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

2 ANALYSIS OF GATE 2016 Mechanical Engineering GATE ME 31-Jan-2016_Morning Session GA 15% Engineering Mechanics 7% MOM 9% Maths 15% TOM 8% Machine design 5% Industrial Engg. 5% Manufacturing Engg. 9% Thermodynamics 12% Heat Transfer 7% Fluid Mechanics 8% : , info@thegateacademy.com Copyright reserved. Web: 1

3 GATE ME 31-Jan-2016_Morning SUBJECT NO OF QUES. Topics Asked in Paper Level of Toughness Total Marks Engineering Mechanics Mechanics of Materials Theory of Machines Machine Design Fluid Mechanics Heat Transfer Thermodynami cs Manufacturing Engg. Industrial Engg. Mathematics GA 1 M: 1 2 M: 3 1 M: 3 2 M: 3 1 M: 2 2 M: 3 1M: 1 2M: 2 1 M: 2 2 M: 3 1 M: 3 2 M: 2 1 M: 4 2 M: 4 1 M: 3 2 M: 3 1 M: 1 2 M: 2 1 M: 5 2 M: 5 1 M: 5 2 M: 5 Free body diagram ;Dynamics Kinematics Stress & Strain;Moment of inertia Statically indeterminate beams Mohr's Circle;Torsion ;Plain Stress & Strain Degree of freedom ;Velocity Analysis Vibrations ;Balancing* Moderate to Difficult Easy to Moderate Easy to Moderate Bearings ;Bolted joint Easy 5 Continuty;Pipe flow;impact of Jet Fluid Kinematics Convection;Heat Exchanger; Radiation Lumped Model Power Engg;Properties of pure sustances;steady Flow energy equation;psychrometry Compressors*;Refregaration Unconventional Machining;Gear Menufacturing;Casting (Cooling) Tool Life;Resistance welding; Shear Angle CPM PERT;Linear Programming; Forecasting Numerical Methods;Complex Variable Probability & Distributions;Limit & Continuty ;Liner algebra Time & Work;Mixtures;Directions Venn Diagram ;Mensuration & Area Clock Moderate 8 Easy to Moderate 7 Moderate 12 Easy to Moderate 9 Easy 5 Easy 15 Easy 15 Total 65 Moderate 100 * Indicates Questions from New Syllabus. Faculty Feedback: Compared to 2015 paper, 2016 paper of GATE (ME) was well balanced with some easy, some moderate and other difficult or conceptual questions. Many MCQs were designed to confuse students and invite errors. A cut-off between is expected for general category. Regarding weightage, Applied Mechanics and Design (Mechanics/SOM/TOM/MD) had an increased weightage while Manufacturing Engineering s weightage was decreased a little bit. : , info@thegateacademy.com Copyright reserved. Web: 2

4 GATE-2016 Question Paper & Answer Keys

5 GATE 2016 Examination Mechanical Engineering Test Date: 31/01/2016 Test Time: 9:00 AM 12:00 PM Subject Name: MECHANICAL ENGINEERING Section: General Aptitude Q.No. 1 [Ans. C] In degrees of comparison Mr. X is taller than Mr. Y is apt. Positive degree tall Comparative degree taller Superlative degree tallest Q.No. 2 : , info@thegateacademy.com Copyright reserved. Web: 3

6 Q.No. 3 [Ans. C] rest is history is an idiomatic expression which means rest is well known Q.No. 4 [Ans. C] Q.No. 5 [Ans. B] M works with twice efficiency as E but worked for half as many days. So in this repsect they will do equal work if their shifts would have been for same timings. But M s shift is for hrs, while E s shift for 12 hrs. Hence E will do twice the work as M. Ratio of contribution of M : E in work, 1 : 2 : , info@thegateacademy.com Copyright reserved. Web: 4

7 Q.No. 6 [Ans. D] The number of student who like to read books or play sports have been shown = = 108 Q.No : , info@thegateacademy.com Copyright reserved. Web: 5

8 [Ans. A] Until the colonial period means pre-colonial origin. Other options can t be inferred. Q.No. 8 [Ans. D] Mirror image of 1 : 20 is 10 : : 30 was the time two and quarter hour back so time now will be 12 : 45 Q.No. 9 [Ans. C] : , info@thegateacademy.com Copyright reserved. Web: 6

9 M 10 km km See the adjoining figure for solution MM = = NM = = MN = (MM ) 2 + (NM ) 2 N M = ( ) + ( ) 2 = Q.No. 10 [Ans. B] : , info@thegateacademy.com Copyright reserved. Web: 7

10 Q.No. 1 Section: Technical [Ans. C] We know that a square matric can be written as, [A] = 1 2 (A + AT ) (A AT ) where 1 2 (A + AT ) is a symmetric and 1 2 (A AT ) is a skew symmetric matix Now, if A = A T or A T = A [A] = 1 2 (A A) + 1 [A ( A)] 2 Symmetric part becomes zero while skew symmetric part is left therefore a square matrix is called a skew symmetric matrix if A T = A Q.No. 2 [Ans. C] Substituting x = 0, we find that function takes a form 0/0 therefore we can use L hopital s rule. log (L + 4x) 4 lim x 0 e 3x = lim 1 x 0 (1 + 4x)3e 3x = 4 (1 + 0)3e o = 4 3 Q.No. 3 [Ans. B] : , info@thegateacademy.com Copyright reserved. Web: 8

11 Q.No. 4 [Ans. *] Range: 99.6 to 99.8 A standard normal curve (as shown in figure) has 68% area in limits 1 to + 1, 95% area is limits 2 to + 2 and 99.7% area is limits 3 to % of the data are within 3 standard deviations of the mean Q.No. 5 [Ans. C] According to Newton-Raphson scheme x n+1 = x n f(x n) f (x n ) x 1 = x 0 f(x o ) ( ) f = 1 (x o ) [3 (1) 2 + 1] = x 1 = 3 4 = 0.75 : , info@thegateacademy.com Copyright reserved. Web: 9

12 Q.No. 6 [Ans. A] While drawing tree body diagram all the supports are removed and forced applied due to those supports are drawn. Further, a fixed support or a clamped support resists force in any direction, as well as any tendency of rotation. : , info@thegateacademy.com Copyright reserved. Web: 10

13 Q.No. 7 [Ans. C] Both made of same material E 1 = E 2 Flexural rigidity = EI a r a l 1 = E 1 I 1 l 2 = E 2 I 2 a 4 l 1 = I 1 12 = π l 2 I 2 64 d4 Given A 1 = A 2 a 2 = πr 2 r = a/ π d = 2a/ π l 1 l 2 = l 1 l 2 = π/3 a 4 12 π a 4 (2a π )4 d = 2r : , info@thegateacademy.com Copyright reserved. Web: 11

14 Q.No. 8 [Ans. B] For stress system shown in figure (a) σ x = p, σ y = p and τ = 0 τ xx = ( σ x + σ y ) + ( σ x σ y ) cos 2θ + τ sin 2θ 2 2 = ( p p ( p) ) + [p ] cos 90 o + 0 = τ yy + τ xx = σ x + σ y τ yy = (σ x + σ y ) τ xy = 0 And τ xy = ( σ x σ y ) sin 2θ τ cos 2θ 2 = [ p (p) ] sin( 90 o ) 0 = p 2 τ xx = 0 τ yy = 0 τ xy = p : , info@thegateacademy.com Copyright reserved. Web: 12

15 Q.No. 9 [Ans. D] Since PQ is a rigid link, the distance between P and Q cannot change (increase or decrease). Therefore, the velocity of Q relative to P i.e., V QP will have only component perpendicular to PQ. Q.No. 10 [Ans. B] Number of degrees of freedom is planar mechanism having n links and j simple hinge joints is given by Gruebler s equation F = 3 (n 1) 2j Q.No. 11 [Ans. *] Range: 99 to 101 M = 1 kg Δ = 1 mm g = 10 m/s 2 w n = g Δ = 10 = 100 rad/s 10 3 : , info@thegateacademy.com Copyright reserved. Web: 13

16 Q.No. 12 [Ans. C] All four bearings are shown in figure. Obviously, straight cylindrical roller bearings can t take thrust load (A) (B) (C) (D) Q.No. 13 [Ans. *] Range: 0.99 to 1.01 Q 2 Q 1 Q 3 Q 1 = Q 2 + Q 3 A 1 V 1 = A 2 V 2 + A 3 V 3 : , info@thegateacademy.com Copyright reserved. Web: 14

17 0.9 = V = 3 10 V = V 3 2 V 2 = 1 m s Q.No. 14 [Ans. B] V = 2xy i y 2 j u = 2xy v = y 2 dx u = dy v dx 2xy = dy y 2 1 dx 2 x = dy y 1 dx 2 x + dy y = 0 1 ln x + ln y = ln c 2 ln x + 2 ln y = 2 ln c xy 2 = c Q.No. 15 : , info@thegateacademy.com Copyright reserved. Web: 15

18 [Ans. *] Range: 67 to o C k = 20 k = o C 0.1 m 0.3 T 1 T 2 ( 0.1 ) = T 2 T 3 ( 0.3 ) (130 T) ( ) = (100 ) (T 30) T = 5T = 8T T = 67.5 o C Q.No. 16 [Ans. B] Grashoff number signifies the ratio of buoyance force to viscous force Q.No. 17 [Ans. C] Ice is converted directly into vapor, if it is heated at constant pressure which is less than triple point pressure. Therefore, direct conversion of ice into vapor does not involve critical point. : , info@thegateacademy.com Copyright reserved. Web: 16

19 Q.No. 18 [Ans. B] Heat transfer rate q = C h (T hi T ho ) = C c (T co T ci ) Let C h = C min ; Then q = C min (T hi T ho ) Also dg = C h dt h = C c dtc If C h < C c D Th > dtc Temperature of hot fluid will change fster compared to that of cold fluid and if the heat exchanger is infinitely long (for q max), temperature T ho would approach T ci Thus q max = C min (T hi T ci ) = C min Δ T max Q.No. 19 [Ans. *] Range: 50 to 54 e = V t1 u 1 V t2 u 2 = [V t1 V t2 ]u.. (1) As the flow velocity is same 150 cos β = 300 cos 65 o β = o V t1 = 300 sin 65 o and V t2 = 150 sin o Substituting in equation (1) and solving we obtain E = 52.8 kj/kg : , info@thegateacademy.com Copyright reserved. Web: 17

20 Q.No. 20 [Ans. C] ϵ = l 2 l 1 l 1 ϵ = (Given) T. S. ln(1 + ϵ) T. S = ln(1.001) = 0.099% Q.No. 21 [Ans. C] α = k ρc Steel α = Copper α = Aluminum α = Copper-Aluminium-Steel Q.No. 22 [Ans. B] In a wire cut EDM process, the wire and the sample both should be electrically conducting in order to make a successful cut : , info@thegateacademy.com Copyright reserved. Web: 18

21 Q.No. 23 [Ans. B] Internal gears are manufactured by shaping process with pinion cutter Q.No. 24 [Ans. C] G01 is used for linear interpolation G03 is used for circular interpolation, anticlockwise M03 is used for spindle rotation, clockwise And M05 is used for spindle stop Q.No. 25 [Ans. D] In PERT network, an activity time is assumed to have Beta distribution while the project duration follows normal distribution. Q.No. 26 [Ans. *] Range: 2 to 2 As two Eigenvalue of given matrix are identical, the eigen vectors resulting from these identical eigen values would be identical/linearly dependent. Therefore, number of linearly independent eigen vectors = n 1 = 2 : , info@thegateacademy.com Copyright reserved. Web: 19

22 Q.No. 27 [Ans. *] Range: 15.9 to 16.1 dr dy dx 4 x dr = r ds F. dr = (yi + 2xj ). dr c Applying Green s theorem (yi + 2xj ). dr c = [ x (2x) y (y)] A Area of the circle = { of radius 4 x units} = π ( 4 x ) 2 = 16 dxdy = dxdy A Q.No. 28 [Ans. C] Let x = 1 t ; then lim x x2 + x 1 x = lim t 0 1 t t 1 (1 t ) 1 + t t 2 1 = lim t 0 t Since the function has 0/0 from now, we can apply L Hopital rule, 1 + t t (1 2t) lim = lim {[ 0] /(1) } t 0 t t t t2 Applying limit now, lim x x2 + x 1 x = 1 2 : , info@thegateacademy.com Copyright reserved. Web: 20

23 Q.No. 29 [Ans. A] Total ways in which three cards can be drawn 52 c 3 Number of ways in which a King, a queen and a jack can be drawn = 4c 1 4c 1 4 c1 The required probability is = Q.No = = [Ans. *] Range: 130 to 135 T T 200 N ma 100 N m 1 g = m = kg 9.81 m 2 g = 200 m 2 = kg ma ma 100 N T 100 = m 1 a T = a : , info@thegateacademy.com Copyright reserved. Web: 21

24 200 N 200 T = m 2 a T = a Equation 1 and a = a a = 100 a = 3.27 m s 2 T = = N Q.No. 31 ma [Ans. *] Range: 19.9 to 20.1 A 30 Meters B V Applying energy conservation, Total energy at A { when the disc is at rest } = { Total energy at B When disc is rolling without slipping } Mgh = 1 2 MV Iω2 = 1 2 MV2 + 1 MR ω2 V 2 + R2 ω 2 = 2gh 2 But Rω = V 3V2 2 = 2gh V2 = 4 3 gh Substituting, g = 10 m s 2 and h = 30 meters, we get V = 4 gh 3 = 20 m s : , info@thegateacademy.com Copyright reserved. Web: 22

25 Q.No. 32 [Ans. *] Range: 2.9 to 3.1 A V 1 Q 2 45 o B 45 o 90 o α = α = α = o V 1 = i + 2j a. b = a b cos θ cos θ = a. b (1) ; cos θ = a b 5 1 θ = V 1 = (AI)w V 1 = i + 2j V 1 = 5 5 = (AI)w (i) V 2 = (BI)w. (ii) (ii) (i) = V 2 5 = BI AI BI sin(71.58) = BI AI = 1.34 V 2 5 = 1.34 V 2 = 2.99 m/s ω I AI sin (45) (i) V 2 : , info@thegateacademy.com Copyright reserved. Web: 23

26 Q.No. 33 [Ans. *] Range: 0.18 to 0.22 Σ x = σ x E μ σ y E ( Δl l ) = σ x E μ σ y E ( 2Δ l ) = P (1 μ) E ( ) = 2 Q.No (1 μ) = μ = [Ans. D] Solid G L J T θ s Hollow G L J T θ H τ s T J = Gθ L = τ R T J = Gθ L τ H : , info@thegateacademy.com Copyright reserved. Web: 24

27 θ = ( TL GJ ) θ s = θ H τ = TR J τ < R τ H > τ S Q.No. 35 [Ans. B] w L y UDL = y R wl 4 8EI = RL3 3EI R = 3wL 8 Q.No. 36 [Ans. *] Range:1.9 to 2.1 : , info@thegateacademy.com Copyright reserved. Web: 25

28 1 m 1 20 mm 20 mm m mm 0.2 m 1 50 mm 2.5 (m 1 )(20)(2.5) + (1)(50)(0.2) = (1)(50)(2.2) M 1 = 2kg Q.No. 37 [Ans. *] Range: 0.49 to 0.51 k M F sin(wt) w = 3k m MF = A Fo/s 1 (1 ( w ) 2 ) + (2ξ ω ) 2 w n ω n ξ = 0, ω n = k m ω = 3 ω n 1 MF = (1 3) = 1/2 2 : , info@thegateacademy.com Copyright reserved. Web: 26

29 Q.No. 38 [Ans. *] Range: 39.9 to 40.1 Q.No. 39 [Ans. *] Range: 2.6 to 2.7 τ w = μ U 1 R 2 = 2μ U 1 R K = 2 : , info@thegateacademy.com Copyright reserved. Web: 27

30 Q.No. 40 [Ans. *] Range: 8.76 to 8.78 Q.No. 41 [Ans. B] a x = u u x + v u y x + y 2 x 2x = x 2 + y 2 (x2 (x 2 + y 2 ) 2 ) = x(x2 + y 2 2x 2 ) 2xy 2 (x 2 y 2 )(x 2 + y 2 ) y (x 2 + y 2 ) x 1 (x 2 + y 2 ) 2 2y = x3 xy 2 (x 2 + y 2 ) 3 : , info@thegateacademy.com Copyright reserved. Web: 28

31 x a x = (x 2 + y 2 ) 2 a y = u v x + v v y x = (x 2 + y 2 ) y (x 2 + y 2 ) 2 2x + y (x 2 + y 2 ) + y 2 ) y 2y ((x2 (x 2 + y 2 ) 2 ) = 2x2 y + yx 2 y 3 (x 2 + y 2 ) 3 y = (x 2 + y 2 ) 2 The velocity components are not functions of time, so flow is steady according to continuity equation, u x + v y = (x2 y 2 ) (x 2 + y 2 ) 2 + (x2 y 2 ) (x 2 + y 2 ) 2 = 0 Since it satisfies the above continuity equation for 2D and incompressible flow. The flow is incompressible. Q.No. 42 [Ans. *] 10.9 to 11.2 T = 1000 k T = 400 k = 0.5 = mn σ = w/m 2 k 4 σ(t 4 1 T 4 2 ) Q = ( 1 1 ) ( 1 2 ) 1 A 1 A 1 F 12 2 A 2 A 1 = A 2 = A F 12 = F 21 = 1 ( Q A ) = ( ) q = w/m 2 q = kw/m 2 Q.No. 43 [Ans. *] Range: 42.0 to 45.0 : , info@thegateacademy.com Copyright reserved. Web: 29

32 d = 0.01/m L = 0.2 m T i = 750 o C T a = 100 o C h = 250 W/m 2 K ρ = 7801 kg/m 3 C p = 473 J/kg/K k = 43 W/m/K T = 300 o C T T a = exp ( hast ) T i T a ρvc p ( ) = exp ( h4t ρdg ) 4(250)T 1.17 = T = sec Q.No. 44 [Ans. *] Range: 84 to 86 h 1 = 100 kj kg V 2 = 200 m s V 1 = 100 m s S.F.E.E o o o dh + d(p. e) + d(k. e) = Q w (h 2 h 1 ) (V 2 2 V 1 2 ) = 0 (h 2 h 1 ) ( ) = 0 h 2 h = 0 h 2 = = 8500 J kg = 85 kj kg Q.No. 45 [Ans. *] Range: 16.9 to 17.1 p v w = p t p v : , info@thegateacademy.com Copyright reserved. Web: 30

33 20 = ( ) = kg w kg da = 17 g w kg da Q.No. 46 [Ans. *] Range: 460 to 470 P 4 = P 3 = P 2 P 3 P 2 P 1 P 4 P 1 = 10 P = 10 P 4 = 1000 kpa p 4 p 3 p 2 = 10 p 3 p 2 p 1 ( P 4 P 3 ) 3 = 10 (1000)3 = (P 10 3 ) 3 P 3 = 1000 = kpa 101/3 Q.No. 47 [Ans. *] Range: 1095 to 1130 Specific Enthalpy = 1054 kj/kg : , info@thegateacademy.com Copyright reserved. Web: 31

34 Q.No. 48 [Ans. *] Range: 2.85 to mm d = 5 mm 2 mm Given, L = 1400 kj kg R = 200 μω ρ = 8000 kg m 3 I = 4000 A J = 0.2 secs H = I 2 RT = = 640 J 1400 kj 1 kg m megget = v ρ = π 4 d2 h ρ = π 4 ( ) 2 h 8000 = h J = kg h = = m Or h = 2.9 mm Q.No. 49 [Ans. B] = 22, t 2 = 0.8 mm, V c = 48 m min, f = 0.4 mm rev tan ϕ = r cos 0.5 cos 22 = 1 r sin sin 22 = 0.57 : , info@thegateacademy.com Copyright reserved. Web: 32

35 r = t 1 = 0.4 t = 0.5 ϕ tan 1 (0.57) = 29.7 Q.No. 50 [Ans. *] Range: 4.9 to 5.1 Q.No. 51 [Ans. *] Range: 14 to 18 n = 0.25, V 2 = 0.5 V 1 VT n = C V 1 T = 0.5 V 1 T ( T 1 ) = 0.5 T 1 = (0.5) = T 2 T 2 T 2 = 16 T 1 Q.No. 52 [Ans. *] Range: 674 to 676 Q.No. 53 [Ans. A] : , info@thegateacademy.com Copyright reserved. Web: 33

36 Q.No. 54 [Ans. D] Q.No. 55 [Ans. *] Range: 39,000 to 41,000 : , info@thegateacademy.com Copyright reserved. Web: 34