CPT-7 / XI-LJ / NARAYANA I I T A C A D E M Y CPT 7 XI-LJ (Date:.0.7) CODE XI-LJ PHYSICS CHEMISTRY MATHEMATICS. (B). (A) 6. (A). (B). (B) 6. (C). (D). (D) 6. (B). (B). (C) 6. (C) 5. (D) 5. (C) 65. (A) 6. (B) 6. (B) 66. (A) 7. (D) 7. (A) 67. (C) 8. (C) 8. (B) 68. (D) 9. (B) 9. (C) 69. (D) 0. (B) 0. (B) 70. (D). (B). (B) 7. (B). (B). (D) 7. (B). (C). (B) 7. (C). (C). (C) 7. (B) 5. (B) 5. (C) 75. (A) 6. (C) 6. (A) 76. (D) 7. (C) 7. (D) 77. (C) 8. (A) 8. (C) 78. (A) 9. (D) 9. (D) 79. (A) 0. (B) 50. (C) 80. (C). (B) 5. (B) 8. (C). (B) 5. (B) 8. (D). (C) 5. (D) 8. (A). (C) 5. (D) 8. (A) 5. (D) 55. (A) 85. (B) 6. (A) 56. (A) 86. (A) 7. (C) 57. (A) 87. (B) 8. (D) 58. (A) 88. (B) 9. (B) 59. (B) 89. (C) 0. (D) 60. (D) 90. (A)
CPT-7 / XI-LJ /. (B) GM W U 0 R Hits & Solutio PHYSICS 6.67 0 00 0 0 0 0 0 6.67 0 J. B α ωf ωi 60 rad / s t 5. At equator the value of g is iiu so it is profitable to purchase sugar at this positio.. Acceleratio due to gravity at height h is give by g 00 R g R h R R h 0 h 9 R. R g g R h 5. If P is the poit where et gravitatioal force is ero the F PA FPB G G ( d ) By solvig d For the give proble d D, earth, oo ad 8 So D D 8 D 9 0 9 D d g d d 6. g g g d R R R R P F PA A B d 8 d F PB 7. (D) Gravitatioal field iside the hollow sphere is equal to ero 8. (C) At the poit P, we have I I 0 (because the gravitatioal field iside a shell is ero). Hece, I I 9. As dv I, if I 0 the V costat. d 0. Potetial icreases by 0 J / kg everywhere so it will be 0 5 5J / kg at P
CPT-7 / XI-LJ / G G G. Net potetial at origi V... r r r G G G 8 O 8. Work doe gh h / R, If h R the work doe gr R / R gr.. ω v v ; R; ω v R R. g ' g d R 5. Distace of ull poit d 6. By the law of coservatio of agular oetu I Iω ( I )ω Iω Agular velocity of syste ω I Iω I Rotatioal kietic eergy ( I I ) ω ( I I ) ω I I ω I Iω I ( Iω Iω ) ( I I ) 7. Iitial agular oetu of the syste Agular oetu of bullet before collisio L Mv...(i) let the rod rotates with agular velocity ω. Fial agular oetu of the syste By equatio (i) ad (ii) 8. L r p ˆi ˆj kˆ ML L ω M ω...(ii) L ML ML Mv ω or ω v / L 0ˆi ˆj ˆ k ˆj kˆ ad the X- ais is give by i 0 ˆj 0kˆ Dot product of these two vectors is ero i.e. agular oetu is perpedicular to X- ais. 9. Rotatioal kietic eergy I ω MR ω 0 (0.5) ( 0 ) 50 J. v M
CPT-7 / XI-LJ / v I R 0. Use d ( v A t ) L E L L. As, E I E L L E 9E E 800% of E [As L L 00%. L L ] gh gh. Velocity at the botto (v) gh. K R g siθ. a k R. (C) g si 0 o g k [As R Gravitatioal potetial at the surface of the earth ad GM R o θ 0 ] Gravitatioal potetial at 0 GM GM Potetial differece 0 R V R Workdoe i shiftig a body of kg fro the surface of the earth to ifiity. GM GM W V Workdoe per kg gr R R 5. (C) Gravitatioal field iside a shell is ero, but for poits outside it, the shell behaves as if whole of its ass is cocetrated at its cetre. 6. Iitial agular oetu of rig I ω MRω If four object each of ass, ad kept getly to the opposite eds of two perpedicular diaeters of the rig the fial agular oetu ( MR R ) ω' By the coservatio of agular oetu Iitial agular oetu Fial agular oetu M MR ω ( MR R ) ω' ω' ω. M G 8. GPE ( U ), use, U et U U U r 9. 0. k R 0 g siθ k g si 0 5 a g R 5 CHEMISTRY
CPT-7 / XI-LJ /. A Half eutraliatio acid salt [ ] [ ] P P P H H H Pk log log k k ati log P a H a Pk a P 0 0 a 5. [ salt] [ acid ] H. B Equilibriu shift i backward directio. D Isoelectric poit is the coditio, whe witterios or sol particles do ot ove uder the ifluece of electric field, i.e., they lose their charge.. C Maiu buffer capacity whe [salt] [acid] P H Pk a 5. C 6. B A buffer of H CO ad HCO is fored. 7. A ph logk a log [Salt] [Acid] log0 log [Sice K a K b 0 Give K b 0 0 K a 0 ] 8. B Miig of NH OH ad (NH ) SO gives a basic buffer. [NH ] poh pk b log ; pk a [NH OH] NH 9.6 pk.7 bnhoh 5.7.7 log a 5 ; where a is illiole of [NH ] obtaied o iig. a 50 illiole of (NH ) SO 5 or, ole of (NH ) SO 0.05 5
CPT-7 / XI-LJ / 9. C (00 V ) ph pk a log ; let V be volue of beoic acid ad thus (00 V ) is V volue of beoate. 00 V.5. log V 00 V V V 00 L 0. B HI H I K a [H ][I ] [HI] K a K w K b 0 [I ] [HI] or [H ] 0 or ph 0 [H ]. B NH HCl i : will give NH NH Cl i : ratio.. D Buffer solutio. B Fid solubility for each separately by S K SP for MS ad ZS. 08S 5 K SP for Bi S ad S K SP for Ag S.. C Let S is the solubility of BaF i a solutio of BaNO, the K SP [Ba ][F ] ; the [F ] S; the [F ] S 5. C Presece of coo io decreases the solubility of salt. 6. A Presece of coo io decreases the solubility of salt. 7. D K SP of PbCl S (0.0) 0 6 I NaCl solutio for PbCl ; K SP [Pb ][Cl ] or, 0 6 [Pb ][0.] [Pb ] 0 M 6
CPT-7 / XI-LJ / 8. C 9. D 50. C S Ag S AgCl S Ag S AgBr S AgCrO K SP Also, S Ag K SP K SP.8 0 0. 0 5 M 5.0 0 7.07 0 7 M. 0 0.8 0 M S 0.8 0.6 0 M AgCrO This will give aiu value. Ksp Ba SO.5 0 0 SO 9.5 0 SO 7 MgCl NaOH Mg(OH) NaCl before 0 0 0 0 reactio 0 0 0 0 Thus, 0 illi-ole of Mg(OH) are fored. The product of [Mg ][OH ] 0 0 is therefore 00 00 5 0 which is ore tha K SP of Mg(OH). Now solubility (S) of Mg(OH) ca be derived by K SP S. 0 S [OH ] S.8 0 K SP 5. B Due to coo io effect 5. B [Mg ][OH ] 5. D 5. D 0 ; 0 [OH ] 0 5 0.0 or poh 5 ad thus ph 9 [OH ] Cα C K b.c K b C.0 0 0.0. 0.0 0 7 M 7
CPT-7 / XI-LJ / 55. A 56. A 57. A 58. A 59. B 60. D MX M X ; K SP (K SP ) / M X M X ; K SP MX M X ; K SP 7 K SP K SP 7 / / MATHEMATICS 6. Give a,b,c are i G.P, the Let these are a, ar, ar Now, give ab, a b ad bc 6, b c ab 6...( ) a b bc 8...( ) b c a ar ar Fro () 6 6. () a ar r ar ar ar Fro () 8 8 (5) ar ar r Fro () ad () r ad a 8 The a b c 0 6. Let A λ, h λ G AH λ a, b, are the roots of ( ) t λt λ 0 t a b t ab 0 or ( ) { ( )} λ ± λ λ t λ ± a: b ( ) : ( ) or ( ) : ( ) 6... 5... as is odd { ( )..... } 6. Multiply ad divide by ( a) i the give series ( ) ( ) 8
CPT-7 / XI-LJ / 65. 66. 00 00 a a a... a α 00 a a a... a α 99 α a r β a H PQ P Q 67. If, y, are i G.P, the log, log y, log are i A.P. 68. Give a a 0 a 9d 9d d /9 the a a d / 7/ 9 d ', d ' (give h 0 ad h ) h0 h 5 7 Now, 6 d ' h7 h 8 8 7 8 The h 7 therefore ah 7 6 7 7 69. If a, a,.. a are i H.P, the,,..., a a a are i A.P. Now, let... a a a a a a d a a a a a a... d aa a a a a 70. d log d log d 5log ( d )( d )( d ) 5 d d 5log 5 5log ( d ) log y y 9
CPT-7 / XI-LJ / 7. Use A.M. > G.M for (a, b, c) The a b c > (abc) /..() Now ab c, a b c, a b c 5 are i A.P. (a, b, c, > 0) The, abc, a b c are also i A.P. abc a b c, (abc ) 0, abc, put abc i equatio () 7. t..5... ( ) ( ) ( )..5.....5... ( )..5... ( ) Give series t..5... ( ) as ( ) 7. Let G, S 5 Ad r is coo ratio a The S r S, r / 5 r < r <, for Ifiite G.P. < < 5 0 < < 0 a, a, a, a are i HP 7.,,, a a a a are i AP λ ( say) a a a a a a a a a a a a a a a a a a a a a a a a a a or a a a a ( a a ) r a a r r 0
CPT-7 / XI-LJ / ( ) ( ) a a a a a a a a a a ar ar r a a a a Hece, is a root of 5 0 a b a a a a... a a ad ab g. g g. g... g. g ab ad h. a b a a a a a a... a b a b g g g g g g ab ab h 75. 76. Give log ( c a) log ( a c) log ( a b c) log ( a c) ( a b c) 77. ( c a) ( a c) ( a b c) ( c a ) ( a c )( a b c ) log log After solvig, ac b a c Hece, a, b, c are i H.P ( )( )( ) tr S say 8 r r ( ) ( )( ) tr S 8 ( )( ) ( ) t S S li li li r t r r ( ) ) r ( ) ( )( ) li r ( )( ) ( ) li ( )( ) 0 78. d,e,f are i G.P. e df d e f λ λ 0 d df f 0 ( d f ) 0 f d Now, substitutig the value of i a b c 0 af b f a b c a b c i.e., c 0 0 or 0 d d d df f d e f a b c,, d e f are i HP
CPT-7 / XI-LJ / i j i ( i ) 79. j i i i j k i j i i i 6 6 80. H ab H, a b a b a b, 8. ( )( ) ( ) ( )( ) H a b a (i) H a b a Siilarly, H b a b (ii) H b a b H a H b H a H b [ ] a a 5 a a 75 i ai [ a a ] 75 900 8. AM of positive ubers G.M. of positive ubers ab ac bc ba ca cb 6 ( ) ( ) ( ) 6 6 6 6 ( a b c ) a b c b c a c a b 6. abc l 6 8. I first 0 iutes, the uber of otes couted is 50 0 500 a a a. Upto ters 000 9 000 0 8. a c b b b b ab bc y ( ab ad y bc) 85.
CPT-7 / XI-LJ / ( ap b) ( bp c) ( cp d ) 0 ( ap b) ( bp c) ( cp d ) 0 b c d p a, b, c, d areig. P. a b c 86. ab a b A G 7 a b ab 7 87. y b y a ad b a y y y a b y 88. Give (y ), (y a), (y ) are i H.P. 89.,, are i A.P. y (y a) y a y a y a a a, y a, a are i G.P. a ( ) d a ( ) d Subtractig : ( ) d d a T ( ) 90.,, a b c are i A.P. b a c b