Aiats Jee Adv-solution

7/28/2019 Aiats Jee Adv-solution

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Test - 3 (Paper - II) (Answers & Hints) All India Aakash Test Series for JEE (Advanced)-2013

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TEST - 3 (Paper - II)

1. (A)

2. (D)

3. (A)

4. (D)

5. (A)

6. (B)

7. (A)

8. (B)

9. (A, B)

10. (A, B, C, D)

11. (A, D)

12. (A, B, C, D)

13. (B, C, D)

14. (A)

15. (A)

16. (B)

17. (D)

18. (6)

19. (2)

20. (1)

PHYSICS CHEMISTRY MATHEMATICS

21. (D)

22. (D)

23. (B)

24. (Deleted)

25. (B)

26. (A)

27. (D)

28. (B)

29. (B, D)

30. (B, C)

31. (C, D)

32. (C, D)

33. (B, C, D)

34. (C)

35. (C)

36. (A)

37. (A)

38. (7)

39. (6)

40. (2)

41. (B)

42. (B)

43. (B)

44. (A)

45. (A)

46. (B)

47. (B)

48. (A)

49. (A, D)

50. (A, B, C, D)

51. (A, B)

52. (A, B, D)

53. (A, C)

54. (B)

55. (D)

56. (C)

57. (D)

58. (3)

59. (6)

60. (1)

ANSWERS


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All India Aakash Test Series for JEE (Advanced)-2013 Test - 3 (Paper - II) (Answers & Hints)

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1. Answer (A)

mg+ FST = B

mg+ 4aS = a2hgh

FST mg

B

Now, m = a2

4a s

hga

2. Answer (D)

3. Answer (A)

2dv

F m av dt

2 dvm a dt v

RM nR

Integrating, we get

20

GH

tR

GM

nR

dvm a dt

v

4. Answer (D)

Mass =3 34 ( )

3 b a

Work done by gravity, while taking a unit mass =

b b

a ag dr gdr

Consider a point at a distance rfrom centre.

g(r) = 3 3

2

4

3

r aG

r

W=

3 3

2

4

3

b

a

G r adr

r

In units of 4G,

2 3 33 2

6

a b b aW

b

Wext =2 3 32 2

6

a b b a

b

5. Answer (A)

In equilibrium vertical line passes through the centre of mass.

Let O be the COMA

B

C

Dl/2 O

P

2l/3

In AOP

32tan2 4

3

l

l

PART - I (PHYSICS)

ANSWERS & HINTS


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6. Answer (B)

The acceleration and velocity of centre Cis given as

Acceleration of point Pw.r.t. surface isC

( )r R Cv

Ca

Ta( )r R

( )r R 2

Ta( )r R

Ca( )r R 2

R

r

22

( )Ca r R r

Net acceleration = C Ca a

= 2( )

R R r

r

= 3R2

7. Answer (A)

34

3 v r

3 vdr

dvr

3

dv dr

v r

1

3

dr dv

r v

vdP

Bdv

dv dP Mg

v B aB

3

dr Mg

r aB

8. Answer (B)

There is no horizontal force, momentum is conserved.

mv1 mv2 = 0

For hinged point,

1 22

Lv v

22 2 21 2

1 1 1

2 2 2 12

mlmgl mv mv

v1

v2

L

m

13

5v gl

v1

= 10 m/s


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9. Answer (A, B)

1 2 2(4 ) 16 P

GM GM E

a a

2 2 2(6 ) 5 P

GM GM E

a a

= 261

900

GM

a

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

11. Answer (A, D)

2 3T r

32 44

22

1 104 10 km

8

r

r

2 2| |

v v =4 410 4 10

21 8

= 410 km/h

1 2( ) =2 1

2 1

v v

r r=

4

4

10

3 10

= 3

vrel = (v2 + v1)

=

4 410 10 42

1 8

= 3 104 km/h

rel =4

4

3 10

5 10

= 3 rad/h5


2

h

gr

PE =2( )

2 2

h hmg r h g

Work by surface tension = (2 ) r h =2

2

rgr

=2

4g


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Work done by gravity =2 21

2 r h g

=2

22 2 2

1 4

2

r g

r g

=22

g

Heat =2

2 24 1

2

r h g

g

=22

g

13. Answer (B, C, D)

10 1N 4500 N

0.002

F

Stress =6 2

4

45005.0 10 N/m

9 10

F

A

Board will break since stress of 5.0 106 N/m2 is more than breaking stress of 3.6 106 N/m2.

14. Answer(A)

In Case - 1, weight of liquid displaced Vg= (10 + 3)g

In Case - 2, weight of water displaced Vg= (10 + 4)g

13

14

specific gravity.

15. Answer(A)

In Case - 1, Vg= 3g ...(i)

In Case - 2, Vg= (W + 1)g ...(ii)

From (i) and (ii), m = 2 kg

In Case - 3, Vg+ 2 m

g g mg

1

2

= or 2

=

16. Answer(B)In reference frame of truck, angular momentum is conserved about an axis fixed to surface of truck

MvR = 22

5 MvR MR

= 22

5

vMvR MR

R

00

57=

5 7

vv v v

v0 is the speed in truck frame, in ground frame velocity is 0 0 05 2

.

7 7

v v v


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17. Answer (D)

The total work done by kinetic friction would be same in all frames. So, in truck frame20

1

7f f iw k k mv

18. Answer (6)

2

2 6

l mlT

mg T= ma

3 3( )

Tmg ma

ml ml ; 0

2

la

T

a

mg

l2 l

2

l

2

l2

2 3( )a g a

l l

5a = 3g

23 6 m/s5

ga

19. Answer (2)

22 sin ( )2

T dm r

22 ( )2

T Ar r

/2

/2

T

A

12 rad/s

r

20. Answer (1)

Taking moments about end point

2(6 ) 182 rg

x A A g

x= 1 m

PART - II (CHEMISTRY)

21. Answer (D)

H2SO4 + 2NaOH Na2SO4 + 2H2O

n = 2

H2C2O4 + 2NaOH Na2C2O4 + 2H2O

n = 2

Let number of millimoles of H2SO4 = a


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And number of millimoles of H2C2O4 = b

Eacids = Ebase

1000(a b) 2 3 0.1

10 (i)

In another experiment, KMnO4 reacts with H2C2O4 only

KMnO4 Mn+2

+7

n = 5C O2 4

2CO2

+3

n = 2

+4

2 2 4 4H C O KMnOE E

1000b 2 4 0.02 5

100 (ii)

Solving (i) and (ii)

b = 2, a = 13

Wt. of H2SO4 = 13 103

98 = 1.274 g

w% of H2SO4 =1.274

100 40%3.185

22. Answer (D)

Effective number of Na+ = 4

Effective number of remaining Cl1 1 13

4 28 2 4

Packing fractionVolume occupied by constituent ions

Total volume

3 3

3

13 4 4r 4 r

4 3 38(r r )

3 3

3

13 16r r

3 38(r r )

23. Answer (B)

H2

O2

+ 2I I2

2 2 2 2 3 4 6I 2S O S O 2I

2 2

1 1 2 2(H O ) (Hypo)

N V N V

2 2H O20 1

N 2 N10

24. Deleted

25. Answer (B)

3Zn2+ + 2K4[Fe(CN)6] K2Zn3[Fe(CN)6]2 + 6K+


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26. Answer (A)

H 2 22 4 2

n 8 n 5

Cu S KMnO Mn Cu SO

2 4Cu S KMnOE E

2 8 x 5

16

x 3.25

27. Answer (D)

3 3r a 5.16 2.23

4 4

28. Answer (B)

md

V (For solution)

m = d V = 1.18 1000 = 1180msolvent = msolution msolute

= 1180 180 = 1000

1m 1

1kg

Tf = Kf m = 1.86 1 = 1.86

Freezing point of solution = 0 1.86 = 1.86C

29. Answer (B, D)

Fact

30. Answer (B, C)

In fcc arrangement, corner atoms do not touch each other. The co-ordination number in fcc arrangement is 12.

31. Answer (C, D)

2 2 4 4BaCl Na SO BaSO 2NaCl

moles 0.04 0.05 0 0

0.01 0.08

0.01 0.08conc. M M

0.7 0.7

= (i1C1 + i2C2)RT

0.01 0.083 2 0.0821 300

0.7 0.7

= 6.685 atm

32. Answer (C, D)

In H2SO5 and H2S2O8, S has highest oxidation state i.e. + 6.

33. Answer (B, C, D)

KI1

ICl

n = 2f

C O2 42

CO2+3

n = 2f

+4+1

Fe O2 3+3

n = 2f

FeSO4+2


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34. Answer (C)

C

X

3rd nearest neighbour is3

XC a 1.22 a2

35. Answer (C)

The number of 1st nearest neighbour is 12.

The number of 2nd nearest neighbour is 6.

36. Answer (A)

o o

o o o o

A B A B B A

A B A B

y y y P y P1

P P P P P

o o

o o oA B

AA A B A

P P 5400P60 30yP y (P P )

oAP 60

oBP 90

37. Answer (A)

When yA = 0.6, P = 69.23

o

AA

A

Py (69.23)(0.6)X 0.6923

60P

38. Answer (7)

Species Oxidation state of underlined atom

S8 0

CrO5 +6O

O O

OCr

O

H2S2O8 +6H O S O O S O H

O O

O O

H2SO5 +6 H O O S O H

O

O

K4[Fe(CN)6] +2 (x + 6(1) = 4 x = +2)

K2Cr2O7 +6

CrO2Cl2 +6

XeO3 +6

[Cr(CO)6] 0

XeO F2 2(+6)

F Xe F

O O


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39. Answer (6)

NO2 NH2

Oxidation stateof N = +3

Oxidation stateof N is 3

40. Answer (2)

2CO C 2COInitially 1 0

1 x 2x

1 x + 2x = 1.5

x = 0.5

2COV 1 0.5 0.5 L

VCO = 2 0.5 = 1 L

2

CO

CO

V 12.0

V 0.5

PART - III (MATHEMATICS)

41. Answer (B)

1 2

1 2

cos cos3

sin sin

1 2 1 2

1 2 1 2

2cos cos

2 23

2cos sin

2 2

1 2cot cot

2 6

1 2

2 6

1 2 3

31 32 =

31 = + 32

1 2 2sin3 sin 3 sin3

sin31 + sin32 = 0


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42. Answer (B)

cos2z+ sin2z= (cosx+ cosy)2 + (sinx+ siny)2

1 = 2cos(x y) + 2

1

cos( ) 2x y

2 11 2sin2 2

x y

2 3sin2 4

x y

3sin

2 2

x y

2sin 32

x y

43. Answer (B)

Ifx> y> 0, then

2 2 2 2sin 2 cossin2 cos

2

x yxy

2 2 2 2

sin2 cos 1

2sin 2 cos

xy

x y

3 3 3 3

2 2 2 2

( )sin2 cos

2sin 2 cos

xy x y x y

x y

44. Answer (A)

(A)

Y

X

No. of solution = 1No. of solution = 1

k> 0 2

2

O

Y

XO

k= 0

No. of solution = 1

Y

XO

k< 0

In all cases no. of solution = 1

45. Answer (A)

sin( 2 ) 1

sin 3

Using componendo and dividendo,sin( 2 ) sin

2sin( 2 ) sin

tan( + )cot = 2 tan( + ) = 2tan

tan( + ) + 2tan = 0


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46. Answer (B)

1 ( )

2 2

A

r a b c a h rs

1A

b ch r

a a

1Bc a

h rb b

A

FE

CB D1Ca b

h rc c

CA B hh h

r r r = 3

a b b c a c

b a c b c a

3 + 6 = 9 (Using AM GM)

47. Answer (B)1 1(2[tan ] 1)([tan ] 2) 0x x

11 [tan ] 22

x

1[tan ] 1x 11 tan 2x

11 tan

2x

tan 1 x<

48. Answer (A)

2 2 2

22

( )

sin2

OA r r s s a

Abcbc

2 2

2

( )OB r s s c

ca

A

CBD

F

O

E

2 2

2

( )OC r s s c

ab

2 2 2OA OB OC

bc ca ab =

2

2 (3 ( ))

r s

s a b c

=2 2 2

2 21

r s

49. Answer (A, D)

cos = cos, cos3

= cos

3

3

= 2n +

3


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= 6n + , nI

cos = cos

cos3

= cos

3 3

2 3 3 3

n

, n I

= 6n +

cos = cos(6n + ) = cos

cos3

= cos

3 3

3

= 2

3 3n

= (6n + 1) + , n I

cos = cos

cos 3

= cos 3 3

cos 3

= cos 3 3

3

=

22

3 3n

, n I

= 6n +2

cos = cos


1 2 3

1 1 1 1

r r r r

=1 1 6 3 2

12 3 6

r=6

11

a = 1 2 3

1 2 2 3 3 1

r r r

r r r r r r

=5

11

Area of the ABC=1 2 3 6

11

r r r

S

b = 2 3 1

1 2 2 3 3 1

8

11

r r r

r r r r r r

51. Answer (A, B)

sin1 + cos1 + tan1

sin1 + cos1 =2

1 1

1tan

4 4


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1 3tan4 2 4

=4

x= cot3

x= cot12

= cot15 = tan75

x= 2 3

2

3x = 4

2 3 2 3 4x x

2 2 3 1x x = 0

Again, (x 2)2 = 3

x2 + 4 4x= 3

x2 4x+ 1 = 0

52. Answer (A, B, D)

1 1 1 1 1 1 2(cos cos cos ) (cos cos cos ) 9x y z

is possible only when

1 1 1cos cos cosx y z x= y= z= 1

and 1 1 1cos cos cos = = = 1

(A) is true, (B) is true

(C) is false, (D) is true

53. Answer (A, C)

Let DOE=

EOA =

AOB =

Since AOB = , ACB = /2

E

D

OC

B

F

A

180

2

FBC= 90 /2

DOC= 2 DBC= 2 (90 /2)

= 180


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Area ofOAB =1 1

sin sin2 2

OA OB

Area ofODC=1

sin(180 )2

OC OD

1 11 1 sin sin

2 2

Ar(OAB) = Ar(ODC)

Here angle BOC= 180 ( + )

Area of the pentagon =

1 1 1 1sin sin sin sin(180 )

2 2 2 2

1sin(180 ( ))

2

1 1 1 1 1

sin sin sin sin sin( )2 2 2 2 2

It is maximum when, = 90, = = 60

3 31

4

sq. unit

54. Answer(B)

55. Answer(D)

Solution of Q54 to Q55

EF+ DF+ DE= a cosA + b cosB + ccosC

= 4RsinA sinB sinC

For a triangleABC

3 3sin sin sin

8A B C

A

CBD

F E

3 34 sin sin sin

2R A B C R

2

Rr

ris the in-radius of incircle ofABC

3 33 3

2EF DF DE R r

But it is given that 3 3EF DF DE r EF + DF + DE= 3 3r

EF= DF= DE

DEFis an equilateral triangle

D = 180 2A = 60, E= 60, F= 60

C= B =A = 60


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ABCis an equilateral triangle.

3a = 18

a = 6

(Area ofABC) =

3

36 9 3 sq. unit4

Area ofDEF=23 3 9 3

94 4 4 4

a

Area of 9 3 44

9Area of

3

ABC

DEF

56. Answer(C)

Vertical line is drawn through the intersection A1 andA4

cosx= tanx

cos2x= sinx sin2x+ sinx 1 = 0

5 1sin

2x

MQ

L P

O

Y

A x1 = cos

A x3 = cot

A x5 = cosec

A x6 = sec

A x2 = sin

A x4 = tan

/2 X

|LM| =1

sinsin

xx

=2 5 1

25 1

=

2( 5 1) 5 1

4 2

=5 1 5 1

12

57. Answer (D)

Vertical line is drawn through the intersection ofA2 andA3

sin x= cotx cos2x+ cosx 1 = 0

5 1cos

2

x

/3 2 /3 3 /3 4 /3 5 /3 6 /3

y= 1

y=

y=

13

13

X

y= 1

O

Y

|PQ| =1

coscos

xx

=2 5 1

125 1

13sin3 1 sin3

3

No. of solutions = 12


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58. Answer (3)

A + B + C=

cos2A + cos2B + cos2C= 1 2cosA cosB cosC

cos2 sin2 + cos2 sin2 + cos2 sin2 + 2cos cos cos. sin sin sin = 1

= (cos2 + sin2) (cos2 + sin2) (cos2+ sin2) ...(i)Solving both the sides of (i) we get

(sin sin sin cos.cos cos)2 = 0

tan tan tan = 1

tan + tan + tan 3(tan tan tan)1/3 = 3

Minimum value is 3.

59. Answer (6)

In any triangleABC,1 2 3

1 1 1 1

r r r r

1 2 3

1 2 3

31 1 13

r r r

r r r

= 3r (Using A.M H.M)

1 2 3 9r r r

r

Minimum of 1 2 3 9r r r

r

amin = 9

Similarly, (r1 r2 r3)1/3

1 2 3

3

1 1 1r r r

Using (GM HM)

1 2 33

27r r r

r bmin = 27

2 2 2 2min mintan cot 3 tan 3cot3 9

a bA A A A

= 3(tan2A + cot2A) 6, Minimum value of the expression is 6.

60. Answer (1)

ABD and APCare similar

CB

A

P

D

b = 3

a = 2

c= 4

AB BD AD

AP PC AC

1 1 11

lcL l bc

L b

similarly L2l2 = ca, L3l3 = ab

(L1L2L3)(l1l2l3) = a2b2c2

2 2 2

1 2 3 1 2 3( )( ) 2 3 4 124 24 24 24

L L L l l l a b c abc

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