Download - Past Year Papers AIEEE 2007
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1 (1) 2 (2) 3 (2) 4 (4) 5 (4) 6 (3) 7 (3) 8 (2) 9 (2) 10 (4)
11 (4) 12 (2) 13 (3) 14 (4) 15 (4) 16 (1) 17 (2) 18 (4) 19 (1) 20 (4)
21 (3) 22 (3) 23 (2) 24 (3) 25 (4) 26 (1) 27 (1) 28 (2) 29 (3) 30 (4)
31 (1) 32 (1) 33 (1) 34 (3) 35 (1) 36 (3) 37 (4) 38 (4) 39 (1) 40 (1)
41 (4) 42 (3) 43 (3) 44 (1) 45 (4) 46 (4) 47 (1) 48 (1) 49 (1) 50 (4)
51 (3) 52 (4) 53 (2) 54 (4) 55 (3) 56 (4) 57 (3) 58 (2) 59 (4) 60 (3)
61 (2) 62 (3) 63 (2) 64 (2) 65 (4) 66 (4) 67 (1) 68 (1) 69 (1) 70 (3)
71 (3) 72 (4) 73 (1) 74 (3) 75 (4) 76 (3) 77 (3) 78 (3) 79 (1) 80 (4)
81 (4) 82 (2) 83 (4) 84 (3) 85 (4) 86 (3) 87 (4) 88 (2) 89 (1) 90 (3)
91 (4) 92 (3) 93 (2) 94 (1,2) 95 (3) 96 (1) 97 (2) 98 (2) 99 (4) 100 (4)
101 (1) 102 (2) 103 (3) 104 (2) 105 (4) 106 (3) 107 (3) 108 (4) 109 (1) 110 (4)
111 (3) 112 (1) 113 (3) 114 (1) 115 (3) 116 (4) 117 (1) 118 (1) 119 (3) 120 (1)
PHYSICS
1. x=(210-2) cos t
Here,a=210-2 m = 2 cm
At t = 0, x = 2 cm, i.e., the object is at positive extreme, so to acquire maximum speed (ie.,
to
reach mean position) it takes1
th4
of time period.
Required timeT
4=
where2
T
= =
T = 2s
So, required timeT 2
0.5s4 4
= = =
2. For given circuit current is lagging the voltage by2
, so circuit is purely inductive and there
is no power consumption in circuit. The work done by battery is stored as magnetic energy in
the inductor.
3.1r 2i 2 j= +
( ) ( )2 21 1r r 2 2 2= = + =
2r 2i 0j= +
2 2r r 2 = =
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Potential at point A is
3 6
A
0 1 0
1q 1 10 10V
4 r 4 2
= =
Potential at pointBis3 6
B
0 2 0
1 q 1 10 10V
4 r 4 2
= =
VA
VB
= 0
4. Required ratioEnergystoredincapacitor
Work doenby thebattery=
2
2
1CV
2Ce
= where C = Capacitance of capacitor,
V = Potential difference
e = emf of battery
( )2
2
1Ce
2 V eCe
= =
1
2=
5. Rise of current in L-R circuit is given by
( )t/0I I 1 e =
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Where0
E 5I 1A
R 5= = =
Now,L 10
2sR 5
= = =
After 2s, ie, at t = 2s
RiseofcurrentI=(1e-1)A
6. Uniformcurrentisowing.Currentenclosedinthe1st amperean path is
2 2
1 1
2 2
I r Ir
R R
=
2
0 0 1 0 1
2 2
1
current .Ir IrB
path 2 r R 2 R
= = =
Magnetic induction at a distance 02
2
.Ir
2 r
=
1 1 2
2 2
2
a.2a
B r r 2 1B R a
= = =
7. There is no current inside the pipe and hence Amperes law can not be applied. So, the
magneticeldatanypointinsidethepipe=0
8. Bindingenergy
( )2
nucleus nucleonsBE M M c= ( )2
0 p nM 8M 9M c= [nucleons=protons+neutrons= p n8M 9M+ ]9. rayemissiontakesplaceduetodeexcitationofthenucleus.Thereforeduring-ray emission,
there is no change in the proton and neutron number.
10. For Vi0, the diode is forward biased and circuit would be as shown in Figure (3) and V
0=
ViHence, the option (4) is correct.
11. Energy of a photon E = h .(i)
Also E = pc ..(ii)
where p is the momentum of a photon
From (i) and (ii), we get
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hh pc p
c
= =
12. We know that,dx
v dt=
dx = v dt
integrating,tx
0 0
dx vdt=
( )t
2 3t2
0 0
0 0
gt ftx v gt ft dt v t
2 3
= + + = + +
2 3
0
gt ftx v t c
2 3
= + + + where c is the constant of integration.
Byquestion,x=0att= 0
0
g f0 v 0 0 0 c
2 3 = + + + c = 0
2 3
0
gt ftx v t
2 3 = + +
At t = 1,0
g fx v
2 3= + +
13. By perpendicular axes theorem, M.I. about perpendicular axis passes through centre of
lamina.
( )2 22 2 2z
M a aa b 2al M M
12 12 12
++= = =
Byperpendicularaxestheorem,
AC BD zI I I+ =
2z
AC
I MaI
2 12 = =
Bythesametheorem2
zEF
I MaI
2 12= = .
IAC
= IEF.
14.0
x x cos t4
=
Acceleration,2
2
d xa
dt=
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2
0x cos t
4
=
2
0
3
x cos t 4
= +
So, A = 2x0
and3
4
=
15. Asshowninthegure,theresultantelectriceldsbeforeandafterinterchangingthecharges
will have the same magnitude but opposite directions.
Also, the potential will be same in both cases as it is a scalar quantity.
16.1
2
T , half life ofY
X = , mean life of Y
X Y
ln2 1=
X Yln2 =
X Y >
X Yt t
X 0 Y 0A A e ; A A e = =
X will decay faster than Y.
17. For Carnot engine using as refrigerator
12
2
TW Q 1
T
=
It is given1
10 =
2
1
T1
T =
2
1
T 9
T 10 =
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2 210
10 Q 1 Q 90 J9
= =
18. Si and Ge are semiconductors but C is an insulator. Also, the conductivity of Si and Ge is more
than C because the valence electrons of Si, Ge and C lie in third, fourth and second orbit
respectively.
19. When E
and B
areperpendicularandvelocityhasnochangesthenqE=qvBie.,E
vB
= .
The two forces oppose each other if v is along2
E BE B i.e.,v
B
=
As E
and B
are perpendicular to each other
2 2
E B EB sin 90 E
BB B
= =
For historic and standard experimentslikeThomsonse/mvalue,ifvisgivenonlyasE/B,
it would have been better from the pedagogic view, although the answer is numerically
correct.
20. V V V
E i j kx y z
=
x 2
V d 20E
x dx x 4
= =
( )2
2
40x
x 4
=
x
10E at x 4 m V / m
9 = =
and is along positive x direction.
21. Wehavetondthefrequencyofemittedphotons.Foremissionofphotonsthetransitionmust
take place from a higher energy level to a lower energy level which are given only in option
(2) and (3).
Frequency is given by,2 2
1 2
1 1hv 13.6
n n
=
For transition from n = 6 to n = 2
1 2 2
13.6 1 1 2 13.6v
h 9 h6 2
= =
For transition from n = 2 to n = 1.
2 2 2
13.6 1 1 3 13.6v
h 4 h2 1
= =
v2
> v1
Hence, option (3) is the correct answer.
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22. Acceleration of the systemF
am M
=+
Force on block of mass m = mamF
m M
=+
.
23. Power of a lens is reciprocal of its focal length.
Power of combined lens is
P = P1
+ P2
=15+5=10D
1 100f cm
P 10 = =
f=10 cm
24. LetTbethetemperatureoftheinterface.Asthetwosectionsareinseries,therateofowof heat in them will be equal
( ) ( )1 1 2 21 2
K A T T K A T T
l l
=
where A is the area of cross-section.
K1A(T
1T)l
2= K
2A(TT
2)l
1
K1T
1l2k
1Tl
2= K
2Tl
1K
2T
2l1
(K2l1
+ K1l2)T = K
1T
1l2
+ K2T
2l1
1 1 2 2 2 1 1 2 1 2 1 2
2 1 1 2 1 2 2 1
K T l K T l K l T K l TT
K l K l K l K l
+ + = =
+ +
25. 1 21 2
0 0
l lL 10log ; L 10log
l l
= =
1 21 2
0 0
l lL L 10log 10log
l l
=
1
2
lL 10log
l
=
1
2
l20dB 10log
l
=
2 1
2
l10
l =
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12
ll
100 =
26. According to Mayers relation,
p vR RC Cm 28
= =
27. Whenachargeparticleentersamagneticeldatadirectionperpendiculartothedirectionof
motion, the path of the motion is circular. In circular motion the direction of velocity changes
at every point (the magnitude remains constant). Therefore, the momentum will change
ateverypoint.Butkineticenergywillremainconstantasitisgiven by 21
mv2
and v2 is the
square of the magnitude of velocity which does not change.
28. The eld at the same point at the same distance from the mutually perpendicular wires
carrying current will be having the same magnitude but in perpendicular directions.
2 2
1 2B B B = +
( )
12 20 21 2
B l l2 d
= +
29. From Rt= R
0(1 + t)
5 = R0
(1 + 50) .(i)
and 6 = R0
(1 + 100) . (ii)
5 1 50
6 1 100
+ =
+
1
200 =
Putting value of in Eq (i), we get
0
15 R 1 50
200
= +
R0
= 4
30. The potential energy of a charged capacitor is given by2Q
U2C
=
If a dielectric slab is inserted between the plates, the energy is given by
2
Q2KC
, where K is the
dielectric constant.
Again, when the dielectric slab is removed slowly its energy increases to initial potential
energy. Thus work done is zero.
31. Since electronic charge (1.6 1019 C) is universal constant. It does not depend on g.
Electronic charge on the moon = electronic charge on the earth
electronicchargeonthemoon
electronicchargeontheearth = 1
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32. In this question distance of centre of mass of new disc from the centre of mass of remaining
disc is R.
Mass of remaining disc
M 3MM
4 4= =
3M MR R 0
4 4 + =
1
3 =
33. The acceleration of a solid sphere of mass M, radius R and moment of inertia I rolling
down (without slipping an inclined plane making an angle with the horizontal) is given by
2 2
2
2
gsin Ia ,whereI MK K
MK1
R
= = =
+
2
gsina
I1
MR
=
+
34. Central forces passes through axis of rotation so torque is zero.
If no external torque is acting on a particle, the angular momentum of a particle is
constant.
35. 2kF 15
a 7.5m / sm 2
= = =
Now, 21
ma kx2
=
212 7.5 10000 x2
=
x2=3 103
x = 0.055 m
x = 5.5 cm
36. Let K be the K.E. at the highest point.
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Then
2
x
1K' mv
2= (v
y= 0 at highest point)
( )21 m ucos2
=
2 2 2 21 1
mu cos K.cos K mu2 2
= = =
K = K.cos260 ( = 60)
21 K
K.2 4
= =
37. In Youngs double slit experiment intensity at a point is given by
2
0I I cos
2
=
. (i)
where = phase difference, l0
= maximum intensity
2
0
Icos
I 2
=
Phase difference2
pathdifference
=
2
6
=
3
= .(ii)
Substitute eqn. (ii) in eqn. (i), we get
2
0
Icos
I 6
=
0
I 3
I 4 = .
38. 1 2k k1
f
2 m
+=
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and 1 2k k1
f ' .2 2f 2 m
+= =
39. For path iaf, Q = 50 cal, W = 20 cal
Byrstlawofthermodynamics
U=QW=5020=30cal.
For path ibf, Q = 36 cal, W = ? Byrstlawofthermodynamics,
Q = U + W
W=QU
Since, the change in internal energy does not depend on the path, therefore U = 30 cal
W=QU=3630=6cal.
40. For a particle to execute simple harmonic motion its displacement at any time t is given by
x(t) = a (cos t + )
where, a = amplitude, = angular frequency, = phase constant.
Let us choose = 0 x(t) = acost
Velocity of a particledx
v a sin tdt
= =
Kinetic energy of a particle is 21
K mv2
=
2 2 21 ma sin t2
=
Average kinetic energy = 2 2 21
ma sin t
2
< >
2 2 21 m a sin t2
= < >
2 2 21 1 1m a sin2 2 2
= < >=
( )221 ma 2 [ 2 ]4
= =
2 2 2ma=
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CHEMISTRY
41.2 2
A B 2 AB+
Ea(forward) = 180 kJ mol1
Ea (backward) = 200 kJ mol
1
In the presence of catalyst :
Ea(forward)=180100=80kJmol1
Ea(backward)=200100=100kJmol1
H = Ea(forward)E
a(backward)
=80100
=20kJmol1
42. Ecell
= 0; when cell is completely discharged.
2
cell .cell 2
Zn0.059E E log
2 Cu
+
+
=
2
2
Zn0.0590 1.1 log
2 Cu
+
+
=
2
2
Zn 2 1.1log 37.3
0.059Cu
+
+
= =
2
37.3
2
Zn10
Cu
+
+
=
43. For acidic buffer,a
ApH pK log
HA
= +
When the acid is 50% ionised, [A] = [HA]
pH = pKa
+ log1
or pH = pKa
Given pKa= 4.5
pH = 4.5
pOH=144.5=9.5
44. 2A+B products
[B]isdoubled,half-lifedidntchangehalf-lifeisindependentofchangeinconc.ofreactant
First order
Firstorderw.r.ttoB
[A] is doubled, rate increased by two times
First order w.r.t A
Hence, net order of reaction = 1 + 1 = 2
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unit for the rate constant = conc.(1n) t1
= (mol. L1)1. s1
= L. mol1 s1
45. 4f orbital is nearer to nucleus as compared to 5f orbital therefore, shielding of 4f is more than
5f.
46. In 4-coordinate complexes Pt, the four ligands are arranged about the central 2-valent
platinumioninasquareplanarconguration.
47. The molecule, which is optically active, has chiral centre, is expected to rotate the plane of
polarized light.
One chiral centre optically active
Two chiral centres, but plane of symmetry within molecule optically inactive
48. Proteins have two types of secondary structures -helix and -plated sheet.
49.
HBr
3 2 3 2 2CH CH CH CH CH CHBr =
2HBr
3 2CH CH CH CHBr
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HBr
3 2 3 3CH CH CH CH CHBr CH =
50. This is carbylamine reaction.
3 2 2 3
2 5 2
CH CH NH CHCl 3KOH
C H NC 3KCl 3H o
+ +
+ +
51.
52.
NO2groupwithdrawelectronfromthering,showsMeffectmakesringelectrondecient,
thus, deactivates ring for electrophilic substitution.
53. NO NO+
NO Totale 14
+
=
2 2 2 2 1 1 1 1 2
x y z1s *1s 2s *2s 2p 2p 2p+ + =
Diamagnetic
Bondorder10 4
32
= =
(NO) Total e = 15
2 2 2 2 2 1 1 1 1
z x y1s *1s 2s *2s 2p 2p 2p+ +
1x y*2p 2p =
Paramagnetic
Bondorder10 5
2.52
= =
Electron is taken away from non-bonding molecular orbital, thats why bond order increases.
54. More the distance between nucleus and outer orbital, lesser will be force of attraction on
them. Distance between nucleus and 5f orbital is more as compared to distance between 4f
orbital and nucleus. So actinides exhibit more number of oxidation states in general than the
lanthanides.
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55. Let the mass of methane and oxygen be m gm. Mole fraction of oxygen,2o
x
mm 32 132
m m 32 3m 3
32 16
= = =
+
Let the total pressure be P.
Partial pressure of O2,
2 2O OP P x=
1 1P P
3 3= =
56. Solution is isotonic.
C1RT = C
2RT
C1
= C2
Density of both the solutions are assumed to be equal to 1.0 g cm3
molality = molarity 5.25%
in 100 g 5.25 g of substance therefore, in 1000g 52.5 g of substance
Hence,52.5 15
M 60= ,
M = molecular mass of the substance
52.5 60M 210
15
= =
57. Isotonic solution have same osmotic pressure.
1 1 2 2C RT, C RT = =
For isotonic solution 1
= 2
C1C
2
1.5 / 60 5.25 / M
V V =
[Where M = molecular weight of the substance]
1.5 5.25
60 M =
M = 210.
58. Let x = solubility
3 3AgIO Ag IO+ +
2
sp 3K [Ag ][IO ] x x x+ = = =
Given Ksp
=1108
8 4
spx K 1 10 1.0 10 mol / lit = = =
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=1.0104283g/lit
41.0 10 283 100
gm / 100ml1000
=
=2.83103 gm/100 ml
59. Activity N
n
0
N 1
N 2
=
n1 1
10 2
=
10 = 2n
Taking log on both sides
log 10 = n log 2
1n 3.32
0.301= =
time=nhalf-life
= 3.3230=99.6days.
60. Chiral conformation will not have plane of symmetry. Since twisted boat does not have plane
of symmetry it is chiral.
61.
Chair form is unsymmetrical and absence of any element of symmetry.
62.2P l Mg
3 2 3 2 EtherCH CH OH CH CH I
A
+
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63. n = 3, l = 0 represents 3s orbital
n = 3, l = 1 represents 3p orbital
n = 3, l = 2 represents 3d orbital
n = 4, l = 0 represents 4s orbital
The order of increasing energy of the orbital is
3s < 3p < 4s < 3d
64. Hydrogen bond is strongest in HF due to higher electro negativity of F.
65. 3(s) (aq) (aq) (aq) 2(g)
2Al 6HCl 2Al 6Cl 3H+ + + +
6 moles of HCl produces = 3 moles of H2
=322.4 L of H2
2
3 22.4
1moleofHClproduces 11.2L ofH2
= =
Since, 2 moles of Al produces 3 moles of H2=322.4LofH
2
1 mole of Al produces2
3 22.433.6L ofH
2
= =
66.( ) ( )24 4 2 4 42NH SO 2H O 2H SO 2NH OH
Strongacid Weakbase
+ + + +
(NH4)
2SO
4on hydrolysis produces strong acid H
2SO
4, which increases the acidity of the soil.
67. In an isolated system where either mass or energy are not exchanged with surrounding for
the spontaneous process, the change in entropy is positive.
68. Isotopes are atoms of same element having same atomic number but different atomic masses.
Neutron has atomic number 0 and atomic mass 1. So loss of neutron will generate isotope.
69. According to Kohlrauschs law, the molar conductivity at innite dilution () for weak
electrolyte, CH3COOH is
3 3CH COOH CH COONa HCl NaCl
= +
So, for calculating the value of3CH COOH
, value of NaCl should also be known.
70. In aqueous solution, basicity order :
Dimethylamine > methylamine > trimethylamine > aniline
2 13
71. When alkyl benzene are oxidized with alkaline KMnO4, the entire alkyl group is oxidized to
COOHgroupregardlessoflengthofsidechain.
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72.
73. Diamagnetic species have no unpaired electrons
2 2 2 2 2 2 2 2 2
2 z x y x yO 1s , s , s , 2p , 2p , 2p , *2p , *2p 1
74. Due to the inert pair effect (the reluctance of ns2 electrons of outermost shell to participate
in bonding) the stability of M2+ ions (of group IV elements) increases as we go down the
group.
75. Br2reactswithhotandstrongNaOHtogiveNaBr,NaBrO3 and H2O.
2 3 26NaOH 3Br 5NaBr NaBrO 3H O
conc.andhot
+ + +
76. Smaller the size and higher the charge more will be polarizing power of cation. So the correct
order of polarizing power is K+ < Ca2+ < Mg2+
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80. For G = HTS
For a spontaneous process G < 0
i.e., HTS < 0
H < TS
TS > H
TS
>
i.e.,179.1 1000
T160.2
>
T > 1117.9 K 1118 K.
MATHEMATICS
81. Since, each term is equal to the sum of the next two terms.
n 1 n n 1ar ar ar + = +
21 r r = +
r2+r1=0
5 1 5 1r r
2 2
=
82. 1 1x 5
sin cosec5 4 2
+ =
1 1x 5
sin cosec5 2 4
=
1 1x 4
sin sin5 2 5
=
[sin-1 x + cos1 x = /2]
1 1x 4
sin cos ....(i)5 5
=
Let 14
cos A5
=
4cosA
5 =
3sinA
5 =
13
A sin5
=
1 14 3cos sin5 5
=
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equation (i) become 1 1x 3
sin sin5 5
=
x 3
5 5 =
x = 3
83. Given n n 4 4 n n 5 55 6 4 5t t 0 C a ( b) C a ( b) 0 + = + = ( ) ( )(4 5n n 4 n n 54 5C a b C a b =
a (n 4)! n 4
b (n 5)!5 5
= =
84. Required number of ways 12 8 44 4 4
C C C=
12! 8!1
8! 4! 4! 4!=
( )3
12!
4!=
85.2x 1 xf(x) 4 cos 1 log(cosx)
2
= + +
f(x)isdened ifx
1 1 12
and cos x > 0
x0 2and x
2 2 2
< <
0 x 4and x2 2