GATE—2017 C
Transcript of GATE—2017 C
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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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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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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