3- Solution Report (3)

Paper Code : 0000CT103116004 CLASSROOM CONTACT PROGRAMME (Academic Session : 2016 - 2017) LEADER & ENTHUSIAST COURSE TARGET : JEE (MAIN) 2017 Test Type : ALL INDIA OPEN TEST Test Pattern : JEE-Main TEST DATE : 19 - 03 - 2017 ANSWER KEY Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Ans. 3 1 3 1 1 1 4 3 4 2 3 3 2 2 2 1 3 1 1 2 Que. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 Ans. 4 3 2 3 4 1 4 1 1 4 1 2 3 3 1 2 4 3 2 1 Que. 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 Ans. 4 2 2 2 4 3 3 2 2 3 4 4 4 3 3 3 4 3 2 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 2 3 3 1 4 4 1 2 1 1 3 3 4 2 1 4 3 1 Que. 81 82 83 84 85 86 87 88 89 90 Ans. 2 2 4 1 2 1 2 3 2 1 1. Ans. (3) HINT – SHEET 2. Ans. (1) Sol. i  dQ  3 12t dt v0 qE0 = max t= 1 sec Sol. 4 R H – 1/4 3  12t2  Rdt 0 = mv0 0 0 v = 2v0 3  12t3 1/4 2 = 3  12 R 0 v2x  v 2  4v20 = 1 27  3R 0 36 4 vx = 3v0 = qE0 t m0 Corporate Office :  CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan)-324005 HS-1/8 +91-744-5156100 [email protected] www.allen.ac.in

3. Ans. (3) ALL INDIA OPEN TEST/LEADER & ENTHUSIAST COURSE/JEE (Main)/19-03-2017 8. Ans. (3) R Rsin t ×2 Sol. 345 × 106 × 2 × 10–2 × 2 × 10–2 = F = mg 2 13800 kg = m t 9. Ans. (4) Sol. R R Sol. p =  gdy  p0 r 10  = 10 × 100  6R2 dh  105 0 S = 2R sin  t  = 104 + 2 × 104 + 105 = 1.3 × 105 Pa  2  10. Ans. (2) Sol. Jc = E0 sin (t – kx) 4. Ans. (1) E0 + id 0 dE = 0 A dE + dt dt – + Sol. – B = 0 × A E0 cos(t– kx) – Jd = 0E0 A 5. Ans. (1) Jc   Jd 0  3GMm 11. Ans. (3) Sol. Ui =  2R Sol.  + x = 2.12 GMm 2 + x = 4.10 Uf = R x = 0.14 12. Ans. (3) GMm mgh 13. Ans. (2) 2R 2 U =  Sol. dH  AT4 dt = 4 10  6.4 106  1.28 108 J dH  5.67 108  4004 2 dt / A 6. Ans. (1) vrelx  2v m = 5.67 × 64 × 64 > 2000 W/m2 Sol. vx = v 14. Ans. (2) Sol. When magnet is divided into two equal vy  tan 45 m parts, the magnetic dipole moment 2v smooth vy = 2v 45° M' = pole strength × 1  M 22 Rmax = v2y 2v2 2g  (pole strength remains same) Also, the mass magnet becomes half, ie, g 7. Ans. (4) m Sol. px = 3NS m' = py = 4NS 2 pz = –5NS Moment of inertia of magnet p = 32  42  52  5 2NS m2 I= 12 HS-2/8 0000CT103116004

New moment of inertia ALL INDIA OPEN TEST/LEADER & ENTHUSIAST COURSE/JEE (Main)/19-03-2017 23. Ans. (2) I'  1  m   2  m2 12  2   2 12 8    Sol. 1   I' = 8 T = 2 I 5 sin   1   4 Now,  MBH   = 53° T'  2  I'   2  I/8  P 2 1  cos 53 1 2 1  20%            M 'BH MBH / 2 PT 4 2 5 5 T  T'  1 24. Ans. (3) 2 T2  T' = Sol. N1  N e1t 01 15. Ans. (2) N e2t 02 16. Ans. (1) N2  17. Ans. (3) N1  2  e1 2 t N2 1 A  1 P  1 q  2r  3 s Sol. A 2P 2q r s 1t = 3n2 = 0.5 + 1.5 + 1 + 1 = 4% 2t = 3n2 × 3  2n2 18. Ans. (1) 4.5 N1  2  en2 1  N2 10 10 Sol. 25. Ans. (4) A 20 C 26. Ans. (1) Req = 20 2R2 42   1 2  20 dH  100  5units Sol. h =   = 0.02m dt 20  19. Ans. (1) 2g 20 Sol. y = 2x – x2 = 0 27. Ans. (4) x=0 x=2 Sol. n  v  275 F = Bi 2 1.95 = B0i × 2 20. Ans. (2) n 1  v  330 21. Ans. (4) 22. Ans. (3) 2 1.95 0000CT103116004 v  55 2 1.95 v = 55 × 3.9 = 214.5 m/s HS-3/8

ALL INDIA OPEN TEST/LEADER & ENTHUSIAST COURSE/JEE (Main)/19-03-2017 28. Ans. (1) 37. Ans.(4) Sol. Equation of motion Sol particles are positively charged and T T hence, coagulation occur at cathode. a2 a1 38. Ans.(3) mg f Formula based. mg – T = ma2 T + f =ma1 39. Ans.(2) Angular acceleration   (T  f)R L.E. : 1 H2  HCOO–  HCOOH + e– mR 2 2 vm = vc + R ; a2 = a1 + R + e–  1 2 also a2 = 2R ( a1 = R) R.E. : CH COOH CH COO– + H2 3 3 29. Ans. (1) Net reaction : CH COOH + HCOO–  3 30. Ans. (4) 31. Ans.(1) CH COO– + HCOOH 3 Z.A d = NA.V Keq = Ka (CH 3COOH)  1.8 105  3 K a (HCOOH) 2.4 104 40 or, 5 = 6A 2 1010 )  A 6 1023  24 2  (100 = 48. 0.06 .log 3  0.0672V  Eº = 1 40 32. Ans.(2) cell [HA] = 20  0.4  0.08M 40. Ans.(1) final 100 11 C H (g) + O (g)  4CO (g) + 3H O(l) 46 22 22  [H+] = Ka.C H = [4 × (–94) + 3 × (–68)] – [–30] + = 4 107  0.08  32 109 10 – 4 × 20 – 3 × 10 and pH = – log (32 × 10–9)1/2 = 3.75 = –650 kcal/mol 33. Ans.(3) 41. Ans. (4) 34. Ans.(3) 42. Ans. (2) As T = T –T = 0, U = 0, H = 0, 43. Ans. (2) fi 44. Ans. (2) but S  0 because volume of gas is increased. 35. Ans.(1) 45. Ans. (4) ln K2  Ea 1  1  46. Ans. (3) K1 R  T2  47. Ans. (3)  T1  48. Ans. (2) or, ln 0.08  Ea 1  1  49. Ans. (2) 0.04 2  300 310  50. Ans. (3)  E = 13020 cal / mol = 13.02 kcal/mol 51. Ans. (4) a 52. Ans. (4) 53. Ans. (4) 36. Ans.(2) 54. Ans. (3) 55. Ans. (3) 2FeS + 11 O2  Fe O + 4SO 2 2 23 2 1 mole 20  1 mole 3 120 6 SO + 2NaOH  Na SO + H O 56. Ans. (3) 2 23 2 57. Ans. (4) 58. Ans. (3) 1 mole 2 mole 59. Ans. (2) 3 3 60. Ans. (3) 2 = 1  400  M M= 10 3 2 1000 3 HS-4/8 0000CT103116004

ALL INDIA OPEN TEST/LEADER & ENTHUSIAST COURSE/JEE (Main)/19-03-2017 61. Ans. (2) 66. Ans. (1) S = (1 – )(1 – 2)+.....+(2017 – )(2017 – 2)  ae,  b2  ends of L.R  a    2017 2017 S =  (n  )(n  2)   (n2  n  1)  tan gents are  e x  1 y  1 n1 n1 aa  Area  2a2  a2e2  2017.2018.4035  2017.2018  2017 e 62  e3  2 S.   2018.4035  10091  62. Ans. (3) 2017  6  even x  y 1 = (odd + even) = odd ×  y=x+1 0 1  cos  S   cos  oddx  1  2017  1 67. Ans. (4)  x  1  x2 1dx 0 1 Reflexive, symmetric but not transitive.  68. Ans. (4) 6 63. Ans. (2) dy  x sin2 y  sin y cos y 2017C0 + 2017C1 + ..... 2017C1008 = 22016 = 2 dx  = 21008  8.32201 = 8(33  1)201 = –8 = 25 cosec2y dy  x  cot y 69. Ans. (1) dx Fixed point of family is (1,–1) Let – coty = v  other bisector is L1 (2,3) B1 dv  v  x y 1   1 x 1 dx 4   cot y.ex  xexdx x + 4y + 3 = 0 P(1,–1) L2  coty = (x – 1) + Ce–x 70. Ans. (2) 64. Ans. (3)  b  cos x2  a x x 2    lim 2 x x  0  x  2,2 x 2 x x2a x2  a sin x2  a Let x2 – a = t  n = 2  (n) = 0 lim b  cos t 65. Ans. (3) t0 t sin(c(t  a)  a) D(0,,0)  b 1  b=1 0 A(2,1,–1) C(2,–1,3) lim 1  cos t  lim 2sin2 t 2 t0 t sin(ct  a(c 1) t0 t sin(ct  a(c 1)) B(3,0,1) sin t 0  c=1 lim 2 sin a(c 1) 2 1   1 1 3  1  5 t0 sin ct  a(c 1) 6 sin t 1  L1 2 1   3 lim 2 2   = 8, –7 t0 sin t 2 Sum = 1 0000CT103116004 HS-5/8

ALL INDIA OPEN TEST/LEADER & ENTHUSIAST COURSE/JEE (Main)/19-03-2017 71. Ans. (1) 2  3  24  1  0 3 1 1 x put in P2 : 16 + 2 sin 2  x 2 sin cos x  cos 2x  cos 3x  cos 4x   0 2 2 1 1  9x x  5  5  0  1 2 2 sin  2 2  3    sin  sin  0 3 x () = (3, 1, –2) it also satisfy P1  = 2 2 sin  9x  76. Ans. (2)  2    0  x  2n ,n  9m m  I p ~p q pq q  ~p (pq) (q  ~p)  x  9 sin  2  T FT T T T T FF F F T 72. Ans. (1) F TT T T T tan1 tan1 tan 1  1  F TF T T T  2   x  2   x  2  77. Ans. (1) tan1  1 x 2 x 2 2   tan1  1       2   x 2x P(2t2,4t) x = 1, –5(reject) Q 73. Ans. (3) x+y+4=0 34 terms so mean of 17th and 18th term is for minimum distance dy  1 median dx P x10+n = 148 + (n–1) (–2) = x17 = 136, x18 = 134  t = –1 hence median = 135  min distance = PQ = 2 2 74. Ans. (3) 78. Ans. (4) B(0,4) Total cases  15C2.2! = 15.14 2x = 3y  (3, 2),(6, 4),(9, 6),(12, 8),(15,10) Fovourable cases = 5 A(–5cos,–4sin) C(5cos,–4sin) 51 Probability = 15.14 = 42 Area  1 10 cos  4  4 sin  79. Ans. (3) 2 cot x  1  cot x  tan x  dA  0     2  2 2  d 6 A max  15 3 cot x  1 1  cot x  tan x   tan x 2   4 4    2 2  75. Ans. (4)  1 cot x  1 tan x  1 tan x 4 44 42 2 8 1  Let point be  3 , 3,6  which also  1  cot x  tan x   1 tan x  1 tan x 8  8 8  4 4 2 2 satisfies both the plane P1 = 0 = P2 HS-6/8 0000CT103116004

ALL INDIA OPEN TEST/LEADER & ENTHUSIAST COURSE/JEE (Main)/19-03-2017 1  x  1  x .... 83. Ans. (4) 2n  2n  2n1  2n 1  cot x  cot  tan  1 tan  x   1 tan  x   1 tan  x  4  x2 6 2 8  8  4  4  2  2  6 9  x2 2 3  x x3 14  x2  3 1  x2  1 tan x  1 tan x  1 tan x ......  1 cot  x   cot x 2 2 4 2 8 8 2n  2n  84. Ans. (1)  1 tan x  1 tan x  ..... terms A B 2 24 4  lim 1 cot  x   cot x  1  cot x 2n  2n  x n   P Q  10 put x   10 10 2 d 80. Ans. (1) d = 10cot; d = 30cot2 B 10cot = 3cot2 P(x,y) y=ƒ(x)   = 30º 85. Ans. (2) 4  r2  4  r (0,2) (r,2) A r3 0 2 C  AC = BC 86. Ans. (1)  P is mid point of AB Let lengths of sides are 2a, 2b, 2c  A(2x.0) & B(0, 2y) = B(a,b,–c), A(–a, b, c), C(–a, –b, –c)  dy y OB.OC cos   c2  a2  b2   a2  b2  c2 dx x |OB||OC|  xy = c  xy = 6 Similarly 81. Ans. (2)  cos   1 2 n 87. Ans.(2)  p1 n(n  1) m(m 1) 2 2    80  t3 |x| |x|1  t3  5|x|  3   3 t dt 6  nt  t       0t n2(n 1)2  n(n 1)(2n 1)  n(n 1)  80  0 4 12 4 n|x||x3|3  |x|2 |x| 5|x|  n=4 |x|  6 82. Ans. (2)  3 11 1 n|x||x3|2  |x| 1  5   2 ah1  2 bh2  2 ch3 1  6 3  h1  2 and h2  2 and h3 2 n|x||x3|2   |x|  1 a b  1 3 6 c  1  1  1  1 a  b  c = 27 n|x|(|x|2 3)  2|x|1 h1 h2 h3 2 15 2 0000CT103116004 HS-7/8

ALL INDIA OPEN TEST/LEADER & ENTHUSIAST COURSE/JEE (Main)/19-03-2017  n|x| 2|x|1 Total number of numbers = 1 2(|x|2 1) (when we use only 0) |x| = t = t > 0 Ans. = 1010– 910 – 1 90. Ans. (1) nt  2t 1 2(t2 1) 3ˆi  2ˆj  5kˆ  (2  µ  2)ˆi  (  3µ  )ˆj    2µ  3)kˆ  2 solutions for |x|  4 solutions. 88. Ans. (3)  2 + µ – 2 = 3 –  + 3µ +  = 2 C  B C  – 2µ – 3 = –5   = 1,  = 4,  = 5 89. Ans. (2) I II III IX X Total number of numbers = 1010 (without any restriction) Total number of numbers = 910 (when we do not use 1) HS-8/8 0000CT103116004