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• 2 PART A : PHYSICS SECTION I Straight Objective Type This section contains 9 multiple choice questions numbered 1 to 9. Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. 1. A freely falling object crosses a T.V. tower of height 102.9 m in three seconds. Find the height above the top of the tower from which it would have started falling. (A) 122.5 m (B) 102.9 m (C) 19.6 m (D) 82.3 m 2. A frame of mass 200 gms, when suspended from a coil spring is found to stretch by 10 cms. A stone of mass 200 gms is dropped from rest on to the pan of the frame from a height 30 cm as shown in Figure. Find the maximum distance moved by frame downwards. (A) 20 cm (B) 10 cm (C) 30 cm (D) 40 cm 3. A plane harmonic acoustic wave y = a sin (ωt − mx) is travelling in a gaseous medium. Find the phase difference between pressure and displacement. π π (A) 0 (B) (C) (D) π 4 2 4. A battery of e.m.f. E and internal resistance r is connected to an external resistance R, the maximum power in the external circuit is 9 watts. The current flowing in the circuit under the conditions is 3 ampere. What is the value of E? (A) 4 V (B) 6 V (C) 8 V (D) 3 V SPACE FOR ROUGH WORK ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(I)/Qns - 2
• 3 5. A single turn circular coil produces at its centre a magnetic induction B when a current is passing through it. It is reshaped into a circular coil of 2 turns and if the same current is passed through it what is the magnetic induction at the centre? (A) 2B (B) 3B (C) 4B (D) 0.5B 6. In two separate setups of Young s double slit experiment fringes of equal width are observed when light of wavelength in the ratio 1 : 2 are used. If the ratio of slit separation in the two cases is 2 : 1 the ratio of distances between the plane of slits and screen are in the ratio (A) 4 : 1 (B) 1 : 1 (C) 1 : 4 (D) 2 : 1 7. Find the number of neutrons generated per unit time in a uranium reactor whose thermal power is 100 MW if the average number of neutrons liberated per fission is 2.5. Each fission releases energy 200 MeV. 18 5 (A) 7.8 × 10 (B) 7.8 × 10 10 12 (C) 7.8 × 10 (D) 7.8 × 10 8. A 1 kg block is executing S.H.M. of amplitude 0.1 m on a smooth horizontal surface under the restoring force of a spring constant 100 N/m. A block of mass 3 kg is gently placed on it as it passes through the mean position. Assuming that the blocks move together, find the amplitude of motion. (A) 4 cm (B) 5 cm (C) 6 cm (D) 3 cm 2 9. A wheel rotates with constant angular acceleration a = 2 rad/sec . If t = 0.5 s 2 after motion begins, the total acceleration of the wheel becomes 13.6 m/s . Determine the radius of wheel. (A) 5.1 m (B) 4.1 m (C) 6.1 m (D) 21. m SPACE FOR ROUGH WORK ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(I)/Qns - 3
• 4 SECTION II Assertion and Reason Type This section contains 4 questions numbered 10 to 13. Each question contains STATEMENT 1 (Assertion) and STATEMENT 2 (Reason). Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct. (A) Statement 1 is True, statement 2 is True; statement 2 is a correct explanation for statement 1. (B) Statement 1 is True, statement 2 is True; statement 2 is not a correct explanation for statement 1. (C) Statement 1 is True, statement 2 is False. (D) Statement 1 is False, statement 2 is True. 10. Statement 1: The trajectory followed by electron, when subjected to a magnetic field acting at right angles to its direction of motion is a parabola. because Statement 2: A charged particle subjected to a magnetic field perpendicular to its direction of motion moves entirely in the plane perpendicular mv to the magnetic field in a circular radius . eB 11. Statement 1: The critical angle for total internal reflection at glass water interface is greater than the critical angle at glass air interface. because Statement 2: The refractive index of glass is greater than the refractive index of water. SPACE FOR ROUGH WORK ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(I)/Qns - 4
• 5 12. Statement 1: A coil of metal wire is kept stationary in a non-uniform magnetic field. An e.m.f is induced in the coil. because Statement 2: There must be a variation in magnetic field with time if the e.m.f is to be generated. 13. Statement 1: The dielectric constant of a conductor is zero. because Statement 2: If a conductor is placed in the electric field the intensity inside the conductor is zero. SECTION III Linked Comprehension Type This section contains 2 paragraphs. Based upon each paragraph, 3 multiple choice questions have to be answered. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct. Paragraph for Question Nos. 14 to 16 A small disc of mass m1 and a thin uniform rod of mass m2 and length l lie on a smooth horizontal plane. The disc is set in motion in horizontal direction and perpendicular to the rod with velocity v after which it elastically collides with the end m 2 of the rod. The ratio of = η. m 1 14. What is the velocity of disc after collision? v 4 η v 4 η v v (A) (B) (C) (D) 4 η 4 η 4 η 4 η SPACE FOR ROUGH WORK ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(I)/Qns - 5
• 6 15. What is the angular velocity of rod after collision? 12v v (A) ω = (B) ω = l 4 η l 4 η 6v 6v (C) ω = (D) ω = l 4 η l η 4 m 2 16. For what ratio of (= η) the disc will reverse its direction of motion? m 1 (A) η > 4 (B) η > 3 (C) η < 4 (D) η < 3 Paragraph for Question Nos. 17 to 19 Figure shows a conducting circular loop of radius a placed in a uniform perpendicular magnetic field B. A metal rod OA is pivoted at the centre O of loop. The other end A of the rod touches the loop. The rod and loop have no resistance. A resistor R is connected between O and a fixed point C on the loop. The rod OA is made to rotate anticlockwise with a small angular velocity ω by an external force. 17. What is the current flowing in the resistance R? 2 2 2 Bωa (A) Bω a (B) Bωa (C) (D) B ωa R 2R R 2R SPACE FOR ROUGH WORK ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(I)/Qns - 6
• 7 18. What is the force on the rod due to magnetic field? 2 3 2 2 2 2 Bωa (A) B ωa (B) B ωa (C) B ωa (D) 2R 2R R 2R 19. Find the torque of external force needed to keep the rod rotating with constant angular velocity ω. 2 4 2 2 2 3 2 4 (A) B ωa (B) B ωa (C) B ωa (D) B ωa 4R 2R 2R R SECTION IV Matrix-Match Type This section contains 3 questions. Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (p, q, r, s) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-p, A-s, B-q, B-r, C-p, C-q and D-s, then the correctly bubbled 4 × 4 matrix should be as follows: p q r s A p q r s B p q r s C p q r s D p q r s SPACE FOR ROUGH WORK ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(I)/Qns - 7
• 8 20. Column I Column II (A) A loaded spring gun of mass M fires a shot of mass m (p) 2.0 with a velocity ω at an elevation θ. The gun is initially at rest on the frictionless horizontal surface. After firing the velocity of the centre of mass of system is (in terms of ω) (B) A body of mass M moving with speed ω makes head on (q) zero collision with another body of mass m initially at rest. If M > m, the speed of mass m is (in terms of ω) (C) Three masses each of mass m are located at the corners (r) 1.0 of an equilateral triangle ABC. They start moving with equal speed ω along the medians and collide at centroid. After collision A comes to rest and B retraces its path. What is the speed of C after collision? (in terms of ω) (D) A particle A undergoes oblique impact with particle B (s) 1.57 that is at rest initially. If their masses are equal the velocity of A after collision, makes an angle with that of B equal to (in radian) 21. Column I Column II Physical Quality Name of units (A) Angle in a plane (p) Radian (B) Solid angle (q) Steradian (C) Electric dipolemoment (r) Coulomb metre (D) Electric field intensity (s) Volt per metre 22. Column I lists the physical quantities associated with photon and Column II lists the formulae for calculating them. Match them properly. Column I Column II (A) The momentum of a moving particle is p and the (p) E/p wavelength of associated matter wave will be 2 (B) The energy of the photon is E and its momentum is p. (q) hν/c The velocity of photon will be (C) A photon in motion of energy E has a mass equal to (r) h/p (D) The mass of photon at rest is (s) zero SPACE FOR ROUGH WORK ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(I)/Qns - 8
• 9 PART B : CHEMISTRY SECTION I Straight Objective Type This section contains 9 multiple choice questions numbered 23 to 31. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 23. Matte in metallurgy is (A) artificially produced oxides (B) artificially produced sulphides (C) natural sulphides (D)none of these 24. In which of the following reactions, H2O2 acts as a reducing agent? (A) PbO2(s) + H2O2(aq) → PbO + H2O(l) + O2(g) (s) (B) Na2SO3(aq) + H2O2(aq) → Na2SO4(aq) + H2O(l) (C) 2KI(aq) + H2O2(aq) → 2KOH(aq) + I2(s) (D)KNO2(aq) + H2O2(aq) → KNO3(aq) + H2O(l) 25. TlI3 is a black coloured sparingly soluble ionic compound. In its aqueous solution, it will give 3+ − (A) Tl and I ions (B) Tl and I ions 3 + − + − (C) Tl , I ions and I2 (D) Tl and I ions 26. Which of the following compounds can be oxidised by MnO2? (A) C6H5CH2OH (B) CH3CH2CH = CHCH2OH (C) C6H5CHOHCH2CH2OH (D) All are correct SPACE FOR ROUGH WORK ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(I)/Qns - 9
• 10 27. Which of the following has the most acidic hydrogen? (A) 3-hexanone (B) 2, 4-hexanedione (C) 2, 5-hexanedione (D) 2, 3-hexanedione 28. Which of the following will be most readily dehydrated in acidic conditions? (A) (B) (C) (D) 29. 20 mL of 0.2 M MnSO4 solution was oxidised by 0.05 N KMnO4. MnO2 is formed as one of the product. Find out the volume of KMnO4 required for this reaction. (A) 160 mL (B) 100 mL (C) 200 mL (D) 250 mL 30. Two separate bulbs contain ideal gases A and B. The density of the gas A is twice that of gas B. The molecular weight of A is half that of gas B. Both the gases are at the same temperature. The ratio of the pressure of A to that of B is (A) 2 (B) 1 (C) 4 (D) 1 2 4 31. For a I order reaction, identify the correct statement. −kt (A) the degree of dissociation is equal to (1 − e ). (B) a plot of reciprocal concentration vs time gives a straight line. (C) the time taken for the completion of 75% reaction is thrice the t1/2 of the reaction. 2 (D)the pre-exponential factor in the Arrhenius equation has dimensions of T . SPACE FOR ROUGH WORK ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(I)/Qns - 10
• 11 SECTION II Assertion-Reason Type This section contains 4 questions numbered 32 to 35. Each question contains STATEMENT 1 (Assertion) and STATEMENT 2 (Reason). Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct. (A) Statement 1 is True, statement 2 is True; statement 2 is a correct explanation for statement 1. (B) Statement 1 is True, statement 2 is True; statement 2 is not a correct explanation for statement 1. (C) Statement 1 is True, statement 2 is False. (D)Statement 1 is False, statement 2 is True. 32. Statement 1: Standard free energy change of a reaction (∆G°) is not affected by catalyst. because Statement 2: Kp of a reaction is also not changed by a catalyst. + + 33. Statement 1: K ion is a weaker acid than Na ion. because Statement 2: E° value of K is less than that of Na. 34. Statement 1: Neopentyl alcohol on acid catalysed dehydration gives 2-methyl-2-butene. because Statement 2: Neopentyl Me C CH carbocation is the stable intermediate. 3 2 35. Statement 1: Pure chloroform does not produce a white precipitate with aqueous AgNO . 3 because Statement 2: Chloroform is not easily miscible with water. SPACE FOR ROUGH WORK ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(I)/Qns - 11
• 12 SECTION III Linked Comprehension Type This section contains 2 paragraphs. Based upon each paragraph, 3 multiple choice questions have to be answered. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct. Paragraph for Question Nos. 36 to 38 Nucleophilic substitution reactions: Due to the electronegativity difference, the δ+ δ− δ+ δ− − C − X bond is highly polarized bond ( − C − X ). Thus the carbon centre of C −X bond becomes prone to attack by a nucleophile. − − R − X + Nu → R − Nu + X These nucleophilic substitution reactions may take place by SN1 and SN2 mechanism. 36. X; X is (A) (B) (C) (D) SPACE FOR ROUGH WORK ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(I)/Qns - 12
• 13 37. Which is SN2 mechanism? − − (A) C2H5Br + OH → C2H5OH + Br − (B) CH CH I NH → CH3CH2NH2 + I 3 2 2 − − (C) (CH3)3 C − Br + OH → (CH3)3C − OH + Br (D)Both A and B 38. Which is the correct statement? (A) Haloalkanes are insoluble in water. (B) CH3 − CH2 − I is more reactive than CH3 − CH2 − Br towards nucleophilic substitution reactions. (C) Haloarenes are less reactive than haloalkanes towards nucleophilic substitution reactions. (D)All are correct. Paragraph for Question Nos. 39 to 41 Consider an aqueous 0.01 M sodium acetate solution. Given: log 1.85 = 0.27, Ka of −5 acetic acid = 1.85 × 10 at 298 K. 39. pH of the solution is (A) 7.0 (B) 8.36 (C) 9.2 (D) 6.0 40. The hydrolysis constant is −10 10 (A) 5.45 × 10 (B) 5.45 × 10 8 −10 (C) 54.5 × 10 (D) 54.5 × 10 41. Degree of hydrolysis is 4 −4 −4 4 (A) 23.4 × 10 (B) 23.4 × 10 (C) 2.34 × 10 (D) 2.34 × 10 SPACE FOR ROUGH WORK ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(I)/Qns - 13
• 14 SECTION IV Matrix-Match Type This section contains 3 questions. Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (p, q, r, s) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-p, A-s, B-q, B-r, C-p, C-q and D-s, then the correctly bubbled 4 × 4 matrix should be as follows: p q r s A p q r s B p q r s C p q r s D p q r s 42. Column I Column II 2+ + (A) Zn | Zn Ag | Ag (p) Redox system − − (B) Pt, Cl2 | Cl Cl | Cl2, Pt (q) Gas electrode P1 atm P2 atm 2+ 2+ (C) Cu | Cu Cu | Cu (r) Concentration cell c1 c2 3+ 2+ − (D) Pt, Fe |Fe OH | O2, Pt (s) ∆G° = − nE°F SPACE FOR ROUGH WORK ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(I)/Qns - 14
• 15 43. Column I Column II (Ion) µ(B.M) 2+ (A) Fe (p) 0 2+ (B) Cu (q) > 1.5 but less than 3 3+ (C) Ti (r) > 3 but less than 6 2+ (D) Zn (s) four unpaired electrons 44. Column I Column II (A) Freon (p) Catalyst (B) SbCl5 (q) Camphor substitute (C) AlCl3 (r) Refrigerant (D) C2Cl6 (s) Lewis acid SPACE FOR ROUGH WORK ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(I)/Qns - 15
• 16 PART C : MATHEMATICS SECTION I Straight Objective Type This section contains 9 multiple choice questions numbered 45 to 53. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 2π 2 2π 2 3π 45. The sum to 14 terms of sin sin sin ... is 7 7 7 (A) 0 (B) 14 (C) 7 (D) 21 2 46. If (a , a + 1) is a point on the angle between the lines 3x − y + 1 = 0, x + 2y − 5 = 0 containing the origin, then (A) a ≥ 1 or a ≤ − 3 (B) a ∈ (0, 1) 1 (C) a ∈ (− 3, 0) ∪ ,1 (D) no real value of a exists 3 47. The sixth term of an A.P. a1, a2, .... is 2. The common difference of the A.P., such that a1 a4 a5 is minimum is given by 2 8 1 2 (A) (B) (C) (D) 3 5 3 9 48. The value of the expression n 1 2 3 4 n is C 2 C C C ... C 2 2 2 2 2 3 2 n 1 (A) ∑n (B) ∑n (C) ∑n (D) 2 SPACE FOR ROUGH WORK ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(I)/Qns - 16
• 17 49. One ticket is selected at random from 100 tickets numbered 00, 01, 02, ..., 99. Suppose S and P are the sum and product of the digits found on the ticket. Then the probability that S = 7 given P = 0 is 2 1 2 1 (A) (B) (C) (D) 3 50 19 19 50. The value of the determinant 2 2cos x sin2x sinx 2 sin2x 2 sin x cos x sinx cos x 0 is (A) 1 (B) 2 (C) 3 (D) 0 51. In a right angled triangle ABC, the bisector of the right angle C, divide AB into A B segments of lengths p, q. Also tan = k. Then p : q is 2 2 1 k k 1 (A) (B) k (C) (D) 1 k 1 2 2 −1 −1 −1 3π 52. If sin x + sin y + sin z = and f (p + q) = f(p) ⋅ f(q) for all p, q ∈ R and f(1) = 1, 2 f 1 f 2 f 3 x y z then the value of x y z is f 1 f 2 f 3 x y z (A) 0 (B) 2 (C) 3 (D) 1 SPACE FOR ROUGH WORK ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(I)/Qns - 17
• 18 53. The range of the function, 3 2 −1 f : [0, 1] → R ; f(x) = x − x + 4x + 2 sin x is (A) [2, 3] (B) [0, 4 + π] (C) [0, 2 + π] (D) [− π − 2, 0] SECTION II Assertion and Reason Type This section contains 4 questions numbered 54 to 57. Each question contains STATEMENT 1 (Assertion) and STATEMENT 2 (Reason). Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct. (A) Statement 1 is True, statement 2 is True; statement 2 is a correct explanation for statement 1. (B) Statement 1 is True, statement 2 is True; statement 2 is not a correct explanation for statement 1. (C) Statement 1 is True, statement 2 is False. (D)Statement 1 is False, statement 2 is True. 54. Statement 1: In a triangle, centroid is the origin and 5 i 4j 2 k is the position vector of the orthocentre, then the position vector of the 5 circumcentre is i 2j k . 2 because Statement 2: S, the circumcentre, G, the centroid and H, the orthocentre are collinear and SG : GH = 1 : 2. SPACE FOR ROUGH WORK ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(I)/Qns - 18
• 19 55. Statement 1: If l, m, n are consecutive positive even integers, then the family of lines lx + my + n = 0 are concurrent at (1, − 2). because Statement 2: Three consecutive positive even integers are in A.P. 3 1 3 56. Statement 1: For any two events A and B, P (A ∪ B) ≥ and ≤ P (A ∩ B) ≤ 4 8 8 7 11 then ≤ P(A) + P(B) ≤ . 8 18 because Statement 2: For any two events E and F, P(E ∪ F) = P(E) + P(F) − P(E ∩ F). 3 5 57. Statement 1: In ∆ABC, cos A = , cos B = , then the value of cos C can be 5 13 7 . 13 because Statement 2: In ∆ABC, tan A + tan B + tan C = tan A tan B tan C. SECTION III Linked Comprehension Type This section contains 2 paragraphs. Based upon each paragraph, 3 multiple choice questions have to be answered. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct. SPACE FOR ROUGH WORK ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(I)/Qns - 19
• 20 Paragraph for Question Nos. 58 to 60 a × b ⋅ c is called scalar triple product of 3 vectors. It is denoted as a b c . Properties: 1) In scalar triple product dot and cross can be interchanged. 2) Value is unaltered for cyclic permutation of vectors. 3) If two vectors are equal, value is zero. Vector triple product a × b × c = a⋅c b a⋅b c 58. If a, b, c are non - coplanar vectors and p, q, r are defined as b×c c×a a× b p = , q = , r = , b c a c a b a b c then a b ⋅p b c ⋅q c a ⋅ r is (A) 0 (B) 1 (C) 2 (D) 3 59. If a, b, c are non-coplanar, non-zero vectors, then a× b × a×c b×c × b×a c × a × c × b is equal to 2 (A) a b c a b c (B) a b c a b c (C) 0 (D) 3a SPACE FOR ROUGH WORK ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(I)/Qns - 20
• 21 1 60. If a, b, c are non-coplanar unit vectors such that a × b × c = b c and if 2 α, β are the angles between a and b and a and c , then α + β is π 2π (A) π (B) (C) 2π (D) 2 3 Paragraph for Question Nos. 61 to 63 Let f(x) be a continuous function defined on the closed interval [a, b]. n 1 1 1 r Then lim ∑ n f n = ∫f x dx. n → ∞ r =0 0 97 97 97 1 2 ... n 61. lim is 98 n→∞ n 98 1 1 1 (A) (B) (C) (D) 100 99 98 100 n 62. 6 πr 1 is lim ∑ sin n→∞ r=1 n n 5 5 5 5 (A) (B) (C) (D) 4 8 16 32 63. lim 1 1 1 1 is ... n→∞ 2n 2 2 2 4n 1 4n 4 3n 2n 1 π π π π (A) (B) (C) (D) 3 6 4 2 SPACE FOR ROUGH WORK ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(I)/Qns - 21
• 22 SECTION IV Matrix-Match Type This section contains 3 questions. Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (p, q, r, s) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-p, A-s, B-q, B-r, C-p, C-q and D-s, then the correctly bubbled 4 × 4 matrix should be as follows: p q r s A p q r s B p q r s C p q r s D p q r s 2 64. The normals drawn at P, Q, R on y = 8x are concurrent at (6, 0). Column I Column II (A) Centroid of ∆PQR (p) (5, 0) 4 (B) Circumcentre of ∆PQR (q) , 0 3 (C) Area of ∆PQR (r) 5 (D) Circumradius of ∆PQR (s) 8 SPACE FOR ROUGH WORK ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(I)/Qns - 22
• 23 65. Column I Column II (A) The number of the points of discontinuity of (p) infinite f(x) = [cos x + sin x], where [ ] is greatest integer function and 0 < x < 2π is 2 (B) Let f(x) = x − |x − x |, x ∈ [− 1, 1]. Then the (q) 0 number of points at which f(x) is discontinuous is (C) The number of points of discontinuity of (r) 1 2n 2 sin x f x = lim is n 2n n→∞ 3 2 cos x (D) The number of points of discontinuity of (s) 4 x 2 f x = x is |x 2| 66. Column I Column II (A) If 3 sin x + 5 cos x = 5, then the value of (p) 0 5 sin x − 3 cos x is (B) The number of values of x for which (q) 4 tan 3x tan 2x = 1 is 1 tan 3x tan 2x (C) With usual notation in ∆ABC, if b + c = 3a, then (r) 2 B C cot cot is 2 2 (D) In ∆ABC, tan A, tan B are the roots of (s) 3 2 2 2 2 2 abx − c x + ab = 0. Then sin A + sin B + sin C is SPACE FOR ROUGH WORK ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(I)/Qns - 23
• 24 SPACE FOR ROUGH WORK ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(I)/Qns - 24
• 25 SPACE FOR ROUGH WORK ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(I)/Qns - 25
• Name: . Enrollment No.: Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose. C. Question paper format: 13. The question paper consists of 3 parts (Physics, Chemistry and Mathematics). Each part has 4 sections. 14. Section I contains 9 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which only one is correct. 15. Section II contains 4 questions. Each question contains STATEMENT-1 (Assertion) and STATEMENT- 2 (Reason). Bubble (A) if both the statements are TRUE and STATEMENT-2 is the correct explanation of STATEMENT-1. Bubble (B) if both the statements are TRUE but STATEMENT-2 is not the correct explanation of STATEMENT-1. Bubble (C) if STATEMENT-1 is TRUE and STATEMENT-2 is FALSE. Bubble (D) if STATEMENT-1 is FALSE and STATEMENT-2 is TRUE. 16. Section III contains 2 paragraphs. Based upon each paragraph, 3 multiple choice questions have to be answered. Each question has 4 choices (A), (B), (C) and (D), out of which only one is correct. 17. Section IV contains 3 questions. Each question contains statements given in 2 columns. Statements in the first column have to be matched with statements in the second column. The answers to these questions have to be appropriately bubbled in the ORS as per the instructions given at the beginning of the section. D. Marking scheme: 18. For each question in Section I, you will be awarded 3 marks if you darken only the bubble corresponding to the correct answer and zero mark if no bubble is darkened. In all other cases, minus one (−1) mark will be awarded. 19. For each question in Section II, you will be awarded 3 marks if you darken only the bubble corresponding to the correct answer and zero mark if no bubble is darkened. In all other cases, minus one (−1) mark will be awarded. 20. For each question in Section III, you will be awarded 4 marks if you darken only the bubble corresponding to the correct answer and zero mark if no bubble is darkened. In all other cases, minus one (−1) mark will be awarded. 21. For each question in Section IV, you will be awarded 6 marks if you darken ALL the bubbles corresponding ONLY to the correct answer. No negative mark will be awarded for an incorrectly bubbled answer. ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(I)/Qns - 26
• 1 ® ® IIT-JEE 2008 STS VIII/PCM/P(I)/SOLNS BRILLIANT S HOME-BASED FULL-SYLLABUS SIMULATOR TEST SERIES FOR OUR STUDENTS TOWARDS IIT-JOINT ENTRANCE EXAMINATION, 2008 PAPER I - SOLUTIONS PHYSICS − CHEMISTRY − MATHEMATICS PART A : PHYSICS SECTION I 1 1. (C) S = ut + at2 2 1 102.9 = u × 3 + × 9.8 × 9 2 u = velocity of body at the top of tower = 19.6 m/s 2 For a freely falling body u = 2gh 2 19.6 = 2 × 9.8 × h h = 19.6 m The height above the top of the tower from which it should have started falling is 19.6 m. 2. (C) When the stone falls on the pan of the frame the impact is completely inelastic. At the instant of impact the stone has a velocity = 2gh = 2g × 30 . ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(I)/Solns - 1
• 2 By principle of conservation of momentum the stone plus frame will have a velocity given by 200 2gh 2gh V= = 200 200 2 1 2gh Kinetic energy of stone and pan = × 400 × 2 4 = 100 gh = 3000 g If the maximum stretching of spring due to impact is x the work done in spring to stretch it from elongation 10 cm to (10 + x) cm must be equal to kinetic energy of stone + frame + the loss of potential energy of (stone + frame) 1 2 1 2 If k is spring constant, work done = k 10 x × k × 10 2 2 200 g k= = 20 g 10 1 × 20 × g [(10 + x)2 − 102] = 3000 g + 400x g 2 2 x − 20x − 300 = 0 Solving, x = 30 cm dy 3. (C) Volumetric strain in a gaseous medium = dx dy Volume strain = = − am cos (ωt − mx) dx stress By Hooke s law volume elasticity of gas E = strain p change in pressure = dy dx dy p= E⋅ dx ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(I)/Solns - 2
• 3 Adiabatic elasticity = Ep = γp ∴ p = γPatmosphere am cos (ωt − mx) π = γp a am sin ω t mx 2 π This shows that pressure is out of phase with displacement. 2 π ∴ phase difference = 2 4. (B) When the power in the external circuit is maximum, the current is maximum i.e., when R = r. E = 3 ampere R R ∴ maximum power = 9 W 2 9 9 i R = 9 or R = = = 1 ohm 2 9 i ∴ E = 3 or E = 6 V 1 1 µ i 0 5. (C) B = 2a When it is reshaped radius = r 2πa = 2 ⋅ 2πr a r= 2 µ ni µ ni µ × i ×2 2µ i 0 0 0 0 B′ = = = = 2a 2r a a 2 × 2 B′ = 4B λ D λ D 1 1 2 2 6. (A) β = = d d 1 2 D λ d 2 2 4 1 2 1 = = × = D 2 λ d 1 1 1 1 2 Ratio = 4 : 1 ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(I)/Solns - 3
• 4 7. (A) Power of reactor = 100 MW = 100 × 106 watt = 100 × 106 J/s 6 100 × 10 Number of uranium atoms splitting per second = 19 6 1.6 × 10 × 200 × 100 23 100 × 10 = 6 3.2 × 10 17 100 × 10 18 Number of nucleus liberated = × 2.5 = 7.8 × 10 3.2 1 2 1 2 8. (B) For mass m: mu = kA , u2 = kA2 2 2 When M is added at mean position, mu = (m + M) V u V = . 4 (K.E.) at mean = (P.E.) at extreme 1 2 1 2 ∴ m M V = kA ′ 2 2 2 u 2 ∴ = kA ′ (P.E = 0at mean) 4 A A′ = = 5 cm 2 2 9. (C) Angular acceleration = 2 rad/s 1 Time = s 2 Initial angular velocity = 0 1 Final angular velocity = 0 + × 2 = 1 rad/s 2 Linear velocity = rω = 1 × r m/s 2 2 Normal acceleration = v = r = r r r Tangential acceleration = r × 2 2 2 Total acceleration = r 2r = r 5 ; r 5 = 13.6 ∴ r = 13.6 = 6.1 m 5 ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(I)/Solns - 4
• 5 SECTION II 10. (D) µ 1 11. (A) Critical angle ic is given by sin ic = µ 2 µ in 4 2 8 When it travels from glass to water = = × = µ 3 3 9 g 1 When it travels from glass to air = µ g sin i′c = 2 3 ∴ ic > i′c 12. (D) There must be a variation in the magnetic field with time so that there is a change in magnetic flux with time which is responsible for induced e.m.f. E 13. (D) The dielectric constant of a medium = where the electric field E is E′ reduced to E′ in the presence of dielectric. If the conductor is placed in the electric field the intensity inside the conductor is zero. Therefore the dielectric constant of the conductor is infinite. SECTION III 14. (A) Since the collision of disc with rod is elastic, linear momentum, angular momentum and energy are conserved. Let v′ and V be the velocities of disc and centre of mass of rod after collision and ω the angular velocity of rod about its centre of mass. m 1 m1v = m1v′ + m2V or V = v v′ m 2 m 1 1 If = η, V = v v′ ... (1) m η 2 2 l l m l m v ⋅ = m v′ ⋅ Iω , where I = 2 1 1 2 2 12 1 lω = 6(v − v′) ... (2) η ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(I)/Solns - 5
• 6 By principle of conservation of energy, 1 2 1 2 1 2 1 2 m v = m v′ m V Iω 1 1 2 2 2 2 2 2 2 2 m 2 From (1) and (2), m1 (v − v′ ) = 4 1 v v′ m 2 v 4 η v′ = ... (3) 4 η 15. (A) From (2) and (3) in the above problem ω= 12v l 4 η 16. (A) The disc will reverse its direction of motion if v′ becomes negative i.e., when 4 < η or η > 4. m 2 >4 m 1 17. (B) The e.m.f between the ends of rotating rod is a E= ∫ dE = ∫ Bω x dx 0 1 2 = Bωa 2 The positive charges of the rod will be pushed towards O by the magnetic field. Thus the rod may be replaced by a battery by e.m.f 1 Bωa 2 2 with the positive terminal towards O. The equivalent circuit diagram is shown. The circular loop forming A to C by a resistanceless path. 2 E Bωa ∴ current in resistance R = i = = R 2R 18. (A) The force on the rod due to magnetic field = Bia 2 2 3 ∴F= B⋅ Bωa B ωa ⋅a = 2R 2R ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(I)/Solns - 6
• 7 19. (A) As the force is uniformly distributed over OA it may be assumed to act at the mid part of OA. The torque is therefore 2 4 a B ωa τ = ia B = in clockwise direction 2 4R 2 4 B ωa To keep the rod rotating at uniform angular speed an external torque 4R in anticlockwise direction is needed. SECTION IV 20. (A) − (q); (B) − (p); (C) − (r); (D) − (s) (A) The motion of the centre of mass of a system of two particles is unaffected by their internal forces irrespective of the actual direction of internal forces. The velocity of centre of mass of system is zero. (B) Let A and B be particles of mass M and m. The Figures below are indicated the situation before collision and after collision. Applying the principle of conservation of momentum, we get 2M v′ = ⋅v if M >> m M m v′ = 2v; v = ω ∴ v′ = 2ω The speed of mass m = 2ω (C) Applying the principle of conservation of momentum, we get the speed of C after collision is ω. (D) Applying the principle of conservation of momentum and Newton s law, we get the velocity of A after impact is at right angles to that of B π 3.14 θ = = = 1.57 rad . 2 2 21. (A) − (p); (B) − (q); (C) − (r); (D) − (s) Refer text book 22. (A) − (r); (B) − (p); (C) − (q); (D) − (s) Refer text book ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(I)/Solns - 7
• 8 PART B : CHEMISTRY SECTION I 23. (B) 4 1 2 0 24. (A) Pb O H O → PbO H O O 2 s 2 2 aq s 2 l 2 g 25. (A) Due to inert pair effect, +1 oxidation state of Tl is more stable in TlI , it is 3 thallium triiodide Tl I 3 . MnO2 26. (A) C6H5CH2OH → C6H5CHO 27. (B) In B, the methylene group is most active due to the presence of C = O group on either side. 28. (C) Conjugated system is formed, which is stable. 29. (A) In the conversion, 2 4 Mn SO → Mn O 4 2 There is a change in oxidation state of Mn by two units. M Equivalent weight of MnSO = 4 2 ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(I)/Solns - 8
• 9 Molecular weight Normality of MnSO4 = Molarity × Equivalent weight M = 0.2 × = 0.4 N M ⁄2 0.05 × V1 = 0.4 × 20 (KMnO4) (MnSO4) 0.4 × 20 V = = 160 mL 1 0.05 PM 30. (C) d = RT dA = 2dB M B M = A 2 d RT d RT A B P = ; P = A B M M A B P d RT M d × M A A B A B = × = P M d RT M ×d B A B A B 2d × M = B B 4 = M 1 B d × B 2 PA : PB = 4 : 1 31. (A) For a I order reaction, 1 a k = log t a x a kt = log a x kt a e = a x kt a x x e = =1 a a x kt Degree of dissociation = =1 e a ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(I)/Solns - 9
• 10 SECTION II 32. (A) 33. (B) 34. (C) Neopentyl carbocation formed undergoes rearrangement leading to the formation of 2-methyl-but-2-ene. 35. (B) SECTION III 36. (B) At high temperature, allylic substitution takes place. 37. (D) 1° - haloalkanes undergo S 2 mechanism. N 38. (D) Haloalkanes are insoluble in water, due to the absence of formation of H-bonding. 1 1 1 39. (B) pH = pK pK log c w a 2 2 2 1 1 1 2 = 14 4.73 log 10 2 2 2 = 7 + 2.365 − 1 = 8.365 K 14 w 10 10 40. (A) K h = = 5.45 × 10 K 5 a 1.85 × 10 K h 41. (C) α = c 10 5.45 × 10 4 = = 2.34 × 10 2 10 SECTION IV 42. (A) − (p), (s); (B) − (q), (r), (s); (C) − (r), (s); (D) − (p), (s) ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(I)/Solns - 10
• 11 43. (A) − (r), (s); (B) − (q); (C) − (q); (D) − (p) − Ion Configuration Unpaired e µ 2+ 6 (A) Fe 3d 4 4.9 2+ 9 (B) Cu 3d 1 1.73 (C) Ti3+ 3d1 1 1.73 2+ 10 (D) Zn 3d 0 0.0 µ = n n 2 If n = 1, µ = 13 = 3 = 1.73 n = 2; µ = 2 4 = 2.83 etc . 44. (A) − (r); (B) − (p), (s); (C) − (p), (s); (D) − (q) ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(I)/Solns - 11
• 12 PART C : MATHEMATICS SECTION I 1 2π 4π 6π 45. (C) The series = 1 cos 1 cos 1 cos ... 14 terms 2 7 7 7 1 2π 4π = 14 cos cos ... 14 terms 2 7 7 2π π 14π cos 13 ⋅ sin 1 7 7 7 = 7 2 π sin 7 1 = 7 ⋅ 0 =7 2 46. (C) 2 From the Figure, it is clear that (0, 0) and (a , a + 1) lie on the same side of both the lines. 2 3a − (a + 1) + 1 > 0 2 1 3a − a > 0 ⇒ a < 0 or a > 3 2 a + 2 (a + 1) − 5 < 0 2 a + 2a − 3 < 0 a ∈ (− 3, 1) Shaded portion is the required region 1 ∴ a ∈ (− 3, 0) ∪ , 1 3 ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(I)/Solns - 12
• 13 47. (A) a1 + 5d = 2 Now, P = a1 a4 a5 = a1 (a1 + 3d) (a1 + 4d) = (2 − 5d) (2 − 2d) (2 − d) 2 3 = 2 [4 − 16d + 17d − 5d ] Consider, S = − 5d3 + 17d2 − 16d + 4 S′ = − 15d2 + 34d − 16 2 8 S′ = 0 ⇒ d = , 3 5 S″ = − 30d + 34 At 2 , S″ = − 20 + 34 = +ve 3 2 ∴d= gives minimum value. 3 48. (B) Expression n 1 3 3 4 = C 2 C C C ... 2 3 2 2 n 1 4 4 = C 2 C C ... 2 3 2 n 1 5 5 n = C 2 C C ... C 2 3 2 2 n 1 n 1 = C 2 C ultimately 2 3 n 1 n 1 n 1 = C C C 2 3 3 n 2 n 1 = C C 3 3 n 2 n 1 n n 1 n n 1 = 6 6 n n 1 = n 2 n 1 6 n n 1 2n 1 = = ∑n2 6 49. (C) S = 7 = (07, 16, 25, 34, 43, 52, 61, 70) P = 0 = {00, 01, 02, ..., 09, 10, 20, 30, ..., 90} S = 7 ∩ P = 0 = {07, 70} P′ S = 7 ∩ P = 0 P′ {S = 7/P = 0} = P′ P = 0 2 = 19 ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(I)/Solns - 13
• 14 50. (B) Applying R1 → R1 − 2R3 sin x, R2 → R2 + 2 cos x R3. we get 2 0 sin x 0 2 cos x sin x cos x 0 2 2 2 = 2 (cos x) + sin x (2 sin x) = 2 (cos x + sin x) = 2 x p 51. (A) = = p 2 sin A 1 2 x = q 2 sin B A B A B 2 cos ⋅ sin p sin B q p 2 2 = ⇒ = q sin A q p A B A B 2 sin ⋅ cos 2 2 π C A B = cot ⋅ tan 2 2 2 =1⋅k q p ⇒ = k q p p 1 k ∴ = q 1 k 52. (B) Put p = 1, q = 1 2 f(2) = (f(1)) = 1 Put p = 2, q = 1 f(3) = f(2) f(1) = 1 ⋅ 1 = 1 −1 −1 −1 3π From sin x + sin y + sin z= , 2 π we get each = 2 ⇒x=y=z=1 1 1 1 x y z ∴ expression = x + y + z − x y z =3−1=2 ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(I)/Solns - 14
• 15 −1 53. (B) f(x) = x3 − x2 + 4x + 2 sin x 2 2 f′(x) = 3x − 2x + 4 + 2 1 x 2 For 3x − 2x + 4, 2 coefficient of x = 3 > 0 Discriminant = 4 − 48 is − ve 2 ∴ 3x − 2x + 4 is always positive. 2 ⇒ is always positive. 2 1 x ⇒ f ′ (x) > 0 for all real x. ⇒ f(x) is increasing. Range [f(0), f(1)] = [0, 4 + π] SECTION II 54. (A) 5 2x 5 = 0 ⇒ x = 3 2 4 2y = 0 ⇒ y=−2 3 2 2z = 0 ⇒ z=−1 3 5 ∴ position vector of circumcentre is i 2j k 2 55. (A) l, m, n are in A.P. ⇒ 2m = l + n ⇒ l − 2m + n = 0 ⇒ l (1) + m (− 2) + n = 0 ⇒ (1, − 2) is a point on lx + my + n = 0 3 56. (A) P(A) + P(B) − P (A ∩ B) ≥ 4 3 1 P(A) + P(B) ≥ 4 8 7 i.e., ≥ 8 ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(I)/Solns - 15
• 16 Also P (A ∪ B) ≤ 1 P(A) + P(B) − P (A ∩ B) ≤ 1 3 P(A) + P(B) ≤ 1 + 8 11 i.e., ≤ 8 4 12 4 12 57. (D) tan C = ⋅ tan C 3 5 3 5 56 16 tan C = tan C 15 5 56 ⇒ tan C = 33 33 cos C = 65 SECTION III 58. (D) The expression a b ⋅ b ×c = ∑ b c a a b c = ∑ = ∑1 a b c =1+1+1=3 59. (A) The expression = a b c a 0 b c a b 0 c a b c 0 = a b c a b c 1 1 60. (A) Given equation is a ⋅ c b a ⋅b c = b c 2 2 Because of the given conditions, 1, 1 a ⋅c = a ⋅b = 2 2 Since a, b, c are unit vectors, the above implies ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(I)/Solns - 16
• 17 Angle between a and c is π . 3 Angle between a and b is 2π . 3 2π π 3π ∴α+β= = = π 3 3 3 61. (C) Required limit 97 97 97 1 1 2 n = Lt ... n → ∞ n n n n 1 1 98 x 1 = ∫x 97 dx = 98 = 98 0 0 1 62. (C) Required limit = ∫sin 6 π x dx 0 Put y = πx. π = ∫sin y dy π 6 0 π /2 = 2 π ∫ sin y dy 6 0 2 5 3 1 π = ⋅ ⋅ ⋅ ⋅ π 6 4 2 2 5 = 16 1 1 1 63. (B) Lt ... 2 2 2 2 n → ∞ 4n 0 4n 1 4n n 1 n 1 1 = Lt ∑ 2 2 n→ ∞ r = 0 4n r n 1 1 = Lt ∑ n→ ∞ r = 0 2 r n 4 n ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(I)/Solns - 17
• 18 1 dx = ∫ 2 dx 0 4 x 1 = sin 1 x 2 0 π π = 0 = 6 6 SECTION IV 64. (A) − (q); (B) − (p); (C) − (s); (D) − (r) Equation to the normal at t is y + xt = 4t + 2t3. It passes through (6, 0) 3 2t − 2t = 0 3 t − t = 0 ⇒ t = 0, 1, − 1 ∴ P is (0, 0) and Q is (2, 4), (2, − 4) 4 (A) Centroid of ∆PQR = , 0 3 (B) Let the equation of the circle be 2 2 x + y + 2gx + 2fy = 0 4g + 8f = − 20 4g − 8f = − 20 Solving, we get g = − 5, f = 0 ⇒ centre (5, 0) 1 (C) Area of ∆PQR = 2 4 2 4 2 1 = 8 8 2 = 8 (D) Radius = 5 2 = 5 ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(I)/Solns - 18
• 19 65. (A) − (s); (B) − (q); (C) − (p); (D) − (r) π (A) cos x + sin x = 2 cos x 4 π f x = 2 cos x 4 [x] is discontinuous at all integral values. π Now, 2 cos x is an integer in 0 < x < 2π 4 π π π π 3π π at x = , , π , 2 2 4 4 2 4 Points of discontinuity are 4 in number. (B) f(x) = x − |x| |1 − x| Since x, |x| |1 − x| are continuous everywhere, f(x) is discontinuous at no point in [− 1, 1]. (C) The function is discontinuous if n 2n 3 − (2 cos x) =0 n 2 (cos x) n = 3 4 cos x = 3 2 4 π x = nπ ± 4 There are infinite number of points of discontinuity. y (D) = 1 if y > 0 |y | = − 1 if y < 0 At x = − 2, h Right limit = Lt 2 h h→ 0 |h| =−2+1=−1 ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(I)/Solns - 19
• 20 h Left limit = Lt 2 h h→0 | h| =−2−1=−3 Right limit ≠ left limit ∴ x = − 2 is the only point of discontinuity. 66. (A) − (s); (B) − (p); (C) − (r); (D) − (r) (A) 3 sin x + 5 cos x = 5 Squaring, 2 2 9 sin x + 25 cos x + 30 sin x cos x = 25 9 (1 − cos2x) + 25 (1 − sin2x) + 30 sin x cos x = 25 2 2 9 cos x + 25 sin x − 30 sin x cos x = 9 2 (5 sin x − 3 cos x) = 9 5 sin x − 3 cos x = 3 (B) tan (3x − 2x) = 1 tan x = 1 π π x = nπ + . This value does not satisfy the equation, because tan 2x = tan = ∞. 4 2 There is no value of x satisfying the equation. (C) From the result, sin B + sin C = 3 sin A B C B C A A 2 sin cos = 6 sin cos 2 2 2 2 B C A cos = 3 sin 2 2 B C B C cos = 3 cos 2 2 B C B C B C B C cos cos sin sin = 3 cos cos 3 sin sin 2 2 2 2 2 2 2 2 B C B C 2 cos cos = 4 sin sin 2 2 2 2 B C cot cot = 2 2 2 ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(I)/Solns - 20
• 21 2 (D) tan A + tan B = c , tan A tan B = 1 ab tan A = cot B = tan (90° − B) A + B = 90°, C = 90° ∴ triangle is right-angled. b sin A = a , sin B = c c 2 2 2 a +b =c a b tan A = , tan B = b a 2 2 2 tan A + tan B = a b = c ab ab 2 2 2 Now, sin A + sin B + sin C 2 2 2 2 2 a b a b c = 1 = 2 2 2 c c c 2 2c = =2 2 c ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(I)/Solns - 21
• 2 PART A : PHYSICS SECTION I Straight Objective Type This section contains 9 multiple choice questions numbered 1 to 9. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1. A loop of string whose linear density is m is whirled at a high angular velocity ω so that it becomes a taut circle of radius R. A kink develops in the whirling string as shown in Figure. What is the tension in the string? 2 2 mω R 2 (A) mω R (B) mrω (C) (D) mω R 2 2. An open organ pipe in which air is at a temperature of 15°C and a sonometer wire of frequency 512 Hz when sounded together produces 5 beats per second with the organ pipe emitting its fundamental note. If a slight reduction in tension of sonometer wire is made it produces resonance between the notes what change in the temperature of air in organ pipe would have produced resonance with the vibrating of sonometer wire with 512 Hz? (A) 20.7°C (B) 5.7°C (C) 15.7°C (D) 3.7°C 3. A gas containing only rigid diatomic molecules is at temperature T. If I is the moment of inertia of the molecule, the angular mean square velocity of a rotating molecule in terms of Boltzmann constant k is 2kT kT 3kT 5kT (A) (B) (C) (D) I I I I SPACE FOR ROUGH WORK ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(II)/Qns - 2
• 3 4. A convex lens of focal length 12 cm lies in a uniform magnetic field 1.2 tesla parallel to the principal axis. A charged particle of mass 20 mg and charge 2 milli coulomb is projected perpendicular to the plane of diagram with a speed of 4.8 m/s. The particle moves along a circle with centre on the principal axis at a distance 18 cm from lens. The radius of the image circle formed by lens is (A) 18 cm (B) 8 cm (C) 4 cm (D) 12 cm 5. What is the equivalent resistance between points a and b of the circuit shown below? (A) 15 Ω (B) 5 Ω (C) 7 Ω (D) 7.5 Ω 6. An air filled parallel plate capacitor is constructed which can store 12 µC of 6 −1 charge when operated at 1200 volt. The dielectric strength of air is 3 × 10 Vm . What is the minimum area of the plates of capacitor? 2 2 2 2 (A) 1 m (B) 0.45 m (C) 1.5 m (D) 1.2 m 2 7. A 40 cm long wire having a mass 3.2 g and area of cross section 1 mm is stretched between supports 40.05 cm apart. In its fundamental note it vibrates with a frequency 220 Hz. What is the Young s modulus of the wire? 11 2 11 2 (A) 1.98 × 10 N/m (B) 2.2 × 10 N/m 11 2 11 2 (C) 3.96 × 10 N/m (D) 8.2 × 10 N/m SPACE FOR ROUGH WORK ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(II)/Qns - 3
• 4 8. The orbital period of a satellite in a circular orbit of radius r about a spherical planet of mass M and mean density ρ, for a low altitude orbit (r = rp) will be 3π π (A) (B) 3πGρ (C) (D) 2Gρ Gρ Gρ 9. It takes one minute for a person standing on an escalator to reach the top from the ground. If the escalator is not moving it takes him 3 minute to walk on the steps to reach the top. How long will it take for the person to reach the top if he walks up the escalator while it is moving? (A) 2 minute (B) 1.5 minute (C) 0.75 minute (D) 1.25 minute SECTION II Assertion and Reason Type This section contains 4 questions numbered 10 to 13. Each question contains STATEMENT 1 (Assertion) and STATEMENT 2 (Reason). Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct. (A) Statement 1 is True, statement 2 is True; statement 2 is a correct explanation for statement 1. (B) Statement 1 is True, statement 2 is True; statement 2 is not a correct explanation for statement 1. (C) Statement 1 is True, statement 2 is False. (D)Statement 1 is False, statement 2 is True. 10. Statement 1: A sail boat can be propelled by air blown at the sails from a fan attached to the boat. because Statement 2: The force applied being internal to the system cannot change the state of motion of the system. SPACE FOR ROUGH WORK ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(II)/Qns - 4
• 5 11. Statement 1: A block of wood is floating in a tank containing water. The apparent weight of the floating block is zero. because Statement 2: The entire weight of the block is supported by the buoyant force (upward thrust) due to water. 12. Statement 1: The velocity, wavelength and frequency all undergo a change when a wave travels from one medium to another medium. because Statement 2: The frequency of a wave does not change when a wave travels from one medium to another. 13. Statement 1: The self induced emf produced by a variable current in a coil always tends to decrease the current. because Statement 2: If there is an increase in current, the self induced emf tend to decrease the current and vice versa according to Lenz s law. SECTION III Linked Comprehension Type This section contains two paragraphs. Based upon each paragraph three multiple choice questions have to be answered. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct. Paragraph for Question Nos. 14 to 16 A 50 µF capacitor initially uncharged is connected through a 300 Ω resistor to a 12 V battery. 14. What is the magnitude of final charge? (A) 500 µC (B) 600 µC (C) 400 µC (D) 300 µC SPACE FOR ROUGH WORK ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(II)/Qns - 5
• 6 15. How long after the capacitor is connected to the battery will it be charged to half the maximum value? (A) 15 millisecond (B) 10.5 millisecond (C) 7.5 millisecond (D) 12.5 millisecond 16. How long will it take the capacitor to be charged to 90% of maximum value? (A) 34.5 millisecond (B) 17.25 millisecond (C) 68 millisecond (D) 90 millisecond Paragraph for Question Nos. 17 to 19 A uniform rod of mass m and length 2l stands vertically on a rough horizontal floor and it is allowed to fall, the slipping has not occurred during the motion. 17. What is the angular velocity of the fall when the rod makes an angle θ with vertical? 3g g (A) 1 cos θ (B) 1 cos θ 2l 2l 2g (C) 1 cos θ (D) 3g l 1 cos θ 3l 18. What is the normal force exerted by the floor on the rod in this position when it makes an angle θ with vertical? 2 2 (A) mg (3 − 4 sin θ) (B) mg (4 − 3 sin θ) mg 2 mg 2 (C) (4 − 3 sin θ) (D) (4 − 3 sin θ) 3 4 19. If slipping occurs at an angle θ = 30°, what is the coefficient of friction between rod and floor? (A) 0.4 (B) 0.3 (C) 0.48 (D) 0.6 SPACE FOR ROUGH WORK ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(II)/Qns - 6
• 7 SECTION IV Matrix-Match Type This section contains 3 questions. Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (p, q, r, s) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-p, A-s, B-q, B-r, C-p, C-q and D-s, then the correctly bubbled 4 × 4 matrix should be as follows: 20. Match the quantities in Column I and Column II correctly. Column I Column II (p) 1 δP (A) The speed of sound in a gas is v. The R.M.S. velocity of the molecules is c. The ratio of v to c γ P will be (B) During an adiabatic process, the pressure P of a (q) γ fixed mass of gas changes by δP and the volume 3 δV changes by δV. The value of will be equal to V 1−γ γ (C) The pressure-temperature relation for an (r) P T = constant adiabatic expansion is (D) If γ denotes the ratio of specific heats of a gas the (s) γ ratio of slopes of adiabatic and isothermal P-V curves at their point of intersection is SPACE FOR ROUGH WORK ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(II)/Qns - 7
• 8 21. Column I Column II 10 (A) A cylinder is released from rest from the (p) g l sin θ 7 top of an incline of inclination θ and length l. If the cylinder rolls without slipping its speed when it reaches the bottom is 4 (B) A solid sphere of mass M and radius R rolls (q) g l sin θ 3 down an inclined plane of length l and inclination θ without slipping from the top. The speed of its centre of mass when it reaches the bottom is v (C) A circular disc rotates in a vertical plane (r) 2 about a fixed horizontal axis which passes through a point X on the circumference of the disc, when the centre of mass moves with speed v the speed of the opposite end of the diameter through X will be (D) A body of mass m slides down a frictionless (s) 2v inclined plane and reaches the bottom with a velocity v. If the same mass is in the form of a ring which rolls down without slipping on identical but rough inclined plane, the velocity of the ring at the bottom will be SPACE FOR ROUGH WORK ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(II)/Qns - 8
• 9 22. Column I Column II (A) The space between the plates of a parallel (p) 0.75 µF plate capacitor of capacity 10 µF having air between the plates is filled with mica of dielectric constant K = 2, what will be the new capacity? (B) Three capacitors, each of capacity 1 µF are (q) 20 µF connected in parallel to this combination at fourth capacitor of capacitance 1 µF is connected in series. The resultant capacity will be (C) A capacitor of capacitance 2 µF is charged to (r) 3 µF a potential difference of 200 V. After disconnecting from battery it is connected in parallel with another uncharged capacitor. The common potential becomes 20 V. The capacitance of second capacitor is (D) In the circuit diagram shown below, what is (s) 18 µF the effective capacitance between P and Q? SPACE FOR ROUGH WORK ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(II)/Qns - 9
• 10 PART B : CHEMISTRY SECTION I Straight Objective Type This section contains 9 multiple choice questions numbered 23 to 31. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 23. As the p - character decreases, the bond angle in hybrid orbitals formed by s and p - atomic orbitals (A) decreases (B) increases (C) doubles (D) remains unchanged 24. In OF2 molecule, the total number of bond pairs and lone pairs of electrons present respectively are (A) 2, 6 (B) 2, 8 (C) 2, 10 (D) 3, 10 25. In electrorefining of copper, some gold is deposited as (A) anode mud (B) cathode mud (C) cathode (D) electrolyte 26. Tautomerism is not exhibited by (A) (B) (C) (D) SPACE FOR ROUGH WORK ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(II)/Qns - 10
• 11 27. Consider the acidity of carboxylic acids: PhCOOH o - NO2C6H4COOH p - NO2C6H4COOH m - NO2C6H4COOH I II III IV Which of the following order is correct? (A) I > II > III > IV (B) II > IV > III >I (C) II > IV > I > III (D) II > III > IV > I 28. Which of the following will not yield tartaric acid? (A) Hydrolysis of glyoxal cyanohydrin. (B) Oxidation of fumaric acid with KMnO4 . (C) Treatment of argol first with Ca(OH)2 , then with CaCl2 and finally with H2SO4 . (D) Heating of cream of tartar. 29. The value of k for the reaction, 2A(g) B(g) + C(g) at 750 K and 10 atm pressure is 2.86. The value of k at 750 K and 20 atm is (A) 28.6 (B) 5.72 (C) 2.86 (D) 11.4 30. When equal volumes of the following solutions are mixed, precipitation of AgCl −10 (Ksp of AgCl is 1.8 × 10 ) will occur only in −4 + −4 − −5 + −5 − (A) 10 M Ag and 10 M Cl (B) 10 M Ag and 10 M Cl −6 + −6 − −10 + −10 − (C) 10 M Ag and 10 M Cl (D) 10 M Ag and 10 M Cl SPACE FOR ROUGH WORK ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(II)/Qns - 11
• 12 − − 31. For the reaction, NH3 + OCl → N2H4 + Cl occurring in basic medium, the coefficient of N2H4 in the balanced equation is (A) 1 (B) 2 (C) 3 (D) 4 SECTION II Assertion - Reason Type This section contains 4 questions numbered 32 to 35. Each question contains STATEMENT 1 (Assertion) and STATEMENT 2 (Reason). Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct. (A) Statement 1 is True, statement 2 is True; statement 2 is a correct explanation for statement 1. (B) Statement 1 is True, statement 2 is True; statement 2 is not a correct explanation for statement 1. (C) Statement 1 is True, statement 2 is False. (D)Statement 1 is False, statement 2 is True. 32. Statement 1: Lithium is the best reducing agent in aqueous solution. because + Statement 2: Hydration energy of Li ion is appreciably high. 33. Statement 1: Equivalent conductance of acetic acid at infinite dilution cannot be experimentally determined. because Statement 2: Acetic acid being an organic acid is not soluble in water. 34. Statement 1: The reaction, PCl5(g) PCl3(g) + Cl2(g) is favoured by temperature rise. because Statement 2: ∆S is positive for the above reaction. SPACE FOR ROUGH WORK ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(II)/Qns - 12
• 13 35. Statement 1: and can be distinguished using neutral ferric chloride. because Statement 2: Phenol is acidic but cyclohexanol is neutral. SECTION III Linked Comprehension Type This section contains 2 paragraphs. Based upon each paragraph, 3 multiple choice questions have to be answered. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct. Paragraph for Question Nos. 36 to 38 Consider a solution of Ba(NO3)2 containing 11.0 g in 100 g of water which boils at −1 −1 373.6 K. (Given: Kb for water = 0.52° molal ; Molar mass of Ba(NO3)2 = 259.34 g mol ) 36. The calculated value of elevation in boiling point of water in the above system is (A) 0.22 K (B) 0.6 K (C) 1.0 K (D) 0.0 K 37. van’t Hoff factor i is (A) 1.0 (B) 2.73 (C) 0.366 (D) 0.0 38. Percentage dissociation of Ba(NO3)2 is (A) 50 (B) 75 (C) 100 (D) 86.4 Paragraph for Question Nos. 39 to 41 The zero (or 18) group of periodic table consists of six gaseous elements namely He, Ne, Ar, Kr, Xe, and Rn. On account of their highly stable ns2p6 configuration in the SPACE FOR ROUGH WORK ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(II)/Qns - 13
• 14 valence shell, these gases have little tendency to undergo any reaction, hence they were called inert gases . However, due to finding of number of reactions of these elements, these are correctly called noble gases . 39. The following noble gas not present in the atmosphere is (A) Ne (B) Xe (C) Rn (D) Ar 40. Xe forms more number of compounds than the other noble gases (A) due to its lower ionization potential. (B) due to its higher electron affinity. (C) due to its electronic structure. (D) none of these 41. Charcoal at 100°C absorbs (A) Ne and Kr (B) He and Ne (C) He and Ar (D) Ar, Kr and Xe SECTION IV Matrix-Match Type This section contains 3 questions. Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (p, q, r, s) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-p, A-s, B-q, B-r, C-p, C-q and D-s, then the correctly bubbled 4 × 4 matrix should be as follows: p q r s A p q r s B p q r s C p q r s D p q r s SPACE FOR ROUGH WORK ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(II)/Qns - 14
• 15 42. Column I Column II (A) KMnO4 (p) Bleaching action 2+ (B) K2Cr2O7 (q) Fe (C) SO2(g) (r) Oxidising agent (D) Cl2(g) (s) Oxidation state = + 7 43. Column I Column II PV (A) Boyle temperature (p) =1 RT (B) NH3(g) (q) van der Waals equation (C) HCl(g) (r) CP − CV ≠ R a (D) Ideal gas (s) bR 44. Column I Column II (A) CHCl3 (p) freon (B) Cl2CF2 (q) white precipitate with alcoholic AgNO3 (C) CH = CH − CH Cl (r) chlorine is least reactive 2 2 (D) CH3 − CH = CHCl (s) responds carbylamine test SPACE FOR ROUGH WORK ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(II)/Qns - 15
• 16 PART C : MATHEMATICS SECTION I Straight Objective Type This section contains 9 multiple choice questions numbered 45 to 53. Each question has four choices (A), (B), (C), (D) out of which ONLY ONE is correct. 2 6 8 45. The radius of a circle that touches the parabola 75y = 64 (5x − 3) at , and 5 5 also x-axis is (A) 1 (B) 2 (C) 3 (D) 5 46. F1, F2, F3, F4 are the faces of a tetrahedron, V , V , V , V are the vectors 1 2 3 4 whose magnitudes are respectively equal to the areas of F1, F2, F3, F4 and whose directions are perpendicular to their faces in outward direction, then V V V V is equal to 1 2 3 4 (A) 1 (B) 2 (C) 3 (D) 0 2 2 2 47. In a triangle ABC, a + b + c = ca + ab 3, then the triangle is (A) equilateral (B) right angled and isosceles (C) with A = 90°, B = 60°, C = 30° (D) with A = 90°, B = 30°, C = 60° π 48. The value of ∫ esec x sec3 x [sin2 x + cos x + sin x + sin x cos x] dx is 0 1 1 (A) 0 (B) e (C) e (D) e e e 2/3 49. The critical points of the function f(x) = (x − 2) (2x + 1) is 1 (A) 2 and 3 (B) 1 and 2 (C) −1 and 2 (D) 1 and − 2 SPACE FOR ROUGH WORK ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(II)/Qns - 16
• 17 sin θ − cos θ 50. The minimum value of 2 +2 is 1 1 1 1 1 1 2 2 2 (A) 2 (B) 2 (C) 2 (D) 1 2 2 2 2 51. If the latus rectum of the ellipse x tan α + y sec α = 1 is 1 , then α is equal to 2 (where 0 < α < π), π π 2π 5π (A) (B) (C) (D) 6 3 3 12 k 2 n cos n 52. The value of Lim , where 0 < k < 1; is n→∞ n 1 (A) 0 (B) 1 (C) infinity (D) does not exist 1 1 1 x x 2 x x 2 x x 2 53. The value of 3 3 4 4 5 5 is x x 2 x x 2 x x 2 3 3 4 4 5 5 x −x (A) 1 (B) 0 (C) 60 (D) 60 SECTION II Assertion-Reason Type This section contains 4 questions numbered 54 to 57. Each question contains STATEMENT 1 (Assertion) and STATEMENT 2 (Reason). Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct. (A) Statement 1 is True, statement 2 is True; statement 2 is a correct explanation for statement 1. (B) Statement 1 is True, statement 2 is True; statement 2 is not a correct explanation for statement 1. SPACE FOR ROUGH WORK ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(II)/Qns - 17
• 18 (C) Statement 1 is True, statement 2 is False. (D)Statement 1 is False, statement 2 is True. 54. Statement 1: The number of common tangents to the circles 2 2 2 2 x + y + 2x + 8y − 23 = 0 and x + y − 4x − 10y + 19 = 0 is 1. because Statement 2: If two circles touch internally, the number of common tangent is 1. 55. Statement 1: Two regular polygons have their number of sides in the ratio 5 : 4, and the difference between their angles is 6°. Then the number of sides are 15 and 12 respectively. because 2n 4 Statement 2: Each interior angle of regular polygon of n sides is right n angle. 1 12 1 3 1 63 56. Statement 1: tan tan tan =π 5 4 16 because 1 1 1 x y Statement 2: tan x tan y=π tan , if x, y > 0 and xy > 1. 1 xy α α 2α 2α 57. Statement 1: If f α = cos i sin cos i sin .... k 2 2 2 2 k k k k α α cos i sin , then the value of Lim f π is 0. k k n n → ∞ because Statement 2: If n is positive, (cos θ1 + i sin θ1) (cos θ2 + i sin θ2) ... (cos θn + i sin θ ) n = cos (θ1 + θ2 + ... + θn) + i sin (θ1 + θ2 + ... + θn). SPACE FOR ROUGH WORK ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(II)/Qns - 18
• 19 SECTION III Linked Comprehension Type This section contains 2 paragraphs. Based upon each paragraph, 3 multiple choice questions have to be answered. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct. Paragraph for Question Nos 58 to 60 From any external point P two tangents can be drawn to a circle with centre C. PT1 and PT2 are the two tangents. θ If the angle between the two tangents is θ, then T PC = T PC = . 1 2 2 2 2 58. From the origin O, the tangents are drawn to the circle x + y + 4x − 8y + 7 = 0, meeting at P and Q, then the circumradius of ∆ OPQ is (A) 5 (B) 3 (C) 2 (D) 2 59. From a point P on the line 4x − 3y = 6; tangents are drawn to the circle 2 2 1 24 . Then P can be x + y − 6x − 4y + 4 = 0, inclined at an angle tan 7 (A) (6, 0) (B) ( − 6, 0) (C) (6, 6) (D) (2, 0) 2 2 60. Tangents are drawn from point P ( − 8, 0) to the circle x + y = 16 meeting at A and B. Then, the area of the quadrilateral PAOB is (A) 16 (B) 16 3 (C) 8 3 (D) 32 SPACE FOR ROUGH WORK ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(II)/Qns - 19
• 20 Paragraph for Question Nos 61 to 63 If E1, E2, E3 ... En are n mutually exclusive and exhaustive events and A is an event which takes place in conjunction with any one of Ei, then the probability of the event A P E P i E E i i Ei happening when the event A takes place is given by P = A n A ∑ P E i P i = 1 E i 61. In a factory, the machines A, B and C produce 25%, 35%, 40% products respectively. Of their total output 5, 4 and 2% are defective. A product is chosen and is found to be defective. The probability that it was manufactured by machine C is 16 25 13 1 (A) (B) (C) (D) 69 69 69 69 62. There are two boxes. First box contains 4 white and 5 black balls. Second box contains 6 white and 5 black balls. One box is selected at random, a ball is chosen and is found to be white. The probability that it has come from 2nd box is 17 23 12 27 (A) (B) (C) (D) 49 49 49 49 63. A letter is known to have come from either MAHARASHTRA or RAJASTHAN. On the postal mark, only consecutive letters RA can be read clearly. The chance that the letter come from MAHARASHTRA is 8 5 12 1 (A) (B) (C) (D) 13 13 13 13 SECTION IV Matrix-Match Type This section contains 3 questions. Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be SPACE FOR ROUGH WORK ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(II)/Qns - 20
• 21 matched with statements (p, q, r, s) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-p, A-s, B-q, B-r, C-p, C-q and D-s, then the correctly bubbled 4 × 4 matrix should be as follows: p q r s A p q r s B p q r s C p q r s D p q r s 64. Column I Column II a b c (A) If a, b, c are positive, then is (p) 3 b c c a a b greater than or equal to −1 −2 −4 3 (B) The sum to ∞ of the product (1 + 3 ) (1 + 3 ) (1 + 3 ) (q) −8 2 (1 + 3 ) ... is 2 3 4 5 6 (C) If (1 − p) (1 + 2x + 4x + 8x + 16x + 32x ) = 1 − p , then (r) 4 p the value of is x (D) If the second, third and sixth terms of an A.P. are (s) 2 the consecutive terms of a G.P., then the common ratio of the G.P. is 65. Column I Column II xf 3 3f x (A) If f (3) = 4, f ′(3) = 1, then Lim is (p) 6 x → 3 x 3 (B) If f (x + y) = f (x) f(y) for all x, y and f(4) = 2 and (q) 1 f ′(0) = 3, then f ′(4) is equal to SPACE FOR ROUGH WORK ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(II)/Qns - 21
• 22 1 (C) When 8 sec θ + cosec θ is minimum, (r) tan θ is equal to 3 cos x cos 2x cos 3x cos 2x cos 3x 1 (s) (D) If f x = 3 4 cos x 3 4 cos x , 2 1 cos x cos x 1 π⁄2 then ∫f x dx is equal to 0 66. Column I Column II (A) A has 5 different books and B has 8 different (p) 14400 books. The number of ways they can exchange their books so that each keeps his initial number of books, is (B) 5 boys and 3 girls are to be seated in a row. (q) 56 The number of ways in which they can be seated in a row, such that no two girls sit together is (C) The number of ways in which 5 prizes can be (r) 1286 given to 4 boys, given that each boy is eligible for all the prizes is (D) There are 3 bags each containing unlimited (s) 1024 number of balls of colour white, black, red and green. The number of ways in which 9 balls can be selected when every coloured variety is represented in the selection is SPACE FOR ROUGH WORK ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(II)/Qns - 22
• 23 SPACE FOR ROUGH WORK ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(II)/Qns - 23
• Name: . Enrollment No.: Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose. C. Question paper format: 13. The question paper consists of 3 parts (Physics, Chemistry and Mathematics). Each part has 4 sections. 14. Section I contains 9 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which only one is correct. 15. Section II contains 4 questions. Each question contains STATEMENT-1 (Assertion) and STATEMENT- 2 (Reason). Bubble (A) if both the statements are TRUE and STATEMENT-2 is the correct explanation of STATEMENT-1. Bubble (B) if both the statements are TRUE but STATEMENT-2 is not the correct explanation of STATEMENT-1. Bubble (C) if STATEMENT-1 is TRUE and STATEMENT-2 is FALSE. Bubble (D) if STATEMENT-1 is FALSE and STATEMENT-2 is TRUE. 16. Section III contains 2 paragraphs. Based upon each paragraph, 3 multiple choice questions have to be answered. Each question has 4 choices (A), (B), (C) and (D), out of which only one is correct. 17. Section IV contains 3 questions. Each question contains statements given in 2 columns. Statements in the first column have to be matched with statements in the second column. The answers to these questions have to be appropriately bubbled in the ORS as per the instructions given at the beginning of the section. D. Marking scheme: 18. For each question in Section I, you will be awarded 3 marks if you darken only the bubble corresponding to the correct answer and zero mark if no bubble is darkened. In all other cases, minus one (−1) mark will be awarded. 19. For each question in Section II, you will be awarded 3 marks if you darken only the bubble corresponding to the correct answer and zero mark if no bubble is darkened. In all other cases, minus one (−1) mark will be awarded. 20. For each question in Section III, you will be awarded 4 marks if you darken only the bubble corresponding to the correct answer and zero mark if no bubble is darkened. In all other cases, minus one (−1) mark will be awarded. 21. For each question in Section IV, you will be awarded 6 marks if you darken ALL the bubbles corresponding ONLY to the correct answer. No negative mark will be awarded for an incorrectly bubbled answer. ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(II)/Qns - 24
• 1 ® ® IIT-JEE 2008 STS VIII/PCM/P(II)/SOLNS BRILLIANT S HOME-BASED FULL-SYLLABUS SIMULATOR TEST SERIES FOR OUR STUDENTS TOWARDS IIT-JOINT ENTRANCE EXAMINATION, 2008 PAPER II - SOLUTIONS PHYSICS − CHEMISTRY − MATHEMATICS PART A : PHYSICS SECTION I 1. (A) As shown in Figure, we assume ACB to be a small unkinked section of rope subtending an angle ∆θ at O. We choose C at the midpoint of an arc. The centripetal force is provided by the sum of two tension forces FA and FB. Fnet = FA + FB In order that the force be directed towards O we must have |FA| = |FB| = F. ∆θ The net inward radial force = 2F sin = F ⋅ ∆θ, where ∆θ is small. 2 Equating this to the required centripetal force, we have 2 (mR∆θ) ω R = F ⋅ ∆θ since, mR (∆θ) is the mass of this section. 2 ∴ F = mω R ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(II)/Solns - 1
• 2 2. (A) Frequency of sonometer wire = 512 Hz Number of beats = 5 Frequency of pipe = 517 or 507 Hz Since the frequency of sonometer is reduced because of reduction in tension and since it produces unison after change in tension with organ pipe the frequency of organ pipe = 507 Hz. Frequency at 15°C = 507 Hz Let the temperature be raised to x°C so that the frequency becomes 512. v 273 15 288 15 = = v 273 x 273 x x v v 507 x 15 = 512 ∴ = 2l v 512 x 507 288 = 512 273 x Solving, x = 20.7°C 3. (A) A diatomic molecule has 5 degrees of freedom. Each degree of freedom has kT energy (3 translational and two rotational). 2 kT Two degrees of rotational motion = 2 ⋅ = kT 2 1 2 ∴ rotational kinetic energy = Iω 2 1 2 Iω = kT 2 ω= 2kT I 2 6 mv 20 × 10 × 4.8 4. (B) Bev = mv −2 or r = = = 4 × 10 m Be 3 r 1.2 × 2 × 10 1 1 1 = v u f ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(II)/Solns - 2
• 3 1 1 1 = v 18 12 v = 36 cm v 36 Magnification = = =2 u 18 −2 Radius of image circle = 2 × 4 × 10 m = 8 cm 5. (C) The current distribution is shown in Figure. Applying Kirchhoff s law, Va − Vb = (Va − Ve) + (Vc − Vd) = 10i1 + 5(i − i1) = 5i1 + 5i ... (1) Va − Vb = (Va − Ve) + (Vc − Vd) + (Vd − Vb) = 10i1 + 5(2i1 − i) + 10i1 = − 5i + 30i1 ... (2) Multiplying equation (1) by 6 and subtracting equation (2), we eliminate i1 5(Va − Vb) = 35i V V a b =7 i Equivalent resistance = 7 Ω R R R R 2R R 10 × 5 5×5 2 × 10 × 5 1 3 2 3 1 2 Aliter: R = = AB R R 2R 10 5 2 ×5 1 2 3 175 = 25 =7Ω ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(II)/Solns - 3
• 4 −6 6. (B) Q = 12 × 10 C, V = 1200 V 6 Q 12 × 10 −8 Capacitance = = = 10 farad V 1200 6 Dielectric strength = 3 × 10 V/m If t is thickness and operating voltage is 1200 V and x is the distance between plates 1200 = 3 × 106 x −4 or x = 4 × 10 m ε A 4 × 10 4 × 10 8 0 Capacitance = or A = t 12 8.85 × 10 2 A = 0.45 m 3 7. (A) l = 40 cm = 0.4 m, m = 3.2 × 10 0.4 1 T 1 T n= = 2l m 2l 3 3.2 × 10 0.4 1 0.4T 220 = 2 × 0.4 3 3.2 × 10 2 8 × 176 T= 3 10 2 T 8 × 176 11 2 Y= = = 1.98 × 10 N/m A × strain 3 6 0.05 10 × 1 × 10 × 40 2 mv GMm 2π r 8. (A) = ,v= r 2 T r 2 2 3⁄2 4π r GMm 2π r m⋅ = or T = 2 2 rT r GM ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(II)/Solns - 4
• 5 4 3 M= πr ρ and r = rp 3 3⁄ 2 2π r 3π p T= = 4 3 Gρ G⋅ πr ρ p 3 9. (C) Let the distance from ground to top be d. The ground is frame at rest and moving escalator is a moving frame. If v1 is the velocity of person for walking d time, t1 = . This is true when escalator is at rest. When it is moving with v 1 velocity v the time taken by person = d (t2). Finally the person walks on v steps while it is moving. Velocity of person with respect to ground = v1 + v d v v v 1 1 1 v 1 1 The time, t3 = or = = = v v t d d d t t 1 3 1 2 t t 3 × 1 = 0.75 minute 1 2 t3 = = t t 3 1 1 2 SECTION II 10. (D) 11. (A) The loss of weight in water is just equal to the total weight of the block. 12. (D) The frequency remains unchanged, the wavelength as well as the velocity undergo a change depending upon the refractive index of the medium. 13. (D) If the change in current is positive (increase), the self induced emf will tend to decrease the current. If the current is decreasing the self induced emf will tend to increase the current. SECTION III − t/RC 14. (B) In charging, q = q0 (1 − e ) q0 = CV −6 q0 = 50 × 10 × 12 = 600 µC ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(II)/Solns - 5
• 6 −6 15. (B) Time constant = RC = 300 × 50 × 10 = 15 millisecond q From the formula, we see that the charge reaches a value 0 in time t such that 2 = 1 − t′/RC e 2 t′ = loge 2 = 0.693 RC t′ = RC × 0.693 = 15 × 0.693 = 10.5 millisecond − t′/RC 90 1 16. (A) Similarly, e =1 = 100 10 t′ = loge 10 = 2.303 × 1 RC t′ = 2.303 × 1 × 15 = 34.5 millisecond 17. (A) The forces acting on the rod are the weight mg acting downwards and normal force N and frictional force, F = µN. The net torque, τ = Iα, where I is the moment of inertia of rod. 4 2 ∴ τ = mgl sin θ = ml ⋅ α 3 3g α= sin θ 4l dω dω dθ dω 3g α= = ⋅ = ω⋅ = sin θ dt dθ dt dθ 4l ω θ ∫ω dω = ∫ 3 g sin θ dθ 4 l 0 0 θ 2 ω 3 g = cos θ 2 4 l 0 2 3g ω = 1 cos θ 2l 3g ω= 1 cos θ 2l ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(II)/Solns - 6
• 7 18. (D) Linear acceleration is lα at right angles to the direction of rod. Resolving it horizontally and vertically, the components are lα cos θ and lα sin θ. The horizontal acceleration of centre of gravity due to force mg and N is 3 2 By Newton s law, mg − N = mlα sin θ = mg sin θ 4 Fmax = µN = mlα cos θ = 3 mg sin θ cos θ 4 3 2 mg 2 ∴ N = mg − mg sin θ = (4 − 3 sin θ) 4 4 19. (A) When θ = 30° slipping starts. At this F has its maximum value. 3 F mg sin θ cos θ max 4 3 sin θ cos θ 3 3 µ= = = = = 0.400 N mg 2 2 13 4 3 sin θ 4 3 sin θ 4 SECTION IV 20. (A) − (q); (B) − (p); (C) − (r); (D) − (s) γP 3P (A) v = ;c= ρ ρ v γ = c 3 γ (B) For adiabatic change PV = constant γ−1 γ dP Differentiating P ⋅ γ ⋅ V +V =0 dV dV 1 dP = ⋅ V γ P (C) PVγ = constant PV = RT Eliminating V between these equations, we get P1 − γ T γ = constant (D) γ 21. (A) − (q); (B) − (p); (C) − (s); (D) − (r) gl sin θ 4gl sin θ (A) In the case of cylinder, v = = 2 2 3 K ⁄R 1 ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(II)/Solns - 7
• 8 gl sin θ (B) Velocity of the sphere is given by 2 2 K ⁄R 1 where K is the radius of gyration of sphere, R is the radius. 10g l sin θ ∴v= 7 (C) Speed of opposite end of diameter = 2v (D) When it rolls, v = gl sin θ When it slides, v = 2gl sin θ 1 Ratio = 2 22. (A) − (q); (B) − (p); (C) − (s); (D) − (r) ε A (A) Capacity of air capacitor = 0 D Kε A If the space is filled with dielectric of constant K = 0 D ∴ capacity = 20 µF (B) Capacitors in parallel capacity = 1 + 1 + 1 = 3 µF 3 When 1 µF is connected in parallel with this C = µF = 0.75 µF 4 (C) Charge given to the first capacitor = 200 × 2 = 400 µC The common potential is 20 volts after connecting it in parallel. ∴ the capacity of the uncharged capacitor connected in parallel = 18 µF (D) When capacitors are connected in parallel, the effective capacity = C1 + C2 + C3 + ..... When capacitors are connected in series, 1 1 1 1 the effective capacity = = ... C C C C 1 2 3 Using the two formulae, we get effective capacity = 3 µF ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(II)/Solns - 8
• 9 PART B : CHEMISTRY SECTION I 3 2 23. (B) s-character increases in the order sp (25%), sp (33%), sp (50%). p-character decreases as sp3 (75%), sp2 (66%), sp (50%). Bond angle increases in the order sp3 (109°28′), sp2 (120°), sp (180°). Thus as p-character decreases, bond angle increases. 24. (B) In OF2 molecule, there are two bond pairs and eight lone pairs (two from O atom and three from each F atom) of electrons. 25. (A) In electrorefining of copper, impure copper is made as anode while pure copper plate is made as cathode. On passing current, impurities like Fe,Zn, Ni, Co dissolve in the solution and Au, Ag, etc., settle down at the bottom as anode mud. 26. (B) 27. (D) 28. (D) • Cream of tartar itself is potassium acid tartarate, and on heating it does not give tartaric acid. • Argol is potassium hydrogen tartarate. 29. (C) As temperature is the same, k is also the same. + − 30. (A) In (A), [Ag ] [Cl ] > Ksp . − − 31. (A) 2NH3 + 2OH → N2H4 +2H2O + 2e − − 2NH3 + OCl → N2H4 + Cl + H2 O. The coefficient of N2H4 is 1. SECTION II 32. (A) 33. (C) 34. (B) This is an endothermic reaction. Hence, the reaction is favoured with increase in temperature. 35. (B) ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(II)/Solns - 9
• 10 SECTION III 1000K w 1000 × 0.52 × 11 b 2 36. (A) ∆T = = = 0.22 K b w M 100 × 259.34 1 2 ∆ T obs b 37. (B) van t Hoff factor, i = ∆ T cal b 0.6 = 0.22 = 2.727 2.73 2+ 38. (D) Ba(NO3)2 Ba + 2NO (3 species) 3 i 1 2.727 1 1.727 α = = = = 0.8635 n 1 3 1 2 i.e., α = 86.35 86.4% 39. (C) 40. (A) 41. (D) SECTION IV 42. (A) − (q), (r),(s); (B) − (q), (r); (C) − (p), (r); (D) − (p), (r) 43. (A) − (s); (B) − (q), (r) ,(s); (C) − (q), (r), (s); (D) − (p) 44. (A) − (s); (B) − (p); (C) − (q); (D) − (r) ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(II)/Solns - 10
• 11 PART C : MATHEMATICS SECTION I 45. (A) Equation of the parabola is 2 75y = 32 (10x − 6) Tangent at (x1, y1) is 75yy1 = 32 (5x + 5x1 − 6) 6 8 Put x 1 = , y = 1 5 5 The equation to the tangent is 3y = 4x 4 Slope = 3 3 Slope of common normal = 4 Equation to the common normal is 6 8 x y 5 5 = = r. 4 3 5 5 Circle touches x-axis ∴ y-coordinate of the centre = radius. 8 r 5 ∴ = r ⇒ r = 4 3 5 Put r = − r in RHS 8 r 5 = r 3 5 ⇒r=1 ∴ one radius is 4 and the other radius is 1 ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(II)/Solns - 11
• 12 46. (D) Let OABC be the tetrahedron, O being the origin. Let OA = a, OB = b, OC = c Then V = 1 a × b 1 2 1 V = b ×c 2 2 1 V = c ×a 3 2 1 V = c a × b a 4 2 1 = c × b c × a a × b 2 ∴ V V V V =0 1 2 3 4 ⇒ V V V V = 0 1 2 3 4 47. (C) We have 2 2 2 a + b + c − ca − ab 3 = 0 2 2 a 3 a b c = 0 2 2 This is possible if a 3 a b = 0 and c = 0. 2 2 3a = 2b; a = 2c k k c = k ⇒a = ; b = ; k is a constant. 3 2 2 3 2 2 2 2 1 1 2 4 k 2 b c = k = k ⋅ = = a 4 12 12 3 A = 90° b 3 sin B = = a 2 ∴ B = 60° ⇒ C = 30° ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(II)/Solns - 12
• 13 π π π sec x 2 sec x 2 48. (C) ∫e tan x sec x dx ∫e sec x dx ∫ e sec x sec 2 x tan x dx 0 0 0 π ∫ e sec x sec x tan x dx 0 π π π sec x 2 sec x sec x = ∫ e tan x sec x dx ∫ e d tan x dx ∫ sec x d e dx 0 0 0 π ∫ e sec x sec x tan x dx 0 π π π sec x 2 sec x = ∫e tan x sec x dx e tan x 0 ∫ tan x e sec x ⋅sec x tan x dx 0 0 π π π sec x sec x e sec x 0 ∫e sec x tan x dx ∫ e sec x sec x tan x dx 0 0 π sec x = e sec x 0 −1 1 =e ( − 1) − e ⋅ 1 1 =− e e 2/3 2 −1/3 49. (B) f ′(x) = (x − 2) ⋅ 2 + (2x + 1) ⋅ ⋅ (x − 2) 3 2 3x 2 2x 1 = 1⁄ 3 3 x 2 2 5x 5 = 1⁄ 3 3 x 2 f ′(x) = 0 ⇒ x = 1 The point at which f ′(x) does not exist is also a critical point. So x = 2 is also a critical point. ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(II)/Solns - 13
• 14 50. (B) Applying A.M. ≥ G.M. sin θ cos θ 2 2 sin θ cos θ ≥ 2 ⋅2 2 1 sin θ cos θ sin θ cos θ 1 2 2 2 ≥ 2 ⋅2 1 π 1 2 sin θ 2 4 ≥2 1 π 1 sin θ 2 4 ≥2 π For minimum value, sin θ =−1 4 1 1 2 ∴ minimum value = 2 51. (D) The equation of the ellipse 2 2 x y = 1. 2 2 cot α cos α 2 Latus rectum = 2b = 1 a 2 2 4b = a 2 4 cos α = cot α 1 π 4 cos α = Q α ≠ sin α 2 2 sin 2 α = 1 1 sin 2 α = 2 n π 2 α = n π + ( − 1) 6 π Put n = 0 ⇒ 2α = 6 π ⇒α = 12 ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(II)/Solns - 14
• 15 π Put n = 1 ⇒ 2α = π − 6 5π = 6 5π α = 12 k 2 n cos n 52. (A) Lim n → ∞ 1 n 1 n 1 2 = Lim cos n n → ∞ 1 k 1 n 1 n 2 cos n is a bounded function and 0 < k < 1. ∴ above limit = 0. 53. (B) Applying R2 → R2 − R3 we get 1 1 1 x x x x x x Det = 2 ⋅ 2 ⋅ 3 ⋅ 3 2 ⋅2 ⋅4 ⋅4 2 ⋅2 ⋅5 ⋅5 2 2 2 x x x x x x 3 3 4 4 5 5 1 1 1 1 1 1 = 4 2 2 2 x x x x x x 3 3 4 4 5 5 =4⋅0=0 SECTION II 54. (B) Centre A of I circle = ( − 1, − 4) Radius r1 of I circle = 1 16 23 = 40 = 2 10 Centre B of II circle = (2, 5) Radius r2 of II circle = 4 25 19 = 10 Distance between the centres = 9 81 = 90 = 3 10 ∴ distance = r1 + r2 The two circles touch externally. Number of common tangents = 3. ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(II)/Solns - 15
• 16 55. (A) Let n1 and n2 be the number of sides of the two regular polygons. n 5 1 = n 4 2 2n 4 2n 4 1 2 Also 90° 90° = 6° n n 1 2 4 4 1 = n n 15 1 2 1 1 1 = n n 60 2 1 1 1 1 = 4 n 60 n 1 1 5 1 1 1 = 4 n 60 1 n1 = 15 4 n = × 15 = 12 2 5 12 3 56. (A) ⋅ >1 5 4 12 3 −1 12 −1 3 1 5 4 ∴ tan + tan =π tan 5 4 12 3 1 ⋅ 5 4 1 63 =π tan 16 1 63 =π tan 16 ∴ L.H.S = π ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(II)/Solns - 16
• 17 π π 2π 2π nπ nπ 57. (D) fn(π) = cos i sin cos i sin ... cos i sin 2 2 2 2 2 2 n n n n n n π 2π ... nπ π 2π ... nπ fn(π) = cos i sin 2 2 n n n n 1π n n 1π = cos i sin 2 2 2n 2n 2 1 2 1 n 1 π n 1 π n n = cos i sin 2 2 2n 2n 1 π 1 π = cos 1 i sin 1 n 2 n 2 π π ∴ Lt f π = cos i sin n n → ∞ 2 2 =0+i⋅1=i SECTION III 58. (A) Centre C is ( − 2, 4) Evidently O, the origin is an external point of the circle. OPC = 90° = OQC. ∴ P and Q lie on the circle on OC as diameter. It is the circumcircle of ∆ OPQ. 2 2 ∴ OC = 2 4 = 20 1 1 Circumradius = 20 = ×2 5 = 5 2 2 59. (C) Centre C is (3, 2) Angle between them 1 24 2α = tan 7 24 tan 2 α = 7 ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(II)/Solns - 17
• 18 2 tan α 24 = 2 7 1 tan α (8 tan α − 6) (3 tan α + 4) = 0 6 3 4 ∴ tan α = = ; ∴ tan α = − , not possible α is acute. 8 4 3 CT1 = Radius = 9 4 4 = 3 ∴ PT2 = 4 ∴ PC = 5 If P is (α, β), then (α − 3)2 + (β − 2)2 = 25 and 4α − 3β = 6. Solving, α = 6 or 0 β = 6 or − 2 P can be (6, 6) 60. (B) Radius OA = 4 PO = 8 4 1 sin θ = = ; θ = 30° 8 2 ∴ A O P = 60° 1 Area of ∆ OAP = OA ⋅ PO sin 60° 2 1 3 = × 4 ×8 × 2 2 =8 3 ∴ area of the quadrilateral = 2 × 8 3 = 16 3 25 61. (A) P(E1) → P (A is production) = 100 35 40 P E = , P E3 = 2 100 100 E − Probability that the product is defective P(E/E1) = 5 100 P(E/E2) = 4 100 ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(II)/Solns - 18
• 19 P(E/E3) = 2 100 P E P E⁄E 3 3 P(E3/E) = P E P E⁄E P E P E⁄E P E P E⁄E 1 1 2 2 3 3 40 2 ⋅ 100 100 = 25 5 35 4 40 2 ⋅ ⋅ ⋅ 100 100 100 100 100 100 16 = 69 62. (D) E1 - I box is chosen; E2 → II box is chosen 1 1 P E = P E = 1 2 2 2 E - ball is white P(E/E1) = 4 ; P E⁄E = 6 2 9 11 P E P E⁄E 2 2 P E ⁄E = 2 P E P E ⁄E P E P E⁄E 1 1 2 2 1 6 6 ⋅ 2 11 11 = = 1 4 1 6 98 ⋅ ⋅ 2 9 2 11 99 6 ×9 54 27 = = = 98 98 49 63. (A) E1 − MAHARASHTRA chosen E2 − RAJASTHAN chosen 1 1 P E = P E = 1 2 2 2 E → Consecutive letters are R and A M A H A R A S H T R A (1) (2) (3) (4) (5) (6) (7) (8) (9) (10) → Favourable = 2 Exhaustive = 10 P(E/E1) = 2 = 1 10 5 ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(II)/Solns - 19
• 20 R A J A S T H A N Favourable = 1 (1) (2) (3) (4) (5) (6) (7) (8) Exhaustive = 8 P(E/E2) = 1 8 1 1 P E ⋅ P E ⁄E ⋅ 1 1 2 5 P E ⁄E = = 1 P E ⋅ P E⁄E P E P E ⁄E 1 1 1 1 1 1 2 2 ⋅ ⋅ 2 5 2 8 1 5 8 = = 1 1 13 5 8 SECTION IV 64. (A) − (q); (B) − (q); (C) − (s); D − (p) 1 a b c a b c 3 (A) Applying A.M. ≥ G.M. ≥3 ⋅ ⋅ b c c a a b b c c a a b 1 Also b c c a a b b c c a a b 3 ≥3 ⋅ ⋅ a b c a b c Also b c c a a b = b a c a b c a b c a b a c c b which is ≥ 2 + 2 + 2 i.e., ≥ 6 a b c 1 ∴ we get ≥ 9× b c c a a b 6 3 i.e., ≥ 2 (B) Upto n terms, n −1 −2 2 product P = (1 + 3 ) (1 + 3 ) ... 1 3 −1 −2 −2 P(1 − 3 ) = (1 − 3 ) (1 + 3 ) −4 = (1 − 3 ) ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(II)/Solns - 20
• 21 Ultimately 1 1 P1 3 = 1 2n 1 3 1 1 P = 1 1 2n 1 1 3 3 3 1 = 1 2 2n 1 3 3 As n → ∞ the above = 2 6 2 3 4 5 1 2x (C) 1 + 2x + 4x + 8x + 16x + 32x = 1 2x 6 1 p = 1 p Comparing; p = 2x is a solution p ⇒ = 2 x a 2d a 5d 3d (D) Now = ; each = =3 a d a 2d d ∴ common ratio = 3 65. (A) − (q); (B) − (p); (C) − (s); (D) − (r) (A) Applying L′ Hospital rule, the required limit f 3 3f ′ x = Lt x → 3 1 4 3 ⋅1 = =1 1 (B) Put y = 0 ∴ f(x + 0) = f(x) ⋅ f(0), ∀x ⇒ f(0) = 1 ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(II)/Solns - 21
• 22 f 4 h f 4 f′ 4 = Lt h → 0 h f 4 f h f 4 = Lt h → 0 h f h f 0 = f 4 Lt h → 0 h = f(4) ⋅ f ′(0) = 2 ⋅ 3 = 6 (C) f(θ) = 8 sec θ + cosec θ f ′(θ) = 8 sec θ tan θ − cosec θ cot θ f ′(θ) = 0 ⇒ tan3 θ = 1 8 1 ⇒ tan θ = 2 For this f ″(θ) is +ve. 1 ∴ tan θ = gives minimum. 2 (D) In the determinant, applying C1 = C1 − (C2 + C3) cos x cos 2x cos 3x we get f x = 0 3 4cos x 0 cos x 1 2 = cos x (3 − 4 cos x) = − cos 3x π π 2 2 ∫f x dx = ∫ cos 3x dx 0 0 π sin 3x 2 = 3 0 1 1 = 0 = 3 3 ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(II)/Solns - 22
• 23 66. (A) − (r); (B) − (p); (C) − (s); (D) − (q) (A) On the whole there are 5 + 8 = 13 books. Then A can select 5 books in 13 C5 ways. This includes 1 way in which he can select his original books. ∴ required number of ways = 13C5 − 1 = 1286. (B) 5 boys can be seated in 5 ways = 120 Number of spaces for girls = 4 + 2 = 6 [4 gaps 2 extremes] 6 3 girls can be arranged in P3 ways = 120 Required number of ways = 120 × 120 = 14400 (C) Any one prize can be given to any one of the 4 boys. For each prize there are 4 choices and there are 5 prizes. Required number of ways = 45 = 1024 (D) The required number of ways = Coefficient of t9 in (t + t2 + ... )4 9 4 4 = Coefficient of t in t (1 + t + ... ) 5 −4 = Coefficient of t in (1 − t) = (4 + 5 − 1)C5 8 8 = C5 = C3 = 56 ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(II)/Solns - 23