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GATE 2014 CIVIL ENGINEERING Memory Based Questions-EVENING SLOT

ENGINEERS INSTITUTE OF INDIA-E.I.I India’s Leading Institute for GATE, IES & PSU’s

Head office: 28B/7 Jiasarai Near IIT New Delhi-16 www.engineersinstitute.com

PH. 011-26514888

India’s Leading Institute for GATE, IES & PSU’s

Head office: 28B/7 Jiasarai Near IIT HauzKhas New Delhi-16

www.engineersinstitute.com

[email protected] PH. 011-26514888

Cell: 0-9990657855

Admission Open for Fresh Batch : IES, GATE & PSU’s 12

nd March 22

nd March

GATE 2014 Civil Engineering: Memory Based Questions

http://www.engineersinstitute.com/

mailto:[email protected]

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PH. 011-26514888

1. The clay mineral primarily governing the swelling behavior of black cotton soil is

(a) kaolinite (b) lllite (c) Halloysite (d) montmorillonite

Ans. (d) Montmorillonite is having a very high activity and it is a very plastic mineral. The presence

of montmorillonite is responsible for this behavior of Black cotton soil.

2. A certain soil has the following properties

Gs = 2.71

n = 40%

w = 20%

Find the degree of saturation of the soil rounded off to the nearest percentage is .

Ans. :

Gs= specific gravity of soil =2.71

n=porosity =40%

w=water content =20%

There is relation :

es=WGs

Where,

e=void ratio

w=water content

Gs=specific gravity

S=degree of saturation

0.41

0.67

en

e

e

0.2 2.710.813

0.67

wGs

e

Since the answer has been asked in % therefore it would be;

S=81.3%

3. Find the Polar moment of inertia Ip( in cm4) at a rectangular section having width b = 2 cm and

depth d = 6cm. ____________

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Ans. Polar moment of inertia is summation of moment of area about x axis and Y-axis. This is used

in study of torsion

Like for as circular section;

4 4

; .64 64

XX YY

d dI I

Ip =polar moment of inertia =

4 4 4

.64 64 32

d d d

Similarly, for rectangular section;

3

12XX

bdI

where b is width of beam and d is depth of beam

Similarly, =

3

12YY

dbI

where , d is depth of beam and (b) is width of beam

34

33

2 6(36)

12

6 2(4 )

12

XX

YY

I cm

I cm

pI polar moment of inertia =40cm

4.

4. The determinant of matrix

0 1 2 3

1 0 3 0;

2 3 0 1

3 0 1 2

is

Ans. The value of determinant does not change due to row transformation

1

2

3

4

0 1 2 3

1 0 3 0

2 3 0 1

3 0 1 2

R

R

R

R

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3 3 2

4 4 2

2

3

R R R

R R R

0 1 2 3

1 0 3 0

0 3 6 1

0 0 8 2

Now we can see that in column 1 we have only one non-zero element.

1 2 3

1 3 6 1

0 8 2

[1( 12 8) 3(4 24)] (88)

5. An isolated 3-h rainfall event on a small catchment produces a hydrograph peak and point of

inflection on the falling limb of the hydrograph are 7 hours and 8.5 hours respectively after the

start of the rainfall. Assuming no losses and no base flow contribution, the time of

concentration (in hours) for this catchment is approximately

(a) 8.5 (b) 7.0 (c) 6.5 (d) 5.5

Ans. Time of concentration is a concept of used in hydrology to measure the response of watershed to

a rain event. It is defined as time needed for water to flow from most remote point to a

watershed outlet. For small catchment time of concentration is approximately equal to lag time

of peak flow this is equal to distance between centroid of rainfall and peak of flow.

Hence, the time of concentration= 5.5 hours.

6. The first moment of area about the axis of bending for a beam cross section is

(a) Moment of inertia (b) section modulus

(c) shape modulus (d) polar moment of inertia

Ans. (b)

Ze or section modulus is defined as first moment of area and is defined as ratio of second of

area or moment of inertia with respect to distance of farthest point from neutral axis.

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7. The two air pollution control devices that are usually used to remove very fine particles from the

gas are

(a) cyclone and venture scrubber (b) cyclone and scrubber

(c) electrostatic precipitator and Fabric filter (d) setting camber and tray scubber

Ans. (c)

Though all the given options are basically used to reduce air pollution but electrostatic

precipitator and fabric filter are used to remove very fine particles

8. A student riding a bicycle on a 5 km one way street takes 40 minutes to reach home. The

student stopped for 15 minutes during this ride. 60 vehicles over took the student (Assume the

number of vehicle overtaken by the student is zero) during the ride and 45 vehicles while the

student stopped. The speed of vehicle stream on that road (in km/hr) is

(a) 7.5 (b) 12 (c) 40 (d) 60

Ans. Velocity of student =4

12 /40 15

km hr

tanding

Vehicle /min (Vehicle/min )

Relative speed of vehicle w.r.t student Relative speed of vehicle w.r.t student smotion

60 45

25 15

12 ( 0)

60 /

x x

x km hr

9. The target mean strength fcm for concrete mix design obtained from the characteristic strength fck

and standard deviation , as define in IS 456-2000 is

(a) 1.35ck

f (b) 1.45ck

f (c) 1.55ck

f (d) 1.65ck

f

Ans. (d) This is based on provision of IS-456 2000. Target mean strength is used for mix

design in laboratory.

10. The static indeterminacy of the two span continuous beams with an internal hinges shown below

is

Ans. (d)

Degree of static indeterminacy =(No. of unknown Reaction) – (No. of equilibrium equation )

No. of unknown reaction= 2 [at hinge]+ 1[at roller support]+1[at second roller support]=4

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No. of equilibrium equation=

0

0

0

x

y

z

F

F

F

And since moment about roller support is zero. Therefore it will provide an additional equation

Degree of static indeterminacy =(4-4)=0

11. A fair (unbiased) coin was tossed Four times in a succession and resulted in the following

outcomes (I) H (II) H (III) H (IV) H. The probability of obtaining a “TAIL’ when the coin is tossed again is

(a) 0 (b) ½ (c) 4/5 (d) 4.5

Ans. (b)

In this case, since the result of earlier trials and new trials are not dependent therefore P(T)=1

2

12. A plane flow has velocity components

1 2

,x y

u vT T

and w = 0 along x, y and z direction

Respectively, where T1 ( 0) and T2 ( 0) are constant. having the dimension of the time.

The given flow is incompressible if

(a) T1=-T2 (b) 21

2

TT

(c) 21

2

TT (d) T1=T2

Ans. (d)

For an incompressible flow the continuity equation must be satisfied

Sine density is not varying continuity equation is given as follows:-

1 2

1 2

0

;

0

1 10

u V W

x y Z

x yu V

T T

u V

x y

T T

Or, 1 2T T

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13. 2 3

5

iz

i

, can be expressed as

(a) -0.5-0.5i (b) -0.5+0.5i (c) 0.5-0.5i (d) 0.5+0.5i

Ans.(b)

2 3

5

iZ

i

In order to eliminate denominator we will have to rationalise denominator

2

2 2

2

2 3 5

5 5

10 2 15 3

( 5 ) ( )

13 13[ 1]

26

i iZ

i i

i i iZ

i

iZ i

Q

Z= -0.5+0.5 i

14. The survey carried out to determine nature feature such as hills rivers, forests and man-made

features such as turns, towns, villages, building roads, transmission lines and canals is classified

as

(a) Engineering survey (b) geological survey (c) land survey (d) topographic survey

Ans. (d) Topographical surveys are used to determine horizontal and vertical location of certain points

by linear and angular measurements and is made determine natural features of a country such as

rivers, Lakes, woods, hills and artificial features such as canals .

Note: Land survey includes a more holistic approach. It also includes cadastral survey which is

used to fix property lines. Therefore best option is topographical survey

15. A single vertical friction pile of diameter 500 mm and length 20 m is subjected to a vertical

compressive load. The pile is embedded in a homogenous sandy strata where angle of internal

friction () = 30°, dry unit weight ( d ) = 20 kN/m

3 and angle of wall friction ( ) = 2 / 3 .

Considering the coefficient of lateral earth pressure (K) = 2.7 and the bearing capacity factor

(Nq) = 25, calculate the ultimate bearing capacity of the pile (in kN).

Ans.

Soil properties

= 30º; d = 20KN/m3.

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Wall friction angle o220

3

Lateral earth pressure coefficient =

K = 2.7

The ultimate load capacity is given by = 9 9up P s sA A

9up = This is due to point bearing resistance

AP = surface area of base.

The frictional resistance = a

2 29 9

9

tan

side resistance area

point bearing = N

(20 20) / (400) /

25

s s s

s

F A K A

A

vertical stress at bottom

KN m N KN m N

N

Total bearing resistance = (400 × 25) KN/m2 = (10000)KN/m

2

Area of bottom 2(0.5)4

Point bearing resistance = 21000 (0.5) 19644

KN

Frictional resistance 1tan

2K

K = coefficient of lateral pressure = (2.7)

o 219 400 (2.7) tan 20 (197) /

2s KN m

Frictional resistance = 197 × × 0.5 × 20 = (6185)

Thus, ultimate bearing capacity = (6185 + 1964) = (8149) KN.

16. Dominating micro-organisms in Active Sludge Reactor process.

(a) Aerobic heterotrops (b) Anaerobic heterotrops (c) Autotrops (d) Phototrops

Ans. (a) Activated sludge reactor is continuously supplied with air. The reactor has aerobic

Heterotrophs.

17. Group-I contains representative stress-strain curve as shown in the figure while group II gives the

list of materials. Match the strain-strain curves with the corresponding materials

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Group –I Group -II P- Curve J (1) Cement paste

Q- curve K (2) coarse aggregate

R- Curve –L (3.) concrete

(a) P-1 Q-3 R-2

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

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

(d) P-3 Q-2 R-1

Ans. (b)

18. The modulus of elasticity E = 5000 ckf where fck is the characteristic compressive strength of

concrete specified in IS456 - 2000 is based on

(a) tangent modulus (b) intir tangenst modulus

(c) secant modulus (d) chora modulus

Ans. (c)

E = 5000 ckf is taken as secant modulus at 1/3

rd of strength of concrete

19. The Muskingum model of routing a flood through a stream reach is expressed as O2 = K0T2 +

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K1 T1 + K201 where K0, K1 and K2 are routing coefficient for the concerned reach I1 and I2 are

the inflows to the reach and O1 and O2 are the outflows from the reach corresponding to times

steps 1 and 2 respectively. The sum of K0, K1 and K2 of the model is

(a) -1 (b) -0.5 (c) 0.5 (d) 1

Ans. (d)

According to Muskingum equation

2 2 1 1 2 1

1 2 1

o

o

O K T K T K O

K K K

This is related to flood routing

20. As per Indian standard soil classification system (IS : 1498 – 1970) an expression for A-line is

(a) 0.73( 20)p L

I W (b) 0.70( 20)p L

I W

(c) 0.73( 10)p L

I W (d) 0.70( 10)p L

I W

Ans. (a)

A line is used for soil classification of fine grained soil. This is given by

Ip=0.73(WL-20)

Soil lying above A-line are classified as clay. Those lying below A- line is known as silt.

21. if f(x) is a continuous real valued random variable defined over the interval ( , ) and its

occurrence is defined by the density function given as

21

21( )

2

x a

bf x eb

where a and b

are the statistical attributes of the random variable {x}. the value of the integral 2

1

21

2

x aa

be dx isb

(a) 1 (b) 0.5 (c) (d) /2

Ans. (b) the density function is symmetric about x=a

1( )

2

1

2

x at

b

dxdt

b

2dx bdt

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20

2

1/2

2

2

tbe dt

b

t x

t x

1/21

2dt x dx

0 11

1/2 2

0

1 1 11/ 2

2 2 2

x xe x dx e x dx

22. A road is being designed for a speed of 110 Km/hr and horizontal curve with super elevation of

8%.If the coefficient of side friction is 0.1. The minimum radius of the curve in (m) required for

safe vehicular movement is

(a) 115 (b) 152.3 (c) 264.3 (d) 528.5

Ans. (d)

If V is speed in km/hr then ruling minimum radius is given by following formulas:-

2

min ;where V is speed in kmph127( )

VR

e f

e=super elevation =8% =(0.08)

f=coefficient of lateral friction =0.10

Ruling minimum Radius =110 110

528.3127 0.18

m

Closest option is 528.5 m

23. Contact pressure for a rigid footing resting on clay at the centre and the edge are respectively

(a) maximum and zero (b) maximum and minimum

(c) zero and maximum (d) minimum and maximum

Ans.(d)

Contact pressure for a rigid footing resting on clay

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Contact pressure is maximum at ends. At centre it is least .

24. Considering the symmetry of a rigid frame as shown below the magnitude of the bending

moment (in kNm) at C (using moment distribution method) is –

(a) 170 (b) 172 (c) 176 (d) 178

Ans. C.

25. The average spacing between vehicles in a traffic steam is 50 m then the density in (veh/km) of

the steam is

Ans. 20 veh/km

Density being asked is vehicles/km.

Density= 1000 1000

20Average spacing of vechicles 50

26. A rectangular channel of 2.5 m width is carrying a discharge of 4m3/s. considering that

acceleration due to gravity as 9.81 m/s2, the velocity of flow ( in m/s) corresponding to the

critical depth at which the specific energy is min. is –

Ans. We know that specific energy is minimum for critical flow

At critical flow

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Froude No.=1

c

V

gy=1

For rectangular channel critical depth =

1/32

q

g

Where 9 is discharge per unit width

2

1/32

4(1.6) /

2.5

(1.6)(0.639 )

9.81

9.8 0.639

2.5 /

c

c

q m s

y m

V gy

V m s

27.

Group-I

P. Anemometer

Q. Hygrometer

R. Pitot tube

S. Tensiometer

Group –II

(1) Capillary potential of the soil water

(2) Fluid velocity at a point in the flowing area

(3) Water pressure content of air

(4) Wind pressure

Ans. anemometer:- this measures the wind pressure.

Hygrometer:- this is used for measuring moisture content of air

Pitot tube- fluid velocity at a specific point

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Tensiometer: capillary potential of soil water

28. The rank of the matrix

6 0 4 4

2 14 8 18

14 14 0 10

is

Ans. Rank=2

6 0 4 4

2 14 8 18

14 14 0 10

Rank of matrix is defined as maximum no. of linearly independent rows or linearly independent

column whichever is less

3 3 12R R R

6 0 4 4

2 14 8 18

2 14 8 18

3 3 2R R R

6 0 4 4

2 14 8 18

0 0 0 0

Rank (A) =2

29. The flexural tensile strength of M25 grade of concrete in N/mm2 as per IS: 456 -2000 is

Ans. Flexural tensile strength according to IS-456: 2000 is =0.7 .ck

f

ckf =characteristic strength in N/mm

2.

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Tensile strength =20.7 25 (3.5) /N mm

30. The integrating factor for the differential equation 1

2 1

k t

o

dpk p k L e is

dt

(a) 1k te

(b) 2k te

(c) 1k te (d) 2k t

e

Ans. (d)

This is a Bernoulli’s equation

Integrating factor = 2 2K dt K te e

When once e we substitute and multiply by integrating factor on both the sides then

2 2 2 1

2 2 1

( )

2 1

( )

1

.

, ( . )

k t k t K K t

o

k t k k t

o

dpe k e p k L e

dt

dor p e K L e

dt

This equation can be now be integrated easily

31. . For the state of stress (in Mpa) shown in figure below the max. shear stress (in MPa) is

Ans. Principle stresses are given by:-

22

1 2/2 2

2; 4

4

x y x y

xy

x y

xy

2 2

1 2

2 4/ (3) (4) 1 5

2

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1 2/ 6, 4

Maximum shear stress = difference of principal stresses 6 4

(5)2 2

units

32. With reference to a standard cartesian (xy) plane, the parabolic velocity distribution profile of

fully developed lamina flow in x direction between two parallel. Stationary and identical plate

that are seperated by height + h 2

8

h dpu

dx

In this equation y = 0 axis lies equidistance between the plates at a distance h/2 form the two

plate. Maximum and average velocity will be ?

Ans :

2 2

1 48

h P yu

hx

Clearly, at y = 0

2

2

max

8

,

8

av

h Pu

x

thus

h Pu

x

Qu

A

Let a unit width of plate be taken.

2

max 2

.

41

av

av

u dAu

A

yu dA

hu

A

Since, we have taken unit width =

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2

max 2

2

0

3max

2

max max max

1

4( 1)

2( 1)

2 4

2 8 3

2 2 2

2 6 3 3

h

av

av

av

h h

yu dy

hu

h

u h hu

h h

u u uh h hu

h h

33. The chaiange of the interestion point of two straight is 1585.60 m and the angle of intersection is

40°. If the radius of a circular curve is 600 m, the tangent distance (in m) and length of the

curve (in m) are –

(a) 418.88 and 1466.08 (b) 218.38 and 1648 .45 (c) 218.38 and 418.88 (d) 418.88 and 218.38

Tangent Distance =1 2

tan ( / 2) 600tan 20 218.88 .Qt Qt R m

Length of curve= 180

Radius of curve

40 600 418.82180

m

34. A circular raft. foundation of 20 m diameter and 1.8 m thick is provided for a tank that applies

a bearing pressure of 110 kPa. on sandy soil with young modulus Es = 30 MPa and = 0.3. The

raft is made of concrete Ec = 30 GPa and = 0.15). Considering the raft as rigid, the elastic

settlement is

(a) 50.96 (b) 53.36 (c) 63.72 (d) 66.71

Ans. (b) Elastic settlement of Rigid footing =

2

5

(1 )0.8

qB

E

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2

3

110 20 (1 0.3 )0.8 53.38

30 10

kPa mmm

kPa

35. A waste water stream (flow = 2 m

3/s) ultimate BOD = 90 mg/l is joining a small river (flow = 12

m3/s), ultimate BOD = 5 mg/l Both water streams get mixed up instaneously, Cross section area

at the river is 50 m2. Assuming the de-oxygenation const. K = 0.25 /day, the BOD (in mg/l) of

the river water, 10 km downstream of the mixing point is

(a) 1.68 (b) 12.63 (c) 15.46 (d) 1.37

Ans. (c)

After mixing, BOD of combined sample:- 2 90 12 5

( ) / 17.14 /14

BOD mg l mg l

Now, there are two ways of assuming velocity of river flow:-

Case-I – No increase in velocity of stream due to waste water mixing

Velocity of flow =12

0.24 /50

m s

Velocity in kmph= (0.864) km /hr

Time to cover 10 km =11.57 hours =0.482 days

BOD Remaining 0.25 0.48217.14 15.18 /e mg l

If stream velocity is considered after taking into account discharge of wastewater

Velocity =

3

2

14 /0.28 /

50

m sm s

m

Velocity in kmph=1.008 kmph

Time to cover 10 km =(9.92)=0.413 days

0.25 0.41317.14 15.46 /BOD e mg l

36. An effluent at a flow rate of 276 m3/d from a sewage treatment plant is to be distinfected. The

laboratory data of disinfection studies with a chlorine doses in mg/l yields the model 0.145t

t oN N e

Where Nt = no. of micro-organism surviving at time t (in min) and No = no. of

micro-organism present intially (at t = 0) the volume of disinfection unit (in m3) required to

achieve a 98% kill of microorganism is -

Ans. Flow rate =(2760m3/day)

0.145t

t oN N e

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Note:- Nt= no. of microorganism surviving at any time (t)

0.145

0.145

0.02

(0.02)

26.98minutes.

t

o o

t

N N e

e

t

Volume of tank=Detention time

Discharge

Detention time =26.98 min.

3 32760

volume of tank = 2698 51.7160 24

m m

37. In a Marshall sample, the bulk specific gravity of mix and aggregate are 2.324 and 2.546

respectively. The sample includes 5% of bitumen (by total weight of mix) of specific gravity

1.1. The theoretical maximum specific gravity of mix is 2.44 the void filled with btumen (VFB)

in the Marshal sample (in %) is -

Ans. volume of air voids =VV=

2.441 2.324100 100 (4.793)%

2.441

t m

m

G G

G

Vb=volume of voids with Bitumen =5

2.324 (10.564)%1.1

Total voids = (4.793)% (10.564)% 15.35%V b

V V

Voids filled with Bitumen =10.564

100 68.78%15.357

38. A tachometer was placed at point P is estimate the horizontal distances PQ and PR. The

corresponding stadia intercepts with the telescope kept horizontal are 0.320 and 0.210 m

respectively. The QPR is measured to be 61°30’30’’. If the stadia multiplication constant = 100 and stadia addition const. = 0.10 m. The horizontal distance (in m) between the points Q

and R is –

33. Ans : According to Tacheometric method:

Horizontal distance = (Ks + C) = D

Where,

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D = Horizontal distance

K = Multiplying constant

C = additive constant

Distance between P and Q = (100 × 0.32) + 0.10 = (32.1)m

Distance between P and R = (100 × 0.210) + 0.1m = 21.1m

Triangle PQR

From Cosine Rule

2 2 2

2

( ) ( ) 2( )( )cos61'30'30'')

830

28.80

QR PQ PR PQ PR

QR

QR m

39. On a section of a highway the space-density relationship is linear and is given by 2

(80 )3

v K

where v = km/hr K = Veh/km the capacity (in veh/h) of the section of the highway would

(a) 1200 (b) 2400 (c) 4800 (d) 9600

Ans. (b) capacity at section of Highway is a function of speed and density

capacity = velocity × Density

Capacity =2

803

K K

2280

3

Kq k

For maximum capacity

0

4,80 0

3

60

dq

dk

kor

k

Density at maximum capacity =(60)

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Speed at maximum capacity = 2

80 60 (40)3

capacity = 40 60 2400

40. A horizontal nozzle of 30 mm diameter discharges a steady jet of water into the atmosphere at

a rate of 15 litre per second. The diameter of inlet to the nozzle is 100 mm. The Jet impinges.

Normal to a flat stationary plate held close to the nozzle end. Neglecting air friction and

considering the density of water as 1000 kg/m3, the force exerted by the jet (in N’) on the plate is ________

Ans : Force on plate = change in momentum of fluid

Force on plate = mass of fluid striking per second × change invelocity.

Force on plate = ( × a × V) (V – 0) = (aV2)

= 1000kg/m3

a = area of nozzle.

V = Velocity of Jet.

Velocity of Jet 3

2

15 1021.2 /

(0.03)4

m s

Force on plate = (aV2) 2 23.14

1000 (0.03) (21.2) (318.2 )4

N

Note: In this question there is no use of diameter inlet of nozzle. The diameter of horizontal

nozzle i.e. outlet only matters.

41. An infinity long slope is made up of a C-

soil having the properties. cohesion (C) = 20 Kpa

and dry unit wt ( d

) = 16kN/m3. The angle of inclination and critical height of the slope are

40° and 5m respectively. To maintain the limiting equilibrium, the angle of internal friction of

the soil (in degree)_________

Ans : To maintain equilibrium safe height of slope.

2

o 2

(tan tan )cos

20

16(tan 40 tan )cos 40

d

CH

angle of slope

Shear angleof soil

H

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PH. 011-26514888

o 2

5

205

16 tan 40 tan cos 40

tan 40 tan 0.426

H m

o

0.84 tan 0.426

tan 0.44 22.44

42. In any given year, the probability of an earthquake greater than magnitude 6 occurring in the

Garhwal Himalayas is 0.04. The average time between successive occurance of such arthquake

in years.

Ans. Average time interval =1 1

25.yearsProbablity 0.04

43. Irrigation water is to be provided to a crop in field to bring the moisture content of the soil from

the existing 18% to the field capacity of the soil at 28%. The effective root zone of the crop is

70 cm. If the densities of the soil and water are 1.3 g/cm3 and 1g/cm

3 respectively, Depth of

irrigation water (in mm)-

Ans. Root zone depth =(70 cm )

d Density of soil =(1.3g/cc)

Available moisture =18%

Field capacity =(28)%

Thus, in order to make current moisture content from 18% to 28% Required moisture =28%-

18% =10%

1.3

70 0.1 9.1 ofwater.1

Depeth cm

Since answer is asked in mm

Depth =(91mm) .

44. A suspension of sand like particles in water with particles of diameter 0.10 mm and below is

falling into a sedimentation tank at 0.01 m3/s, g = 9.8 m/s

2, Specific gravity of particles = 2.65,

and kinematic viscosity of water = 1.0105 × 10–2

cm2/s. The min. surface area (in m

2 required

for this setting tank to remove particles of size 0.06 mm and above with 100% efficiency is Ans. Discharge =0.1 m

3/s

Surface overflow rate =

3

2

0.1 /m s

Am

23




PH. 011-26514888

Whore, A is surface area in (m2). According to stokes haw:- settling velocity =

21

18

sG gd

v

3 2

6

1.65 9.81 (0.06 10 ) 0.1

18 (1.010 10 ) A

231.2A m

45. Find shear force, moment and axial stress at A. Assume pully to be frictionless.

Ans. axial stress = 250

(1250) /0.2 0.2

KNKN m

46. Closest meaning- as a women i have no. country

(a) Women have no country

(b) Women are not citizen of any country

(c) Women’s’ solidarity knows no national boundaries

(d) Women’s of all countries have earn legal rights-

Ans. (c)

47. An observes counts 240 veh/hr at a specific highway location Assuming that the vehicle arrives

at the location is poisson distributed, the probabality of having one vehicle arriving over a 30

second time in the interval is- Ans. According to passion distribution.

( )

( )n t

n

t eP n

L

=average vehicle flow or arrival rate in vehicle/ time.

t=duration of time interval over which vehicles are counted.

24




PH. 011-26514888

Since we are being asked with vehicles in 30 seconds, therefore we to find vehicles per second.

240 1

3600 15

1 130

15

2

130

15(1)

(1) 2. 0.270

n

e

PL

P e

48. A person suffering from Alzheimer’s disease... short term memory loss. (a) Experienced (b) has experienced

(c) is experience (d) experiences

Ans. (d)

49. X is 1km northeast of Y. Y is 1 km southeast of Z, W is 1 km west of Z, D is 1 km south of W.

Q is 1 km east of D. Distance between x and Q in km

(a) 1 (b) 2 (c) 3 (d) 2

50. 0

1lim

x

is equal to

(a) log x (b) 0 (b) x log x (d)

Ans. (a)

0

1limx

x

This is 0

0

from and hence we can apply L’ Hospital Rule

0 0

( 1)1

lim lim (log )

( )x x

dx

x d xd

d

51. The beam of an overall depth 250 mm (shown below) is used in a building subjected to two

different thermal environments. Temperature at top and bottom surface of the beam is 36°C and

72°C, 5150 10 per ºC. The vertical deflection of the beam (in m) at its mid span is

Ans : Temperature at bottom = (72ºC)

Temperature at top = (36ºC)

This means the average change in temperature = (36ºC)

Strain 6

4150 10 36(27 10 )

2 2

T

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PH. 011-26514888

According to theory of Bending

1

M E FyI R

f straincurvature

R Ey y

At bottom fibre y = h/2 2

2

TT

Curvatureh h

2 2 2

8 8 8

L L T TLDeflection

R h h

In the question

= 15 × 10-5

T = 36º

L = 13m h = 0.25

2 515 10 36 13 13, (45.6)

8 8 0.25

TLThus Deflection downward mm

h

52. In a group of four children, Som is younger to Riaz. Shiv is elder to Ansu. Ansu is youngest in

the group. Which of the following statements is required to find the eldest child in the group?

Statement -

1. Shiv is younger to Riaz

2. Shiv is elder to Som

A. Statement 1 by itself determines the eldest child

B. Statement 2 by itself determines the eldest child

C. Statement 1 & 2 are both reqquired to determine the eldest child.

D. Statement 1 & 2 are not sufficient to determine the eldest child.

53. A venture meter having a throat diameter of 0.1 m is used to estimate the flow rate of a horizontal

Pile having a diameter of 0.2 m. For a observed pressure difference of 2m of water head and

Coefficient of discharge equal to unity, assuming that the energy losses are negligible, flow rate

(in m3/s) is

(a) 0.500 (b) 0.150 (c) 0.050 (d) 0.015

Ans. Throat diameter =(0.1m)

Pipe diameter =(0.2m)

Pressure difference =(2m)

For venture meter discharge, we can use the direct formulae:-

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PH. 011-26514888

2

2 2

31 2

2 2 2 2

1 2

(0.1) (0.2)4

2 2 9.81 2 0.05 /

0.2 0.14 4

a aQ gh m s

a a

54.

The tension ( in kN) in a 10 m long cable (neglecting self wt)...............

(a) 120 (b) 75 (c) 60 (d) 45

Ans.

According to symmetry.

2Tcos =12.

60 5

754

T KN

55. A pre-timed four phase signal has critical lane flow rate for the first three phases as 200,187 and

210 veh/hr. with saturation flow rate of 1800 veh/hr/ lanes . The lost time is given as 4 second

27




PH. 011-26514888

for each phase, If the cycle length is 60 sec, the effective green time (in second) of the fourth

phase is-

Ans. This is related to Webster method.

11

1

22

2

33

3

200

1800

187

1800

210

1800

qy

S

qy

S

qy

S

1 2 3 4 40.33167y y y y y y

Total no. of phase =4

Time lost per phase =4

Total lost time =16 seconds.

According to Webster method, the optimum signal cycle is:-

1.5 5

1o

LC

y

Where, L= total lost time per cycle

4

4

1.5 16 560

1 0.33167

0.185

y

y

Effective green time = 4

1 2 3 4

( ) (60 16) 0.18515.755sec .

0.33167 0.185

oC L yonds

y y y y

56.

Group-I

P. Pressure meter

Q. Static cone penetration test

R. Standard penetration test

S. Vane shear test

Group –II

1. Menards modulus (En)

2. No. of blows

3. Skin resistance

4. untrained cohesion

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PH. 011-26514888

Ans. Pressure meter test is conducted to measure at rest horizontal Eauth pressure. Pressure Louis

menard developed it.

Pressure meter is related to menard modulus.

Static cone penetration test is recognized as a standard field test to collect data about bearing

capacity and flectional resistance of soil.

Standard penetration test is used to measure no. of Blows of resistance into soil for 150 mm to

450mm.

Vane shear test can be used in field as well as laboratory to measure undrained cohesion.

57. Water is flowing at a steady rate through a homogeneous and saturated horizontal soil strip of 10 m length. The strip is being subjected to a constant water head of 5m at the beginning and 1m at

the end. If the governing equation of flow in the soil is

2

20(

d Hx

dx distance along the soil

strip) value of H(in m) at the middle of the strips- Ans. This is question of mathematics

2

20

d H

dx

dHK

dx

H kx C

At 0, 5 5x H c

At 0, 1 1 10 5x H K

K 0.4

0.4 5H x

At x=5m or at the midpoint

0.4 5 5 3H m

58. The population of new city 18.5 million and is growing at 20% annually. How many years would

it take to double. at the growth rate

(a) 3-4 year (b) 4-5 year (c) 5-6 year (d) 6-7 year

Ans. (a) Since population gets doubled

(20) 10(1 0.20)

2 (1.20)

3.8

n

n

n years

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PH. 011-26514888

59. A prismatic beam has plastic moment capacity of Mp, the collapse loan P of the beam-

(a) 2 pM

L (b)

4 pM

L (c)

6 pM

L (d)

8p

M

L

Ans.

Let us drew Bending moment Diagram:-

3

2A B

PV V

4

2 3 2A

P L LV L P

4 7

6 2 6

3

B

A

P P PV

PV

Bending moment diagram.

Maximum moment = 6

p

PLM

6 pM

PL

60. The axial load in (kN) in the rod PQ for the arrangement below is – R

30




PH. 011-26514888

Ans :

This question is though incomplete and thus needs an assumption. The assumption in this case

would be to assume vertical rod (PQ) as rigid.

The deflection at Q can be assumed to be equal to (zero).

Let the axial load at (Q) be equal to (A) KN.

Let us see deflection at (Q)

At (Q) deflection due to A KN = 3(4)

3

A

EI

Due to 160KN

Deflection at 3 2160 (2) 160 (2)

23 2

QEI EI

The deflection due to A KN is in upward direction. The deflection due to 160KN will be in

downward direction.

3 2 3160(2) 160(2) (2) (4)

3 2 3

64 160 8 160 4 3

50

A

EI EI EI

A

A KN

Thus, Tension in PQ = 50KN

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PH. 011-26514888

61. Surface water treatment plant operates round the clock with a flow rate of 3.5 m3/min. The

water temperature 15°C and jar testing indicated an alum dosage of 25 mg/l, Gt = 4×104 for

Producing optimum result. The alum req. for 30 days (in kg) is - Ans. The use of Gt in this question is not there.

Quantity of alum in 30 days=

33.5 10 60 24 30 25 10 3780 .kg

62. _________ is the key to their happiness. They are satisfied with what they have

(a) Contentment (b) ambition (c) perseverance (d) hunger

Ans.(a) contentment is acknowledgement and satisfaction of reaching capacity.

Answer of Few Questions is pending due to lack of data.

ALL Questions is UNDER REVIEW and UPDATED Questions will be available Very soon

Any Query or error report us immediately [email protected]

mailto:[email protected]

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