Jee Main 2013 Test8

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Test - 8 (Paper-I) (Answers & Hints) All India Aakash Test Series for JEE (Main)-2013

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TEST - 8 (Paper-I)

ANSWERS

1. (2)

2. (3)

3. (2)

4. (4)

5. (4)6. (3)

7. (3)

8. (1)

9. (3)

10. (4)

11. (4)

12. (3)

13. (1)

14. (3)

15. (1)

16. (2)

17. (2)

18. (3)

19. (2)

20. (1)

21. (1)

22. (1)

23. (1)

24. (1)

25. (2)

26. (4)

27. (4)

28. (4)

29. (4)

30. (1)

PHYSICS CHEMISTRY MATHEMATICS

31. (3)

32. (4)

33. (3)

34. (3)

35. (2)36. (1)

37. (1)

38. (1)

39. (3)

40. (4)

41. (3)

42. (4)

43. (1)

44. (1)

45. (2)

46. (2)

47. (4)

48. (1)

49. (3)

50. (2)

51. (2)

52. (3)

53. (2)

54. (2)

55. (1)

56. (1)

57. (2)

58. (4)

59. (1)

60. (3)

61. (1)

62. (3)

63. (1)

64. (2)

65. (3)66. (1)

67. (3)

68. (1)

69. (1)

70. (1)

71. (4)

72. (4)

73. (2)

74. (1)

75. (1)

76. (2)

77. (3)

78. (1)

79. (2)

80. (1)

81. (4)

82. (1)

83. (1)

84. (4)

85. (1)

86. (3)

87. (4)

88. (1)

89. (1)

90. (4)

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All India Aakash Test Series for JEE (Main)-2013 Test - 8 (Paper-I) (Answers & Hints)

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1. Answer (2)

45

P

30

30

6090

Q R

Light ray will fall normally on PR

Using Snell's law

1 sin45 = sin30

1 1

22=

2 =

2. Answer (3)

Spherical aberration is inability of mirror to focus all

the rays coming parallel to principal axis at a single

point. So it exists in concave mirror. Chromatic

aberration is due to behavior of light. It does not exist.

3. Answer (2)

01/2

0 (4 )( 1 min)

= =

t NP R t

0

1/20 ( )

( 2 min)=

=t N

Q R t

After four minutes, the number of nuclei ofPand Q

are same and equal to0

4

N.

So number of nuclei ofRpresent at this time is

0 0 00

3 94

4 4 2

+ =

N N NN

4. Answer (4)i4

i3 = 0

i2

i1

15 V

25

25

25

75

75

Given circuit

Can be looked upon as

i1

= i2

+ i4

i1 = 0.2 Ai2

= i1

= 0.1 A

PART - A (PHYSICS)

5. Answer (4)

For no average deviation ( = ( 1)A)

(1.48 1)1

(1.64 1)3 + (1.48 1)2

= 0

1

+ 2

= 4

6. Answer (3)

observerfish fish bird

(as seen by bird) object

v v v

= +

=1

5 3 104

+

=

55m/s

4

7. Answer (3)

2

> (as it is converging the rays)

and 1

= (No deviation)

8. Answer (1)

238 234 4

92 90 2U U He +

234 234 0

90 91 1X Y +

9. Answer (3)

M m1

+ m2

Rest1

GP 2

GP

Using conservation of linear momentum

1 2 0+ =G GP P

1 2+ =G GP P P

=

h

P

So,1

2

1

=

10. Answer (4)

=

D

d

As water

< air

So, if d is decreased, D is increased or is

increased original can be restored.

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11. Answer (4)

y

x

vcos

vsin

v

O

vsin

vcosv

(see the figure)

So, cos2 sin2Iv v i v j = + G

10cos2 10sin2Iv i j= + G

12. Answer (3)

hc

k W=

=1242eV nm

2 eV 1.105 eV400nm

=

Linear momentum = 2mv mk =

= 31 192 9.1 10 1.105 10 1.6

= 5.67 1025 kg m/s

Now =mV

rqB

B = 35.8 106 T

13. Answer (1)

A

B

Y = A.B

(AND gate)

A

B

So, ifA = 0, B = 1

Y= 0

and ifA = 1, B = 1

Y= 1

14. Answer (3)

WhenA is at positive potential and B is at negative

potential. Diodes are in forward bias.

So, Req

= 8

WhenA is at negative potential and B is at positive

potential

Reverse bias

Req

= 19

15. Answer (1)

Population covered = P d2 (P= population density)

2=d Rh

On solving Population covered

= 76.8 lakhs

16. Answer (2)

Least count = 1 M.S.D. 1 V.S.D.

= 1 M.S.D. 18

20M.S.D.

=1

10M.S.D.

= 0.1 mm

17. Answer (2)

2

Dimensionless

x

t

So,2 1L T

18. Answer (3)

2

0( )+ =I I I (Initially)

04

=I

I

When one slit is covered

2

4

+ =

II I

099

4 16 = =

II I

19. Answer (2)

K.E. = |E|

= 3.4 eV

2 (K.E.) =

h

m

20. Answer (1)

cos2

= R

f R

R> 2f

CBC

A

f 30

R F

AB = R f

BC= (R f)sin30

2

=

R fBC

2>

fBC

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21. Answer (1)

1

2

90

Here > critical angle

1 2

1

sin

>

Now, 1sin(90 ) = 1 sin

2 21 1 2sin cos = =

Since, > critical angle

2 2

1 2sin <

22. Answer (1)

xyQ

z=

1

2

= + +

Q x y z

Q x y z

= 0.48%

23. Answer (1)

( )f a z b= (Moseley's law)

forK

line b = 1

2 2

1 1

1

( 1)= = c

f a z

2 2

2 2

2

( 1)= = c

f a z

2

1 2

2

2 1

( 1)

( 1)

=

z

z

2

2

2

( 1)

4 (11 1)

=

z

z2 = 624. Answer (1)

2

2

4=

Ag

T

2 = +

A

A

g T

g T

A and Tare least and same in options 1, 2 and 4but n is highest for option (1)

25. Answer (2)

= i+ e A

(for same deviation there are two values of i) the

second value ofifor same deviation is the angle of

emergence for first value ofi

So, 23 = 15 + 35 A

A = 27

26. Answer (4)

Lateral displacement can never be greater than

thickness.

27. Answer (4)

Intensity of characteristic X-ray is independent ofapplied voltage.

28. Answer (4)

Photoelectric emission will not take place for < 0,

so no stopping potential.

29. Answer (4)

Image will not move if the mirror moves parallel to

itself.

30. Answer (1)

Eg 1 eV (for semiconductor)

PART - B (CHEMISTRY)

31. Answer (3)

Enol form is

CH CH CH CH CH CH CH OH3

In this there are three double bonds hence possible

isomers are 23 = 8.

32. Answer (4)

O O

I2OH

CHI3+

O IC +

In the above reaction

O

CH I2and

O

CHI2

formed as intermediate products hence allO

,

O

CH I2and

O

CHI2give iodoform test.

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33. Answer (3)

OH

O

O

OOH

(A)

(No elimination at bridge head carbon)

So, (A) gives positive DNP test.

34. Answer (3)

CH N N2

+

O

CH N N2

O

+O

O CF CO H3 3

O

35. Answer (2)

CHO

CHONO

2

NO2

OH

Conc.

(Cannizaroreaction)

Conc. H SO2 4

CH2COO

NO2

NO2

O HO

O

NO2

NO2

36. Answer (1)

OH

OH

O

OHO

O

O

OH

OH

Molar mass = Mg (M 44)g

O

+ CO2

O

O

O + H O2

(M 62)g

37. Answer (1)

Glycosidic linkage is ether linkage.

38. Answer (1)

Amylose is a polymer of maltose.

39. Answer (3)

At isoelectric point the net charge on amino acid is

zero. So, for glycine it is H N CH COO3 2

+

40. Answer (4)

Fact

41. Answer (3)

O C NHO C NH2

O rB

HOBrBr Br

C N Br

HO

C N Br

O

OBr

D O2

ND2OH

O C N

42. Answer (4)

In aldol reaction

Generation of nucleophile is an acid/base reaction

Attack of nucleophile is nucleophil ic addition

reaction.

In this reaction :

CH C H + CH C H3 3

+1 3

O O

OH

OH

H C CH CHO3 2C

H

20

So, it is a redox reaction.

43. Answer (1)

KOH

conc.HCHO + Ph CHO (A) + (B)

Here HCHO is oxidized by the attack of OH andPhCHO is reduced by the hydride shift.

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44. Answer (1)

Hinsberg reagent test : Primary amines react with

ArSO2Cl forming products soluble in KOH.

45. Answer (2)

Dil

H SO2 4

OO

CH O3 O

O

CH OH +3

O HO

H HOO

OH O

+

46. Answer (2)

dil.H SO

2 4

O OO

OO

OHOH Tautomerize

H

OH

Since compound does not have alcoholic group hence

does not give victor meyer test.

47. Answer (4)

Polyamides have carbon chain hence have van der

Waal attraction force and due to amide linkage it has

dipole-dipole interaction and H-bonding

48. Answer (1)

Fact

49. Answer (3)

Fact

50. Answer (2)

Vulcanised rubber have C S C linkage which

is thio ether linkage.

51. Answer (2)

In Stephen's reaction Sn oxidizes as

Sn Sn . Hence n = 2.+2 +4 factorof SnCl2

52. Answer (3)

Highest boiling point is in propan1ol due to inter-

molecular hydrogen bonding.

53. Answer (2)

CH CH COOH3 2 PCl

3

Br2

H2O

Br Br

Cl OH

O O

54. Answer (2)

In esterification higher the leaving aptitude of 'X' in

O

R C X higher will be the rate of the reaction.

So, rate is highest in

O

R C Cl.

55. Answer (1)

Zwitter ion is formed in sulphonation of aniline.

H SO2 4

NH2

NH3HSO

4

+NH

3

SO3

+

zwitter ion

56. Answer (1)O

CH COCl3

NH2

NH C CH3

(lessreactive)

(morereactive)

H N C CH3

O

+

Reactivity towards electrophilic substitution

57. Answer (2)

HCOOH is a reducing agent. Hence gives Tollen's

test positive.

58. Answer (4)

COOH

COOH

NH3

O

O

O

O

NH4

NH4

+

+

59. Answer (1)

N N + NH2

+

N N NH2

Coloured due to extended conjugation.

60. Answer (3)

Primary structure of protein decides the amino acid

sequence and quaternary form is biologically active

form of protein.

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61. Answer (1)

, ,AB AC ADJJJG JJJG JJJG

are coplanar.

9 5 2AB i j k= +

JJJG

11 4AC i j k= + +JJJG

7 14 ( 2)AD i j k k= + + JJJG

9 5 2

11 4 1 0

7 14 2k

=

k= 1

62. Answer (3)

Let us assume, 2 3 5A i j k= + +

G

B ai bj ck = + +G

We know

. | || | cosA B A B= G G G G

. | || |A B A BG G G G

2 2 22 3 5 4 9 25a b c a b c + + + + + +

2 2 22 3 5

38

a b ca b c

+ ++ +

2 2 2 100

38a b c+ +

2 2 2 50 .

19a b c+ +

63. Answer (1)

M

M

M

M

M

M

1

2

3

4

5

6

SM

TWThFSa

A B

Total ways = (7)6

Favourable cases = 6 33 1 6C

Probability =

6 3 33

6 6

6 20 6

7 7

C

=

PART - C (MATHEMATICS)

64. Answer (2)

2 22x y y c+ =

2 4 0dy dy x ydx dx

+ =

(4 1) 2dy

y xdx

=

2

4 1

dy x

dx y

=

...(i)

2

4 1

dx x

dy y

=

...(ii)

2 (4 1)

dx dy

x y=

1 1log log (4 1) log

2 4e e ex y c= +

2log log (4 1) 4loge e ex y c= +

{ }2 4log log (4 1)e ex c y= x2 = k(4y1)

65. Answer (3)

New S.D. =10

5| 2 |

=

66. Answer (1)

A = (1, 2, 3), B = ( 1, 2, 3), C= (1, 2, 3)

Let the plane is

a(x 1) + b(y 2) + c(z+ 3) = 0 ...(i)

B will satisfy (i)

2a + 6c= 0 ...(ii)

Cwill satisfy (i)

4b + 6c= 0 ...(iii)

From (ii) and (iii)2a = 4b = 6c

a = 3c,2

2

cb =

33 ( 1) ( 2) ( 3) 0

2

cc x y c z + + + =

3

3 3 3 3 02

yx z + + + =

3

3 3

2

yx z+ + =

6x+ 3y+ 2z= 6

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1.1 2 3

x y z+ + =

This plane intersects the co-ordinate axes at

(1, 0, 0), (0, 2, 0) and (0, 0, 3).

3

1 0 01

0 2 0 1 units6

0 0 3

V= =

67. Answer (3)

( ) ( )a i b i GG

Let a i R =GG

( )R b i= G G

( )R b i= G G

{ } ( )a i b i = GG

{ } ( )b a i i = G G

{ } ( ) ( )b i a b a i i = G GG G

(( ) ( ) )b a i b i a i = G GG G

( ) ( )( )b a b i a i = G GG G

Let1 2 3 a a i a i a k = + +

G

1 2 3 b b i b j b k = + +

G

1 1( )a b a b= GG

...(i)

Similarly

( ) ( )a j b j GG

2 2a b a b=

GG

...(ii)

and ( ) ( )a k b k GG

3 3a b a b= GG

...(iii)

On adding (i), (ii) and (iii)

( ) ( ) ( ) ( ) ( ) ( )a i b i a j b j a k b k + + G GG G G

1 1 2 2 3 33 ( )a b a b a b a b= + +GG

3a b a b= G GG G

2a b= GG

68. Answer (1)

The equation of normal at P(x, y) is

1( )

( , )

Y y X x dy

x y

dx

=

xintercept =x+ yM(where( ),

=dy

Mdx x y

)

yintercept = = +x

y yM

HM = 2

22

ab

a b=

+

1 1 1.a b

+ =

1 11

xx YMy

M

+ =+ +

11

M

x YM x yM+ =

+ +

M+ 1 =x+ Y.M

M(1 y) =x 1

1

1

dy x

dx y

=

( 1) ( 1) constantx dx y dy+ = (x 1)2 + (y 1)2 = c

69. Answer (1)

The given triangle is equilateral triangle

AB = BC= CA = 2 26 6 6 2+ = units

So circumcenter is = (1, 4, 3)In an equilateral triangle circumcentre coincides with

centroid.

70. Answer (1)

Number of elements in the sample space = 3 3 6.

Favourable cases are

(2, 1, 1), (2, 1, 5), (2, 3, 3), (2, 5, 1), (2, 5, 5)

(4, 1, 3) (4, 3, 1), (4, 3, 5), (4, 5, 3), (6, 1, 1), (6, 1, 5)

(6, 3, 3), (6, 5, 1), (6, 5, 5)

14 73 3 6 27

P= =

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71. Answer (4)

Form the equation, whose roots are 1, 1, 3, 3

(t 1) (t+ 1) (t 3) (t+ 3) = 0

(t2 1) (t2 9) = 0

t4

10t2

+ 9 = 0So the required differential equation will be

4 2

4 210 9 0

d y d y y

dx dx + =

1 4

2 3

0 9 9

10 0 10

k k

k k

+ + = =

+ +

72. Answer (4)

73. Answer (2)

The tree diagram of the given problem will be as

follows.

R611

B511

R(4/

8)

R(6/

11)

B (5/11)

R(5/1

1)

B (6/11)

R(7/11)

B (4/11)

B (4/8)

B(3/8)

R(6/11)

B (5/11)

B (5/8)

So, Total probability is

6 4 5 6 4 6 5 3 4 5 5 5

11 8 11 11 8 11 11 8 11 11 8 11

= + + +

120 144 60 125 449.

11 11 8 968

+ + += =

74. Answer (1)

Case I :n = 2m

Average1

(1 2 3...2 )2

mm

= + +

1 2 1 2(1 2 )

2 2 2

m mm

m

+= + =

1

2m= +

Mean deviation =

1 32 ... terms

2 22

m

m

+

2 (1 ( 1) 1)2

2

mm

m

+ =

2 4

m n= =

Case II :n = 2m +1

Average { }1

1 2 3...(2 1)(2 1)

mm

= + + ++

= (m + 1)

Mean deviation =2(1 2 3.... ) ( 1)

(2 1) (2 1)

m m m

m m

+ + +=

+ +

2 1

4

n

n

=

75. Answer (1)

A (90

)

(1, 2, 3) (3, 3, 3)

C

B

A1

B1

x + y + z 3 = 0

9090

4 1 36 41AB = + + = units

2 1 1 1 1 6 9cos(90 ) .

3 4 1 36 3 41

+ + = =

+ +

9sin

3 41 =

41 42cos 14

123AC AB

= = = units

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76. Answer (2)

2| | ( ) ( )a b c a b c a b c + = + + G G GG G G G G G

2 2 2| | | | | |a b c= + +GG G

2 2 2a c b c a b + G GG G G G

...(i)

2| | ( ) ( )b c a b c a b c b+ = + + G G G GG G G G G

2 2 2| | | | | | 2 2 2b c a b c c a a b= + + + G G GG G G G G G

...(ii)

2| | ( ) ( )a c b a c b a c b+ = + + G G GG G G G G G

2 2 2| | | | | | 2 2 2a c b a b b c a c = + + + G G GG G G G G G

...(iii)

On adding (i), (ii) and (iii).

2 2 2| | | | | |a b c b c a a c b+ + + + + G G GG G G G G G

2 2 22(| | | | | | ) 2( )a b c a b b c c a= + + + + G G GG G G G G

1 2 32(1 1 1) 2(cos cos cos )= + + + +

1 1 16 2

2 2 2

=

36 2 9.2

= =

{1

= 2

= 3

= 120}

77. Answer (3)

The planesx+ 2y+ 3z 6 = 0 and

x+ y z+ 1 = 0 are perpendicular to each other, so

image is

x + 2y + 3z 6 = 0

78. Answer (1)Plane isx+ y+ z 6 = 0

S(P) = a + 2a + 4 6 = 3a 2

S(Q) = 2a + ( a) + 3 6 = a 3

S(P) S(Q) > 0

(3a 2) (a 3) > 0

+ + + + + + +

2/3 3

( ,2 / 3) (3, )a

79. Answer (2)

3log loge ex xdy x x dx ydx=

3log ( log )e ex xdy x x y dx=

3 log

log

e

e

x x ydy

dx x x

=

2

log+ =

e

dy yx

dx x x

dyP y Q

dx+ =

I.F. = P dx

e

=1

logedx

x x

e

= log (log )e e xe

= loge x

2(log ) loge ey x x x dx c = +

3 3

(log ) log3 9

e e

x xy x x c = +

80. Answer (1)

The equation of the plane is

(2x 3y+ 5z 7) + k(x + y + 2z 4) = 0

(2 + k)x+ (k 3)y+ (2k+ 5) z (4k+ 7) = 0

and

(2 + k) (1) + (k 3) (1) + (2k+ 5) (1) = 0

4k+ 4 = 0

k= 1

So, the equation of the plane isx 4y+ 3z 3 = 0

81. Answer (4)82. Answer (1)

Probability of getting success =6 1

6 6 6=

Case I :3 success and 2 unsuccess.

3 25

4 5

1 5 250

6 6 6C

=

Case II : 4 success and 1 unsuccess.

4 154 5

1 5 256 6 6

C =

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Case III: 5 success

55

5 5

1 1

6 6C

=

Total probability = 5250 25 1

6

+ +

= 5 4276 46

6 6=

=23

648

83. Answer (1)

Let us assume the equation of the line is

x y z

a b c= =

1 1 1

2 3 4 0

a b c

=

1(3c 4b) + 1 (2c 4a) + 1 (2b 3a) = 0

3c 4b + 2c 4a + 2b 3a = 0

7a 2b + 5c= 0

7a + 2b 5c= 0 ...(i)

3 9 / 2 0

1 2 1 0

a b c

=

93(2 ) ( ) 0

2c b c a =

9 96 3 0

2 2

c ac b + =

12c 3b 9c+ 9a = 0

9a 6b + 3c= 0

3a 2b + c= 0 ...(ii)

From (i) and (ii)

2 10 7 15 14 6

a b c= =

+

8 22 20

a b c= =

4 11 10

a b c= =

84. Answer (4)

| ( ) | | | | | | |a b c a b c = G GG G G G

| || | sin | | | | | |a b c a b c = G GG G G G

| | | | | | sin sin | | | | | |a b c a b c = G GG G G G

sin sin 1 =

sin = 1 and sin = 1

= /2 and = /2

,a bGG

are perpendicular and a bGG

is

perpendicular to cG

, ,a b cGG G are coplanar.

85. Answer (1)

| |

u vx

u v

+=

+

G GG

G G =3

u v+G G

| |

u wy

u w

+=

+

G GG

G G =3

u w+G G

| |

v wz

v w

+=

+

G GG

G G =2

v w+G G

[ ]x y y z z x G G G G G G

= 2[ ]x y zG G G

=

2

3 3 2

u v u w v w + +

G G G G G G

=21 [ ]

18u v u w v w + + +G G G G G G

=24 [ ]

18uv wG G G

=22 [ ]

9uv wG G G

2

[ ] 2

9[ ]

x y y z z x

u v w

=

G G G G G G

G G G

86. Answer (3)

(1) is true (2) is false.

7/29/2019 Jee Main 2013 Test8

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87. Answer (4)

Putx+ 2y = t

21

dy dt

dx dx + =

13

2 2 1

dttdx

t

+=

2 61

2 1

dt t

dx t

+= +

4 5

2 1

dt t

dx t

+=

2 1

4 5

t dt dx t

=+

1 (4 5) 7

2 (4 5)

tdt x c

t

+ = +

+

1log | 4 5 |

2 4et t x c

7 + = +

7

log | 14 5 | 2 2 '4

et t x c x c + = + = +

7

2 log | 4 8 5 | 2 '4

ex y x y x c+ + + = +

4 8 7log | 4 8 5 | 8 "ex y x y x c+ + + = +

4 8 7log | 4 8 5 |ex y x y k + + + =

(1) is false and (2) is true false

88. Answer (1)

a1a

2+ b

1b

2+ c

1c

2= 1 4 9 = 12 < 0

(a1x

1+ b

1y

1+ c

1z+ d

1) (a

2x

1+ b

2y

1+ c

2z

1+ d

2)

= (1) (10) > 0

(1) is false and (2) is true

89. Answer (1)

41 2 3( )

x cxy c c e c e += + +

41 3

cx xy k e c e e= +

1 2xy k e k e= +

1 2( )xy k k e= +

xy ke=

xdy kedx

=

Sody

ydx

=

So, this is a differential equation of order = 1,

(1) is true and (2) is true and (2) is the correct expla-

nation of (1).

90. Answer (4)

x + y= 10

2 3 = 18x + y

(0, 10)

(0, 6)

(9, 0) (10, 0)

We find that the feasible region of empty set.

(1) is false, (2) is true.

Jee Main 2013 Test8

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