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ME GATE Paper 2008 www.gateforum.com Answer keys 1 C 2 B 3 B 4 A 5 C 6 D 7 D 8 D 9 B 10 D 11 B 12 B 13 B 14 C 15 A 16 B 17 18 B 19 20 D 21 D 22 B 23 B 24 C 25 26 A 27 D 28 C 29 A 30 31 D 32 B 33 B 34 B 35 B 36 37 D 38 39 D 40 A 41 B 42 43 A 44 C 45 C 46 A 47 A 48 C 49 C 50 D 51 A 52 53 54 C 55 56 D 57 C 58 D 59 60 C 61 C 62 63 B 64 D 65 66 67 B 68 B 69 D 70 71 72 73 74 A 75 A 76 77 78 79 80 81 82 D 83 B 84 85 Explanation:- −2 1 1 3 x −2 x 3 1 3. Lt , Applying L-Hospital’s Rule , we get, Lt 3 = x →8 x−8 x →8 1 12 3 1 1 1 1 4. Required probability = 4C3 = 2 2 4 1 2 4 5. Given matrix is 3 0 6 1 1 P Let λ1 , λ2 and λ3 be the Eigen values of the above matrix Let λ1 = 3 ( Given) Now, λ1 + λ2 + λ3 = sum of diagonal elements = 1 + P ∴ λ2 + λ3 = P + 1 − 3 = P − 2 6. Given vector is F = ( x − y ) ˆ + ( y − x ) ˆ + ( x + y + z ) k i j ˆ ∂F ∂F ∂F Divergence = + + = 1+1+1 = 3 ∂x ∂y ∂z 8. Both young’s Modulus and shear Modulus are required as linear strain will be calculated by young modulus. Change in diameter can be calculated from Poisson’s ratio which depends on young’s modulus and shear modulus. © All rights reserved by GATE Forum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission. Discuss this questions paper at www.gateforum.com. 1 ME GATE Paper 2008 www.gateforum.com 10. Let W be the weight of counterweight. P Q R • 2⋅0 Taking moment about Q 0.5 75 × 2 = W × 0.5, W = 300Kg 75Kg W 11. Grubler’s criterion is applied to mechanism with only single degree of freedom. Given conditions satisfy Grubler’s criterion i.e. 3l − 2j − 4 = 0 where, l = link, j = No. of joints 13. Since the final temperature is same as that of initial temperature µCP 0.001 × 1 × 103 14. Prandtl Number, Pr = = K 1 Given δ = Hydroxynamic Boundary layer = 1 δt = Thermal boundary layer = ? δ 1 = Pr 3 ⇒ δt = 1 δt 18. Job with higher Processing time will be taken first since it will minimize the total holding cost. 21 21. I= ∫ ∫ xydxdx (0.1) 00 2 1 2 2 y2 x x2 = ∫ 2 xdx = ∫ 2 dx = 4 = 1 0 0 0 0 • 0 (2.0) ∂F ˆ ∂F ˆ ∂F 22. Gradient will ∇f = ˆ i +j +k ∂x ∂y ∂z ˆ ˆ ∇f = 2xˆ + 4yj + k i Now ∇f at the po int (1,1, 2 ) ˆ ˆ ˆ ∇f = 2i + 4j + k ˆ ˆ Directonal derivative of f in the direction 3i + 4k is = (2i + 4j + k ) . (3i − 4j) ˆ ˆ = 6 − 16 −10 = = −2 3 + ( −4 ) 5 5 2 2 © All rights reserved by GATE Forum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission. Discuss this questions paper at www.gateforum.com. 2 ME GATE Paper 2008 www.gateforum.com 28 f = yx lnf = xlny differentiating with respect to x 1 ∂f ∂f = lny ⇒ = flny f ∂x ∂x ∂ ∂f ∂x = ∂y ( flny ) = ∂y y lny ∂y ∂ ∂ x ( ) 1 = yx + lnyxyx −1 ⇒ yx −1 ( xlny + 1) y Now x = 2, y = 1 ∂2 f So =1 ∂x∂y 29. y′′ + 2y′ + y = 0 ⇒ D2 + 2D + 1 = 0 i.e. (D + 1) = 0, 2 D = −1, − 1 So solution will be y= ( C1 + C2 ) e− x Now given, y=0 at x=0 and y=0 at x=1 So we get C1 = C2 = 0 y = constant y ( 0.5) = 0 32. Let FS be the shear stress π T = × fs × d3 ⇒ fs = 51MPa, ft = Tensile stress = 50MPa 16 2 ft f Maximum principal stress, σmax = + t + fs2 82MPa 2 2 34. At node P TPQ cos 45º + TPR cos 60º +F = 0........ (1) F P TPQ sin 45º = TPR sin 60º.................. (2 ) TPQ TPR from these two equations 45º 60º we can find out TPQ and TPRin terms of F. Now, At node Q. Q 45º 30º R TQR = TPQ cos 45º TQR ooo On solving we get, TQR = 0.63F 35. Given spring system forms a parallel combination © All rights reserved by GATE Forum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission. Discuss this questions paper at www.gateforum.com. 3 ME GATE Paper 2008 www.gateforum.com Keq = K1 + K2 = 4000 + 1600 = 5600N / m 1 K Natural frequency f = = 10Hz 2π m G.d14 G.d24 39. K1 = & K2 = 8D13.n1 8D23.n2 d1 = d2 = 2mm dia of spring wire G = 80GPa n1 = n2 = 10 D1 = 20mm, D2 = 10mm 3 K1 D2 1 ∴ = = ⇒ K2 = 8K1 K2 D1 8 56. Direction of heat flow is always normal to surface of constant temperature. dT So, for surface P , =0 dx From energy conservation, heat rate at P = Heat rate at Q dT dT 0.1 × 1 × = 0.1 × 2 × dy P dx Q dT ∴ = 20 K / m dy 63. Riser takes care of solidification/contraction in liquid state and phase transition. So volume of metal compensated from the riser = 3% + 4% = 7% 67. Heat supplied by power source = Heat required melting efficiency × transfer efficiency × welding power = cross sectional area × welding speed × 10 .5×.7×2×103 = 5 × 10 × V ⇒ V = 14 mm / s © All rights reserved by GATE Forum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission. Discuss this questions paper at www.gateforum.com. 4 ME GATE Paper 2008 www.gateforum.com 2 R 3 − Ri3 41. Torque carrying capacity, T = µw o2 3 R o − Ri2 ( w = P × π R 02 − Ri2 ) Given R0 = 50mm, Ri = 20mm, P = 2MPa and µ =0.4 So, T=196NM 45. Given mc = 2mh Mass flow rate Th,i ch = 2cc specific heat So, we get Tc,o Heat capacity Hot fluid = Heat capacity Cold fluid Th,o Tc,i ∴ LMTD = ∆T1 = Th,i − Tc,o 20 = 100 − Tc,o ⇒ Tc,o = 80º C 82. τs = 250MPa. V = 180m / min, F = 0.20mm / rev r = 0.5, α = rake angle = 7º φ = shear angle r cos α tan φ = ⇒ φ = 28º 1 − r sin α Now shear force wt1τs Fs = w = depth of cut = 3mm, t1 = feed = 0.02mm sin φ ∴ Fs = 320KN 83. From Merchant’s theory 2φ + λ − α = 90º ∴ λ = Friction Angle = 90º +7º −2 × 28º = 41º µ = tan λ = .87 Form Merchant circle Fs FC = R cos ( λ − α ) ....... (1) and R = ........ (2 ) cos ( φ + λ − α ) FS cos ( λ − α ) R = Resultant force ∴ FC = ,FC = 565N cos ( φ + λ − α ) © All rights reserved by GATE Forum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission. Discuss this questions paper at www.gateforum.com. 5

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