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IIT-JEE 2011 PAPER-2 SOLUTIONS ( T.I.M.E) SOLUTIONS

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IIT-JEE 2011     PAPER-2  SOLUTIONS ( T.I.M.E)   SOLUTIONS Powered By Docstoc
					             MODEL SOLUTIONS TO IIT JEE 2011
                                                       Paper II
                                                            PART I
                               1        2        3           4   5          6            7         8
                               C        D        B          A   D           C            A         B

                                    9                  10              11                    12
                                 C, D            A, B, C, D       A, B, D                A, C, D

                                        13       14         15    16        17 18
                                        8         6         4    7          6    8
                                               19                             20
                                        A – r, s, t                    A – p, r, s
                                         B – p, s                       B – r, s
                                         C – r, s                        C–t
                                        D – q, r                      D – p, q, r, t


                       Section I                                 6.    Aromatic primary amines form diazonium salt
                                                                       with NaNO2 and HCl at low temperature which
1.   [Co(NH3)6]Cl3, Na3[Co(ox)3], [K2Pt(CN)4]               &          couples with β–naphthol to form coloured azo
     [Zn(H2O)6](NO3)2 are diamagnetic.                                 dye.


                              PO2 × [H+ ]4                       7.    ∆Tf = i × Kf × m
                     0.06
2.   Ecell = E0 +         log                                                                     0. 1
              cell     4        [Fe2+ ]2                                     = 4 × 1.86 ×              × 10
                                                                                                  329
                               0.1 × (10 −3 )4                             = 0.023
         = 1.67 + 0.015 log
                                                                       F.P = –2.3 × 10 °
                                                                                      –2
                                   (10 −3 )2                                             C
         = 1.67 – 0.015 × 7
                                                                 8.    The structure given is that of β–D–glucose.
         = 1.57 V

                                                                                              Section II
                 RCH2OH
3.                                                               9.    Cu
                                                                            +2
                                                                                 is reduced to Cu
                                                                                                       +1       –     –
                                                                                                            by CN & SCN
                   H+
                                    O          OCH2R
         O
                                                                 10. In (B), (C) and (D), X–(CH2)4–X is converted to a
4.   CuS & HgS are insoluble in dilute mineral acids.                diamine which can form condensation polymer
       2+      2+
     Cu & Hg ions belong to group II of qualitative                  with adipic acid. In (A), X−(CH2)4–X is converted
     analysis.                                                       to a diol which gives polyester.

5.   Haematite is Fe2O3. Oxidation state of Fe is +3                             0.693
                                                                 11. K =
     Magnetite is Fe3O4. Oxidation state of Fe is +2                              t1
                                                                                   2
     and +3.
    K increases with increase of temperature and                               CH3
    hence half life decreases.
       t1                                           18. ClCH2        CH2       C     CH2       CH3
                 100
    t = 2 log          = 8t 1
        0 .3      0 .4       2                                                 H
              -kt
                                                                                          (2 isomers)
    R = R0e                                                     H      CH 3

12. MnO4− → Mn (acid medium)
                     2+                                                                  CH3
                                                         CH3    C     C        CH2
    MnO4− → MnO2 (Neutral and aqueous mediums)
                                                                Cl     H
                                                                                (2 different chiral
                          Section III                                           carbon - 4 isomers)
                                                                           CH3

13. A truncated octahedrous has 8 hexagonal and 6        CH3    CH2        C       CH2     CH3 (1 isomer)
    square faces. (36 edges and 24 vertices)
                                                                           Cl
                                                                           CH2Cl
14. There are six C – H bonds that can involve in
    hyperconjugation.                                           CH2        C       CH2    CH3 (1 isomer)
                                                         CH3

15. PCl5 + SO2 → POCl3 + SOCl2                                             H
    PCl5 + H2O → POCl3 + 2HCl                                              Section IV
    PCl5 + H2SO4 → SO2Cl2 + 2POCl3 + 2HCl
    6PCl5 + P4O10 → 10POCl3                         19. (A) is intramolecular aldol condensation
                                                        (B) involves Grignard reagent addition. To
16. [Cl ] from CuCl = 10−
       –                      3                         carbonyl compound.
                    +    –                              (C)     involves   nucleophilic   addition and
    Ksp(AgCl) = [Ag ] [Cl ]
              1.6 × 10 −10                              dehydration.
                              = 1.6 × 10
        +                                  –7
    [Ag ] =                                             (D) involves dehydration and intramolecular
                    10 −3
                                                        Friedel–Craft reaction.

17. Millimoles of Cl = 30 × 0.01 × 2
                          –
                                                    20. (A) involves transition from solid to gas phase
                      = 0.6
                                                        with the absorption of heat.
                            0 .6
    Vol. of 0.1 M AgNO3 =        = 6 mL                 (B) is exothermic and a gaseous product is
                            0 .1
                                                        formed from a solid.
                                                        (C) involves association
                                                        (D) white phosphorous is converted to the
                                                        polymeric red allotropic form as heating. Different
                                                        solids are considered as different phases.
                                                             PART II

                                  21     22        23         24   25       26         27        28
                                  B      B         C         A     C        D          C         A

                                       29               30          31                     32
                                       B, C         A, C           C, D                A, B, D
                                         33      34          35     36        37       38
                                         2        7          5     5          4         4

                                              39                                  40
                                            A – p, t                      A – p, r, t
                                            B – p, s                       B – p, r
                                            C – q, s                       C – q, s
                                            D – q, r                       D – r, t


                        Section I                                                   µ0N          i
                                                                         −dB =              dr.
                                                                                   (b − a ) 2r
                     GMm                                                                   µ N.i      b
21. Escape KE =
                      r
                                                                         B=   ∫   dB = 0
                                                                                        2(b − a ) a
                                                                                                   ln

               GM
    and V =
                r                                                  25. At θc, 100% reflected.
                                                                       (B or C) But C is correct.
    Eliminate r
                                                                       Θ At θ = 0, T + R = 100,
22. Phasor:                                                            R≠0

       A              Resultant
                                                                   26. 0.01 V = 0.01 V’ + 0.2 V’’
                   π/3                                                                            2h
                                                                         But time of fall =          =1
                         A                                                                         g
           ∴ Required                                                    ∴ V’ = 100, V’’ = 20
                                                                         ∴ V = 500
                4π 
    A sin  ωt +    
                3                                                27. If non-conservative induced electric field this
                                                                       pattern is possible.
23. L.C = 0.01 mm
    diameter = 2.70 mm                                                               m
                                                                   28. T = 2π          holds.
                 1                                                                   K
    % error =
               2.70                                                      At equilibrium position,
     ∆ρ 3∆dia ∆m
        =        +                                                       kx0 = QE
     ρ     dia      m                                                    ∴ Equilibrium shifted.
                3
    % error =       + 2 = 3.1%
               2.70                                                                                      Ι
                                                                                                Section IΙ
24. Take dr at r                                                   29.
                       N.dr                                                            R
    No. of turns =
                      (b − a)                                                        ZA           A
                                                                                                 XC
                                                                  2u sin 60° 2 × 10 × 3
                 R                                    35. T =               =           = 3 s
                                                                      g         10 × 2
                ZB             XB
                                C
                                                                          1
                                                            (u cosθ) T =
                                                                              2
                                                                            aT + 1.15
                                                                          2
            1
      XB =                                                  (u = 10 m s− )
                  A
       C       < XC                                                     1
           Cω
                                                            ⇒ a = 5 m s−
                                                                          2

      ⇒ ZB < ZA
      ⇒ ΙB > ΙR and VC > VC
         R
              A      A    B                                  st
                                                      36. 1 surface
                                                            µ    µ     µ − µ1
                                                          − 1+ 2 = 2
30. Linear momentum in the horizontal direction is           u     v     R
    zero before and after collision ⇒ CM is                          7 7 
                                                                        − 1
    stationary.     for  conservation of    angular
                                                                    +  =
                                                               1      4     4 
                                                          −
    momentum, ω of ring increases after collision ⇒         (− 24 ) V1        6
    slip ⇒ friction towards left.                           ⇒ v1 = 21 cm
                                                             nd
                                                            2 surface
31.                   d
                                                            (Θ R = ∞ for 2 surface)
                                                                          nd

         +q1           •            +q2                       7
                                                               
          (or −q1)    E=0           (−q2)
                                                             − +
                                                                4     4
                                                                          =0
                                                                21 3 v 2
      W agent = ∆U = q(V2 − V1)
                                                            ⇒ v2 = 16 cm ⇒ x = 18 − 16 = 2 cm
              = 1 × (VB − VA)
              = (VB − VA)
                                                             1                 1
                                                               mv = µmg S + kS
                                                                  2                 2
                                                      37.
                                                             2                 2
32. VdA + VdB = 2VdF = 0
                                                            m = 0.18 kg, µ = 0.1,
    ∴ dA + dB = 2dF
                                                            g = 10 m s− , S = 0.06 m
                                                                       2
    Obviously dA < dF and dB > dF                               2
                                                            ⇒ v = 0.12 + 0.04 = 0.16
                                                                  0.16 = 0.4 m s−
                                                                                      1
                                                            ⇒v=
                                      Ι
                             Section IΙΙ
                                                                            4
                                                                        =     ⇒N=4
                                                                          10
    4           7      4 7
                  −1     −                                 2    2
                                                      38. Z = R + XC
                                                                      2

33. 3 − 1 = 4        + 3 4
                                                          ⇒ XC = 1.25 R − R
                                                              2           2   2
    v − 24        6      ∞
                                                                          2
    ⇒ v = 16 from top surface of liquid                          = 0.25 R
    ⇒ 2 cm from bottom surface of liquid                  ⇒ XC = 0.5 R
                                                                   1          1     1
                                                          ⇒C=          =          =
             1242 eV nm                                           XCω 0.5 R × 500 250 R
34. hυ =                = 6.21 eV
               200nm                                                    1
                                                            τ = RC =       s = 4 ms
                                                                       250
      KE = 6.21 − 4.7 = 1.51 eV
      kq                                                                    Section IV
         = 1.5 V for stopping electrons.
       r

      ∴q=
                     ( ) = Ne
              1.5 × 10 −2                             39. Knowledge based.
                         9
                9 × 10
      ∴ N = 1 × 10
                     7                                40. Knowledge based.
                                                            PART III

                                 41        42        43      44    45 46            47   48
                                 B         D         A       A     B   C            D    C
                                      49               50           51                52
                                   A, D              C, D         A, B, D        A, B, C, D

                                           53    54         55      56      57      58
                                           9      2         3       1       2       9

                                                59                             60
                                           A – p, r, s                      A–q
                                            B – r, t                        B– p
                                             C–r                            C–s
                                             D–r                            D–s
                                                                                     2
                            Section I                                    = f(sin(sin(x )
                                                                                2      2
                                                                         = sin (sin(x )
                                                                                             2
41. Let α denote the common root                                         gogof(x) = gog(x )
                                                                                                2
    α + bα − 1 = 0                                                                   = g(sin(x)
      2
                                                                                                   2
    α +α+b=0                                                                         = sin(sin (x ))
      2

                                                                         ∴ sin (sin(x ) = sin sin (x )
                                                                                 2      2              2
    bα − 1 = α + b
                                                                         ⇒ sin(sin(x ) (sin sin(x ) −1) = 0
                                                                                       2             2
    α (b − 1) = b + 1
                                                                                       2                 2
         b +1                                                            ⇒ sin(sin(x ) = 0 or sin (sin(x ) = 1
    α=                                                                             2              2
         b −1                                                            ⇒ sin(x ) = 0 sin (x )
                                                                              2
    (b + 1) + b(b − 1) − (b − 1) = 0
           2      2             2                                        ⇒ x = nπ
    4b + b(b − 1) = 0
             2
                                                                         ∴x=±       nπ n = {0, 1, 2 ……..}
         2
    4+b =0
    b =−3
     2
                                                                                                 1       a b
    b=±I        3                                                  44. for non-singular matrics, ω       1 c ≠0
                                                                                                    ϖ2   ω 1
42. Let the circle be                                                    ⇒ 1 - cω - aω - acω ≠ 0
                                                                                              2
     2    2
    x + y + 2gx + 2fy + c =                                              (1 - cω) (1 - cω) ≠ 0
    Passed thro (−1, 0) → − 2g + c = −1(1)                                      1             1
                                                                         ⇒a≠        =ω and c ≠   =ω
                                                                                       2            2
    Passes thro (0, 2) →4f + c = − 4→(2)                                       ω             ω
         g2 + f 2 − c = g                                                ∴ a = c =ω and b = ω pr ω
                                                                                                  2


    f −c=0                                                               Hence there are two such matrices.
     2

     2
    f =c
    4f + f = − 4
          2                                                        45. Let the point be ‘θ’
    (f + 2) = 0 ⇒ f = −2
            2                                                          Normal at θ is
    ⇒ c = − 4 − 4f = 4                                                   ax      by
                                                                              +       = a2 + b2
                  5                                                     sec θ tan θ
    2g = 5, g =
                  2                                                    Normal passes thro (9, 0)
    Circles is x + y + 5x − 4y + 4 = 0
                2    2                                                   9a
                                                                              = a2 + b2 (1)
    (−4, 0) satisfies the equation                                      sec θ
                                                                       a sec θ = 6
                                                                                6
43. fogogo f(x)                                                        sec θ =
              2
    = fogog(x )                                                                 a
    = fog(sin(x)
                 2                                                     (1) reduces
                 a                                                     48.
      9a×
                     2   2
                   =a +b
                 6                                                                                               A
            2                                                                                                        2
      3a                  2               2                                                                      (t , 2t)
                  =a +b
       2
                  = a + a (e − 1)
                          2               2   2

      3
        =1+e −1
             2
                                                                                            P
      2
           2                                                                 0 (0, 0
        =e
                  3
      e=                                                                      OP 1
                  2                                                               =
                                                                              PA 3
                                                                             Let P be (x, y)
46.
                                                  y                               t2      2t
                                                                             X=      ,Y=
                                                                                  4       4
                                                                             Lows of P is
                                                               1             4X = 4Y
                                                                                     2
                                                          x=
                                                               2                 2
                                                                             ⇒Y =X
                                  −1 0                             x
                                                      1   2
                                                                                                      Section II

      f(x) = f(1 −x)                                                           (            ) (
                                                                       49. P E ∩ F + P F ∩ E =         )   11
                                                                                                           25
                                       1
      ⇒ curve is symmetrical about x =

                  2
                                       2                                                (
                                                                             Also P E ∩ F =       )   2
                                                                                                      25
      R1 =       ∫ xf ( x ) dx                                                      (           ) (        ) (
                                                                             But P E ∩ F = P E ∩ F = P E ∪ F             )
                  1
                                                                                             2    23
           2                                                                 ∴ P(E∩F) = 1 −     =
                                                                                            25 25
      =    ∫ (−1 + 2 − x) f (−1 + 2 − x) dx                                  P(E∪F) = 1 −
                                                                                          2
                                                                                             =
                                                                                               23
           1
      2
                                                                                          25 25
                                                                             P(E∪F) = P(E ∩ F ) + P(F ∩ E ) + P(E ∩ F)
      ∫ (1− x) f (1 − x) dx
      1                                                                      (refer figure)
                                                                                          + P(E ∩ F )
           2                                                                     23 11
                                                                             ∴       =
      =    ∫ (1 − x) f ( x) dx                                                   25 25
           −1
                                                                             ∴ P(E ∩ F ) =     = P(E ) × P(F )
                                                                                            12
      2                           2                                                                                      E
                                                                                            25                                           F
      ∫   f ( x ) dx −            ∫    xf ( x ) dx
                                                                             From the options                                E∩F   F∩E
      −1                          −1                                                  3            4
      = R2 − R1                                                              P(E) = and P(F) =
                                                                                      5            5                                E∩P
      2R1 = R2                                                               or
                                                                                       4           3
47. Which is of the form 1α                                                  P(E) = and P(F) =
                                                                                      5            5
          lim ( f ( x )−1) g( x )
      ex →0
                                              1                                       b−x
               (1+ x log(1+ b 2 )−1)                                   50. Let y =
      = e                                     x                                      1 − bx
             x log(1+ b 2 )                                                  ⇒ (1 – bx) y = b – x
      =    e      x                                                          ⇒x(1 – by) = b – y
      = elog(1+b
                              2
                                  )                                                 b−y
                                                                             ⇒ x=
                                                                                    1 − by
      ∴ 1 + b = 2b sin θ
                      2                       2
                                                                             -1
              1+ b                    2                                      f (x) = f(x)
      sin θ =                                                                        (1 − bx )(−1) + (b − x )b
           2
               2b                                                            f’(x) =
      Since b > 0                                                                           (1 − bx )2
               π
      ∴θ=±
               2
                 (
               − 1 − b2  )   1                                      1  0 
                              (      )
     f’(b) =              = 2                                          
                    22   b −1                                 M = 1 =  0  ⇒ 2 + 3 + c3 = 12
               1 − b                                              1 12 
                                                                     
               − 1 + b2       1                                                 ⇒ c3 = 7.
     f’(0) =            =
                   1      f ' (b )                              A1 + b 2 + c 3 = 0 + 2 + 7 = 9


51. Normal at ‘t’ is                                        54. Origin lies on the side 2x – 3y - < 0
                    3
    y + xt = 2t + t
                    3                                           The smaller intersection Is on 2x – 3y – 1 > 0
    6 + 9t = 2t + t
    t − 7t − 6 = 0
     3

    t = − 1, t = − 2, t = 3                                                                        6       2x – 3y = 1
    Normal are
    Y − x = − 3,
    And y + 3x = 33
                                                                             − 6                            6
    And y − 2x = −12

       π−                                                                                       − 6
52. f  − =0
       2 
       −π +          −π 
     f       = − cos    =0                                  All points ∈ S lies inside the circle, except
        2            2 
                                                                5 3        1 1
     (A) is true                                                 ,  and  ,  satisfies the inequality
                                                                 2 4      8 4
     f(x) is continuous at x = 0
         -                                                      2x – 3y – 1 < 0.
     f’(0 ) = 0
         +                                                      ∴ The remaining 2 points lie on the smaller
     f’(0 ) = 1
                                                                intersection.
     (B) is true
     F(x) is continuous at x = 1
         -                                                  55. Let A = ai + bj + ck
     f’(1 ) = 1
                                                                                                       2
         +
     f’(1 ) = 1                                                  P =i+j+k
                                                                  1                                P1 = 0
     f(x) is differentiable at x = 1.                            P2 = i + wj + w k
                                                                                             2         2
                                                                                                   P2 = 0
     (c) is true.
                                                                                                       2
                                                                 P3 = i + w j + wk
                                                                                 2
      22 11                                                                                        P3 = 0
           =     =1
      14      7
                                                                Them x = A, P1 , y = A, P2 z = A, P3
           −3                −π 
      x=       lies in  − ∞,    
                                                                                               A  P1 + P2 + P3 
                                                                                                2    2     2   2
            2                2                                     2
                                                                 x + y +z
                                                                             2           2                      
                                                                                             =                  
          ⇒ f(x) is differentiable there, since f(x) is a            2
                                                                 a +b +c
                                                                             2           2
                                                                                                       A
                                                                                                         2

     linear function in that interval
                                                                =3+0+0=3
     (D) is true.

                                                            56. Let y(x) = n as g (x) = u
                          Section III
                                                                ∴ The given equation becomes
                                                                du + udu = udu
         a1 a2 a3 
                                                                  dy
                                                                        + u = v ⇒ ue = e +c
                                                                                      u   u
53. M =  b1 b2 b3                                             ⇒
                                                                    du
        c c c 
         1 2    3                                                       g(x)   g(x)
                                                                i.e y(x) e = e + c
       0   − 1                                              y(0) = 0 as g(0) = 0 ⇒ C = 0
         
                                                                ∴ y(2) e = e + 0 ⇒ y (2) = 1
                                                                          g(2)  g(2)
     M 1  =  2  ⇒ a2 = - 1; b2 = 2; c2 = 3
      0  3 
         
                                                            57. x − 4x + 12x + x − 1 = 0
                                                                 4       3           2
       1  1
                                                            f(0) < 0, f(1) > 0
     M − 1 =  1  ⇒ a1 + 1 = 1 ⇒ a1 = 0
                                                                ∴ one root in (0, 1)
       0   − 1
                                                            f (x) = 4x − 12x + 24x + 1
                                                                 1          3      2

                                                                     f (x) = 12(x − 2x +2) > 0
                                                                      11           2
                             C1 – 3 = - 1 ⇒ c1 = 2
                                                                         1
                                                                     ⇒ f (x) is increasing
          ∴ it has exactly one root                             ∴ x ∈ ( − ∞, 0] ∪ [ z, ∞)
          ⇒ f(x) may have almost 2 distinct roots
          Since real roots are even in number                                cos θ         sin θ      cos θ
                                                                     1
          f(x) has 2 distinct real roots                (c)                  − sin θ       cos θ      sin θ`
                                                                cos3 θ
                                                                             − cos θ − sin θ cos θ
58.   (r − c) × b = 0                                                               0          0       2 cos θ
      r = c +mb                                                          1
                                                               =                 − sin θ     cos θ      sin θ
                                                                    cos3 θ
      0 = r . a = c . a + m b .a                                                 − cos θ − sin θ        cos θ
      = −4 + m, m = 4                                       2 cos θ
                                                        =                = 2 sec 2 θ ∈ [2, ∞ )
      r = c + 4b                                                3
                                                            cos θ
                               2
      r . b = c .b + 4 b
                                                                 3
      =1+8=9
                                                        (d)     x2   (3x – 10)
      ∈ (−∞, −1]U[1, ∞)                                                  5          3
                                                               = 3 x 2 − 10 x 2
                           Section IV                                              3
                                                                    15 2
                                                               f’(x) =  x − 15 x
                 2iz                                               2
59. (a). (a) Re 
                                  
                                                                     x 
                 1 − z2                                      = 15 x  − 1 ≥ 0
                                                                      2 
                   
              2i                                             If x ≥ 2.
          Re       
              1 − z
             z
                   
                                                                                −1 + 3 1
                                                    60. (a) cos θ =                    =
                2i                                                               4     2
          = Re                                                             π
               z−z                                           ∴θ=
                                                                             3
                    1
          = Re                                                                                        π 2π
                (− Ιm z )                                      ∴ required angle is π -                  =
                                                                                                      3   3
               1
          =
            − Im z
                                                                x

                                                                ∫ (f ( x) − 3x )dx = a
                                                                                           2
                        x -1
                                                        (b)                                    − x2
      (b) Put y = 3
                                                                a
                8
             < y                                               f(x) – 3x = - 2x ⇒ f(x) = x
          −1 3 ≤ 1                                                 π π
             1 − y2                                            ∴ f  =
                                                                   6 6
               8
           ⇒     xy ≤ 1 − xy 2
               3
                                                                π
                                                                    log[sec πx + tan πx ]7 6 = π
                                                                                         5
              8                                         (c)
           ⇒ y ≤ 1− y2        if 1 −y ≥ 0
                                     2                        log 3                        6
              3
          3y + 8y − 3 ≤ 0 i.e y∈ [−1, 1]
            2
                                                                    1 
          3y +ay − y − 3 ≤ 0
            2
                                                        (d)     Arg       = Arg(z − 1)
                                      1                           z − 1
          (3y − 1) (y +3) ≤ 0 y ∈ − 3, 
                                      3                      Maximum value = π.
              1         1
          y∈ 0,  = 3− ≤
                       1

              3         3
              x≤0
           8
             y ≤ y 2 − 1 if 1− y ≤ 0
                                2
           3
          3y − 8y − 3 ≥ 0
             2

          3y − 9y + y − 3 ≥ 0
             2

          (3y +1) (y− 3) ≥ 0          y≥3
                             ⇒3 − ≥3
                                  x 1

               x ≥ 2 (r, t)

				
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posted:4/11/2011
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Description: IIT-JEE 2011 PAPER-2 SOLUTIONS ( T.I.M.E) SOLUTIONS