IIT JEE 2010 Solution Paper 1 English

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IIT JEE 2010 Solution Paper 1 English by Resonance

IIT JEE 2010 Solution Paper 1 English by Resonance

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  • 1. Date : 11-04-2010 Duration : 3 Hours Max. Marks : 252 PAPER - 1 Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose. INSTRUCTIONS A. General : 1. This Question Paper contains 32 pages having 84 questions. 2. The question paper CODE is printed on the right hand top corner of this sheet and also on the back page of this booklet. 3. 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. 5. The answer sheet, a machine-gradable Objective Response Sheet (ORS), is provided separately. 6. Do not Tamper / mutilate the ORS or this booklet. 7. Do not break the seals of the question-paper booklet before instructed to do so by the invigilators. B. Filling the bottom-half of the ORS : 8. The ORS has CODE printed on its lower and upper Parts. 9. Make sure the CODE on the ORS is the same as that on this booklet. If the Codes do not match, ask for a change of the Booklet. 10. Write your Registration No., Name and Name of centre and sig with pen in appropriate boxes. Do not write these anywhere else. 11. Darken the appropriate bubbles below your registration number with HB Pencil. C. Question paper format and Marking scheme : 12. The question paper consists of 3 parts (Chemistry, Mathematics and Physics). Each part consists of four Sections. 13. For each question is Section–I, 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) will be awarded. 14. For each question is Section–II, you will be awarded 3 marks if you have darken only the bubble corresponding to the correct answer and zero mark if no bubbles are darkened. Partial marks will be answered for partially correct answers. No negative marks will be awarded in this Section. 15. For each question is Section–III, you will be awarded 3 marks if you have darken only the bubble corresponding to the correct answer and zero mark if no bubbles are darkened. In all other cases, minus one (–1) will be awarded. 16. For each question is Section–IV, you will be awarded 3 marks if you darken the bubble corresponding to the correct answer and zero mark if no bubbles is darkened. No negative marks will be awarded for in this Section.
  • 2. Useful Data : Atomic Numbers : Be 4; N 7; O 8; Al 13 ; Si 14; Cr 24 ; Fe 26; Fe 26; Zn 30; Br 35. 1 amu = 1.66 × 10–27 kg R = 0.082 L-atm K–1 mol–1 h = 6.626 × 10–34 J s NA = 6.022 × 1023 me = 9.1 × 10–31 kg e = 1.6 × 10–19 C c = 3.0 × 108 m s–1 F = 96500 C mol–1 RH = 2.18 × 10–18 J 40 = 1.11 × 10–10 J–1 C2 m–1 RESONANCE Page # 2
  • 3. PART- I CHEMISTRY SECTION - I Single Correct Choice Type This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct. 1. The species which by definition has ZERO standard molar enthalpy of formation at 298 K is : (A) Br2(g) (B) Cl2(g) (C) H2O(g) (D) CH4(g) Ans. (B) Sol. Hf° (Cl2,g) = 0, As Hf° of elements in their standard state is taken to be zero. 2. The bond energy (in kcal mol–1) of a C–C single bond is approximately : (A) 1 (B) 10 (C) 100 (D) 1000 Ans. (C) Sol. EC – C  100 KCal/mole. 3. The correct structure of ethylenediaminetetraacetic acid (EDTA) is : (A) (B) (C) (D) Ans. (C) RESONANCE Page # 3
  • 4. Sol. Structure of EDTA is : CHEMISTRY 4. The ionization isomer of [Cr(H2O)4Cl (NO2)]Cl is : (A) [Cr(H2O)4(O2N)]Cl2 (B) [Cr(H2O)4Cl2](NO2) (C) [Cr(H2O)4Cl(ONO)]Cl (D) [Cr(H2O)4Cl2(NO2)].H2O Ans. (B) Sol. The ionisation isomer for the given compound will be obtained by exchanging ligand with counter ion as : [Co(H2O)4Cl2](NO2). 5. The synthesis of 3-octyne is achieved by adding a bromoalkane into a mixture of sodium amide and an alkyne. The bromoalkane and alkyne respectively are : (A) BrCH2CH2CH2CH2CH3 and CH3CH2C CH (B) BrCH2CH2CH3 and CH3CH2CH2C CH (C) BrCH2CH2CH2CH2CH3 and CH3C CH (D) BrCH2CH2CH2CH3 and CH3CH2C CH Ans. (D) NaNH CH3 – CH – CH 2 – CH 2 – Br Sol. CH3–CH2–C  C–H  2  CH3–CH2–C  C    2         CH3–CH2–C  C–CH2–CH2–CH2–CH3 3–octyne 6. The correct statement about the following disaccharide is : (A) Ring (a) is pyranose with -glycosidic link (B) Ring (a) is furanose with -glycosidic link (C) Ring (b) is furanose with -glycosidic link (D) Ring (b) is pyranose with -glycosidic link Ans. (A) Sol. Ring (a) is pyranose with -glycosidic linkage and ring (b) is furanose with -glycosidic linkage. RESONANCE Page # 4
  • 5. CHEMISTRY HBr 7. In the reaction OCH3   the products are :  (A) Br OCH3 and H2 (B) Br and CH3Br (C) Br and CH3OH (D) OH and CH3Br Ans. (D) Sol. 8. Plots showing the variation of the rate constant (k) with temperature (T) are given below. The plot that follows Arrhenius equation is : (A) (B) (C) (D) Ans. (A) Sol. k = Ae–Ea/RT So, variation will be RESONANCE Page # 5
  • 6. SECTION - II CHEMISTRY Multiple Correct Correct Type This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE OR MORE may be correct. 9. Aqueous solutions of HNO3, KOH, CH3COOH, and CH3COONa of identical concentrations are provided. The pair (s) of solutions which form a buffer upon mixing is (are) : (A) HNO3 and CH3COOH (B) KOH and CH3COONa (C) HNO3 and CH3COONa (D) CH3COOH and CH3COONa Ans. (C,D) Sol. (C) HNO3 + CH3COONa mixture can act as buffer solution if volume of HNO3 solution taken is lesser than volume of CH3COONa solution because of following reaction : CH3COONa + HNO3  CH3COOH + NaNO3 (D) CH3COOH + CH3COONa - mixture will act as buffer. 10. Among the following, the intensive property is (properties are) : (A) molar conductivity (B) electromotive force (C) resistance (D) heat capacity Ans. (A,B) Sol. Molar conductivity and electromotive force (emf) are intensive properties as these are size independent. 11. The reagent (s) used for softening the temporary hardness of water is (are) : (A) Ca3(PO4)2 (B) Ca (OH)2 (C) Na2CO3 (D) NaOCl Ans. (B,C) Sol. (B) HCO3– + OH–  H2O + CO32– and CaCO3 will get precipitated. (C) CO32– + Ca2+  CaCO3, CaCO3 (white) precipitated. RESONANCE Page # 6
  • 7. CHEMISTRY 12. In the reaction the intermediate (s) is (are) : (A) (B) (C) (D) Ans. (A,C) Sol. NaOH     It is ortho-para directing. 13. In the Newman projection for 2, 2-dimethylbutane X and Y can respectively be : (A) H and H (B) H and C2H5 (C) C2H5 and H (D) CH3 and CH3 Ans. (B,D) CH3 CH3 | | Sol.  CH3 – C – CH2 – CH3 ,  CH3 – C – C 2H5 | | CH3 CH3 RESONANCE Page # 7
  • 8. SECTION - III CHEMISTRY Comprehension Type This section contains 2 Paragraphs. Based upon the first paragraph 3 multiple choice questions and based upon the second paragraph 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 Question Nos. 14 to 16 Copper is the most noble of the first row transition metals and occurs in small deposits in several countries, Ores of copper include chalcanthite (CuSO4.5H2O), atacamite (Cu2Cl(OH)3), cuprite (Cu2O), copper glance (Cu2S) and malachite (Cu2(OH)2CO3). However, 80% of the world copper production comes from the ore chalcopyrite (CuFeS2). The extraction of copper from chalcopyrite involves partial roasting, removal of iron and self-reduction. 14. Partial roasting of Chalcopyrite produces : (A) Cu2S and FeO (B) Cu2O and FeO (C) CuS and Fe2O2 (D) Cu2O and Fe2O2 Ans. (A) Sol. 2 CuFeS2 + 4O2  Cu2S + 2FeO + 3SO2 15. Iron is removed from chalcopyrite as : (A) FeO (B) FeS (C) Fe2O3 (D) FeSiO3 Ans. (D) Sol. Iron is removed in the form of slag of FeSiO3 FeO + SiO2 FeSiO3 16. In self-reduction, the reduction species is : (A) S (B) O2– (C) S2– (D) SO2 Ans. (C) Sol. S2– acts as reducing species in self reduction reaction 2Cu2O + Cu2S 6Cu + SO2 RESONANCE Page # 8
  • 9. Paragraph for Question Nos. 17 to 18 CHEMISTRY The concentration of potassium ions inside a biological cell is at least twenty times higher than the outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. A simple model for such a concentration cell involving a metal M is : M(s) | M+ (aq ; 0.05 molar) || M+ (aq ; 1 molar) | M(s) For the above electrolytic cell the magnitude of the cell potential | Ecell | = 70 mV. 17. For the above cell : (A) Ecell < 0 ; G > 0 (B) Ecell > 0 ; G < 0 (C) Ecell < 0 ; Gº > 0 (D) Ecell > 0 ; Gº < 0 Ans. (B) Sol. M (s) | M+ (aq, 0.05 M) || M+ (aq, 1 M) | M(s) Anode : M (s)  M+ (aq) + e– Cathode : M+ (aq) + e–  M (s) _____________________________________ M+ (aq) |c M+ (aq) |a M (aq) |a Ecell = E°cell – 0.0591 log 1 M (aq) |c 0.0591  0.05  =0– log 1  1    = + ve = 70 mV and hence G = – nFEcell = – ve. 18. If the 0.05 molar solution of M+ is replaced by a 0.0025 molar M+ solution, then the magnitude of the cell potential would be : (A) 35 mV (B) 70 mV (C) 140 mV (D) 700 mV Ans. (C)  0.0591  0.0025  0.0591  0.05  Sol. Ecell = log   =  log  1  1  1  20   0.0591 = 70 mV + log 20 = 140 mV. . 1 RESONANCE Page # 9
  • 10. SECTION - IV CHEMISTRY (Integer Type) This section contains TEN questions. The asnwer to each question is a single-digit integer, ranging from 0 to 9. The correct digit below the question number in the ORS is to be bubbled. 19. The concentration of R in the reaction R  P was measured as a function of time and the following data is obtained : [R](molar ) 1.0 0.75 0.40 0.10 t(min.) 0.0 0.05 0.12 0.18 The order of the reaction is : Ans. 0 C0  C 1 0.75 0.25 Sol. K= = = =5 t 0.05 0.05 0.75  0.40 0.35 K= = =5 0.07 0.07 So, reaction must be of zero order. 235 142 90 20. The number of neutrons emitted when 92 U undergoes controlled nuclear fission to 54 Xe and 38 Sr is : Ans. 3 Sol. 21. The total number of basic groups in the following form of lysine is : Ans. 2 Sol. – NH2 and  COO are two basic groups. RESONANCE Page # 10
  • 11. CHEMISTRY 22. The total number of cyclic isomers possible for a hydrocarbon with the molecular formula C4H6 is : Ans. 5 Sol. , , , , 23. In the scheme given below, the total number of intramolecular aldol condensation products formed from 'Y' is: 1. O 3 Y NaOH() 1.   aq 2. Zn, H2O 2. heat   Ans. 1 31. O NaOH Sol. 2. Zn / H2O       24. Amongst the following, the total number of compounds soluble in aqueous NaOH is : Ans. 5 RESONANCE Page # 11
  • 12. CHEMISTRY Sol. , , , & are soluble in aqueous NaOH. 25. Amongst the following, the total number of compounds whose aqueous solution turns red litmus paper blue is : KCN K2SO4 (NH4)2C2O4 NaCl Zn(NO3)2 FeCl3 K2CO3 NH4NO3 LiCN Ans. 3 Sol. Basic solutions will convert red litmus blue. , their aqueous solution will be basic due to anionic hydrolysis. 26. Based on VSEPR theory, the number of 90 degree F–Br–F angles is BrF5 is : Ans. 8 Sol. 27. The value of n in the molecular formula BenAl2Si6O18 is : Ans. 3 Sol. BenAl2Si6O18. The value of n = 3 by charge balancing. (Be2+)3 (Al3+)2 (Si6O18)12– [2 × n + 2 (+3) + (–12) = 0  n = 3]. 28. A student performs a titration with different burettes and finds titre values of 25.2 mL, 25.25 mL, and 25.0 mL. The number of significant figures in the average titre value is : Ans. 3 25 .2  25.25  25.0 75.45 Sol. Average titre value = = = 25.15 = 25.2 mL 3 3 number of significant figures will be 3. RESONANCE Page # 12
  • 13. PART-II MATHEMATICS SECTION - I Single Correct Choice Type This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct. 29. Let p and q be real numbers such that p  0, p3  q and p3  – q. If  and  are nonzero complex numbers satisfying  +  = – p and 3 +  3 = q, then a quadratic equation having and as its roots is     (A) (p3 + q) x2 – (p3 + 2q)x + (p3 + q) = 0 (B) (p3 + q) x2 – (p3 – 2q)x + (p3 + q) = 0 (C) (p3 – q) x2 – (5p3 – 2q)x + (p3 – q) = 0 (D) (p3 – q) x2 – (5p3 + 2q)x + (p3 – q) = 0 Ans. (B) Sol. Product = 1 2  2 (   ) 2  2 Sum = =   Since 3 +  3 = q  – p (2 +  2 – ) = q q q (( + )2 – 3 ) = – p2 + = 3 p p   2 2  p 3  q  p    3p p3  2q  3  p   Hence sum =  = 3    (p 3  q) p q  p3  2q  so the equation is x2 +  3  x+1=0  p q     (p3 + q) x2 – (p3 – 2q)x + (p3 + q) = 0 2 2 30. Let f, g and h be real-valued functions defined on the interval [0, 1] by f(x) = e x + e–x , 2 2 2 2 g(x) = xe x + e – x and h(x) = x 2 e x + e – x . If a, b and c denote, respectively, the absolute maximum of f, g and h on [0, 1], then (A) a = b and c  b (B) a = c and a  b (C) a  b and c  d (D) a = b = c Ans. (D) RESONANCE Page # 13
  • 14. Sol. Clearly f(x) = e x 2 + e–x 2 MATHEMATICS 2 2 1 f(x) = 2x ( e x – e – x )  0 increasing fmax = f(1) = e + e  2 2 2 2 2 1 g(x) = x e x + e – x  g(x) = e x + 2x2 e x – 2x e – x > 0 increasing  gmax = g(1) = e + e 2 2 2 2 2  2 2 2  h(x) = x2 e x + e – x h(x) = 2x e x + 2x3 e x – 2x e – x = 2x  e x  x 2 e x  e  x  > 0   1 hmax = h(1) = e + , e  so a = b = c 31. If the angle A, B and C of a triangle are in arithmetic progression and if a, b and c denote the lengths of the a c sides opposite to A, B and C respectively, then the value of the expression sin 2C + sin 2A is c a 1 3 (A) (B) (C) 1 (D) 3 2 2 Ans. (D) a c 2 b Sol. sin 2C + sin 2A = (a cos C + c cos A) = = 2 sin B = 2 sin 60º = 3 c a 2R R x y z 32. Equation of the plane containing the straight line = = and perpendicular to the plane containing the 2 3 4 x y z x y z straight line = = and = = is 3 4 2 4 2 3 (A) x + 2y – 2z = 0 (B) 3x + 2y – 2z = 0 (C) x – 2y + z = 0 (D) 5x + 2y – 4z = 0 Ans. (C) Sol. Direction ratio of normal to plane containing the straight line i ˆ j ˆ k ˆ 3 4 2 = 8 ˆ – ˆ – 10 k i j 4 2 3 ˆ x 0 y 0 z0 Required plane 2 3 4 =0  – 26x + 52y – 26z = 0  x – 2y + z = 0 8 1  10 RESONANCE Page # 14
  • 15. MATHEMATICS 33. Let  be a complex cube root of unity with  1. A fair die is thrown three times. If r1, r2 and r3 are the numbers obtained on the die, then the probability that r1  r2  r3 = 0 is 1 1 2 1 (A) (B) (C) (D) 18 9 9 36 Ans. (C) Sol. r1  r2  r3 = 0 ; r1, r2, r3 are to be selected form {1, 2, 3, 4, 5, 6} As we know that 1 +  + 2 = 0  form r1, r2, r3 , one has remainder 1, other has remainder 2 and third has remainder 0 when divided by 3.  we have to select r1, r2, r3 form (1, 4) or (2, 5) or (3, 6) which can be done in 2C1 × 2C1 × 2C1 ways value of r1, r2, r3 can be interchanged in 3! ways. ( 2 C1 2 C1 2 C1 )  3! 2 required probability = = 666 9  34. i j i i j i j Let P, Q, R and S be the points on the plane with position vectors – 2 ˆ – ˆ , 4 ˆ , 3 ˆ + 3 ˆ and – 3 ˆ + 2 ˆ respectively. The quadrilateral PQRS must be a (A) parallelogram, which is neither a rhombus nor a rectangle (B) square (C) rectangle, but not a square (D) rhombus, but not a square Ans. (A) Sol. PQ = 36  1 = 37 = RS, PQ  PS PS = 1 9 = 10 = QR 1 slope of PQ = , slope of PS = – 3 6 PQ is not  to PS So it is parallelogram, which is neither a rhombus nor a rectangle x 1 35. The number of 3 × 3 matrices A whose entries are either 0 or 1 and for which the system A  y  = 0 has     z   0   exactly two distinct solutions, is (A) 0 (B) 29 – 1 (C) 168 (D) 2 Ans. (A) RESONANCE Page # 15
  • 16. Sol. a1x + b1y + c1z = 1 MATHEMATICS a2x + b2y + c2z = 0 a3x + b3y + c3z = 0 No three planes can meet at two distinct points. So number of matrices is 0 x 1 tn (1  t ) 36. The value of lim dt is x 0 x3 0  t4  4 1 1 1 (A) 0 (B) (C) (D) 12 24 64 Ans. (B) xn (1  x ) 1 1 1 Sol. lim 4 2 = lim = x 0 ( x  4)  3 x x 0 4 3 12 × SECTION - II Multiple Correct ChoiceType This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE OR MORE may be correct. 1 x 4 (1  x ) 4 37. The value(s) of dx is (are) 0  1 x 2 22 2 71 3 (A) (B) (C) 0 (D) 7 105 15 2 – – Ans. (A) 1 1 1 x 4 (1  x ) 4 x 4 [(1  x 2 )  2 x ] 2 x 4 [(1  x 2 )2  4 x(1  x 2 )  4 x 2 ] Sol. = = 1 x 2 0 1 x2 0 1 x2 0    1 4x 2  4x 6  x 4 (1  x 2 )  4 x   6 4 5  dx = x  x  4x   dx  = 0    1 x2      1 x2   Now on polynomial division of x6 by 1 + x2 , we obtain 6 1  4  x  x 4  4 x 5  4 ( x 4  x 2  1)    dx = 6  4x 5  5x 4  4x 2  4  =  x  dx     1  x 2  1 x 2     RESONANCE Page # 16
  • 17. MATHEMATICS 1  x 7 4 x 6 5 . x 5 4x 3 x 1 1 4 4  1 12 =  7  6  5  3  4 x   4 tan x 0 =    1   4  – 4   =    5 –      0 7 6 3 4 7 6        1 22 =   3 –  =  7 7 –  x 38. Let f be a real-valued function defined on the interval (0, ) by f(x) = n x + 1  sin t dt. Then which of the 0  following statement(s) is (are) true? (A) f(x) exists for all x  (0, ) (B) f(x) exists for all x  (0, ) and f is continuous on (0, ), but not differentiable on (0, ) (C) there exists  > 1 such that |f(x)| < |f(x)| for all x  (, ) (D) there exists  > 1 such that |f(x)| + |f(x)|   for all x  (0, ) Ans. (B, C) 2 Sol. f(x) = n x + 1  sin t dt 0  1 f(x) = + 1 sin x x 1 cos x f(x) = – 2 + x 2 1  sin x (A) f is not defined for x = – + 2n , n   2 so (A) is wrong (B) f(x) always exist for x > 0 (C) | f| < | f | Since f > 0 and f>0 f < f x 1 + 1 sin x < n x + 1  sin x dx x 0  LHS is bounded RHS is Increasing with range  So there exist some  beyond which RHS is greater than LHS (D) | f | + | f|  b is wrong as f is M & its range is not bound while  is finite RESONANCE Page # 17
  • 18. MATHEMATICS 39. Let A and B be two distinct points on the parabola y2 = 4x. If the axis of the parabola touches a circle of radius r having AB as its diameter, then the slope of the line joining A and B can be 1 1 2 2 (A) – (B) (C) (D) – r r r r Ans. (C, D) Sol. A(t12, 2t1), B (t22 , 2t2)  t 2  t 22 centre of circle  1 , t1  t 2     2   2 2 | t1 + t2 | = r , slope of AB = t  t = ± 1 2 r  40. Let ABC be a rectangle such that ACB = and let a, b and c denote the lengths of the sides opposite to  6 A, B and C respectively. The value(s) of x for which a = x2 + x + 1, b = x2 – 1 and c = 2x + 1 is (are) (A) – 2  3   (B) 1 + 3 (C) 2 + 3 (D) 4 3 Ans. (A, B) ( x 2  1)2  ( x 2  x  1)2  (2x  1)2 Sol. cos =  6 2( x 2  x  1)( x 2  1) 3 ( x 2  1)2  ( x 2  3 x  2)( x 2  x ) = 2 2( x 2  x  1)( x 2  1) 3 ( x 2  1)2  ( x  1)( x  2)x( x  1) = 2 2( x 2  x  1)( x 2  1) x 2  1  x( x  2) 3 = 3 (x2 + x + 1) = 2x2 + 2x – 1 x2  x  1    ( 3 – 2) x2 + ( 3 – 2) x + ( 3 + 1) = 0 on solving  x 2 + x – (3 3  5 ) = 0 x= 3 + 1, – (2  3 ) RESONANCE Page # 18
  • 19. MATHEMATICS 41. Let z1 and z2 be two distinct complex numbers and let z = (1 – t) z1 + tz2 for some real number t with 0 < t < 1. If Arg(w) denotes the principal argument of a nonzero complex number w, then (A) |z – z1| + |z – z2| = |z1 – z2| (B) Arg (z – z1) = Arg (z – z2) z  z1 z  z1 (C) z  z z  z = 0 (D) Arg (z – z1) = Arg (z2 – z1) 2 1 2 1 Ans. (A, C, D) Sol. (A) |z – z1| + |z – z2| = |z1 – z2| AB + BC = AC (B) Arg (z – z1) – Arg (z – z2) =  z z 1 z1 z1 1 (C) =0 z2 z2 1 z  z1 z  z1 0 z1  z 2 z1  z 2 0 =0 z2 z2 1 z  z1 z  z1 z1  z 2 z1  z 2 =0 (D) Arg (z – z1) = Arg (z2 – z1) SECTION - III Comprehension Type This section contains 2 Paragraphs. Based upon the first paragraph 3 multiple choice questions and based upon the second paragraph 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. RESONANCE Page # 19
  • 20. MATHEMATICS Paragraph for Question Nos. 42 to 44 Let p be an odd prime number and Tp be the following set of 2 × 2 matrices : a b Tp = A  c a : a, b, c  {0, 1, 2,....., p – 1}            42. The number of A in Tp such that A is either symmetric or skew-symmetric or both, and det (A) divisible by p is (A) (p – 1)2 (B) 2 (p – 1) (C) (p – 1)2 + 1 (D) 2p – 1 Ans. (D) a b  Sol. A = c a where a, b, c  {0,1,2,....p – 1}   Case- I A is symmetric matrix  b=c  det(A) = a2 – b2 is divisible by p  (a – b) (a + b) is divisible by p (a) a – b is divisible by p if a = b, then 'p' cases are possible (p – 1) (b) a + b is divisible by p if a + b = p, then 2 × 2 = (p – 1) cases are possible Case - II A is skew symmetric matrix if a = 0, b + c = 0, then det (A) = b2  b2 can never be divisible by p. So No case is possible Total number of A is possible = 2p – 1 43. The number of A in Tp such that the trace of A is not divisible by p but det (A) is divisible by p is [Note : The trace of matrix is the sum of its diagonal entries.] (A) (p – 1)(p2 – p + 1) (B) p3 – (p – 1)2 (C) (p – 1)2 (D) (p – 1)(p2 – 2) Ans. (C) Sol. a2 – bc  p a can be chosen in p – 1 ways (a  0) Let a be 4 & p = 5 so a2 = 16 and hence bc should be chosen such that a2 – bc  p Now b can be chosen in p – 1 ways and c in only 1 (one be is chosen) RESONANCE Page # 20
  • 21. Explanation : If b=1c=1 MATHEMATICS b=2c=3 b=3c=2 b=4c=4 Hence a can be chosen in p – 1 ways and then b can be chosen in p – 1 ways c can be chosen in 1 ways so (p – 1)2 44. The number of A in Tp such that det (A) is not divisible by p is (A) 2p2 (B) p3 – 5p (C) p3 – 3p (D) p3 – p2 Ans. (D) Sol. As = Total cases – (a  0 and |A| is divisible by p) – (a = 0 and |A| is divisible by p) = p3 – (p –1)2 – (2p – 1) = p3 – p2 – bc  p since b & c both we coprime to p  one of them must be zero. If b = 0  c can be chosen in {0,1,....p – 1} If c = 0  b can be chosen in {0,1,....p – 1} Paragraph for Question Nos. 45 to 46 x2 y2 The circle x2 + y2 – 8x = 0 and hyperbola = 1 intersect at the points A and B. 9 4 – 45. Equation of a common tangent with positive slope to the circle as well as to the hyperbola is (A) 2x – 5 y – 20 = 0 (B) 2x – 5y+4=0 (C) 3x – 4y + 8 = 0 (D) 4x – 3y + 4 = 0 Ans. (B) Sol. Let equation of tangent to ellipse sec  tan  x– y 1 3 2 2sec x – 3 tan y = 6 It is also tangent to circle x2 + y2 – 8x = 0 8 sec  – 6 =4 4 sec 2   9 tan2   (8sec – 6)2 = 16 (13sec2 – 9)  12sec2 + 8sec – 15 = 0 RESONANCE Page # 21
  • 22. 5 3 5 MATHEMATICS  sec = and – but sec  6 2 6 3 5  sec = –  tan =  slope is positive 2 2 Equation of tangent = 2x – 5y+4=0 46. Equation of the circle with AB as its diameter is (A) x2 + y2 – 12x + 24 = 0 (B) x2 + y2 + 12x + 24 = 0 (C) x2 + y2 + 24x – 12 = 0 (D) x2 + y2 – 24x – 12 = 0 Ans. (D) Sol. x2 + y2 – 8x = 0 x2 y2 =1  4x2 – 9y2 = 36 9 4 –  4x2 – 9(8x – x2) = 36 13x2 – 72x – 36 = 0 13x2 – 78x + 6x – 36 = 0 (13x + 6) (x – 6) = 0 6 x=– and x = 6 13  But x > 0  x=6  A(6, 2 ) and B (6, – 2 )  Equation of circle with AB as a diameter x2 + y2 – 12x + 24 = 0 SECTION - IV (Integer Type) This section contains TEN questions. The asnwer to each question is a single-digit integer, ranging from 0 to 9. The correct digit below the question number in the ORS is to be bubbled. RESONANCE Page # 22
  • 23. MATHEMATICS 47. The number of all possible values of , where 0 <  < , for which the system of equations (y + z) cos 3 = (xyz) sin 3 2 cos 3 2 sin 3 x sin 3 = y + z (xyz) sin 3 = (y + 2z) cos 3  + y sin 3 have a solution (x0, y0, z0) with y0 z0  0, is (y + z) cos 3 = (xyz) sin 3 2 cos 3 2 sin 3 x sin 3 = y + z (xyz) sin 3 = (y + 2z) cos 3  + y sin 3 Ans. 3 Sol. Let xyz = t t sin3 – y cos 3 – z cos3 = 0 ..........(1) t sin3 – 2y sin 3 – 2z cos3 = 0 ..........(2) t sin3 – y (cos 3 + sin3 ) – 2z cos 3 = 0 ..........(3) y0. z0  0 hence homogeneous equation has non-trivial solution sin 3 – cos 3 – cos 3 D = sin 3 – 2 cos 3 – 2 cos 3 =0 sin 3 – (cos 3  sin 3) – 2 cos 3 sin3cos3(sin 3 – cos3) = 0 sin3= 0 or cos3= 0 or tan3 = 1 Case - I sin3= 0 From equation (2) z = 0 not possible Case - II cos3 = 0, sin3 0 t. sin3 = 0 t = 0 x = 0 From equation (2) y = 0 not possible Case- III tan3= 1 3 = n+ ,nI  4  n x. y. z sin3= 0 + ,nI  3 12  =  5 9 x = 0 , sin3 0 , , 12 12 12  = Hence 3 solutions RESONANCE Page # 23
  • 24. MATHEMATICS 48. Let f be a real-valued differentiable function on R (the set of all real numbers) such that f(1) = 1. If the y-intercept of the tangent at any point P(x, y) on the curve y = f(x) is equal to the cube of the abscissa of P, then the value of f(–3) is equal to Ans. 9 Sol. Y – y = m (X – x) y-intercept (x = 0) y = y – mx dy Given that y – mx = x3 x – y = – x3 dx  dy y = – x2 dx x  – 1 dx 1 Intergrating factor e  x = – x 1 1 x3 solution y . = . (  x 2 ) dx f(x) = y = – + cx x x 2   3 Given f(1) = 1 c= 2  x3 3x f(x) = – + f(–3) = 9 2 2   n 49. The number of values of  in the interval  – 2 , 2  such that   for n = 0, ±1, ± 2 and      5 tan = cot 5 as well as sin 2 = cos 4is Ans. 3 Sol. tan = cot 5 sin  cos 5 =  cos 6= 0 cos  sin    6 = (2n + 1)  2   = (2n + 1) ; n  12 RESONANCE Page # 24
  • 25. 5   5 MATHEMATICS  = – ,– , – , , , .........(1)   12 4 12 12 4 12 sin2 = cos4  sin2 = 1 – 2 sin2 2  2sin22 + sin2– 1 = 0 1 sin2 = – 1, 2  2= (4m – 1) , p + (–1)p   2 6   p q = (4m – 1) , + (–1)p ; m, p  I  4 2 12    5 , , ...........(2) 4 12 12  = – From (1) & (2) 5   – 4 , 12 , 12       Number of solution 1 50. The maximum value of the expression is sin   3 sin  cos   5 cos 2  2 Ans. 2 1 1 2 Sol. f() = 2 2 = = sin   3 sin  cos   5 cos  1  cos 2 3 5(1  cos 2) 6  3 sin 2  4 cos 2  sin 2  2 2 2 2 f()max = =2 65  ˆ – 2ˆ i j 2ˆ  ˆ  3k i j 51. If a and b are vectors in space given by a = ˆ and b = , then the value of     5 14 2a  b. a  b a – 2b is       Ans. 5 RESONANCE Page # 25
  • 26. ˆ – 2ˆ MATHEMATICS i j 2ˆ  ˆ  3k i j ˆ Sol. = , b =  a  5 14 |a | = 1 , |b | = 1   a .b = 0   (2a  b) . (a  b)  (a – 2b)         = (2a  b) . a .(a – 2b) b – (b.(a – 2b) a            = (2a  b) .  (a. a – 2a . b ) b – (b. a – 2b. b ) a              = (2a  b) . (1 – 0) b – (0  2) a      = (2a  b) . b  2a      = 2 (a . b)  4(a . a)  b. b  2(b. a) = 0 + 4 + 1 + 0 = 5        x2 y2 52. The line 2x + y = 1 is tangent to the hyperbola = 1 . If this line passes through the point a2 b2 – of intersection of the nearest directrix and the x-axis, then the eccentricity of the hyperbola is Ans. 2 Sol. x = a/e b2 y = –2x + 1 e2 = 1 + a2 2a ( 4a 2 – 1) 0=– – +1 =1+ e a2 a 1 1 = e2 = 1 + 4 – e 2 a2  4 e = 2a e2 = 5 – e2 RESONANCE Page # 26
  • 27. MATHEMATICS c2 = a2 m2 – b2  e4 – 5e2 + 4 = 0  1 = 4a2 – b2  (e2 – 1)(e2 – 4) = 0  1 + b2 – 4a2 = 0 e2 – 1  0 e=2 53. If the distance between the plane Ax – 2y + z = d and the plane containing the lines x –1 y–2 z–3 x–2 y–3 z–4 = = and = = is 6 , then |d| is 2 3 4 3 4 5 Ans. 6 x –1 y – 2 z –3 Sol. Equation of plane is 2 3 4 =0 3 4 5 x – 2y + z = 0 ..........(1) Ax – 2y + z = d ..........(2) A –2 1 Compare = =  A = 1 1 –2 1 d Distance between planes is  6 1 1 4  |d| = 6 54. For any real number, let [x] denote the largest integer less than or equal to x. Let f be a real valued function defined on the interval [–10, 10] by  x – [ x] if [ x ] is odd, f(x) = 1  [ x ] – x if [ x ] is even  10 2 Then the value of 10  f ( x) cos x dx is –10 Ans. 4 RESONANCE Page # 27
  • 28.  { x} , 2n  1  x  2n MATHEMATICS Sol. f(x) = 1  { x } , 2n  x  2n  1  Clearly f(x) is a periodic function with period = 2 Hence f(x) . cos x is also periodic with period = 2 10 2 1 2 f ( x ) cos( x ) dx = 2 f ( x ) cos( x ) dx = 2  ((1  {x})  {x}) cos( x ) dx 10 10 0 0   1 1  x sin x cos x   2  = 22 (  x cos x ) dx = – 22  2  = – 2 2   =4  2   0 0     2 2 55. Let  be the complex number cos + i sin . Then the number of distinct complex numbers 3 3 z 1  2 z  2 1 z satisfying = 0 is equal to  2  1 z Ans. 1 2 2 Sol.  = cos + i sin 3 3 R1  R1 + R2 + R3 z  2 1  2 z z  2 1 1 z  2 1 =0  z =0 z 1 z 1 1 z z=0 1  2 0 z  2 –  1 – 2 =0 0 1–  z   – 2 (z + 2 – )(z +  – 2) – (1 – )(1 – 2 ) = 0 z2 = 0 only one solution RESONANCE Page # 28
  • 29. MATHEMATICS k –1 56. Let Sk, k = 1, 2,...., 100, denote the sum of the infinite geometric series whose first term is k ! 100 1 100 2 and the common ratio is . Then the value of k 100 ! +  (k 2 – 3k  1) Sk is k 1 Ans. 3 100 2 Sol.  (k k 2 – 3k  1) Sk for k = 2 |k2 – 3k + 1) Sk| = 1 100 k –1 k – 1 1  (k – 2)! – (k – 1)! k 3 100 1 1 1 1  (k – 3)!  (k – 2)! – (k – 2)! – (k – 1)! k 3 100  1 1    (k – 3)! – (k – 1)!  k 3     1 1 1  1 1  1 1  1 1  1 1  S = 1  1 –    –    –    –    –   .... 2!   1! 3!   2! 4!   3! 5!   4! 6!  94! 96!   –       1 1   1 1   1 1  1 1 +  95! – 97!    96! – 98!    97! – 99 !  = 2 – 98! 99!   –       100 2 1 1 E= +3– 100! 98! 99.98!  – 100 2 100 100 2 100 = +3– = +3– =3 100! 99! 100 .99! 99! RESONANCE Page # 29
  • 30. PART-III PHYSICS SECTION - I Single Correct Choice Type This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D) four choices (A), (B), (C) and (D) out of which ONLY ONE is correct. 57. A block of mass m is on inclined plane of angle . The coefficient of friction between the block and the plane is  and tan > . The block is held stationary by applying a force P parallel to the plane. The direction of force pointing up the plane is taken to be positive. As P is varied from P1 = mg(sin – cos) to P2 = mg(sin + cos), the frictional force f versus P graph will look like : (A) (B) (C) (D) Ans. (A) Sol. P1 = mgsin – mgcos P2 = mgsin + mgcos Initially block has tendency to slide down and as tan > , maximum friction mgcos will act in positive direction. When magnitude P is increased from P1 to P2, friction reverse its direction from positive to negative and becomes maximum i.e.mgcos in opposite direction. RESONANCE Page # 30
  • 31. PHYSICS 58. A thin uniform annular disc (see figure) of mass M has outer radius 4R and inner radius 3R. The work required to take a unit mass from point P on its axis to infinity is : P 4R 3R 4R 2GM 2GM GM 2GM (A) 7R 4 2 5   (B)  7R 4 2 5   (C) 4R (D) 5R  2  1 Ans. (A) Sol. W ext = U – UP Gdm  W ext = 0 – .1  – –  x  M 2rdr rdr P 2GM W ext = G    7R 2 = 16R 2  r 2 7R 2 16R 2  r 2 x  2GM zdz 2GM = = [Z] 3R 2 7R  z 7R 2 4R r dr 4R 2GM  W ext = 16R 2  r 2  7R 2    3R  2GM W ext = 4 2R – 5R 7R 2  2GM W ext = 4 2 –5 . 7R 2  59. Consider a thin square sheet of side L and thickness t, made of a material of resitivity . The resistance between two opposite faces, shown by the shaded areas in the figure is : t. L (A) directly proportional to L (B) directly proportional to t (C) independent of L (D) independent of t Ans. (C) RESONANCE Page # 31
  • 32. l PHYSICS Sol. R= A L  R= = tL t Independent of L. 60. A real gas behaves like an ideal gas if its (A) pressure and temperature are both high (B) pressure and temperature are both low (C) pressure is high and temperature is low (D) pressure is low and temperature is high Ans. (D) Sol. At low pressure and high temperature inter molecular forces become ineffective. So a real gas behaves like an ideal gas. 61. Incandescent bulbs are designed by keeping in mind that the resistance of their filament increases with increase in temperature. If at room temperature, 100 W, 60 W and 40 W bulbs have filament resistances R100, R60 and R40, respectively, the relation between these resistance is : 1 1 1 1 1 1 (A) R  R 40 R 60  (B) R100 = R40 + R60 (C) R100 > R60 > R40 (D) R  R 60 R 40  100 100 Ans. (D) V2 1 100 Sol. 100 = R'100 = V 2 R'100  where R’100 is resistance at any temperature corresponds to 100 W V2 1 60 60 = R' 60 = V 2 R' 60  V2 1 40 40 = R' 40 = V 2 R' 40  From above equations we can say 1 1 1 R'100 > R' 60 > R' 40 . So, most appropriate answer is option (D). RESONANCE Page # 32
  • 33. PHYSICS 62. To verify Ohm's law, a student is provided with a test resistor RT, a high resistance R1, a small resistance R2, two identical galvanometers G1 and G2, and a variable voltage source V. The correct circuit to carry out the experiment is : G1 G1 R2 R1 G2 G2 RT R1 RT R2 (A) (B) V V R1 R2 G1 G1 G2 G2 RT RT (C) R2 (D) R1 V V Ans. (C) Sol. To verify Ohm's law one galvaometer is used as ammeter and other galvanometer is used as voltameter. Voltameter should have high resistance and ammeter should have low resistance as voltameter is used in parallel and ammeter in series that is in option (C). 63. An AC voltage source of variable angular frequency  and fixed amplitude V connected in series with a capacitance C and an electric bulb of resistance R (inductance zero). When  is increased : (A) the bulb glows dimmer (B) the bulb glows brighter (C) total impedence of the circuit is unchanged (D) total impedence of the circuit increases Ans. (B) Vrms Sol. irms = 1 R2  2 2  c when  increases, irms increases so the bulb glows brighter RESONANCE Page # 33
  • 34. PHYSICS 64. A thin flexible wire of length L is connected to two adjacent fixed points carries a current  in the clockwise direction, as shown in the figure. When system is put in a uniform magnetic field of strength B going into the plane of paper, the wire takes the shape of a circle. The tension in the wire is : BL BL (A) BL (B) (C) (D) BL  2 4 Ans. (C) Sol.  d  d B L 2Tsin   = dlB 2T = RdlB T = BR = .  2  2 2   SECTION - II Multiple Correct Choice Type This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE OR MORE may be correct. 65. One mole of an ideal gas in initial state A undergoes a cyclic process ABCA, as shown in the figure. Its pressure at A is P0. Choose the correct option(s) from the following : (A) Internal energies at A and B are the same (B) Work done by the gas in process AB is P0V0 n 4 P0 T0 (C) Pressure at C is (D) Temperature at C is 4 4 Ans. (A, B) RESONANCE Page # 34
  • 35. PHYSICS Sol. f U= nRT, , where f,n,R are constants. Also temperature T is same at A & B. 2   UA = UB  Vf 4 V0 Also, W AB = nRT0 n  V  = nRT n V0 = nRT0 n4 = P0V0 n4  0  i    So, answers are (A) & (B). 66. A ray OP of monochromatic light is incident on the face AB of prism ABCD near vertex B at an incident angle of 60º (see figure). If the refractive index of the material of the prism is 3 , which of the following is (are) correct ? (A) The ray gets totally internally reflected at face CD (B) The ray comes out through face AD (C) The angle between the incident ray and the emergent ray is 90º (D) The angle between the incident ray and the emergent ray is 120º Ans. (A, B, C) RESONANCE Page # 35
  • 36. PHYSICS 45º Q Sol. 60º By refraction at face AB : 1.sin60º = 3 . sinr1 So, r1 = 30º This shows that the refracted ray is parallel to side BC of prism. For side 'CD' angle of incidence will be 45º, which can be calculated from quadrilateral PBCQ. By refraction at face CD : 3 sin45º = 1 sinr2 3 So, sin r2 = 2 which is impossible. So, there will be T.I.R. at face CD. Now, by geometry angle of incidence at AD will be 30º. So, angle of emergence will be 60º. Hence, angle between incident and emergent beams is 90º 67. A few electric field lines for a system of two charges Q1 and Q2 fixed at two different points on the x-axis are shown in the figure. These lines suggest that : (A) |Q1| > |Q2| (B) |Q1 | < |Q2| (C) at a finite distance to the left of Q1 the electric field is zero (D) at a finite distance to the right of Q2 the electric field is zero Ans. (A, D) RESONANCE Page # 36
  • 37. PHYSICS Q1 Sol. Q2 From the diagram, it can be observed that Q1 is positive, Q2 is negative. No. of lines on Q1 is greater and number of lines is directly proportional to magnitude of charge. So, |Q1| > |Q2| Electric field will be zero to the right of Q2 as it has small magnitude & opposite sign to that of Q1 . 68. A student uses a simple pendulum of exactly 1m length to determine g, the acceleration due to gravity. He uses a stop watch with the least count of 1sec for this and records 40 seconds for 20 oscillations. For this observation, which of the following statement(s) is (are) true ? (A) Error T in measuring T, the time period, is 0.05 seconds (B) Error T in measuring T, the time period, is 1 second (C) Percentage error in the determination of g is 5% (D) Percentage error in the determination of g is 2.5% Ans. (A, C) t 40 Sol. Since, t = nT. So, T= = or T = 2 sec. n 20 Now, t = n T t T and t T = 1 T So, = T = 0.05 40 2  T = 2 g  Time period, T 1 g g T So, = 2 g or 2 T g = T  g So, percentage error in g = g × 100 T 0.05 = 2 = – 2 T 2 × 100 × 100 = 5%. RESONANCE Page # 37
  • 38. PHYSICS 69. A point mass of 1kg collides elastically with a stationary point mass of 5 kg. After their collision, the 1 kg mass reverses its direction and moves with a speed of 2 ms–1. Which of the following statement(s) is (are) correct for the system of these two masses ? (A) Total momentum of the system is 3 kg ms–1 (B) Momentum of 5 kg mass after collision is 4 kg ms–1 (C) Kinetic energy of the centre of mass is 0.75 J (D) Total kinetic energy of the system is 4 J Ans. (A, C) Sol. Since collision is elastic, so e = 1 Velocity of approach = velocity of separation So, u=v+2 .............(i) By momentum conservation : 1 × u = 5v – 1 × 2 u = 5v – 2 v + 2 = 5v – 2 So, v = 1 m/s and u = 3 m/s Momentum of system = 1 × 3 = 3 kgm/s Momentum of 5kg after collision = 5 × 1 = 5 kgm/s 2 1  m1u  So, kinetic energy of centre of mass = (m + m2) 2 1 m m    1 2   2 1  1 3  = (1 + 5)  2  6   = 0.75 J 1 Total kinetic energy = × 1 × 32 2 = 4.5 J. RESONANCE Page # 38
  • 39. PHYSICS SECTION–III Paragraph Type This section contains 2 paragraphs. Based upon the first paragraph 3 multiple choice questions and based upon the second paragraph 2 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 Question Nos. 70 to 72 When a particle of mass m moves on the x-axis in a potential of the form V(x) = kx2, it performs simple m harmonic motion. The corresponding time period is proportional to , as can be seen easily using k dimensional analysis. However, the motion of a particle can be periodic even when its potential energy increases on both sides of x = 0 in a way different from kx2 and its total energy is such that the particle does not escape to infinity. Consider a particle of mass m moving on the x-axis. Its potential energy is V(x) = x4 ( > 0) for |x| near the origin and becomes a constant equal to V0 for |x|  X0 (see figure) 70. If the total energy of the particle is E, it will perform periodic motion only if : (A) E < 0 (B) E > 0 (C) V0 > E > 0 (D) E > V0 Ans. (C) Sol. When 0 < E < V0 there will be acting a restoring force to perform oscillation because in this case particle will be in the region |x|  x0 . 71. For periodic motion of small amplitude A, the time period T of this particle is proportional to : m 1 m 1 (A) A (B) (C) A (D)    A  m A m Ans. (B) Sol. V = x4 1 T.E. = m2A2 = A4 (not strictly applicable just for dimension matching it is used) 2 2A 2 1 m 2 = T m A   RESONANCE Page # 39
  • 40. 72. The acceleration of this particle for |x| > X0 is : PHYSICS V0 (A) proportional to V0 (B) proportional to mX 0 V0 (C) proportional to mX 0 (D) zero Ans. (D) dU Sol. F=  dx as for |x| > x0 V = V0 = constant dU =0 dx   F = 0. Paragraph for Question Nos. 73 to 74 Electrical resistance of certain materials, known as superconductors, changes abruptly from a nonzero value TC (B) to zero as their temperature is lowered below a critical TC (0) temperature T C (0). An interesting property of superconductors is that their critical temperature becomes smaller than TC (0) if they are placed in a magnetic field, i.e., the critical temperature TC (B) is a function of the O B magnetic field strength B. The dependence of TC (B) on B is shown in the figure. 73. In the graphs below, the resistance R of a superconductor is shown as a function of its temperature T for two different magnetic fields B1 (solid line) and B2 (dashed line). If B2 is larger than B1, which of the following graphs shows the correct variation of R with T in these fields? R R B2 B1 (A) B2 B1 (B) O T O T R R B1 B1 B2 B2 (C) (D) O T O T Ans. (A) RESONANCE Page # 40
  • 41. PHYSICS Sol. As the magnetic field is greater, the critical temperature is lower and as B2 is larger than B1. Graph ‘A’ is correct. 74. A superconductor has TC (0) = 100 K. When a magnetic field of 7.5 Tesla is applied, its TC decreases to 75 K. For this material one can definitely say that when : (A) B = 5 Tesla, TC (B) = 80 K (B) B = 5 Tesla, 75 K < TC (B) < 100 K (C) B = 10 Tesla, 75 K < TC (B) < 100 K (D) B = 10 Tesla, TC (B) = 70 K Ans. (B) Sol. For B=0 TC = 100 k B = 7.5 T TC = 75 k For B = 5T 75 < TC < 100 SECTION–IV Integer Type This Section contains 10 questions. The answer to each question is a single-digit integer, ranging from 0 to 9. The correct digit below the question number in the ORS is to be bubbled. 75. A piece of ice (heat capacity = 2100 J kg–1 ºC–1 and latent heat = 3.36 × 105 J kg–1) of mass m grams is at –5 ºC at atmospheric pressure. It is given 420 J of heat so that the ice starts melting. Finally when the ice- water mixture is in equilibrium, it is found that 1 gm of ice has melted. Assuming there is no other heat exchange in the process, the value of m is : Ans. 8 gm Sol. S = 2100 J kg–1 ºC–1 L = 3.36 × 105 J kg–1 420 = m S Q + (1) × 10–3 × L 420 = m s(5) + 3.36 × 102 420 – 336 = m(2100) × 5 1 m= 125 × 1000 = 8 gm. 76. A stationary source is emitting sound at a fixed frequency f0, which is reflected by two cars approaching the source. The difference between the frequencies of sound reflected from the cars is 1.2% of f0. What is the difference in the speeds of the cars (in km per hour) to the nearest integer ? The cars are moving at constant speeds much smaller than the speed of sound which is 330 ms–1. Ans. 7 Sol. Let speed of cars are V1 and V2  v  frequency received by car f1 =  v  v  f0 1    RESONANCE Page # 41
  • 42.  v  PHYSICS  v  v1  frequency reflected by car f2 =   v  v  f0  v    1    v  v 2 v  v1  f = f2’ – f2 =  v  v  v  v  f0 2 1     ( v  v 2 )(v  v1 )  ( v  v 1 )( v  v 2 )  f =   ( v  v 1 )( v  v 2 )  f  0   2v( v 2  v 1 ) f0 2( v 2  v 1 ) f0 f = ( v  v )(v  v ) 1 2 v  2( v 2  v 1 ) f0 1 .2 Given = f v 100 0  v2 – v1 = 7.126 Answer in nearest integer is 7. 77. The focal length of a thin biconvex lens is 20cm. When an object is moved from a distance of 25cm in front m 25 of it to 50cm, the magnification of its image changes from m25 to m50. The ratio m is : 50 Ans. 6 Sol. When object distance is 25. 1 1 1 = v u f – 1 1 1 = v = 100 cm. v (–25 ) 20 –  v 100 m25 = = = – 4. u – 25 When object distance is 50. 1 1 1 100 = u= cm v (–50) 20 3 –  100 3 2 m 25 –4 m50 = =– m 50 = 2 = 6. – 50 3 3 – Alternate : f m 25 20  25 30 m 50 = f = 5 =6 20  50 RESONANCE Page # 42
  • 43. PHYSICS 78. An -particle and a proton are accelerated from rest by a potential difference of 100V. After this, their p de-Broglie wavelength are  and p respectively. The ratio , to the nearest integer, is :  Ans. 3 Sol. P1 = 2m(100 eV ) h h P =  = 2m(100 eV ) 2( 4m)(100 eV )  P  = 8 p  The ratio , to the nearest integer, is equal to 3.  79. When two identical batteries of internal resistance 1each are connected in series across a resistor R, the rate of heat produced in R is J1. When the same batteries are connected in parallel across R, the rate is J2. If J1 = 2.25 J2 the value of R in  is : Ans. 4 Sol. 2 i= 2R 2  2  J1 =   R 2R RESONANCE Page # 43
  • 44. PHYSICS   1 1  eq = 1 1 = 1 1  1 2 req = i= 1 =  2 R 2R  1  2 2  2  J2 =   R  1  2R  9 Given J1 = J 4 2 2 2  2  9  2   R=  R 2R 4  1  2R     2 3 = 2R 1  2R   2 + 4R = 6 + 3R  R = 4. 80. Two spherical bodies A (radius 6 cm) and B (radius 18 cm) are at temperature T1 and T2 respectively. The maximum intensity in the emission spectrum of A is at 500 nm and in that of B is at 1500 nm. Considering them to be black bodies, what will be the ratio of the rate of total energy radiated by A to that of B ? Ans. 9 Sol. (m)B = 3(m)A  TA = 3TB E1 = 4 (6)2 TA4 = 4(6)2 (3TB)4 E2 = 4 (18)2  TB4 E1 E 2 = 9. 81. When two progressive waves y1 = 4 sin (2x – 6t) and y2 = 3 sin  2x – 6t –  are superimposed, the    2 amplitude of the resultant wave is : Ans. 5 Sol. Aeq = A 1  A 2  2A 1A 2 cos  2 2 Aeq = 4 2  3 2  2( 4)(3) cos  2 Aeq = 5. RESONANCE Page # 44
  • 45. PHYSICS 82. A 0.1 kg mass is suspended from a wire of negligible mass. The length of the wire is 1m and its cross-sectional area is 4.9 × 10–7 m2. If the mass is pulled a little in the vertically downward direction and released, it performs simple harmonic motion of angular frequency 140 rad s–1. If the Young's modulus of the material of the wire is n × 109 Nm–2, the value of n is : Ans. 4 k yA /  yA (n  10 9 )  ( 4.9  10 7 ) Sol. n = = = = 140 n = 4. m m m 1 0.1   83. A binary star consists of two stars A (mass 2.2 MS) and B ( mass 11 MS) is the same mass of the sun. They are separated by distance d and are rotating about their centre of mass, which is stationary. The ratio of the total angular momentum of hte binary star to the angular momentum of star B about the centre of mass is : Ans. 6 c.m. A B Sol. 2.2 Ms 5d d 6 6 11 Ms  5d   5d   d d (2.2Ms )      (11Ms )      Total angular momentum about c.m.  6  6   66 Angular momentum of B about c.m. =  dd = 6. (11 Ms )       66 6 84. Gravitational acceleration on the surface of a planet is g, where g is the gravitational acceleration on the 11 2 surface of the earth. The average mass density of the planet is times that of the earth. If the escape speed 3 on the surface of the earth is taken to be 11 kms–1, the escape speed on the surface of the planet in kms–1 will be : Ans. 3 4 GM (G) R 3   Sol. g= = 3 ; g R R2 2  R 4 g'  '   R'   2   R'  6 R' 3 6 GM (G) ()) R 3     =    =     = Given, = Ve = = 3 g    R 3  R  11 R 22 R   R     Ve  R  ; Ve = 3 km/hr. RESONANCE Page # 45