AIEEE_2011_S-Series_Solution T.I.M.E INSTITUTE by thekingmaker

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									                      SOLUTIONS & ANSWERS FOR AIEEE-2011
                                 VERSION – S

               PART A − CHEMISTRY                                 Sol:    A1B5        = A2 B5
                                                                                  2
1.   Ans:   Acetaldehyde
                                                               10. Ans:   AlCl3
     Sol:   Acetaldehyde reduces tollen’s reagent to
            metallic silver on warming.                           Sol:    Fajan’s rules.
                                                                            3+
                                                                          Al is the smallest cation and it has high
2.   Ans:   0 .086                                                        charge.

     Sol:   Mole fraction of methanol                          11. Ans:   Al2O3 < MgO < Na2O < K2O
                moles of methanol                    5 .2
             =                        =                           Sol:    K2O is more basic than Na2O. Al2O3 is
                     total moles                        1000
                                                5 .2 +                    amphoteric.
                                                          18
              = 0.086
                                                               12. Ans:   pentagonal bipyramid
3.   Ans:   2, 2, 2–Trichloroethanol
                                                                  Sol:    IF7 is pentagonal bipyramidal.
     Sol:   2Cl3C – CHO + NaOH
                                                               13. Ans:   2–Pentanone
                  → Cl3C – CH2OH + Cl3C – COONa
                                                                                      O
4.   Ans:   32 times
                                                                  Sol:
                                                                          CH3 C           CH2 CH 2 CH3
     Sol:                          C
            2 times increase for 10°
             5
                                        C
            2 = 32 times increase for 50°                                             ketoform
                                                                                            OH
5.   Ans: a for Cl2 > a for C2H6 but b for Cl2 < b for
          C2H6                                                                        CH3 C        CH       CH2 CH 3

     Sol:   ‘a’ is a measure of attraction between the                                          enol form
            molecules and ‘b’ the size of the
            molecules.                                         14. Ans:   2, 4, 6–Tribromophenol
                          –1       –1
6.   Ans:   38.3 J mol         K                                  Sol:    Phenol forms 2, 4, 6–tribromophenol when
                                                                          treated with a mixture of KBr, KBrO3 and
                                        V2                                HCl.
     Sol:   ∆S = 2.303 nR log
                                        V1
                                                               15. Ans:   804.32 g
              = 2.303 × 2 × 8.314 × log 10
                        –1
              = 38.3 J K                                                                  W2   1
                                                                  Sol:    ∆Tf = Kf ×         ×
                                                                                          M2   W1
7.   Ans:   1.8 atm
                                                                                          W2   1
                                                                          6 = 1.86 ×         ×
     Sol:   CO2 + C → 2CO                                                                 62   4
            a–x       2x                                                  W 2 = 800 g
                                                                          Wt. of glycol required is more than 800 g
            a = 0.5 atm
            a + x = 0.8 atm                                                        i −1
            x =0.3 atm                                         16. Ans:   α=
                                                                               ( x + y − 1)
                     2
                    pCO            (0.6)2
            Kp =               =          = 1.8 atm
                    pCO2             0 .2                         Sol:    i = 1 – α + nα; n = x + y
                                                                                  i −1
                                                                          α=
8.   Ans:   743 nm                                                              x + y −1


                                                                          BF6 −
              1      1    1                                                 3
     Sol:        =      −                                      17. Ans:
             355    680   λ
            λ = 743 nm
                                                                  Sol:    Boron cannot form BF6 − since boron has
                                                                                              3
9.   Ans:   A2 B5                                                         no available d–orbitals.
18. Ans:   CH3CH2CH(Cl)CO2H                                      Sol:   Sulphur exhibits oxidation state lower than
                                                                        +4 in its compounds.
    Sol:   Presence of Cl having –I effect on the
           α–carbon makes 2–chlorobutanoic acid              30. Ans: The stability of hydrides increases from
           the strongest acid among the given                         NH3 to BiH3 in group 15 of the periodic
           compounds.                                                 table.

19. Ans:   Ethyl ethanoate                                       Sol:   Stability of hydrides decreases from NH3
                                                                        to BiH3.
    Sol:   CH3COCl + C2H5ONa
                        → CH3COOC2H5 + NaCl                                   Part – B – Mathematics

20. Ans:   2
               nd
                                                             31. Ans:   −5

    Sol:   RNA contains β–D–ribose while DNA                     Sol: |a| = |b| = 1 a. b = 0
           contains β–D–2–deoxyribose.                                (2a − b). ((a × b) × (a + 2b))
                                                                      = (2a − b) ×
21. Ans:
               7
           4f 5d 6s
                        1   2                                              [(a. a) b − (a. b) a + (2b. a) b − (2b. b)]
                                                                      (2a − b). (b − 2a) = − 5
    Sol:   The outer electronic         configuration   of
                     7  1   2
           64Gd is 4f 5d 6s
                                                             32. Ans:    − 144
22. Ans:   2.82 BM
                                                                        (1 − x − x + x ) = (1 − x) (1 − x )
                                                                                       2        3 6                   6   2 6
                                                                 Sol:
                                                                        = (1 − 6x + ..− 20x … − 6x ) x
                                                                                           3       5
    Sol:   There are two unpaired electrons is
                  2–
                                                                              (1 − 6x + 75x − 20x ….)
                                                                                      2      4     6
           [NiCl4] hence the paramagnetic moment
           is 2.82 BM.                                                  = 120 − 300 + 36
                    2           3                                       = 156 − 300 = − 144
23. Ans:   sp , sp, sp
                                                             33. Ans:    β ∈ (1, ∞)
    Sol:     −          +
           NO3 – sp , NO2 – sp and NH+ – sp
                                2                   3
                                     4
                                                                 Sol:   If 1 + ai is root (a, real)
                                                                        Then (1 + i a) + α (1 + I a) + β = 0
                                                                                        2
24. Ans:   a vinyl group
                                                                        2a + aα = 0 ⇒ α = − 2 a ≠ 0
                                                                        1−a +α+β=0
                                                                              2
    Sol:   Formation of HCHO in ozonolysis shows
                                                                        1−a +β=0
                                                                              2
           the presence of CH2 = CH – group.
                                                                        β = a + 1 > 1 ∴ β ∈ (1, ∞)
                                                                              2

25. Ans:   The complex is an outer orbital complex
                                                             34. Ans:    ~ (Q ↔ (P ∧~R))
    Sol:   [Cr(NH3)6]Cl3 is not an outer orbital
           complex.                                              Sol:   The given statement is
                                                                        (P ∧~R) ↔ Q ≡ Q ↔(P ∧~R)
                                    +
26. Ans:   p(H2) = 2 atm and [H ] = 1.0 M                               ∴ The required negative is
                                                                         ~ [Q ↔ (P ∧~R)]
           2H + 2e → H2
                   +        –
    Sol:
                  0.0591     [H+ ]2                                        d2 y   dy  −3
           ECl =         log                                 35. Ans:    − 2  
                     2        [H2 ]                                        dx   dx 
                                                                                
                    + 2
           [H2] > [H ]
                                                                         d2 x          d    dx 
27. Ans: Availability of 4f electrons results in the             Sol:         2
                                                                                  =        
                                                                                            dy 
                                                                                                
         formation of compounds in +4 state for all                      dy           dy       
         the members of the series                                                           
                                                                                       d  1 
                                                                                  =
    Sol:   All the lanthanoids does not exhibit +4                                    dy  dy 
           oxidation state.                                                               dx 
                                                                                             
                                                                                           −1            d    dy 
28. Ans:   Neutral FeCl3                                                          =             2
                                                                                                    .            
                                                                                       dy             dy    dx 
                                                                                          
    Sol:   Neutral FeCl3 solution gives violet colour                                  dx 
           with phenol.
                                                                                           −1           d2 y  dx 
                                                                                  =             2
                                                                                                             .
                                                                                                                 
                                                                                                                  
29. Ans: The oxidation state of sulphur is never                                        dy            dx 2  dy 
         less than +4 in its compounds                                                     
                                                                                        dx 
                                d2 y   dy  −3                   ∴ f(x) has maximum at x = π
                            = − 2                                And minimum at x = 2π
                                dx   dx 
                                     
                                                                              a⋅c
36. Ans:      Statement-1 is true, Statement-2 is true;   41. Ans: c −            b
                                                                              a.b
             Statement -2 is not a correct explanation
             for Statement-1.
                                                             Sol:      b × c = b×d
   Sol:      A (1, 0, 7)      B,(1, 6, 3)                            a⋅d = 0
              x y −1 2 − 2
                =       =                                                (
                                                                     b× c −d = 0      )
              1     2       5
             P (λ, 2λ + 1, 3λ + 2)                                                (
                                                                     b and c − d are collinear)
             drs (λ - 1, 2λ + 1, 3λ - 5)                                      (
                                                                     b = k c −d           )
             ∴ λ - 1 + 2 (2λ + 1) + 3(3λ - 5) = 0
             14λ - 14 = 0 ⇒ λ = 1
                                                                             (
                                                                     a ⋅ b =k c − c − a ⋅ d   )   ]
             P (1, 3, 5) is mid point of A and B                    k [c − c ]
           Statement-1 is true
                                                                             a⋅b
           Statement-2 is also true but                             k=
           statement-1does not follow from 2                                 a⋅c
                                                                             ac
                                                                     b c−d=      b
37. Ans:     P(C|D) ≥ P(C)                                                  a ⋅b
                                                                            a⋅c
                     P(CD)                                             d=c−      b
   Sol: P(C | D) =
                      P(D )
                                                                            a.b

                  P(C )                                   42. Ans:     Statement-1 is true, Statement-2 is true;
                =
                  P(D )                                               Statement -2 is not a correct explanation
                ≥ P(C)                                                for Statement-1.

                                                             Sol:     A = (x, y) y – x ∈ z
               1
38. Ans:      0, 2                                                  B = (x , y) x = αy for rational α
                                                                    A : x – x = 0 ∈ z ⇒ (x, x) ∈ A reflexive
   Sol: 1 -P ≥
                5           31                                        y–x∈z⇒x–y∈z
                            32                                                     ⇒ (y, x) ∈ A symmetric
                31                                                    y – x ∈ z and z – y ∈ z ⇒ z – x ∈ z
   P ≤1-
       5
                32                                                    ∴ (x, z) ∈ A transitive
                                                                      A is equivalence relation
      1
   ≤                                                                  Statement – 1 is true
     32                                                               B: x = 1, x ⇒ (x, x) ∈ B reflexive
        1  1                                                                       1
   P≤     = 0,                                                      x = αy ⇒ y =     x     ∴ (y, x) ∈ B
        2    2                                                                     α
   Choice (3)                                                                                     symmetric
                                                                      x = αy and y = αz ⇒ x = α z
                                                                                                   2

39. Ans:      π log 2                                                 ∴ (x, z) ∈ B transitive
                                                                      B is equivalence relation
                        1                                             Statement – 2 is true but I does not
                            log (1 + x )
   Sol:      I= 8       ∫     1+ x2
                                                                                                  follow from 2.
                        0
                    π
                                                          43. Ans:     Statement-1 is true, Statement-2 is true;
                        4                                             Statement -2 is not a correct explanation
                    ∫
             = 8 Log(1 + tan θ) dθ                                    for Statement-1.
                    0
             = π log 2                                       Sol:     if AB = BA
                                                                           T    T  T
                                                                      (AB) = A B
40. Ans:      local maximum at π and local minimum at                 ⇒ AB is symmetric
             2π                                                       Statement-2 is true
                                                                           T    T T    T
                                                                    (ABA) = A B A
                                                                    Take A = I and B = some non – symmetric
   Sol:       f’(x) = x sin x                                         ∴ABA always
                       2x cos x + sin x                               ∴A(BA) and (AB)A are symmetric
              f’’(x) =
                            2 x                                       Statement-1 is true nut does not depend
           f’(x) = 0 ⇒ x = nπ, n ∈ Z                                  on Statement-2
           ie., x = π, 2π in 0 5π   (
                                    2
                                            )
           f’’(π) < 0 and f’’(2π) > 0
44. Ans:       a =c                                                             Point on the parabola corresponding to t = 1 is
                                                                                    1 1
                                                                                ⇒  , 
   Sol:    Two circle should touch each other                                      4 2
                      a                                                                             1 1
          Centres are  ,0  and (0,0)                                                                 − +1
                      2                                                                                     3 2
                                                                                ∴ shortest distance = 4 2   =
          ∴ also second circle passes through (0, 0)                                                    2      8
          ∴c=a⇒ a =c
                                                                             51. Ans:     21 months
45. Ans:      Does not exist                                                    Sol:      Total savings = 11040
                                                                                          Savings in the first 2 months = 400
                    sin( x − 2)                                                           Hence, savings in the next n months
   Sol:       lim        2
              x →2    ( x − 2)                                                                                          = 10640
                                                                                          We have
          Limit does not exist
                                                                                           n
                                                                                             [ 400 + (n − 1)40] = 10640
              3                                                                            2
46. Ans:        ≤A≤1                                                                      [200 + (n−1) 20] n = 10640
              4
                                                                                          200n + 20 n − 20 n = 10640
                                                                                                        2

                                                                                          20n + 180 n − 10640 = 0
                                                                                               2
                     2               4
   Sol: A = sin x + cos x
                4       2
          = cos x – cos x + 1                                                               n2 + 9n − 532 = 0
                                         2
                          1  3                                                                  9 ± 81 + 2128
              =  cos2 x −  +                                                            n=
                          2  4                                                                       2
                  3                                                                         − 9 ± 2209     −9 ± 47
              ∴     ≤ A≤ 1                                                                    =          =
                  4                                                                              2            2
                                                                                         = 19
47. Ans:      Statement-1 is true, Statement-2 is false.                        Therefore, answer is 21 months

   Sol:    P is (-2, -2) and Q (-1, 2) since R bisect                        52. Ans:     4
           ∠POQ, PR ¨RQ = OP : OQ
           = 4 + 4 : 1+ 4 = 8 : 5                                                                 25a + 26b
                                                                                Sol: Median =
           ∴ Statement 1 is true                                                                      2
           But statement 2 is false.                                                              51a
                                                                                               =
                                                                                                   2
48. Ans:      (-∞, 0)                                                                   Numerical value of the sum of the derivation
                                                                                                     1 3 5            49 
   Sol: |x| − x > 0                                                                            = 2a + + + .... +         
        ⇒|x| > x                                                                                     2 2 2             2 
        ⇒ x ∈ (-∞, 0)                                                                                     2a × 252
                                                                                                      =            = 252 a
                                                                                                              2
               2
49. Ans:
               3                                                                                                             25 2 a
                                                                                        Mean derivation about median =
                                                                                                                              50
                                             3
   Sol: The angle is sin−
                                         1

                                             14                                          25 2 a
                                                                                                = 50
                         1 + 4 + 3λ                         3                             50
          ∴                                         =
                (1 + 4 + λ )(1+ 4 + 9)
                                 2                          14
                                                                                |a| =
                                                                                         50 × 50
                                                                                                 =4
          14 (3λ + 5) = 9 × 14 (5 + λ )
                             2                          2                                25 × 25
            2                2
          9λ + 30λ + 25 = 9λ + 45
                             2                                               53. Ans:     (1, 1)
          ⇒ 30λ = 20 ⇒ λ =
                             3                                                  Sol: (1 + ω) = A + Bω
                                                                                                  7

                                                                                     (− ω ) = A + B ω
                                                                                          2 7

                                                                                     − ω = A + Bω
                                                                                         14
           3 2
50. Ans:
                                                                                     − ω = A + Bω
                                                                                         2
             8
   Sol: Slope of                     the     line       perpendicular   to           1 + ω = = A + Bω
   y – x = 1is (-1)                                                                     ∴A=1 B=1
   Hence t = 1                                                                          ∴ (1, 1)
             3                                                                                   3     1
54. Ans:       square units                                                  ∴p= −                 ,q=
             2                                                                                   2     2

                                                                   57. Ans:    Statement-1 is true, Statement-2 is true;
                                                                               Statement -2 is a correct explanation for
                                                                               Statement-1.
                                          1
                                                                      Sol:     x1 + x2 + x3 +x4 = 6
                                       y= x
                                                                             xI ≥ 0
                                                                                            9
                                                                             no. of ways = C3
                 A                 x=c                                       S2 is true
                                                                             S1 is true
                                                                             S1 follows from S2

                                                                                  3x + 5y − 32 = 0
                                                                                        2            2
   Sol:     y=x                                                    58. Ans:
               1     2
            y=    ⇒x =1
               x                                                              x2            y2
                                                                      Sol:             +         =1
                  ⇒ x = 1 (x > 0)                                             a2            b2
               1
            y=   ,x=e⇒x=e                                                     9              1
               x                                                                       +         =1
                                                                              a2            b2
                     e
                             1                                               1                 9
   ∴area A =         ∫  x − x  dx
                                                                            b   2
                                                                                       = 1−
                                                                                                 a
                     1

                         e −1
                          2                                                            1                 a2 − 9
                 =
                          2
                              − log e
                                                                                   (
                                                                              a 1− 2
                                                                               1
                                                                                             5
                                                                                                 )   =
                                                                                                          a2
                         e2 − 3                                                      3
                                                                             a −9=
                                                                              2
                 =
                           2                                                         5
                                        1 2 e2 − 3 3                                 3 32
                                                                                        =
                                                                              2
            Required area =               .e −    =                          a =9+
                                        2      2    2                                5     3
                                                                                      3 32 3 32
                                                                             b =a ×      =    × =
                                                                              2   2
55. Ans:     2                                                                        5     3 5       5
                                                                             Equation of the ellipse is
            4 k 2                                                              x    y2
                                                                                  +    =1
   Sol:     k 4 1 =0                                                          32 32
            2 2 1                                                             3     5
                                                                                3x + 5y − 32 = 0
                                                                                  2    2
           4(4 – 2) – k (k -2) + 2 (2k – 8) = 0
                  2
           = 8 – k + 2k + 4k – 16 = 0
               2
          ⇒ - k + 6k – 8 = 0                                                                kT 2
            2                                                      59. Ans:           I−
           k – 6k + 8 = 0                                                                    2
          ⇒ (k – 4) (k – 2) = 0
          ⇒ k = 2, 4                                                                  dv ( t )
           ∴k=2                                                       Sol:                     = − k (T − t)
                                                                                       dt

56. Ans:    p= −
                         3
                           ,q=
                               1                                                  V(t) =         ∫ − k (T − t ) dt
                         2     2
                                                                                   k (T − t )2
                                                                                               +C
                 sin(p + 1) x + sin x                                                  2
   Sol: f(x) =                        ,x<0                                        t = 0, V(t) = I
                          x
              = q, x = 0                                                                         kT 2
                                                                                  ⇒I=                 +C
                                                                                                  2
            x + x2 − x
                               ,x>0
                     3                                                                   kT 2
                x2                                                                C=I−
          is continuous.                                                                   2
                                                                                  Therefore,
                                                      x
          ⇒ p + 1 + 1 = q = lim                                                                  k (T − t )2     kT 2
                                       x →0    3                                                           +I−
                                              x 2  x + x2 + x                   V(t) =
                                                                                                     2            2
                                                              
                                       1                                                                          kT 2
                                   =                                              ⇒ V(T) = 0 + I −
                                       2                                                                           2
                                                                  = 5 × 10− × 2 × 1.50
                                                                                5
                               kT 2
                         =I−
                                2                                 = 0.15 mV

60. Ans:   7                                           64. Ans:    Wave moving in −x direction with speed
         dy                                                         b
   Sol:     = y+3
         dx                                                         a
          dy
               = dx
                                                                  y(x, t) = e− ( a x + b t )
         y+3                                                                               2
                                                          Sol:
        log (y + 3) = x + c
                                                                  This is of the form y(x, t) = f(x + vt), where
        ∴y+3=ce
                      x

        x=0 y=2 ⇒ c=5                                                      b
                                                                  v=            travels in negative x direction.
        ∴y=5e −3
                  x
                                                                           a
        ∴ y (log 2) = 5 e       −3
                          log 2

                     =5×2−3=7
                                                                  πv 4
                                                       65. Ans:
                                                                   g2
                 PART – B – PHYSICS
                                                                               v2
             1          1                                 Sol:    Rmax =
                × 15 =    m s−
                              1
61. Ans:                                                                       g
            152        15
                                                                  Area = π(Rmax)
                                                                                      2


           1      1   1                                                        πv 4
   Sol:      +      =                                                    =
           v − 2 .8 0 . 2                                                      g2
              1 15
           ⇒     =
              v 2 .8                                               π
                                                       66. Ans:
               2 .8                                                2
           v=
               15
           v    1                                         Sol:   Particle 1 is at equilibrium position (φ = 0).
             =
           u 15                                                                                     π
                                                          Particle 2 is at maximum position.  φ = 
           v2        1                                                                              2
            2
                 =
           u    15 2
                                                       67. Ans:   Statement – 1 is false, Statement-2 is
           1 1 1
             + =                                                  true.
           v u f
              dv                                          Sol:    If υ ⇒ 2υ,
                     v2
           ⇒ dt = − 2                                             V0’ > 2V0, well known result
              du     u                                            ⇒ Statement 1 is wrong.
              dt                                                  Statement 2 is true.
           dv   v 2 du
              = 2.
           dt   u dt                                   68. Ans:    45°
                   1          1
                      × 15 =    m s−
                                                                    [           ] [         ]
                                    1
              =                                                      ˆ           ˆ
                15  2        15                           Sol:    µ1 N × K1 = µ2 N × K 2 .        But    plane     of
                                                                  separation need to be XY.
62. Ans:   20 min
                                                       69. Ans:   372 K and 310 K
                     N0
   Sol:    N=
               2t / T1 / 2                                              T2 1
                                                          Sol:    1−      =
           N0       N             log 3                                 T1 6
              = t /020 ⇒ t 2 = 20
            3   2   2             log 2                                 T2 − 62 1
                                                                  1−           =
              2       N         20(log 3 − log 2)                         T1     3
           N0 = t /020 ⇒ t1 =
              3 21                    log 2                       T2 5
                                                                    =
                        20                                        T1 6
           t2 − t1 =         (log 3 − log 3 + log 2)
                       log 2
                     = 20 min                                     T2 − 62 2
                                                                         =
                                                                     T1    3
63. Ans:   0.15 mV

   Sol:    E = Bλv
             T2     5                                                q0
                  =                                      i.e. q’ =
           T2 − 62 4                                                     2
                                                          q0
           4T2 = 5T2 − 310                                     = q0 cosωt
                                                           2
                                                                     π
           T2 = 310 K                                    ⇒ ωt =
                                                                     4
           ⇒ T1 = 372 K                                        π
                                                         t=      LC
                                                               4

70. Ans:    108.8 eV                          75. Ans: Statement-1 is true, Statement-2 is true
                                                       and Statement -2 is not the correct
            13.6 Z2               1
   Sol:               = 13.6 × 9 1 −                 explanation of statement - 1
              n2                  9
                                     8            Sol:   Statement-1 is true, Statement-2 is true
                       = 13.6 × 9 ×
                                     9                   and Statement -2 is not the correct
                       = 108.8 eV                        explanation of statement - 1

            2.7 × 10 Ω
                      6
71. Ans:                                      76. Ans:   0.052 cm

           V = V0(1 − e− )
                              t/RC
   Sol:                                                         1
                                                  Sol:   LC =      = 0.01 mm
           120 = 200(1 − e− )
                               t/RC
                                                              100
                      2                                  Reading = PSR × pitch + CSR × LC
           e−
               t/RC
                    =
                      5                                     = 0 + 52 × 0.01
             t/RC
           e       = 2.5                                    = 0.52 mm
               t                                           = 0.052 cm
                  = 0.4 × 2.5 × 2.303
           RC
           ⇒ R = 2.7 × 10 Ω
                             6                            n1T1 + n2T2 + n3T3
                                              77. Ans:
                                                             n1 + n2 + n3
            1 Mv 2 (γ − 1)
72. Ans:                                          Sol:   P1V = n1KT1
            2     R
                                                         P2V = n2KT2
                                                         P3V = n3KT3
   Sol:    Volume is constant                             1        3          3      3
                  R                                         mv 2 = KT1 × n1 + KT2n2 + K − T3n3
           Cv =                                          2         2          2      2
                (γ − 1)                                             3
                 1                                               = K(n1 + n2 + n3)T
           KE = Mv 2                                                2
                 2                                           n1T1 + n2T2 + n3T3
           ∆Q = nCv ∆θ = 1 × Cv ∆θ                       T=
                                                                 n1 + n2 + n3
                    KE 1 Mv 2 (γ − 1)
           ∴ ∆θ =     =
                    Cv 2     R                78. Ans:   −6 ε0a

                                                                2
73. Ans: 0.4π mJ                                  Sol:   V = ar + b
                                                               dV
                                                         E= −       = −2ar
           E = T.8π(r2 − r1 )
                          2     2
   Sol:                                                         dr
                    25       9                            2      Q
             = 8πT  4 − 4 
                                                       4πr .E =
                    10    10                                     ε0
             = 8 × 16 × π × 0.03 × 10−                   Q = −4πr .2ar.ε0
                                          4                          2

             = 0.4π mJ                                       − 8πar 3ε0
                                                         ρ=
                                                                4 3
            π                                                     πr
74. Ans:      LC                                                3
            4                                              = −6 ε0a
   Sol:    q’ = q0 cosωt
                                              79. Ans:    Statement 1 is true. Statement 2 is true.
              q 2                                        and statement 2 is the correct explanation
           E= 0
              2C                                         for statement – 1.
            E 1 q0 2
             =
            2 2 2C
   Sol:    Statement 1 is true. Statement 2 is true.               ⇒ −2.5t = 2 v   [ ]       0
                                                                                             6.25
           and statement 2 is the correct explanation
                                                                      2 6.25
           for statement – 1.                                      t=
                                                                        2 .5
            2                                                         2 × 2 .5
80. Ans:      g                                                     =          =2
            3                                                           2 .5

                                                                   3.6 × 10− m
                                                                               3
   Sol:    mg − T = ma                                  86. Ans:
               mR 2 a
           TR =      .                                                   1
                  2 R                                      Sol:    P0 +    ρv12 + ρgh
                                                                        2
                   3
           ⇒ mg =    ma                                                    1
                   2                                               = P0 + ρv 22
                                                                           2
                 2
           ⇒a=     g                                               ⇒2gh = (v2 − v1 )
                                                                               2     2
                 3                                                             2     2
                                                                   ⇒ 2gh + v1 = v2 ;
                                                                   v1 = 0.4 m s− , h2 = 0.2 m
                                                                                 1
               µ0Ι
81. Ans:                                                           ⇒ v2 = 2.0396 m s−
                                                                                        1
               π2R
                                                                                             2      d12 v1
                                                                   A1v1 = A2v2 ⇒ d2 =
                        Ι     µ                                                                      v2
   Sol:    B=             Rdθ 0 sin θ
                       πR    2πR                                                       v1
                                                                   ⇒ d2 = d1.
                                                                                       v2
                                 dθ           •
               •           R
                                          •                                                   0 .4
                                                                   = 8 × 10− ×
                                                                               3
                   •
                       •              •                                                     2.0396
                           •     •
                                                                   ≅ 3.6 × 10− m
                                                                                   3

                   µ0 Ι    π/2

                2π R ∫0
           =           sin θdθ
                       2                                87. Ans:    v∝x
                µ0Ι
           =                                               Sol:    T cosθ = mg
                π2R                                                T sinθ = F
                                                                            F
82. Ans:    Increases by 0.2%                                      tanθ =
                                                                           mg
           R∝λ
                       2                                           x   F
   Sol:                                                              =
           R’ ∝ λ’
                   2
                                                                   2λ mg
              ∝ (1.001) λ
                        2 2
                                                                   F∝x
            ∆R
            R
                = 0.002                                            ∫ vdv ∝ ∫ xdx
                                                                   v ∝x
                                                                    2     2
           ∴ 0.002 × 100
           = 0.2%                                                  v∝x

                                                                               1/ 2
83. Ans:   First increases and then decreases.                      M+m
                                                        88. Ans:       
                                                                     M 
   Sol:    Angular momentum is conserved.
           Ι decreases ω increases then Ι increases        Sol:    Mv1 = (M + m) v2
           ω decreases.                                            v1 M + m
                                                                      =
                                                                   v2     M
84. Ans:    8.4 kJ
                                                                    1
                                                                      (M + m)v 22 = 1 KA 22
   Sol:    ∆U = mC∆T                                                2               2
              = 4184 × 20 × 0.1                                     1         1
              = 8.4 kJ
                                                                          2
                                                                      Mv1 = KA12
                                                                    2         2
                                                                    1         1
85. Ans:    2s                                                        Mv12 = KA12
                                                                    2         2
                                                                                                         2
           dv
              = −2.5 v                                                  A12         M M+m 
   Sol:                                                            ⇒           =          
           dt                                                           A 22       M+m  M 
           dv
              = −2.5 dt                                                                M+m
            v                                                                  =
                                                                                        M
                                         1/ 2   90. Ans:   more than 3 but less than 6.
                     A1  M + m 
            ∴          =       
                     A2  M                       Sol:    τ = Fr = 40t − 10t
                                                                                2

                                                                τ
                                                           α = = 4t − t
                                                                         2
              −9 Gm                                             Ι
89. Ans:
                 r
                                                            dω                           t3
                                                                = 4 t − t 2 ⇒ ω = 2t 2 −
            Gm               G4 m                           dt                           3
   Sol:                  =                                 (Θ At t = 0, ω = 0)
             x   2
                             (r − x )2                     At t = 6 s. ω again become zero
            (r − x )2         =4                                dθ          t3       2t 3 t 4
                     2                                     ω=      = 2t 3 −    ⇒ θ−      −
              x                                                 dt          3         3    12
           r − x = 2x                                      ∴ θ in 6 s = (144 − 108) = 36 rad
                r                                                   θ    36
           x=                                              ⇒N=         =     = 5.72rotation.
                3                                                  2π 2π
               −Gm G4 m
          V=         −
                 r       2r
                 3        3
           = −
                Gm
                     (3 + 6 )
                   r
              −9 Gm
           =
                  r

								
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