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12th ABCD (Date: 06-11-2011) Review Test-6
PAPER-2
Code-A
ANSWER KEY
CHEMISTRY
SECTION-2
PART-A
Q.1 C
Q.2 A
Q.3 B
Q.4 C
Q.5 C
Q.6 C
Q.7 D
Q.8 C
Q.9 B
Q.10 A,B,C,D
Q.11 A,B,C,D
Q.12 C,D
Q.13 A,B,C,D
PART-B
Q.1 (A) P,
(B) R,S
(C) Q,R,S
PART-C
Q.1 0007
Q.2 0005
Q.3 0040
Q.4 0004
Q.5 9260
MATHS
SECTION-3
PART-A
Q.1 A
Q.2 Bonus
Q.3 C
Q.4 C
Q.5 D
Q.6 D
Q.7 B
Q.8 A
Q.9 C
Q.10 A,C,D
Q.11 A,B,D
Q.12 A,B,C
Q.13 B,C,D
PART-B
Q.1 (A) Q
(B) P
(C) S
PART-C
Q.1 0003
Q.2 0001 or 0006
Q.3 0086
Q.4 0040
Q.5 0003
PHYSICS
SECTION-1
PART-A
Q.1 D
Q.2 D
Q.3 A
Q.4 B
Q.5 D
Q.6 C
Q.7 A,C,D
Q.8 A,B
Q.9 A,B
Q.10 B,D
Q.11 A,C,D
Q.12 A,C,D
Q.13 A,B
PART-B
Q.1 (A) R
(B) P,R
(C) R
PART-C
Q.1 0004
Q.2 0021
Q.3 0002
Q.4 0001
Q.5 5880
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PHYSICS
Code-A Page # 1
PART-A
Q.1
[Sol. La
= Lb
mvar = 3mv
br
b
a
v
v= 3 : 1 ]
Q.2
[Sol. =E
hc=
1.1
1240nm ~ 1100 nm infra red. ]
Q.3
[Sol. AB = CD = T
T
= v
wave]
Q.4
[Sol. Fbottom
= L2 g h
hF
side= g
2
h hL
2
Lgh2
=6
1 ghL2
h =3
L= 20 cm ]
Q.6
[Sol.c
QiR = 0
i
Q= RC = constant. ]
Q.8
[Sol. (A) no net change in p. (B) no net change in p
(C) p along x-axis. (D) p along x-axis. ]
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Q.9
[Sol. = ( 1)A = (1.5 1) 180
9.0 =
400
rad.
dt
dN=
hc
= 834
9
1031063.6
102212.7
= 8 1018 /sec.
py = dtdN
h
sin
= 8 1018 9
34
10221
1063.6
400
= 6 1011 N ]
Q.10
[Sol. =f
v=
850
340=
5
2= 0.4 m
L =4
,4
3.......... = 10 cm, 30 cm, 50 cm etc.
l + e =4
3l = 2 cm ]
Q.12
[Sol. (A) EKE
L+ E
LE
N= E
KE
W= E
K. K
rray
= EKEM + EMEN ]
Q.13
[Sol.1
2
3
l v =2
4
5
l v
1
2
l
l
=6
5]
PART-B
Q.1
[Sol. Central maxima is at the point wherex = 0.
C
I
Here x = ( 1)t d sin = 0
sin= 1d
t)1( (Not possible)
So central maxima does not exist in part A & C. ]
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Code-A Page # 3
PART-C
Q.1
[Sol.X
4
4
9V
1
x = P.d. across 4 = 4V ]
Q.2[Sol. Q = U + W
Q W = UmL PV =U1 540 4200 105 (VS 10
3) = U
VS
=P
nRT=
M
m RT
= 31018
1
3
25
105
373=
216
373m3
= 2.268000 105
216
373= U
= 105
216
37368.22 ~ 21 105 J ]
Q.3
[Sol. k BjBiBB321
4 109 v1 kBjBiBk45sinj45cos 321 = iF1
321BBB2
1
2
1
0
kji
=
2
B2
B 23 i + j2B1
k2
B1
B1
= 0 kBjB32
2F
= q 2 104 kBjBi32
= q 2 104 jBkB32
B2
= 0
F2
= 4 105 = 4 109 2 104 B3
B3
= 2T ]
Q.4
[Sol. Initially, only m2oscillates
m1 m2 t =2
T=
k
m2 =
25
next both m1
and m2
oscillate about their CM.
t' =2
'T=
k
=
k)mm(
mm
21
21
= 210
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Code-A Page # 4
m2
=50
k= 3kg
)m3(
3m
1
1
150 =
200
1
4m1
+ 3 + m1 m
1= 1 kg ]
Q.5
[Sol. B = 0ni B =
0
Ni N = i
B
0
L = N 2r = ir2B
0
=
2104
1001.024.1168.07
= 5880 m ]
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Code - AA Page # 1
PART-A
Q.1
[Sol. Dissociation of water will be maximum when [H+] = [OH] at 25C
[H+] = [OH] = 107 pH = 7 ]
Q.2
[Sol. Compouds which produce H+ ion in solution is called Bronsted Acid
(A) H[Co(CO)4
] H+ + [Co(CO)4
]
(B) LiAlH4 Li+ [AlH4]
It provides hydrides [H]
in the solution. ]
Q.3
[Sol. (I)
MeS Br
NGPS
KOHaq
N
MeS OH
It is formed by SNNGP path
(II)
MeS Br
2S
KOHaq
N
MeS OHS
N2 path ]
Q.4
[Sol. Mass of neutron is greater than that of proton. ]
Q.5
[Sol. M(s) + O2
(g) I MO
2(s) S = a
M(l) + O2
(g) II MO
2(s) S = b
M(g) + O2(g)
III MO2(s) S = c
|a| < |b| < |c|
i.e.
M(g) + O2(g) MO
2(s) minimum entropy change
M(g) + O2(g) MO
2(s) maximum entropy change
We know that
G = H + TS
G = (S)T + H
y = mx + C
where y = G, m =S, x = T, C = H
M(s) + O2
(g) I MO
2(s) m
I=S =(a) = a
M(l) + O2
(g) II MO
2(s) m
II=S =(b) = b
M(g) + O2(g)
III MO2(s) m
III=S =(c) = c
|c| > |b| > |a| mIII
> mII
> mI
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Code - AA Page # 2
(C) III
III
G
(KJmol )1
T(C)
]
Q.6
[Sol. (a)
D H
H
HH
Cl
KOH.alc
)major(CHCHD 2 + CH2 = CH2
(b)
H
D
HH
Cl
H
KOH.alc
)major(CHCHD 2 + CH2 = CH2
There is free rotation along CC bond & CH bond fission take place more easily in comparison of
CD bond fission. ]
Paragraph for question nos. 7 to 9
[Sol.
CH3 CH3
O3Zu/H O2
CH3CH CH CH2 2
CH3CH CH CH = CCH2 2
(A) (B)
CH+
H O/H2 NaOH,
NaOH, CH2CH3
CH OH2
COONa
+
Aldol reaction
CannizarosReaction
Electrophilic Acidreaction
(C)
(D)
(E)
CH3CH CH CHCH2 2 2
CH3CH CH = CHCH2 2
CH3CH CH2
CH3CH CH = CCH2
OHCH3
CH3
CH3
CH3
Acid d-Base reactionEllimination reaction
H /
O /Zn, H O3 2
NaOH, NaOH,
Aldol reaction
+ CHCH2 (F)
CH3CH2CH = C
CH3O = CH
O O
O
O O
O
CH CH3 2 CH CH = CH2
Sol.7 In complete paragraph hetrolytic fission takes place so no any free radical reaction
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Code - AA Page # 3
Sol.8
OOO
O
CH CH3 2 CH CH = CH2
CH CH3 2 CH CHCH2
CH CH3 2 CH CH2 CH CH3 2 CH CH2
CH CH CH CH3 2 2
CH CH3 2 CH CH2
CH3
CH3
O3
O O
O
(A) (B) (C) (D)
O O
+
+
+CH CHCH3
CH3
CH3CH3
CH3
O O O
O OO OO O
CH CH CH CH CH CH2 2 3CH
Combination of A&B or C&D Combination of B&C Combination of A&D
Sol.9 So three ozonides
E & F both give aldol condensation. ]
Q.10
[Sol. Using phenolphthalein,
1 32CONa
M 10 = 2.5 5
1= 0.5
32CONaM = 0.05 M
Using methyl orange
2 32CONa
M 10 + 1 3NaHCO
M 10 = 7.5
5
1
2 0.05 10 + 103NaHCO
M = 1.5
3NaHCOM = 0.05 M
Mass of32CONa
M in 10 ml solution = 10 0.05 106 gm 103 = 0.053 gm
Mass of NaHCO3
in 10 ml solution = 10 0.05 84 103 gm = 0.042 gm ]
Q.12
[Sol. A2(g) 1
K2A (g)
B3(g) 2
K3 B (g)
r1
= k1[A
2]
r2
= k2[B
3] r
1= r
2 k
1= k
2
10RT
200005
RT
14000
T = K314.8
1200
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Code - AA Page # 4
Initial moles of A2
and B3
are same so
22 BAPP and Rate of disappearance of AA
2and B
3is also same
so1
P
P
timeanyatB
A
3
2
Pi= atm237.0
314.8
12000821.0
100
2
so P can't be less than 0.237 Total moles are increasing.
1P
P
B
A
Ans. (C) and (D) ]
PART-B
Q.1
[Sol. (A) 64OP
3
+ 6H2O 33POH43
(B) 4P0
+ 3NaOH (aq.) + 3H2O
3
PH3
+22
PONaH31
(Disproportionation reaction)
(C)33
POH43
3PH
3+ 33POH3
5
(Disproportionation reaction) ]
PART-C
Q.1
[Sol. AgBr Ag+ + Br
G = 80104(100)
= 76
G =2.303 RT logKP
1000 76 =2.303 PKlog4.1303.2314.8
7600314.8
log KP
=7600
4.1761000
log KSP =1.4 10logK
sp=14 or K
sp= 1014
Solubility = 107
Ans. = 7 ]
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Code - AA Page # 5
Q.2
[Sol. [Zn(gly)2] Td [M(AB)
2]
M
A
BAB
[Optically Active]
[Mn BrCl (PH3) (H
2O)] Td [Mabcd]
M
a
bcd
[Optically Active]
[Pt(NH3) (H
2O) FCl] Square planar [Mabcd]
M
a b
cd
[Optically inactive]
[RhCl(CO)(PPh3) (NH
3)] Square planar [Mabcd]
M
a b
cd[Optically inactive]
[Be4O(CH
3COO)
6] Symmetrical
or
[Be4O(OAC)
6] Be
OAC
Be
OAC
Be
OAC
OA
C
Be
OAC
OAC
O
[Fe(EDTA)] Oh Optically active
[Cr(NH3)3FCl Br] Oh [Ma
3bcd]
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Code - AA Page # 6
M
ac
d
[Optically active]
a
b
a
[Co(en)3]3+ Oh [M(AA)
3]3+
M
AA
A
[Optically active]
A
A
A
5 Molecules show optical isomerism. ]
Q.3
[Sol. (i) In presence of NaCl, Ist reaction shifts backward and IInd reaction shifts forwards. So solubility of
AgCl may increase or decrease.
(ii) SAgCl
= [Ag+] + [AgCl2]
= ]Cl[k]Cl[
k2
1
(iii) For minimum solubility
]Cl[ddS = 22
1 k]Cl[d
k = 0
[Cl] =08.0
102
k
k 10
2
1
= 5 105
31
2
2
]Cl[
k2
]Cl[d
Sd
> 0
(iv) If complex formation is not occurring, then only reaction 1 will occur and solubility will decrease as
[Cl] increases.
13 + 27 = 0040 Ans. ]
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Code - AA Page # 7
Q.4
[Sol.
Na CO
(aqueous)2 3
SrCl2
CaCl2
BaCl2
Hg(NO)2 3 2
HgCl2CuCl2
Pb(CHCOO)3 2
NHCl4
BaCO3(white)
HgCO3(Yellow)
HgCO 3HgO3(Reddish Brown)
Cu(OH) 2CuCO
(Blue)2 3
Pb(CH) 2PbCO
(White)
2 3
(NH ) CO
(Soluble)4 2 3
SrCO
(White)3
CaCO
(White)3
]
Q.5
[Sol. (a) C C = C C
C
(b)
(c) (d) CH2
= CH2
]
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PART-A
Q.1
[Sol. I = 2
2
12
1
dx1xx
xtan.......(1) put x =
t
1
I =
2
2
12
1
dt1tt
t
1
tan.........(2)
Now, (1) + (2) gives
2I =
2
2
122
2
3
2
1x
dx
2
I =
2
2
1
1
3
1x2tan3
24
=
0332
=36
2 A. Ans. ]
Q.2
[Sol. P(E) = P( R R W W B or R R R W W or W W W R R)
= !2!2
!5
53
1
+ !2!3
!5
53
1
+ 53!2!3
!5
Simplifying P(E) = 53
50 k = 50 Ans.]
Q.3
[Sol. )x(fLim
2x
=h
(sinh)sinLim
0h = 1 and )x(fLim
2x
= 1h
)1(sinLim
0h
.
So, )x(fLim2
x does not exist. Ans.]
Q.4
[Sol. 1, x, y G.P. Let x = k, y = k 2
and x, y, 3 A.P. 2y = x + 3 2k2 = k + 3
2k2 k 3 = 0 k = 1,2
3.
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Code-A Page # 2
x + y = k2 + k =4
1
2
1k
2
.
x + y|max
=4
15
4
14 at k =
2
3. Ans.]
Q.5
[Sol. As, ar.(ABC) = ar. (ACD) + ar. (BCD)
2
1(3) (4) =
2
1(4) (CD) sin 60 + 30sinCD3
2
1
A
CB
D
60
30
b=4
a=3
c=5
CD =334
24
=
39
)334(24 = )334(
13
8 . Ans.]
Q.6
[Sol. Possible cases :
(i) 0 0 0 0 0 3 !5!6
(when digit 0 comes at first place then number of arrangement =!4
!5)
Number of six digit numbers with 0 0 0 0 0 3 =!5
!6
!4
!5= 1
(ii) Similarly for 0 0 0 2 2 1
!2!2
!5
!2!3
!6= 30
(iii) 1 1 1 1 1 2 !5
!6= 6
Total = 1 + 30 + 6 = 37. Ans.]
Paragraph for question nos. 7 to 9
Sol.
(i) No American together
A1A
2A
3A
4B, C, D, E
For A1
we have four (B, C, D, E) favourable cases out of total cases 7.
Hence, probability =7
4(say) A
1and B are paired and the remaining
are, A2A
3A
4C D E
for A2, favourable cases 3 out of total cases 5
Hence, probability =5
3say AA
2and C are paired and the remaining are
Now, A3
A4
D, E
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for A3, favourable cases 2 out of total 3.
Hence probability =3
2.
P (No two Americans together) =35
8
3
2
5
3
7
4 (B)
(ii) P (delegates of the same country form both pairs)
A1A
2A
3A
4E, B, C, D
= 1151
73
[For A1
we have 3 favourable and of 7 and two A2
only 1 favourable out of 5 are for the remaining, no
constraints.]
EBCD
=35
3(B)
(iii) P (delegates of the same country not forming any pair
+ forming both pair + forming exactly one pair) = 1
P (forming exactly one pair) = 1 353
358 =
3524 (C). Ans.
Alternatively: A1, A
2, A
3, A
4, B, C, D, E
n(S) =!4)!2(
!84 = 105
(i) The probability that no two delegates of the same country are paired =105
CCC1
2
1
3
1
4
=
105
234=
35
8
(ii) The probability that delegates of the same country form two pairs =
105
33=
35
3
(As A1, A
2, A
3, A
4can be paired in 3 ways and B, C, D, E can be paired in 3 ways.)
(iii) The probability that exactly two delegates of the same country are paired together
=105
34C2
4 =
35
24Ans.]
Q.10
[Sol. We have
g(x) =
1,0xif,0
1,0x,1x
sin1xx
sinx22
Clearly, g(x) is differentiable x R.(As sum and product of differentiable functions is also differentiable function.)
Now, g '(x) =
1,0x,0
1,0x,1x
sin)1x(21x
cosx
sinx2x
cos
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Clearly, g'(x) is discontinuous at both x = 0 and x = 1.
Also, g(0) = 0 = g(1) (Given)
So, Rolle's theorem is applicable for g(x) in [0, 1].
Q.11
[Sol.
(A) P(A/B) = P(B/A) P(A) = P(B)Now, P(A B) = P(A) + P(B) P(A B) P(A B) = P(A) + P(B) P(A B)
P(A B) = 2P(A) 1 > 0; P(A) >21 True
(B) P(B) =4
3; P(A/ B) =
2
1
)B(P
)BA(P =
2
1 P(A B) =
4
3
2
1=
8
3
Now, P(A) + P(B) P(A B)= P(A B) 1
P(A) +4
3
8
3 1
P(A) 8
5
P(A)]max.
=8
5 True
(C) P(AC BC)C = C4443)aa()aa( = (a4)
C = a1
+ a2
+ a3
= P(A + B)
P(AC BC) = (a1
+ a3
+ a4)C = a
2= P(AB)
Hence P(AC BC) + P(AC BC) = P(A) + P(B)] =6
23
3
1
2
1 =
6
5.
(D) Given, A is subset of B.
Now, P(B/A) =)A(P
)AB(P =
)A(P
)A(P= 1.
S
A
B
Q.12
[Sol. A vector coplanar with kji2a
, kjib
and perpendicular to k6j2i5c
will be
along cba
= acbbca
= kji29kji18 = kj39k9j27
Vector will be along k
j
3 .This vector will lie in the plane which will be parallel to it.
Normal of plane will be perpendicular to vector. Ans.]
Q.13
[Sol. Equating the coefficient of x20, we get
220 a20 = 1 20 20 12a (C)
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put, x =2
1, we get
0 1020
q2
p
4
1b
2
a
0q2
p
4
1b
2
a1020
b2
a and 0q
2
p
4
1
a + 2b = 0 (B)
b =2
1220 20
.
put, x = 0 we get
1 b20 = q10
1
2020 20
2
12
= q
10
10
20
20
q2
121
10
20q
2
1 q =
4
1
Using, 0q2
p
4
1
04
1
2
p
4
1
p =
1
Hence B, C, D. Ans.]
PART-B
Q.1
[Sol.
(A) We know that | adj A | = | A |2 for a 3 3 matrix
Given adj A = KAT |adj A| = |KAT| = K3 | A | (|AT| = | A | ) K3 | A | = | A |2
K3 = | A | ; Now det A = 1 (1 4) 2(2 4) + 2 (4 + 2) = 27 k3 = 27K = 3.
(B) Given, f (x) = (2x + 1)50 (3x 4)60
f ' (x) = 220(2x + 1)48 (3x 4)58 (2x + 1)(3x 4)(3x 1)
1
2
1
3
4
3
+ +
Sign scheme of f '(x)
Least positive integer is k = 2
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(C) By applying continuity and differentiability at x = 1, we get a = 3, b =1.
Hence, (2a + b) = 2 (3) 1 = 5. Ans.]
PART-C
Q.1
[Sol. Line L1
is parallel to vector jin1
and line L2
is parallel to vector k2j2in2
. As, normal of
plane is perpendicular to both lines L1
and L2
, so n
= normal vector of plane
=211
011
kji
nn21
= j2i2 .P(x,y,z)
N(1,1,5)
n = 2i 2j ^ ^
So, equation of plane is 2(x 1) 2(y + 1) + 0(z 5) = 0 x + y = 0 ........(1)
So, distance of plane from 0,8,2M =2
82 = 3. Ans.]
Q.2 Let f(x) be a differentiable function satisfying
x
0
x
0
dtxttandtttan)t(f)x(f = 0
2
x .
Find the number of solutions of the equation f(x) = 0.
[Ans. 1]
[Sol. x
0
x
0 KingUsing
dtxttandtttan)t(f)x(f
= 0
x
0
x
0
dtttandtttan)t(f)x(f = 0
Differentiate w.r.t. x
f '(x) + 1)x(f tan x = 0
1)x(f
)x('f
+ tan x = 0
integrate w.r.t. x
ln 1)x(f + ln (sec x) = C
f(0) = 0 C = 0.
Hence
1xcos
1)x(f
f(x) = cos x 1 f(x) = 0 cos x = 1
Hence only solution in
2
,2
is x = 0.
Hence number of solution is 1.
Note that domain of f(x) is
2
,2
Ans.]
7/30/2019 Rt Solutions-06!11!2011 XII ABCD Paper II Code A
19/19
MATHEMATICS
C d A P # 7
Q.3
[Sol. We have, f '' (x) = 12x 4
f ' (x) = 6x24x + cAs, f ' (1) = 0 c = 2 f ' (x) = 6x24x 2 f (x) = 2x32x22x + As, f (1) = 0 = 2Hence, f (x) = 2(x3x2x + 1)
or f (x) = 2(x
1)2
(x + 1).Now, M (x = 2, y = 6) and f ' (2) = 14.
So, the equation of normal at M is (y 6) =14
1(x 2)
For x-intercept, put y = 0
we get x = 86 Ans.]
Q.4
[Sol. Let O be the centre of polygon
Area of rectangle = 4 OA1A
2= 6 .........(1)
and Area OA1A2 = n1 area of polygon .........(2) A
1
A2
Ak + 1
Ak
O
(1) and (2)4
6= 60
n
1
n = 40. Ans.]
Q.5
[Sol. Using tan1 + tan1 + tan1 = tan1
1
In L.H.S. we get
tan1 2)c)(bx()bx(ax)ax(c1
bxaxcbxaxc
where c =
8
x
x
1.
Now, 1 = ax
8
x
x
1+ abx2 + bx
8
x
x
1 1 = 222 x
8
bbabxx
8
aa x R
1 = (a + b) + x2
8
ba
ab
On comparing, we geta + b = 1 ..........(1)
and ab =8
ba =
8
1.......(2) (Using (1))
Now, a2 + b2 + 2ab = 1 a2 + b2 +4
1= 1
Hence, 4(a2 + b2) = 3. Ans.]