FIITJEE JEE (Advanced) 2023 Past Paper PDF - Part 1 - Paper 2

FIITJEE ALL INDIA TEST SERIES JEE (Advanced)-2023 PART TEST – I PA...

FIITJEE ALL INDIA TEST SERIES JEE (Advanced)-2023 PART TEST – I PAPER –2 TEST DATE: 20-11-2022 ANSWERS, HINTS & SOLUTIONS Physics PART – I Section – A 1. D Sol.  = I (about ICOR) 2  4  1  m 2   N1 ICOR mg      m     5  2  12  2    COM 6g N2  5   3g    acm    2 5 mg  N2   maCM  cos  mg acm 3mg  4  mg  N2    5 5 13mg  N2  25 2. A Sol.   t    cons tant  F cos  N = F sin t F v mdv   Fcos t  N  mg   dt N F sin  0 T  mdv   F  cos t   sin t   mg dt 0 0 mg F T   sin t   cos t 0  mgT  f = N  Also T  2 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com AITS-PT-I (Paper-2)-PCM(Sol.)-JEE(Advanced)/2023 2      F  sin   cos     mgT  2 2   F 1     mg 2 mg  F. 2(1   ) 3. B Sol. F.B.D. of rod F1 N1  0  0 (about pivot) mg f1  N1  …(i) F2 O 2 mg F.B.D. of cylinder  N1  O C  0  f1  f1R  f2R y f2 C f1  f2 …(ii)  mg (say f)   Also 0  0 x N2 N1  mg   N2  N2  N1  mg  3mg N2  …(iii) 2  x 0 (on the cylinder) F  f 1  cos    N2 sin  3mgsin   f 2 1  cos   Also f1 max  f2max  N1  N2  3mgsin  mg  f  N1   2 1  cos   2 3 sin   1  cos  4. A Sol. F.B.D. of block is as shown (  90°) N N = F sin  + mg cos  Fcos f = F cos  – mg sin  for No sliding f  N 1 mgsin +f Fcos   mgsin   F sin   mgcos   3 sin  mg  cos   cos    sin    Fsin + mgcos 3 F  3  sin  cos    0 ( no sliding even when F ) 3 tan   3    60 For  > 90° N  F sin   mgcos  FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com  3 AITS-PT-I (Paper-2)-PCM(Sol.)-JEE(Advanced)/2023 f  mgsin   F cos  f  N    120  Range    [60°, 120°] 5. A, B, D Sol. Applying WET on bead from A to B W N + W F = K m u O  F R  x   R  x     v 2  u2  R+x conservative 2  P 4Fx Fx v 2  u2   u0  2 N+F x m m mv 2 v Also N  F  R R–x m 4Fx  m N  u2   F  u2  u20   F R m  R  m m 4Fx N   2  1 u02  F   2  1 F R  R  m  N  F  4  2  1  1      N  F 1  4 1  2  ˆi    6. A, D Sol. As PA + PB = 10  = Constant Y The bead will move along the elliptical path in vertical plane. Such that 2a = 10  and 2ae = 8  a = 5, b = 3 A B  Equation of ellipse is x x2 y2  1  5 2  3 2 P 3/2 2 r0  1   y     3 at  0, 3  ( y = 0, y = ) y  25 25   r0  3 m 2  3 2T0 cos   mg  r0  2g   cos     5 T0   T0 31 mg T0 . v  2g 30 mg 7. A, C Sol. Angular momentum of system is conserved about the axis passing through the pivots   Li (Just before collision) = L f (Just after collision)  2  mv0R =  mR2   5m  R2    5  FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com AITS-PT-I (Paper-2)-PCM(Sol.)-JEE(Advanced)/2023 4 v0   3R Now, applying conservation of mechanical energy after the collision 1 2  3R  mR2   5m  R2  2  mgR  5mg 2 5  8 mR2 v2 7mgR  3  02  2 9R 8 21R 7g Also v 0  2gh  h   8 12R 8. A, D Sol. For u to be minimum, it just grazes the cylinder at two points as shown. From COME mu2 mv 2 v   mgh 2 2  u2  v 2  2gh   2 2 u0 u  v  2gR 1  cos   h v 0 2Rsin sin 2 Rg Also 2R sin   v 2  v2  …(2) g cos  From (1) and (2)  1  u2  Rg 2  2cos    cos   d 2 1 For u min. d   u  0  cos   2   = 45 u02  Rg 2  2 2  Also ucos   v cos  tan2 0  [3  2 2] 9. A, D Sol. Just after the jump by the frog. 5v0 (w.r.t. to A) According to COLM of (A + frog) (4v 0 – v) m – mv = 0 v = 2v 0 37° K A B V Now consider the motion of blocks in their v0 A C.M. v0 COM frame. m m t=0 At any time t 2K 2K 2k vBC  vB  v CM  v 0 cos  t  ,   m vB  v CM  v 0 cos t vB  v0  v 0 cos t FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com  5 AITS-PT-I (Paper-2)-PCM(Sol.)-JEE(Advanced)/2023 Also the frog is in projectile motion from ground frame 3v0  3v  2v 12v02  6v 0  R  2 0  0  T    g  g g  g  2v0 R Also frog land on B T Then  0   vB dt  R 0 T  0  R   v 0  cos t  1 dt 0 v  0  R  v 0 T  0 sin T  12v 20 6v 20 m  2K 6v 0   0    v0 sin   g g 2K  m g   6v02  2  0  3   . g   10. B, C dm Sol. F  mg  v 0 also m = y dt dy F  yg  v0  dt = yg  v 02 dE Heat loss per unit time = Fv 0  dt d 1 y   yg  v 20 v 0    y  v 20  yg  dt  2 2 v 30 ygv 0   v03    ygv 0 2 v 30 . 2 Section – B 11. 2 Sol. Observing the motion in COM frame. Then system will look like v/2 v/2   v/2 v/2 v/2 Just before the collision Just after the collision After the collision, ring starts sliding and then starts pure rolling on wedge as shown. R = 2v (pure rolling) FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com AITS-PT-I (Paper-2)-PCM(Sol.)-JEE(Advanced)/2023 6 Applying COAM of ring in centre of mass frame (after the collision) v about point P on wedge  2  v 2 mR    mR   mRv   mR  v  2 P v v mR  3mRv   v   2 6 v Therefore the velocity of wedge is (in centre of mass frame) when pure rolling starts after the 6 first collision. Then they again collides and ring again starts pure rolling on wedge and velocity of v nd wedge is (in centre of mass frame) after 2 collision. 18 The process goes on and there will be infinite collisions and finally wedge comes at rest in centre of mass frame. v  Velocity of wedge = (in ground frame) 2 12. 5   Sol. v  u  gt   g    v u also s   v ave  t   t  2  t     v  u  gt …(i)  u   2s v v u  …(ii) s t t=0 Squaring and multiplying        v.v  u.u  2u.v  v.v  u.u  2u.v   4g2s2   2 s  100  2u.v 100  2u.v  4g2    s is max. when u.v  0  u  v 100  smax   5m 2 10  13. 6 Sol. If it strike again at same point, then the horizontal component of their velocities are same after the first collision. v0cos v2  v0 v1 v1  v0sin  Just before collision Just after collision As e = 1 vsep = vapp v2cos = v0sin FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com  7 AITS-PT-I (Paper-2)-PCM(Sol.)-JEE(Advanced)/2023 3  v 2  v 0 tan    40  = 30 m/s 4 2v 2 t0  = 6 sec. g 14. 9 Sol. Tmax  T1 v Tmin = T2 mu2 T1  mg  …(1) l T2 + mg 2 mv l v T2  mg  …(2) l From COME (of pendulum + earth system) 1 1 T mu2  mv 2  mg(2l) 2 2 l mg  u2  v 2  4gl …(3) T1 T1 also 2 …(4) T2 u from (1), (2) (3) and (4) mg u  11gl , v  7 gl  v   9gl , mv 2 T  9mg K=9 l 15. 2 Sol.  V0 2R/3 P V R/3 C.O.M. Applying COLM v mv0  m  2m  v v  0 3 Applying COAM about point P 2R  MR2  v R mv0     2m 0  mR2  3    2  3 2v  R  0 3 R  2 v FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com AITS-PT-I (Paper-2)-PCM(Sol.)-JEE(Advanced)/2023 8 16. 3  3  /2 2v 0 cos    Sol. T  2  h   v0 gcos     2 t=0 2v    t=T  0  4cos2    3  g  2  /2 Also Rsin = V0sin2T /2 Rsin 2v 2 R sin   0  2sin  cos   2 1  cos    3  gcos(/2) g As v02  2gh R  h 8 cos   2cos   1  h  0 2cos2   cos   0 cos   2cos   1  0 1 cos   2   K = 3. 3 2nd Method:     v0 [90 (/2)] /2   2     90    180  2 3   90   90– 2  Now  > 0  v0  < 60° /2  max  radian 3 /2 17. 6 Sol. Applying impulse momentum theorem. N v0/2 3  Ndt  mv 0 2 30° 30°  v0   Ndt  m  2  v  30° 30° v0 v  v  0 1  3 2   f = N v03/2 Just before the impact Also v 02  2gH and FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com  9 AITS-PT-I (Paper-2)-PCM(Sol.)-JEE(Advanced)/2023 2gH 2   2gh 1  3   4  1  3  l v 1 h Also   a 2 3 30° 30° 2  1 H 1 Just after the impact h 1    1    2 4 2 a = g/2 [1 + 3 ] H  6 4h 18. 8 Sol. In equilibrium, the centre of mass of the system must lies on the vertical line passing through hinge.  XCM = 0 O L  2R  m1  m2  0 2    x L2 2R    R  2  R  L = 2R m2 m1  2R   m2R YCM  L m1  m2 m1 y  L2  R2    4  L       4,   2  L  R 22   = 8. FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com AITS-PT-I (Paper-2)-PCM(Sol.)-JEE(Advanced)/2023 10 Chemistry PART – II Section – A 19. A Sol. PCl5  PCl3  g   Cl2  g Dd  , n=2 n  1 d D 1       d D Thus, 1   will linearly increases with increase of  .  d 20. A So Ho Sol. nK   R R o S  8R 21. B 2x 2x 2x  16x Sol. Change in potential energy     2x     2x  8 8 8 14x 7x   8 4 22. A 2.303 100 Sol. K 298  log t1/2 50 2.303 100 K 308  log t1/2 25 K 298 log 2 1    0.5 K 308 log 4 2 23. A, B Sol. Possible transition 3 2 and 4 2 36  3 2  5R 16  4 2  3R 24. A, B, C Sol. Molecular mass of A 2B4  100 2A  4B 2A  4B  100,  5  given  2A A 10 A  10,B  20,   1: 2 B 20 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com  11 AITS-PT-I (Paper-2)-PCM(Sol.)-JEE(Advanced)/2023 A 2B4  g   A 2  g  2B2  g 1   2 At 10 min. V.D. = 25 Molecular mass = 50 100 50  1  2 1 1  %B2  by mole    100  50% 2 2 1 %A 2  by mole    100  25% 4 At t = 20 min. 100 40  1  2   0.75 d  A 2B 4  0.5  0.25 0.25    dt 10 10 d  A 2  0.75  0.50 0.25    dt 10 10 Ratio of rate = 1 : 1 25. A, C, D Sol. 2HI  H2  I2 At equilibrium 4  2x x  3y x 2N2  6H2  4NH3 At equilibrium 4.25  2y x  3y 4y Given x = 1.0 M 4y = 0.5 M 5 Using these value K C1  32 45 K C2  6 5  32  K C2     125  26. B, C, D Sol. SO2 1 p  p  1 p  d  SO3 1 p  p  2  p  d  XeO 4 4  p  d  ClO3 2  p  d  FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com AITS-PT-I (Paper-2)-PCM(Sol.)-JEE(Advanced)/2023 12 27. A, B, D Sol. H H C C nodal plane H H Equitorial plane F F Axial H F H F C S C S H F F H F F Axial plane Equitorial Axial More stable form due Less stable due to more to bent rule. nodal repulsion. Its nodal plane lies in axial. plane lies in equitorial 28. C, D Sol.  AlCl3 n has (3C – 4e) bond not (3C – 2e) B2H6  Two (3C – 2e) bond H H H H H B B Al H H H H H H H H H B B B Be H H H H H H 6-(3C-2e) bond 4-(3C-2e) bond Section – B 29. 5 Sol. x = 9, y = 3, z = 2 In one electron system all orbitals of a shell are degenerate (same energy level). In case of many electron system different orbitals of a shell are non-degenarate. Hence, in second excited state only three p-orbitals (2p) are degenerate. H-atom 1s 2s 2p 3s 3p 3d n=1 n=2 n=3 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com  13 AITS-PT-I (Paper-2)-PCM(Sol.)-JEE(Advanced)/2023 H-ion 1s 1s 2s 2p 1s 2s 2p Ground state 1st excited state 2nd excited state 30. 6  Sol. (i) Borax  NaBO2  B 2 O3  (ii) Borax  NH4 Cl  BN  B2 O3  NaCl  H2 O (iii) Borax  HCl  H3 BO3 (iv) Borax  H2 O  Na B  OH 4   H3 BO3  (v) 3B2 H6  6NH3  2B3 N3 H6  12H2 THF (vi) B3 N3 H3 Cl3  3LiBH4  B3 N3H6  3LiCl  3BH3 (vii) LiH  B 2 H6  2LiBH4  sp3 (viii) K 3  AlF6   BF3  3KBF4  AlF3  sp3 Except (vii) and (viii) all options can give sp2 hybridised Boron centre as one of the product. 31. 6 2   Sol. Lone pair in  TeBr6  , IF2  , SNF3 and  XeF3 are 1, 2, 0 and 3 respectively. Hence, x = 6. Diamagnetic species are Li2 , C2 , O22 , K 2 O, N2 O 4 , B3 N3 H6. Note: Be2 does not exist so we can’t say it is paramagnetic or diamagnetic species. Hence, y = 6. xy  6. 2 32. 4 Sol. In container (I) In container (II) 2SO2  g  O2  g  2SO3  g  KMnO4 will oxidized only SO2 2x1 x1 0 Moles of O.A. Moles of R.A.  2x1  2a x1  a 2a 2 5 0.2  4 Moles of SO 2 Total moles nT   3x1  a   2 5 x1  a 1 x1 Given  a Mole of  SO2 equilibrium  2 3x1  a 4 3 2 2SO3  g  2SO2  g   O2  g  2a    20  10  2x1 2x1 x1  nT   1  1 y  KP  2   atm 2x1 = 2  2x1  2a   x1  a   20    20    20  x1  1.0  nT   nT  FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com AITS-PT-I (Paper-2)-PCM(Sol.)-JEE(Advanced)/2023 14 2 y 80  2  1 Value of  4     x 20 5 5 1 x  KC 2  mol /   8 80   5 33. 6 Sol. a = 4 i, iv, vi, vii b = 2 ii, ix v and viii forms  M.O. d = 2 iii, x a2  b2  2d 24  6 4 4 34. 9 Sol. I, II, III and IV are correct due to bent rule and V is incorrect. Correct order in case of V is II > III > IV > I = V. x=4 y=1 2x + y = 9 35. 8 3  5.8  Sol. K SP of Mg  OH 2 is    10 3 .4  4  1012  58  2 Mg  OH 2  s   Mg  2OH x  0.01 2x  x  0.01  4x 2   4  10 12 4x 2  0.01  4  1012 x  105 mole /  Mg2  in container II  1.0  105   NH4    For basic buffer pOH  pK b  log  4  NH OH pOH  5  log5 = 5.70 OH  2  106 4  1012 Mg2     4  1012  1.0 M 1.6  10 20  Al3     2  103 M   8  1018 x  105 y  1.0 z  2  103 yz 200 Value of   8. 25x 25 36. 5 Sol. Statement 4, 11 and 12 are incorrect and other statements are correct. x = 9, y = 3. FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com  15 AITS-PT-I (Paper-2)-PCM(Sol.)-JEE(Advanced)/2023 Mathematics PART – III Section – A 37. A 1 tan1 x Sol. By differentiating, f   x   2 f  x  0 1 x 1 x2 1 f  x   1  tan1 x  2e tan x 38. A a   f  x   f  x   dx  g  a  3 Sol. ; f(x) = 2x + 6x – 8 1 3 f x 1  101   g  x   5 dx  2 ln  22  2 39. D 2 x 2  2x  cot x   ex  Sol.  e  sin x   ln  sin x  dx 2 ex Let = t   ln  t  dt  t  ln t  1  c sin x 40. B Sol. x2 + y2 – ax = 0  x2 + 2xyy – y2 = 0 2xy 2xy Orthogonal curve : x 2   y2  0 ; y  2 y x  y2 y P By solving, x 2  y 2  c Passing through (1, 1) so x 2 + y2 = 2y For shortest distance, Q = (1, 0) Q D = PQ – r = 2  1 41. A, B, D Sol. f(x) = f(x)(tan x – 1)(tan3 x – tan2 x + 3 tan x + 1)    g(x) = tan3 x – tan2 x + 3 tan x + 1 = 0 has exactly one root in x    ,  which lies in  2 2    x   , 0  4      So, f(x) = 0 at x  and k where k    , 0  4  4  42. A, C Sol. f  x   f 1  x   x y -1 -1 y = f (x) (x, f (x)) 1 (f(x), x) y = f(x) (x, f(x)) x 1 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com AITS-PT-I (Paper-2)-PCM(Sol.)-JEE(Advanced)/2023 16 43. A, B 1  Sol. fg = fg + fg by solving g  x   e x  2 g(x) is discontinuous at one point at x = 2 44. B, D Sol. Graph of y = f(x)  2 x –1 0 1 45. A, D 3/ 4  1  2 cos t  3 sin t Sol. f  x   x  2  dt 0 cos t  2   cos t  2    3  2x ; x0 1  2 x 3 f  x    x  3u du  f  x     x  ; 0x3 0  3 2  3  x2 ; x3  46. B, C x2  x  1  1  f(x) Sol.  ,3  x  R  x2  x  1  3  Maxima at x = –1 and minima at x = 1 2 x –1 1   2 Section – B 47. 1 Sol.  6x f  x   4x 2   2f  x  dx   2x dx   fx dx   1 c   3f  x   2x   x  f  x  2 2  x2  f  x  2  x2  f  x   2 2 x  f  x 2 Substitute f  x   4x  3f  x  48. 2 16  4  4 Sol.  2 8    x  4 x  4  dx  3  0  4 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com  17 AITS-PT-I (Paper-2)-PCM(Sol.)-JEE(Advanced)/2023 49. 5  n  n2  mr  1  1   n n2 r 1  Sol. lim    2 2  m   lim 2   = lim   2 2  2 2 dx   r 1  n  r  n2  k 2  m     r 1 n  r 0 n  x n n  k 1      m      1  n n  r  tan1 x 2 = lim   2 2 tan1      dx  n  r 1 n  r  n   0 1 x2 32 50. 7  x 2  1  x 2  9  |f(|x|)| Sol. f x  x  R – {–3, 1}  x  1  x  3  Graph of y = |f(|x|)| Discontinuity at x =  1 Non-differentiable at x = 0, 1, 3 –3 –1 1 3 51. 8 4 4 4  /8 dx dx 1 1 Sol. I     I  2 2 dx  64  2 dx 0 2  sin 4x 0 2  sin 4x 0 4  sin 4x 0 4  sin 4x  /8 3  /8 2 128 1 I  64  dx ; 3x = t  I   dt 7  cos8x 3  8t  0 0 7  cos   3 52. 4 Sol. f(x) = k(x + 1)2(x – 1)2(x – 5)2 + 4  f(2) = 81k + 4 = 85  k = 1 f(x) = (x + 1)2(x – 1)2(x – 5)2 + 4  f(x) = 2(x + 1)(x – 1)(x – 5)(3x2 – 10x – 1) 10  x 2  9  19     lim  2  3 x 10  x  5x  6  4 3 53. 4  1   2022 1  x 2022  2021  1 2023 1  x 2023  2021  1  1  x x 1 Sol. lim 2021  x 0 2022 x 2021  = 2021 = 43  47 number of divisors = 2  2 = 4 54. 2  y2  3  1 Sol. Area = 2   y  dy   4  3 1 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com