PROGRESS TEST-5 RBS-1801 & 1802 JEE MAIN PATTERN

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Theodore Collins
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1 PRGRESS TEST-5 RBS-80 & 80 JEE MAI PATTER Test Date:

2 [ ] PT-V (Main) RBS-80-80_ q. Potential at this point, V re E r 4 0 Work done = qv = 4E joules. (B). (A). As W F ds cos qe ds cos 4 0.E cos60, E 0 /C. (D) dv dx 4. E 0 x 0 0V/m x PHYSICS 5. (A) K r 4q K q r r r and r r 0 6. Hence, r = 0 cm (B) q q q q q q U i 0 U f = 0 4 a q q q q q q 4 a W ext = U = 0 (D) V V 7. E iˆ ˆ j K yiˆ xj ˆ x y 0 E 0 E x E y ( Ky) ( Kx) = Kr i.e., E r (B)

3 PT-V (Main) RBS-80-80_ [ ] 8. For coherent source I max. = 4I for incoherent source I = I + I = I Ratio = (B) 9. D, d D / d 4 (C) 0. (D). (C). D D n n d d n n n 84 n (D). Velocity of image = v 0 vm = 0 A cost Acos t t = (D) 4. I 4I 0 cos I 0 4I 0 cos cos 4 Path difference x 4 t or 0.5t t 4 4 min (C) 5. For refraction at spherical surface v R v R R (B)

4 [ 4 ] 6. d v dt is the tangential acceleration. (D) 7. Initial velocity of boy with respect to bus = 0 ms acceleration of boy with respect to bus = ms s ut at 48 0t t t 0t 96 0 s t t 8t 96 0 (t 8)(t ) 0 t 8s and s (A) Boy PT-V (Main) RBS-80-80_ a = ms Bus u=0 ms 48 m 8. t w v sin sin sin 0 v w=5m = 80 = 50 (A) 9. Horizontal component of velocity of A is 0 cos 60 or 5 m/s which is equal to the velocity of B in horizontal direction. They will collide at C if time of flight of both the particles are equal i.e. t A = t B or h = usin h h gtb g g u sin g (0) 0 (C) 0. r 4sint ˆi 4cost ˆj x 4sin t = 5 m

5 PT-V (Main) RBS-80-80_ [ 5 ] y 4cos t x y 6 sin 6cos t x y 6 (C) v. T vb 6 or vt vb ms 0.5. (A) (A) a 6î 8ĵ a r = 8 and a t = 6 r andr. (B) 4. (B) 5. (C) 6. mg mg a = 0.4 g m/s m (B) 7. Reading reduces when the lift starts accelerating downwards and then original value is restored as lift moves with constant velocity. Apparent weight mg a, where a is acceleration of lift. (C) 8. f F F 7 (A) f a =.5 ms 6 kg F 9. Friction is static so a 0 m/s, f T cos cos 60 0 (C) mg 0. T cos, T = mg cos T sec T (B)

6 [ 6 ] PT-V (MAI) RBS-80-80_ (A). (A). (B) CHEMISTRY utput of C per hour = 44 g =.4 litre at S.T.P. Reduction of C per hour = ml = L Fraction of time the converter has to be operated = C output rate C reduction rate Gas with higher critical temperature and higher inversion temperature can be easily liquefied. Volume of H Volume of Cylinder m 5 L w PV RT m w g wh Reaction : Fe(s) HCl (aq.) FeCl (aq.) H 56 g g Mass of Pure Iron % Purity = g Mass of pure iron Mass of Impure iron (D) Mass of HS g 00 Zn H S ZnS H 4 4 Reaction : mol mol.4l at STP 98 g Volume of H produced = L 98

7 PT-V (MAI) RBS-80-80_ [ 7 ] 5. (B) 6. (A) umber of equivalents of KMn 4 = umber of equivalents of FeS 4 + umber of equivalents of FeC 4 x 5 = + = 4 7. (A) 8. (C) 4 x 5 In H 4,.S. of is. There is total loss of ten moles of e, each atom is losing 5 mole e, therefore, the.s. of nitrogen changes from to +. The.S. of nitrogen in compound Y is (D) 4 Fe Mn Fe Mn n 5 Fe 7 Fe Cr Fe Cr n 6 Fe Since 5 : (D) 8a a TC 0 PC 7 7Rb 7b TC 8a 7b 0 P 7Rb a 7 C 8b 0 R b = (A) Phenol and Alcohol are functional group isomer of each other 4. (C) B.L B. as delocalisation of C C bond increases B. decreases B.L, increases Delocalisation of C C -hydrogen

8 [ 8 ] PT-V (MAI) RBS-80-80_ (A) Resonance increases dipolemoment. CH CH me me me me me me 44. (C) Resonance o due to SIR no resonance CH CH CH CH H CH CH CH CH H H CH C CH CH CH CH CH H CH CH CH CH CH CH CH CH CH CH CH CH 45. (C) CH 8 Kcal/mole H Kcal/mole 6 Kcal/mole

9 PT-V (MAI) RBS-80-80_ [ 9 ] 46. (D) 47. (B) 5 CH Me 4 CH C -Formyl-4-ethyl pent-4-ene--nitrile C 4 5 CH 6 C 48. (A) 49. (C) 50. (C) 5. (D) 5-Formyl benzene-,-dicarbonitrile Because both the oxygen make p p backbonding with carbon. Acidic strength Stability of conjugate base Stability of A Rectangular shape so (a) and (b) are not equal. H C H H C Inter-molecular H-Bond Intramolecular H-Bond H 5. (C) Solubility = II > I (i) & (ii) Bond order increases.

10 [ 0 ] PT-V (MAI) RBS-80-80_ (C) + 80º Ȯ. 54. (B) 0º sp HSi HSi Ȯ. SiH lone pair Ȯ. sp 0º Vacant d orbital Ȯ... (due to Back Bonding) CH CH CH ovacant orbital (o back bonding) P HSi SiH SiH (o back bonding due to large size of atoms) 55. (D) As electronegativity of halogen attached with sulphur increases, suphur becomes more electron deficient and hence its tendency of get electrons from oxygen through p d bonding also increases i.e. extent of p d bonding increases and hence, bond order also increases. 56. (A) Temperature > 00K 57. (D) 58. (A) 59. (D) 60. (B) Cl Cl Be Be Cl Cl dimer Among B, C, b orbitals filled before b, indicating s p intermixing. IF IF I 5

11 PT-V (MAI) RBS-80-80_ [ ] 6. (B) MATHEMATICS x e 6. (D) 6. (A) 64. (C) 65. (A) 66. (C) f(x) cot x ; f(x), 4 b b 0 8 b =, cos x x0 x (x lim )cos x In cos x In y = lim 0 x sec x e 67. (A) a sin bx lim x 0 x cos x e ab 68. (B) dy dy sin y cos ec x cot x ; dx dx dy dy sin y cos y cos ecx cot x cos ec x dx dx e dy x dx dy dx

12 [ ] PT-V (MAI) RBS-80-80_ (C) d x x d x x dx 0 e e dy dy (x 8x 6) at x dy g'( 4) 70. (D) bvious 7. (D) lim h0 h f '( h ) hf '( h ) f( h ) f( h ) h 7. (C) 7. (A) 74. (A) 75. (C) 76. (B) h'(x) f ' x g(x) cos x (g(x) xg(x) sin x) x Q f(x) x x Q f( x) x dy dy x dx dx 6y 8 y x ; 6y ; x 6 y. x x x a 0; is equivalent to x + 4x + = 0 a = (B) x (a )x a 0 xr 4(a ) 4 9a 0 ; a 4a 4 9a 0 a 5a 4 0 (a ) (a 4) 0

13 PT-V (MAI) RBS-80-80_ [ ] 78. (C) dy dx sin x y cos x ( sin x) (x x) cos x sin x x ; y x sin x y y x 79. (C) 80. (C) 8. (B) 8. (D) 8. (C) 84. (B) 85. (C) 86. (A) 87. (D) T = S 88. (C) 89. (A) x y 6 b (0 / ) p = q = 0 60 L L 90. (D) Clearly ; a 8 4 Length of latus rectum = 4 4a 4 8

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