with a road of grade –3%. The length of the vertical curve is 400 m measured along the horizontal. The vertical point of curvature (VPC) is located on the road of grade +5%. The difference in height between VPC and vertical point of intersection (VPI) (in m, round off to the nearest integer) is ______________

Solution:

Height difference between

VPI & VPC = 5% of 200 m = 10 m

QUESTION: 45

If the section in the figure turns from fully-elastic to fully-plastic, the depth of neutral axis (N.A.), decreases by

All dimensions are in mm

Solution:

For fully elastic case,

For fully plastic case,

y = Equal area axis = 5

N.A reduces by = 18.75 – 5 = 13.75 mm

*Answer can only contain numeric values

QUESTION: 46

A portal frame shown in figure (not drawn to scale) has a hinge support at joint P and a roller support at joint R. A point load of 50 kN is acting at joint R in the horizontal direction. The flexural rigidity. EI, of each member is 10^{6} kNm^{2}. Under the applied load, the horizontal displacement (in mm, round off to 1 decimal place) of joint R would be __________

Solution:

For reaction

∑M_{p }= 0

–W × 5 + 50 × 10 = 0

When unit load at R is acting in the direction of 50kN load, then reaction at R = 2 (downward)

Alternatively:

*Answer can only contain numeric values

QUESTION: 47

A box measuring 50 cm × 50 cm × 50 cm is filled to the top with dry coarse aggregate of mass 187.5 kg. The water absorption and specific gravity of the aggregate are 0.5% and 2.5, respectively. The maximum quantity of water (in kg, round off to 2 decimal places) required to fill the box completely is _________

Solution:

Volume of the box = 0.5 × 0.5 × 0.5

= 0.125 m^{3}

Mass of aggregate = 187.5 kg

G_{agg} = 2.5

Volume of aggregate

Volume of empty space = 0.125 – 0.075

= 0.05 m^{3}

Water absorption = 0.5%

Volume of water absorbed

Total volume of water that can be filled

= 9.375 × 10–4 + 0.05

= 0.0509 m^{3}

Mass of water = 50.94 kg

*Answer can only contain numeric values

QUESTION: 48

A wastewater is to be disinfected with 35mg/L of chlorine to obtain 99% kill of microorganisms. The number of micro-organisms remaining alive (N_{t}) at time t, is modelled by Nt = N0e^{–kt}, where N0 is number of microorganisms at t = 0, and k is the rate of kill. The wastewater flow rate is 36m^{3}/h, and k = 0.23 min-If the depth and width of the chlorination tank are 1.5 m and 1.0m, respectively, the length of the tank (in m, round off to 2 decimal places) is _________

Solution:

For 99% kill of mircoorganision

–0.23t = –4.605

t = 20.022 min

Volume of tank req. = Q.t

QUESTION: 49

The cross-section of a built-up wooden beam as shown in the figure (not drawn to scale) is subjected to a vertical shear force of 8kN. The beam is symmetrical about the neutral axis (N.A.) shown, and the moment of inertia about N.A. is 1.5 × 10^{9} mm^{4}. Considering that the nails at the location P are spaced longitudinally (along the length of the beam) at 60 mm, each of the nails at P will be subjected to the shear force of

All dimensions are in mm

Solution:

Shear force in nail at

*Answer can only contain numeric values

QUESTION: 50

A staff is placed on a benchmark (BM) of reduced level (RL) 100.000 m and a theodolite is placed at a horizontal distance of 50m from the BM to measure the vertical angles. The measured vertical angles from the horizontal at the staff readings of 0.400m and 2.400 m are found to be the same. Taking the height of the instrument as 1.400 m, the RL (in m) of the theodolite station is _________

Solution:

2x = 2.8

x = 1.4m

H.O.I = 100 + 1.4

= 101.4m

RL of theodelite station = 101.4 – theodolite hight

= 101.4 – 1.4

= 100m

*Answer can only contain numeric values

QUESTION: 51

A 0.80 m deep bed of sand filter (length 4m and width 3m) is made of uniform particles (diameter = 0.40 mm, specific gravity = 2.65, shape factor = 0.85) with bed porosity of 0.the bed has to be backwashed at a flow rate of 3.60 m^{3}/min. During backwashing, if the terminal settling velocity of sand particles is 0.05 m/s, the expanded bed depth (in m, round off to 2 decimal places) is ______

Solution:

Expanded bed depth = D_{e} = D[(1−n)/(1−n_{e})]

D = Depth of bed = 0.8m; n = Bed porosity = 0.4

ne = porosity of expanded bed =(v_{b}V_{s})0.22

V_{s} = terminal velocity of settling particles during back wasting = 0.05m/s

V_{b} = backwash velocity

= QBL(Q=Backwashflowrate)

[Q = 3.6 m^{3}/min B = 3m L = 4m]

⇒ V_{b} = 3.6/(60×3×4)

= 0.005m/s

⇒ n_{e} = (0.005/0.05)0.22

= 0.60255

∴ D_{e} = 0.8[(1−0.4)/(1−0.60255)]

= 1.21m/s

*Answer can only contain numeric values

QUESTION: 52

A reinforced concrete circular pile of 12m length and 0.6 m diameter is embedded in stiff clay which has an undrained unit cohesion of 110 kN/m^{2}. The adhesion factor is 0.The Net Ultimate Pullout (Uplift) Load for the pile (in kN, round off to 1 decimal place is) is

Solution:

p = perimeter ℓ = length

= 0.5 × 110 × 12 × π(0.6)

= 1244.07 kN

QUESTION: 53

A survey line was measured to be 285.5m with a tape having a nominal length of 30m. On checking, the true length of the tape was found to be 0.05 m too short. If the line lay on a slope of 1 in 10, the reduced length (horizontal length) of the line for plotting of survey work would be

Solution:

Measured length = 285.5 m

Nominal length of tape = 30 m

Slope = 1 in 10

The tape is 0.05 m too short

Actual length of tape = 30 – 0.05 = 29.95 m

Actual length measured

Now slope correction

∵ Slope = 1 in 10

⇒ Length to be plotted

= Actual length measured + correction

= 285.024 + (-1.42512)

= 283.599 m

Hence, option (d) is correct.

QUESTION: 54

The rigid-joined plane frame QRS shown in the figure is subjected to a load P at the joint R. Let the axial deformation in the frame be neglected. If the support S undergoes a settlement of the vertical reaction at the support S will become zero when β is equal to

Solution:

Using slope deflection method,

If reaction at S is equal to zero

.....(i)

From equilibrium of joint

.....(ii)

QUESTION: 55

Which one of the following is NOT a correct statement?

Solution:

y = x^{1/x}

⇒ x = e

Thus point x = e is the critical point for y = x^{1/x}

Now at x = e, dy/dx changes its sign from (+ve) to (-ve). Thus point (x = e) is point of global maxima.

y = x^{3} has neither global minima nor global maxima, it only have saddle point at x = 0

y | x |; attains its minimum value at x = 0; so x = 0 is the global minima for y = f(x)

QUESTION: 56

A rectangular open channel has a width of 5m and a bed slope of 0.001. For a uniform flow of depth 2m, the velocity is 2m/s. The Manning’s roughness coefficient for the channel is

Solution:

For a rectangular channel

Width of channel = 5 m

Depth of flow = 2 m

Bed slope = 0.001

Velocity V = 2 m/sec.

From manning’s

*Answer can only contain numeric values

QUESTION: 57

The hyetograph of a storm event of duration 140 minutes is shown in the figure.

The infiltration capacity at the start of this event (t = 0) is 17mm/hour, which linearly decreases to 10 mm/hour after 40 minutes duration. As the event progresses, the infiltration rate further drops down linearly to attain a value of 4mm/ hour at t = 100 minutes and remains constant thereafter till the end of the storm event. The value of the infiltration index, ϕ (in mm/hour, round off to 2 decimal places), is _______

Solution:

Depth of infiltration = Area of hyetograph above Horton’s curve

Now, assuming

*Answer can only contain numeric values

QUESTION: 58

Consider a laminar flow in the x-direction between two infinite parallel plates (Couette flow). The lower plate is stationary and the upper plate is moving with a velocity of 1 cm/ s in the x-direction. The distance between the plates is 5mm and the dynamic viscosity of the fluid is 0.01 N-s/mIf the shear stress on the lower plate is zero, the pressure gradient, (in N/m^{2} per m, round off to 1 decimal place) is______

Solution:

Velocity of plate, V = 1 cm/sec

Distance between the late = 5 mm

Dynamic viscosity of fluid = 0.01 N-S/m^{2}

Shear stress at lower plate = 0

Pressure gradient

We know that, in case of couette flow, shear stress (τ) is given by

At lower plate, y = 0; τ = 0 [Given]

*Answer can only contain numeric values

QUESTION: 59

A granular soil has a saturated unit weight of 20 kN/m^{3} and an effective angle of shearing resistance of 30°. The unit weight of water is 9.81 kN/m^{3}. A slope is to be made on this soil deposit in which the seepage occurs parallel to the slope up to the free surface. Under this seepage condition for a factor of safety of 1.5, the safe slope angle (in degree, round off to 1 decimal place) would be ____

Solution:

*Answer can only contain numeric values

QUESTION: 60

Two water reservoirs are connected by a siphon (running full) of total length 5000 m and diameter of 0.10 m, as shown below (figure not drawn to scale).

The inlet leg length of the siphon to its summit is 2000 m. The difference in the water surface levels of the two reservoirs is 5m. Assume the permissible minimum absolute pressure at the summit of siphon to be 2.5m of water when running full. Given: friction factor f = 0.02 throughout, atmospheric pressure = 10.3 m of water, and acceleration due to gravity g = 9.81 m/s^{2}. Considering only major loss using Darcy- Weisbach equation the maximum height of the summit of siphon from the water level of upper reservoir, h (in m round off to 1 decimal place) is _____

Solution:

d = 0.1 m

Length of siphon = 5000 m

Length of siphon upto summit = 2000 m

Friction Factor, f = 0.02

Acceleration due to gravity, g = 9.81 m/sec^{2}

Applying Energy equation between point 1 and 3 to get

[From Darcy Weisback equation

Now applying energy equation between 1 and 2 to get

QUESTION: 61

Sedimentation basin in a water treatment plant is designed for a flow rate of 0.2 m^{3}/s. The basin is rectangular with a length of 32m, width of 8m and depth of 4m. Assume that the settling velocity of these particles is governed by the Stokes’ law. Given: density of the particles = 2.5 g/cm^{3}; density of water = 1 g/cm^{3}; dynamic viscosity of water = 0.01 g/(cm.s); gravitational acceleration = 980 cm/s^{2}. If the incoming water contains particles of diameter 25 μm (spherical and uniform) the removal efficiency of these particles is

Solution:

Flow rate = 0.2 m^{3}/sec

Dimension of tank = 32m × 8 m × 4 m

Density of particles = 2.5 g/cc

Density of water = 1 g/cc

Dynamic viscosity of water = 0.01 g/cm-S

Diameter of particle = 25 μm

We know that

Over flow rate of tank

= 7.8125 × 10^{-4 }m/sec

And settling velocity of particle (v_{s}),

Now, % removal efficiency

Hence option (b) is correct.

*Answer can only contain numeric values

QUESTION: 62

A square footing of 4m side is placed at 1 m depth in a sand deposit. The dry unit weight (γ) of sand is 15 kN/m^{3}. This footing has an ultimate bearing capacity of 600 kPa. Consider the depth factors; d_{q} = d_{γ} = 1.0 and the bearing capacity factor: N_{γ} = 18.This footing is placed at a depth of 2m in the same soil deposit. For a factor of safety of 3.0 per Terzaghi’s theory, the safe bearing capacity (in kPa) of this footing would be ________

Solution:

Side of square footing = 4 m

Depth of footing = 1 m

Unit weight of soil = 15 KN/m^{3}

Ultimate bearing capacity = 600 KPa

Depth factors, d_{q} = d_{γ} = 1

N_{γ} = 18.75

According to terzaghi, the ultimate bearing capacity of square footing is given as

At depth of footing = 1 m

q_{u} = 1.3CN_{C} + qN_{q}d_{q} + 04B_{γ}N_{γ}d_{γ}

For sand, C = 0, q = _{γ}D_{f} = 15 × 1 = 15KN m^{2}

600 = 0 + 15 × N_{q} × 1 + 0.4 × 4 × 15 × 18.75 × 1

⇒ Nq = 10

Now at depth of footing at 2m

q_{u} = 1.3 CN_{C} + qN_{q} + 0.4B_{γ}N_{γ}d_{γ}

q_{u} = 0 + (2 × 15)10 × 1 + 0.4 × 4 × 15 × 18.75 × 1

q_{u }= 750 KP_{a}

∵ We know that

q_{nu} = q_{u} - _{γ}D_{f}

q_{nu} = 750 – 15 × 2

q_{nu} = 720 KPa

and safe bearing capacity q_{safe}

QUESTION: 63

Consider two funct In Which one of the following is the

correct expression for ?

Solution:

Putting value of ϕ in (i)

Assuming y constant and differentating ψ w.r.t. x.

Puttting value of from (ii) in equation (iii)

*Answer can only contain numeric values

QUESTION: 64

A 16 mm thick gusset plate is connected to the 12 mm thick flange plate of an I-section using fillet welds on both sides as shown in the figure (not drawn to scale). The gusset plate is subjected to a point load of 350 kN acting at a distance of 100 mm from the flange plate. Size of fillet weld is 10 mm.

The maximum resultant stress (in MPa, round off to 1 decimal place) on the fillet weld along the vertical plane would be ___________

Solution:

Given Data :

Thickness of gusset plate (t) = 16 mm

Point load (P) = 350 KN

Eccentricity (e) = 100 mm

Direct shear stress, q = p/2ht

And bending stress on the extreme edge of weld (f)

⇒

For checking the safety

Stress = (√60^{2 }+ 3 × 50^{2})

= 105.36 N/mm^{2}

QUESTION: 65

A one-dimensional domain is discretized into N sub-domains of width Δx with node numbers i = 0, 1, 2, 3, ...., N. If the time scale is discretized in steps of Δt, the forward-time and centered-space finite difference approximation at nth node and n^{th }time step, for the partial differential equation

Solution:

Given differential equation

Using forward time finite difference

Using centred space finite difference

Putting (i) and (ii) in PDE

So, option (d) is correct.

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### Syllabus - Civil Engineering, GATE - Practice

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- Practice Test: Gate Civil Engineering(CE) 2020 Paper: (Session I)
Test | 65 questions | 180 min

- Practice Test: Gate Civil Engineering(CE) 2019 Paper: (Session II)
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- Practice Test: Gate Civil Engineering(CE) 2018 Paper: (Session I)
Test | 65 questions | 180 min

- Practice Test: Gate Civil Engineering(CE) 2018 Paper: (Session II)
Test | 65 questions | 180 min

- Practice Test: Gate Civil Engineering(CE) 2020 Paper: (Session II)
Test | 65 questions | 180 min