aiats_jeeadvanced_test-6 (1)

8/9/2019 aiats_jeeadvanced_test-6 (1)

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Test - 6_Code - C_(Paper - II) (Answers & Hints) All India Aakash Test Series for JEE (Advanced)-2015

1/7

PHYSICS

1. (B)

2. (B)

3. (C)

4. (B)

5. (B)

6. (D)

7. (D)

8. (B, D)

9. (A, B, D)

10. (C, D)

11. (A, D)

12. (D)

13. (A)

14. (B)

15. (C)

16. (A)

17. (A)

18. (6)

19. (4)

20. (3)

CHEMISTRY

21. (C)

22. (C)

23. (B)

24. (C)

25. (B)

26. (C)

27. (A, B, C, D)

28. (A, B, C, D)

29. (A, B)

30. (B, D)

31. (A, C)

32. (C)

33. (C)

34. (C)

35. (A)

36. (C)

37. (B)

38. (6)

39. (5)

40. (4)

MATHEMATICS

41. (B)

42. (B)

43. (A)

44. (C)

45. (A)

46. (B)

47. (A, B, C, D)

48. (A, C, D)

49. (A, C)

50. (A, C)

51. (C, D)

52. (A)

53. (D)

54. (A)

55. (B)

56. (B)

57. (D)

58. (2)

59. (4)

60. (9)

ANSWERS

TEST - 6 (Paper-II) - Code-C

All India Aakash Test Series for JEE (Advanced)-2015

Test Date : 01-02-2015

Click here for Code-D Solution


2/14

All India Aakash Test Seriesfor JEE (Advanced)-2015 Test - 6_Code - C_(Paper - II) (Answers & Hints)

2/7

1. Answer (B)

2. Answer (B)

3. Answer (C)

r

r

r(1 cos) =x ...(i)

rsin= y ...(ii)

From (i) & (ii)

Momentum =

2

2

qB yx

x

4. Answer (B)

5. Answer (B)

F= eE= 2

r dB

edt

6. Answer (D)

At = 0

z=

i= 0

7. Answer (D)

8. Answer (B, D)

9. Answer (A, B, D)

10. Answer (C, D)

11. Answer (A, D)

qvBl

C i= avBl

mg ilB= ma

mg avl2B2= ma

2 2 constant

( )

mga

m vl B

12. Answer (D)

PART - I (PHYSICS)

ANSWERS & HINTS

13. Answer (A)

21

2qEr mv (i)

2qEr

vm

2

2 1

B E

mvN F F N

r (ii)

2

2 1 0

E B

mvN N F F

r (iii) [Given]

2

2

9

qB rEm

14. Answer (B)

i= 0.5 Wb

f= 2 Wb

= 1.5 Wb

15. Answer (C)

16. Answer (A)

17. Answer (A)

18. Answer (6)

When iis max V2C

= V3C

Common potential =1 1 2 2

1 2

C V C V

C C

=0 0

73 2

2

3 2

C V C V

C C

= 2V0

2 20 01

(3 ) 4 62

U C V CV

19. Answer (4)

qvBl

C (i)

i

aBlC

(ii)

Now, kx ilB= ma (iii)

kx CaB2l2= ma


3/14


3/7

kx= (CB2l2+ m)a

2 2

ka x

CB l m

2 2

2 m B l C

T

k

Time =4

T

21. Answer (C)

w

h

a

KK

K

Here Ka is the ionization constant of parent acid and

H

is the weakest acid so, its Kh is

highest.

22. Answer (C)

Nitro group stabilizes phenoxide ion formed and

steric inhibition of resonance of NO2stop stablizationof phenoxide ion by resonance. So, p-nitrophenol is

most acidic.

23. Answer (B)

Bulky groups at equitorial positions brings most

stable conformer.

24. Answer (C)

is crystalizes as ionic solid by keeping

negative terminal at five membered ring and positive

terminal on seven membered ring in order to maintain

aromaticity.

25. Answer (B)

O

O

O HH O

OH

OHH

IIIII

I

IV

20. Answer (3)

B= 012 cos 30 cos 604 3

2

i

l

B= 0 3 3 1il

PART - II (CHEMISTRY)

From I to II to III it is a clockwise movement hence,

configuration at Cis R.

O

O

O HH O

OH

OHH

III

I

II

*

H

From I to II to III it is a anticlockwise movement

hence, configuration at C* is S.

26. Answer (C)

Molecule as per Newman projection is

H

CH3

H

H3C

Cis(1, 4) dimethylcyclohexane

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

2HCl D O HOD DCl

In solution available electrophiles are H+and D+. So,

probable products are

OD

H

,

OD

D

,

OD

H

,

OD

D



4/14


4/7

29. Answer (A, B)

Claisen rearrangement mechanism is completely

concerted hence, no crossover product formation is

observed.

30. Answer (B, D)

LiAlH4is not able to reduce unconjugated C C .

31. Answer (A, C)

In case of t-butylbenzene because of the stability of

, electrophile NO2substitute t-butyl group.

32. Answer (C)

33. Answer (C)

Basic strength of amines is also dependent on steric

hinderence of the group bonded to N and hydration.

34. Answer (C)

O H O

H H

Cl Cl

D D

NH2

OH

D

35. Answer (A)

36. Answer (C)

(P)

OO O

ClHO

(i) LiAlH4

(ii) H O2

OH

OHHO

DBE = 2

(Q)

OO

HO

C NNaBH

4

H O2

OHO

HO

C N

DBE = 5

(R) Cold alk.

KMnO4

OH

OH

OHOH

OH

HO

DBE = 1

(S) mCPBA

O

O

O

DBE = 4

37. Answer (B)

In aqueous medium down the group nucleophilicity

increases, across the period nucleophilicity

increases. Anions are better nucleophile than neutral

ones and nucleophilicity decreases as steric

hinderence increases.

38. Answer (6)

Cl

H H

Cl

H

Cl

H

Cl

H

Cl

Cl

HCl

Cl Cl

H

Cl H

Cl

H Cl

H

Six sterioisomers are formed.

39. Answer (5)

(1), (3), (5), (7) and (9) show Walden inversion in SN2.

40. Answer (4)

Hydrogen from phenolic group, COOH, SH and

C

NH

C

OO

are removed from NaOH.


5/14


5/7

41. Answer (B)

Length of sub-normal = length of sub-tangent

1 dy

dx

If 1dy

dx, equation of the tangent y 4 =x 3

yx= 1, area of OAB=1 1

1 12 2

If 1dy

dx, equation of the tangent is y 4 = x+ 3

y+x= 7, area =

1 497 7

2 2

42. Answer (B)

v=x3and the percent error in measuring edge =

100dx

Kx

The percent error in measuring volume = 100dv

v

Now, 23dv

xdx

dv= 3x2dx

2

3

3 3

dv x dx dx

v xx

3

100 100 3 dv dx

Kv x

43. Answer (A)

f(x) = (x 2)(x 3) x3

= (x 2)(x 3) x< 3

Graph of f(x) isy

0 2 3 x

5

2

Clearly f(x) is increasing in 5

, 3,2

44. Answer (C)

log 1 1

x x

x x

x

e e

I e e dx e

=

log 11

x

x x

x

ee e dx

e

= log 1 log 1 x x xe e e C

= exlog(ex+ 1) log(1 + ex) +x+ C

= 1 log 1x xe e x C

45. Answer (A)

5

4

3 44

2 cos 4

1

x

x dx

I

e

Put 4

x t

2 cos

1

t

tI

e

2cos( )

1

tt

I

e

Adding

2 2 cos 0

I t I= 0

46. Answer (B)

0 x< 1

y= [x+ 1] = 1

So required area =

1

0

ydy

=

13

2

0

2

3

y

= 2 sq. unit3

PART - III (MATHEMATICS)


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6/7


Here dy y

dx a

Sub-tangent = constant

y

ady

dx

Sub-normal =2dy y y

y ydx a a

Equation of tangent at (x1, y

1) is

11 1 ( )

yy y x x

a

This meets thex-axis at a point given by

1

1 1 [ ]

y

y x xa

x=x1 a

The curve meets the y-axis at (0, C)

(0, )

c

dy c

dx a

Equation of normal 1

( 0)y c xc

a

ax+ cy= c2

48. Answer (A, C, D)

Graph of f(x)

y

21x

When is evenn

y

21x

When is oddn

49. Answer (A, C)

2

3

2 2

1

1

xx x

I e dx

x

=

2

3 3

22 2 2

1

1( 1)

x x x

e dx

xx

=32

2 2

1

1( 1)

x x

e dx

xx

= ( ) ( ) x

e f x f x dx where2

1( )

1

f x

x

2

( )

1

x

x ee f x c c

x

f(x) =2

1

1x

50. Answer (A, C)

Let f(x) = 33 x

Clearly f(x) is increasing in [1, 3]

The least value of function m= f(1) = 33 1 2

& maximum value M= f(3) = 33 3 30

Therefore,

3

2

1

(3 1) 2 3 (3 1) 30x dx

So

2

3

1

4 3 2 30x dx

51. Answer (C, D)

Since the curve

1

2

y ax bx passes through thepoint (1, 2)

2 = a+ b ...(i)

By observation the curve also passes through (0, 0)

14

2

0

8A ax bx dx

2

8 16 83 2

a b

2

13

ab ...(ii)

Solving (i) & (ii) we get a= 3, b= 1


7/14


7/7

52. Answer (A)

h(x) = f(x) a((f(x))2+ a(f(x))3)

h(x) = f(x) 2a f(x) f(x) + 3a(f(x))2f(x)

h(x) = f(x) [3a[f(x)]2 2af(x) + 1]

Now h(x) increases if f(x) increases and

3a(f(x))2 2af(x) + 1 > 0 x R

3a> 0 & 4a2 12a0

a> 0 & a[0, 3]

a[0, 3]

53. Answer (D)

h(x) is non-monotonic functions if

3a(f(x))2 2a f(x) + 1 changes sign for which D0 or

4a2 12a0

a(, 0] [3, )54. Answer (A)

Put 1 +x= t6

To get3

61

t

I dtt

=3

1 16

1

t

dtt

=2 1

6 1 1

t t dt t

= 1 11 1

6 62 32 1 3 1 6(1 ) 6ln (1 ) 1x x x x c

55. Answer (B)

Let 2 2

2

xt

x

2

2 1

2

x

x t

by C & D

2

2

12

1

tx

t

2 2

2 2 22

(1 )2 (1 )2 82

(1 ) 1

t t t t t dx dt dt

t t

2

2 2 2

1 1 8

2 1 (1 )

t tt dt

t t

=

2

2 24

1 1

tdt

t t

2 22

1 1

dt dt

t t

11log 2 tan1

tt C

t

where

2

2

xt

x

56. Answer (B)

57. Answer (D)

58. Answer (2)

1

0

(2 )I xf x dx , Put 2x= t

2

dtdx

So

2

0

1( )

4I tf t dt

Apply by parts to solve the integral.

59. Answer (4)

x1

22 2 32 2

00 0

8 4(2 ) 4 4

3 3 3

yA xdy y y dy

60. Answer (9)

Differentiating the curve we get

6 8 0dy

x ydx

( , )

3 3

4 2a b

dy a

dx b

2a

b ...(i)

Also (a, b) lies on the curve

So

3a2 4b2= 72 ...(ii)

Solving (i) & (ii) we get a= 6

& b= 3

Out of (6, 3) & (6, 3), (6, 3) is closer to line

So a= 6 & b= 3

b a= 9


8/14

Test - 6_Code - D_(Paper - II) (Answers & Hints) All India Aakash Test Series for JEE (Advanced)-2015

1/7

PHYSICS

1. (D)

2. (B)

3. (B)

4. (C)

5. (B)

6. (B)

7. (A, D)

8. (C, D)

9. (A, B, D)

10. (B, D)

11. (D)

12. (B)

13. (C)

14. (D)

15. (A)

16. (A)

17. (A)

18. (3)

19. (4)

20. (6)

CHEMISTRY

21. (C)

22. (B)

23. (C)

24. (B)

25. (C)

26. (C)

27. (A, C)

28. (B, D)

29. (A, B)

30. (A, B, C, D)

31. (A, B, C, D)

32. (C)

33. (A)

34. (C)

35. (C)

36. (B)

37. (C)

38. (4)

39. (5)

40. (6)

MATHEMATICS

41. (B)

42. (A)

43. (C)

44. (A)

45. (B)

46. (B)

47. (C, D)

48. (A, C)

49. (A, C)

50. (A, C, D)

51. (A, B, C, D)

52. (A)

53. (B)

54. (A)

55. (D)

56. (D)

57. (B)

58. (9)

59. (4)

60. (2)

ANSWERS

TEST - 6 (Paper-II) - Code-D

All India Aakash Test Series for JEE (Advanced)-2015

Test Date : 01-02-2015

Click here Code-C Solution


9/14

All India Aakash Test Seriesfor JEE (Advanced)-2015 Test - 6_Code - D_(Paper - II) (Answers & Hints)

2/7

1. Answer (D)

At = 0

z=

i= 0

2. Answer (B)

F= eE=2

r dBe

dt

3. Answer (B)

4. Answer (C)

r

r

r(1 cos) =x ...(i)

rsin= y ...(ii)

From (i) & (ii)

Momentum =

2

2

qB yxx

5. Answer (B)

6. Answer (B)

7. Answer (A, D)

qvBl

C i= avBl

mg ilB= ma

mg avl2B2= ma

2 2 constant

( )mga

m vl B

8. Answer (C, D)

9. Answer (A, B, D)

10. Answer (B, D)

11. Answer (D)

12. Answer (B)

i= 0.5 Wb

f= 2 Wb

= 1.5 Wb

PART - I (PHYSICS)

ANSWERS & HINTS

13. Answer (C)

14. Answer (D)

15. Answer (A)

21

2qEr mv (i)

2qEr

vm

2

2 1

B E

mvN F F N

r (ii)

2

2 1 0

E BmvN N F F

r (iii) [Given]

22

9

qB rE

m

16. Answer (A)

17. Answer (A)

18. Answer (3)

B= 0

12 cos 30 cos 604 3

2

i

l

B= 0 3 3 1il

19. Answer (4)

qvBl

C (i)

i

aBl

C

(ii)

Now, kx ilB= ma (iii)

kx CaB2l2= ma

kx= (CB2l2+ m)a

2 2

ka x

CB l m

2 2

2 m B l C

Tk

Time =4

T


10/14


3/7

20. Answer (6)

When iis max V2C

= V3C

Common potential =1 1 2 2

1 2

C V C V

C C

21. Answer (C)

Molecule as per Newman projection is

H

CH3

H

H3C

Cis(1, 4) dimethylcyclohexane

22. Answer (B)

O

O

O HH O

OH

OHH

IIIII

I

IV

From I to II to III it is a clockwise movement hence,

configuration at Cis R.

O

O

O HH O

OH

OHH

III

I

II

*

H

From I to II to III it is a anticlockwise movement

hence, configuration at C* is S.

23. Answer (C)

is crystalizes as ionic solid by keeping

negative terminal at five membered ring and positive

terminal on seven membered ring in order to maintain

aromaticity.

=0 0

73 2

2

3 2

C V C V

C C

= 2V0

2 20 01 (3 ) 4 62U C V CV

PART - II (CHEMISTRY)

24. Answer (B)

Bulky groups at equitorial positions brings most

stable conformer.

25. Answer (C)

Nitro group stabilizes phenoxide ion formed andsteric inhibition of resonance of NO

2stop stablization

of phenoxide ion by resonance. So, p-nitrophenol is

most acidic.

26. Answer (C)

w

h

a

KK

K

Here Ka is the ionization constant of parent acid and

H

is the weakest acid so, its Kh is

highest.

27. Answer (A, C)

In case of t-butylbenzene because of the stability of

, electrophile NO2substitute t-butyl group.

28. Answer (B, D)

LiAlH4is not able to reduce unconjugated C C .

29. Answer (A, B)

Claisen rearrangement mechanism is completely

concerted hence, no crossover product formation is

observed.



11/14


4/7


2HCl D O HOD DCl

In solution available electrophiles are H+and D+. So,

probable products are

OD

H

,

OD

D

,

OD

H

,

OD

D

32. Answer (C)

O H O

H H

Cl Cl

D D

NH2

OH

D

33. Answer (A)

34. Answer (C)

35. Answer (C)

Basic strength of amines is also dependent on steric

hinderence of the group bonded to N and hydration.

36. Answer (B)

In aqueous medium down the group nucleophilicity

increases, across the period nucleophilicity

increases. Anions are better nucleophile than neutral

ones and nucleophilicity decreases as steric

hinderence increases.

37. Answer (C)

(P)

OO O

ClHO

(i) LiAlH4

(ii) H O2

OH

OHHO

DBE = 2

(Q)

OO

HO

C NNaBH

4

H O2

OHO

HO

C N

DBE = 5

(R) Cold alk.

KMnO4

OH

OH

OHOH

OH

HO

DBE = 1

(S) mCPBA

O

O

O

DBE = 4

38. Answer (4)

Hydrogen from phenolic group, COOH, SH and

C

NH

C

OO

are removed from NaOH.

39. Answer (5)

(1), (3), (5), (7) and (9) show Walden inversion in SN2.

40. Answer (6)

Cl

H H

Cl

H

Cl

H

Cl

H

Cl

Cl

HCl

Cl Cl

H

Cl H

Cl

H Cl

H

Six sterioisomers are formed.


12/14


5/7

41. Answer (B)

0 x< 1

y= [x+ 1] = 1

So required area =

1

0

ydy

=

13

2

0

2

3

y

=2

sq. unit3

42. Answer (A)

5

4

3 44

2 cos 4

1

x

x dx

I

e

Put 4

x t

2 cos

1

tt

I

e

2cos( )

1

tt

I

e

Adding

2 2 cos 0

I t I= 0

43. Answer (C)

log 11

x xx x

x

e eI e e dx

e

=

log 11

x

x x

x

ee e dx

e

= log 1 log 1 x x xe e e C

= exlog(ex+ 1) log(1 + ex) +x+ C

= 1 log 1x xe e x C

44. Answer (A)

f(x) = (x 2)(x 3) x3

= (x 2)(x 3) x< 3

Graph of f(x) is

y

0 2 3 x

5

2

Clearly f(x) is increasing in 5 , 3,2

45. Answer (B)

v=x3and the percent error in measuring edge =

100dx

Kx

The percent error in measuring volume = 100dv

v

Now,

23

dvx

dx

dv= 3x2dx

2

3

3 3

dv x dx dx

v xx

3

100 100 3 dv dx

Kv x

46. Answer (B)

Length of sub-normal = length of sub-tangent

1 dy

dx

If 1dy

dx, equation of the tangent y 4 =x 3

yx= 1, area of OAB=1 1

1 12 2

If 1dy

dx, equation of the tangent is y 4 = x+ 3

y+x= 7, area =

1 497 7

2 2

PART - III (MATHEMATICS)


13/14


6/7

47. Answer (C, D)

Since the curve

1

2 y ax bx passes through the

point (1, 2)

2 = a+ b ...(i)By observation the curve also passes through (0, 0)

14

2

0

8A ax bx dx

2

8 16 83 2

a b

2

13

ab ...(ii)

Solving (i) & (ii) we get a= 3, b= 1

48. Answer (A, C)

Let f(x) = 33 x

Clearly f(x) is increasing in [1, 3]

The least value of function m= f(1) = 33 1 2

& maximum value M= f(3) =3

3 3 30

Therefore,

3

2

1

(3 1) 2 3 (3 1) 30x dx

So

2

3

1

4 3 2 30x dx

49. Answer (A, C)

2

3

2 2

1

1

xx x

I e dx

x

=

2

3 3

22 2 2

1

1( 1)

x x x

e dx

xx

=32

2 2

1

1( 1)

x x

e dx

xx

= ( ) ( ) x

e f x f x dx where2

1( )

1

f x

x

2 ( )

1

x

x ee f x c c

x f(x) = 2

1

1x

50. Answer (A, C, D)

Graph of f(x)

y

21x

When is evenn

y

21x

When is oddn


Here dy y

dx a

Sub-tangent = constanty

ady

dx

Sub-normal =2dy y y

y ydx a a

Equation of tangent at (x1, y

1) is

11 1 ( )

yy y x x

a

This meets thex-axis at a point given by

1

1 1 [ ]

y

y x xa

x=x1 a

The curve meets the y-axis at (0, C)

(0, )

c

dy c

dx a

Equation of normal 1

( 0)y c xc

a

ax+ cy= c2


14/14


7/7

52. Answer (A)

Put 1 +x= t6

To get3

61

t

I dtt

=3 1 1

6 1

t

dtt

=2 1

6 1 1

t t dt t

= 1 1

1 1

6 62 32 1 3 1 6(1 ) 6ln (1 ) 1x x x x c

53. Answer (B)

Let 2 2

2

xt

x

2

2 1

2

x

x t

by C & D

2

2

12

1

tx

t

2 2

2 2 22

(1 )2 (1 )2 82

(1 ) 1

t t t t t dx dt dt

t t

2

2 2 2

1 1 8

2 1 (1 )

t tt dt

t t

=

2

2 24

1 1

tdt

t t

2 22 1 1

dt dt

t t

11log 2 tan

1

tt C

t

where

2

2

xt

x

54. Answer (A)

h(x) = f(x) a((f(x))2+ a(f(x))3)

h(x) = f(x) 2a f(x) f(x) + 3a(f(x))2f(x)

h(x) = f(x) [3a[f(x)]2 2af(x) + 1]

Now h(x) increases if f(x) increases and

3a(f(x))

2

2af(x) + 1 > 0x R

3a> 0 & 4a2 12a0

a> 0 & a[0, 3]

a[0, 3]

55. Answer (D)

h(x) is non-monotonic functions if

3a(f(x))2 2a f(x) + 1 changes sign for which D0 or

4a2 12a0

a(, 0] [3, )

56. Answer (D)

57. Answer (B)

58. Answer (9)

Differentiating the curve we get

6 8 0dy

x ydx

( , )

3 3

4 2a b

dy a

dx b

2a

b ...(i)

Also (a, b) lies on the curve

So

3a2

4b2

= 72 ...(ii)Solving (i) & (ii) we get a= 6

& b= 3

Out of (6, 3) & (6, 3), (6, 3) is closer to line

So a= 6 & b= 3

b a= 9

59. Answer (4)

x1

22 2 32 2

00 0

8 4(2 ) 4 4

3 3 3

yA xdy y y dy

60. Answer (2)

1

0

(2 )I xf x dx , Put 2x= t

2

dtdx

So

2

0

1( )

4I tf t dt

Apply by parts to solve the integral.

aiats_jeeadvanced_test-6 (1)

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