ce gate 2015 morning session

7/21/2019 Ce Gate 2015 Morning Session

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8 t h F e b ru a r y, 2 0 15Exam Solutions (Civil)

(Morning Session)



Office Address : F-126, Katwaria Sarai, New Delhi - 110 016.

Telephone : 011-41013406, Mobile : 8130909220, 9711853908, 7838813406

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

I E S M A S T

E R1. A function f(x) is linear and has a value of 29 at x = – 2 and 39 at x = 3. Find its value at x = 5.

(A) 59 (B) 45 (C) 43 (D) 35

Sol. (c)

f (x) is linear

Let f (x) = ax + b

f (– 2) = 29

– 2a + b = 29 ...(i)

f (3) = 39

3a + b = 39 ...(ii)from (i) and (ii)

a = 2, b = 33

f (x) = 2x + 33

f (5) = 2(5) + 33 = 43

2. Extreme focus on syllabus and studying for tests has become such a dominant concern of Indian

students that they close their minds to anything______to the requirements of the exam.

(A) related (B) extraneous (C) outside (D) useful

Sol. (b)

3. The Tamil version of ______John Abraham-starrer Madras Cafe _____cleared by the Censor Board

with no cuts last week but the film’s distributors______no takers among the exhibitors for a release in

Tamil Nadu____this Friday.

(A) Mr. was, found, on (B) a, was, found, at (C) the, was, found, on (D) a, being, find at

Sol. (c)

4. Select the pair that best expresses a relationship similar to that expressed in the pair:

Children : Pediatrician

(A) Adult : Orthopaedist (B) Females : Gynaecologist

(C) Kidney : Nephrologist (D) Skin : Dermatologist

Sol. (b)

5. If ROAD is written as URDG then SWAN should be written as

(A) VXDQ (B) VZDQ (C) VZDP (D) UXDQ



Office Address : F-126, Katwaria Sarai, New Delhi - 110 016.Telephone : 011-41013406, Mobile : 8130909220, 9711853908, 7838813406


I E S M

A S T E

R

Sol. (b)

R O A DU R D G

The alphabets are shifted by ‘+3’

S W A N

V Z D Q

6. The exports and imports (in crores of Rs.) of a country from the year 2000 to 2007 are given in the

following bar chart. In which year is the combined percentage increase in imports and exports the

highest?

120

110100

90

80

70

60

50

40

30

20

10

0

2000 2001 2002 2003 2004 2005 2006 2007

Sol. (2006)

YEAR PERCENTAGE

2001110 90

100 22.22%90

=

2002 0

130 110100 1818

110

=

2003 00

2004 0

150 130100 15.38

130

=

2005 0

160 150100 6.67

150

=

2006 0

220 160100 37.5

100

=

2007 0

230 220100 4.54

220

=

The year 2006 shows the highest growth.






I E S M A S T

E R7. Choose the most appropriate equation for the function drawn as a thick line, in the plot below.

(0, –1)

(2, 0) =

y

(A) x = y – |y| (B) x = – (y – |y|) (C) x = y – |y| (D) x = – (y + |y|)

Sol. (b)

y = 0 at x = 0

y = – 1 at x = 2

By options we can can see that these points satisfies the only equation

x = (y – |y|)

8. Most experts feel that in spite of possessing all the technical skills required to be a batsman of the

highest order, he is unlikely to be so due to lack requisite temperament. He was guilty of throwing away

his wicket several times after working hard to lay a strong foundatin. His critics pointed out that until

he addressed that problem, success at the highest level will continue to elude him.

Which of the statement(s) below is/are logically valid and can be inferred from the above passage?

(i) He was already a successful batsman at the highest level

(ii) He has to improve his temperament in order to become a great batsman

(iii) He failed to make many of his good starts count.

(iv) Improving has technical skills will guarantee success

(A) (iii), and (iv) (b) (ii) and (iii) (C) (i), (ii) and (iii) (D) (ii) only

Sol. (b)

9. Alexendar turned his attention towards India, since he had conquered Persia.

Which one of the statements below is logically valid and can be inferred from the above sentence?

(A) Alexander would not have turned his attention towards India has he not conquered Persia.

(B) Alexander was not ready to rest on his laurels and wanted to march to India.

(C) Alexander was completely in control of his army and could command it to move towards India.(D) Since Alexander’s kingdom extended to Indian borders after the conquest of Persia, he was keen

to move further

Sol. (a)

10. The head of a newly formed government desires to appoint five of the six selected members P, Q, R,

S, T, and U to portfolios of Home, Power, Defense, Telecom and Finance. U does not want any portfolio





I E S M

A S T E

R

if S gets one of the five R wants either Home or Finanace or no portfolio. Q says tht if S gets either

Power or Telecom, then she must get the other one. T insists on a portfolio if P gets one.

Which is the valid distribution of portfoliors?

(A) P-Home, Q-Power, R-Defense, S-Telecom, T-Finance

(B) R-Home, S-Power, P-Defense, Q-Telecom, T-Finance

(C) P-Home, Q-Power, T-Defense, S-Telecom, U-Finance

(D) Q-Home, U-Power, T-Defense, R-Telecom, P-Finance

Sol. (b)

We have been given, R wants flome or finance or no portfolio. Hence, option A & D are eliminated since

R has been given defence and Telecom in those. Option c is eliminated since U has given a portfolio

white S has also given one. [ U does not want any portfolio if S gets one]

Option (b) satisfies the conditions stated for the portfolios.

Section Name Civil Engg. and Maths

1. The penetration value of a bitumen sample tested at 25°C is 80. When this sample is heated to 60°C

and tested again the needle of the penetration test apparatus penetrates the bitumen sample by d mm.

The value of d CANNOT be less than ______ mm.

Sol. (8 mm)

Relationship between penetration & temperature is given byP = AT + K

where P = penetration

A = temperature susceptibility

K = constant

However, value of A varies considerably between 0.0015 to 0.06

Hence for simplicity, we will simply take d not to be less then

80

10= 8mm

2. In an unconsolidated undrained triaxial test, it is observed that an increase in cell pressure from 150

kPa to 250 kPa leads to a pore pressure increase of 80 kPa. It is further observed that, an increaseof 50 kPa in deviatoric stress results in an increase of 25 kPa in the pore pressure. The value of

Skempton’s pore pressure parameter B is:

(A) 0.5 (B) 0.625 (C) 0.8 (D) 1.0

Sol. (c)

U = 3 1 3B A






I E S M A S T

E Rduring cell pressure applicatin, 1 = 3

hence, 1 = 3

U = 3B

80 = B(100)

B = 0.8

3. In a two-dimensional steady flow field, in a certain region of the x-y plane the velocity component in

the x-direction is given by vx = x2 and the density varies as1

x = Which of the following is a valid

expression for the velocity component in the y-direction, v y?(A) vy = – x/y (B) vy = x/y (C) vy = –xy (D) vy = xy

Sol. (c)

Continuity equation for steady flow is given by

u v

x y

= 0

21 1x v

x xx y

= 0

vx1y

= 0

v

xy

= –1

v

y

= – x

v = –xy + f(x)

4. In a closed loop traverse of 1 km total length, the closing errors in departure and latitude are 0.3 mand 0.4 m respectively. The relative precision of this traverse will be

(A) 1:5000 (B) 1:4000 (C) 1:3000 (D) 1:2000

Sol. (d)

Relative precision =1

p/e





I E S M

A S T E

R

where,

p = Perimeter of traverse

e = Closing error

e = 2 20.3 0.4 0.5 =

Relative precision =1 1

1000 20000.5

=

5. Two triangular wedges are glued together as shown in the following figure. The stress acting normal

to the interface, n is ______MPa.

100 MPa

100 MPa100 MPa

100 MPa

45°

n

Sol. (0)

Mohr circle for the given condition is

2 = 90°

(100, 0)

(0, 100)

(–100, 0)

(0, –100)

Here, normal stress is zero at 45° to the principal plane.

6. Total Kjeldahl Nitrogen (TKN) concentration (mg/L as N) in domestic sewage is the sum of the con-

centrations of

(A) organic and inorganic nitrogen in sewage

(B) organic nitrogen and nitrate in sewage






I E S M A S T

E R(C) organic nitrogen and ammonia in sewage

(D) ammonia and nitrate in sewage

Sol. (c) Kjehldahl Nitrogen = organic nitrogen + free ammonia

7. Which of the following statements CANNOT be used to describe free flow speed (uf ) of a traffic stream?

(A) uf is the speed when flow is negligible

(B) uf is the speed when density is negligible

(C) uf is affected by geometry and surface conditions of the road

(D) uf is the speed at which flow is maximum and density is optimum

Sol. (d)

Vf

K j

qmax

K jK/2 j

qmax

Flow is negligible at free flow speed.

Density is negligible or almost zero at free flow speed.

8. The development length of a deformed reinforcement bar can be expressed as (1/k) s bd( / ). From

the IS: 456-2000, the value of k can be calcualted as ______.

Sol. (6.4)

2y0.87f

4 = d bdL

Ld =y

bd

0.7f

4 1.6

=S

bd6.4

(For deformed bar, increase bd by 60%)





I E S M

A S T E

R

9. Which of the following statements is TRUE for degree of disturbance of collected soil sample?

(A) Thinner the sampler wall, lower the degree of disturbance of collected soil sample

(B) Thicker the sampler wall, lower the degree of disturbance of collected soil sample

(C) Thickness of the sampler wall and the degree of disturbance of collected soil sample are unrelated

(D) The degree of disturbance of collected soil sample is proportional to the inner diameter of thesampling tube

Sol. (a)

10. Solid waste generated from an industry contains only two components, X and Y as shown in the table

below:

1 1

2 2

X cY c

3Component Composition Density

(% weight) (kg / m )

Assuming (c1 + c2) = 100, the composite density of the solid waste ( ) is given by:

(A)1 2

1 2

100

c c

(B) 1 2

1 2

100c c

(C) 1 1 2 2100(c c ) (D)

1 2

1 1 2 2

100c c

Sol. (a)

CompereComposition(% weight)

Density

(Kg/m )3

X

Y

C

C

1

2

1

2

Assumng C1 + C2 = 100

Volume of X =1

1

C

Volume of Y =2

2

C

Combined density =1 2C C

Volume of X Volume of Y

=1 2

1 2

100

C C






I E S M A S T

E R11. Which of the following statement is NOT correct?

(A) Loose sand exhibits contractive behavior upon shearing

(B) Dense sand when sheared under undrained condition, may lead to generation of negative porepressure

(C) Black cotton soil exhibits expansive behavior

(D) Liquefaction is the phenomenon where cohesionless soil near the downstream side of dams or sheet-piles loses its shear strength due to high upward hydraulic gradient

Sol. (d)

The phenomenon where cohesionless soil near the d/s side of dams or sheet piles loses its shear

strength due to high upward hydraulic gradient is called Piping.

12. A circular pipe has a diameter of 1m, bed slope of 1 in 1000, and Manning’s roughness coefficient

equal to 0.01. It may be treated as an open channel flow when it is flowing just full i.e., the water level

just touches. The crest.The discharge in this condition is denoted by Qfull. Similarly, the discharge when

the pipe is flowing half-full, i.e. with a flow depth of 0.5 m is denoted by Qhalf. The rato Qfull/Qhalf is

(A) 1 (B) 2 (C) 2 (D) 4

Sol. (c)

Given diameter, d = 1 m

Slope, S =1

1000n = 0.01

full

half

Q

Q =

2/3 1/2

full

2/3 1/2

half

AR S

n

AR S

n

=

2/3full

2/3half

AR

AR

=

2/32 2

full2/32 2

half

d d/ d4 4

d d d

8 8 2

=2

1





I E S M

A S T E

R

13. Consider the following probability mass function (p.m.f.) of a random variable X.

q if X 0p(x,q) 1 q if X 1

0 otherwise

=

= =

If q = 0.4, the variancew of X is _______.

Sol. (0.24)

x 0 1

p(x) q q

E(x) = i ii x p 0 q 1 1 q 0.61 q

E(x2) = 2 2 2i ii

i

x p 0 q 1 1 q 0.61 q

V(x) = E(x2) – [E(x)]2 = 0.6 – 0.36 = 0.24

14. Which of the following statements is FALSE?

(A) Plumb line is along the direction of gravity

(B) Mean Sea Level (MSL) is used as a reference surface for establishing the horizontal control

(C) Mean Sea Level (MSL) is a simplification

(D) Geoid is an equi-potential surface of gravity

Sol. (b)

Mean sea level (MSL) is used as a reference surface for establishing the vertical control.

15. Which of the following statements is TRUE for the relation between discharge velocity and seepage

velocity?

(A) Seepage velocity is always smaller than discharge velocity

(B) Seepage velocity can never be smaller than discharge velocity

(C) Seepage velocity is equal to the discharge velocity

(D) No relation between seepage velocity and discharge velocity can be established

Sol. (b)

VS =V

n

where VS is discharge velocity

V is seepage velocity

n = VV; n 1

V

Hence SV V






I E S M A S T

E R16. For steady incompressible flow through a closed-conduit of uniform cross-section the direction of flow

will always be

(A) from higher to lower elevation

(B) from higher to lower pressure

(C) from higher to lower velocity

(D) from higher to lower piezometric head

Sol. (d)

Flow takes place from high total head to low total head. Since cross-section is uniform, velocity is

constant.

Flow will take place from high total piezometric head to low total piezometric head.

17. The two columns below show some parameters and their possible values:

Parameters Value

P–Gross Command Area I–100 hectares/cumec

Q–Permanent Wilting Point II–6 °C

R–Duty of canal water III–1000 hectares

S–Delta of wheat IV–1000 cm

V – 40 0cm

VI – 0.12

Which of the following options matches the parameters and the values correctly?

(A) P-I, Q-II, R-III, S-IV (B) P-III, Q-VI, R-I, S-V

(C) P-I, Q-V, R-VI, S-II (D) P-III, Q-II, R-V, S-IV

Sol. (b)

GCA is a hectares. Hence P-III.

Duty has units of ha/cumec. R-I.

Delta of wheat cannot be 1000 cm, so 40 cm [ S - V]

18. Consider the following statements for air-entrained concrete

(i) Air-entrainment reduces the water demand for a given level of workability

(ii) Use of air-entrained concrete is required in environments where cyclic freezing and thawing isexpected

Which of the following is TRUE?

(A) Both (i) and (ii) are True (B) Both (i) and (ii) are False

(C) (i) is True and (ii) is False (D) (i) is False and (ii) is True





I E S M

A S T E

R

Sol. (a)

An air-entraining agent introduces air in the form of minute bubbles that occupy upto 5% of the volume

of concrete distributed uniformly throughout cement paste. Thus, for the same amount of water/cement

ratio, we get higher workability.

Air air-entrained improves freezing and thawing resistance since cracks in the concrete during freezing

and melting of water is avoided because water can take up the air voids present is concrete.

19. Workability of concrete can be measured using slump, compaction factor and Vebe time. Consider thefollowing statements for workability of concrete.

(i) As the slump increases, the Vebe time increases

(ii) As the slump increases, the compaction factor increases

Which of the following is TRUE?

(A) Both (i) and (ii) are True (B) Both (i) and (ii) are False

(C) (i) is True and (ii) is False (D) (i) is False and (ii) is True

Sol. (d)

Slump is the subsidence of the specimen recorded in mm. As slump increases, water content increases

Workability increases

Compaction factor = Mass of partially compacted concrete/Mass of fully compacted concrete

CF will also increase

Vee bee time is the time required for transforming by vibration, a concrete specimen in the shape of a conical frustum into a cylinder. Hence, as workability increases, Vee bee time decreases.

20. For what value of p the following set of equations will have no solution?

2x + 3y = 5

3x + py = 10

Sol. (4.5)

Given system of equations has no solution if

2

3=

3 5

p 10

2

3=

3 1

p 2

p =9

2






I E S M A S T

E R21. For the beam shown below, the stiffness coefficient K22 can be written as

A, E, I

L

2

3

1

Note: 1, 2 and 3

are the d.o.f

(A) 2

6EI

L(B) 3

12EI

L(C)

3EI

L(D) 2

EI

6L

Sol. (b)

3

12EI

3

12EI

2

6EI

2

6EI

By giving unit displacement in (2) direction without giving displacement in any other direction, the

force developed is

K22 = 3

12EI

22. For the beam shown below, the value of support moment M is ____kN-m.

M3 m 1 m 1 m 3 m

20 kN

EI EI EI

Internal hinge

Sol. (5 kN-m)10

1m

10 15kN m

2

=

3m





I E S M

A S T E

R

23. A fine grained soil has 60% (by weight) silt content. The soil behaves as semi-solid when water content

is between 15% and 28%. The soil behaves fluid-like when the water content is more than 40%. The

‘Activity’ of the soil is

(A) 3.33 (B) 0.42 (C) 0.30 (D) 0.20

Sol. (0.3)

Activity =PI

C(%clay fraction)

=40 28

40

= 0.3

24. The integral2

1

x 2

xx dx with x2 > x1 > 0 is evaluated analytically as well as numerically using a single

application of the trapezoidal rule. If I is the exact value of the integral obtained analytically and J is

the approximate value obtained using the trapezoidal rule, which of the following statements is correct

about their relationship?

(A) J > I (B) J < I

(C) J = I (D) Insufficient data to determine the relationship

Sol. (a)

By Trapezoidal Rule

b

a

f(x)dx =h

2[( y

0+ y

n) + 2 (y

1 + y

2 + ....+ y

n-1)] –

where is maximum absolute error and

=2

1

h(b a)max f "( ) [a,b]

12

So > 0

I = J –

J > I

25. Consider the singly reinforced beam shown in the figure below:






I E S M A S T

E R At cross-section XX, which of the following statements is TRUE at the limit state?

(A) The variation of stress is linear and that of strain is non-linear

(B) The variation of strain is linear and that of stress is non-linear

(C) The variation of both stress and strain is linear

(D) The variation of both stress and strain is non-linear

Sol. (b)

26. The composition of an air-entrained concrete is given below :

Water 184 kg/m3

Ordinary Portland Cement (OPC) 368 kg/m3

Sand 606 kg/m3

Coarse aggregate 1155 kg/m3

Assume the specific gravity of OPC, sand and coarse aggregate to be 3.14, 2.67 and 2.74 respectively.

The air content is _______ liters/m3.

Sol. (50.32)

c fa caair

c fa ca

M M MV V

= 1.0

V368 606 1155 184

V 1.03.14 1000 2.67 1000 2.74 1000 1000

V0.11719 0.22696 0.42153 0.184 V 1.0

VV = 0.05032

VV = 0.05032 × 1000 = 350.32 ml

27. The directional derivative of the field u(x, y, z) = x2 – 3yz in the direction of the vector ˆ ˆ(i j 2k) at

point (2, –1, 4) is ______.

Sol. (–5.715)

u (x, y, z) = x2

– 3yz

u = ˆ ˆ2xi 3zj 3yk

at(2,–1,4)u| = ˆ ˆ ˆ4i 12j 3k

Directional derivative = ˆ ˆ ˆi j kˆ ˆ ˆ4i 12 j 3k

6





I E S M

A S T E

R

=4 12 6 14

6 6

= –7 6

3 = –5.715

28. A short reach of a 2 m wide rectangular open channel has its bed level rising in the direction of flow

at a slope of 1 in 10000. It carries a discharge of 4 m 3/s and its Manning’s roughness coefficient is

0.01. The flow in this reach is gradually varying. At a certain section in this reach, the depth of flow

was measured as 0.5 m. The rate of change of the water depth with distance, dy/dx, at this section

is _____(use g = 10 m/s2).

Sol. (3.1×10 –3)

Adverse slope = – 1

,10000

Q = 4m3/s,

n = 0.01

y = 0.5 m

dy

dx=

0 f 2r

S S

1 F

Fr =

V Q

gy 2xy gy=

=4

1.7882 0.5 10 0.5

=

Q =2/3 1/2

f

1 AR S

n

1/2f S =

2/3 2/3

Q n 4 0.01

AR 2 0.52 0.5

2 1

=

Sf

= 6.922 × 10 –3

dy

dx=

3

32

16.922 10

10000 3.196 101 1.788

=

29. An earth embankment is to be constructed with compacted cohesionless soil. The volume of the

embankment is 5000 m3 and the target dry unit weight is 16.2 kN/m3. Three nearby sites (see figure

below) have been identified from where the required soil can be transported to the construction site.






I E S M A S T

E RThe void ratos (e) of different sites are shown in the figure. Assume the specific gravity of soil to be

2.7 for all three sites. If the cost of transporation per km is twice the cost of excavation per m3 of borrow

pits, which site would you choose as the most economic solution? (Use unit weight of water =

10 kN/m3)

(A) Site X (B) Site Y (C) Site Z (D) Any of the sites

Sol. (a)

VsWs

ss

s w

WV

G

Vs = 316.2 5000m

2.7 10

Volume of solids will remain constant.

also, V = Vv + Vs = eVs + Vs = Vs(1 + e)

V = 3000(1 + e)Vx = 3000 × 1.6 = 4800 m3

Vy = 3000 × 1.7 = 5100 m3

Vz = 3000 × 1.64 = 4920 m3

Let cost of excavation be / 3m

Px = 3000 1.6 140 2

= 5080





I E S M

A S T E

R

Py = 5100 80 2

= 5260

Pz = 4920 100 2

= 5120

Hence Px is least.

Hence total cost for site x is least.

30. In a catchment, there are four rain-gauge stations, P, Q, R and S. Normal annual precipitation values

at these stations are 780 mm, 850 mm, 920 mm and 980 mm, respectively. In the year 2013, stations

Q, R and S, were operative but P was not. Using the normal ratio method, the precipitation at station

P for the year 2013 has been estimated as 860 mm. If the observed precipitation at stations Q and

R for the year 2013 were 930 mm and 1010 mm respectively, what was the observed precipitation (in

mm) at station S for that year.

Sol. (1093.433)

X

X

P

N =31 2 m

1 2 3 m

PP P P1...

N N N Nm

P

P

P

N =QS R

S Q R

PP P1

N N N3

860

780=

S1 P 930 1010

3 980 850 920

3.3076 = SP1.0941 1.0978

980

PS = 1093.433 mm

31. A sign is required to be put up asking drivers to slow down to 30 km/h before entering Zone Y(see

figure). On this road, vehicles require 174 m to slow down to 30 km/h (the distance of 174m includes

the distance travelled during the perception-reaction time of drivers). The sign can be read by 6/6 vision

drivers from a distance of 48 m. The sign is placed at a distance of x m from the start of Zone Y so

that even a 6/9 vision driver can slow down to 30 km/h before entering the zone. The minimum valueof x is _____ m.

Sign

Direction of vehicle movement

Road

Start of Zone Y

Zone Yx






I E S M A S T

E RSol. (142)

For a 6/6 person driver can see the board from distance 48 m.

For a 6/9 person driver can see the board from distance6

48 32m9

Vehicle require 174 m to slow down to 30 km/hr. So minimum distance of X from the start of y zone

= 174 – 32

= 142 m

32. A non-homogenous soil deposit consists of a silt layer sandwiched between the fine-sand layer at top

and clay layer below. Permeability of the silt layer is 10 times the permeability of the clay layer andone-tenth of the permeability of the sand layer. Thickness of the silt layer is 2 times the thickness of

the sand layer and two-third of the thickness of the clay layer. The ratio of equivalent horizontal and

equivalent vertical permeability of the deposit is _____.

Sol. (10.9684)

Z1

Z2

Z3

fine sand

silt

clay

1

2

3

22 3 3

KK 10K or K

10 ; 2 1 1 2

1K K or K 10K

10 ; 2

2 1 1

ZZ 2Z or Z

2 ; 2 3 3 2

2 3Z Z or Z Z

3 2

KX =1 1 2 2 3 3

1 2 3

K Z K Z K Z

Z Z Z

=

2 2 22 2 2

22 2

10K Z K 3K Z Z

2 10 2Z 3

Z Z

2 2

=2 2 2 2 2 2

2

5K Z K Z 0.15K Z

3Z

= 2 2

6.15K 2.05K

3





I E S M

A S T E

RKZ =

22 2

2 2 2

2 2 2

Z 3Z Z

2 2Z Z 1.5Z 10

2 10K K K

=2

22

2

3Z0.1869K

Z 11 15

K 20

=x 2

z 2

K 2.05K

K 0.1869K

= 10.9684

33. In a survey work, three independent angle X, Y and Z were observed with weights WX, WY, WZ

respectively. The weight of the sum of angles X, Y and Z is given by

(A)

x Y Z

1

1 1 1

W W W(B)

x Y Z

1 1 1

W W W

(C) X Y ZW W W (D) 2 2 2X Y ZW W W

Sol. (a)

x y z

1

1 1 1

W W W

34. A square footing (2m × 2m) is subjected to an inclined point load, P as shown in the figure below. The

water table is located well below the base of the footing. Considering one-way eccentricity, the net safe

load carrying capacity of the footing for a factor of safety of 3.0 is ____kN.

The following factors may be used :

Bearing capacity factors : qN 33.3, N 37.16 , Shape factors : qs sF F 1.314 ; Depth factors

: qd dF F 1.113 ; Inclination factors : qi iF 0.444, F 0.02

P

GL

30°1m

2m

0.85mUnit weight = 18 kN/m

3

Cohesion = 0Friction angle = 35°






I E S M A S T

E RSol. (450.037)

qns = nu

1q

3

qnu = C q

1CN q N 1 B N

2 (for normal case)

For inclined loading, net safe load carrying capacity

qns = q 1 qs qd qp s d p

1 1q N F F F B N F F F

3 2

qns =

1 118 1 33.3 1 1.314 1.113 0.444 2 18 37.16 1.314 1.113 0.02

3 2

=2397.09

132.364 kN/ m3

=

For one way eccentricity, we will have to take reduced area of footing

reduced area of footing tomake the load concentric

= 2×1.7m =3.4m2 2

0.85 m

0.85 m

Load carrying capacity = 132.364 × 3.4 = 450.037 kN

35. Two reservoirs are connected through a 930 m long, 0.3 m diameter pipe, which has a gate valve. The

pie-entrance is sharp (loss coefficient = 0.5) and the valve is half-open (loss coefficient = 5.5). The

head difference between the two reservoirs is 20 m. Assume the friction factor for the pipe as 0.03

and g = 10 m/s2. The discharge in the pipe accounting for all minor and major losses is ____ m3/s.

Sol. (0.1413)

1

2

3

20 md = 0.3m

f = 0.03

l = 930 m

The loss will be due to





I E S M

A S T E

R

1. Entry (minor)

2. Head loss in pipe (major)

3. Valve half open

4. Exit loss (minor)

Total loss = 20 m

20 =2 2 2 20.5 v 0.03 930 v 5.5 v v

2g 2 g 0.3 2g 2g

20 × 2 × 10 =2

2 2 20.03 930 v0.5v 5.5v v

0.3

400 = 2 2 2 20.5v 93v 5.5v v

100v2 = 400

v = 2 m/s

Discharge, Q = 2d v4

= 2 30.3 2 m s4

= 0.1413 m3/s

36. Consider a primary sedimentation tank (PST) in a water treatment plant with Surface Overflow Rate(SOR) of 40 m3/m2/d. The diameter of the spherical particle which will have 90 percent theoretical

removal efficiency in this tank is _____ m . Assume that settling velocity of the particles in water is

described by Stoke’s Law.

Given : Density of water = 1000 kg/m3, Density of particle = 2650 Kg/m3, g = 9.81 m/s2. Kinematic

viscosity of water (v) = 1.10×10 –6 m2/s.

Sol. (22.573 m )

% removal =s

s

V100

V

sV = 0.9 × Vs

= 0.9 × overflow rate

sV =0.9 40

m / s86400

2s d

18

=

0.9 40

86400






I E S M A S T

E Rd = s

0.9 40 18

86400 G 1 g

d(m) =60.9 40 18 1.1 10

86400 1.64 9.81

= 2.2576 × 10 –5 m

= 22.576 m

37. Consider the following differential equation :

y

x cosydx xdy

x

= y

y sinxdy ydx

xWhich of the following is the solution of the above equation (c is an arbitrary constant) ?

(A) x y

cos cy x

(B) x y

sin cy x

(C) y

xycos cx

(D) y

xysin cx

Sol. (c)

x(ydx + xdy)y

cosx

= y (xdy – ydx) siny

x

ydx xdy

xdy ydx

=

y ytan

x x

Let y = v·x

dy = v dx + xdv

2

2

vxdx vxdx x dv

vxdx x dv vxdx

= v tan v

xdv 2vdx

xdv

= v tan v

1 +2v dx

x dv= v tan v

2v dx

x dv= v tan v – 1

dx2

x=

1tanv dv

v

Integrating both sides





I E S M

A S T E

R

2 log x = log |sec v | – log v + log c

x2 =csecv

v

2 yx .

x= c sec y/x

xy cosy

x= c

38. On a circular curve, the rate of super elevation is e. While negotiating the curve a vehicle comes to

a stop. It is seen that the stopped vehicle does not slide inwards (in the radial direction). The coefficient

of side friction is f . Which of the following is true :

(A) e f (B) f e 2f (C) e 2f (D) none of the above

Sol. (a)

fN

N

mgmgcos

mg sin

fN mg sin

f mg cos mg sin

fcos sin

f tan

f e

or e f

39. The quadratic equation 2x 4x 4 0 is the solved numerically, starting with the initial guess 0x 3 .The Newton-Raphson method is applied once to get a new estimate and then the Secant method is

applied once using the initial guess and this new estimate. The estimated value of the root after the

application of the Second method is ______.

Sol. (7/3)

f(x) = x2 – 4x + 4






I E S M A S T

E Rf' (x) = 2x – 4

x0 = 3

f (3) = 1, f' (3) = 2

By Newton Rapshon method

x1 =0

00

(x ) 1 5x 3

f '(x ) 2 2

f (5/2) =25 1

10 44 4

By secant method

x2 = x1 –1 0 1

1 0(x x )f(x )f(x ) f(x )

= 5 1

2 414

352 1

=5 1

2 6

=7

3

40. A 20 m thick clay layer is sandwiched between a silty sand layer and a gravelly sand layer. The layer

experiences 30 mm settlement in 2 years.

Given :

TV =

2

10

Ufor U 60%

4 100

1.784 0.933log forU 60%100 U

where TV is the time factor an U is the degree of consolidation in %

If the coefficient of consolidation of the layer is 0.003 cm 2/s, the deposit will experience a total of 50

mm settlement in the next ____ years.

Sol. (4.43)

Clay20 m

Silty sand

Gravelly sand





I E S M

A S T E

R

v2

C t

H= Tv

Tv =

2

22

cm0.003 2 86400 365 sec

sec

20cm100

2

Tv = 20.1892 (U )4

Tv < 0.2826

U < 60%

U = 0.4908

Total consolidation =30

61.12 mm0.4908

Degree of consolidation for 50 mm settlement

=50

U 0.818 81.8%61.12

Tv = 0.60835 = Cvt/H2

t = 2

4

0.60835 10 sec

0.003 10

= 6.43 yr

Additional number of years = 4.43 yrs

41. Consider the singly reinforced beam section given below (left figure). The stress block parameters for

the cross-section from IS : 456-2000 are also given below (right figure) . The moment of resistance

for the given section by the limit method is _____ kN-m.

300mm

200mm

4-12Fe415

M25

d

xu

0.42

0.36f xck u

xu, max

for Fe415

=0.48d






I E S M A S T

E RSol. (42.7708)

Ast =2 24 12 452.38 mm

4

=

xu,max = 0.48×300

= 144 mm

C = T

0.36 f ck bxu = 0.87 f y Ast

xu =y st

ck

0.87 f A90.73mm

0.36 f b =

xu < xu,lim, it is under reinforced section

MOR = 0.36 f ck b xu (d – 0.42 xu)

= 42.7708 × 106 N-mm

= 42.7708 kN-m

42. A water tank is to be constructed on the soil deposit shown in the figure below. A circular footing of

diameter 3 m and depth of embedment 1 m has been designed to support the tank. The total vertical

load to be taken by the footing is 1500 kN. Assume the unit weight of water as 10 kN/m 3 and the load

dispersion pattern as 2V:1H. The expected settlement of the tank due to primary consolidation of the

clay layer is _____ mm.

2m

6m

10m

Dense Sand

Normallyconsolidated

clay

Silty Sand

Sand

Bulk unit weight=15 kN/m3

Saturated unit weight =18kN/m3

GWT

GL

Saturated unit weight = 18 kN/mCompression index = 0.3Initial void ratio = 0.7

Coefficient of consolidation = 0.004 cm /s

3

2

Sol. (53.236)





I E S M

A S T E

R

1500 kN

3m

1m2m

6m

5m 2V

1H 10m

6m 3m 6m

2V

1H

Dense sand

NCC

Ysat=18 kN/m3

3

C =0.3

e =0.7C =0.004 cm /s

C

0

v

Silty sand

t=15 kN/m

sand

=18 kN/m

3

3sat

GL

3w 10kN/m =

H =C 0 0

100 0

C Hlog

1 e

H0 = thickness of clay layer = 10 m.

0 = effective stress before application of load at the centre of clay layer.

0 = 15 2 18 10 6 18 10 5

= 118 kN/m2

= increase in effective stress due to the structural weight at the centre of clay layer.

=

2

2

15008.488 kN/m

3 124

=

H = 10

0.3 10 118 8.488log

1 0.7 118

= 0.053236 m

= 53.236 mm






I E S M A S T

E R43. The smallest and largest Eigen value of the following matrix are :

3 2 2

4 4 6

2 3 5

(A) 1.5 and 2.5 (B) 0.5 and 2.5 (C) 1.0 and 3.0 (D) 1.0 and 2.0

Sol. (d)

For eigen values A I 0

3 2 2

4 4 6 0

2 3 5

23 ( 20 4 5 18) 2(20 4 12) 2( 12 8 2 ) 0

3 24 5 2 0

1, 1, 2

Smallest eigen value = 1

Largest eigen value = 2

44. The concentration of Sulfur Dioxide (SO2) in ambient atmosphere was measured as 30 330 g m .

Under the same conditions, the above SO2 concentration expressed in ppm i s_____

Given : P/(RT) = 41.6 mol/m3, where P =Pressure; T = Temperature; R = Universal gas constant;

Molecular weight of SO2 = 64.

Sol. (0.0113)

1 m3 of air has 30 mg SO2

106 m3 of air has 3OgSO2

3Og SO2 =30

64

mole SO2

From PV = nRT

V =nRT n

P P / RT= =

3

30 / 64 mole

41.6 mol/m





I E S M

A S T E

R

Volume of SO2 =330

0.01127m64 41.6

=

Concentration of SO2 in PPM = 0.0113 ppm

45. The drag force FD on a sphere kept in a uniform flow field depends on the diameter of the sphere D;

flow velocity, V fluid density and dynamic v iscosity, . Which of the following options represents the

non-dimensional parameters which could be used to analyze this problem ?

(A)

DF

andVD VD

(B)

D

2

F VDand

VD

(C)

D2 2

F VD

andV D (D)

D3 3

F

andV D VD

Sol. (c)

46. The acceleration-time relationship for vehicle subjected to nonuniform acceleration is,

dv

dt= t

0v e

where, v is the speed in m/s, t is the time in s, and are parameters, and v0 is the initial speed

in m/s. If the accelerating behavior of a vehicle, whose driver intends to overtake a slow moving vehicle

ahead, is described as,

dv

dt = v

Considering 22m s , 10.05s and 2dv1.3m s

dt at t = 3 s, the distance (in m) travelled by the

vehicle in 35 s is _____

Sol. (900.832m)

dv

dt= t

0V e

dV = t0V e dt

V = t

0V eC

at t = 0, V = V0

V0 = 0V

C

C =0

0

VV






I E S M A S T

E RC =

V = t

0V e

Distance travelled in t0 sec

= 0t0 0 02 2

t V Ve 0 1

x = 0t0 02

t Ve 1

t 3 sec

dV

dt =

= 30V e 1.3 =

0V = 3

1.3

e

x = 0t0

2 3

t 1.3e 1

e

Putting the values-

x =

35 0.05

23 0.05

2 35 1.3 e 1

0.05 0.05 e

= 1400 – 499.168 = 900.832 m

47. The 4-hr unit hydrograph for a catchment is given in the table below. What would be the maximum

ordinate of the S-curve (in m3/s) derived from this hydrograph ?

Time (hr)

Unit hydrograph ordinate (m /s)3

0 2 4 6 8 10 12 14 16 18 20 22 24

0 0.6 3.1 10 13 9 5 2 0.7 0.3 0.2 0.1 0

Sol. (22)





I E S M

A S T E

R3

Unit HydrographTime S curve(hr) ordinatesordinate (m /s)

0 0 0

2 0.6 0.6

4 3.1 3.1

6 10 10.6

8 13 16.1

10 9 19.6

12 5 21.1

14 2 21.6

16 0.7 21.8

18 0.3 21.9

20 0.2 22

22 0.1 22

24 0 22

Hence, maximum ordinate is 22.

48. A tapered circular rod of diameter varying from 20 mm to 10 mm is connected to another uniform

circular rod of diameter 10 mm as shown in the following figure. Both bars are made of same material

with the modulus of elasticity, E = 2×105 MPa. When subjected to a load P 30 kN, the deflection

at point A is _____ mm.






I E S M A S T

E Rd = 20mm1

2m

1.5m

d =10 mm2

A

P = 30 kN

Sol. (15)

d =20 mm1

A

2m

1.5m

B

d =10 mm2

P 30 kN=

Total elongation, = AB BC

= 2

55

30 1.5 30 20.01 0.020.01 2 102 10 44

= 2 55

30 1.5 4 30 2 4

0.01 0.02 2 100.01 2 10

= 9 + 6

= 15 mm





I E S M

A S T E

R

49. Consider the following complex function :

f(z) = 2

9

z 1 z 1

Which of the following is one of the residues of the above function ?

(A) –1 (B)9

16(C) 2 (D) 9

Sol. (–9/4)

f(z) = 2

9

z 1 z 1

poles are z = 1, –1

Res {f(z);z = 1} = 2

9 9

41 1 =

Res {f(z);z = –1} = g 1

1!

where g(z) =

9

z 1

g(z) = 2

9

z 1

= – 2

9

1 1

=9

4

50. For formation of collapse mechanism in the following figure, the minimum value of Pu is cMP/L. MP and

3MP denote the plastic moment capacities of beam sections as shown in the figure. The value of is

_____.

2m

1m

Pu

MP

1m

3MP

Sol. (13.33)

Assuming L = 4m






I E S M A S T

E RMechanism 1

3MP MP

Pu

MPMP

L/2 L/4 L/4

P P P P

L3M M M M P

4 = 0

u

LP

4 = 6 MP

Pu = P24M

L C = 24

Mechanism 2

3MP

MP

L/4

Pu

3L/4

MP

MP

L

4 =

3L

4

= 3

P P4M 2M =PL

4

PP

4M2M

3

=

PL

4

P10M

3

=

PL

4





I E S M

A S T E

R

P = P40M 40C

3L 3

Load is smaller obtained out of two mechanisms =P40M

3L

C = 40/3 = 13.33

51. In a region with magnetic declination of 2°E, the magnetic Fore bearing (FB) of a line AB was measured

as N79°50É. There was local attraction at A. To determine the correct magnetic bearing of the line,

a pint O was selected at which there was no local attraction. The magnetic FB of line AO and OA were

observed to be S52°40É and N50°20´, respectively. What is the true FB of line AB ?

(A) N81°50É (B) N82°10É (C) N84°10É (D) N77°50É

Sol. (c)

Magnetic Declination, 2 E =

Magnetic F.B of AB = N 79°50' E 79°50

To find local attraction at Stn A

As Stn O is free from local attraction

Hence F.B of OA will be correct

Correct F.B of OA = N 50°20 W 309 40

Correct B.B of OA = 129°40'

Obs F.B of AO = Obs B.B of OA = S52° 40'E = 127°20'Error = M.V – T.V = 127°20' – 129°40' = – 2°20'

Correction = + 2°20'

Local attraction @ Stn A = 2 20 2 20 E

Magnetic F.B of AB = N 79°50'E

2 E, = L.A = 2°20'E

T.B of F.B of AB = N 79°50' E + 2° + 2° 20' = N84° 10'E

52. A hydraulic jump is formed in a 2 m wide rectangular channel which is horizontal and frictionless. The

post-jump depth and velocity are 0.8 m and 1 m/s, respectively. The pre-jump velocity is _____ m/s.

(use g = 10m/s2

).

Sol. (4.828 m/s)

y2 = 0.8 m

v2 = 1 m/s

2r F = 2 1

gD 0.8 210

2

=






I E S M A S T

E R2r

F = 0.3535

y1 = 2

22r

y1 1 8F

2

= 20.81 1 8 0.3535

2

= 0.165685

V1y1 = V2y2

V1 =1 0.8

0.165685

= 4.828 m/s

53. A bracked plate connected to a column flange transmit a load of 100 kN as shown in the following

figure. The maximum force for which the bolts should be designed is ____ kN.

75 75

75

75

600

100 kN

All dimensions are in mm

Sol. (156.2 kN)

Fd =100

20 kN5

FT =2

Mr

r =

3

2 6

100 0.6 75 2 10

4 1075 2

=

21000

10

= 141.42 kN

= 45°

F1 = 2 2 2 20 141.42 cos4520141.42

= 156.2 kN





I E S M

A S T E

R

54. For the 2D truss with the applied loads, shown below, the strain energy in the member XY is _____

kN-m. For member XY, assume AE = 30 kN, where A is cross-section area and E is the modulus of

elasticity.

3m

3m

3m

10 kN

X

7m

Sol. (5)

Truss members carry axial force only, hence strain energy will be only due to axial stress.

Strain energy, U =2P L

2AE

when AE = 30 kN (given), L = 3m

Calculation for force P is xy member.

AM = 0

RB × 3 – 10 × 9 – 5 × 3 = 0

RB = 35 kN

H A = 10 kN

BM = 0

R A × 3 + 10 × 9 = 0

R A = – 30

(opposite to the direction assumed)

10 kN

5 kN

3m

3m

3m

HA

R A RB

3m

AB

45°

45°

45°

x y

O´

O

cutting a section 0-0'' from xy as shown.






I E S M A S T

E RH =10 A

R =30 kN A R =35 kNB

Fxy

horizontalF = 0

Fxy = 10 kN

Now, U =10 10 3

2 30

= 5 kN-m

55. Two beams are connected by a linear spring as shown in the following figure. For a load P as shown

in the figure, the percentage of the applied load P carried by the spring is _____.

L

EI

P

K = 3EI/(2L )S

3

EI

Sol. (25%) P

R

R

KS

R

R





I E S M

A S T E

R

Compression of spring = 3 3

S

P R L RL R

3EI 3EI K

=

3 3 3P R L RL R 2L

3EI 3EI 3EI

=

P R R = 2R

P = 4R

R =P

4

% force carried by spring = 25%

ce gate 2015 morning session

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