4- Solution Report (4)

Paper Code : 1001CT102116063 CLASSROOM CONTACT PROGRAMME (Academic Session : 2016 - 2017) JEE (Main + Advanced) LEADER COURSE (SCORE-I) & ENTHUSIAST COURSE (SCORE-II) ANSWER KEY TEST DATE : 12-03-2017 Test Type : FULL SYLLABUS Test Pattern : JEE-Main Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Ans. 4 4 2 4 3 2 1 2 1 3 2 2 2 1 4 3 2 3 4 3 Que. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 Ans. 2 4 4 2 4 4 4 3 4 2 3 1 4 4 2 3 3 3 2 3 Que. 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 Ans. 2 1 2 1 4 4 2 1 1 2 4 3 4 4 4 3 4 4 3 3 Que. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 Ans. 2 3 1 3 3 2 1 2 2 2 4 2 3 4 1 2 3 2 3 1 Que. 81 82 83 84 85 86 87 88 89 90 Ans. 3 3 1 2 1 3 4 2 3 1

Paper Code : 1001C T102116063 CLASSROOM CONTACT PROGRAMME (Academic Session : 2016 - 2017) JEE (Main + Advanced) LEADER COURSE (SCORE-I) & ENTHUSIAST COURSE (SCORE-II) Test Type : FULL SYLLABUS Test Pattern : JEE-Main TEST DATE : 12 - 03 - 2017 SOLUTION 1. Ans. (4) P3  90°– Sol. F = F – mg =Vg – mg = r2hg – mg P1 P2 net b 2. Ans. (4) Sol. mT = constant 3. Ans. (2) Sol. Flux linked with both loops is same, so emf Intensity of light emitted from P induced is same for both loops. 3 4. Ans. (4) Conceptual I1  I0 cos2  5. Ans. (3) 2 Sol. Between 2 – 3sec both x and y position are not Intensity of light transmitted from last polaroid changing. P2 = I1cos2 (90° – ) 6. Ans. (2) Sol. Some of the characteristics of an optical fibre P= I0 cos2 sin2  2 2 are as follows (i) This works on the principle of total internal P = I0 (2 sin  cos )2 reflection. 28 (ii) It consists of core made up of glass/silica/ P2 = I0 sin2 2 8 plastic with refractive index n , which is 1 8. Ans. (2) surrounded by a glass or plastic cladding Sol. For permanent magnet we prefer a material with refractive index n2 (n2 > n1). The refractive index of cladding can be either changing abruptly or gradually changing with high retentivity (so as to make a stronger (graded index fibre). magnet) and high coercivity (so that (iii) There is a very little transmission loss magnetization any not be wiped out easily). for through optical fibres. electromagnet we prefer high saturated (iv) There is no interference from stray electric magnetism low coercivity and least possible and magnetic field to the signals through area of hysteresis loop so that electromagnet optical fibres. develops high magnetization, is easily 7. Ans. (1) demagnetized and energy loss in a magnetization cycle is least. Therefore, P is Sol. Let initial intensity of light is I0. So intensity suitable for making permanent magnet and Q of light after transmission from first for making electromagnet. polaroid = I0 . 2 Corporate Office :  CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan)-324005 HS-1/7 +91-744-5156100 [email protected] www.allen.ac.in

Target : JEE (Main + Advanced) 2017/12-03-2017 9. Ans. (1) 15. Ans. (4) 10. Ans. (3) Sol. From Newton's law of cooling Sol. F = mg sin  ms  1  2    1  2  0  mg tan   t   2   mg dy  mg  2x in the first case  50  40    50  40  0  dx 40 5   2   ms  a= xm 2 ......(i) a =  x in second case  45  41.5   45  41.5    2  5   2  0 1 ......(ii) = 2 11. Ans. (2) From (i) and (ii) Sol. For constant pressure process. 0 = 33.3ºC Work done (W) = nRT. 16. Ans. (3) Sol. Flux going in pyramid = Q 20 Which is divided equally among all 4 faces  T   25   Flux through one face = Q  nR  80 Now, for above process f = 6, 17. Ans. (2) So CP 1  f  R  4R Sol. Behaviour of lens is opp. If outside medium is  2   denser than lens medium  Heat absorbed = nC T 18. Ans. (3) P = n × 4R × 25 100 J Sol. A1  1N1 nR A2 2N2 12. Ans. (2)  N0e 2 n2 n2 1 2 n2 2 2 Sol. At equilibrium dU  0 2 2 dr  N0e 4 n2  1  2 12A 6B 4 2 r13  r7  0 19. Ans. (4) r   2A 1/ 6 Sol. La = Lb  B mvar = 3mvbr   13. Ans. (2)  va =3:1 vb R0 R0 20. Ans. (3) Sol. R0 R0 R0 R0= R/3 v(m/s) R0 R0 R0 R0 7R0 = Req 10 R0 3 Sol. R0 t(s) R0 1s t 2s 14. Ans. (1) 1 1 210 10t  135 Sol.  h  h 6.63 1034 mv 2Km 2  54 1.6 1019  9.11031 2 t = 12s  (3) HS-2/7 1001CT102116063

21. Ans. (2) Leader Course (Score-I) & Enthusiast Course (Score-II)/12-03-2017 28. Ans. (3) Sol. Induced e.m.f. = Bv = 0.125 V Sol : geff = 5g current l = 10  e  10  0.125V R 2  P  5gR 0.5T 29. Ans. (4) Sol : Use concept of series combination. + 10V 0.25m 30. Ans. (2) – 0.5N 2 V Q Sol. vv Force Bl fb  4  4  x = 0.5  10  0.125V  0.25  0.5N(given) v  1 1  v  x  2  4    4  =    x      x  Solving V = 16 m/s. 22. Ans. (4)  vx ( x << ) 42 L C Sol.  2 cos  = v0  31. Ans. (3) v=?  L v0 Sol. Metal hydrides are less dense than the parent  2 sin  = v metal because the crystal lattice has expanded  v = v0 tan  through the inclusion of hydrogen. 23. Ans. (4) Sol. y  A  B B 32. Ans. (1) (4) 33. Ans. (4) 24. Ans. (2) 34. Ans. (4) Sol. 40 × 10–3 = mvp2 v = 50 × 10–3 × vp2 × 20 Sol. 2KMnO4  K2MnO4  MnO2  O2 4KMnO4  4KOH  4K2MnO4  O2  2H2O 4 = v 2  0.2 m/s = v = 20 cm/s 2MnO2  4KOH  O2  2K2MnO4  2H2O 100 p p KMnO4  H2SO4 (con)  K2SO4  Mn2O7  H2O 35. Ans. (2) 25. Ans. (4) Sol : Resistance in box is 30  26. Ans. (4) Sol. 1msd = 200 × 0.005 = 1mm 2r = 4 × 1 + 25 × 0.005 – 5 × 0.005 Sol. 6PCl5  P4O10 10POCl3 = 4.1 POCl3  3H2O 3HCl  H3PO4 r = 2.05 mm 36. Ans. (3) (2) Sol. Due to synergic bonding between Pt and ethylene, C2H4 is distorted and bond angles 27. Ans. (4) change from free C2H4. C 60 103 103 37. Ans. (3) Sol. W  J/s J/s cop 601.2 1.2 No of units consumed  103   4  30  Sol. (1). H PO  one P–H bond.  1.2  33    = 100 units (2). H4P2O7  no P–H bond. 1000 (3). H P O  two P–H bond. Cost of energy = (6) (100) = 600 Rs 42 5  (4) (4). HPO  no P–H bond 42 6 1001CT102116063 HS-3/7

Target : JEE (Main + Advanced) 2017/12-03-2017 38. Ans. (3) 45. Ans. (4) Sol. K [Co(C O ) ] : diamagnetic, NO[PF ] : Sol. Gold number is minimum amount of protective 3 2 43 6 colloid when can protect 10 ml standard gold sol from coagulation when 1 ml of 10% NaCl NO [PF6] is added. dia dia [NMe4]O3 : NMe4 O3 , H[BF4] : 46. Ans. (4) dia Para 47. Ans. (2) H BF4 dia 0.0591 1 2 [Cu2 ] 39. Ans. (2) Sol. E= E0cell – log . cell 10 2 2– 48. Ans. (1) O HO O OH Sol. H solution < 0 ; Exothermic B Sol. 1 B 3 Temperature  ; Solubility  for X, O OH HO O 49. Ans. (1) 40. Ans. (3) Sol. U = 2RT Sol. Hydrated aluminium chloride is formed in (1) MPS M and (2) 41. Ans. (2) U = 3RT RMS M Sol. Rn1 = n12 = 1   n1 = 1,2, 3 etc. 8RT Rn2 n22 4 n2 24 6 Uav = M Among the first four orbits n and n can be 1 12 50. Ans. (2) 51. Ans. (4) and 2 or 2 and 4.  Energy difference can be : E = 10.2 eV or E = 2.55 eV. Sol. H /Pd-BaSO reduced alkyne into alkene. 21 42 24 42. Ans. (1) 52. Ans. (3) Sol. K depends only on temperature and mode of Sol. Compound CH3 can not exhibit P CH3 representation so  will change on changing pressure and PH2 > PN2 . geometrical isomerism. 43. Ans. (2) 53. Ans. (4) Sol : Suniverse : Ssystem + Ssurrounding Sol. Refer NCERT = 100 (0.4 – 0.3) + (75–80) 54. Ans. (4) = 5kJ/K Sol. Internal cannizzaro reaction then esterification 44. Ans. (1) 55. Ans. (4) Sol : 2C6H5COOH  (C6H5COOH)2 O Sol. Product is CH3 – C – CH3 t =0 1 0 Degree of unsaturation = 1. t 1–x x/2 i = 1 – x + x/2 It can undergo aldol condensation. i  Tf  0.504 56. Ans. (3) kbm 57. Ans. (4) 1– x  0.504 Sol. More is the stability, lesser will be heat of 2 hydrogenation. x = 0.992 HS-4/7 1001CT102116063

Leader Course (Score-I) & Enthusiast Course (Score-II)/12-03-2017 58. Ans. (4) 66. Ans. (2) Sol : Refer NCERT Ellipse and circle touches each other at (2,0) 59. Ans. (3)  Length of common chord = 0 Sol : Refer NCERT 67. Ans. (1) 60. Ans. (3) ,  are roots of x2 + x +  = 0 is 61. Ans. (2)  +  = – ...(1) 1  icos   1  i cos   1  2i cos  and  =  ...(2) 1  2i cos  1  2i cos  1  2i cos   from (1) & (2)  = 1,  = –2 1 2 cos2   3i cos  Now ||y + 2| – 1| < 1 = 1 4cos2  –1 < |y + 2| – 1 < 1 0 < |y + 2| < 2 3cos   –2 < y + 2 < 2 and X  –2 is a real number only if 1 4cos2  = 0 i.e. if cos = 0 y  (–4, 0) – {–2} 68. Ans. (2) i.e. if  = (2n + 1)/2, n  I 1 1 1  A  (r  1) D a b c So (b) is correct alternative.  A  (p  1) D ,  A  (q  1) D , 62. Ans. (3) 1  1  (p  q) D or p–q= ba and so on ab abD 1 1  2 2 m  Pi .x i   1  c b ˆi a c ˆj b a kˆ u     bcD caD abD 2  m 2 1 4 Pi  xi  3  6  1 Consider   c b  ac  ba = 0 u.v abcD abcD abcD 63. Ans. (1) 69. Ans. (2)  6!  Let the equation of plane be  3!3! 26 1  20 a(x + 1) + b(y – 3) + c(z + 2) = 0 63 –3a + 2b + c = 0 plane passes through (0, 7, –7) 64. Ans. (3) a + 4b – 5c = 0  1  cos 3  a = b = c  2 4   plane is x + y + z = 0 log perpendicular line on which image lie is   E = 1  x1  y2  z1  41cos 4    111   foot of perpendicular lies on plane  1  1   2 1  2  2 2  ( + 1) + ( + 2) + ( – 1) = 0 = log 1   = log 2     2  2  4 3 41 1 2 1 2   1 4 5  2 1 foot of perpendicular is  3 , 3 , 3  21  2 2  = log =1 22  & image point is  1 , 2 , 7  .  3 3 3  65. Ans. (3) 70. Ans. (2) 4sin 9sin 21sin 39sin 51sin 69sin 81 DAB  ADC 180º sin 54  ADP  DAP 90º (because DQ and AS = 4 sin9 cos9 . sin39 cos 39 sin 21 cos 21 are angle bisectors of angle A&D) sin 54 = sin18 . sin78 sin 42  DPA  90º  SPQ  90º . Similiarly, 2 sin54 PSR 90º , SRQ  90º , PQR  90º .So = sin18 (cos36  cos 60) 1 PQRS is a rectangle. =8 4 sin 54 1001CT102116063 HS-5/7

Target : JEE (Main + Advanced) 2017/12-03-2017 71. Ans. (4) 77. Ans. (3) a             Since 32 –  < 100 < 32 b [ b c a]b [b c a]c 2 =       a.b  |  |2   a.c    b.c a  sin–1 (sin 100) = 100 – 32 [b c a] b b a b         tan–1 (tan 100) = 100 – 32 = [a b c][(a  c  b  a)b  (| b |2 b  c)a] cos–1 (cos 100) = 32 – 100 72. Ans. (2) cot–1 (cot 100) = 100 – 31 3 3 78. Ans. (2)  sin2 sin2 2d  2 sin2 sin2 2d The given differential equation can be written 3 0 as y5 xdx + ydx – xdy = 0 . Multiplying by 3  x3/y5 , we have   8 sin4  cos2 d  24 sin4  cos2 d x4 dx  x3  ydx  xdy   0 00 y3  y2    /2 Integrating, we get x5/5 + (1/4) (x/y)4 = C  48 sin4  cos2 d 0  48.3.1.1 .   3 79. Ans. (3) 6.4.2 2 2 12 73. Ans. (3)      Area = ex  (1 x) dx + ex  (x  1) dx 01 Let side of square and cube = x  A  x2  dA  2x dx  dx  1 dt dt dt 2 Now V  x3  dv  3x2 dx  dv  6 dt dt dt x2 74. Ans. (4) ƒ(x + 2y) = 2yƒ(x) + xƒ(y) – 3xy + 1  x2 1  x2  2  ƒ(0) = 1  2 0  2  let y = 0, then ƒ(x) = x + 1 = e x x + e x   x  1  = e1 1   e1 1 1 75. Ans. (1)  1 2 – 1+ e2 2 2 –   2   Minimum distance will be along the common = e2 – 2 normal of parabola y2 = 8x is y = mx – 4m – 2m3 80. Ans. (1) should passes through centre of circle x2 – 2p x + p  4 = 0 i.e. (–2,–8) p5 p5 m=1 foot of normal is (am2, – 2am) = (2, –4) Minimum distance between curves 4 2 2 3 2 76. Ans. (2)  f(0) > 0, f(2) < 0, f(3) > 0 Range :   ,   //////////////////////////2///////////// p4  4  1/2 f(0) > 0  p  5 > 0...... (1) Non derivable at x  1 p  24 f(2) < 0  p  5 < 0..........(2) 2 4p  49 f(3) > 0  p  5 > 0..........(3) I and II are wrong  49 , 24   4  Intersection of (1) (2) & (3) gives HS-6/7 1001CT102116063

Leader Course (Score-I) & Enthusiast Course (Score-II)/12-03-2017 81. Ans. (3) 87. Ans. (4) Sum of roots = sum of diagonal elements f(1–) f(1) and f(1+)  f(1)  2 = 1 + 2 + k  k = –1 and product of roots = value of the determinant. – 2 + log (b2 – 2)  5 2 12 3 0 < b2 – 2  128 2 < b2  130  = 2 2 1  2k  24 88. Ans. (2) 30 k f(x) = 2 has 3 solutions x = –3, 1/2, where  (2 < <3). 82. Ans. (3) Now f(x) = –3 has no solution nC1.21 + nC2.22 + nC323 + .... + nCn2n f(x) = 1/2 has 2 solutions (1 + 2)n = nC0 + nC121 + nC2.22 + .... + nCn2n f(x) =  has 2 solutions (3n – 1) = nC1.21 + nC2.22 + ....... + nCn.2n So, f(f(x)) = 2 has 4 solutions 83. Ans. (1) 89. Ans. (3) Let d1 & d2 are the distance of point (1, 2) from f(x) is symmetrical about the line x = 7. the bisector B & B . Let x1, x2, x3, x4 and x5 are the real and distinct 12 387 4 x1  x5  7 , d= =5 roots of f(x) = 0. Then x3 = 7, 2 15 4  6  14 16 x2  x4  7 . d2 = 5 =5 2 d1 < d2 S = x1 + x2 + x3 + x4 + x5 = 35 B1 is an acute angle bisector S/7 = 5 84. Ans. (2) 90. Ans. (1) As altitude from A is fixed and orthocentre lies Replacing even numbers by zero and odd on altitude hence x + y = 3 is required locus. numbers by one, we have 85. Ans. (1) Curve represent a triangle formed by x = 0, 100 y = 0, x + y = . |A| = 0 1 0 = 1 86. Ans. (3) 001 f(a2) – 3f(a) = 0  (a4 + a2 + 1) –3(a2 + a + 1) = 0 which is an odd number and hence |A| can not (a2 + a + 1) (a2– a + 1) –3(a2 + a + 1) = 0 be zero. Hence A is invertible for all n  N. (a2 + a + 1) (a2 – a – 2) = 0 1001CT102116063 HS-7/7