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(a) | | 2 c − | a | =12 2 (b) | | 2 c + | a | =30 2 (c) | a× b + c × a| = 48 3 (d) a ⋅ b =− 72 44. The tangent PT and the normal PN to the parabola y 2 = 4ax at a point P on it meet its axis at points T and N, respectively. The locus of the centroid of the triangle PTN is a parabola whose ⎛ 2a (a) vertex is , ⎝ ⎜ 3 0 ⎞ ⎠ ⎟ (b) directrix is x = 0 (c) latus rectum is 2 a (d) focus is (a, 0) 3 45. Let f : R R be such that f (2x – 1) = f (x) for all x ∈ R. If f is continuous at x = 1 and f (1) = 1, then (a) f (2) = 1 (b) f (2) = 2 (c) f is continuous only at x = 1 (d) f is continuous at all points 46. The angle of intersection between the curves y = [|sinx| + |cosx|] and x 2 + y 2 = 10, where [x] denotes the greatest integer x, is (a) tan –1 (3) (b) tan –1 (–3) −1 (c) tan ( 3 ) (d) tan −1 ( 1/ 3) 47. If the equation x 2 + y 2 – 10x + 21 = 0 has real roots x = a and y = b then (a) 3 x 7 (b) 3 y 7 (c) –2 y 2 (d) –2 x 2 48. If [x] denotes the greatest integer x, then the value of lim | | [cos x x ] is x →0 (a) 0 (b) 1 (c) –1 (d) does not exist 49. If A, B are two events such that P(A ∪ B) ≥ 3 4 and 1 3 ≤PA ( ∩B) ≤ then 8 8 (a) PA ( ) + PB ( ) ≤ 11 (b) PA ( ) ⋅PB ( ) ≤ 3 8 8 (c) PA ( ) + PB ( ) ≥ 7 (d) None of these 8 50. If cosx and sinx are solutions of the differential equation a d 2 y a dy 0 + a y 2 1 + 2 = 0, where a 0 , a 1 , a 2 dx dx are real constants then which of the following is/are always true? (a) Acosx + Bsinx is a solution, where A and B are real constants. ⎛ π⎞ (b) Acos x+ ⎝ ⎜ 4⎠ ⎟ is a solution, where A is a real constant. (c) Acosx sinx is a solution, where A is real constant. ⎛ π⎞ (d) Acos⎜ x+ Bsin x ⎝ ⎠ ⎟ + ⎛ ⎜ ⎝ − π⎞ ⎟ is a solution, 4 4⎠ where A and B are real constants. SECTION 3 (MAXIMUM MARKS : 12) • This section contains TO paragraphs. • ased on each paragraph, there are TO questions. • Each question has FOUR options (a), (b), (c) and (d). ONLY ONE of these four options is correct. • For each question, darken the bubble corresponding to the correct option in the ORS. • For each question, marks will be awarded in one of the following categories : Full Marks : +3 If only the bubble corresponding to the correct option is darkened. Zero Marks : 0 In all other cases. PARAGRAPH 1 Consider the circle x 2 + y 2 = 9 and the parabola y 2 = 8x. They intersect at P and Q in the first and the fourth quadrants, respectively. Tangents to the circle at P and Q intersect the x-axis at R and tangents to the parabola at P and Q intersect the x-axis at S. 51. The ratio of the areas of the triangles PQS and PQR is (a) 1 : 2 (b) 1 : 2 (c) 1 : 4 (d) 1 : 8 52. The radius of the circumcircle of the triangle PRS is (a) 5 (b) 3 3 (c) 3 2 (d) 2 3 PARAGRAPH 2 A f ai r d i e i s t o s s ed r ep eat ed l y u n t i l a s i x i s o b t ai n ed . L et X d en o t e t h e n u m b er o f t o s s es r eq u i r ed . 53. The probability that X ≥ 3 equals (a) 125 216 (b) 25 36 (c) 5 36 (d) 25 216 54. The conditional probability that X 6 given X 3 equals 125 25 5 (a) (b) (c) (d) 25 216 216 36 36 CHEMISTRY TODAY | APRIL ‘17 77
SOLUTIONS PAPER-I 1. (a) : The photon of energy 10.2 eV excites the electron from n = 1 to n = 2 as E 2 – E 1 = – 3.4 eV – (– 13.6 eV) = 10.2 eV The electron returns to the ground state in less than a microsecond and releases a photon of energy 10.2 eV. As the ionisation energy is 13.6 eV, the second photon of 15 eV energy ionises the atom by ejecting an electron and the balance of energy (15 eV – 13.6 eV = 1.4 eV) is retained by the ejected electron. 2. (b): The rod will rotate about A. Therefore, A from conservation of mechanical energy, Decrease in gravitational potential energy = increase in rotational kinetic energy v c C about A or mg l 1 2 = I 2 2 A ω or mg l ml 2 1 2 = ⎛ 2 2⎝ ⎜ ⎞ 3 ⎠ ⎟ ω ω 2 3 = g …(i) l Centripetal force of COM of rod in this position is m l 2 3 ω = mg (towards A) 2 2 Let F be the force exerted by the hinge on the rod upwards. Then mg F− mg = 3 5 ∴ F = mg 2 2 or force exerted by the rod on the hinge is 5 2 mg downwards. 3. (a) : It is clear from figure (0, a ) ( m 3 )2 kg that coordinates of centre of mass C, mx mx mx XCM = 1 1+ 2 2 + 3 3 m1+ m2 + m3 6 = × 0 + 2 × 0 + 2 × a a = 6+ 2+ 2 5 ∴ = ⎛ ⎝ ⎜ a a⎞ ( XCM , YCM ) , ⎠ ⎟ 5 5 Hence OC a a = ^ i + ^ j 5 5 Angle made by OC with x-axis −1⎛ y⎞ = ⎝ ⎜ ⎟ ⎠ = −1⎛ a/ 5⎞ tan tan ⎝ ⎜ ⎟ 5⎠ = 45 x a/ ° ( m 1 ) 6 kg ( a, 0) O (0, 0) ( m 2 )2 kg 4. (a) : Let v v ̂i v ̂ r = rx + ry j and v ̂ m = 5 i (in 1st case) vrm = ( vrx − vm)^ i + v ^ ry j vry Case (i) : tan90°= or v rx = 5 m s –1 vrx − 5 Case (ii) : v = ( 5̂i− 10 ̂i) + v ̂j ∵ v 10̂i rm ry ( m = ) v ry tan60°= or v 5 − 10 ry =−5 3 v ̂i ̂ r = 5 −5 3 j v −1 r = 10 m s ⎛ − ⎞ − ∠ φ= tan 1 5 3 ⎜ ⎟ or φ= 120° ⎝ 5 ⎠ 5. (c) ^ ^ 6. (c, d) : Here, E= E0 jB , = B0 j y v If θ = 0°, then due to magnetic E B force path is circular but due to electric force qE 0 () q will have x accelerated motion along y-axis. +q So <strong>com</strong>bined path of q will be a helical path with variable pitch. So (a) and (b) are wrong. If θ = 10° then due to vcosθ, path is circular and due to qE 0 and vsinθ, q has accelerated motion along y-axis so <strong>com</strong>bined path is a helical path with variable pitch. So (c) is correct. If θ = 90° then F B = 0 and due to qE 0 motion is accelerated along y-axis. So (d) is correct. 7. (a, b, c) : Here, t = 8 hours As N = N 0 e –λt N t or = e −λ N0 0.0039 = e –λ8 or e λ8 1 = 0 . 0039 e λ8 = 256 or e λ8 = 2 8 Taking natural logarithm on both sides, we get 8λ = 8ln2 or λ = ln2 per hour Option (c) is correct. ln2 T 12 / = = 1 hour λ Option (a) is correct. 1 1 Mean time, τ = = hour λ ln2 Option (b) is correct. ⎛ 1 ⎜ ⎞ ⎠ ⎟ ⎛ N = N ⎝ ⎜ ⎞ ⎝ 8 1 ⎠ ⎟ 2 1 8 7 ( 10) = × 10 = = 5 2 × 10 2 2 Option (d) is incorrect. 8. (a, b, c) 9. (b, c, d) 10. (b, d) 78 CHEMISTRY TODAY | APRIL ‘17
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Volume 26 No. 4 April 2017 Managing
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(R) K[Co(NH ) (NO ) 4 ] 1. The volu
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24. m-Bromoaniline can be prepared
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(a) (b) (c) (d) CH CH C CH CH CH C
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20. (d) : CH 2 5 alk. KMnO 4 ( A) C
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41. (a) : According to Gibbs’-Hel
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5. Compound 'X' C 7 H 8 O, is insol
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(c) (d) O HC 2 O O HC 2 O O CH CH C
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4. As 2 S 3 sol carries a negative
Page 21 and 22: 15. Cl CHONa + EtOH 2 5 Heat Possib
Page 23 and 24: (c) If θ = 10°, the charge underg
Page 25 and 26: (c) (d) NH 2 O heat NH HO 2 C CO 2
Page 27 and 28: (c) Dil. HCl followed by reaction w
Page 29 and 30: 33. Which of the following reaction
Page 31 and 32: Balz-Schiemann reaction : tetrauoro
Page 33 and 34: 26. (c) : HC 3 HC 3 -H O CH 2 3 H +
Page 35 and 36: 12. At equimolar concentration of F
Page 37 and 38: (a) 1s 2 , 2s 2 2p 6 , 3s 2 3p 6 3d
Page 39 and 40: Fraction of M present as M 2+ = 88
Page 41: 26. (b) : According to Gay-Lussac
Page 47 and 48: CONCEPT MAP PERIODICITY IN PROPERTI
Page 49 and 50: Class XII This specially designed c
Page 51 and 52: JEE MAIN / JEE ADVANCED / PETs Only
Page 53 and 54: CHEMISTRY MUSING SOLUTION SET 44 24
Page 55 and 56: Isothermal reversible expansion :
Page 57 and 58: (a) H 2 O/H + , BH 3·THF/H 2 O 2·
Page 59 and 60: 22. H 2 can be obtained from 28. Th
Page 61 and 62: 7. Predict the product for the foll
Page 63 and 64: 7. (d) : Cyclopropane is under seve
Page 65 and 66: 46. If the first and the (2n - 1) t
Page 67 and 68: 1 (a) 6 2 ms − (b) 4 2 (c) 7 m s
Page 69 and 70: tension, viscous force, weight but
Page 71: (a) HN 2 SO 3 H (b) (NH 2 ) 2 CO (c
Page 75 and 76: 11. (a, d) : As, F 1 = k 1 x, F 2 =
Page 77 and 78: Given that, at equilibrium, ⎛ n
Page 79 and 80: 2 ∫ 53. (3) : (| x− 2| + [ x])
Page 81 and 82: percentage change in v = 1 2 perce
Page 83 and 84: Again, c 2 a 2 b 2 = + + 2a⋅b
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