2- Solution Report (2)

Paper Code : 1001CT102116064 CLASSROOM CONTACT PROGRAMME (Academic Session : 2016 - 2017) JEE (Main + Advanced) LEADER COURSE (SCORE-I) & ENTHUSIAST COURSE (SCORE-II) ANSWER KEY TEST DATE : 21-03-2017 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 2 3 1 3 4 1 4 4 2 4 4 4 2 3 3 1 4 2 1 Que. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 Ans . 4 4 3 1 2 4 2 2 2 4 1 1 3 2 2 3 1 4 4 2 Que. 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 Ans . 2 3 4 1 4 3 3 4 2 2 2 1 2 4 2 3 2 4 3 4 Que. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 Ans . 2 4 3 2 2 4 3 2 1 4 3 3 2 3 3 4 4 2 2 2 Que. 81 82 83 84 85 86 87 88 89 90 Ans . 2 4 3 2 4 3 3 1 2 2

Paper Code : 1001C T102116064 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 : 21 - 03 - 2017 1. Ans. (4) SOLUTION Momentum conservation ucos Mu  M  Mu' u    u' Angular momentum conservation usin M.u   Mu'   M2 . 2 2 12 ucos u  u'   6 h  2u sin   h  2u2 sin2  4H u sin  g g For elastic collision     u'  u 2 2. Ans. (2) Surface Area = 2 [ab + bc + ca] On solving  2u ;   12u = 2 [ 1.5 × 1.5 + 1.5 × 1.0 + 1.5 × 1.0] 5 5 = 2 [ 2.25 + 1.50 + 1.50] = 2 [ 2.2 + 1.5 + 1.5] now KE of upper half part = 2 × 5.2 = 10.4 cm2  1  M   Vcm 2  1 .Icm .2 3. Ans. (3) 2  2  2 t.u1 cos 1  5  5/u1 cos = t  1 M .  2u  3u 2 + 1 M .  2  12u 2 2 2  5 5  2 2 2  5  10  t. u2 cos 2  10 ; t = u2 cos2  Now 12 5 10 Mu 2  t t   25 2m  5. Ans. (3) F  P   15m  15 m g For the given situation disc will perform t t2 2h 2t translatory motion in radius l. Hence case is F  15 0.210  7.5N like simple pendulum (Refer to H.C. V. exercise 22 S.H.M.) 4. Ans. (1) 6. Ans. (4) For speed to be zero. v 1 K (S  )2  mg(S )  2 MM S2 + 2 - 2S  = 2mg .S K u u’  KS2–2 KS  2mgS  K2  0 Corporate Office :  CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan)-324005 HS-1/13 +91-744-5156100 [email protected] www.allen.ac.in

Target : JEE (Main + Advanced) 2017/21-03-2017 2K  2mg  4(K  mg)2  4K2 k 10. Ans. (2) S= at some distance from centre inside core 2K  G 4  r3 3   m  S = K  mg   K 22  m2g 2  2Kmg  K 22  3 r   2 K F =      K  mg   2mgK  m2g2 ma = – 4 Gmr S= a = – 4 Gr K so   4 G  2 Now maximum speed will be at equilibrium T position 1 mv2  1 K  mg 2  mg    mg  or T = 2 . 1  2 2  K  K  4 G G     1 now time for A to B 2 G 1 mv2  1 m2 g 2  mg  m2 g 2 11. Ans. (4) 2 2K K 1 mv2  mg  m2g 2 GMm 2 2K For energy in radius r = 2r =nk i v = 2g  mg 2 where n is integer k is constant energy , now K GMm = nk 2R 2 GMm = (n-1) k time of free fall is g  3R   2  and clearly option 4 is wrong. 2 7. Ans. (1) GMm diameter = 1 × pitch + L.C×C.S.R so solving k = 6R = 1.5 mm + 1.5mm  76  2.64mm now this implies that for R 100 max 8. Ans. (4) GMm  1 GMm  Rmax  3R 2Rmax 6R particle speed = –v × slope w 12. Ans. (4) = 2×(2/1) at 3 sec . Let R  1 . l 4 Ka 9. Ans. (4) 2K K 2K = nCV t  Vf PdV E F Q 2K D K C 2K Vi C= U W A B nΔT nT nT =C +  . Vf VdV K V nt Vi R/4 R/2 R/4 R 1 V 2  Vi 2 1 2 f =   nT =  R  1 PfV f  PiVi R 1 2 nT = R  1  1  R   1 Req = R  1     1 2    2 2  ka  2    1   HS-2/13 1001CT102116064

Leader Course (Score-I) & Enthusiast Course (Score-II)/21-03-2017 13. Ans. (4) Also, angle rotated by wheel in given time, av   it  1  t2 4A 2 4A h = 2.5 (7)  1 272  6rad  343.7 P1 av 2 A A i.e. wheel is 16.3° short of complete revolution  at that instant , so, desired colour is green. h 17. Ans. (1) a Finally system behaves as shown v P0  1 2v2  P1  2 gh  1 2   av 2 K =1, A =5m2, d=1m 2 2 A   P0  1   av 2   gh  P1  O  1   av 2 C1 C2 2  4A 2  A C3   K=3 C1 = 30(10)   A = 10 m2 K=2 1   d= 1m A = 5 m2 d= 1m C1 =300  v2   a2v2   gh  2 gh   a2v2   a2v2 C2 =50 32 A2 A2 2 A2 C1 =100 v2 1  a2  a2 .v 2  a 2v 2   gh  32 A2 A2 2 A2   (150 )(300 450  100  8.851011 F (net = ) 3gh v= 1  17a2 18. Ans. (4) 32 A2 V = y3+2 14. Ans. (2) y F = (P0 – Paverage). lh B F =  2T   gh  h  d 2  h  2T 2T 2 x  gd  gd 2 A so F = z 15. Ans. (3) V y ln 2 n2  n(2)T0  E= iˆ  3y2 ˆj R half time =  rate constant R / T0 VA = 2 volt 16. Ans. (3) VB = 10 volt [V = y3+2] current in bulbs B = 1A q(V –V )= 1 mv2  1 (8)  1 2V 2 1 BA 2 2 2 max current in B2 = 2A  V = 2 m/s For same illumination current should be same So, velocity of ball before collision = (2m/s)j in both bulbs i.e. i = I (l - e–t/) l = 2 (l - e–t/)  e–t/ = 2 So, velocity of ball after collision = –(1.5m/s)j  t= ln2 = L ln 2 = 20 ln 2 = 7 seconds R 2  change in momentum = m VF Vi  7N.S  j 1001CT102116064 HS-3/13

Target : JEE (Main + Advanced) 2017/21-03-2017 Net force = (–7) /(0.1) = (–70N)j 22. Ans. (4) from FBD of ball during collision B= E  104  3.31013T qE C 3108  Fnet 23. Ans. (3) Fwall Let Ec be the amplitude of carrier wave and Es F Fnet = wall –qE is signal amplitude Fwall = Fnet + qE = (70+6) = 76 N then EC   Vmax  Vmin  , Es   Vmax  Vmin  [E at top face = 3y2 = 3(2)2 = 12 N/C]  2   2  19. Ans. (2) E = mE  4 = m6  m= 2 8V CC R 3 + Ans. 24.(1)   - 2V   .eff vB    2iˆ  3 ˆj  4kˆ  . 3iˆ  4 ˆj   |  | 32volt 25. Ans. (2) Xi 1  Li Xf Ci  E VLED 82 1 i = 20mA = = CF  Lf RR R  6V  300   0n2A 1  1 K  20mA K0A =  =   Also reverse voltage across red diode is 2V 0A 2A K  1 r  which is fine for LED with reverse breakdown  0rn voltage of 3V. 20. Ans. (1) [Using 00  1 ] & [2A2 n2  C2 ] C2 water level y 26. Ans. (4) dy dr r  h =app observerhi  1 dy  tan 1 1    i  0 y2 1 4   21. Ans. (4) X Y Z X .Y Y.Z R  X .Y  Y .Z dR  dr 4r 2 000 1 1 0 010 1 1 0 dR   r2 dr (r2  r1) 4 r1 r2 4r1r2 001 1 1 0  R   R  011 1 0 0 100 1 1 0 Rate of melting is max when power dissipated 110 0 1 0 in sphere is max. Using maximum power 101 1 1 0 transfer theorem, 111 0 0 1 R = r. of battery i.e. (r2  r1)  2 4r1r2  From truth table, answer is AND gate. HS-4/13 1001CT102116064

Leader Course (Score-I) & Enthusiast Course (Score-II)/21-03-2017   8r1r2  8(200)  160 (in SI)  k  78 12  22 = k  25 r2  r1 10 100 k 78  3 25 k 22 Also,   1  10  5 31. Ans. (1)  16 8 27. Ans. (2) Average atomic mass of carbon = 0.988 × 12 + 0.0118 × 13 + 0.0002 × 14 = 12.0122 d 32. Ans. (1) 33. Ans. (3) Considering gaussian surface In H2O strength of H-bonding is greater than in NH3.  qin  E 4d2  2ed3 34. Ans. (2) 0 0R3 35. Ans. (2) E  2edk 36. Ans. (3) R3 For equilibrium of charge, Ag(CN) – is more stable than Ag(NH3)2 2 kee   k2ed  e 37. Ans. (1) (2d)2  R3  1 2d R3 38. Ans. (4) 4d 2 R3 8   d3   d  R/2 a 3 a3 2 2 28. Ans. (2) rc + ra =  180 = When light enters medium of refractive index a= 360 = 120 3 pm. c 3 , its speed decreases, to .   wavefront at point P is option (2). Closest distance between two cation 29. Ans. (2) ynth  n1D ; n = 2, 4, 6, 8..... = a  120 3 pm 2d 39. Ans. (4) ymth  m2 D ; m= 2, 4, 6, 8...... 40. Ans. (2) 2m (Na  )    (SO42 ) 2d m y  ynth mth d from central fringe.  1000(2.6103)  260Scm2mol1 2 0.001 30. Ans. (4) 41. Ans.(2) M k 3 CH3 EDG [strongly activating] EWG +M>-I O - M - I [Deactivating] L k 12 K 78 N C A  hc  H O  k    Ek  EL  [weakly activating] H EDG [moderately activating] B O +M>-I D  hc   Ek  Em  Rate of EAS C < B < D < A    k  1001CT102116064 HS-5/13

Target : JEE (Main + Advanced) 2017/21-03-2017 42. Ans.(3) HBr ,ROOR  Br CH3 CH3 Br (1) light / Heat + Intermediate : CH3 Br Stereochemistry of addition – (Syn + Anti) ++ Regiochemistry of addition – Markovnikov Br CH3 Br CH3 Br CH3 43. Ans. (4) SOCl2 / Py SN 2 Intermediate – Free Radical PBr3  SN 2 OH NaH Stereochemistry of addition – (syn + Anti) Br H2 Regiochemistry of addition – Antimarkovnikov O (2) Br2H2O OH + Br O Na  I Br CH3 CH3 ICl Br OH AlCl3 Intermediate 44. Ans. (1) CH3 + CH3 HO O OH Br HO Br OH Stereochemistry of addition – Anti OH - ketohexofuranose Regiochemistry of addition – Markovnikov CH2OH like HO (3)   (1)B2H6 -THF CH3 HO OH has 7(c) atoms hence not (2)H2O2 -OH- HO H OH OH hexose sugar. + OH HOH2C O HO H HO OH HO OH Stereochemistry of addition – Syn is -Aldohexopyranose. Regiochemistry of addition – Antimarkovnikov OH O OH (4) HBr OH OH OH HS-6/13 -Ketohexopyranose 1001CT102116064

Leader Course (Score-I) & Enthusiast Course (Score-II)/21-03-2017 45. Ans.(4) Cl (4) Cl CH3 - MgCl (1) Mg,diethyl ether O MgCl CH3 Cl  O O MgCl CH3 CH3 Na  H2O O O MgBr 46. Ans. (3) OH H + Na  H 1 + 2 H2  OH + Na  O Na + (2) H  1 H2  (3) 2 1001CT102116064 H2O H OH + 2Na  2Na CH3  COOH  Na  CH3COO N   1 H 2 2 a 47. Ans. (3) CH3CHO NaOH  CHI 3   HCOO  I2 O CH3 - MgCl PhCHO Tollen'sReagent  PhCOO  Ag  CH3CHO Tollen 's Reagent CH3COO  Ag  CHO O MgCl H OH HO H CH3 H2O H OH Br2 H2O OH H OH CH2OH glucose HS-7/13

Target : JEE (Main + Advanced) 2017/21-03-2017 COOH (2) H & H Br H OH Br HO H H OH  enantiomers H OH Cl Br CH2OH (3) &  identical Fructose Br2H2O No Reaction Br Cl O (4) & 2(P,4H4DN5P)  chain isomers NO2 50. Ans. (2) N - NH - NO2 O OH yellow/ orange/red I = H3C H+ H3C H3C O 2(P,4H4DN5P) Chiral H OO H3C NO2 H3C + H3C N - NH - NO2 H enantiomers yellow/ orange/red 48. Ans. (4) (a)NaBH4 OH O (b) H2O II = O H3C O Chiral (a)DIBAL-H H3C Cl (b)H2O Ch3 H no racemization under acidic condition 49. Ans. (2) O O (1) O O H3C N2H4 OH H or  & H3C H  chain isomers H3C HS-8/13 1001CT102116064

Leader Course (Score-I) & Enthusiast Course (Score-II)/21-03-2017 51. Ans. (2) 63. Ans. (3) The correct of order of bond energy in f x  2  x4  3 4x  23 7x2  halogens is I < F < Br < Cl 22 22 Let h x  x4  7x2  4x  23 52. Ans. (1) Bond order in the four cases is 1, 1.33, 1.5 and   x2  4 2   x  22  3 2 respectively. h(x)  3 53. Ans. (2) Range of h(x) is [3,) KNO3  KNO2  O2  Range of f(x) is (2, 3] 54. Ans. (4) 64. Ans. (2) 55. Ans. (2) y = f(x) 56. Ans. (3) It has two geometric isomers and two optical x=1 y=2 isomers. 57. Ans. (2) y=1 58. Ans. (4) x 59. Ans. (3) 2 CuSO4 + 4KI  K2SO4 + Cu2I2 + I2 x=0 60. Ans. (4) 1 Mg3N2  6H2O  3Mg(OH)2  2NH3  A= 2  x3  3x2  3x 1 0 NaNO3  4Zn  7NaOH  1  A  x3  3x2  3x 1 dx 4Na2ZnO2  NH3  2H2O 0 CaNCN  3H2O  CaCO3  2NH3 A=  x4  x3  3x2  1  1 61. Ans. (2) 4 2 4 x 0 65. Ans. (2) x coty y dy + ln siny dx + ln cosx dyy tan x dx=0 f  5x  3y   5 f  x  3 f  y  53   d  x ln sin y   d  y ln cos x   0 Given 53 x ln sin y + y ln cos x = c Which satisfies section formula for abscissa on (sin y)x . (cos x)y = c L.H.S. and ordinate on R.H.S. Hence f(x) must be linear function 62. Ans. (4) let f(x) = ax + b f(0) = b = 1  f(x) = 2x + 1 f'(0) = a = 2 period of sin (2x + 1) is  66. Ans. (4) x = 1  12k . 1211 Ck C11 1 k 1 k n 12 2   12.K 12CK . 11CK1  122 x=1 C11 K 1 K 1 K 1 x=0 x=2  122. 22! 11! 11!  12. 21.19.17......3 212 .6  p  6 11! 1001CT102116064 HS-9/13

Target : JEE (Main + Advanced) 2017/21-03-2017 67. Ans. (3) 71. Ans. (3) We have, z = 0 for the point where the line  a = 0 and y = bx2 + cx + d is symmetric intersects the curve about x   c 2b x  2  y 1  01 3 2 1 x = k =  c k  c  0 2b 2b  x = 5 and y = 1 Putting these values in xy = c2  5 = c2  c   5 a c k 0 68. Ans.(2) 2b 1 72. Ans. (3) y Put x  y  tan1 x  2  cot 1 x  1     0  2    1  1     tan1 x 1 tan1 x  2  0 1  f  y  1  dy  y  y2   0  1  0  1  dy  1  tan1 x  2 f y  dy  f y    tan1  x  tan 2  73. Ans.(2)    y  y  y2      1 0  1    1   log2 1 6x  x2  8  0 y y y dy 0  z   Putting z  f    f z  dz  1 1 6x  x2 8 1 6x  x2 8  0  x2  6x 8  0  x  2x  4  0 69. Ans.(1) 2 x4. D C  xy y x Now f '(x) = x2 + 2x + 2 > 0  x  R A B  f(x) is strictly increasing in [2, 4] sin 2 1 f  x  x3  x2  2x In ABD  2x 2R1 3 x  25sin cos , a  f 2  8  4  4  32 In ACD sin 2  1 33 2 y 2R2 b  f 4  64 16  8  136 y  50 sin cos 33 tan  x  1  y  50. 1 . 2  20 y2 55 a  b  56 and x = 10 74. Ans. (3) D.R's of normal to plane x + y + z1 =0 and Area of rhombus = 2xy= 400 x + ky + 3z1 = 0 is (1, 1, 1) and (1, k, 3) respectively 70. Ans.(4)  D.R. of normal to a plane perpendicular to |z|2 -|z| - 2 < 0 given planes   z  2  z 1  0  z  2 iˆ ˆj kˆ Now z2  zsin  z 2  zsin  z 2  z  4 2  6  1 1 1  iˆ3  k   ˆj 2  kˆ k 1 1k 3 HS-10/13 1001CT102116064

 1 2  2    1 3 Leader Course (Score-I) & Enthusiast Course (Score-II)/21-03-2017 3  k 2 k 1 77. Ans. (4)  2  4  6  3  2k  k 2x3  3x2  x  3  2x  1  1  3 x2  x  2 x 1  4 10  4  2  12  8     2 x 2 3 d 2x3  3x2  x 3  2  1  3 dx   2   2  8  6  2  2k  2  k  3  x2  x  x 12  x  22 3  A = 2, B = 1, C = 3  2  4   4 k   10   4 k  k  5 A + B + C = 0 3 3 33 2 78. Ans. (2) PA × PB = (PT)2 75. Ans. (3) where PT = length of tangent (PT)2 = (1)2 + 32  2(1) + 4(3)  8 = 16 g(x) = x |x|3 has 4 repeated roots P(A) P(B) = 16 g\"(x) is cont. and diff. at x = 0  consider f(x) = xp sin 1 x0 x x =0 0 h p sin 1 AM  GM f ' 0   lim h  0 if p > 1 PA  PB  8 h0 h  f '0  0 for p > 1 79. Ans. (2) n 1 n  n 1 n!  f ' x  p x p1 sin 1  x p2 cos 1 x  0  Tn  n2 1 n  n n 1 n 1 n xx Sn  n n 1 0 x=0 T10  10110  101.  b  a  9 p h p1 sin 1  h p2 cos 1 S10 10 11 110 80. Ans. (2) f \" 0   lim h h  0 Since f\"(x) > 0 h0 h if p > 3  f ' x is always increasing sin 1  cos 1 x x     f \"x   p p 1 x p2  x p4 2x p3 for x  0 = 0 for x = 0  f \"(x) to be continuous p 4,  g'(x) =2f '(2x33x2)×(6x2 6x) + f'(6x2 4x3  3)(12x12x2) 76. Ans. (4) = 12(x2x) (f '(2x33x2)f '(6x24x33)) =12x(x1)][f '(2x33x2)f '(6x24x33))] lim ax2  bx  c  1 For increasing g'(x) > 0 2 Case-I x < 0 or x > 1 x1 2x 12  f(2x33x2) > f '(6x24x33) 2  2x33x2 > 6x24x33  ax2  bx  c  1 2x 12  f ' xisincreasing  2  ax2  bx  c  2x2  2x  1 2 a  2,b  2, c  1   x 12  x  1   0  x   1 2  2  2  x  2 x  2  x  1  3  1   2   2  lim  4  6  x   , 0  1,  x2 x  2 2 1001CT102116064 HS-11/13

Case II : If 0 < x < 1 Target : JEE (Main + Advanced) 2017/21-03-2017 f '(2x33x2) < f '(6x24x33) 83. Ans. (3)  x  12  x  1   0 AI 0  2  1 1 2 1 0 2 1 0 2 1 0 2  x   , so there is no solution x1 1     2   2  0    0 1  2  0   2      0 2   Hence the values are x    1 , 0   1,  1  2   2 1 4  2  0  2  1   2  4  4  3  2  0 81. Ans. (2) 2  4  4  3  42  4  3  2  0 0 x  3  52  6 1  5 1 1 6 A3  5A  I  A  I  cos x > x  cos x  1 a = 5, b = 1 or a = 1, b = 5 x a+b= 6  /6 cos x  /6 84. Ans. (2) 0x dx  1.dx 0 I  ; x p q pq (pq)p ~q q~q [(pq)p] (q~q) 63 2 TT T T F F F cos x < x TF F T T F F FTF T F F F x  /2 x  /2  1 dx   cos cos dx  J  FF F T T F F x  /3 x  /3 6 Given compound statement is always false. So it is a contradiction. 82. Ans. (4) ax + by + c = 0 85. Ans. (4) a + c = 2b  a  2b + c = 0 x = 1, y = 2 n1 = 10, n2 = 10 average m1 = 60, m2 = 40 (1, 2) = ,   1  4,  2  6 (x  1)2 + (y + 2)2 =   x2+y22x + 4y + 5   =0 Standard deviation of combined series it is orthogonal to x2+y2  4x  4y  1 = 0 4 8  5 1      n1 12 2 2 n1  n2 2 n1n2 m1  m2  8  n1  n2 2  n2     128  7  10 16 10  36  10 1060  40 2 10 10 10 102  8 18 100  126  11.2 HS-12/13 1001CT102116064

Leader Course (Score-I) & Enthusiast Course (Score-II)/21-03-2017 86. Ans. (3) 89. Ans. (2) Problem is same as arranging 8 things out of 2 cases arise (i) P & P are in same pair & one 12 8! which 5 identical i.e. 5! which gives total loses p 1 11 number of ways of selecting block and 8! (ii) P1 & P2 are in different pairs & one loses distributing them away 3 children i.e. 5! 3!. 10  1 1 1 1  5  p  11   2  2  2  2   11 87. Ans. (3)  c For a b to be greatest  must be 6  a  required probability = 11 . b  wpeirthpebndiccu.lar to both and c i.e. collinear 90. Ans. (2)   iˆ ˆj kˆ Region where 2 tangents to two different branches can be drawn. b c  1 4 6 = 2iˆ  2 ˆj  kˆ y 2 7 10 y = 4x/5 y = 4x/5    2iˆ  2 ˆj  kˆ a 3 88. Ans. (1) x The normal at the extremities of focal chord meet at right angle. So orthocentre is the point of intersection of normals.    If P at12 , 2at1 ,Q at22 , 2at2 then t1t2 = 1,.  (1, 6), (1, 3) But from (1, 6) 2 tangents to circle can be drawn Point of intersection of normals  Ans. (1, 3)  h  a t12  t22  t1t2  2 k  at1t2 t1  t2  1001CT102116064 HS-13/13