May 2011 - Career Point
May 2011 - Career Point May 2011 - Career Point
XtraEdge for IIT-JEE 80 MAY 2011
IIT-JEE 2011PAPER-II (PAPER & SOLUTION)Time : 3 Hours Total Marks : 240Instructions : [Each subject contain]Section – I :Section – II :Multiple choice questions with only one correct answer. +3 marks will be awarded for correct answerand -1 mark for wrong answer. [No. of Ques. : 8]Multiple choice questions with multiple correct answer. +4 marks will be awarded for correct answerand No Negative marking for wrong answer. [No. of Ques. : 4]Section – III : Numerical Response Question (single digit Ans. type) +4 marks will be awarded for correct answerand No Negative marking for wrong answer. [No. of Ques. : 6]Section – IV : Column Matching type question +2 marks for each correctly matched row and No Negative markingfor wrong answer. [No. of Ques. : 2]CHEMISTRYSECTION – ISingle Correct Choice TypeThis section contains 8 multiple choice questions. Eachquestion has 4 choices (A), (B), (C) and (D), out ofwhich ONLY ONE is correct.1. Oxidation states of the metal in the mineralshaematite and magnetite respectively, are(A) II, III in haematile and III in magnetite(B) II, III in haematite and II in magnetite(C) II in haematite and II, III in magnetite(D) III in haematite and II, III in magnetiteAns. [D]+ 3Sol. Haematite is F e2O3+ 2+ 3Magnetite is Fe 3 O 4 or FeO. Fe2O32. The following carbohydrate isH OHH OHOHOOHH OHH H(A) a ketohexose (B) an aldohexose(C) an α-furanose (D) an α-pyranoseAns. [B]Sol. Aldohexose3. The major product of the following reaction isRCH 2 OHH + (anhydrous)O(A) a hemiacetal (B) an acetal(C) an ether(D) an esterAns.Sol.[B]OR- CH 2 –OH⊕H (anhydrous)OOCH 2 –RAcetal4. Amongst the compounds given, the one thatwould form a brilliant colored dye on treatmentwith NaNO 2 in dill. HCl followed by addition toan alkaline solution of β -naphthol isN(CH 3 ) 2NHCH 3(A)(B)Ans.Sol.(C)[C]H 3 CNH 2(D)NaNO 2 + HCl (dil.)CH 3 NH 2 CH 30–5°CCH 3Alkaline solutionN = NHOColoured dyeCH 2 NH 2N 2 ClOH5. The freezing point (in °C) of a solution containing0.1 g of K 3 [Fe(CN) 6 ] (mol. wt. 329) in 100 g ofwater (k f = 1.86 K kg mol –1 ) is(A) –2.3 × 10 –2 (B) –5.7 × 10 –2(C) –5.7 × 10 –3 (D) –1.2 × 10 –2XtraEdge for IIT-JEE 81 MAY 2011
- Page 32 and 33: KEY CONCEPTPhysicalChemistryFundame
- Page 34 and 35: ⎛ n ⎞ ⎛ n ⎞Correction term
- Page 36 and 37: Kinetic Isotope Effects :The kineti
- Page 38 and 39: (m / 32) 1Moles fraction of O 2 ==(
- Page 40 and 41: `tà{xÅtà|vtÄ V{tÄÄxÇzxá1 Se
- Page 42 and 43: Students' ForumExpert’s Solution
- Page 44 and 45: MATHSCOMPLEX NUMBERMathematics Fund
- Page 46 and 47: MATHSMATRICES &DETERMINANTSMathemat
- Page 48 and 49: aBased on New PatternIIT-JEE 2012Xt
- Page 50 and 51: dxO12. The time of crossing the riv
- Page 52 and 53: 6. Choose the INCORRECT statement f
- Page 54 and 55: 7. If the equation 3x 4 - 16x 3 + 3
- Page 56 and 57: Based on New PatternIIT-JEE 2013Xtr
- Page 58 and 59: 17. A stone is projected from level
- Page 60 and 61: 16. Half a mole of photon is used t
- Page 62 and 63: 17. Ifsin 3θcos 2θ2-147137= 0Then
- Page 64 and 65: IIT-JEE 2011PAPER-I (PAPER & SOLUTI
- Page 66 and 67: 9. Amongst the given options, the c
- Page 68 and 69: 22. The total number of alkenes pos
- Page 70 and 71: 1 2 × 8∆U = × [V - 0]22 2 + 81
- Page 72 and 73: shown in the figure. We use the sig
- Page 74 and 75: 40. A block is moving on an incline
- Page 76 and 77: MATHEMATICSSECTION - ISingle Correc
- Page 78 and 79: Ans.Sol.[A,D]r = xiˆ + yj ˆ + zk
- Page 80 and 81: ⎡323 21× ⎤=⎡32 1⎤1 C2C21 C
- Page 84 and 85: Ans. [A]Sol. ∆T = k f × m × i
- Page 86 and 87: SECTION - IVMatrix match TypeThis s
- Page 88 and 89: Ans.[D]m 1 = 0.01 kgvH = 5mm 2 = 0.
- Page 90 and 91: 31. A series R-C circuit is connect
- Page 92 and 93: P3PBAColumn-I(A) Pipe closed at one
- Page 94 and 95: 21/ x45. If lim [1 + xl n(1 + b )]
- Page 96 and 97: So f(x) is monotonically decreasing
- Page 98 and 99: (C)25 / 6πI = x dxn ∫sec πl 327
- Page 100 and 101: Subscription Offer for Students'Xtr
- Page 102: XtraEdge for IIT-JEE 100 MAY 2011
IIT-JEE <strong>2011</strong>PAPER-II (PAPER & SOLUTION)Time : 3 Hours Total Marks : 240Instructions : [Each subject contain]Section – I :Section – II :Multiple choice questions with only one correct answer. +3 marks will be awarded for correct answerand -1 mark for wrong answer. [No. of Ques. : 8]Multiple choice questions with multiple correct answer. +4 marks will be awarded for correct answerand No Negative marking for wrong answer. [No. of Ques. : 4]Section – III : Numerical Response Question (single digit Ans. type) +4 marks will be awarded for correct answerand No Negative marking for wrong answer. [No. of Ques. : 6]Section – IV : Column Matching type question +2 marks for each correctly matched row and No Negative markingfor wrong answer. [No. of Ques. : 2]CHEMISTRYSECTION – ISingle Correct Choice TypeThis section contains 8 multiple choice questions. Eachquestion has 4 choices (A), (B), (C) and (D), out ofwhich ONLY ONE is correct.1. Oxidation states of the metal in the mineralshaematite and magnetite respectively, are(A) II, III in haematile and III in magnetite(B) II, III in haematite and II in magnetite(C) II in haematite and II, III in magnetite(D) III in haematite and II, III in magnetiteAns. [D]+ 3Sol. Haematite is F e2O3+ 2+ 3Magnetite is Fe 3 O 4 or FeO. Fe2O32. The following carbohydrate isH OHH OHOHOOHH OHH H(A) a ketohexose (B) an aldohexose(C) an α-furanose (D) an α-pyranoseAns. [B]Sol. Aldohexose3. The major product of the following reaction isRCH 2 OHH + (anhydrous)O(A) a hemiacetal (B) an acetal(C) an ether(D) an esterAns.Sol.[B]OR- CH 2 –OH⊕H (anhydrous)OOCH 2 –RAcetal4. Amongst the compounds given, the one thatwould form a brilliant colored dye on treatmentwith NaNO 2 in dill. HCl followed by addition toan alkaline solution of β -naphthol isN(CH 3 ) 2NHCH 3(A)(B)Ans.Sol.(C)[C]H 3 CNH 2(D)NaNO 2 + HCl (dil.)CH 3 NH 2 CH 30–5°CCH 3Alkaline solutionN = NHOColoured dyeCH 2 NH 2N 2 ClOH5. The freezing point (in °C) of a solution containing0.1 g of K 3 [Fe(CN) 6 ] (mol. wt. 329) in 100 g ofwater (k f = 1.86 K kg mol –1 ) is(A) –2.3 × 10 –2 (B) –5.7 × 10 –2(C) –5.7 × 10 –3 (D) –1.2 × 10 –2XtraEdge for IIT-JEE 81 MAY <strong>2011</strong>