NARAYANA IIT ACADEMY INDIA

SEC:LT-IIT Dt: 27-01-2018 Time: 3Hrs JEE MAINS QP Max.Marks: 360

MATH KEY

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 2 1 3 4 3 3 2 1 3 1 3 4 1 2 2

16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 2 1 4 4 3 3 1 1 3 2 1 2 2 1 3

CHEMISTRY KEY

31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 3 3 1 3 1 1 2 1 2 2 3 4 3 2 4

46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 2 3 3 4 4 4 1 1 2 2 3 1 3 4 3

PHYSICS

61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 A C A D D A B B D B C B A C B 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 B A B A A A C A C C C D C B A

MATHS SOLUTIONS

1. 3 1 3 3; 21 1 21 21 7n n nT nc T T n C nc n

2. 1 22 1 2 1 ..... 2 1 255nn C n C n C

0 1 21 2 1 2 1 .... 2 1 1 2562

n C n C n C n

2 82 2 4n n

3. 94864 = 4 2 22 7 11

Required number = 1 4 1 2 1 2 1 1 232

4. 2 2 1 1 1351000 5 2 3 7 17

Sum of proper diverors = 3 1 1 3 2 22 1 3 1 5 1 7 1 17 1 1 357000

2 1 3 1 5 1 7 1 17 1

4 31 8 18 1 357000 89291

5. 2 2 1 15 1 4 2 5 4 2c c c c

10 12 40

6. 3 310 4 1 117c c

7. No of diagonals = 3 12 12 3

542 2

n n

8. Conceptual

9. Conceptual

10. Conceptual

11. 6 7 8 315 6 7 ......30P

21 26 316 11 165 10 15 20 25 30P

Exponent of P in 5 = 137

12. Recived no of hignglass = 3 14nc n n n G

3 110 10 10 6 50c c

13. All are distinct - 56 6c

3 distinct 2 alike 3 15 3 30c c ; 2 distinet 3 alike = 1 22 .5 20c c

2 alike, 2alike 1 distinct = 2 13 4 12c c

3alike 2alike = 1 12 .2 4c c

Total = 6 30 20 12 4 72

14. 20x y z w t 0t

5 1 420 5 1 24c c

15. Btal no of ways = 7! 210

2!3!2!

16. Conceptual

17. 12!n s 4!9!n A 155

P A

18. 3

3

1 32 20xc xxc

19. 81n S 22 6 5!n A P 528

P A

20. Conceptual

21. 64 63n S 224n A 224 164 63 18

P A

22. Conceptual

23. Conceptual

24. 4,5,6A 3 16 2

P A

25. Conceptual

26. 1 555rT c 11

5 10.r r

y x

0,10,20,30,40,50r will give the terms which are free from radicals

27. 1 2 1 3 2 39 4 12z z z z z z

3 3 1 2 2 2 1 3 1 1 2 3 12z z z z z z z z z z z z

1 2 3z z z 1 2 3z z z = 12 1 2 3 2z z z

28. 2 11 2 3 .... nS n

2 1

2 1

2 .... 11 1 ....

n n

n n

S n nS n

1

nS

29. Conceptual

30. Conceptual

CHEMISTRY SOLUTIONS

31. Conceptual

32. NC Ag CN and CN Ag NC

33. In the presence of strong ligands the electronic configuration of 2Co is 6 12gt eg . The

electron in higher energy 1eg will be given easily to get stability.

34. ( ) 223 3 4

4Zn NH Zn NH++ é ù+ ë û

23 4

423

Zn NHK

Zn NH

2

4 42 933 4

1 13 10 10

Zn

K NHZn NH

= 143.33 10

35. In 2

2 6Ni H O

the 3NH can attack at 6 coordination sites. But as the number of 2H O

molecules substituted increases the probability of the coordination sites for the attack decreases. So k values decreases

36. Conceptual

37. Due to increase in negative charge density on central metal, electron transfer through back bonding from metal to CO increases, thus causing increase in M – C bond strength and decrease in C – O bond strength. So M – C bond stretching frequency increases with increase in negative charge.

38. Conceptual

39. Conceptual

40. no. of moles of AgCl formula 28.7 0.2143.5

0.1 mole complex then 0.2 moles of tree Cl

41. Conceptual

42. .(I) Fe+2 3d6 4s0 0 unpaired e-

II)Fe+2 3d5 4s0 1 unpaired e-

III) 3Cv 3d3 4s0 3 unpaired e-

IV) 2Ni 3d8 2 unpaired e-

CN Strength of ligands

43. A) 3-

2 4 3Co C O EAN=36

B) 4( ) EAN=36Ni CO

C) 4

2 6EAN=32Ti H O

D) 4

6EAN=36Fe CN

44. conceptual

45. conceptual

46. 24NiCl is square planar

47. Conceptual

48. 3 :Fe Blood red, 2 :Co Blue

49. AgCl + K2Cr2O7 + H2SO4 No gas.

50. Conceptual

51. To convert the covalent compound into a mixture of ionic compounds

52. NaCN and 2Na S decomposes into HCN and H2S

53. 238NV ml

732 32 700P

2

700 38 273350 760NV at STP ml

222400 28ml gm N

2 38 273 28 2 38 273760 22400 760

gm

% Nitrogen = 28 2 38 273 100 6.25%22400 760 0.546

54.

2

2

224 2 2 6

Blue6 4Zn

CaCoCl H O Co H O Cl

Addition of 2Ca shifts equilibrium left, and addition of 2Zn shifts right

IIA group elements preferably forms complex with O-donar ligands where as zinc

resembles transition elements and forms stable complexes with not only O-donar but

also S-donar ligands.

55. Conceptual

56. Increasing in polarizing power order 2 2 2 2Mg Zn Cd Hg is a reflection of

decreasing nuclear shielding and consequent increase in power of distortion in the

sequence filled p-shell < filled d-shell < filled f-shell.

57. conceptual

58. 3 3 3 32 2NaCH CHCl CH Cl CH CH CH

59. 4 3 4Al C givenCH

60. Due to antiperiplanar requirement conjugated diene is form

PHYSICS SOLUTIONS

61. Current capacity of galvanometer Ig=25 ×4×10-4=10-2 Resistance of galvanometer G=50 Ω Range of voltmeter=25 V From formula resistance to be connected in series

R=(V/Ig - G R=2450 Ω

62.

Magnetic field due to current in wire 1 at point P distant r from the wire is

direction of magnetic field is perpendicular to the plane of paper, inward The force exerted due to this magnetic field on current element i2dl is dF = i2dlBsin90

63.

Current is coming out of the paper. By Fleming's left hand rule force on the conductor is BIL as angle between direction of current and magnetic field is 90°, component of force parallel to inclined plane is BIlcosθ

From figure BIlcosθ=mgsinθ

64. τ=NIABsinθτ=(1)(10)(0.01)(0.1)sin90=0.01 Nm

65. Conceptual

66. Magnetic moment M=AI=πr2(2qf) Angular momentum=2mr2 ω But ω=2πf Tahe ratio of M and L

67. use formula r=mv/qB

68. First case :Protons released from rest thus no magnetic force act on it, Proton moves along with direction of electric field. Thus electric field direction is along west direction Now F= eE and Fma0 eE=ma0

Second case: Proton moves with initial velocity projected towards north moves along west. Thus proton experience electric filed and magnetic field

Now j × n must be -i It will be possible is n = -k or direction of magnetic field is down ward direction Now above equation reduced to

69. Conceptual

70. Given A =1.2×10-3 m2 N1 = 20000 and N2 = 300 length l = 0.3m dI = 2-(-2) = 4 dt = 0.25 sec From the formula for mutual inductance

Emf E = M( dI/dt) E = 30.14×10-4 × (4/0.25) E = 4.82×10-2 V

71. USe formula E = Blv

72. USe E = -L(dφ/dt)

73. Use formula Φ = M(dIx/dt)

74. E = NABω

75. Use E = -dφ/dt to get equation then substitue t = 3

76. E = -dφ/dt RI= -dφ/dt 2×I = 8/0.2 I = 20 Amp In 1 second charge flown = 20 colunlomb ∴ in 0.2 second = 4 coulomb

77. In case of full wave rectifier, Fundamental frequency=2 × mains frequency=2×50=100Hz

78. Barrier potential does not depend in diode design while barrier potential depends upon temperature, doping density and forward biasing

79. Conceptual

80. Conceptual

81. Conceptual

82. Conceptual

83. Conceptual

84. Conceptual

85. Body A gets 5 sec. thus displacement in fifth seconds S=(a1 /2 )( 2× 5 - 1)=a1 (9/2) --(1) Body gets 3 sec, thus displacement in third seconds S=(a2 /2) ( 2× 3 - 1)=a2 (5/2) --(2) from equation (1) and (2) we get (9a1/2)= (5a2/2) a1 : a2=5:9

86. v=√(µgr) v=√(0.2×100×9.8)=14 m/s

87. Formula for acceleration along the inclined plane is

For solid sphere I=(2/5)mr2 and θ=30° on substituting the values we get

88. Work ∝ (radius)2 Thus W ∝ R2 W2∝(2R)2 ∴ W2 ∝ 4R2 on taking ratio W2=4W

89. Let frequency of first A=n Frequency of last tuning fork L=2n difference d=6 Number of tune forks N=56 From formula L=A+(N-1)d 2n=n + (56-1)6 n=330 Frequency of last tuning fork=2n=2×330=660

90. W=PΔV W=105× (0.091)×10-6 W=0.0091J W is positive because the system is expanding