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2. 2. 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. 3. 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. 4. 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. 5. 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. 6. 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. 7. 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. 8. 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. 9. 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. 10. 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. 11. 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. 12. 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. 13. 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. 14. 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. 15. 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. 16. 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. 17. 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. 18. 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. 19. 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. 20. 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. 21. 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. 22. 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. 23. 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. 24. 24 SPACE FOR ROUGH WORK ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(I)/Qns - 24
25. 25. 25 SPACE FOR ROUGH WORK ◊ Brilliant Tutorials Pvt. Ltd. IIT/STS VIII/PCM/P(I)/Qns - 25
26. 26. 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
27. 27. 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
28. 28. 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
29. 29. 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
30. 30. 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
31. 31. 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
32. 32. 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
33. 33. 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
34. 34. 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
35. 35. 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
36. 36. 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
37. 37. 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
38. 38. 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
39. 39. 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
40. 40. 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
41. 41. 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
42. 42. 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
43. 43. 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