Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our User Agreement and Privacy Policy.

Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our Privacy Policy and User Agreement for details.

Successfully reported this slideshow.

Like this document? Why not share!

No Downloads

Total views

1,739

On SlideShare

0

From Embeds

0

Number of Embeds

16

Shares

0

Downloads

39

Comments

0

Likes

2

No embeds

No notes for slide

- 1. QUESTIONS & SOLUTIONS OF IIT-JEE 2009 Date : 12-04-2009 Duration : 3 Hours Max. Marks : 240 Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose. INSTRUCTIONS PAPER - 1 A. General 1. This booklet is your Question Paper containing 60 questions. 2. The Question Paper CODE is printed on the right hand top corner of this page and on the back page of this booklet. 3. Each page of this question paper contain half page for rough work (except front and back page). No additional sheets will be provided for rough work. 4. Blank paper, clipboard, log tables, slide rules, calculators, cellular phones, pagers, and electronic gadgets in any form are not allowed to be carried inside the examination hall. 5. Fill in the boxes provided below on this page and also write your Name and Roll No. in the space provide on the back page of this booklet. 6. The answer sheet, a machine-readable Objective Response Sheet (ORS), is provided separately. 7. DO NOT TAMPER WITH/ MULTILATE THE ORS OR THE BOOKLET. 8. Do not open the seals of question-paper booklet before being instructed to do so by the invigilators. B. Filling the ORS : 9. Write your Roll No. in ink, in the box provided in the Upper part of the ORS and darken the appropriate bubble UNDER each digit of your Roll No. with a good quality HB pencil. C. Question paper format : D. Marking scheme Read the instructions printed on the back page of this booklet. Name of the Candidate Roll Number I have read all the instructions and -------------------------------- shall abide by them. Signature of the Invigilator -------------------------------- Signature of the Candidate
- 2. PART- I CHEMISTRY SECTION - I Straight Objective Type This section contains 8 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which ONLY ONE is correct. 1. Among the electrolytes Na2SO4, CaCl2, Al2(SO4)3 and NH4Cl, the most effective coagulating agent for Sb2S3 sol is : (A) Na2SO4 (B) CaCl2 (C) Al2(SO4)3 (D) NH4Cl Sol. (C) Most effective coagulating agent for Sb2S3 is Al2(SO4)3 because of high charge. 2. The term that corrects for the attractive forces present in a real gas in the vander Waals equation is : an2 an2 (A) nb (B) (C) – (D) –nb V2 V2 Sol. (B) an 2 Correction factor for attractive force in to the real gas is given by . V2 3. The Henry's law constant for the solubility of N2 gas in water at 298 K is 1.0 × 105 atm. The mole fraction of N2 in air is 0.8. The number of moles of N2 from air dissolved in 10 moles of water of 298 K and 5 atm pressure is : (A) 4 × 10–4 (B) 4.0 × 10–5 (C) 5.0 × 10–4 (D) 4.0 × 10–6 Sol. (A) PN = KH × x N2 2 1 x N2 = × 0.8 × 5 = 4 × 10–5 per mole 105 In 10 mole solubility is 4 × 10–4 . RESONANCE Page # 2
- 3. 4. The reaction of P4 with X leads selectively to P4O6. The X is : (A) Dry O2 (B) A mixture of O2 and N2 (C) Moist O2 (D) O2 in the presence of aqueous NaOH Sol. (B) In presence of N 2 P4 + 3O2 P4O6. Here nitrogen acts as diluent. Note : In dry O2 following reactions may take place. P4 + 3O2 P4O6. P4O6 + 2O2 P4O10. In moist O2 the P4O6 gets hydrolysed forming H3PO3. P4O6 + 6H2O 4H3PO3. In presence of NaOH. P4 + 3OH– + 3H2O PH3 + 3H2PO2– 5. Among celluose, poly(vinyl chloride), nylon and natural rubber, the polymer in which the intermolecular force of attraction is weakest is : (A) Nylon (B) Poly(vinyl chloride) (C) Cellulose (D) Natural Rubber Sol. (D) The natural rubber has intermolecular forces which are weak dispersion force (van-der-waal forces of attraction) and is an example of an elastomer (polymer). 6. The correct acidity order of the following is : (A) (III) > (IV) > (II) > (I) (B) (IV) > (III) > (I) > (II) (C) (III) > (II) > (I) > (IV) (D) (II) > (III) > (IV) > (I) Sol. (A) – H More stable conjugate base (carboxylate anion, –ve charge always on oxygen atom). RESONANCE Page # 3
- 4. – H More stable conjugate base (carboxylate anion, –ve charge always on oxygen atom). + I and hyperconjugative effect of – CH3 group decreases stability of benzoate anion. – H Less stable conjugate base (phenoxide ion is less resonance stabilized than benzoate anion). –Cl group exhibits – I effect. – H Less stable conjugate base (phenoxide ion is less resonance stabilized than benzoate anion). 7. Given that the abundances of isotopes 54Fe, 56Fe and 57Fe are 5%, 90% and 5%, respectively, the atomic mass of Fe is : (A) 55.85 (B) 55.95 (C) 55.75 (D) 56.05 Sol. (B) 54Fe 5% 56Fe 90% 57Fe 5% Av. atomic mass = x1A1 + x2A2 + x3A3 = 54 × 0.05 + 56 × 0.9 + 57 × 0.05 = 55.95 RESONANCE Page # 4
- 5. 8. The IUPAC name of the following compound is : (A) 4-Bromo-3-cyanophenol (B) 2-Bromo-5-hydroxybenzonitrile (C) 2-Cyano-4-hydroxybromobenzene (D) 6-Bromo-3-hydroxybenzonitrile Sol. (B) 2-Bromo-5-hydroxybenzonitrile (–CN group gets higher priority over –OH and –Br) SECTION - II Multiple Correct Answer Type This section contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which ONE OR MORE is/are correct. 9. The correct statement(s) about the compound H3C(HO)HC–CH=CH–CH(OH)CH3 (X) is(are) : (A) The total number of stereoisomers possible for X is 6. (B) The total number of diastereomers possible for X is 3. (C) If the stereochemistry about the double bond in X is trans, the number of enantiomers possible for X is 4. (D) If the stereochemistry about the double bond in X is cis, the number of enantiomers possible for X is 2. Sol. (A, D) (X) has configurations (1) R Z R (2) S Z S (3) R Z S (4) R E R (5) S E S (6) R E S RESONANCE Page # 5
- 6. 10. The compound(s) formed upon combustion of sodium metal in excess air is(are) : (A) Na2O2 (B) Na2O (C) NaO2 (D) NaOH Sol. (A, B) combustion 6Na + 2O2 (from air) 2Na2O + Na2O2 11. The correct statement(s) regarding defects in solids is(are) : (A) Frenkel defect is usually favoured by a very small difference in the sizes of cation and anion. (B) Frenkel defect is a dislocation defect. (C) Trapping of an electron in the lattice leads to the formation of F-center. (D) Schottky defects have no effect on the physical properties of solids. Sol. (B, C) Frenkel defect is a dislocation defect. Trapping of an electron in the lattice leads to the formation of F-center. 12. The compound(s) that exhibit(s) geometrical isomerism is(are) : (A) [Pt(en)Cl2] (B) [Pt(en)2]Cl2 (C) [Pt(en)2Cl2]Cl2 (D) [Pt(NH3)2Cl2] Sol. (C, D) (A) [Pt(en)Cl2] : exist only in one form. (B) [Pt(en)2]Cl2 : exist only in one form (en is symmetrical lignad). (C) [Pt(en)2Cl2]Cl2 : RESONANCE Page # 6
- 7. (D) [Pt(NH3)2Cl2] : Pt is in +2 oxidation state having 5d8 configuration. Hence the hybridisation of complex is dsp2 and geometry is square planar. SECTION - III Comprehension Type This section contains 2 groups of questions. Each group has 3 multiple choice questions based on a paragraph. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which ONLY ONE is correct. Paragraph for Question Nos. 13 to 15 A carbonyl compound P, which gives positive iodoform test, undergoes reaction with MeMgBr followed by dehydration to give an olefin Q. Ozonolysis of Q leads to a dicarbonyl compound R, which undergoes intramolecular aldol reaction to give predominantly S. 1. MeMgBr 1. O 3 1. OH P Q R S 2. Zn, H2O 2. 2. H , H2O 3. H2SO4 , 13. The structure of the carbonyl compound P is : (A) (B) (C) (D) Ans. (B) 14. The structures of the products Q and R, respectively, are : (A) , (B) , (C) , (D) , Ans. (A) RESONANCE Page # 7
- 8. 15. The structure of the product S is : (A) (B) (C) (D) Ans. (B) Sol. (13, 14 & 15) (1) MeMgBr 2H SO , 4 ( 2 ) H , H2O OH O / Zn,H O 3 2 int ramolcular aldol condensation Paragraph for Question Nos. 16 to 18 p-Amino-N, N-dimethylaniline is added to a strongly acidic solution of X. The resulting solution is treated with a few drops of aqueous solution of Y to yield blue coloration due to the formation of methylene blue. Treatment of the aqueous solution of Y with the reagent potassium hexacyanoferrate(II) leads to the formation of an intense blue precipitate. The precipitate dissolves on excess addition of the reagent. Similarly, treatment of the solution of Y with the solution of potassium hexacyanoferrate(III) leads to a brown coloration due to the formation of Z. RESONANCE Page # 8
- 9. 16. The compound X is : (A) NaNO3 (B) NaCl (C) Na2SO4 (D) Na2S Ans. (D) 17. The compound Y is : (A) MgCl2 (B) FeCl2 (C) FeCl3 (D) ZnCl2 Ans. (C) Sol. (16 to 17) + S2– (X) + + 6Fe3+ (Y) 6Fe2+ + NH4+ + 2H+ + (methylene blue). The compound X is Na2S and Y is FeCl3. 18. The compound Z is : (A) Mg2[Fe(CN)6] (B) Fe[Fe(CN)6] (C) ) Fe4[Fe(CN)6]3 (D) ) K2Zn3[Fe(CN)6]2 Sol. (B) Fe3+ + [Fe(CN)6]3– Fe[Fe(CN)6] (brown colouration) (Z). SECTION - IV Matrix - Match Type This section contains 2 questions. Each question contains statements given in two columns, which have to be matched. The statements in Column-I are labelled A, B, C and D, while the statements in Column-II are labelled p, q, r, s and t. Any given statement in Coloumn-I can have correct matching with ONE OR MORE statement(s) in Coloumn-II. The appropriate bubbles corresponding to the answers to these questions have to be drakened as illustrated in the following example. If the correct matches are A-p, s and t ; B-q and r; C-p and q; and D-s and t; then the correct darkening of bubbles will look like the following : RESONANCE Page # 9
- 10. 19. Match each of the compounds in Column I with its characteristic reaction(s) in Column II. Column I Column II (A) CH3CH2CH2CN (p) Reduction with Pd-C/H2 (B) CH3CH2OCOCH3 (q) Reduction with SnCl2/HCl (C) CH3–CH=CH–CH2OH (r) Development of foul smell on treatment with chloroform and alcoholic KOH. (D) CH3CH2CH2CH2NH2 (s) Reduction with diisobutylaluminium hydride (DIBAL-H) (t) Alkaline hydrolysis Ans. (A) – p, q, s, t ; (B) – p, s, t ; (C) – p ; (D) – r Sol. Pd C / H 2 (A) CH3CH2CH2CN CH3CH2CH2CH2NH2 (p) SnCl / HCl 2 CH3CH2CH2CHO (q) DIBAl H CH3CH2CH2CHO (s) OH CH3CH2CH2COO (t) H2O 2Pd C / H (B) CH3CH2OCOCH3 2CH3—CH2 – OH (p) DIBAl H CH3—CHO (s) OH CH —COO (t) H2O 3 2 Pd C / H (C) CH3–CH=CH–CH2OH CH3CH2CH2CH2OH (p) 3 CHCl (D) CH3CH2CH2CH2NH2 CH3CH2CH2CH2NC (r) KOH RESONANCE Page # 10
- 11. 20. Match each of the diatomic molecules in Column I with its property/properties in Column II. Column I Column II (A) B2 (p) Paramagnetic (B) N2 (q) Undergoes oxidation (C) O2– (r) Undergoes reduction (D) O2 (s) Bond order 2 (t) Mixing of 's' and 'p' orbitals Ans. (A) - p, r, t ; (B) - s, t ; (C) - p, q, r ; (D) - p, r, s Sol. (A) B2 1s2 1s 2 2s 2 2s 2 2px1 = 2py1 Bond order = 1 Paramagnetic (B) N2 1s2 1s 2 2s 2 2s 2 2px2 = 2py2 2pz2 Bond order = 3 Diamagnetic (C) O2– 1s2 1s 2 2s 2 2s 2 2pz2 2px2 = 2px 2 2px 2 = 2py 1 Bond order = 1.5 Paramagnetic (D) O2 1s2 1s 2 2s 2 2s 2 2pz2 2px2 = 2px 2 2px 1 = 2py 1 Bond order = 2 Paramagnetic RESONANCE Page # 11
- 12. PART-II MATHEMATICS SECTION - I Single Correct Choice Type This section contains 8 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which ONLY ONE is correct. 15 2m 1 21. Let z = cos + i sin . Then the value of m ( z m 1 ) at = 2º is 1 1 1 1 (A) (B) (C) (D) sin 2º 3 sin 2º 2 sin 2º 4 sin 2º Sol. (D) Given that z = cos + i sin = ei 15 15 15 Im (z2m–1) = Im (ei)2m–1 = Im e i(2m–1) = sin+ sin3+ sin5 + .......... + sin29 m 1 m 1 m 1 29 15 2 sin sin 2 2 sin(15) sin (15) = = = 1 2 sin sin 4sin2º 2 22. The number of seven digit integers, with sum of the digits equal to 10 and formed by using the digits 1, 2 and 3 only, is (A) 55 (B) 66 (C) 77 (D) 88 Sol. (C) There are two possible cases Case 1 : Five 1’s, one 2’s, one 3’s 7! Number of numbers = = 42 5! Case 2 : Four 1’s, three 2’s 7! Number of numbers = = 35 4! 3! Total number of numbers = 42 + 35 = 77 RESONANCE Page # 12
- 13. 23. Let P(3, 2, 6) be a point in space and Q be a point on the line r ( ˆ ˆ 2k ) ( 3ˆ ˆ 5k ) . Then the value i j i j ˆ ˆ of for which the vector PQ is parallel to the plane x – 4y + 3z = 1 is 1 1 1 1 (A) (B) – (C) (D) – 4 4 8 8 Sol. (A) Given OQ = (1–3) ˆ + ( –1) ˆ + (5 +2) k , OP = 3ˆ 2ˆ 6k (where O is origin) i j ˆ i j ˆ PQ = (1–3– 3) ˆ + (–1–2) ˆ + (5 +2–6) k i j ˆ = (–2 – 3) ˆ + (–3) ˆ + (5 –4) k i j ˆ PQ is parallel to the plane x – 4y + 3z = 1 –2–3 –4 + 12 + 15 – 12 = 0 8 = 2 = 1 4 x x 24. Let f be a non-negative function defined on the interval [0, 1]. If 1 ( f ( t ))2 dt = 0 f ( t) dt, 0 x 1 and 0 f(0) = 0, then 1 1 1 1 1 1 1 1 (A) f < and f > (B) f > and f > 2 2 3 3 2 2 3 3 1 1 1 1 1 1 1 1 (C) f < and f < (D) f > and f < 2 2 3 3 2 2 3 3 Sol. (C) x x Given 1 – ( f ' ( t ))2 dt = f (t )dt, 0 x 1 0 0 Apply Leibnitz theorem, we get 1 – ( f ' ( x ))2 = f (x) 1 – (f(x))2 = f2(x) (f(x))2 = 1–f2(x) dy f ' (x) = 1 – f 2 (x) = 1– y2 , where y = f(x) dx dy = dx 1– y2 Integrating both sides sin –1 (y) = x + c f (0) = 0 c=0 RESONANCE Page # 13
- 14. y = ± sin x y = sin x = f(x) given f(x) 0 for x [0, 1] it is known that sinx < x, x R+ sin < f < and sin < f < 1 1 1 1 1 1 1 1 2 2 2 2 3 3 3 3 25. Let z = x + iy be a complex number where x and y are integers. Then the area of the rectangle whose vertices are the roots of the equation zz 3 zz 3 350 is (A) 48 (B) 32 (C) 40 (D) 80 Sol. (A) z z (z2 + z 2 ) = 350 2(x2 + y2) (x2 – y2 ) = 350 (x2 + y2) (x2 –y2) = 175 Since x, y I , the only possible case which gives integral solution, is x2 + y2 = 25 ......... (1) x2 – y2 =7 ..........(2) From (1) and (2) x2 = 16 ; y2 = 9 x= 4; y= 3 Area = 48 26. Tangents drawn from the point P(1, 8) to the circle x2 + y2 – 6x – 4y – 11 = 0 touch the circle at the points A and B. The equation of the circumcircle of the triangle PAB is (A) x2 + y2 + 4x – 6y + 19 = 0 (B) x2 + y2 – 4x – 10y + 19 = 0 (C) x2 + y2 – 2x + 6y – 29 = 0 (D) x2 + y2 – 6x – 4y + 19 = 0 Sol. (B) For required circle, P(1, 8) and O(3, 2) will be the end points of its diameter. (x – 1) (x – 3) + (y – 8) (y – 2) = 0 x2 + y2 – 4x – 10y + 19 = 0 RESONANCE Page # 14
- 15. 27. The line passing through the extremity A of the major axis and extremity B of the minor axis of the ellipse x2 + 9y2 = 9 meets its auxiliary circle at the point M. Then the area of the triangle with vertices at A, M and the origin O is 31 29 21 27 (A) (B) (C) (D) 10 10 10 10 Sol. (D) Equation of auxiliary circle is x2 + y2 = 9 (1) x y Equation of AM is + =1 3 1 (2) 12 9 on solving (1) and (2), we get M , 5 5 1 Now, area of AOM = . OA × MN = square unit 27 2 10 1 28. If a, b, c and d are unit vectors such that (a b ) . (c d) = 1 and a . c , then 2 (A) a, b, c are non-coplanar (B) b, c, d are non-coplanar (C) b, d are non-parallel (D) a, d are parallel and b, c are parallel Sol. (C) Let angle between a and b be 1 , c and d be 2 and a b and c d be . (a b ) . (c d) = 1 sin 1 . sin 2 . cos = 1 1 = 90º, 2 = 90º , = 0º a b, c d, (a b) | | (c d) a b = k (c d) so a b = k(c d) and (a b) . c = k(c d) . c and (a b) . d = k (c d) . d [a b c ] 0 and [a b d] 0 a, b, c and a, b, d are coplanar vectors so options A and B are incorrect Let b || d b d as (a b ) . (c d) = 1 ( a b ) . (c b ) = ± 1 [a b c b] = ± 1 [c b a b] = ± 1 RESONANCE Page # 15
- 16. c . [b (a b)] 1 c . [a (b . a) b] = ± 1 c . a 1 ( a . b 0 ) which is a contradiction so option 'C' is correct Let option 'D' is correct d = ± a and c = ± b as (a b) . (c d) 1 (a b ) . (b a) = ± 1 which is a contradiction so option 'D' is incorrect Alternatively option 'C' and 'D' may be observed from the given figure SECTION - II Multiple Correct Choice Type This section contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which ONE OR MORE is/are correct. x2 a a2 x 2 29. Let L = lim 4 , a > 0. If L is finite, then x 0 x4 1 1 (A) a = 2 (B) a = 1 (C) L = (D) L = 64 32 Sol. (A, C) 1 1 x 2 1 2 1 x 4 x2 a a . 1 . 2 . .... x2 2 a 2 2 a4 4 a a2 x 2 L = lim 4 , a > 0 = lim 4 x 0 4 x 0 x x x2 1 x4 x2 . 3 ...... lim = x 0 2a 8 a 4 4 x Since L is finite 2a = 4 a=2 1 lim L = x 0 = 8 . a3 1 64 RESONANCE Page # 16
- 17. 30. Area of the region bounded by the curve y = ex and lines x = 0 and y = e is e 1 e x (A) e – 1 (B) n (e 1 y) dy (C) e – e dx (D) n y dy 1 1 0 Sol. (B, C, D) 1 Shaded area = e – e x dx = 1 0 e Also n( e 1 y) dy 1 put e+1–y=t – dy = dt 1 e e = n t dt = n t (–dt) = e 1 n y dy 1 1 sin4 x cos 4 x 1 31. If + = , then 2 3 5 2 sin8 x cos 8 x 1 (A) tan2x = (B) + = 3 8 27 125 1 sin8 x cos 8 x 2 (C) tan2 x = (D) + = 3 8 27 125 Sol. (A, B) sin4 x cos 4 x 1 sin4 x (1 sin 2 x )2 1 + = + = 2 3 5 2 3 5 sin4 x 1 sin 4 x 2 sin2 x 1 + = 2 3 5 6 5 sin4x – 4 sin2x + 2 = 5 25 sin4x – 20 sin2x + 4 = 0 (5 sin2x – 2)2 = 0 2 3 sin2x = , cos2x = 5 5 tan2x = 2 3 sin 8 x cos 8 x and + = 1 8 27 125 RESONANCE Page # 17
- 18. A 32. In a triangle ABC with fixed base BC, the vertex A moves such that cos B + cos C = 4 sin2 . If a, b and c 2 denote the lengths of the sides of the triangle opposite to the angles A, B and C respectively, then (A) b + c = 4a (B) b + c = 2a (C) locus of points A is an ellipse (D) locus of point A is a pair of straight lines Sol. (B, C) A cos B + cos C = 4 sin2 2 BC BC A 2 cos cos = 4 sin2 2 2 2 A BC A 2 sin cos 2 2 sin 2 = 0 2 B C B C A cos =0 as sin 0 2 2 2 – 2 cos B C B C cos + 3 sin sin =0 2 2 2 2 – cos B C 1 tan tan = 2 2 3 (s a)(s c ) (s b)(s a) 1 . = s(s b) s(s c ) 3 sa 1 = 2s = 3a b + c = 2a s 3 Locus of A is an ellipse SECTION - III This section contains 2 groups of questions. Each group has 3 multiple choice questions based on a paragraph. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which ONLY ONE is correct. Paragraph for Question Nos. 33 to 35 Let be the set of all 3 × 3 symmetric matrices all of whose entries are either 0 or 1. Five of these entries are 1 and four of them are 0. RESONANCE Page # 18
- 19. 33. The number of matrices in is (A) 12 (B) 6 (C) 9 (D) 3 Sol. (A) Case I : All three diagonal elements are 1 No. of matrices = 3C1 = 3 Case II : Two diagonal elements are zero & one element is one No. of matrices = 3C1 . 3C1 = 9 Total matrices = 3 + 9 = 12 x 1 34. The number of matrices A in for which the system of linear equations A y = 0 has a unique solution, z 0 is (A) less than 4 (B) at least 4 but less than 7 (C) at least 7 but less than 10 (D) at least 10 Sol. (B) x 1 A y = 0 z 0 For unique solution det (A) 0 Case I det(A) = = 1 – a2 – b2 – c2 + 2abc 0 Here a, b, c is selected from 1, 0, 0. (No case is possible) Case II (i) det(A) = = 2abc – c2 0 Here a, b, c are selected from 1, 1, 0. (2 cases are possible) (ii) det(A) = = 2abc – b2 0 Here a, b, c are selected from 1, 1, 0. (2 cases are possible) (iii) det(A) = = 2abc – a2 0 Here a, b, c are selected from 1, 1, 0. (2 cases are possible) Hence there are exactly 6 matrices for unique solution . Hence option B is correct RESONANCE Page # 19
- 20. x 1 35. The number of matrices A in for which the system of linear equations A y = 0 is inconsistent, is z 0 (A) 0 (B) more than 2 (C) 2 (D) 1 Sol. (B) x 1 A y = 0 z 0 Case I : 1 a b x 1 y = 0 a 1 c a, b, c are selected from 1, 0, 0 b c 1 z 0 x + ay + bz = 1 ax + y + cz = 0 bx + cy + z = 0 (i) If a = 1 , b = c = 0 then x + y = 1 Inconsitent system of equation x +y=0 (ii) If a = 0 = c, b = 1 then x + z = 1 y=0 Inconsitent system of equation x+z=0 (iii) If c = 1, a = b = 0 then x = 1, z = 0, y = 0 Case II : 1 a b x 1 y a 0 c 0 (i) = a, b, c are selected from 1, 1, 0 z b c 0 0 x + ay + bz = 1 ax + cz = 0 bx + cy = 0 Clearly, In all three cases, solutions are possible so system is consistent. RESONANCE Page # 20
- 21. 0 a b x 1 (ii) a 1 c y = 0 b c 0 z 0 ay + bz = 1 ax + y + cz = 0 bx + cy = 0 Clearly , b = 0, a= c= 1 gives y=1 x+y+z=0 Inconsistent system y=0 More than 2 matrices are possible. Hence option B is correct Paragraph for Question Nos. 36 to 38 A fair die is tossed repeatedly until a six is obtained. Let X denote the number of tosses required. 36. The probability that X = 3 equals 25 25 5 125 (A) (B) (C) (D) 216 36 36 216 Sol. (A) 5 5 1 P(X = 3) = . . = 25 6 6 6 216 37. The probability that X 3 equals 125 25 5 25 (A) (B) (C) (D) 216 36 36 216 Sol. (B) 5 5 P(X 3) = . .1= 25 6 6 36 38. The conditional probability that X 6 given X > 3 equals 125 25 5 25 (A) (B) (C) (D) 216 216 36 36 Sol. (D) 5 5 1 5 6 1 1 . . . ..... 6 6 6 6 P((X 6) / (X > 3)) = P(( X 3) /( X 6)) . P( X 6) = = 25 P( X 3) 5 3 1 5 4 1 . . ...... 36 6 6 6 6 RESONANCE Page # 21
- 22. SECTION - IV Matrix - match Type This section contains 2 questions. Each question contains statements given in two columns, which have to be matched. The statements in Column - I are labelled A, B, C and D, while the statements in Column - II are labelled p, q, r, s and t. Any given statement in Column - I can have correct matching with ONE OR MORE statement(s) in Column - II. The appropriate bubbles corresponding to the answers to these questions have to be darkened as illustrated in the following example : If the correct matches are A – p, s and t ; B – q and r ; C – p and q ; and D – s and t; then the correct darkening of bubbles will look like the following. 39. Match the statements/expressions in Column - I with the open intervals in Column - II Column - I Column - II (A) Interval contained in the domain of definition of (p) – , 2 2 non-zero solutions of the differential equation (x – 3)2 y + y = 0 (B) Interval containing the value of the integral (q) 0, 2 5 ( x – 1) ( x – 2) ( x – 3) ( x – 4) ( x – 5) dx 1 5 (C) Interval in which at least one of the points of local (r) , 8 4 maximum of cos2x + sinx lies (D) Interval in which tan–1 (sinx + cosx) is increasing (s) 0, 8 (t) (– , ) RESONANCE Page # 22
- 23. Sol. (A) (x – 3)2 . y + y = 0 dy y =– dx ( x 3)2 dy dx = y ( x 3)2 1 n y = + n C ( x 3) 1 y = Ce x 3 , C 0 x R – {3} (hence correct options p, q, s) Aliter Given differential equation is homogeneous linear differential equation and has x = 3 as a singular point hence x = 3 cannot be in domain of solution. 5 (B) Let I = ( x 1) ( x 2) ( x 3) ( x 4) ( x 5) dx 1 Let x – 3 = t dx = dt 2 I= (t 2)(t 1) t( t 1) (t 2) dt , 2 Integrand is an odd function I=0 (hence correct options are p, t) Aliter 5 Let = ( x 1) ( x 2) ( x 3) ( x 4) ( x 5) dx .......(1) 1 b b Using f ( x) dx = f (a b x) dx a a 5 = (5 x) (4 x) (3 x) (2 x ) (1 x ) dx ........(2) 1 On adding (1) and (2), we get 2 = 0 =0 RESONANCE Page # 23
- 24. (C) f(x) = cos2x + sin x f(x) = – 2 cos x sin x + cos x = cos x (1 – 2 sin x) = 0 (hence correct options are p, t) Sign scheme for first derivative 5 Points of local maxima are , 6 6 Clearly option (p, q, r, t) are correct. Aliter y = cos2x + sin x 2 5 1 y= – sin x 4 2 For y to be maximum. 2 1 sin x = 0 2 1 sin x = 2 x = n + (–1)n , n 6 Hence option (p, q, r, t) are correct (D) y = tan–1 (sin x + cos x) dy cos x sin x = dx 1 (sin x cos x )2 Clearly by graph cos x > sin x is true for option(s) Ans. (A) (p, q, s), (B) (p, t), (C) (p, q, r, t), (D) (s) RESONANCE Page # 24
- 25. 40. Match the conics in Column - I with the statements/expressions in Column - II. Column - I Column - II (A) Circle (p) The locus of the point (h, k) for which the line hx + ky = 1 touches the circle x2 + y2 = 4 (B) Parabola (q) Points z in the complex plane satisfying |z + 2| – |z – 2| = 3 (C) Ellipse (r) Points of the conic have parametric 1– t2 representation x = 3 , 1 t 2 2t y= 1 t 2 (D) Hyperbola (s) The eccentricity of the conic lies in the interval 1 x < (t) Points z in the complex plane satisfying Re(z + 1)2 = |z|2 + 1 1 1 Sol. (p) =2 h2 + k2 = locus is a circle. h k2 2 4 (q) z 2 – z – 2 3 and 2 –(–2) = 4 > 3 locus is a hyperbola. 1 t2 2t (r) x= 3 1 t 2 , y= 1 t 2 Let tan =t x= 3 cos 2 y = sin 2 x2 Hence + y2 = 1 locus is an ellipse 3 (s) Eccentricity x = 1 Parabola , 1< x < Hyperbola (t) z = x + iy Re(z + 1)2 = |z|2 + 1 (x + 1)2 – y2 = x2 + y2 + 1 2x = 2y2 x = y2 Hence locus is a parabola Ans. (A) (p), (B) (s, t), (C) (r), (D) (q, s) RESONANCE Page # 25
- 26. PART-III PHYSICS SECTION - I Single Correct Choice Type Thus section contains 8 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out which ONLY ONE is correct. 41. A disk of radius a/4 having a uniformly distributed charge 6C is placed in the x-y plane with its centre at (–a/2, 0, 0). A rod of length a carrying a uniformly distributed charge 8C is placed on the x-axis from x = a/4 to x = 5a/4. Two point charges –7C and 3C are placed at (a/4, –a/4, 0) and (–3a/4, 3a/4, 0), respectively. Consider a cubical surface formed by six surfaces x = ± a/2, y = ± a/2, z = ±a/2. The electric flux through this cubical surface is 2C 2C 10C 12C (A) (B) (C) (D) 0 0 0 0 Sol. (A) From Gauss's law Qin (8C / 4 ) – 7C (6C / 2) 2C 0 = 0 =– . 0 42. The x-t graph of a particle undergoing simple harmonic motion is shown below. The acceleration of the particle at t = 4/3 s is 1 x(cm) 0 t(s) 4 8 12 –1 3 2 2 2 3 2 (A) cm/s2 (B) cm/s2 (C) cm/s2 (D) – cm/s2 32 32 32 32 RESONANCE Page # 26
- 27. Sol. (D) 2 T = 8 second., A = 1 cm, x = A sint. = 1sin t. 8 2 2 2 a = – 2x = – sin t cm/s2 8 8 2 4 2 3 2 At, t = second. a= – sin =– cm/s2. 3 8 3 32 43. Three concentric metallic spherical shells of radii R, 2R, 3R, are given charges Q1, Q2, Q3, respectively. It is found that the surface charge densities on the outer surfaces of the shells are equal. Then, the ratio of the charges given to the shells, Q1 : Q2 : Q3, is (A) 1 : 2 : 3 (B) 1 : 3 : 5 (C) 1 : 4 : 9 (D) 1 : 8 : 18 Sol. (B) The charge distribution on the surfaces of the shells are given. As per the given condition. Q1 Q1 Q2 Q1 Q2 Q3 Q1 Q2 Q3 = = . 4R 2 4 ( 2R )2 4 (3R )2 1 3 5 44. A ball is dropped from a height of 20 m above the surface of water in a lake. The refractive index of water is 4/3. A fish inside the lake, in the line of fall of the ball, is looking at the ball. At an instant, When the ball is 12.8 m above the water surface, the fish sees the speed of ball as [Take g = 10 m/s2] (A) 9 m/s (B) 12 m/s (C) 16 m/s (D) 21.33 m/s Sol. (C) x v 2 10 ( 20 – 12 .8) 4 x' = n , v' = n = = 16 m/s. rel rel 1 3 45. Look at the drawing given in the figure which has been drawn with ink of uniform line-thickness. The mass of ink used to draw each of the two inner circles, and each of the two line segments is m. The mass of the ink used to draw the outer circle is 6m. The coordinates of the centres of the different parts are: outer circle (0, 0), left inner circle (–a, a), right inner circle (a, a), vertical line (0, 0) and horizontal line (0, –a). The y- coordinate of the centre of mass of the ink in this drawing is RESONANCE Page # 27
- 28. a a a a (A) (B) (C) (D) 10 8 12 3 Sol. (A) m1y1 m2 y 2 m3 y 3 m4 y 4 m5 y 5 ycm = m1 m2 m3 m4 m5 6m(0) m(a ) m(0 ) m( a) m(– a ) a ycm = = . m m m m 6m 10 46. A block of base 10 cm × 10 cm and height 15 cm is kept on an inclined plane. The coefficient of friction between them is 3 . The inclination of this inclined plane from the horizontal plane is gradually increased from 0º. Then (A) at = 30º, the block will start sliding down the plane (B) the block will remain at rest on the plane up to certain and then it will topple (C) at = 60º, the block will start sliding down the plane and continue to do so at higher angles (D) at = 60º, the block will start sliding down the plane and on further increasing , it will topple at certain Sol. (B) Angle of repose 0 = tan–1 = tan–1 3 = 60º 5 2 tan = = . < 45º. 15 / 2 3 Block will topple before it starts slide down. 47. Two small particles of equal masses start moving in opposite directions from a point A in a horizontal circular orbit. Their tangential velocities are v and 2v, respectively, as shown in the figure. Between collisions, the particles move with constant speeds. After making how many elastic collisions, other than that at A, these two particles will again reach the point A? RESONANCE Page # 28
- 29. v A 2v (A) 4 (B) 3 (C) 2 (D) 1 Sol. (C) Since masses of particles are equal, collisons are elastic, so particles will exchange velocities after each collision. The first collision will be at a point P and second at point Q again and before third collision the particles will reach at A. 48. The figure shows certain wire segments joined together to form a coplanar loop. The loop is placed in a perpendicular magnetic field in the direction going into the plane of the figure. The magnitude of the field increases with time. 1 and 2 are the currents in the segments ab and cd. Then, (A) 1 > 2 (B) 1 < 2 (C) 1 is in the direction ba and 2 is in the direction cd (D) 1 is in the direction ab and 2 is in the direction dc Sol. (D) Current 1 = 2 , Since magnetic field increases with time So induced net flux should be outward i.e. current will flow from a to b RESONANCE Page # 29
- 30. SECTION - II Multiple Correct Choice Type This section contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which ONE OR MORE is/are correct. 49. A student performed the experiment of determination of focal length of a concave mirror by u-v method using an optical bench of length 1.5 meter. The focal length of the mirror used is 24 cm. The maximum error in the location of the image can be 0.2 cm. The 5 sets of (u, v) values recorded by the student (in cm) are : (42, 56), (48, 48), (60, 40), (66, 33), (78, 39). The data set(s) that cannot come from experiment and is (are) incor- rectly recorded, is (are) (A) (42, 56) (B) (48, 48) (C) (66, 33) (D) (78, 39) Sol. (C, D) 1 1 1 1 1 1 |u| |f | or |v| = | u | | f | v u f |v| |u| |f | ( 42) ( 24 ) For |u| = 42, |f| = 24 ; |v| = = 56 cm so (42, 56) is correct observation 42 24 For |u| = 48 or |u| = 2f or |v| = 2f so (48, 48) is correct observation For |u| = 66 cm ; |f| = 24 cm ( 66 ) ( 24 ) |v| = 36 cm which is not in the permissible limit 66 24 so (66, 33), is incorrect recorded For |u| = 78, |f| = 24 cm (78 ) ( 24 ) |v| = 32 cm which is also not in the permissible limit. 78 24 so (78, 39), is incorrect recorded. 50. If the resultant of all the external forces acting on a system of particles is zero, then from an inertial frame, one can surely say that (A) linear momentum of the system does not change in time (B) kinetic energy of the system does not changes in time (C) angular momentum of the system does not change in time (D) potential energy of the system does not change in time RESONANCE Page # 30
- 31. Sol. (A) (A) Since there in no resultant external force, linear momentum of the system remains constant. (B) Kinetic energy of the system may change. (C) Angular momentum of the system may change as in case of couple, net force is zero but torque is not zero. Hence angular momentum of the system is not constant. (D) Potential energy may also change. 51. Cv and Cp denote the molar specific heat capacities of a gas at constant volume and constant pressure, respectively. Then (A) Cp – Cv is larger for a diatomic ideal gas than for a monoatomic ideal gas (B) Cp + Cv is larger for a diatomic ideal gas than for a monoatomic ideal gas (C) Cp / Cv is larger for a diatomic ideal gas than for a monoatomic ideal gas (D) Cp . Cv is larger for a diatomic ideal gas than for a monoatomic ideal gas Sol. (B, D) 5 3 For monoatomic gas, Cp = R, Cv = R. Cp – Cv = R 2 2 7 5 For diatomic gas Cp = R, Cv = R. Cp – Cv = R 2 2 Cp – Cv is same for both Cp + Cv = 6R (for diatomic) Cp + Cv = 4R (for mono) so (Cp + Cv)dia > (Cp + Cv)mono Cp 7 Cv 5 = 1.4 (for diatomic) Cp 5 Cv 3 = 1.66 (for monoatomic) 35 2 (Cp) (Cv) = R = 8.75 R2 (for diatomic) 4 15 2 (Cp) (Cv) = R (for monoatomic) 4 so (Cp . Cv)diatomic > (Cp . Cv)monoatomic 52. For the circuit shown in the figure [Current Electricity] (A) the current I through the battery is 7.5 mA (B) the potential difference across RL is 18 V (C) ratio of powers dissipated in R1 and R2 is 3 (D) if R1 and R2 are interchanged, magnitude of the power dissipated in RL will decrease by a factor of 9 RESONANCE Page # 31
- 32. Sol. (A, D) 6 1 .5 6 16 Req = 7.5 2 = +2= k 5 5 V 24 3 = R = eq 16 2 × 5 mA = × 5 = 7.5 mA for potential difference across R1 V1 = 7.5 × 2 = 15 V for potential difference across R2 V2 = 24 – 15 = 9 V 2 2 V1 V2 : (15 )2 92 25 for power : PR1 : PR 2 = = : = R1 R2 2 6 3 2 PRL V2 92 = = 54 mW RL 1 .5 If R1 and R2 are interchanged ( 2) (1.5 ) 3 R´ = R1 || R2 = 2 1.5 = 3 .5 R´ V´L = R2 R´ × 24 V = 3 V V ´2 L 32 Now power dissipated in RL is P´L = = = 6 mW RL 1 .5 SECTION–III Comprehension Type This section contains 2 groups of questions. Each group has 3 multiple choice questions based on a paragraph. Each question has 4 choices (A),(B),(C) and (D) for its answer, out of which ONLY ONE is correct. Paragraph for Question Nos. 53 to 55 2 Scientists are working hard to develop nuclear fusion reactor. Nuclei of heavy hydrogen, 1 H , known as deuteron and denoted by D, can be thought of as a candidate for fusion reactor. The D-D reaction is 2 2 1 H 1 H 3 He n energy . In the core of fusion reactor, a gas of heavy hydrogen is fully ionized into 2 2 deuteron nuclei and electrons. This collection of 1 H nuclei and electrons is known as plasma. The nuclei move randomly in the reactor core and occasionally come close enough for nuclear fusion to take place. Usually, the temperatures in the reactor core are too high and no material wall can be used to confine the plasma. Special techniques are used which confine the plasma for a time t0 before the particles fly away from the core. If n is the density (number/volume) of deuterons, the product nt0 is called Lawson number. In one of the criteria, a reactor is termed successful if Lawson number is greater than 5×1014 s/cm3. e2 It may be helpful to use the following: Boltzman constant k = 8.6×10–5 eV/K ; = 1.44 × 10–9 eVm. 4 0 RESONANCE Page # 32
- 33. 53. In the core of nuclear fusion reactor, the gas becomes plasma because of (A) strong nuclear force acting between the deuterons (B) Coulomb force acting between the deuterons (C) Coulomb force acting between deuterons-electrons pairs (D) the high temperature maintained inside the reactor core Sol. (D) the high temperature maintained inside the reactor core 54. Assume that two deuteron nuclei in the core of fusion reactor at temperature T are moving towards each other, each with kinetic energy 1.5 kT, when the separation between them is large enough to neglect Coulomb potential energy. Also neglect any interaction from other particles in the core. The minimum temperature T required for them to reach a separation of 4 × 10–15 m in the range. (A) 1.0 × 109 K < T < 2.0 × 109 K (B) 2.0 × 109 K < T < 3.0 × 109 K (C) 3.0 × 109 K < T < 4.0 × 109 K (D) 4.0 × 109 K < T < 5.0 × 109 K Sol. (A) From energy conservation 3 1 e2 2 kT 2 4 0 r (1.44 10 9 ) 1 T= = 1.39 × 109 K (8.6 10 5 ) 3 4 10 15 × 55. Results of calculations for four different designs of a fusion reactor using D-D reaction are given below. Which of these is most promising based on Lawson criterion ? (A) deuteron density = 2.0 × 1012 cm–3, confinement time = 5.0 × 10–3 s (B) deuteron density = 8.0 × 1014 cm–3, confinement time = 9.0 × 10–1 s (C) deuteron density = 4.0 × 1023 cm–3, confinement time = 1.0 × 10–11 s (D) deuteron density = 1.0 × 1024 cm–3, confinement time = 4.0 × 10–12 s Sol. (B) nt0 > 5 × 1014 s/cm3 for deuteron density = 8.0 × 1014 cm–3, confinement time = 9.0 × 10–1 s nt0 = 7.2 × 1014 s/cm3 Paragraph for Question Nos. 56 to 58 When a particle is restricted to move along x-axis between x = 0 and x = a, where a is of nanometer dimension, its energy can take only certain specific values. The allowed energies of the particle moving in such a restricted region, correspond to the formation of standing waves with nodes at its ends x = 0 and x = a. The wavelength of this standing wave is related to the linear momentum p of the particle according to p2 the de-Broglie relation. The energy of the particle of mass m is related to its linear momentum as E = . 2m Thus, the energy of the particle can be denoted by a quantum number ‘n’ taking values 1,2,3,......., (n = 1, called the ground state) corresponding to the number of loops in the standing wave. Use the model described above to answer the following three questions for a particle moving in the line x = 0 to x = a. Take h = 6.6 × 10–34 J s and e = 1.6 × 10–19 C. RESONANCE Page # 33
- 34. 56. The allowed energy for the particle for a particular value of n is proportional to : (A) a–2 (B) a–3/2 (C) a–1 (D) a2 Sol. (A) h nh n =a and p= = 2 2a p2 n2h2 1 Energy E = = E 2m 8ma2 a2 57. If the mass of the particle is m = 1.0 × 10–30 kg and a = 6.6 nm, the energy of the particle in its ground state is closest to : (A) 0.8 meV (B) 8 meV (C) 80 meV (D) 800 meV Sol. (B) n2h2 E= 8 ma 2 (1)2 ( 6.6 10 34 )2 For ground state n = 1 E1 = 30 = 8 meV 8 10 ( 6.6 10 9 )2 1.6 10 19 58. The speed of the particle, that can take discrete values, is proportional to : (A) n–3/2 (B) n–1 (C) n1/2 (D) n Sol. (D) nh P= = mv 2a vn Section–IV Matrix – Match Type This section contains 2 questions. Each question contains statements given in two columns, which have to be matched. the statements in Column–I are labelled A,B,C and D, while the statements in Column-II are labelled p,q,r,s and t. Any given statement in Column-I can have correct matching with ONE OR MORE statement(s) in Column-II. The appropriate bubbles corresponding to the answers to these questions have to be darkened as illustrated in the following examples : If the correct matches A–p,s and t; B–q and r; C–p and q ; and D–s and t; then the correct darkening of bubbles will look like the following. RESONANCE Page # 34
- 35. p q r s t A p q r s t B p q r s t C p q r s t D p q r s t 59. Six point charges, each of the same magnitude q, are arranged in different manners as shown in Column–II. In each case, a point M and a line PQ passing through M are shown. Let E be the electric field and V be the electric potential at M (potential at infinity is zero) due to the given charge distribution when it is at rest. Now, the whole system is set into rotation with a constant angular velocity about the line PQ. Let B be the magnetic field at M and be the magnetic moment of the system in this condition. Assume each rotating charge to be equivalent to a steady current. Column–I Column–II (A) E = 0 (p) Charges are at the corners of a + Q regular hexagon. M is at the – – + centre of the hexagon. PQ is perpendicular to the plane of + the hexagon. P – (B) V 0 (q) P Charges are on a line perpendicular to PQ at equal intervals. M is the – + – + – + midpoint between the two innermost M charges. Q (C) B = 0 (r) + Q Charges are placed on two coplanar – + insulating rings at equal intervals. M is the common centre of the rings. PQ is M perpendicular to the plane of rings. – – P + (D) 0 (s) – + – Charges are placed at the corners of a M rectangle of sides a and 2a and at the P Q mid–points of the longer sides. M is at – + – the centre of the rectangular. PQ is parallel to the longer sides. RESONANCE Page # 35
- 36. P (t) + – Charges are placed on two coplanar, identical insulating rings at equal + + M– – intervals. M is the mid–point between Q the centres of the rings. PQ is perpendicular to the line joining the centres and coplanar to the rings. Ans. : (A) (p), (r), (s); (B) (r), (s); (C) (p), (q), (t) ; (D) (r), (s) Sol. (p) By symmetry E=0,V =0,B=0 + – Q – + and µ = NA but effective = 0, So µ = 0 + P – (q) P E 0,V=0 – + – + – + Since effective = 0 B = 0 and µ = 0 M Q (r) + Q E = 0 (By Symmetry) – + V 0 (since distances are different) – M – P B 0 (since Radius is different) + µ0 (s) – + – E = 0 (By symmetry) M P Q V 0 (since distances are different) – + – B0 µ0 P (t) + – E 0, V = 0, µ = 0, B = 0 and given each rotating charge to be equivalent to a steady current so B = 0 so (t C) + + M– – and if it was not given that each rotating charge to be Q equivalent to steady current then B 0. RESONANCE Page # 36
- 37. 60. Column II shows five systems in which two objects are labelled as X and Y. Also in each case a point P is shown. Column I gives some statements about X and and/or Y. Match these statements to the appropriate system(s) from Column II. Column I Column II (A) The force exerted (p) Block Y of mass M left on a by X on Y has a fixed inclined plane X, slides magnitude Mg. on it with a constant velocity. (B) The gravitational (q) Two ring magnets Y and Z, potential energy of each of mass M, are kept in X is continuously frictionless vertical plastic increasing, stand so that they repel each other. Y rests on the base X and Z hangs in air in equilibrium. P is the topmost point of the stand on the common axis of the two rings. The whole system is in a lift that is going up with a constant velocity. (C) Mechanical energy (r) A pulley Y of mass m0 is fixed of the system X + Y to a table through a clamp X. is continuously A block of mass M hangs from decreasing. a string that goes over the pulley and is fixed at point P of the table. The whole system is kept in a lift that is going down with a constant velocity. (D) The torque of the (s) A sphere Y of mass M is put weight of Y about in a nonviscous liquid X kept point P is zero. in a container at rest. The sphere is released and it moves down in the liquid. (t) A sphere Y of mass M is falling with its terminal velocity in a viscous liquid X kept in a container. Ans. : (A) (p), (t); (B) (q), (s), (t); (C) (p), (r), (t); (D) (q) RESONANCE Page # 37

No public clipboards found for this slide

×
### Save the most important slides with Clipping

Clipping is a handy way to collect and organize the most important slides from a presentation. You can keep your great finds in clipboards organized around topics.

Be the first to comment