NARAYANA IIT ACADEMY · 67. use formula r=mv/qB 68. ... 78. Barrier potential ... Frequency of last...
Transcript of NARAYANA IIT ACADEMY · 67. use formula r=mv/qB 68. ... 78. Barrier potential ... Frequency of last...
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