Aieee 2012 Nst3 Solns

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Brilliant Tutorials Pvt. Ltd. AIEEE/PET/PMC/Obj/Solns - 1

1. (3) Theory

2. (3) Theory

3. (4) Q0 = C0 V0

= 2.0 10 6 20

= 40 10 6 Coulomb

C = 0.1 part in 2

= 1 part in 20 = 5%

V = 1 part in 100 = 1%

Q = 5% + 1% = 6%

Q = Q0 + Q

= 40 10 6 6% Coulomb

= 40 2.4 10 6 Coulomb

4. (2) tan 37 =3

5

Also tan 37 =10.8

v

3 10.8

5 v =

v = 18 km/hr

51818

=

= 5 m/s

37

rmv

mv

rv

5. (3)

BRILLIANTS

PROGRESSIVE TEST

FOR STUDENTS OF

OUR ONE/TWO-YEAR POSTAL COURSES

TOWARDSALL INDIA ENGINEERING ENTRANCE EXAMINATION, 2012

AIEEE 2012PT3 (ONE YEAR)/PT6 (TWO YEAR)

SOLNS

PHYSICS

MATHEMATICS

CHEMISTRY

SOLUTIONS

PART A : PHYSICS


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6. (2)

Q

1000NQa

Qx

PX

300 N

T

14T T

Denote

x = displacement

a = acceleration

T = tension

Constraint equation is

xP + 4xQ = constant

aP + 4aQ = 0

P Qa 4a = (i)

For P,

T 300 = P300

ag

(ii)

For Q,

1000 4T = Q1000

ag

(iii)

Put T from (ii), in (iii) and aP from (i) in

(iv)

1000 4 (300 + 120 aQ) = 100 aQ

2Q

20a m/ s

58

=

Required SQ

2 2Q

1 1 20 40a t (2) metres

2 2 58 58

= = =

where ve sign indicates that SQ is

upwards as aQ is downwards (+ ve)

7. (3)

8. (4)

9. (4) When S is opened at t = 0 s

q0 = CV

and q = q0t/RCe for t > 0

S

RC

During t = 2 ms, 0q

q4

=

t/RC4 e =

tR

(ln 4)C =

3

6

2 10R

2.303 0.6021 1 10

=

= 1.45 k


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10. (4) AB is isobaric expansion

Temperature increases

UB UA > 0

AC is isothermal expansion

Temperature remains constant

UC UA = 0

AD is adiabatic expansion

Temperature decreases

UD UA < 0

11. (3)D 2

R2 2

= = = 1 m

I = MR2 = 1 (1)2 = 1 kg-m2

= 2

= 2 3.14 10

31.4

= 2 rad/s

L = I

= (1) (2)

= 2 kg-m2/s

12. (1) For fundamental frequency s2

=

= 2s

52.87 10 vf

s

= =

52.87 10v

s

=

= 2 2.87 105 cm/s

v

=

2v

=

v

2 =

5 22.66 (5.74 10 ) =

= 87.6 1010 Pa

13. (2) Power absorbed by blackbody

= AT4 = 30 103 W

(5.67 10 8) 4 (0.03)2 T4

= 30 103

T4 = 4.68 1013

T = 2600 K

14. (4)K.E. imparted to water

Pt

=

K.E

Volume of water

=

Volume of watert

The first term viz;K.E.

Volume

v2

where v is velocity

The second term viz;Volume

t

v

P v2 v = v3

Flux v

P 3

As = 2. we have P = 8P

15. (2) At entering surface, angle of incidence= 0

Deflection occurs at second surfaceonly.

By geometry, i = 30

1 sin i = 2 sin r

1.55sin r (sin30 )

1

=

1.552

= = 0.77

D = r i = sin 1 (0.77) 30


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16. (3)4

3

r3g = 6rv

2 2 3

2

(2 10 ) (1.47 10 ) (9.8)

9 0.21 10

=

= 152.4 kg m 1 s 1

17. (4)

18. (4) If d sin = ( i) t, the path differ-rence for central (bright) fringe will

be zero

Bright fringe will be at O

But, if d sin > ( 1) t, it will be above O

and if d sin < ( 1) t, it will be below O

19. (1) For real image u = u1

v = 2 u1

f = 20 cm

1 1 1

v u f+ =

1 1

1 1 1

2 u u 20 =

u1 = 30 cm

For virtual image u = u2

v = 2 u2

f = 20 cm

2 2

1 1 1

2u u 20

=

u2 = 10 cm

required distance = u1 u2 = 30 10

= 20 cm

20. (2) For a:

Potential of outer sphere

02

Kq KqV

R R= +

where q is to be found

But V2 = 0 ( earthed)

q = q0

For b:

Potential of inner sphere

01

KqKqV

r R

= +

where q is to be found

But V1 = 0 ( earthed)

q = q0r

R

For c:

Potential of inner sphere

0 01

Kq KqV

r R=

01 1

Kqr R

=

For d:

Potential of outer sphere

0 02

q r KqKV

R R R

= +

0Kq r1R R

=

21. (4)

22. (3) The emf induced in conductor ab

E = Bv = (0.3) (0.4) (5)

= 0.6 V

The conductor ab is regarded as

voltage source with R1 and R2 in

parallel as shown

total 6 4R 0.66 4

= + +

= 3


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1 1R I

2R

a b

r 0.6=

ab0.6

3=I = 0.2 A

14

0.210

= I = 0.08 A

Current through R2 = Iab I1

= 0.20 0.08

= 0.12 A

23. (1) Work done by electric field

= gain in kinetic energy

(qE) (2y0) =2 21 m (2v) v

2

(qE) 2y0 =21 m 3v

2

2

0

3 mvE4 qy

=

Rate of work done by electric field

at P

3

0

3 mvF v (qE) v

4 y

= = =

24. (2)

25. (2) 21 = 100 25 = 75C

13 = 25 20 = 5C for water

In steady state, 2 1KA ( ) dQd dt

=

dQ 0.9 8 75

dt 20

= = 27 cal/s

Heat absorbed by water per second

= m (1 2)

= m (5)

where m is rate of f low of water

5 m = 27

m = 5.4 gm/sec

26. (3)

27. (2) To resonate on a circular orbit, a

wave must circle back on itself insuch a way that crest (say C1) falls

on crests and trough falls on trough

(say C2)

circumference = 2R = NN

where N = 1, 2, 3,

1C

2C

( )2 2 2N

N 2 2

p N hK.E

2m 8 R m= =

as in theory of

Bohrs model of H-atom

Use h = 6.63 10 34 J-s

R = 0.5 10 9 m

m = 9.1 10 31 kg

(K.E)N = 2.44 10 20 N2 J

= 0.153 N2 eV

28. (4)

29. (1)

30. (3)


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31. (4) Given f(x) = sin x and = 2

g(x) x 1

range of f = [ 1, 1] and

domain of g = (1, )

go f is not defined.

32. (3) Area of hexagon

= 6 OA1 A2

= 1 2 1 2

16 OA OA sin A OA

2

= 3 1 1 sin 60

= 3

3 sq. units2

i 3

2A e

i 5 3

6A e

5A

i 4 3e

4A

1A

3A

i 2 3e

ie

O

33. (3)=

m

i 0

10 20

i m i

Coefficient of xm in (1 + x)10 (1 + x)20

is 30Cm. For maximum value, m = 15.

34. (2) Any line through P( 2, 1) is

y + 1 = m (x + 2).

It touches the circle if

2

2m 1 41 m 0, .

31 m

= =

+

the equation of PB is

4y 1 (x 2)

3+ = +

4x 3y 5 0 + =

A point on PB is ( 5, 5). Its image

by the line y = 1 is P( 5, 3). So the

equation of the incident ray is

+ = +

+

3 1y 3 (x 5)

5 2

+ + =4x 3y 11 0

35. (1) We have for all x (x 0),

f(x3) = x5

Differentiating w.r.t. x,

f (x3) 3x2 = 5x4

=3 25

f (x ) x3

= =25

f (27) (3) 15

3

36. (1)

x x

x 0

e e 2x 0lim form

x sin x 0

+ =

x x

x 0

e e 2lim

1 cos x

( using L-Hospitals rule )

+= = =

x x x x

x 0 x 0

e e e elim lim 2

sin x cos x

37. (4) Area

b

x 1f(x)dx

==

PART B: MATHEMATICS


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= + 2

b 1 2

b2 2

1

b 1 1 1 x 1

= + + = +

2

2

d xf(x) x 1

dx x 1

= + = +

38. (4) f(x) is continuous in the interval

[ 1, 1], but f(x) is not differentiable

at x = 0. Hence mean value theorem

is not applicable. So no C can be

found.

39. (3) Period of

x

[x]

2

is1.2

Period of sin 1 ({x}) is 1.

Period of ( )1sin cos x is 2, whereas

( )1 2sin cos (x ) is non-periodic.

40. (2) (a, 2a) is an interior point of y2 16 x = 0,

if (2a)2 16a < 0 a2 4a < 0

V (0, 0) and (a, 2a) are on the same

side of x 4 = 0

a 4 < 0 a < 4

Now, a2 4a < 0 0 < a < 4.

41. (3) (a + b) x + (a b) y = 2a + 3b

a (x + y 2) + b (x y 3) = 0

b(x y 2) (x y 3) 0

a

+ + =

It passes through the intersection

point of lines + = =x y 2 and x y 3

= = 5 1

x , y2 2

42. (1) cot A tan A = 2 cot 2A

cot 5 tan 5 2 tan 10 4 tan 20 8 cot 40

= 2 cot10 2 tan10 4 tan 20

8 cot40

2 (2 cot 20 ) 4 tan 20 8 cot 40=

= 4 (2 cot 40 ) 8 cot 40

= 0

43. (3) Out of 22 players, 4 of them being

excluded, so selection is out of

remaining 18. Also 6 is always

included so only 4 more is to be

selected out of remaining 12.

number of ways12

4C 495= =

44. (2) The equation of line is

x cos + y sin = p.

y x cot p cosec . = +

Now,2 2 2 2

c a cot b=

2 2 2 2 2p cosec a cot b =

2 2 2 2 2a cos b sin p . =

45. (3) 2

2

2

5a 2 a 1

5b 2 b 1

5c 2 c 1

+

+

+

2

2

2

5a a 15a 2 1

5b 2 1 5b b 1

5c 2 1 5c c 1

= +

= +

2

2

2

a a 1

0 5 b b 1

c c 1

= 5 (a b) (b c) (c a)

46. (3) = + + + 2 2

x y z (a b)(a b )(a b )

= + + + +2 2 2

(a b)(a ab( ) b )

= + +2 2

(a b)(a ab b )

= +3 3

a b

47. (2)n n n n

(A ) (A ) ( A) A , = = = since n is

odd.


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48. (2) Let , be the roots of ax2 + bx + c = 0

+ = =b c

,a a

Given that + = +

2 2

1 1

+ + =

2

2

( ) 2

( )

=

2 2b c b c

2a a a a

= 2 2

3 2

bc b c2

aa a

= 2 2 2bc ab 2 ca

+ =2 2 2ab bc 2 ca

+ = + =2 2 2

2 2 2

ab bc b bc2 2

acca ca a

49 (1) We have100

50

100!C

50!50!=

The exponent of 7 in 50! Is

+ = +

250 50

7 17 7

= 8

The exponent of 7 in 100! Is

+ = + =

2100 100

14 2 167 7

exponent of 7 in100

50C is 16 2(8) = 0

50. (1)17 9 3

3 2 2 2 24 4 2

+ = + +

23

22

= +

(1)

15 1517 3

and 3 3 2 24 2

+ =

( )

51015

11 10

3

T C 2 ,2

= which is a

positive integer.

51. (4)x 1

x sin {x}Lt

x 1

x 0

(1 x )sin {1 x}Lt

1 x 1 +

=

x 0

(1 x) sin[1 x]Lt

x +

= =

x 1

x sin {x}Lt

x 1 +

x 0(1 x) sin {1 x}Lt

1 x 1+ +=

+

x 0

(1 x )sin xLt 1

x

+= =

x 1

x sin {x}Lt

x 1

does not exist.

52. (2) Let P = (x1, y1)

AP = 2 AB

B is the midpoint of AP

1 1x y 3

B ,2 2

+ =

B lies on the circle

2 2

1 1 1x x y 3

4 3 02 2 2

+ + + =

2 21 1 1

x 8x (y 3) 0 + + =

the locus of P is x2 + 8x + (y 3)2 = 0

53. (4) The points P = (a cos , b sin ) and

D2

+

= ( a sin , b cos )

Centre C = (0, 0) CP2 + CD2 = a2 + b2

54. (4) The abscissae of the points of

intersection of the given curves are


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connected by

42 2

2

c

x ax+ =

x4 a2 x2 + c4 = 0

As x1, x2, x3, x4 are the roots of this

equation,

x1 + x2 + x3 + x4 = 0,

x1 x2 x3 x4 = c4

Similarly, y1 + y2 + y3 + y4 = 0

y1 y2 y3 y4 = c4

55. (4)3 3

4 4 4 4n

1 2 1Lt ....

2nn 1 n 2

+ + +

+ +

3 3 3

4 4 4 4 4 4n

1 2 nLt ...

n 1 n 2 n n

= + + +

+ + +

n 3

4 4n r 1

rLt

n r =

=

+

3

n

4n r 1

r

1 nLt

n r1

n

=

=

+

13

40

x 1dx log 241 x

= =+

56. (2) Clearly, such a point P is the point of

intersections of two perpendicular

tangents to the parabola y2 = 8x.

Hence P must lie on the directrix

x + a = 0 or x + 2 = 0 x = 2.

the point P is ( 2, 0).

Statement 2 is false, as every

parabola is symmetric about its axis.

57. (4) 4t 0

1 cos (1 cos t)

Lt t

2

2 4t 0

1 cos (1 cos t) (1 cos t)Lt

(1 cos t) t

=

= =

2

2x 0

1 1 1 cos ax a

Lt2 4 2x

=1

8

58. (3) Let us consider a fourth order matrix.

Then

0 1 8 15

1 0 5 12A ,

8 5 0 7

15 12 7 0

=

which is

a skew symmetric matrix. So

statement 1 is false.

Now, 2A (1 5 8 12 7 15) ,= + a

perfect square.

59. (1) Given b2 = a c and

2 (log 2b log 3c) log a log 2b =

log 3c log a+

22b 3c

log log3c 2b

=

3cb

2 =

Now,2

b 3b 9ca

c 2 4= = =

a is the largest side.

Now,

2 2 2b c a

cosA2 bc

+ =

22 29c 81c c

4 163c

2 c2

+ =

, negative

A > 90 The triangle is obtuse.

60. (4) We have f(x + f(y)) = f(x) + y x, y R.

Putting y = 0, we get f(x + 1) = f(x)

f(x) is periodic with period 1.

4 1

0 0f(x)dx 4 f(x)dx 4 = = I.


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61. (1) As per de Broglies equation =h

mv,

1

mv. Velocity being equal for all

particles. Wavelength of the sub-

atomic particle will be large if its

mass is small.

62. (4) C C sigma bond has free rotation

around its axis.

63. (2) = 1 2 3 4 5 6

2 2 2CH CH CH CH C CH

1-hexene-6-yne.

64. (3) It is not a redox reaction all others

are autooxidation and reduction

reaction.

65. (3) NaOH + HCl NaCl + H2O.

NaCl is salt of strong acid with strong

alkali.

1 103 M NaOH + 1 103 M HCl

Neutral.

Number of millimoles of acid and

alkali = 10 0.1 = 1

pH = 7.

66. (2) + + (aq) (aq) 2 (aq)H OH H O

H of strong acid with strong base

= 13.7 kcal.

Acidic nature of acid decreases with

lower enthalpy of neutralization.

67. (4)

68. (3)

69. (4) Bond order of nitrogen is 3, with

smaller bond length and greater

bond energy.

70. (3)

71. (3)

72. (1) T = I Kf m

1.1 = i 1.86 0.2

i = =

1.12.95

1.86 0.2

i = 1 + (n 1)

2.95 = 1 + (3 1)

= 0.975 or 97.5%

73. (4) log

=

a1

1

E4K 320 300

K 2.303 2 320 300

Ea = 0.602 2.303 2 320 300

20

= 13.3 kcal

74. (1) +0 2(S) (aq)Mg Mg 2e Oxidation,

anode; E0 = 2.37 V

++

0(aq) (S)2Ag 2e 2Ag Reduction,

cathode; E0 = 0.80 V.

n = 2

= 0 0 0

cell cathode anodeE E E

= 0.80 ( 2.37) = 3.17 V

G0 = nFE0

= 2 96500 3.17

= 611.8 kJ

75. (1) In BCC lattice,

radius of the species (r)

=edge length(a) 3

4

= =428 1.732r 185 pm.4

76. (3)

77. (1)

PART C: CHEMISTRY


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83. (3) D Nitro group most powerful I

effect electron withdrawing

group in the benzene ring,

decreases the electron density

Least reactive to electrophilic

substitution reaction.

C Chlorine has electron with-drawing I effect in addition,

due to + M effect increases

the electron density.

B Methyl group by + I and hyper-

conjugation increases electron

density highly.

84. (2)

85. (4)

86. (1)

87. (3)88. (1)

89. (2)

90. (3)

78. (4) Greater the charge/size of cation, greater will be its polarizing power of an anion. Be +2

has the highest value, K+ has the least value.

79. (2)

80. (1) La+3 ion is bigger than Lu+3 ion. Bigger the size of the cation, greater will be the basic

nature M(OH)3.

81. (1)

Coen

trans

en

Cl

Cl

Coen

Cl

cisen

Cl

Co

Cl

en

Cl

en

Mirror

Optical isomersGeometricalisomerism

82. (3) Greater the distance between the atoms or groups, more is the stability of the

conformer.

F

H

H

OH

H

H

FH

HHOH

F

H

H

H

H

HO

Anti/Staggered Fully eclipsed Skew/Gauche

Most stable Least stable Stable moderate

Aieee 2012 Nst3 Solns

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