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GENERAL APTI TUDE (GA) (Q.1 – 5) : Car r y One M ar k Each

1. I f I were you, I __________ that laptop. I t ’s muchtoo expensive.

(a) won’t buy (b) shan’t buy

(c) wouldn’t buy (d) would buy

2. He turned a deaf ear to my request . What doesthe under l ined phrasal verb mean?

(a) ignored (b) appreciated

(c) twisted (d) returned

3. Choose the most appropr iate set of words fromthe opt ions given below to complete the fol lowingsentence.

_________ _________ is a wi l l , _________ is a way.

(a) Wear, there, their

(b) Were, their, there

(c) Where, there, there

(d) Where, their, their

4. (x % of y) + (y % of x) is equivalent to _______.

(a) 2 % of xy (b) 2 % of (xy/100)

(c) xy % of 100 (d) 100 % of xy

5. The sum of the digits of a two digit number is 12.I f the new number formed by reversing the digitsis greater than the or iginal number by 54, findthe or iginal number.

(a) 39 (b) 57

(c) 66 (d) 93

(Q. 6 – 10) : Car r y Tw o M ar k s Each

6. Two finance companies, P and Q, declared fixedannual rates of interest on the amounts investedwith them. The rates of interest offered by thesecompanies may differ from year to year. Year-wiseannual r at es of i n t er est of f er ed by t hesecompanies are shown by the l ine graph providedbelow.

2000

79 9.5 10

98

6

4

86.5

7.5886.5

2001 2002 2003 2004 2005 2006

P Q

I f the amounts invested in the companies, P andQ, in 2006 are in the rat io 8 : 9, then the amountsr ecei ved af t er one year as i n t er est s f r omcompanies P and Q would be in the rat io:

(a) 2 : 3 (b) 3 : 4

(c) 6 : 7 (d) 4 : 3

7. Today, we consider Ashoka as a great r ulerbecause of the copious evidence he left behind inthe form of stone carved edicts. Histor ians tendto cor relate greatness of a king at his t ime withthe avai labi l i ty of evidence today. Which of thefol lowing can be logical ly infer red from the abovesentences?

(a) Emper or s who do not l eave si gn i f i can tsculpted evidence are completely forgot ten.

I NSTRUCTI ONS1. Total of 65 quest ions car rying 100 marks, out of which 10 quest ions car rying a total of 15 marks are in

General Apt i tude (GA)2. The Engineer ing Mathemat ics wi l l car ry around 15% of t he t ot al mar k s, the General Apt i tude sect ion

wil l car ry 15% of t he t ot al mar k s and the r emai n i ng 70% of t he t ot al mar k s.3. Types of Quest i ons

(a ) M u l t i p le Choi ce Quest i ons (M CQ) car rying 1 or 2 marks each in al l papers and sect ions. Thesequest ions are object ive in nature, and each will have a choice of four opt ions, out of which the candidatehas to mark the cor rect answer(s).

(b) Numer i cal Answ er Quest i ons of 1 or 2 marks each in al l papers and sect ions. For these quest ionsthe answer is a real number, to be entered by the candidate using the vir tual keypad. No choices wil lbe shown for these type of quest ions.

4. For 1-mar k mult iple-choice quest ions, 1/3 mar k s wil l be deducted for a wrong answer. L ikewise, for 2-mar k s mult iple-choice quest ions, 2/3 marks wil l be deducted for a wrong answer. There is no negat ivemarking for numer ical answer type quest ions.

GATE – 2016 CE : CI VI L ENGI NEERI NG

Set - 2No. of Quest i ons : 65 Maxi mum Mar k s : 100

2 SOLVED PAPER – 2016 (SET - 2)

f(x) = 2 22 cos cos3

.....4 1 3

x x

sin sin 2 sin 3

.....1 2 3

x x x .

The convergence of the above Four ier ser ies atx = 0 gives

(a)2

21

16n n

(b)

1 2

21

1

12

n

n n

(c)

2

21

182 1n n

(d)

1

1

1

2 1 4

n

nn

4. X and Y are two random independent events. I tis known that P(X ) = 0.40 and P(X YC) = 0.7.Whi ch one of t he fol l owi ng i s t he val ue ofP(X Y ) ?

(a) 0.7 (b) 0.5

(c) 0.4 (d) 0.3

5. What is the value of 2 200

l im ?xy

xy

x y

(a) 1 (b) – 1

(c) 0 (d) L imit does not exist

6. The kinemat ic indeterminacy of the plane t russshown in the figure is

(a) 11 (b) 8

(c) 3 (d) 0

7. As per IS 456-2000 for the design of reinforcedconcrete beam, the maximum al lowable shearst ress (c max) depends on the

(a) grade of concrete and grade of steel

(b) grade of concrete only

(c) grade of steel only

(d) gr ade of concr et e and per cent age ofreinforcement

8. An assembly made of a r igid arm A-B-C hingedat end A and suppor ted by an elast ic rope C-D atend C is shown in the figure. The members maybe assumed to be weight less and the lengths ofthe respect ive members are as shown in thefigure.

(b) Ashoka pr oduced st one car ved edict s t oensure that later histor ians wil l respect him.

(c) Stat ues of k ings ar e a r eminder of t hei rgreatness.

(d) A king’s greatness, as we know him today, isinterpreted by histor ians.

8. Fact 1 : Humans are mammals.

Fact 2 : Some humans are engineers.

Fact 3 : Engineers bui ld houses.

I f the above statements are facts, which of thefol lowing can be logical ly infer red ?

I . Al l mammals bui ld houses.

I I . Engineers are mammals.

I I I .Some humans are not engineers.

(a) I I only (b) I I I only

(c) I , I I and I I I (d) I only

9. A square pyramid has a base per imeter x, andthe slant height is half of the per imeter. What isthe lateral sur face area of the pyramid?

(a) x2 (b) 0.75 x2

(c) 0.50 x2 (d) 0.25 x2

10. Ananth takes 6 hours and Bharath takes 4 hoursto read a book. Both star ted reading copies of thebook at the same t ime. After how many hours isthe number of pages t o be read by Ananth, twicethat t o be read by Bharath? Assume Ananth andBharath read al l the pages with constant pace.

(a) 1 (b) 2

(c) 3 (d) 4

TECH NI CAL SECTI ON

(Q.1 – 25) : Car r y One M ar k Each

1. The spot speeds (expressed in km/hr ) observedat a road sect ion are 66, 62, 45, 79, 32, 51, 56, 60,53, and 49. The median speed (expressed inkm/hr) is _________

(Note : answer wi th one decimal accuracy)

2. The opt imum value of the funct ion

f(x) = x2 – 4x + 2 is

(a) 2 (maximum) (b) 2 (minimum)

(c) – 2 (maximum) (d) – 2 (minimum)

3. The Four ier ser ies of the funct ion,

f(x) = 0, – < x 0

= – x, 0 < x <

in the interval [– , ] is

SOLVED PAPER – 2016 (SET - 2) 3

Under the act ion of a concentrated load P at C asshown, the magnitude of tension developed in therope is

(a)3P

2(b)

P

2

(c)3P8

(d) 2 P

9. As per I ndian standards for br icks, minimumacceptable compressive st rength of any class ofburnt clay br icks in dry state is

(a) 10.0 MPa (b) 7.5 MPa

(c) 5.0 MPa (d) 3.5 MPa

10. A construct ion project consists of twelve act ivit ies.The est imated durat ion (in days) requir ed tocomplete each of the act ivi t ies along with thecorresponding network diagram is shown below.

Act ivi ty Durat ion (days)

A Inaugurat ion 1

B Foundat ion work 7

C Structural construct ion-1 30

D Structural construct ion-2 30

E Br ick masonry work 25

F Plaster ing 7

G Floor ing 25

H Electr ificat ion 7

I Plumbing 7

J Wood work 7

K Color ing 3

L Handing over funct ion 1

1 2 3

4 6 8

10 11 12

5 7 9

A B K L

C

J

D I

E F

G H

Total floats (in days) for the act ivi t ies 5-7 and11-12 for the project are, respect ively,

(a) 25 and 1 (b) 1 and 1

(c) 0 and 0 (d) 81 and 0

11. A str ip foot ing is rest ing on the surface of a purelyclayey soi l deposit . I f the width of the foot ing isdoubled, the ult imate bear ing capacity of the soi l

(a) becomes double

(b) becomes half

(c) becomes four-t imes

(d) remains the same

12. The relat ionship between the specific gravity ofsand (G) and the hydraul ic gradient (i ) to ini t iatequick condit ion in the sand layer having porosityof 30% is

(a) G = 0.7 i + 1 (b) G = 1.43 i – 1

(c) G = 1.43 i + 1 (d) G = 0.7 i – 1

13. The r esu l t s of a consol i dat i on t est on anundisturbed soi l , sampled at a depth of 10 mbelow the ground level are as fol lows :

Saturated unit weight : 16 kN/m3

Pre-consol idat ion pressure : 90 kPa

The water table was encountered at the ground level.Assuming the unit weight of water as 10 kN/m3,the over-consol idat ion rat io of the soi l is

(a) 0.67 (b) 1.50

(c) 1.77 (d) 2.00

14. Profi le of a wei r on permeable foundat ion isshown in figure I and an elementary profi le of‘upstream pi le only case' according to Khosla'stheory is shown in figure I I . The upl i ft pressureheads at key points Q, R and S are 3.14 m, 2.75 mand 0 m, respect ively (refer figure I I ).

10 m 5 m 25 m

P

Weir

Gate

Floor

1 m

3 m

5 m

Fi gu r e I

40 m

S

4 m

5 mR

Q

Fi gu r e I I

What is t he upl i ft pressure head at point Pdownstream of the weir (junct ion of floor and pileas shown in the figure I )?

(a) 2.75 m (b) 1.25 m

(c) 0.8 m (d) Data not sufficient


15. Water table of an aquifer drops by 100 cm overan area of 1000 km2. The porosity and specificretent ion of the aquifer mater ial are 25% and 5%,respect ively. The amount of water (expressed inkm3) drained out from the area is _________

16. Group I contains the types of fluids while Group I Icon t ai ns t he shear st r ess - r at e of shearrelat ionship of different types of fluids, as shownin the figure.

Rat e of sh ear

Sh

ear

stre

ss

Yie

ld s

tres

s

2

3

4

5

1

Gr oup I Gr oup I I

P. Newtonian fluid 1. Curve 1

Q. Pseudo plast ic fluid 2. Curve 2

R. Plast ic fluid 3. Curve 3

S. Dilatant fluid 4. Curve 4

5. Curve 5

The cor r ect mat ch bet ween Gr oup I andGroup I I is

(a) P-2, Q-4, R-1, S-5

(b) P-2, Q-5, R-4, S-1

(c) P-2, Q-4, R-5, S-3

(d) P-2, Q-1, R-3, S-4

17. The atmospher ic layer closest to the ear th sur faceis

(a) the mesosphere (b) the st ratosphere

(c) the thermosphere (d) the t roposphere

18. A water supply board is responsible for t reat ing1500 m3/day of water. A set t l ing column analysisindicates that an over flow rate of 20 m/day wil lpr oduce sat i sfact or y r emoval for a dept h of3.1 m. I t is decided to have two circular set t l ingt ank s i n par al l el . The r equ i r ed di amet er(expressed in m) of the set t l ing tanks is ________

19. The hardness of a ground water sample was foundto be 420 mg/L as CaCO3. A softener containingion exchange resins was instal led to reduce thet ot al har dness t o 75 mg/L as CaCO3 befor esupplying to 4 households. Each household getst r eat ed water at a r at e of 540 L /day. I f t heefficiency of the softener is 100%, the bypass flowrate (expressed in L/day) is _________

20. The sound pressure (expressed in Pa) of thefaintest sound that a normal healthy individualcan hear is

(a) 0.2 (b) 2

(c) 20 (d) 55

21. I n the context of the IRC 58-2011 guidel ines forr igid pavement design, consider the following pairof statements.

I . Radius of relat ive st i ffness is direct ly relatedt o modul us of elast i ci t y of concr et e andinversely related to Poisson's rat io

I I . Radius of relat ive st i ffness is direct ly relatedto thickness of slab and modulus of subgradereact ion.

Whi ch one of t he fol l owing combinat ions i scor rect?

(a) I : True; I I : True (b) I : False; I I : False

(c) I : True; I I : False (d) I : False; I I : True

22. I f the total number of commercial vehicles perday ranges from 3000 to 6000, t he minimumpercentage of commercial t raffic to be surveyedfor axle load is

(a) 15 (b) 20

(c) 25 (d) 30

23. Opt imal fl ight planning for a photogrammetr icsurvey should be car r ied out consider ing

(a) only side-lap

(b) only end-lap

(c) either side-lap or end-lap

(d) both side-lap as well as end-lap

24. The reduced bear ing of a 10 m long line is N30E.The depar ture of the l ine is

(a) 10.00 m (b) 8.66 m

(c) 7.52 m (d) 5.00 m

25. A circular curve of radius R connects two straightswith a deflect ion angle of 60. The tangent lengthis

(a) 0.577 R (b) 1.155 R

(c) 1.732 R (d) 3.464 R

(Q. 26 – 55) : Car r y Tw o M ar k s Each

26. Consider the fol lowing l inear system.

x + 2y – 3z = a

2x + 3y + 3z = b

5x + 9y – 6z = c

This system is consistent i f a, b and c sat isfy theequat ion

(a) 7a – b – c = 0 (b) 3a + b – c = 0

(c) 3a – b + c = 0 (d) 7a – b + c = 0


27. I f f (x) and g(x) ar e t wo pr obabi l i t y densi t yfunct ions,

f(x) =

1 : 0

1 : 0

0 : otherwise

xa x

ax

x aa

g(x) =

: 0

: 0

0 : otherwise

xa x

ax

x aa

Which one of the fol lowing statements is t rue?

(a) Mean of f(x) and g(x) are same; Var iance off(x) and g(x) are same

(b) Mean of f(x) and g(x) are same; Var iance off(x) and g(x) are different

(c) Mean of f(x) and g(x) are different ; Var ianceof f(x) and g(x) are same

(d) Mean of f(x) and g(x) are different ; Var ianceof f(x) and g(x) are different

28. The angle of intersect ion of the curves x2 = 4yand y2 = 4x at point (0, 0) is

(a) 0 (b) 30

(c) 45 (d) 90

29. The area between the parabola x2 = 8y and thestraight l ine y = 8 is _________.

30. The quadrat ic approximat ion of f(x) = x3 – 3x2 – 5at the point x = 0 is

(a) 3 x2 – 6x – 5 (b) – 3x2 – 5

(c) – 3x2 + 6x – 5 (d) 3x2 – 5

31. An elast ic isot ropic body is in a hydrostat ic stateof st ress as shown in the figure. For no change inthe volume to occur, what should be i ts Poisson'srat io?

y

x

z

(a) 0.00 (b) 0.25

(c) 0.50 (d) 1.00

32. For the st ress state (in MPa) shown in the figure,the major pr incipal st ress is 10 MPa.

55

5

The shear st ress is

(a) 10.0 MPa (b) 5.0 MPa

(c) 2.5 MPa (d) 0.0 MPa

33. The por tal frame shown in the figure is subjectedto a uniformly dist r ibuted ver t ical load w (perunit length).

L

P

R

L/2

S

w

Q

The bending moment in the beam at the joint‘Q’ is

(a) zero (b)2L

(hogging)24

w

(c)2L

(hogging)12

w(d)

2L(sagging)

8w

34. Consider t he st ructural system shown in t hefigure under the act ion of weight W. All the jointsare hinged. The proper t ies of the members interms of length (L), area (A) and the modulus ofelast ici ty (E) are also given in the figure. Let L,A and E be 1 m, 0.05 m2 and 30 106 N/m2,respect ively, and W be 100 kN.

90°P

A, E

A, EL

2A, E45°45°

45°45°

Q R

90°S

A, E

W

A, E

5


Which one of the fol lowing sets gives the cor rectvalues of the force, st ress and change in lengthof the hor izontal member QR?

(a) Compr essi ve for ce = 25 k N ; St r ess =250 kN/m2; Shor tening = 0.0118 m

(b) Compr essi ve for ce = 14.14 kN ; St r ess =141.4 kN/m2; Extension = 0.0118 m

(c) Compr essi ve for ce = 100 k N ; St r ess =1000 kN/m2; Shor tening = 0.0417 m

(d) Compr essi ve for ce = 100 k N ; St r ess =1000 kN/m2; Extension = 0.0417 m

35. A haunched (varying depth) reinforced concretebeam is simply suppor ted at both ends, as shownin the figure. The beam is subjected to a uniformlydist r ibuted factored load of intensity 10 kN/m.

The design shear force (expressed in kN) at thesect ion X-X of the beam is ______

10 kN/m5 m X

600 mm400 mm

effect ive span = 20mX

36. A 450 mm long plain concrete pr ism is subjectedto the concentrated ver t ical loads as shown in thefigure. Cross sect ion of the pr ism is given as150 mm 150 mm. Consider ing l inear st ressdistr ibut ion across the cross-sect ion, the modulusof rupture (expressed in MPa) is ________

11.25 kN 11.25 kN

P

Q RS

150 mm 150 mm150 mm

37. Two bolted plates under tension with alternat ivearrangement of bolt holes are shown in figures 1and 2. The hole diameter, pi tch, and gauge lengthare d, p and g, respect ively.

Pd

Figure 1

P

p

P

Figure 2

gP

Which one of the fol lowing condit ions must beensured to have higher net tensi le capacity ofconfigurat ion shown in Figure 2 than that shownin Figure 1 ?

(a) p2 > 2gd (b) 2 4p gd

(c) p2 > 4gd (d) p > 4gd

38. A fixed-end beam is subjected to a concentratedload (P) as shown in the figure. The beam hastwo different segments having different plast icmoment capacit ies (M p, 2M p) as shown.

2L/3

2M P

LL

M P

P

The minimum value of load (P) at which the beamwould col lapse (ult imate load) is

(a) P7.5 ML

(b) P5.0 ML

(c) P4.5 ML

(d) P2.5 ML

39. The act ivi ty-on-ar row network of act ivi t ies for aconstruct ion project is shown in the figure. Thedurat ions (expressed in days) of the act ivi t ies arement ioned below the ar rows.

10 20 40 50

30 70

80 90

W3

T5

Q3

P2

R4

U3

S3

V2

X2

The cr i t ical durat ion for this construct ion projectis

(a) 13 days

(b) 14 days

(c) 15 days

(d) 16 days

40. The seepage occurr ing through an ear then dami s r epr esent ed by a f l ownet compr i si ng of10 equipotent ial drops and 20 flow channels. Thecoefficient of permeabil ity of the soi l is 3 mm/minand the head loss is 5 m. The rate of seepage(expressed in cm3/s per m length of the dam)through the ear then dam is _________


41. The soi l profi le at a si te consists of a 5 m thicksand layer under lain by a c- soi l as shown infigure. The water table is found 1 m below theground level. The ent ire soi l mass is retained bya concrete retaining wall and is in the act ive state.The back of the wall is smooth and ver t ical. Thetotal act ive ear th pressure (expressed in kN/m2)at point A as per Rankine's theory is _________.

1 m

4 m

3 m

A

bulk = 16.5 kN/m3

Sand sat W = 19 kN/m , = 9.81 kN/m ,3 3

= 32°

c- soil

sat W = 18.5 kN/m , = 9.81 kN/m3 3

c = 25 kN/m , = 242 °

42. OM C-SP and M DD-SP denot e t he opt i mummoisture content and maximum dry densi t yobtained from standard Proctor compact ion test ,respect ively. OMC-MP and MDD-MP denote theopt imum moisture content and maximum drydensi t y obt ained fr om t he modi f i ed Pr oct orcompact ion test , respect ively. Which one of thefol lowing is cor rect?

(a) OMC-SP < OMC-MP and MDD-SP < MDD-MP

(b) OMC-SP > OMC-MP and MDD-SP < MDD-MP

(c) OMC-SP < OMC-MP and MDD-SP > MDD-MP

(d) OMC-SP > OMC-MP and MDD-SP > MDD-MP

43. Water flows from P to Q through two soil samples,Soi l 1 and Soi l 2, having cross sect ional area of80 cm2 as shown in the figure. Over a per iod of15 minutes, 200 ml of water was observed to passthrough any cross sect ion. The flow condit ionscan be assumed t o be st eady st at e. I f t hecoefficient of permeabil i ty of Soi l 1 is 0.02 mm/s,the coefficient of permeabil ity of Soi l 2 (expressedin mm/s) would be ________

Soil 1 Soil 2P Q

150 mm 150 mm

600

mm

300

mm

44. A 4 m wide st r ip foot ing is founded at a depth of1.5 m below the ground sur face in a c- soi l asshown in the figure. The water table is at a depthof 5.5 m below ground sur face. The soil proper t iesare: c = 35 kN/m2, = 28.63, sat = 19 kN/m3,bulk = 17 kN/m3 and w = 9.81 kN/m3. The valuesof bear ing capaci ty factors for di fferent aregiven below.

N N N

15 12.9 4.4 2.5

20 17.7 7.4 5.0

25 25.1 12.7 9.7

30 37.2 22.5 19.7

c q

4 m

1.5 m

5.5 m

Using Terzaghi 's bear ing capacity equat ion anda factor of safety Fs = 2.5, the net safe bear ingcapaci t y (expressed in kN/m2) for local shearfai lure of the soi l is __________

45. A square plate is suspended ver t ical ly from oneof i ts edges using a hinge suppor t as shown infigure. A water jet of 20 mm diameter having avelocity of 10 m/s str ikes the plate at its mid-point,at an angle of 30 with the ver t ical. Consider g as9.81 m/s2 and neglect the self-weight of the plate.The force F (expressed in N) required to keep theplate in i ts ver t ical posit ion is _________

Water jet

20 mm

30° Plate

200 mm

F


46. The ordinates of a one-hour unit hydrograph atsixty minute interval are 0, 3, 12, 8, 6, 3 and0 m3/s. A two-hour storm of 4 cm excess rainfal loccurred in the basin from 10 AM. Consider ingconstant base flow of 20 m3/s, the flow of the r iver(expressed in m3/s) at 1 PM is _________

47. A 3 m wide rectangular channel car r ies a flow of6 m3/s. The depth of flow at a sect ion P is 0.5 m. Af l at -t opped hump i s t o be pl aced at t hedownstream of the sect ion P. Assume negl igibleenergy loss between sect ion P and hump, andconsider g as 9.81 m/s2. The maximum height ofthe hump (expressed in m) which wil l not changethe depth of flow at sect ion P is _________

48. A penstock of 1 m diameter and 5 km length isused to supply water fr om a reservoi r t o animpulse turbine. A nozzle of 15 cm diameter isfixed at the end of the penstock. The elevat iondifference between the turbine and water levelin the reservoir is 500 m. Consider the head lossdue to fr ict ion as 5% of the velocity head availableat the jet. Assume unit weight of water = 10 kN/m3

and accelerat ion due to gravity (g) = 10 m/s2. I fthe overal l efficiency is 80%, power generated(expressed in kW and rounded to nearest integer)is _______________

49. A t racer takes 100 days to t ravel from Well-1 toWell-2 which are 100 m apar t . The elevat ion ofwater sur face in Wel l -2 is 3 m below that inWel l -1. Assuming porosi ty equal t o 15%, thecoefficient of permeabil i ty (expressed in m/day)is

(a) 0.30 (b) 0.45

(c) 1.00 (d) 5.00

50. A sample of water has been analyzed for commonions and results are presented in the form of abar diagram as shown.

3.30

2.65 4.10 6.35 6.85meq/L

3.90 6.75

0

0

Ca2+ Mg2+ Na+ K +

Cl –SO4

2–HCO3–

meq/L

The non-carbonate hardness (expressed in mg/Las CaCO3) of the sample is

(a) 40 (b) 165

(c) 195 (d) 205

51. A noise meter located at a distance of 30 m froma point source recorded 74 dB. The reading at adistance of 60 m from the point source wouldbe _________

52. For a wast ewat er sampl e, t he t h r ee-daybi ochemi cal oxygen demand at i ncubat i ontemperature of 20C (BOD3day, 20C) is est imatedas 200 mg/L. Taking the value of the first orderBOD react ion rate constant as 0.22 day-1, the five-day BOD (expressed in mg/L) of the wastewaterat incubat ion temperature of 20C (BOD5day, 20c)would be _________

53. The cr i t ical flow rat ios for a three-phase signalare found to be 0.30, 0.25, and 0.25. The totalt ime lost in the cycle is 10 s. Pedestr ian crossingsat this junct ion are not significant. The respect iveGreen t imes (expressed in seconds and roundedoff to the nearest integer) for the three phasesare

(a) 34, 28, and 28 (b) 40, 25, and 25

(c) 40, 30, and 30 (d) 50, 25, and 25

54. A motor ist t ravel ing at 100 km/h on a highwayneeds to take the next exit , which has a speedlimit of 50 km/h. The sect ion of the roadway beforethe r amp ent r y has a downgrade of 3% andcoefficient of fr ict ion (f) is 0.35. I n order to enterthe ramp at the maximum al lowable speed l imit ,the braking distance (expressed in m) from theexit ramp is _________

55. A tal l tower was photographed from an elevat ionof 700 m above the datum. The radial distancesof the top and bot tom of t he tower fr om thepr incipal points are 112.50 mm and 82.40 mm,respect ively. I f the bot tom of the tower is at anelevat ion 250 m above the datum, then the height(expressed in m) of the tower is _________


ANSWERSGener al Apt i t ude (GA)

1. (c) 2. (a) 3. (c) 4. (a) 5. (a) 6. (d) 7. (d)

8. (b) 9. (d) 10. (c)Techn i cal Sect i on

1. (54.5) 2. (d) 3. (c) 4. (a) 5. (d) 6. (a) 7. (b)

8. (b) 9. (d) 10. (c) 11. (d) 12. (c) 13. (b) 14. (b)

15. (0.2) 16. (c) 17. (d) 18. (6.909) 19. (1774.29) 20. (c) 21. (b)

22. (a) 23. (d) 24. (d) 25.( 60R tan

2

)26. (b) 27. (b) 28. (d)

29. (85.33) 30. (b) 31. (c) 32. (b) 33. (a) 34. (c) 35. (65)

36. (12) 37. (c) 38. (a) 39. (c) 40. (500) 41. (69.5) 42. (b)

43. (0.04545) 44. (298.48) 45. (7.85) 46. (60) 47. (0.20) 48. (6570) 49. (5m/day)

50. (a) 51. (67.979) 52. (276.19) 53. (34,28,28) 54. (92.32) 55. (120.4)

EXPL ANATI ONS

Gener al Apt i t ude (GA)

4. (x% of y) + (y% of x)

100 100x y

y x = 2100

xy = 2% of xy

Opt ion (a) is cor rect .

5. Let the two digit number be xy

Now sum of the digit of a two digit number is 12

x + y = 12 ...(i )

Now according to quest ion,

(10y + x) – (10x + y) = 54

9y – 9x = 54

y – x = 6 ...(i i )

Now solving equat ion (i ) and (i i ) we get ,

x = 3 and y = 9

Hence the or iginal number is 39.

Opt ion (a) is cor rect .

6. Amount invested by Company P in 2006 = ` 8x

and amount invested by Company Q in 2006 = ` 9x

I nterest from Company P

=

8 61009 4100

x

x

`

`

= 43

Opt ion (d) is cor rect .

Techn i cal Sect i on

1. Median speed calculat ion = 32, 45, 49, 51, 53, 56

60, 62, 66, 79

Median speed = 53 56

2

= 54.5 km/hr

6. Kinemat ic indeterminacy

No. of joints = 7

No. of unrest rained displacement possible

= 7 2 = 14

No. of rest raints = 2 + 1 = 3

Therefore, degree of freedom = 14 – 3 = 11

7. VA

HAHD

A D

LP

RigidarmL

Rope

B C

D

L L

VD

MA = 0VD 2l – P.l = 0

VD = P2


Also, Fy = 0

VA = P – VD = P P

P2 2

Fx = 0H A + H D = 0 ...(1)

MC = 0VA . l + H A . l = 0

H A = – VA = P2

...(2)

By 1 & 2,Component of force ‘P’ in the direct ion of rope

= P

2

Therefore, tension in rope = P

2

9. As per BIS : 1077-1957 & 1910 code specificat ion,minimum compr essive st r engt h or cr ushingstrength of br icks should be 3.5 N/mm2

10.

1 2 3

Act ivi ty

5 7

4 6 8

9

11 12A B(1) (7)

C

D(30)

(7)E F

G HI

(25)

(7)(7)(25)

(7)J

K L(1)(3)

T = 38E T = 63E T = 70E

T = 81E

T = 81L

T = 80E

T = 80L

T = 77E

T = 77LT = 70E

T = 70L

T = 63E

T = 63L

T = 38E

T = 38L

T = 0E T = 1E T = 8E

(30)

,5 7

11 12 ,

F = (63 – 38) – 25 = 0T

F = (81 – 80) – 1 = 0T

–

–

11. Bear ing capacity of clayey soi ld does not dependupon the size of foot ing.

Therefore, the bear ing capacity remains same.

12. i = G 11 e

n = 30%

e = 1

nn

= 0.3 30.7 7

= 0.428

Therefore, G = i (1 + e) + 1

= i (1 + 0.428) + 1

= 1.43 i + 1

13. m m

r sat = 16 kN/m310m

r w = 10 kN/m3

Pre-consol idat ion pressure = 90 kPa

Effect ive st ress = r sat . 10 – r w . 10

= 16 10 – 10 10

= 60 kN/m2

Over-consol idat ion rat io

= Pre-consol idat ion pressure

Current effect ive st ress

= 90 330 2

= 1.5

14. R = 2.75

1004

= 68.75%

P = 100 – R

= 31.25%

P = Pressure head to point P

100total head

31.25 = 1004h

h = 1.25 m

15. H = 100 cm

Area of aquifer = 1000 km2

Porosity = 25%

SR = 5%

Total volume of water reduced

= 3101000 km

100

= 1 km3

= 109 m3

Also, SR = Rvolume of water rate (V )total volume of water

9510

100 = Volume of water rate

Therefore, VR = 5 107 m3

SR + SY = n

5% + SY = 25%

SY = 20%

Volume of water drained = 20% of total water

= 92010

100

= 2 108 m3

= 0.2 km3

16. P : Newtonian fluid – curve 2

Q – Pseudoplast ic fluid – curve 4

R – Plast ic fluid – curve 5

S = Dilatent fluid – curve 3


17. Figure

Attitude

90 km

60 km

10 km

500 km Exosphere

Thermosphere

Mesosphere

StratosphereTroposphere

0–100 –50 50 100 150 200

Temperature (°C)

18. Q = 1500 m3/day

Over flow rate = 20 m/day

Depth of tank = 3.1 m

Area of set t l ing tank required (A)

= Q

O.F.R.

= 2150075 m

20

Since provision of two tanks is there

Therefore area of each tank

= 27537.5 m

2

Therefore, diameter of set t l ing tank (D)

=

Area 37.5 4

4

= 6.909 m

19. Hardness = 420 mg/l as CaCO3

Reduced hardness = 75 mg/l as CaCO3

Total hardness removed = 345 mg/l as CaCO3

Total quant i ty of t reated water = 540 4

= 2160 1/day with 75 mg/l hardness

Total hardness for supply water

= 2160 75 mg/day

= 162000 mg/day

Volume of water required not to be t reated

= 162000

420mgday

= 385.71l trday

Therefore, volume of water required to be treated

= 2160 – 385.71

= 1774.29 l t rs/day

20. The faintest sound we can hear is 0.00002 Pa andloudest round we can hear is 20 Pa

21. l = 3

2

E

12 1

h

Where l = radius of rel. st i ffness in cm

E = modulud of elast ici ty of

cement concrete in kg/cm2

H = slab thickness

= Poission's rat io

K = Modulus of subgrade react ion

22. Total number of commercial vehicles per day

= 3000 to 6000

Minimum % of commercial t raffic to be surveyedfor axle load = 15

23. For opt i mal f l i gh t pl ann i ng for aphotogrammater ic survey, both side-lap as wellas end lap are considered.

24. L

D

L = 10 m

= N. 30 E

Depar ture = L sin

= 10 sin 30

= 10 (1/2) = 5 m

25. Radius (R)

Deflect ion, L = 60

Tangent length = R tan2

= 60R tan

2

30. R tan 30 = 0.577 R

31.

x

y

z

x = E E E

yx z

y = E E E

y x z

x = E E E

yz x


V = V X Y Z

V X Y Z

VV

= x + y + z

= 1 2

E x y z

Therefore, V = 0, either (1 – 2 ) = 0

Or, x + y + z = 0

Therefore, 1 – 2 = 0

= 1

0.52

32. x + y = 1 + 2

5 + 5 = 10 + 2

2 = 0

max = 1 2

2

= 10 0

2

= 5 MPa

33. Q Rw

SH S

H P L

P

L/2

wL2

wL2

FX = 0

H P – H S = 0

Therefore,

H P = H S = 0

MQ = PH0

2l

Bending moment at Q = 0

34.

A,EA,E

A,EA,E

RQ

P

S

45° 45°45°45°

2A,E

W

W

At joint S, FX = 0

FSQ = FSR

FY = 0

FSQ cos 45 + FSR cos 45 = W

FSQ + FSR = W 2

FSQ = FSR = W 2

Similar ly at point P we get ,

FSQ = SRF W 2

Now at joint Q,

FX = 0

FQP cos 45 + FQS cos 45 + FQR = 0

QRW W

F2 2 = 0

FQR = – W kN

= – 100 kN

Stress in QR = QRF

2 0.05 = 1000 kN

Shor tening due to compressive st ress

= 100 2 1

2 0.05 30 1000

= 0.0471 m

35.

100 kN

5m10 kN/m

100 kN20 m

Cross sect ional area at sect ion X-X of the beam is

Depth (m) = 600 400600 5

10

= 600 – 100 = 500 m

Shear force at sect ion X-X

= 100 – (10 5)

= 50 kN

Moment at sect ion X-X

= M u = 100 5 – 10 5 52

= 500 – 125

= 375 kN-m

Design shear force at sect ion X-X = Vu

= MV tanu

d

=

375200

0.55010000

= 65 kN


36.

Q RSP

11.25 kN 11.25 kN

450 mm 450 mm 150 mm

Compressive st rength of pr ism

= 150 mm 150 mm

Maximum moment

= 11.25 (350 + 75) – (11.25 75)

= 11.25 (225 150)

= 1687.5 kN-mm

Sect ion modulus = Iy

=

3.3

362

b h

h

= 2 2150 150

24 24bh

Modulus of rupture = M2

= 6

21.6875 10

24150 150

= 2

N120

mm = 12 MPa

37. Condit ion is : The loss due to second bolt is lessthan the gain in st rength due to stagger ing ofthe second bolt

2

4pg

> d,

i .e., p2 > 4 gd

38. Number of hinges for the mechanism

= Ds + 1 = 2 + 1 = 3

Case I : When hinges are at A, B & C

P2L3

2MP + 2.MP + MP = 2

L.3

p

5MP = 2

P.L.3

P = P

15M

2L

= PM7.5

L

Case I I : Hinges are formed at A, B & D

P2L3

P P P2M M ( ) M = 2P

L.3

PM 35.5

L 2 = P

P = P8.25. ML

Therefore,23 =

4.L

3

= 2.

Collapse load is PM7.5

L40. N d = 10,

N f = 20

K = 3 mm/min,

h = 5 m

Q = N

K. .N

f

dh

=

35 201000

60 10

= 500 cm3/s

41. I n c- soi l

Ear th pressure Pa = 2a v ak c k

K a = 1 sin 241 sin 24

= 0.4217

Now at point A

v = 1 4 3b sat sat

Ear th pressure Pa = 2a v ak c k

= 0.4217 (16.5 + 4(19 – 9.81) + 3(18.5 – 9.81))

+ 7 9.81 – 2 25 0.4217

= 69.65 kN/m2

42. Because energy impar ted to the mould in case ofstandard proctor t est is less as compared tomodified proctor test. Therefore, more compact ionis obtained for modified proctor test and hencehigh maximum dry density.

Si m i l ar i l y t he moi st u r e cont en t at wh i chmaximum compact ion is obtained will also be lessfor modified proctor test .


43. h = 600 – 300 mm

Q = 200 ml/15 minutes

= 200 1000

1000 15 60

= 1000

5 900= 0.222 cm3/sec

A = 80 cm2

V = QA

= 1000

5 900 80

= 0.0277 mm/sec

V = K i

V = 300300

k

Therefore, K = 1 2

1 2

1 2

H HH HK K

2

1K

= 72 – 50

K 2 = 0.04545 mm/sec

44. qu = CcN 8D N 0.5 B Nf q

qu = 2

35 17.7 17 1.5 7.4 0.5 4 17 53

qu = 771.7 kN/m2

qnu = qu – D f

= 771.7 – 17 1.5

qnu = 746.2 kN/m2

qnet = Fnuq

= 746.2

2.5= 298.48 kN/m2

45.

100 mmJet

30°

100 mm

F

Jet

30°

W cos 30f °

W sin 30f °

Fy

Force exer ted by water jet on plate

= 2 2P 20 (10)4

= 2

20P (100)

4 100

= 10 = 31.4 N

Wf sin 30 100 = F 200

Therefore, F = 1

W sin 302 f

= W 31.44 4

f

= 7.85 N

46. Ti me Or di nat e of 1 h r UH (m 3/sec)

Of fset Or di nat e (m 3/sec)

Or di nat e of 2 h r UH (m 3/sec)

Or di nat e of DRH 4cm r ai n fal l excess(m 3/sec)

Or di nat e of f l ood hydr ogr aph (m 3/sec)

10 : 00 0 - 0 0 20

11 : 00 3 0 3 6 26

12 : 00 12 3 15 30 50

1 : 00 8 12 20 40 60

2 : 00 6 8 14 28 48

3 : 00 3 6 9 18 38

4 : 00 0 3 3 6 26

5 : 00 0 0 0 20

Ordinate of 1 : 00 pm = 60 m3/sec


47. E1 = E2min + Zmax

E2min =

1 32 32

qg

=

1 326 3 39.81 2

= 1.1123 m

E1 = 2

22

qy

gy

= 2

22

0.52 9.81 0.5

= 1.3155 m

Zmax = 1.3155 – 1.1123

= 0.20 m

48. Appl y ener gy equat i on of f r ee su r face ofreserveoir,

500 = 21head loss

2v

g

= 2 21 10.05

2 2v v

g g

v1 = 2 10 500

1.05

v1 = 97.59secm

Water power = 21

12

mv

= 3

2

110

2 0.15 97.594

= 8212.178

Therefore, 0 = S.P.W.P.

0.8 = S.P.

8212.178 S.P. = 6570 kW

49. Time = 100 days

Distance = 100 m

n = 15%

h = 3m

VS = 100100

= 1m/day

VS = kni

i = Lh

= 3

100

kn = svi

= 100

m/day3

k = 100

0.153

= 5m/day

50. Ca2+ = 2.65 meq/1

Mg2+ = 4.10 – 2.65

= 1.45 meq/1

Na+ = 6.35 – 4.10

= 2.25 meq/1

K + = 6.85 – 6.35

= 0.5 meq/1

HCO3– = 3.30 meq/1

SO42– = 3.9 – 3.3

= 0.6 meq/1

Cl – = 6.75 – 3.90

= 2.85 meq/1

Total hardness (mg/l as CaCO3) = Total meq/1

Equat ion weight of CaCO3 in mg

= (2.65 + 1.45) 50

= 205 mg/l as CaCO3

Alkal ini ty mg/l as CaCO3 = 3.30 50

= 165 mg/1 as CaCO3

Non-car bonat e har dness = Total har dness –Carbonate hardness = 205 – 165 = 40 mg/l asCaCO3

51. L 2 = 11

2

RL 20 log

R

= 60

74 20 log30

= 74 20 log0.5

= 69.979 dB

52. BOD5 = 0.22 50L 1 e (1)

BOD3 = 0.22 30L 1 e (2)


From 1 & 2,

5

3

BODBOD

= 0.22 5

0.22 31

1

e

e

BOD5 = 0.667

2000.483

= 276.19 mg/l

53. Y = y1 + y2 + y3

= 0.30 + 0.25 + 0.25 = 0.80

L = 10 sec

Opt imum cycle t ime = 1.5L 5

1 Y

Therefore, C0 = 100 sec

Now, green t imes are calculate dby

G1 = 10C L

yy

= 0.30

(100 10)0.80

= 34 sec

G2 = 20C L

yy

= 0.25100 10

0.80

= 28 sec

G3 = 30C L

yy

= 0.25100 10

0.80

= 28 sec

54. Total energy lost between point 1 & 2 = work doneby fr ict ional force

2 21 10.03

2 2mv mv mg sce = f · (mg) · s

289.815 = 3.1392s

s = 92.32 m

55. Relief distance,

d = r – r 0

= 112.5 – 82.40

= 30.1 mm

d = H

rhh

30.1 = 112.5700 250

h

h = 120.4 mm

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