Past Year Papers AIEEE 2007

7/31/2019 Past Year Papers AIEEE 2007

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485Past 5 Years Papers: Solved

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 = =

AIEEE 2007 SOLUTION


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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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492 AIEEE

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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504 AIEEE

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

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Past Year Papers AIEEE 2007