13 11 2011 xiii vxy paper i code a

7/30/2019 13 11 2011 Xiii Vxy Paper i Code A

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13th VXY (Date: 13-11-2011) Review Test-4

PAPER-1

Code-A

ANSWER KEY

PHYSICS

SECTION-2PART-A

Q.1 B

Q.2 C

Q.3 B

Q.4 D

Q.5 B

Q.6 C

Q.7 B

Q.8 C

Q.9 A

Q.10 A

Q.11 A

Q.12 D

Q.13 B

PART-B

Q.1 (A) Q,R,S,T(B) P,R,S

(C) Q,R,T

(D) P,R,T

PART-C

Q.1 0001

Q.2 0010

Q.3 0113

Q.4 0375

CHEMISTRY

SECTION-1PART-A

Q.1 A

Only V-Group Batch

Q.2 B

Only XY Batch

Q.2 B

Q.3 D

Q.4 B

Only V-Group Batch

Q.5 COnly XY Batch

Q.5 D

Q.6 B

Q.7 D

Q.8 D

Q.9 A

Q.10 A

Q.11 A

Q.12 C

Only XY Batch

Q.13 C

Only V-Group Batch

Q.13 D

PART-B

Only V-Group Batch

Q.1 (A) PQRT (B) PQS

(C) QRT (D) PQR

Only XY Batch

Q.1 (A) S (B) Q

(C) P (D) R

PART-C

Q.1 0287

Q.2 4302

Only XY Batch

Q.3 0006

Only V-Group Batch

Q.3 0004

Q.4 0002

MATHS

SECTION-3PART-A

Q.1 B

Q.2 D

Q.3 B

Q.4 A

Q.5 C

Q.6 D

Q.7 D

Q.8 B

Q.9 C

Q.10 B

Q.11 D

Q.12 C

Q.13 D

PART-B

Q.1 (A) R(B) T

(C) Q

(D) P,Q,T

PART-C

Q.1 0003

Q.2 0009

Q.3 0004

Q.4 0012


2/20

CHEMISTRY

Code-A Page # 1

PART-A

Q.1

[Sol. C + O2

CO2

mole x x x

C + 2O2

1 CO

mole y 2

y

ywt. of C : 12 (x + y) = 2 ....(1)

wt. of O2

: 32 (x +2

y) = 20 ....(2)

x + y = 1

x +2

y

8

3

2

y , y =

4

3, x =

4

1

mass ratio of CO and CO2

=x44

y28=

14

43

44

28

=

11

21Ans. ]

Only V-Group Batch

Q.2

[Sol. Ph

OH

/H

Ph

C=C

H

H

Me

additionAnti

2BrH HBr Br

Br BrH H

Me Me

Ph Ph

(B)

Ph

H H

Me

4

2

CCl

Br

additionAnti

H HBr Br

Br BrH H

Me Me

Ph Ph

(C)

(B) and (C) are diastereomers ]

Only XY Batch

Q.2

[Sol. Optical purity =100

5.125.87 100 = 75 %

Racemic mixture = 25 %

Optical purity = 100pure

mix


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CHEMISTRY

Code-A Page # 2

mix

=100

7530 = 22.5 ]

Q.3

[Sol. K4[Fe(CN)

6]

CN NC G.I.

K4

[Fe(CN)6

] 6 0 [Ma6

] 1

K4[Fe(CN)

5(NC)] 5 1 [Ma

5b] 1

K4[Fe(CN)

4(NC)

2] 4 2 [Ma

4b

2] 2

K4[Fe(CN)

3(NC)

3] 3 3 [Ma

3b

3] 2

K4[Fe(CN)

2(NC)

4] 2 4 [Ma

2b

4] 2

K4[Fe(CN)(NC)

5] 1 5 [Mab

5] 1

K4[Fe(NC)

6] 0 6 [Mb

6] 1

________________________

Total isomers = 10

K3[Fe(CN)

6] Same as K

4[Fe(CN)

6]

Total isomers = 10 ]

Q.4

[Sol. G = nRT ln

1

2

P

P

= nRT ln

2

1

V

V(P

1V

1= P

2V

2)

= P1V

1ln

2

1

VV = 105 1.2 103 ln

21

=84 J Ans. ]

Only V-Group Batch

Q.5

[Sol. (A)

OH

3 AlcoholMarkonikoff's product with rearrangement.

(B)HO CH3

No rearrangement

Markonikoff rule followed


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CHEMISTRY

Code-A Page # 3

(C) OH

No rearrangement

Anti Markonikoff product. ]

Only XY Batch

Q.5

[Sol. (I)

NH

lone pair is responsible for aromaticity

(II)N

MoreElectronegativity & localized lone pair

(III) NH2Delocalized more electronegativity

(IV)

CH NH2 2

lone pair localized lone pair ]

Q.6

[Sol.

O

O

H

Ha

O

O Ob O = O

c

B.O. : a 1

b 1.5 Bond lengthoderBond

1

Hence bond length order

c

2 )OH(d)O(d)O(d 22OO3OO2OO

]


5/20

CHEMISTRY

Code-A Page # 4

Paragraph for question nos. 7 to 8

[Sol.

(i)kf

kbA B

For t teq dt

]B[d

dt

]A[d (Always)

(ii)dt

]B[ddt

]A[d

kf[A]k

b[B] .....(A)

At eq. kf[A] = k

b[B] = 0 .....(B)

kf 0.63 0.4 = 0

kf=

6.0

4.03= 2

dt

]B[d= 2[A]3 [B] .....(C) ]

Paragraph for question nos. 9 to 11

[Sol. FeSO4excess KCN

(A)

K [Fe(CN) ]4 6+2

oxidation(B)

K [Fe(CN) ]3 6+3

(C)

K [Fe(CN) a ]3 3 3

3CN

+3a

4CN

+4b

(D)K [Fe(CN) b ]3 2 4

(i) K3

[Fe(CN)6]

Total possible geometrical and structural isomers = 10

(ii) K3[Fe(CN)

2b

4]

CN NC b G.I.

2 0 4 [Ma4b

2] 2

1 1 4 [Ma4bc] 2

0 2 4 [Ma4c

2] 2

_____________________

Total isomers = 6

_____________________

(iii) ])CN(Fe[K2

64

EAN = 262 + 6 2 = 36 ]


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CHEMISTRY

Code-A Page # 5

Q.12

[Sol. M ofNO2

>CN but SIR will work onNO2

therefore it can not showM on Phenyl ring. ]

Only XY Batch

Q.13

[Sol. ST-1 : The dipole moment of O2F

2is not zero : True

O

O

F

Fa

Open book like structure.

D.M. () 0ST-2 : All atoms are lying in the same plane : False

O

O

F

F

Open book like structure

3D structure

Hence maximum 3 atoms are lying in the same plane. ]

Only V-Group Batch

Q.13

[Sol. ST-1 : False :

Because electrolysis of aq. solution of magnesium chloride gives H2

gas at cathode and

Cl2

gas at anode.

ST-2 : True :

MgCl2

. 6H2O HCldry

MgCl

2(anhydrous) + 6H

2O ]

PART-B

Only V-Group Batch

Q.1

[Sol. (A) nCH2=CH2 CH2CH 2 n

Polymerisation

(B) CO2H

OO

C

nHOCH2CH OH + nHO C2 2

OCH2CH C2

Dacron

n

copolymerisation

H ,


7/20

CHEMISTRY

Code-A Page # 6

(C) Polymerisation(1,4 addition reaction)

[CC = CC]

C

n

(D)n

nCH =2 CHCH=CH +2 CH 2 CH=CHCH CH CH2 2 nCH =CH2

CN

CNcopolymerisation

]

Only XY Batch

Q.1

[Sol. F 7.75

Cl 6.25

Br 3.25

I 1.0

Nucleophilicity order of Xin polar aprotic solvent.

F> Cl> Br> I (Xis not solvated in DMF)

Since stability of Xin free state

F< Cl< Br< I ]

PART-C

Q.1

[Sol. 10C(g) + 4H2 (solid) H

f

HfC

10H

8(s) = (10 715.5 + 4 436)(5 617 + 6 345.5 + 8 413 + 72)

= 365 kJ mol1

C10

H8

+ 12O2 10CO

2+ 4H

2O(l) H =5157 kJmol1

5157 = 10 (393.5) + 4(286) HfC

10H

8(s)

or HfC

10H

8(s) = 78 kJmol1

|RE| = 36578 = 287 kJmol1 ]

Only XY Batch

Q.3

[Sol. 2H2SO

4H

2O + O H

2S

2O

8(X)

H2S

2O

8:

H OSOOSOH:

:

:

:

:

:

:

:

O

O

O

O

sp3

sp3

sp3

sp3

sp3

sp3 ]

Only V-Group Batch

Q.3

[Sol. Process

(i) Roasting / Calcination

(ii) Reduction

(iii) Roasting / Calcination

(iv) Roasting / Calcination

(v) Reduction

(vi) Roasting / Calcination ]


8/20

CHEMISTRY

Code-A Page # 7

Q.4

[Sol. Rate = k' [Sugar]a [H+]b Here conc. of H+ remain constant duration reaction.

Rate = k' [sugar]a

k' = k[H+]b

50

50010b = 101

b = 1 Ans]


9/20

PHYSICS

Code-A Page # 1

PART-A

Q.1

[Sol. E

=2/

k

[sin 45 + sin 45] =

k22]

Q.2

[Sol. m (g + a) cos 60

qE cos 30 = max.

30

30

60

Rcosec30

=2h

qE

mamg

h

an

=2

)ag(

m2

3qE

2h =2

1a

nt2

t =

m2

3qE

2

ag

h4

]

Q.3

[Sol.

)TT2(AK00

=dt

d

dt

d=

dx

KAdt

dt

d

x1k

dx

0

= AA T

T0

dT

0k

ln

2

0

x1

dt

d

= A(T

TT0)

dt

d=

AK0

(TT0) = )TT2(

AK0

0

TT0

= 2T0

ln3T ln3

T =3n1

T)13n2(0

]

Q.4

[Sol. Qene

= 100

E =0

2

= 200 20

= 4 1040

= 35.4 108 C ]


10/20

PHYSICS

Code-A Page # 2

Q.5

[Sol. kxf = Ma

a

f f

a'

f = Ma'

a' = aR

fR =2

1MR2

M

f= a

M

f2

M

f3= a

a' =M

f3

M

f2=

M

f]

Q.6

[Sol. QE = MgQE

mg

45

d

QV= w

Q =V

wd]

Q.8

[Sol.dt

dQ=

L

kA(T

2T

1)

65 = 31018

3.1k

(TT

1) = 3104

3.11

(13T)

Tglass = 13T = 0.2T = 12.8

3.1

501018 3 = k (12.80.2)

k =6.12

50010183

=14

1]

Q.9

[Sol.

R2

)RR(3)RRRR(4R2

1

2

1

2

1

=100

5

13

2

)RR(R

)RRRR(

1

21

21

=

20

1


11/20

PHYSICS

Code-A Page # 3

20

19=

)RR(R

)RRRR(

3

2

1

21

21

57(RR1

+ R2) = 40 (R1

2 + RR1

+ R2)

40R1

217RR117R2 = 0

R

R1

=80

1716028917 =

80

300917 ]

Q.10

[Sol. Idisc

=2

1M(R2 + R

12)

IRing

= MR2

2

21

22

MR

)RR(M2

1MR

=20

1

1

2

1

2

21

2

R

)RR(

= 20

1

10

19= 1 +

2

21

R

R

R

R1

=10

9]

Q.11

[Sol.

r

k2

)sin2(R

k

r

k2

=20

1 a

20

1sin =20

1

sin =20

19= 2a100

10

400

361

= 2a100

100

36100 + 361a2 = 100 400

361a2 = 100 39 a =361

39100]


12/20

PHYSICS

Code-A Page # 4

PART-B

Q.1

[Sol.(A)After first collision,

v

v

v

pure rotation, after 1/2 rotation

elastic collision

only translation.

(B)

v

vm 2m m

only rotation

(C)

/2

/2

/2

/2

after 1/4 rotation.

/2 /2

/2 /2only translation.

(D)

/2

/2

only rotation. ]

PART-C

Q.1

[Sol. AdE

=0

dV

E 4r2 =0

2n drr4rk

=0

k4

3x

r3x

E = )r()3n(

k 1n

0

n + 1 = 2 n = 1 ]


13/20

PHYSICS

Code-A Page # 5

Q.2

[Sol. For second lens,

f = 30 cm

3030x

x 22x

u = 30x

v = 22x

30

1

u

1

v

1

x22

1

x30

1

=

30

1

x52x660

x22x302

=

30

1

240 = 660 + x252x

x252x + 420 = 0

x242x10x + 420 = 0

x = 42, 10 cm ]

Q.3

[Sol. 36 103 = msT

= 103

0.09 TT = 400C

v = v0

(1 + rT) =9

1000(1 + 5 105 400)

=9

1000(1 + 2 102) =

9

1000(1.02) =

9

1020= 113.33 cc ]

Q.4

[Sol. T =2/mg

I2

=2/mg

3

m

2

2

=

30

2=

4

1 =

4

15m = 3.75 m = 375 cm ]


14/20

MATHEMATICS

Code-A Page # 1

PART-A

Q.1

[Sol. We have kxxtan22

32 21

R

2cottan,As

11

tan1 (x2 + x + k) =4

x2 + x + k = 1 x2 + x + (k1) = 0

For required condition, put D > 0 14(k1) > 0 54k > 0 k

13 11 2011 xiii vxy paper i code a

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