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AIEEE 2010-Solution

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PREVIOUS YEAR PAPERS CBSE BOARD EXAM AIEEE BITSAT ISAT VITEEE IIT-JEE STUDY MATERIAL PHYSICS CLASS XI XII SAMPLE PAPERS KEY SOLUTIONS ANSWERS QUESTIONS

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									                      SOLUTIONS & ANSWERS FOR AIEEE-2010
                                 VERSION – A

                        [PHYSICS, CHEMISTRY & MATHEMATICS]
                PART – A – PHYSICS                                    Ans:     Statement – 1 is true, Statement-2 is
                                                                               false.
1.   The initial shape of the wavefront of the ----                   Sol:     hυ = KEmax + φ
                                                                               ∴ if hυ increases, KEm,ax increases
     Ans:   Planar
                                                                               ∴ Stopping potential increases
     Sol:   Initially parallel, cylindrical beam will have                     Photoelectrons have various speeds.
            planar wavefront.
                                                                 7.   Statement – 1:
2.   The speed of light in the medium is
                                                                      Two particles moving in the same direction do
     Ans:   Minimum on the axis of the beam                           not ------

     Sol:                                                             Ans:     Statement -1 is true, Statement – 2 is true;
                                                                               Statement-2 is not the correct explanation
                                 Ιr < ΙC (µr < µC)                             of Statement-1
                                ΙC
                                                                      Sol:     Statement – 1 is true because the energy
                                             (µr < µC)                         considered is not kinetic energy alone.
                                                                               Statement – 1 is correct. Statement -2 is
                                                                               correct but not the explanation for
                                                     wavefront
                                                                               Statement – 1.
3.   As the beam enters the medium ------

     Ans:   Diverge                                              8.   The figure shows the position – time (x − t) graph
                                                                      of ---
     Sol:   From the above diagram the beam on
            coming out into the medium, diverges.                     Ans:     0.8 Ns

4.   The speed of daughter nuclei -----                               Sol:     Impulse = change in momentum
                                                                               = mv2 − mv1
                2 ∆m                                                           = 0.4 × (−1) − 0.4 × 1 = −0.8 Ns
     Ans:   c
                  M
                                                                 9.   Two long parallel wires are at a distance 2d
     Sol:   Mass lost = (∆m)                                          apart. They carry -------
                                   2
            Energy released = (∆m)c
                                                                      Ans:
            By conservation of momentum and energy
                             1M 2                                                                  B
            each has energy      v
                             2 2
                                                                                        •                  •
                                          2(∆m )
                                                                         X                                            X’
                    v = (∆m ) c 2 ⇒ v = c
                1M 2 1
            ∴
                2 2    2                    M                                               d          d

                                                                      Sol:
5.   The binding energy per nucleon for the ----------

     Ans:   E2 > E1
                                                                                            Bmax
     Sol:   In radioactive decay, the parent nucleus                                               B                   Bmax
            decays to a more stable daughter nuclei.
                                                                                •                                 •
            ∴ E2 > E1                                                               A              A       B
                                                                        Bmax
6.   Statement – 1                                                                                         Bmax
     When ultraviolet light is incident on a photocell,
     its stopping potential -----
                                                                               Hence graph (2) is correct.
10. A ball is made of a material of density ρ where ---     13. The respective number of significant figures for
    ---                                                         the numbers ------

    Ans:                                                        Ans:   5, 1, 2

                                              oil           14. The combination of gates shown below ------

                                                                Ans:   OR gate
                                              water

                                                                Sol:   A.B = A + B
    Sol:   Since ρwater is greater than ρoil, ρoil shoule              OR gate
           be above water.
           ρ > ρoil it should sink in oil and float in      15. If a source of power 4 kW produces 10
                                                                                                                   20
                                                                                                                        photons
           water.                                               / second, the ----
           Hence Answer 3.
                                                                Ans:   X - rays
11. A thin semicircular ring of radius r has positive
    charge q ----                                                                                       4000
                                                                Sol:   Energy of a photon =                    J
                                                                                                        1020
                        q                                                  4000            1
    Ans:    −                     ˆ
                                  j                                    =             ×        eV
                    2
                2π ε0r        2                                           10     1.6 × 10 −19
                                                                                20

                                                                       = 250 eV
    Sol:                                                                   1242
                                                                       ⇒        nm ≅ 5 nm (X- rays)
                                                                            250
                                                        •
                                                        •
                    •
                    •                                       16. A radioactive nucleus (initial mass number A and
                                          θ                     atomic number Z) emits ------

                                                                       A−Z−4
                                      θ             E           Ans:
                              E                                         Z−8
           Taking symmetrical elements of charge as
                                                                Sol:   ZX
                                                                            A
                                                                                →Z −8 Y A −12 + 3α + 2β+
           shown the sinθ components cancel out.
           The cosθ components add upto                                No. of neutrons = A − 12 − (Z − 8)
              π / 2 K dq                                                          A−Z−4
                                                                       ∴ Ratio =
           2
             0  ∫    r2
                         cos θ                                                      Z−8
                        π/2      q   cos θ
           = 2
                    ∫
                    0
                              K. rdθ 2
                                 πr   r
                                                            17. Let there be a spherically symmetric charge
                                                                distribution with charge -----
                1    q                     π/2
           = 2.
              4πε0 πr 2                   ∫ coθ dθ
                                          0                            ρ0r      5 r 
                 q
                                                                Ans:
                                                                       4ε 0     3 − R 
           = − 2 2ˆ    j                                                              
              2π ε0r
                                                                Sol:
12. A diatomic ideal gas is used in a Carnot engine
    as the working substance ----
                                                                                                       dx
    Ans:   0.75                                                                                x
                                                                                                   r
    Sol:   In the adiabatic part of the cycle
           T1V1γ − = T2 V2γ −
                  1           1

           = T2 (32 V1)γ −
                           1

               T
           ∴ 1 = (32)γ − = (32) − = (32)                                       5 x
                            1     7/5 1       2/5
                                                                       ρx = ρ0  −  4πx dx
                                                                                        2
               T2
                                                                               4 R
           = 4 ⇒ T1 = 4 T2
                                                                       Total charge upto r is
               T −T     3
           η = 1 2 = = 0.75
                 T1      4
                                                                     y [1 − K t ] = x [K t − 1] + constant
                                                                      2                2 2         2    2 2
                         5 x 
                         r
            ρr =
                     ∫ρ0  −  4πx dx
                                                                                      [ ]
                                  2

                     0 4 R                                                       x 2 K 2t 2 − 1
                                                                                    [1− K t ] + constant
                                                                      2
                                                                     y =                     2 2
                                 x 2
                                 r5
            = 4πρ0
                             ∫ −  x dx
                             04 R                                  = −x + current = x + constant
                                                                               2                        2

                                                                     ∴ y = x + constant
                                                                        2   2
                                  x3 
                                         r
                                            x4  
                                                 r
                     5
            = 4πρ0                 −         
                   4              3 0  4 R 0 
                   
                                                   21. Let C be the capacitance of a capacitor
                                                              discharging through ------
                    5 r3      r4 
            = 4πρ0        −       
                   4 3 4 R
                                                                   1
                                                              Ans:
            Gauss’s law is                                           2
                      4πρ0  5 r 3     r4 
                                     −                               Q = R0 e−
                 2                                                                       t/RC
            E.4πr =                                         Sol:
                       ε0  4 3 4 R 
                                                                   Q12 Q02 1
                                                                        =   .
                   ρ0r  5 r                                        2C   2C 2
            ⇒E=             −
                   4ε 0  3 R 
                                                                                      Q0
                                                                     ⇒ Q1 =
                                                                                          2
18. In a series LCR circuit R = 200 Ω and the voltage                         Q0
                                                                     ∴               = Q0 .e − t1 / τ
    and the frequency of the main supply ------                                2
                                                                      1
    Ans:    242 W                                                              = e −t1 / τ
                                                                          2

                                                                       2 = e t1 / τ ln 2 = 1
                                                                                           t
    Sol:    Since the lag by removing the capacitance                                                         ---(1)
                                                                                            τ
            is equal to the lead by removing the
                                                                     Q2 2 Q0 1
                                                                           2
            inductor XC = XL.                                            =   .
                                                                     2C 2C 4
            The circuit is in resonance condition.
                                            V 2 (220 )2
                                                                                   Q0
                                                                     Q2 =
            Power dissipated is                =                                    2
                                            R    200
                                                                              Q0
            = 242 W                                                  ∴           = Q0 e − t 2 / τ
                                                                               2
                                                                     ∴ 2 = et 2 / τ
19. In the circuit shown below, the key K is closed at
                                                                                        t2
    t = 0. --------                                                  ∴ ln2 =
                                                                                         τ

            V               V (R1 + R2 )                             ∴
                                                                              t1 ln 2 1
                                                                                 =      =
    Ans:       at t = 0 and              at t = ∞                             t2   ln 2   2
            R2                 R1R2

    Sol:    At the instant of switching on there is no    22. A rectangular loop has a sliding connector PQ of
            current through L. Therefore current at           length l and -----
                     V
            t = 0 is
                     R2                                                                      Bλv     2Bλv
                                                              Ans:        Ι1 = Ι2 =              ,Ι=
            At t = ∞, VL = 0                                                                 3R      3R
                    R1R2        V (R1 + R2 )
            Reff =          ⇒Ι=                               Sol:   Motional emf, E = Blv
                   R1 + R2         R1R2
                                                                                                                       R
                                                                     Reffective (external) = R || R =
                                                                                                                       2
20. A particle is moving with velocity -------
                                                                     Internal resistance = R
             2       2                                                                      R 3R
    Ans:    y = x + constant                                         Total resistance = R +    =
                                                                                             2   2
    Sol:            i(
            v = K y ˆ + xˆj        )                                 ∴Ι=
                                                                                E
                                                                                    =
                                                                                      2E
                                                                                         =
                                                                                           2 Bλv
                                                                              3R  3 R     3R
            dr
               = Ky ˆ + Kx ˆ
                      i     j                                                    
            dt                                                                2 
                                                                                Ι Bλv
            ⇒ r = Kyt ˆ + Kxtˆ + C
                        i     j                                      Ι1 = Ι2 = =
                 2           2 2 2      2 2 2
                                                                                2 3R
            ⇒ r = K y t + K x t + constant
              2   2     2 2 2   2 2 2
            (x + y ) = K y t + K x t + constant
           Aliter                                                                   V2
                                                                          ac =
                                                                                    R
           ΙR + Ι1R = Blv ---- (i)
                                                                                         V2
           ΙR + Ι2R = Blv ---- (ii)                                       (ac)X = −         cos θ ˆ
                                                                                                  i
                                                                                         R
           (i) − (ii) ⇒ Ι1R −Ι2R = 0 ⇒ Ι1 = Ι2
                                                                                         V2
           Ι = Ι1 + Ι2 = 2Ι2                                              (ac)Y = −         sin θˆ
                                                                                                 j
           ∴ (ii) ⇒ 3 Ι2 R = Blv                                                         R
                      Bλv                                                                V2           V2
           ⇒ Ι2 =                                                         ∴ ac = −          cos θ ˆ −
                                                                                                  i      sin θ ˆ
                                                                                                               j
                      3R                                                                 R            R
                       Bλv     2 Bλv
           Ι1 = Ι2 =       ;Ι=
                       3R       3R                          26. A small particle of mass m is projected at an
                                                                angle θ -----
23. The equation of a wave on a string of linear mass
                                                                               mv 0gt 2       ˆ
    density 0.04                                                 Ans:     −             cos θ k
                                                                                  2
    Ans:   6.25 N
                                                                 Sol:                               ˆ
                                                                          L = r × p ⇒ L is in the − k direction
                         2πt   2πx 
    Sol:   y = 0.02 sin      −                                          x = tv cos θ ˆ
                                                                                    0  i
                         0.04 0.50 
                                                                                           1     
           Compare with y = A sin (ωt − kx)                               y =  tv 0 sin θ − gt 2  ˆ
                                                                                                    j
                                                                                           2     
                  2π             2π
           ⇒ω=        and k =                                                                              gt 
                 0.04          0.50                                       r = v 0 t cos θ ˆ + t v 0 sin θ −  ˆ
                                                                                          i                      j
                ω    0 .5                                                                                  2
           ∴v=     =      = 12.5 m s−
                                      1
                k 0.04                                                    v = v ˆ+v ˆ
                                                                                   ix    jy
                        T                                                 = v0 cosθ ˆ + (v0 sinθ − gt) ˆ
                          ⇒T=v µ
                              2                                                     i                  j
           But v =
                        µ
                                                                          p = mv = mv 0 cos θ ˆ + m(v 0 sin θ − gt )ˆ
                                                                                                i                   j
           ∴ T = (12.5) × 0.04
                           2

                                                                          L = r×p
           = 6.25 N
                                                                                                         gt  
                                                                          = v 0 t cos θ ˆ + t v 0 sin θ −  ˆ ×
                                                                                         i                     j
24. Two fixed frictionless inclined planes making an                                                     2
    angle -------                                                              [
                                                                             mv cos θ ˆ + m(v sin θ − gt )ˆ
                                                                                     0     i            0        j   ]
    Ans:   Zero                                                           =
                                                                                                                    gt  ˆ
                                                            mv 0 t cos θ(v 0 sin θ − gt )k − tmv 0 cos θ v 0 sin θ − k
                                                                                         ˆ
    Sol:   Vertical acceleration of A and B are g.                                                                  2
           Hence relative vertical acceleration of A
           w.r.t B is zero.
                                                                               [
                                                                         = mv t sin θ cos θ − mv gt cos θ k
                                                                                2
                                                                                    0
                                                                                                        2
                                                                                                                0
                                                                                                                   ˆ     ]
                                                                             2                   mv 0gt 2  ˆ
                                                                          − mv 0 t sin θ cos θ −          k
25. For a particle in uniform circular motion, the                          
                                                                                                    2    
    acceleration ------
                                                                                   mv 0gt 2       ˆ
                                                                          = −               cos θ k
                                                                                      2
               V2           V2
    Ans:   −      cos θ ˆ −
                        i      sin θ ˆ
                                     j
               R            R
                                                            27. Two identical charged spheres are suspended by
                                                                strings of
    Sol:
                                  Y
                                                                 Ans:     2
                                          V
                                                                 Sol:
                                              P(R, θ)
                                   R
                                      θ
                                                        X                                              15° 15°
                                                                                                   l


                                                                          q2                  •                     •
                                                                        4πε0d2                              d
                                                                                              mg
                                   q2                                                       6 b x −7
           l sin 15° =                                                                  1=      .
                                4πε0d       2                                              12 a x −13
           l cos 15° = mg                                                                 b 6
                                                                                        =    .x
                                                                                          2a
                                       q2
           tan 15° =                                   ---- (i)                                         1/ 6
                               mg d2 4 πε0                                                   2a 
                                                                                        ∴x=                  at equilibrium
                             σ                                                             b 
                                       0 .8  g
           In liquid g’ = g 1 −  = g1 −   =                                         U(x) = ∞ = 0
                               ρ     1 .6  2
                                                                                                                  a                   b
           ε = ε0 K                                                                     Uat equilibrium =            12 / 6
                                                                                                                              −            6/6
                                                                                                             2a                  2a 
           Again tan 15°                                                                                                         
                      q2                        2q2                                                          b                   b 
           =                       =                       ------(ii)
                       2
               mg' d 4 πε              mgd2 4πε0K                                           ab2       b2    b2
                                                                                        =         −      =−
                                                                                            4a2       2a    4a
                                                 2q2                 q2
           From (i) and (ii)                                 =                                     b2  b2
                                            4πε0Kmgd2             mgd2 4πε0             ∴ D = 0 − −   =
                                                                                                   4a  4a
               2                                                                                      
           ⇒     =1 ⇒ K = 2
               K
                                                                              30. Two conductors have the same resistance at
28. A point P moves in counter-clockwise direction                                0 °C but their temperature
    on a circular path as shown ----
                                                                                 Ans:
    Ans:
                                                                                 Sol:   In series
                  3                                                                     R0 = R1 + R2
    Sol:   S=t +s
                        ds       2                                                      Rt = R1’ + R2’
           Speed v =        = 3t
                         dt                                                             = R1 + R1α1t + R2 + R2α2t
           At t = 2 s, v = 12 m s−
                                   1                                                    = (R1 + R2) + t[R1 α1 + R2 α2] ------ (i)
                 v 2 (12)2                                                              But Rt = R0 + R0 αt
                     =        = 7.2 m s−
                                        2
           ac =                                                                         = (R1 + R2) + (R1 + R2)αt ----- (ii)
                  r      20
                                                                                        From (i) & (ii)
           Tangential acceleration,
                 dv                                                                     α=
                                                                                             (R1α1 + R2α2 )
           at =
                 dt
                    = 6t                                                                        (R1 + R2 )
                                                                                            α1 + α 2
           At t = 2 s, at = 6 × 2 = 12 m s−
                                           2
                                                                                        =            (Θ R1 = R2 )
                                                                                               2
           ∴a=         ac 2 + a t 2 =            (7.2)2 + (12)2
           =    51.84 + 144                                                             In parallel
                                                                                              R
           =    195.84                                                                  R0 =
                                                                                              2
           ≅ 14 m s−
                           2
                                                                                              R[1 + α1t ]R[1 + α 2 t ]
                                                                                        Rt =
                                                                                             R{[1 + α1t ] + [1 + α 2t ]}
29. The potential energy function for the force
                                                                                            R(1 + α1t )(1 + α 2t )
    between two atoms -----                                                             =                                             ----(ii)
                                                                                             [2 + α1t + α2t]
    Ans:
                b2  b2
           0 − −   =                                                                  But Rt =
                                                                                                  R
                                                                                                      [1 + αt ] --- (ii)
                4a  4a                                                                          2
                                                                                      From (i) & (ii) (1 + αt)
                                                                                          2[1 + α1t ](1 + α 2t )
                                                                                        =
    Sol:   U(x) =
                       a
                               −
                                   b
                                                                                           [2 + α1t + α2t]
                                                                                                  [                               ]
                       12
                 x      x6
                 dU(x )                                                                        2 1 + α1t + α 2 t + α1α 2t 2
                                                                                        αt =                                −1
           F= −
                   dx
                        = −[−12 ax− + 6bx− ]
                                   13     7
                                                                                                     [2 + α1t + α2t]
           = 12ax− − 6 bx−
                   13        7                                                              α1t + α 2t + α1α 2 t 2
                                                                                        =
           At equilibrium, F = 0 ⇒ 0                                                          (2 + α1t + α2t )
           = 12ax− − 6bx−                                                                   t(α1 + α 2 + α1α 2t )
                   13       7
                                                                                        =
                   −13
           ∴ 12ax = 6bx     −7
                                                                                               2 + (α1 + α 2 )t
                       α1 + α 2 + α1α 2t                                                                                              F
             ⇒α=                         ; At t = 0,
                        2 + (α1 + α 2 )t
                    α1 + α 2                                                                                        HBF4
             α=
                       2                                                                                            heat

                    PART B − CHEMISTRY                                                         –4        3                                             3
                                                                                   35. If 10 dm of water is introduced into a 1.0 dm
31. In aqueous solution the ionization constants                                        flask at 300 K, ..................
     .........                                                                                                    –3
                                                                                        Ans : 1.27 × 10                mol
                                         −
     Ans : The concentration of H and HCO3 are
                                                        +
                                                                                        Sol : PV = nRT
                                                                                                              3170 (Pa ) × 1 × 10 −3 (m3 )
             approximately equal.
                                                                                                    n=
                                         + −                                                                 8.314 (J K −1 mol−1) × 300(K )
     Sol : H2CO3                    H + HCO3                                                                            –3
                                                                                                     = 1.27 × 10             mol
                −
             HCO3                       H + CO3 −
                                            + 2

             Since the k2 value is very low compared to                            36. From amongst the following alcohols the one that
                                        +         −
             that of k1, the H obtainable from HCO3 is                                 would react fastest with..............

             negligibly small.
                                                                                       Ans : 2–Methylpropan–2–ol

32. Solubility product                      of       silver         bromide   is
             –13
                                                                                                               CH3
    5.0 × 10 .......
                                                                                       Sol :
                                                                                                    CH3 C              OH        2-Methylpropan-2-ol
     Ans : 1.2 × 10
                           –9
                                g                                                                                                     alcohol)
                                                                                                                                    (3°
                                                                                                           CH3
                                        +        –                                                  Order of reactivity of alcohols                with
     Sol : ksp(AgBr) = [Ag ] [Br ]
                                                                                                    con.HCl/ZnCl2 (Lucas reagent) is
                    5 × 10 −13
                –
             [Br ] =
                                       –11
                               = 1 × 10 moles/L                                                     3° > 2° > 1°
                       0.05
             ∴ No. of moles of KBr = 10
                                        –11

                                  –11       –9
                                                                                   37. If sodium sulphate is considered to be completely
             Wt of KBr = 120 × 10 = 1.2 × 10 g                                          dissociated into cations and anions in aqueous
                                                                                        solution, the change ................
33. The correct sequence which shows decreasing
    order of ...............                                                            Ans : 0.0558 K
               2–      –            +            2+           3+
     Ans : O        > F > Na > Mg                     > Al                              Sol : ∆Tf = i × kf × m
                                                                                                  = 3 × 1.86 × 0.01
     Sol : For isoelectronic species the radii                                                    = 0.0558 K
           decreases with increase in atomic
           number.
                                                                                   38. Three reactions involving H2PO− are given
                                                                                                                     4

34. In the chemical reactions, ................                                        below:..................


     Ans : benzene diazonium chloride and                                              Ans : (ii) only
           fluorobenzene
                                                                                                    H2PO− act as H donor in reaction (ii).
                                                                                                                             +
                                                                                       Sol :            4
                    NH2                                       N2Cl
                                                                                   39. The main product of the following reaction is .....
                                 NaNO2
     Sol :                                                                              Ans : H5C6                           H
                                HCl, 278K
                                                                                                                C       C
                                                              (A)                                        H                    CH(CH3)2
     Sol :                                                                                    z2
                                         CH3                                    Sol : E α
                                                                                              n2
                                                             +
                                                           H                                    9
                  CH 2     CH        CH         CH3                                     ELi2 + =   × EHe+
      C6H5                                                 H2 O
                                                                                                4
                                                                                                  9
                                                                                               = × − 19.6 × 10 −18 J atom−1
                           OH
                                                CH3                                               4
                                                                                                           –17      –1
                                                      CH 3                                     = –4.41 × 10 J atom
              C6H 5      CH2        CH          CH

                          (2°carbocation)                                   43. On mixing, heptane and octane form an ideal
              1, 2 −migration                                                   solution. At 373 K, the vapour pressures ...........
                     
                →
                 of hydride ion
                                            CH 3                                 Ans : 72.0 kPa
                                                                 −H+
             C6H5       CH         CH2      CH        CH3 →
                                                                                              25
                                                                                 Sol : nA =       = 0.25
                   (more stable                                                              100
                   2°benzylic carbocation)                                                    35
                           H5C6                                                         nB =      = 0.31
                                           H                                                 114
                                  C    C                                                     0.25
                                           CH(CH3)2                                     xA =      = 0.45
                               H                                                             0.56
                                                                                                          0
                                                                                        p = p 0 . x A + p B .x B
                                                                                              A
40. The energy required to break one mole of Cl – Cl
                              –1
    bonds in Cl2 is 242 kJ mol ...........                                                = 105 × 0.45 + 45 × 0.55
                                                                                          = 72 kPa
    Ans : 494 nm
                                                                            44. Which one of the following has an optical
                                                                                isomer? .........
                      242 × 103                       –1
    Sol : E =                            J molecule
                                    23
                      6.02 × 10                                                  Ans : [Co(en)3]
                                                                                                     3+

                      h×c
             E=
                       λ
                                                                                                     3+
                                                                                 Sol : [Co(en)3]          is chiral.
             ∴λ
                  6.626 × 10 −34 (Js) × 3 × 108 (ms −1)                     45. Consider the following bromides:............
             =
                      242 × 103 
                                   ( J molecule −1)                           Ans : B > C > A
                      6.02 × 1023 
                                  
                               –6
             = 0.494 × 10           m                                           Sol : Order of SN1 reactivity is related to the
             = 494 nm                                                                 relative stability of carbocation formed by
                                                                                      ionisation (B) gives allylic secondary
41. 29.5 mg of an organic compound containing                                         carbocation,      (C)    gives    secondary
    nitrogen was digested according to Kjeldahl’s                                     carbocation and (A) gives primary
    method ................                                                           carbocation on ionisation.

     Ans : 23.7                                                             46. One mole of a symmetrical alkene on ozonolysis
                                                                                gives two moles of an aldehyde ................
                           14 × ( V1 − V2 ) N1 × 100
     Sol : % of N =
                                   w × 1000                                      Ans : 2–butene
                      14 × (20 − 15 ) × 0.1 × 100
                  =                               = 23.7                                                                     ozonolysis
                           0.0295 × 1000                                         Sol : CH3 − CH = CH − CH3    →
                                                                                                   2 − butene

42. Ionisation energy of He+ is 19.6 × 10
                                                       –18
                                                             J atom .
                                                                       –1                                                  2CH3 − CHO
                                                                                                                              Ethanal
    The energy ...............
                                                                                                                       Molecular mass : 44 u
                             –17           –1
    Ans : –4.41 × 10               J atom
47. Consider the reaction:                             52. For a particular reversible reaction at
    Cl2(aq) + H2S(aq) → S(s) + 2H (aq) + 2Cl (aq)          temperature T, ∆H and ∆S were found to be
                                 +          –

    ...................                                    ...........

    Ans : A only                                            Ans : T > Te

    Sol : Slow step is the rate determining step.           Sol : At equilibrium, ∆H = Te ∆S
          According to A; rate = K[Cl2][H2S]                      ∴ ∆G = ∆H – T∆S
                                  K[Cl 2 ][H 2 S]                        = ∆S (Te – T)
          According to B; rate =
                                     [H + ]                       ∆G will be negative when T > Te.

                                                       53. Percentages of free space in cubic close packed
48. The Gibbs energy for the decomposition of Al2O3
                                                           structure and in body centered ..............
    at 500°C is as follows:..................

                                                           Ans : 26% and 32%
    Ans : 2.5 V

                                                           Sol : For ccp and bcc percentages of free
    Sol : ∆G = –nFE
          3                                                      space are 26% and 32% respectively.
             × 966 × 10 3 (J) = 6 × 96500 × E
           2
                                                       54. The polymer containing strong intermolecular
          E = 2.5 V
                                                           forces e.g. hydrogen bonding ...............
49. The correct order of increasing basicity of the
                                                            Ans : nylon–6, 6
    given conjugate bases ..................

                                                            Sol : Nylon–6,6 is a fibre having strong
                  −             −   −     −
    Ans : RCO O < HC ≡ C < N H2 < R                               intermolecular forces due to hydrogen
                                                                  bonding.
    Sol : Acidic strength of the corresponding
          conjugate acid is                            55. At 25° C, the solubility product of Mg(OH)2 is
                                                                      –11                     2+
          CH3 – COOH > CH ≡ CH > NH3 > CH4                 1.0 × 10 . At which pH, will Mg ions start
          Hence the basicity of the conjugate base         precipitating ................
          must be the reverse.
                                                           Ans : 10
50. The edge length of a face centered cubic cell of
                                                                                               2+          – 2
    an anionic substance is 508 pm............             Sol :   k sp[Mg(OH)2 ] = [Mg ] [OH ]

                                                                                  10 −11
                                                                   ∴ [OH ] =
                                                                           – 2                        –8
    Ans : 144 pm                                                                               = 10
                                                                                      10 − 3
                                                                       –         –4
                                                                   [OH ] = 10
    Sol : 2(r(+) + r(–)) = a
                                                                   pOH = 4
                         508
          r(+) + r(–) =      = 254                                 pH = 10
                          2
          r(–) = 254 – 110 = 144 pm
                                                       56. The correct order of E0 2 +              values with negative
                                                                                         M     /M
51. Out of the following, the alkene that exhibits          sign for the four successive elements ...............
    optical isomerism is .............
                                                            Ans : Mn > Cr > Fe > Co
    Ans : 3–methyl–1–pentene
                                                            Sol : Mn > Cr > Fe > Co
                            H
                                                                  Standard reduction potential values of
                                                                     2+
            CH3       CH2   C       CH   CH2                      Mn /Mn = –1.18 V
                                                                    2+
    Sol :                                                         Cr /Cr = –0.91 V
                            CH3                                     2+
                                                                  Fe /Fe = –0.44 V
                                                                     2+
                                                                  Co /Co = –0.28 V
                   3-Methyl-1-pentene
             It contains a chiral carbon atom.
57. Biuret test is not given by ................                            if w ≠ 0
                                                                            ie x R0 ⇒ 0 Rx ie; R is not symmetric
     Ans : carbohydrates                                                    ∴ R is not an equivalence relation.
                                                                             m S p ⇒ mq = pn
                                                                             n   q
     Sol : Biuret test is             not        answered   by
           carbohydrates.                                                   ∴    m Sm     exists by the definition so S is
                                                                                 n n
                                                                            reflexive .
58. The time for half life period of a certain reaction
                                                                             m S p ⇒ mq = pn ⇒ pn = mq ⇒           p Sm
    A → Products is 1 hour. When the initial
                                                                             n   q                                 q n
    concentration of the reactant ‘A’,..................
                                                                            ∴ S is symmetric.

      Ans : 0.25 h                                                                                     ⇒ mq = pn and ps = qr
                                                                            Again,       m Sp , pS r
                                                                                         n q q s
      Sol : For a zero order reaction, t 1 α a
                                                     2                      ie; mq.ps = pn.qr ⇒ ms = nr ⇒          mSr
                             → 1.0 mol L ; t 1 = 1 hour
                        –1               –1                                                                        n s
             2.0 mol L
                                                     2                      ∴ S is transitive
             0.5 mol L
                        –1
                             → 0.25 mol L ;
                                            –1                              ∴ S is an equivalence relation but is not
                                         t 1 = 0.25 hour                    an equivalence relation.
                                                 2
                                                                 62. The number of complex numbers z such that
59. A solution containing 2.675 g of CoCl3.6NH3                      |z − 1| = |z + 1| = …..
                                    –1
    (molar mass = 267.5 g mol ) is passed through                    Ans:    1
    a cation exchanger. ..................
                                                                     Sol:   z is a point equidistant from 3 given
     Ans : [Co(NH3)6]Cl3                                                    points.
                                                                            ∴ z is the centre of the circle passing
                                     4.78                                   through 1, −1, i.
     Sol : No. of moles of AgCl =          ≅ 0.03
                                    143.5                        63. If α and B are the roots of the equation
                                                                     x − x + 1 = 0, ……
             i.e., 0.01 moles of the compound gives                    2

             0.03 moles of AgCl
             ∴ No. of moles of Cl per unit = 3
                                 –                                   Ans:   1
             ∴Formula of the complex is [Co(NH3)6]Cl3
                                                                            −ω, −ω
                                                                                     2
                                                                     Sol:
                                                                            α      = (−ω)
                                                                              2009        2009
60. The standard enthalpy of formation of NH3 is                                   = −ω
                                                                                        2007
                                                                                              .ω
                                                                                                2
                 –1
                                                                                   = −ω
    –46.0 kJ mol . If the enthalpy of formation of H2                                   2
                                  –1
                                                                            β      = (−ω )
                                                                              2009      2 2009
    from its atoms is –436 kJ mol and that of N2 is
                                                                                   = −ω
                –1                                                                      4018
    –712 kJ mol . ..........
                                                                                   = −ω       ×ω
                                                                                        4017


      Ans : +352 kJ mol
                             –1                                                    = −ω
                                                                            −ω − ω = −(ω + ω) = 1.
                                                                                2            2

                                                                 64. Consider the system of linear equations :
      Sol : N2 + 3H2 → 2NH3                                          x1 + 2x2 + x3 = 3
            2 × –46 = +712 + 3 × +436 – (6 × N – H)                  2x1 + 3x2 + x3 = 3
                                –1
            N – H = +352 kJ mol                                      ………………………

                                                                     Ans:   No solution.
               PART – C -MATHEMATICS
                                                                                1 2 1
61. Consider the following relations :                                               
    R = {(x, y)|x, y are real numbers and …..                        Sol:   A = 2 3 1 ⇒ A = 0
                                                                                3 5 2
                                                                                     
     Ans:    S is an equivalence relation but R is not
             an equivalence relation.                                             3 2 1
                                                                                         
                                                                            Ax1 = 3 3 1 ⇒ Ax1 ≠ 0.
     Sol:    x Ry = x = wy ⇒ x Rx                                                  1 5 2
                                                                                         
             ∴ R is reflexive
             xRy ⇒ x = wy and y Rx ⇒ y = w’x                                ∴ The given system has no solutions.
                            1
             where w ’ =      , this is possible only
                           w
                                                                                      [300 + (n − 1)(− 2)]
65. There are two urns.            Urn A has 3 distinct red                         n
    balls …….                                                            ∴ 3150 =
                                                                                    2
                                                                                  = n[150 − n + 1]
    Ans:   108
                                                                         3150 = 151n − n
                                                                                          2
    Sol:   A ⇒ 3 distinct red balls
                                                                         ∴ n − 151n + 3150 = 0
                                                                             2
           B ⇒ 9 distinct blue balls
                                                                                252      50
           3
               C2 × 9C2 = 3 × 36 = 108.                                  ⇒n=          or
                                                                                  2       2
                                                                            n = 25
66. Let f : (−1, 1) → R be a differentiable function                     ∴ Total time = 25 + 9 = 34
    with ………                                                             = 34 mts.

    Ans:   −4                                                 70. The equation of the tangent to the curve
                                                                          4
    Sol:   g(x) = [f(2f(x) + 2)]
                                     2                            y = x + 2 , ……
                                                                          x
           g’ (x) = 2f(2f(x) + 2) × 2f’ (x)
           g ‘(0) = 2f(2f(0) + 2) × 2f’(0)                        Ans:    y=3
                   = 4 × 1 × f(2 − 2)
                   = 4 f(0)                                                        4
                   = −4.                                          Sol:   y=x+
                                                                                   x2
67. Let f : R → R be a positive increasing function                      dy            8
                                                                            = 0 ⇒ 1− 3 = 0
              f (3 x )                                                   dx           x
    with lim           = 1 …..
         x →∞ f (x )                                                     ⇒x=2
                                                                         ∴y=3
    Ans:   1                                                             ∴ Equation of tangent y = 3.

                         f (3 x )                             71. The area bounded by the curves y = cos x and
    Sol:   Given Lim              =1
                          f (x )
                                                                  y = sin x ……
                        x →∞
           since f(x) is an increasing function,
                f (2x )                                           Ans:    4 2 −2
            Lim         is also equal to 1.
           x →∞ f (x )
                                                                  Sol:
68. Let p(x) be a function defined on R such that
    p’(x) = p’(1 − x), …….

    Ans:   21

    Sol:   f(x) = p(x) + p(1 − x)                                        Required area =
           f’(x) = p’(x) − p’(1 − x) = 0(given)                           π                     5π
           ∴ f’(x) = 0                                                    4                      4
           ⇒ f(x) = k constant
           when x = 0, p(0) + p(1) ⇒ k = 42
                                                                          ∫ (cos x − sin x )dx + ∫ (sin x − cos x)dx
                                                                          0                      π
           p(x) + p(1 − x) = 42                                                                  4
                1              1                                                  5π
                    p(x )dx + p(1 − x ) = 42
           ∴
                ∫              ∫                                                   2

                0
                    1
                               0                                              +
                                                                                  ∫ (cos x − sin x )dx
                                                                                  5π
           ∴2
                    ∫ p(x )dx = 42
                    0                                                    =
                                                                                   4

                                                                              2 − 1+ 2 2 + 2 − 1
                1                                                        = 4 2 −2.
                    p(x )dx = 21 .
           ∴
                ∫
                0                                             72. Solution of the differential equation
                                                                  cos x dy = y (sinx − y) dx, ……
69. A person is to count 4500 currency notes. …..
                                                                  Ans:    secx = (tanx +c)y
    Ans:   34 minutes
                                                                  Sol:   Consider dy = y(sinx − y)dx
    Sol:   In the first 9 minutes the person counts                                dy
                                                                         consider      = y sin x − y 2
           9 × 150 = 1350 notes                                                    dx
           Total left notes = 4500 − 1350                                 dy
                             = 3150                                          = y tan x − y 2 sec x
                                                                          dx
           He counts in A.P with d = (−2) and a = 150
            dy                                                               −3
                − y tan x = − y 2 sec x                               ∴c=       . ie; equation of line k becomes
            dx                                                                4
            −1 dy 1                                                   4x − 3y = −3.
                    + tan x = sec x
            y 2 dx y                                                  ∴ The distance between them
                 1   dz −1 dy                                                           20 − (− 3 )
           z=      ⇒   =                                                            =
                 y   dx y 2 dx                                                               16 + 1
              dz                                                                            23
           ∴      + z tan x = sec x                                                 =            .
              dx                                                                            17
           ∴ I. F e
                    logsecx
                            = secx
                                                          77. A line AB                     in       three-dimensional   space
                           ∫ sec
                                   2
           ∴z                          x = tan x + C
                secx
                       =
                                                              makes……..
            sec x
                  = tan x + C                                 Ans:     60°
              y
           ∴ secx = y(tanx +C).                               Sol:    cos 45 + cos 120 + cos θ = 1
                                                                            2                    2           2

                                                                       1 1
        ρ             ρ                                                 + + cos2 θ = 1
            j ˆ           i j ˆ
73. Let a = ˆ − k and c = ˆ − ˆ − k . ……                               2 4
                                                                                   3 1
                                                                      cos θ = 1 −   =
                                                                         2
    Ans:     −i + j − 2k                                                           4 4
                                                                                 1
    Sol:   (a × b) + c = 0                                            ∴ cosθ =
                                                                                 2
           a × (a × b) + a × c = 0                                    ∴ θ = 60°.
           (a . b)a − (a . a)b + a × c = 0
            3 j − 3k − 2b − 2i − j − k = 0                78. Let S be a non-empty subsets of R. …..
           ∴ 2b = −2i + 2 j − 4k
           ∴ b = −i + j − 2k                                  Ans: There is a rational number x ∈ S such that
                                                              x ≤ 0.
                   ρ            ˆ
74. If the vectors a = ˆ − ˆ + 2k , …….
                       i j                                    Sol: The negation of the given statement is –
                                                              ‘There is no rational number x ∈ S such that
    Ans:   (−3, 2)                                            x > 0.’ The equivalent statement is given above.


                                                          79. Let cos (α + β ) =
    Sol:   a.c = 0                                                                      4
                                                                                          and . …..
           ⇒ λ − 1 + 2µ = 0                                                             5
           ⇒ λ + 2µ = 1 -------(1)
                                                                       56
           b.c = 0                                            Ans:
           ⇒ 2λ + 4 + µ = 0                                            33
           ∴ 2λ + µ = −4 ----(2)
           ∴ Solving λ = −3 and µ = 2                         Sol:    tan2α = tan(α + β + α − β)
                                                                                tan(α + β ) + tan(α − β )
                                                                            =
75. If two tangents drawn from a point P to the                                1 − tan(α + β) tan(α − β )
              2
    parabola y = 4x are at right angles, ….                                            3 5
                                                                                          +
    Ans:   x = −1                                                                   = 4 12
                                                                                         3 5
                                                                                      1− ×
    Sol:   Locus of p is directrix of y = 4x
                                                2                                        4 12
           ∴ x = −1                                                                   56
                                                                                    =    .
                                                                                      33
                              x y
76. The line L given by        + = 1 passes ……..                                2       2
                                                          80. The circle x + y = 4x + 8y + 5 ……
                              5 b

    Ans:    1                                                 Ans:     −35 < m < 15

           x y                                                Sol:    Perpendicular distance from
    Sol:     + = 1 passes through (13, 32)                            (2, 4) < Radius
           5 b
                                                                       6 − 16 − m
           ⇒ = −20                                                                <5
           ∴ Equation is 4x − y = 20. It is parallel to                     25
           x y                                                            −10 − m
             + =1                                                     =           <5
           c 3                                                              5
                                                                      = |10 − m| < 25
           −25 < 10 + m < 25                             84. The number of 3 × 3 non-singular matrices……
           − 35 < m < 15.
                                                             Ans:    at least 7
81. For two data sets, each of size 5……                                         1 1 0
                                                                                     
                                                             Sol: Consider  0 1 0  . The 1 on the non
            11                                                                  0 0 1
    Ans:                                                                             
            2
                                                                  diagonal position can be shifted to 5 more
                                                                  positions. Further we can consider
                  n1σ12 + n2σ22 + n1d12 + n2d22
    Sol:   σ2 =                                 ,                  1 1 0
                             n1 + n2                                        
                                                                   0 0 1  . ∴ at least 7 matrices are there.
                             n1 x1 + n2 x 2                        1 0 0
                            x=                                              
                                n1 + n2
           since n1 = n2 we get                          85. Let f : R → R be defined by……
               σ 2 + σ22 + d12 + d22          x + x2
           σ2 = 1                          x= 1
                         2                      2            Ans:   −1
                                              2+4
           d1 = (2 − 3) = 1
             2          2
                                           x=      =3        Sol:   Since function has local minimum it must
                                               2
                                                                    be continuous at x = −1
           d2 = (4 − 3) = 1
             2          2
                                                                    ∴ Lim f (x ) = Lim f (x )
                   4 + 5 + 1 + 1 11                                     x → −1+                 x → −1−
           ∴ σ2 =               =   .
                        2         2                                 1=k+2
                                                                    ∴ k = −1.
82. An urn contains nine balls of which…..
                                                         86. Four numbers are chosen at random……
            2
    Ans:                                                     Ans:   Statement 1 is true, Statement 2 is false.
            7

    Sol:   Three balls without replacement can               Sol: If four chosen numbers form an AP, the
                         3× 4×2                              common differences can be ± 1, ±2, ±3, ±4, ±5 or
           be done in = 9                                    ±6. (e.g. 1, 7, 13, 19 is an AP with common
                            C3                               difference 6)
                            2                                ∴ Statement 2 is not true.
                        =     .
                            7
                                                                         10                                 10
83. For a regular polygon, let r and R be the…..         87. Let S1 =   ∑ j (j − 1)10 C j , S2 = ∑ j 10C j …..
                                                                         j =1                               j=1
                                              r   2
    Ans:   There is a regular polygon with      =            Ans:   Statement 1 is true, Statement 2 is false.
                                              R 3
                                                                                10
    Sol:   Let n sided regular polygon is inscribed in
           a circle. From the figure it is clear that        Sol:   S1 =        ∑ j(j − 1)
                                                                                j =1
                                                                                                 10
                                                                                                      Cj

                                                                                10
                                                                    S2 =        ∑j
                                                                                j =1
                                                                                        10
                                                                                          Cj

                                                                                10

                            R
                                      2π                            S3 =        ∑j
                                                                                j =1
                                                                                       2 10
                                                                                               Cj
                                      n
                                  r

                                                                                       ∑ (j                 )
                                                                                        10
                                                                    S1 − S3 =                   2
                                                                                                    − j − j2 ×10 C j
                                                                                        j =1
                 π r                                                                    10
           ∴ cos  =
                 n R                                                            =−     ∑j
                                                                                          j=1
                                                                                                    10
                                                                                                       Cj
           There an possible integer value
                                  1 1     3                                   = − S2
           corresponding to        ,  and                           S1+ S2 = S3.
                                  2 2     2
                                                                                  j( j − 1)
                                                                          10!
           But cosθ =
                        2  π        2
                          ⇒ = cos −1                               j! (10 − j)!
                        3  4        3                                     10!                             8!
                                                                                         = 9 × 10 ×
           ⇒ n is not an integer.                                    (j − 2)! (10 − j) !            (j − 2)! (10 − j)!
            10                                   10                        90. Let A be a 2 × 2 matrix with non-zero……..
           ∑j =1
                   j(j − 1) 10C j = 90           ∑
                                                 j=1
                                                          8
                                                              C j− 2
                                                                               Ans:   Statement 1 is false, Statement 2 is true.
                  = 90 × 2 .
                                         8

.                                                                                             a b
                                                                               Sol:   Let A =       
88. Statement 1 : The point A (3, 1, 6) is the mirror                                         c d
    image…….                                                                          Given |A| = 1
                                                                                      ad − bc = 1 ------(1)
    Ans:   Statement 1 is true, Statement 2 is true;
           Statement 2 is a correct explanation for                                        a b a b  1 0
                                                                                      A =  c d   c d  =  0 1
                                                                                       2
           statement 1                                                                                        
                                                                                                             
                                                                                       a2 + bc (a + d)b 
    Sol:   A (3, 1, 6)                                                                                  
           B = (1, 3, 4)                                                               (a + d)c bc + d2 
                                                                                                        
           Midpoint of AB is (2, 2, 5)
           2 −2 + 5 = 5                                                                          
                                                                                      a2 + bc = 1
                                                                                                  (1)
           Statement 2 is true                                                        d + bc = 1
                                                                                       2
                                                                                                 
           D. R’s of AB are [2, −2, 2] or [1, −1, 1]
           ⇒ which represent the D.R’s of normal                                      (a + d)b = 0
                                                                                      (a + d)c = 0
                                                                                                    ( 2)
             to the plane x − y + z = 5                                                           
           ⇒ Statement 1 is true                                                      Case 1
           We used statement 2 to prove                                               b = 0 and c = 0
           statement 1.                                                                    a 0
                                                                                      A=   0 d
                                                                                                
89. Let f : R → R be a continuous function……                                                   
                                                                                      Using (1)
    Ans:   Statement 1 is true, Statement 2 is true;                                         ±1 0 
           Statement 2 is a correct explanation for                                   A=    0 ± 1
                                                                                                   
           statement 1
                                                                                      It is obvious that for a given A, Trace (a)
                                                                                      can be different from zero.
                                             ⇒ f (x ) > 0
                                 1
    Sol:   f(x) =                                                                     Therefore, statement 1 is not true.
                        ex + 2e− x
                             −1
                                                 [e   x
                                                          − 2e − x     ]                             OR
                        (e                   )
           f’(x) =
                             x
                                 + 2e   −x 2                                                                      1 0
                                                                                      Take the 2 × 2 unit matrix 
                                                                                                                  0 1  as A.
                                                                                                                       
           f’(x) = 0                                                                                                  
                                                                                                   2
             x    2                                                                   |A| = 1 and A = I
           e = x                                                                      However, Trace (A) ≠ 0
                  e
                                                                                      Statement 1 is not true.
                   2x        1
           ⇒e =2⇒x=            log2
                             2
           Checking the sign of f’(x) as x crosses
            1
              log 2, we note that f(x) is maximum at
            2
                1
           x = log 2 .
                2
                                             1
           Maximum value of f(x) =
                                                1
                                         2 + 2×
                                                2
                                         1
                                    =
                                        2 2
           Statement 2 is true
              1       2 1.414
                        ==       = 0.3535
            2 2      4       4
           Since f(x) is continuous in R,
           f(x) has to assume all values between
           0 and 0.3535
                   1
           Since      is a number lying between o and
                  3
           0.3535, statement 1 is also true.

								
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