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NAITS_NFT_IIT_SOLUTION_MERGE

VIEWS: 6 PAGES: 28

AIEEE IIT-JEE SAMPLE PAPERS NARAYANA TUTORIALS TEST PAPER KEY SOLUTIONS ANSWERS QUESTIONS STUDY MATERIAL

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									This Key & Hints paper contains 28 Pages                       NFT_6 PAPER I & II DT : 16-01-11



                            NARAYANA
                            I   I   T         A     C      A     D    E      M     Y

                        ALL INDIA TEST SERIES
                    NARAYANA FULL TEST – 6 PAPER – I&II
                                          ANSWERS
                                           PAPER – I
MATHS                               PHYSICS                            CHEMISTRY
1.     (A)                          24.    (A)                         47.       (A)
2.     (C)                          25.    (D)                         48.       (D)
3.     (C)                          26.    (C)                         49.       (D)
4.     (A)                          27.    (A)                         50.       (A)
5.     (D)                          28.    (B)                         51.       (A)
6.     (B)                          29.    (C)                         52.       (C)
7.     (A), (B), (C), (D)           30.    (B), (C), (D)               53.       (A), (B), (C), (D)
8.     (B), (C)                     31.    (A), (C)                    54.       (A), (B), (C), (D)
9.     (A), (B), (C)                32.    (A), (B), (D)               55.       (B), (C), (D)
10.    (A, (B)                      33.    (A), (B), (C)               56.       (A), (C)
11.    (A)                          34.    (A)                         57.       (A)
12.    (D)                          35.    (D)                         58.       (C)
13.    (A)                          36.    (D)                         59.       (B)
14.    (D)                          37.    (D)                         60.       (A)
15.    (C)                          38.    (D)                         61.       (C)
16.    (C)                          39.    (D)                         62.       (D)
17.    (D)                          40.    (C)                         63.       (B)
18.    (C)                          41.    (B)                         64.       (A)
19.    (D)                          42.    (D)                         65.       (C)
20.    (A)                          43.    (A)                         66.       (A)
21.    (D)                          44.    (C)                         67.       (A)
22.    (C)                          45.    (B)                         68.       (A)
23.    (A)                          46.    (D)                         69.       (C)




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This Key & Hints paper contains 28 Pages          NFT_6 PAPER I & II DT : 16-01-11


                                        PAPER - II
MATHS                            PHYSICS                         CHEMISTRY
1.  (D)                          23.  (D)                        45. (B)
2.  (C)                          24.  (C)                        46. (A)
3.  (A)                          25.  (B)                        47. (A)
4.  (B)                          26.  (C)                        48. (C)
5.  (B)                          27.  (D)                        49. (C)
6.  (C)                          28.  (D)                        50. (A)
7.  (B)                          29.  (A)                        51. (B)
8.  (C)                          30.  (C)                        52. (D)
9.  (B)                          31.  (D)                        53. (B)
10. (C)                          32.  (C)                        54. (B)
11. (D)                          33.  (A)                        55. (B)
12. (A)                          34.  (D)                        56. (A)
13. (A)                          35.  (D)                        57. (B)
14. (C)                          36.  (A)                        58. (B)
15. (D)                          37.  (C)                        59. (A)
16. (D)                          38.  (D)                        60. (A)
17. (C)                          39.  (C)                        61. (B)
18. (D)                          40.  (A)                        62. (A)
19. (C)                          41.  (C)                        63. (D)
20. (A) – (p)                    42.  (A) – (p), (q), (r),       64. (A) – (q)
    (B) – (r)                         (B) - (p), (q), (r),           (B) – (p)
    (C) – (q), (s),                   (C) – (p), (s)                 (C) – (r)
    (D) – (p), (r), (s)               (D) - (p), (s)                 (D) – (s)
21. (A) – (s)                    43.  (A) – (q)                  65. (A) – (r)
    (B) – (r)                         (B) – (p)                      (B) – (p), (s)
    (C) – (q)                         (C) – (r)                      (C) – (r)
    (D) – (p)                         (D) – (p)                      (D) – (q), (s)
22. (A) – (q)                    44.  (A) – (q)                  66. (A) – (p), (s)
    (B) – (s)                         (B) – (p)                      (B) – (p)
    (C) – (p)                         (C) – (r)                      (C) – (q), (r)
    (D) – (r)                         (D) – (p)                      (D) – (q)




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This Key & Hints paper contains 28 Pages                                     NFT_6 PAPER I & II DT : 16-01-11


                               HINTS AND SOLUTION
                                                            PAPER - I

                                               MATHEMATICS
1.     (A)
Since PQ is of fixed length                                                    R

          1
PQR       | PQ || RP |sin                                        P(4,4)
          2                                                      
                                             R(0, 7)
       This will be maximum,
       if sin  = 1 and RP is
       maximum.
                                                                           3x + 4y + 5 = 0
       Since line y = mx + 7                                      Q
       rotates about (0, 7).
       If PR ' is perpendicular        y = mx + 7
       to the line then RR ' is
       maximum value of PR.
                      40 4
        m               
                      47 3
2.     (C)
       The common roots must be roots of the equation 2x2 + (r – q) = 0. So their sum is
       0. Then the third root of the first equation must be 5 and that of second equation
       is 7.
3.     (C)
       Let, a  tan 2  and b  tan 2 
       We have to determine the minimum value of
             a  1          b  1
                       2                   2

                                              , where a, b  0 .
               b                   a
                        a  1           b  1
                                   2                   2
                                                              a 2 1 b2 1  a b
       We have,                                                  2  
                               b                  a           b b a a     b a
                                               1/ 4
                a 2 1 b2 1 
                                                                 1/ 2
                                                         a b
             4 . . .                                4 .           8
                b b a a                                b a
4.     (A)
       We have,    A  B   A  B  A
                               2       2                     3
                                                                  A2 B  AB2  B3
            A  B   A  B  0
               2           2


       As det (A - B)  0. det(A2 + B2) = 0.
5.     (D)
       In order to have an equilateral triangle, the sides must be face diagonals. There
       are three face diagonal coming out of each vertex. Any two of such three face

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       diagonals are the sides of an equilateral triangle. Hence each vertex is the vertex
       of three distinct equilateral triangles.
                                            8     1
       Thus the required probability =         
                                           C3 7
                                                     f ( x) f ( x)           f ( x) f ( x)
                                                                           a                   a
       (B) f ( x) f ( x)  f ( x)  1                             1     ( f ( x))4  1dx   dx
                                        4
6.
                                                    ( f ( x)) 4  1        0                    0

                                   a
             tan 1 (( f ( x)) 2        2a
                                   0

            tan 1 ( 3 )  tan 1 (1)  2a
             
               2a
            3 4
       
          a
       24
                                     3 1 5 1
7.     (a, b, c, d)        P( A  B)     
                                     8 2 8 4
                             A C  P( B)  P( A  B) 1
                             B 
                           P                        
                                        P( B)         2
                             A  P ( A)  P ( A  B ) 1
                           P C                     
                            B          P( B C )       4
                             A  P( A  B) 1
                           P              
                            B     P( B)       2
                             A  1  P( A  B) 3
                               C
                           P C  
                            B                   
                                    1  P( B)     4
                             B  P( B)  P( A  B) 2
                           P C                     
                            A         1  P( A)       5
                                                         
8.     (b, c)                    Obviously A  C and B  D will be non-zero collinear
       vectors
                                                                    
        ( A  C ) . ( B  D )  0 and similarly              ( A  B), (C  D) will be non-zero
       collinear vectors
                                   
            Hence       ( A  B)  (C  D)  0
9.     (A), (B), (C)
       We can write,
       m + (m + 1) + ……+ (m + k) = 1000
          k 1
              2m  k  1000
          2 
            2m  k  k 1  2000          (1)
       Since (2m + k) – (k + 1) = 2m – 1 = odd integer

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         One of two terms on the left hand side is even and other is odd. Moreover it is
        clear that
        2m + k > k + 1.
        Since 2000 = 24.53, its odd factor only 1, 5, 25 and 125. For odd k + 1 we need
        consider only 1, 5 and 25 and for odd (2m + k) we can have only 125. Hence,
        2m + k = 2000;                k+1=1                m = 1000, k = 0
        2m + k = 400; k + 1 = 5                     m = 198, k = 4
        2m + k = 80; k + 1 = 25                     m = 28, k = 24
         2m + k = 125; k + 1 = 16             m = 55, k = 15
10.      (a, b)
        Given circles touch externally at real axis. So, the                          Y
        centre C of the desired circles lies on real axis, which
        has radius r. thus CC1 = CC2 = 1 + r.
                                                                                          C1
              C1C3 = C2C3 = 1
                                                                                          C3
              CC3 = 2 – r                                                         C
                                                                                                            X
                                                                           (–1, 0)                 (3, 0)
       (1 + r)2 = 12 + (2 – r)2  r = 2/3
                                           2     1                                    C2
So, the centre of desired circle is at 1   
                                           3     3
           2 7
or t = 3   .
           3 3
So, required equation of circles are
              1 2            7 2
         z   and z  
              3 3            3 3
11.      (A)
                  1
         Let z   r
                  z
                           3
                      1       1         1       1        1
                   z    z  3  3  z    z  3  3 z   2  3r
                             3                   3
        Now
                      z      z          z      z         z
         r 3  3r  2  0
          r  2  r  1  0
                             2


        This implies r  2
12.     (D)
        Though | x  1| is non-differentiable at x = 1.
        | x  1| (x - 1) is differentiable at x = 1, for which LMVT is applicable.
13.     (A)
                      b                b
        Given that  | g  x  | dx   g  x  dx
                      a                a

         y = g(x) cuts the x-axis at least once, then f(x)g(x) changes the sign at least
                                  b
        once in (a, b), hence  f  x  g  x  dx can be equal to zero.
                                  a


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14.    (D)
       If f '  x  exists, then f '  x  may or may not be continuous.
       For example,
                   2     1
                   x sin   ;       x0
       f x            x
                  0,                 x0
                  
       has a derivative at x = 0. but f '  x  is discontinuous at x = 0.
                             1
15.    (C)      x i 1  xi 
                           ln 2
                        y           1                  dy
                                   ln 2    ln 2dx   y
            i.e.   x  1   x  
                           
                        y      
              ln y   x ln 2  c it passes through (0, 2)
            c  ln 2
            y  21 x
                                x y
               xi 1              y
16.    (C)            2              2
                xi                x
                        y
            y  
                        x
            xy  c

                 1    1                    1        1 1
17.    (D)            c                          
                xi 1 xi                   y        x c
                                        x
                                           y
                                                        c
              x 2 y 1  y ( x  c)    ln y  ln x       ln c1
                                                        x
                               c
                           
              y  c1 xe       x




18.    (C)
       Equation of tangent at point P(t) on the parabola y2 = 8x is
           ty = x + 2t2               (1)
       Equation of chord of contact of the circle x2 + y2 = 4 is,
           x + y = 4                (2)
       Q(, ) lies on (1)
       Hence, t =  + 2t2            (3)
                          
            x  y   2t   4  0
                   t       
                              y
        2  yt  2     x    0
                              t
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                                            y         2
          For point of concurrency, x =      and y 
                                            t         t
           locus is y2 + 2x = 0

  19.     (D)
          Required point will lies on the director circle of the given circle as well as on
          directrix of parabola.
           x1  y1  8 and x1 + 2 = 0
                 2     2
                                                   point =  2,  2
  20.     (A)
          Let circum centre be (h, k)
           h = /2; k = /2   = 2h and  = 2k
          Also t =  + 2t2 or
                 2 + 8 = 0
          So therefore locus is, x – 2y + 4 = 0.
  21.     (D)
              a
                b
            c
                                            3
                                 
                              1/3

               3                                 4
  22.     (C)
          If a  1 , then       1
                        
                                       2 3 1/6
                   2 2 3 3 3     
                                         2 3 
                           6             2 .3 
            2  3  
                         max     1
                               432
  23.     (A)
                  27 is essentially true.
                       3



                  27
                     3
                                  is possible only
                  27
                       3


                                  1
          Whence,      
                                  4
                                                  3 3     9
          a  b              =    
                                                  4 16   16

                                        PHYSICS
24.     (A)
                       4s         28R 3Po
               2R
         Wgas    Po   4r dr 
                              2
                                              24Rs
                R 
                         r            3
25.     (D)
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                                                                v

                                                3
                                                R          ER
                                            ’
                                            v
       COAM  mvR = mv3R                        …(i)
                  1        GMm 1           GMm
       COE         mv2          mv '2        …(ii)
                  2          R     2        3R
                  3 GM
        v
                  2 R
26.    (C)
       Let total mass is 3m, from impulse momentum principle, we get
       p  p  3mg  m 4i  16j  2m ˆ  ˆ  3m v ˆ  v ˆ  30mj
        t 3     t 2
                                 ˆ  ˆ      i j        i     j   ˆ
                                                                    2x      2y   
              v 2x  2 m / s and v 2 y  4 m / s
       Now u x  v 2x  2 m / s and v 2y  u y  g  2  u y  24 m / s
               ˆ
        v  2i  24ˆ  j
27.    (A)
       For standard clock
                                 2                                           2
        of minute hand =            ;               of hour hand =
                               60min                                     12  60 min
               2   2  rad 2  11  rad
       rel                     
               60 12  60  min 60  12  min
       Relative angular displacement = 2
                                                           2         60 12
        Time after which they come together =                  min         min  65.45min
                                                         211          11
                                                         60 12
        Watch will gain time by 0.45 min  27 s
28.        (B)
       A = A0 e–t and I = i0e–t/
                A A0et A0 t   1  
       Given,                   e
                 i i0e t/    i0
                                    i
       for time independent value of .
                                    A
             1
          0
             
               1
       =
               
                 1
       Tmean = =  = 0.4 s
                 




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29.    (C)                                                                    P
       Net force on the system is zero hence
displacement of the centre of mass of the system will                lo               lo
be zero. Let x be the shift in P, then                                             3
          l                       l                 
        m  o 1  T   x   m  o 1  3T   x                 P
          2                       2                 
                 l T                                                     x
        x= o
                    2
        (C)
30.    (B), (C), (D)
       Since water falls at maximum distance from wall so the hole should be made at h.
       For toppling, av 2  h  r 2 2hg  r
        a2gh  h  r 2 2hgr
                r 3
        h
                 a
        (B), (C), (D)
31.    (A), (C)
       Let Q be the charge on the sphere at t = t and Ro be the initial separation between centre of
       the sphere and charge q. So at t = t, potential of sphere
                    Q           q
            V                            0
                4 o R 4 o  R o  vt 
                    qR
        Q
                  R o  vt 
            dQ          Rqv
              i
                     R o  vt 
                                 2
            dt
        (A), (C)
32.    (A), (B), (D)
                            CE
       Initially, Q A  Q B   , when dielectric is inserted.
                             2
                     2CE              2CE CE CE
       Q 'A  Q 'B      ,  Q =                 
                      3                 3      2       6
                                             2
                               CE        CE
       Work done by battery =      E 
                                 6         6
                           2     2       2
                         Q'    Q     7CE
       U A  U f  U i  A  A 
                          2C 2C       72
33.    (A), (B), (C)
       For isothermal process Q = W.
       Curve 3 corresponds to isothermal process, and Curves 1 and 2 correspond to isobaric
       processes.
       For isobaric process
            W = pV = RT (n = 1)

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                          i2                                            W      2
            Q = CpT =         RT    ( i = degrees of freedom)            
                           2                                             Q i2
                        2     W 2
       For curve 1,                        i = 5 (diatomic gas) (given tan1 = 2/7)
                       i2 Q 7
                          2    W 2
       For curve 2,                        i = 3 (monatomic gas) (given tan 2 = 2/5)
                        i2 Q 5
34.    (A)
          pm  B, pm  iA
       Since statement-2 is true.  with deformation area reduces, so does the p m and  .
        (A)
35.    (C)
       Conceptual.
36.    (D)
                       Ring


                     
         ma
                  mg
       Using work energy theorem.
           Wpeudo force + Wgravity = KEf – KEi
       mg
          Lsin  mgL 1  cos   0 (1)
        3
           3 1  cos    sin 
       Solving we get,  = 60o.
        (D)
37.    (D)
                       q  1 1  q  r2  r1 
           V1  V2         
                     4o  r1 r2  4o r1r2
                                qr
        V  k  say  
                          4o r1  r1  r 
      Obviously r is not constant. So equipotential surfaces differing by K volt (say) are not
      equi-spaced. So statement – 1 is wrong but statement – 2 is correct.
38.   (D)
       dy                          dy
           tan 60o  3, v p   v     20 3   v 3
       dx                          dx
       v = 20 cm/s
       (D)
39.   (D)
      From graph, A = 4  10-2 m,  = (5.5 - 1.5) = 4  10-2 m
       K = 2/ = 50,  = Kv = 50  (1/5) = 10
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       Hence equation of wave is y = (4  10-2)sin(10t + 50x + )
       At x = 0, y = 2 2 102        2 2 102  4 102 sin 
                     1         3
        sin          ,  ,
                      2       4 4
       As the particle is moving up at t = 0, x = 0, hence  = /4
        (D)
40.    (C)
       Conceptual.
41.    (B)
            r  dB                                                                             FE sin
       E                                                                        FE
            2  dt
       FE  qE                                           A                                              B
                                                                                  FE cos
                     r              q  L qB L
       FE cos   q Bo cos    Bo   o                             L/2
                    2              2 2       4                                  
                                                                                            r
           F cos  qBo L
       a E                                                                O
              m          m4
       Hence velocity at end A
                         qB L           qBo
            vA  2  o  L  L
                          m4             2m
       Time taken to reach at end A
                  2s      2  L  4m       2m
            t                        2
                  a           qBo L        qBo
42.    (D)
                                         qB L                   N
        N  Fm  FE cos 45o  qv A B  o                                    L/2
                                           4
                qBo             2m qBo L 9                      A                                                B
        qL            Bo 2               qBo L                                          L/2
                 2m            qBo      4    4                      Fm
                                                          FE                             o
                                                                                       45
                                                               FEcos45o                         O


43.    (A)
                                             qBo
                                     mL
                             mv              2m  L
       Radius of curvature =    
                             qB               2m  4
                                  q  Bo  2
                                              qBo
44.    (C)
          mvsin 
                   vsin   1.2  107 m / s (Charge of -particle = 3.2  10 C)
                                                                             -19
       R
             qB
          2m                    PqB
       P      vcos   vcos          9  106 m / s
           qB                    2m
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        V        vsin 2   vcos 2  1.5  107 m / s
45.    (B)
       m y v y  m  v   v y  2.715  105 m / s
        TE released during an -decay of the nucleus X is,
                             1       1
             E  KE y  KE  my v2  m v  4.7 MeV
                                  y
                                          2

                             2       2
46.    (D)
                                                E
       Mass lost during -decay, m                0.005U
                                               931
        mass of nucleus X,
             m x  m y  m  m  225.038U
        mass defect in nucleus X,
                                                    
          md  92mp   225  92 mn   mx U  1.895U
                                     
                                          m  931
        BE per nucleon in nucleus X = d           7.84MeV
                                            225



                                          CHEMISTRY
47.    (A)
        Tb  i  m  K b
       2.08 = i  1  0.52
       i = 4 (so number of particles are four)
       Hence complex will be [Fe(CN)6]3
       EAN of Fe = 26 – 3 + 12 = 35
48.    (D)
       Oxidation number of M = (2/x)
       Let y atoms be in +2
              2   y   3100  y   2
       So
                          100               x
       2y + 300 – 3y = 200/x
       y + 300 = 200/x
       300 – (200/x) = y
                 3x  2 
        y  100           
                 x 
                                                                1 x 
       (100 – y) = 100 – 300 + (200/x) = -200 + (200/x) = 200        
                                                                 x 
           100  y         1 x        3x  2  2 1  x 
       so            200        100         
              y            x           x   3x  2 
49.    (D)
       (A)  stability increases with size
       (B)  stability increases with increase in ionic character
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       (C)  solubility decreases with increase in size
       (D)  solubility increases due to less lattice energy with increase in size of cation.
50.    (A)
       Flourine forms H2F2
             H 2 F2  OH   HF2 (acidic salt)
51.    (A)
       OMe is e donating.
        NO 2 is e withdrawing (having major role at ortho and para position as compared to
       meta)
52.    (C)
       IV forms aromatic ion
       III form antiaromatic ion
       II’s cation has less strain than I’s cation
53.    (A), (B), (C), (D)
       Theoretical
54.    (A), (B), (C), (D)
       Due to inert pair effect down the group stability of lower oxidation state increases and
       stability of higher oxidation state decreases.
55.    (B), (C), (D)
                           dU an 2
       (A) For real gas              i.e. internal energy is volume dependent also.
                           dV V 2
       (B) Z > 1  Vreal > Videal so forces of repulsion are dominant.
       (D) For ideal gas U = 0 but for real gas U  0 .
56.    (A), (C), (D)
                                                 O
                                                                                OD
                                                        
                                               
       Me  CH 2  CHO  Me  CD2  C  H  Me
                                     OD / D2 O               CN / D2 O
                                                                          CD2   C CN
                                                                                H
57.    (A)
       For hydrogen low molecular density causes repulsion at all pressure.
58.    (C)
          SO3 is S3O9
               O           O
                       S
                   O       O
           O                 O
                   S       S
                       O
               O            O
59.    (B)
       Both statements are independently true.
60.    (A)
       Mg3C2 + H2O  Mg(OH)2 + C3H4
       CaC2 + H2O  Ca(OH)2 + C2H2
       Be2C + H2O  Be(OH)2 + CH4
61.    (C)
       Al2O3 + 3C + N2  2AlN + 3CO
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62.    (D)
       Theoretical.
63.    (B)
       Cryolite is Na3AlF6
64.    (C)
       Half cell will be
       2Ag s   2Cl  aq   2AgCl s   2e
       Hg2Cl2 s   2e  2Hg l   2Cl  aq 
       Both are metal insoluble salt.
65.    C
       E o  E Oxd  E o
         cell
               o
                       Red

                               0.0591
       Eo  Eo  
        Oxd  Ag/Ag
                                      log K sp AgCl  s 
                                  1
                                 0.0591
       Eo
        Red    E o 2 /Hgl  
                  Hg 2
                                        log K sp Hg 2Cl2  s 
                                    2

                              0.0591            0.0591
       Eo  Eoxd  Eo  
        cell
             o
                    Red              log1010         log1018
                                 1                 2
                = 0.0591  (9) + 10  0.0591 = 0.0591

66.    (A)
       G  nFE  2  96500   0.0455  8780J / mol .


                               HINTS AND SOLUTION
                                                   PAPER - II


                                         MATHEMATICS
1.     (D)
                          sin x 
                               2     n

       f  x   lim       n
                   3
                n 

                       cos x 
                               2   n

                   4
                    
       If x  n  , then f(x) does not exist and so it is discontinuous.
                    6
                             n
                         3
                          
                      4                1                                 0
       f    lim                 lim           1;        f  0   lim         0
          3  n   3   1 
                          n     n               n                               n
                                    n 
                                            1                        n 
                                                                            3
                                    1                                 1
                    4 4                 3                             4




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                      1
       f    lim          
          2  n   3 
                          n

                      0
                    4
2.     (C)
           RBS  RAS  90o
       and ASB  ARB  90o

       So equation of circle with RS as a diameter is same circle with AB as diameter.
        Equation of required circle


                                            S               B(1, 1)


                                                                         y = 3x + 4
                                                                P


                                           A              R
                                                Q                   A = (1/3, 5)

                                                      y = 3x  6

            x  1  x     y  1 y  5   0
                         1
                          
                        3
           3x 2  3y 2  4x  18y  16  0
3.    (A)
      Equation of the required plane is
      1(x - 1) + 2(y - 2) + 3(z - 3) = 0
       x + 2y + 3z = 14
4.    (B)
      We can assume that, circumcentre of ABC is at                                  A
      origin. If R is the circumradius,
      BC = 2R sin A = R.
      Also, if z1, z2, z3 are the complex number
      representing A, B, C respectively, then                                         H
      z1 + z2 + z3 represents H and M is
       z 2  z3                                                             B
                                                                                          M
                                                                                                C
                . If z represents T, then
           2
       z  z1  z 2  z3 z 2  z3
                                       z  z1                                      T
                2               2
      AT = 2|z1| = 2R = 2BC
6.   (C) B  ( I  A) ( I  A) 1
       B T  ( I  AT ) 1 ( I  AT )  ( I  A) 1 ( I  A),
       BB T  ( I  A) ( I  A) 1 ( I  A) 1 ( I  A)
              ( I  A) ( I  A) 1 ( I  A) 1 ( I  A)  I

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       (As ( I  A).(I  A)  ( I  A)(I  A))
7.     (B)
                          2 tan x
           1  tan 2 x 
                          tan 2x
                                         2k  
                                  2 tan  n        
                     2 2              2 1 
                           k
        1  tan  n           
                       2  1  tan  2  
                                           k 1

                                        n       
                                        2 1 
       It follows the product to
                       2               2 
             2n tan n           2n.tan  n       
                     2 1              2  1   2n
                  2n 1                 2 
             tan  n        tan  2  2n  1 
                                                  
                  2 1 
8.     (C)
       10999 = 2999 .5999
       So the number of positive divisor of 10999 is
            = (999 + 1)(999 + 1) = 10002                       (1)
       Similarly, the number of positive divisor of
            10998 is = 9992 .
       So the number of positive number which divide 10999 but not 10998 is equal to
            = 10002 – 9992 = (1000 + 999)(1000 – 999) = 1999
9.     (B)
       (1 + x)n = n Co  n C1x  n C 2 x 2  ......  nC n x n
       On integrating,
       1  x 
                  n 1
                         1      x 2 n x3
                                     C2  ......
                               n Co x  n C1
          n 1                   2         3
       Multiplying with x and differentiating,
       1  x            x  n  11  x   1
                  n 1                          n
                                                                        x2           x3
                                                   2. n Co x  3. n C1     4. n C2     ......
                            n  1                                     2            3
                                                                                   2n  n  3   1
       Put x = 1  2                  Co    C1    C2    C3   ....... =
                                   n         3 n      4 n      5 n
                                             2        3        4                        n 1
10.    (C)
       Period of sin 1  sin x  is 2 and that of tan 1  tan x  is . Hence required period is 2.
                   12                x 6
       2tanx + x =       tan x =  
                    5                 2 5
       From the graph only two solutions are possible.




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            O               /2                   3/2       2     (12/5, 0)




11.    (D)
       Statement-1 is false as consider the function f  x   max 0, x 3 which is equivalent to
               0;       x0
       f x   3
               x ;      x0
       Here, f(x) is continuous and differentiable at x = 0.
       However, statement-2 is obviously true.
12.    (A)
       R1: Point P(x, y) is nearer to (1, 0) than to x  1.
              x  1        y 2  | x  1|
                        2
       
        y 2  4x
        point P lies inside parabola y 2  4x .
       R2: Point P(x, y) is nearer to (0, 0) than to (8, 0).
        | x |  | x  8 |  x < 4.
        Point P is towards left side of line x = 4.
       The area of common region of R1 and R2 in first quadrant is the area bounded by x = 4 and
       y2 = 4x, and x  0 , y  0 .
                                x=4 2                                      x = 2y
                                       y = 4x
                                                                                  y=4




       This area is equal to the area bounded by x  2 y and y = 4. Now the area bounded by
       x2 y
                                                          4
                              x2 
                               4            x3 
       and y = 4 is A    4  dx  16    0  16   64   16  
                                                        1            16 32
                        0
                               4            12        12             3  3
13.    (A)
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       Image of (4, 3) in the line x + y = 3
       is
                                                                            (1, 2)       (0, 1)
       (0, 1).
       Now points (0, 1), (1, 2) and (6, y)                                   
       are collinear.                                          (6, y)           (4, 3)
        y = 17.



                          x
                  1
14-16. f  x   1  
                  x
                             1
       f(x) is defined if 1    0  x   , 1   0, 
                             x
                                                  
                         1   1         x x 1 
                               x

       Now, f '  x   1   ln 1         2 
                         x    x  1 1 x 
                                            x     
               1   1            1 
                        x

             1   ln 1          
               x    x  1 x 
                           1     1
       Let     g(x) = ln  1   
                           x  1 x
                       1      1   1          1
        g ' x            2                              (1)
                         1 x  x  1 2
                                          x  x  1
                                                     2
                    1
                         x
       For x > 0, g'  x   0
        g(x) is monotonically decreasing for x   0,  .
        g(x) > lim g  x   g(x) > 0
                   x 

       And since g(x) > 0  f '  x   0
       Now, for x   , 1 , g'  x   0
        g(x) is monotonically increasing for x   , 1 .
        g  x   lim g  x   g  x   0  f '  x   0
                   x 
       Hence, f(x) is monotonically increasing in its domain.
                              x
                      1
       Also,    lim 1    e
               x 
                      x
                    x                           x
              1                       1
       lim 1    1 and lim 1    
       x 0 
           
               x                x 1  x
        Ranger of f(x) is 1,  ~ e


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                                             y = f(x)

                                                         y=e

               x = 1
                                     O

                        /2
17-19. f  x     sin x cos t f  t  dt  sin x
                        0
                                      /2
          f  x    sin x  cos t f  t  dt  sin x
                                         0
                                                                                                   /2
          f  x   Asin x  sin x  f  x    A 1 sin x where A    cos tf  t  dt
                                                                                                   0
                               /2
                                                             A  1    /2
                                                                                        A 1 
          A    cos t  A  1 sin tdt                                sin 2tdt        
                               0
                                                                2        0               2 
                                           2 
          A                     f x        sin x
                    2                     2 
               2 
                      sin x  2  sin x  2  
               2 
              | 2   | 1  1    3
                /2                              /2
                                                        2 
                     f  x  dx  3              2    sin xdx  3
                0                                0        
                2                 4
                   3  
              2                 3
20.      (A) – (p) ; (B) – (r) ; (C) – (q), (s), (t) ; (D) – (p), (r), (s).
         (A) Lines must be non-parallel and coplanar
                                     
              b  d  0 and [a  c , b , d ]  0
         (B) Lines must be coincident
                               
             b  d  0 and (a  c )  b  0
         (C) Either lines are skew or parallel
                                                     
             [ a  c , b , d ]  0 or (a  c )  b  0 and b  d  0 ,
             (and if skew then lies in set of parallel planes)
         (D) Either intersecting or parallel
                                           
             [a  c , b , d ]  0, b  d  0 or b  d  0
21.      (A) – (s), (B) – (r), (C) – (q), (D) – (p)

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                                                   x3 x5         
                                               x    .......  x
                            sin x  x
                                       lim                      
                                                     3! 5!
       (A)   f  0   lim                                  2
                                                                     0
                       x 0 x sin x     x 0              x
                   1                                 1 
                                                              1
                           1        1                           dx   1
       (B)   lim                      ......                 ln  3
             n  n
                         n2 n4                  n  2n  0 1  2x 2
       (C)    = 0  k  1
       (D)    f  x max  6



                                                      y=x-1

                 (0, 0)



22.    (A) – (q), (B) – (s), (C) – (p), (D) – (r)
       (A) xy = c2 = ve ( c is purely imaginary number)
        If x is positive, then y must be negative real.
            1 3
       
               2
       (B) 2b = a + c
       Now, (a + c - b)s = kb2
             b a  b  c           3b 2                3
                           kb 2         kb 2  k 
                   2                   2                 2
       (C) r     3  0   4  0       23
                          2            2




                               7  4cos   0                                      
                                                                                          2
                                                       2 30  4sin   0                      185  56cos   16 30 sin 
                                                  2
       (D) distance =

                                                        56                 
                                                                                  2
        maximum distance = 185                                     16 30            17
                                                                2




                                                  PHYSICS
23.    (D)
       Charge is proportional to lines of force.
24.    (C)
       Event 1                                                       Event 2
        v  v1  2v 2                                                      v
                                                                     2v  4  2v1  4v 2
        v  v 2  v1                                                       2

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                     2                               v
             v2      v                        v       v2  v1
                     3                               2
                        2   4                          5
         Fdt  2  v  v                       v2  v
                        3   3                          6
                                                               5  1  4
                                                  Fdt  4  v  v   v
                                                               6  2  3
       Impulse imparted is same in both events so area in F-t should be same. So (C) may be a
       possible solution.
25.    (B)
       When displaced by (dx) towards right, then
                   K
                              dx cos 
                           
                                   dx
       Elongation of left spring = dx cos 
                 (kdxcos)cos          kdx

        Restoring force =  Kdx  Kdx cos2   K 1  cos2  dx
                                                  
                                 K
        Acceleration =  1  cos 2  dx
                                 m              
                  K
        =           1  cos 2    2 rad / s
                  m               
26.    (C)
       Fmax for no sliding between cone and plank is
                                1
              (m + M)g =  2 10 = 5 N
                                4
       For toppling of cone
                    N
               ma
                             H/4
                        r          fr
                            mg

                T           T
            mF      H                           r
       i.e.         mgr  F  4  M  m  g      F4
           Mm 4                                H
        Toppling will occur before sliding.
        Fmax for equilibrium = 4 N
27.    (D)
                                            P  0.99  50 kV  20 mA 
       Average rate of rise of temperature                                2o C / sec
                                           t ms 1kg   495J kg C 
                                                                   1o 1




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                                                              hc 12400
       The minimum wavelength of the X-rays emitted  min               0.248 Å
                                                              eV 50 103
28.    (D)
             D           2 D         dy2 2
              and y 2                     v
              d            d            dt   d
29.    (A)
        2       1
            1  2
        v 2R      R
       v
       After reflection from the spherical surface the ray become parallel to optic axis. Hence
       after reflection from silvered surface it will retrace its initial path and image coincides
       with object.
30.    (C)
                                                           R      17R
        P  V 2 so PV2 = constant,  C  CV                   
                                                          1 x     6
                                            17R                   17      17
        Heat supplied Q = nCT = n                4To  To   nRTo  Po Vo
                                              6                    2       2
31.    (D)
       V = i(G + H)  I = V/(G + H)  50  10-6 = 50/(100 + R)  R = 106 
        (A) is correct.
             i G
       S  G  0.5A  (B) is correct.
            i  ig
       (C) is also correct, the incorrect option is (D).
32.    (C)
       Energy of individual m is not maximum but the total energy associated with m is
       maximum.
33.   (A)
34.    (D)
      Only those images appear in the photograph which are focused on to the film by the lens.
      The lens does not form an image of the scratches on itself, so they do not appear in the
      photograph. So statement - 1 is wrong.
      However, they spoil the sharpness of the photograph by scattering some of the light
      passing through the lens. So statement - 2 is correct.
35.    (D)
       In resonance condition when the energy across the capacitor is maximum, energy stored in
       the inductor is zero, vice versa is also true. Hence statement-1 is false.
36.    (A)
       Conceptual.
37.    (C)




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                                            mg


                                                                      v = vo - r

                                                         
                                                                          vo
             t=0                                 kmg
                                                         t when rolling starts
                                        
       Impulse momentum principle kmgt = m(vo  r)  kgt = vo r                       (1)
                                                            2 2               2
       Angular impulse momentum principle kmgrt = mr    k gt  r (2)
                                                            5                 5
                                       2v o         2v
       From equation (1) and (2), t         , v  o due East
                                      7 k g         7
38.    (D)
       In ground frame: work done by friction = change in KE
             1        12       
        WG   Mv2   Mr 2  2   0                  (3)
             2        25       
                                        2v o         2v                                   1
       From equation (1) and (2), t           , v  o , substituting in (3), we get WG  Mvo  2

                                       7 k g          7                                  7
       In    Belt    frame:     Work        done        by    friction   =      change   in    KE:
             12                   1             1
        WB   Mr 2  Mr 2  2  Mv o   Mv o
                                         2              2

             25                   2             7
39.    (C)
            ˆ            ˆ
       a  3i  4ˆ  a z k
                 j
       The acceleration of the block in the x-y plane for the block not to slip in that plane is
         32  42  5m/ s2
        Friction required = 5  1 = 0.9  1 [g + az]
                         5         5  10
        10  a z           az          10  4.44 m / s2
                        0.9          9
40.    (A)
       N z  m  g  a z   15 N
        fs  max  N z  0.6  1`5  9 N ,
                                      which is greater than force required to provide the same
       acceleration as that of the box along the x-y plane.
        Friction = 1  5 = 5 N
        R  N2   friction   152  52  250
                                    2
              z




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41.      (C)                                                    z
         With respect to the box
                                                                                        y
             Max = 3 N, may = 4 N
          Net pseudo force in x-y plane =
           32  42  5N
         Friction will try to oppose this force, but
         (fs)max = 0.2  1  (10 + 5) = 3 N                                 max
               In X-Y plane                                                                 mg
                                                                                  may maz
              Fnet = 5 – 3 = 2 N                                                                           x
                                                                     g =10m/s2
          This force will be acting at an angle 53o in
      anticlockwise direction from the (-)ve x-axis.                       F.B.D. of the block

                                                          ˆ      ˆ
          a block w.r.t.box  2cos53o ˆ  2sin 53o ˆ  1.2i  1.6j
                                        i             j


42.     (A) – (p), (q), (r), (t), (B) – (p), (q), (r), (t), (C) – (p), (s), (D) – (p), (s)
        Conceptual
43.     (A) – (q), (B) – (p), (C) – (r), (D) – (p)
        Use concept of field of view.
44.      (A) – (q), (B) – (p), (C) – (r), (D) – (p)
         (A) In case of resistance (r) acceleration of rod, mg sin   iBl  ma
                               Bl
          a  g sin   i
                               m
                    Blv                        B2 l 2                              g sin 
         Also i            a  g sin               v  when a = 0; v                     = constant.
                     R                         mR                                  B2 l 2 
                                                                                            
                                                                                     mR 
                                                                    dq          dv
         (B) In case of capacitance C: q  C  Blv   i                 BCl        BlC  a
                                                                     dt         dt
                                                                                           g sin 
          Again from Newton's law: mg sin    BlCa  Bl  ma  a                                  = constant
                                                                                         B2l 2C 
                                                                                        1  m 
                                                                                                   
                                                        di      dx
         (C) In case of inductance only (L): L  Bl                     Li = Blx , i  (Bl/L)x
                                                        dt       dt
                 Again mg sin  - ilB = ma                                                   a = (g sin )
           B l 
               2 2
                                   B l 
                                      2 2
                  x  a    mL  x   g sin  
            mL                         
          equation of SHM.
         (D) As induced emf in the rod is equal to external cell, so the current in the circuit is zero
         and hence it acceleration with g sin .




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                                            CHEMISTRY
               –
45.    (A) I is preferentially oxidized to I2 farming a violet color layer.
                                           1
46.    (A) Moles of complex =
                                         269.5
                           –                  1
            Moles of Cl formed = 2               = 0.0074
                                            269.5
       So, moles of HCl in 1 liter solution = 0.0074
       If V is volume in litre of Ca(OH)2 then V  0.1  2 = 0.0074
       V=0.037 Litre.
47.   (A)       NH 4Cl   NH 4  Cl  130 ----------- (1)
                                



         NaOH   Na    HO   220            ----------- (2)
         NaCl   Na   Cl   100             ----------- (3)
        [(1) + (2)] – [(3)]
        NH 4   HO   250    NH 4OH
                                  10   1
       ' 'of NH 4OH                 
                                 250 25
                        1  1
        HO    C 
                          4 104
                       100 25
        pH  10  2log 2  10.6
48.    (C) (A) In NOCl it is NO+ hence B.O. is 3
       (B) NCl3 hydrolysis to give NH3.
       (C) In PF3Cl2 two chlorine & one Fluorine are equatorial & two fluorine are axial
         (D)
                       O       O

                       S       S     O
                   O
                       OH      OH
49.   (C) Intermediate formed clearing nucleophilic substitution is more stable because of
      having cyclopentadienyl anion which is aromatic
               F
                               OMe

                               F

                       F



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                                  
50.    (A)   CRT so, C=               4 104 M
                                  RT
                           w×1000
       So, Mol . wt=                     5 × 104 gm/mole
                       molarity × volume
       Mol wt of glucose = 180, on forming starch 18 gm of H2O is lost so average wt = 162
                5 104
       So,             310 units.
                 162
51.    (B) – OMe (– I) (+ R), Dominant effects (+R)
             – CH3 (+ I) (Hyper congregation)
             - Cl, NO2, (– I) strong.
52.    (D) HCl is halides
53.    (B) Only B is unsymmetrical
54.    (B) In water extent of Hydrogen bonding is more Hence boiling point is
             more but strength of H-bonding is weak in H2O then HF.
                         R
55.    (B) Cp – Cv =              M=Molar mass
                         M
                                R
             0.125 – 0.075 =               M=40
                                M
56.    (B) Energy increases as the orbit numbers increases
57.    (B) conceptual
58.    (B) Meq of H3PO4 = 300  0.1 = 30
             Meq of NaOH = 50  0.1 = 5
             H3 PO4 + NaOH = NaH2 PO4 + H2O
               30        5             0      0
               25        0             5      5
             Solution is buffer
                                5
             pH = P ka1 + log       3 + log 1/5
                                25
                                   2.3010
59.    (A) At point ‘C’ a buffer of H2 PO  & HPO4  exists.
                                          4
                                                 2



60.    (A) At Point ‘D’ only Amphiprotic salt Na2HPO4 remains.
                      P ka2  P ka3   8  13
             So, pH                         10.5
                            2           2
61.    (B)


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                  Et                                Et
                           O                                                      O                  O
                                                             O
                                   EtO                                    H            OMe or H  C  OEt
                                                         (-)



                  Et
                                    H                                Et
                           O
                                                                              O
                                    C       OMe

                                    -O                                        C       H
                                                                              O

62.    (A)
                  Et                            Et                                             Et
                           O                              O                                              O
                               R3 N                                                       Me
                                                         (–)                                                    Me
                               H                                 H                    O
                       O                                 O                                      CHO O

63.    (D) Loss of H2O is by E-1 CB mechanism



                                                                                                     O                          OH
64.    (A)                     Br    
                                        Mg/Ether
                                                                          MgBr                               
                                                                                                         
                                                                                                             LiAlH4 /H 2 O

                       O                                             O
             MeO                                         MeO




                                                                 OMgBr                              OH
       (B)                     MgBr
                       O
             Me

                                    Me
       (C)                     Br        
                                          Mg/ether
                                                                                     MgBr                               
                                                                                                                          LiAlH4
                                                                                                                                
                       O                                                          O
             MeO                                                 MeO                                     O                      OH



                               Br         ether 
                                         
                                           Mg
                                                                              MgBr
       (D)                 O                                              O
             Me                                          Me
                                                                                                         HO

65.    (A) – (r)
       (B) – (p), (s)

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       (C) – (r)
       (D) – (q), (s)
66.    (A) – (p), (s)
       (B) – (p)
       (C) – (q), (r)

       (D) – (q)




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