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PREVIOUS YEAR PAPERS CBSE BOARD EXAM AIEEE BITSAT ISAT VITEEE IIT-JEE STUDY MATERIAL PHYSICS CLASS XI XII SAMPLE PAPERS KEY SOLUTIONS ANSWERS QUESTIONS
IIT-JEE 2010 Answers by (Division of Aakash Educational Services Ltd.) PAPER - 2 : CODES Q.No. 0 1 2 3 4 5 6 7 8 9 01. C D A C D C D C C D 02. A A C B C A C B A C 03. D C D C B D A D C B 04. B C D A D B D D D A 05. C B C D A C C A B D 06. D D B D C D B C D C 07. 7 2 7 2 7 2 2 3 2 3 08. 2 6 2 7 2 6 2 7 2 6 09. 2 3 2 6 3 2 7 6 7 7 10. 6 2 6 3 2 3 6 2 3 2 11. 3 7 3 2 6 7 3 2 6 2 12. B B B B B B B B B B 13. A C A C A C A C A C 14. D B D B D B D B D B 15. B B B B B B B B B B 16. C A C A C A C A C A 17. B D B D B D B D B D 18. A(r,s) A(p,s) A(r,s) A(p,s) A(r,s) A(p,s) A(r,s) A(p,s) A(r,s) A(p,s) B(t) B(p,q,r,t) B(t) B(p,q,r,t) B(t) B(p,q,r,t) B(t) B(p,q,r,t) B(t) B(p,q,r,t) C(p,q) C(p,q) C(p,q) C(p,q) C(p,q) C(p,q) C(p,q) C(p,q) C(p,q) C(p,q) D(r) D(p) D(r) D(p) D(r) D(p) D(r) D(p) D(r) D(p) 19. A(p,s) A(r,s) A(p,s) A(r,s) A(p,s) A(r,s) A(p,s) A(r,s) A(p,s) A(r,s) B(p,q,r,t) B(t) B(p,q,r,t) B(t) B(p,q,r,t) B(t) B(p,q,r,t) B(t) B(p,q,r,t) B(t) C(p,q) C(p,q) C(p,q) C(p,q) C(p,q) C(p,q) C(p,q) C(p,q) C(p,q) C(p,q) D(p) D(r) D(p) D(r) D(p) D(r) D(p) D(r) D(p) D(r) 20. A B C B D B D A B B 21. C C A D B C A D C B 22. B A B A D D C D A D 23. D B D C B D B B D C 24. B D B D C A B C D D 25. D D D B A B D B B A 26. 0 4 0 4 0 1 4 3 4 3 27. 1 3 1 0 4 3 1 0 1 3 28. 4 3 4 3 3 4 0 3 0 0 29. 3 1 3 3 1 3 3 4 3 4 30. 3 0 3 1 3 0 3 1 3 1 31. C D C D C D C D C D 32. A C A C A C A C A C 33. B A B A B A B A B A 34. D C D C D C D C D C 35. C A C A C A C A C A 36. A B A B A B A B A B 37. A(q,r) A(t) A(q,r) A(t) A(q,r) A(t) A(q,r) A(t) A(q,r) A(t) B(p) B(p,r) B(p) B(p,r) B(p) B(p,r) B(p) B(p,r) B(p) B(p,r) C(p,s,t) C(q) C(p,s,t) C(q) C(p,s,t) C(q) C(p,s,t) C(q) C(p,s,t) C(q) D(q,r,s,t) D(r) D(q,r,s,t) D(r) D(q,r,s,t) D(r) D(q,r,s,t) D(r) D(q,r,s,t) D(r) 38. A(t) A(q,r) A(t) A(q,r) A(t) A(q,r) A(t) A(q,r) A(t) A(q,r) B(p,r) B(p) B(p,r) B(p) B(p,r) B(p) B(p,r) B(p) B(p,r) B(p) C(q) C(p,s,t) C(q) C(p,s,t) C(q) C(p,s,t) C(q) C(p,s,t) C(q) C(p,s,t) D(r) D(q,r,s,t) D(r) D(q,r,s,t) D(r) D(q,r,s,t) D(r) D(q,r,s,t) D(r) D(q,r,s,t) 39. C D A B B B B B B D 40. A A C D B A C D A B 41. D C D C D B A C C D 42. D B B A D D D B B A 43. B D B B A C B A D B 44. B B D D C D D D D C 45. 6 8 6 8 6 3 8 2 8 2 46. 3 4 3 6 8 4 3 6 3 4 47. 8 2 8 4 2 8 6 4 6 6 48. 4 3 4 2 3 2 4 8 2 8 49. 2 6 2 3 4 6 2 3 4 3 50. C D C D C D C D C D 51. A B A B A B A B A B 52. B C B C B C B C B C 53. D C D C D C D C D C 54. B A B A B A B A B A 55. C B C B C B C B C B 56. A(p,r) A(r,s,t) A(p,r) A(r,s,t) A(p,r) A(r,s,t) A(p,r) A(r,s,t) A(p,r) A(r,s,t) B(q,s,t) B(q,r,s,t) B(q,s,t) B(q,r,s,t) B(q,s,t) B(q,r,s,t) B(q,s,t) B(q,r,s,t) B(q,s,t) B(q,r,s,t) C(p,r,t) C(p,q) C(p,r,t) C(p,q) C(p,r,t) C(p,q) C(p,r,t) C(p,q) C(p,r,t) C(p,q) D(q,s) D(q,r,s,t) D(q,s) D(q,r,s,t) D(q,s) D(q,r,s,t) D(q,s) D(q,r,s,t) D(q,s) D(q,r,s,t) 57. A(r,s,t) A(p,r) A(r,s,t) A(p,r) A(r,s,t) A(p,r) A(r,s,t) A(p,r) A(r,s,t) A(p,r) B(q,r,s,t) B(q,s,t) B(q,r,s,t) B(q,s,t) B(q,r,s,t) B(q,s,t) B(q,r,s,t) B(q,s,t) B(q,r,s,t) B(q,s,t) C(p,q) C(p,r,t) C(p,q) C(p,r,t) C(p,q) C(p,r,t) C(p,q) C(p,r,t) C(p,q) C(p,r,t) D(q,r,s,t) D(q,s) D(q,r,s,t) D(q,s) D(q,r,s,t) D(q,s) D(q,r,s,t) D(q,s) D(q,r,s,t) D(q,s) Though every care has been taken to provide the answers correctly but the Institute shall not be responsible for error, if any. (Division of Aakash Educational Services Ltd.) Regd. Office: Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-110075. Ph.: (011) 47623456. 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VaranasI: (0542) 2314975 INFORMATION CENTRES AT: Chennai: 0442-6209607, 9380029607 Hamirpur: 9318989643/44 Hyderabad: 040-66102991, 9396775000 Indore: 0731-4239679, 9302184721 / 23 / 24 Kalu Sarai: (011) 46080497/98, 9312230103 / 303 Kanpur: 0512-2503594, 9358547135, 9335151568 Kota: 0744-3237788, 9352027261/2 Patna: (0612) 6531267 E-mail: iitjee@aakashiitjee.com | Website: www.aakashiitjee.com | SMS Aakash to 56677 TOP RANKERS ALWAYS FROM AAKASH Downloaded from www.icbse.com DATE : 11/04/2010 Regd. Office : Aakash Tower, Plot No.-4, Sec-11, MLU, Dwarka, New Delhi-110075 Ph.: 011-47623456 Fax : 011-47623472 Time : 3 hrs. Solutions Max. Marks: 252 to IIT-JEE 2010 PAPER - 2 (Code - 5) Instructions : 1. The question paper consists of 3 parts (Chemistry, Mathematics and Physics) and each Part consists of four Sections. 2. For each question in Section I, you will be awarded 5 marks if you have darkened only the bubble corresponding to the correct answer and zero mark if no bubbles are darkened. In all other cases, minus two (–2) mark will be awarded. 3. For each question in Section II, you will be awarded 3 marks if you have darkened the bubble corresponding to the correct answer and zero mark if no bubbles are darkened. No negative marks will be awarded for incorrect answers in this Section. 4. For each question in Section III, you will be awarded 3 marks if you have darkened only the bubble corresponding to the correct answer and zero mark if no bubbles are darkened. In all other cases, minus one (–1) mark will be awarded. 5. For each question in Section IV, you will be awarded 2 marks for each row in which you have darkened the bubble(s) corresponding to the correct answer. Thus, each question in this section carries a maximum of 8 marks. There is no negative marks awarded for incorrect answer(s) in this Section. Downloaded from www.icbse.com IIT-JEE 2010 (Paper-2) PART–I : CHEMISTRY SECTION - I Single Correct Choice Type This section contains 6 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. O (1) NaOH/Br2 1. In the reaction H3C C O T , the structure of the product T is NH2 (2) C Cl O H3C C O NH (A) O C (B) C CH3 O O H3C NH H3C C O (C) (D) NH C) C td . sL i ce O lS e rv n a Answer (C) u c a ti o d Hints : s hE ka Aa O O f no O C i si o NaOH + Br2 Cl H3C C H3C NH2 C v NH2 (D i NH H 3C 2. Assuming that Hund’s rule is violated, the bond order and magnetic nature of the diatomic molecule B2 is (A) 1 and diamagnetic (B) 0 and diamagnetic (C) 1 and paramagnetic (D) 0 and paramagnetic Answer (A) Hints : Molecular orbital diagram for B2 where Hund’s rule is violated. π2p2 σ1s 2 σ * 1s 2 σ2s 2 σ * 2s 2 X π2p0 Y ∴ Bond order = 1 and diamagnetic. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 011-47623417/23 Fax : 47623472 (2) Downloaded from www.icbse.com IIT-JEE 2010 (Paper-2) 3. The species having pyramidal shape is (A) SO3 (B) BrF3 (C) SiO2– 3 (D) OSF2 Answer (D) Hints : Pyramidal S O F F 4. The complex showing a spin-only magnetic moment of 2.82 B.M. is (A) Ni(CO)4 (B) [NiCl4]2– (C) Ni(PPh3)4 (D) [Ni(CN)4]2– Answer (B) Hints : Ni2+ is sp3 hybridized and metal ion is connected with weak ligands. ∴ Ni2+ ⇒ d 8 two unpaired electrons sp3 td .) sL ∴ μs = 2 (2 + 2) B.M. i ce = 8 B.M. = 2.83 B.M. lS e rv n a 5. The compounds P, Q and S u c a ti o d s hE Oka f Aa COOH OCH3 io no C (D i vi s O HO H3C P Q S were separately subjected to nitration using HNO3/H2SO4 mixture. The major product formed in each case respectively, is O COOH OCH3 C (A) O HO H3C NO2 NO2 O2N O COOH OCH3 C (B) O HO NO2 H3C NO2 NO2 Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 011-47623417/23 Fax : 47623472 (3) Downloaded from www.icbse.com IIT-JEE 2010 (Paper-2) O NO2 COOH OCH3 C (C) O HO H3C NO2 NO2 O COOH OCH3 C (D) O NO2 HO H3C NO2 NO2 Answer (C) Hints : COOH COOH HNO3/H 2SO4 HO HO NO2 OCH3 HNO3/H2SO4 OCH3 H3C H3C NO2 td .) O O NO2 sL i ce C C lS e rv HNO3/H2SO4 n a a ti o O O d u c hE The packing efficiency of the two-dimensional squares ka unit cell shown below is 6. fA a no v i si o (D i L (A) 39.27% (B) 68.02% (C) 74.05% (D) 78.54% Answer (D) Hints : Area covered by particle Packing efficiency = Total area 2 × πr 2 2 × πr 2 π = = = (2 ) 2 a 2 2r 4 Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 011-47623417/23 Fax : 47623472 (4) Downloaded from www.icbse.com IIT-JEE 2010 (Paper-2) SECTION - II Integer Type This section contains a group of 5 questions. The answer to each of the questions is a Single-digit Integer, ranging from 0 to 9. The correct digit below the equation number in the ORS is to be bubbled. 7. Among the following, the number of elements showing only one non-zero oxidation state is O, Cl, F, N, P, Sn, Tl, Na, Ti Answer (2) Hints : F will only exhibit –1 oxidation state except zero. and Na will exhibit +1 oxidation state. 8. The total number of diprotic acids among the following is H3PO4 H2SO4 H3PO3 H2CO3 H2S2O7 H3BO3 H3PO2 H2CrO4 H2SO3 Answer (6) Hints : H2SO4, H3PO3, H2CO3, H2CrO4 and H2SO3 and H2S2O7 will behave as dibasic acid. 9. One mole of an ideal gas is taken from a to b along two paths denoted by the solid and the dashed lines as shown in the graph below. If the work done along the solid line path is Ws and that along the dotted line path is Wd, then the integer closest to the ratio Wd/Ws is 4.5 4.0 a 3.5 td .) sL 3.0 i ce e rv P (atm.) 2.5 n a lS a ti o 2.0 1.5 d u c s hE ka 1.0 0.5 o f Aa b i on vi s 0.0 (D i 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 V (lit.) Answer (2) Hints : Solid line path work done (Ws) is isothermal beacuse PV is constant at each point & dash line path work done (Wd) is isobaric. V2 Total work done on solid line path (Ws) = 2.303 nRT log V1 V2 = 2.303 PV log V1 5.5 = 2.303 × 2log 0.5 ≈ 4.6 l atm Total work done on dash line path (Wd) = 4 × 1.5 + 1 × 1 + 0.5 × 2.5 = 8.255 l atm. Wd 8.25 = ≈ 2(closest integer) Ws 4.6 Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 011-47623417/23 Fax : 47623472 (5) Downloaded from www.icbse.com IIT-JEE 2010 (Paper-2) 10. Total number of geometrical isomers for the complex [RhCl(CO)(PPh3)(NH3)] is Answer (3) Hints : Rh Cl(CO)(PPh3)(NH3)] is a square planar complex with four different ligands and hence it will have three geometrical isomers a b a b a c Rh Rh Rh d c c d d b 11. Silver (atomic weight = 108 g mol–1) has a density of 10.5 g cm–3. The number of silver atoms on a surface of area 10–12 m2 can be expressed in scientific notation as y × 10x. The value of x is Answer (7) Hints : 4 3 Volume of 1 Ag atom = πr 3 4 3 108 ∴ πr = cm3 3 6.023 × 10 × 10.5 23 ∴ r = 1.6 × 10–8 cm r = 1.6 × 10–10 m ∴ Number of Ag atoms in 10–12 m2 td .) sL πr2 ×n= 10–12 m2 i ce –12 lS e rv n= 10 n a 3.14 × (1.6 × 10–10 )2 u c a ti o d 10 × 10 7 s hE n= ka 8 f Aa io no i vi s n = 1.25 × 107 ∴ The value of x = 7 (D SECTION - III Paragraph Type This section contains 2 paragraphs. Based upon each of the paragraphs 3 multiple choice questions have to be answered. Each of these questions has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Paragraph for questions 12 to 14 The hydrogen-like species Li2+ is in a spherically symmetric state S1 with one radial node. Upon absorbing light the ion undergoes transition to a state S2. The state S2 has one radial node and its energy is equal to the ground state energy of the hydrogen atom. 12. The state S1 is (A) 1s (B) 2s (C) 2p (D) 3s Answer (B) Hints : S1 state is 2s In 2s, one radial node is present Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 011-47623417/23 Fax : 47623472 (6) Downloaded from www.icbse.com IIT-JEE 2010 (Paper-2) 13. Energy of the state S1 in units of the hydrogen atom ground state energy is (A) 0.75 (B) 1.50 (C) 2.25 (D) 4.50 Answer (C) Hints : Energy in S1 state –13.6 × 32 9 E= 2 = – 13.6 eV 2 4 E( H1 ) = –13.6 eV [Ground state] 1 So energy = 2.25 × energy of e– in ground state in 1H1. 14. The orbital angular momentum quantum number of the state S2 is (A) 0 (B) 1 (C) 2 (D) 3 Answer (B) Hints : S2 state is 3p orbital Orbital angular momentum of 3p is 1. Paragraph for questions 15 to 17 Two aliphatic aldehydes P and Q react in the presence of aqueous K2CO3 to give compound R, which upon treatment with HCN provides compound S. On acidification and heating, S gives the product shownd.) t below : L es H3C rvic Se OH H3C al ti o n d u ca hE O O ka s Aa 15. The compounds P and Q respectively are f CH3 io no CH (D i vi s H and H3C H (A) H3C C C O O CH3 CH H and H H (B) H3C C C O O H3C CH2 H H3C H CH C and C (C) CH3 O O H3C CH2 H H H CH C and C (D) CH3 O O Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 011-47623417/23 Fax : 47623472 (7) Downloaded from www.icbse.com IIT-JEE 2010 (Paper-2) 16. The compound R is O O H3C H3C C C H C H C H3C (A) H3C (B) CH2 CH OH H3C OH CH3 O CH3 O CH C CH C H3C CH H H3C CH H (C) (D) CH2 CH OH H3C OH 17. The compound S is CH3 O O CH H3C C C H3C CH H C H (A) (B) H3C CH2 CH2 CN CN CH3 CN CN CH H3C CH CH OH .) H3C CH OH (D) H3C C L td ce s (C) CH2 rvi Se CH2 OH n al OH ti o u ca Hints (15-17) : 15. Answer (B) h Ed 16. Answer (A) a k as fA 17. Answer (D) io no O i vi s CH3 (D H3C 1. K2CO3 C C H 2. H H C H H3C C H3C H C (Q) CH2 (P) O 3. H + O OH (R) HCN OH H3C CN CH O C C H3C H3C CH Hydrolysis C OH H2C HO H3C O H CH2 Δ OH (S) OH H3C C O H3C H2C O Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 011-47623417/23 Fax : 47623472 (8) Downloaded from www.icbse.com IIT-JEE 2010 (Paper-2) SECTION - IV Matrix Type This section contains 2 questions. Each question has four choices (A), (B), (C) and (D) given in Column I and five statements (p), (q), (r), (s) and (t) in Column II. Any given statement in Column I can have correct matching with one or more statement(s) given in Column II. For example, if for a given question, statement B matches with the statements given in (q) and (r), then for that particular question, against statement B, darken the bubbles corresponding to (q) and (r) in ORS. 18. All the compounds listed in Column I react with water. Match the result of the respective reactions with the appropriate options listed in Column II. Column I Column II (A) (CH3)2SiCl2 (p) Hydrogen halide formation (B) XeF4 (q) Redox reaction (C) Cl2 (r) Reacts with glass (D) VCl5 (s) Polymerization (t) O2 formation Answer : A(p, s), B(p, q, r, t), C(p, q), D(p) Hints : CH3 CH3 Polymerise Si (A) (CH3)2SiCl2 + H2O Si—OH O O OH CH3 CH3 td .) n sL i ce (B) 6XeF4 + 12H2 O ⎯⎯ 4Xe + 2XeO3 + 24HF + 3O 2 → lS e rv n a (C) Cl2 + H2O ⎯⎯ HCl + HOCl → u c a ti o d (D) VCl5 + H2O ⎯⎯ VOCl3 + HCl → s hE ka o f Aa i on Note : Vanadium in (+V) oxidation state from only fluoride. Existence of VCl5 is doubtful. i vi s (D 19. Match the reactions in Column I with appropriate options in Column II. Column I Column II NaOH/H2O (A) N2Cl + OH N=N OH (p) Racemic mixture 0°C O OH OH CH3 C H2SO4 (B) H3C – C — C – CH3 H3C C (q) Addition reaction CH3 CH3 CH3 CH3 O OH 1. LiAlH4 (C) C + CH (r) Substitution reaction 2. H3O CH3 CH3 Base (D) HS Cl S (s) Coupling reaction (t) Carbocation intermediate Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 011-47623417/23 Fax : 47623472 (9) Downloaded from www.icbse.com IIT-JEE 2010 (Paper-2) Answer : A(r, s), B(t), C(p, q), D(r) Hints : (A) N≡ N + O—H Cl– OH N=N O H N=N O ∴ Coupling reaction and electrophilic substitution reaction. OH OH OH OH2 td .) sL i ce (B) +H e rv CH3 — C — C — CH3 CH3 — C — C — CH3 CH3 CH3 CH3 CH3 n a lS u c a ti o d s hE ka f Aa O CH3 no OH i—oC — C si CH3 CHv CH3 — C — C — CH3 (D i 3 CH3 CH3 CH3 Carbocation intermediate O OH O + LiAlH4 H H3O (C) C Nucleophilic addition C H C H CH3 CH3 CH3 Racemic mixture Nucleophilic H Base S substitution S (D) S Cl Cl B Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 011-47623417/23 Fax : 47623472 (10) Downloaded from www.icbse.com IIT-JEE 2010 (Paper-2) PART–II : MATHEMATICS SECTION - I (Single Correct Choice Type) This section contains 5 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. x 20. Let f be a real-valued function defined on the interval (–1, 1) such that e f ( x ) = 2 + ∫ t + 1 dt , for all −x 4 0 x ∈ (–1, 1), and let f –1 be the inverse function of f. Then (f –1)′ (2) is equal to 1 1 1 (A) 1 (B) (C) (D) 3 2 e Answer (B) Hints : We have, x e − x f ( x ) = 2 + ∫ t 4 + 1 dt x ∈ (–1, 1) 0 Differentiating w.r.t. x, we get e −x (f ' ( x ) − f ( x )) = x +1 4 td .) sL i ce ⇒ f ' ( x ) = f ( x ) + x 4 + 1e x lS e rv n a ∵ f –1 is the inverse of f u c a ti o d s hE ∴ f –1′(f(x)) = x ka f Aa ⇒ f –1′(f(x)) f ′(x) = 1 io no (D i vi s ⇒ f −1 ' ( f ( x ) ) = 1 f '(x) ⇒ f −1 ' ( f ( x ) ) = 1 f ( x ) + x4 + 1 ex at x = 0, f(x) = 2 1 1 f ' ( 2) = −1 = 2+1 3 4 1 21. A signal which can be green or red with probability and respectively, is received by station A and then 5 5 3 transmitted to station B. The probability of each station receiving the signal correctly is . If the signal received 4 at station B is green, then the probability that the original signal was green is 3 6 20 9 (A) (B) (C) (D) 5 7 23 20 Answer (C) Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 011-47623417/23 Fax : 47623472 (11) Downloaded from www.icbse.com IIT-JEE 2010 (Paper-2) Hints : From the tree-diagram it follows that S 4 1 46 P (BG ) = 5 5 80 R P ( BG | G ) = 10 5 = 3G 1 3 1 16 8 4 4 4 4 P ( BG ∩ G ) = 5 4 1 AG AR AR AG ∴ × = 3 1 3 1 1 3 3 8 5 2 4 4 4 4 4 4 1 4 1 4 2 = 1 × 80 = 20 BG BR BG BR BG P ( G | BG ) = P (BG ) 2 46 23 22. Let S = {1, 2, 3, 4}. The total number of unordered pairs of disjoint subsets of S is equal to (A) 25 (B) 34 (C) 42 (D) 41 Answer (D) Hints : Let A ∩ B = φ, A, B ⊂ S The number of elements in A : 0 Choices for B : 24 The number of elements in A : 1 Choices for B : 38 The number of elements in A : 2 Choices for B : 3 Total number of possible subsets is 41 td .) sL i ce 23. For r = 0, 1,..., 10, let Ar , B r and C r denote, respectively, theScoefficient e rv al of x r in the expansions of ti o n ) ca 10 (1 + x)10, (1 + x)20 and (1 + x)30. Then ∑ Ar ( B10Br − C10 Arduis equal to E r =1 sh a ka (B) A10 ( B10 − C10 A10 ) of A (C) 0 2 (A) B10 – C10 (D) C10 – B10 i on Answer (D) i vi s (D10 r Hints : Ar = Coefficient of x in (1 + x)10 = Cr Br = Coefficient of xr in (1 + x)20 = 20C r Cr = Coefficient of xr in (1 + x)30 = 30C r 10 10 10 ∑ A (B r =1 r 10 Br − C10 Ar ) = ∑ Ar B10 Br − ∑ Ar C10 Ar r =1 r =1 10 10 = ∑ Cr Cr − ∑ C r 10 20 20 10 30 10 C10 C10 Cr r =1 r =1 10 10 = ∑ C10− r Cr − ∑ C10 − r 10 20 20 10 30 10 C10 C10 Cr r =1 r =1 10 10 = C10 ∑ C10 − r . Cr − C10 ∑ 20 10 20 30 10 10 C 10− r Cr r =1 r =1 = C10 20 ( 30 ) C10 − 1 − C10 30 ( 20 C10 − 1 ) = C10 − C10 = C10 − B10 30 20 Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 011-47623417/23 Fax : 47623472 (12) Downloaded from www.icbse.com IIT-JEE 2010 (Paper-2) 24. If the distance of the point P(1, –2, 1) from the plane x + 2y – 2z = α, where α > 0, is 5, then the foot of the perpendicular from P to the plane is ⎛8 4 7⎞ ⎛ 4 4 1⎞ ⎛ 1 2 10 ⎞ ⎛2 1 5⎞ (A) ⎜ , , − ⎟ (B) ⎜ , − , ⎟ (C) ⎜ , , ⎟ (D) ⎜ , − , ⎟ ⎝3 3 3⎠ ⎝3 3 3⎠ ⎝3 3 3 ⎠ ⎝3 3 2⎠ Answer (A) Hints : Distance of point P from plane = 5 1− 4 − 2 − α ∴ 5= 3 P(1, –2, 1) α = 10 Foot of perpendicular x −1 y + 2 z −1 5 = = = 1 2 −2 3 8 4 7 x= ,y = ,z=− 3 3 3 Thus the foot of the perpendicular is ⎛8 4 7⎞ ∴f⎜ , ,− ⎟ ⎝3 3 3⎠ td .) sL 25. Two adjacent sides of a parallelogram ABCD are given by i ce AB = 2 i + 10 j + 11k and AD = − i + 2 j + 2 k . lS e rv ti o na The side AD is rotated by an acute angle α in the plane of the parallelogram so that AD becomes AD ′. If AD ′ d uca α is given by makes a right angle with the side AB, then the cosine of the angle E sh 8 a ka 1 of A (C) 17 4 5 (A) (B) (D) 9 9 io n 9 9 Answer (B) (D i vi s Hints : AB = 2 i + 10 j + 11k AD = − i + 2 j + 2 k D C Angle 'θ' between AB and AD α θ A B −2 + 20 + 22 8 AB . AD = cos ( θ ) = = = AB AD (15)( 3 ) 9 17 ⇒ sin ( θ ) = 9 α + θ = 90 17 cos ( α ) = cos ( 90 − θ ) = sin ( θ ) = 9 Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 011-47623417/23 Fax : 47623472 (13) Downloaded from www.icbse.com IIT-JEE 2010 (Paper-2) SECTION - II (Integer Type) This Section contains 5 questions. The answer to each question is a single-digit integer, ranging from 0 to 9. The correct digit below the question no. in the ORS is to be bubbled. 26. Let f be a function defined on R (the set of all real numbers) such that f '(x) = 2010 (x – 2009) (x – 2010)2 (x – 2011)3 (x – 2012)4, for all x ∈ R. If g is a function defined on R with values in the interval (0, ∞) such that f(x) = ln (g(x)), for all x ∈ R, then the number of points in R at which g has a local maximum is Answer (1) f (x ) Hints : g ( x ) = e , ∀x ∈R f (x) ⇒ g '( x ) = e .f ' ( x ) ⇒ f′(x) changes its sign from positive to negative in the neighbourhood of x = 2009 ⇒ f(x) has local maxima at x = 2009 So, the number of local maximum is one. 27. Two parallel chords of a circle of radius 2 are at a distance 3 + 1 apart. If the chords subtend at the center, π 2π angles of and , where k > 0, then the value of [k] is k k [Note : [k] denotes the largest integer less than or equal to k] td .) sL Answer (3) i ce lS e rv π n a Hints : Let θ = 2k u c a ti o d s hE x ka cos θ = 2 o f Aa θ x n i si o 2 v (D i C 3 + 1− x ⇒ cos2θ = 2 2 2θ 3+1–x 3 + 1− x ⇒ 2cos2 θ − 1 = 2 ⎛ x2 ⎞ 3 + 1− x ⇒ 2⎜ ⎟ − 1= ⎝ 4 ⎠ 2 ⇒ x2 + x − 3 − 3 = 0 −1 ± 1 + 12 + 4 3 x= 2 −1 ± 13 + 4 3 = 2 −1 + 2 3 + 1 = = 3 2 Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 011-47623417/23 Fax : 47623472 (14) Downloaded from www.icbse.com IIT-JEE 2010 (Paper-2) 3 π ∴ cos θ = ⇒θ= 2 6 π π ∴ Required angle = = 2θ = k 3 ⇒k = 3 28. Let k be a positive real number and let ⎡ 2k − 1 2 k 2 k⎤ ⎡ 0 2k − 1 k ⎤ ⎢ ⎥ ⎢ ⎥ A=⎢2 k 1 −2k ⎥ and B = ⎢1 − 2k 0 2 k ⎥. ⎢ ⎥ ⎢ ⎥ ⎢ −2 k 2k ⎣ −1 ⎥ ⎦ ⎢− k ⎣ −2 k 0 ⎥⎦ If det (adj A) + det(adj B) = 106, then [k] is equal to [Note : adj M denotes the adjoint of a square matrix M and [k] denotes the largest integer less than or equal to k] Answer (4) Hints : |A| = (2k + 1)3, |B| = 0 ⇒ det (adj A) ≠ det (adj B) = (2k + 1)6 = 106 ⇒ k = 9 td .) sL . i ce 2 ⇒ [k] = 4 lS e rv n a u c a ti o Ed 29. Consider a triangle ABC and let a, b and c denote the lengths of the sides opposite to vertices A, B and C h respectively. Suppose a = 6, b = 10 and the area of as triangle is 15 3. If ∠ACB is obtuse and if r denotes ak the the radius of the incircle of the triangle, then r2 of A is equal to i on vi s (D i Answer (3) Hints : 3 sin C = and C is given to be obtuse. 2 2π ⇒ C= 3 30. Let a1, a2, a3,......, a11 be real numbers satisfying a1 = 15, 27 – 2a2 > 0 and ak = 2ak – 1 – ak – 2 for k = 3, 4, ...., 11. a1 + a2 + .... + a11 2 2 2 If = 90, 11 a1 + a2 + .... + a11 then the value of is equal to 11 Answer (0) Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 011-47623417/23 Fax : 47623472 (15) Downloaded from www.icbse.com IIT-JEE 2010 (Paper-2) SECTION-III (Paragraph Type) This section contains 2 paragraphs. Based upon each of the paragraphs 3 multiple choice questions have to be answered. Each of these question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Paragraph for questions 31 to 33 x2 y 2 Tangents are drawn from the point P(3, 4) to the ellipse + = 1 touching the ellipse at points A and B. 9 4 31. The coordinates of A and B are ⎛ 8 2 161 ⎞ ⎛ 9 8⎞ (A) (3, 0) and (0, 2) (B) ⎜ − 5 , 15 ⎟ and ⎜ − 5 , 5 ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ 8 2 161 ⎞ ⎛ 9 8⎞ (C) ⎜ − 5 , 15 ⎟ and (0, 2) ⎜ ⎟ (D) (3, 0) and ⎜ − , ⎟ ⎝ ⎠ ⎝ 5 5⎠ Answer (D) Hints : Figure is self explanatory P D (3, 4) B td .) sL i ce F A lS e rv (3, 0) n a u c a ti o d s hE ka 32. The orthocentre of the triangle PAB is f Aa ⎛ 8⎞ ⎛ 7 25 ⎞ isio no ⎛ 11 8 ⎞ ⎛ 8 7⎞ ⎟ iv ⎝ 5 8 (D (A) ⎜ 5, ⎟ (B) ⎜ , (C) ⎜ , ⎟ (D) ⎜ , ⎟ ⎝ 7⎠ ⎠ ⎝ 5 5⎠ ⎝ 25 5 ⎠ Answer (C) Hints : Equation of AB is P(3, 4) 8 y −0 = 5 ( x − 3) = 8 ( x − 3) 9 −24 –9, 8 − −3 5 5 5 B A (3, 0) 1 ⇒ y = − ( x − 3) 3 ⇒ x + 3y = 3 ...(i) Equation of the straight line perpendicular to AB through P is 3x – y = 5 Equation of PA is x – 3 = 0 Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 011-47623417/23 Fax : 47623472 (16) Downloaded from www.icbse.com IIT-JEE 2010 (Paper-2) ⎛ −9 8 ⎞ 8 The equation of straight line perpendicular to PA through B ⎜ , ⎟ is y = . ⎝ 5 5⎠ 5 Hence the orthocentre is ⎛ 11, 8 ⎞ ⎜ 5 5⎟ ⎝ ⎠ 33. The equation of the locus of the point whose distances from the point P and the line AB are equal, is (A) 9x2 + y2 – 6xy – 54x – 62y + 241 = 0 (B) x2 + 9y2 + 6xy – 54x + 62y – 241 = 0 (C) 9x2 + 9y2 – 6xy – 54x – 62y – 241 = 0 (D) x2 + y2 – 2xy + 27x + 31y – 120 = 0 Answer (A) Hints : 1 Equation of AB = y − 0 = − ( x − 3) 3 x + 3y – 3 = 0 |x + 3y – 3|2 = 10[(x – 3)2 + (y – 4)2] (Look at coefficient of x2 & y2 in the answers) td .) sL i ce Paragraph for question 34 to 36 lS e rv Consider the polynomial f(x) = 1 + 2x + 3x2 + 4x3. Let s be the sum na all distinct real roots of f(x) = 0 and let a ti o of t = |s|. c d u s hE ka 34. The real number s lies in the interval o f Aa (B) ⎛ −11, − 3 ⎞ sion (A) ⎛ − 1 , 0 ⎞ (C) ⎛ − 3 , − 1 ⎞ ⎛ 1⎞ 4 ⎟ i vi ⎜ ⎜ 4 (D) ⎜ 0, ⎟ ⎜ 4 ⎝ ⎟ ⎠ ⎝ D ⎠ ⎝ 2⎟⎠ ⎝ 4⎠ ( Answer (C) Hints : f ′(x) = 2(6x2 + 3x + 1) Δ = 9 – 24 < 0 Hence f(x) = 0 has only one real root. ⎛ 1⎞ 3 4 f ⎜ − ⎟ = 1− 1+ − > 0 ⎝ 2⎠ 4 8 ⎛ 3⎞ 6 27 108 f ⎜ − ⎟ = 1− + − ⎝ 4⎠ 4 16 64 64 − 96 + 108 − 108 = <0 64 ⎛ 3 −1 ⎞ ⎛ −3 −1 ⎞ f(x) changes its sign in ⎜ − , ⎟ , hence f(x) = 0 has a root in ⎜ , ⎟ . ⎝ 4 2⎠ ⎝ 4 2⎠ Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 011-47623417/23 Fax : 47623472 (17) Downloaded from www.icbse.com IIT-JEE 2010 (Paper-2) 35. The area bounded by the curve y = f(x) and the lines x = 0, y = 0 and x = t, lies in the interval ⎛3 ⎞ ⎛ 21 11 ⎞ (A) ⎜ , 3 ⎟ (B) ⎜ , ⎟ ⎝4 ⎠ ⎝ 64 16 ⎠ ⎛ 21 ⎞ (C) (9, 10) (D) ⎜ 0, ⎟ ⎝ 64 ⎠ Answer (A) Hints : 1 3 2 t 4 ∫ f ( x )dx < ∫ f ( x )dx < ∫ f ( x )dx 1 0 0 0 1 t 3 ∫ f ( x )dx = x + x + x3 + x 4 2 2 4 1 2 13 3 ∫ f ( x )dx = 16 > 4 0 3 4 525 ∫ f ( x )dx = 256 < 3 0 td .) sL i ce 36. The function f′(x) is lS e rv n a c a ti o (A) Increasing in ⎜ −t , − ⎟ and decreasing in ⎜ − , t ⎟ Edu ⎛ 1⎞ ⎛ 1 ⎞ ⎝ 4⎠ ⎝ 4 sh ⎠ a ka of A ⎛ 1⎞ i on ⎛ 1 ⎞ i vi s (B) Decreasing in ⎜ −t , − ⎟ and increasing in ⎜ − , ⎝ 4⎠ (D ⎝ 4 t⎟ ⎠ (C) Increasing in (–t, t) (D) Decreasing in (–t, t) Answer (B) Hints : Figure is self explanatory 1 S –3 –1 1 3 4 2 2 4 Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 011-47623417/23 Fax : 47623472 (18) Downloaded from www.icbse.com IIT-JEE 2010 (Paper-2) SECTION - IV (Matrix Type) This Section contains 2 questions. Each question has four statements (A, B, C and D) given in Column I and five statements (p, q, r, s and t) in Column II. Any given statement in Column I can have correct matching with one or more statement(s) given in Column II. For example, if for a given question, statement B matches with the statements given in q and r, then for that particular question, against statement B, darken the bubbles corresponding to q and r in the ORS. 37. Match the statements in Column-I with the values in Column-II. Column I Column II (A) A line from the origin meets the lines (p) –4 x − 2 y −1 z +1 = = and 1 −2 1 8 x− 3 = y + 3 = z −1 at P and Q 2 −1 1 respectively. If length PQ = d, then d2 is (B) The values of x satisfying (q) 0 ⎛3⎞ tan−1( x + 3) − tan−1( x − 3) = sin−1 ⎜ ⎟ are ⎝5⎠ (C) Non-zero vectors a, b and c satisfy a ⋅ b = 0 , (r) 4 td .) sL (b − a) ⋅ (b + c ) = 0 and 2 | b + c | = | b − a | . i ce If a = μb + 4c , then the possible values of lS e rv n a μ are c a ti o (D) Let f be the function on [–π, π] given by du E(s) 5 s h ⎛ 9x ⎞ ⎛x⎞ fA a ka f(0) = 9 and f ( x ) = sin ⎜ ⎟ sin ⎜ 2 ⎟ o ⎝ 2 ⎠ ⎝ ⎠ i on vi s π (D i ∫ f ( x ) dx is 2 for x ≠ 0. The value of π −π (t) 6 Answer → A(t), B(p, r), C(q), D(r) Hints : x y z (A) Equation of line passing through origin is = = a b c 2 1 −1 ∴ 1 −2 1 =0 a b c ⇒ a(–1) – b(3) + c(–5) = 0 –a – 3b – 5c = 0 a + 3b + 5c = 0 ... (1) Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 011-47623417/23 Fax : 47623472 (19) Downloaded from www.icbse.com IIT-JEE 2010 (Paper-2) 8 −3 1 3 Also, 2 −1 1 = 0 a b c ⎛2⎞ ⎛ 10 ⎞ a( −2) − b ⎜ ⎟ + c ⎜ ⎟=0 ⎝ 3⎠ ⎝ 3 ⎠ 2b 10c 2a + − =0 3 3 3a + b – 5c = 0 ... (2) Now, a + 3b + 5c = 0 3a + b – 5c = 0 a b c = = −20 20 −8 a b c = = 5 −5 4 Equation of line is x y z = = ... (2) 5 −5 +4 x − 2 y −1 z +1 td .) = = sL i ce ... (3) −2 e rv 1 1 8 n a lS a ti o x− 3 = y + 3 = z −1 c Now, ... (4) d u 2 −1 1 s hE ka Point on (2) is (5λ, –5λ, + 4λ) f Aa Point on (3) is (2 + k1, 1 – 2k1, –1 + k1)o n v i si o ⎛8 (D i ⎞ Point on (4) is ⎜ + 2k2 , − 3 − k2 , 1 + k2 ⎟ ⎝3 ⎠ On solving, 2 + k1 + 1 – 2k1 = 0 –k1 + 3 = 0 k1 = 3 P ≡ (5, –5, 2) Again for Q, 8 + 2k2 − 3 − k 2 = 0 3 1 k2 − =0 3 1 k2 = 3 Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 011-47623417/23 Fax : 47623472 (20) Downloaded from www.icbse.com IIT-JEE 2010 (Paper-2) ⎛ 10 −10 4 ⎞ Q ≡⎜ , , ⎝ 3 3 3⎟ ⎠ 2 2 2 ⎛5⎞ ⎛5⎞ ⎛ 2⎞ PQ = ⎜ ⎟ +⎜ ⎟ +⎜ ⎟ ⎝3⎠ ⎝3⎠ ⎝3⎠ 54 = 3 54 PQ2 = d2 = =6 9 ⎛ x +3−x +3⎞ −1 ⎛ 3 ⎞ (B) tan−1 ⎜ ⎜ 1 + ( x 2 − 9) ⎟ = tan ⎝ 4 ⎠ ⎟ ⎜ ⎟ ⎝ ⎠ 6 3 = x −8 4 2 3x2 – 24 = 24 3x2 = 48 x = ±4 ⎛ a − μb ⎞ (C) (b − a ) ⋅ ⎜ b + ⎜ ⎟=0 ⎝ 4 ⎟ ⎠ (b − a) ⋅ (4b + a − μb ) = 0 (4 – μ)b2 – a2 = 0 ... (1) Also, 2 b + a − μb = | b − a| td .) sL i ce 4 (4 − μ)b + a lS e rv 2 n a a ti o 4 d u c (4 − μ)2 b 2 a2 s hE + = b2 + a 2 ka 4 4 f Aa 3a 2 (4 − μ)2 − 4 2 io no 4 = 4 ⋅b (D i vi s 3a2 = ((4 – μ)2 – 4)b2 ... (2) From equations (1) and (2), 3(4 – μ) = (4 – μ)2 – 4 (4 – μ)2 – 3(4 – μ) – 4 = 0 ⇒ μ = 0, 5 μ = 5 is not admissible. (D) f(0) = 9, ⎛ 9x ⎞ sin ⎜ ⎟ ⎝ 2 ⎠ f(x) = x sin 2 ⎛ x ⎞⎛ 3x ⎞ = ⎜ 3 − 4 sin2 ⎟ ⎜ 3 − 4 sin2 ⎝ 2 ⎠⎝ 2 ⎟ ⎠ x 3x x 3x = 9 − 12 sin − 12 sin2 + 16 sin2 ⋅ sin2 2 2 2 2 2 = 9 – 6(1 – cos x) – 6(1 – cos 3x) + 4(1 – cos x)(1 – cos 3x) = 1 + 6 cos x + 6 cos 3x – 4 cos x – 4 cos 3x + 4 cos x cos 3x Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 011-47623417/23 Fax : 47623472 (21) Downloaded from www.icbse.com IIT-JEE 2010 (Paper-2) π 9x sin ∫ 2 2 I = π x −π sin 2 π ∫ 1 + 2 cos x + 2 cos 3x + 2(cos 4x + cos x ) 4 = π 0 4 = ×π π =4 38. Match the statements in Column-I with those in Column-II. [Note: Here z takes values in the complex plane and Im z and Re z denote, respectively, the imaginary part and the real part of z.] Column I Column II 4 (A) The set of points z satisfying (p) An ellipse with eccentricity 5 | z – i | z || = | z + i | z || is contained in or equal to (B) The set of points z satisfying (q) The set of points z satisfying Im z = 0 | z + 4 | + | z – 4 | = 10 is contained in td .) or equal to ce sL (C) If | w | = 2, then the set of points (r) The set of points vi satisfying | Im z | ≤ 1 er z a lS 1 a ti o n z =w − is contained in or equal to d uc hE w s (D) If | w | = 1, then the set of points fA a ka (s) The set of points z satisfying | Re z | ≤ 2 o i n is contained in or equal itoo iv s 1 z =w + w (D (t) The set of points z satisfying | z | ≤ 3 Answer → A(q, r), B(p), C(p, s, t), D(q, r, s, t) Hints : (A) z is equidistant from the points i| z | and –i| z |, whose perpendicular bisector is Im (z) = 0. (B) Sum of distance of z from (4, 0) and (–4, 0) is a constant 10, hence locus of z is ellipse with semi-major axis 5 and focus at (±4, 0). ae = 4 4 ∴ e= 5 1 5 (C) | z | ≤ |w | + = <3 w 2 1 (D) | z | ≤ |w | + =2 w ∴ Re z ≤ | z | ≤ 2 Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 011-47623417/23 Fax : 47623472 (22) Downloaded from www.icbse.com IIT-JEE 2010 (Paper-2) PART–III : PHYSICS SECTION - I Single Correct Choice Type This section contains 6 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct. 39. A biconvex lens of focal length 15 cm is in front of a plane mirror. The distance between the lens and the mirror is 10 cm. A small object is kept at a distance of 30 cm from the lens. The final image is (A) Virtual and at a distance of 16 cm from the mirror (B) Real and at a distance of 16 cm from the mirror (C) Virtual and at a distance of 20 cm from the mirror (D) Real and at a distance of 20 cm from the mirror Answer (B) 16 cm Hints : 15 cm Object is placed at O. O B I A First refraction through lens 6 cm 10 cm .) 1 1 1 − = V −30 +15 30 cm td L 20 cm 10 cmvi r ce s 30 cm ⇒ V = +30 cm e n al S ⇒ Real image is formed at A. ti o 20 cm d u ca E as h This image acts as virtual object for plane mirror. Image is formed 20 cm before mirror (B) This acts as virtual object for lens. ak fA o Second refraction through lens i on vi s 1 1 1 (D i − = V +10 +15 ⇒ V = +6 : Real image Distance from mirror is 16 cm. 40. A uniformly charged thin spherical shell of radius R carries uniform surface charge density of σ per unit area. It is made of two hemispherical shells, held together by pressing them with force F (see figure). F is proportional to F F 1 2 2 1 2 1 σ2 1 σ2 (A) ε σ R (B) ε σ R (C) (D) 0 0 ε0 R ε0 R 2 Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 011-47623417/23 Fax : 47623472 (23) Downloaded from www.icbse.com IIT-JEE 2010 (Paper-2) Answer (A) Hints : By dimensional analysis ⎡ 1 q 2 ⎤ ⎡ 1 σ 2 A2 ⎤ [F ] = ⎢ 2 ⎥ =⎢ 2 ⎥ ⎣ 4 πε0 r ⎦ ⎣ ε0 R ⎦ ⎡ 1 2 2⎤ = ⎢ε σ R ⎥ ⎣ 0 ⎦ 41. A hollow pipe of length 0.8 m is closed at one end. At its open end a 0.5 m long uniform string is vibrating in its second harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the wire is 50 N and the speed of sound is 320 ms–1, the mass of the string is (A) 5 grams (B) 10 grams (C) 20 grams (D) 40 grams Answer (B) Hints : For hollow pipe, fundamental frequency v 320 f = = 4l 4 × 0.8 For string in 2nd harmonic 1 T 1 Tl 1 50 × 0.5 f = = = l μ l m 0.5 m Equating we get, m = 0.01 kg = 10 g td .) sL i ce e rv 42. A Vernier calipers has 1 mm marks on the main scale. It has 20 equal divisions on the Vernier scale which match with 16 main scale divisions. For this Vernier calipers, the least countSis l (C) 0.1 mmio na (A) 0.02 mm (B) 0.05 mm at uc (D) 0.2 mm Ed sh Answer (D) ka Hints : f Aa 20 VSD = 16 MSD io no ⇒ 5 VSD = 4 MSD (D i vi s LC = 1 MSD – 1 VSD 4 1 = 1MSD − MSD = MSD = 0.2 mm 5 5 43. A block of mass 2 kg is free to move along the x-axis. It is at rest and from t = 0 onwards it is subjected to a time-dependent force F(t) in the x direction. The fore F(t) varies with t as shown in the figure. The kinetic energy of the block after 4.5 seconds is (F)t 4N 4.5 s t O 3s (A) 4.50 J (B) 7.50 J (C) 5.06 J (D) 14.06 J Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 011-47623417/23 Fax : 47623472 (24) Downloaded from www.icbse.com IIT-JEE 2010 (Paper-2) Answer (C) Hints : Area under the curve of F – t graph provides change in momentum of particle since Fdt = d(mV) ⇒ ∫ Fdt = Δ mV ⇒ p = 4.5 kg ms–1 p2 4.52 Kinetic energy = = = 5.06 J 2m 2 × 2 44. A tiny spherical oil drop carrying a net charge q is balanced in still air with a vertical uniform electric field of strength 81π × 105 Vm −1 . When the field is switched off, the drop is observed to fall with terminal velocity 2 × 10–3 ms–1. Given 7 g = 9.8 ms–2, viscosity of the air = 1.8 × 10–5 Ns m–2 and the density of oil = 900 kg m–3, the magnitude of q is (A) 1.6 × 10–19 C (B) 3.2 × 10–19 C (C) 4.8 × 10–19 C (D) 8.0 × 10–19 C Answer (D) Hints : During equilibrium in presence of electric field qE = mg 4 3 ⇒ qE = πr ρg …(i) 3 When the drop descends with constant velocity mg = 6πηrV td .) sL i ce 4 3 πr ρg = 6πηrV e rv ⇒ lS 3 n a a ti o Putting the values, we obtain d u c hE 3 r = × 10−5 m 7 ka s f Aa Putting back in equation (i) io no q ≈ 8× 10–19 C (D i vi s SECTION - II Integer Type This section contains 5 questions. The answer to each of the questions is a single-digit integer, ranging from 0 to 9. The correct digit below the question number in the ORS is to be bubbled. 45. Image of an object approaching a convex mirror of radius of curvature 20 m along its optical axis is observed to 25 50 move from m to m in 30 seconds. What is the speed of the object in km per hour? 3 7 Answer (3) Hints : Convex mirror forms virtual image of object behind the mirror. By Mirror formula 1 1 1 + = v u f Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 011-47623417/23 Fax : 47623472 (25) Downloaded from www.icbse.com IIT-JEE 2010 (Paper-2) For first object 1 1 1 ⇒ + = + 25 / 3 u +10 O I ⇒ u = – 50 m For second object 10 cm 1 1 1 ⇒ + = + 50 / 7 u +10 ⇒ u = – 25 m 25 Speed = m/s 30 25 36 = × km/h = 3 km/h 30 10 1 46. A diatomic ideal gas is compressed adiabatically to of its initial volume. If the initial temperature of the gas 32 is Ti (in kelvin) and the final temperature is aTi, the value of a is Answer (4) Hints : td .) sL TV γ – 1 = constant i ce lS e rv n a a ti o 7 γ= 5 d u c s hE 2 ka ⎛ V ⎞5 Aa 2 TiVi = α Ti ⎜ i ⎟ f no 5 ⎝ 32 ⎠ io ⇒ α=4 (D i vi s dN (t ) dN (t ) 47. To determine the half life of a radioactive element, a student plots a graph of ln versus t. Here is dt dt the rate of radioactive decay at time t. If the number of radioactive nuclei of this element decreases by a factor of p after 4.16 years, the value of p is 6 5 dN (t ) 4 dt 3 n 2 1 2 3 4 5 6 7 8 Years Answer (8) Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 011-47623417/23 Fax : 47623472 (26) Downloaded from www.icbse.com IIT-JEE 2010 (Paper-2) Hints : Equation for radioactive decay is – λt N = N0 dN – λt = – λ N0 dt dN ln = + ln(λN0 ) – λt dt 3–4 ⇒ Slope = – λ = (from graph) 6–4 1 ⇒ λ= year –1 2 0.693 Half life = = 2 × 0.693 1.386 λ 4.16 years is approximately 3 half lives ⇒ Nuclei will decay by a factor of 23 = 8 .) Lt d 48. At time t = 0, a battery of 10 V is connected across points A and B in the given circuit. If the capacitors have s i ce no charge initially, at what time (in seconds) does the voltage across them become 4 V? [Take : ln 5 = 1.6, ln 3 = 1.1] lS e rv a i on t2μF ca 2MΩ d u s hE ka f Aa A io no B (D i vi s 2MΩ 2μF Answer (2) Hints : Resistance of circuit = 1 MΩ Capacitance of circuit = 4 μF Time constant = RC = 4 s Voltage across capacitor at time 0 = 0 V Voltage across capacitor at time ∞ = 10 V In general –t V =10(1 – α 4 ) Putting V = 4 V We get t = 2 s Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 011-47623417/23 Fax : 47623472 (27) Downloaded from www.icbse.com IIT-JEE 2010 (Paper-2) 49. A large glass slab (μ = 5/3) of thickness 8 cm is placed over a point source of light on a plane surface. It is seen that light emerges out of the top surface of the slab from a circular area of radius R cm. What is the value of R? Answer (6) Hints : Light will appear only from a circular region because rays going out at an angle greater than θc will suffer TIR. 1 3 sin θc = = μ 5 R From figure R sin θc = 8 cm R + 82 2 Equating R 3 = R +8 2 2 5 Solving R = 6 cm SECTION - III Paragraph Type This section contains 2 paragraphs. Based upon each of the paragraphs 3 multiple choice questions have to be ) . answered. Each of these questions has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct. td sL Paragraph for questions 50 to 52 i ce lS e rv The key feature of Bohr’s theory of spectrum of hydrogen atom is the quantization of angular momentum when o a an electron is revolving around a proton. We will extend this to na general rotational motion to find quantized ti rotational energy of a diatomic molecule assuming it to be ca E du rigid. The rule to be applied is Bohr’s quantization ashquantization condition its rotational energy in the nth level condition. A diatomic molecule has moment of inertia l. ByAak of 50. Bohr’s i on (n = 0 is not allowed) is 1 ⎛ h ⎞ 2 1 ⎛ h (D⎞ i vi s 2 ⎛ h ⎞ 2 ⎛ h ⎞ 2 2 (A) ⎜ ⎟ ⎜ ⎟ (B) n n⎜ (C) ⎟ n ⎜ (D) ⎟ n ⎝ 8π I ⎠ 2 2 ⎝ 8π I ⎠ 2 ⎝ 8π I ⎠ 2 ⎝ 8π I ⎠ 2 Answer (D) Hints : nh L= 2π L2 n 2 h 2 KE = = 2I 8π2I 51. It is found that the excitation frequency from ground to the first excited state of rotation for the CO molecule is 4 close to × 1011 Hz. Then the moment of inertia of CO molecule about its centre of mass is close to (Take h = π 2π × 10–34 J s) (A) 2.76 × 10–46 kg m2 (B) 1.87 × 10–46 kg m2 (C) 4.67 × 10–47 kg m2 (D) 1.17 × 10–47 kg m2 Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 011-47623417/23 Fax : 47623472 (28) Downloaded from www.icbse.com IIT-JEE 2010 (Paper-2) Answer (B) Hints : 4 (22 – 12 )h 2 h× × 1011 = π 8π 2I 3h ⇒ I= = 1.87 × 10 –46 kg m2 32π × 10 22 52. In a CO molecule, the distance between C (mass = 12 a.m.u.) and O (mass = 16 a.m.u.), where 5 1 a.m.u. = × 10 –27 kg , is close to 3 (A) 2.4 × 10–10 m (B) 1.9 × 10–10 m (C) 1.3 × 10–10 m (D) 4.4 × 10–11 m Answer (C) Hints : m1m2 2 I = 1.87 × 10–46 = r m1 + m2 12 × 16 5 1.87 × 10 –46 = × × 10–27 r 2 12 + 16 3 td .) sL i ce ⇒ r = 1.3 × 10–10 m lS e rv a ti n Paragraph for questions 53oto 55 When liquid medicine of density ρ is to be put in the eye, it isEd u ca sh done with the help of a dropper. As the bulb on the top of the dropper is pressed, a drop forms at the openingaof the dropper. We wish to estimate the size of the drop. k f isa We first assume that the drop formed at the opening A spherical because that requires a minimum increase in its no i si o surface energy. To determine the size, we calculate the net vertical force due to the surface tension T when the radius v (D i of the drop is R. When this force becomes smaller than the weight of the drop, the drop gets detached from the dropper. 53. If the radius of the opening of the dropper is r, the vertical force due to the surface tension on the drop of radius R(assuming r << R) is 2πr 2T 2πR 2T (A) 2 πrT (B) 2 πRT (C) (D) R r Answer (C) Hints : dF = Tdl r Vertical force = ∫ Tdl sin θ r R =T× × 2πr R 2πr 2T = R Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 011-47623417/23 Fax : 47623472 (29) Downloaded from www.icbse.com IIT-JEE 2010 (Paper-2) 54. If r = 5 × 10–4 m, ρ = 103 kgm–3, g = 10 ms–2, T = 0.11 Nm–1, the radius of the drop when it detaches from the dropper is approximately (A) 1.4 × 10–3 m (B) 3.3 × 10–3 m (C) 2.0 × 10–3 m (D) 4.1 × 10–3 m Answer (A) 2πr 2T 4 Hints : For equation = πR 3ρg R 3 1/ 4 ⎛ 2 r 2T ⎞ ⇒ R =⎜ ⎜ 3 ρg ⎟ ⎟ ⎝ ⎠ 55. After the drop detaches, its surface energy is (A) 1.4 × 10–6 J (B) 2.7 × 10–6 J (C) 5.4 × 10–6 J (D) 8.1 × 10–6 J Answer (B) Hints : Surface energy = 4πR2T SECTION - IV (Matrix Type) This section contains 2 questions. Each Question has four statements (A, B, C and D) given in Column I and five statements (p, q, r, s and t) in Column II. Any given statement in Column I can have correct matching with one or more statement(s) given in Column II. For example, if for a given question, statement B matches with the statements given in q are r, then for that particular question, against statement B, darken the bubble corresponding to q and r in the ORS. 56. You are given many resistances, capacitors and inductors. These are connected to a variable DC voltage source (the first two circuits) or an AC voltage source of 50 Hz frequency (the next three circuits) in different ways as d. ) shown in Column II. When a current I (steady state for DC or rms for AC) flows through the circuit, the corresponding voltage V1 and V2. (indicated in circuits) are related as shown s Lt in Column I. Match the two i ce Column I Column II lS e rv n a a ti o V1 V2 uc h Ed 6 mH 3 μF s (A) I ≠ 0, V1 is proportional to I fA a ka (p) o i on V (D i vi s V1 V2 6 mH 2Ω (B) I ≠ 0, V2 > V1 (q) V V1 V2 6 mH 2Ω (C) V1 = 0, V2 = V (r) V V1 V2 6 mH 3 μF (D) I ≠ 0, V2 is proportional to I (s) V V1 V2 1 kΩ 3 μF (t) V Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 011-47623417/23 Fax : 47623472 (30) Downloaded from www.icbse.com IIT-JEE 2010 (Paper-2) Answer A(r, s, t), B(q, r, s, t), C(p, q), D(q, r, s, t) Hints : (p) V1 = 0, V2 = V ( XL = 0) (q) XL = 0, V1 = 0, V2 = V (r) XL = ωL = 2πfL = 1.884 Ω V1 = IXL V2 = IR V1 < V2 (s) XL = 1.884 Ω 1 10 XC = = × 103 Ω ωL 3 π XC >> XL V2 = IXC, V1 = IXL (t) R = 1 kΩ 10 XC = × 103 Ω 3π ⇒ XC > R ⇒ V2 > V1 td .) Also, V2 = IXL, V1 = IR. sL rv i ce Se 57. Two transparent media of refractive indices μ1 and μ3 have a solid lens shaped transparent material of refractive al tion Match them to the ray diagrams shown in index μ2 between them as shown in figures in Column II. A ray traversing these media is also shown in figures. In Column I different relationships between μ1, μ2 and μ3 are ca u given. Column II. h Ed s Column I fA aka Column II io no (D i vi s (A) μ1 < μ2 (p) μ3 μ2 μ1 μ2 μ1 (B) μ1 > μ2 (q) μ3 (C) μ2 = μ3 (r) μ3 μ μ1 2 (D) μ2 > μ3 (s) μ μ2 μ1 3 (t) μ3 μ2 μ1 Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 011-47623417/23 Fax : 47623472 (31) Downloaded from www.icbse.com IIT-JEE 2010 (Paper-2) Answer A(p, r), B(q, s, t), C(p, r, t), D(q, s) Hints : (p) μ2 = μ3, as there is no deviation μ1 < μ2, as ray bends towards normal. (q) μ1 > μ2, as ray bends away from normal μ2 > μ3, for similar reasons (r) Similar to (p) (s) Similar to (q) (t) μ1 > μ2, as ray bends away from normal μ2 = μ3, as there is no deviation td .) sL i ce lS e rv n a u c a ti o d s hE ka f Aa io no (D i vi s Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 011-47623417/23 Fax : 47623472 (32) Downloaded from www.icbse.com