iit jee model paper answer.pdf
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Max Marks:360
KEY & HINTS
PHYSICS1) 1 2) 2 3) 2 4) 1 5) 1 6) 2
7) 2 8) 2 9) 4 10) 2 11) 4 12) 1
13) 1 14) 3 15) 3 16) 3 17) 4 18) 3
19) 2 20) 3 21) 2 22) 3 23) 1 24) 3
25) 2 26) 1 27) 4 28) 4 29) 3 30) 2
CHEMISTRY31) 1 32) 1 33) 1 34) 1 35) 3 36) 4
37) 2 38) 1 39) 3 40) 1 41) 3 42) 1
43) 3 44) 3 45) 4 46) 2 47) 1 48) 3
49) 1 50) 4 51) 3 52) 1 53) 4 54) 4
55) 4 56) 3 57) 1 58) 3 59) 3 60) 3
MATHEMATICS61) 2 62) 1 63) 3 64) 2 65) 1 66) 4
67) 4 68) 3 69) 4 70) 3 71) 1 72) 3
73) 2 74) 3 75) 3 76) 4 77) 3 78) 3
79) 4 80) 3 81) 3 82) 1 83) 3 84) 4
85) 2 86) 1 87) 3 88) 1 89) 1 90) 4
HINTSPHYSICS
1. VS=4, MS=2.7= MS+VS x LC=2.7 + 0.04=2.74 ]
3. Spring force does not change instantaneouslyThus for 1 1 0;m a aFor 2m 2 2pS
F m a ….(i) instantaneously after 2F is withdrawn
Initially 2 2 0pSF F m a
2a
pSF
2 2 0Fsp F m a ii
From (i) and (ii) 22 0
2
Fa am
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4. WD k
21 21 .6 02 6m mmgl v vm m
1 273
v v gl
5. sin60tan30tan60
oo
o
eV
2tan30 tan 30tan60
oo
oe
6.1 2 2
2cme eV gh
221 2
max 2 4cm e e hV
hg
8.
12 rad/hr1
22 rad/hr8
2
1 1
2 3
T RT R
2
1
R 4R 4
2R 4x10 km
411
2 RV 2 x10 km/hr1h
422
2 RV x10 km/hr8h
At closest separation4
rel4
V toline joining x10 km/hr rad/hr.lengthof line journing 3x10 km 3
]
9.When source is at origin, the observer receives the sound emitted by the source,when it was at P.
Such that 50t 1cos200t 4
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0
s
v (v) 90x200v 50v v cos 2004
V=96 Hz
10. 3 3L2 2L
0y A sinkxsin t2 3
02 L 5A sin x sin t2L 2 6
05A sin t6
11. Comment for discussion. In the ideal case that we normally consider eachcollision transfers twice the magnitude of its normal momentum. On the faceEFGH, it transfers only half of that.
13. Let x mole of the gas dissociate at 1000 KNo. of mole of diatomic gas molecule =1-xNo. of moles of monatomic gas molecules=2 x XEnergy of diatomic molecules = energy of monatomic molecule
5 31 x RT 2x x RT2 2
x 5/11
Now new no.of moles = (1-x)+2x=1+x=(16/11)
P= nRTV
Pressure initially at 300K= i300RP
V
Pressure finally at 1000K =
f
1 x R x1000 16 RP 100V 11 V
]
15. 13 3 27 1x x 3x+39=-x+27x=-3
a b
27SoV V (x 13)
17
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16. iEA
2
2
kr iER r
kiEr
17. 4 4 1I' I4 G 4 G 5
(given) where G=16
4 3I" I4 163
1I" I13
or 1 5I" 51' I '13 13
i 4xG 4x i G 16W5 5
In second case (i-i’) x 16= 4 i '3 eq
2 4 4R2 4 3
i '4i 4i '3
13i ' 124i i ' i
3 13
1 1i i ' i x0.65mA 0.05mA13 13
]
19. Introducing two equal and opposite current 1I and also 2I between A & C.
Force on ABCA closed loop zeroForce on ADCA closed loop zeroForce on extra 1I & 2I
1 2F I I lB=IlB
20. 2 21 1Li C2 2
128
3
C 9x10 12x3i 12 10L 2.5x10 5
4i 7.2x10 A ]22. maxV V 2 2WA
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24. 2eV
1
hc 8eV 2 1T 2T
If 1l is the wavelength corresponding to maximum intensity at 1 2 2T & T at T ;Then 2 1 /2 (by wein’s displacement Law)
2 1
hc 2hc 16ev
2ev maxhcK.E. 14ev
26. longest minEfor Balmer longest is for n 2 3 for Lyman longest is for n 1 2 longest Balmer> longest Lyman
CHEMISTRY:31. 3 2 2 22 2 2AgClO Cl AgCl ClO O 50. The nitrogen has 7 electrons while the oxygen has 8 electron. The MO electronic
configuration of NO molecule is 2 * 2 2 * 2 2 2 2 * 11 . 1 . 2 . 2 . 2 . 2 . 2 . 2z x y ys s s s p P P P .
Bond order is 2.5. When NO molecule is converted in to NO ion, the electron in* MO is removed. Therefore, bond order increases, 2.5 to 3, bond length
decreases and becomes diamagnetic. The N-O bond length in NO is 106 pm.MATHEMATICS
61) Conceptual
62)
63)
64) Since x, y, z are in GP,
Hence,
\
are in AP.
are in HP.
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65)
66) Conceptual
67)
68) =
69)
0 1 1 0 0 0
| | 0 1 0
0 1
A a b c d c d cd
ab cd a b ab
70)
71) Reflexive but not Symmetric
72)
No. of non negative inetegral solutions
Total no. of solutions = 16 x 21 = 336
73) a = p(A getting 6)
b = p(B getting 7)
74)
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75) Adjust the series and apply the formula
76)
The tangent at
----- (1)
The normal at
----- (2)
(1), (2) are identical ;
Eliminating
77)
78)
79)
80) Given that
Differentiable with respective x,
81)
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82)
83) For continuity,
This is possible only when
For to exist ; f(x) will not be differentiable if ;
84) Equation of pair of tangents,
angle between them,
85) Passing through
86) Common tangents to and are
87) Slope of tangent = 1/2 and slope of PS = 4/3
89) Let the three given points lie on the line , where l, m and n are constants, them
; ; For
i.e are roots ; Them
90) ; ;
; ;
Required line passing through .
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