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Solutions

The following solutions are for questions in the sequence as given in Set – B of the JEE Screening Paper.

  • 1. (C); H1 = (As = 0)
    H2 = ;
    Þ H2 =(3/2)H1 & hence the required ratio is 2/3.

    2. (C); Magnetic moment ‘M’ = (qw /2p )(p r2)
    angular momentum ‘L’ = m    w r2
    Þ M/L = q/2m

    3. (D); n fundamental = (1/2L)(T/m )1/2
    m 1 = p r (2r)2 , m 2 = p r r2 ; L1 = L, L2 = 2L
    n 1/n 2 = (L2/L1) (m 2/m 1)1/2 = 1

    4. None of the given choices is correct. The correct dimensional formula for (1/2)e 0E2 which is simply energy density (i.e. energy per unit volume ) is ML-1T-2.

    5. (B)’ vsound = (g RT/M)1/2
    As     g (=cP/cv) and T are same for both the gases, required ratio equals (m2/m1)1/2.

    6. (C); For an ideal gas in an isobaric process (D V/VD T) = 1/T, hence d = 1/T.

    7. (A); During phase changes, temperature remains constant otherwise it increases linearly.

    8. (A); While falling down v = -(2gh)1/2. Just before impact v1 = -(2gd)1/2.
    Just after impact v2 = + {(2g(d/2)}1/2 ; afterwards, v =

    9. (D); f1 = f (V/(V - )), f2 = f (V/(V - ))
    Þ f1/f2 = (V - )/(V - ) = 323/306 = 19/18

    •
    1.  
    2. (A); at (tangential acceleration) = La . Hence, net normal reaction = ma L. The bead starts slipping, when centrifugal force m(a t)2L > m m a L Þ t > (m /a )1/2.
  • 11. (D); TVg -1 = constant
    Þ T1/T2 = (V2/V1)g -1 = (L2/L1)g -1, where g monoatomic = 5/3

    12. (C); toppling begins when FL = Mg(L/2)
    Þ L = Mg/2

    13. (C); R¢ = 2R [As R µ mq4 ; where m and q are mass and charge of electron respectively.
    1/    l = R¢ [], Here nf = 2 ; ni = 3, as maximum wavelength corresponds to minimum energy of transition.

    •
    1. (B); let C1 =(e 0(A/2)k1)/(d/2) ; C2 = (e 0(A/2)k2)/(d/2) ;
      C3 = (      e 0Ak3) /(d/2); then, C = (C1 + C2)C3/(C1 + C2 + C3)
      equating ‘C’ to (k      e 0A/d), we get 1/k = {1/(k1 + k2)} + 1/2k3

    15. (B); = E(2p r) = e induced = (p a2)(dB/dt)

  • 16. (A); En = (-13.6 eV)/n2 ; KEn = |En|
    PEn = 2En
    •

    • 17. (B); Every emergent ray is parallel to corresponding incident ray.

    1. (A); Imax = (Ö I1 + Ö I2)2
      Imin = (      Ö I1 - Ö I2)2

    19. (B); T = Wien’s constant/l min

  • 20. (D); A concave lens behaves as diverging when its refracting index is greater than that of surrounding medium.

    21. (D); i2(t) B(t) µ i2 (t)i1(t) µ (1–e–t/t )e–t/t
    At t = 0, i1(t).i2 = 0

    Also i1(t).i2 (t) ® 0 as t ® ¥

  • 22. (A); sin a m/sin r = n1/n2 . . . (i)
    r +     q c =p /2 . . . .(ii)
    where critical angle     q c = sin- 1(n2/n1);
    Þ a m = sin- 1((n1/n2)cosq c)
    = sin    - 1[(n1/n2 cos(sin- 1(n2/n1))]
    •
  • 23. (A); T = 2p (l /geff)1/2 ; geff = || = g cos q
  • 24. (A); Wisobaric > Wisothermal > Wadi.
    Þ W2 > W1 > W3.
    •
  • 25. (D); In similar D ABI & D MNI
    MN/AB = IQ/IP
    Þ MN = (3L/L)(d) = 3d
  • 26. (D); N1/N0 = . . .(i)
    N2/N0 = . . .(ii)
    Þ N1/N2 =
    putting N1/N2 = 1/e we obtain (    l 1 - l 2)t = 1
    Þ t = 1/(l 1 - l 2) = 1/(10l - l ) = 1/9l .

    27. (C); Object lies slightly beyond the focus of the objective. Therefore intermediate image is real, inverted & magnified.

    28. (D); l min = 12400/V = 12400/(80 ´ 1000) » 0.155 A0 Þ continuous X-ray having wave length l ³ 0.155 A0 will be emitted. Also energy of the incident electron is sufficient to knock out an electron from K-shell. Hence, characteristic X-rays will also be emitted.

    29. (C); Since magnetic force = q()
    for +ve charge q = +q
    for –ve charge q = -q
    therefore both will be accelerated along –y axis.

    30. (B); Qq/4p e 0a + Qq/4p e 0Ö 2a + q2/4p e 0a = 0
    Þ a = -2q/(2+ Ö 2).

    31. (C); P = F. v
    since F =     r Av2
    P     µ v3.

    32. (D); Ixx’ = (3/2)mR2 since m = r L & R = L/2p
    IXX    ¢ = 3r L3/8p 2

    33. (B); Since no net torque is acting along the axis of rotation & no dissipative force is present, hence angular momentum & energy will be conserved.

    •
  • 34. (A); A1v1 = A2v2 ; where A is area of cross section
    & v is velocity of efflux.
    Þ (L2)(2gy)1/2 = (p R2)(2g(4y))1/2
    Þ R = L/(2p )1/2
  • 35. (B); At the mid-point of the line joining the conductors, = 0. As we come close to wires magnitude of B increases. The direction of magnetic field on opposite sides of a wire will be opposite.
    •

    • 36. (A); Undergoes dehydration easily as the product obtained is conjugated and is more stable.

    37. (B); D H = (reactant) =( –110.5 – 241.08) – (–393.5) = 41.2 kJmol–1-

    38. (A); PhCH=CHCH3 will exhibit geometrical isomerism.

    39. (D); Kc= = , (R in L.atm.K–1 mole–1).

    40. (C); C6H5–COOH + SOCl2 ¾ ¾ ® C6H5COCl + SO2 + HCl

  • 41. (B); For M+ + X– ¾ ® M + X, = 0.44 – 0.33 = 0.11V is positive, hence reaction is spontaneous.

    42. (B); Effective nuclear charge decreases from F– to N3– , hence the radii follows the order:
    F–     < O2– < N3–.

    • 43. (A); px orbital being dumbell shaped, number of nodal planes = 1

  • 44. (B); Zn(Hg), HCl cannot be used when acid sensitive group like –OH is present, but NH2NH2, OH– can be used.

    45. (C); –CH3 is the best nucleophile because carbon has the least electronegativity among the among the given options.

    •

    • 46. (C); Þ T2 = 2T1

  • 47. (B); 3d54s1 system is more stable than 3d44s2, hence the former is the ground state configuration.

    48. (D); C + O2 ¾ ¾ ® CO2, CO2 + C ¾ ¾ ® 2CO, The CO so produced reduces Fe2O3 to Fe via Fe3O4. and FeO.

    •

    • 49. (D); \ Cr is in +6 oxidation state.

    50. (B); (PV)Observed / (PV)Ideal < 1 Þ Vobs < Videal, Vobs < 22.4 litre.

    51. (D); Aliphatic amines are more basic than aromatic amines.

    52. (D); At initial stage concentration of each product will increase and hence Q will increase.

    53. (D); As unit of k is sec–1 , reaction is of first order r = K[N2O5] \ [N2O5] = = 0.8

  • 54. (D); In propyne the terminal hydrogen is acidic and reacts with ammonical AgNO3, but in propene there is no acidic hydrogen.
    •

    • 55. (C); Mass of 1L water vapour = 0.6 g , V = = 0.6 cm3.

    57. (C); Number of P–O–P bonds = 3 in trimetaphosphate which is the most stable one among all the cyclic metaphosphates.

     

  • 58. (A); (A) and (B) are almost equally least stable among the given olefins, but adsorption of A will occur easily on the nickel surface making hydrogenation easier.

    59. (D); Rate of reaction will be R – I > R – Br > R – Cl > R – F. As I– is the best leaving group among halide ions.

    60. (D); Structure of SF4 is T.B.P. with one lone pair , while CF4 is tetrahedral with no lone pair and XeF4 is octahedral with 2 lone pairs and is square planar.

    •

    • 61. (B); N is sp, sp2 and sp3 hybridised in and , respectively.

  • 62. (B); In –CH2– group is flanked on both sides by electron - withdrawing groups and hence is most acidic.

    63. (A); Non-metallic oxides are acidic and acidic character decreases with decreasing non metallic character.

    64. (C); NH3 does not react with CaO

    65. (A); S, Se, Te cannot undergo hydrogen bond formation because of their larger size and lower electronegativity values.

    66. (C); Assertion is correct but reasoning is incorrect because the energy of 2s orbital is less than 2p orbital.

    67. (C); Assertion is correct but reasoning is incorrect because 1-butene reacts with HBr in presence of peroxide to form n – butyl bromide via secondary free radical.

    •

    • 68. (B); Q = 0 since W = PD V = 0 ´ D V = 0 (D E = 0 in an isothermal process).

    69. (A); c µ , mc µ , frequency of collision µ c,where c = rms velocity

     

  • 70. (A); Due to +M effect of , is stabilized, hence its intermediate carbocation is more stable than the one in benzene.
    •

    • 71. (C); f(q ) = sinq ´ 2 sin2q cosq = sin22q ³ 0 " q Î R

    72. (B); The given normal is y = -x + k

    Now normal to the parabola y2 =12x of slope -1 is

    y = –x + 6 +3 ( y = mx –2am – am3)

    Þ y = –x + 9. Thus k = 9.

  • 73. (D); nCr +2.nCr-1 + nCr-2 = nCr + nCr-1 + nCr-1 + nCr-2

    = n+1Cr + n+1Cr-1 = n+2Cr .

    • 74. (B); Given a < b and c < 0 < b

    Since a + b = –b and a b = c

    a + b < 0 and a b < 0 i.e. a and b are of opposite signs.

    Þ As a < b Þ a < 0 < b and |a | > b ( since a + b < 0 ).

    Alternatively: Let f(x) = x2 + bx +c Þ f(0) = c < 0
  • Also vertex of this parabola lies towards the left of the origin

    • Þ b < |a |.

    75. (C); For the given f :R ® R

    If f is onto, |f| will be into.

    If f is one-one, |f| can be many-one.

    If f is differentiable |f| can be non differentiable.

    But if f is continuous, then |f| is also continuous.

    76. (D); 2x + 2y =2 Þ 2y = 2 – 2x Þ y = log2(2 –2x)

    y will be real if 2 – 2x > 0 Þ 2x < 2

    Þ – ¥ < x < 1.

  • 77. (B); x2 + y2 =1 Þ 2x +2yy¢ = 0

    Þ 1 + yy² + (y¢ )2 = 0 .

    78. (A); We have

    Þ 1 ³ M1/2 Þ M £ 1Þ 0 < M £ 1.

    79. (D); For non-trivial solutions =0

    Þ 1(2) + k(-k+1) –1( k+1) =0 Þ k2 –1= 0 Þ k = ± 1.

    • 80. (C); RQ = 5, OR = OQ = 5 (O being the origin)

    Þ Ð ROQ = Þ Ð RPQ =

    81. (B); 2ac sin(p /2–B) = 2ac .

    82. (C); = .

  • 83. (D); Given

    Þ Þ 16 r2 – 16r +3 = 0

    Þ (4r –1)(4r –3) = 0 Þ r = 1/4, 3/4 .

    • 84. (B); g(2) =

    Þ Þ 0 £ g(2) < 2 .

    85. (A); 2(r + R) = 2(s –c) tan+ c = 2s –2c + c = 2s – c =a + b.

  • 86. (C); Odd digits = 3, 3, 5, 5, Even digits = 2, 2, 8, 8, 8

    We have to fill nine places so that odd digits occupy the even places.

    Þ Total number of different numbers = .

    • 87. (A); arg(-z) – arg(z) = arg= arg(-1) = p

    88. (D); S º (13/2, 1). Slope of line PS = .

  • Equation of the required line: y +1 = – i.e., 2x + 9y + 7= 0.

    • 89. (B); Let O be the foot of the tower of height h and a be the angle of elevation

    Þ OA = h cota = OB = OC Þ O is the circumcentre.

    90. (C); = = 0 +2 = 2.

  • 91. (D); Since the triangle is equilateral, incentre of the triangle is the same as the centroid
    º (1, 1/ ).

    92. (D); The statement ‘S’ is true as their derivatives cosx and –sinx respectively are < 0 " x Î
    (    p /2, p ) and the statement ‘R’ is false as from the sign of df/dx, sign of d2f/dx2 can not be checked.

    93. (C); f(x) =

    • f¢ (x) = ex(x –1)(x –2) Þ f¢ (x) < 0 Þ 1 < x < 2

    So f is decreasing in (1, 2).

    94. (A); Using 2g1g2 + 2f1f2 = c1 + c2

    we get 0 + 2.k.k = 6 + k Þ 2k2 –k – 6 = 0

    Þ (2k + 3)( k –2) = 0 Þ k =2 , –3/2 .

    95. (B);We have Þ Þ

    Also, Þ Þ .

  • Alternatively: The sides are taken in cyclic order. Hence the vector area of triangle ABC will be equal Þ .
    •

    • 96. (D); slope of the normal at (3, 4) is tan(3p /4) = -1

    Þ slope of the tangent at (3, 4) is 1 Þ f¢ (3) =1.

    97. (A); Þ are collinear.

  • Now normal vectors of P1 and P2 are and .

    Thus the angle between the planes is zero.

    •

    • 98. (A); f(0) =1, f(0– ) < 1 and f(0+ ) < 1

    Þ f has local maximum at x = 0.

  • 99. (A); Sinceare coplanar

    Þ their linear combinations will also be coplanar

    Þ = 0.

    • 100. (D); Let f(x) = (x –a)(x –b) –1

    Þ f(a) = –1 < 0 and f(b) = –1 < 0

    Þ x1 Î ( – ¥ , a) and x2 Î ( b , ¥ ) .

    101. (A); |zi|2 = 1 Þ , i = 1, 2, 3.

    Þ |z1 + z2 +z3| = = =1.

    102. (C); Let a , a 2 be the roots

    a + a 2 = –, a 3 =1. Hence a =1, w or w 2.

    But a ¹ 1 as p > 0 Þ a = w or w 2 Þ Þ p =3.

    103. (C); y2 = k. Directrix: x –

    Comparing with x –1 = 0, we get Þ 32 – k2 = 4k

    Þ k2 + 4k –32 = 0 Þ ( k+ 8)( k –4) = 0 Þ k =4, –8 .

    104. (B); Let f(x) = ex –1 –x

    f¢ (x) = ex –1 > 0 " x Î (0, 1)

    g(x) = loge(1+x) – x

    g¢ (x) = = < 0

    so g(x) is decreasing Þ loge(1+x) < x.

  • Alternatively: Using graphs of y = ex, y = 1+x, y = ln(1+x), y = sinx, y = x and y = lnx it can be shown that ln(1+x) < x is the only correct choice.

    • 105. (B);

    = –= = .

     

     

     

     

     

     

     

     

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