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AKASH MULTIMEDIA 19

AIEEE-2008 MATHS, CHEMISTRY & PHYSICS

88. (2) Directions: Question No. 89 and 90 are baseson the followign paragraph.Consider a block of conducting material of

resistivity ’ ’S shown in thefigure. Current ‘|’ enters

at ‘A’ and leaves from ‘D’. We applysuperposition principle to find voltage ’ V ’%

developed between ‘B’ and ‘C’. The calculationis done in the following steps:i) Take current ‘|’ entering from ‘A’ and assumeit to spread over a hemispherical surface in theblock.ii) Calculate field E(r) at distance ‘r’ form A by

using Ohm’s law E = j,S where j is the currentper unit area at ‘r’.iii) From the ‘r’ dependence of E(r), obtain thepotential V(r) at r.iv) Repeat (i), (ii) and (iii) for curent ‘|’ leaving‘D’ and siuperpose results for ‘A’ and ‘D’.

I I

A B C D

a b a

V%

89. (3) Choosing A as origin,

2

|E j

2 r� S � S

Q

(a b)

C B 2a

| 1 | 1 1V V dr

2 2 (a b) ar

�

S S ¨ ·

� � � � �© ¸

Q Q �ª ¹

°

B C

| 1 1V V

2 a (a b)

S ¨ ·

� � �© ¸

Q �ª ¹

90. (3) 2E

2 r

S�

Q

91. (3) 1 1 1

v u f� � = constant

92. (3) m1u

1 + m

2u

2 = (m

1 + m

2)v v = 2/3 m/s

Energy loss = 1

2(0.5)s (2)2–

1

2(1.5)s

22

3¥ ´¦ µ§ ¶

= 0.67 J

93. (3) Capillary rise h = 2Tcos

gr

R

S.

As soap solution has lower T, h will be low.

94. (1) 2k mv

r r�

mv2 = k (independent of r)

hn

2¥ ´

¦ µ§ ¶Q

= mvr r� t n and T = 1

2mv2 is

independent of n.

95. (1) y = 0.005 cos ( x t)B �C

comparing the equation with the standard form,

x ty Acos 2

T

¨ ·¥ ´

� � Q¦ µ© ¸§ ¶Mª ¹

2 /Q M � B and2 / TQ � C

2 / 0.08 25.00B � Q � Q C � Q

96. (4) 40 1 2N N AM 2.4 10 H

l�

N

� � Q s

97. (3)

A B C0 0 00 1 11 0 11 1 1

98. (2) x1 (t) =

21at

2 x

2(t) = vt x

1–x

2 =

1

2at2 – vt

99. (1)

100.(1) 2

0

0 if r R

E(r) Qif r R

4 r

�«

®� ¬

r®

QF

101.(3)

2 1P P

5 0 22 10 1V V1 1 12 10 1

� �

� �

� �

P2

P1

1082V5V

28 i 18

2 1P P2 1

V V| 0.03from P P

10

�

� � m

102. (3) 7

60 i 4 10 100B 5 10 T

2 R 2 4

�

�

N Q s

� � s � s

Q Q

southward

103. (2) For diamagnetic materials r 1F � and Ur<1

104. (4) Diameter = M.S.R. + C.S.R × L.C. + Z.E.

= 3 + 35 × (0.5/50) + 0.03 = 3.38 mm

105.(1) U = U1 + U

2

1 1 2 2 1 2

1 1 2 2 2 1

(P V P V )T TT

(P V T P V T )

�

�

�

P

AKASH MULTIMEDIA18

AIEEE-2008MATHS, CHEMISTRY & PHYSICS

PART-C : PHYSICS

Directions : Questions No. 71, 72 and 73 arebased on the following paragraph.

Wave property of electrons implies that they willshow diffraction effects. Davisson and Germerdemonstrated this by diffracting electrons fromcrystals. The law governing the diffraction froma crystal is obtained by requiring that electronwaves reflected from the planes of atoms in acrystal interfere constructively (see figure).

i

d

OutgoingElectrons

IncomingElectrons

Crystal plane

71. (2) 2d cos i = nM 2d cos i = h

2meV

v = 50 volt

i

72. (4) 2d cos i = dBnM

73. (4) Diffraction pattern will be wider than theslit.

74. (3)

esc

2GM 2G 10MV 10 11 110 km / s

R R /10

s

� � � s �

75. (1) 21 2 TVg Vg kvS �S �

1 2T

Vg( )v

k

S �S

� �

76. (2) 55 R 55 8

R 22020 80 2

s

� � � � 8

77. (1)

n

cm n

xk .xdxdmx dx.x L

Xdm dm x

k dxL

¥ ´¦ µM § ¶

� � �

¥ ´¦ µ§ ¶

°° °° ° °

=

Ln 2

n

Ln 1

no

o

kx

(n 2)Lx(n 1)kx

n 2(n 1)L

�

�

¨ ·

© ¸�

© ¸

�¨ ·© ¸

�© ¸© ¸ �� ª ¹

© ¸

© ¸

© ¸ª ¹

cm

L 2L 3L 4L 5LX , , , , ,.....

2 3 4 5 6�

78. (2) 1 RT

n4x M

H

� 1 RT

xn4 M

H

� x Tt

79. (3) F = qvB

B = F/qv = MC–1T–1

80. (4)

22 2

2cm

a 2 ma ma 2l l m ma

2 6 2 3

¥ ´� � � � �¦ µ

§ ¶

81. (1) P = mv = 3.513×5.00z 17.6

82. (4) Approximate mass = 60kg

Approximate velocity = 10m/s

Approximate 1

KE 60 100 3000J2

� s s �

KE range � 2000 to 5000 joule

83. (3) 0 0 0

1 2

A A 18AC’

d d d 2d 4d9 183 6

F F F

� � �

��

C’ = 40.5PF

84. No option is correct

RTv

M

H

�

1 1 2

2 2 1

74V M 5

5V M 323

s

H

� �

H

s

22

460 21 460 5 2 2v 1420

V 25 8 21

s s

� � � �

s

85. (4)

86. (2) g = GM/r2

87. (4) As liquid 1 floats above liquid 2, 1 2S � S

The ball is unable to sink into liquid 2, 3 2S � S

The ball is unable to rise over liquid 1,1 3S � S

Thus, 1 3 2S � S � S

AKASH MULTIMEDIA 3


22. Let

11 sin 1

10 1

x if xf x x

if x

«

� x®

� �¬

®�

Then which one of the following is true?

1) f is neither differentiable at x = 0 nor at x =1

2) f is differentiable at x = 0 and at x=1

3) f is differentiable at x = 0 but not at x=1

4) f is differentiable at x =1 but not at x = 0

23. The first two terms of a geometric progressionadd up to 12. The sum of the third and the fourthterms is 48. If the terms of the geometric pro-gression are alternately positive and negative,then the first term is1) –4 2) –12 3) 12 4) 4

24. Suppose the cubic x3-px+q has three distinctreal roots where p > 0 and q > 0. Then whichone of the following holds?

1) The cubic has minima at 3

pand maxima at

3

p�

2) The cubic has minima at 3

p� and maxima

at 3

p

3) The cubic has minima at both 3

p� and

3

p

4) The cubic has maxima at both 3

pand

3

p�

25. How many real solutions does the equationx7 + 14x5 + 16x3 + 30x – 560 = 0 have ?

1) 7 2) 1 3) 3 4) 5

26. The statement pm (qmp) is equivalent to

1) pm (pmq) 2) pm (p� q)

3) pm (p� q) 4) pm (pjq)

27. The value of 1 15 2cot cos tan

3 3ec� �

¥ ´�¦ µ§ ¶

is

1) 6/17 2) 3/17 3) 4/17 4) 5/17

28. The differential equation of the family of circleswith fixed radius 5 units and centre on the liney=2 is

1) 222 25 2x y y� � � �a

2) 222 25 2y y y� � � �a

3) 2 222 25 2y y y� � � �a

4) 2 222 25 2x y y� � � �a

29. Let 1 1

0 0

sin cosx xI dx and J dx

x x� �° °

Then which one of the following is true ?

1) 2

3I � and J > 2

2) 2

3I � and J < 2

3) 2

3I � and J > 2

4) 2

3I � and J < 2

30. The area of the plane region bounded by the

curves 22 0x y� � and 23 1x y� � is equal to

1) 5/3 2) 1/3 3) 2/3 4) 4/3

31. The value of sin

2sin

4

xdx

xQ¥ ´

�¦ µ§ ¶

° is

1) log cos4

x x cQ¥ ´

� � �¦ µ§ ¶

2) log sin4

x x cQ¥ ´

� � �¦ µ§ ¶

3) log sin4

x x cQ¥ ´

� � �¦ µ§ ¶

4) log cos4

x x cQ¥ ´

� � �¦ µ§ ¶

AKASH MULTIMEDIA4


32. How many different words can be formed byjumbling the letters in the word MISSISSIPPIin which no two S are adjacent?1) 8 .6 4C . 7

4C 2) 6 . 7 . 8 4C3) 6 . 8 .7 4C 4) 7 . 6 4C . 8

4C

33. Let a, b, c any real numbers. Suppose that thereare real numbers x, y, z not all zero such that

x=cy+bz, y=az+cx, z=bx+ay. Thena2+b2+c2+2abc is equal to1) 2 2) –1 3) 0 4) 1

34. Let A be a square matrix all of whose entriesare integers. Then which one of the followingis true?

1) If det 1A � p , then 1A� exists but all its en-tries are not necessarily integers.2) If det 1A x p , then 1A� exists and all its en-tries are non-integers.3) If det 1A � p , then 1A� exists and all its en-tries are integers.4) If det 1A � p , then 1A� need not exist.

35. The quadratic equations x2–6x+a=0 and x2–cx+6=0 have one root in common. The otherroots of the first and second equations are inte-gers in the ratio 4 : 3. Then the common root is1) 1 2) 4 3) 3 4) 2

PART-B : CHEMISTRY

36. The organic chloro compound, which showscomplete stereochemical inversion during a S

N2

reaction, is

1) (C2H

5)

2CHCl 2) (CH

3)

3CCl

3) (CH3)

2CHCl 4) CH

3Cl

37. Toluene is nitrated and the resulting product isreduced with tin and hydrochloric acid. Theproduct so obtained is diazotised and thenheated with cuprous bromide. The reactionmixture so formed contains

1) mixture of o - and p-bromotoluences

2) mixture of o- and p-dibromobenzenes

3) mixture of o- and p-bromoanilines

4) mixture of o-and m-bromotoluenes

38. The coordination number and the oxidation stateof the element ‘E’ in the complex[E(en)

2(C

2O

4)]NO

2 (where ‘en’ is ethylene

diamine) are, respectively,

1) 6 and 2 2) 4 and 2

3) 4 and 3 4) 6 and 3

Coordination no. = 6 and Oxidation no. = +3

39. Identify the wrong statements in the following :

1) Chlorofluorocarbons are responsible forozone layer depletion

2) Greenhouse effect is responsible for globalwarming

3) Ozone layer does not permit infrared radiationfrom the sun to reach the earth

4) Acid rains is mostly because of oxides ofnitrogen and sulphur.

40. Phenol, when it first reacts with concentratedsulphuric acid and then with concentrated nitricacid, gives

1) 2, 4, 6- trinitrobenzene 2) o-nitrophenol

3) p-nitrphenol 4) nitrobenzene

41. In the following sequence of reactions, thealkene affords the compound ‘B’CH

3CH = CHCH

3 3 2O H O

ZnA B}}m }}}m

The compound B is

1) CH3CH

2CHO 2) CH

3COCH

3

3) CH3CH

2COCH

34) CH

3CHO

42. Larger number of oxidation states are exhibitedby the actinoids than those by the lanthanoids,the main reason being1) 4f orbitals more diffused than the 5f orbitals2) lesser energy difference between 5f and 6d

than between 4f and 5d orbitals3) more energy difference between 5f and 6d

than between 4f and 5d orbitals4) more reactive nature of the actinoids than

the lanthanoids

43. In which of the following octahedral complexesof Co (at. no. 27), will be magnitude of 0% bethe highest ?1) [Co(CN)

6]3– 2) [Co(C

2O

4)

3]3–

3) [Co(H2O)

6]3+ 4) [Co(NH

3)

6]3+

AKASH MULTIMEDIA 5


44. At 800C, the vapour pressure of pure liquid ‘A’is 520 mm Hg and that of pure liquid ‘B’ is 1000mm Hg. If a mixture solution of ‘A’ and ‘B’ boilsat 800C and 1 atm pressure, the amount of ‘A’in the mixture is (1 atm = 760 mm Hg)1) 52 mol percent 3) 34 mol percent3) 48 mol percent 4) 50 mol percent

45. For a reaction 12 A 2Bm , rate of disappearance

of ‘A’ is related to the rate of appearance of ‘B’by the expression.

1) < > < >d A d B1

dt 2 dt� � 2)

< > < >d A d B1

dt 4 dt�

3) < > < >d A d B

dt dt� � 4)

< > < >d A d B4

dt dt� �

46. The equilibrium constants 1PK and

2PK for thereactions X 2YU and Z P Q�U , respectivelyare in the ratio of 1 : 9. If the degree ofdissociation of X and Z be equal then the ratioof total pressure at these equilibria is1) 1 : 36 2) 1 : 1 3) 1 : 3 4) 1 : 9

47. Oxidising power of chlorine in aqueous solutioncan be determined by the parameters indicatedbelow :

diss eg hyd

1H H H2

2 g g g aq

1Cl Cl Cl Cl

2

22 2

%

% %

� �

}}}}m }}}m }}}m .

The energy involved in the conversion of

2 g

1Cl

2 to gCl� (using the data,

2diss ClH2% = 240

kJmol–1, eg ClH2% = – 249 kjmol–1,

hyd ClH2% = – 381 kjmol–1) will be

1) + 152 kJmol–1 2) – 610 kJmol–1

3) – 850 kJmol–1 4) + 120 kJmol–1

48. Which of the following factors is of nosignificance for roasting sulphide ores to theoxides and not subjecting the sulphide ores tocarbon reduction directly ?

1) Metal sulphides are thermodynamically morestable than CS

2

2) CO2 is thermodynamically more stable than

CS2

3) Metal sulphides are less stable than thecorresponding oxides

4) CO2 is more volatile than CS

2

49. Bakelite is obtained from phenol by reacting

with

1) (CH2OH)

22) CH

3CHO

3) CH3COCH

34) HCHO

50. For the following three reactions a, b, and c,

equilibrium constants are given :

a) CO(g)

+ H2O

(g) U CO

2(g) + H

2(g) ; K

1

b) CH4(g)

+ H2O

(g) U CO

(g) + 3H

2(g); K

2

c) CH4(g)

+ 2H2O

(g) U CO

2(g) + 4H

2(g); K

3

Which of the following relations is correct ?

1) 1 2 3K K K� 2) 2 3 1K K K�

3) K3 = K

1K

24) 3 2

3 2 1K .K K�

51. The absolute configuration of

HO2C CO2H

OHHHHO

is

1) S, S 2) R, R 3) R, S 4) S, R

52. The electrophile, E� attacks the benzene ringto generate the intermediate T -complex islowest energy ?

1)

NO2

H E

2)

HE

3)

HE

NO2

4) HE

NO2

53. B - D - ( + ) - glucose and C -D(+) - glucose are1) conformers 2) epimers3) anomers 4) enantiomers

54. Standard entropy of X2, Y

2 and XY

3 are 60, 40

and 50 JK–1 mol–1, respectively. For the reaction,

2 2 3

1 3X Y XY

2 2� m , H% = – 30 kJ, to be at

equilibrium, the temperature will be

1) 1250 K 2) 500 K

3) 750 K 4) 1000 K

AKASH MULTIMEDIA6


55. Four species are listed below

i) 3HCO� ii) H3O+ iii) 4HSO� iv) HSO

3F

Which one of the following is the correctsequence of their acid strength ?1) iv < ii < iii < 1 2) ii < iii < i < iv3) i < iii < ii < iv 4) iii < i < iv < ii

56. Which one of the following constitutes a groupof the isoelectronic species ?

1) 22 2C ,O ,CO,NO� � 2) 2

2 2NO ,C ,CN ,N� � �

3) 2 22 2 2CN ,N O ,C� � � 4) N

2, 2O� , NO+, CO

57. Which one of the following pairs of species havethe same bond order ?1) CN– and NO+ 2) CN– and CN+

3) 2O� and CN– 4) NO+ and CN+

58. The ionization enthalpy of hydrogen atom is1.312 x 106 J mol–1. The energy required toexcite the electron in the atom from n = 1 to n =2 is1) 8.51 × 105 Jmol–1 2) 6.56 × 105 Jmol–1

3) 7.56 × 105 Jmol–1 4) 9.84 × 105 Jmol–1

59. Which one of the following is the correctstatement?1) Boric acid is a protonic acid2) Beryllium exhibits coordination number of

six.3) Chlorides of both beryllium and aluminium

have bridged chloride structures in solidphase

4) B2H

6.2NH

3 is known as ‘inorganic benzene’

60. Given 3O

Cr / CrE 0.72V,

�� 2

O

Fe / FeE 0.42V

�� . The

potential for the cellCr | Cr3+ (0.1M) | Fe2+(0.01 M) Fe is1) 0.26 V 2) 0.399 V3) – 0.339 V 4) – 0.26 V

61. Amount of oxalic acid present in a solution canbe determined by its titration with KMnO

4

solution in the presence of H2SO

4. The titration

gives unsatisfactory result when carried out inthe presence of HCl, because HCl1) gets oxidised by oxalic acid to chlorine2) furnishes H+ ions in addition to those from

oxalic acid3) reduces permanganate to Mn2+

4) Oxidises oxalic acid to carbon dioxide andwater

62. The vapour pressure of water at 200C is17.5 mm Hg. If 18 g of glucose (C

6H

12O

6) is

added to 178.2g of water at 200C, the vapourpressure of the resulting solution will be

1) 17.675 mm Hg 2) 15.750 mm Hg

3) 16.500 mm Hg 4) 17.325 mm Hg

63. Among the following substituted silanes the one

which will give rise to cross linked siliconepolymer on hydrolysis is

1) R4Si 2) RSiCl

3

3) R2SiCl

24) R

3SiCl

64. In context with the industrial preparation ofhydrogen from water gas (CO + H

2), Which of

the following is the correct statement ?

1) CO and H2 are fractionally separated using

differences in their densities

2) CO is removed by absorption in aqueousCu

2Cl

2 solution

3) H2 is removed through occulsion with Pd

4) CO is oxidised to CO2 with steam in the

presence of a catalyst followed by absorption

of CO2 in alkali

65. In a compound atoms of element Y from ccplattice and those of element X occupy 2/3rd of

tetrahedral voids. The formula of the compoundwill be

1) X4Y

32) X

2Y

3

3) X2Y 4) X

3Y

4

66. Gold numbers of protective colloids A, B, C and

D are 0.50, 0.01, 0.10 and 0.005, respectively.The correct order of their protective powers is

1) D < A < C < B 2) C < B < D < A

3) A < C < B < D 4) B < D < A < C

67. The hydrocarbon which can react with sodium

i liquid ammonia is

1) CH3CH

2CH

2Cy CCH

2CH

2CH

3

2) CH3CH

2C y CH

3) CH3CH = CHCH

3

4) CH3CH

2C y CCH

2CH

3

AKASH MULTIMEDIA 7


68. The treatment of CH3MgX with CH

3Cy C–H

produces

1) CH3 – CH = CH

22) CH

3Cy C–CH

3

3) 3 3

H H | |CH C C CH� � �

4) CH4

69. The correct decreasing order of priority for the

functional groups of organic compounds in the

IUPAC system of nomenclature is

1) – COOH, – SO3H, – CONH

2, – CHO

2) – SO3H, – COOH, – CONH

2, – CHO

3) – CHO, – COOH, – SO3H, – CONH

2

4) – CONH2, – CHO, – SO

3H, – COOH

70. The pKa of a weak acid, HA, is 4.80. The pK

b

of a weak base, BOH, is 4.78. The pH of an

aqueous solution of the corresponding salt, BA,

will be

1) 9.58 2) 4.79 3) 7.01 4) 9.22

PART-C : PHYSICS


Wave property of electrons implies that they willshow diffraction effects. Davisson and Germerdemonstrated this by diffracting electrons fromcrystals. The law governing the diffraction froma crystal is obtained by assuning that electronwaves reflected from the planes of atoms in acrystal interfere constructively (see figure).

i

d

OutgoingElectrons

IncomingElectrons

Crystal plane

71. Electrons accelerated by potential V are

diffracted from a crystal. If d = 1o

A and i = 300,

V should be about (h = 6.6 × 10–34 Js,

me = 9.1 × 10–31 kg, e = 1.6 × 10–19 C)

1) 2000 V 2) 50 V

3) 500 V 4) 1000 V

72. If a strong diffraction peak is observed whenelectron are incident at an angle 'i' from thenormal to the crystal planes with distance 'd'between them (see figure) de Brogliewavelength dBM of electrons can be calculatedby the relationship (n is an integer)1) dBdsin i n� M 2) dBd cos i n� M

3) dB2dsin i n� M 4) dB2d cos i n� M

73. In an experiment, electrons are made to passthrough a narrow slit of width 'd' comparable totheir de Broglie wavelength. They are detectedon a screen at a distance 'D' from the slit (seefigure).

d

D

y = 0

Which of the following graphs can be expectedto represent the number of electrons 'N' detectedas a function of the detector position 'y'(y = 0 corresponds to the middle of the slit) ?

1) d

y

N2)

d

y

N

3) d

y

N 4) d

y

N

74. A planet in a distant solar system is 10 timesmore massive than the earth and its radius is 10times smaller. Given that the escape velocityfrom the earth is 11 kms–1, the escape velocityfrom the surface of the planet woudl be1) 1.1 kms–1 2) 11 kms–1

3) 110 kms–1 4) 0.11 kms–1

75. A spherical solid ball of volume V is made of amaterial of density 1S . It is falling through aliquid of density 2 2 1( )S S � S . Assume that theliquid applies a viscous force on the ball that isproportional to the square of its speed v, i.e.,F

viscous = – kv2(k>0). The terminal speed of the

ball is

1) 1 2Vg( )kS �S

2) 1VgkS

3) 1VgkS

4) 1 2Vg( )kS �S

AKASH MULTIMEDIA8


76. Shown in the figure below is a meter-bridge setup with null deflection in the galvanometer.

R558

G

20 cm

The value of the unknown resistor R is1) 13.75 8 2) 22083) 110 8 4) 558

77. A thin rod of length 'L' is lying along the x-axiswith its ends at x=0 and x=L. Its linear density

(mass/length) varies with x as n

xkL

¥ ´§ ¶ , where n

can be zero or any positive number. If theposition x

CM of the centre of mass of the rod is

plotted against 'n', which of the following graphsbest approximates the dependence of x

CMonn?

1)

L

L/2

O n

XCM

2) L/2

O n

XCM

3)

L

L/2

O n

XCM

4)

L

L/2

O n

XCM

78. While measuring the speed of sound byperforming a resonance column experiment, astudent gets the first resonance condition at acolumn length of 18 cm during winter.Repeating the same experiment during summer,she measures the column length to be x cm forthe second resonacne. Then1) 18 > x 2) x > 543) 54 > x > 36 4) 36 > x > 18

79. The dimension of magnetic field in M, L, T andC (Coulomb) is given as

1) MLT–1C–1 2) MT2C–2

3) MT–1C–1 4) MT–2C–1

80. Consider a uniform square plate of side ‘a’ andmass ‘m’. The moment of inertia of this plateabout an axis perpendicular to its plane andpassing through one of its corners is

1) 25

ma6

2)21

ma12

3) 27

ma12

4) 22

ma3

81. A body of mass m = 3.513 kg is moving alongthe x-axis with a speed of 5.00 ms–1. Themagnitude of its momentum is recorded as1) 17.6 kg ms–1 2) 17.565 kg ms–1

3) 17.56 kg ms–1 4) 17.57 kg ms–1

82. An athlete in the olympic games covers adistance of 100m in 10s. His kinetic energy canbe estimated to be in the range1) 200J–500J 2) 2×105J–3×103) 20,000J–50,000J 4) 2,000J–5,000J

83. A parallel plate capacitor with air between theplates has a capacitance of 9 pF. The separationbetween its plates is 'd'. The space between theplates is now filled with two dielectrics. One ofthe dielectrics has dielectric constant k

1 = 3 and

thickness d3

while the other one has dielectric

constant k2 = 6 and thickness2d

3. Capcitance

of the capacitor is now

1) 1.8 pF 2) 45 pF

3) 40.5 pF 4) 20.25 pF

84. The speed of sound in oxygen (O2) at a certain

temperature is 460 ms–1. The speed of sound inhelium (He) at the same temperature will be(assumed both gases to be ideal)1) 460 ms–1 2) 500 ms–1

3) 650ms–1 4) 330 ms–1

85. This question contains Statement-1 and Statement-2. Of the four choices given after the statements,choosethe one that best describes the twostatements.Statement-I: Energy is released when heavynuclei undergo fission or light nuclei undergofusion. and

Statement-II : For heavy nuclei, binding energyper nucleon increases with increasing Z whilefor light nuclei it decrease wth increasing Z.

1) Statement –1is false, Statement –2 is true.

2) Statement – 1is true, Statement – 2 is true;Statement–2 is correct explanation forStatement–1.

3) Statement –1 is true, Statement –2 is true;Statement –2 is not a correct explanation forStatement –1.

4) Statement –1 is true, Statement –2 is False.

AKASH MULTIMEDIA 9


86. This question contains Statement–1 andStatement–2. Of the four choices given after thestatements, choose the one that best describesthe two statements.Statement-1 : For a mass M kept at the centreof a cube of side ‘a’, the flux of gravitationalfield passing through its sides is 4Q GM.andStatement-II : If the direction of a field due toa point source is radial and its dependence onthe distance ‘r’ for the source is given as 1/r2,its flux through a closed surface depends onlyon the strength of the source enclosed by thesurface and not on the size or shape of thesurface1) Statement – 1is false, Statement – 2 is true2) Statement – 1is true, Statement –2 is true;

Statement –2 is correct explanation forStatement–1

3) Statement – 1 is true, Statement –2 is true;Statement -2 is not a correct explanation forStatement -1

4) Statement –1 is true, Statement –2 is False.

87. A jar filled with two non mixing liquids 1 and 2having densities 1S and 2S respectively. A solidball, made of a material of density 3S , isdropped in the jar. It comes to equilibrium inthe position shown in the figure.

Which of the following is true for 1 2,S S and 3S ?1) 3 1 2S � S � S

2) 1 3 2S � S � S

3) 1 2 3S � S � S

Liquid 1 1S

3S

Liquid 2 2S4) 1 3 2S � S � S

88. A working transistor with its three legs marked P,Q and R is tested using a multimeter. Noconduction is found between P and Q. Byconnecting the common (negative) terminal of themultimeter to R and the other (positive) terminalto P or Q, some resistance is seen on themultimeter. Which of the following is ture for thetransistor ?1) It is an npn transistor with R as base2) It is a pnp transistor with R as collector3) It is a pnp transistor with R as emitter4) It is an npn transistor with R as collectorDirections: Question No. 89 and 90 are baseson the followign paragraph.

Consider a block of conducting material of



developed between ‘B’ and ‘C’. The calculationis done in the following steps:i) Take current ‘|’ entering from ‘A’ and assume

it to spread over a hemispherical surface inthe block.

ii) Calculate field E(r) at distance ‘r’ form A by

using Ohm’s law E = j,S where j is the currentper unit area at ‘r’.

iii) From the ‘r’ dependence of E(r), obtain thepotential V(r) at r.

iv) Repeat (i), (ii) and (iii) for curent ‘|’ leaving‘D’ and siuperpose results for ‘A’ and ‘D’.

I I

A B C D

a b a

V%

89. V% measured between B and C is

1) | |

a (a b)

S S�

Q Q �

2) | |

a (a b)

S S�

�

3) | |

2 a 2 (a b)

S S�

Q Q �

4) |

2 (a b)

S

Q �

90. For current entering at A, the electric field at adistance ‘r’ from A is

1) 2

|

8 r

S

Q

2) 2

|

r

S3) 2

|

2 r

S

Q

4) 2

|

4 r

S

Q

91. A student measures the focal length of convexlens by putting an object pin at a distance ‘u’from the lens and measuring the distance ‘v’ ofthe image pin. The graph between ‘u’ and ‘v’plotted by the student should look like

1)

v(cm)

u(cm)O

2)u(cm)

v(cm)

O

3)

v(cm)

O u(cm)

4)

v(cm)

O u(cm)

AKASH MULTIMEDIA10


92. A block of mass 0.50 kg is moving with a speedof 2.00 m/s on a smooth surface. It strikesanother mass 1.00 kg and then they movetogether as a single body. The energy loss duringthe collision is1) 0.16 J 2) 1.00 J 3) 0.67 J 4) 0.34 J

93. A capillary tube (A) is dropped in water.Another identical tube (B) is dipped in a soapwater solution.Which of the following shows the relativenature of the liquid columns in the two tubes?

1)

A B

2)

A B

3)

A B

4)

A B

94. Suppose an electron is attracted towards the

origin by a force kr

where 'k' is a constant and

'r' is the distance of the electron from the origin.By applying Bohr model to this system, theradius of the nth orbital of the electron is foundto be 'r

n' and the kinetic energy of the electron

to the 'Tn'. Then which of the following is true?

1) Tn independent of n, nr nt

2) n n1T , r nn

t t

3) 2n n

1T , r nn

t t 4) 2

n n21T , r nn

t t

95. A wave travelling along the x-axis is describedby the equation y(x,t) = 0.005 cos( x t)B �C . Ifthe wavelength and the time period of the waveare 0.08 m and 2.0s, respectively, then B andC in appropriate units are

1) 25.00 ,B � Q C � Q 2) 0.08 2.0,B �

Q Q

3) 0.04 1.0,B � C �

Q Q

4) 12.50 ,2.0Q

B � Q C �

96. Two coaxial solenoids are made by winding thininsulated wire over a pipe of cross-sectional areaA = 10cm2 and length = 20 cm. If one of thesolenoid has 300 turns and the other 400 turns,

their mutual inductance is 7 10( 4 10 Tm A )� �

N � Q s

1) 44.8 10 H�

Q s 2) 54.8 10 H�

Q s

3) 42.4 10 H�

Q s 4) 52.4 10 H�

Q s

97. In the circuit below, A and B represent two in-puts and C represents the output.

B

A

C

The circuit represents

1) AND gate 2) NAND gate

3) OR gate 4) NOR gate

98. A body is at rest at x = 0, At t = 0, it startsmoving in the positive x-direction with aconstant acceleration. At the same instantanother body passes through x = 0 moving inthe positive x-direction with a constant speed.The position of the first body is given by x

1(t)

and the position of the second body is x2(t) after

the same time interval. Which of the followinggraphs correctly describes (x

1–x

2) as a function

of time 't'?

1)O t

(x1 – x2

2)O t

(x1 – x2

3) O

(x1 – x2

4) O t

(x1 – x2

99. An experiment is performed to find the refractiveindex of glass using a travelling microscope. Inthis experiment distances are measured by

1) a vernier scale provided on the microscope

2) a standard laboratory scale

3) a meter scale provided on the microscope4) a screw gauge provided on the microscope

AKASH MULTIMEDIA 11


100. A thin spherical shell of radius R has chargeQ spread uniformly over its surface. Which ofthe following graphs most closely represents theelectric field E(r) produced by the shell in therange 0 rb � d , where r is the distance fromthe centre of the shell ?

1)

E(r)

O R r2)

E(r)

O R r

3)

E(r)

O R r

4)

E(r)

O R r

101. A 5V battery with internal resistance 28 and a2V battery with internal resistance 18 areconnected to a 108 resistor as shown in thefigure.

P2

P1

5V 2V108 1828

The current in the 108 resistor is

1) 0.27A P2 to P

1

2) 0.03A P1 to P

2

3) 0.03A P2 to P

1

4) 0.27A P1 to P

2

102. A horizontal overhead powerline is at a heightof 4m from the ground and carries a current of100 A from east to west. The magentic fielddirectly below it on the ground is

7 10( 4 10 Tm A )� �

N � Q s

1) 2.5×10–7T southward

2) 5×10–6T northward

3) 5×10–6T southward

4) 2.5×10–7T northward

103. Relative permitivity and permeability of amaterial are rF and rN , respectively. Which ofthe following values of these quantities areallowed for a diamagnetic material?

1) r r0.5, 1.5F � N �

2) r r1.5, 0.5F � N �

3) r r0.5, 0.5F � N �

4) r r1.5, 1.5F � N �

104. Two full turns of the circular scale of a screwgauge cover a distance of 1mm on its mainscale. The total number of divisions on the cir-cular scale is 50. Further, it is found that thescrew gauge has a zero error of –0.03 mm.While measuring the diameter of a thin wire, astudent notes the main scale reading of 3 mmand the number of circular scale divisions inline with the main scale as 35. The diameter ofthe wire is

1) 3.32 mm 2) 3.73 mm

3) 3.67 mm 4) 3.38 mm

105. An insulated container of gas has two cham-bers separated by an insulating partition. Oneof the chambers has volume V1

and containsideal gas at pressure P

1 and temperature T

1. The

other chamber has volume V2 and contains ideal

gas at pressure P2 and temperature T

2. If the par-

tition is removed without doing any work onthe gas, the final equilibrium temperature of thegas in the container will be

1) 1 2 1 1 2 2

1 1 2 2 2 1

T T (P V P V )P V T P V T

�

�

2) 1 1 1 2 2 2

1 1 2 2

P V T P V TP V P V

�

�

3) 1 1 2 2 2 1

1 1 2 2

P V T P V TP V P V

�

�

4) 1 2 1 1 2 2

1 1 1 2 2 2

T T (P V P V )P V T P V T

�

�

AKASH MULTIMEDIA12


PART-A : MATHEMATICS

1) 2 2) 2 3) 3 4) 1 5) 2

6) 3 7) 4 8) 3 9) 4 10) 4

11) 4 12) 4 13) 4 14) 4 15) 3

16) 1 17) 3 18) 4 19) 2 20) 4

21) 1 22) 1 23) 2 24) 1 25) 2

26) 2 27) 1 28) 3 29) 2 30) 4

31) 3 32) 4 33) 4 34) 3 35) 4

PART-B : CHEMISTRY

36) 4 37) 1 38) 4 39) 3 40) 2

41) 4 42) 2 43) 1 44) 4 45) 2

46) 1 47) 2 48) 1 49) 4 50) 3

51) 2 52) 2 53) 3 54) 3 55) 3

56) 2 57) 1 58) 4 59) 3 60) 1

61) 3 62) 4 63) 2 64) 4 65) 1

66) 3 67) 2 68) 4 69) 2 70) 3

PART-C : PHYSICS

71) 2 72) 4 73) 4 74) 3 75) 1

76) 2 77) 1 78) 2 79) 3 80) 4

81) 1 82) 4 83) 3 84) - 85) 4

86) 2 87) 4 88) 2 89) 3 90) 3

91) 3 92) 3 93) 3 94) 1 95) 1

96) 4 97) 3 98) 2 99) 1 100) 1

101) 3 102) 3 103) 2 104) 4 105) 1

123456789012345678901234123456789012345678901234123456789012345678901234AIEEE 2008 ANSWERS

PART-A : MATHEMATICS1. (2) BD = AB = 7 + x

Also AB = x tan 600 = 3x

3 7x x= � �

7

3 1� �

�

x

7 3

3 12

AB � �

2. (2) 1

( ) 2

P A B

P B

�

� , 2

( ) 3

P A B

P A

�

�

Hence ( ) 3

( ) 4

P A

P B� (But P(A) = 1/4)

1

3P B� �

3. (3) A = {4, 5, 6}, B = {1, 2, 3, 4}.

Obviously ( ) 1P A B� �

4. (1) 4a

aee� �

12 4

2a¥ ´

� �¦ µ§ ¶

a = 8/3

5. (2) Vertex is (1, 0)

x=2

(2, 0)O

6. (3) Centre (–1, –2)

Let , )B C is the required point

11

2

B �

� �

and 02

2

C �

� �

; 3, 4B � � C � �

7. (4) Function is increasing 3

( )4

yx g y

�

� �

8. (3) Put –i in place of i Hence 1

1i

�

�

9. (4) T = {(x,y) : x–y�1}as 0�1 T is a reflexive relation

If 1 1x y y x� � � � �

=T is symmetrical alsoIf x – y = I

1 and y – z = I

2

Then x – z = (x – y) + (y – z) = I1 + I

2�I

=T is also transitive.Hence T is an equivalence relation

Clearly 1 ( , )x x x x Sx � � �

=S is not reflexive

.. ..AIEEE 2008 HINTS AND SOLUTIONS

AKASH MULTIMEDIA 13


10. (4) Slope of bisector = k–1

Middle point 1 7

,2 2

k �¥ ´� ¦ µ§ ¶

Equation of bisector is

7 ( 1)( 1)

2 2

�¥ ´� � � �¦ µ

§ ¶

ky k x

Put x = 0 and y = –4

4k� � p

11. (4) y = vx, dy dv

v xdx dx

� �

1dv

v x vdx

� � � dx

dvx

� �

= v = log x + c logy

x cx

� � �

Since, y(1) = 1, we havey = xlog x + x

12. (4) Mean of a, b, 8, 5 10 is 6

8 5 106

5

a b� � � �

� � �a + b = 7 ... (1)

Given that variance is 6.8=Variance

2 2( 6) ( 6) 4 1 166.8

5

a b� � � � � �

� �

�a2 + b2 = 25

a2 + (7 – a)2 = 25 (using (1))

�a2 – 7a + 12 = 0

=a = 4, 3 and b = 3, 4

13. (4) ˆ ˆ( )a b c� M �

G

ˆˆ ˆ2ˆˆ ˆ22

i i ki j k

¥ ´� ��B � �C � M

¦ µ§ ¶

2M � B and 2M �

and 2M � C

1�B � and 1C �

14. (4) Since 8a b�

G

G

7c b� �

G

G

= aG and b

G

are like vectors and bG

and cG are

unlike� aG and c

G will be unlike

Hence, angle between aG and c

G = Q

15. (3) Equation of line passing through (5,1,a) and(3,b,1) is

5 1

2 1 1

x y z a

b a

� � �

� � � M

� �

If line crosses yz-plane i.e., x = 0

x = 2M + 5 = 0 5 / 2�M �

Since, 17

(1 ) 12

y b� M � � �

5 17(1 ) 1

2 2b� � � � b = 4

Also, 13

( 1)2

z a a� M � � � �

5 13( 1)

2 2

� �

� � �a a 6a� �

16. (1) 1 2 3

2 3

x y z

k

� � �

� � and

2 3 1

3 2

x y z

k

� � �

� � Since lines intersect in a

point

2 3

3 2 0

1 1 2

k

k �

�

=2k2 + 5k – 25 = 0 k = –5, 5/2

17. (3) 1 1 1

( ) ...1 2

P nn

� � � �

1 1(2) 2

1 2P � � �

Let us assume that

1 1 1

( ) ...1 2

P k kk

� � � � � is true

1 1 1 1

( 1) ... 11 2 1

P k kk k

= � � � � � � � �

�

has to be true

L.H.S.> 1 11

1 1

k kk

k k

� �

� �

� �

Since ( 1)k k k� � ( 0)k� r

1 1 11

1 1

k k kk

k k

� � �= � � �

� �

Let ( ) 1 1P n n n n� � � �

Statement –1 is correct (2) 2 3 3P � s �

If ( ) ( 1) ( 1)P k k k k� � � � is true

AKASH MULTIMEDIA14


Now ( 1) ( 1)( 2) 2P k k k k� � � � � �

Since ( 1) 2k k� � �

( 1)( 2) 2= � � � �k k k

Hence Statement –2 is not a correct explanationof Statement -1.

18. (4) a b

Ac d

¥ ´�¦ µ§ ¶

so that

22

2

1 0

0 1

a bc ab bdA

ac cd bc d

¥ ´� � ¥ ´� �¦ µ ¦ µ

§ ¶� �§ ¶

�a2+bc = 1 = bc + d2 and (a + d)c=0=(a + d)b.Since xA I , 1A x , a = –d and hence det

1

1

bc bA

c bc

�

�

� �

= – 1 + bc – bc = –1

Statement 1 is trueBut tr. A = 0 and hence statement 2 is false.

19. (2) 0 0

( 1)n n

n n nr r r

r r

r C r C C� �

� � �¤ ¤

1 11

0 0

2 2� �

�

� �

� � � �¤ ¤n n

n n nr r

r r

nr C C n n

r

=2n–1(n + 2)Statement-1 is true

( 1) n r n r n rr r rr C x r C x C x� � �¤ ¤ ¤

11

0 0

�

�

� �

� � �¤ ¤n n

n r n rr r

r r

n C x C x nx (1+x)n–1+ (1+x)n

Substituting x = 11( 1) 2 2n n

rr C n n�

� � �¤Hence Statement-2 is also true and is a correctexplanation of Statement-1.

20. (4) Given statement r =~ p qj

Statement-1 1 ( ~ ) (~ )r p q p q� � � �

Statement-2 :

2 ~ ( ~ ) ( ) (~ ~ )r p q p q q p� j � � � �

From the truth table of r, r1 and r

2

r = r1

Hence Statement-1 is true and Statement-2 is false

21. (1) 1 2 3 4 5 6x x x x x� � � � �

5+6–1C5–1

= 10C4

22. (1) 0

(1 ) (1)’ 1 lim

h

f h ff

hm

� �

�

0

1(1 1)sin 0

1 1’ 1 lim

h

hh

fhm

¥ ´

� � �

¦ µ

§ ¶� �

�

0

1lim sinm

¥ ´� ¦ µ

§ ¶h

h

h h

0

1’ 1 lim sin

hf

hm

¥ ´� � ¦ µ

§ ¶

= f is not differentiable at x = 1

Similarly, 0

( ) (0)’ 0 lim

h

f h ff

hm

�

� �

0

1( 1)sin sin(1)

1’ 0 lim

m

¥ ´

� �

¦ µ

§ ¶�

� �

h

hh

fh

� f is also not differentiable at x = 0

23. (2) Let a, ar, ar2, ....

a + ar = 12 ... (1)

ar2 + ar3 = 48 ... (2)

dividing (2) by (1), we have

2 (1 )4

( 1)

��

�

ar r

a r2 4r� �

if 1r x �

= r = –2 also, a = –12 (using (1))

24. (1) f(x) = x3 – px + q

Now for maxima/minima f'(x) = 0

�3x2 – p = 0

2

3

px� �

/ 3p�

/ 3p

2

3

px= � p

25. (2) x7 + 14x5 + 16x3 + 30x – 560 = 0

Let f(x) = x7 + 14x5 + 16x3 + 30x

� f'(x) = 7x6 + 70x4 + 48x2 + 30 > 0 � x

=f(x) is an increasing function � x

26. (2) pm (qmp) = ~ ( )p q p� m

~ (~ )p q p� � � since ~p p� is always true

~ ( )p p q p p q� � � � m �

AKASH MULTIMEDIA 15


27. (1) Let 1 15 2

cot cos tan3 3

E ec� �

¥ ´� �¦ µ

§ ¶

1 13 2cot tan tan

4 3E � �

¥ ´¥ ´ ¥ ´� � �¦ µ ¦ µ¦ µ§ ¶ § ¶§ ¶

1

3 24 3cot tan

3 21 ,

2 3

E �

¥ ´¥ ´�

¦ µ¦ µ� � ¦ µ¦ µ

¦ µ�¦ µ§ ¶§ ¶

1 17 6cot tan

6 17E �

¥ ´� � �¦ µ

§ ¶

28. (3) 2 2

2 25x h y� � � �.... (1)

2 2 2 0dy

x h ydx

� � � �

( ) 2dy

x h ydx

� � � �

substituting in (1), we have

2

2 22 2 25

dyy y

dx¥ ´

� � � �¦ µ§ ¶

2 222 ’ 25 2y y y� � � �

29. (2)

11 1 13 / 2

0 0 0 0

sin 2 2

3 3� � � � �° ° °

x xI dx dx xdx x

x x

2

3I� �

1 1

0 0

cos 1xJ dx dx

x x� �° °

1

02 2x� � 2J= b

30. (4) Solving the equations we get the points ofintersection (–2,1) and (–2, –1). The boundedregion is shown as shaded region.

The required area =

12 2

0

2 (1 3 ) ( 2 )� � �° y y dy

11 32

0 0

2 42 (1 ) 2 2

3 3 3

yy dy y

¨ ·

� � � � � s �© ¸

ª ¹

°

y

x

(-2, -1)

(-2, 1)

x+2y2=0

x+3y2=1

(1, 0)

31. (3)

sinsin 4 4

2 2sin sin

4 4

Q Q¥ ´� �¦ µ

§ ¶�

Q Q¥ ´ ¥ ´� �¦ µ ¦ µ

§ ¶ § ¶

° °x

xdxdx

x x

2 cos cot sin4 4 4

x dx¥ ´Q Q Q¥ ´

� � �¦ µ¦ µ§ ¶§ ¶

°

cot4

dx x dxQ¥ ´

� � �¦ µ§ ¶

° °

sin4

x In x cQ¥ ´

� � � �¦ µ§ ¶

32. (4) Other than S, seven letters M, I, I, I, P, P, I

can be arranged in 7!

7.5.32!4!

�

Now four S can be placed in 8 spaces in 8C4

ways.Desired number of ways = 7.5.3. 8C

4= 7. 6C

4.8C

4

33. (4) The system of equations x–cy–bz=0,cx–y+az=0 and bx+ay–z=0 have non-trivial

solution if

1

1 0

1

c b

c a

b a

� �

� �

�

�1(1–a2) + c(–c–

ab)–b(ca+b)=02 2 2 2 1a b c abc� � � � �

34. (3) Now det 1A � p and 1 1

( )det( )

A adjAA

�

�

� all entries in 1A� are integers

35. (4) Let B and 4C be roots of x2–6x+a=0 and

,3B C be the roots of x2–cx+6=0, then

4 6B � C � and 4 aBC �

3 cB � C � and 3 6BC �

we get 2 8aBC � � �

So the first equation is x2 –6x +8=0 � x = 2, 4

If 2B � and 4 4C � then 3 3C �

If 4B � and4 2C � , then 3 3 / 2C � (non-integer)

=common root is x = 2

AKASH MULTIMEDIA16


PART-B : CHEMISTRY36. (4) For S

N2 reaction, the C atom is least hindered

towards the attack of nucleophile in the case of(CH

3Cl).

37. (1)

CH3 CH3NO2

CH3

NO2

Nitration}}}}m

Sn/HClSn/HCl

CH3NH2

CH3

NH2NaNO2/HCl

NaNO2/HCl

CH3N2

+ClCH3

N2+Cl

CuBrCuBr

CH3Br

CH3

Br

38. (4) E

en

en

ox NO2

Coordination no. = 6 and Oxidation no. = +3

39. (3) Ozone layer does not allow ultravioletradiation from sun to reach earth.

40. (2)

OH

2 4.Conc H SO}}}}}m

OH

SO3H

3.Conc HNO}}}}m

OH

NO2

41. (4)

H3C – CH == CH – CH

3 H3C CH CH CH3

O O

O

H2O/Zn

O

CH3CH(B)

42. (2) Being lesser energy difference between 5fand 6d than 4f and 5d orbitals.

43. (1) CH2 is stronger ligand hence 0% is highest.

44. (4) .0 0t A A B BP P X P X� �

760 = 520 XA + 0

B AP 1 X� AX 0.5� �

Thus, mole % of A = 50%

45. (2) 1

A 2B2m

< > < >2d A d B

dt 2dt

�

� �

< > < >d A d B1

dt 4 dt

�

� �

46. (1) X 2YZZXYZZ

1 0 (1 – x) 2x

1

2 1

1p

2x PK

1 x 1 x¥ ´

�¦ µ

§ ¶� �

Z P Q�ZZXYZZ

1 0 0(1 – x) x x

2

122

p

PxK

1 x 1 x¥ ´

�¦ µ

§ ¶� �

1 1

2 2

4 P P1 1

P 9 P 36

s

� � �

47. (2) For the process aq2 g

1Cl Cl

2�

m

diss

1H H

2% � % of Cl

2 + egCl% + hydCl% –

= 240

349 3812

� � � = – 610 kJmol–1

48. (1) Metal slphides are more stable.49. (4)

AKASH MULTIMEDIA 17


O

OH OH

+HCHO

CH2OH

CH2OHPolymerize}}}}m

CH2

CH2

n

50. (3) Equation (c) = equation (a) + equation (b)thus K

3 = K

1. K

2

51. (2)

HO2C CO2H

OHHHHO

1 2

Both C1 and C

2 have R- configuration.

52. (2) – NO2 is electron withdrawing which will

destabilize T - complex

53. (3) B - D - ( + ) - glucose and C -D(+) - glucoseare anomers.

54. (3) 2 2 3

1 3X Y XY

2 2� m

1reaction

3 1S 50 40 60 40Jmol

2 2�

¥ ´% � � s � s � �¦ µ

§ ¶

G G T S% � % � %

at equilibrium G% = 0

H T S% � %

330 10 T 40s � s � T = 750 K

55. (3) (iv) > (ii) > (iii) > (i) HSO3F is most acidic

56. (2) 22 2NO ,C ,CN ,N� � � all have fourteen electrons.

57. (1) Both are isoelectronic and have same order.

58. (4) 6 6

2 1 2

1.312 10 1.312 10E E E

12

¥ ´s s% � � � � � �

¦ µ§ ¶

= 9.84 × 105 Jmol–1

59. (3)

Cl

Cl

Al Al

Cl

Cl

Cl

Cl

Cl

Br Be Be

Cl Cl

ClCl Cl

60. (1) As 3O

Cr / CrE 0.72V

�� and 2

O

Fe / FeE 0.42V

��

2 Cr + 3Fe2+ m 3Fe + 2Cr3+

Ecell

= E0Cell

–

23

32

Cr0.0591log

6 Fe

�

�

= (–0.42 + 0.72)–

2

3

0.10.0591log

6 0.01

2

3

0.10.05910.30 log

6 0.01� �

= 2

6

0.0591 100.30 log

6 10

�

�

� = 40.05910.30 log10

6�

Ecell

= 0.2606 V

61. (3) HCl being stronger reducing agent reducesMnO

4– to Mn2+ and result of the titration becomes

unsatisfactory.

62. (4) 0

ssolute

s

P PX

P

�

� = s

s

17.5 P 0.1

P 10

�

�

ss

s

17.5 P0.01 P

P

�

� � = 17.325mm Hg

63. (2) R Si Cl

Cl

Cl

2H O}}}m R Si OH

OH

OH

Condensationpolymerization}}}}}m

| | Si Si | | O O | |R Si O Si R | | O O | |

� � � �

Si Si | |

¨ ·

© ¸

© ¸

© ¸

© ¸

© ¸

© ¸

© ¸

© ¸

© ¸

© ¸ª ¹

64. (4) CO + H22H O

}}}mCO2+2H2

KOH

K2CO3

65. (1) Number of atoms of Y = 4; X = 2

83s

Formula of compound will be X4Y

3

66. (3) Higher the gold number lesser will be theprotective power of colloid

67. (2)

3Na / Liq.NH3 2 3 2CH CH C CH CH CH C C Na

2�

%� y }}}}}m y

It is a terminal alkyne, having acidic hydrogen.

68. (4) CH3 – MgX + CH

3 – Cy C – H m CH

4

69. (2) – SO3H, – COOH, – CONH

2, – CHO

70. (3) It is a salt of weak acid and weak base

w a

b

K KH

K�

�

¨ · �ª ¹ pH = 7.01

AKASH MULTIMEDIA18


PART-C : PHYSICS


Wave property of electrons implies that they willshow diffraction effects. Davisson and Germerdemonstrated this by diffracting electrons fromcrystals. The law governing the diffraction froma crystal is obtained by requiring that electronwaves reflected from the planes of atoms in acrystal interfere constructively (see figure).

i

d

OutgoingElectrons

IncomingElectrons

Crystal plane

71. (2) 2d cos i = nM 2d cos i = h

2meV

v = 50 volt

i

72. (4) 2d cos i = dBnM

73. (4) Diffraction pattern will be wider than theslit.

74. (3)

esc

2GM 2G 10MV 10 11 110 km / s

R R /10

s

� � � s �

75. (1) 21 2 TVg Vg kvS �S �

1 2T

Vg( )v

k

S �S

� �

76. (2) 55 R 55 8

R 22020 80 2

s

� � � � 8

77. (1)

n

cm n

xk .xdxdmx dx.x L

Xdm dm x

k dxL

¥ ´¦ µM § ¶

� � �

¥ ´¦ µ§ ¶

°° °° ° °

=

Ln 2

n

Ln 1

no

o

kx

(n 2)Lx(n 1)kx

n 2(n 1)L

�

�

¨ ·

© ¸�

© ¸

�¨ ·© ¸

�© ¸© ¸ �� ª ¹

© ¸

© ¸

© ¸ª ¹

cm

L 2L 3L 4L 5LX , , , , ,.....

2 3 4 5 6�

78. (2) 1 RT

n4x M

H

� 1 RT

xn4 M

H

� x Tt

79. (3) F = qvB

B = F/qv = MC–1T–1

80. (4)

22 2

2cm

a 2 ma ma 2l l m ma

2 6 2 3

¥ ´� � � � �¦ µ

§ ¶

81. (1) P = mv = 3.513×5.00z 17.6

82. (4) Approximate mass = 60kg

Approximate velocity = 10m/s

Approximate 1

KE 60 100 3000J2

� s s �

KE range � 2000 to 5000 joule

83. (3) 0 0 0

1 2

A A 18AC’

d d d 2d 4d9 183 6

F F F

� � �

��

C’ = 40.5PF

84. No option is correct

RTv

M

H

�

1 1 2

2 2 1

74V M 5

5V M 323

s

H

� �

H

s

22

460 21 460 5 2 2v 1420

V 25 8 21

s s

� � � �

s

85. (4)

86. (2) g = GM/r2

87. (4) As liquid 1 floats above liquid 2, 1 2S � S

The ball is unable to sink into liquid 2, 3 2S � S

The ball is unable to rise over liquid 1,1 3S � S

Thus, 1 3 2S � S � S

AKASH MULTIMEDIA 19


88. (2) Directions: Question No. 89 and 90 are baseson the followign paragraph.Consider a block of conducting material of



developed between ‘B’ and ‘C’. The calculationis done in the following steps:i) Take current ‘|’ entering from ‘A’ and assumeit to spread over a hemispherical surface in theblock.ii) Calculate field E(r) at distance ‘r’ form A by

using Ohm’s law E = j,S where j is the currentper unit area at ‘r’.iii) From the ‘r’ dependence of E(r), obtain thepotential V(r) at r.iv) Repeat (i), (ii) and (iii) for curent ‘|’ leaving‘D’ and siuperpose results for ‘A’ and ‘D’.

I I

A B C D

a b a

V%

89. (3) Choosing A as origin,

2

|E j

2 r� S � S

Q

(a b)

C B 2a

| 1 | 1 1V V dr

2 2 (a b) ar

�

S S ¨ ·

� � � � �© ¸

Q Q �ª ¹

°

B C

| 1 1V V

2 a (a b)

S ¨ ·

� � �© ¸

Q �ª ¹

90. (3) 2E

2 r

S�

Q

91. (3) 1 1 1

v u f� � = constant

92. (3) m1u

1 + m

2u

2 = (m

1 + m

2)v v = 2/3 m/s

Energy loss = 1

2(0.5)s (2)2–

1

2(1.5)s

22

3¥ ´¦ µ§ ¶

= 0.67 J

93. (3) Capillary rise h = 2Tcos

gr

R

S.

As soap solution has lower T, h will be low.

94. (1) 2k mv

r r�

mv2 = k (independent of r)

hn

2¥ ´

¦ µ§ ¶Q

= mvr r� t n and T = 1

2mv2 is

independent of n.

95. (1) y = 0.005 cos ( x t)B �C

comparing the equation with the standard form,

x ty Acos 2

T

¨ ·¥ ´

� � Q¦ µ© ¸§ ¶Mª ¹

2 /Q M � B and2 / TQ � C

2 / 0.08 25.00B � Q � Q C � Q

96. (4) 40 1 2N N AM 2.4 10 H

l�

N

� � Q s

97. (3)

A B C0 0 00 1 11 0 11 1 1

98. (2) x1 (t) =

21at

2 x

2(t) = vt x

1–x

2 =

1

2at2 – vt

99. (1)

100.(1) 2

0

0 if r R

E(r) Qif r R

4 r

�«

®� ¬

r®

QF

101.(3)

2 1P P

5 0 22 10 1V V1 1 12 10 1

� �

� �

� �

P2

P1

1082V5V

28 i 18

2 1P P2 1

V V| 0.03from P P

10

�

� � m

102. (3) 7

60 i 4 10 100B 5 10 T

2 R 2 4

�

�

N Q s

� � s � s

Q Q

southward

103. (2) For diamagnetic materials r 1F � and Ur<1

104. (4) Diameter = M.S.R. + C.S.R × L.C. + Z.E.

= 3 + 35 × (0.5/50) + 0.03 = 3.38 mm

105.(1) U = U1 + U

2

1 1 2 2 1 2

1 1 2 2 2 1

(P V P V )T TT

(P V T P V T )

�

�

�

P

These Solutions Prepared by Srichaitanya Educational Institutions,Vijayawada.

These Solutions Prepared by Srichaitanya educational Institutions, Vijayawada

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