GATE : , Copyright reserved. Web:

Size: px

Start display at page:

Download "GATE : , Copyright reserved. Web:www.thegateacademy.com"

Tiffany Rice
6 years ago
Views:

1 GATE-2016 Index 1. Question Paper Analysis 2. Question Paper & Answer keys : , info@thegateacademy.com Copyright reserved. Web:

Engineering 15% RCC 5% Transportation Engineering 6% Structural Analysis 3% Steel Structure

2 ANALYSIS OF GATE 2016 Civil Engineering SOM 5% CMM 6% Mathematics 13% Water Resource Engineering 11% General Aptitude 15% Surveying 5% Environmental Engineering 10% Geotechnical Engineering 15% RCC 5% Transportation Engineering 6% Structural Analysis 3% Steel Structure 3% Mechanics 3% : , info@thegateacademy.com Copyright reserved. Web: 1

3 GATE CE 6-Feb 9 AM-12 PM SUBJECT NO OF QUESTION Topics Asked in Paper Level of Toughness Total Marks Enviormental Engineering Structural Analysis Water Resources Engineering SOM/Solid mechanics R.C.C. Geomatics/Surve -ying Engineering Geotechnical Engineering 1 M: 4 2 M: 3 1 M: 1 2 M: 1 1 M: 3 2 M: 4 1 M: 1 2 M: 2 1 M: 1 2 M: 2 1 M:3 2 M:1 1 M: 3 2 M: 6 BOD,Air Pollution;Sludge Disposal ; Domestic wastewater Moderate Deflection of beam;truess & Strain; Tough 3 Unit hydrographs; flood estimation; fluid routing 10 Easy 11 Struess;Deflection of Beams Easy 5 Concrete Technology;shear; compression and torsion by limit state methods; Leveling;coordinate system; map projections;theodolite traversing; Earth Pressure; Consolidation; Shear Strength;permeability& seepage;compaction; Easy 5 Moderate 5 Moderate 15 Transportation Engineering * Construction Materials and Management Engg. Mechanics Steel Structure 1 M: 2 2 M: 2 1 M: 0 2 M: 3 1 M: 1 2 M: 1 1 M: 1 2 M: 1 Traffic characteristics, Theory of traffic flow; pavements;gis Remote sensing *Construction Management ;Analysis of beam sections at transfer and service loads Trusses and Frames; Kinematics of Rectilinear Motion Plastic Analysis ; Design of Tension Member ;Beams Easy 6 Tough 6 Tough 3 Moderate 3 Engineering 1 M: 5 Numerical Method ;Matrices;Complex Mathematics 2 M: 4 Variable; Probability ; Limit & Continuty. Moderate 13 GA 1 M: 5 Time & Work ; Mixtures,Directions ; Venn 2 M: 5 Diagram ; Mensuration& Area, clock. Easy 15 Total 65 Moderate 100 * Indicates Questions from New Syllabus. Faculty Feedback: Few questions came from New Syllabus; General Ability was pretty easy; many question from Geotechnical Engg,Water Resources Engg, Construction Materials and Management was scoring subject, qualifying is easy but scoring is tough. Practice previous question papers will be beneficial. : , info@thegateacademy.com Copyright reserved. Web: 2

4 GATE-2016 Question Paper & Answer Keys

5 Q.No. 1 Section: General Aptitude [Ans. D] until itself is negative so it can t take one more negative i.e., does not. Hence, Option (D) is the right answer Q.No. 2 [Ans. A] paraphrase means a restatement of a text, passage or a rewording of something written or spoken. Q.No. 3 [Ans. A] figurative means representing by a figure or resemblance or expressing one thing in terms normally denoting another with which it may be regarded as analogous. Q.No. 4 [Ans. C] From given codes Relftaga carefree : , info@thegateacademy.com Copyright reserved. Web: 3

6 Otaga careful Fertaga careless From these codes, clearly known that care means taga, from given alternatives, option C is correct. Q.No. 5 [Ans. D] From given data, 64 cubic blocks of one unit Sizes are formed No. of faces of the Cube is 6 No. of corners of the Cube is 8 After removing one Cubic block from Each corner, The resulting surface area of the body = 6 (4) = 96 sq. Units : , info@thegateacademy.com Copyright reserved. Web: 4

7 Q.No. 6 [Ans. B] Total No. of razors Elegance type from all four quarters = = 10,2510 Total No. of razors of Smooth type from all four quarters = = 8,0059 Total No. of razors of Soft type from all four Quarters = = 7,2186 Total No. of razors of Executive type from all four Quarters = = 3,9286 The revenue of the company in that year of 4 different types of razors Elegance = 10, = 49,20,480 Smooth = 8, = 50,43,717 Soft = 7, = 56,30,508 Executive = 3, = 67,96,132 The Executive of razors contributes the greatest revenue of the company in that year. Q.No. 7 [Ans. D] If seventeen languages were not an indication of the nation s diversity, nothing else is. If nothing else is so the best inference is option D. : , info@thegateacademy.com Copyright reserved. Web: 5

8 Q.No. 8 [Ans. D] Q.No. 9 [Ans. B] f (x) = 2x 7 + 3x 5 for x = 1 the equation is satisfied. The factor is (x 1). Q.No. 10 [Ans. B] load cycles for failure exponentially Eg: load = x Cycles for failure = y x 2 = y2, x 3 = y3, x 4 = y4 By elimination procedure from options, Option (A) Load = 40units, it says that, the load is halved, it takes 10,000 cycles for failure = (100)2 40 units = 10,000 cycles, it is not Option (D) Load = units means it is more than 80 units, so it is not. Option (C) Load = units : , info@thegateacademy.com Copyright reserved. Web: 6

3 (80 units) = 60.01 units 4 From the given relation 3 (80 units) = (10)4 3 = (100) 1 (100) 1/3 = 100 4.64 = 464 4 It is not Option (B) only possible At the load of 46.

9 3 (80 units) = units 4 From the given relation 3 (80 units) = (10)4 3 = (100) 1 (100) 1/3 = = It is not Option (B) only possible At the load of units, the failure will happen in 5000 cycles. Q.No. 1 Section: Technical [Ans. *] Range: to Q.No. 2 [Ans. C] Q.No. 3 [Ans. D] Q.No. 4 [Ans. A] : , info@thegateacademy.com Copyright reserved. Web: 7

10 Q.No. 5 [Ans. B] Q.No. 6 [Ans. A] ΣF H = 0 H B = D ΣM c = 0 V B 2 L = 0 V B = 0 ΣV = 0 V c = P : , info@thegateacademy.com Copyright reserved. Web: 8

Q.No. 7 [Ans. D] Under No circumstances even with the shear reinforcement shall the nominal shear stress, τ V should exceed τ c,max. This prevents the failure of section by diagonal compression Q.No. 8 [Ans.

11 Q.No. 7 [Ans. D] Under No circumstances even with the shear reinforcement shall the nominal shear stress, τ V should exceed τ c,max. This prevents the failure of section by diagonal compression Q.No. 8 [Ans. *] Range: 99.9 to As per IS 800, the design bending strength of laterally unsupported beam as governed by lateral torsional buckling is: M d = β b. Z p. f bd β b = Z e Z P ; for semi compact section So, M d = Z e Z P Z P. f bd = Z e. f bd = = 100 kn-m Q.No. 9 [Ans. C] Bull s trench kiln is used to manufacture bricks Q.No. 10 [Ans. B] Tri calcium silicate (C 3 S) hydrates and hardens rapidly and is largely responsible for initial set and early strength : , info@thegateacademy.com Copyright reserved. Web: 9

12 Q.No. 11 [Ans. B] Consolidate undrained test (CU) test, drainage is permitted in consolidation stage. So, pore water pressure = 0. However, in shear stage, No drainage is permitted. So, pore water pressure develops Q.No. 12 [Ans. D] Soil is plastic in range of 26% to 48%. So, plastic limit = 26 %. Liquid limit =48% Since 35% <LL<50%... So, CI. Q.No. 13 [Ans. D] P(a) = Ka. γ. z 2c K a = 0 Z e = QC γ. K a So, max depth of unsupported excavation = 2Z e = 4c γ K a : , info@thegateacademy.com Copyright reserved. Web: 10

13 Q.No. 14 [Ans. *] Range: 21.5 to 21.7 Let area of the catchment be a hectares So, = A A = 21.6 ha Q.No. 15 [Ans. B] : , info@thegateacademy.com Copyright reserved. Web: 11

Q.No. 16 [Ans. *] Range: 13.6 to 13.7 F 1 = 10 y 2 y 1 = 1 2 + 1 2 1 + 8 F 1 2 y 2 = 1 y 1 2 + 1 1 + 8 100 2 = 13.65 Q.No. 17 [Ans.

18 [Ans. A] Crown corrosion is caused due to H 2 S being released from waste water Q.No. 19 [Ans. B] Given, D= 0.45m, L= 7.5m No.

14 Q.No. 16 [Ans. *] Range: 13.6 to 13.7 F 1 = 10 y 2 y 1 = F 1 2 y 2 = 1 y = Q.No. 17 [Ans. A] Pre-Cursors to photochemical oxidants are NOX, VOCs and sunlight photochemical smooth is formed when pre-cursor pollutants undergo reaction in sunlight Q.No. 18 [Ans. A] Crown corrosion is caused due to H 2 S being released from waste water Q.No. 19 [Ans. B] Given, D= 0.45m, L= 7.5m No. of fabric filter bags, N = A t A b A t = Total area of filter = 300m 2 2 A b = Area of one bag = πdl = π = 10.6 m 2 N = 300 = : , info@thegateacademy.com Copyright reserved. Web: 12

15 Q.No. 20 [Ans. A] Biological Nitrogen Removal Nitrifies and Denitrifies Q.No. 21 [Ans. B] Voids filled with bitumen Bitumen content : , info@thegateacademy.com Copyright reserved. Web: 13

16 Q.No. 22 [Ans. *] Range: 1199 to 1201 Flow, q = 3600 = 3600 = 1200 veh/hr t H 3 Q.No. 23 [Ans. C] Min No. of satellites needed for a GPS to determine the position precisely = 4 Q.No. 24 [Ans. C] Passive Remote sensing uses sun as a source of EM energy and records the energy that is naturally radiated or reflected from objects Q.No. 25 [Ans. A] RL of bottom of beam = = m : , info@thegateacademy.com Copyright reserved. Web: 14

17 Q.No. 26 [Ans. A] Q.No. 27 [Ans. B] Q.No. 28 [Ans. B] Q.No. 29 [Ans. C] : , info@thegateacademy.com Copyright reserved. Web: 15

18 Q.No. 30 [Ans. A] Q.No. 31 [Ans. *] 2 x 1 EI. dy dx = 10x2 2 + c 1 0 y 0.5 EI. dy 10 = 10y dx 3 y3 + c 1 a + y = 0.5, dy dx = 0 0 = (0.5)3 + c 1 c 1 = dy dx = dy x=1 dx x=0 5 + c 1 = : , info@thegateacademy.com Copyright reserved. Web: 16

19 C 1 = So, dy dx = C 1 max EI = = Q.No. 32 [Ans. D] By Principal of superposition δ max 1 = δ max 2 and θ max 1 = θ max 2 : , info@thegateacademy.com Copyright reserved. Web: 17

20 Q.No. 33 [Ans. A] At any joint of a truss, if there is no external force, and if two members are collinear then third member carries the zero force So, F = 0 in FT, TG, HU, MP, PL : , info@thegateacademy.com Copyright reserved. Web: 18

21 Q.No. 34 [Ans. D] Taking moments about A = 0 R Bh 8 + P 1.5 = 0 R Bh = 1.5 P 8 ΣF H = 0 R Ah = R Bh = 1.5 P 8 ΣF v = 0 R AV = P : , info@thegateacademy.com Copyright reserved. Web: 19

Q.No. 35 [Ans. B] As per IS 456 : 2000, Min tensile reinf, Ast bd = 0.85 f y Ast = 85 bd = 0.85 250 400 f y 415 = 204.81 mm 2 205 mm 2 (Ast) max = 0.4 % of bd = 0.4 250 400 = 4000 mm2 100 Q.No. 36 [Ans.

22 Q.No. 35 [Ans. B] As per IS 456 : 2000, Min tensile reinf, Ast bd = 0.85 f y Ast = 85 bd = f y 415 = mm mm 2 (Ast) max = 0.4 % of bd = = 4000 mm2 100 Q.No. 36 [Ans. *] Range: 9999 to E l = E s 1 + θ Where, E s = 5000 fck = = MPa θ = Creep cofficient = 1.5 E l = = MPa : , info@thegateacademy.com Copyright reserved. Web: 20

23 Q.No. 37 [Ans. B] P L 2 P L 2 MP θ MP L 2 θ = L 2 ϕ θ = ϕ By virtual work method Work done by external force = work done by internal moments M P θ + M P (2θ) = P. L/2θ P U = 6M P L ϕ MP : , info@thegateacademy.com Copyright reserved. Web: 21

24 Q.No. 38 [Ans. B] Max force carried by plates P = A g f y = γ mo 1.1 = kn Load carried by each weld = P = kn 2 For minimum length of weld Strength of weld = Load carried by weld f u l w t = γ mw l w (10 0.7) 410 = l w = mm 105 mm : , info@thegateacademy.com Copyright reserved. Web: 22

25 Q.No. 39 [Ans. *] Range: 37 to E F G H I t E = t O + 4t L + t P 6 E F: t E = F G: t E = G H: t E = H I: t E = = 62 6 = 49 6 = 43 6 = 73 6 t E = = days 6 Q.No. 40 [Ans. *] Range: 41 to 43 n= 0.7, S = 40% V = 100 m 3 n = V v V V v = n v = = 70 m 3 S = V w V v V w = = 28 m 3 V w + V a = V v 28 + V a = 70 V a = 42 m 3 : , info@thegateacademy.com Copyright reserved. Web: 23

26 Q.No. 41 [Ans. A] W Pa P H 3 = 2m 2m K a. γ. H = 40 ka. γ = 40 H So, Pa = 1 2. Ka. γ. H2 = 20 H = 120 kn Taking moment about P = 0 Pa 2 = W 2 Pa = W = 120 kn Ka. γ. H = 40 : , info@thegateacademy.com Copyright reserved. Web: 24

27 Q.No. 42 [Ans. *] Range: 36 to 38 S = C r. H o log σ c 1 + e o σ + C c. H o log σ o + Δσ 1 + e o σ c Where, C r = 0.05, C c = 0.50, H o = 1m Far clay, e o = = σ o = 60 kpa, σ o = ( ) 0.5 = kpa [ γ clay = G + e 1 + e γ w = 10 = kn/m 3 ] Δσ o = = 50 kn/m 2 S = log log = m = m = mm : , info@thegateacademy.com Copyright reserved. Web: 25

28 Q.No. 43 [Ans. *] Range: 65.3 to 65.6 : , info@thegateacademy.com Copyright reserved. Web: 26

29 Q.No. 44 [Ans. *] Range: 100 to 110 CD test: ϕ = 22, C = 15 KPa CU test: σ 1 U = (σ 3 U)tan 2 (45 + ϕ 2 ) + 2c tan (45 + ϕ 2 ) σ = ( ) tan 2 ( ) tan( ) σ = 50 tan tan(56 ) σ = 50 (1.484) σ 1 = = kn/m 2 So, Deviator Stress = σ 1 σ 3 = = kn/m 2 : , info@thegateacademy.com Copyright reserved. Web: 27

30 Q.No. 45 [Ans. *] Range: to m 40 m γ w H 2 = 10 5 = 50 kn/m 2 γ w H 1 = = 650 kn/m 2 γ w H γ w(h 2 H 1 ) = = 250 kn/m2 3 P = 1 2 ( ) ( ) 40 2 = = kn/m : , info@thegateacademy.com Copyright reserved. Web: 28

31 Q.No. 46 [Ans. B] Flow is normal to bedding plane K avg = Στ Σ τ = 1 20 = 2 m/day ki Head difference i = = = 5 Length = 1 12 Seepage discharge q = K avg i A = = 0.5 m 3 /day/m : , info@thegateacademy.com Copyright reserved. Web: 29

32 Q.No. 47 [Ans. B] For Rectangular channel, y c = ( q2 g ) Where, q = Q B = 16 4 = 4 m2 /s y c = [ (4)2 1/3 10 ] = 1.17 m By Manning s method Q = 1 n A (R)2 3 (S) = (4) y n (y n ) 2 3 (0.001) 1 2 y n = 1.28 m y n > y c & s o < s c so mild slope 1 3 : , info@thegateacademy.com Copyright reserved. Web: 30

33 Q.No. 48 [Ans. *] Range: 126 to 128 A 5 m 2. 5 m B m F H = 1 2 C (5)2 = 125 kn 1000 kn F H acts a distance = 5 = 1.67 m from the base 3 F v weight of water enclosed or supported ( actual or imaginary) by the curved surface = ρg volume of portion ABC = [ π ] = [ 25 π ] 6 = : , info@thegateacademy.com Copyright reserved. Web: 31

34 = 22.7 kn F R = F H 2 + F V 2 = (125) 2 + (22.7) 2 =127kN Q.No. 49 [Ans. *] Range: 2.29 to y = 2 m B For hydraulically efficient channel, B = 2 3. y = = 4 3 = 2.31 m Q.No. 50 [Ans. *] Range: 4.4 to 4.6 A P H = 4.2 P OH = 9.8 [OH ] = mol/l B [OH] = mol/l P OH = 9.8 log 10 (2) = 9.5 P H = 4.5 Q.No. 51 [Ans. *] Range: 8000 to 8020 A 1 So, ln(1 η 1 ) = A 2 ln(1 η 2 ) 5600 ln (1 0.96) = A ln (1 0.99) A = m 2 : , info@thegateacademy.com Copyright reserved. Web: 32

Q.No. 52 [Ans. *] Range: 0.29 to 0.31 BOD 2 = Lo (1 e k 2 ) 100 = Lo (1 e 2k ) (i) Also 155 = Lo (1 e 4k ) (ii) (i)/(ii) 100 155 = 1 e 2k 1 e 4k 1 e 4k = 1.55 1.55 e 2k e 4k 1.55 e 2k + 0.

35 Q.No. 52 [Ans. *] Range: 0.29 to 0.31 BOD 2 = Lo (1 e k 2 ) 100 = Lo (1 e 2k ) (i) Also 155 = Lo (1 e 4k ) (ii) (i)/(ii) = 1 e 2k 1 e 4k 1 e 4k = e 2k e 4k 1.55 e 2k = 0 let e 2k = x x x = 0 x = 0.55 e 2k = 0.55 K = 0.3 day 1 Q.No. 53 [Ans. B] n. l2 W me,60 = 2R = 2 (5) = 0.5 m W ps = V 9.5 R = = 0.89 m W me,80 = 2 (5) = 0.50 m 80 W ps,80 = = 1.19 m : , info@thegateacademy.com Copyright reserved. Web: 33

36 Q.No. 54 [Ans. *] Range: 2.9 to 3.1 Q.No. 55 [Ans. *] Range: to T R 10.5 S 16.5 T T BM Δ TST, tan 16.5 = T T x (1) ΔTRT, tan(10.5 ) = TT x + 60 = T T + 2 x + 60 From (1) and (2) x tan 16.5 = (x + 10) tan(10.5 ) 2 x = (x + 10) x = m (2) So, T T = = m So, RL of top of tower = = m P Q 60 m x : , info@thegateacademy.com Copyright reserved. Web: 34

Q. 1 Q. 5 carry one mark each.

GATE 016 General Aptitude - GA Set-5 Q. 1 Q. 5 carry one mark each. Q.1 Out of the following four sentences, select the most suitable sentence with respect to grammar and usage. (A) I will not leave the