NARAYANA. XI-REG (Date: ) Physics Chemistry Mathematics

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NARAYANA I I T / N E E T A C A D E M Y. XI-REG (Date: ) Physics Chemistry Mathematics

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CPT-5 / XI-REG / 0..07 / Hits & Solutio CODE NARAYANA I I T / N E E T A C A D E M Y XI-REG XI-REG (Date: 0..07) Physics Chemistry Mathematics. 3. 6.. 3 3. 6. 3. 33. 63.. 3. 6. 5. 35. 65. 6. 3 36. 66. 7. 3 37. 67. 8. 38. 68. 9. 39. 69. 0. 0. 70.. 3. 7... 7. 3. 3. 3 73... 7. 3 5. 3 5. 75. 6. 6. 76. 7. 7. 77. 8. 8. 78. 9. 9. 3 79. 0. 50. 80. 3. 5. 8.. 5. 8. 3. 53. 3 83.. 5. 8. 3 5. 55. 3 85. 3 6. 56. 86. 7. 57. 87. 3 8. 58. 88. 3 9. 59. 89. 30. 60. 90. Page

CPT-5 / XI-REG / 0..07 / Hits & Solutio Page Hits & Solutio PART A: PHYSICS K. F, F mr r 3. wghw mghm W 5. Relative desity of the body air Wair Wwater Where Wair 5N (weigh of the body i air) Ad Wwater N (weight of the body i water) 5N R. Dblock 5 5N N. loss of weight i liquid 5 3 R. D L loss of weight i water 5 3 GMm 6. Total eergy of the system is E a GMm Which is coserved. So, KE PE a At positio r, orbital eed of the satellite is V. GMm The, KE mv ad PE r So, GMm GMm mv V GM r a r a. Apt wt of st body Apt wt of d body 8 m g Fb 8 g 5.6 36 m g Fb 36 g d bodies are i equilibrium apt wts are equal 8 36 8 36 56 g d g 36 ; 8 5 36 d 36 36 36 3 3 ; d.8 g / cc. d 3. accordig to the law of coservatio of agular mometum, r r r r K r K. r

CPT-5 / XI-REG / 0..07 / Hits & Solutio. F upthrust force V g a g a 000 3.75 N 800 F T W ma 3.75 T 0 ; w Mass of block desity of block T.75N 5. Vdbg V dg VDg 6. The tesio o deser here is upwards ad o lighter here is dowwards. Vb g T Vb g F.B.D of heavier body Lighter T w F.B.D of lighter body F B T F Heavier F B T mg mg 50 0 6 800 50 0 g T 6 liquid g i Vb g T Vb g 50 0 6 00 g T 50 0 6 liquid g ii Substract eqs. (ii) from (i), we get 6 T 500 00g T 0.5N 7. P0 P0 PQ P0 h g g h g g PQ P0 0 gh 8. W TE3 T. E R 9. ta r v cos t 0. W k. E K. E. mvr mvr () r a e r a() e GMm GMm mv mv (3) r r solve for v Page 3 R

CPT-5 / XI-REG / 0..07 / Hits & Solutio R. g g R h 3. W mv K. E. da L 3 6. T R dt m 8. Gravitatioal potetial eergy at ay poit at a distace from the cetre of the earth is 9. Gravitatioal field is a coservative field. Therefore, work doe i takig a mass from oe poit to aother i this field depeds oly o its ed poits but idepedet of the path followed. 3. For precipitatio reactio: Q. I P K PART B: CHEMISTRY 3 0 0 9 B QIP Ca F.50 K, precipitate will be formed 3. For precipitatio to occur K <Q 0 0 9.50 a Q K 3. Catios of strog bases ad aios of strog acids do ot hydrolyse, rather they exists as hydrated ios i water. 35. Catios ad aios of weak acids ad weak bases gives correodig acids or bases by hydrolysis. 36. a NaCl cotais Na + catio of strog base NaOH ad Cl aio of strog base which do ot hydrolyse. 39. BaSO s Ba aq SO aq S At Eqm. Solubility K S S 5 5 S K. 0.050 M 0. For Zr PO S 3 3 K 7 K 3 3 69 S Page

CPT-5 / XI-REG / 0..07 / Hits & Solutio 3 /5 3 K.0 5 /5 5 0 0 3 3 08 Ni OH Ni OH. S. S S NaOH Na OH 0. 0. K Ni OH 5 S 0. 0 5 0. 0 S S 3 S 0 M. For salt of weak acid ad strog base 3 ph 7 log C log K a 7 log 0. log.350 7 0.303 3 7.8697 7.938 3.35 0 gl AgCl 6. Molarity of 5 3.50 5 0 3.5 K S 0 0 5 0 7. Metals have low ioizatio eergy. Ioizatio eergy of H- atom is 3 kj mol while that of Li, it is 50kJ mol oly 8. H or Q is ot a isotopes of hydroge. 9. H e, p, o H e, p, 3 H e, p, 53. The very first method of removig hardess of water used : 57. 0 volume H O 0 LO at STP from L H O H O H O O 3 g.l. LO 3g H O 0 3 0 LO g H O 30.357 g H O. Stregth of HO 30.357 gl 58. KO is superoxide K O. BaO is peroxide is Ba O MO ad NO are dioxides O M O M 59. s aq Page 5 MX M aq X s s 3 K s. s s 0 s or s 0 3 3

CPT-5 / XI-REG / 0..07 / Hits & Solutio or, M X solid M X aq s 0 M M 0 M 60. aq K s s 56 s. Solubility product, 5 K K 56 56 5 s /5 PART C: MATHEMATICS 6. Give that 5 P 3 P 6. 63. 6.! 5! 3.!!! 5. 3 0 3 5! 3.! 3 0 3 0 =. P 5 P 3 3! 5.! 3! 3! 5 3 5 3! 5 56! 5! 30800. 56 r 6! 5 r 3! 56! 30800. 5! 50 r! 5 r! 56.55 30800 5 r 30800 5 r 0 56.55 r 0 0 K K PK K K 0. K. K! K K! K Page 6

CPT-5 / XI-REG / 0..07 / Hits & Solutio Page 7 0 K! K! K 0 65. sum=! d d d3... d 3! 9 933 66. 3 5 6 7 First arrage ay 3 cosoats at eve places i p3 ways. Now the ewly created four odd places ca be filled by the remaiig letters which icludes 3 vowels ad cosoats, which ca be doe i p ways. Hece the required umber of permutatio. p p 3 576. 67. There are total 9 places out of which are eve ad rest 5 place are odd. wome ca be arraged at eve places i! Ways ad 5 me ca be placed i remaiig 5 places i 5 p5 ways. Hece the required umber of permutatio. 5! P5 0 880 68. Total umber of arragemet =8!=030 Number of arragemet whe best ad worst papers are together = 7!! 0080 Numbers of arragemet i which best ad worst papers are ot together =030 0080=300 69. Ay 5 soldiers ca stad i oe row ad the remaiig 5 soldiers ca stad i other row. 0 5 Required umber of arragemets: p5 p5! 757600 70. A passeger from ay statio may purchase ticket for ayoe of the other 9 statios. Therefore total umber of differet tickets 90 90 7. Whe L ad T are fixed as first ad last letters of the word, the we have oly 6 letters to be arraged. Hece required umber of permutatios 6! 7. 90!!! M, M, T, T, H, C, S A, A, E, I 7 Whe all the vowels are together the 7 8 8!! Required umber of permutatios =0960!!! 73. T,T,T N,N,.N,G,G.G,R,I,I,I There are total 3 letters of which 3 are T s, 3 are G s, 3 are I s. Also there are 6 eve places ad 7 odd places. 3Is (i.e vowels) ca be arrages i the 6 eve places i 6 P3 0 ways. 3! 0! Now, we are left with 0 places i which 0 letters ca be filled up i 3! 3! 3! 6800 The total umber of permutatios i which vowels occupy eve places=06800 336000 7. Number of umbers of digits i which repetitio allowed 9000 9000 Number of umbers of digits i which repetitios is ot allowed 9987 536 Hece the required umbers of digits =9000 536 6 75. Total umber of 3digit umbers i which there is at least oe digit is 9 900 68 5 76. First letter ca be posted i letter boxes i ways. Simillarly secod letter ca be posted i letter boxes i ways ad so o. Hece a the 5 letters ca be posted i 0 77. Total umber of arragemet =6!=70

CPT-5 / XI-REG / 0..07 / Hits & Solutio Numbers of arragemet i which two ladies are together =5! 0 Number of arragemets i which two ladies are ever together =70 0=80 78. Assumig Sahara, Ambai ad Mahidra as a sigle persoality, the there are total (0 3)+=8 perso, which ca be arraged i 7! Ways. Therefore required umber of arragemets=7! 0080 79. A perso ca be chose out of 8 people i 8 ways to be seated betwee Musharraf ad Mamoha. Now cosider Musharraf, Mamoha ad the third perso,sittig betwee them as a sigle persoality, we ca arrage them i 7! Ways but Musharraf ad Mamoha ca also be arraged i ways. 8 7! 8! 80. Required umbers of permutatio = C x C x C x... 3 5 3 3 3 x 8. 8, x 6 6 8 ; ax 6 8 8 Now put x = a = 3 3 8. 3 83. 8. C!!! C!!! / p / q 85. I 3 0 0 0 0 0 0 C0 C... C9 C0 C... C 0 0 0 0 0 C 0 C... C 0 C0 9 0 C0 x y, if p or q divide r the umber of to ratioal terms are k else L.C.M. {p, q} ad [.] is greatest iteger fuctio k = L.C.M. {5, 0} = 0 5 divides 5 umber of terms are 86. r = 3, = 0 umber of terms 87. 5 5 0 C C 03 3 = 3 mm mx x... x x... mm m 3m m 3, mm 3 6 m 5 m m 7m -m + = - m = 88. 0 x a x a C C x a... k, where k = Page 8

CPT-5 / XI-REG / 0..07 / Hits & Solutio 3 3 x x x x 5 5 / / 5 x(x ) x C x(x ) Highest power is 7 89. Put x = i formula give i hit, the 5 3 3 5 3 6 7 C 3 C C 3 C0... 3 6 6 90. I f, f ' : 0 f ', 0 f 6 6 6 6 6 I f f ' C0 C C C 6 I + f + f = iteger f + f = iteger betwee 0, ad f + f = I 8 5 5 I = (99) - = 97 0 f f ' Page 9