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File 02 Iit Jee 09 Paper 02 Pcm
File 02 Iit Jee 09 Paper 02 Pcm
File 02 Iit Jee 09 Paper 02 Pcm
File 02 Iit Jee 09 Paper 02 Pcm
File 02 Iit Jee 09 Paper 02 Pcm
File 02 Iit Jee 09 Paper 02 Pcm
File 02 Iit Jee 09 Paper 02 Pcm
File 02 Iit Jee 09 Paper 02 Pcm
File 02 Iit Jee 09 Paper 02 Pcm
File 02 Iit Jee 09 Paper 02 Pcm
File 02 Iit Jee 09 Paper 02 Pcm
File 02 Iit Jee 09 Paper 02 Pcm
File 02 Iit Jee 09 Paper 02 Pcm
File 02 Iit Jee 09 Paper 02 Pcm
File 02 Iit Jee 09 Paper 02 Pcm
File 02 Iit Jee 09 Paper 02 Pcm
File 02 Iit Jee 09 Paper 02 Pcm
File 02 Iit Jee 09 Paper 02 Pcm
File 02 Iit Jee 09 Paper 02 Pcm
File 02 Iit Jee 09 Paper 02 Pcm
File 02 Iit Jee 09 Paper 02 Pcm
File 02 Iit Jee 09 Paper 02 Pcm
File 02 Iit Jee 09 Paper 02 Pcm
File 02 Iit Jee 09 Paper 02 Pcm
File 02 Iit Jee 09 Paper 02 Pcm
File 02 Iit Jee 09 Paper 02 Pcm
File 02 Iit Jee 09 Paper 02 Pcm
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  • 1. IITJEE 2009 SOLUTIONS 22 SOLUTIONS TO IIT-JEE 2009 CHEMISTRY: Paper-II (Code: 04) PART – I SECTION – I Single 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 ONLY ONE is correct. Note: Questions with (*) mark are from syllabus of class XI. 1. In the following carbocation, H/CH3 that is most likely to migrate to the positively charged carbon is H H 1 2 + 4 5 H3C C C 3 C CH3 HO H CH3 (A) CH3 at C–4 (B) H at C–4 (C) CH3 at C–2 (D) H at C–2 Sol.: Migrating tendency of hydride is greater than that of alkyl group. Migration of hydride from second carbon gives more stable carbocation (stabilized by +R effect of OH group and +I effect of methyl group). Correct choice: (D) 2. The spin only magnetic moment value (in Bohr Magneton units) of Cr(CO) 6 is (A) 0 (B) 2.84 (C) 4.90 (D) 5.92 3d 4s 4p Sol.: Cr(CO)6 XX XX XX XX XX XX d2sp3 hybridisation CO is a strong field ligand; it forces the unpaired electrons of Cr to pair up. As the complex does not have any unpaired electron, its magnetic moment is zero. Correct choice: (A) *3. The correct stability order of the following resonance structures is + – + – – + – + H2C=N=N H2C–N=N H2C–NN H2C–N=N (I) (II) (III) (IV) (A) (I) > (II) > (IV) > (III) (B) (I) > (III) > (II) > (IV) (C) (II) > (I) > (III) > (IV) (D) (III) > (I) > (IV) > (II) Sol.: The structure with complete octet of all the atoms and having greater number of covalent bonds are more stable than the others. The structure with –ve charge on more electronegative atom is more stable than the one in which –ve charge is present on less electronegative atom. Correct choice: (B) 4. For a first order reaction A  P, the temperature (T) dependent rate constant (k) was found to follow the equation 1 log k = –(2000) + 6.0. The pre-exponential factor A and the activation energy Ea, respectively, are T (A) 1.0 × 106 s–1 and 9.2 kJ mol–1 (B) 6.0 s–1 and 16.6 kJ mol–1 6 –1 –1 (C) 1.0 × 10 s and 16.6 kJ mol (D) 1.0 × 106 s–1 and 38.3 kJ mol–1 Ea 1 Sol.: log k = log A – ; log k = 6.0 – (2000) ; log A = 6.0 ; A = 106 s–1 2.303RT T Ea = 2000  Ea = 2000 × 2.303 × 8.314 = 38.29 kJ mol –1. 2.303R Correct choice: (D) SECTION – II Multiple Correct Choice Type This section contains 5 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. 5. The nitrogen oxide(s) that contain(s) N-N bond(s) is(are) (A) N2O (B) N2O3 (C) N2O4 (D) N2O5 Brilliant Tutorials Pvt. Ltd. Head Office: 12, Masilamani Street, T. Nagar, Chennai-600 017 Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016 Ph.: 2653 7392/93/94/95 Fax: 2653 7396
  • 2. IITJEE 2009 SOLUTIONS 23 Sol.: N2O, N2O3 and N2O4 contain N–N bonds. Their structures are as given below: O O O 175 pm O NNO 114 pm NN 121 pm NN 121 pm 113 pm 119 pm 186 pm O O O (N2O) (N2O3) (N2O4) Correct choice: (A), (B) and (C) 6. The correct statement(s) about the following sugars X and Y is(are) CH2OH CH2OH O CH2OH H H H O HOH2C O H H HO H H O H H O OH H OH H O H HO CH2OH OH H H HO OH H HO H OH OH H H OH X Y (A) X is a reducing sugar and Y is a non-reducing sugar. (B) X is a non-reducing sugar and Y is a reducing sugar. (C) The glucosidic linkages in X and Y are  and , respectively. (D) The glucosidic linkages in X and Y are  and , respectively. Correct choice: (B) and (C) *7. In the reaction, 2X + B2H6  [BH2(X)2]+ [BH4]– the amine(s) X is(are) (A) NH3 (B) CH3NH2 (C) (CH3)2NH (D) (CH3)3N Sol.: B2H6 with lower amine such as NH3, CH3NH2 and (CH3)2NH undergoes unsymmetrical cleavage while with large amine such as (CH3)3N, C5H5N it undergoes symmetrical cleavage. Correct choice: (A), (B) and (C) *8. Among the following, the state function(s) is(are) (A) Internal energy (B) Irreversible expansion work (C) Reversible expansion work (D) Molar enthalpy Sol.: Internal energy and molar enthalpy are state functions. Correct choice: (A) and (D)  *9. For the reduction of NO3 ion in an aqueous solution, E° is +0.96 V. Values of E° for some metal ions are given below V2+(aq) + 2e–  V ; E° = –1.19 V Fe3+(aq) + 3e–  Fe ; E° = –0.04 V Au3+(aq) + 3e–  Au ; E° = +1.40 V – Hg (aq) + 2e  Hg 2+ ; E° = +0.86 V  The pair(s) of metals that is(are) oxidized by NO3 in aqueous solution is(are) (A) V and Hg (B) Hg and Fe (C) Fe and Au (D) Fe and V o o o Sol.: E cell = E cathode – E anode o o  For the reduction to occur E cathode should be greater than E anode . Therefore, V, Fe and Hg can be oxidized by NO3 . Correct choice: (A), (B) and (D) SECTION  III 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. Brilliant Tutorials Pvt. Ltd. Head Office: 12, Masilamani Street, T. Nagar, Chennai-600 017 Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016 Ph.: 2653 7392/93/94/95 Fax: 2653 7396
  • 3. IITJEE 2009 SOLUTIONS 24 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 10. Match each of the reactions given in Column I with the corresponding product(s) in Column II. Column I Column II (A) Cu + dil. HNO3 (p) NO (B) Cu + conc. HNO3 (q) NO2 (C) Zn + dil. HNO3 (r) N2O (D) Zn + conc. HNO3 (s) Cu(NO3)2 (t) Zn(NO3)2 Sol.: (A) – (p), (s) ; (B) – (q), (s) ; (C) – (r), (t) ; (D) – (q), (t) 11. Match each of the compound in Column I with its characteristic reaction(s) in Column II. Column I Column II Br (A) (p) Nucleophilic substitution O (B) OH (q) Elimination CHO (C) (r) Nucleophilic addition OH Br (D) (s) Esterification with acetic anhydride NO2 (t) Dehydrogenation Sol.: (A) – (p), (q), (t) ; (B) – (p), (s), (t) ; (C) – (r), (s) ; (D) – (p) SECTIONIV Integer Answer Type This section contains 8 questions. The answer to each of the X Y Z W questions is a single digit integer, ranging from 0 to 9. 0 0 0 0 The appropriate bubbles below the respective question 1 1 1 1 numbers in the ORS have to be darkened. For example, if the 2 2 2 2 correct answers to question numbers X, Y, Z and W (say) are 6, 0, 9 and 2, respectively, then the correct darkening of 3 3 3 3 bubbles will look like the following: 4 4 4 4 5 5 5 5 6 6 6 6 7 7 7 7 8 8 8 8 9 9 9 9 12. The coordination number of Al in the crystalline state of AlCl 3 is Sol.: In the crystalline state of AlCl3, the Cl– ions form space lattice with Al3+ occupying octahedral voids. The co-ordination number of Al in the crystalline state of AlCl3 is 6. Brilliant Tutorials Pvt. Ltd. Head Office: 12, Masilamani Street, T. Nagar, Chennai-600 017 Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016 Ph.: 2653 7392/93/94/95 Fax: 2653 7396
  • 4. IITJEE 2009 SOLUTIONS 25 13. The oxidation number of Mn in the product of alkaline oxidative fusion of MnO 2 is Sol.: 4KOH + 2MnO2 + O2  2K2MnO4 + 2H2O The oxidation number of Mn in the product formed in the above reaction is 6. 14. The total number of  and  particles emitted in the nuclear reaction 238 92 U  214 82 Pb is  1  238 214 4 0 Sol.: 92 U 82 Pb + 6 2 He + 2  total number of particles emitted is 8. *15. The dissociation constant of a substituted benzoic acid at 25°C is 1.0 × 10 –4. The pH of a 0.01 M solution of its sodium salt is 1 Sol.: pH of sodium salt of weak acid = (pKw + pKa + log C) 2 1 (14 + 4 – 2) = 8. = 2 16. The number of water molecule(s) directly bonded to the metal centre in CuSO4.5H2O is Sol.: The number of water molecules directly bonded to the metal centre in CuSO 4.5H2O is 4. H2O 2+ OH2 Cu SO2.H2O 4 H2O OH2 *17. At 400 K, the root mean square (rms) speed of a gas X (molecular weight = 40) is equal to the most probable speed of gas Y at 60 K. The molecular weight of the gas Y is 3RTX Sol.: Urms of X = MX 2RTY Ump of Y = MY 3RTX 2RTY = MX MY 2TY M X 2  60  40 MY = = = 4. 3TX 3  400 *18. The total number of cyclic structural as well as stereo isomers possible for a compound with the molecular formula C5H10 is Me Me Me Me H H Me Sol.: Me Et Me H H H Me Me H 7 cyclic isomers are possible for the compound having molecular formula C5H10. *19. In a constant volume calorimeter, 3.5 g of a gas with molecular weight 28 was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K due to the combustion process. Given that the heat capacity of the calorimeter is 2.5 kJ K–1, the numerical value for the enthalpy of combustion of the gas in kJ mol–1 is Sol.: Heat absorbed by calorimeter = 2.5 × (298.45 – 298) kJ Heat evolved by combustion of 3.5 g gas = 2.5 × 0.45 kJ 2.5  0.45 Heat evolved by combustion of 1 mol gas = = 9 kJ mol–1. 3.5 28 Heat of combustion at constant volume = 9 kJ mol–1. Brilliant Tutorials Pvt. Ltd. Head Office: 12, Masilamani Street, T. Nagar, Chennai-600 017 Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016 Ph.: 2653 7392/93/94/95 Fax: 2653 7396
  • 5. IITJEE 2009 SOLUTIONS 26 SOLUTIONS TO IIT-JEE 2009 MATHEMATICS: Paper-II (Code: 04) PART – II SECTION – I Single 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 ONLY ONE is correct. Note: Questions with (*) mark are from syllabus of class XI. *20. The normal at a point P on the ellipse x2  4 y 2  16 meets the x-axis at Q. If M is the mid point of the line segment PQ, then the locus of M intersects the latus rectums of the given ellipse at the points  3 5 2  3 5 19   1  4 3 (A)   ,  (B)   ,  (C)   2 3 ,   (D)   2 3 ,    2 7  2 4   7  7        y 7  4x 2 Sol.: M  cos , sin     y2 1 2  49 b2 P (4cos , 2sin) x  4 1   2 3 a2 M x x 2  12 Q (3cos , 0) 1 y2  49 Correct choice: (C) 21. A line with positive direction cosines passes through the point P2,  1, 2 and makes equal angles with the coordinates axes. The line meets the plane 2 x  y  z  9 at point Q. The length of the line segment PQ equals (A) 1 (B) 2 (C) 3 (D) 2 x  2 y 1 z  2 Sol.: Equation of the line through P is   r ;  Q be r  2, r 1, r  2 1 1 1 So, 2r  2  r 1  r  2  9  r  1  PQ  3 Correct choice: (C) *22. The locus of the orthocentre of the triangle formed by the lines 1  px  py  p1  p   0 , 1  q x  qy  q1  q  0 and y  0 , where p  q , is (A) a hyperbola (B) a parabola (C) an ellipse (D) a straight line Sol.: Equation of the line AM: x  pq …(i) A (pq, (1+p) (1+q)) q Equation of the line BN: y  0  x  p  …(ii) N q 1 Point of intersection of these lines will be orthocentre. O q y  pq  p    pq . Hence locus of orthocentre x  y  0 which q 1 B M C is straight line. (–p, 0) (–q, 0) Correct choice: (D) *23. If the sum of first n terms of an A.P. is cn2 , then the sum of squares of these n terms is (A)   n 4n 2  1 c 2 (B)   n 4n 2  1 c 2 (C)   n 4n 2  1 c 2 (D)   n 4n 2  1 c 2 6 3 3 6 Sol.: Sn  cn2  Tn  Sn  Sn 1  c[n2  n  1 ]  Tn  c2n 1  2 T n 2  c2 4n 2  4n  1    n 4n 2  1 c 2 Required sum  c 2 [ 4 n  4n  1]  2 3 Correct choice: (C) Brilliant Tutorials Pvt. Ltd. Head Office: 12, Masilamani Street, T. Nagar, Chennai-600 017 Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016 Ph.: 2653 7392/93/94/95 Fax: 2653 7396
  • 6. IITJEE 2009 SOLUTIONS 27 SECTION – II Multiple Correct Choice Type This section contains 5 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. *24. The tangent PT and the normal PN to the parabola y 2  4ax at a point P on it meet its axis at points T and N, respectively. The locus of the centroid of the triangle PTN is a parabola whose  2a  (A) vertex is  , 0  (B) directrix is x  0  3  (D) focus is a, 0 2a (C) latus rectum is 3 Sol.: ST  SN y  PS is a median.  PG : GS  2 :1 at , 2at 2 P  2a  at 2 2at   G  ,   h, k   3 3    G T  at , 0 2 x 3k 9k O S(a, 0) N 2a  at , 0 2  3h  2a  at2 and t  3h  2a  2 2a 4a 4a  2a  Locus of (h, k) is, y 2  x   3  3   2a  , Focus  a, 0 a 4a Vertex   , 0  , Directrix  x   0 ; Latus rectum   3  3 3 Correct choice: (A) and (D)  6 m  1  cosec   m   4 *25. For 0    2 , the solution(s) of cosec   m 1 4      4  2 is (are)    5 (A) (B) (C) (D) 4 6 12 12 6 m  1  cosec   m   4 Sol.:  cosec   m 1 4      4  2   m   m  1     sin           6   4   4   6  m  1   cot   m   4  2    m 1  sin   m    m  1   4 2   cot   4      4   . sin    m 1    4   4    3  1  5  cot   cot     4  cot   tan   4  sin 2    and  2  2 12 12 Correct choice: (C) and (D) For the function f x   x cos , x  1 , 1 26. x (A) for at least one x in the interval [1, ), f x  2  f x  2 (B) lim f x   1 x  (C) for all x in the interval [1, ), f x  2  f x  2 (D) f x  is strictly decreasing in the interval [1, ) 1 1 f x   cos  x sin  2  cos  sin 1 1 1 1 Sol.: x  x x x x x lim f x   1 x   1  1  1  1  1  1 1 f x    sin    2   2 sin  cos   2   2 sin   2 sin  3 cos   3 cos  0,   [1, ) 1 1 1 1 1 1 1 1  x  x  x x x  x  x  x  x x x x x x x Brilliant Tutorials Pvt. Ltd. Head Office: 12, Masilamani Street, T. Nagar, Chennai-600 017 Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016 Ph.: 2653 7392/93/94/95 Fax: 2653 7396
  • 7. IITJEE 2009 SOLUTIONS 28 f x   lim f x   1 x  f x  2   f x  f x    1  f x  2  f x  2 x2 x Correct choice: (B), (C) and (D) *27. An ellipse intersects the hyperbola 2 x2  2 y 2  1 orthogonally. The eccentricity of the ellipse is reciprocal of that of the hyperbola. If the axes of the ellipse are along the coordinate axes, then (A) Equation of ellipse is x2  2 y 2  2 (B) The foci of ellipse are  1, 0 (C) Equation of ellipse is x  2 y  4 2 2  (D) The foci of ellipse are  2 , 0  1 y Sol.: x2  y2  , e 2 …(i) 2 x2 y2 (x1, y1) 2  1 …(ii) M1 a b2 P 1 e x 2 M2 b 2  a 2 1 e 2    1  a2  b 2  a 2 1     2 2 a2 b2  2 dy dy x1 From equation (i)  2 x  2 y 0   at P dx dx y1 x2 2y2 From equation (ii)  2   1  x2  2y2  a2 …(iii) a a2 dy dy x 2x  4 y 0;   1 at P  m1 m2  1 dx dx 2y1 x1  x1 1   1  x1  2y1 . 2 2 From equation (i)  x1  y1  2 2 y1 2 y1 2 1 1 2 y1  y1  2 2  y1  and x1  1 2 2  a 2  2, b 2  1 2 2 Equation of ellipse is x 2  2 y 2  2 and focus  1, 0 Correct choice: (A) and (B)   1   sin x dx, n  0, 1, 2, .......... , then sin nx 28. If I n  x  10 10 (A) I n  I n  2 (B)  m 1 I 2 m 1  10 (C) I m 1 2m 0 (D) I n  I n 1  a a   f x  dx    f x  f  x dx sin nx Sol.: In  dx Using sin x 0 a 0 Now I n2  I n    sinn  2 x  sin nx cosn  1x sin x   dx  2 cosn  1x dx   [sinn  1x] 0  0 ; I 1  , I 0  0 2   dx  2 sin x sin x n 1 0 0 0 10 10  I m 1 2 m 1  10 , I 1 2m 0 Correct choice: (A), (B) and (C) Brilliant Tutorials Pvt. Ltd. Head Office: 12, Masilamani Street, T. Nagar, Chennai-600 017 Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016 Ph.: 2653 7392/93/94/95 Fax: 2653 7396
  • 8. IITJEE 2009 SOLUTIONS 29 SECTION  III 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. 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 29. Match the statements/ expressions given in Column I with the values given in Column II. Column I Column II *(A) Root(s) of the equation 2 sin   sin 2  2 2 2  (p) 6  6 x   3x   (B) Points of discontinuity of the function f x     cos  , where [y] denotes (q)   4 the largest integer less than or equal to y (C) Volume of the parallelopiped with its edges represented by the vectors  ˆ (r) i  ˆ, i  2 ˆ and i  ˆ  k ˆ j ˆ j ˆ j 3      (D) Angle between vectors a and b where a , b and c are unit vectors satisfying  (s)     2 a b  3c  0 (t)  Sol.: A-q, s 2 sin2   sin2 2  2  0      2 sin2   4 sin2  cos2   2  0  2 1  cos2   4 1  cos2  cos2   2  0 1  2  2 cos2   4 cos2   4 cos4   2  0  2 cos4  1 cos2   0  cos2   0 or cos 2   2        n  or   n    or   2 4 2 4 6x  3x     B-p, r, s, t When is integer then function will be discontinuous if cos   0 i.e., , , ,    6 3 2  ˆ ˆ  ˆ  ˆ ˆ ˆ C-t Let a  i  j ; b  i  2 ˆ ; c  i  j  k j 1 1 0    Volume of parallelopiped  [a b c ]  1 2 0  2  1   1 1     D-r Since a, b , c are unit vectors     a  b  c 1    Now a  b   3 c  a  b . a  b           3c .  3c   2  2   2      1  a  b  1  1  2a . b  3  a . b   2a . b  3 c 2   1   1   a . b cos   , where  is angle between vector a and b  cos     2 2 3 Brilliant Tutorials Pvt. Ltd. Head Office: 12, Masilamani Street, T. Nagar, Chennai-600 017 Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016 Ph.: 2653 7392/93/94/95 Fax: 2653 7396
  • 9. IITJEE 2009 SOLUTIONS 30 30. Match the statements/ expressions given in Column I with the values given in Column II. Column I Column II   (p) 1 (A) The number of solutions of the equation xesin x  cos x  0 in the interval  0,   2 (B) Value(s) of k for which the planes kx  4 y  z  0 , 4 x  ky  2 z  0 and (q) 2 2 x  2 y  z  0 intersect in a straight line *(C) Value(s) of k for which x  1  x  2  x  1  x  2  4k has integer (r) 3 solution(s) (D) If y  y  1 and y0  1 , then value(s) of yln 2 (s) 4 (t) 5 Sol.: A-p xe sin x  cos x  0 y xe sin x  cos x xesin x cos x x O  2 B-q, s For non trivial solution k 4 1 k  4 0 1  4 k 2 0  4 k 2 0 R1  R1  2 R3 2 2 1 2 2 1  k  4k  4 18  2k   0  k  4k  4  2  0  k  4 and 2 C-q, r, s, t x 1  x  2  x  1  x  2  4k Case I : x  2 4 x  4k x  k , therefore k  2, 3, 4, 5 Case II: 2  x  1 or 1  x  1, 1  x  2 No integral solution on exists. Case III: For x  2 x  k ,  k  2, 3, 4, 5 D-r y  y 1   dy dy dy   y 1   dx   dx dx y 1 y 1  y 1  y 1  log y  1  x  logc  log x   ex  c  c  y  ce x  1 y0  1  1  c 1  c  2  y  2e x  1  yln 2  2e ln 2 1  4 1  3 Brilliant Tutorials Pvt. Ltd. Head Office: 12, Masilamani Street, T. Nagar, Chennai-600 017 Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016 Ph.: 2653 7392/93/94/95 Fax: 2653 7396
  • 10. IITJEE 2009 SOLUTIONS 31 SECTIONIV Integer Answer Type This section contains 8 questions. The answer to each of the X Y Z W questions is a single digit integer, ranging from 0 to 9. The 0 0 0 0 appropriate bubbles below the respective question numbers in the ORS have to be darkened. For example, if the correct answers to 1 1 1 1 question numbers X, Y, Z and W (say) are 6, 0, 9 and 2, 2 2 2 2 respectively, then the correct darkening of bubbles will look like the 3 3 3 3 following: 4 4 4 4 5 5 5 5 6 6 6 6 7 7 7 7 8 8 8 8 9 9 9 9 *31.   The smallest value of k, for which both the roots of the equation x2  8kx  16 k 2  k  1  0 are real, distinct and have values at least 4, is Sol.: D0 8k 2  64k 2  k  1  0  k 1   f 4  0  16  32k  16 k 2  k  1  0  1  2k  k 2  k  1  0  k 2  3k  2  0  k  1 or k  2 Least value of k = 2 x 32. Let f : R  R be a continuous function which satisfies f x    f t  dt . Then the value of 0 f ln 5 is Sol.: f x  f x   f x   e x ;  f 0  0    0  f x   0 f ln 5  0  p x   33. Let px  be a polynomial of degree 4 having extremum at x = 1, 2 and lim1  2   2 . Then the value of p2 is x 0  x   P x   Sol.: Given lim1  2   2 . Let Px   x 2  ax3  bx 4 x 0 x  P x   2 x  3ax2  4bx3 P1  2  3a  4b  0 …(i); P2  4  12a  32b  0 …(ii) . So Px   x 2  x 3  x 4  P2  4  8  4  0 1 1 From (i) and (ii)  a  1, b  4 4 Alternative Solution: Px   x 2 x  x     1 P x   2 xx  x    x 2 x    x 2 x   P1  2  1    1    0      4 P2  42  2    16  4    0 P2  42  2    4   16  0 Brilliant Tutorials Pvt. Ltd. Head Office: 12, Masilamani Street, T. Nagar, Chennai-600 017 Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016 Ph.: 2653 7392/93/94/95 Fax: 2653 7396
  • 11. IITJEE 2009 SOLUTIONS 32 *34. Let ABC and ABC be two non-congruent triangles with sides AB  4, AC  AC  2 2 and angle B  30 . The absolute value of the difference between the areas of these triangles is A Sol.: If AB  4, AC  2 2 , B  30 Let lengths of 3rd side be l (BC and BC  ) So using cosine formula cos 30    42  l 2  2 2 2 4 2 2 2 2 2. 4.l 30° p B C D C 3 l 8 2   l 2  4 3l  8  0  l  2 3  2 . So, BC  2 3  2 and BC   2 3  2 (as mentioned in figure) 2 8l So difference between area of these triangles  ADBC  BC  AD  AB sin 30  2 1 2 1   . 2 . 2 3  2  2 3  2 = 4 sq. unit 2   *35. The centres of two circles C1 and C2 each of unit radius are at a distance of 6 units from each other. Let P be the mid-point of the line segment joining the centres of C1 and C2 and C be a circle touching circles C1 and C2 externally. If a common tangent to C1 and C passing through P is also a common tangent to C2 and C, then the radius of the circle C is Sol.: In triangle CMC2 CC2 2  CM 2  MC2 2 M C r r  1 2 2    r  1  2 8 2  r 8 1 1 1 8 C1 (0, 0) P(3, 0) C2 (6, 0) 8 *36. Let x, y, z  be points with integer coordinates satisfying the system of homogeneous equations: 3x  y  z  0 , 3x  z  0 and 3x  2 y  z  0 . Then the number of such points for which x2  y 2  z 2  100 is Sol.: The point satisfying the system of given equation will be , 0, 3 where   I . Now  2  9 2  100  3    3  Number of total points are seven. x 37. If the function f x   x3  e 2 and g x   f 1 x  , then the value of g 1 is ex 2 f x   x 3  e x 2 ; g x   f x  ; f x   3x 2  ; f 0  ; g  f x   x 1 1 Sol.: 2 2 g  f x f x  1  g  f 0 f 0  1  g 1 f 0  1  g 1  1  g 1  2 1  2 38. The maximum value of the function f x   2 x3  15x2  36x  48 on the set A  x | x2  20  9 x is   Sol.:   A  x | x 2  20  9 x  x 2  9 x  20  0  x  4x  5  0  4 x5   f x   2 x 3 15x 2  36x  48 ; f x   6 x 2  30x  36  6 x 2  5x  6  6x  2x  3  0 , x [4, 5] So f 5 would be maximum.  f 5  2125 1525  365  48  7 Brilliant Tutorials Pvt. Ltd. Head Office: 12, Masilamani Street, T. Nagar, Chennai-600 017 Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016 Ph.: 2653 7392/93/94/95 Fax: 2653 7396
  • 12. IITJEE 2009 SOLUTIONS 33 SOLUTIONS TO IIT-JEE 2009 PHYSICS: Paper-II (Code: 04) PART – III SECTION – I Single 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 ONLY ONE is correct. 39. Photoelectric effect experiments are performed using three different metal plates p, q and r having work functions p = 2.0 eV, q = 2.5 eV and r = 3.0 eV, respectively. A light beam containing wavelengths of 550 nm, 450 nm and 350 nm with equal intensities illuminates each of the plates. The correct I-V graph for the experiment is [Take hc = 1240 eV nm] I I p (A) q (B) p q r r V V I I r q (C) p (D) r q p V V 1240 Sol.: E1   2.25eV 550 1240 E2   2.76 eV 450 1240 E3   3.54 eV 350 In case of plate p, all three wavelengths are capable of ejecting electrons In case of plate q, only two wavelengths are capable of ejecting electrons In case of plate r, only one wavelength is capable of ejecting electrons Correct choice: (A) *40. A piece of wire is bent in the shape of a parabola y  kx2 (y –axis vertical) with a bead of mass m on it. The bead can slide on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the x-axis with a constant acceleration a. The distance of the new equilibrium position of the bead, where the bead can stay at rest with respect to the wire, from the y-axis is a a 2a a (A) (B) (C) (D) gk 2 gk gk 4 gk Sol.: macos  mg sin y a  g tan a dy a tan  ;  g dx g a a a 2kx  ; x g 2kg N Correct choice: (B) ma   x mg Brilliant Tutorials Pvt. Ltd. Head Office: 12, Masilamani Street, T. Nagar, Chennai-600 017 Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016 Ph.: 2653 7392/93/94/95 Fax: 2653 7396
  • 13. IITJEE 2009 SOLUTIONS 34 *41. A uniform rod of length L and mass M is pivoted at the centre. Its two ends are attached to two springs of equal spring constants k. The springs are fixed to rigid supports as shown in the figure, and the rod is free to oscillate in the horizontal plane. The rod is gently pushed through a small angle  in one direction and released. The frequency of oscillation is 1 2k 1 k 1 6k 1 24k (A) (B) (C) (D) 2 M 2 M 2 M 2 M L L ML2 Sol.: k    2     2 2 12 k 2 ML2 L    2 12 6k M 1 6k     T  2  f  M 6k 2 M Correct choice: (C) *42. The mass M show in the figure oscillates in simple harmonic motion with amplitude A. The amplitude of the point P is k1 k2 M P k1 A k A k1 A k2 A (A) (B) 2 (C) (D) k2 k1 k1  k2 k1  k2 Sol.: Let x1 and x2 are the amplitudes of points P and Q respectively x1  x2  A k1 k2 Q P k1x1  k2 x2 M k2 A  x1  k1  k2 Correct choice: (D) SECTION – II Multiple Correct Choice Type This section contains 5 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. 43. Under the influence of the Coulomb field of charge +Q, a charge – q is moving around it in an elliptical orbit. Find out the correct statement(s). (A) The angular momentum of the charge – q is constant (B) The linear momentum of the charge – q is constant (C) The angular velocity of the charge – q is constant (D) The linear speed of the charge – q is constant Sol.: Force is always directed toward the centre of +Q, hence net torque on the charge –q is zero.   As F  0 so p will change. Since moment of inertia is changing,  will not be constant. Correct choice: (A) Brilliant Tutorials Pvt. Ltd. Head Office: 12, Masilamani Street, T. Nagar, Chennai-600 017 Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016 Ph.: 2653 7392/93/94/95 Fax: 2653 7396
  • 14. IITJEE 2009 SOLUTIONS 35 *44. A sphere is rolling without slipping on a fixed horizontal C plane surface. In the figure, A is the point of contact, B is the centre of the sphere and C is its topmost point. Then, B A     (A) VC  VA  2 VB  VC      (B) VC  VB  VB  VA         (C) | VC  V A | 2 | VB  VC | (D) | VC  V A | 4 | VB |       Sol.: If VB  V then VC  2V and V A  0 Correct choice: (B, C) *45. The figure shows the P-V plot of an ideal gas taken through P a cycle ABCDA. The part ABC is a semi-circle and CDA is A 3 half of an ellipse. Then, B (A) the process during the path A  B is isothermal 2 D (B) heat flows out of the gas during the path B  C D 1 C (C) work done during the path A  B  C is zero (D) positive work is done by the gas in the cycle ABCDA 0 V 1 2 3 Sol.: Process A  B is not isothermal as it is not rectangular hyperbola During process BCD W < 0 and U < 0  Q < 0 During A  B  C Area under the curve is positive So non zero positive work is done Whole cycle is clockwise so positive work is done by the gas. Correct choice: (B, D) *46. A student performed the experiment to measure the speed of sound in air using resonance air-column method. Two resonances in the air-column were obtained by lowering the water level. The resonance with the shorter air-column is the first resonance and that with the longer air-column is the second resonance. Then, (A) the intensity of the sound heard at the first resonance was more than that at the second resonance (B) the prongs of the tuning fork were kept in a horizontal plane above the resonance tube (C) the amplitude of vibration of the ends of the prongs is typically around 1 cm (D) the length of the air-column at the first resonance was somewhat shorter than 1/4th of the wavelength of the sound in air Sol.: For a longer air column, absorption of energy is more.  Due to end correction, l  e  4 Correct choice: (A, D) 47. Two metallic rings A and B, identical in shape and size A B but having different resistivities A and B, are kept on top of two identical solenoids as shown in the figure. When current I is switched on in the both the solenoids in identical manner, the rings A and B jump to heights hA and hB, respectively, with hA > hB. The possible relation(s) between their resistivities and their masses mA and mB is(are) (A) A > B and mA = mB (B) A < B and mA = mB (C) A > B and mA > mB (D) A < B and mA < mB Brilliant Tutorials Pvt. Ltd. Head Office: 12, Masilamani Street, T. Nagar, Chennai-600 017 Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016 Ph.: 2653 7392/93/94/95 Fax: 2653 7396
  • 15. IITJEE 2009 SOLUTIONS 36 Sol.: Due to induced currents in the rings they will be repelled by the magnetic field of the solenoids. Now since change of flux is same in both so induced emf will be same but induced current will be different as resistivity is not same. The ring with lesser resistivity will get higher impulse. Given that hA > hB This is possible if  A   B and mA = mB or mA < mB Correct choice: (B, D) SECTION  III 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. 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 48. Column II gives certain systems undergoing a process. Column I suggests changes in some of the parameters related to the system. Match the statements in Column I to the appropriate process (es) from Column II. Column–I Column–II (A) The energy of the system is increased (p) System: A capacitor, initially uncharged Process: It is connected to a battery (B) Mechanical energy is provided to the system, which (q) System: A gas in an adiabatic container fitted with an is converted into energy of random motion of its parts. adiabatic piston Process: The gas is compressed by pushing the piston (C) Internal energy of the system is converted into its (r) System: A gas in a rigid container mechanical energy Process: The gas gets cooled due to colder atmosphere surrounding it. (s) System: A heavy nucleus , initially at rest (D) Mass of the system is decreased Process: The nucleus fissions into two fragments of nearly equal masses and some neutrons are emitted (t) System: A resistive wire loop Process: The loop is placed in a time varying magnetic field perpendicular to its plane Sol.: (A) – (p, q, t) ; (B) –(q) ; (C) – (s) ; (D) – (s) Brilliant Tutorials Pvt. Ltd. Head Office: 12, Masilamani Street, T. Nagar, Chennai-600 017 Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016 Ph.: 2653 7392/93/94/95 Fax: 2653 7396
  • 16. IITJEE 2009 SOLUTIONS 37 49. Column I shows four situations of standard Young’s double slit arrangement with the screen placed far away from the slits S1 and S2. In each of these cases S1P0 = S2P0, S1P1 –S2P1 = /4 and S1P2 – S2P2 = /3, where  is the wavelength of the light used. In the cases B, C and D, a transparent sheet of refractive index  and thickness t is pasted on slit S2. The thicknesses of the sheets are different in different cases. The phase difference between the light waves reaching a point P on the screen from the two slits is denoted by (P) and the intensity by I(P) . Match each situation given in Column I with the statement(s) in Column II valid for that situation. Column–I Column–II P2 S2 P1 P0 (A) (p) (P0) = 0 S1 P2 S2 P1 P0 (B) ( – 1) t = /4 (q) (P1) = 0 S1 P2 S2 P1 P0 (C) ( – 1) t = /2 (r) I(P1) = 0 S1 P2 S2 P1 P0 (D) ( – 1) t = 3/4 (s) I(P0) > I (P1) S1 (t) I(P2) > I (P1) Sol.: (A) –(p,s); (B) –(q); (C) – (t); (D) –(r,s,t) SECTIONIV Integer Answer Type This section contains 8 questions. The answer to each of the X Y Z W questions is a single digit integer, ranging from 0 to 9. The 0 0 0 0 appropriate bubbles below the respective question numbers in the ORS have to be darkened. For example, if the correct answers to 1 1 1 1 question numbers X, Y, Z and W (say) are 6, 0, 9 and 2, 2 2 2 2 respectively, then the correct darkening of bubbles will look like the 3 3 3 3 following: 4 4 4 4 5 5 5 5 6 6 6 6 7 7 7 7 8 8 8 8 9 9 9 9 50. A solid sphere of radius R has a charge Q distributed in its volume with a charge density   kr a , where k and a are R 1 constants and r is the distance from its centre. If the electric field at r  is times that at r = R , find the value of a. 2 8 R/2 R/2 a 3  r a 3  4k  R   R Sol.: At r  ; qencl  4r 2 drkra  4k      2 0  a  30   a3 2  At r = R Brilliant Tutorials Pvt. Ltd. Head Office: 12, Masilamani Street, T. Nagar, Chennai-600 017 Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016 Ph.: 2653 7392/93/94/95 Fax: 2653 7396
  • 17. IITJEE 2009 SOLUTIONS 38 a 3 4k R 4 1 Ra 3 Ra 3 ; encl 2  1 encl 2 ; q q' q' encl     a3 4 0 r 8 4 0 R 2 R2 8 R2 2a  3  32 ; a35 ; a=2 51. A steady current I goes through a wire loop PQR having shape of a right angle triangle with PQ = 3x, PR = 4x and QR = 5x.   I  If the magnitude of the magnetic field at P due to this loop is k  0  , find the value of k.  48x  0 I   Q Sol.: Bp  sin 370  sin 530 4r 0 I 3 4 7 0 I   I  5x     =   k 0   3x 4 4 x 3 5 5 48x  48x  r 5 370 530  k=7 P R 4x *52. Three objects A, B and C are kept in a straight line on a m 2m m frictionless horizontal surface. These have masses m, 2m and m, respectively. The object A moves towards B A B C with a speed 9 m/s and makes an elastic collision with it. Thereafter, B makes completely inelastic collision with C. All motions occur on the same straight line. Find the final speed (in m/s) of the object C. Sol.: Let after collision velocity of block A and B be vA and vB respectively. 9m  mvA  2mvB vB  vA  9  vB = 6 m/s When B collides with C, let the combined mass moves with velocity v 3mv  2mvB 26 v  4 m/s 3 *53. Two soap bubbles A and B are kept in a closed chamber where the air is maintained at pressure 8 N/m2. The radii of bubbles A and B are 2 cm and 4 cm, respectively. Surface tension of the soap-water used to make bubbles is 0.04 N/m. Find the ratio nB/nA, where nA and nB are the number of moles of air in bubbles A and B, respectively. [Neglect the effect of gravity.] 4S 4  0.04  100 Sol.: For bubble A, PA  P0   8  16 N/m2 RA 2 4S 4  0.04  100 For bubble B, PB  P0   8  12 N/m2 RB 4 4 3 PB RB nB 3 12  64   6 n A P 4 R 3 16  8 A A 3 Brilliant Tutorials Pvt. Ltd. Head Office: 12, Masilamani Street, T. Nagar, Chennai-600 017 Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016 Ph.: 2653 7392/93/94/95 Fax: 2653 7396
  • 18. IITJEE 2009 SOLUTIONS 39 *54. A light inextensible string that goes over a smooth fixed pulley as shown in the figure connects two blocks of masses 0.36 kg and 0.72kg. Taking g = 10 m/s2, find the work done (in joules) by the string on the block of mass 0.36 kg during the first second after the system is released from rest. Sol.: Let the two blocks move with acceleration a and tension in the string be T. 0.72  0.36 10 2  0.72  0.36 1 5 a 10  m/s2; T   10  4.8 N; S  3.3312  m 0.72  0.36 3 0.72  0.36 2 3 WT = 8 J *55. A cylindrical vessel of height 500 mm has an orifice (small hole) at its bottom. The orifice is initially closed and water is filled in it up to height H. Now the top is completely sealed with a cap and the orifice at the bottom is opened. Some water comes out from the orifice and the water level in the vessel becomes steady with height of water column being 200 mm. Find the fall in height (in mm) of water level due to opening of the orifice. [Take atmospheric pressure = 1.0  105 N/m2, density of water = 1000 kg/m3 and g = 10 m/s2. Neglect any effect of surface tension.] Sol.: P  gh  P0 200 P  1000 10   105 1000 500mm P  105  2000  9.8  104 N/m2 P PV f  P0Vi  9.8 104  A  300  10 104  A  500  H  1000 1000 H H = 206 mm 200mm Height fallen = 206 – 200 = 6 mm *56. A 20 cm long string, having a mass of 1.0 g, is fixed at both the ends. The tension in the string is 0.5 N. The string is set into vibrations using an external vibrator of frequency 100 Hz. Find the separation (in cm) between the successive nodes on the string. n T Sol.: Let the string vibrates with n loops  100 2l  n  100 0.5  100 ; n=4 2  20 1  10 3 20  10 2 Let S be the separation between two successive nodes nS = 20 ; S = 5 cm *57. A metal rod AB of length 10x has its one end A in ice at 00C and the other end B in water at 1000C. If a point P on the rod is maintained at 4000C, then it is found that equal amounts of water and ice evaporate and melt per unit time. The latent heat of evaporation of water is 540 cal/g and latent heat of melting of ice is 80 cal/g. If the point P is at a distance of x from the ice end A, find the value of . [Neglect any heat loss to the surrounding.] Sol.: Let H1 amount of heat flows per unit time form P to A and H2 amount of heat flows per unit time from P to B. P(4000C) 00C ice H1 H2 1000C water A B x (10–)x KA400  0 KA400  100 H1  ; H2  x 10   x Let m be the mass of ice melting or water evaporating per unit time KA400 KA300  m80 … (i) ;  m540 … (ii) x 10   x 410    4 Dividing (i) and (ii)    =9 3 27 Brilliant Tutorials Pvt. Ltd. Head Office: 12, Masilamani Street, T. Nagar, Chennai-600 017 Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016 Ph.: 2653 7392/93/94/95 Fax: 2653 7396
  • 19. IITJEE 2009 SOLUTIONS 40 BREAK UP 1 (LEVEL OF DIFFICULTY) BREAK UP 1 (LEVEL OF DIFFICULTY) Brilliant Tutorials Pvt. Ltd. Head Office: 12, Masilamani Street, T. Nagar, Chennai-600 017 Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016 Ph.: 2653 7392/93/94/95 Fax: 2653 7396
  • 20. IITJEE 2009 SOLUTIONS 41 BREAK UP 1 (LEVEL OF DIFFICULTY) Brilliant Tutorials Pvt. Ltd. Head Office: 12, Masilamani Street, T. Nagar, Chennai-600 017 Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016 Ph.: 2653 7392/93/94/95 Fax: 2653 7396
  • 21. IITJEE 2009 SOLUTIONS 42 Brilliant Tutorials Pvt. Ltd. Head Office: 12, Masilamani Street, T. Nagar, Chennai-600 017 Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016 Ph.: 2653 7392/93/94/95 Fax: 2653 7396
  • 22. IITJEE 2009 SOLUTIONS 43 BREAK UP 2 ( XI-XII) BREAK UP 2 ( XI-XII) Brilliant Tutorials Pvt. Ltd. Head Office: 12, Masilamani Street, T. Nagar, Chennai-600 017 Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016 Ph.: 2653 7392/93/94/95 Fax: 2653 7396
  • 23. IITJEE 2009 SOLUTIONS 44 BREAK UP 2 ( XI-XII) Brilliant Tutorials Pvt. Ltd. Head Office: 12, Masilamani Street, T. Nagar, Chennai-600 017 Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016 Ph.: 2653 7392/93/94/95 Fax: 2653 7396
  • 24. IITJEE 2009 SOLUTIONS 45 Brilliant Tutorials Pvt. Ltd. Head Office: 12, Masilamani Street, T. Nagar, Chennai-600 017 Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016 Ph.: 2653 7392/93/94/95 Fax: 2653 7396
  • 25. IITJEE 2009 SOLUTIONS 46 BREAK UP 3 (TOPICWISE/PARTWISE) Brilliant Tutorials Pvt. Ltd. Head Office: 12, Masilamani Street, T. Nagar, Chennai-600 017 Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016 Ph.: 2653 7392/93/94/95 Fax: 2653 7396
  • 26. IITJEE 2009 SOLUTIONS 47 BREAK UP 3 (TOPICWISE/PARTWISE) Brilliant Tutorials Pvt. Ltd. Head Office: 12, Masilamani Street, T. Nagar, Chennai-600 017 Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016 Ph.: 2653 7392/93/94/95 Fax: 2653 7396
  • 27. IITJEE 2009 SOLUTIONS 48 BREAK UP 3 (TOPICWISE/PARTWISE) Brilliant Tutorials Pvt. Ltd. Head Office: 12, Masilamani Street, T. Nagar, Chennai-600 017 Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016 Ph.: 2653 7392/93/94/95 Fax: 2653 7396

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