CPT XI-IC (Date: ) PHYSICS CHEMISTRY MATHEMATICS 1. (A) 61. (C) 2. (C) 31. (D) 32. (D) 62. (A) 3. (A) 63. (B) 4. (C) 33. (C) 34.
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1 CPT-0 / XI-IC / / Hints & Solution NARAYANA I I T / N E E T A C A D E M Y CPT - 0 XI-IC (Date: ) CODE XI-IC-M PHYSICS CHEMISTRY MATHEMATICS 1. (A) 1. (D) 61. (C). (C). (D) 6. (A). (A). (C) 6. (B) 4. (C) 4. (B) 64. (C) 5. (D) 5. (B) 65. (D) 6. (C) 6. (B) 66. (A) 7. (D) 7. (C) 67. (B) 8. (C) 8. (B) 68. (C) 9. (C) 9. (A) 69. (B) 10. (B) 40. (C) 70. (C) 11. (D) 41. (A) 71. (C) 1. (D) 4. (A) 7. (C) 1. (C) 4. (D) 7. (B) 14. (C) 44. (B) 74. (A) 15. (D) 45. (B) 75. (D) 16. (C) 46. (A) 76. (B) 17. (B) 47. (C) 77. (B) 18. (D) 48. (C) 78. (D) 19. (C) 49. (C) 79. (D) 0. (A) 50. (D) 80. (C) 1. (B) 51. (A) 81. (A). (B) 5. (D) 8. (A). (D) 5. (D) 8. (C) 4. (C) 54. (B) 84. (B) 5. (C) 55. (A) 85. (A) 6. (C) 56. (A) 86. (B) 7. (A) 57. (A) 87. (D) 8. (A) 58. (A) 88. (A) 9. (B) 59. (D) 89. (D) 0. (C) 60. (A) 90. (A) 1 P a g e N A R A Y A N A I I T / N E E T A C A D E M Y
2 CPT-0 / XI-IC / / Hints & Solution 1. (A) Hints & Solution PHYSICS The figue shown is an ellipse which is taced out by a planet aound the sun. The closet point is P called Peihelion and the fathest point is A which is called Aphelion. B P S ' S A b C a. (C) / T / 4 4 T K K. (A) 4. (C) K 4 Gien, T = 84 y, 10 a 8710 m a T y m.0010 y m 4 4 Fom conseation of angula momentum, L= m = constant (fo planet at all position) (i) Gien, min max 110 ms 1 10 ms 4 max 410 km, min? Fom equation (i), m max 4 1 min = m min max (at maximum distance, elocity is minimum and ice-esa) max min min max O 4 min max min 4 max km P a g e N A R A Y A N A I I T / N E E T A C A D E M Y
3 CPT-0 / XI-IC / / Hints & Solution 5. (D) Gaitational potential at some height h fom the suface of the eath is gien by..(i) R h And acceleation due to gaity at some height h fom the eath suface can be gien as ' g 6. (C) R h (ii) Fom equations (i) and (ii), we get ' R h.(iii) g J kg and g ' 6.0 ms Radius of eath, R = 6400 km Substitute these alues in equations (iii), we get R h 9 10 R h 6 h.610 m h 600 km Gaitational potential at cente, i.e., centoid of tiangle G, T A B C Also, 7. (D) Gm Gm Gm T AG BG CG L L AG BG CG T Gm AG BG CG 9Gm Gm L L P a g e N A R A Y A N A I I T / N E E T A C A D E M Y o Gm L L G A m m m B C Wok done in lifting a paticle fom the suface of the eath R 1 R h is gien by, W 1 e m e m 1 R h R 1 to height h
4 CPT-0 / XI-IC / / Hints & Solution e e mh m h R R h R R h O Using 10. (B) e 11. (D) 1. (C) 14. (C) 15. (D) 16. (C) W g mh R h 1 R e 1 R mgh h 1 R e W 1 G M R o mghr R h Escape elocity doesn t depend on angle of pojection. d Edx, g R T g g0 g h 1 T 17. (B) T 6400 c R c R 4 P a g e N A R A Y A N A I I T / N E E T A C A D E M Y
5 CPT-0 / XI-IC / / Hints & Solution 18. (D) 19. (C) T doesn t depend on mass of satellite. T 8h w w e 1. (B). (D) 4. (C) s 4 6h w w s e m G m m Gm 1 1 n n 1 T T Gm n 1 1 L L 5. (C) 6. (C) 7. (A) 9. (B) cente suface 0 R R R R K E KE e 0 escape 0 Due to esistance E KE E KE Initially el max h h 0. (C) T 5 P a g e N A R A Y A N A I I T / N E E T A C A D E M Y
6 CPT-0 / XI-IC / / Hints & Solution CHEMISTRY 1. (D) Ag + ion fom a complex with NH [Ag(NH ) ] +. (D) 6 Ionic poduct of M(OH) Kionic > Ksp of Fe(OH) 4. (B) Because of common ion effect 5. (B) s Ksp [ OH ] 7. (C) Ksp Ag Cl Ksp x x 0.1 Ksp 0.1x Ksp x (C) A B A B Ksp A B 5 Ksp x x 108x 4. (A) Ksp1 of MgF Mg F Ksp of CaF Ca F 1 15 Ksp Mg 1 4 Ksp Ca 44. (B) Fe(OH) is a weak base hence solubility incease as the solution becomes acidic 45. (B) Ksp Zn S 10 S H S Ka 10 H S H 0.1 ph P a g e N A R A Y A N A I I T / N E E T A C A D E M Y
7 CPT-0 / XI-IC / / Hints & Solution 47. (C) H O is good oxidizing agent and weak educing agent, its bleaching action is due to oxidizing natue 5. (D) NaH Na H H - ion gie hydogen at anode 55. (A) Atomic hydogen is highly enegetic and hence its is stongest educing agent 61. (C) y 6. (A) c 5 y 5 5 c 5 10 C. y. C.c y T +1 = MATHEMATICS y Fo finding coefficient of y, 10 = 1 = 9 = 5 C.c 5! = c 10c!! x k 5 x T +1 = 5 C. (x ) 5- k. x 5 C.x 10-. (k /x ) Fo finding coefficient of x, 10 = 1 = 9 = 5 C.k = 70 10k = 70 k = 7 k = 6. (B) n is een use the fomula n Cn/ 64. (C) ax bx T +1 = C. ax 11 fo find coefficient of x 7, 1 bx = C.a x b x 1 ax bx = 7 T +1 = 11 C.(ax) bx = 15 = = C.a x b.x = a b C x.a b C 1 x = 7 P a g e N A R A Y A N A I I T / N E E T A C A D E M Y
8 CPT-0 / XI-IC / / Hints & Solution 11 6 a C 5 5 b Coefficient of x 7, 11 = 7 = 18 = a C a 1 = C6 6 b b 65. (D) Multiple and diided 1 x and simplify. 66. (A) Coefficient of nd, d and 4 th tems in the expansion of (1 + x) n = n C 1, n C, n C ae in A. P. n C = n C 1 + n C n n n C1 C C = Afte soling, we get, n = (B) (1 + x 4 )(1 x) 8 = ( C0 1 x C x... C8 1 x Coefficient of x ! = C !! x 4 ) (C) n = 4096, find n and hence geatest coefficient. 69. (B) n C + n C + n C n + n C n = 440 ( n C + n C ) = 440, n = (C) n n1 x x 1 x n We can wite S x x x 1 x n n C n coefficient of x = 71. (C) Putting x = i, we get (1 + i + i ) 48 = a o + a 1 i + a i +. + a 98 i 96 i 48 = (a o a + a 4..) + i(a 1 a +. ) Equating the eal pats, we get a o a + a 4 a a 96 = 1 7. (C) We hae, 1 + x + x + x = 1 + x + x (1 + x) = (1 + x)(1 + x ) (1 + x + x + x ) 11 = (1 + x) 11 (1 + x ) 11 = ( C 1 x + 11 C x + 11 C x + 11 C 4 x 4 +.)( C 1 x + 11 C x ) Thus, coefficient of x 4 in (1 + x + x + x ) 11 = (1)( 11 C ) + ( 11 C )( 11 C 1 ) + ( 11 C 4 )(1) = 55 + (55)(11) + 0 = 990 n 8 P a g e N A R A Y A N A I I T / N E E T A C A D E M Y
9 CPT-0 / XI-IC / / Hints & Solution 7. (B) (1 + x)(1 x) n = [ (1 x)](1 x) n = (1 x) n (1 x) n+1 coefficient of x n in (1 + x)(1 x) n = (1) n n+1 C n (1) n = (1) n ( n 1) = (1) n (1 n) 74. (A) Numbe of tems of in (x + y + z) n = numbe of non-negatie integal solution of p + q + = n = n+ C = 1 n n (D) As = 15, fo a fie digit numbe eithe do not use 0 o. When 0 is not used, the numbe of such numbes is 5! And when is not used, the numbe of such numbes is 5! 4!. Requied numbe of ways = 5! + (5! 4!) = (B) Since = 18, the 5 digit numbe will be diisible by if eithe 0 o is note used. When 0 is not used, fo the unit place we hae choices (, 4 o 8) and fo the emaining place we hae 4! Choices. When is not used : In this case if 0 is used at the unit s place, the numbe of choices is 4! If 0 is not used at the unit s place, then unit s place can be filled up in ways and the emaining places in (4!!) ways. Thus, equied numbe of numbes = (4!) + 4! + (4!!) = (B) I + E + 1A = 6 Set-I = I + E + 1A = 6 Set-II = NTRMPT = 6 Set-III six lettes aanged 6!/! Ways of these 7 gaps 6 lettes of set I ae 7 P 6 aanged in!! ways. 78. (D) 7! 5!! = 6.6! = (D) n. n! = [(n + 1) 1]n! = (n + 1)! n! 80. (C) Weaing of coats = 4 P Weaing of waist coats = 5 P Weaing of caps = 6 P 81. (A) owels occupy een places to P ways emaining 4 places occupy by consonants in 4! P 4! 8. (A) 6 P + 6 P 5 P = 0 9 P a g e N A R A Y A N A I I T / N E E T A C A D E M Y
10 CPT-0 / XI-IC / / Hints & Solution 8. (C) TRAC; E E N N; 7!! 84. (B) 4 x x x x = 5!/! = 60 5 x x x x = 5!/(!) = (A) MADHUR : ADHMRU. 5! + 0.4! + 0.! + 0.! + 1.1! + 1 = (B). 18! 87. (D) Total aangement with the lettes of the wod DEIL = 5! = 10 Numbe of aangement stating with D = 4 = 4! Numbe of aangement end with L = 4 = 4! Numbe of aangement that begin with D and end with L is 6 Numbe of aangements equied = 10 ( ) = (A) ( )(1111)! = 9,4 89. (D) NNN = One unit Remaining = 8 units, total 9 units Total pemutations E s come togethe 90. (A) 6!!!! 10 P a g e N A R A Y A N A I I T / N E E T A C A D E M Y
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