GATE—2017 C

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GATE—2017 CIVIL ENGINEERING SESSION–2 QUESTION AND DETAILS SOLUTION | 1

8010009955, 9711853908 [email protected], [email protected]. : E-mail:

Regd. office : Phone : F-126, (Upper Basement), Katwaria Sarai, New Delhi-110016 011-41013406

Q.1 - Q. 5 Carry one mark each.1. The VPI (vertical point of intersection) is

100 m away (when measured along thehorizontal) from the VPC (vertical point ofcurvature). If the vertical curve is parabolic thelength of the curve (in meters and measuredalong the horizontal) is_____________

Ans. (200)

l1 l2lP.TP.C

For flat vertical curve we assume that 1 2l l

l = 1 2l l = 100 + 100 = 200 m2. Consider a rigid retaining wall with partially

submerged cohesionless backfill with asurcharge. Which one of the following diagramsclosely represents the Rankine’s active earthpressure distribution against this wall?

(a) (b)

(c) (d)

Ans. (b)Pressure diagram for partially submerged soilwith surcharge

But no option is matching.

If the water table is at ground table then

Option b can be the answer

3. The most important type of species involved inthe degradation of organic matter in the caseof activated sludge process is

(a) autotrophs

(b) heterotrophs

(c) prototrophs

(d) photo-autotrophs

Ans. (c)4. The infiltration capacity of a soil follows the

Horton’s exponential model f = c1 + c2 e–kt.During an experiment, the intial infiltrationcapacity was observed to be 200 mm/hr. Aftera long time the infiltration capacity wasreduced to 25 mm/h, if the infiltration capacityafter 1 hour was 90 mm/h the value of thedecay rate constant k (in h–1 up to two decimalplaces) is______

Ans. (0.99)

GATE—2017Civil Engineering Questions and Details Solution

Session-2

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2 | GATE—2017 CIVIL ENGINEERING SESSION–2 QUESTION AND DETAILS SOLUTION



f = C + .e 1–ktC2

f1

f0

C1 = f0 C2 = f1 – f0

= 25 mm/hr = 200 – 25= 175

f = 90 mm/hr after 1 hrHence, 90 = 25 + 175 × e–k×1

e–k =90 25 0.3715

175

Taking loge on both side–x loge e = loge (0.3714)

–x = –0.990

x = 0.99

5. Let G be the specific gravity of soil solids, w

the water content in the soil sample, w the

unit weight of water, and d the dry unit weightof the soil. The equation for the zero air voidsline in a compaction test plot is

(a) wd

G1 Gw

(b) w

dGGw

(c)w

dw

G1

(d)

wd

w

G1

Ans. (a)

d = a s w(1 ) G1 e

for zero air voids a 0 ... (1)

and S + ac = 1 where ac = air content

for zero air voids ac = 0

Hence, S 1 ... (ii)

from (1) and (ii)

d = s w

s

(1 0) GwG1

S

s wd

S

G1 wG

6. A two-faced fair coin has its faces designatedas head (H) and tail (T). This coin is tossedthree times in succession to record the followingoutcomes: H.H.H. If the coin is tossed onemore time, the probability (up to one decimalplace) of obtaining H again given the previousrealizations of H, H and H. would be________

Ans. (0.5)

Probability of getting (H.H.H) = 1 1 12 2 2

= 18

Probability of getting (H.H.H.H)

= 1 1 1 12 2 2 2

= 1

16 Given condition is that (H.H.H) is already

realized

Conditional probability of getting next H after(H.H.H)

=

11618

= 0.5

7. Consider the following statements related tothe pore pressure parameters. A and B:

P. A always lies between 0 and 1.0

Q. A can be less than 0 or greater than 1.0

R. B always lies between 0 and 1.0

S B can be less than 0 or greater than 1.0

For these statements, which one of the follow-ing options is correct

(a) P and R (b) P and S

(c) Q and R (d) Q and S

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Ans. (c)Value of B lies between 0 and 1.0

for completely saturated soi, B = 1

for completely dry soil, B = 0

Value of A may be as large as 2 to 3 for veryloose saturated fine sand and it can be lessthan zero for over consolidated clay.

8. During a storm event in a certain period, therainfall intensity is 3.5 cm/hour and the

index is 1.5 cm/hour. The intensity ofeffective rainfall (in cm/hour up to one decimalplace) for this period is _________

Ans. (2.0)Rainfall intensity = 3.5 cm/hr

index = 1.5 cm/hr

Intensity of effective rainfall = 3.5 – 1.5

= 2 cm/hr

9. While aligning a hill road with a ruling gradientof 6%, a horizontal curve of radius 50 m isencountered. The grade compensation (inpercentage up two equal places) to be providedfor this case would be______

Ans. (1.5)R = 50

Grade compensation = 30 R 30 50

R 50

= 80 1.650

(Grade compensation)max = 75 75R 50

= 1.5

The compensated gradient should begreater than 4%.

Compensated gradient = 6 – 1.5 = 4.5

Hence, (grade compensation)max = 1.5

10. Let the characteristic strength be defined asthat value, below which not more than 50% ofthe results are expected to fall. Assuming astandard deviation of 4 MPa, the target meanstrength (in MPa) to be considered in the mixdesign of a M25 concrete would be

(a) 18.42 (b) 21.00

(c) 25.00 (d) 31.58

Ans. (c)Since characteristic strength has beendefined as the value below which not morethan 50% of results are likely to fall.

Hence, Target mean strength

= Characteristic strength

= fck

= 25 N/mm2

11. Following observations have been made for theelevation and temperature to ascertain thestability of the atmosphere.

Elevation (in m) Temperature (in C)10 15.560 15.0

130 14.3

The atmosphere is classified as

(a) Stable (b) Unstable

(c) Neutral (d) Inverse

Ans. (c)12. A sheet pile has an embedment depth of 12 m

in a homogeneous soil stratum. The coefficientof permeability of soils is 10–6 m/s. Differencein the water levels between the two sides ofthe sheet pile is 4m. The flow net is constructedwith five number of flow lines and eleven numberof equipotential lines. The quantity of seepage(in cm3/s per m. up to one decimal place)under the sheet pile is__________

Ans. (1.6)K = 10–6 m/s

H = 4 m

N f = 5 – 1 = 4

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Nd = 11 – 1 =10

Q =f

D

kHNN

= 6 410 m/s × 4m× per meter width10

= 1.6 × 10–6 m3/s/m width = 1.6 cm3/s/meter width

13. In a material under a state of plane strain, a10× 10 mm square centred at a point getsdeformed as shown in the figure.

10 mm

10 mm

0.004 mm

y

x0.0005 rad2

If the shear strain xy at this point is expressed

as 0.001 k (in rad.) the value of k is(a) 0.50 (b) 0.25(c) –0.25 (d) –0.50

Ans.(d)Shear strain in an element is positive when theangle between two positive faces (or two negativefaces) is reduced.The strain is negative when the angle betweentwo positive (or two negative) faces increase.

2

3

1

4

y

x

Face 2 & 3 are +ve face

Face 1 & 4 are –ve face.

Angle between 1 & 4 is increased by

0.0005 rad.

xy = – 0.0005 = 0.001 K

K = –0.5

14. The divergence of the vector field V = x2 i +2y3 j + z4 k at x = 1, y = 2, z = 3 is _____

Ans. (134)

V = 2 3 4x i 2y j z k

Divergence ( ) = 2 3 2(x )i (2y )j (z )kx y x

= 2x + 6y2 + 4z3

[Divergence (V)](x = 1, y = 2, z = 3)

= 2 × 1 + 6 × 22 + 4 × 33

= 134

15. The safety within a roundabout and theefficiency of a roundabout can be increasedrespectively by

(a) increasing the entry radius and increas-ing the exit radius.

(b) increasing the entry radius and decreas-ing the exit radius.

(c) decreasing the entry radius and increas-ing the exit radius.

(d) decreasing the entry radius and decreas-ing the exit radius.

Ans. (c)

16. Given that the scope of the construction workis well-def ined with al l i ts drawings,specifications, quantities and estimates. Whichone of the following types of contract would bemost preferred?

(a) EPC contract

(b) Percentage rate contract

(c) Item rate contract

(d) Lump sum contract

Ans. (d)

Scope of construction work is well-definedwith all its drawings, specification quantitiesand estimates, then lump sum contract isused.

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17. As per Noise pollution (Regulation and Control)Rules 2000 of India, the day time noise limitfor a residential zone expressed in dB (A) Leqis

(a) 55 (b) 65

(c) 75 (d) 85

Ans. (a)As per Noise pollution (Regulation andcontrol) Rules 2000 of India.

Area code

Area/zone Limits in dB Leq

Day time Night time

ABCD

IndustrialCommerical areaResidential areaSilence Zone

75655550

70554540

18. The method of orientation used, when the planetable occupies a position not yet located onthe map is called as

(a) traversing (b) radiation

(c) levelling (d) resection

Ans. (d)Resection method of orientation isemployed when the plane table occupies aposition not yet plotted on the drawingsheet.

19. Consider the following simultaneous equation(with c1 and c2 beings constants):

3x1 + 2x2 = c1

4x1 + x2 = c2

The characteristic equation for thesesimultaneous equations is

(a) 2 4 5 0 (b) 2 4 5 0

(c) 2 4 5 0 (d) 2 4 5 0

Ans. (a)3x1 + 2x2 = c1

4x1 + x2 = c2

3 24 1

= [A]

[A] [I] = 0

3 42 1

= 0

23 4 8 = 0

2 4 5 = 020. If a centrifugal pump has an impeller speed of

N (in rpm) discharge Q (in m3/s) and the totalhead H (in m), the expression for the specificspeed Ns of the pump is given by

(a)0.5

s 0.5NQNH

(b)0.5

sNQN

H

(c)0.5

s 0.75NQNH

(d) s 0.75NQN

H

Ans. (c)

NS = 3/4N QH

=0.5

0.75NQH

21. For a broad gauge railway track on a horizontalcurve of radius R ( in m) the equilibrium cante required for a train moving at a speed of V(in km per hour) is

(a)2Ve 1.676

R (b)

2Ve 1.315R

(c)2Ve 0.80

R (d)

2Ve 0.60R

Ans. (b)

Equilibrium cant = 2GV

127 RV in kmph and R in m.For B.G track G = 1.676

Cant =21.676 V

127 R m

e =21.319 V cm

RThe unit of e is not specified in question but optionb is matching in cm units.

22. Consider the frame shown in figure.

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If the axial and shear deformations in differentmembers of the frame are assumed to benegligible the reduction in the degree ofkinematical indeterminacy would be equal to

(a) 5 (b) 6

(c) 7 (d) 8

Ans. (b)

(x, y, ) (x, y, )

(x, y, )(x, y, )

( ) ( )

()

( ) ( )

(, x)

() ( ), x

Total degree of freedom = 3 + 3 + 3 + 3 +1 + 1= 14

When axial and sheardeformations are negligible,degree of freedom = 1 + 1 + 1 +1+2 + 2 = 8

Hence, reduction in degree of kinematicalindetermiancy = 14 – 8 = 6

23. Let w = f(x, y), where x and y are functions of

t. Then according to the chain rule dwdt

is

equal to

(a)dw dx dw dtdx dt dy dt

(b)w x w yx t y t

(c)w dx w dyx dt y dt

(d)dw x dw ydx t dy t

Ans. (c)w = f (x, y)

dwdt

=w dx w dyx dt y dt

24. For a construction project the mean andstandard deviation of the completion time are200 days and 6.1 days respectively. Assumenormal distribution and use the value ofstandard normal deviate z = 1.64 for the 95%confidence level. The maximum time required(in days) for the completion of the project wouldbe_________

Ans. (210)

Z = s ET T

1.64 = sT 2006.1

Ts = 210 days

25. The plate load test was conducted on a clayeystrata by using a plate of 0.3m × 0.3mdimensions, and the ultimate load per unit areafor the plate was found to be 180 kPa. Theultimate bearing capacity (in kPa) of a 2mwide square footing would be

(a) 27 (b) 180(c) 1200 (d) 2000

Ans. (b)In plate load test bearing capacity of claydoes not depend upon size of footing.

qup = quf (for clay)

Ultimate bearing capacity of a 2m widesquare footing = 180 kPa

26. Two identical concrete piles having the plandimensions 50cm × 50cm are driven into ahomogeneous sandy layer as shown in thefigures. Consider the bearing capacity factor

Nq for 30 as 24.

Dry Sand = 18 kN/m

3

= 30°20m

QP1

Saturated Sand = 19 kN/msat

3

= 30°

20m

QP2

If QP1 and QP2 represent the ultimate pointbearing resistances of the piles under dry andsubmerged conditions, respectively, which oneof the following statements is correct?(a) QP1 > QP2 by about 100%(b) QP1 < QP2 by about 100%(c) QP1 > QP2 by about 5%(d) QP1 < QP2 by about 5%

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Ans. (a)

QP1 = qqN 0.4B N

= 18 × 20 ×Nq + 0.4 × 0.5 × 18 × N

= q18 (20 N 0.2N )

QP2 = qqN 0.4B N

= (19 – 10) × 20 × Nq + 0.4 × 0.5

× (19 – 10)N

= 9 (20 Nq + 0.2 N )

(Assuming 3w 10 kN / m )

1

2

P

P

QQ = 2

Hence, 1 2P PQ Q by about 100%.

27. Consider the portal frame shown in the figureand assume the modulus of elasticity,E = 2.5 × 104 MPa and the moment of mertia,I = 8 × 108 mm4 for all the members of theframe.

2000 kNPE,

1650 kN/m 2m

Q

R

4m

4m

E, s

E,

The rotation (in degrees, up to one decimalplace) at the rigid joint Q would be________

Ans. (1.0)

Moment at Q = MQ = 2000 × 2– 221650

2

= 700 kNm anticlockwise

2 m

4 m

1650 kN/m

2000 kNP

Q E,I

R

S

4 m

KQR = QS4EI EI 4EI EIK4 1m 4 1m

KQ = QR QS2EIK K1m

MQ = Q QK

Q =6

Q4 8

Q

M 700 10 1000K 2 2.5 10 8 10

= 0.0175 rad

= 0.0175 × 180

= 1.003 = 1.0

28. A 1m wide rectangular channel carries adischarge of 2m3/s. The specific energy-depthdiagram is prepared for the channel. It isobserved in this diagram that corresponding toa particular specific energy. The subcriticaldepth is twice the supercritical depth. Thesubcritical depth (in meters, up to two decimalplaces) is equal to____________

Ans. (1.07)Let, Subcritical depth = y

then, super critcal depth = y2

Specific energy at subcritical depth

= Specific energy at super-critical depth

2

2 2Qy

2gB y =

2

22

y Q2 y2gB

2

yy2

=2

2 2 2Q 1 4

2gB y y

y2

=2

2 22 3

2 9.81 1 y

y = 1.069 1.07

29. For a given water sample, the ratio betweenBOD5-day. 20°C and the ultimate BOD is 0.68.The value of the reaction rate constant k (onbase e) (in day–1, up to two decimal places)is__________.

Ans. (0.23)

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Let BOD5day = L0 (1 – e–5k)

Ultimate BOD = L0

5k

0

0

L (1 e )L

= 0.68

e–5k = 0.32

K =1 1n5 0.32

l

= 0.2279

0.23

30. The tangent to the curve represented byy = x ln x is required to have 45° inclinationwith the x-axis. The coordinates of the tangentpoint would be

(a) (1, 0) (b) (0, 1)

(c) (1, 1) (d) 2, 2

Ans. (a)Target is having inclination of 45° withx-axis

dydx

= tan 45

d(x nx)

dxl

= 1

xnxx

l = 1

At x = 1, y = 1× n 1 0 l

(1, 0)

31. Consider the following statements:

P. Walls of one brick thick are measured insquare meters.

Q. Walls of one brick thick are measured incubic meters.

R. No deduction in the brickwork quantity ismade for openings in walls up to 0.1m2

area.

S. For the measurement of excavation fromthe borrow pit in a fairly uniform ground,deadmen are left at suitable intervals.

For the above statements, the correct optionis

(a) P – False; Q – True: R – False: S – True

(b) P – False; Q – True: R – False: S –False

(c) P – True; Q – False: R – True: S – False

(d) P – True; Q – False: R – True: S – True

Ans. (d)

Brick masanry is measured in volume for thick-ness more than single bricks. For masonry withsingle bricks it is measured in square metres.

No deduction is made for following.

Opening each up to 1000 sq. cm (0.1 sq.meter).

Ends of beams rafters etc up to 500 sq. cmor 0.05 sq. m in section.

Bed plata wall plate bearing of balcony andthe like up to 10cm depth bearing of floor androof slabs an not deducted from masonry.

When the ground is not uniform levels shall betaken before the start, after site clearance andafter the completion of the work and the quantityof excavation in cutting computed from theselevels.

32. An observer standing on the deck of a shipjust sees the top of a lighthouse. The top ofthe lighthouse is 40m above the sea level andthe height of the observer’s eye is 5m abovethe sea level. The distance (in km. up to onedecimal place) of the observer from thelighthouse is_____________

Ans.(33.0)

0.0673 d12 = 5 and, 0.0673 d2

2 = 40

d1 = 8.62 km, d2 = 24.38 km

Distance of observer from the lighthouse

= d = d1 + d2

= 33 km

33. Two prismatic beams having the same flexualrigidity of 1000 kN-m2 are shown in the figures.

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6 kN/m

4m1

120 kN

21m 1m

If the mid-span deflections of these beams aredenoted by 1 and 2 (as indicated in thefigures). the correct option is

(a) 1 2 (b) 1 2

(c) 1 2 (d) 1 2

Ans.(a)w = 6 kN/m

L = 4m

1

1 = 45 wL

384 E =

45 6 4384 10000 m

= 20 mm

P = 120 kN

L = 2m

L

L = 3PL

48E = 3120 2

48 10000 m

= 20 mm

1 = 2

34. The composition of a municipal solid wastesample is given below:

Component Percent by Moisture Content Energy ContentMass (%) (kJ / kg. on as discarded basis)

Food Waste 20 70 2500Paper 10 4 10000

Cardboard 10 4 8000Plastics 10 1 14000Garden 40 60 3500

TrimmingsWood 5 20 14000

Tin Cans 5 2 100

The difference between the energy content ofthe waste sample calculated on dry basis andas discarded basis (in kJ/kg) wouldbe_________

Ans. (3870)

%Bymass

(2)

MoistureContent %

(3)

drycontent%

(4)= 100 – (3)

energyas

discarded(5)

(kJ/kg)

Component(1)

Food wastePaperCard boardPlasticsGarden trimmingWood Tin Cars

201010104055

70441

60202

30969699408098

2500100008000

1400035001400

10

Total dry mass

(6)

Total energy as

discarded (7)

(2) (4)100 (2) (5)

100

69.69.69.9164

4.9

500100080014001400700

5

= 60 = 5805 kJ/kg

Energy on dry basis: It will be the total energywhen whole mass is dry.

In current situation only 60% mass is dry

energy corresponding to 60% dry mass= 5805 kJ/kg

Energy corresponding to 100% dry mass

= 5805 10060

= 9675 kJ/kg

So, energy as on dry basis = 9675 kJ/kg

Energy based on as discarded basis = 5805kJ/kg

So, difference = 9675 – 5805 = 3870 kJ/kg

35. Two plates of 8 mm thickness each areconnected by a fillet weld of 6mm thicknessas shown in the figure.

100mmP

6

50 mmP

100mm

The permissible stresses in the plate and theweld are 150 MPa and 110 MPa. respectively.Assuming the length of the weld shown in thefigure to be the effective length the permissibleload P(in kN) is___________

Ans.(60 )

100mmP

50 mmP

100mm

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1 0 | GATE—2017 CIVIL ENGINEERING SESSION–2 QUESTION AND DETAILS SOLUTION



Maximum load taken by plate

= 150 × 50 × 8 = 60 kN

Maximum load taken by the weld

= KS leff

= 0.7 × 6 × (100 + 50 + 100)×110 = 115.5 kN

Permissible load = minimum of above two

= 60 kN

36. For the construction of a highway a cut is tobe made as shown in the figure.

Potentialshear surface

Point A

4m2m

The soil exhibits c 20 kPa , 18 , andthe undrained shear strength = 80 kPa. Theunit weight of water is 9.81 kN/m3. The unitweights of the soil above and below the groundwater table are 18 and 20 kN/m3, respectively.If the shear stress at Point A is 50 kPa, thefactors of safety against the shear failure atthis point, considering the undrained anddrained conditions respectively, would be

(a) 1.6 and 0.9 (b) 0.9 and 1.6

(c) 0.6 and 1.2 (d) 1.2 and 0.6

Ans. (a)

Case-I : Undrained condition

F.o.S = resisting shear strength

Acting shear stress

=8050 = 1.6

Case-II : Drained condition

F.o.S =tan c

Acting shear stress

= [2 18 4(20 9.81)] tan18 20

50

= 0.9

37. A municipal corporation is required to treat 1000m3/day of water. It is found that an overflowrate of 20 m/day will produce a satisfactoryremoval of the discrete suspended particles ata depth of 3m. The diameter (in meters,rounded to the nearest integer) of a circularsettling tank designed for the removal of theseparticles would be_______

Ans. (8)

Q =3m1000

day

Overflow Rate = 20m/day

H = 3m

Surface Area of circular Tank

=1000

20= 50m2

Assuming diameter to be ‘d’

Hence, 2d4 = 50

d =50 4

= 7.978

8m

38. Consider the following second-order differentialequation:

2y 4y 3y 2t 3t

The particular solution of the differential equa-tion is

(A) – 2 – 2t – t2 (B) – 2t – t2

(C) 2t – 3t2 (D) – 2 – 2t – 3t2

Ans. (a)

y 4y 3y = 2t – 3t2

f(D) = D2 – 4D + 3

P.I. =2(2t 3t )

f(D)

= 22

1 (2t 3t )D 4D 3

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=

21 1 3ttD1 D 23 13

(1 + D + D2 +....)2 23t 1 D Dt t ...

2 3 3 9

232tt2

= 23tt 1 3t 3

2

–

21 3t 1 11 83 2 3 3

= – 2 – 2t – t2

39. Two cars P and Q are moving in a racing trackcontinuously for two hours. Assume that noother vehicles are using the track during thistime. The expressions relating the distancetravelled d (in km) and time t (in hour) for boththe vehicles are given as

P: d = 60t

Q: d = 60t2

Within the first one hour, the maximum spaceheadway would be

(a) 15 km at 30 minutes

(b) 15 km at 15 minutes

(c) 30 km at 30 minutes

(d) 30 km at 15 minutes

Ans. (a)

P: d = 60t

Q: d = 60t2

Space Headway (S) = 60t2 – 60t

For space headway to be max.,

dsdt = 0

120t – 60 = 0

60(2t – 1) = 0

t =12

= 30 minutes

Maximum space Headway

=21 160 60

2 2

=1 160 602 4

= 30 – 15

= 15 km

40. The culturable command area of a canal is10,000 ha. The area grows only two crops-ricein the Kharif season and wheat in the Rabiseason. The design discharge of the canal isbased on the rice requirements, which has anirrigated area of 2500 ha, base period of 150days and delta of 130 cm. The maximumpermissible irrigated area (in ha) for wheat,with a base period of 120 days and delta of 50cm. is_________

Ans.(5199.97) For Rice,

CCA = 10000 ha

B.P. = 150 days

= 130cm = 1.3 m

D =B.P. 8.64

=150 8.64

1.3

= 996.923 ha/m3/s

Area irrigated = 2500 ha

Design Discharge

= 2500

996.923

= 2.5077 m3/s

For, wheat,

B.P. = 120 days

= 50 cm

D =8.64 120

0.5

= 2073.6 hect./m3/s

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Area irrigated = Duty × Design Discharge(Q)

= 2073.6 × 2.5077

= 5199.97 ha.

41. Consider the following definite integral:

211

20

sin xdx

1 x

The value of the integral is

(a)3

24

(b)3

12

(c)3

48

(d)3

64

Ans. (a)

I =1 1 2

20

(sin x) dx1 x

Put sin–1 x = t

2

dx

1 x= dt

=/2

2

0

t dt

=

/23

0

t3

=3

24

42. If 1 5

A6 2

and 3 7

B8 4

. ABABT is equal to

(a)38 2832 56

(b)3 4042 8

(c)43 2734 50

(d)38 3228 56

Ans. (a)

A =1 5 3 7

[B]6 2 8 4

ABT =1 5 3 86 2 7 4

=3 35 8 20

18 14 48 8

=38 2832 56

43. Consider the three prismatic beams with theclamped supports, P, Q and R as shown inthe figures.

P

80 N

E.I.

8 m

8 m

QE.I.

20 N/m

R640 N-m

E.I.

8 mGiven that the modulus of elasticity. E is 2.5 ×104 MPa. and the moment of inertia. I is 8 × 108

mm4, the correct comparison of the magnitudesof the shear force S and the bending moment Mdeveloped at the supports is

(a) P Q R P Q RS S S , M M M

(b) P Q R P Q RS S S , M M M

(c) P Q R P Q RS S S , M M M

(d) P Q R P Q RS S S , M M M

Ans.(c)

E, I

8m

20 N/mE, I

8m

E, I

8m

P 640 N-m80 N Q R

E = 2.5 × 104 MPa

= 8 × 108 mm4

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For P,8m

80N

V

vf = 0

V 80N

Shear force at support = Sp = 80N

Moment at support = 80 × 8 = 640 Nm.

Mp = 640 Nm

for Q E, I

8m

20 N/m

V

vf = 0

V 20 8 160N

Shear force at support = SQ = 160N

Moment at support = MQ = 820 82

= 640 NmFor R,

640 N-m

V

vf = 0

V = 0 = SR

Moment at support = MR = 640 NmHence, by above values

P Q RS S S & P Q RM M M

44. The radii of relative stiffness of the rigid pave-ment P and Q are denoted by IP and IQ re-spectively. The geometric and material proper-ties of the concrete slab and underlying soilare given below.

PavementConcrete Soil

PQ

LL

BB

h0.5h

EE

K2K

Length ofSlab

Breadthof Slab

Thicknessof Slab

Modulus ofElasticity

Poisson’sRatio

SubgradeReactionModulus

The ratio (up to one decimal place) of P QI / I

is _____.

Ans. (2)

l =

1/43

2Eh

12 (1 )k

P

Q

ll =

1/43QP P

3Q PQ

KE hE Kh

=

1/43

3E h 2kE K(0.5h)

= [16]1/4 = 2

P

Q

ll = 2

45. The analysis of a water sample produces thefollowing results

2

2

24

3

Ion milligrampermilli equivalent Cncentrationfor the ion mg L

Ca 20.0 60Mg 12.2 36.6Na 23.0 92K 39.1 78.2C 35.5 71

SO 48.0 72HCO 61.0 122

l

The total hardness (in mg/L as CaO3) of thewater sample is _____.

Ans. (300)Hardness will be due to multivalent cationsonly

Ca2+ = 60 mg/L equivalent weight = 20(given)

mg2+ = 36.6 mg/L equivalent weight =12.2 (given)

miliequivalent of

Ca2+ = 60 320

mg2+ = 36.6 312.2

Total hardness (mg/L as CaCO3)

= ( miliequivalent of Ca2+ + miliequivalent ofmg2+) × equivalent weight of CaCO3

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= 100(3 3)

2

= 300 mg/L as CaCO3

46. Consider a square-shaped area ABCD on theground with tis centre at M as shown in thefigure. Four concentrated vertical loads of P =5000 kN are applied on this area, one at eachcorner.

P

B

C

A

D

4 m

M

P

P

P

4 m

The vertical stress increment (in kPa up toone decimal place) due to these loads ac-cording to the Boussinesq’s equation at a point5 m right below M is _____.

Ans. (190.84)Due to one corner load

2 =5/2

2 23P 1

3 z r1z

r = 2 22 2 2 2 m

z = 5 m

P = 5000 kN

2 =5/2

5 23 5000 12 5 2 21

5

= 47.709 kN/m2

So due to four load vertical stress

= z9

= 4 × 47.709

= 190.84 kN/m2

47. Two towers, A and B standing vertically on ahorizontal ground appear in a vertical aerialphotograph as shown in the figure.

BA

p

The length of the image of the tower A on thephotograph is 1.5 cm and of the tower B is 2.0cm. The distance of the top of the tower A (asshown in the arrowhead is 4.0 cm and thedistance of the top of the tower B is 6.0 cm.as measured from the principal point p of thephotograph. If the height of the tower B is 80m the height (in meters) of the tower A is_____.

Ans. (90)

A B

P

for Tower B,

Radial distance of top of tower = r = 6cm

Length of image = d = 2cm

Height of tower = 80m = h2

Flying height above datum = H – h1

d = 2

1

rhH h

H – h1 = 2rhd

=6 80

2

= 240 m

for Tower A,

both on same plane

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d = 2

1

rhH h

r = 4 cm

d = 1.5 cm

H – h1 = 240 m

h2 = 1d H hr

=

1.5 2404

h2 = 90 m

48. Water is pupped at steady uniform flow rateof 0.01 m3/s though a horizontal smooth circu-lar pipe of 100 mm diameter. Given that theReynold number is 800 and g is 9.81 m/s2,the head loss (in meters upto one decimalplace) per km length due to friction would be_____.

Ans. (66.1)Q = 0.01 m3/s

d = 100 mm

Re = 800

g = 9.81 m/s2

h f =2f V

2gdl

Since Re is less than 2000 the flow will belaminar flow

Hence, f =e

64R

=64 0.08

800

h f =2

5f Q

12.1dl

=2

50.08 (0.01)

12.1 (0.1) l

= 0.0661l

Hence head loss per km length = 0.0661× 1000 = 66.1 m

49. A 2 m long axially loaded mild steel rod of 8mm diameter exhibits the load-displacement(P - ) behavior as shown in the figure.

14000

12000

10000

8000

6000

4000

2000

0

Axia

l Loa

d, P

(kN

)

Displacement 5(mm)0 1 2 3 4 5 6 7 8 9 10

Assume the yield stress of steel as 250 MPa.The complementary strain energy in N-mm)stored in the bar up to its linear elastic behav-ior will be _____

Ans. (15707.96)

Complementryenergy

Strainenergy

The area enclosed by the inclined line andthe vertical axis is called complementarystrain energy

2.5 mm

=l =

2.52000 =

1800

y = 250 MPa

Complementry strain energy = strain energy

= y1 f vol. of bar2

= 21 1250 8 20002 800 4

= 15707.96 Nmm

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50. A simply supported rectangular concrete beamof span 8 m has to be prestressed with a forceof 1600 kN. The tendon is of parabolic profilehaving zero eccentricity at the supports. Thebeam has to carry an external uniformly dis-tributed load of intensity 30 kN/m. Neglectingthe self-weight of the beam, the maximum dip(in meters upto two decimal place) of the ten-don at the mid-span to balance the externalload should be _____.

Ans.(0.15)

8m

30 kN/m

Prestressing force = 1600 kN

e

Mmax =2wl

8

=230 8

8

= 240 kN-m

Shape of tendon profile will follow the shape ofbending moment diagram and for equillibriumof the section having maximum BendingMoment.

Mmax = Pe

240 kN-m = 1600 kN × e

3240kN 10 mm

1600kN

= e

e = 150 mm

= 0.15 m

51. A catchment is idealized as a 25 km × 25 kmsquare. It has five rain gauges, one at eachcorner and one at the centre, as shown in thefigure.

G1

G4

G2

G3

G5

During a month the precipitation at thesegauges is measured as G1 = 300 mm. G2 =285 mm, G3 = 272 m, G4 = 290 mm and G5= 288 mm. The average precipitation (in mmup to one decimal place) over the catchmentduring this month by using the Thiessen poly-gon method is _____.

Ans.(287.375)

G1

G4 G3

G2B

D

CA

For G1 area is AG1B

For G2 area is BG2C

For G3 area is CG3D

For G4 area is DG4A

For G5 area is ABCD

Take 1AG B ,

B

A

G1

252

252

AB =25 22

= 17.67 km

Area of ABCD =225

2

= 312.5 km2

Area of AG1B = area of AG4D

= Area of CG3D

= area of BG2C

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=625 312.5

4

= 78.125 km2

Average presipitation

= 1 1 2 2 3 3 4 4 5 5

1 2 3 4 5

G A G A G A G A G AA A A A A

= (300 285 272 290) 78.125 288 312.5

625

= 287.375 mm

52. The figure shows a U-tube having a 5 mm × 5mm square cross-section filled with mercury(specific gravity = 13.6) upto a height of 20 cmin each limb (open to the atmosphere).

50 cm

Mercury20 cm

If 5 cm3 of water is added to the right limb, thenew height (in cm up to two decimal places)of mercury in the LEFT limb will be _____.

Ans.(20.74)

50 cm Mercury

20 cm

S gravity = 13.6

p 21

5cm 5cm

(0.2+x)

Mercury 21

5cm 5cm

(0.2–x)

water

Height of water column

=35cm 1000

5mm 5mm

= 20cm = 0.2m

Since, P2 = P1

Hence, w w0.2 (0.2 x) 13.6

= w(0.2 x) 13.6

0.2 + 13.6 × 0.2 – 13.6x

= 13.6 × 0.2 + 13.6x

0.2 = 27.2x

x = 7.35 × 10–3m

= 0.735cm

New height = 20.735

53. Group I gives a list of test methods and testapparatus for evaluating some of the proper-ties of Ordinary Portland Cement (OPC) andconcrete Group II gives the list of these prop-erties.

Group I

P. Le Chatelier test

Q. Vee-Bee test

R. Blaine air permeability test

S. The Vicat apparatus

Group-II1. Soundness of OPC2. Consistency and setting time of OPC3. Consistency or workability of concrete

4. Fineness of OPC

The correct match of the items in Group I withitems in Group II is

(a) P-1, Q-3, R-4, S-2

(b) P-2, Q-3, R-1, S-4

(c) P-4, Q-2, R-4, S-1

(d) P-1, Q-4, R-2, S-3

Ans.(a)(1) Le chatelier Test : This test is used to mea-

sure the soundness of OPC due to lime. Lime &Magnesia are two primary compounds respon-sible for soundness of cement.

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(2) Vee Bee Test : It is one of the methods ofmeasuring the workability of concrete.

(3) Blaine Air Permeability: It is used to measurefineness of cement.

(4) The Vicat Apparatus: It is used to measuresetting time and consistency of concrete.

54. Following are the statements related to thestress paths in a triaxial testing of soils.

P. If 1 3 , the stress point lies at theorigin of the p-q plot

Q. If 1 3 , the stress point lies on the p-axis o the p-q plot

R. If 1 3 , both the stress points p andq are positive.

For the above statements, the correct combi-nation is(a) P-False; Q-True; R-True(b) P-True; Q-False; R-True(c) P-False; Q-True; R-False(d) P-True; Q-True; R-False

Ans.(a)

1 3q

2

1 3 p2

p-q plot

When 1 3

q = 0, p = 122

= 1

So stress point lies on p-axis of p-q plot

If 1 3

q = 1 3

2

so q > 0

p = 1 3

2

so p > 0

so, stress point p & q are positive.

55. A hollow circular shaft has an outer diameterof 100 mm and inner diameter of 50 mm. If theallowable shear stress is 125 MPa, the maxi-mum torque (in kN-m) that the shaft can resistis _____.

Ans. (23)

allowable = 2N125

mm

Tmax.

D = 100mmo

D = 50mm i

max.TJ = allowable

R

Tmax. = 4 4125 100 5050 32

= 23009711.82 N-mm

= 23.00 × 10+6 N-mm

= 23 kN-m

General Aptitude1. What is the value of x when

x 2 2x 416 381 14425 5

?

(a) 1

(b) –1

(c) –2

(d) Cannot be determined

Ans. (b)x 2 2x 416 381

25 5

= 144

2x 4 2x 44 5

5 3

=14481

2x 44

3

=14481

x 24

3

=14481

x4 16

3 9

=129

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x4

3

=34

x4

3

=14

3

x = –1

2. There was no doubt that their work was thorough

Which of the words is closest in meaning tothe underlined word above?

(a) pretty (b) complete

(c) sloppy (d) haphazard

Ans. (b)3. Four cards lie on a table. Each card has a

number printed on the one side and a color onthe other. The faces visible on the cards are 2,3, red and blue.

Proposition. If a card has an even value on oneside then its opposite face is red.

The cards which MUST be turned over to verifythe above proposition are

(a) 2, red (b) 2, 3, red

(c) 2, blue (d) 2, red, blue

Ans. (a)Card with visible face 2 should have red onthe other side

Card with visible face 3 may have any coloron other side

Card with visible face red should have evennumber on other side

Card with visible face blue may have anynumber on other side.

Hence, To verify the above proposition,cards with face 2 and Red must be turned.

4. The event would have been successful if you_____ able to come.

(a) are

(b) had been

(c) have been

(d) would have been

Ans. (b)

5. Two dice are thrown simultaneously. The prob-ability that the product of the numbers appear-ing on the top faces of the dice is a perfectsquare is

(a) 1/9 (b) 2/9

(c) 1/3 (d) 4/9

Ans. (b)

Favourable outcomes = {(1, 1), (1, 4), (2,2) (3, 3), (4, 1), (4, 4), (5, 5), (6, 6)}

No. of favourable outcomes = 8

Total outcomes = 36

Probability = 8 236 9

6. Bhaiclung was observing the pattern of peopleentering and leaving a car service centre. Therewas a single window where customers werebeing served. He saw that people inevitablycame out of the centre in the order that theywent in. However, the time they spent insideseemed to vary a lot some people came outin a matter of minutes while for others it tookmuch longer.

From this what can one conclude?

(a) The centre operates on a first-come-first-served basis., but with variable servicetimes depending on specific customerneeds.

(b) Customers were served in an arbitraryorder since they took varying amounts oftime for service completion in the centre.

(c) Since some people came out within a fewminutes of entering the centre, the systemis likely to operate on a last-come-first-served basis.

(d) Entering the centre early ensured that onewould have shorter service times andmost-people attempted to do this.

Ans. (a)

7. The points in the graph below represent thehalts of a lift for duration of 1 minute, over aperiod of 1 hour.

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543210

0 5 10 15 20 25 30 35 40 45 50 55 60

Floo

r num

ber

Time (min)

Which of the following statements are correct ?

(i) the elevator never moves directly from anynon-ground floor to another non-groundfloor over the one hour period.

(ii) The elevator stays on the fourth floor forthe longest duration over the one hourperiod.

(a) only i (b) only ii

(c) Both i and ii (d) Neither i nor ii

Ans. (d)

The elevators has moved from 2nd to 5th

floor between time 25 min to 30 min.

Elevation stayed at 4th floor for

= 3 + 4 + 3 + 2 + 3 + 2 + 2 = 19

= 19 min

Elevation stayed at ground floor for

= 1 + 2 + 1 + 1 + 1 + 1 + 2 + 1

+ 2 + 1 + 1 + 1 + 1 + 2 + 1 + 1 +1

= 21 min

Hence, (i) and (ii) both are false.

8. A map shows the elevations of Darjeeling,Gangtok, Kalimpong, Pelling, and SiliguriKalimpong is at a lower elevation than Gangtok.Pelling is at a lower elevation than GangtokPelling is at a higher elevation than Siliguri .Darjeeling is at a higher elevation than Gangtok.

Which of the following statements can be inferred from the paragraph above ?

(i) Pelling is at a higher elevation thanKalimpong

(ii) Kalimpong is at a lower elevation thanDarjeeling

(iii) Kalimpong is at a higher elevation thanSiliguri

(iv) Siliguri is at a lower elevation thanGangtok.

(a) Only ii (b) Only ii and iii

(c) Only ii and iv (d) Only iii and iv

Ans. (c)

K < G, P < G, P > S, G < D

if K < G and G < D

Hence, K < D

If S < P and P < G

Hence, S < G

Hence, (ii) and (iv) are correct

9. P, Q, R, S, T and U are sealed around acircular table. R is seated two places to theright of Q. P is seated three places to the leftof R. S is seated opposite U. If P and U nowswitch seats which of the following must nec-essarily be true.

(a) P is immediately to the right of R

(b) T is immediately to the left of P.

(c) T is immediately to the left of P or P isimmediately to other right of Q.

(d) U is immediately to the right of R or P isimmediately to the left of T.

Ans. (c)Q

PS or U

V or S

Before switching of seats of P and U

R T

Q QU US P

P S

Ist possibility 2nd possibility

R RT T

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10. Budhan covers a distance of 19 km in 2 hoursby cycling one fourth of the time and walkingthe rest. The next day he cycles (at the samespeed as before) for half the time and walksthe rest (at the same speed as before) andcovers 26 km in 2 hours. The speed inkm/h at which Budhan walks is

(a) 1 (b) 4

(c) 5 (d) 6

Ans. (d)Let C = Speed by cycling (km/h)

W = Speed of walking (km/h)

2 3C W 24 4

= 19

C + 3W = 38 ... (1)

And,2 2C W2 2

= 26

C + W = 26 ... (2)

From (i) and (ii): W = 6 km/h

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