Aiats Iitjee 2011 Test-8 P-i Solution

8/18/2019 Aiats Iitjee 2011 Test-8 P-i Solution

1/21

Test - 8 (Paper - I) (Answers & Hints) All India Aakash Test Series for IIT-JEE 2011

1 / 2 11 / 2 11 / 2 11 / 2 11 / 2 1

TEST - 8 (Paper - I)

CHEMISTRY MATHEMATICS PHYSICS

A N S W E R S

29. (C)

30. (B)

31. (A)

32. (D)

33. (C)34. (C)

35. (A)

36. (D)

37. (B, C)

38. (A, C, D)

39. (A, B, C)

40. (A, C)

41. (A, B, C, D)

42. (B)

43. (A)

44. (A)

45. (B)

46. (D)

47. (1)

48. (1)

49. (6)

50. (2)

51. (5)

52. (2)

53. (3)

54. (2)

55. (1)

56. (3)

1. (A)

2. (A)

3. (D)

4. (B)

5. (D)6. (C)

7. (B)

8. (B)

9. (A, C)

10. (A, C)

11. (A, C, D)

12. (A, C)

13. (A, C, D)

14. (C)

15. (B)

16. (C)

17. (C)

18. (C)

19. (7)

20. (3)

21. (2)

22. (5)

23. (5)

24. (8)

25. (0)

26. (8)

27. (5)

28. (6)

57. (B)

58. (D)

59. (A)

60. (C)

61. (B)62. (C)

63. (D)

64. (B)

65. (A, B, C)

66. (A, D)

67. (A, B, D)

68. (A, B, C, D)

69. (A, B, C)

70. (D)

71. (A)

72. (A)

73. (A)

74. (B)

75. (2)

76. (4)

77. (3)

78. (3)

79. (6)

80. (3)

81. (5)

82. (5)

83. (2)

84. (1)


2/21

All India Aakash Test Series for IIT-JEE 2011 Test - 8 (Paper - I) (Answers & Hints)

2 / 2 12 / 2 12 / 2 12 / 2 12 / 2 1

PART - I (CHEMISTRY)

ANSWERS & HINTS

1. Answer (A)

CR H HS SH

S

SRH

(i) Base

(ii) R X′

O

CR

O

R′

Hydrolysis S

SR′

R

2. Answer (A)

CH3

Br

(CH ) COK3 3

(CH ) COH3 3

CH2

(i) BH THF3⋅

(ii) O H O2H/ 2

CH2

OH3. Answer (D)

Both SN 2 and SN Ar reaction will take place simultaneously.

4. Answer (B)

It is an example of Beckmann rearrangement.

5. Answer (D)

NaBH4 being a less reactive hydride donor only reduces keto group in the presence of ester.

6. Answer (C)

OH

NH2

More nucleophilic site.

7. Answer (B)

Because enol form is stabilized by strong intramolecular hydrogen bonding.

O

OH

8. Answer (B)

CHO(i) O3

(ii) ZnH O⋅ 2 CHO

(i) OH/H O2

(ii) H O3

+

OH

(i) H+

(ii) ∆

CHO

CHO


3/21


3 / 2 13 / 2 13 / 2 13 / 2 13 / 2 1

9. Answer (A, C)

10. Answer (A, C)

I

+ Ag+

+ AgI

∴ This reaction does not occur due to antiaromatic intermediate.

I I etc.

∴ Due to partial double bond character in C—I bond dissociation of C—I bond is very difficult.

11. Answer (A, C, D)

CH3

H

NH2

CH3

CH3

X

CH3

CH3

CH3

Y

ZZ

Y

12. Answer (A, C)

[X]gives positive

Baeyer’s test

COOH

COOH

∆

[Y]give positive test

with DNP

O


In sucrose free hemiacetal linkage is not present and hence it is not capable of exhibiting mutarotation, osazoneformation or reducing behavior.

14. Answer (C)

15. Answer (B)

16. Answer (C)

17. Answer (C)

18. Answer (C)


4/21


4 / 2 14 / 2 14 / 2 14 / 2 14 / 2 1

Solution of Q.16 to Q18

As the reaction proceeds through free radical mechanism, it will involve the formation of most stable free radical.

(i) —CH—CH CH2

H

O2

h ν —CH—CH CH

2

More stable resonancestabilized free radical

(ii)O

2

h ν

OOH

(iii) O2

h ν

O—O—HHO H O

19. Answer (7)

O

O

H

(i) NaOH

O

HO

O

OH

O

H

OHMe

Seven dehydrated products possible from here.

20. Answer (3)

If has three basic sites

∴ Protonation at three N will lead to the development of +3 charge.

21. Answer (2)

It has α-D-Glucosidic linkage and β-D-Fructosidic linkage.

22. Answer (5)

COOH

COOH

∆O

Five memberedring


5/21


5 / 2 15 / 2 15 / 2 15 / 2 15 / 2 1

23. Answer (5)

C C H,H

H,

H H

and

OH

are not sufficiently acidic and hence the acid base

reactions of these acids with aq. NaOH shifts in backward direction.

24. Answer (8)

It has three stereocenters, therefore 8 stereoisomers are possible.

25. Answer (0)

Due to resonance stabilization C—Br bonds are very strong in aryl halides and hence no AgBr is precipitated in thegiven example.

26. Answer (8)

27. Answer (5)

CH3

28. Answer (6)

⇒ 3 × 2 × 1 = 6.

PART - II (MATHEMATICS)

29. Answer (C)

Consider A as origin and

A

B b( )

C c( )D d( )

(origin)

E F

( ), ( ), ( ), ,2 2

+

r r r

r r r r r c b d

B b C c D d E F

∴ ( )λ + + + = µuuur uuur uuur uuur uuur

AB AD CD CB EF

⇒ ( ) ( )2

b d d c b c b d c µ

λ + + − + − = + −r r r r r r

r r r

⇒2 ( ) ( )

2b d c b d c

µλ + − = + −

r r r r r r

⇒14

λ=

µ

30. Answer (B)

{H, T } → {1, 2} i.e. H = 1, T = 2Choices of

coefficients a,b,c

a 121

2

b 1

1,21,2

1,2

c 112

2

Coefficients are obtained by tossing three coins

Total number of quadratic equations obtained = 8

No. of quadratic equation ax 2 + bx + c = 0 having non-real roots is root 7.

∴ Required probability =78


6/21


6 / 2 16 / 2 16 / 2 16 / 2 16 / 2 1

31. Answer (A)

A(x 1, y 1, z 1), B (x 2, y 2, z 2), C (x 3, y 3, z 3), D(x 4, y 4, z 4)

S R

C

Q P

A

B

D Let plane is ax + by + cz + d = 0

∴1 1 1

2 2 2

+ + += −

+ + +

ax by cz d AP

PB ax by cz d

2 2 2

3 3 3

+ + += −

+ + +

ax by cz d BQ

QC ax by cz d

3 3 3

4 4 4

+ + += −

+ + +

ax by cz d CR

RD ax by cz d

4 4 4

1 1 1

+ + += −

+ + +

ax by cz d DS

SA ax by cz d

∴ . . . 1=AP BQ CR DS

PB QC RD SA

32. Answer (D)

Required probability =5

3(2!)

5!

C =

16

33. Answer (C)

The equation ax 2 + by 2 + cz 2 + 2hxy + 2fyz + 2gzx = 0 ... (i)

represents a pair of planes if 0=

a h g

h b f

g f c

⇒

2 2 2

3 2 0+ + + + + =px y qz xy yz zx represents pair of perpendicular planes

∴

3 12 2

3 1 1 0

11

2

=

p

q

and p + 1 + q = 0 ... (ii)

⇒ 4pq – 4p – 9q + 5 = 0

Using (i) & (ii),

4p 2

– p – 14 = 0 i.e. p = 2,

7

4−p = 2, q = –3

p – q = 5

34. Answer (C)

We have,

3/

0lim 15

3→

+ +=

x x x x

x

a b c

⇒ 0

3 3lim .

3 15

x x x

x

a b c

x e →

+ + − =


7/21


7 / 2 17 / 2 17 / 2 17 / 2 17 / 2 1

⇒ log( ) 15abc e =

⇒ abc = 15 ... (i)

∴ Number of favourable ways for choice of a,b,c satisfying eq. (i) = 6

∴ Required probability6 1

216 36= =

35. Answer (A)

Let F (x 1,y 1,z 1) be foot of perpendicular drawn, from origin O to variable plane.

∴ Equation of variable plane is x 1(x – x 1) + y 1(y – y 1) + z 1(z – z 1) = 0

2 2 21 1 1 1 1 1xx yy zz x y z + + = + +

According to question,

2 2 22 2 2 2 2 2 2 2 21 1 1 1 1 1 1 1 1

1 1 116

x y z x y z x y z

x y z

+ + + + + + + + =

∴ Locus of F (x 1,y 1,z 1) is

12 2 2 2

2 2 21 1 1

1 1 1( ) 16x y z

x y z

− + + = + +

∴ λ = 1

36. Answer (D)

p = Probability of getting a black ball (success) =B

B Y +

q = Probability of getting a yellow ball (failure) =Y

B Y +

Probability (player P 2 wins) = qp + q 4p + .....

= 31

qp

q −

= 2 2 2( )

3 3

Y B Y

B B Y Y

+

+ +

∴ λ = 1, µ = 3, t = 3

∴ λ + µ + t = 7

37. Answer (B, C)

1 2 32 3 4a b c r ur vr wr − + = = + +r

r r r r r r

= ( ) ( ) ( )u v w a u v w b u v w c − + + + − + − + +r

r r

Q , ,a b c r

r r are non-coplanar

∴ 2, 3, 4u v w u v w u v w − + = + − = − − + + =

Solving1 1

, , 32 2

u v w = − = =

∴ u + v = 0 , 2u + 4v + w = 4


8/21


8 / 2 18 / 2 18 / 2 18 / 2 18 / 2 1


We have,

( ) ( )a b c d × × ×r r

r r

= (( ). ) (( ). )a b d c a b c d × − ×r r r r

r r r r

= [ ] [ ]d a b c a b c d −r r r r

r r r r

∴ 4 5c d +r

r = [ ] [ ]d a b c a b c d −r r r r

r r r r

⇒ [ ] 4, [ ] 5d a b a b c = = −r r r

r r r

Also ( ) ( )a b c d × × ×r

r r

= – ( )( )( )c d a b × × ×r r

r r

= – ( )[ ] [ ]b c d a c d a b −r r r r

r r r r

= [ ] [ ]d a c b b c d a −r r r r

r r r r

∴ 2 3a b +r

r = [ ] [ ]d a c b b c d a −r r r r

r r r r

[ ]d a c r

r r

= 3, [ ]b c d =r r

r

–2

39. Answer (A, B, C)

, andr r r

x y z are non-zero non-coplanar vectors

=×r r r

x y z , =×r r r

y z x , =×r r r

z x y

= 0x z y z z x ⋅ ⋅ = ⋅ =r r r r r r

Also | | = | | | | = | |x y z x y z × ⇒r r r r r r

... (i)

Similarly | | | | = | |y z x r r r

... (ii)

| | | | = | |x z y r r r

... (iii)

From (i) & (ii),

2| | | | = | |z y z r r r

⇒ 2| |y r

= 1

⇒ | |y r

= 1

Similarly | | | | = 1x z =r r

Also Q x y z × =r r r

z x y z z ⋅ × = ⋅r r r r r

[ ] 1x y z =

r r r


9/21


9 / 2 19 / 2 19 / 2 19 / 2 19 / 2 1

40. Answer (A, C)

Required probability1

1.3

4

n n

n

C −=

Also for n = 3,

Required probability3 2

13

.3 27644

C = =

41. Answer (A, B, C, D)

Given ˆ ˆ ˆ2 3 4OA i j k = + +uuur

ˆ ˆ ˆ3 4 2OB i j k = + +uuur

ˆ ˆ ˆ4 2 3OC i j k = + +uuur

6AB BC AC = = =

uuur uuur uuur

∴ ∆ABC is equilateral triangle

∴ Circumcenter of ∆ABC coincides with centroid (Incenter, orthocentere) of triangle

∴ Circumcenter of ∆ABC has the position vector ˆ ˆ ˆ3( )i j k + +

Orthocentre (incenter) of ∆ABC has the position vectors ˆ ˆ ˆ3( )i j k + +

The vector ˆ ˆ ˆ( )i j k + + makes equal angle with vector ,OA OB and OC respectively which is1 9cos

29-q= .

Position vector of normal to the plane of triangle is – ˆ ˆ ˆ3( )i j k + + .

Position vector of centroid is ˆ ˆ ˆ3( )i j k + + so both are along the same line hence perpendicular drawn from origin

passes through centroid of ∆ABC.

2 3 4

3 4 2 27

4 2 3

OA OB OC = = −

uuur uuur uuuur

27OA OB OC = uuur uuur uuuur

42. Answer (B)

We have,

l + m + n = 0,

3lm – 4mn + 3nl = 0

⇒ m (3l – 4n ) + 3nl = 0

⇒ (l + n ) (4n – 3l ) + 3nl = 0

⇒

2

4 4 3 0n n

l l

+ − =


10/21


1 0/2110/2110/2110/2110/21

⇒12

n

l = or

32

n

l = −

l = 2n , m = –3n 2n = –3l ,3n

m −=

2 3 1l m n

= =− 2 1 3

l m n = =

− −

∴ Direction ratio of two lines are < 2, –3, 1 > and < –2, –1, 3 >

4 3 3 1cos( )

714 14

− + +θ = =

tan 4 3θ =

( )1tan 4 3−θ =

43. Answer (A)

Direction cosine of two lines are2 3 1 2 1 3

, , and , ,14 14 14 14 14 14

− − −< > < >

Let2 3 1 2 3 1

, , , , ,14 14 14 14 14 14

A A− − −

′

and2 1 3

, ,14 14 14

B − −

A m B

N

A′

O∴ 1,OA OB = =uuur uuur

O is origin

M (mid point) of AB is 2 20, ,

14 14

M −

N (mid point) of A′B is 2 2 1 , ,14 14 14

N −

OM and ON are internal and external angle bisector of ∠AOB respectively

Direction cosines of line OM 1 1

0, ,2 2

−< >

Direction cosines of line ON 2 1 1

, ,6 6 6

− −<

44. Answer (A)

( 2, 4)( 2 / 4)

( 4)P N S

P N S P S

= == = =

=

=

1 3.

4 361 1 1 1 1 3 1 1

. . .2 6 4 12 8 216 16 1296

+ + +

=432

2197


11/21


1 1/2 11 1/2 11 1/2 11 1/2 11 1/2 1

45. Answer (B)

( 4, )( 4 / ).

( )P S N even

P S N even P N even

= == = =

=

=( 4, 2) ( 4, 4)

( 2) ( 4) .....

P S N P S N

P N P N

= = + = =

= + = +

=

1 3 1 14 36 16 1296

1 1 1......

4 16 64

+

+ + +

=1 1 1

3. . 34 36 144

+

=3 2

4 33 4 1

4 .3

+

46. Answer (D)( 2, 1, 4)

( 2 / 4, = 1)( 4, 1)

P N D S P N S D

P S D

= = == = =

= =

=

1 3.

4 361 3 1 2 1 1

. . .4 36 8 108 16 1296

+ +

=1

1 11

6 44

+ +

=144169

47. Answer (1)

Consider vertex A as origin of the parallelogram

Let AB b =uuur r

D d( )

M

(origin)A ( 0 ) B b( )

C c( )N

AC c =uuur

r

AD d =uuur r

∴ M and N are mid points of sides BC and CD

∴2

b c AM

+=

r r

uuuur

,2

c d AN

+=

r r

uuur

Q ABCD is a parallelogram

∴ 2b d c + =r r

r

Now 3AM AN AC λ + µ =uuuur uuur uuur

⇒ 32 2

b c c d

c

+ +λ + µ =

r r

r r

r


12/21


1 2/2112/2112/2112/2112/21

⇒ 3 02 2c b

λ λ − µ + µ − + =

r r

⇒ 32λ

+ µ = and λ = µ

⇒ 1λ

=µ

(∴ b r

and c r

are non-collinear vectors)

48. Answer (1)

Given ˆˆ ˆ 1a b c = = = . ˆ. 1a x =r

, 3ˆ.2

b x =r , 2x =

r

ˆˆ ˆa b c x + + = r

2ˆˆ ˆ. . .a x b x c x x + + =r r r r

⇒ 3 ˆ1 . 42c x + + =

r

⇒3ˆ.2

c x =r

⇒1 32cos cos

2− α =

β

⇒34

α=

β

⇒

4

3

β=

α

⇒2

13

β + =

α

49. Answer (6)

Given planes containing the line L are x + y – z + 2 = 0 and x – y + z + 4 = 0

Required plane containing the line L (passing through intersection of given plane’s) is

(x + y – z + 2) + λ (x – y + z + 4) = 0

Q It passes through point P (1, 2, –1)

∴ 6 + λ(2) = 0⇒ λ = –3

Required plane is x – 2y + 2z + 5 = 0

Lengths of intercept on coordinates axis of required plane are a = 5,52

b = ,52

c =

Perpendicular distance of required plane from origin,53

d =

∴ 6a b c

d

+ +=


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1 3/2113/2113/2113/2113/21

50. Answer (2)

Red2

Yellow2

Green2

Whitex

Required probability =(2)(1) (2)(1) (2)(1) ( 1)

(6 )(5 )

x x

x x

+ + + −

+ +

⇒15

=2

2

6

11 30

x x

x x

− +

+ +

⇒ 4x 2 – 16x = 0

⇒ 4x (x – 4) = 0

⇒ x = 4 (Q x > 0)

∴21( 7)

3x − = 3

∴ No. of positive divisor of 3 is 2

51. Answer (5)

Equation of six face’s of parallelopiped are x = –1, y = 3, y = 2, y = 1, z = 7, z = 4

∴ Length of diagonal of parallelopiped = 16 1 9+ +

⇒ k = 26

⇒2 1

55

k −=

52. Answer (2)l + m + n = 0, 2lm + 3mn + nl = 0

m (2l + 3n ) – nl = 0

2l 2 + 3n 2 + 6nl = 0

2

2 6 3 0l l

n n

+ + =

... (i)

From (i),

3 32l n − += or3 3

2l

n

+= −

Similarly,

1 32

m

n

−= or

1 32

m

n

+=

∴ 23 3 1 3

l m n = =

− − 2(3 3) 1 3

l m n = =

− + +

∴ d ′ ratio of straight lines are 3 3,1 3,2< − − > and (3 3),1 3,2< − + + >

If θ is an acute angle between two straight lines


14/21


1 4/2114/2114/2114/2114/21

∴2

cos( )13

θ =

∴3 1

tan tan2

λ + θ = =

λ

⇒ 2λ =

53. Answer (3)

ABC is equilateral triangle such |AB | = |BC | = |AC | = l A

B C

O

1 2 0 ° θ

P

Consider Circumcentre O as origin

Point P lies on the circumference of circle

Let OA OB OC r = = =uuur uuur uuur

(r = radius of circle)

2 2 2 2( 1)PA PB PC l + + = λ +uuur uuur uuur

2 2 2 2( 1)OA OP OB OP OC OP l − + − + − = λ +uuur uuur uuur uuur uuur uuur

⇒ ( )2 26 2 ( 1)r OA OP OB OP OC OP l − ⋅ + ⋅ + ⋅ = λ +uuur uuur uuur uuur uuur uuur

⇒

226 – 0 ( 1)

3

l l

= λ +

⇒ λ +1 = 2

⇒ λ = 1

∴ 10OQ =

21 1 10+ + µ =

∴2 1

33

µ +=

[Here 2 2 2cos( ) cos(120 ) cos(240 )OA OP OB OP OC OP r r r ⋅ + ⋅ + ⋅ = θ + + θ + + θuuur uuur uuur uuur uuur uuur

( )2 cos cos(120 ) cos(240 )r = θ + + θ + + θ

= 0]

54. Answer (2)

w r

= ( ) ( ) ( ) ( ) ( ) ( )a b r c b c r a c a r b × × × + + × × + × × ×r r r

r r r r r r r r r

= 3[ , , ] [ , , ] [ , , ] [ , , ]a b c r r a b c r b c a r c a b − + +r r r r

r r r r r r r r r r r r

= 3[ , , ] [ , , ]a b c r a b c r −r r

r r r r r r

w r

= 2[ ]a b c r r

r r r

w r

=2 r λ

r

(λ = [a b c ] given)


15/21


1 5/2115/2115/2115/2115/21

Note that: , ,a b c r

r r

are non-coplanar vectors and r r

is any arbitrary vector

We can write

r xa yb zc = + +r

r r r

... (i)

( ) ( )r a y b a z c a × = × + ×

r r r r r r

∴[ , , ]

( ) [ ][ ]

r a b r a b z a b c z

a b c × . = ⇒ =

r r r

r r r r r r

r r r

Similarly[ , , ]

[ ]

r b c x

a b c =

r r r

r

r r and

[ , , ]

[ ]

r c a y

a b c =

r r r

r r r

Put in (i)

[ , , ] [ , , ] [ , , ]

[ ] [ ] [ ]

r b c r c a r a b r a b c

a b c a b c a b c = + +

r r r r r r r r r

r r r r

r r r r r r r r r

55. Answer (1)Given

r

b andr

c are orthogonal unit vectors

0,b c b c a ⋅ = × =r r

r r r

⇒ sin(90 )a b c b c = × = °r

r r

[ ]a b c a b b c a b c + + + + = r r r r

r r r r r r

= 1 {Q , anda b c r

r r are mutually perpendicular unit vectors}

56. Answer (3)

Required probability = 3 14. .27.6.5

C

p

q =

435

∴ p = 4 , q = 35 {∴ p, q have no common factor except ± 1}

∴ q – 8p = 3

PART - III (PHYSICS)

57. Answer (B)

z y

x 2 ms

–1

2 ms –1

2 ms –1

I

M

N

2 ms –1

2 ms –1

2 ms –1

O

v yrel = 0, v zrel = 0, ˆ4= −xrel v i


16/21


1 6/2116/2116/2116/2116/21

58. Answer (D)

M ′ M

N ′N

P I O O ′

20 cm21 cm

The incident rays upon concave mirror should be normally.

the shift1 2

1 13 3

′ = − = − = µ

t OO t t

21 21 20 3 cm3 3

t t

PO R t ′ = − = ⇒ − = ⇒ =

59. Answer (A)

of water 4of air 3

µ= = µ =

µa w

x

y

60. Answer (C)

0( ) ( ) air0

2 / 3

=

µ =

= = = = µ +

∫ ∫ ∫ ∫

x l

x a x x

dx dx dx x t dt

v v v l

20 0

air air0

723 2 6

x l

x

l x x v l v

=

=

µ µ= + =

61. Answer (B)

0

t N e N

−λ=

N = N 0e –λt

1 1

10

11 1 1 1 1t

N e e e e p

N e

−λ −λτ −λ⋅ −λ∆ = − = − = − = − = = −

1/22 −λ= t p e

2log 1 1(2)

2⋅−λ −

λ= = =e

e

∴ 1

2

12 1

p

p e

= −

62. Answer (C)

( ) 1 −λ = − λ ⇒ = + ⇒ = − − λ λ∫ ∫

t t

dN dN AA N dt c N e

dt A N

2 log 220050 [1 ] sec.2 2

t e

e t −

= − ⇒ =


17/21


1 7/2117/2117/2117/2117/21

63. Answer (D)

0 0 0 0 12

12 2 2 2 4

µ µ µ µ= = ⋅ = ⋅ = ⋅

π πH

H

I e e v e v B

r r T r r r

[ ] [ ]

2

1 1 1 1

3

[ ] 32 2 2H H H H e e

M I r v r v r e v r = × π = × = × = ×

30

1

16 H

B

M r

µ =

π

64. Answer (B)

Saturation current depends upon intensity of incident light whereas cut off voltage depends upon frequency ofincident light

(h ν = W + e ⋅ Vc)


Wavelength and cut off voltage will not change

where as intensity2

1∝

r

66. Answer (A, D)

Final image will be at ∞ when

f eff = 30 cm,

1 3 21 20 cm

2R

f R

= − × ⇒ =

1 1 1 30 cm20 60 eff eff f

f = − ⇒ =

1 3/2 21 30 cm

9/8 20f

f

= − × ⇒ =

67. Answer (A, B, D)

1ˆ ˆu ai bj = +

r

2ˆ ˆu ci dj = +

r

1 2| | | | 1u u = =r r

∴ a 2 + b 2 = c 2 + d 2

Snell’s law µ1sini = µ2sinr

2 2 2 21.5 2

a c

a b c d × = ×

+ +

4,

3a

c =

Now, a 2 + b 2 = 1, c 2 + d 2 = 1

∴ 16d 2 – 9b 2 = 7


18/21


1 8/2118/2118/2118/2118/21

68. Answer (A, B, C, D)

Since z is same

Hence λ1 = λ2 = λ3


4.25 eV = W A + T A …(i)

1

2A

AmT λ = …(ii)

4.70 eV = W B + T B …(iii)

T B = T A – 1.50 eV …(iv)

1

2B

B mT λ = …(v)

λB = 2λA …(vi)

Solving T A = 2.00 eV, W A = 2.25 eV, W B = 4.20 eV

70. Answer (D)

Using relation2

x π

∆φ = × ∆λ

1 2

23π

∆φ =S S

1 3

83π

∆φ =S S

2 3

8 2 23 3π π

∆φ = − = πS S

71. Answer (A)

Let R be the amplitude of each wave at point P .2R

120°

R

R ′2 = R 2 + (2R )2 + 2 × R + 2R ×cos2

3π

R ′2 = 3R 2

I P = 3I

72. Answer (A)

73. Answer (A)

Solution of Q. 72 & 73

The total angular momentum is µωr 2, where µ is reduced mass, ω is angular velocity of the nucleus as well asthe charged particle about the common centre of mass.

⇒ µωr 2 =2nh

π

Also,( ) ( ) 2

2

k Ze e r

r = µω

⇒ r =2 2

2 24

n h

kZe π µ


19/21


1 9/2119/2119/2119/2119/21

or r = 0 e r m

z µ

Here, µ =Mm

m m + =

5 1840 105 1840 10

e e

e e

m m

m m

× ×

× +

µ =9200921 e

m

orbital radius of charged particle

r ′ =M

r M m

+ =

r

m

µ

r ′ =10 e m

µ × 0 e

r m

Z µ

=0 0.52

10 10 2

r

Z = ×

⇒ r ′ = 0.026 Å

Also, r =0 e r m

Z µ =

0.529200

2921

e

e

m

m

×

×

Now, r 1 = (2)2 r = 4r

74. Answer (B)

2

213.6 eVn

e

Z E

m n

µ= × ×

4 920013.6 eV

4 921

= ×

75. Answer (2)

2 2 21 | |2 2

dU nh U kr F mv kr mvr

dr = ⇒ = ⇒ = ⇒ =

π

12 2 22 2

2 2( )

4

mv m n h r r n

k k mr = = × ⇒ ∝

π

76. Answer (4)

1 cm

2 cmC O ′ P

I

30

O

I ′

20 cm

10 cm 30 cm

From the ray diagram, it is clear that R = 40 cm ⇒ f = 20 cm

⇒ 5f

= 4


20/21


20/2120/2120/2120/2120/21

77. Answer (3)

PHPH

,E/ce

h h h h

pe p p λ = = λ = =

9

PH 8

c 1 13.33 10

E E 3 10e h h p

p c

−λ = λ ⇒ = ⇒ = = = ××

78. Answer (3)

11

1 1 5E 13.6 13.6 eV

4 9 36hc ′

= − = × = λ

22

1 3E 13.6 1 13.6 eV

4 4hc ′

= − = × = λ

1 1

2 2

E 5 27 1,

E 27 5λ

= = = = =λ

c a c

31 1

2 2p 5b 3p 27

× = = = ⇒ = ⇒ =

E b c c d E a

79. Answer (6)

f 0 = 50 cm, f e = 5 cm, D = 25 cm = v e , u 0 = 200 cm

For objective,0

1 1 1 1 1 1,

200 50− = − =

−v u f v

0200

cm3

v =

00

0

200/3 1

200 3

v

m x u = = = − =−

For eyepiece,1 1 1 1 1 1 25

cm25 5 6

−− = ⇒ − = ⇒ =

−e

e

u v u f u

25 |X| 1 16

25/6 | | 3 6 3−

= = = = ⇒ = =− × ×

e e

e

v m Y

u Y N

80. Answer (3)

22 1

1,

2= ∆ = − = +

λ λ

h hc MV E E E MV

∆∆ ≅ ⇒ = =λ λhc h E E MV

c

1927 1

8

10.2 1.6 105.44 10 kg ms

3 10

−− −× ×= = ×

×

81. Answer (5)

2

1 2 1 1 1

1 1 1 1 1 1 3,

150 3/2 2= + ⇒ = + = −

−

F

F F F F F F

F 1 = 50 cm, F 2 = 75 cm

|F 1| ~ |F 2| = 25 cm


21/21


21/2121/212 1/2121/2121/21

82. Answer (5)2

222

2

1sin

4( – 1)

4

= = ⇒ =µ µ

+

r a c r

a r a /2

O

r

c

222

6Area painted 64[ 1]

π= π =µ −

a r

2

2 2

6Fraction

56 4[ 1]

r

a

π π π= = =

µ −

83. Answer (2)

At mid-point x = I 0 = 4I ′ (I 1 = I 2 = I ′)

At point P ,

2 2 21 24 cos 4 cos tan 4 cos .2 2P

d I I I d I y

D

φ π π ′= ′ = ′ × × θ = λ λ

2 04 cos2

I y x I x

x

π ′ = ′ = =

2x

x ′ =

84. Answer (1)

7 7Required distance

2 2D

x d

λ= =

7 (0.5 m)(5000 Å)1 mm

2 (0.875 mm)= × =

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Aiats Iitjee 2011 Test-8 P-i Solution