CODE # JEE-2015 : Advanced Paper – 2 Answers and Explanations Physics Chemistry Mathematics 1 6 11 A 21 2 31 C 41 8 51 D 2 3 12 B,D 22 6 32 A 42 4 52 A,D 3 2 13 A,C 23 6 33 B 43 2 53 B,C,D 4 2 14 D 24 3 34 A 44 9 54 A,B 5 1 15 A,B 25 9 35 B,C 45 7 55 A,B,D 6 2 16 B,C 26 8 36 C,D 46 9 56 A,C 7 7 17 A,D 27 4 37 C 47 4 57 A,B 8 4 18 A,C 28 4 38 D 48 9 58 C,D 9 D 19 A,C 29 B 39 A 49 B,C 59 A,B,C 10 A,B,C 20 D 30 B,C,D 40 B 50 B 60 C,D PART - I : PHYSICS 1. 6 dI = 2. 3 The phase diagram for amplitudes will be 2 dm r 2 3 A dm = 4πr 2 ⋅ ρ dr for sphere A 60° R A 6 5 8 πk 5 r dr = 8πk ⋅ R = 4πk R IA = ∫ 3R 3R 6 9 0 R IB = 8πk ∫ 3R5 r 0 IB 3 6 = = IA 5 10 n=6 9 A 120° A 10 5 dr = 8πk ⋅ R = = 4π k R 15 3R5 10 Ares = J3 A ∴ Ires = 3 I0 3. 2 A = A0 e–t/τ A dA = R = 0 e− t / τ τ dt Both sources have same A0 Rρ at (t = τ) R Q ... 1 ... = τQ e−2τ / τ Rρ 2 ⋅ = τρ e−2τ / 2τ ; RQ e 4. 2 sini =n, sinr1 ∴ sinr1 = 6. 2 sini n λ= h p nh 2π mv rn = sini dr cosr1 1 = − 2 dn n At second surface nh P = mv = 2πr n sinr2 1 = sin θ n h 2πrn 2πn2r0 = = p n nz ∴ sin θ = nsinr2 λ = 2nπ dθ dr = sinr2 + ncosr2 2 ∴ cos θ dn dθ here dθ 1 dr2 ∴ dn = cos θ sinr2 + ncosr2 dn r0 z 3.h mvr = 2π n=3 z=3 r2 = A - r1 λ= dθ 1 −dr1 ∴ dn = cos θ sinr2 + ncosr2 dn 7. 7 6πr0 ,P=2 z For tension to be zero, = ncosr2 sini 1 sinr2 + cos θ cosr1n2 GMm 1 1 2Gm2 − = l2 9 T6 l2 = cosr2 sini 1 sinr2 + cos θ cosr1n m= 8. 4 3 1 = 2 + =2 2 2 × 3 2 E(t) = A2e–αt log E(t) = 2 log A – αt 100 × 1 A 5. 1 7M ,k=7 288 ∆E ∆A = 2 100 + α∆t × 100 E ∆ ∆t × 100 = 1.5% T 8 2 6 10 12 ∴ ∆t = B 4 1.5 × 5 100 Solving we get C 9. D CA = There is a wheatstone bridge between A & B whose equivalent is 2Ω. There is again a wheat stone bridge between A & C whose equivalent is 4.5 D= ∆E × 100 = 4% E 2 εA /1 d/2 C = 4 εA /2 A d/2 2 4 = 6.5 =1A 2 + 4.5 2 CA = ... 2 ... 2 εA /2 =C d CA = 2C, CB = 4C, CC = C ∴ equivalent capacitance Option C: 4C 4×2 7C =C+ C = C + 3 = 4 2 + 3 ∴ C2 7 ∴ C =3 1 P2 V2 P1V1 = T2 T1 P2 × 3V1 P1V1 = 4T1 T1 4 P1 . 3 i.e. work done by gas is stored in potential energy of spring = ½kx2. Plus the work done v against atmospheric pressure = P1x ∴ work done by gas = P1xA + ½kx2 V2 = V1 + Ax = 3V1 ∴ {Ax = 2V1} P2A = P1A + kx ∴ P2 = 10. A, B, C T 1 ,P 1 ,V 1 P1 1–n T 2 ,P 2 ,V 2 Ideal monoatomic P2A = P1A + kx P1V1 = µRT1 P2V2 = µRT2 ∴ 4 AP1 = P1A + kx 3 ⇒ 1 P1A = kx 3 work done by gas = = 1 (kx)x + P1xA 2 P1V1 P2 V2 = T1 T2 = 1 1 P1A x + P1xA 2 3 3 ∴ P2 = P1 2 = an equilibrium P2A = P1A + kx → 7 7 7 P1Ax = P1 2V1 = P1V1 6 6 3 Change in intend energy = ncr(T2 – T1) = ncr3T1 P1V1 = nRT1 3 P, 2 A = P1A + kx P1V1 9R 9 × = P1V1 R 2 2 ∆v = ∆Q – ∆W 1 P1A = kx 2 further V1 + Ax = V2 = 2V1 ∴ Ax = V1 ∴ nT1 = ∴ Energy stored = ∴ ∆Q = 1 1 1 V (kx) × x = × P1A × 1 2 2 2 A P1V1 4 Change in intend energy = ncr(T2 – T1) = ncr2T1 But, P1V1 = nRT1 1 2 kx + P1xA 2 = 9 7 21 + 14 P1V1 + P1V1 = (P1V1 ) 2 3 6 35 (P1V1 ) 6 = 11. A P1V1 3 × 2 × R = 3P1V1 R 2 ∴ option (B) is correct. ∴ (nT1)×2cr = ... 3 ... 94 U → 140 54 Xe + 38 Sr + x + y Q = kxe + ksr + kx + ky = ksr + kxe + 4 = 8 : 5 (234) – 236(7.5) = 1989 – 1770 = 219 ∴ kxe + ksr = 219 – 4 = 215 mev ∴ Mass number = 236 – (140 + 94) = 2(X ∪ Y) Atomic number = 92 – (54 + 38) = 0 (x ∪ Y) Only option is (A). 236 92 12. B, D dp 4πr2 = r (4π r2dr) ρ ρ at depth r will be directly proportional to r. ∴ dp = – krdr. Integration n1 σ1 P1 L1 0 ∫ dp = –k rdr ] R r P L2 P = k (R2 – r2) Where k is some cosnt. solving we get, option (B) & (C) are correct σ2 P2 n2 Since the rope is in tension force on P will be upward, force on Q will be downward. ∴ they will more in opposite direction if revosed. 17. A, D ev dB = ∴ VP .VQ < 0 {D} Since the 2 spheres are in equilibrium, ∴ T = Net force due to wt & liquid = 6πnrV vd = ev w I IB ne wd ⇒ v = ne d if B, n & I remain same 1 v 2 = d1 ; d v1 d2 for d1 = 2d2 v2 = 2v1 for d1 = d2 v1 = v2 n1 | Vp | ∴ n | V | =1 2 q v∝ | Vp | n2 ∴ | V | = n {B} q 1 18. A, C if I & w remain same 13. A, C By dimensional analysis. Only option A & C are correct. v∝ Pa 14. D Electric field inside cority = 3ε 0 B n v 2 B2 n1 = × v1 B1 n2 Only option is (D). if 15. A, B (A) (B) (from stress strain graph) Tenside strength is the stress at which metal breaks. B1 = 32 & n1 = 2n2 v2 = 2v1 B = 2B2 & n1 = n2 v2 = 0.5 v1 if (A) (C) 19. A, C 16. B, C The situation is similar to that of earth where pressure keeps on increasing as we go down. c i p for S1 p+d p sin c = Let us there a shell at r distance, the pressure on 2 sides is (p + dp) & P. the difference in pressure dp can be found by dpA = net attractive force on shell dpA = drx (ρ) ... 4 ... 2 5 for S1 in water & S2 in cos c = 16 3 15 , 1 5 K HX 10 −4 = = 10 −3 K HY 10 −1 45 2 × 4 5 4 sin im = 3 KHX = 10–3KHY in KHX – log KHY = – 3 (– log KHX) – (– log KHY) = 3 pKHX – pKHY = 3 9 sin im = 16 & 16 sin im = 3 15 sin im = 8 15 × 5 8 9 16 92 U238 → 82Pb 206 + x ( 2 He 4 ) + y ( −1 eo ) 92 = 82 + 2x – y 2x – y = 10 238 = 206 + 4x + 0 4x = 32 x=8 y=6 (A) for S1 in air and S2 in 1 × sin im = 25. 9 4 15 3 (for S1) 4 Pf × V 9RT = Pi × V 1RT & Pf =9 Pi 4 8 15 sin im = × 15 5 8 → 26. 8 5 [Fe(H2 O)2 (C2 O 4 )2 ] + 3MnO 4− + 24H+ 2− 3 sin im = 4 (C) 5Fe3 + + 10H2 O + 10 CO2 + 3Mn2 + n Rate H+ 24 = H = =8 − Rate MnO 4 nMnO 3 20. D if NA1 < NA2 we will take smaller numerical question. PART - II : CHEMISTRY + − 4 + H → heat 27. 4 21. 2 HO CH3 + CH3 CH3 CH3 22. 6 1, 2 methyl → shift 23. 6 3B 2H6 + 6CH3 OH → 6(CH3 O)3 B + 6H2 CH3 24. 3 ∧HX = x C = 0.01 M α1 = + ∧HY = 10x C = 0.1 M ∧HX 0 ∧HX α2 = aq.dil. → kMnO (excess) 4 HO CH3 OH OH CH3 28. 4 α1 = 0.1 α2 KHX = Cα12 = 0.01 α12 KHX = 0.01(0.1 α2)2 KHY = 0.1 α 22 KHX = 10–4 α 2 KHY = 10–1 α 22 2 CH3 OH ∧HY 0 ∧HY α1 X 1 = = α 2 10x 10 CH3 CH3 29. B 30. B, C, D As oxygen chemically participates with metal and metal donates electrons to antibonding π-orbitrals of oxygen. ... 5 ... 31. C 35. B, C OH CH3 → 32. A O CH3 C –H Cl Cl H C lO 2 (ii) H C lO 3 (iii) O Cl O H C lO 4 (iv) For questions no. 37 and 38. O C C –H NH → CH OH OH 36. C, D → CH2 CH3 O H C = O + H 2N H CH3 O HO (i) O3 (ii) Zn, H2 O O O CH (i) CH CH2 Pd − BaSO → H 4 2 C H 2O H H CH3 N –H CH + N → 2 C CH3 − (i) B H , (ii) H O ,OH → 6 2 2 CH COCH3 CH3 CH (ii) 2+ + Hg ,H → + CH2 = CH − CH3 , H → High Pr essure, ∆ 33. B CH3 HC CH3 CH3 C C CH3 + (i) C2H5 MgBr,H2 O (ii) H , heat → OOH (M ajor) O radical initiator → 2 37. C + NH2 34. A N = NCl NaNO / HCl → 0° C 38. D – + 2 39. A 40. B O– N=N OH → CH3 COOH + NaOH → CH3 COONa + H2 O Initial 0.2 0.1 0 Final 0.1 0 0.1 0.1 −5 pH = − log(2 × 10 ) + log 0.1 = – [0.3010 – 5] = 4.6990 ... 6 ... PART - III : MATHEMATICS lim F (x ) x →1 G 41. 8 42. 4 Coeff of ×9 in (1 + x) (1 + x2) L (1 + x100) = Coeff of x9 in (1 + x) (1 + x2) L (1 + x9) f (1) f ( f ( x )) 1 x2 y2 + = 1 , e = F1 (2, 0), F2 (–2, 0) 2 9 5 F1 (2, 0) 2F2 (–4, 0) P1 : y2 = 8x P2 : y2 = –16x Tangent to P1 (2t2, 4t) : ty = x +2t2 , it passes through (–4, 0) 0 = –4 + 2t2 ⇒ t2 = 2 m1 = 46. 9 47. 4 Tangent to P2 at (–4t2, 8t) ty = – x + 4 + 2, it passes through (2, 0) f ( f ( x )) 1 14 S = 4i$ + 3j$ + 5k αk +1 − αk = const (r ) k =1 3 ∑ α4k −1 − α4k −2 = 12r =4 3r k =1 43. 2 ( ) −e0 cos an 48. 9 0 m a →0 a ( ).(− sinan ).(nan−1 ) −e = cos an e mam−1 a→0 44. 9 ( ) −1 1 3 ax + tan x α = ∫9 + .e dx 2 + 1 x 0 9x + tan−1 x =e 1 0 =e loge (1 + α ) = 9 + F (x ) = x ∫ −1 9+ 3π 4 d=9 −1 49. B, C h(x) = f(x) – 3g (x) h(–1) = f(–1) – 3g (–1) = 3 h (0) = 6 – 3 = 3 h(2) = 3 h(–1) = h(0) = h(2) ∴ h’(x) = 0 has one root in (–1, 0) & one root in (0, 2) 3π 4 f ( t ) dt ⇒ F (1) = 1 ∫ f ( x ) dt = 0 −1 x 1 −1 −1 7 [2a + 6d] 6 2 = 11 11 2a + 10d] [ 2 ⇒ 7 (2a + 6d) = 6 (2a + 10d) 14a + 41d = 12a + 60d 2a = 18d 29 = 18d ⇒ a = 9d T7 = a + 6d = 15d 130 < 15d < 140 2 m = 2n G (x ) = x →1 x. = 12 m22 = 2 45. 7 (x ) f (x ) 1 1 ⇒ f = 7 14 2 ∑ αk +1 − αk 1 0=–2+4+2 ⇒ t = 2 = lim x →1 G' = lim → 2 lim = F' ( x ) x=–4+3+5=4 y=4–3+5=6 z=–4–3+5=–2 1 1 ⇒ = t2 = 2 2 t m1 e (x ) = lim ∫ t f (f (t )) df ⇒ G (1) = ∫ x. f (f (t )) dt = 0 ... 7 ... {h'' ( x ) ≠ 0} 50. B 53. B, C, D f(x) = (7tan6x – 3 tan2x) sec2x π 4 π 4 0 0 54. A, B ∫ f ( x ) dx = ∫ (7 tan 6 ) x − 3 tan2 x sec 2 xdx 55. A, B, D 56. A, C 1 Let tan x = t = ∫ (7t 6 0 ) 1 − 3t 2 dt = t7 − t3 = 0 0 π 4 π 1 ( 57. A, B ) 7 3 ∫ xf ( x ) dx = x f ( x ) dx 02 − ∫ t − t dt 0 0 1 x1 − x2 < 1 ( x1 − x2 )2 − 4x1x2 α2 <1 n3 1 n1 + 2 n1 + n2 n3 + n4 58. C, D n1 n −1 n2 n1 1 + = . 1 . n1 + n2 n1 + n2 n1 + n2 n1 + n2 − 1 3 −4 <1 n1 1 = n1 + n2 3 <5 1 < α2 5 α2 < II 1 II = P = R 1 n1 1 n3 3 + 2 n1 + m 2 n3 + n4 52. A, D 1 I 1 n3 2 n3 + n4 51. D α2 n3 R n1 B P (R ) = t8 t 4 1 1 1 =0− − = − + = 8 4 8 8 4 0 1 n1 R n2 B −1 5 or α > 1 59. A, B, C f(x) = x.F(x) f(2) = 2. F(2) 5 1 F’(x) < 0 for x ∈ ,3 2 therefore F(2) < F(1) ⇒ f(2) < 0 f’(x) = F(x) + xF’(x) < 0 for x ∈ (1, 3) f’(1) = F(1) + F’(1) < 0 1 1 1 1 or ∴− < α < − <α< 2 2 5 5 1 1 2 2 1 − 4α > 0 4α − 1 < 0 − 2 < α < 2 60. C, D ... 8 ...
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