civil enginering morning session gate 2014

8/12/2019 Civil Enginering Morning Session Gate 2014

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1. Ranjan was not happy that Sajan decided to do the project on his own on observing his unhappiness,

sajan explained to Rajan that he preferred to work independently

Which one of the statement below is logically valid and can be inferred from the above

sentences ?

1 mark

(a) Rajan has decided to work only is group.

(b) Rajan and Sajan were formed into a group against their wishes.

(c) Sajan decided to give into Rajans request to work with him.

(d) Rajan had believed that Sajan and he would be working together.

Sol. (d)

2. I f y = 5x2+3 than the tangent at x = 0, y = 3

1 mark

(a) passes through x = 0, y = 0

(b) has a slope +1

(c) is paralled to x-axis

(d) has a slope of 1

Sol. : (c)

y =5x2+ 3

dy

dx=10x

tangent at x =0, y = 3 is equal to zero

tangent is parallel to x-axis.

3. A student is required to demonstrate a high level of comprehension the subject, espesially in the

social sciences.

The word closest in meaning to comprehension is

1 mark

(a) understanding (b)Meaning

(c) Concentration (d) Stability

Sol. (a)

4. Choose the most appropirate word from the option given below to complete the following sentences,


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one of his biggest ................... was his ability to forgive

1 mark

(a) vice (b) virtues

(c) choices (d) strength

Sol. (b)

5. A boundary has a fixed daily cost of Rs 50,000 whenever it operates and variable cost of Rs. 8000

Q, where Q is the daily production in tonnes. What is the cost of production in Rs. per tonne for

a daily production of 100 tones.

1 mark

Sol. :

Total cost for 100 tonn

=8000 100 + 50000

=8,50,000

Cost per tonn =850000

Rs.8500/tonn.=100

6. Find the odd one in the following group ALRVX, EPVZB, ITZDF, OYEIX

(a) ALRVX (b) EPVZB

(c) ITZDF (d) OYEIX

2 marks

Sol. (d)

7. One percent of the people of country X are taller than 6ft, 2 percent of the people of country Y

are taller thatn 6ft. There are thrice as many people in country X as in country Y. Taking both

countries together. What % of people taller than 6ft.

(a) 3 (b)2.5

(c) 1.5 (d)1.25

2 marks

Sol. (d)

Let the population of x = Nxand population of y = Ny

Nx= 3Ny


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Total no. of people taller than 6ft =yx

2NN 1

100 100

% of people taken than 6ft =

y y

y y

3N 2N100

1003N N

=1.25

8. The sum of eigen value matrix (M) is

When [ M] =

215 650 795

655 150 835

485 355 550

2 marks

(a) 915 (b) 1355

(c) 1640 (d) 2180

Sol. (a)

9. The smallest angle of a triangle is equal to two third of the smallest angle of a quadrilateral. The

ratio between the angle of the quadritateral is 3 : 4 : 5 : 6. The largest angle of the triangle is

twice its smallest angle what is the sum, in degrees of the second largest angle of the triangle

and the largest angle of the quadritateral.

2 marks

Sol. :

3

2

1

3

5

4

6 18 360=

20=

largest angle of quadrilateral =120

smallest angle of quadrilateral = 60

Smallest angle of triangle = 2

40=3 203

Largest angle of triangle =2 40 = 80

Third angle of triangle =60


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Sum of largest angle of quadrilateral and second largest angle of triangle.

=120 + 60 = 180

10. The monthly rainfall chart based on 50 years of rainfall in Agra in shown in the following figure

which of the following are true ? (K percentile is the value such that K % of data fall below that

value)

800

700

600

500

400

300

200

100

Feb Mar Apr May une uly Aug Sep Oct Nov Dec

650

600

400

200

100

50

Rainfall (mm)

650

(i) On average it rains more in J uly than in Dec.

(ii) Every year, the amount of rainfall in August is more than that in J anuary.

(iii) J uly rainfall can be estimated with better confidence than Feb. rainfall.

(iv) In Aug, there is at least 500 mm of rainfall.

(a) (i) and (ii) (b) (i) and (iii)

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

Sol. (b)

11. Some of the non-toxic metal normally found in natural water are :

(a) Arsenic, Lead, Mercury (b)Calcium, Sodium, Silver

(c) Cadmium, curomium, copper (d) Iron, Mangnese, Magnesium

Sol. (d)

12. An incompressible homogeneous fluid flowing steadily in a variable dia pipe having the large and


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14. The minimum value of 15 minutes peak hour factor on a section of a road is

(a) 0.1 (b) 0.2

(c) 0.25 (d) 0.33

1 mark

Sol. : (c)

15 min peak hr factor is used for traffic intersection design

PHF =

15

V/4V

V =Peak hourly volumeveh.

inhr

V15 =Max 15 Min Vol. within the peak hr (veh.)

Max value is 1.0 & min value is 0.25

Normal range is 0.7 0.98

Ans is 0.25

15. the ultimate collapse load (Pu ) is terms of plastic moment Mpby kinematic approach for a propped

cantilever of length L with P acting at its mid span as shown in fig would be

L/2 L/2

(a) P =p2M

L(b)P =

p4M

L

(c)p6M

L(d)

p8M

L

1 mark

Sol. : (c)

2

L/2 L/2

MP

Pu


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L

2 =

External work done =load defleciton

= uP

= uL

P2

Internal work done = p pM M 2

= p3M

By principle of virtual work,

u

LP

2

= p3M

uP =p

6M

L

16. for a saturated cohesive soil, a triaxial test yeilds the angle of interval friction ( ) as zero. The

conducted test is

(a) Consolidated Drained (CD) test

(b) Consolidated undrain (CU) test

(c) Unconfined compresion (UC) test

(d) Uncosolidated undrain (UU) test

1 mark

Sol. : (d)

17. The degree of static indeterminacy of a rigid jointed frame PQR supported as shown is

1 mark


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y

R

EI

Q

90

S

Px

EI

145

Sol. : (d) (Unstable)

18. The action of negative friction on the pile is to

(a) Increase the ultimate load on the pile

(b) Reduce the allowable load on the pile

(c) Maintain the working load on the pile

(d) Reduce the settlement

1 mark

Sol.(b)

19. In a beam of length L four influence line diagram for shear at a section located at a distance

of L /4 from the left support marked (P, Q, R, S) are shown below the correct, I LD is .....

(a)

0.75

L/4 3 L/4

P0.25

(b)

0.6

L/4

0.6

3 L/4

Q

(c)

0.5

L/4 3L/4

0.5R

(d) L/4 3 L/4S

A (P), B (Q) C (R) D (S)

Sol. : (a)


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y

x

L/4

3L/4

xL

4

=y

3L

4

x =y

3

also x+y =1

y

y3

=1

4y

3=1

y =3

4

x = 14

Ans is (a)

20. A conventional flow duration curve is a plot between 1-marks

(a) Flow and % time flow is exceeded

(b) Duration of flooding and ground level elevation

(c) Duration of water supply in a city and proportion of area recurring supply exceeding this

duration.

(d) Flow rate and duration of time taken to empty of a reservoir at that flow rate.

1-marks

Sol. (a)

21. The following statements are related to temperature stress developed in concrete pavement slab

with for edge (without any restrain)

P : The temperature stress will be zero during both day and night time if the pavement slab

is considered weight less.


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Q. : The temp stress will be compressive at the bottom of the slab during night time if the self

weight of the pavement slab is considered.

R : The temp stress will be compressive at the bottom of the slab during day time if the self

weight of the pavement slab is considered.

1 mark

The Ture statements is/are :

(a) P only (b) Q only

(c) P and Q only (d) P and R only.

Sol. : (c)

22.x

x sinxlim

x

is equal to

1 marks

(a) (b) 0

(c) 1 (d)

Sol. : (a)

x

x sinxlim

x

=x

sinxlim 1

x

=1

23. A steel section is subjected to a combination of shear and bending action. The applied shear force

is V and shear capacity of the section is Vs. For such a section, high shear force (as per IS 800

- 2007) is defined as

1 mark

(a) V > 0.6Vs (b) V > 0.7Vs

(c) V > 0.8 Vs (d) V > 0

Sol. : (a)

Clause 9.2.1 IS 800:2007

24. The dimension for kinematic viscosity is :

(a)L

MT(b) 2

L

T


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(c)

2L

T (d)ML

T

Sol. : (c)

SI unit is m2/s

25. In reservoir with an uncontrolled spillway the peak of the plotted outflow hydrogroph

1 mark

(a) Lies outside the plotted inflow hydrograph.

(b) Lies on the recession limb of the plotted inflow hydrograph.(c) Lies on the peak of the inflow hydrograph.

(d) is higher than peak of the plotted inflow hydrograph.

Sol. : (b)

inflow hydrograph

outflow hydrograph

Recession limb

26. For a sample of water with the ionic composition shown below, the Carbonate and Non-carbonate

hardness concentration (in mg/l as CaCO3) respectively are.

meq/l

meq/l

0 4 5 7

Ca2+ Mg

2+Na

+

HCO3 SO42

3.5 7

2 marks

(a) 200 and 500 (b) 175 and 75

(c) 75 and 175 (d) 50 and 200

Sol. :

Carbonate hardness =3.5 103g eq [if NCH is present sodium alkalimitry will be

absent i.e. NaHCO3absent]

=3.5 10350g

las CaCO3


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=175 mg/l as CaCO3

Non carbonate hardness = total hardness - carbonate hardness

Total hardness = 5 50 mg/l as CaCO3[total hardness is due to Ca2+and Mg2+]

= 250 mg/l as CaCO3

NCH = 250 175 = 75 mg/l as CaCO3

27. The perception - reaction time for a vechile travelling at 90 km/h, given the co-efficienc of

longitudinal friction of 0.35 and the stopping sight distance of 170 m(assume g = 9.81 m/s2) is

...... seconds.

2 marks

Sol. :

SSD =2V

Vt2g

170 =

25

905 18

90 t18 2 9.81 0.35

t =3.1594 sec.

28. A traffice surveying conducted on a road yield an average daily traffic count of 5000 vehicle. The

axle load distribution on the same road is given in the following table.

Axle Load(tones)

Frequency ofTraffic (f)

18141086

1020351520

the design period of the road is 15 years. The yearly Traffice growth rate is 7.5% the load safety

factor (LSF) is 1.3. I f the vehicle damage factor (VDF) is calculated from he above data, the design

traffic (In million standard axle load MSA) is

2 marks

Sol. :

Calculation of vehicle damage factor :


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VDF =

4 y 4 4 4

3 51 2 41 2 3 4 5

s 3 s s s

1 2 3 4 5

W WW W WV V V V VW W W W W

V V V V V

where Ws= standard axle load = 80 kN = 8.2 tonn

VDF =

4 4 4 4 418 14 10 8 6

10 20 35 15 208.2 8.2 8.2 8.2 8.2

10 20 35 15 20

=4.988.

N =15365 5000 (1.075) 1

4.9880.075

=237.758 MSA

Note : that LSF is used to calculate streses not for MSA calculation.

29. Mathematical idealization of a crane has three bar with their vertices arranged as shown with

load of 80 kN hanging vertically. The co-ordinate of the vertices are given in parenthesis. The

forces in members QR is ..................

2 marks

P (0,4)

80kN22.84

104.03

(3,0)

R

55.13(1, 0)

Q

x

y

Sol. :


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P (0,4)

80 kN22.84

104.03

(3,0)R

55.13

(1, 0)

Q

R

RR

3m

4m

1m

Q RR R 80

80 3 = RQ 2

RQ= 120 kN

RR= 40 kN

tan =1 1 4

sin , cos4 17 17

tan =3 4

Sin , cos4 5 5

120

A

A40

P

80

120

FPR

FQR

taking moment of all forces about P.

pM 0

120 1, + FQR 4 =0

FQR = 30 kN

30. The reinforced concrete section, the stress at extreme fibre in compression is 5.8 MPa. The depth

of Neutral Axis in the section is 58 mm and grade of concrete is M25. Assuming Linear elastic

behavior of the concrete, the effective curvature of the section (in per mm) is


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(a) 2 106 (b) 3 106

(c) 4 106 (d)5 1062 mark

Sol. :

5.8 MPa

58 mm

M25 conrature

M f E

I y R , E = 5000 25 25000 N/mm2

26

2

1 f 5.8N / mm4 10 per mm

R yE 58mm 25000N /mm

curvature = 4 106per mm

31. An isolated three-phase traffic signal is designed by websters method. The critical flow ratio for

three phase are 0.2, 0.3 and 0.25 respectively and lost time per phase is 4 second. The optimum

cycle length (in sec.) is ........

2 mark

Sol. :

Total lost time in a cycle L = 4 3 = 12 sec.

C =1.5L 5

1 y

:

1.5 12 592sec1 (0.2 0.3 0.25)

32. Group - I Group - II

P : Curve J (i) No apporent heaving of soil around the footing.

Q Curve K (ii) Rankine passive zone develops imperfectly

R : Curve L (ii i) Well defined slip surface extends to ground surface.


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Load

K L

Settlement

(a) P -Q, Q - 3, R - 2

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

(c) P - 1, Q-2, R - 3

Sol. : (d)

L General sheer failre

K local shear failre

J Punching sheer failre

33. 16 MLD of water is flowing through a 2.5 km long pipe of diameter 45 cm. The chlorine at the

rate of 32 kg/d is applied at the entry of this pipe so that disinfected water is obtained at the exit.

These is a proposal to increase the flow through the pipe to 22 ML D from 16 mLD. Assume the

dilution co-efficient n = 1. The minimum amount of chlorine (in kg per day) to be applied to

achieve the same degree of disinfection for the enhanced flow.

(a) 60.5 (b) 4.4

(c) 38 (d) 23.27

2 mark

Sol. : In the disinfection process we have the relationship,

tCn =K

where t = time required to kill all organism.

c = cocentration of disinfactant

n = dilution coefficient

k = constant

n1 1t C =n

2 2t C

in our case n = 1

1 1 2 2t C t C

t1 =1

L

v , L = length of pipe ; V1= velocity of flow

t1 =1

L

Q / A


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t1

=1

LA

Q

C1 =1

1

W

Q , where W1= weight of disinfactant per day ; Q1= discharge per day

1

1 1

WLA

Q Q =

2

2 2

WLA

Q Q

2W =22

121

QW

Q =

222

32kg/day16

= 60.5 kg/day

Ans. (a)

34. A levelling is carried out to established the reduced level (RL) of point R with respect to the bench

mark (BM) at P. The staff reading taken are given below

Staff station BS IS FS RL

P 1.655

Q 0.95 1.5

R 0.75

I f RL of P is + 100 m then RL (m) of R is

(a) 103.355 (b) 103.155

(c) 101.455 (d) 100.355 2 mark

Sol. : (c)

station BS FS

P 1.655

Q 0.95 1.5

R 0.75

RL of P =100

RL of Q =100+1.655+1.5

=103.155

RL of R =103.155 (0.95 + 0.75)

=103.155 1.70

RL of R =101.455

35. The flow net constructed for a dam is shown in the figure below. Taking coefficient of permeability

as 3.8 106 m/s, the quantity of flow (in cm3/sec) under the dam per m is


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50 m

6.3 m

1.6 m

17.2 m

9.4 m

2 mark

Sol. :

Q =

f

d

NKH

N

Nd =10

Nf =3

H =6.3 m

K =3.810-6m/s

Q =6 m 3

3.8 10 6.3ms 10

=3

6 m /s7.182 10

m

=3

6 6 cm /s7.182 10 10

m

Q =7.1823

cm /s

m

36. The tension and shear force (both in kN) in each both of the joint as shown below respectively

are


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4

53

P = 250 kNu

(a) 30.3 and 20 (b)30.33 and 25

(c) 33.33 and 20 (d)33.33 and 25

2 mark

Sol. : tan =3

4

cos =4

5

u4Pu

P cos5

Pu

u

3PuP sin

5

sin =3

5

Tension in each bolt =u4P

5 6

=4 250

6 5

= 33.33 kN

Shear in each bolt =u3P

5 6= 25 kN

Ans. (d)

37. An incompressible fluid is flown at steady rate in horizontal pipe. From a section the pipe divides

into two horizontal parallel pipes of diameter (d1and d2) that run, for a distance of L each and

then again join back to a pipe of the original size. For both the pipes, assume the head loss due

to friction only and the Darcy-weilbach friction factor to be same. The velocity ratio betweenbigger and smaller branched pipe is.

Sol. :


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B

D

C

d1

A

Q1

Q2d2

L

L

1d = 24d

21

1

flv

2gd=

22

2

flv

2gd

21

1

v

d =

22

2

v

d

2122

v

v=

1

2

d

d = 4

1

2

v

v =2

Ans. 2

38. A box of weight 100 kN shown in the figure to be lifted without swinging. If all the forces are

coplanar, the magnitude and direction ( ) of force F w.r.t. x axis is

30

45

40 kNF

90 y

100 kN

X dir

(a) F = 56.389 kN and = 28.28

(b) F= 56.389 kN and = 28.28(c) F = 9.055, = 1.1414

(d) F = 9.055, = 1.1414

Sol. : For no swinging HorizontalF 0


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3045

40 kN F

x-axis

90kN

100 kN

90 cos 30 = 40 cos 45 + F cos

49.658 F cos

F cos from option (A) = 56.389 cos 28.28 = 49.658 kN

Ans. (a)

39. The speed-density (v k) relationship on a single lane road with unidirectional flow is

v = 70 0.7 K, where v is in km/hr and k is in veh/km. The capacity of the road (veh/hr) is

2 mark

Sol. :

u 70 0.7 K

km veh

hr km

100 =kJ

v =70f

u

k

Capacity =f jV K

4=

km Veh70 100hr km

4

= 1750 veh/hr

Ans. 1750 veh/hr

40. For a cantilever beam of span 3m as shown a concentrated load of 20 kN applied at the free end

causes a vertical displacement of 2 mm at a section located at a distance of 1m from the fixed

end. I f a concentrated vertically downward load of 10 kN is applied at the section located at a

distance of 1 m from the fixed end (with no other load on the beam) the maximum vertical

displacement in the same beam (in mm) is


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2mm

20 kN

1m2

2 mark

Sol. :

20kN

2mm

1m 2m

1m10kN

2m

20 =10 2 (From Betts Therorem

=1 mmAns = 1mm

41. For a beam cross section W = 230 mm, effective depth = 500 mm, the number of rebars of 12

mm diameter required to satisfy minimum tension reinforcement requirement specified by

IS -456:2000 (assume grade of steel as Fe500) is ..............

Sol. :

stminA

bd=

y

0.85

f

Ast =20.85 230 500mm

500

n =

st2

A

d

4

= 2

0.85 2302nos

(12)4


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42. A venturimeter having diameter of 7.5 cm at the throat and 15 cm at the enlarged end is installedin a horizontal pipline of 15 cm diameter. The pipe carries incompressible fluid at steady rate of

36 l/s. The difference of pressure head measured in terms of the moving fluid in between the

enlarged and the throat of the vent is observed to be 2.45 m. Taking the g = 9.8, the coefficient

of discharge of venturimeter (correct upto 2 decimal) is .......

2 mark

Sol. :

15 cm

7.5 cm

1 2

21

Q =30 l/s

1 21 2

P PZ Z

=2.45 m = h

g =9.81

Q =1 2

d2 21 2

a aC 2gh

a a

Cd =1 2

2 21 2

Q

a a2gh

a a

=

3 3

2 2

4 4

30 30 m / s

(0.15) (0.075)4 2 9.81 2.45

(0.15) (0.075)

Cd =0.95

43. A particles moves along a curve whose parametric equation are x = t3+ 2t, y = 3e2tand z

= 2 sin (5t), whre x, y and z show variation of the distance covered by the particles in (cm) with

time (t) (in second). The magnitude of the acceleration of the particle (in cm/s2) at t = 0 is ......

2 mark


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Sol. :

x =t3+2t

y =3 e2t

z =2 sin (5t)

dx

dt= 3t2+ 2

ax=2

2

d x

dt=6t

dydt

=3 e2t (2) = 6e2t

ay =2

2

d y

dt=12 e2t

dz

dt=5 2 cos (5t)

2

2

d z

dt=10 5 sin (5t) = 50 sin 5t

az =

2

2

d z

dt =50 Sin 5t

a

= x y z a i a i a k

a

at t =0 = 0i 12j 0k

a

= 12j

Meg. of acceleration at t = 0 = 12 cm/s2

Ans =12 cm/s2

44. The full data are given for laboratory sample 0 175kPa , 0 0 0e 1.1, 300kPa, e 0.9 .I f thickness of the clay specimen is 25 mm the value of coeffcient of volume compressibility is

...... 104m2/kN.

2 mark

Sol. :

mv =0 v

0

e

1 e a

1 e


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=(1.1 0.9)

125

2.1

=0.2

2.1 125

=7.619 104m2/kN

Ans is = 7.619

45. A given cohesionless soil has emax= 0.85, emin= 0.5. In the field, the soil is compasted to a mass

density of 1800 kg/m3at w/c of 8%. Take the mass density of water as 1000 kg/m3and Gs= 2.7.

The relative density (in %) of the soil is

(a) 56.43 (b) 60.25

(c) 62.87 (d)65.41

2 mark

Sol. : (d)

emax =0.85

emin =0.5

field =1800 kg/m3at w/c = 8%

w =1000 kg/m3

Gs =2.7

Relative density = Dr= ?

= wG 2.7 10001 w 1.08=

1 e 1 e

1 + e =2.7 1000 1.08

1800

e =0.62

Dr =Relative density = maxmax min

e e100e e

=0.85 0.62

1000.85 0.5

=65.714%

46. I f the following equaiton establishes equilibrium in slightly bent position,


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2

2d ydx

+ py 0EI

the mid-span deflection of a member shown in figure is

P

M

yEI

P

NL

y

x

[a is the amplitude contant for y]

(a) y =1 2

1 acosP L

x

(b) y =1 2

1 asinP L

x

(c) y = asinn x /L (d)y = acosn / Lx

2 mark

Sol. : (c)

47. Three rigid bucket are of idential height and base area. Further assume that each of these

buckets have negeligiuble mass and are full of water. The weight of water in these bucket are

denoted by W1, W2, W3respectively. Also let the force of water on the base of the bucket be F 1,

F2, F3respectively. Which of the following option are correct.

hhh

All these bucket have the same area

(a) W2= W1= W3and F2> F1> F2

(b) W2> W1> W3and F2> F1> F3

(c) W2= W1= W3and F1= F2= F3

(d) W2

> W1

> W3

and F2

= F1

= F3

2 mark

Sol. (d)

Bucket identical height

identical Base area


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h h h

AA

1 2 3

2 1 3W W W

Force on the base in each case will be equal to = w hA

Hence F1= F2= F3

48. A horizontal jet of water with its cross section area 0.0028m2hits a fixed vertical plate with a

velocity of 5 m/s. After impact the jet split symmetrically in a plane parallel to the plane of the

plate. The force of impact (in N) of the jet on the plate is

(a) 90, (b)80

(c) 70 (d)60

2 mark

Sol. : (c)

5m/s

area of jet =0.0028m2

Force on plate = waV V

= 2wav = 1000 0.0028 (5)

2

=70 N

49. A rectangular beam of 230 mm width and effective depth = 450 mm, is reinforced with 4 bars

of 12 mm diameter. The grads of cencrete is M 20, grade of steel is Fe 500. Given that for M20

grade of concrete, the ultimate shear strength uc = 0.36 N/mm2for steel percentage of = 0.25,

and 2uc 0.48N / m for steel percentage =0.5. For a factored shear force of 45 kN, the diameter

(mm) of Fe500 steel 2 legged stirrups to be used at spacing of 325 mm should be

(a) 8 (b)10


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(c) 12 (d)16

2 marks

Sol. (a)

230

450

4 12

uc =0.36 N/mm2 [For M 20] for

stA 100 0.25bd

uc =0.48 N/mm2 [For M 20] for

stA 100 0.5bd

Factored SF =45 kN = Vu

We have to calculate the dia of Fe500 2-L egged stirrup to be used at a spacing of 325 mm c/

c.

v =uV

bd =

45 1000

230 450

= 0.4348 N/mm2

% tensile steel =

24 (12)4 100

230 450

=0.437 %

c =0.12

0.36 (0.437 0.25)0.25

=0.45 N/mm2

Since v c 0

Min shear reinforcement is required

Min sheer reinforcement is given by

sv

v

A

bS =0.4

0.87fy

Asv =v0.4 (S )(b)

0.87fy


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since we limit fyto 415 N/mm2hence,

Asv =22 ( )

4

=

20.4 325 230 82.814mm0.87 415

=7.26 mm

adopt =8 mm

50. A straight 100 m long raw water gravity main is to carry water from intake to the jackwell of

a water treatmented plant. . The required flow of w =0.25 m3/s. Allowable velocity through main

= 0.75 m/s. Assume f = 0.01, g = 9.81. The minimum gradient (in cm/100 length) required to

be given to this main so that water flow without any difficulty

Sol. :

Q =0.25 m3/s

Allowable velocity =0.75 m/s

f =0.01

g =9.81

2d

4

=

2Q 0.25 1mV 0.75 3

d =0.6515 m

2flv

2gd=hf=

20.01 100 (0.75)m

2 9.81 0.6515

= 0.044 m = 4.4 cm

Min gradent =fh 4.4cm

100m

l

hence ans is 4.4

51. Group I Group II

P Alidade

Q Arrow 1. Chain Survey

R Bubble tube 2. Levelling

S Stedia hair 3. Plant table surveying

4. Theodolite

(a) P 3 Q = 2, R1, S4.

(b) P2, Q4, R3, S-1

(c) P1, Q2, R4, S-3

(d) P3, Q1, R2, S-4


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Group I is 100/ ibs fruit colum group II is method of surv. Match.

Sol. (d)

P 3

Q 1

R 2

S 4

52. A traffic office impose on an average 5 number of penalties daily on traffic violators. Assume that

the number of penalties on different day is independent and follows a poisson distribution. The

probability that there will be less than 4 penalties in a day is .....

Sol. :

Mean = 5

P (x < 4) =p (x = 0) + p (x =1) + P (x = 2) + P (x = 3)

=5 0 5 1 5 2 5 3e s e 5 e 5 e 5

0! 1! 1! 3!

=5 25 125e 1 5

2 6

=5 118e

3

=0.265

53. The potable water is prepared from turbid surfaec water by adopting the foil treatment square.

(a) Turbid surface water Coagulation Flocculation Sedimentation Filteration disinfection storage and supply

(b) Turbid surface water disinfection Flocculation Sedimentation F ilteration Coagulation storage and supply

(c) Turbid surface water Filteration Sedimentation disinfection Flocculation Coagulation

(d) Turbid surface water Sedimentation Flocculation Coagulation disinfection Filteration.

Sol. (a)

54. A rectangular channel flow have bed slope of 0.0001 width = 3m coefficient n = 0.015, Q = 1 m3/

sec given that normal depth of flow ranges between 0.76 m and 0.8 m. The minimum width of


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throat (in m) that is possible at a given section while ensuring that the prevailing normal depth

does not exceed along the reach upstream of the contraction is approximately, equal to (assumenegligiable loss)

(a) 0.64 (b) 0.84

(c) 1.04 (d) 1.24

Sol. (b)

3mS = 0.00010

n = 0.015

Q = 1 m3/s

Normal depth of flow between 0.76 m to 0.8 m.

If prevailing normal depth of flow is not excreded, there must not be chocking of the section or

there must be just chocking.

Thus the width of the section should be such that for the prevailing sp. energy there should be

critical flow at the contracted section

i.e.

1/3

23 q2 g

=Ec= E initial

1/32

min

Q

B3

2 g

=E initial

Let is now calculate Einitial

Q =

2/3 1/2

0

1

AR Sn

1 =2/3

1/21 3y(3y) (0.0001)0.015 3 2y

y =0.78 m

E initial =y +2

2

q

2gy


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=

2

2

13

0.782 9.81 (0.78)

=0.7893 m

1/32

Q

Bmin3

2 g

=0.7893

2/3

2/31/3min

3 (1)

2g B=0.7893

Bmin =0.836 m

55. For the truss shown below, the member PQ is short by 3 mm. The magnitude of the vertical

displacement of joint R in mm is ............

QP

R

3m

4m 4m

Sol.

PQ is short by 3 mm

We have to find out vertical displacement of joint R in mm

R = u( )

Let in apply unit load at R as shown below


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R

1/21/2

uPR

R

uPR sin =1

2

uPQ+ uPR cos =0

uPQ = uPR cos

=1

.cos2sin

uPQ =1

cot2

=1 4/3

2

=2

3

R = PQ PQ2

u ( 3)3

= 2 mm upwards.

56. The probability density function of evaporation E on any day during a year in watershed is given

by

f(E) =

10 E 5 mm/day

50 otherwise

The probability that E lies in between 2 and 4 mm/day in a day in watershed is (in decimal)

Sol.

f(E) =

10 E 5 mm/day

50 otherwise


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P(2 < E < 4) =

4

2

f(E)dE = 4

42

2

1 1dE = E5 5

= 1

4 25

= 2/5

57. While designing for a steel column of Fe250 grade the base plate resting on a concrete pedestal

of M20 grade, the bearing strength of concrete (N/mm2) in LSM of design as per IS 456:2000 is

1 marks

Sol.

Permissible bearing stress

=0.45 fck

=0.45 20= 9 N/mm2

58. A long slope is formed in a soil with shear strength parameter C = 0, 34 .= F irm strate lies

below the slope and it is assumed that water table may occassionally rise to the surface, with

seepage taking place parallel to the slope. Use sat = 18 kN/m3 and w = 10 kN/m

3. The

maximum slope angle (in degree) to ensure the factor of safety 1.5. Assuming a potential failuresurface parallel to the slope would be

(a) 45.3 (b)44.7

(c) 12.3 (d)11.3

Sol.: (d)

3sat 18kN/m=

3w 10kN/m=

C 0=34=

F.O.S when water surface rises to surface is given by


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F.O.S = subsat

tantan

tan =sub

sat

tan

F.O.S

tan =sub

sat

tan

1.5

tan = 18 10 tan3418 1.5

=11.30

59. The amount of CO2generated in kg while completly oxidising one kg of CH4is .......

1 marks

Sol.( )

4 2 2 2O16g 44g

CH 2O CO 2H

16 g of CH4 when completely oxidised leads to 44 g of CO2

1 kg of CH4when completely oxidised leads to44

116

= 2.75 kg CO2

60. Given J =

13 2 1and K then=2 4 2 2

1 2 6 1

product KTJ K is

Sol.(a)

J =

3 2 12 4 21 2 6

, k =

121

KTJ K = 13 2 1

1 2 1 2 4 2 21 2 6 1


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= 1

6 8 1 21

=6 + 16 + 1

=23

61. With reference to the conventional cartesion (X, Y) the vertices of a triangle have x1, y1= 1, 0,

x2, y2= 2, 2, and x3, y3= 4, 3, the area of triangle is

(a)3

2(b)

3

4

(c)4

5(d)

5

2-1 marks

Sol. : (a)

A

B

a

c

C

(2, 2)

(4,3)

(1,0)

y

x

b

area of triangle is A = p(p a)(p p)(p c)

when P =a b c

2

a = 2 2(4 2) (2 1) 5

b = 2 2(4 1) (3) 3 2

c = 2 2(2 1) (2) 5

p =5 5 3 2 3

52 2


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A = 3 3 3 35 5 3 22 2 2 2

=

23 2 3 3

5 522 2 2

62. The degree of disturbances of a sample collected by sampler is expressed by a term called the area

ratio. I f outer diameter and inner dia of sample are D0and Direspectively, the area ratio is-

(a)

2 2

0 i2

i

D D

D

(b)

2 2

i 02

i

D D

D

(c)

2 2

0 i2

0

D D

D

(d)

2 2

i 02

0

D D

D

Sol. : (a)


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Vikas AhlawatGATE - 2013 & ESE - 2012

Landmark

Achievement

of

invery

of

in

ESE-2012

All India

1st RankFirst Year

IES Master

Divey Sethi

AIR - 34 (CE)

Bharat Bhushan

AIR - 35 (CE)

Manohar

AIR - 3 (CE)

Swapnil Chaurasia

AIR - 6 (CE)

Abh ina v Garg

AIR - 20 (CE)

Amit esh Pr iyadar shi

AIR - 43 (CE)

Nilesh Kumar

AIR - 105 (CE)

Sumit Garg

AIR - 114 (CE)

Vishal Mehta

AIR - 139 (CE)

Bharat Bhushan

AIR - 159 (CE)

and many more....

Inderjeet Kaushik

AIR2CE

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Rohit Sharma

AIR16ME

Nilesh Kumar

AIR25CE

AIR47

CE

Hemant Kumar

AIR58

AIR60

AIR61

AIR63

CE CE ME CE

Vivek Nandan Avinash Mishra Ranjeet Kumar Manish Kanodia

AIR68

CE

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AIR77

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CE CE

Raj at Kr. Shar ma Ro hi t J or wal

AIR95

CE

Vivek Kushwaha

AIR88

CE

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AIR96

CE

Manish Garg

AIR111

CE

Divey SethiKr.

AIR116

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AIR134

CE CE CE

Rakesh Uday Roopesh D

AIR147

CE

AIR155

ME

Rahul Meena

AIR159

CE

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And the list conti nues.....

AIR161

CE

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CE

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AIR162

CE

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