2015-1 SOLVED PAPER - 2015

GENERAL APTITUDEQUESTION 1 TO 5 CARRY ONE MARK EACH

1. Select the pair that best expresses a relationship similar tothat expressed in the pair:Children : Pediatrician(a) Adult : Orthopaedist(b) Females : Gynaecologist(c) Kidney : Nephrologist(d) Skin : Dermatologist

2. If ROAD is written as URDG, then SWAN should be writtenas:(a) VXDQ (b) VZDQ(c) VZDP (d) UXDQ

3. Extreme focus on syllabus and studying for tests has becomesuch a dominant concern of Indian students that they closetheir minds to anything _______ to the requirements of theexam.(a) related (b) extraneous(c) outside (d) useful

4. A function f(x) is linear and has a value of 29 at x = –2 and 39at x = 3. Find its value at x = 5.(a) 59 (b) 45(c) 43 (d) 35

5. The Tamil version of ______ John Abraham-starrer MadrasCafe _____ cleared by the Censor Board with no cuts lastweek, but the film’s distributors _______ no takers amongthe 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

QUESTION 6 TO 10 CARRY TWO MARKS EACH

6. The exports and imports(in crores of Rs.) of a country fromthe year 2000 to 2007 are given in the following bar chart. Inwhich year is the combined percentage increase in importsand exports the highest ?

Duration: 3 hrs Maximum Marks: 100

INSTRUCTIONS

1. There are a total of 65 questions carrying 100 marks.

2. This question paper consists of 2 sections, General Aptitude (GA) for 15 marks and the subject specificGATE paper for 85 marks. Both these sections are compulsory.

3. The GA section consists of 10 questions. Question numbers 1 to 5 are of 1-mark each, while question numbers 6 to10 are of 2-marks each. The subject specific GATE paper section consists of 55 questions, out of which questionnumbers 11 to 35 are of 1-mark each, while question numbers 36 to 65 are of 2-marks each.

4. Questions are of Multiple Choice Question (MCQ) or Numerical Answer type. A multiple choice question will havefour choices for the answer with only one correct choice. For numerical answer type questions, the answer is a numberand no choices will be given.

5. Questions not attempted will result in zero mark. Wrong answers for multiple choice type questions will result in

NEGATIVE marks. For all 1 mark questions, 13

mark will be deducted for each wrong answer. For all 2 marks questions,

23

mark will be deducted for each wrong answer..

6. There is NO NEGATIVE MARKING for questions of NUMERICAL ANSWER TYPE.

2015SET - 1

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020406080

100120140

2000 2001 2002 2003 2004 2005 2006

ExportsImports

7. Alexander turned his attention towards India, since he hadconquered Persia.Which one of the statements below is logically valid andcan be inferred from the above sentence ?(a) Alexander would not have turned his attention towards

India had 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 movefurther.

8. Choose the most appropriate equation for the function drawnas a thick line, in the plot below.

(0, – 1)

(2, 0) x

y

(a) x = y – |y| (b) x = – (y – |y|)(c) x = y + |y| (d) x = – (y + |y|)

9. The head of a newly formed government desires to appointfive of the six selected members P,Q,R,S,T, and U to portfoliosof Home, Power, Defense, Telecom, and Finance. U does notwant any portfolio if S gets one of the five. R wants eitherHome or Finance or no portfolio. Q says that if S gets eitherPower or Telecom, then she must get the other one. T insistson a portfolio if P gets one.Which is the valid distribution of portfolios ?(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

10. Most experts feel that in spite of possessing all the technicalskills required to be a batsman of the highest order, he isunlikely to be so due to lack of requisite temperament. Hewas guilty of throwing away his wicket several times afterworking hard to lay a strong foundation. His critics pointed

out that until he addressed this problem, success at thehighest level will continue to elude him.Which of the statement(s) below is/are logically valid andcan 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 his technical skills will guarantee success.(a) (iii) and (iv) (b) (ii) and (iii)(c) (i), (ii) and (iii) (d) (ii) only

TECHNICAL SECTION

QUESTION 11 TO 35 CARRY ONE MARK EACH

11. Total Kjeldahl Nitrogen (TKN) concentration (mg/L as N) indomestic sewage is the sum of the concentrations of :(a) Organic and inorganic nitrogen in sewage(b) Organic nitrogen and nitrate in sewage(c) Organic nitrogen and ammonia in sewage(d) Ammonia and nitrate in sewage

12. For the beam shown below, the value of the support momentM is _____ kN-m.

20 kN

3 m 1 m

EI

Internal hinge

EI EI

3 m1 mM

13. The penetration value of bitumen sample tested at 25°C is80. When this sample is heated to 60°C and tested again, theneedle of the penetration test apparatus penetrates thebitumen sample by d mm. The value of d CANNOT be lessthan _____ mm.

14. Consider the singly reinforced beam shown in the figurebelow:

L L/2 L

PX

P

X

At cross-section XX, which of the following statements isTRUE at the limit state ?

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(a) The variation of stress is linear and that of strain isnon-linear

(b) The variation of strain is linear and that of stress isnon-linear

(c) The variations of both stress and strain is linear(d) The variation of both stress and strain is non-linear

15. For the beam shown below, the stiffness coefficient K22 canbe written as

Note: 1, 2 and 3are the d.o.f

1

2

3L

A.E.I

(a) 26EIL

(b) 312EI

L

(c)3EIL

(d) 2EI

6L16. For what value of p the following set of equation will have

no solution ?2x + 3y = 53x + py = 10

17. Consider the following probability mass function (p.m.f) ofa random variable X.

( )q if X 0

p x,q 1 q if X 10 otherwise

=ìï= - =íïî

If q = 0.4, the variance of X is _________.18. A circular pipe has a diameter of 1m, bed slope of 1 in 1000,

and Manning’s roughness coefficient equal to 0.01. It maybe treated as an open channel flow when it is flowing justfull, i.e., the water level just touches the crest. The dischargein this condition is denoted by Qfull. Similarly, the dischargewhen the pipe is flowing half-full, i.e., with a flow depth of0.5m, is denoted by Qhalf.The ratio Qfull/Qhalf is:(a) 1 (b) 2(c) 2 (d) 4

19. For steady incompressible flow through a closed-conduit ofuniform 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

20. Workability of concrete can be measured using slump,compaction factor and Vebe time. Consider the followingstatements for workability of concrete:(i) As the slump increases, the Vebe time increases(ii) As the slump increases, the compaction factor

increasesWhich 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

21. In a closed loop traverse of 1 km total length, the closingerrors in departure and latitude are 0.3 m and 0.4 m,respectively.The relative precision of this traverse will be:(a) 1 : 5000 (b) 1 : 4000(c) 1 : 3000 (d) 1 : 2000

22. Which of the following statements is TRUE for degree ofdisturbance 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 the sampling tube.23. The development length of a deformed reinforcement bar

can be expressed as (1/k) (fss / tbd ) .From the IS:456-2000, the value of k can be calculatedas _________.

24. Which of the following statements is NOT correct ?(a) Loose sand exhibits contractive behavior upon

shearing(b) Dense sand when under undrained condition, may lead

to generation of negative pore pressure(c) Black cotton soil exhibits expansive behavior(d) Liquefaction is the phenomenon where cohesionless

soil near the downstream side of dams or sheet-pilesloses its shear strength due to high upward hydraulicgradient

25. In an unconsolidated undrained triaxial test, it is observedthat an increase in cell pressure from 150 kPa to 250 kPaleads to a pore pressure increase of 80 kPa. It is furtherobserved that, an increase of 50 kPa in deviatoric stressresults in an increase of 25 kPa in the pore pressure. Thevalue of Skemptons’s pore pressure parameter B is:(a) 0.5 (b) 0.625(c) 0.8 (d) 1.0

26. 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 ?

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(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

27. A fine-grained soil has 60% (by weight) silt content. Thesoil behaves as semi-solid when water content is between15% and 28%. The soil behaves fluid-like when the watercontent is more than 40%. The ‘Activity’ of the soil is(a) 3.33 (b) 0.42(c) 0.30 (d) 0.20

28. Solid waste generated from an industry contains only twocomponents, X and Y as shown in the table belowComponent Composition Density

( %weight) (kg/m3)X c1 r1Y c2 r2

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

(a) 1 2

1 2

100c cæ ö

+ç ÷r rè ø(b)

1 2

1 2100

c cæ ör r

+ç ÷è ø

(c) ( )1 1 2 2100 c cr + r (d)1 2

1 1 2 2100

c cæ ör rç ÷r + rè ø

29. The two columns below show some parameters and theirpossible values.ParametersP – Gross Command AreaQ – Permanent Wilting PointR – Duty of canal waterS – Delta of wheatValueI –100 hectares/cumec II – 6° CIII – 1000 hectares IV – 1000 cmV – 40 cm VI – 0.12Which of the following options matches the parameters andthe 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

30. Which of the following statements CANNOT be used todescribe 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

optimum31. 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

32. 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 of the Geoid(d) Geoid is an equi-potential surface of gravity

33. In a two-dimensional steady flow field, in a certain region ofthe x-y plane, the velocity component in the x-direction is

given by vx = x2 and the density varies as 1x

r = .

Which of the following is a valid expression for the velocitycomponent in the y-direction, vy ?(a) vy = – x/y (b) vy = x/y(c) vy = –xy (d) vy = xy

34. The integral 2

1

x2

xx dxò with x2 > x1 > 0 is evaluated analytically

as well as numerically using a single application of thetrapezoidal rule. If I is the exact value of the integral obtainedanalytically and J is the approximate value obtained usingthe trapezoidal rule, which of the following statements iscorrect about their relationship ?(a) J > I(b) J < I(c) J = I(d) Insufficient data to determine the relationship

35. Two Triangular wedges are glued together as shown in thefollowing figure. The stress acting normal to the interface,sn is __ MPa.

100 Mpa 100 Mpa

sn

45°

100 Mpa

100 Mpa

QUESTION 36 TO 65 CARRY TWO MARKS EACH36. Consider the following differential equation:

( ) ( )x yx ydx xdy cos y xdy ydx siny x

+ = -

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

(a) x ycos cy x

= (b)x ysin cy x

=

(c)yxycos cx

= (d)yxysin cx

=

2015-5 SOLVED PAPER - 2015

37. In a survey work, three independent angles X,Y and Z wereobserved with weights WX,WY,WZ respectively. The weightof the sum of angles X,Y and Z is given by:

(a)X Y Z

1 1 11/W W W

æ ö+ +ç ÷

è ø(b)

X Y Z

1 1 1W W W

æ ö+ +ç ÷

è ø

(c) WX + WY + WZ (d) 2 2 2X Y ZW W W+ +

38. Two reservoirs are connected through a 930 m long, 0.3 mdiameter pipe, which has a gate valve. The pipe entrance issharp (loss coefficient= 0.5) and the valve is half-open (losscoefficient = 5.5). The head difference between the tworeservoirs is 20 m. Assume the friction factor for the pipe as0.03 and g =10 m/s2. The discharge in the pipe accountingfor all minor and major losses is _________ m3/s.

39. The quadratic equation x2 – 4x + 4 = 0 is to be solvednumerically, starting with the initial guess xo = 3. The Newton-Raphson method is applied once to get a new estimate andthen the Secant method is applied once using in the initialguess and this new estimate. The estimated value of theroot after the application of the secant method is___________.

40. The acceleration –time relationship for a vehicle subjectedto non-uniform acceleration is,

( ) to

dV V edt

-b= a-b

Where, V is the speed in m/s, t is the time in s, a and b areparameters, and V0 is the initial speed in m/s. If theaccelerating behavior of a vehicle, whose driver intends toovertake a slow moving vehicle ahead, is descried as,

( )dv Vdt= a -b

Considering a = 2 m/s2, b = 0.05 s–1 and 2dV 1.3m / sdt

= at

t = 3s, the distance (in m) travelled by the vehicle in 35 sis _______

41. A square footing (2 m × 2 m) is subjected to an inclined pointload, P as shown in the figure below. The Water table islocated well below the base of the footing. Considering oneway eccentricity, the net safe load carrying capacity of thefooting for a factor of safety of 3.0 is _________ kN.The following factors may be used:Bearing capacity factors: Nq = 33.3, Ng = 37.16; Shape factors: Fqs = Fgs=1.314; depth factors: Fqd = Fgd = 1.113; Inclinationfactors: Fqi = 0.444, Fgi = 0.02

0.85 m

30°

2 m

P

GL

1 m

Unit weight = 18 kN/m3

Cohesion = 0Friction angle = 35°

42. A tapered circular rod of diameter varying from 20 mm to10 mm is connected to another uniform circular rod of diameter10 mm as shown in the following figure. Both bars are madeof same material with the modulus of elasticity, E = 2 × 105

MPa. When subjected to a load P= 30p kN, the deflection atpoint A is ________ mm.

1.5 m

d = 20 mm1

d = 10 mm2

P = 30 kNp

A

2 m

43. An earth embankment is to be constructed with compactedcohesionless soil. The volume of the embankment is 5000 m3

and the target dry unit weight is 16.2 kN/m3. Three nearlysites (see figure below) have been identified from where therequired soil can be transported to the constructed site.The void ratios(e) of different sites are shown in the figure.Assume the specific gravity of soil to be 2.7 for all threesites. If the cost of transportation per km is twice the cost ofexcavation per m3 of borrow pits, which site would youchoose as the most economic solution ?( use unit weight ofwater = 10 kN/m3)

Site Xe = 0.6

Site Ye = 0.7

Site Ze = 0.64

Constructionsite

100 km

140 km

80 km

(a) Site X (b) Site Y(c) Site Z (d) Any of the sites

2015 -6 SOLVED PAPER - 2015

44. In a catchment, there are four rain – gauge stations, P,Q,R,and S. Normal annual precipitation values at these stationsare 780 mm, 850 mm, 920 mm, and 980 mm, respectively. Inthe year 2013, stations Q,R, and S, were operative but P wasnot. Using the normal ratio method, the precipitation atstation P for the year 2013 has been estimated as 860 mm. Ifthe observed precipitation at stations Q and R for the year2013 were 930 mm and 1010 mm, respectively, what was theobserved precipitation (in mm) at station S for that year ?

45. A bracket plate connected to a column flange transmits aload of 100 kN as shown in the following figure. The maximumforce for which the bolts should be designed is _____ kN.

75

75

75 75All dimensions are in mm

600

100 kN

46. For the 2D truss with the applied loads shown below, thestrain energy in the member XY is _______ kN-m. Formember XY, assume AE=30kN, where A is cross section areaand E is the modulus of elasticity.

5 kN

3 m

3 m

3 m

X

10 kN

Y

3 m

47. The concentration of Sulphur Dioxide (SO2) in ambientatmosphere was measured as 30 mg/m3. Under the sameconditions, the above SO2 concentration expressed in ppmis ___.Given: P/(RT) = 41.6 mol/m3; where, P = Pressure;T = Temperature; R= universal gas constant; MolecularWeight of SO2 = 64.

48. On a circular curve, the rate of superelevation is e. Whilenegotiating the curve a vehicle comes to a stop. It is seenthat the stopped vehicle does not slide inwards (in the radialdirection). The coefficient of side friction is f. Which of thefollowing is true:(a) e £ f (b) f < e < 2f(c) e ³ 2f (d) none of the above

49. 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 coarseaggregate to be 3.14, 2.67 and 2.74, respectively. The aircontent is ______ litres/m3.

50. Consider a primary sedimentation tank (PST) in a watertreatment plant with Surface Overflow Rate (SOR) of 40 m3/m2/d. The diameter of the spherical particle which will have90 percent theoretical removal efficiency in this tank is________ mm. Assume that settling velocity of the particlesin water is described by stokes’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

51. The smallest and largest Eigen values of the following matrixare:

3 2 24 4 62 3 5

-é ùê ú-ê úê ú-ë û

(a) 1.5 and 2.5 (b) 0.5 and 2.5(c) 1.0 and 3.0 (d) 1.0 and 2.0

52. A hydraulic jump is a 2m wide rectangular channel which ishorizontal and frictionless. The post-jump depth and velocityare 0.8 m and 1m/s, respectively. The pre-jump velocity is___________ m/s.(use g = 10 m/s2)

53. The drag force, FD on sphere kept in a uniform flow fielddepends on the diameter of the sphere, D; flow velocity, V,fluid density, r ; and dynamic viscosity, m. Which of thefollowing options represents the non-dimensionalparameters which could be used to analyze this problem ?

(a) DFand

VD VDm

r (b)D

2F VDandVD

rmr

(c)D2 2

F VDandV D

rmr (d)

D3 3

F andVDV Dm

rr54. The directional derivative of the field u(x, y, z) = x2– 3yz in

the direction of the vector ( )ˆ ˆ ˆi j 2k+ - at point (2,–1,4) is

_________.55. A water tank is to be constructed on the soil deposit shown

in the figure below. A circular footing of diameter 3 m anddepth of embedment 1m has been designed to support the

2015-7 SOLVED PAPER - 2015

tank. The total vertical load to be taken by the footing is1500 kN. Assume the unit weight of water as 10 kN/m3 andthe load dispersion pattern as 2V:1H. The expected settlementof the tank due to primary consolidation of the clay layer is______mm.

Silty Sand

Sand

Normally Consolidated clay

Bulk unit weight = 15 kN/m3

Saturated unit weight = 18 kN/m3

Saturated unit weight = 18 kN/m3

Compressed index = 0.3Initial void ratio = 0.7Coefficient of consolidated = 0.004 cm /s

2

Dense sand

6 m

10 m

2 mGL

GWT

56. The 4-hr unit hydrograph for a catchment is given in thetable below. What would be the maximum ordinate of theS-curve (in m3/s) derived from this hydrograph?

Time(hr) 0 2 4 6 8 10 12 14 16 18 20 22 24

Unit hydrograph ordinate (m3/s)

0 1 3 10 13 9 5 2 1 0 0 0 0

57. Consider the singly reinforced beam section given below(leftfigure). the stress block parameters for the cross- sectionfrom IS: 456-2000 are also given below (right figure).The moment of resistance for the given section by the limitstate method is ______ kN-m.

300 mm

200 mm

M25

4-12 F

Fe415

0.42xu

0.36 Xfck u

X , max = 0.48dufor Fe415

Xu

d

58. A 20m thick clay layer is sandwiched between a silty sandlayer and a gravelly sand layer. The layer experiences 30 mmsettlement in 2 years.

( )

2

V

10

U for U 60%T 4 1001.781 0.933log 100 U for U 60%

ì p æ öï £ç ÷= í è øï - - <î

Where Tv is the time factor and U is the degree ofconsolidation in %

If the coefficient of consolidation of the layer is 0.003 cm2/s, thedeposit will experience a total of 50 mm settlement in thenext ________ years.

59. Two beams are connected by a linear spring as shown in thefollowing figure. For a load P as shown in the figure, thepercentage of the applied load P carried by the springis ________.

K = 3EI/(2L )S3

EI

L P

EI

60. A short reach of a 2 m wide rectangular open channel has itsbed level rising in the direction of flow at a slope of 1 in10000. It carries a discharge of 4 m3/s and its Manning’sroughness coefficient is 0.01.The flow in this reach is gradually varying. At a certain section inthis reach, the depth of flow was measured as 0.5 m. The rate ofchange of the water depth with distance, dy/dx, at this sectionis ___(use g = 10 m/s2)

61. A non-homogeneous soil deposit consists of a silt layersandwiched between a fine-sand layer at top and a claylayer below.Permeability of the silt layer is 10 times the permeability ofthe clay layer and one-tenth of the permeability of the sandlayer.Thickness of the silt layer is 2 Times the thickness of thesand layer and two-third of the thickness of the clay layer.The ratio of equivalent horizontal and equivalent verticalpermeability of the deposit is ________.

62. For formation of collapse mechanism in the following figure,the minimum value of Pu is cmp/L. mp and 3mp denote theplastic moment capacities of beam sections as shown in thisfigure. The value of c is________.

3Mp

2m

1m Pu

mp

1m

63. Consider the following complex function:

( )( ) ( )2

9f zz 1 z 2

=- +

Which of the following is one of the residues of the abovefunction ?(a) –1 (b) 9/16(c) 2 (d) 9

2015 -8 SOLVED PAPER - 2015

64. A sign is required to be put up asking drivers to slow downto 30 km/h before entering Zone Y (see figure). On this road,vehicles require 174 m to slow down to 30 km/h( the distanceof 174 m includes the distance travelled during the perception– reaction time of drivers). The sign can be read by 6/6vision drivers from a distance of 48 m. The sign is placed ata distance of x m from the start of Zone Y so that even a 6/9vision driver can slow down to 30 km/h before entering thezone. The minimum value of x is ________ m.Direction of vehicle movement.

Sign

Road

Start of Zone Y

Zone Yx

65. In a region with magnetic declination of 2°E, the magneticFore bearing(FB) of a line AB was measured as N79°50¢E.There was local attraction at A. To determine the correctmagnetic bearing of the line, a point O was selected at whichthere was no local attraction. The magnetic FB of line AOand OA were observed to be S52°40¢E and N50°20¢W,respectively. What is the true FB of line AB ?(a) N81°50¢E (b) N82°10¢SE(c) N84°10¢E (d) N77°50¢E

2015-9 SOLVED PAPER - 2015

GENERAL APTITUDE

1. (b) First part is the community of people second part isthe doctor particular to those people.

2. (b) Each alphabet in code is 3 letters from thecorresponding alphabet of the word.

R : R + 3 = UO : O + 3 = RA : A + 3 = DD : D + 3 = G

\ S : S + 3 = VW : W + 3 = ZA : A + 3 = DN : N + 3 = Q

3 (b) extraneous-meaning : irrelevant or unrelated to asubject being dealt with.

4. (c) Let f(x) = ax + bGiven f(x) = 29 at x = –2\ –2a + b = 29 ..........(1)Given f(x) = 39 at x =3\ 3a + b = 39 ..........(2)Subtracting (2) from (1)Þ 5a = 10\ a = 2Substitute in (1) or (2)Þ b = 33\ f(x) = 2x + 33\ At x = 5

f(x) = 2 × 5 + 33 = 435 (c)6. 2006

From the graph itself it can be seen that there is asudden increase in import and export in 2006In 2006

% increase in export = 100 70

70-

= 42.8%

% increase in import = 120 90

90-

= 33.3%

7. (a)8. (b) From the graph, when y = – 1, x = 0

Substitute in the option given.At y = –1

(a) y – | y | = –1 –1 = –2(b) – (y – | y |) = – (–1 –1) = 2(c) y + | y | = –1 + 1 = 0(d) – (y + | y |) = + (–1 + 1) = 0

\ Answer is (b)

9. (b) U does not want any portfolio if S gets one of five\ (c) is ruled out.R wants home or defence or no portfolio\ (a) and (d) are also ruled out\ Answer is option (b)

10. (b)

TECHNICAL SECTION

11 (c) T.KN = Ammonia (60%) + Organic Nitrogen (40%)12. 5

The beam is symmetrical\ Take half of beam to left. The same moment M will

be there on other end.

M3 m 1 m

10 kN

10 12

M ´= = 5 kNm

13. 814. (b) Variation of strain Variation of stress

15. (b) 26EI

LD

312EI

LD

2

3

1

22 312 E IK =

L16. 4.5

2x + 3y = 5 ...(1)3x + py = 10 ...(2)Ax = B

Þ2 3 53 10

é ù é ùé ù =ê ú ê úê úë ûë û ë ûxyp

2015 -10 SOLVED PAPER - 2015

Augmented matrix, [A/B]

[A/B] = 2 3 53 10p

é ùê úë û

= 2 3 50 2 – 5 5p

é ùê úë û R2 ® 2R2– - 3R1

There is no solution if p(A/B) ¹ p(A)\ 2p – 9 = 0

Þ p = 9/2 = 4.5Shortcut methodMultiply first equation with 3 and second equationwith 2Þ 6x + 9y = 15

6x + 2py = 20From the equation it is clear that, there is no solutionif 9 = 2pÞ p = 9/2 = 4.5

17. 0.24

0( , ) 1 – 1

0 otherwise

=ìï= =íïî

q if xp x q q if x

When q = 0.4

0.4 0( , ) 0.6 1

0 otherwise

=ìï= =íïî

if xp x q if x

0 1( ) 0.4 0.6=

xp X x

Variance of x, V(X) = E(X2) – [E(X)]2

E(X) = Sxi pi = (0 × 0.4) + (1 × 0.6) = 0.6E(X2) = Sxi

2 pi = (02 × 0.4) + (12 × 0.6) = 0.6

V(X) = 0.6 – (0.6)2

= 0.6 – 0.36= 0.24

18. (c) Discharge, Q = 1n A R2/3 S1/2

Qfull = 2 / 3

2 1/ 214 4

DD Sn

pæ öæ öç ÷ç ÷è øè ø

Qhalf = 1n .

2 / 32 1/ 2

8 4DD Spæ öæ ö

ç ÷ç ÷è øè ø

full

half2=

QQ

19. (b) Flow will be from high pressure to low pressure.20. (d) As slump increases, Vebe time increases.

21. (d) 2 2e l d= +l = 0.3, d = 0.4

\ 2 20.3 0.4e = +

= 0.25 = 0.5 m

Relative precision 0.5

1000= = 1:2000

22. (d) When the thickness of sampler wall is less, thedegree of disturbance of collected sample is lower.

23. 6.4

4fs

=t

std

bdL

For deformed bars, tbd is increased by 60%

\ 4 ( 1.6)fs

=´ t ´d

st

bdL

6.4st st

bd bdkfs fs

= =t t

\ k = 6.424. (d) Liquefaction is due to cyclic load and not due to

high hydraulic gradient.25. (c) Skempton’s pore pressure parameter, B

3

D=

DsUB

80 80250 –150 100

= = = 0.8

26. (a) Both statements are true

27. (c) Activity = pI

C

Ip = WL – Wp= 40 – 28 = 12

\ Activity 12

100 60=

- = 0.3

28. (a) Composite density, r

1 2 1 2

1 2

c c c c+= +

r r r

Þ1 2

1 2

100 c c= +

r r r

\ r = 1 2

1 2

100c cé ù

+ê úr rë û

2015-11 SOLVED PAPER - 2015

29. (b) Gross command area – 1000 hectaresPermanent wilting point – 0.12Duty of canal water – 100 hectares/cumecDelta of wheat – 40 cm

30. (d) uf is the speed at which flow is negligible and densityis also negligible.

31. (b) Seepage velocity, sVVn

=

n(porosity) is always less than 1\ Vs > V (always)

32. (b) Mean sea level (MSL) is used as a reference surfacefor establishing the vertical control.

33. (c) Continuity equation

( ) ( ) 0¶ ¶r + r =¶ ¶x yv vx y

Þ1

( ) . 0¶ ¶ æ ö+ =ç ÷è ø¶ ¶ yx vx y x

Þ 1 0æ ö¶+ =ç ÷¶ è ø

yvy x

Þ –1æ ö¶ =ç ÷¶ è ø

yvy x

Þ vy = – xy34. (a) The approximate value obtained by trapezoidal

method (J) is always greater than analytic method (I)J > I

35. 0 Normal stress, s n

s n = cos22 2

s + s s - s+ qx y x y

100 100 100 (–100) cos (2 45)2 2- -

= + ´

2000 02

= + ´ = 0

36. (c) Given x(y dx + x dy) cos yx

= y(x dy – y dx) sin yx

Þ x(y dx + x dy) cos yx + (– sin y

x ) y(x dy – y dx)

= 0Divide by x

Þ (y dx + x dy) cos yx + (– sin y

x)

( – )y xdy ydxx

= 0

Þ (y dx + x dy) cos yx + – sin y

xæ öç ÷è ø

( – ) 0=xdy ydx xy

x

Þ ( ) cos 0yd xyx

æ ö =ç ÷è ø

Integrating

xy. cos y cx

=

37. (a)38. 0.1413

Total loss = difference in head = 20m.

Entry loss = 20.5

2vg

Loss due to value = 25.5

2vg

Exit loss = 2

2vg

Friction loss = 2

2f Lv

g d

\2 2 2 20.5 5.5 20

2 2 2 2v v v fLvg g g g d

+ + + =

2 0.03 9300.5 5.5 1 202 0.3vg

´æ ö+ + + =ç ÷è ø

Þ2

1002vg´ = 20 Þ v = 2 m/s

q = 24

d vp´ ´

2(0.3) 24p

´ ´ = 0.1413 m3/s

39. 2.333f(x) = x2–4x + 4Given, x0 = 3f¢ (x) = 2x – 4Newton – Raphson method

x1 = x0 – 0

0

( )'( )

f xf x

13 – 2.52

= =

Secant methodx0 = 2.5, x1 = 3

2015 -12 SOLVED PAPER - 2015

x2 = x1 – 1 01

1 0( )

( ) – ( )x x

f xf x f x

-

= 3 – (3 – 2.5) . (3)

(3) – (2.5)f

f f

= 0.53 – 1

1– 0.25´

= 3 – 0.667= 2.333

40. 900.83Given,

dVdt = (a – bV0) e–bT

Integratingò dV = ò(a–bV0).e–bt.dt

Þ–

0( – ).C

–

tV eV

ba b= +

b

At t = 0, V = V0

V0 = 0( – )–

Va bb + C

00

( – )VC V

a b= +

b

0 0–V Vb + a b=

b

Þ C = ab

Þ V = –0( )

.–

tVe ba - b a

+b b

= –

0– ( – )te Vba a bb

Integrating

ò V = x = 0–0 0–12

( )( )tt Ve ba a - b

+b b

At t = 3

–30( ) 1.3dV V e

dtb= a -b =

Þ a – bV0 = –31.3

e b

\ x = 0–0–3 2

1.3 ( –1)tte

eb

ba

+b ´b

a = 2, b = 0.05

\ x = –35 0.05

2 –3 0.052 35 1.3( –1)0.05 (0.05) ( )

´

´´

+e

e

= 1400 – 499.17 = 900.83m

41. 450 qsafe 3nuq

=

qnu = cNc + qNq + 0.5 Bg Ng – 8D c = 0 (Cohesion = 0)qnu = q (Nq – 1) + 0.5 Bg Ng

1( 1) 0.5

3 g g g gé ù= - ´ + + g ´ ´ ´ë ûq qs qd qp s o pq N F F F B N F F F

1 1= 18×(33.3 –1)×1.314×1.113 0.444 +

3 2é +êë

× 2 ×

18 × 37.16 × 1.314 × 1.113 × 0.02 ùû

397.033

= = 132.364 kN/m2

0.85 0.852

2

Reduced area of footing = 2 × 1.7 = 3.4 m2

Load carrying capacity = 132.364 × 3.4 = 450 kN

42. 1520 mm

30 p

30 p

10 mm

1PLAE

D =1 2

4.

PLd d E

=p

6

54 30 2 10

20 10 2 10´ p´ ´

=p´ ´ ´ ´

= 6 mm

2015-13 SOLVED PAPER - 2015

2PLAE

D =

30 p

30 p6

54 30 1.5 10

10 10 2 10´ p´ ´

=p´ ´ ´ ´

= 9 mm

\ D = D1 + D2 = 15mm43. (b) V a l + e

\ 1 1 1 1= = =

+ + + +yx z

x y z

VV VVe e e e

gd = .

1wGeg+

Þ 16.2 = 2.7 10

1 e´+

Þ 1 + e = 1.67\ e = 0.67

\5000

1 .67 1 .6 1 .64 1 .7yx zVV V

= = =+ + + +

Þ 35000 1.6 4790.41.67xV m´

= =

35000 1.64 4910.181.67yV m´

= =

35000 1.7 5089.81.67zV m´

= =

Let cost of excavation per unit m3 be C.\ Cost of excavating and transporting from eachsiteCx = C × 4790.42 + (2 × C × 140) = 5070.4 CCy = C × 4910.18 + (2 × C × 80) = 5070.1 CCz = C × 5089.82 + (2 × C × 100) = 5289.8 C

44. 1076.2

13

s p Q R

s p Q R

P P P PN N N N

é ù= + +ê úê úë û

Þ1 860 930 1010

980 3 780 850 920sP é ù= + +ê úë û

Ps = 1076.20 mm

45. 156.20

100F 5D

Pn

= = = 20 kN

75

75 2

75 75

75 q

75

75

2( . )F tP d r

r=

S

2100 600 75 2

4 (75 2)´ ´

=´

= 141.42 kN

2 2 2 cosR D t D tF F F F F= + + ´ ´ q

2 2 75(20) (141.42) 2 20 141.4275 2

= + + ´ ´ ´

400 19999.616 3999.96= + += 156.204 kN

46. 5

3m

10 kN5 kN

E

C

x

A B

y

D3m

3m

3m

RA × 3 + 10 × 9 = 0Þ RA = –30 kNRB = 35 kNTaking joint A

10 kN

30 kN

10 kN

30 kN

2015 -14 SOLVED PAPER - 2015

Taking joint B

10 2

10 kN 35 kNB

Taking joint xFx = 10 kN

2

2 E

F LUA´

=

210 32 3

´=

´ = 5 kNm

10

10 230

47. 0.0133Given concentration of SO2 = 30 mg/m3

106 m3 of air has 30gSO2 3064

= mol SO2

/nRT nV

p p RT= =

330 / 6441.6 /

molmol m

= = 0.0113m3

Conc. of SO2 in ppm = 0.0113 ppm48. (a) Since stopped vehicle does not

slide inside,

p

w

F N

q

w.sinq

q

F > w.sinqF = P cosqP = w.fwhere,

P = centrifugal forcef = coeff of lateral friction

Þ P.cosq > w sin qwf. cosq > w sin qf > tan q

Þ f > eÞ e < f

49. 51

1c s aw a

c s a

M M MV V+ + + + =

r r r

Þ368 606

3.14 1000 2.67 1000+

´ ´

1155 184 12.74 1000 1000

+ + + =´ aV

Þ 0.117 + 0.227 + 0.421 + 0.184 + Va = 1\ Va = 0.051

= 51 l/m3

50. 22.58

% removal ' 100= ´s

s

VV = 90%

\ Vs¢= 0.9 Vs

=

3 2

–4

40 / /40 /

3600 244.63 10 /

sV m m d

m s

m s

ì =ïï =í

´ï= ´ïî

2( –1)18s

gV G dv

=¢

\18

( –1)sV vdg G´ ´=

¢

–4 –6(0.9 4.63 10 ) 18 1.10 102.6509.81 –11000

´ ´ ´ ´ ´=

æ öç ÷è ø

= 22.58 mm

51. (d)3 –2 24 –4 62 –3 5

Aé ù

= ê úê úë û

Characteristic equation|A – l I| = 0

3 –2 24 –4 – 62 –3 5 –

- ll

l

Þ l3 – 4l2 + 5l – 2 = 0(l –1) (l2– 3l + 2) = 0(l –1) (l –1) (l –2) = 0l = 1, 1, 2

Smallest and largest are 1 and 2.

2015-15 SOLVED PAPER - 2015

52. 4.94B = 2m, y2 = 0.8m, V2 = 1 m/s

22

2=

VF

gy

1 0.3510 0.8

= =´

212

2

1 1– 1 82 2

yF

y= + +

Þ 21 1 1 1 8 (3.5)0.8 2 2

-= + + ´

y = 0.203

\ y1 = 0.8 × 0.203 = 0.162 mQ = AV2= B.y2 × V2 = By1 × V1Þ 0.8 × 1 = 0.162 × V1Þ V1 = 4.94 m/s

53. (c)2

2 2 22

3 2

( / )

( )æ öæ öç ÷ç ÷è ø è ø

DF kg m sV D kg m m

m s

= (dimensionless parameter)

ReVDr =m (reynold’s number) which is dimensionless

54. –5.72u(x, y, z) = x2 – 3yz

Ñu = u u ui j kx y z

¶ ¶ ¶+ +

¶ ¶ ¶= i. 2x + j.(–3z) + k(–3y)At (2, –1, – 4)Ñu = i × 4 + j (–3 × 4) + k (–3×–1)= 4i – 12j + 3k

ar

= i + j –2 k

| | 1 1 4 6a = + + =r

Directional derivative = Ñu. a

= (4i – 12j + 3k) .( – 2 )

6i j k+

4 12 66

- - = – 5.72

55. 53.236

2

6

10

1

Settlement, DH = 001

cCH

e+ Log

0

0

æ ös + Dsç ÷sè ø

s0 = (15 × 2) + (18 – 10) × 6 + (18 – 10) × 5= 118 kN/m2

Ds0 = 2

1500

(3 6 6)4p

+ + = 8.48 kN/m2

\ DH = 0.3 10

1 0.7´

+ 10118 8.488log

118+æ ö

ç ÷è ø

= 0.0532 m= 53.2 mm

56. 22 ATime UHO S-curve addition S0 0 02 0.6 0.64 3.1 0 3.16 10 0.6 10.68 13 3.1 16.1

10 9 10 19.612 5 16.1 21.114 2 19.6 21.616 0.7 21.1 21.818 0.3 21.6 21.920 0.2 21.8 2222 0.1 21.9 2224 0 22 22

\ Maximum ordinate of S curve is 22

57. 42.82

Ast = 244

dp´ ´

= 4×4p

× ( 12 )2

= 453 mm2

0.36 fck. b. xu = 0.87 fy. Ast

Þ0.87 .0.36 .

y stu

ck

f Ax

f b=

0.87 415 4530.36 25 200

´ ´=

´ ´= 90.86 mmxu.max = 0.48d= 0.48 × 300 = 144 mm\ xu < xu max\ it is U.R section (under reinforced)Moment of resistance,MV = 0.87 ´ fy ´ Ast ´ (d – 0.42 xu)= 0.87 ´ 415 ´ 453 (300 – 0.42 × 90.86)= 42.82 kNm

2015 -16 SOLVED PAPER - 2015

58. 4.43

2.v

vC t

Td

=

20.003 (2 86400 365)

20 1002

´ ´ ´=

æ ö´ç ÷è ø

= 0.189

2 0.1894p

´ =u

Þ u = 0.49 < 60%

Consolidation 30

0.49= = 61.22 mm

For 50 mm settlement

50 0.81761.22

u = = = 81.7%

TV = 1.784 – 0.933 log10(100 – u)= 0.608

Tv = 2 0.608vC td

=

\ t = 20.608

v

dC

´

= 2

–40.608 100.003 10

´

´ seconds

–2

–40.608 10 1

60 60 24 3650.003 10´

= ´´ ´ ´´

Extra years = 6.43 – 2= 4.43 years

59. 33.33

RK

D = 323

R LEI

= ´ 332EIKL

é ù=ê úë û

D = 32

3RLEI

3 3 323 3 3PL RL RLEI EI EI

- =

3 3 2 13 3 3

æ ö= +ç ÷è ø

PL RLEI EI =

3RLEI

Þ3P R=

\ % of applied load carried by spring = 1 1003

´

= 33.3%

60. 0.0032

01

10000S =

q = 4 m3/s, n = 0.01, y = 0.5m

Rate of change of water depth, dydx

02

–

1–= f

r

S Sdydx F

2 / 3 1/ 21fQ AR S

n=

4 = 2 / 3

1/ 21 2 0.5(2 0.5)0.01 2 1 fS´æ ö´ ´ ´ ´ç ÷+è ø

\ Sf = 6.92 × 10–3

rVFgy

= QA gy

=

4(2 0.5) 10 5

=´ ´ ´ = 1.79

\

–3

2

1 – 6.92 1010000

1– (1.79)dydx

´=

= 3.2 × 10–3 = 0.003261. 10.967

fine sandsilt

Clay

k1 d1

k2 d2

k3 d3

Given, k2 = 10 k3 and 2 11

10k k=

d2 = 2d1 and d2 = 23 d3

\ k1 = 10k2, k3 = 21

10k

d1 =12

d2 , 3 232

d d=

1 1 2 2 3 3

1 2 3

+ +=

+ +Hk d k d k d

kd d d

2015-17 SOLVED PAPER - 2015

2 2 2 2 2 2

2 2 2

1 1 3(10 )2 10 2

1 32 2

k d k d k d

d d d

æ ö´ + + ´ç ÷è ø=

+ +

2 2

2

(5 1 3/ 20)3

k dd

+ +=

= 212360

k ´ = 2.05k2

kv = 1 2 3

31 2

1 2 3

d d ddd d

k k k

+ +

+ +

2 2 2

2 22

2 2 2

1 32 2

1 32 2

11010

+ +=

+ +

d d d

d ddk k k

2

2

2

(3)1 30120 2

=æ ö+ +ç ÷è ø

ddk

= k2 3

1 20 30020

æ öç ÷ç ÷+ +ç ÷ç ÷è ø

260321

k= = 0.187 k2

\ H

V

kk

= 2

2

2.0510.9675

0.187kk

=

62. 13.33Mechanism I

3mp.q + mp.(2 q) + mp q = 4uLP ´ ´q

mp

3 mp mp

pu

2q

q q

Þ 6mp.q = 4uLP ´ ´q

Þ Pu. = 24 pmL

Mechanism II

mp

3 mp mp

pu

q + f

q f

1.f = 3q

Þ f = 3q

3mp.q + mp (q + f) + mp f = pu. 4L

× f

3mp.q + mp (q + 3q) + mp –3q = pu.3q. 4L

Þ 10mp q = pu. 3 .4L

q

Þ40P3

= pu

mL

13.33= pmL

\ c = 13.33

63. (a) 29( )

( –1)( 2)f z

z z=

+

z = 1 is simple pole, z = –2 is pole of order 2Residues at z = 1 and –2

[Res f (z)]z =1 = 21

9lim( 1)( 1)( 2)z

zz z®

-- +

= 29 9

9(1 2)=

+ = 1

[Res f(z)]z = –2 = 11! –2

limz®

ddz

2

29( 2)

( 1) ( 2)z

z z

é ù+ê ú

- +ê úë û

= 2–2

9lim( 1)z z®

-

-–99

= = –1

2015 -18 SOLVED PAPER - 2015

64. 142

For a 6/6 person, driver can see from 48 m.

For a 6/9 person, driver can see from 48 × 69 = 32m

Vehicle requires 174m to slow to 30 kmph.

Minimum distance, x = 174 – 32 = 142m

65. (c) d = 2° E

Magnetic FB of AB = N 79° 50¢ E

= 79° 50¢

Corrected FB of O A = N 50° 20¢W

= 309°40¢

\ Correct BB of OA = 129° 40¢

Observed FB of AO = Observed BB of OA

= 552°40¢ E

= 127° 20¢

Error = MV – TV = – 2° 20¢

\ correction = 2°20¢

TB of FB of AB = N 79°50¢ E + d + correction

= N 79°50¢ E + 2 + 2°20¢

= N 84°10¢ E