NPT-2_14.11.10_H_S PAPER-I_II

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					                                                                                               CODE–NPT-2-PAPER-I&II


                                                                                                    CODE        NPT-2




                                        NARAYANA
                                        I   I     T             A     C        A         D    E       M         Y

                      NARAYANA PART TEST - 2
                                     PAPER – I & II -:- Date: 14.11.2010

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


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


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CODE–NPT-2-PAPER-I&II



                                   HINTS & SOLUTION
                                       PAPER - I
                                      CHEMISTRY
1.      Mg+1 3s1
        Al+1  3s2
        Si+1  3s2 3p1.
2.      Pseudohalide ions contain two or more electronegative atom, having similar
        property of halide Ion, Example CN–, NC–, SCN–, OCN–. Dimer of pseudohalide
        are pseudohalogen. Example (CN)2, (SCN)2, (OCN)2.
3.      [IBrCl]– is sp3d hybridized with linear shape.
4.      In graphite each carbon is left with one spare electron in its p orbital. This
        electron then overlaps with each other to form a  bond. Hence C—C distance in
        graphite is shorter (1.42 Å) than that of diamond (1.54 Å).
                              OH                    Two Boron are sp2 and
                              B                     Two Boron are sp3
                        O           O               hybridized in borax.


       2Na
             +
                 HO     B         O B OH     8H2O
5.

                        O           O
                              B
                              OH

                        Structure of Borax
             MnCl4  paramagnetic (5 unpaired e– ), sp3
                         2
6.      1.
                             2
        2. Cu  NH3 4  paramagnetic (1 unpaired e–), dsp2
                          
        3.  CoF6  paramagnetic (4 unpaired e–), sp3d2.
                   3


                                          
7.      (A) cis form of Co  en 2 Cl2  can show optical isomerism
                                       
                                              
        (B) trans form of Co  en 2 Cl2  does not show optical isomerism, cannot have
                                           
            optical is omerism Co  NH3 3 Cl3 
                                               
                                  
        (D) 1. Co  en 2 Cl2  Geometrical, only cis form optical
                               
            2. Co  NH3 3 Cl3  Geometrical, No optical
                                   
            3. Co  H2O4 Cl2  Cl Geometrical, no optical.
                                   
8.        n  n  2
          6.92 BM
        n=6
        chromium is in zero oxidation state.
        Cr24  4s1 3d 5 .

 2
                                                                       CODE–NPT-2-PAPER-I&II




9.    Ma2b2, M(AA)b2 complex can show geometrical isomerism
      M(AA)2, Ma3b type complex can not show geometrical isomerism.
10.   All these processes are not involved during extraction of aluminium metal from
      bauxite ore.
11.   4NaCl  K 2Cr2O 7  6H 2SO 4  4NaHSO 4  2KHSO 4 3H 2 O  2CrO 2Cl 2
                                     
                                                                            chromyl chloride 
                                                                            deep red 
                       
      CrO2Cl2  4NaOH  Na 2CrO4  2NaCl  2H2O
                                   yellow 
      Na 2CrO4   CH3COO2 Pb  PbCrO4  2CH3COONa
                                
                                                yellow ppt 
      heavy metal does not show this test like Hg, Pb, Sn, Sb etc.
12.   A, C, D methods are correctly matched. Van Arkal method is for Nickel
      purification.
13-15 MnSO4 + 2Na2CO3 + 2KNO3  Na2MnO4 + 2KNO2 + Na2SO4 + 2CO2
        (A)                                    (B)
      2Na2MnO4 + 2H2SO4  2NaMnO4 + 2Na2SO4
                                       (C)
      MnSO4 + 2NaOH  Mn(OH)2 + Na2SO4
                                   (white)
      Mn(OH)2 + [O]  MnO(OH)2
                               (D)
      MnSO4 + (CH3COO)2Pb  PbSO4 + (CH3COO)2Mn
                                        white ppt.
16.   2KClO3 + 4HCl  2KCl + 2ClO2 + Cl2 + 2H2O
      Euchlorine (ClO2 + Cl2)
17.   Pb3O4 + 8HCl  3PbCl2 + Cl2 + 4H2O
      2AgClO3 + Cl2  2AgCl + 2ClO2 + O2
18.   2ClO 2  2KOH  KClO2 + KClO3 + H2O




                                  MATHEMATICS
      xx      3
21.       < sinx < x  x > 0
       6
                                              x  x3 
                                            f         
           x  x3 
                     > f (sin x) > f(x)  
                                                   6  f (sin x)
       f                                                      1
           6                                 f (x)        f (x)
                                              f (sin x)
      According to Sandwitch theorem lim                 1
                                         x     f (x)
           4
22.   I =  p(x) dx
           1
           4
      I=    p(5  x) dx
           1


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CODE–NPT-2-PAPER-I&II


             4
        2I =  (p(x)  p(5  x) dx
             1
        Now p(x) = p(5–x)
         p(x) + p(5 – x) = C
        C=1+7=8
                 4
         2I =  8dx = 24  I = 12.
                 1

                f (x)  3x  1 
23.     (x) =                  1  f is odd.
                  2  3x  1 
24.     y3 – 3y + x = 0
        x = –y3 + 3y < –2
         y3 – 3y – 2 > 0
        (y + 1)2 (y – 2) > 0
        y>2
         dy       1
                        0
         dx 3(1  y 2 )
                           1
25.     f(x) = cot–1x + ln x
                           2
                      1      1
        f(x) = –          
                   1  x 2x
                         2


           2x  1  x 2     (1  x) 2
        =                  
            2x(1  x 2 ) 2x(1  x 2 )
        f is increasing  fmax = f( 3 )
26.     Df = {–3, –2, –1, 1, 2, 3}
                   1 5          
        Range =  , ,1  2 
                   9 4          
                 dy dx        1 1
27.     D.E. is 2  2  y    dy = 0
                  y x         y x
             dy dx
                
             y2 x 2                     1 1 y2
                      y dy  0  – ln      =C
              1 1                       y x 2
                
               y x
              xy   y2
         ln         =C
             xy 2
        1

         (x ln x  2x  2x
                               2
28.                                ) dx
        0

           1      2 3  12  8 7
        =    1              
           4      3      12      12
29.     f(x) = g(x) = h(x) = p(x) = x




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                                                                         CODE–NPT-2-PAPER-I&II


      1                            1      1
             dx            x                 x 2 dx
30.     (1  x 2 )n  (1  x 2 )n  2n  (1  x 2 )n 1
       0                                0
                                  0
      In = 2–n + 2n In – 2n In+1
       2nIn+1 = 2–n + (2n – 1) In
      Putting n = 1
              1           1
      2I2 =       + I1 =     + /4
              2           2
             1 
      I2 = 
             4 8
              1 1 x2
31.   lim
      x 0       x
                   x2    1              |x|
      = lim                = lim
         x 0
               1 1 x 2 x   x 0
                                  x (1  1  x 2 )
                  1
      L.H.D. =
                    2
                   1
      R.H.D. =
                    2
              x
32.   f(x) =  e t sin(x  t) dt
              0
                  x
      f (x) =  et cos(x  t) dt
                  0
                            x
      f (x) = ex –  e t sin(x  t) dt
                            0

      f (x) + f(x) = ex
34.   f(x) – x2 = (x – 1) (x – 2) (x – 3)
      f(x) = x2 + (x – 1) (x – 2) (x – 3)
      f(4) = 16 + 6 = 22
35.   f(x) = x2 + (x – 1) (x – 2) (x – 3)
      f (x) = 2x + (x – 2) (x – 3) + (x –1) (x – 3) + (x – 1) (x – 2)
      f (2) = 4 – 1 = 3.
36.   For domain
      0  [2x]  1 & 0  [2-x]  1  x  (-1, 1)
37.   For x (0, 1)
      [2x] = 1 & [2-x] = 0  f(x)  {/2}
      f(0) = 0.
38.                   /2



              O




                                                                                           5
CODE–NPT-2-PAPER-I&II



        (A) f(x) = e x 3x  2 (3x2 – 3)
                                 3
39.
        = 3 e x 3x  2 (x2 – 1)  0 x(–, –1]  f is increasing
               3



        Range = (0, e4]   = 4


        (B)                                                         y = (1 - x)2
                   y = x2

                                                          y = 2x (1 - x)


             ab ab
        (C)           = /4
              8     2
             b          3
        m = tan    4m = 3
             a          4
        (D) f (f–1(x)) = x
                                                                                   1
        f  (f–1(x)) . (f–1(x)) = 1  (f–1(x)) =
                                                f (f (x))                         1


         d –1                 1          1     1
           f (x)|x=–2 =                     
        dx              f (f ( 2)) f ( 1) 14
                             1

                                               
                        dx        1                        dx
40.     (A)        e1x  e3x  e2
                  1
                                                e  1
                                                            e1x
                                                        x 1

                   
             1      dt
              2  t
        =                put t = x – 1
             e 0 e  et
                   
             1 e t dt   
              2  2t
        =             = 2
             e 0 e  1 4e
                           x

                            (1  sin 2t)
                                             1/ t
                                                    .dt                                              sin 2x
                                                                                              lim
        (B) lim            0
                                                           = lim (1  sin 2x)      1/x
                                                                                         =e   x 0      x
                                                                                                              = e2
                  x 0                   x                         x 0
                           5
        (C) I =  {x  0.5}dx
                           1
        Let x - 0.5 = t
        dx = dt
            4.5                      4                         1
        =  {t}dt =   {t}dt = 4  {t}dt = 2
            0.5                      0                         0
                       100 
        (D) I             
                           0
                               (tan x  tan 2x  ... + tan 10 x) dx
                       
        = 100  (tan x  tan x  ...  tan10x)dx = 0
                       0




 6
                                                 CODE–NPT-2-PAPER-I&II



                                    PHYSICS
41.   The period of oscillation is
               2            2 
          T         2       
                            K 6
       K = 288 N/m
      The force needed is
          F = Kx = 288 × 2 × 10–2 = 5.76 N.
                 T         500
42.       v                   50 m / s
                          0.2
               1
          P  Va 22
               2
               2v
      and  
                 
      putting the values
          P = 197 W.
                P                5
43.       I          
              4r  2
                          4 (20 102 ) 2
      Also, I = 222a2v
                        I
       A
                 2  v
                     2 2


      Putting the values, we get
          A = 1.15 × 10–6 m.
44.   Given
          A1V1 = A2V2
       L2 2gy  R 2 2g  4y
                 L
       R            .
                 2
                       k
45.        2f 
                      m
       K  (2f ) m2


                      1
      Total energy  KA 2  0.5J
                      2
                1      1 1     1         1
          A                     10       .
                K 2f 0.1 2 10         2 10
46.   We have from F = 0
         N1 + 2 N2 = mg
         N2 = 1 N1




                                                                   7
CODE–NPT-2-PAPER-I&II


             2N 2
        A            N2

                           N1
                      
                      1N1 B
             Mg
        And from  = 0
            (mg cos )  (1N1 ) sin   (N1 ) cos 
                         2
        eliminating N1 and N2 we have
                    1  1 2 3
            tan               .
                       21      2
47.     Power = (energy in unit length) × wave speed
                1M 2
            P        v max  C
                2
              1M 2 2
                  AC
              2
                1 45 103 2                   40
            3               A (2 30)2
                2 15                     (45 103 /15)
                     T
            c
                   M/
         A = 4.88 × 10–4 m.
48.     Due to symmetry, the oscillation of the two masses are the same. Consider one of
        them and write down the equation of motion
               d2x
            m 2  Kx  K '(x  x)  (K  2K ')x
                dt
        where x is the displacement from the respective equilibrium position. Then the
        angular frequency of oscillation is
                    K  2K '
            
                       m
        and the period of oscillation is
                         m
            T  2              .
                      K  2K '
49.     The equation of motion are
            f = ma
                        mr 2
             fr  I        
                         2
                 f     3
         a              1.7 m / s 2
                 m 1.8
                   2f
                     17 rad / s .
                   mr



 8
                                                                    CODE–NPT-2-PAPER-I&II


50.   As the two ends of the strings are fixed
           n  2 ;  length of string.
      Let wavelength corresponding to 160, 240, 320 Hz be 1, 2 and 3 respectively
      then
           n1 = (n + 1) 2 = (n + 2) 3 = 200
           160 1 = 240 2 = 320 3
       n=1
      and 1 = 100 cm, 2 = 67 cm and 3 = 50 cm.
51.   For successive resonaces, the air column differ in height by half wavelength.
                 
           d
                 2
           d = 48.45 – 15.95 = 80.95 – 48.45 = 32.50 cm
            = 2d = 65 cm.
           v =   v = 330 m/s.
          
      As  16.25 cm and 16.25 cm – 15.95 cm = 0.3 cm
          4
      Hence uppermost antinode is located at 0.3 cm above top of pipe.
52.        V2 = n2 (8ax – x2 – 12a2)
               dv
       2V            x 2 (8a  2x)
               dx
            d2x                                          d2X
              2
                    n 2 (x  4a)                         2
                                                               n 2 X
            dt                                           dt
      where X = x – 4a
       motion is SHM with  = n and mean position at x = 4a.
      for V = 0 x = 2a, 6a
       amplitude = 6a – 4a = 2a.
      Time taken to move from x = 4a to x = 6a i.e. from mean position to extreme end
         T 
                 .
         4 2n
53.   In case of ring f2 = 0.
          there is no slipping anywhere hence friction between both cylinders and rod is
      same.
                                     N2            N1


                                  A              B

                                          f1            f1
                                  A              B
                                          f2            f2

                            2f1                         F
      For rod F – 2f1 = Ma …(i)
      For cylinder
                 f
          a cm  1
                m
                a      a f1
      but a cm            …(ii)
                2      2 m
                                                                                      9
CODE–NPT-2-PAPER-I&II


        from (i) and (ii)
                   F
            a           3 m / s2
                mM
                 m
        and f1  a  1.5 N .
                 2
                                                                         Mg     3Mg
54.         f1 is constant but N1 & N2 are variable. They vary between      and     .
                                                                          4      4
        Hence  is variable.
                     f    1.5
           min  1            0.2 .
                   N min  Mg
                           4
                              F 2L
55.        V  2as  2                  3.46 m / s .
                           mM 3
56.     The T is as shown.

                  B

                         A
                m 2        13
          IA        m 2  m          2

                 12        12
                1 2
          IB  m
                3
                       17
         I  IA  IB  m 2 .
                       12
57.         C.M. of B raised through   (1  cos )
                                     2
            C.M. of A raised through (1  cos )
         p.e. of system  mg (1  cos )  mg (1  cos )
                             2
             3
           mg (1  cos ) .
             2
58.     From energy principle
             17 2  d  3mg
                           2

                  m                (1  cos )  constant
             24       dt       2
        differentiating w.r.t. time and rearranging
             d 2     18 g sin      18g
                2
                                        [for small ]
             dt       17             17
                      17
         T  2           .
                      18g
59.          F + f = ma




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                                                                    CODE–NPT-2-PAPER-I&II




                 h



                     f
          Fh – f.R = I
          a = R
                        I 
              F h        
          f 
                       mR 
                        I
                 R
                      mR
              I
     for h        friction will be backward
             mR
               I
     for h         friction will be zero
              mR
               I
     for h         friction will be forward.
              mR

                                     PAPER - II
                                    CHEMISTRY
1.   Borax is Na2B4O7. 10H2O
     The steps are
                                                              CoO
                                                
     Na2B4O7.10H2O  Na2B4O7  NaBO2 + B2O3  Co (BO2)2
                     (blue)
2.   The reaction is
     [Fe(CO)5 + 2NO  [Fe(CO)2(NO)2] + 3CO
     NO acts as a three electron donor. Therefore three CO groups are replaced by two
     NO groups.
3.   If Xenon (+VI) disproportionate to give Xe (zero), the other Xenon is expected to
     be Xenon (+VIII)
     2Na[HXeO4] + 2NaOH  Na4XeO6 + Xe + O2 + 2H2O
4.   Information based.
5.   The salt is soluble in water when hydration energy is more than lattice enthalpy.
6.    Ma 5b , M  AA 3  type complex do not show geometrical isomerism.
                          
7.   XeOF4 is square pyramidal
              O
       F               F

               Xe

     F                   F


8.   Electrolysis of liquid H2 may not give any result. As the boiling points of H2
     (20.4K) & D2 (24.6K) are different, distillation of liquid hydrogen will leave
     behind small amount of liquid D2.
                                                                                     11
CODE–NPT-2-PAPER-I&II



9.      BrF5 + 3H2O  5HF + HBrO3
        In ClF3, Cl is sp3d hybridized. It is T-shaped
                 F


                 Cl
                                 F
                 F
        Equatorial Cl – F bond distance is shorter than the other two.

10.     CO2 and SO2 are acidic oxides and can be absorbed by bases like CaO, NaOH.
        CO gets absorbed by ammonical CuCl. Both CaO and P4O10 absorbs H2O.
11.     Ferrocene is a  bonded organometallic substance.
        In metal carbonyls, CO acts as a pi- acceptor.
12.     Mass of Mg2+ in 100 mL water = 0.216 mg
        Mass of Mg2+ in 106 mL (106 g) water = 0.216 × 104 mg
                                                  0.216 104
        Mass of Mg2+ in 106 mL (106 g) water =                g
                                                      103
                             2.16
        Moles of Mg2+ =            0.09
                              24
        Moles of CaCO3 = 0.09
        Mass of CaCO3 par 106 g H2O = 0.09 × 100
        = 9 g.
13.     It is a cyclic structure. Two oxygens of each tetrahedron are shared by the other.
        Number of Si atoms = 3
        Number of O atoms = 9
        Silicte in (Si3O9)-6.
14.     The structure of isomers are
                        Py                        Py
                              CH 3NH2                                           Py
                                                          CH 3NH2
        Ox                                                          CH 3NH2
                       Co               Ox      Co
                                                                               Co       Ox
                             CH 3NH2                     Py
                                                                      Py
                  Py                          CH 3NH2
                                                                              CH 3NH2
15.     4FeCr2O4 + 16NaOH + 7O2  8Na2CrO4 + 2Fe2O3 + 8H2O
        n-factor of O2 = 4
        n-factor for FeCr2O4
        Fe+2  Fe+3 + e 
        2Cr+3  2Cr+6 + 6e 
        n-factor = 6 + 1 = 7.
16.      Ti (zero)  Ar 3d2 4s2
        Ti (II)  Ar 3d2
        Ti (III)  Ar 3d1
        Ti (IV)  Ar 3d0
        Maximum number of unpaired electron are observed in Ti (0) and Ti+2.


12
                                                                         CODE–NPT-2-PAPER-I&II



                                              1 1
      Maximum spin multiplicity = 2S + 1 = 2    + 1 = 3.
                                              2 2
17.   When dry iodine reacts with ozone it forms I4O9. It is yellow in colour.
                        O      O
       Solid SO3            S
                        O      O
18.    or  SO3      O             O
       Molecule         S      S
                           O      O
                      O
19.   Oleum H2S2O7
           O
      O          O
            Cr        CrO5
      O          O
      Sulphurous acid H2SO3  ln + 4 oxidation state
      Marshall acid H2S2O8
      Caro acid H2SO5
      Sulphuric acid H2SO4
      Chromic acid H2CrO4.

                                        MATHEMATICS
20.   sin51x (sinx)49
      = (sinx . cos50x (sinx)49
      + cosx sin50x (sinx)49
           d
              ((sin x)50 sin 50x)
      = dx
                     50
21.   It is obvious from the given functional relation that degree of polynomial is 1.
      Putting
      f(x) = ax + b
      we get a = 2, b = -3. i.e., f(x) = 2x - 3.
                                    f (x)  f (0)
22.   According to L.M.V.T                           | f (C1 ) |1
                                           x
       |f (x)|  1
                f (x)  f (0)
      Again                    f (C 2 ) 1
                      x
       |f(x)  |x|
23.   0  {x} < 1
      0  sin {x} < sin 1
         1         1
                       
       sin1 sin{x}
       1 
       sin{x}   {1, 2, 3, ... }
                 
            f (a  h) g(a  h)  f (a) g(a)
24.   lim
      h 0                  h
               f (a  h)  f (a)                 g(a  h)  g(a)
      = lim                      g(a  h)  lim                  f (a)
         h 0          h                    h 0        h
                                                                                          13
CODE–NPT-2-PAPER-I&II



                 f (a  h)  f (a)
        = lim                      g(a)  g(a) f (a)
            h 0         h
25.     f (x) = 2x ln2 – 2x
        f  (x) = 2x (ln2)2 – 2
        f (x) = 2x (ln2)3 > 0
        f(0) = ln2)2 – 2 < 0
        As f (x) is increasing and f  (0) < 0
        f (x) has only one root i.e., f (x) has at the most two roots.
        f (0) = ln 2 > 0
        f (1) = 2ln2 – 2 < 0
        f (4) = 16 ln2 – 8 > 0
         f (x) = 0 has exactly two roots
         f(x) = 0 has at the most three roots
                  f (0)  0 f (4)  0
        again                            f(x) = 0 has three roots.
                  f (1)  0 f (5)  0
                     f (x)    if f (x)  Q     x       if x  Q
26.     If [f(x)] =                       
                    1  f (x) if f (x) Q 1  (1  x) if x  Q
27.    Clearly from graph
            / 4         / 2
                                                        cotx               tanx
       I =  tan x dx   cot x dx
              0                / 4
                                            
       = [ln | sec x |]0 / 4  [ln | sin x |] // 2
                                                  4
                                                          O       /4      /2
       = 2 ln 2
       (d) obviously the period of tanx and
         cotx is 
28.     f (x) = (x – 1) (x – 2)2  x = 1, 2
        Around x = 1, f(x) changes sign from –ve to +ve f(x) is minimum at x = 1, f(1) =
           17                                  4              4
              f(x) changes sign around x = 2, , x = 2, x =     are the point of inflexion
           12                                  3              3
                  4 4         112
        f(2) =  , f             .
                  3 3           81
29.    (A)
        f  x  x  2 x
                         1
        f '  x   1
                          x
        f  x  is strictly increasing function.
        So least value is f  0   0
        (B)
                 x 1          2
        f x          1
                 x 1        x 1
                     2
        f 'x             0
                  x  1
                          2


        f  x  is strictly increasing function.
14
                                                                              CODE–NPT-2-PAPER-I&II


      So least value is -1
                     1 x  x2          2x
      (C) f  x                1
                     1 x  x 2
                                     1 x  x2

      f '  x   2
                            
                       x2  1
                                  0
                                           
                                              
                                2
                     1 x  x2
      So f  x  is strictly decreasing function
      Least value  f 1  1
      (D)
      f  x   x4  2x2  5
      f '  x   4x3  4x  4x x2  1                
      f '  x   0  x  1,0,1
      f  1  f 1  4
      f 0  5
      f  2  f  2  13
      So, least value = 4.
30.   (A) f ''  x   0  f '  x  is increasing function
      g'  x   f '  4  x   f '  2  x 
      If g'  x   0  f '  2  x   f '  4  x   2  x  4  x or x  1
      (B) f '  x   3  x  1 x  1  f '  x   0 has roots x  1 1
                                                                        ,
      f  x   0 will have exactly one real root if f  1 f 1  0
       a  2a  2  0  a  2 or a  2
      (C) f '  x    sinx  a2  0x  R
       a2  sin xx  R
                                                                        1
      (D) f '  x   2e x  ae  x  2a  1  2e  x  e x  a   e x  
                                                                        2
      f '  x  0
       ex  a  0x  R  a  0

              1  x 1
       2
31.     x
      1/2 
            x   e x dx

                    1  x 1
           2
      =      x 1  2  e x dx
           1/2 
                    x 
                     1 2         2         1
                x                    x
      I = xe         x
                            e            x
                                               dx
                      1/2       1/2

           3
       I = e5/ 2 .
           2

                                                                                               15
CODE–NPT-2-PAPER-I&II



                     x3        x 1
32.     f(x) =  2
                px  qx  r x  1
                     3x 2 ,     x 1
         f (x) = 
                    2px  q, x  1
                  6x x  1
        f (x) =            . If f(x) exist xR then
                  2p x  1
        p+q+r=1
        2p + q = 3
        2p = 6
         p = 3, q = -3, r = 1
        |p| + |q| + |r| = 7


                                           
                                                n
33.     If n is even               3  1  ( 3  1) n = even integer
                       n
         3  1 = odd integer
              
        { 3  1}n = 1 - ( 3  1) n

                                                         
                                            n                   n
        If n is odd                3 1                 3 1       = even integer
                       n
         3  1 = even odd
              

         3 1   3 1
                       n                            n




         L = 0, M = 1.
        e 3                        1

         f         (x) dx =  tf (t) dt
               1
34.
         2                          0
        1                                       1

         tf (t) dt  | tf (t) |   f (t) dt
                                        1
                                        0
        0                                           0

        = e + 3 -  et  t 2  t 
                                                     1

                                                    0

        = e + 3 - (e + 2 - 1) = 2
35.     Domain = [-1, 1]
                1                              (x  1)
        f(x) = (sin-1 x + cos-1x + tan-1x) +
                                            (x  1) 2  9
                      3 2        47
        fmax = f(1) =        =     =M
                      4 13       52
                                                                                             A
                                  1                                                  B
36.    Required area = 4 – 4 ×      × /2
                                  2
       = 4 – .

                                                                                         C   D




 16
                                                                       CODE–NPT-2-PAPER-I&II


37.   The given sum has 63 terms, each of which equals either [x] or [x+1] . Moreover
      63 × 8 < 546 < 63 × 9  [x] = 8.
                 49                 50 
      i.e.,  x       = 8 and  x  100  = 9
              100                      
       8.50  x < 8.51
                              [100x]
       [100x] = 850                 5
                                170
38.   Here f(x) is divisible by (x – 16)
       f(2x) is divisible by (x – 8)  f(x) is divisible by (x – 8)
       f(2x) is divisible by x – 4  f(x) is divisible by (x – 2)
      i.e., f(x) = (x – 2) (x – 4)(x – 8) (x – 16) (x)
      Putting in the given equation we get (2x) = (x) = C = 1
       f(x) = (x – 2) (x – 4) (x – 8) (x – 16)

                                       PHYSICS
39.   Let the elastic constant of the rubber pad be K, then
                         K 8 9.8
          Kx  mg                    98 s 2
                         m x 0.1
                                                K
       natural frequency of system              9.9 s 1
                                                m
      Hence when the motor is rotating at a rate
            60  9.9
                         94.5 rpm
           2      2
      resonance will take place and the motor will exhibit the largest vertical vibration.
               f
40.       a   g
              m
                                              

                                                    V
                                   f
                (mg)R 2g
                          
               I mR 2 / 2       R
          V = V0 + at1
                V V
       t1  
                 a g
           = 0 + t2
                 R
       t2  
                 2g
      cylinder stops at t1 = t2
                         2V
      which gives R         .
                          
41.   After striking with P, length of pendulum will be (L–x).
          T  length
                        1
          T'  L x        2
                  
          T  L 
                                                                                        17
CODE–NPT-2-PAPER-I&II


                                                 1
                       T T ' T T  L  x  2 5T
        required time                       (given)
                        2 2 2 2 L              8
                       15L
        or solving x      .
                        16
42.        mg – N = m2x  N = mg – m2 x
           N = 0, when mass leaves the contact.
                                      N

                                    mg
                                                     x
                                                       e
                                                      mna
                                                      oii n
                                                     psto
                  g    8
         x          2
                  2
                      4
            v = speed when mass leaves the contact.
                                      82
              A2  x 2  2 1 
                                     164
                               8     v2     g    42     g2 
        Maximum height  2                        1     
                              4 2g 42 2g  164 
              22     g
                   2  2.13m .
               g     8
                 3
43.          n      1.5s1
                 2
                   2           2
        also       x               0.01    0.16 cm
                               8 
            V = n = 1.5 × 0.16 = 24 cm/s.
        The wave equation is
                              
             y  2sin  3t  x 
                             8 
        For x = 4 cm at t = 0
                            4 
             y  2sin  3       2cm .
                             8 
44.     Detector moves away from source and source moves away from detector till the
        semi–circle is completed.
                                 v o
                                 s s
                                  c                  v o
                                                      c
                                                     0 s

                                                    



         f < f
        After semi–circle is completed they are moving towards each other


                                         vcos v0cos
                                          s

                                               

         f > f.
18
                                                               CODE–NPT-2-PAPER-I&II


45.   Upthrust = Net weight
          3
             LAdg  (d1  d 2 )LAg
          2
                          3
       d1  d 2  d
                          2
      Also d2 – d1 = d
                          d
      Solving d1  .
                          4
                                 C  Vs 330  33
46.        in front of source                    0.9 m .
                                   n0       330
                       C  V0          330  66
          n '  n0                330           440 Hz .
                       C  Vs          330  33
                          C  Vs 330  66
           reflected                        0.6 m .
                              n'       440
47.   Let as is the acceleration of system. From FBD we have
                                          
                                                    as
                               V        2R
                                                T
                                       R T               m
                                                
                                         F1         F2
                                        ta )
                                      (s tic       in tic
                                                 (k e )
          F2 – F1 = (M + m) as …(i)
                       1
          F1R = I = MasR …(ii)
                       2
                1
          F1  Ma s …(iii)
                2
          F2 =  (mg – T sin ) …(iv)
      Putting F1 and F2 in (i)
                               1
           (mg – T sin )  Ma s  (M  m)a s …(v)
                               2
      For plate
          Tcos – F1 = Mas
                      3
       T cos   Ma s (using (iii))
                      2
                3Ma s
          T            …(vi)
               2cos 
      Putting (vi) in (v) and solving
          a s  2.08 m / s 2
             V2
         S        0.98 m .
             2a s
             2s
          t     0.98 s .
             v0



                                                                                19
CODE–NPT-2-PAPER-I&II


48.     Decrease in P means height of mercury column will increase.
        So point A will move up.
        Point B and C will remains stationary. Point D will move down as level of
        mercury in container will go down.

49.     Conceptual.

50.         PA = pB = atm
                         H
            A
                    h
              B
             1             1
              PVA  gh  2 …(i)
                   2
                              V
             2             2
                      A
        Also AVA  V3
                      2
        or VB  2VA  2 2gH …(ii)
        from (i) and (ii)
            1                1
              (2gH)  gh  (8gH)
            2                2
         h  3H  3m .

51.     Acceleration when block B is in liquid
                                m
                 m A g  (m Bg  B g)
                                 2
            a1                         …(i)
                        mA  mB
        acceleration when B is out of liquid
                 (m B  m A )g
            a2                 …(ii)
                   mA  mB
        Given a1 = a2
           mB = 12 kg
         solving mA = 9 kg.

                 F
52.          
                AY
           A  FA  A  A B  YB      m 1 1
                                                       2
                                                            m
                                   r     
           B  FB  B  A A  YA      3  2r   3r  36r
                                                               2


           m        1
              2
                 2 given
          36r      6r
         m = 6 kg.

53.     Consider a point distance x above bottom of the string
                        Mgx
                 T
           v                 gx
                       M/
20
                                                                   CODE–NPT-2-PAPER-I&II


      The time taken by pulse to travel from x  to x is
                          x
          t1  2       2
                     g    g
      The time taken for a body to fall from rest through distance               x is
             2(  x)
      t2               .
                   g
          t1  t 2

       solving x          2 m.
                       9

54.   The final steady state achieved is as shown
                                       
                                       ’        ’
                                                   ’
                                                    


      Using impulse momentum principle
           J12 = I( – )                          …(i)
           J21 + J23 = I( – )                    …(ii)
           J32 = I( – )                          …(iii)
      (i) + (iii) – (ii) gives I(3 – ) = 0
                  
      or  ' 
                   3
                                                  1
                                                    (3I2 )
       ratio of spin k.e. before and after is  2           9.
                                                  1
                                                    (3I ' )
                                                          2

                                                  2
55.       pA = pB
      Pressure of air in tube = (75 + 15 + x) cm of Hg
          = (90 + x) cm of Hg
                 h
          C
                 x
          A     B



      Final Volume = xA.
      As, Pi Vi = Pf Vf
       Patm ( A)  (90  x)Ax
           75  (90  x)x
       x = 15 cm.
      Force to be applied = (pA – pC) A = x gA = 1.02 N.




                                                                                    21
CODE–NPT-2-PAPER-I&II



               3a
56.                a cos Kx
               2
         O
        /2                1
                         0C     D0
                                 2
                 A      B



                   0m 0m 0m
                  1c 1c 1c
                              
             Kx       or x 
                   6          12
            BO1 = O1C = 5 cm
        So OA = 5 cm.
             
                OA  AB  BO1  20 cm
             2
        AB = 10 cm = BC
          = 40 cm, L = 60 cm
         It is second overtone.
57.     M.I. of the ring w.r.t. pivot P is I = 2MR2.
        Conservation of angular momentum about P gives
            2MR2 – 2mR (V – 2R) = 0
                      mV
         
                  R(M  2m)
        The velocity of bug at point x w.r.t. table is
                           MV
             V  2R               4 m/s.
                         M  2m




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