past year papers aieee_2010

8/6/2019 Past Year Papers AIEEE_2010

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360 AIEEE

1 (1) 2 (2) 3 (3) 4 (3) 5 (4) 6 (1) 7 (2) 8 (3) 9 (2) 10 (3)

11 (4) 12 (3) 13 (2) 14 (2) 15 (2) 16 (3) 17 (3) 18 (1) 19 (3) 20 (1)21 (4) 22 (3) 23 (1) 24 (1) 25 (4) 26 (4) 27 (4) 28 (1) 29 (4) 30 (1)

31 (4) 32 (3) 33 (1) 34 (4) 35 (1) 36 (3) 37 (3) 38 (2) 39 (2) 40 (1)

41 (4) 42 (3) 43 (2) 44 (3) 45 (2) 46 (4) 47 (1) 48 (4) 49 (1) 50 (1)

51 (4) 52 (3) 53 (3) 54 (3) 55 (3) 56 (2) 57 (2) 58 (4) 59 (2) 60 (3)

61 (3) 62 (2) 63 (3) 64 (4) 65 (4) 66 (2) 67 (1) 68 (2) 69 (2) 70 (4)

71 (1) 72 (1) 73 (1) 74 (1) 75 (3) 76 (4) 77 (3) 78 (3) 79 (2) 80 (2)

81 (2) 82 (2) 83 (3) 84 (4) 85 (4) 86 (3) 87 (3) 88 (1) 89 (1) 90 (3)

PHYSICS

1. For a parallel cylindrical beam, wave front will be planar.

2. Since medium0 2

cV

I =

+ and I is inversely proportional to r

For a cylindrical beam ( ) 0 2medium

cI I,

V= = +

medium0 2

cV

I =

+ . Thus speed will be minimum at the axis as intensity is maximum

3. As intensity is maximum at axis. will be maximum and speed will be minimum on the axis of the beam.

Beam will converge.

4. By Conservation of the momentum

1 2M M0 V V2 2

=

V1 = V 2 .(i)

2 2 2

1 2

1 M 1 Mmc . V . V2 2 2 2 = + .(ii)

2 21

Mmc V2

=

2

21

2 mc VM

=

12 mV c

M=

5. After decay, the daughter nuclei will be stable hence binding energy per nucleon will be morethan that of their parent nucleus.

AIEEE 2010 SOLUTION


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361Past 5 Years Papers: Solved

6. By the equation

0 0hv hv eV== +

00

hc hceV= +

0 maxas V K

This proves that statement 1 is true and 2 is false.

7. If the particles moving in same direction lose all their energy, nal momentum will becomezero, whereas initial momentum is not zero.

8. Because of impulse direction of velocity changes as can be detected from the slope of the graph.

Initial velocity 2 1 m / s2

= =

Final velocity 2 1 m / s2

= =

iP 0.4N s=

iPj 0.4N s=

( )f iJ P P 0.4 0.4 0.8N s J impulse= = = =

J 0.8N s=

9. In region 1, i.e., on the left of the rst wire eld will remain negative, and as the positionchanges from to A, eld increase in magnitude from zero to large value. As one movesfrom A to B, eld changes sign from positive to negative, becoming zero at mid point as onemove in region 3, from B to +, eld decreases from a large value to zero.

10. As oil > , ball must sink in oil alone.

As water < , ball mus t oat in water.

11. By symmetry the cosine component cancels so,

Linear charge density qr

=

( ) ( )2K.dq E dEsin j sin jr= =

( )2K qr E d sin jrr= ( )2

0

K q sin jr

=

( )2 20

q j2 r

=


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362 AIEEE

12. For adiabatic expansion

y 1 y 1

1 1 2 2T V T V =

( ) ( )

y 1 7 1 2/51 2 5

2 1

T V32 32 4T V

= = = =

1

2

T 11 1 0.75T 4

= = =

13. By the rules to nd out signi cant gures the following gures have the respective signi cantdigits.

23.023 5 0.0003 1

32.1 10 2

14.

15. P = nh v

3 20 344 10 10 6.63 10 v = v

174v 10 Hz6.63= v

this is range of X-rays.

16. When positive beta decay takes place, a proton is transformed into a neutron and a positronis emitted.

0p n e+ + +

no. of neutrons initially was A Z

no. of neutrons after decay

(A Z) 3 2 (Due to alpha particles) + 2 1 (Due to positive beta decay)

The no. of proton will reduce by 8.[As 3 2 (due to alpha particles) 2+ (Due to positive beta decay)]

Hence atomic number reduces by 8.

17. 05 r(r)4 R =

By Gauss Law

enc

0

qE.dA =


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r 22 0

0

(r)4 r drE.4 r

=

r 200

0

5 r 4 r dt4 R =

r r2 30 0 0

2

0

5 14 r dr r dr4 RE.4 r

=

3 4

2 0 5 r 1 rE.r4 3 R 4 =

22 0

0

5 rE.r r4 3 R

=

0

0

r 5 rE4 4 R

=

18. When capacitance is taken out, then

LXtan30

R =

L

200X

3 =

When inductance is taken out, then

CXtan30

R =

C

200X

3 =

While all three are present,

{ L C0 X X = =thus power dissipated ms msP E .i cos0

=

( )2

220 220 220 220 242W200200 0

= = =+

19. At t = 0, no cu rrent ows through inductor

So,2

VI

R=

At t = , inductor behaves as a conductor

So, ( )( )

1 2

1 2

VI

R RR R

=

+


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364 AIEEE

20. dx dyKy ; Kxdt dt

= =

dy x

dx y=

ydy = xdx

2 2y x cons tan t = +

21. Since

( )22 0 2t /RC 2t/RC0

qqU e U e2C 2C

= = =

t/RC0q q e=

On substituting values of charge, when charge becomes 14

times, energy becomes 116

times.

So, t 1 = one half life, while t 2 = 4 half lives.22. A moving conductor is equivalent to a battery of emf = v B l (motion emf) Equivalent circuit

1 2l l l= +

Applying Kirchoffs law

1I R IR vBl 0 ......(1)+ =

2I R IR vBl 0 ......(2)+ =

Adding (1) and (2)

2IR IR 2vBl+ =

2vBlI3R

=

1 2

vBlI I3R

= =

23.

20.500.04v 12.5m / s

2k 0.040.50

= = = =

Tv =


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( )22T v 12.5 0.04= = = 6.25 N

24.( )

2

A alongveticalgsin 60 =

( )2

B Alongvertical gsin 30 =

( )2

A/B alongvertical

3 1 gg 4.9m / s4 4 2

= = = 25. For a particle in uniform circular motion,

2vaR

= towards centre of circle

( )2v

a cos i sin jR = 26. Gravitation torque

( )F dist.= = mgx

( )0mg v cos dldt

=

2

0

tl l ldt mgv cos 2= =

( )l m r v= (20 t mgv cos k2=

27. When charged sphere are in air

EF T sin=

mg T cos=

()EF mg tan ....... i = When in liquid,

ET 'sin F ' =


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366 AIEEE

BT 'cos F mg + =

BT 'cos mg F =

( ) ( )E BF ' mg F tan ....... ii = By (1) and (2)

E

E B

F mgF ' mg F

=

B

1 1KF 111

R.Dmg

= =

( )

1K 211 1.6 / 0.8

= =

28. S = t 3 + 5

tangential speed dsvdt

=

3v 3t 0= + at t = 2 sec v = 12 m/s

Tangential acceleration Tdvadt

=

a T = 6tat t = 2 s

a T = 12 m/s2

Centripetal acceleration2

RvaR

=

( )2 2R

12 144a 7.2m / sec20 20

= = =

Total acceleration ( ) ( )2 2totala 12 72= += 14 m/s 2

29. ( ) 12 6a bU x x x= At equilibrium position, F net = 0

( )dU x0

dx =

13 7

12a 6b 0x x

+ =


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1/62aX 0andXb

= =

thus 12 61/6 1/6a bD

2a 2ab b

=

2 2 2 2

2

a.b b b b2a 4a 4a4a

= = =

30. In series combination

eff 0 0 0 2 0R R R t R t R= + + +

( )0 0 1 22R R t= + +

1 2eff 0R 2R 1 t2

+ = +

1 2eff 2 +

=

In parallel combination,

( ) ( )eff 0 1 0 21 1 1

R R 1 t R 1 t

= +

+ +

( ) ( )1 2eff 0

1 11 t 1 t

R R = +

( )0

eff 1 2

RR

2 t=

+

1

0 1 2eff

RR 1 t

2 2

+ =

( )1 20 tR 12 2

+ = +

1 2eff 2

+ =

CHEMISTRY

31. Since K 2 < < K 1 Conc. of H + and 3HCO

are approximately same.

32. [Ag + ] = 0.05, [Br ] = x MK

sp= S 2=[Ag + ] [Br ]


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368 AIEEE

5 1013 = 0.05 x

x = 10 11 M

Solubility of KBr is (120 10 11 ) 1.2 10 9 g/L

33. 2 2 3O F Na Mg Al + + +> > > > .

Decreasing ionic radii with increasing effective nuclear charge for isoelectronic species.(Generally if element is carrying ve charge size is more,if +ve charge small size neutral is inbetween for the successeive elements.

34.

(conversion of aniline to benzene diazonium chloride is Diazotization reaction)

35. PV = nRT (ideal gas equation)

n = no moles has to calculated i.e., n =PV/RT

3170 10 3 = n 8.314 300

3

331.7 10n 1.27 108.314 3

= =

36. Alcohols which give more stable carbocation is more reactive with Lucas reagent (Anhy. ZnCl 2 + conc. HCl)

Reactivity of alcohols 3 0>2 0>1 0. 3 0 form ppt immediately, 2 0 form after 5min, 1 0 will notrespond r very slow.

37. Tf

= i Kf m

i for Na 2SO 4 ionize as a 2Na+ and SO 4

-2 i.e., is i= 3(100% ionisation)

f 0.01T 3 1.86

1 =

T f = 0.0558 K

38.


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Bronsted acid base theory. Conjugate base conjugate acid

39.

Dehydration of alcohols gives alkenes. Above case major is according to resonance and minoris sytzeff rule

40. hcE =

Ev =h v h is plank constant

3

23242 10E J / atom

6.023 10=

3 34 8

23

242 10 6.6 10 3 106.023 10

=

26 236

3

19.8 10 6.023 10 0.494 10242 10

= =

= 494 nm.

41. 1.4NV%NW

= : the amount of acid absorbed by NH 3 is 5ml.

( )3

1.4 0.1 20 15 70023.7

29.529.5 10

= = =

42.2 2

2He He

2Li Li

E ZE Z

+ +

+ +

= Energy directly proportional to the z

2

18

Li

19.6 10 4E 9+

=

2

18Li

9E 19.6 104+

=


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370 AIEEE

= 4.41 10 17 J/atom Energy of orbit of Li +2 is 4.41 10 17 J/atom.

43. 0 0A A B BP X P X P= +

heptane25n 0.25

100= = (pressure exerted by heptanes)

octane35n 0.307

114= = (pressure exerted by octane) by the addion of the these will give the

total pressure.

0.25 0.307P 105 45

0.25 0.307 0.25 0.307= +

+ + = 47.127 + 24.84 = 71.96

= 72 kPa

44.

In order to show optically active, the objects should not be superimposed on each other and

it exists as enantiomeric pair ( ) 33Co en+

45. Formation of carbocation is rate determining step in S N1 reaction. Hence, alkyl halide whichgives more stable carbocation is more reactive towards S N1 reaction

B. stabilized by resonance, c is stabilised by to 2 +I groups, A has of +I group

46. 32

(i)O3 3 3(i i) Zn H OCH CH CH CH CH CHO =

44 amu47. Rate depends only on slow step. For slow step activation energy is more. Hence, it takes more

time, which in turn rate of the reaction increase.

48. 2 3 22 4Al O Al O3 3

+

G = 966 kJ/mol

4e are involved G = nFE i.e., E= -G/nF

E = potential difference


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966 10 3 = 4 96500 E

966E V 2.5V

4 965= =

2.5 V potential difference is required.

49. 2R COO HC C NH R

increasingbasicstrength < < T e

53. S pace occupied in fcc and bcc lattices are 74% and 68% space left out 26% and 32%.

54. Nylon 6, 6 involves amide linkage therefore, it will also have very strong intermolecularhydrogen bonding between

To show hydrogen bond H should attach with EN element like O,N,F, which will partiallyattached with adjacent EN element group of two polyamide chains.


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372 AIEEE

55. Ksp = S3= [Mg +2 ] [OH ] 2

1 1011 = 0.001 [OH ] 2

[OH ] = 10 4 M

pOH = 4, pH = 10 pH= pH+ poH

56. Mn > Cr > Fe > Co Reduction potential value are give of each element more is the negativemore is the tendency to get reduce

20Mn /MnE 1.18+ =

20Cr /Cr

E 0.91+ =

20Fe /FeE 0.44+ =

20CO /CO

E 0.28+ =

57. Biuret test is only given by amides. Carbohydrates are not amides and hence, it does notgive biuret test.(this test used to identify the peptide bonds) The biuret reagent is made of potassium hydroxide(KOH) and hydrated copper (II) sulfate, together with potassium sodiumtartrate. The reagent turns from blue to violet in the presence of proteins, blue to pink whencombined with short-chain polypeptides

58.12

a 2K 12t 2 1

= = =

in zero order reaction half life is directly proportional to conc of reactant.

Hence,

0 1C C 0.5 0.25t 0.25h

K 1 = = =

59. Moles of complex = 2.675 0.01267.5

=

Moles of AgCl precipitated = 4 78 0.033143.5

=

It means 3Cl are released by one molecule of complex [Co(NH 3) 6]Cl 3 the chloride ionswhich are outside the co-ordination sphere can form white ppt.

60. 2 2 31 3N H NH2 2+

3NHH 46kJ =

3 2 21 3NH N H2 2

+

( )N H 1 346 3 H 712 4362 2=

N HH 352kJ / mol =


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MATHEMATICS

61. R is not an equivalence relation because 0 R 1 but 1 R 0, S is an equivalence relation.

62. We have,

|z 1| = |z + 1| = |z i|. So z is at equal distance from the points (1, 0) and (0, 1) and (1,0). Hence z is the circumcentre of the triangle formed by the vertices (1, 0) and (0, 1) and(1, 0) which is unique.

63. and are roots of the equation x 2 x + 1 = 0

+ = 1, = 1

+ = =1 3i 1 3i 1 3ix x or2 2 2

x = , 2

Thus, = 2, then =

= , then = 2 where 3 = 1

2009 2009 2009 2 2009( ) ( ) + = +

Since 2009 leaves 2 as remainder, we have 2009 2009 2009 2 2009 2 4( ) ( ) 1 + = + = =

64. The given system of linear equations can be put in the matrix form as

1

2

3

x1 2 1 32 3 1 x 33 5 2 1x

=

Observe that the elements in the rst two columns are in AP. Then the determinant of thematrix will be zero. Hence the given system of equations has no solution.

65.

Two balls from urn A and two balls from urn B can

be selected in 3 92 2C C ways.

= 3 36 = 108

66. We have,

f : (1, 1) Rf(0) = 1 f(0) = 1

g(x) = [f(2f(x) + 2)] 2

g(x) = 2[ f(2f(x) + 2)] f (2f(x) + 2) 2f (x)

g(0) = 2[ f(2 f(0) + 2)] f (2 f(0) + 2) 2f(0)


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374 AIEEE

= 2[f(0)] f (0) 2f (0)

= 2 1 1 2 1 = 4

67. We have, f : R R. xf(3x)

lim 1f(x) =

Given that the function is increasing and positive. Hence we have ( ) ( ) ( )0 f x f 2x f 3x< < < .The question should have given that domain also is positive.

Now, we have ( ) ( ) ( ) f (2x) f (3x)0 f x f 2x f 3x 1f (x) f (x)

< < <

past year papers aieee_2010

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