INDIA XI_IC_SPARK ime : 3 Hours JEE-MAIN CP -5 Date: 5-0-8 Max Marks : 360 KEY SHEE PHYSICS.. 3. 4. 5. 6. 7. 8. 9. 0... 3. 4. 5. 6. 7. 8. 9. 0... 3. 4. 5. 6. 7. 8. 9. 30. CHEMISRY 3. 3. 33. 34. 35. 36. 37. 38. 39. 40. 4. 4. 43. 44. 45. 46. 47. 48. 49. 50. 5. 5. 53. 54. 55. 56. 57. 58. 59. 60. MAHEMAICS 6. 6. 63. 64. 65. 66. 67. 68. 69. 70. 7. 7. 73. 74. 75. 76. 77. 78. 79. 80. 8. 8. 83. 84. 85. 86. 87. 88. 89. 90.
XI-IC-Spark_CP-5 [5/0/8] PHYSICS. Z e Z e x x e x 4 K= 4 Hence wave velocity 4m / sec along +x-axis direction K. General wave equation along the +ve x-axis y A sin t kx 0 A for initial phase 0 particle at X=0 approaches the mean. Below the mean. Hence 0 6 y= sin t kx 6 y sin t kx 6 3. Velocity of the transverse pulse V 0 00m / s 0.00 Speed of the transverse pulse is almost constant. 0.0 V 00 ime taken 4. ransverse wave velocity V 5. 00 0.0 00 N mg sin 00 m.0 sin 300 m 0 CD AB V AB V CD 6. V 7. A.s r.s V r VA rb VB ra y A sin kx t A y cos kx t Amplitude A frequency Page
8. XI-IC-Spark_CP-5 [5/0/8] he phase difference between two particles having their relative velocity zero is x x 9. 0. Pulse moves relative to the string l V0t at t V0 V0 al a V0 / M a g /3.. Put t = sec y e x with centre of symmetry at x = d dm. R sin d Rd R dm.v / r d R 3. d Vw R R Also speed of string is R he velocity of disturbance w.r.t. ground R R R. As shown in the curve, if wave is moving Along x axis, V p is positive. y A 60 x s Vp 4. tan V p V tan Vw he intensity of sound is proportional to (amplitude) x (frequency) 00 a 3, n 50 50 I 3 50 9 4 a 4, n 75 : 4 I 4 75 6 9 Page 3
XI-IC-Spark_CP-5 [5/0/8] 6. I a f 3 8 I a f Intensity of sound reduces by 0 % through the first slab, therefore transmitted intensity is 80 %. hrough the second slab, it again decrease by 0 % of 80, i.e., by 6. hus, the total decrease in intensity is 0 + 6 = 36% 7. SL SL SL 0log0 8. 400 SL 0 log0 0log0 0 3dB 0 R v M v M 7/5 4 v M 5/3 5 9. Maximum particle velocity = 4 (wave velocity) 5. I P 0log0 I P y 0 4 f y 0 f 4 f 0. y0 y 5sin t A 5 4 y sin t cos t y sin t cos t A 88 4 A 5 A 4 y sin t cos t It can be written as,.. y sin t 4 Amplitude m At t 0, y sin m 4 y y y y 3 a sin t 45 a sin t a sin t 45 a[sin t 45 sin( t 45 )] a sin t a[sin t cos 45 ] a sin t a sin t (a) Amplitude = a Page 4
XI-IC-Spark_CP-5 [5/0/8] (b) he motion will be at 45 phased difference with the first one (c) Energy (amplitude)² a ER 3 E a 3. 4. 5. A particle oscillating under a force F kx bv is a damped oscillator. he first term kx represents the restoring force and second term bv represents the damping force As v L Y vl Y r Y Y and v r v / r stress Y 0 stress = 07 N / m v L / v 00 6. he given equation of wave is x t t y 5sin cm / s cos 0.04 0.04 4 he particle velocity of wave is vp dy x t 5 cm / s cos dt 0.04 0.04 4 cos has maximum value + x t, 0.04 4 For maximum particle velocity cos 7. 8. v p max 5 cm / s 5cm / s.5m / s 0.04 Amplitude of foscillationssatany instant t is A A 0e bt /m When t 00, A A0 3 Let the amplitude be A at t = 00 A0 A 0e 00h /m 3 And A ' A 0e 00h/m From (i) e 00b/m 3 A' A From (ii), e 00b/m or A'= 0 A0 9 9 A= 9..(i).(ii) he trajectory of motion in the particle will be a circle of radius a Page 5
XI-IC-Spark_CP-5 [5/0/8] CHEMISRY 3. 3. 33. 34. 37. Maximum branching, maximum octane number. Alkene contains minimum two carbons HI / P CH 3 CH CH COOH CH 3 CH CH CH 3 Reaction is Wolf-Kishner reduction he first structure contains two six membered rings. he second structure contains one fivemembered ring and one seven-membered ring and the third structure contains one four- membered ring and one seven-membered ring Reactivity of six, seven and eight membered ring compounds is almost same but reactivity of four and five membered ring compounds is more reactive than five,and six membered ring compounds. MAHEMAICS 6. Eliminating t from the given two equation, we have x y 6 48, Whose eccentricity is e 6. 48. 6 Let the asymtotes be x 3 y 0 and x y 0 It will pass through centre (,). Hence, 8, 5 he equation of the hyperbola is. ( x 3 y 8)( x y 5) 0 It passes through (.4), (4+-8) (+8-5)+ 0 It passes through (, 4) (4+-8) (+8-5)+ 0 40 Hence, equation of hyperbola is ( x 3 y 8)( x y 5) 40 63. Page 6
XI-IC-Spark_CP-5 [5/0/8] Let CP r be inclines to transverse axis at an angle so that P is (rcos, r sin ) and P Lies on the hyperbola. It gives cos sin r a b Replacement by 900, we have sin cos r a b cos sin sin cos r r a b a b cos sin X r r a b a b 64. r CP r a CQ b a b Normals ar p( ), Q( ) are ax cos + by cot = a + b ax cos + by cot = a + b where and these pass through (h, k) herefore, ah cos bk cot a b and ah sin bk tan a b Eliminating h, we have bk cot sin tan cos a b sin cos 65. a b k b Let a pair of tangent be drawn from point (x, y) to hyperbola x y 9 hen chord of contact will be xx yy 9..(i) But the given chord of contact is x 9..(ii) As Eqs. (i) and (ii) represent the same line, these equation should be identical and hence Page 7
XI-IC-Spark_CP-5 [5/0/8] x y 9 x, y 0 0 9 herefore, the equation of pair of tangent drawn from (, 0) to x y 9 is x y 9 0 9 x. y.0 9 x y 9 8 x 9 (using SS ) 8 x 8 y 7 x 8 x 8 9 x 8 y 8 x 9 0 66. Equation of tangent to hyperbola x y 4 at any point x, y is xx yy 4 Comparing with x 6 y or 4 x 6 y 4, we have x 4 and y 6 4, 6 is the required point of the contact 67. he given hyperbola is x y 4 a, b, e 3 b herefore, the required area a e a 3 3 3 68. Any tangent to y = 8x = mx m if x mx has equal roots m 3 4m m. m Hence tangent is y = x +. 69. CS p Page 8
XI-IC-Spark_CP-5 [5/0/8] Coordinate of S is either h p, k p or h p, k p 70. If eccentricities of ellipse and hyperbola are e and e Foci ( ae, 0) and ( ae 0) Here, ae = ae a e a e b b a a a a a b a b 44 8 5 b 9 5 5 b 6 74. ( x )( x )( x ) ( x )( x ) x3 x x 75. Equation of tangent at P is xh a yk a put x in equation x y a b b h yk y k yk a 4 y a b4 b h k a 4 ka 4 a4 y y a b 4h bh h Page 9
XI-IC-Spark_CP-5 [5/0/8] ka 4 y y b h a 4 a h h ka b (a h ) k h b a 76. k k k b b ix a ib ix ix ix a ib ix ix x ix a ib x x x a and b x x Now, x can be written as x x x x x x x x b b a a n b a b a a b 77. z z 3 z 3 3z 6 w i w 8 i 78. w w 8 As, a b 0, a b ab But z a b..(i) Page 0
XI-IC-Spark_CP-5 [5/0/8] From Eq. (i) we get z ab z a b a b ab z z a b z a b z a b, as z is positive a b z 79. We have, z z z z z z z z z z z z z z z z z z z and z z z z z z z z z z z z z 80. z k Let z x iy We have, z a z i 0 x i y x y 0 on equating real and imaginary parts and x x 4 0 x x x 5 or x x 5 0 Since, x is real D B 4 AC 0 4 0 0 5 5 8. Given, z Let 5 z 5 z 5 0 0 Locus of, i.e, 5 z is the circle having centre at and radius 0. a = 0 Again, z z z z cos 0 z z cos 0 z z Page
XI-IC-Spark_CP-5 [5/0/8] z cos 4 cos 4 cos i sin z z : z a :0 8. We have, z z z z z z z z z z z z z 3 z 3 So, the maximum value of z is 3. 83. We have, z z z z z z z z z z 0 z z z z 85. z z 3 i a ib c id On taking argument both sides, we get tan tan 87. b d tan tan a c 3 b d tan n, n Z a c 6 Let z r cos i sin rei Given expression rei.e i i e rei i e r Since, imaginary part of given expression is zero, we have r sin sin 0 r r 0 r re i r or sin 0 z 0 arg z Page
88. XI-IC-Spark_CP-5 [5/0/8] 0 k k i cos k 0 k k i sin i cos k 0 0 i k k k i cos i sin i e k k 0 i k i e k th i (sum of roots of unit ) i 0 i We have, 89. sin We have, abcd cos i sin / abcd cos i sin Or abcd cos i sin..(i) [using De-Moivre s theorem] cos i sin abcd On adding Eqs. (i) and (ii), we get abcd cos abcd 90. (ii) Let cot p, then p cot m =e mi cot p pi mi i cot. e. i cot pi m m e mi e i e 0 Page 3