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GATE- Mechanical_ME_- 2008 Exam Paper

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GATE- Mechanical_ME_- 2008 Exam Paper Powered By Docstoc
					                            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.

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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




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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



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           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




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                                                            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 )




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