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					    The Key & Hint paper contains 28 Pages                        NFT-7 PAPER I & II Dt : 13-02-2011




                    NARAYANA
                    I     I     T          A     C      A     D         E     M      Y
                  ALL INDIA TEST SERIES - 2011
                              NARAYANA FULL TEST – (PAPER I & II)



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




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    The Key & Hint paper contains 28 Pages                   NFT-7 PAPER I & II Dt : 13-02-2011



                                          PAPER – II KEY

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




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     The Key & Hint paper contains 28 Pages                               NFT-7 PAPER I & II Dt : 13-02-2011



                                                HINTS & SOLUTION
                                                      PAPER - I
CHEMISTRY
1.      Blue colour of solution is due to presence of ammoniated electrons which is
        paramagnetic in nature
         M  M  e
            
            +                           +
         M + y NH3              [M(NH3)y]
        –                                –
        e + x NH3               [e(NH3)x] Ammoniated electron.
2.       (D)
                        H
        H               N       H
                B           B
                N           N
        H               B       H
                        H
3.      (C)




                                                         –      +
                Ph–CH–COOH                  Ph–CH–COO +H; 1.5 × 10–5            ...(i)
                   Cl                          Cl
                                            +     –
                                                                          –3
                    HCN                     H + CN           ; 4.5 × 10        ...(ii)
                    –       +                                   1       –3
                CN+H                        HCN                       ...(iii)
                                                             ; 4.5 × 10
                                    –                                     Adding (i) and (iii)
                                                     –
         Ph–CH–COOH+CN                      Ph–CH–COO +HCN; k = k1 × k2
            Cl                                 Cl
                                                                               1
                                                           1.5  105               3.3  103
                                                                          4.5  103
4.      Since the reaction is redox reaction so n factor will be number of moles of e– required
        for 1 mole of compound.
5.      (C)
        Conceptual
6.      Because nucleophilicity is a kinetic property but basicity is a thermodynamic property.

7.                    
         Cl2  H 2 O  HOCl  HCl
        (X)
                                      1
                       
         HOCl  AgNO3  AgCl  HNO3  O2
                                      2
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      The Key & Hint paper contains 28 Pages                                        NFT-7 PAPER I & II Dt : 13-02-2011



          AgCl  2NH 4 OH  [Ag(NH3 ) 2 ] Cl   2H 2O
                           
                                         Soluble complex
                     
          2HCl  Mg  MgCl2  H 2 
                                                               (Y)
9.       (A), (B)
               CH3                       O CH3                                       CH3          O CH3
         Ph      CH NH2      OH C C C2H5                                   Ph        CH    NH C C C2H5 (R R)
                                              H                                                 H (A)



                                                                           CH3              O C2 H 5
                                                                      Ph   CH NH C C CH3 (R S)
                                                                                                H     (B)

10.       (A), (B) & (D)
         In A, B & D compounds are cyclic planar having delocalization of  e  & they are
         following Huckel number rule.

                                                                            +
                                                                                –
                                                                                      both part are
                            N Sp2
                            ..                                                        aromatic
                            not  e– = 6
12.       (A) (C) & (D)
         [Mn(CO)5 ]  EAN  25  0  10  35
         Can gain one e– to get stable electronic configuration gain of e– is reduction so
         behaves as oxidising agent
         Complex dimerizes to get stability

                      pKa1  pKa 2
14.       (A) pI 
                           2

15.       (A)
                                          +
                                          NH3–CH2–COOH
                                 2
                              =
                             H




                                              Cationic
                            p




          H2N–CH2–COOH p         H
                                 =
                                     9
                                                           –
                                         H2NCH2COO                   will move
                                           Anionic                   towards anode


                                          pK a1  pK a 2
         Isoelectric point (PI) 
                                                2
                  1.9  3.7 5.6
                               2.8
                      2      2
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16.       (B)
         Conceptual
17.       (A)
         Iron exists in +2 state in haemoglobin because it releases two H+ ion
18.       (D)
                         
         In [Fe(CN) 6 ]4
         Fe2+    3d6
                 –
         Since CN is strong field ligand so pairing will occur
                        x x x xxx
         But In [Fe(CN)6]3–    d2Sp3
          Fe3+    3d5

         After pairing
                                    2   3
                                   d Sp but weakly paramagnetic
                 x x x xxx

19.       (A)—p, q; (B)—q, r, s, t; (C)—p; (D)—r, t

20.       (A)—p, q, r, s; (B)—q, s; (C)—p, r, s; (D)—r, s, t




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MATHS

21.       (C)
                f(x) is periodic with period 2
                 2n                    2
                 
                 2n
                       [f (x)]dx  2n  [f (x)]dx
                                       0

                         1            1
                                                    
                 2n  [f (x)]dx   [f (2  x)]dx 
                        0             0            
                [ f (1  x)  f (1  x)  0
                      1            1
                                                  
                 2n  [f (x)]dx   [ f (x)]dx  x  x  1
                      0            0             
           f (x)  f (2  x)  0 ]
                      1             1
                                                        
                  2n  [f (x)]dx   ( 1  [f (x)]dx 
                      0             0                  
                  2n [1]  2n

22.       (D)


                                                            f–1(x)

                                       f(x)                                   2
                                 –2          –




                                 –2          –                              2




              1
         2.     (2)2  4 2 sq units
              2
23.       (A)
         1 + 2 + 3 + …. 8 = 36
         Hence number of balls is boxes are 1, 2, 3, ….8
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         It can be done in 8! Ways
                                           8
                                               C4  2!
         Required probability is
                                                8!
                       8! 2!     1
                             
                       4! 4!8! 12  4!
24.      (A)
         Divide denominator and numerator by cos2 x we have
               cos x  x sin x 
              
                                dx
                                
                    cos 2 x
          I
                  x         x 
                      1         
                cos x  cos x 
                   x
         Put           t
                cos x
                 cos x  x sin x 
               
                                  dx  dt
                                  
                      cos 2 x
                  dt
          I
              t (1  t)
              1      1 
                         dt
               t (1  t) 
           log t  log (1  t)  logC
                 Ct         Cx 
                      log 
                            cos x  x 
          log
                1 t                   

25.      (B)
         Total no. of correct combination of the question is 24  1  15.
         The student will answer the question correctly either in 1st attempt or in 2nd attempt or
               in 3rd attempt.
          The required probability is
           1 14 1 14 13 1 1 1 1 3 1
             .  . .     
          15 15 14 15 14 13 15 15 15 15 5
26.      (B)
         Put tan   t
                 9t 5 + 7t 3
         I=ò 3               dt
                (t + t )1/ 2
         Multiply No. & Den. By t3
                 9t 8 + 7t 6
         I = ò 9 7 1/ 2
                (t + t )
         Put t  t 7  
              9


                      du
          I
                       
           2  C
           2 t9  t7  C

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         = 2 tan9 q + tan7 q + C
          2.(tan )7 / 2 sec   C
27.      (A)
         Consider the expansion of (a  b) n
         Put a = 1 + 3x & b = 2x
         Given term is the co-efficient of xk in (1  3x  2x) n
          n Ck

28.      (C)
         Other points is reflection of (1, 1) in line y = x tan  / 6




                                                         (1, 1)
                                                     
                                                     6




29.      (A, C)
         Consider f (x)  x 3  ax 2  bx  2
           2
          , ,   N  two of them are 1 & other is 2
                                           f ''(0)
                a  (     )  4 
                                              2
          b  (     )  5  f '(0)
         Hence f (x)  x 3  4x 2  5x  2
         Again 1 is a repeated root of f(x) = 0
              f (1)  0
30.      (A, B, C, D)
           x  [y]  {z}  1.1                            (1)
          [x]  {y}  z  2.2                             (2)
          {x}  y  [z]  3.3                             (3)
          (1) + (2) + (3)
                  2(x  y  z)  6.6
                  x  y  z  3.3                        (4)
          (4)  (1)
          {y}  [z]  2.2  {y}  0.2 &[z]  2
          (4) – (2)
          {x}  [y]  1.1  {x}  0.1, [y]  1
          (4) – (3)
          [x]  {z}  0       [x]  0 & {z}  0


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           x  0.1
          
           y  1.2
          z  2
          
31.       (B, D)
          b2  ac
                   a 2  c2  b2
          cos B 
                         2ac
            a 2  c2  b 2
          
                 2b2
                             a2  c2 1
                   cos B           
                              2ac     2
                       1 a2  c2             1
           cos B                  cos B   1
                       2     2ac             2
                       1
           cos B 
                       2
          B (0,  / 3)
32. (B, C, D)
          det (M  I)  det (M  I) T
                     det (MT  I)
                              det (M  I), if n is even
           det (  M  I)  
                               det (M  I), if n is odd
33.      (C)
         OP = 2OD = 4R cos B cos C
                        OP          OQ          OR
                                                      12R
                    cos B cos C cos A cos C cos A cos B
34.       (A)
                        D         c cos C     E
                                                  2R
                    C




                                                       co
                   s
                co




                                                         sA
             sB




                                                           o  c
           co




                 –A                                    –B s C
         2R




                2                                      2
          P             X        c cos C      Y                   Q
                    PQ = 2ccosC
                    PQ
                      2
                    DE
                    PQ RQ PR
                          6
                    DE EF DF
35.       (B)

         Orthocentre of ABO is C and Orthocentre of ACO is B.

36.      (A)
         z  a  bt  ct 2  z  a  t (b  ct)
         Put z  x  iy
         a  a1  ia 2
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          b  b1  ib2
          c  c1  ic2
         Comparing Real & imaginary parts
          x  a1  t (b1  c1t)
          y  a 2  t (b2  c2 t)
          If c2  b2  0 or c1  b1  0
                  y  a2
                         k i.e. straight line or b or c are real or purely imaginary
                  x  a1
37.      (A)
         Again if b1  0 & c2  0
               x  a1  c1t 2
         and    y  a 2  b2 t
                                         2
                             y  a2 
               x  a1  c1 
                             b     
                                     2
                                 2
                               b
               (y  a 2 )      2
                                  (x  a1 )
                               c1
38.      (C)
                                   b2  (b  b) 2
         Length of latus Ractum is 2 
                                   c1  2 (c  c)

39.      (A) —r; (B)—p; (C)—q; (D)—s
         (A)
                        é      1ù          é     1ù
                sin - 1 êx 2 + ú + cos - 1 êx 2 - ú
                        ë      2û          ë     2û
               is defined only if
                é 2 1ù
                êx + 2 ú = 0 or 1
                ë         û
         (B)
                                    1
                f (x) + f (- x) =
                                  1+ x2
         (C)   Square & add given equals
                24(sin A cos B + cos Asin B) = 12
                                1
                 sin C =
                                2
                            p 5p
                 C= ,
                            6 6
                         5                        1
               If c          then A   sin A 
                          6              6          2
                 3sin A + 4cos B < 6
                                                        
                  i.e. impossible  only solution is
                                                        6


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40.      (A)–q, r, s; (B) r, s, t; (C) –r; (D) – t
         (A) The two circles must be intersecting then  (0,4)
         (B) 1  2sin 2 x  a sin x  2a  7
                                 a4
                   sin x 
                                  2
                        a4
                  1         1
                         2
                  2a 6
                       
                         sin 4 x
         (C)      I1           dx
                       0
                           x2
                                    
                     sin 4 x       4sin 3 x cos x
                                                dx
                    x 0 0                  x
                   
                       (3sin x  sin 3x) cos x
                                              dx
                   0
                                  x
                                             
                      3 sin 2x  1 sin 4x  sin 2x
                       x dx  2 
                      20          0
                                         x
                                                  dx

                    3   1
                   I2  (I2  I2 )
                    2   2
                    I2
                  
                    2
         (D)          This is possible when
                       
                                         1
                   cos 2                  There exist four values of  , so there are four possible
                                          
         triplets.



PHYSICS
41. (D)
          dNB
               1NA   2 NB
           dt
                                       3N 
          0  1 (2N 0 )e t 0   2  0 
                                       2 
                 1  4 1 
          t0      ln
                 1  3 2 
                          
42.      (A)
                            dB
           E.dl  A dt
                                                              R 2K
                      E2 x  d  R K ; E 
                                 2        2       2

                                                            2 x2  d2
                                     
                                                  qR 2
                      Wext   qE.dx                  K
                                     0
                                                    4

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43.      (D)
         PV = c
          dP    dV                  P
                    0  dP =  0 dV
           P     V                   V
         Restoring force F = kdx + S0 dP
                    k  S0 P0 
          a                  dx
                        m      
                       k  P0 S 0
                                  .
                           m
44.      (A)
         x = 0; t  2 ( 2 –1)d .
45.      (C)
         Velocity of efflux is 2g
                       1
          s y  u y t  gt 2 ; s x  u x t
                       2
46.      (B)
                 CV
          q 
                  2
                               2
                     CV
          Wb  qV      ; Wb  (U f  Ui )  H
                      2
                          1
         Loss of energy = CV 2 .
                           4
47.      (C)
           Ndt = 30sin 37m
            Ndt =10 cos 37m
                                                 37       37
             1 4 4
            .
             3 3 9
48.      (B)
                                             4R
         Dipole moment of ring P  (R)         4R 2
                                              
            I
          4R 2 E – fR  mR 2                            …(i)
          Fx  ma cm
         f = ma                                           …(ii)
         a = R                                           …(iii)
         f = 2RE
49.      (B, D)
              4       R 1 2
         E  G r  ;   at
              3       2 2
               R
          t
               a



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                    6
          T4           .
                  4G
50.       (A, C, D)
                 V1 + V2 = 6
                       V             3        9
                 3V1  2  6 ; V1  Volt; V2  Volt
                        3            2        2
         From figure (i)
                       3          9
                 IR1  and IR 2 
                       2          2
                       R
                R1  2        ….(i)
                       3
         From figure (ii)
                 8I       I
                    3   R2
                  3       3
                 R2 = 24 and R1 = 8
                     3
                 I A                                                 Fig. (ii)
                    16
51.       (A, D)
52.       (B, C)
                               




         r  Rsin  ; x  R cos 
         dq  (2R 2 )sin d
               
         d I  dq  R 2 sin d
               2
                   0 dI r 2
         dB 
               2(r 2  x 2 ) 2/3
                      2
         B   dB  0R
                      3
               q              4
                L ;   R 4
              2m              3
53.      (B)
54.      (B)
55.      (A)
Sol.     Fcos   1  5
         Fsin  – mg  ma P / T
          a P / T  6m / s 2
          a P / G  62  52  61 m / s2
           VP/ T t 6  6  2  12 m/ s
56.      (C)
57.      (B)

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58.      (C)
Sol.     J cos 37 +   Ndt  mvcm
                                                    37°


                                                          f
         Vcm = V0
         J × R – fdtR = I
              2V0
         =
               R

59.      (A) p, r (B) p, r (C) q (D) s
         Consider two equations
                1
         eVs  mvmax  hv  0
                     2
                                                              …(1)
                2
                                             int ensity
         No of photoelectrons ejected/sec.               …(2)
                                                 hv
         (A) As frequency is increased keeping intensity constant.
                                   1
              | VS | will increase, m(v 2 ) will increase and saturation current will decrease.
                                         max
                                   2
         (B) As frequency is increased and intensity is decreased.
                                   1
              | VS | will increase, m(v 2 ) will increase and saturation current will decrease.
                                         max
                                   2
         (C) It work function is increased photo emission may stop.
         (D) If intensity is increased and frequency is decreased saturation current will
         increase.

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




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                                             PAPER - II
CHEMISTRY
                              [In – ]
1.      pH = pk In  log
                              [InH]
                             75
        5 = pk In  log          pkIn = 4.523
                             25
                                10
        pH = 4.523 + log              = 4.523 – .954 = 3.56
                                90
           0    0
         PA PB
2.                  1  PA  PB  2
                                0       0

          2    2
           0       0
         PA 3PB
                      1 atm PA  3PB  4 atm
                                      0     0

          4     4
                 0         0        0
               PA 3PB 4PC
        and                          1  PA  3PB  4PC = 8 atm.
                                                 0     0    0

                8        8        8
         PA  3PB = (8 – 4 × 0.8) atm = 4.8 atm.
                 0        0


3.

           + O–H       OH               OH          OH              OH                      OH

                                             H           Cl2                          –
                               +                                             Cl




4.      Conceptual
5.      Conceptual
6.                         3A(g)            2B(g)  2C(g)
        t=0                P0
                                                  2x 2x
        t = 20             P0 – x
                                                   3 3
                                                  2P0 2P0
         t               0
                                                   3   3
        4P0
             4            P0 = 3 atm
          3
             x
        P0   3.5                                            t50% = 20 is the half life
             3
        x = 1.5                                                t75% = 2 × 20 = 40 min
                                                               t87.5% = 3 × t50% = 3 × 20 = 60
                                                                       2
                                                               t99% =  t 99.9%
                                                                       3
                                                                   2              400
                                                               = 10  t 50% 
                                                                   3               3


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7.
                                       O
                                                                 O
                 O                     C – Cl                        C=O
                              +        CH2                       O   CH2 – Cl
                 OH
                                       Cl

                                                                                O


                CHO

8.              CHO           Gives Cannizaro reaction with NaOH
                CHO                                         CH2OH
                               + NaOH
                 CHO                                        COONa


                CH2OH              +                CH2 OH
                                   H
                C – ONa                             C – OH
                O                                   O
                                                        –H2O
                                                           CH2
                                                                     O
                                                            C
                                                            O
9.       Conceptual.
                                                
10.       A  r, Nucleophilic substitution by N 3 ion and reduction.
          B  p,
                   + O Tautomerism          + O–Na H SO
          R – CH – N               R – CH = N       2  4
                                                         R – CHO
                       O   OH                 O
             H
                     O
                                             
          C  s, R–C–Cl  R  N  C  O  R  NH2
                         NaN3
                          NaCl
                                           H2O



         D  q, t
11.      (A)   5AgNO3(aq.) + 3I2 (excess) + 3H2O
                              5
                    HIO3  5AgI  5HNO3
                   It is disproportionation and redox reaction. AgI is insoluble in H2O.
                              6                                          7         4
         (B)       3K 2 MnO4 (aq.)  2CO2 (g)  2KMnO4  MnO2  2K 2CO3
                   It is dis proportionation and redox reaction. MnO2 is insoluble in water.
                     0             12                    6–6           4
                                    
         (C)                          
                   2C Na 2Cr2O7  Cr2O3  Na 2CO3  CO
                   Redox reaction Cr2O3 (green pigment) is insoluble in water.
                         2                     0          1
         (D)       CuCl2 (aq.)  Cu(s)  Cu 2 Cl 2 (s)


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                                               
12.       3I 2 (s)  6OH –       5I – (aq)  IO3 (aq)  3H 2O(l)

          G = 5 × – 50 + (–123.5) + 3 × (–233)
                    – 0 – 6 × (–150) = – 172.5 kJ/mol
                                              25
          G       = –172.5 kJ/mole = –          300  2.3 10 –3 log K
                                              3
         log K = 30

             30       10 5  10 –1
         10         =
                       [OH – ]6

                    (OH – )  10 –6 ; pOH = 6
                    pH = 14 – 6 = 8.
13.      Conceptual
14.      Molality (experimental)
              Tf       0.29
         =                  = 0.150 mole/1000 gm
              Kf        1.86
                                         moles of solute
         Molality (theoretical) =                           1000
                                        wt of solvent in gm
              1.1 1000
         =            = 0.0412 mole/1000 gm
              267 100
         Number of moles of solute particles produced by
                                   0.156
         1 mole of solute =              4.
                                  0.0412
15.      Conceptual.




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16.




17.      Conceptual.
18.      Conceptual.
                      400          400
19.       log10 P          10        10  9
                       T            400
          P  109 mm Hg
          x  9.

MATHS
             2
20. A(2t1 , t1 )
         B(2t 2 , t 2 )
                    2

         Again P(–1, 0), A & B are collinear
                   t 2  t1 2
                                     t2
                    2
                                     2

                2(t 2  t1 ) 2t 2  1
          (t1  t 2 )  –2t1t 2                   ....(i)
         Let centroid is (h, k)
                        2(t1  t 2 )
               h=                                  ....(ii)
                              3
                         t1  t 2
                          2
                                2
               k =                                 ....(iii)
                             3

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         from (i), (ii), & (iii) eliminate t1 & t2
                3h2 + 2h = 4k
         hence locus is 3x2 + 2x = 4y.
21.      iz  z
          i(x + iy) = (x – iy)
          (i – 1) (x + y) = 0
         which represents the line y = – x
          foot of perpendicular is (2, –2).
22.      | cos x  sin x |  2
                cos x  sin x
                               1
                       2
                      
          | y| .
                      4
23.      Let sin2x = a
         cos2y = b
                a 3  3ab  b3  1  0
                a 3  b3  ( 1)3  3(a)(b)( 1)
                  (a  b  1) ((a  b) 2  (a  1) 2  (b  1) 2  0
                either a + b = 1, or a = b = –1
                sin 2 x  cos 2 y  1 or cos 2 y  sin 2 x  1 (Im possible)
                sin 2 x  sin 2 y either x  n  y
          sin x   sin y x  n  y. x  y  n

24.      C (3, 2)
         r=2
                  (1, –1) is an external point hence two tangents are possible with slopes such that
                                5
                  one them is      & slope of other is not defined
                               12
          their equations are x = 1 & 12y – 5x + 17 = 0
25.      a=2
         for 3y  x  3
         use c2 = a2m2 + b2
                 5
         b2 
                 3
                       1 7
          e                .
                        2 3
                      2
26.      Let A(at1 , 2at1 )
         B(at 2 , 2at 2 )
                2
                         2
          t 2 = - t1 -
                         t1
         for AB to be shortest t1   2
           t2  2 2

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          t1t 2  4
         Hence AOB is a right angle
          Mid pt of AB is circum centre

27.      f (x) = 4x3 – 4 = 0 has only one real root
          f(x) is decreasing in (– , 1) & increasing in (1, )
         but f(1) = – 4 i.e. min. value
         so f(x) = 0 has only two real root of opposite sign.

28.      if        bc, ca, ab – G.P.
         then      bncn, cnan, anbn – G.P. n R.
                  bn cn  a n b n
         and                       cn a n
                         2
             a n  cn
                        bn
                 2
                           2(b n )2 .a n cn
         and      cn a n  n n
                           b (a  cn )
                         cn a n
           bn  2
                       a n  cn


29.      xf(x) is a polynomial of degree 5
                  f(x) is a polynomial of degree 4 such that coefficient of x4 is 1
         Again, f(x) – x = (x–1) (x–2) (x–3) (x–4)
         (A)       f(5) = 29
                         x  f (x)
         (B)       lim               lim(x  1)(x  2)(x  4) = 2
                    x 3   x 3        x 3


                        f (x)  x        (x  1)(x  3)(x  4)
         (C)       lim             lim                        1
                   x 2 x  2x
                           2        x 2           x
         (D)       f (2)  3


30.      (A)       The point (1, , 1) lies on the plane
                   i.e. 2    6  0
                  4
                   The line is also perpendicular to the plane
                    2  6  12  0
                    3
                       2 1 1
         (B)         1 3 1  0
                        1 1
                    3

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                  For   1, the three planes passes through same line.
                     1
                      a b 1
         (C)          1 b c 0
                      a 1 c
                   1  ab  bc  ca  2abc
                     1  ab  bc  ca
                                      (a 2 b2c2 )1/ 4
                             4
                      abc
                           abc
                        2
                   abc  4
                   1     1 1
         (D)               1  0
                      2   2 1
                   (  ) (  1) (  1)  0

                     1 and   1
                   1
          sin 2 x 
31.               x 2  sin 2 y
             sin x
           2
               x
         LHS  1 and RHS  1.
                                                           3
           sin 2 y  1  sin y   1  y                 ,
                                                          2 2
                              1
         and      sin 2 x          …..(1)
                              x2




                                                            x
         There are four values of x satisfying equation (1)
               ordered pairs = 8
32.         ˆ 0
         r 
         and               ˆ                      ˆ
                  2r  (r )  3r  (r  )  r 
                                          ˆ
                        ˆ              ˆ                                  ˆ
                   2((r )r  (r  r ) )  3 ((r ) r  (r  r ))  r 
                                                    ˆ              ˆ
                            ˆ
                   2 | r |2   3| r |2   r 
                                           ˆ        ˆ
                                 ˆ            ˆ
                  3 | r |2   r   2 | r |2 
                           ˆ
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                   9 | r |4  | r |2 2 | r |4
                                  1
                   | r |
                                   7
                1
33.       In   1.(1  x 3 )n dx
                0

                                              1
         In = (1  x )  x   3n  (1  x 3 ) n 1  x 3dx
                            3 n        1
                           0
                                              0

                    1
          In  3n  (1  x 3 )n 1  (x 3  1  1)dx
                    0


          In  3n  –In  In 1 

                   (1  3n)I n  3n  I n 1  0

                 n=4
34.       Tn = (n 2 + 3n + 3) ×(n + 1) !
          = (n 2 + 4n + 4) ×(n + 1) ! - (n + 1) ×(n + 1)!
          Þ Tn = (n + 2) ×(n + 2)!- (n + 1) ×(n + 1)!
          Tn = 1.1!- 0.0!
          T2 = 2.2!- 1.1!
          T3 = 3.3!- 2.2!
         .
         .
         .
         .
          Tn +2 = (n + 2) × + 2)!- (n + 1) (n + 1)!
                           (n
         Adding all the above equation
                        n
          Þ         å
                    n =- 1
                             (n 2 + 3n + 3) (n + 1)! = (n + 2) (n + 2)!
                        n

                        å     é(n 2 + 3n + 3) (n + 1)!ù
                              ë                       û
          Þ         n =- 1
                                                          = (n + 2) = 7
                                       (n + 2)!
          Þ        n=5
35.       f (x) =| cos | x ||
                            ì 2x ,                x³ 0
          g(x) = x + | x |= í
                            î 0 ,                 x<0

                                                              –2                              –2




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                                     4
36.       log 2 (x 2  1)                    4
                              log 2 (x 2  1)
                           1
         and        y2       2
                           y2
                                                  4             1
                    log 2 (x 2  1)                     & y2  2
                                           log 2 (x  1)
                                                    2
                                                               y

                   log 2 (x 2  1)  2        & y  1

                   the point are               
                                             3,1 ,                        
                                                      3, –1 , – 3,1 & – 3, 1 .             
37.       A1OA3  2 / 7
          A3OA 7  4 / 7
          A1OA 7  6 / 7
                                                                           A4

                                                                                  A3


                                                                                       A2
                                                            0        /7


                                                                                  A1

                                                                           A14
         Consider
                                                                Aj



                                                      R


                                                      
                                                 0                               Aj
                                                                R
                    R 2  R 2  (A i A j ) 2
          cos  
                              2R 2
                              (A i A j ) 2
           cos   1 
                               2R 2
                   (Ai A j )2  (1  cos )2R 2
                    (A1A 3 ) 2  (A1A 7 ) 2  (A 3A 7 ) 2
                  cos 2   cos 4   cos 6 
           2R 2 1        1      1    
                      7         7        7 
                  1
           2R 2 3  
                  2
           7R 2
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38.      if K = 49, then both roots are 7
         If K = 13, then roots are 1, 13
          there are 2 values of K

PHYSICS
39. V = –2x + 10
     dx
          10  2x
     dt
     x            t
         dx
                 dt
      10  2x 0
     0
    x = 5 (1–e–2t)
    for x  5, t 
                               2
                      2R 
40.      I0 = IC  M       
                       
                      4MR 2              4
         IC = MR 2            MR 2 1  2 
                         2
                                       
41.      Level of water will rise till volume of water entering into container/s = volume of
         water
         emerging through hole/s then level will become constant.

42.       t(  1)  
                 
          t           2 .
             1.5  1

43.      In S.H.M Umax = mechanical energy (M.E)
               Kmax = M.E – Umin.
               U = 10 + 2x2
                      dU
               F= –         4x
                       dx
         Force constant of S.H.M = 4N/m
         Mean position is at x = 0
               Umin = 10J
               Kmzx = 18– 10 = 8J
               Umax = 18 = 10 + 2x2
          amptitude = 2m
                 2radian / s
                     2
               T         s
                     
44.            at t = 0, VC = 0, VL  
               at t  VC  , VL  0
                                                
          at t = 0, I         and at t , I     .
                             R2                  R1


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45.      VA – 1 × 1 = V0                                    C

               VA = 2V
          V0 = 1V                                          I        3
                                                            0
               VB – 2I1 + 6 = V0                          I    6v
                                                                2
                                                      1
          I1 = 0.5A                                     1A          I
               I3 = I1 + I2 = 1.5A                2v
                                                     A             – B
                                                                    4v
                                                                            1



               V0 – 2 × 1.5 = VC
          VC = – 2V.
46.         E is perpendicular to the line joining A and B
                Line joining A and B will be equipotential line and work done by E on the
         particle from A to B = 0.

47.      For the determination of flux of magnetic flux through outer coil consider I flowing in
         outer coil find flux through inner coil
              I a 2
          o
                2b
                            a 2
           MI  M  0
                            2b

48.      (A–q,s); (B–p); (C–r); (D–t)

49.      Using conservation of angular momentum about 0
                                    before collision just after collision
                                             O                   O
                                      x
                                m                           P
                                     v0
                   ML2        
          mV0 x        mx 2                                    …(i)
                   3          
                                          L
                                              M
         Final momentum pf  mx              dy  y             …(ii)
                                          0
                                              L
         (i) is (ii) 
                      ML  mV0 x
         pf   mx 
                       2  ML2
                                  mx 2
                               3
                                   ML 
                              mx 
                                         mV0 x
                                     2 
          p  p f  pi                         mV0
                                 ML2
                                      mx   2

                                  3
            ML            mV0 ML2
                  mV0 x 
          2                   3
                   ML2
                         mx 2
                     3

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                                2L
         For p  0 x 
                                 3
                      2L
          p  0        x
                       3
                     2L
         p  0 x 
                      3
         About 0 resultant external torque on ball + rod system is zero for any value of x. So
         angular momentum is conserved about 0.


50.      If pv x  constant then
                R       R
         C          
               1 x 1
          3R       R      R
                      
                   1 x 1
           2     7
                 5
          x2

                        r

51.       E4r   4x 2dx
                2

                        0

            Cx    n

                        r

          E4r   4x 2dxcx n
                2

                        0
                n 1
          Er
         If E  r 2  n  1

52.          Peak value of I  3 2A
          R.M.S value = 3A

53.

                                    O
                             2              A
                                        R
                             3 1m

                                S



                         4R 2
         Ratio 
                   2R 2 (1  cos c )
                       2
                
                  1  cos c

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      The Key & Hint paper contains 28 Pages                   NFT-7 PAPER I & II Dt : 13-02-2011



                    3
          sin  c 
                   2
                  1
         cos c 
                  2
         Ratio = 4

               dU
54.       F         m 2 r
                dr
          mv 2
                m 2 r
            r
         V  r                                           …(i)
                 nh
         mvr 
                 2
                nh
         V                                               …(ii)
               2mr
         (i) & (ii) 
                        1
             nh  2
                            1
          r         r  n2
             2m 
                   
          x2


                            wt. of solid in air  wt. of solid in liquid
55.      R.D of liquid 
                            wt. of solid in air  wt. of solid in water
              15  9 6
                     2
              15  12 3

          dy
56.            Particle velocity
          dt
           dy 
                    2  3 m/s
           dt  max
                        3
         Wave velocity  m/s
                        1
                      6
          Ratio       2
                      3




                                                                                           27 Of 28
      The Key & Hint paper contains 28 Pages   NFT-7 PAPER I & II Dt : 13-02-2011



         M     q
57.         
         L 2m
               q
         M        I
              2m
            q mR 2
                     
           2m 2
              qR 2 
         M
                4
          n  4.




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