CIVIL ENGINEERING · 2020-01-07 · CIVIL ENGINEERING F-126, Katwaria Sarai, New Delhi - 110 016...

CIVIL ENGINEERING

F-126, Katwaria Sarai, New Delhi - 110 016 011-41013406, Mobile: 8130909220, 9711853908Ph:

Web: www.iesmaster.org | E-mail: [email protected]

Detailed SolutionSET - A

IES M

ASTER

1. When the deposit of efflorescence is morethan 10% but less than 50% of the exposedarea of the brick, the presence ofefflorescence is(a) Moderate (b) Slight(c) Heavy (d) Serious

Ans. (a)Sol.

Nil

Slight

Moderate

Heavy

Serious

Deposit of efflorescence is

Deposit of salt of brick area is covered with thin.

of exposed area of the bricks. But

Deposit of Salt of exposed area of the bricks. But

of salts

imperceptible.< 10

10-50%not showing

powdering and flaking of surface.

> 50% not

showing powdering and flaking of surface.Heavy deposits showing powdering and/or flaking of surface.

2. Mohs scale is used for stones to determine(a) Flakiness index (b) Durability(c) Strength (d) Hardness

Ans. (d)Sol. Mohs scale is used to determine hardness of

stone.

3. Which of the fol lowing conditions arerecommended for using sulphate resistingcement?1. Concrete to be used in foundation and

basement, where soil is not infested withsulphate.

2. Concrete used for fabrication of pipeswhich are likely to be buried in marshyregion or sulphate bearing soils.

3. Concrete to be used in the constructionof sewage treatment works

(a) 2 and 3 only (b) 1 and 2 only(c) 1 and 3 only (d) 1, 2 and 3

Ans. (a)Sol. Sulphate resisting cement should to be used

in foundation and basement, where soil isinfected with sulphates. Hence, statement 1is incorrect.

4. Which one of the following cements is adeliquescent?(a) Quick setting portland cement(b) White and coloured cement(c) Calcium chloride cement(d) Water repellent cement

Ans. (c)

Sol. Calcium chloride (CaCl2) is widely consideredto be one of the most versatile and widelyused chemicals due to its unique physicaland chemical deliquescent (means it canabsorb enough moisture to convert to liquidbrine) and hydroscopic properties which lendit to perform well in many industrialapplication.Extra rapid hardening or calcium chloridecement is a deliquescent.

5. Consider the following data for concrete withmild exposure:Water- cement ratio = 0.50Water = 191.6 litreThe required cement content will be(a) 561 kg/m3 (b) 472 kg/m3

(c) 383 kg/m3 (d) 294 kg/m3

Ans. (c)Sol. Water content = 191.6 litre

Water cement ratio = 0.50

Cement content = 3191.6 383.2 kg / m0.50

CIVIL ENGINEERING




IES M

ASTER

For mild exposure condition, the minimumcement content = 300 kg/m3

383.2 kg/m3 > 300 kg/m3

6. The strength of a fully matured concretesample is 500 kg/cm2. When cured at anaverage temperature of 20°C in day, 10°C innight, datum temperature T0 is –11°C. IfPlowman constants A is 32 and B is 54, thestrength of identical concrete at 7 days will benearly(a) 333 kg/cm2 (b) 312 kg/cm2

(c) 272 kg/cm2 (d) 243 cm2

Ans. (a)Sol. Maturity of concrete at the age of 7 days

time temperature= 7 × 12 × [20 – (–11)]+7 × 12 × [10 – (–11)]= 4368°C hrNow, A = 32, B = 54 there by% of strength of concrete at maturity of 4368°Chr

= 104368A Blog 66.5%1000

Strength at 7 days = 66.550100

= 332.7 kg/cm2

7. A sample of concrete is prepared by using500 g of cement with water cement ratio of0.55 and 240 N/mm2 intrinsic strength of gel.The theoretical strength of concrete of fullhydration will be nearly.(a) 148 N/mm2 (b) 126 N/mm(c) 104 N/mm2 (d) 82 N/mm2

Ans. (c)Sol. Cement (c) = 500 gms

Water cement ratio = 0.55INtrinsic strength of gel = 240 N/mm2

Gel space ratio = 0

0.657C0.319C W

W0 = 0.55 × 500 = 275 nl

Gel space ratio = 0.657 500 0.756

0.319 500 275

Theoretical strength of concrete

= 240 × (0.756)3 = 103.72 N/mm2

8. The cement and water slurry coming on thetop and setting on the surface is called(a) Crazing(b) Efflorescence(c) Sulphate deterioration(d) Laitance

Ans. (d)Sol. Laitance is defined as cement and water slurry

coming on top and setting on the surface.

9. Polymer concrete is most suitable for(a) Sewage disposal works(b) Mass concreting works(c) Insulating exterior walls of an air-

conditioned building(d) Road repair works

Ans. (a)Polymer concrete have high sulphate and acidresistance properties, therefore it is mostsuitable for sewage disposal work.

10. Which one of the following limes will be usedfor finishing coat in plastering and whitewashing?(a) Semi hydraulic lime(b) Kankar lime(c) Magnesium/Dolomitic lime(d) Eminently hydraulic lime

Ans. (c)Sol.

CIVIL ENGINEERING




IES M

ASTER

Class Lime name WorkClass A

Class B

Class C

Class D

Case E

Eminently hydraulic limeSemi hydraulic limeFat lime (non hydraulic lime) Magnesia/ Dolomite originKankar lime

Setting under waterMasonry constructionWhitewashing and final coat for plasteringFinishing work of plastering and whitewashingMasonry mortar

11. Which one of the following light weight elementwill be added to enhance the protectiveproperties for x-ray shielding mortars?(a) Sodium (b) Potassium(c) Lithium (d) Calcium

Ans. (c)Sol. In x-ray shielding mortar, the protective

property will be enhanced by addition of lightweight elements like hydrogen, lithium,cadmium etc.

12. Which one of the following stone is producedby moulding a mixture of iron slag and portlandcement?(a) Imperial stone (b) Garlic stone(c) Ransom stone (d) Victoria stone

Ans. (b)

13. When a round bar material with diameter of37.5 mm, length of 2.4 m, Young’s modulusof 110 GN/m2 and shear modulus of42 GN/m2 is stretched for 2.5 mm, its bulkmodulus will be nearly

(a) 104 GN/m2 (b) 96 GN/m2

(c) 84 GN/m2 (d) 76 GN/m2

Ans. (b)Sol. E = 110 GN/m2

G = 42 GN/m2

EG

2 1

11042

2 1

0.309

Hence,

Bulk modulus (K) = E

3 1

110k

3 1 2 0.309

k = 96.25 GN/m2

14. A punch of 20 mm diameter is used to puncha hole in 8 mm thick plate. If the force requiredto create a hole is 110 kN, the average shearstress in the place will be nearly(a) 41 MPa (b) 320 MPa(c) 220 MPa (d) 140 MPa

Ans. (c)

Sol. Surface area of hole = d t

= 20 8

= 502.65 mm2

Force = 110 kN

Average shear stress ( avg ) = 110 1000

502.65

avg = 218.83 MPa

15. A reinforced concrete circular section of50,000 mm2 cross-sectional area carries 6reinforcing bars whose total area is 500 mm2.If the concrete is not to be stressed morethan 3.5 MPa and modular ratio for steel andconcrete is 18, the safe load the column cancarry will be nearly.(a) 225 kN (b) 205 kN(c) 180 kN (d) 160 kN

https://iesmaster.org/im/online-registration

CIVIL ENGINEERING




IES M

ASTER

Ans. (b)Sol. Total area of circular column = 50000 mm2

Total area of reinforcing bars = 500 mm2

Modular ratio = 18Safe load carring capacity = Area of column× PermissibleStress + (M – 1) × Area of steel × Permissiblestress= 50000 × 3.5 + (18 – 1) × 500 × 3.5= (50000 + 8500) × 3.5= 204750 N= 204.75 kN 205 kN

16. The strain energy U stored due to bending ofthe cantilever beam due to point load at thefree end will be

(a)2 3W

6 EIl

(b)2 2W

6 EIl

(c)3 3W

36 EIl

(d)2 3W

36 EIl

Ans. (a)Sol.

x

l, EI

W

Strain energy “U” due to bending = 2xM dx

2EIBending moment at x

W

x

Mx

M 0

Mx + Wx = 0

Mx = – Wx for 0 x

2

0

Wx .dxU

2EI

22

0

WU x dx2EI

2 3

0

W xU2EI 3

2 3WU6EI

17. A steel bar 2 m long, 20 mm wide and 15 mmthick is subjected to a tensile load of 30 kN.If Poisson’s ratio is 0.25 and Young’s modulusis 200 GPa, an increase in volume will be(a) 160 mm3 (b) 150 mm3

(c) 140 mm3 (d) 130 mm3

Ans. (b)Sol.

2 m

30 kN30 kN

15 mm

20 mmy

z

x

E = 200 GPA = 200 × 103 MPa

0.25

3

xP 30 10A 20 15

x y z100 MPa, 0, 0

CIVIL ENGINEERING




IES M

ASTER

x y zv 1 2

E

3100 1 2 0.25

200 10

4v 2.5 10

v v. Volume

42.5 10 2000 20 15

3v 150 mm

18. A bolt is under an axial thrust of 9.6 kNtogether with a transverse force of 4.8 kN. Iffactor of safety is 3, yield strength of boltmaterial is 270 N/mm2 and Poisson’s ratio is0.3, its diameter as per maximum principalstress theory will be nearly

(a) 13 mm (b) 15 mm

(c) 17 mm (d) 19 mm

Ans. (a)

Sol. Axial stress on bolt = PA

3

22

9.6 10 12223.1dd

4

Average shear stress = 2V 6111.54A 2d

A per maximum principal stress theory

ymax

fF.O.S

...(1)

For given stress condition

/2

2x y x y 2

max xy2 2

2 2

max 1.2072 2 2

From (1)

2701.2073

212223.1 2701.207

3d

d 12.8 mm

19. In a material the principal stresses are 60MN/m2, 48 MN/m2 and –36 MN/m2 when the

values of E = 200 GN/m2 and 1 0.3,m

the

total strain energy per unit volume will benearly(a) 43.5 KNm/m3 (b) 35.5 KNm/m3

(c) 27.5 KNm/m3 (d) 19.5 KNm/m3

Ans. (d)Sol. Given,

21 60 MN / m

22 48 MN / m

23 36 MN / m

0.3

Strain energy per unit volume (u)

CIVIL ENGINEERING




IES M

ASTER

2 2 21 2 3 1 2 2 3 3 1

1u 22E

2 2 23

1u 60 48 362 200 10

2 0.3 60 48 48 ( 36) ( 36)(60)

3 3MN.m kN mu 0.0195 19.5

m m

20. At a point in a two dimensional stress system,the normal stress on two mutuallyperpendicular planes are xx yyand andshear stress xy. One of the principal stresswill become zero when the value of shearstress xy is

(a) xx yy( ) (b) xx yy

(c) xx yy (d) xx yy

Ans. (d)Sol.

yy

xx

xy

2xx yy xx yy 2

major /minor xy 02 2

2 2

xx yy xx yy 2xy2 2

2 22 2yy xx yy yy xx yy 2xx xx

xy4 4 2 4 4 2

xy xx yy

21. The deflection of the closed coil helicalspring is

(a)2

3WR n8Cd

(b)3

464 WR n

Cd

(c)3

2128 WR n

Cd(d)

2

264 WR n

CdWhere:W is the axial loadR is radius of the coiln is the number of turns of coilC is the modulus of rigidityd is the diameter of the wire of the coil

Ans. (b)

Sol. Strain energy = 2T L

2GJT = (W.R)

L = 2 R n

J = 4d

32

2 2 3

4 4

WR . 2 R.n 32W .R .nUd Cd2.C

32

3

4U 64WR .nw Cd

22. A closely-coiled helical spring of round steelwire 5 mm in diameter having 12 completecoils of 50 mm mean diameter is subjected toan axial load of 100 N. if modulus of rigidityis 80 GPa, the deflection of the spring will be(a) 36 mm (b) 32 mm(c) 28 mm (d) 24 mm

Ans. (d)

Sol.3

464WR .n

C.d

https://iesmaster.org/ese-mains-conventional-question-practice-program

CIVIL ENGINEERING




IES M

ASTER

3

43

64 100 25 12

80 10 5

24 mm

23. A hollow shaft of external and internaldiameters as 100 mm and 40 mm respectivelyis transmitting power at 120 rpm. If theshearing stress is not to exceed 50 MPa, thepower the shaft can transmit will be(a) 100 kW (b) 120 kW(c) 140 kW (d) 160 kW

Ans. (b)Sol.

100 mm = d0

40 mm = d i

Maximum torque which can be tranmittedwithout exceeding allowable shear stress

0max

allowable4 40 i

dT2

(d d )32

50 = max

4 4

100T2

(100 40 )32

Tmax = 9.566 kN-mMaximum power which can be transmitted

Pmax = Tmax .w = 2 N9.56660

Pmax = 2 1209.566

60

Pmax = 120.211 kW

24. A circular beam of 100 mm diameter issubjected to a shear force of 30 kN. Themaximum shear stress will be nearly(a) 5.1 MPa (b) 6.3 MPa(c) 7.5 MPa (d) 8.7 MPa

Ans. (a)Sol.

100 mm

max avg43

max = avg2

4 4 V3 3 d

4

max = 3

2

4 30 103 100

4

max = 5.09 MPa

25. A cantilever beam AB as shown in figure issubjected to a point load of 12 kN over aspan of 6 m with E = 2 × 105 N/mm2 and Ixx= 6 × 107 mm4. The deflection at the free endwill be

W

A Bl

(a) 80 mm (b) 72 mm(c) 64 mm (d) 56 mm

Ans. (b)

Sol. B

W

Al, EI B

B = 3 3 3 3

5 7w (12 10 ) (6 10 )3EI 3 2 10 6 10

l

B = 72 mm

CIVIL ENGINEERING




IES M

ASTER

26. A floor has to carry a load of 12 kN/m2. Thefloor is supported on rectangular joists each100 mm wide, 300 mm deep and 5 m long.If maximum stress in the joist should notexceed 8 MN/m2, the centre to centre distanceof joists will be(a) 430 mm (b) 400 mm(c) 360 mm (d) 320 mm

Ans. (d)Sol.

C

Floor

C = centre-to-centre distance between twojoistsHence, load on each joist will be udl andintensity of udl(w) = 12C kN/mMaximum bending moment in joist

(Mmax) = 2W

8l

Mmax = 2312C 5 10

8

100mm

300mm

X-section of joist

max = 23

max2

M 12C 5 10z bd8

6

8 MN/m2 = 23

212C 5 10

100 30086

C = 0.32 m

C = 320 mm

27. A simply supported wooden beam 100 mmwide, 250 mm deep and 3 m long is carryinga uniformly distributed load of 40 kN/m. Themaximum shear stress will be

(a) 2.4 MPa (b) 2.8 MPa

(c) 3.2 MPa (d) 3.6 MPa

Ans. (d)

Sol.

A B40 kN/m

3 m

100 mm

250 mm

X-section

Maximum shear force will occur at supports

Vmax = W 40 32 2

l = 60 kN

For rectangular beam, maximum shear stress

max

max = maxavg

V3 32 2 bd

=

33 60 102 100 250

max = 3.6 MPa

28. A simply supported of span 8 m carries auniformly distributed load of 24 kN/m run overthe whole span. The beam is propped at themiddle of the span. The values ofE = 200 × 106 kN/m2 and I = 20 × 10–5 m4.The amount by which the prop should yield inorder to make all three reactions equal will benearly

(a) 20 mm (b) 15 mm

(c) 10 mm (d) 5 mm

CIVIL ENGINEERING




IES M

ASTER

Ans. (b)Sol.

4 m 4 m

A

CB

w = 24 kN/m

Flexural rigidity

6 5 42

kN"EI" 200 10 20 10 .mm

= 40000 kN-m2

The amount by which prop yield in order tomake all reactions equal

HA l

w

l

RA RC RB

By using equilibrium equation

x AF 0 H 0 y A C BF 0 R R R w w 2w

Let support “C” yield by " " amountdownward

w

udl

2l

RC

RC

4 4

udlw 25 5 w

384 EI 24 EI

RC

3 3C CR 2 R48EI 6EI

C

34C

udl RR5 w

24 EI 6EI

Now, given situation is all reactions becomesequal when support “C” yield by " "

RA = RC = RB

C3R 2w

C2wR

3

4 35 w 2w24 EI 3 6EI

47 w72 EI

w = 24 kN/m

4 m

EI = 40000 kN-m2

47 24 4 0.01493 m72 40000

14.93 mm

29. A cantilever beam ACB has end A fixed andsubjected to a point load P at free end B. Thepoint C is mid-point of AB and the moment ofinertia of AC is twice that of CB. The deflectionat the free end will be

A B2I C I

P

(a)3P

3 EIl

(b)3P

48 EIl

(c)35 P

96 EIl

(d)39 P

48 EIl

H-21, Sector 63, - 201305 | Call 0120 415 1100Email: [email protected] | Website: iesmaster.org

Noida

New DelhiEmail: [email protected]

22 Departments...

Batches Starts

11February

Assistant Engineer Examination-2019

UPPSC

692 Posts

Click here for more information

https://iesmaster.org/uppsc-ae-classroom-coaching

CIVIL ENGINEERING




IES M

ASTER

Ans. (d)Sol.

l/2 l/2A

2I CB

PI

l/2l/2 CA B

Pl

Pl2

BMD

A1 CA B

Pl2EI

Pl2EI

A2M/EI diagram

Pl4EI

From area moment theorem,

tBA = = 1 1 2 2A x A x

2x = 23 2 3

l l

A2 = 21 P

2 2EI 2 8EI

Pl l l

A1 = 21 P P 3 P

2 2EI 4EI 2 16 EI

l l l l

1x =

P P2 12EI 4EIP P2 3 22EI 4EI

l ll l

l l

1x = 79

l

Hence, = 23 P 7 P

16 EI 9 8EI 3

l l ll

= 3 33 P 9 P

16 EI 48 EI

l l

30. A beam of uniform cross-section, simplysupported at ends carries a concentrated loadW at mid-span. If the ends of the beam arefixed and only load P is applied at the mid-span such that the deflection at the centreremains the same, the value of the load P willbe(a) 6 W (b) 4 W(c) 2 W (d) W

Ans. (b)Sol. From standard results

l/2

W

l/2

3W48EI

l/2

P

l/231 P

4 48EI

For

P = 4W

31. Two wheels loads 200 kN and 80 kN spacedat 2 m apart move on the span of girder of16 m. If any wheel load can lead the other,the maximum bending moment that can occurat a section of 6 m from the left end will be(a) 1050 kNm (b) 990 kNm(c) 870 kNm (d) 750 kNm

Ans. (b)Sol. Maximum bending moment

Case I200 kN80 kN

2 m

BA

6 m 10 mC

CIVIL ENGINEERING




IES M

ASTER

ILD for moment

4 m

10 6 1516 4

15y2

1y4 =

1546

y1 = 15 56 2

Mc = 15 5200 804 2

= 950 kN-m

Case II200 kN

2 m

80 kN

BA

6 m 10 mC

154

4 m

y = 31

8 m 10 m

1y 1548

10

y1 = 15 8 3

4 10

Mc = 15200 3 804

= 990 kN-m

32. A beam of length l is fixed at its both endsand carries two concentrated loads of W each

at a distance of 3l

from both ends. The fixed

end moment at A will be

(a)W3

l(b)

2W9

l

(c)6W15

l(d)

4W27

l

Ans. (b)Sol.

P

a MBMA b

MA = 2

2Pab

(a b)

MB = 2

2Pa b

(a b)

A B

W W

l/3 l/3 l/3

(FEM)A =

2 2

2 2

2 2W W3 3 3 3

l l l l

l l

(FEM)A = 2W

9

l

–ve sign means anticlockwise.

33. The natural frequency of a mass m at the endof the cantilever beam of negligible mass withusual notations will be

(a)1/2

31 3EI

2 mL

(b)1/2

31 6EI

mL

(c)1/2

31 6EI

2 mL

(d)1/2

31 3EI

mL

CIVIL ENGINEERING




IES M

ASTER

Ans. (a)Sol.

Mass = m

Natural circular frequency

nkm

Stiffness = 33EIL

n 33EImL

Natural cycle frequency nnf 2

= 3

1 3EI2 mL

34. The simple osci l lator under idealizedconditions of no-damping, once excited willoscillate indefinitely with constant amplitudeat its natural frequency f will be

(a)1 m

2 k(b)

1 km

(c)1 k

2 m(d)

1 mk

Ans. (c)

Sol. Natural frequency nf2

nkm

1 kf2 m

35. A cable carrying a load of 10 kN/m run ofhorizontal span is stretched between supportsof 100 m apart. If the supports are at samelevel and central dip of 8 m, the ratio ofgreatest and least tensions in the cable willbe

(a) 1.05 (b) 1.35(c) 1.65 (d) 1.95

Ans. (a)Sol.

10 kN/m

8 m

100 m

HA HB

VBVA

C1. Reaction

AM 0

B100V 100 10 100 0

2

BV 500 kN

yF 0 VA + VB – 10 × 100 = 0

VA + 500 – 1000 = 0

AV 500 kN

2. Horizontal thurstIn cable equation of condition is bendingmoment at every point on cable is zero

CM 0

10 kN/m

8 m50 m HB

VB

B B50H 8 V 50 10 50 02

2

B50H 8 500 50 10 02

HB = 1562.5 kN

https://iesmaster.org/master-gate/

CIVIL ENGINEERING




IES M

ASTER

3. Maximum and minimum tension in the cable

2 2max A AT V H

2 2maxT 500 1562.5 1640.55 kN

Tmin = H = Tc = 1562.5 kNratio of max and min tension in cable

= max

min

T 1640.55 1.049 1.05T 1562.5

36. A cable is suspended between two points75 m apart at the same level. It carries auniformly distributed load of 12.5 kN perhorizontal meter. If the maximum tension inthe cable is limited to 1000 kN, the minimumcentral dip will be nearly(a) 14 m (b) 12 m(c) 10 m (d) 8 m

Ans. (c)Sol.

HA HB

VA VB

75 m

C

12.5 kN/m

yc = central drip

1. Reaction

yF 0 VA + VB = 12.5 kN …(i)

AM 02

B(75)V 75 12.5 0

2

VB = 468.75 kNVA = 468.75 kN

2. Horizontal thrust

Bending moment at every point on cableis zero.

HB

12.5 kN/m VB

yc

752

CM 0

2

B c B12.5 75 75H y V 0

2 2 2

2

B c12.5 75 75H y 468.75 0

2 2 2

HByc = 8789.062 kNGiven maximum tension = 1000 kN

Tmax = 2 2A AV H

1000 = 2

2

c

8789.062(468.75)y

780273.4375 = 2

c

8789.062y

yc = 8789.062883.33

cy 9.949 m

37. A tie bar 50 mm × 8 mm is to carry a load of80 kN. A specimen of the same quality steelof cross-sectional area is 250 mm2. For amaximum load of 125 kN carried by thespecimen, the factor of safety in the designwill be(a) 3.0 (b) 2.5(c) 2.0 (d) 1.5

CIVIL ENGINEERING




IES M

ASTER

Ans. (b)Sol. For cross-sectional area 250 mm2,

Maximum load = 125 kN3

allowable12 10 500 MPa

250

For tie bar

P 80 1000 200 MPaA 50 8

allowable 500F.OS 2.5200

38. Hanger connections are made when(a) Beam as well as girder is meeting at

different level. A plate or hanger isinterposed between the beam and thegirder and finally inter-connected bymeans of angle cleats or bolts and rods.

(b) Beam as well as girder is meeting at samelevel. A plate is interposed between thebeam and the girder.

(c) The beams are meeting at different levels.A hanger is interposed between thebeams and finally inter-connected bymeans of angle cleats or bolts and rods.

(d) The girders are meeting at same level. Aplate is interposed between the girdersand finally inter-connected by means ofbolts and rods.

Ans. (c)

39. The splicing of a column becomes necessary,where(a) The available length of structural steel

section is less than the required length ofthe column.

(b) Section remains same throughout at allfloors.

(c) Only riveted columns are to be designed.(d) Splices should be designed to carry axial

loads only.

Ans. (a)Sol. Column sections can be splices in the

following cases:(i) When the length of column is more than

the length of column section available.(ii) In case of multistory buildings, the section

of column required for the various storeysmay be different.

40. A tie bar 50 mm × 8 mm is to carry a load of80 kN. A specimen of same quality steel ofcross-sectional area is 250 mm2. If themaximum load carried by the specimen is125 kN, the gauge length will be(a) 133 mm (b) 126 mm(c) 113 mm (d) 106 mm

Ans. (c)

Sol. Gauge length g = 5.65 Area

= 5.65 50 8

= 113 mm

41. The strength of a column depends on whichof the following imperfections?

1. The material being isotropic andhomogeneous.

2. Geometric variations of columns

3. Eccentricity of load

(a) 1, 2 and 3 (b) 2 and 3 only

(c) 1 and 3 only (d) 1 and 2 only

Ans. (b)

Sol. As per Euler’s theory.2

cr 2EIP

Actual critical load will deviate from this loadif assumptions made by Euler are not met inpractical condition

CIVIL ENGINEERING




IES M

ASTER

Assumptions:

(a) It is valid for long columns

(b) EI is uniform and material is isotropic

(c) Load is purely axial

(d) Axis of the shaft is perfectly straight whenunloaded

Hence 2 & 3 are not meeting the criteria

Option (b) is correct

42. Which of the following types of failures occurin the beam-column connections?

1. Failure by lateral-torsional buckling

2. Failure by combined instability in both theprincipal directions

3. Failure by combined twisting and bendingon the torsionally weak sections

4. Failure by combined twisting and bendingwhen plane of bending does not containthe shear centre

(a) 1, 2 and 3 only (b) 1, 3 and 4 only

(c) 1, 2 and 4 only (d) 2, 3 and 4 only

Ans. (b)

Sol. Types of failure occur in the beam columnconnection are:

1. Failure by instability in plane of bendingwithout twisting.

2. Failure by instability in one of the principaldirection.

3. Failure by lateral torsional buckling.

4. Failure by combined twisting and bendingon these torsionally weak sections.

5. Failure by combined twisting and bendingwhen plane of bending does not containshear centre.

Hence, statement-2 is wrong.

43. In a design of beam columns, the values ofplastic section ratio b 1 , the plasticsectional modulus Zpz = 3948812 mm3, theyield stress fy = 250 N/mm2 and criticalmoment of Mcr = 16866 × 106 N.mm. Thenon-dimensional lateral torsional slendernessratio will be nearly

(a) 0.141 (b) 0.242

(c) 0.323 (d) 0.424

Ans. (b)Sol. Non dimensional lateral torsional slenderness

ratio

= b PZ y

cr

Z fM

= 61 3948812 250 0.2419 0.242

16866 10

44. As per Indian Railway Board, the impact factori (also known as coefficient of dynamicaugment, CDA) in steel girders for single trackspan is

(a)80.15 1.0

6 L

(b)60.75 1.0

8 L

(c)60.15 1.0

8 L

(d)80.75 1.0

6 L

where, L is spanAns. (a)

Sol. Impact factor = 80.15 1.0

6 L

https://iesmaster.org/master-talent-reward-exam/



CIVIL ENGINEERING




IES M

ASTER

45. An ISHB 300 with plastic section modulus of921.68 × 103 mm3, flange width of 250 mm,the yield stress of 250 N/mm2 is embedded ina pocket base to develop its strength withM25 concrete in design of beam-column. Therequired depth will be nearly

(a) 475 mm (b) 425 mm

(c) 375 mm (d) 325 mm

Ans. (a)Sol. The compressive force in concrete =

ck0.45 f d b2 2

= 0.112 fckbdThe moment of resistance of concrete (M)

= 22 C d3

= ck22 0.112 f bd d3

= 0.149 fckbd2

0.149 fckbd2 = 0

y pz

m

f z

= moment of resistance

of the steel section

d = 1/2

y pz

ck

f z1.1 0.149 f b

For plastic and compact section 1

d = 1/2

y pz

ck

f z0.164 f b

d = 1/23250 921.68 10

0.164 25 250

= 474.13 475 mm

46. In beam-columns or eccentric loaded columns,an elastic critical stress in compression fcc is

(a)E

(b)2

2E

(c) 2E

(d)2E

Where, E = Modulus of elasticity of steel

= Slenderness ratio in the plane of bendingAns. (b)Sol. For columns

Elastic critical sresss = 2

2E

47. A square of side a is placed such that itsdiagonal is horizontal. The shape factor ofthe square will be

a

(a) 3.2 (b) 2.0(c) 1.5 (d) 1.0

Ans. (b)Sol.

a2

a2

a2

x x45°

a

fy

fy

yp

y y

.SM SS.F.M .z z

CIVIL ENGINEERING




IES M

ASTER

1 2

a aA 1 a 2 2S y y 2a2 2 3 32

3aS3 2

43

xx

max

aI a12z

ay 6 22

3

3

a3 2S.F. 2a

6 2

48. Which one of the following is the correctassumption made in evaluation of fully plasticmoment?(a) The upper and lower yield stresses and

the modulus of elasticity have differentvalues in compression and tension.

(b) The material is homogeneous andisotropic in both the elastic and plasticstates.

(c) There will be resultant axial force on thebeam.

(d) Some layers of the material are not freeto expand and contract longitudinally andlaterally under stress.

Ans. (b)Sol. 1. The stress-strain curve is idealised to two

straight line.2. Plane section before bending remains

plane after bending.3. Material is homogenous, isotropic and

stress-strain curve is same both in tensionand compression.

4. Rotation of any magnitude can take placeat plastic hinge, but the bending momentremains constant at fully plastic value MP.

5. Effect of axial load and shear on fullyplastic moment capacity is neglected.

6. The deformations are small and equationsof statical equilibrium are same as thosefor undeformed structure.

49. As per IS-456 : 2000, cracking of concrete intension zone cannot be avoided but can belimited by

1. Adhering to the codal requirements ofminimum steel area

2. Proper and prolonged curing of concrete

3. Increasing water cement ratio to increaseworkability

(a) 1 and 2 only (b) 1 and 3 only

(c) 2 and 3 only (d) 1, 2 and 3

Ans. (a)

Sol. Minimum steel area will limit the crack width.Proper and prolonged curing will increase thetensile stress and delay the loading.

Increase in water cement ratio will increasecrack width due to higher shrinkage and creep.

50. Which of the following assumptions are madewith respect to Euler ’s theory applied tocolumns?

1. The section of the column is uniform

2. The length of the column is very largecompared to the lateral dimensions

3. The direct stress is large when comparedwith the bending stress

(a) 1, 2 and 3 (b) 1 and 3 only


Ans. (d)

Sol. Assumptions of Euler’s theory:

CIVIL ENGINEERING




IES M

ASTER

(a) It is valid for long columns(b) EI is uniform and material is isotropic(c) Load is purely axial(d) Axis of the shaft is perfectly straight when

unloadedHence, 1 & 2 are assumptionsOption (d) is correct

51. A rectangular beam with b = 200 mm andeffective depth d = 300 mm is subjected tolimit state shear of 80 kN and torsionalmoment of 6 kNm. The equivalent value ofshear will be(a) 128 kN (b) 116 kN(c) 104 kN (d) 92 kN

Ans. (a)Sol. b = 200 mm

d = 300 mmVu = 80 kNTu = 6 kNm{Already factored shear as given as limit stateis mentioned}Equivalent shear,

Ve = uu

1.6 TVb

= 1.6 680 128 kN

0.2

52. As per IS-456 : 2000, the value of maximumcompression strain in concrete in axialcompression for limit state of collapse is(a) 0.001 (b) 0.002(c) 0.003 (d) 0.004

Ans. (b)Sol. As per IS-456:2000, clause 39.1 (a), the

maximum compressive strain in concrete inaxial compression is taken as 0.002.

53. The positive bending moment coefficient atthe middle of the end-span of a continuousone way slab is

(a) 2dw w L10 12

l (b) 2dw w L9 10

l

(c) 2dw w L12 16

l (d) 2dw w L9 12

l

Where, wl = Live load ; wd = Dead loadAns. (a)Sol. As per Table 12 of IS-456:2000, bending

moment coefficients for dead load is 1

12

and for live load is 1

10 near middle of end

span.

So, bending moment = 2 2d l

1 1w w12 10

l l

= 2w wd10 12

l l

54. Which one of the following are the generaldesign requirements of retaining wall?1. The factor of safety against sliding should

be atleast 1.52. The factor of safety against overturning

should be atleast 2.03. The bearing pressure at toe should be

less than the bearing capacity of the soil4. The length of retaining wall to be cast in

one go should not exceed 10 m otherwisecracks may develop

(a) 1, 2 and 3 only (b) 1, 3 and 4 only(c) 1, 2 and 4 only (d) 2, 3 and 4 only

Ans. (a)Sol. As per IS 14458 (Part-2): 1997, retaining wall

for hill area.Guidelines: Design of retaining/breast walls,clause no. 5.2.Option 1, 2 and 3 are correct.

https://iesmaster.org/testseries/cil

CIVIL ENGINEERING




IES M

ASTER

55. Which of the following are the desirableproperties for efficient functioning in designfor movement joint of water tank?1. The joint should accommodate repeated

movement of the structure without loss ofwater-tightness

2. The design should provide for exclusionof grit and debris which would preventthe closing of the joint

3. The material used in the construction ofmovement joints should not slump undulyin hot weather or become brittle in coldweather

(a) 1, 2 and 3 (b) 1 and 2 only(c) 1 and 3 only (d) 2 and 3 only

Ans. (a)Sol. As per IS 3370 (Part 1): 2009 concrete

structures for storage of liquids, clause number10.2, 1, 2 and 3 are correct.

56. A simply supported beam having 200 mmwidth and 450 mm effective depth supports atotal uniformly distributed load of 2,00,000 N.The nominal shear stress will be nearly(a) 0.8 N/mm2 (b) 1.1 N/mm2

(c) 1.8 N/mm2 (d) 2.2 N/mm2

Ans. (b)Sol. Width, b = 200 mm

Effective depth, d = 450 mmTotal uniformly distributed load = 200000 N

Maximum shear force (Vu) = 200000

2

= 100000 N

Nominal shear stress, v = uVbd

= 2100000 1.11N/mm200 450

ORw

l

200 mm

450 mm

maxw 200000V 100000N2 2

Nominal shear stress

maxavg

V 100000.bd 200 450

2avg. 1.1N / mm

57. Which of the following are correct for cover toreinforcement?

1. The reinforcement shall have a minimumclear cover of 20 mm or diameter of suchbar whichever is more.

2. At each end of reinforcing bar not lessthan 25 mm nor less than twice thediameter of such bar.

3. Increased cover thickness may beprovided when surface of concrete isexposed to the action of harmfulchemicals.

(a) 1, 2 and 3 (b) 1 and 2 only


Ans. (a)Sol. In case of mild exposure cover = 20 mm

Also, in any case cover shall not be less thanmaximum bar diameter, here statement (1) iscorrect.

CIVIL ENGINEERING




IES M

ASTER

Statement (2) is correct as per SP-34-1987

Increased cover thickness is provided whenconcrete is exposed to the action of harmfulchemicals.

Statement (3) is correct.

58. A beam of size 250 mm width and 460 mmeffective depth is subjected to limit statemoment of 146 kNm. If M20 grade concreteand Fe415 steel are used, the area of steelrequired will be

(a) 435 mm2 (b) 935 mm2

(c) 1100 mm2 (d) 1235 mm2

Ans. (c)Sol. Width, b = 250 mm

Effective depth, d = 460 mm

Mu = 146 kNm

Mu,lim = 0.138fckbd2

= 0.138 × 20 × 250 × 4602 × 10–6 = 146 kNm

Section has to be designed as a balancedsection,

Ast = ck2

y ck

f 4.6M1 1 bd2f f bd

= 1096.43 mm2 1100 mm2

59. Air permeability method is used to determine(a) Soundness of cement(b) Setting time(c) Fineness of cement(d) Resistance of cement

Ans. (c)Sol. Fineness of cement is measured by air

permeability method.

60. Which of the following assumptions are correctfor the lateral torsional buckling of an I-sectionbeam?

1. The beam is initially distorted

2. Its behaviour is elastic

3. It is loaded by equal and opposite endmoments in the plane of the web


(c) 1 and 3 only (d) 1, 2 and 3

Ans. (b)Sol. The lateral torsional buckling of an I-section

is considered with the following assumptions:

1. The beam is initially undistorted.

2. Its behaviour is elastic (no yielding).

3. It is loaded by equal and opposite endmoments in the plane of the web.

4. The loads act in the plane of the webonly (there are no externally applied lateralor torsional loads).

5. The beam does not have residualstresses.

6. Its ends are simply supported verticallyand laterally.

61. In an excavation of 3,000 cub.mtr of commonearth for a canal project, 6 men can beeffectively employed on the job. If an outputof a man is taken as 100 cub.mtr per day, theduration of excavation activity will be

(a) 5 days (b) 6 days

(c) 7 days (d) 8 days

Ans. (a)Sol. Duration of excavation activity

= 3000

6 100 = 5 days.

62. The project plan for construction1. Clearly defines project’s scope of work. It

breaks down project objectives into clear,identifiable, attainable and verifiable goals.

CIVIL ENGINEERING




IES M

ASTER

2. Identifies critical activities, thus enablingmanagement of projects by exceptions.

3. Provides the basis for coordinating theefforts of clients, consultants, architects,designers, quality surveyors, specialists,suppliers, contractors and project staff.

(a) 1 and 2 only (b) 1, 2 and 3(c) 1 and 3 only (d) 2 and 3 only

Ans. (b)Sol.

Planning/Design

Determine budget

Risk management plan

Procurement plan

Communication plan

Human resource plan

Stakeholder management plan

Project management plan

Project cost

Estimate cost

Support plans

Quality plan

Prepare work break down structure

Project kick off meeting

Defining scope of work

Role assignment

Project scheduling

Define activities

Estimate resource required

Estimate activity duration

Develope schedule

Sequence activities

63. Which one of the following techniques is notcovered in Project Network Analysis?

(a) Critical Path Method

(b) Program Evaluation and Rev iewTechniques

(c) Procedure Network Analysis

(d) Measurement Book

Ans. (d)

Sol. Measurement Book (M.B.): Measurement ofall works and supplies are recorded in aspecial type of note book.

64. Which of the following statements are correctfor Network Critical Path?

1. The path of critical activities, which linksthe start and end events is critical path.

2. It is the path of activities having zero float.

3. It is the path of events having zero slack.

4. The sum of the duration of the criticalactivities along a critical path gives theduration of the project.

(a) 1, 2, 3 and 4 (b) 1, 2 and 3 only

(c) 1 and 4 only (d) 2, 3 and 4 only

Ans. (c)Sol. Critical path has activities having minimum

floats.

It is not necessary that path of events havingzero slack becomes critical.

65. Independent float is an amount of time bywhich the start of an activity may be delayedwithout affecting

1. the preceding or the following activity

2. the start of a following activity

3. the completion of the project

(a) 1 only (b) 2 only

(c) 3 only (d) 1, 2 and 3

Ans. (d)Sol. Independent float is the minimum amount of

time by which an activity can be delayed.

It does not effect start of succeeding activityi.e. following activity.

It does not effect latest completion ofpreceding activity.

It only affects concerned activity.

Since, it does not effect EST of succeedingactivity it does not effect completion of project.

CIVIL ENGINEERING




IES M

ASTER

66. Consider the following activity for the totalproject :

Immediate DurationActivity Pr edecessors (Days)A 10B 9C A 9D A 8E B 7F B 11G D, E 5

The total project duration for the critical pathwill be(a) 25 days (b) 23 days(c) 21 days (d) 19 days

Ans. (b)Sol.

10,0

2

3

4

10,10

9,12

A10B

9

D8

E5G

5

C9

23,23

F11

18,18

7

Project completion time = 23 days

67. By performing which of the following functionsthe construction manager can achieve theproject goals?

1. Envisioning the task ahead

2. Setting targets and monitoringperformance

3. Motivating the work force

4. Building the line supervisors team


(c) 1, 3 and 4 only (d) 1, 2, 3 and 4

Ans. (d)Sol. The construction manager can play the most

vital lead role in achieving the project goalsby performing the fol lowing functionseffectively and efficiently, these functions aredetailed in the following sub-sections:

Envisioning the task ahead.

Enabling the individuals to perform thetask systematically and efficiently.

Setting targets and monitoringperformance.

Providing resources support.

Implementing a sound incentive scheme.

Communicating feedback.

Motivating the work force.

Building the line supervisors team.

Creating safe working environments.

Abiding by professional ethics.

Leadership makes all the difference.

68. The cost of the machine is Rs 20,00,000 andif it is purchased under instalment basis; thecompany has to pay 25% of the cost at thetime of purchase and the remaining amountin 10 annual equal instalments of Rs 2,50,000each. If rate of interest is 18%, compoundedannually the present worth of the machinewill be(a) Rs 17,01,000 (b) Rs 16,22,500(c) Rs 15,43,000 (d) Rs 14,64,500

Ans. (b)Sol.

0 1 2 3 4 5 6 7 8 9 10

5L2.5L 2.5L 2.5L 2.5L 2.5L 2.5L 2.5L 2.5L 2.5L 2.5L

P = 10

101.18 15,00,000 2,50,000

1.18 0.18

P = 16,23,521.

CIVIL ENGINEERING




IES M

ASTER

69. Which of the following relations are correctfor determining different components of a bidprice?

1. Bid price = Direct cost + Indirect cost +Mark up amount.

2. Direct cost = Project overheads +Common plant and equipment cost +Common work men cost.

3. Mark up amount = Profit + Contingency +Allowances for risks + General overheads.

(a) 1, 2 and 3 (b) 1 and 2 only


Ans. (c)Sol. Bid price = Base cost + Mark up amount

Base cost = Direct cost + Indirect cost

Direct cost = Labour cost + Plant andEquipment cost + materials cost +Subcontractor cost

Mark up cost = Profit + Contingency + Generaloverheads + Allowance for risks

70. Resource smoothing is

(a) An optimization and economical utilizationof resources.

(b) An adjustment of resources withoutaffecting project duration.

(c) A gradual increase in resources.

(d) A gradual decrease in resources.

Ans. (b)Sol. Resources smoothing: In this total project

duration is not changed but some of theactivity start times are shifted by their availablefloat so that more or less uniform demand isgenerated.

71. In PERT technique, the time estimate ofactivities and probability of their occurrencefollow

(a) Binomial distribution(b) Normal distribution(c) Poisson distribution

(d) -distribution

Ans. (d)Sol. In PERT analysis time estimates of activities

and probability of their occurrence follow distribution.

72. Indirect cost due to accidents includes(a) Legal charges(b) Medical expenses for the injured(c) Compensation amount to the injured(d) Over time payment to make up the loss

of timeAns. (d)Sol. Cost of an accident = Direct cost + Indirect

costDirect cost include worker’s compensationpayment, medical expenses and legalservices.Indirect cost includes:

Training replacement employees

Accident investigations

Implementation of corrective measure

Lost productivity

Repairs of damaged equipment andproperty

Cost of absenteecism

73. An oil of specific gravity 0.9 contained in avessel. At a point the height of oil is 40 m andfor the density of water = 1000 kg/m3, thecorresponding height of water at the point willbe(a) 28 m (b) 32 m(c) 36 m (d) 40 m

https://iesmaster.org/testseries/gate-2020

CIVIL ENGINEERING




IES M

ASTER

Ans. (c)Sol. Given: Gs = 0.9; h = 40 m

Pressure at that point = oil gh

= 0.9 × 1000 × 9.81 × 40 = 353160 N/m2

Corresponding height of water at that pointwill be

hwater = w

Pressure at heat point

hwater = 353160 m

9.81 1000

waterh 36 m

74. When speed changes in case of centrifugalpump, which of the following points arecorrect?1. The shape of the velocity triangle will

remain same2. Various angles will remain same3. Magnitude of velocity wil l change

proportionately(a) 1 and 2 only (b) 1 and 3 only(c) 2 and 3 only (d) 1, 2 and 3

Ans. (b)Sol. Upon change in speed of centrifugal pump

Shape of velocity triangle will remain sameAngles 1 2and will change.

Magnitude of velocity wil l changeproportionately

u2

V2

Vr1V1

u1

Vr2

2

75. Which one of the following is the use of flownet analysis in fluid mechanics?(a) To determine the streamlines and

equipotential lines(b) To determine downward lift pressure

above hydraulic structure(c) To determine the viscosity for given

boundaries of flow(d) to design the hydraulic structure

Ans. (a)Sol. The flow net provides a simple graphical

techniques for studying two dimensionalirrotational flows especially in the cases wheremathematical relations for steam function andpotential function are either not available orare rather difficult to solve.

76. A jet propelled aircraft is flying at a speed of1100 km/hour at t = 20°C, k = 1.4 and R =287 J/kg K. The Mach number at a point onthe jet will be nearly(a) 0.3 (b) 0.5(c) 0.7 (d) 0.9

Ans. (d)Sol.

Velocity of aircraft = 1100 1000 m

3600 s

= 305.55 m/s

Velocity of sound = kRT

= 1.4 287 20 273

= 343.114 m/s

Mach number = Velocity of aircraftVelocity of sound

= 305.55 0.89343.114

77. When the drag force becomes equal to theweight of the body, the acceleration ceasesand the net external force acting in the bodybecomes

CIVIL ENGINEERING




IES M

ASTER

(a) Zero and the body will move at constantspeed

(b) Light and the body will move forward(c) Zero and the body will move fast(d) High and the body will move at constant

speedAns. (a)Sol. When drag force becomes equal to the weight

of the body, the acceleration ceases and thenet external force acting in the body becomeszero and the body will move at constantspeed.

78. Which of the following statements is correctregarding flow in open channel?(a) The curve for kinetic energy is a parabola(b) The curve for potential energy is a

parabola(c) Specific energy is asymptotic to the

vertical axis(d) At critical depth the specific energy is

maximumAns. (a)Sol. Statement (b) wrong, as potential energy

will vary linearly w.r.t. depth (y).Statement (c) wrong, as specific energyis asymptotic to horizontal axis.Statement (d) wrong, at critical depth thespecific energy is minimum.Statement (a) Correct, as kinetic energywill vary parabolically w.r.t. velocity.Hence, most appropriate answer is (a).

79. Which one of the following statement is correctregarding critical state of flow through achannel section?(a) Specific energy is a minimum for a given

discharge(b) Specific energy is a maximum for a given

discharge

(c) The Froude number is greater than two(d) The discharge is a minimum for a given

specific forceAns. (a)Sol. Expression of specific energy

E = 2vy

2g

As, v = QA

E = 2

2Qy

2gA

For E to be minimum at constant Q,

Tdy

dA

y

dEdy = 0

dEdy =

2–2Q d1 A

2g dy

dEdy =

2

3Q –2 dA12g dyA

dA can be written as Tdy. Therefore dA T,dy

dEdy =

2

3Q T1–gA

2

3Q T1–gA = 0

2

3Q T 1gA

CIVIL ENGINEERING




IES M

ASTER

2

2Q TA

gA

= 1

2v TgA = 1

vAgT

= 1

For a rectangular channel, we know thatA yT

vgy = 1 r

vF 1gy

Thus, when the specific energy is minimumfor a given discharge, flow will be critical flowand depth of flow will be called as criticaldepth yc and velocity of flow is called as criticalvelocity vc.

80. Which one of the following statement is correctregarding centrifugal pumps?(a) The discharge is fluctuating and pulsating(b) It is used for large discharge through

smaller heads(c) The efficiency is low(d) It runs at low speed

Ans. (b)Sol. Centrifugal pumps are used for pumping large

discharge through smaller heads.

81. A hydraulic press has a ram of 300 mmdiameter and a plunger of 45 mm diameter.When the force applied at the plunger is 50N, the weight lifted by the hydraulic press willbe nearly(a) 2133 N (b) 2223 N(c) 2316 N (d) 2406 N

Ans. (b)

Sol. By Pascal law

WRam

A

a

P

F

P

Pressure intensity = F Wa A

Weight lifted (W) = AFa

= 2

2

(D)4F

d4

= 230050

45

= 2223 N

82. Hydraulic efficiency of Francis turbine is(a) Directly proportional to velocity of whirl at

inlet and inversely proportional to net headon turbine

(b) Directly proportional to velocity of whirl atinlet and net head on turbine

(c) Inversely proportional to velocity of whirlat inlet and net head on turbine

(d) Inversely proportional to velocity of whirlat inlet and directly proportional to nethead on turbine

Ans. (a)Sol. Hydraulic efficiency of a Francis turbine

( h ) = 1w 1V ugH

1n w n1V andH

83. A turbine develops 7225 kW power under ahead of 25 m at 135 rpm. The specific speedof the turbine will be nearly(a) 245 rpm (b) 225 rpm(c) 205 rpm (d) 185 rpm

CIVIL ENGINEERING




IES M

ASTER

Ans. (c)Sol. P = 7225 kW, H = 25 ; N = 135 rpm

Specific speed (Ns) = 5/4N P(H)

= 5/4

135 7225(25)

= 205

84. Which one of the following is an example ofbodies where both drag and lift forces areproduced?(a) Hydrofiles(b) A tall chimney exposed to wind(c) Flow of water past a bridge pier(d) Motion of aeroplanes, submarines,

torpedoesAns. (d)

85. The relative humidity h is a measure of air’scapacity, at its existing temperature, to absorbfurther moisture, and is defined by the relation

(a)s

e 100e

(b) se 100e

(c) s2e 100e

(d)s

2e 100e

where; e = vapour pressurees = saturation vapour pressure

Ans. (a)Sol. Relative humidity

= Water vapour pressure100

Saturation vapour pressure

= s

e100e

The relative humidity is the ratio of watervapour pressure (e) and saturation vapourpressure (es) of present temperature.

86. Which one of the following is not a majordeterrent in water harvesting through watertanks?

(a) Deforestation mainly due to populationpressure in the catchments of tanksystems

(b) Siltation

(c) Lack of maintenance and repairs andbreaches of tank embankments

(d) Shallow depth of water tanks

Ans. (a)Sol. Major deterrent in water harvesting through

water tank.

1. Siltation.

2. Lack of maintenance and repair andbreaches of tank embankment.

3. Shallow depth of water tank.

87. Which one of the following is not a basicrequirement for any well screen?

(a) Resistance to corrosion, incrustation anddeterioration

(b) Enough structural strength to preventcollapse

(c) Suitability for excessive movement of sandinto the well

(d) Minimum resistance to flow of water intothe well

Ans. (c)Sol. Basic requirements for any well screen

1. Resistance to corrosion, incrustation anddeterioration.

2. Enough structural strength to preventcollapse.

3. Minimum resistance to flow of water intothe well.

4. Control sand pumping.

88. Which one of the following methods is not thecategory of Geophysical methods of sub-surface investigation?

https://iesmaster.org/testseries/ssc-je

CIVIL ENGINEERING




IES M

ASTER

(a) Electrical resistivity method

(b) Electric logging

(c) Gamma-ray logging

(d) Electrical response surveying

Ans. (d)Sol. Earth exploration method are:

1. Gamma ray logging

2. Electric logging/well logging

3. Electrical resistivity method

4. Seismic refraction method

89. In which one of the following industries, thewater requirement in kilo litres per unit ofproduction is very high?

(a) Paper industry

(b) Steel industry

(c) Sugar industry

(d) Fertilizer industry

Ans. (a)Sol. According to GOI manual

Industry Unit of production

Water requirementin kilolitres per unit

PaperSteel

Sugar

Fertilizer

TonneTonne

Tonne (cane crushed)

Tonne

200 – 400200 – 250

1 – 2

80 – 200

90. In drip irrigation system, which one of thefollowing emitters is not based on definitionsby American Society of Agricultural Engineers(ASAE)?(a) Emitter(b) Pulsating emitter(c) Long path emitter(d) Multi-outlet emitter

Ans. (b)

Sol. Emitters are small micro-irrigation dispensingdevices designed to dissipate pressure anddischarge of small uniform or trickle of waterat a constant rate (as per ASAE definition).In pulsating emitter, discharge will not beconstant.

91. A Persian wheel with an average dischargeof 230 litre/minute irrigates 1 hectare wheatcrop in 50 hours. The average depth ofirrigation will be nearly(a) 4 cm (b) 5 cm(c) 6 cm (d) 7 cm

Ans. (d)Sol. Q × time = Area × Avg. Depth of irrigation

Avg. Depth of irrigation

= 3

24

230 10 50 60 1010

= 6.9 cm

92. Which one of the following is not the maincause for soil salinity and sodicity?

(a) Irrigation mismanagement

(b) Poor land levelling

(c) Use of heavy machinery, resulting in nosoil compaction

(d) Leaching without adequate drainage

Ans. (c)Sol. Main causes of soil salinity and sodility:

1. Irrigation mismanagement

2. Leaching without adequate drainage

3. Poor land levelling

4. Deforestation

5. Accumulation of air borne or water bornesalts in soil.

93. Which one of the following is not the majorfactor influencing seepage from a canal?

CIVIL ENGINEERING




IES M

ASTER

(a) Characteristics of the soil traversed by thecanal system

(b) Area wetted by the canal(c) Location of the canal(d) Frequencies of canal usage

Ans. (c)

94. Which of the following are the causes of failureof weirs?1. Rupture of floor due to uplift2. Rupture of floor due to suction caused by

standing wave3. Scour on the upstream and downstream

of the weir(a) 1 and 2 only (b) 1 and 3 only(c) 1, 2 and 3 (d) 2 and 3 only

Ans. (c)Sol. A weir may fail due to following reasons.

1. Piping2. Rupture of floor due to uplift3. Rupture of floor due to section caused by

standing wave.4. Scour at the upstream and downstream

side of the weir floor.5. Erosion of apron protection.6. Deterioration of cutof and subsequent loss

of containment.

95. Which of the following are the principal factorsinfluencing the choice of a particular methodof lining?

1. Availability and cost of the material at thesite or within easy reach

2. Velocity of flow in the channel

3. Cost of maintenance


(c) 1, 2 and 3 (d) 2 and 3 only

Ans. (c)Sol. Factors influencing choice of a method of

lining:

1. Availability and cost of material.

2. Velocity of flow.

3. Cost of maintenance.

4. Climatic condition.

5. Size of canal.

6. Position of water table.

7. Imperviousness.

96. Which of the following are the objectives forriver training?

1. High flood discharge may pass safely andquickly through the reach

2. To make the river course stable andreduce bank erosion to minimum

3. To check flow through canal

4. To provide a sufficient draft for navigationas well as good course for it


(c) 1, 2 and 4 only (d) 2, 3 and 4 only

Ans. (c)Sol. Objectives of river training are

(1) High flood discharge may pass safely andquickly through the reach.

(2) Sediment load including bed andsuspended load may be transportedefficiently.

(3) To make the river course stable andreduce bank erosion to minimum.

(4) To provide a sufficient draft for navigationas well as good course for it.

(5) To fire direction of flow through certaindefined reach.

CIVIL ENGINEERING




IES M

ASTER

97. The transition region between unsaturatedzone and saturated zone is called

(a) Capillary fringe

(b) Water table

(c) Yadose water zone

(d) Confining bed

Ans. (a)

Sol. Capillary fringe treated as a boundarycondition separating the water table (saturatedzone) from unsaturated zone.

98. Which of the following chemical parametersare associated with the organic content ofwater?

1. Biological Oxygen Demand (BOD)

2. Chemical Oxygen Demand (COD)

3. Total Organic Carbon (TOC) and TotalOxygen Demand (TOD)


(c) 2 and 3 only (d) 1, 2 and 3

Ans. (d)

Sol. BOD, COD, TOC and TOD all are associatedwith organic content of water.

COD is related to both inorganic and organiccontent of water.

99. When chlorine is dissolved in water, it reactsto form hypochlorous acid and hypochloriteions. At pH < 5, chlorine exists in water as

(a) Elemental or molecular chlorine

(b) Remains in the form of hypochlorous acid

(c) Remains in the form of hypochlorite ions

(d) Remains in the form of both hypochlorousacid and hypochlorite ions

Ans. (a)

Sol. pH 52 2Cl H O HOCl HCl

At pH < 5, chlorine does not react with waterand remains in elemental or molecular chlorineform.

100. Reactive substances are(a) Unstable under normal conditions. They

can cause explosions and/or liberate toxicfumes, gases, and vapors when mixedwith water

(b) Eeasily ignited and burn vigorously andpersistently

(c) Liquids with pH less than 2 or greaterthan 12.5, and those that are capable ofcorroding metal containers

(d) Harmful or fatal when ingested orabsorbed

Ans. (a)Sol. Reactive substances are unstable under

normal conditions and react violently with airand water. They cause explosions or formtoxic vapors.

101. The noise value of sound wave depends upon:1. The frequency of sound waves2. The intensity of sound waves3. The time of exposure of sound waves(a) 1 and 2 only (b) 1 and 3 only(c) 2 and 3 only (d) 1, 2 and 3

Ans. (a)Sol. The noise levels of sound waves depends

upon

Intensity of sound

Frequency of sound wave

Time period of sound waveTime of exposure of sound will not affect noiselevel rather it will effect on human ears.

CIVIL ENGINEERING




IES M

ASTER

102. Which of the following type of treatments willbe used for neutralization of alkaline effluent?(a) Lime stone treatment(b) Caustic lime treatment(c) Carbon dioxide treatment(d) Hydrochloric acid treatment

Ans. (d)Sol. Lime stone (CaCO3), caustic lime (CaO) both

will produce alkalinity in medium.Whereas carbon dioxide will form carbonicacid.

2 2 2 3CO H O H CO

2 3 3H CO H HCO

It will give H+ for neutralization but will alsoproduce alkalinity in medium.Hydrochloric acid (HCl) on the other hand willbe best option for neutralization of alkalineeffluent.

103. Flocculation is the process of(a) Gently mixing the water and coagulant

allowing the formation of large particlesof floc

(b) Removing relatively large floating andsuspended debris

(c) Flow, which is slowed enough so thatgravity will cause the floc to settle

(d) Mixture of solids and liquids collected fromthe settling tank are dewatered anddisposed of

Ans. (a)Sol. Flocculation is the process of gently mixing

the water and coagulant allowing the formationof large particles of floc.

104. In solid waste management, waste utilizationis achieved by

(a) Recover, reclamation and reproduce

(b) Reuse, reclamation and recycling

(c) Recover, recycling and reproduce

(d) Reuse, reproduce and recycling

Ans. (c)Sol. Conceptual.

105. The frequency range for hearing the soundby a human ear is in the range of(a) 20 Hz – 200 kHz(b) 10 Hz – 20kHz(c) 20 Hz – 20 kHz(d) 10 Hz – 20 Hz

Ans. (c)Sol. The human ear can detect a wide range of

frequencies. Frequencies from 20 Hz to 20kHz are audible to human ear.Any sound with a frequency below 20 Hz isknown as an infrasound and any sound witha frequency above 20 kHz is known asultrasound.

106. Physiological responses accompanyingresponse and other noise exposures include:1. A vascular response characteristic by

peripheral vasoconstriction, changes inheart beat rate and blood pressure

2. Various glandular charges such asincreased output of adrenaline evidencedby chemical changes in blood

3. Slow, deep breathing(a) 1 and 2 only (b) 1 and 3 only(c) 2 and 3 only (d) 1, 2 and 3

Ans. (d)Sol. Physiological responses accompanying a

response and other noise exposures includes:1. A vascular response characteristic by

peripheral vasoconstriction, changes inheart rate and blood pressure.

CIVIL ENGINEERING




IES M

ASTER

2. Various glandular changes such asincreased output of adrenaline evidencedas chemical changes in blood duringcirculation.

3. Slow, deep breathing.

4. A change in the electrical resistance ofskin with changes in activity of the sweatglands.

5. Brief changes in skeletal muscle tension.

107. Electrostatic precipitators are used for removalof

1. Gaseous contaminants

2. Liquid contaminants

3. Particulate contaminants


(c) 3 only (d) 1, 2 and 3

Ans. (c)

Sol. Electrostatic precipitators are used for removalof particulate contaminants only.

108. Which one of the following type of ecology isdealt with autecology?

(a) Synecology

(b) Community ecology

(c) Ecosystem ecology

(d) Individual species ecology

Ans. (d)

Sol. Ecology is often broadly div ided intoautecology and synecology. Autecology dealswith the ecological study of one species oforganism. Thus, an autecologist may study tothe l i fe history, population dynamics,behaviour, home range and so on, of a singlespecies, such as the Indian bull frog, or maizeborer. Synecology deals with the ecologicalstudies of communities or entire ecosystems.

109. A soil sample has a porosity of 40%, and thespecific gravity of solid is 2.70. if the soil is50% saturated, the unit weight will be nearly

(a) 22 kN/m3 (b) 20 kN/m3

(c) 18 kN/m3 (d) 16 kN/m3

Ans. (c)Sol. n = 40% = 0.4, Gs = 2.7, S = 50% = 0.5

n 0.4 0.4 2e1 n 1 0.4 0.6 3

Unit weight s wt

G 1 w1 e

es 2 / 3 0.5es wG w 0.213G 2.7

3t

2.7 9.81 1.123 17.84 kN / m1 2 / 3

18 kN/m3

110. Oven dry mass of a pat of clay is 10.8 gmand mass of mercury displaced on immersionis 84.2 gm. If the specific gravity of solids is2.72 and the density of the mercury is 13.6 g/cm3, the shrinkage limit of the soil will benearly

(a) 12% (b) 15%

(c) 18% (d) 21%

Ans. (d)Sol. Mass of mercury displaced = 84.2 gm

Density of mercury = 13.6 g/cm3

Dry mass of a pat of clay = 10.8 gm

Specific gravity of solids = 2.72

Volume of displaced mercury = 84.213.6

= 6.19 cm3

Volume of clay = 6.19 cm3

Dry density of clay = 310.8 1.745 g / cm6.19

CIVIL ENGINEERING




IES M

ASTER

s wd

G1 e

2.72 11.7451 e

e = 0.56

At shrinkage limit clay will be 100% saturated(s = 1)

es = wGe 0.56w 0.206 20.6%G 2.72

111. The suitability number of a backfill for D50 =1 mm, D20 = 0.5 mm and D10 = 0.08 mm willbe nearly

(a) 16 (b) 18

(c) 20 (d) 22

Ans. (d)

Sol. D50 = 1 mm, D20 = 0.5 mm, D10 = 0.08 mm

The grain size distribution of the backfillmaterial is an important factor that controlsthe rate of densification

Brown (1997) has defined a quantity calledthe suitability number for rating backfill as

N 2 2 250 20 10

3 1 1S 1.7D D D

Where, D50, D20 and D10 are the diameters(in mm) through which respectively 50, 20,10% of the material passes.

N 2 2 23 1 1S 1.7 21.7 221 0.5 0.08

112. The porosity of a soil n is

(a)e

1 e (b)e

1 e

(c)e 1

e

(d)e 1

e

where: e = Void ratio

Ans. (a)

Sol.en

1 e

113. A coarse-grained soil has a void ratio of 0.78and specific gravity as 2.67. The criticalgradient at which a quick sand conditionoccurs will be

(a) 0.62 (b) 0.74

(c) 0.82 (d) 0.94

Ans. (d)Sol. Gs = 2.67, e = 0.78

Critical gradient ic = G 11 e

= 2.67 1 1.67 0.941 0.78 1.78

114. Which of the following assumptions of theRankine theory of lateral earth pressure arecorrect?1. The soil mass is semi-infinite,

homogeneous, dry and cohesionless.2. The ground surface is a plane which may

be horizontal or inclined.3. The wall yields about the base and thus

satisfies the deformation condition forplastic equilibrium.

(a) 1 and 2 only (b) 1 and 3 only(c) 1, 2 and 3 (d) 2 and 3 only

Ans. (b)Sol. Assumptions in Rankines theory

1. Soil is semi-inf inite, homogenous,isotropic, dry and cohesionless.

2. The ground surface (backfill soil) ishorizontal.

3. The wall yields about the base and thussatisfies the deformation condition forplastic equilibrium.

CIVIL ENGINEERING




IES M

ASTER

115. The ratio of the horizontal stress to the verticalstress is called coefficient of

(a) Active earth pressure

(b) Passive earth pressure

(c) Earth pressure

(d) Plastic earth pressure

Ans. (c)

Sol. Earth pressure = Horizontal stressVertical stres

116. A bed consists of compressible clay of 4 mthickness with pervious sand on top andimperv ious rock at the bottom. In aconsolidation test on an undisturbed specimenof clay from this deposit 90% settlement wasreached in 4 hours. The specimen was 20mm thick. The time for the building foundedover this deposit to reach 90% of its finalsettlement will be

(a) 91 years (b) 82 years

(c) 73 years (d) 64 years

Ans. (c)Sol. For deposit H1 = 4 m

For clay specimen H2 = 20 mm 10 mm

2

= 0.01 m

Time t2 = 4 hour

Since settlement is same in both cases.

121

tH =

222

tH

12

t4

= 24

(0.01)

t1 = 64 × 104 hours

= 464 10 years

365 24

= 73 years

117. A 30 cm square bearing plate settles by 8mm in the plate load test on cohesionless soilwhen the intensity of loading is 180 kN/m2.The settlement of a shallow foundation of 1.5m square under the same intensity of loadingwill be nearly(a) 30 mm (b) 26 mm(c) 22 mm (d) 18 mm

Ans. (c)Sol. BP = 30 cm = 0.3 m, SP = 8 mm

Bf = 1.5 mSettlement of foundation

2f f P

P P P

S B (B 0.3)S B (B 0.3)

fS8

= 21.5 (0.3 0.3)

0.3 (1.5 0.3)

Sf = 22.22 mm

118. When the observed value of N exceeds 15,the corrected penetration number Nc as perTerzaghi and Peck recommendation in the siltyfine sands will be

(a) R115 (N 15)2

(b) R115 (N 15)2

(c) R115 (N 15)2

(d) R115 (N 15)2

where, N = Penetration number, and NR =Recorded value

Ans. (c)Sol. As per Terzaghi and peck, 1948

Corrected penetration number

Nc = R115 (N 15)2

CIVIL ENGINEERING




IES M

ASTER

119. A canal of 4 m deep has side slopes of 1 : 1.The properties of the soil are c = 15 kN/m2, = 15°, e = 0.76 and G = 2.7. Taylor’s stabilitynumber for that sudden drawdown = 0.136.The factor of safety with respect to cohesionin the case of sudden drawdown will be

(a) 0.64 (b) 1.43

(c) 2.22 (d) 3.01

Ans. (b)Sol. H = 4m , c = 15 kN/m2, e = 0.76. G = 2.7

Sn = 0.136

In case of sudden drawndown, soil will besaturated.

sat = w(G e)(1 e)

=

2.7 0.76 9.811 0.76

= 19.28 kN/m3

Taylor’s stability number

Sn = c

CF H

Fc = n

CS H =

15 1.430.136 19.28 4

120. The stability or shear strength of finegrainedsoils can be increased by draining them withthe passage of direct current through them.This process is known as

(a) Electro-osmosis

(b) Zeta potential

(c) Electro-chemical hardening

(d) Consolidation

Ans. (a)Sol. The stability or shear strength of fine-grained

soils can be increased by draining them withthe passage of direct current through them.This process is known as electro-osmosis.

121. The combined correction for curvature andrefraction for a distance of 3400 m will benearly

(a) 0.2 m (b) 0.4 m

(c) 0.6 m (d) 0.8

Ans. (d)Sol. Combined correction = –0.0673 × d2

= –0.778 m

122. A 100 m tape is suspended between the endsunder a pull of 200 N. If the weight of thetape is 30 N, the correct distance betweenthe tape ends will be nearly

(a) 100.5 m (b) 100.3 m

(c) 100.1 m (d) 99.9 m

Ans. (d)

Sol. Correction for sag, Cs = 2

2w

24 Pl

Cs = 2

230 10024 200

Cs = –0.094 m

Correct length = M.L + Cs

= 100 – 0.094 = 99.906 m

123. In horizontal distance measurement, the basicformula for distance in stadia tacheometry hasan additive constant. An anallatic lens isinserted in the tacheometer to make thisadditive constant zero. This lens is

(a) Convex lens inserted between objectglass and diaphragm

(b) Plano-convex lens between object glassand diaphragm

(c) Plano-convex lens between diaphragmand eye piece

(d) Convex lens inserted between diaphragmand eye piece

https://iesmaster.org/results/ese

CIVIL ENGINEERING




IES M

ASTER

Ans. (b)Sol. Anallatic lens is plano-convex lens between

objective and diaphragm.

124. If the LMT is 8h 12m 16s AM at 38° 45 Wlongitude, the GMT will be

(a) h m s11 12 16 AM

(b) h m s10 47 16 AM

(c) h m s9 29 46 AM

(d) h m s5 29 46 AM

Ans. (b)

Sol. LMT at 38° 45 w = 8 hr min 16 sec

We know that when we travel towards easttime is added and in west vice-versa.

For 15 0 change in longitude time changes1 hr.

GMT = 38 458hr 12 min15 sec

15

= 8 hr 12 min 16 sec + 2.5833 hr

= 8 hr 12 min 16 sec + 2 hr 35 min

= 10 hr 47 min 16 sec

125. A section line AB appears to be 10.16 cm ona photograph for which the focal length is 16cm. The corresponding line measures 2.54

cm on a map, which is to a scale 1

50,000 .

The terrain has an average elevation of 200m above mean sea level. The flying altitudeof the aircraft above mean sea level duringphotograph will be

(a) 1800 m (b) 200 m

(c) 2200 m (d) 2400 m

Ans. (c)

Sol. Sphoto = ab fAB H h

Smap = 21 2.54 10

50,000 AB

AB = 2.54 × 10–2 50,000 = 1270 m

Sphoto = 10.16 161270 H 200

H = 2200 m

126. If backsight and foresight distances arebalanced

1. The difference in elevation between twopoints can be directly calculated by takingdifference of the two readings.

2. No correction for the inclination of the lineof sight is necessary


(c) Both 1 and 2 (d) Neither 1 nor 2

Ans. (c)Sol. If back sight and foresight distances are

balanced then no correction is required forcollimation, curvature and refraction.

Staff readings will be correct and heightdifference can be determined as difference ofstaff readings.

127. A railway curve of 1350 m radius is to be setout to connect two tangents. If the designspeed is 110 kmph and the rate of change ofacceleration is 0.3 m/s3, the shift of the circularcurve will be nearly

(a) 0.18 m (b) 0.16 m

(c) 0.14 m (d) 0.12 m

Ans. (b)

Sol. Length of transition curve = 3v

CR

CIVIL ENGINEERING




IES M

ASTER

=

3511018 70.44 m

0.3 1350

Shift = 2 2sL 70.44

24 R 24 1350

= 0.153 m = 0.16 m

128. A theodolite is called a transit theodolite, whenits telescope can be revolved through acomplete revolution about its

(a) Vertical axis in an inclined plane

(b) Horizontal axis in an inclined plane

(c) Vertical axis in a horizontal plane

(d) Horizontal axis in a vertical plane

Ans. (d)Sol. Transiting is the process of turning the

telescope over the horizontal axis through180° in a vertical plane, making it upside downand pointing in the opposite direction. It isalso called reversing or plunging.

129. Stalactites and stalagmites are features of

(a) Stream erorsion developed in limestoneregion by specific chemical reaction

(b) Ground water deposition in caves formedby precipitation from dripping water richin calcium carbonate

(c) Marine erorsion and deposition formedalong coastal regions by selective erorsionfollowed by deposition by waves

(d) A centripetal drainage in which streamsfrom different directions flow towards acommon central basin

Ans. (b)Sol. Cave features are usually formed by slow

moving ground water that has a high calciumcarbonate content. Chemical changes insidethe cave make the minerals harden and form

deposits, such as stalactites (which hangsfrom ceiling) and stalagmites (which rise upfrom the ground).

130. Which of the following statements withreference to isogonic line are correct inmagnetic declination?

1. It is drawn through the points of samedeclination

2. It does not form complete great circle

3. It radiates from north and south magneticregions and follows irregular paths


(c) 2 and 3 only (d) 1, 2 and 3

Ans. (d)Sol. Isogonic are the lines joining points of equal

declination at a time of observation.

They do not form complete great circles.

The radiate from south and north magneticregion and they are quite irregular neargeographic poles.

131. Mountains resulting from the depression orelevation of blocks of the earth crust on alarge scale due to faulting and these elevatedstructures are commonly called

(a) Fault block mountains

(b) Volcanic mountains

(c) Relict mountains

(d) Residual mountains

Ans. (a)

132. A little gap is left between the head of theglaciated valley and the mass of the glacierice. This gap is known as

(a) Bergs-chrund (b) Arete

(c) Horn (d) Cirque

Ans. (a)

CIVIL ENGINEERING




IES M

ASTER

Sol. Bergs-chrund or Rimaye is a Clevasse thatforms where moving glacier ice separates fromstagnant ice.It is often a serious obstacle for mountaineers.Arete is narrow ridge of rock that separatestwo valleys.A horn results when glacier erode three ormore aretes, usually forming a sharp edgedpeak.Cirque are concave, circular basin carved bythe base of glacier as it erodes the landscape.

133. The sight distance available on a road to adriver at any instance depends on

1. Features of the road ahead2. Height of the driver’s eye above the road

surface3. Height of the object above the road

surface(a) 1 and 2 only (b) 1 and 3 only(c) 2 and 3 only (d) 1, 2 and 3

Ans. (d)Sol. Sight distance on road depends on :

(i) The characteristics/features of roadaheads.

(ii) Height of object above the road surface.

(iii) Height of driver’s eye above the roadsurface.

134. Consider the following data:

Design speed = 96 kmph

Speed of overtaken vehicle = 80 kmph

Reaction time for overtaking = 2 sec

Acceleration = 2.5 kmph/sec

The safe overtaking sight distance on a two-way traffic road will be nearly

(a) 646 m (b) 556 m

(c) 466 m (d) 376 m

Ans. (a)Sol. Design speed = 96 kmph

Speed of overtaken vehicle = 80 kmph

Reaction time for overtaking = 2 sec

Acceleration = 2.5 kmph/sec

OSD = 0.278V × t + 0.278 × V × T + 2S

+ Vd × T × 0.278

where, V = Speed of overtaken vehicle inkmph

Vd = design speed or speed of vehicle inopposite direction (kmph)

T = Time taken in operation of overtaken

S = space headway

S = 0.2 × V + 6.1 = 0.2 × 0.80 + 6.1 = 22.1m

T = 4S 22.1 4

5a 2.518

= 11.28 sec

OSD = 0.278 × 80 × 2 + 0.278 × 80 × 11.28

+ 2 × 22.1 + 0.278 × 96 × 11.28

= 44.48 + 250.87 + 44.2 + 301.04

OSD = 640.59 m

Nearest answer is (a)

135. Which one of the following statements iscorrect?

(a) The ratio of load on wheel to contact areaor area of imprint is called as contactpressure

(b) The ratio of load on wheel to contactpressure is called as rigidity factor

(c) The value of rigidity factor is more thanthree for an average tyre pressure of 7kg/cm2

(d) Rigidity factor does not depend upon thedegree of tension developed in walls oftyres

CIVIL ENGINEERING




IES M

ASTER

Ans. (a)

Sol. Contact pressure = Load on wheel

Contact area of imprint

Rigidity factor = Constant pressure

Tyre pressure

The value of rigidity factor is unit (1.0) foran average tyre pressure of 7 kg/cm2

Rigidity factor depends upon the degreeof tension developed in the walls of tyres.

136. Which one of the following is not the correcttype of critical load position in pavement slabdesign for the load on the pavement surface?(a) Interior loading(b) Edge loading(c) Eccentric loading(d) Corner loading

Ans. (c)Sol. Westergaard considered three load positions

as critical i.e. interior, edge and corner.

137. Which of the following statements are correctregarding Westergaard’s concept fortemperature stresses?1. During the day, the top of the pavement

slab gets heated under the sun light whenthe bottom of the slab becomes hot.

2. During summer season as the meantemperature of the slab increases, theconcrete pavement expands towards theexpansion joints

3. Due to frictional force at the interface,compressive stress is developed at thebottom of the slab as it tends to expand


(c) 1 and 3 only (d) 1, 2 and 3

Ans. (b)Sol. During the day the top of the pavement slab

gets heated under the sunlight when thebottom of slab still remains relatively colder.

138. Which of the following advantages are correctregarding Poly-centric shape tunnel?1. It can be conveniently used for road and

railway traffic2. The number of centres and lengths3. It can resist external and internal pressure

due to its arch action(a) 1 and 2 only (b) 2 and 3 only(c) 1 and 3 only (d) 1, 2 and 3

Ans. (c)Sol. Polycentric tunnels are those tunnel which

has more than one centre and flat base.

It can be used for road and railway traffic.

It can resist external and internal pressurewith their arch shape.

139. Which one of the following statement is correctregarding Journal friction?(a) Caused due to the wave action of rails(b) The amount does not depend upon the

type of bearing(c) For roller bearings, it varies from 0.5 to

1.0 kg per tonne(d) For coupled boxes, it lubricates by hard

grease from 0.5 kg to 1.0 kg per tonneAns. (c)Sol. Journal friction : It is the resistance

considered which are independent of speed.It depends on the type of bearing, the lubricantused, the temperature and condition ofbearing. In the case of roll bearing, it variesfrom 0.5 kg to 1.0 kg per tonne.

140. For the construction of a 640 m long B.G.railway track by using a sleeper density of M+ 5, and the length of each rail is 12.8 m, thenumber of sleepers required will be

(a) 1000 (b) 900

(c) 800 (d) 700

Ans. (b)

https://iesmaster.org/results/gate

CIVIL ENGINEERING




IES M

ASTER

Sol. Sleeper density = M + 5

No. of sleeper in 12.8 m length of rail

= 12.8 + 5 = 17.8 18

No. of sleeper in 640 m,

Long BG rail = 18 640

12.8 = 900

141. Which one of the following statement is correctregarding ballast used for railway tracks?

(a) The minimum depth of ballast for B.G.section is 20 cm - 25 cm

(b) The quantity of stone ballast required forone metre length of track is 0.53 m3 forB.G. section

(c) For M.G. section the width of ballast is1.83 m

(d) The minimum depth of ballast for N.G.section is 10 cm

Ans. (a)Sol. Minimum depth of ballast for SWR and LWR

is 200 mm and 250 mm respectively.

Quantity of stone ballast for BG per metervaries from 0.964 m3 to 2.616 m3.

Width of ballast for MG is

1.83

2.29

Ballast for MG track

142. Which one of the following statement iscorrect?(a) The radius of transition raises from infinity

to a selected minimum in order to attainful l super-elevation and curvaturegradually

(b) The compound curve is an arc of circle

(c) The radius of transition curve is constantfor entire length

(d) The horizontal curves are providedwhenever there is a change in gradient

Ans. (b)Sol. Transition curve is provided to decrease

the radius infinity at the straight road todesirable radius at the starting of circularcurve, in order to attain full super-elevationand curvature gradually.

Vertical curves are provided wheneverthere is a change in gradient.

143. A cross-over occur between two Metre Gaugeparallel tracks of same crossing number 1 in12 with straight intermediate portion betweenthe reverse curves and the distance betweenthe centres of tracks is 3.5 m. If the value ofG is 1 m, the intermediate straight distancewill be nearly(a) 12 m (b) 15 m(c) 18 m (d) 21 m

Ans. (c)Sol.

CLCL

D =

3.5

m

A

B

C

FE

Crossing number = 12

cot = 12

From ABC ,

cos = G

BC

CF = BF – BC = (D – G) – Gsec

CIVIL ENGINEERING




IES M

ASTER

From CEF ,

tan = CFFE

FE = CFcot

= D G Gsec cot

= 2D G G 1 tan N

= 2D G N G 1 N

DE = 23.5 1 12 1 1 12 18 m

144. Which one of the following is the correctstandard for provision of curves on railwaytrack?

(a) Cant excess on B.G. shall not be allowedto exceed 105 mm

(b) Minimum radius of vertical curves forgroup A, Broad Gauge track is 4000 m

(c) The minimum value of superelevation

according to Railway Board is 1 th

10 of

gauge

(d) The speed potential of curve is given by

formula E = 2V

127 R where E is

superelevation in mm.

Ans. (b)Sol.

Cant excess 75 mm

1 1th to th

10 12 is the range of maximum

superelevation.

Correct equation is E = 2GV

127 R .

Directions:Each of the next Six (06) items consists of twostatements, one labelled as the ‘Statement (I)’ andthe other as ‘Statement (II)’. You are to examinethese two statements carefully and select theanswers to these items using the codes givenbelow:Codes:(a) Both Statement (I) and Statement (II) are

individually true and statement (II) is thecorrect explanation of Statement (I).

(b) Both Statement (I) and Statement (II) areindividually true but Statement (II) is NOTthe correct explanation of Statement (I).

(c) Statement (I) is true but Statement (II) isfalse.

(d) Statement (I) is false but Statement (II) istrue.

145. Statement (I): Finer grinding of cement resultsin early development of strength.Statement(II): The finer the cement, the higheris the rate of hydration.

Ans. (a)146. Statement (I): Pozzolana is added to cement

to increase early strength.Statement (II): It reduces the heat ofhydration.

Ans. (d)Sol. Pozzolana lower the rate of development of

strength but ultimate strength is comparablewith ordinary portland cement.

147. Statement (I): Coarser the particles, less isoptimum moisture contant.

Statement (II): The specific surface area ofcoarser particle is less.

Ans. (a)

Sol. Coarser particles have less optimum moisturecontent because specific surface area ofcoarser particles are less.

CIVIL ENGINEERING




IES M

ASTER

148. Statement (I): A reverse curve consists of twoarcs with their centres of curvature on oppositesides of the curve.

Statement (II): Superelevation can beprovided conveniently at the intersection pointof the two arcs.

Ans. (c)

Sol. There will be no super-elevation at theintersection of two reverse curve.

A

D B

EC

FO1

O2

The outertrack AB will be higher than DE forone curve whereas in another curve the trackEF will be higher than BC.

149. Statement (I): The counter interval dependsupon the nature of the ground whether it isundulating or flat.Statement (II): In a hilly terrain or undulatingground a smaller interval is adopted, otherwise

the contours will come too close for plottingdue to the steep slope.

Ans. (c)Sol. Contour interval depends on following factors:

1. Scale of map

2. Purpose of map

3. nature of ground

4. Time

5. Fund

For flat ground, countour interval is small, butfor steep slope contour interval is large.

150. Statement (I): Geodetic survey cannot bedone for works requiring high precision.

Statement (II): The curvature of earth isaccounted for measurements in Geodeticsurvey.

Ans. (d)Sol. Geodetic survey accounts for curvature of

earth whereas plane surveying considers earthas flat.

For survey works requiring high precision hasto be done by Geodetic survey.

CIVIL ENGINEERING · 2020-01-07 · CIVIL ENGINEERING F-126, Katwaria Sarai, New Delhi - 110 016...

Documents

Transcript of CIVIL ENGINEERING · 2020-01-07 · CIVIL ENGINEERING F-126, Katwaria Sarai, New Delhi - 110 016...