Aieee Sample

8/13/2019 Aieee Sample

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SAMPLE TEST PAPERfor

AIEEE2010

Time: 3 Hours Marks: 432

INSTRUCTIONS

1. The Test Booklet consists of 90 questions. The maximum marks are 432.

2. It is mandatory to fill the Test Idand Student Idon the OMR Sheet.

3. There are three parts in the question paper.

The distribution of marks subjectwise in each part is as under for each correct response.

Part AChemistry (144 marks) Question No. 1 to 24 consist FOUR (4)marks each andQuestion No. 25 to 30 consist EIGHT (8)marks each for each correct response.

Part BMathematics (144 marks)Question No. 31 to 32 and 39 to 60 consist FOUR (4)marks each and Question No. 33 to 38 consist EIGHT (8)marks each for correct response

Part CPhysics (144 marks) Question No. 61 to 84 consist FOUR (4) marks each andQuestion No. 85 to 90 consist EIGHT (8)marks each for correct response

4. Candidates will be awarded marks as stated above in instructions No. 2 for correct responseof each question. 1/4 (one fourth)mark will be deducted for indicating incorrect response ofeach question. No deductionfrom the total score will be made if no responseis indicatedfor an item in the answer sheet.

5. Use Blue/Black Ball Point Pen only for writing particulars/marking responses on the AnswerSheet. Use of pencil is strictly prohibited.

6. No candidate is allowed to carry any textual material, printed or written, bits of papers, pager,mobile phone, any electronic device, etc. except the Admit Cad inside the examinationhall/room.

7. Rough work is to be done on the space provided for this purpose in the Test Booklet only.This space is given at the bottom of each pages.

8. On completion of the test, the candidate must hand over the Answer Sheet to the Invigilatoron duty in the Room/Hall. However, the candidates are allowed to take away this TestBooklet with them.

PIE EDUCATION, Corporate Office: 44A/2, Sarvapriya Vihar, New Delhi - 16. Ph: 01141828585. Fax: 01126520144

Head OfficeAXIS 117/N/581, Kakadeo KanpurVisit us : www.pieeducation.com


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PartA : CHEMISTRYUseful Data :

Atomic Mass : H = 1, C = 12, N = 14, O = 16, Na = 23, S = 32, Cl = 35.5, K = 39, Fe = 56, Cu = 63.5.

1. Number of atoms in 558.5 gram Fe (atomic weight of Fe = 55.85 g mol1

) is

(a) twice that in 60g carbon(b) 6.023 1022

(c) half that in 8g He

(d) 558.5 6.023 1023

2. Alum helps in purifying water by

(a) forming Si complex with clay partiles(b) sulphate part which combines with the dirt and removes it(c) coagulating the mud particles(d) making mud water soluble

3. A square planar complex is formed by hybridization of which atomic orbitals?

(a) s, px, py, d (b) s, pyz x, py, 2 2x yd

(c) (d) s, p2x y xs, p , p , d

y, d , dz xy

4. Polymer formation from monomers starts by(a) condensation reaction between monomers(b) coordinate reaction between monomers(c) conversion of monomer to monomer ions by protons

(d) hydrolysis of monomers

5. The type of isomerism present in nitropentamine chromium (III) chloride is(a) optical (b) linkage(c) ionization (d) polymerization

Space for r ough work




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6. Arrangement of (CH3)3 C , (CH3)2 CH , CH3 CH2 when attached to benzyl or an unsaturated groupin increasing order of inductive effect

(a) (CH3)3 C


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11. Units of rate constant of first and zero order reactions in terms of molarity M unit are respectively(a) sec1, Msec1 (b) sec1, M(c) Msec1, sec1 (d) M, sec1

12. In XeF2, XeF4, XeF6the number of lone pairs on Xe are respectively(a) 2, 3, 1 (b) 1, 2, 3

(c) 4, 1, 2 (d) 3, 2, 1

13. In which of the following species the interatomic bond angle is 10928?(a) NH3, (BF4)

1 (b) (NH

44) , BF3

(c) NH3, BF4 (d) (NH1

2) , BF3

14. For the reaction A + 2B C, rate is given by R = [A] [B]2then the order of the reaction is(a) 3 (b) 6(c) 5 (d) 7

15. RNA is different from DNA because RNA contains(a) ribose sugar and thymine(b) ribose sugar and uracil

(c) deoxyribose sugar and thymine(d) deoxyribose sugar and uracil

Space for r ough work




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16. Which of the following are arranged in an increasing order of their bond strengths?

(a)2

2 2 2O O O O +< < 050. c




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Here points (3) and (3) lie on the circle

Z

B C

-3 3

3iA

| Z 3i | 3 2 =

AB = 3 2

AC = 3 2 BC = 6

Also, AB2+ AC2= BC2

2

BAC =

Z 3

Z 3 4

= + Arg

51. c

Let P (3, 4) and C being the centre of circle, C (4, 3). The line which is farthest from the circle is theline perpendicular to CP.

4 31

3 4

+=

+Slope of CP =

Slope of line perpendicular to CP = 1 Equation of required liney 4 = 1(x 3)or, x + y 7 = 0

52. d

As number of common tangents = 3

Both circles touch each other externally. Distance between centres = sum of their radii

5 3 9 k = + 9 k = 4

k = 5

53. b

= (a + b c) (a b + c)= (2s 2c) (2s 2b)or, = 4(s c) (s b)

or, s(s a)(s b)(s c) 4(s b) (s c) =

or,s(s a)

4(s b)(s c)

=

or,A 1

tan

2 4

=

12

84tan A1 15

116

= =

54. b




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2(at ,2at)

2y 4ax=( 4, 0)

A

P

AP = 10

or,2 2 2(at 4) (2at) 10+ + = (i)

Also equation of tangent at point Pty = x + at2

As it passes through (4, 0)2

0 = 4 + atat2= 4 (ii)from (i) and (ii)(4 + 4)2+ 4a(4) = 100

4a = 9

55. a

56. a

( )2

3 16 4 + = 1 = 0 or 6.

57. c

58. a

( ) ( ) ( ) ( ) ( )P A B 1 A B 1 A B 1 P A P B = = = 0.8 = 1 0.7 P(2)

2

7 P(2) = .

59. a

2a 3b 5c 0+ =rr r

) ( )b a 5 c a = r r r 3 ( r 5AB AC3

=uuur uuur

and must be parallel since there is common point A. The points A, B, C must

be collinear .

ACuuur

ABuuur

Hence (a) is the correct answer.

60. d

Distance =5 2 7

50 5 2

+= .

61. b

g2

L In equilibrium Mg = kx + A

Let it be given a small downward push y and released then

( ) gy2

LAyxkMgFr

+++= ( )ygAk + =

M

gAk + 2 =

62. c

63. b




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q q

O(a, 0) (a, 0)

a

KQq2U i=

( )22f xaKQqa2

xa

KQq

xa

KQqU

=

+

+=

( )

=

=22

2

22if xaa

xKQqa2

a

1

xa

aKQq2UU U =

=3

2

a

KQqax2 (since )

222axa

64. d

o

Using the relations E = and V = E.L, we get

t

q

t

L

L

q

t

EL.L.

EX

2

2 =

=

=

65. d

If all points are at same potential, there cannot exist an electric field, as

dr

dVE =

A B C

DHence, (a) is not correct.

In a uniform electric field, the lines of forceare parallel and equidistant (solid lines).

The equipotentials (dotted lines), being

perpendicular to lines of force, are also

equidistant and parallel.

The distance between A and B is same as that between C and D. But, A and B

will have some potential difference, whereas C and D being on the same

equipotential, will have no potential difference. Hence, (b) is not correct.

Points on an equipotential are at the same potential. Hence, (c) is not correct.

66. aLet temperature of interface be T. Since two section of rod are in series, rate of heat flow in them will beequal.

2

22 )(

l

TTAk

1

11

l

)TT(Ak =

1221

212121

lklk

TlkTlk

++

T =

67. b

U = 0a2

kqQ

a

kqQ

a

kq2=++

68. a




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x = 2 5t + 6t2

v =dt

dx= -5 + 12 t

Initial velocity means velocity at t = 0

vinitial = 5 m/s

69. cMotion parameters i.e. the initial velocity and acceleration are same for both the bodies. Therefore, they willfall by the same distance in a given time, maintaining their initial separation.

70. a

( ) ( ) ( )y4g2Rgy2L 22 = Asvs= A vc c

2

L R =

71. dThe force exerted by the ceiling on the spring is the tension in the spring.

72. c

In gamma-decay, the atomic and mass number do not change.

73. cMass defect = 4 4.0026 - 15.9994 = 0.011Energy released by oxygen nuclei = 0.011 931 = 10.24 MeV

74. b

From work-kinetic energy theorem

s.Fmv2

1 2 =

m

Fs2v =

m

1 V

75. d

Both the observes are inertial observers. They will measure different values for work or

kinetic energy. However, acceleration measured by both of them will be the same

76. b

The flux linking with the coil at any instant t is given as

= NBA cos t

tsinNBAdt

d=

=E

Therefore, the maximum value of emf is

E = NBAmax Hence, answer (b) is correct.77. c

Momentum of the electron will increase. So the wavelength (= h/p)of electrons will decrease

and fringe width decreases as

78. d

Due to the motion of the loop, an electric field will be induced both in AD and BC.

79. a

2nt = 400 t n = 200

2= 50oI amp.

R.m.s. current = 2/Io = 50 amp.80. b




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81. c

82. d

The division of current I into the two parallel branches will be as shown,

V = IRA

3

I2 VB= 1.5 R = IRB

3

I VC= 3R = IR

V = VB= VA CB83. c

84. c

3

3 )y2= 5(sin 3t + cos 3t) = 10 sin (3t +

85. d

86. b

87. a

For the cyclic process U = 0

( )CACABCAB W010WWW ++=++ W = JGiven Q = 5 J

From first law of thermodynamics 5 = 10 + 0 + CAW

5WCA =88. c

Heat lost = Heat gained

( ) ( 2V222

1V1

11

TTCRT

VP

TTC.RT

VP

=2

22

1

11

2211

T

VP

T

VP

VPVP

T +

+

=)

89. d

P = P1+P2= [15 + 5]D = 10D

m10

1

P

1 Focal length of the combination f = = = 10cm

90. d

3

2

f

f

2

1 = (1)

30

1

f

1

f

1

21 = (2)Solving equations (1) and (2),

cm (concave lens)15f2= f1= 10 cm (convex lens)

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