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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
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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
]
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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 ,
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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 ]
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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]
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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 ]
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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
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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]
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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 ]
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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 ]
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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