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IIT-JEE 2011 SOLUTIONS PAPER-I (TIME ) SOLUTIONS

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IIT-JEE 2011 SOLUTIONS  PAPER-I (TIME ) SOLUTIONS Powered By Docstoc
					             MODEL SOLUTIONS TO IIT JEE 2011
                                                              Paper I
                                           1   2         3       4         5         6          7
                                           A   D         A       B         D         C          C

                                           8             9             10                   11
                                     A, B, D         A, D            A, C, D              A, B, C

                                               12        13     14         15        16
                                               D         B      B          A         C

                       17              18           19          20              21              22        23
                       9                   5        5            4              6               5         7

                       Section I                                                            O

1.   27 Al + 4 He → 30 P + 1n( Y )                                                          C
     13      2      15     0
     ↓                 ↓                                                                            N   CH2              Br
     30 Si + 1H( X ) 30 Si + 0 e( Z )
     14      1       14      1                                                             C

                                 +                                                         O
2.   Since the mobility of K is nearly same as that of
        +
     Ag which it replaces, the conductance will                                      2–                          2–
                                                                      4.   [NiCl4]   is tetrahedral, [Ni(CN)4]        is square
     remain as more or less constant and will                                                 2+
                                                                           planar & [Ni(H2O)6] is octahedral.
     increase only after the end point.
                                                                      5.   Ba(N3)2 → Ba + 3N2
                   O
                                                                           Very pure N2 is produced by the thermal
3.                 C                                                       decomposition of Barium or sodium azide.
                                     KOH
                           NH                                                     2 × 1.15 × 1000
                                                                      6.   M=
                 C                                                                     1120
                                                                                = 2.05 M
                 O
                  O
                                                                      7.   o-hydroxy benzoic acid: pKa = 2.99
                   C                                                       p–hydroxy benzoic acid: 4.58
                                Br             CH2 Cl
                                                                           p–toluic acid = 4.34
                           NK                                              p–nitrophenol = 7.15
                 C
                                                                                                 Section II
                 O
                                                                      8.   Adsorption is always exothermic.
                                                                           Chemisorption is more exothermic than
                                                                           physisoprtion and it requires activation energy.
9.    A, D
                                                                                       Section IV
10. Cassiterite contain 0.5 – 10% of metal as SnO2
    the rest being impurities of pyrites of Fe, Cu &        17. Maximum no. of electron with n = 3 is 18, of
    wulframite.                                                 which are having −1 spin
                                                                                   2
11. In a, b & c all atoms are in the same plane.
                                                            18. 3Na2CO3 + 3Br2 → 5NaBr + NaBrO3 + 3CO2
                          Section III
                                                            19.
       H3C                         dil.H2SO4                                  Br
      H3C  C       C        CH         HgSO 4
12.                                                                                                         alc.KOH
       H3C        (P)                                                         C    CH2      CH 2      CH3
                                                            CH3 CH 2
       H 3C                        (i) NaBH4
      H3C   C      C        CH3
                                   (ii) dil.acid
        H3C        O
                   OH                                             CH3 CH           C     CH 2      CH2      CH3
       H 3C                                                                                               (2 isomers)
      H3C   C      C        CH 3
        H3C
                      H      (Q)
                                                                  CH3 CH2          C     CH      CH2      CH3
                                                                                                          (2 isomers)
                   OH
     H3C
                                         H+
13. H3C  C         C        CH3
                                       (-H2 O)                                          C       CH2    CH 2     CH3
        H3C                                                            CH3 CH2
                      H                                                                                         (1 isomer)
       H 3C
                                   1, 2-shift
      H3C   C      C        CH3
                                   of -CH3
        H3C           H
                   (2°)                                                 6.626 × 10−34 × 3 × 108
                                                            20. hυ =
                  CH3                                                   300 × 10 −9 × 1.6 × 10 −19
      H3C                          -H   +

              C   C        CH3
                                                                       = 4.14 eV
      H 3C
                  H
      CH3                  CH3 ozonolysis                         E = hυ – hυ0
              C    C                                              For Li, Na, K and Mg hυ0 values are less than
       CH3                 CH3
                                                                  4.14 eV
                                                 O
                                   2
                                                 C          21. Let the number of glycine units be x
                                        H 3C          CH3       Total mass of the hydrolysed products
                                                                             = 796 + 9 × 18 = 958
                                                                                        75 x × 100
                                                                % by mass of glycine =             = 47
                                               1                                           958
14. 2AgNO3 + Cu → Cu(NO3)2 +                     Ag
                                               2                ∴x=6
      The metal dipped is Cu.
                                                            22. S4O2− has sulphur atoms with oxidation states
                                         2+
      Blue colour due to formation of Cu
                                                                   6
                                                                  ‘0’ & +5.
15. The solution in which the metal is dipped in
    AgNO3.                                                                                               2.24
                                                            23. Vol. of 0.1 mole at 0.32 atm =                =7L
                                                                                                         0.32
16. The deep blue colour due to formation of
              2+
    [Cu(NH3)4] and the white precipitate of AgCl
                                            +
    dissolve due to formation of [Ag(NH3)2]
                                                               PART II

                                    24          25        26      27         28        29         30
                                    D           D         D      A           A         A          B

                                    31                    32                 33                   34
                                  B, D           A, B, C, D           A, B, C, D                  A

                                                35        36     37          38        39
                                                C         B      D           C         B

                            40      41               42          43               44              45         46
                            3        5               6            3                9              4          1

                            Section I                                          1        1    1 
                                                                                 = Z2R 2 − 2 
                                                                                               
                                                                              λ2       2    4 
24. x component of area vector is
                                                                                         5
      a2           E 0a 2
            ∴φ =                                                             λ2 = λ1 × 36 × 22
        2               2                                                                3
                                                                                       16
                    2                                                                        0
25. T sinθ = mω Lsinθ
                                            T                                     = 1215 A
               324                      L
      ω=                                    θ
            0 .5 × 0 .5                                                        X   52 + 1
                                                                       30.       =
        = 36                                                                  10 48 + 2
                                    mg
                                                                                                   Section II
26. Position (1): Let charge on 2 µF be Q. Then
                 Q2                                                    31.
      energy =      µJ
                 4                                                                electron
                                                                                                        •B
      Position (2): Ceq = 10 µF
                                            Q2
      Total charge = Q. ∴ Energy =             µJ
                                            20
      Loss % = 80%
                                                                                   proton
          1
27. n =     (Θ 22.4 λ = 1 mole)
          4                                                                         2πm
                                                                             T=         , different for them.
                       1 3                                                           qB
    W = nCV∆T = × R.∆T
                       4 2
    T1V1γ − = T2V2γ − ⇒
             1             1                                                  dQ
                                                                       32.       is same for A and E and both are maximum.
    T2 = 4T1 ∴ ∆T = 3T1                                                       dt
                                                                                                λ 
                                                                           Thermal resistances      are as below.
               f           v                                                                   KA 
28. f’ =            × 1 + 0 
               vs          c 
          1−                                                                                         RB
               c 
         8 × (320 + 10 )                                                                               RC
      =
            320 − 10                                                                         RA                  RE
                                                                                                       RD

      1      1   1                                                              1          4         1       4
29.      = R 2 − 2 
                                                                           RA =    , RB = , RC = , RD =
      λ1    2   3                                                               8          3         2       5
                                                                                   1
                                                                             RE =      . So C is also correct.
                                                                                  24
                             1                                                  1/ 3
    (Θ Eq.R (RC, RB, RD) is    .                                     q2              4+ 2 
                            4                                   ⇒a=                          
                                                                     r                16 πε0 
    D also correct Θ RB parallel                                                             
    RD is equal RC                                              ⇒N=3
    Note: It is assumed that there is no radiation
    loss.                                                 41. Fd = mg sinθ − µmg cosθ
                                                              Fu = mg sinθ + µmg cosθ
33. On interconnecting V same                                 Fu = 3Fd
       Q     Q                                                       1
    ⇒ A = B ∴ (B) correct.                                    ⇒µ=
       R A RB                                                        2
               σR                                                               1
          V=      is standard formula                         ∴ N = 10 µ = 10 × = 5
               ε0                                                               2

    ∴ (C) correct
                            V        V                    42.                  L
    V same ⇒ EA =             , EB =                                    x x x x x x x x
                           RA        RB
                                                                                       B
    ∴ (D) correct.
                                                                            −−−−−−−−−−

34. Under case (B) eventhough the disc is free to                    µ0 Ι
    rotate, it will not rotate. ∴ Cases (A) and (B) are         B=
                                                                      L
    identical in all respects.
                                                                             µ0 πr 2gΙ
                                                                φcoil = Bπr =
                                                                            2
                                                                                 L
                       Section III                                dφ         E
                                                                −    = E, i = ; M = iπr
                                                                                        2
                                                                  dt         R
               e2 
35. [N] = L−   = Force × area
            3                                                   ⇒N=6
               ε0 
               
                                                                      MgL
                                                          43. ∆λ =
36. Find ω by the given formula                                        AY
          c                                                     L’ = L + ∆λ
    λ=
         ω                                                          L'  Mg 
                                                                ⇒      = 1 +      
         2π                                                         L         AY 
                                                                            L'−L
                                                                α=
37. As ball goes up, x is positive and increasing, v is              (L × 40 − L'×30 )
    positive and decreasing. Symmetrical for ball                                Mg
                                                                   =
    coming down.                                                                 Mg  
                                                                      AY 40 − 1 +     30 
                                                                                   AY  
                                p2                              Solving, we get M ≅ 3 kg
38. At x = 0, E = K.E =
                                2m
    ∴ E varies as p
                       2
                                                                          2                2     
                                                          44. N × 10− = 2  mr 2 + md2  + 2  mr 2  ;
                                                                     4
    ∴ E1 = 4E2                                                             5                5    
                                                                             4 2
                                                                                 × 10− = 2 2 × 10− m
                                                                                      2           2
39. Starting from positive positions (i.e. in air),             Here d =
                                                                              2
    amplitude in air will be more than amplitude in
                                                                  5             1
    water. Momentum is negative for downward                        × 10− m; m = kg
                                                                         2
                                                                r=
    journey. Also water produces damping.                         2             2
    ⇒ graph is spiralling in.                                   ⇒N=9


                       Section IV                         45.
                                                                            a = 0.3


40. U =
           q2
          4πε0a
                [
                4 + 2 + 2a r
                          2
                            ]                                                   α
                                                                                    2N
                                                                                             f2

                                                                                                    f1
                           dU
    For equilibrium,          =0
                           da
         f1R − f2R         a                     ∴ M = mN0 = 10− × 10 kg
                                                              25    19
    α=               ,α=
                                                     = 10− × 10 × 10 mg
              2                                           25    19   6
           mR              R
       f1 − f2              2 − f1                   = 1 mg
    ⇒          = a = 0 .3 =        ,m=2
          m                  m
    ⇒ f1 = 1.4, f2 = 0.8 ⇒ µ.2 = 0.8 ⇒ µ = 0.4

                           s−
                      10    1
46. A0 = |−λN0| = 10
             1010
    ∴ N0 =
               λ
         = 10 × 10 = 10
             10   9     19
                                                                       PART III

                                                   47   48        49       50        51            52              53
                                                   C    B         B       A          C             D               C

                                              54                   55                 56                              57
                                           QUESTION
                                                                   B,D               A,D                           B,C
                                          INCORRECT

                                                        58        59      60         61            62
                                                        B         D       D          A             B

                                 63                64        65           66              67                       68              69
                                 7                  9        8            6                1                       2                5

                                 Section I                                                     (b − 1)3 1
                                                                                          =            +
                                                                                                  3      3
        ρ
47. Let v = Ai + B j + C k                                                                     1
                                                                                                         ( x − 1)3 
                                                                                                                                    1

                                                                                               ∫
                                                                                     R2 = ( x − 1) dx =       2
      a×b. v = 0                                                                                                    
                                                                                         b
                                                                                                         3 b
                                                                                                                   
       1     1 1
                                                                                                           1                      (b − 1)3 
                                                                                                                                           
       1     −1 1 = 0                                                                R1 − R2 =               gives = 0 −                   
       A     B       C
                                                                                                           4                      3 
                                                                                                                                           
        2        0 2                                                                 2(b − 1)3 1 1
                                                                                               + =
      ⇒ 1        −1 1 = 0                                                               3        3 4
                                                                                             1
           A     B       C                                                           ⇒b=        satisfies the above equation.
                                                                                             2
      ⇒ 2 (− C − B) + 2 (B + A) = 0
      ⇒A=C
                                                                               49. y + 2 = m (x – 3)
      For the vector in (C),
                                                                                     and m =               3
      A=C
                             (
      Projection of 3i − j + 3k on C       )                                         y+2=                 3 (x – 3)
        3 + 1− 3    1                                                                y-       3 x+2+3 3 =0
      =          =
            3       3
                                                                                                               ln 3
                                                                                                                              x sin x 2
48.
                     y
                                                                               50.            Let I =          ∫      sin x 2 + sin (ln 6 − x 2 )
                                                                                                                                                    dx
                                                                                                               ln 2

                                                                                     Set x = t
                                                                                            1
                                                                                     dx =      dt
                                                                                           2 t
                                                                                     x=        ln 2 , t = ln2
                                               x                                     x=        ln 3 , t = ln3
                 0       b        1                                                                ln 3
                                                                                                                  t sin t         1
                                                                                     ⇒I=             ∫                          ×
                                                                                                          sin t + sin(ln 6 − t ) 2 t
                                                                                                                                     dt
                                                                                                 ln 2
                                                                                              ln 3
                                                    b                                     1                 sin t dt
                                                                                               ∫
             b
                              ( x − 1)        3                                    =
                                                                                                     sin t + sin (ln 6 − t )
             ∫ (x − 1) dx = 
                      2
      R1 =                                                                               2
                                                                                              ln 2
             0
                                      
                                          3    
                                                   0
              ln 3                                                                  But if
          1                 sin(ln 6 − t )
      =
          2      ∫      sin (ln 6 − t ) + sin t
                                                dt                                       The question is reframed in the following
                                                                                    manner;
                     ln 3
                 1
      2I =
                 2      ∫ dt                                                        “ Let M and N be two nonsingular skew-
                     ln 2                                                           symmetric matrices…………”
         1     3
      =    log
         2     2                                                                    The solution is
           1                                                                               2    2      -1      -1        -1 T
      I = log 3                                                                         M N (M N) (MN )
           4     2                                                                     2    2            -1     -1 T    T
                                                                                    = M N (-MN) (N ) M
                                                                                       2    2       -1     -1 -1     T -1
                                                                                    = M N (-N M ) (N ) (-M)
      log x log 2                                                                      2    2    -1     -1        -1
51.        =                                                                        = M N N M (-N) M
      log y log 3                                                                        2    2     -1
                                                                                    =-M N N M N M=-M NM N M
                                                                                                           -1    -1            2   -1   -1
                                                        k
      ⇒ log x = k log 2 ⇒ x = 2                                                                                               2
                                                                                                                      = - M N (N M) M
                                                                                                                                       -1
                              k
      log y = k log 3 ⇒ y = 3                                                                                                 2
                                                                                                                      = - M N(M N) M
                                                                                                                                      -1

      (2 )log 2 = (3 )log 3
          k +1                     k +1                                                                                       2
                                                                                                                      =-M NN M M
                                                                                                                              2
                                                                                                                                 -1    -1

                 k +1               k +1                                                                              =-M
      2log 2            = 3log 3
                                    2                                   2
                                                                                    Choice (c)
      ⇒ ( k +1) (log2) = (k+1) (log 3)
      ⇒
       k = −1
                                                                                            x 2 y2
      ⇒ x0 = 2− = 12                                                                           +   =1
               1
                                                                                55. For
                                                                                            4    1
                                                                                                3
                    π                                                           e=            and focus = (± 3 ,0)
52. P = θ : sin  θ −  = cos θ                                                               2
                    4        
                                                                                                 x2          y2                   2
                            3π                                                  ∴ for            2
                                                                                                         −         = 1, e =
      ⇒P = θ : cos θ = cos     − θ                                                           a           b2                   3
                             4                                                          2
                                                                                        b            1
                           3π                                                   ⇒       2
                                                                                                =
      ∴ P = θ : θ = 2nπ ±     − θ                                                   a            3
                           4      
                                                                                    Substitution                 ( 3,0), a    2          2
                                                                                                                                  = 3, b = 1
                     3π 
      = θ : θ = nπ +                                                              Focus (2, 0)
                     8 
                                                                                                                                      x2
                 π        π                                                     ∴ required hyperbola is                              − y2 = 1
      Q = θ : θ − = 2nπ ±  − θ                                                                                                    3
                 4        2    
                     3π                                                       56. j − k = i + j + k  (i + j + 2k)
      = θ : θ = nπ +    
                     8                                                            & k − j = i + j +2k  (i + 2j +k)
      ∴P=Q                                                                          ∴ j − k, k − j coplanar with the given vectors
                                                                                             1 −1 0           1 0 0
      a10 − 2a8                     α10 − β10 − 2 α 8 − β8  (               )       and ± 1 1 2 = ± 1 1 3 ≠ 0
53.
         2a9
                               =
                                                (   9
                                                2α −β           9
                                                                    )                               1        2     1          0 1 −1
                                                                                    (A) and (D) true.
                                     α8 (α 2 − 2) − β8 (β2 − 2)
                                =
                                                 2(α9 − β9 )
                                                                                57. f (x +y) = f (x) +f(y)
                                     α8 ,6α − β8 6β                                 ⇒ f(x) = kx
                                =
                                            (
                                           2 α9 − β9        )                       f(x) is continuous ∀ x ∈R
                                     6(α9 − β9 )                                    and f’ (x) = k
                                =                           =3
                                     2(α9 − β9 )


                                     Section II

54. Skew symmetric matrix of order 3 is singular. Its
    inverse does not exist. Therefore there is
    NO SOLUTION TO THIS QUESTION.
                                                           a + 48 + 7c = 0
                    Section III                            a+6+c=0
                                                           Solving a = 1 and c = - 7.
                                                           ∴ The equation is x + 6x – 7 = 0
                                                                               2
    For problems (58) and (59)
                                                              1 1 α+β 6
                                  3                        ∴    + =         = <1
                                           1                  α β       αβ    7
                        W         5                  ∞               n
                                                           1   1
                   1                                ∑ α + β
                                                     
                                                     
                                                            
                                                            
                                                                         (is an infinite geometric progression) =
                   2                                n =0
                         R                 1
                                  2                         a            1
          H                                                     =                =7
                                  5        2               1− r              6
                                                                     1−
                                                                             7

                               1
                                           1                                      Section IV
                              10
          T             2W                                    π
                                                    63. θ =
                                       3                      n
                  1          2R                 1
                                                           1         1         1
                  2                   10        3               =         +
                                                         sin θ sin 2θ sin 3θ
                 1R+1W       3         2                   1         1         1
                                       3                        −         =
                             5                           sin θ sin 3θ sin 2θ
                                                            sin 3θ − sin θ       1
58. Probability                                                             =
                                                             sin θ sin 3θ     sin 2θ
      1 3      2 1 3        1 1 3 2
    =  × 1+ × +       × 1+   × + ×                      2 cos 2θ         1
      2 5      5 2 10      10 3 5 3 
                                                                    =
                                                           sin 3θ      sin 2θ
     1 3 1 3          1 2                             ⇒ 2 sin2θ cos2θ = sin3θ
    =
     2 5 + 5 + 10 + 30 + 5 
                                                       sin4θ −sin3θ = 0
     1 46 23                                                        7θ      θ
    = ×      =                                          ⇒ 2 cos         sin = 0
     2 30 30                                                         2      2
                                                                     7π       π
                                                        ⇒ 2 cos          sin     =0
                   1 3         1                                   2n      2n
                        ×1+ 25 × 
                  2 5                                   7π              π
                                                            =  (2k + 1) 
                                2
59. Probability =
                         23                              2n              2
                         30                                          7
                12                                         ⇒n=
             =                                                     2k + 1
                23
                                                           For positive integral values of n
                                                           k = 0 or 3
60. 7(a + b+ c) = 0 , ⇒ a + b + c = 0                      put k = 0, n = 7
             also 2a + b + c = 1
             ⇒ a = 1, b + c = −1
    ∴ 7a + b + c = 7 − 1 = 6                                      sm
                                                                         5n
                                                                             [6 + (5n − 1) (a2 − 3)]
                                                    64. m = 5n;       = 2
                                                                  sn       n
                                                                             [6 + (n − 1) (a2 − 3)
                                           1    3                          2
61. x − 1 = 0 and Im(ω) > 0 ⇒ ω = −          +i
     3
                                           2    2               9 − a2 + 5n (a2 − 3) 
                                                          =5                         
    a = 2 ⇒ 2 + 8b + 7c = 0                                     9 − a2 + n (a2 − 3 ) 
    14 + 7b + 7c = 0                                            9 − a2 + 5 (a2 − 3) (9 − a2 ) + 10(a2 − 3 )
                                                           ∴                       =                        ,   is
    (taking first and third columns)                              9 − a2 + a2 − 3    9 − a2 + 2 (a2 − 3 )
    Solving we get b = 12 and c = − 14
                                                           independent of n
                           3     1     3
    ∴ The equation in 2 + 12 + −14                          4a2 − 6 9a2 − 21
                          ω     ω    ω                             =
                                                              6      a2 + 3
    = 3ω + 1 + 3ω
                    2
                                                           (2a2 − 3) (a2 + 3) = (3a2 − 7)
    = 3 (ω + ω ) + 1 = − 2
                2
                                                           2a2 − 24a2 + 54 = 0
                                                             2
    Correct choice (a)
                                                           a2 − 12a2 + 27 = 0
                                                            2
                   st        rd
62. b = 6. Taking 1 and 3 columns we get                   a2 ≠ 3, a2 = 9
      a −5 + a −4 + a −3 + a −3 + a −3 + 1 + a8 + a10                68. Extremum point of latus rectum are (2,4) and
65.
                              8                                          (2,−4)
                                                                                                           1 
                       (             ( )               )
                                                       1
                                                                         ∴ Area of triangle so formed with  ,2 
                     ≥ a−5a− 4 a−3 13 a8a10            8   =1                                              2 
      ⇒ minimum value of sum = 8.                                                  1
                                                                           ∆1 =      (y1 −y2) (y2−y3) (y3−y1)
                                                                                  8a
          x                                                                   1
                                                                                  (4+4) (−4 −2) (2 −4) = 6
          ∫                                                                =
                                       3
66. 6 f ( t )dt = 3x f(x) – x                                                8 .2
          1
                                               2
                                                                           Eqn. of tangent at (2,4) is y = x +2 _______(1)
      6f(x) = 3xf’(x) + 3f(x) – 3x                                         Eqn. of tangent at (2, −4) is −y = x +2 _____(2)
                       2
      f(x) = xf’(x) – x                                                                       1 
             dy                                                            Eqn. of tangent at  ,2  is y = 2x +1 _____(3)
      y= x       − x2                                                                         2 
             dx
         dy              dy y                                              Tangent at the extremities of latus rectum
       x    − y = x2 ⇒       − =x                                          intersect directrix at (−2, 0).
         dx              dx x
                  1                                                        Point of intersection of (1) and (2) is (1,3) and of
                ∫
      I. F = − dx
                  x                                                        (2) and (3) is (−1, −1)
                e                                                                  1 −2 0
              1        1                                                         1
      ∴
              x        ∫
                = x dx = x + c
               y.
                       x
                                                                           ∆2 = 1 1
                                                                                 2
                                                                                             3 = −3 = 3
                                                                                   1 −1 −1
      y = x(x + c) = f(x)
                                                                               ∆1 6
      f(1) = 2 ⇒ 2 = 1 + c ⇒ c = 1                                         ∴     = =2
      ∴ f(x) = x(x + 1) ⇒ f(2) = 6                                             ∆2 3


                                  sin θ                            69.
67. f(θ) = sin tan
                            -1
                                           
                                  cos 2θ                                                        3,2
                                           
                                  sin θ 
               tan tan−1
                        
                                          
                                  cos 2θ 
                                          
      =
                                 −1 sin θ
              1 + tan2 (tan
                                      cos 2θ                                                  −5
                                                                                             3,
                    sin θ                                                                      2
                    cos 2θ                             sin θ                                    5
                                                                           minimum of 2 z − 3 + i
      =                               =
                                                                                                2
              cos 2θ + sin θ     2                 2
                                           cos θ − sin2 θ + sin2 θ
                  cos 2θ                                                              5   
                                                                                  = 2  + 2 − 2 = 5
                                     = tan θ                                          2   
         d(f ( θ))
      ⇒            =1
        d(tan θ)

				
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Description: IIT-JEE 2011 SOLUTIONS PAPER-I (TIME ) SOLUTIONS