NFT_9 _20.03.11_ Paper-I_II_H_Sol by nehalwan

VIEWS: 2 PAGES: 24

More Info
									                     NARAYANA
                     I   I   T        A      C   A   D      E      M        Y


                   ALL INDIA TEST SERIES
             NFT – 9_PAPER – I & II (20.03.2011)

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



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




          2

   Narayana IIT Academy, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 • Ph.: (011) 42707070, Fax : (011) 41828319
AITS –NFT-9 SOLUTION                                                                                NARAYANA
                                                                                                    GROUP OF EDUCATIONAL INSTITUTIONS




                                   HINTS AND SOLUTION
                                                        PAPER – I


                                             MATHEMATICS
1.         A all the couples sit together
           B  9 couples sit together
                       p(A  B) (10  1)! 210 / (19)! 1
           p(A/B) =                                     
                          p(B)      (11  1)! 29 / (19)! 5
                  1      1 
                 x 4  x 7  dx
2.         I=               
                        4     2
                  6 3  6
                       x     x
                      4 2
           Put 6 – 3  6  t
                    x x
                     1 dt       6x 6  4x 3  2
                    12  t
           I=                                 +C
                                     6x 3
3.         Let ln x – 3 = t
                         tn    0         
            lim                form  = –1
                                         
                            m
               t  0 ln(cos t) 0


                      nt n 1
            lim                = –1
               t 0  m tan t

                                 n
            n – 1 = 1 &   1  n = m = 2.
                                m
                                             t2 t 
4.         Normal to the parabola at point  ,  will be
                                             16 8 
                     1        1
           y + tx = t  t 3 , if it is common normal to circle, then
                     8 16
           33 1          1
              t  t  t 3  t3 = 64 t
            8      8 16
            t = 0,  8
                                     65  1
           So shortest distance is
                                       8




                                                                                                                          3

     Narayana IIT Academy, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 • Ph.: (011) 42707070, Fax : (011) 41828319
     NARAYANA
     GROUP OF EDUCATIONAL INSTITUTIONS
                                                                                                       AITS –NFT-9 SOLUTION



5.              =  – ,  = 2 + ,  = 3 – 
               given expression is equal to
                      3                          
               sin     2sin      3sin     4sin
                      2 2                2       2 2  2                                                       
                                                  
               = – cos  2sin  3cos  4sin
                            2        2       2        2
                                 
              = 2 sin  cos   2 1  sin   5
                        2         2
                         x
6.            Let y = , z  x.r
                          r
                                          1
              x + y + z = xyz  x[r +  1 ] = x3
                                          r
                              1
               x2 = r +  1  x2  3
                               r
                                            38
               x4 + x2 + 7  19  4                2
                                         x  x2  4
                            sin x
7.           Let f(x) =
                               2
             f(x) is increasing
                          x
                   1
                                   2
             and  sin dx 
                   0
                           2       
                        2
             So L <  M
                        

                                            
                       sin 2x       0x
                                             2
                      
                       0            
                                       x
                                    2
8.            f(x) = 
                       sin 2x   x  3
                                            2
                                  3
                       0               x  2
                                   2
              So f(x) is periodic with fundamental period 2 and range [–1, 1].
9.            Put x = 1, 2, 3, 4 in f(x), f (x) , f(x) f(x) respectively
              We get,
              f (2) + f (3) + f (4) = 0
              12 f(1) + 3f (2) + f (3) = 0
              18 f(1) + 2f (2) – f (3) = 0
              6f(1) – f  (4) = 0
              We get f(1) = f (2) = f (3) = f(4) = 0
              So f(x) is constant.

             4

      Narayana IIT Academy, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 • Ph.: (011) 42707070, Fax : (011) 41828319
AITS –NFT-9 SOLUTION                                                                                 NARAYANA
                                                                                                     GROUP OF EDUCATIONAL INSTITUTIONS




10.        Slope of OC > slope of OB                                                                   y = f-1(x)
              x      f 1 (x)                                                                  B(x, f-1(x)            y=x
                  
            f (x)       x                                                                                           y = f(x)
                  –1
           f(x) f (x) < x2 x (0, 1).                                          C(f(x), x)
                                                                                                            A(x, f(x)

11.         Let sin x = t then
            f(t) = t3 + 3t + 6
            f (t) = 3t2 + 3 = 0
            f(t) is increasing and f(–1) > 0
12.         Statement 1 is wrong and Statement 2 is correct.
13.         xn+2 = 2xn+1 + 3xn
                 x                3
             n 2  2 
                 x n 1         x n 1
                                 xn
                           x
            If L = lim n  2
                      n  x
                             n 1

                         3
            L=2+            L = –1, 3
                         L
            L = 3.

14.         Statement 2 is not true.
15.         Let x2 = t
            f(t) = t2 + (1 – 2a) t + a2 – 1 ... (1)
            (*) will have exactly two real roots
            If either (1) has two equal positive roots or one positive, one negative roots.
                                     5
            for this a (–1, 1)    .
                                     4
16.         (*) has exactly 3 distinct real solutions if (1) has one positive root and one zero root for it a
            = 1.
17.         Equation (*) has no solution if either (I) has no real solution or (I) has only negative
            solution.
                                     5 
            For it a (–, –1)   ,  
                                     4 
18.         Let Ei  bag contain 'i' no of red balls
            A  ball drawn from the bag is red
                                            n
            p(Ei) i2  p(Ei) = i2,        p(E )  1
                                           i 0
                                                    i


                            6i 2
             p(Ei) =
                      n(n  1) (2n  1)


                                                                                                                           5

      Narayana IIT Academy, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 • Ph.: (011) 42707070, Fax : (011) 41828319
 NARAYANA
 GROUP OF EDUCATIONAL INSTITUTIONS
                                                                                                   AITS –NFT-9 SOLUTION


                          n                            n
                                                                 6i3
          p(A) =        p(E i ) p(A/Ei) =
                       i 0
                                                       n 2 (n  1) (2n  1)
                                                      i 0

             3(n  1)
          =
            2(2n  1)
                        P(E1 )P(A / E1 )
19.       P(E1/A) =
                             P(A)
                      6           1
                                .
            n(n  1) (2n  1) n           4
          =                         = 2
                    3(n  1)          n (n  1) 2
                   2(2n  1)
                        5
20.       p(En /A) =
                        9
          p(E n ) p(A / E n ) 5
                              
                p(A)            9
                   6n 2        2(2n  1) 5
                                          n = 5.
             n(n  1) (2n  1) 3(n  1) 9
                              A

21.                     -B
                  F
                                  P
                                                E
                                  B A
                          Q       C
                                                      -A
                                          R

                      -C             
             B                D                        C
          A = R, B = P, C = Q
22.       In ABD
               C         AD        BD
                            
           sin(  ) sin B sin(  B)
                  Csin B         Csin(  B)
          AD =            , BD 
                   sin              sin 
          Now in  BQD
              QD         BD
                      
           sin(  C) sin C
                      2R sin(  B)sin(  C)
            QD =
                               sin 
          Now in APC
             b         AP
                               AP = 2R sin( – A)
           sin B sin(  A)
          Now PC = AD – AP – QD
          = 2acos


         6

  Narayana IIT Academy, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 • Ph.: (011) 42707070, Fax : (011) 41828319
AITS –NFT-9 SOLUTION                                                                             NARAYANA
                                                                                                 GROUP OF EDUCATIONAL INSTITUTIONS




                2a cos 
        Now              = 2R1  R1 = 2Rcos
                 sin A
            R1
               2 cos 
            R
         Area of ABC 2R 2 sin Asin Bsin C  1
23.                       2                sec2 
         Area of PQR 2R 1 sin Psin Qsin R  4


                                                  PHYSICS
                    1       1                                                                            V
24.     When x  , y                                                                                 A 1, 1 m
                    2       4                                                                           2 4
        Equation of tangent
                      1
         yx
                      4
        Perpendicular distance of tangent from origin
                        1
                00
                        4      1
                                 m
                     2        4 2
        Also V cos  = 4 m/s
                     dy            1
             tan        2x  2    1
                     dx            2
             = 45º
             V  4 2 m/s
        Angular momentum about origin

                    
              (1) 4 2    
                           1 
                          4 2
                                  1 J s .
25.     Distance of object from first focus = a – f1
        Distance of image from second focus = b – f2
            x1 x2 = f1 f2
            (a – f1) (b – f2) = f1 f2
             f1 f 2
                  1.
             a b
26.     Circuit after shifting the switch is in steady state.
27.          = 120 cm                                                            
                                                                                  P
                                                                           0
                                                                            P
                 6                                                                  0m
                                                                                    3c
                     3  3.6 m .                                        2
                  2                                                                                                        x
                                                                          
                                                                          P
                                                                         – 0
                                                                          2              0m 3c
                                                                                        3c    0m
                                                                                            0m
                                                                                           3c




                                                                                                                       7

  Narayana IIT Academy, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 • Ph.: (011) 42707070, Fax : (011) 41828319
 NARAYANA
 GROUP OF EDUCATIONAL INSTITUTIONS
                                                                                                   AITS –NFT-9 SOLUTION



28.       Let liquid in the left column goes down by h1 and in right column it goes up by h2. So
                 r 2 h1  9r 2 h 2
                h1 = 9h2
          Also wgh =  g(h1 + h2)
                wh = 10 w (9h2 + h2)
                h = 100 h2
                           h
                 h2            1 cm .
                        100
                                dU                     a
29.             F=0                    0  x0 
                                 dx                   2b
          Frestoring = 2a (x0 + x) – 4b (x0 + x)3
                                                3x 
                = 2ax0 + 2ax – 4bx03  1  
                                                x0 
                                   2
                = 2ax – 4bx0 (3x)
                                             a 
                m(acc) = 2ax – 4b   (3x)  4ax
                                             2b 
                           4a          a
                             2         .
                           m           m
                 dN B
30.                        1 N A   2 N B  0 at t = t0
                   dt
                                       3N
           NA  2 NB  2  0
                           1          1      2
                                    t        3N0
          Also 1 (2N0 )e  1 0   2
                                                2
                              3 2
                e  1 t 0 
                              41
                         1      4
                 t0         n 1.
                       1 3 2
31.       For field lines
                     ax dy
          tan            
                     ay dx
          y dy = x dx
          y2 – x2 = c2
          equipotenital surface is perpendicular to field lines
                   dx ax
                         
                   dy ay
           xy = c.
32.       For V = 0, x = amplitude
                                E
          Amplitude 
                                A
         8

  Narayana IIT Academy, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 • Ph.: (011) 42707070, Fax : (011) 41828319
AITS –NFT-9 SOLUTION                                                                             NARAYANA
                                                                                                 GROUP OF EDUCATIONAL INSTITUTIONS




        For x = 0, V = Vmax
                      E
             Vmax 
                      B
             Vmax = A
                    E
                          A
              B 
                    E      B
                    A
                      B
             T  2
                      A
        Dimensions of Ax2 = dimensions of BV2
                                   A [LT 1 ]2
        Dimensional formula for                 [T 2 ] .
                                   B      [L2 ]
                     P                                                P
33.          tan   1
                     V1                                                            (V, 1
                                                                                 X 1 P)
             YZ = V1
             OY = 2V1
                                                                                     
                                 1                                    O                                                V
        Area of triangle OXY  (2V1 )(P1 )  nRT .                              Z       Y
                                 2
34.     Internal forces occur in pair and torque of a couple is independent of the origin
                                                               A
                                                                          F
                                                       O

                                               F           B
               OA  F  OB  F  BA  F .
35.     If we let a free electron absorb the photon completely
             1          hc
               mv 2 
             2          
                    h
             mv 
                    
         v  2c [not possible].
36.     Nuclear force has to stablize the nucleus.
37.     At the location of discrete charges electric field blows.
                y1      1mm
38.     (B)           
             195cm 5cm                                                                                                 y
                                                                                                                       1
                                                                                                                  y
             y1  39 mm
                                                                                                                  2



                y2       1mm
                      
             190cm 10cm
             y2 = 19 mm
        width = 20 mm = 2 cm.

                                                                                                                       9

  Narayana IIT Academy, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 • Ph.: (011) 42707070, Fax : (011) 41828319
 NARAYANA
 GROUP OF EDUCATIONAL INSTITUTIONS
                                                                                                   AITS –NFT-9 SOLUTION



39.       (B)
                         D 5 107  2
          Fringe width                2 103
                          d
                      4
                5 10 m
                                         2  10 2
          Number of fringes                        40
                                         5  10 4
                           D
40.       (B) (  1)t         y2
                           d
                                
                  (1.5  1)t          19 103
                             2 103
                 t = 38 × 10–6 m
                 t = 38 m.




41.       Initial energy
                                                                           C
                                                 i                    q
                                                                      –1
                                                              d
                                                              2       q1
                                                 R                         B
                                                                       q
                                                                      +2
                                                              d        q
                                                                      –2
                                                                           A
                           2         2
                   Q        (60)
                                 300 J .
                   0 A 2  6
                 2
                     d
42.       Using KVL
                q2       q
                      1  iR  0
               0 A 0 A
                 d       2d
          Also q1 + q2 = Q
               dq1 dq 2
          and              0
                dt      dt
                             A
          So, q 2  2q1   0  R i
                             d 
                      dq       dq
          Where i  1   2
                       dt       dt
                    Q
           q1  1  e500 t   20  20 e500 t
                     3           
                    2Q Q 500t
          and q 2        e        .
                      3     3
43.       As t  
                     2Q
               q2         40 C .
                      3
        10

  Narayana IIT Academy, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 • Ph.: (011) 42707070, Fax : (011) 41828319
AITS –NFT-9 SOLUTION                                                                              NARAYANA
                                                                                                  GROUP OF EDUCATIONAL INSTITUTIONS




44.     (D)
        Magnetic field due to outer solenoid
             B = 0nkt
        Magnetic field due to inner solenoid
             B = 0(n) [2kt]
        Magnetic flux linked with cross–section of outer solenoid
             = (R2) [30nkt] + (4R2 – R2) 0nkt
             = 6R2 0nkt.
45.     (B)
        Flux of magnetic field enclosed in radius r
             = R2[30 nkt] +  (r2 – R2) 0 nkt
             = 2R2 0 nkt + r2 0 nkt = 0 nkt [2R2 + r2]
        If induced electric field is E
             E.2r =  0 nk [2R2 + r2]
                   nk
              E  0 [2R 2  r 2 ] .
                    2r
                   qBr qEt
46.     (A) V           
                    m       m
                         0 nk
                               [2R 2  r 2 ]t
                  E        2r
              r t
                  B            0 nkt
                2      2     2
             2r = 2R + r
              r R 2.

                                               CHEMISTRY
47.     (B)
                                                       Cl                             MgCl

           Cl

                               Na ether
                                    Wurtz reaction
                                                                    
                                                                      Mg/ether
                                                                               

                         Br

                                                       Cl                             MgCl
                                                             I                          II 
                                                               D




                                                                            
                                                                             C2 H5 OD
                                                                                     



                                                               D
50.     (C)
                                                                                                                       11

  Narayana IIT Academy, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 • Ph.: (011) 42707070, Fax : (011) 41828319
 NARAYANA
 GROUP OF EDUCATIONAL INSTITUTIONS
                                                                                                   AITS –NFT-9 SOLUTION



                                     13.8 1
          No of K2 CO3 mole =            
                                     138 10
                                   1                                                  1
          no of mole of C =           , Carbon which converted into K2Zn3[Fe(CN)6]2 =    mole
                                  10                                                  20
           12 mole of C required 3 mole of Zn.
                                              3 1
           1/20 mole of C required                 mole of Zn.
                                             12 20
                            3 1
          mass of Zn =            65.5  0.819 gram 0.82 g
                           12 20
51.       (B)
                                
          CH4 + 2O2  CO2 + 2H2O
                 0
          at 127 C the no of gaseous mole will remain same because H2O will be in vapour form.
          These for the pressure change will be due to change in temperature
           P P2             1           P
            1
                                 
           T1 T2        300 K 400 K
                400 4
           P         1.33 atm
                300 3
52.       (D)
          Concentration of |H+| at anode = 10-8 M
           concentration of |H+| at cathode = 10-7 M
                            1                        1
          because pH =  pkw  pka  pkb  [14  4.74  4.74] = 7.
                            2                        2
                 +      -7
          so |H | = 10 M
                   0.059         108
          E=0-             log 7 = E = 0.059 V
                     1           10
56.       Through SPM only migration of solvent particle is permissible and hence none will be
          able to make a mixture.
57.       Activity = rate of decay
                    = N
                        0.693
                    =             N
                          t1/ 2
59.       (B)
          Succinic acid is dicarboxylic acid and thus SOCl2 dehydrate it.
60.       (B)
                                                                                     1       3
          The minimum energy required to excite H-atom is 13.6 ev   1    13.6  ev
                                                                                     4       4
                                     13.6 ev
          which is higher than               .
                                         4
                   RT02         RT02 M         RT0 M
61.        Kb                           
                 1000 lv 1000 Lv                 L 
                                            1000  v 
                                                 T 
                                                  0

        12

  Narayana IIT Academy, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 • Ph.: (011) 42707070, Fax : (011) 41828319
AITS –NFT-9 SOLUTION                                                                             NARAYANA
                                                                                                 GROUP OF EDUCATIONAL INSTITUTIONS




                Lv
         S          ; 10.5  nearly constant for Trouton's rule 
                 T0
         Kb  T0.M
        M = constant for isomeric liquids
         Kb  Tb
              R  320  320
62.     Kb                  3.20
               1000  32 R
        Tb = i.Kb Cm                                            XY2             X2+ + 2Y–
                         0.01
        = 2  3.20                                              1                 0       0 … initially
                     100 /1000
        = 0.64                                                   0.5               0.5     1 … After dissociation
                                                                          2
                                                                 i=        2
                                                                          1

                H v 50  32  2
63.     Sv                      10
                 Tb       320
                                 1
64.     esterification 
                         steric hindrance
65.
                                                       O

                                                           O
        HO       (CH 2)4      COOH 
                                    H 2SO4
                                           

                                   
        TlI 3 exists as Tl 1 and I3
69.




                                                                                                                      13

  Narayana IIT Academy, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 • Ph.: (011) 42707070, Fax : (011) 41828319
 NARAYANA
 GROUP OF EDUCATIONAL INSTITUTIONS
                                                                                                      AITS –NFT-9 SOLUTION




                                     HINTS AND SOLUTION
                                                       PAPER - II

                                             MATHEMATICS
1.         Let  be purely imaginary root, then
           a2  b  c  i = 0 ... (i)

            a 2  b  c  i = 0
            a2 - b + c - i = 0 ... (ii)
                                                           i
           (i) + (ii) a2 + c = 0, (i) – (ii)   =            a = b2 c
                                                           b
2.         abc = 1 in 1 ways
           abc = 2, 3, 5, 7, 11 in 15 ways
           abc = 4, 9 in 12 ways
           abc = 8 in 10 ways
           abc = 6, 10 in 18 ways
           So, total number of solution is 56
           abc
3.                    (abc)1/3
               3
                1     1
                  
               abc 27
            1  1   1 
             1   1    1
            a  3   c 
                  1 1 1 1 1 1                1                         3        3       1
           =1+                           1+                 3
                                                                                         (using G.M  H.M.)
                  a b c ab bc ca abc                                   abc   3 2 2 2
                                                                              a bc     abc
                                3
                       1       64
            1  3          27
                      abc 
              2
4.         4y - 3y + 6x + 1 = 0
                  3y 6x 1
           y2 -             
                   4     4     4
                  3  6x 1 9
                      2
           
           y                
                 8       4     4 64
           If normals at 3 point are con current then centroid lie on the axis of parabola.
5.         3x + 4y + 1 = 0
                      3        1
            y =  x  
                      4        4
           If it is normal to y2 = -x, then



         14

     Narayana IIT Academy, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 • Ph.: (011) 42707070, Fax : (011) 41828319
AITS –NFT-9 SOLUTION                                                                                NARAYANA
                                                                                                    GROUP OF EDUCATIONAL INSTITUTIONS




                    3     3 
                                                                   3
            1
                       
            4     2  4   4  4 
           1 3 27
                       64 = 96 + 27
           4 8 296
               64
          =
              123
6.        Equation of normal at P
                  dx                                                                                   P(x, y)
          Y-y=        X  x
                   dy
                                                                       dy
          Mid point of ON = (x, 0)  x + y                                 2x                                  N
                                                                       dx                O (0, 0)              C
              ydy                                                                                            ydy 
                  x  x2 - y2 = c                                                                      x      ,0
                                                                                                             dx 
              dx
7.        f (x) = 6x2 + 2ax + b
          If f(x) = 0 has 3 distinct
          real roots, then f (x) = 0
          must have 2 distinct real roots for it
          4a2 - 24b > 0  a2 > 6b
           a  3, b  1 and for a = 3, b = 1, f(x) = 0
          has 3 distinct real roots.
                  4               4

                   e dt         e
                          2
                     cos t                 cos2 t
                                                    dt
                  y           xy                        0     
8.         lim                                            form 
           x 0               x                          0     
           = lim 0  0  ecos       = ecos y
                                          (x  y)
                                      2                        2


              x 0                 
           1  tan     (3 tan   tan 3 )
9.                   3
           1  tan        1  3 tan 2 
            3tan4 - 6 tan2 + 8 tan - 1 = 0
           tan tan = -2
10.        Both statement are right and I is due to II.
11.        Statement 1 in not true for example if a = 3, b = 5, c = 7
                                  7
           then 2r = 3 , R =          , ac = 2 .
                                   3
12.        Statement II is not true. As we also need the conditions h2 - ab > 0 and a + b = 0.
                   
                           m  (x  m)3 ,         m  x  m 1
13.        f(x) = f (x)  
                           m  1  (x  m  1) , (m  1 x  m  2
                                               3
                   
                   
           mI
           f(x) is strictly increasing and continuous every where
14.        Equation of ellipse can be rewritten as
                         1   13   12x  5y  2 
                      2                    2               2                     2
               1 
            x    y                      
               5      3   15       13       
                                                                                                                         15

     Narayana IIT Academy, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 • Ph.: (011) 42707070, Fax : (011) 41828319
 NARAYANA
 GROUP OF EDUCATIONAL INSTITUTIONS
                                                                                                                  AITS –NFT-9 SOLUTION



                      1 1
          So focus is  ,   , directrix 12x + 5y - 2 = 0
                      5 3
                            13
          and eccentrics is    .
                            15
                                                         12 9
                                                            2
                                                          5 3     19        1
15.       Distance between focus and directrix will be               = ae  
                                                            13    195       e

                19
          a=
                 56
          Required locus is auxiliary circle of the ellipse.
16.       Major axis is perpendicular to directrix and passes through focus, so it is 5x - 12y - 5 = 0.
                                 3
                   e 1
17.          A r    . , So                            A1 ,     A2 ,   A3 , ... are in H.P.
                    r
                                                                                           1                       
                                                         9                           9
                                    e          1        1 e                                             1
18.          Ar . A r 1 . A r 2    .                                             .                        
                                       r(r  1)(r  2) 2                             r(r  1) (r  1)(r  2) 
                                                                           9                 9
                             n
                                                                   1 e  1 1 e 
          So, lim
                  n 
                         
                         r 1
                                         A r . A r 1 . A r 2       .       
                                                                   2    1.2 4   
                                                         r
                              3 
                                     3
             
                 1                              3
                           r
                 e          e e   k r
                          e
                 Ar
19.       e
                              
                              
                               1
                     n                         3

          So, lim  k  r     k  1  (e e3  1)1
              n                 1 k 1
                   r 1
                            1
                                  k
20.       (A) No. of term is 6 + 4 - 1C4-1 = 9C3 = 84
          (B) 72011 = -7 (1-50)1005 = -7 + 7 × 1005C1 × 50 - 7×1005C2 × 502
          = 351743 + multiple of 100
                                               100  r        r
          (C) Tr+1 = 1001Cr 7 3 115
          Tr+1 will be rational
          if r = 5, 20, 39, 50 ... 995
          So number of rational terms is 67
          (D) 22011 = 4[128]287
          = 4[1 + 127]287
          So remainder is 4.
                             1 4x 2  1
21.       (A) f (x) = 4x - 
                             x       x
             +           –                 –             +
                                 O

                 -1/2            O             1/2



        16

  Narayana IIT Academy, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 • Ph.: (011) 42707070, Fax : (011) 41828319
AITS –NFT-9 SOLUTION                                                                             NARAYANA
                                                                                                 GROUP OF EDUCATIONAL INSTITUTIONS




        (B) f  (x) = 2 - e-x
               –                   +
                        ln (1/2)

        (C) f (x) = - sinx + ex
        (D) f (x) = sinx (–ecosx - e-cosx)
22.     (A) 2x + y = 0, 4y + 3z = 24 and 12x – 7z = 24. Solving we get the point intersection as
            (– 12, 24, –24) whose distance from origin is 36.
                                              1 2 1
        (B) For no unique solution  = 0  3              3           0
                                                        2 1        2
              1(9)   (3)  2 (9)  0                    9    81
        (C) Let the centre be at origin and the vertex at ( 2, 0) . The remaining vertices will be
             corresponding       to     the    complex        numbers                2, 2 2 , 2 3 , ...., 28
                      2         2 
                cos     i sin     . The product of distances
                       9          9 
              = | 2  2 || 2  2 2 | ...  ( 2) 8 |1   | |1   2 | ... | |   8 |  16  9  144
        (D) a  b 3i  (72  56 3i) . ei / 3  (36  28 3i) ( 1  3i)
            a = 36 – 84 = –48, b = 36 + 28 = 64. (a + b)2 = 256.




                                                                                                                      17

  Narayana IIT Academy, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 • Ph.: (011) 42707070, Fax : (011) 41828319
 NARAYANA
 GROUP OF EDUCATIONAL INSTITUTIONS
                                                                                                   AITS –NFT-9 SOLUTION




                                                      PHYSICS
                              V0  V0 sin t  V0 cos t V0  V0 2 sin(t  45º )
23.       Current i                                    
                                          R                        R
                                 T
                                i dt  V 2 .
                                         2

                  i rms        0
                                     T
                                             0


                                 dt0
                                         R
                                                                                                   y
24.       For all the waves to meet at same point their optical
                                                                                         1
                                                                                                        ( ,y 2
                                                                                          =
          path should be same. Let (x, y) be a point on surface.                                        x ) =
                  x  2 (1  x)2  y2  (2) (1)                                                                             x
                                                                                                        m
                                                                                                        1
                  2 (1  x)2  y2  (2  x)
                  4[1  x 2  y 2  2x]  4  x 2  4x
                3x 2  4y 2  4x  0 .
25.       i + e – A = 60º
          i + e = 120º
          and i  30 = e
          i = 75º, e = 45º
          i = 45º, e = 75º.
                     (2n  1)v
26.             f
                      4(  e)
                          v
                 1 e 
                         4f
                         3v           e
                 2 e         2        3
                         4f         1 e

                    2    (3.6  2.34)m and      1    (3.6  3.22) m  e  0.06m  0.6 r
           r = 0.1 m
             A = 100 cm2.
27.       Equation of resultant wave is
                                     3Kx  4Ky       3Kx  4Ky      
              Z  z1  z 2  2A cos             cos             t 
                                         2               2          
                                      3Kx  4Ky
          Equation of wavefront is               t  C (C is constant)
                                          2
                                                     y


                                                 C’            
                                                 4K
                                                                            x
                                                          C’
                                                          3K
          So at any given instant of time the equation of wave front will be

        18

  Narayana IIT Academy, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 • Ph.: (011) 42707070, Fax : (011) 41828319
AITS –NFT-9 SOLUTION                                                                             NARAYANA
                                                                                                 GROUP OF EDUCATIONAL INSTITUTIONS




              3Kx + 4 Ky = C
        Wave travels perpendicular to the wave front also with increase in time both x and y
        should increase, as
                       C'
                               4       4
              tan   3K  ,   tan 1 .
                       C' 3            3
                       4K
28.      E  3.4 1.51  1.89 ev
         hc
              2.24  1.89 =4.13 ev
         
              12400
                  Å
               4.13
          3000 Å.
        Since radiation emitted increases by 16 times, temperature must have become doubled
               T    2T 
                6000 Å.
29.     When the oil is poured and fraction of ice in the water decreases. Volume of ice melted
        into water > volume of water displaced by ice
        So water level rises
        Overall volume of ice will decrease as it melts. So oil's upper level falls.

30.     Let temperature of interface is 
                   0                 70  
                               
                1 1 1            1      1    1 
              4K  R  2R  4  2K   2R  3R 
                                                
               70  
                 
              1       1
              R      6R
               6  70  
             7  6  70
              = 60ºC.
                      1 1
31.          F  A   
                       y1 y2 
                             1 1
              30  (0.1)(2)   
                              y1 y 2 
               y1  y 2
                         150
                y1 y 2
                y1 y 2     1      100
                             m       cm
               y1  y 2 150       150
              y1 = 3 – y2

                                                                                                                      19

  Narayana IIT Academy, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 • Ph.: (011) 42707070, Fax : (011) 41828319
 NARAYANA
 GROUP OF EDUCATIONAL INSTITUTIONS
                                                                                                   AITS –NFT-9 SOLUTION



             (3  y 2 )(y 2 ) 100
                             
                   3           150
             3y2 – y22 = 2
             y22 – 3y2 + 2 = 0
             y2 = 1 cm, 2 cm,  y1 = 2 cm, 1 cm.
32.     Distance traveled in last second of upward motion is equal to distance traveled in first
        second of downward motion.
33.     Power factor is zero.
34.     Same amplitude increases contrast.
35.     Young's modulus depends on material.
36. to 38.
                  di                                                    C             C
        At t = t0      0 , charges on capacitors are as shown
                  dt                                                      ( Vq ( C + ) C +
                                                                    V q –C - ) –2 V q 2 V q
                                                                   C -                     0        0         0        0


               CV0  q 2CV0  q
                                    0                                       L
                   C            C
             2CV0 + q = CV0 –q                                         di
                                                                          =0
                                                                       dt
                            CV0
                  q
                             2
                                                   C                   C
                                             C0
                                             3V        C0
                                                       3V          C0
                                                                   3V   C0
                                                                        3V
                                                     –         –
                                               2           2       2      2
                                                               L

          Conserving energy
                                                       2                 2
                  1          1          1  3V  1  3V  
                     CV02  C(2V0 )2  C  0   C  0   Li 2
                  2          2          2  2  2  2  2 max
                  5CV02        9V 2 1
                           C. 0  L i 2 max
                     2          4    2
                   10  9        1 2
                            CV0  L i max
                                2
                  
                   4             2
                               C
                  i max          V0 .
                               2L
39. to 41.
39.     (A)
        For shortest time, life should move with maximum acceleration and maximum retardation.
                     3Mg  Mg
             a max               2g
                        M
                                   Mg
             (retardation)max        g
                                    M
40.     (B)
        For accelerated motion
            T = 3Mg
            V = 2gt
        20

  Narayana IIT Academy, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 • Ph.: (011) 42707070, Fax : (011) 41828319
AITS –NFT-9 SOLUTION                                                                             NARAYANA
                                                                                                 GROUP OF EDUCATIONAL INSTITUTIONS




             P = 6 Mg2t
        For retardation
             T=0
             P=0
41.     (C)
        For acceleration
             Fnet = 2 Mg
             v = 2 gt
             P = 4 Mg2t
        For retardation
             Fnet = – Mg
             v = V0 – gt
             P = – MgV0 + Mg2t
42.     For AB        PT2 = constant  PV2/3 = constant
                    R      R     3R 3R 9R
             C                       
                   5         2    2    1     2
                     1 1 
                   3         3
             T = < 0
             Q < 0, U < 0, W < 0
        For BC, C = CP
             U > 0, Q > 0, W > 0
        For BC, C = CP
             U > 0, Q > 0, W > 0
        For CA T = constant, VA > VC
             W > 0, Q > 0, U = 0
        For ABCA
             U = 0, W > 0, Q > 0.
43.     Force due to surface tension can balance weight of h height of water column by forming
        meniscus.
44.     R  g2  f 2  C  60m
                                                           
        Angle moved by A in 10 s = 20 /60 =                 = 60º
                                                           3
                    60
         Vav A        6 m/s
                    10
                    0  2
         Vav B             m/s
                       2
                    change in velocity 2 
         a av A                        m/s2
                           Time         10 5
                    2  0 
         a av B          m/s2.
                     10       5
        Also, since B moves with constant acceleration, average acceleration will be same as
        instantaneous acceleration which is a constant.


                                                                                                                      21

  Narayana IIT Academy, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 • Ph.: (011) 42707070, Fax : (011) 41828319
 NARAYANA
 GROUP OF EDUCATIONAL INSTITUTIONS
                                                                                                      AITS –NFT-9 SOLUTION




                                                    CHEMISTRY
45.       (C)
                                                     O        O                              O         O
              O          OH
                                                                  ONa                                        OC2H5
                              CH3    
                                      I2 / NaOH
                                           
                                                                         H
                                                                         CH 2 CH 2 OH, 
                                                                                         

                                                                                                 i) RO
                                                                                                 ii) CH3-Br


                                                                                             O        O

                                                                                                        OC2 H5
                                                                                                     CH3

46.       Suppose weight of each layer is w.
          Total weight of phenol = w  0.7 + w  0.1 = 0.8 w
          Total weight of H2O = w  0.30 + w  0.90 = 1.20 w
           weight of phenol x 0.8w 2
                                 
            Weight of H 2O  y 1.20w 3

                                                    10                                                            9
47.       Molar mass of XY2   a  2b                 100       Molar mass of X 3Y2   3a  2b                  180
                                                    0.1                                                          0.05
                   a  40, b  30

          Weight of 0.02 moles of X2Y is = 0.2  2  40  1 30 =22

48.       I : BrF4+ : Trigonal bipyramidal structure and see saw shape
          II : Bond order of O2 increases by the removal of e, as bond order increases
          III : Br3 and Br3 are of different shape and structure.

          IV : In PCl5, hybrid orbital of P is sp3d z2

51.       (C)
                                                         x
          Freundlich adsorption isotherm is                 kp1/n ;1  n  
                                                         m
53.       (A)
          Buna-S is a copolymer of butadiene and styrene while all the other are homopolymers.
54.       (A)

        22

  Narayana IIT Academy, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 • Ph.: (011) 42707070, Fax : (011) 41828319
AITS –NFT-9 SOLUTION                                                                             NARAYANA
                                                                                                 GROUP OF EDUCATIONAL INSTITUTIONS




        Both statement (1)and (2) are correct and statement (2) is correct explanation of statement
        (1) .
56.     (A)

                                        
        K 2 Cr2 O 7  4NaCl  6H 2SO 4 

        2KHSO 4  4NaHSO 4  2CrO 2Cl 2  3H 2O
                                        Orange  red fumes


               
                 
58-60. 2AgNO3  2Ag  2NO2  O2 
               A

                     
        AgNO3  HCl  AgCl   HNO 3
                                      B

                         
        2AgNO3  K2CrO4  Ag2CrO4  2KNO3
                                               C Red

        AgNO3  KI  AgI   KNO3
                               D

        AgI is insoluble in dil NH3 and in dil HNO3 but readily soluble in KCN and in Sodium
        thiosulphate. AgCl is also soluble in sodium thiosulphate.

        AgX  2Na 2S2O3  Na 3 Ag S2O3 2   NaX
                                           

        As 2 CrO 4 is soluble in dil HNO3 and in NH3 sol^.

        2Ag2CrO4  2H                 4Ag  Cr2O7  H2O
                                                  2


                                                             
        Ag 2CrO4  4NH3                 2 Ag  NH3 2   CrO4 .
                                                      
                                                              2




65.     (A) Mg(OH)2(s)                      Mg2+ + 2OH
                             s              2s
                ksp = 4s3 = 4  10-12
                s = 10-4
                [OH] = 2  10-4

                 pH = 10 + log 2
        (B) Basic buffer is formed. Hence,
                                     1
                pOH = Pkb  log        = 4 – log 2
                                     2

                                                                                                                      23

  Narayana IIT Academy, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 • Ph.: (011) 42707070, Fax : (011) 41828319
NARAYANA
GROUP OF EDUCATIONAL INSTITUTIONS
                                                                                                  AITS –NFT-9 SOLUTION



                pH = 10 + log 2
         (C) A salt of weak acid and weak base is formed in which ka = kb
                hence, pH = 7 or
                           1      1     1
                pH =         Pkw  Pka  Pkb
                           2      2     2

         (D) H3PO4                  H+ + H 2 PO4
                      
                 H2 PO4                      
                                    H   HPO4
                    
                 HPO4                       
                                    H   PO4

                                                  Pka2  Pka3       8.102  12.5
                PH of 0.01 M Na2HPO4 =                          =
                                                      2                  2
                      20.602
                 =            10.301
                         2
                Hence, q and r.




       24

 Narayana IIT Academy, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 • Ph.: (011) 42707070, Fax : (011) 41828319

								
To top