rt solutions-08!05!2011 xii abcd paper i code b
TRANSCRIPT
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12th ABCD (Date: 08-05-2011) Review Test-1
PAPER-1
Code-B
ANSWER KEY
CHEMISTRY
SECTION-1
PART-A
Q.1 A
Q.2 B
Q.3 A
Q.4 D
Q.5 A
Q.6 C
Q.7 D
Q.8 A or C or A,C
[XII (ABCD) except D1, D2]
Q.8 B [XII D1, D2 Students]
Q.9 C
Q.10 A
Q.11 C
Q.12 B
Q.13 CQ.14 C
Q.15 A
PART-C
Q.1 7048
Q.2 5646
Q.3 3375
Q.4 0034
Q.5 0600
Q.6 3018
Q.7 0630
Q.8 0280
PHYSICS
SECTION-2
PART-A
Q.1 D
Q.2 B
Q.3 A
Q.4 A
Q.5 C
Q.6 B
Q.7 A
Q.8 A
Q.9 A
Q.10 A
Q.11 B
Q.12 D
Q.13 C
Q.14 A
Q.15 D
PART-C
Q.1 0125
Q.2 0072
Q.3 0060
Q.4 0006
Q.5 0090
Q.6 0036
Q.7 0009
Q.8 0030
MATHS
SECTION-3
PART-A
Q.1 A
Q.2 A
Q.3 C
Q.4 A
Q.5 D
Q.6 B
Q.7 D
Q.8 C
Q.9 D
Q.10 B
Q.11 C
Q.12 A
Q.13 B
Q.14 C
Q.15 A
PART-C
Q.1 0010
Q.2 0002
Q.3 0002
Q.4 0004
Q.5 0045
Q.6 0001
Q.7 0000
Q.8 0006
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PART-A
Q.1
[Sol. ]
Q.3
[Sol. Pideal
> Preal
]
Q.5
[Sol. ]
Q.6
[Sol.
I Cl
3
1
D
3
2
2
Cl
Cl
x
x
y
x > y
Molecule is polar and planar
BothCl I Cl are equal
Equatorial I
Cl bond has more s-character than axial I
Cl bond.Lone pair electron are present in equatorial position so ICl equatorial bond is longer than expected
because %s decreases in that bond.]
Q.7
[Sol. (A)Plane of symmetry
(B)Plane of symmetry / centre of symmetry
(C)Plane of symmetry
and other form of these compound either has plane of symmetry or centre of symmetry so cna't show
optical isomerism. ]
[NOTE: This question is NOT for D1 and D2 batch students]
[NOTE: This question is NOT for D3 new students]
Q.8
[Sol. 'N' shell = 4th orbit]
[IMPORTANT: This question is ONLY FOR D1 and D2]
[IMPORTANT: This question is ONLY FOR D3 new student]
Q.8
[Sol. Cv,m
=54
80
Tn
qv
= 4 cal/k mol
qp
= nCp,m
T = 4 (4 + 2) 5 = 120 cal.]
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Q.9
[Sol. q = 0 w = U
U =f(T,V)
dU = dVV
UdT
T
U
TV
= nCv,m
dT + dVV
an2
2
U = nCv,m
2
1
2
1
V
V2
T
T
2
V
dVandT
= nCv,m
(T2T
1)n2a
12 V
1
V
1]
Q.10
[Sol. dq = dUdw = 0
dU = dw
or, nCv,m
dT + dVV
an2
2=PdV
or, nCv,m
dT = dVnbV
nRTdV
V
anP
2
2
or, Cv,m
2
1
2
1
V
V
T
TnbV
dVR
T
dT
or, Cv,mln 1
2
T
T
=
Rln nbV
nbV
1
2
or, ln1
2
T
T=
m,vC/R
2
1
2
1
m,v nbV
nbVln
nbV
nbVln
C
R
or, T2 (V
2nb)R/Cv,m
= T
1(V
1nb)R/Cv,m
If 'nb' is negligible m,vm,v C/R11
C/R
22 VTVT ]
Q.11
[Sol. (i) & (viii) ]
Q.12
[Sol. (i), (iv), (vi) & (viii) ]
Q.13
[Sol. (iii) & (vii)]
Q.15
[Sol. Due to the Back bonding]
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PART-C
Q.1
[Sol. General formula = (SiO3)n2n
Actual formula = S23
O70
48 ]
Q.2
[Sol. (a) 5
(b) Stereoisomer = 2n = 26 = 64
where n stereogenic area
(c) Six (threebond and one monocyclic ring one bicyclic ring) ]
Q.3
[Sol. S = 3000 1.25 600
1503000
300
600ln
= 2625 + 750 = 3375 cal]
Q.5
[Sol. w =P (V2V
1) = )m/N1040( 26
36m10
4.2
108
6.3
108
= +600J Ans.]
Q.6
[Sol.
SOS bonds = 3
SS bonds = zero
lone pairs = 18 ]
Q.7
[Sol. Mass = 20 gm
Volume = 200 mlml/gm1.0
200
20
volume
massC
Length of tube (l) = 10 dm
Optical rotation = 30
=l
C
=101.0
30
= 30
(a) If solution is diluted to one litre then new volume
V = 1000 ml
= ?
=l
C
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30 =
101000
20
= 6
(b) Specific rotation does not changed with change in conc. or volume or length etc. therefore it is 30.]
Q.8
[Sol.nk
q = qAB
+ qBC
+ qCD
+ qDA
= nRTB )TT(C.n
V2
VlnnRT)TT(Cn
V
V2ln DAm,v
0
0cBCm,v
0
0
= nR ln2 (TBT
C) = 1 2 0.7 200 = 280 cal/mol]
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PART-A
Q.1
[Sol. (1.51)
21 R
1
R
1=
F
1
134
5.1
21 R
1
R
1
= F
1
5.0
5.0 4 =
F
'F F' = 4F ]
Q.3
[Sol. T = (T + 1) (0.01) T = (T + 1) (0.99)
99.0
1
T
1T
=
99
100
991
1T
1
1 T = 99 K
Alternative : max
T = 0.99 max
(T + 1)
T = 0.99 T + 0.99 T = 99 K ]
Q.4
[Sol. TB
= 4T0
TD
= 5T0
TE
= 4T0
TF
= 3T0
]
Q.5
[Sol.1
d I=
u
x
d = Hx + dI= Hx
1
1 ]
Q.6
[Sol. Obj at 2f image at 2f, inv. ]
Q.7
[Sol.
t
t
A
]
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Q.8
[Sol.
5
du
dv=
2
2
u
v=
2
2
30
50=
9
25
du = dv 25
9=
25
91mm = 0.36 mm ]
Q.9
[Sol. Q = 0 as no temperature difference
U 0 as body is melting.W 0 ]
Q.10
[Sol. p
=1200
104.56
W = mL
M
W=
M
mL=
psay % of fat melted = k
k
pL
M
M3.0 k =
3
6
101503.01200
104.5
10% ]
Q.11
[Sol.u
u
1=
R
1
v
=
u
1
R
)1(
R0 u =ve v =ve v
I
v0 u = +ve (v can be +ve orve) ]
Q.13
[Sol.v3
4
1
=10
13
4
v3
4=
30
1
v =40
2v
1+
803
4
=
10
3
41
=
30
1
2v
1=
30
1
60
1=
60
1 v
2= 60 ]
Q.14
[Sol. Temp. falls as PV2 = CTV = CGraph is a rectangular hyperbola ]
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PART-C
Q.1
[Sol. 8103
= 8 109
8103
n
= 10
n =8
10=
4
5]
Q.2
[Sol. (p0
+ hg) v0 = (p0hg)v
(H + 8) 4 = (H8) 5
4H + 32 = 5H40
72 = H ]
Q.3
[Sol. Beam is parallel to basemm deviation
=
2sin
2sin 3 =
2
60sin
2
60sin
2
60sin =
2
3
2
60 = 60
= 60 ]Q.4
[Sol. u =(302t)
2t 202t 302t
20 30
v = 202t
v
1
u
1=
F
1
t220
1
+
t230
1
=
5
1
t100t4600
t4502
=
5
1
25020 = 600 + 4t2100t
4t2
80 t + 350 = 0 t = 4
1400160040 = 4
1440= 6.465 sec.]
Q.5
[Sol. NS = 15
5.1
11 =
3
15= 5 cm
Mirror is at 505 = 45 cm
Image is 45 cm behind the mirror.Final image =- 45 + 505 = 90 cm ]
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Q.6
[Sol. Initial magnification = 3
3u
v
1
1 v1
= 3u1
111
11
11
11 u4
3
u3u
)u3(u
vu
vu
f .......(i)
In second case m = 2.
2u
v
2
2 v2
= 2u2
3
u2
vu
vu 2
22
22
f .....(ii)
21 u3
2u
4
3 12 u
8
9u
It is given that the shift of the object = 6 cm
cm36484
3
u4
3
f 1 ]Q.7
[Sol.3/4
vv SI =
1
vv S0
v0
4cm/sv
I=
3
v4 0
vIv
f= 16 =
3
4v
0+ 4 = 16
34 v
0= 12 v0 = 9 cm/s ]
Q.8
[Sol.
= 45O
45 =t 45 = 1.5 t
t =5.1
45= 30 sec. ]
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PART-A
Q.1
[Sol. )x(fLim0x
= 0
functionboundedaiseand0xLim x12
0x
k = 0
Now, f ' (0) =h
)0(f)h0(fLim
0h
=
h1
0hehLim
= 0 f ' (0) = 0 Ans.
Note that f (x) is discontinuous at x = 2n
1
l,
3n
1
land so on.]
Q.2
[Hint: f (x) is derivable x R ]
Q.3
[Sol. Let all the 3 quadratic equations have real roots. Hence b2 4ac ; c2 4ab and a2 4bc a2b2c2 64a2b2c2, which is not possible. Hence at least one of the quadratic equation must haveimaginary roots. Maximum possible real roots can be 4 C.
e.g. a, b, c 1, 5, 6 give
3,
2 as roots of x2
+ 5x + 6 = 0
and 1,5
1as roots of 5x2 + 6x + 1 = 0.
But 6x2 + x + 5 = 0 has imaginary roots. Ans.]
Q.4
[Sol. Select 4 gaps EEEEE for BKRP in 6C4 ways and arrange them in 4 ! ways.
Hence total 6C4 4!. Ans.]
Q.5
[Sol. (24 sin x)3/2 = 24 cos x
24 (sin x)3/2 = cos x 24 sin3x = cos2x = 1sin2xput sin x = t, we get
24t3 + t21 = 0
(3t1)
0
21t3t8
= 0 t =3
1
t =3
1i.e. sin x =
3
1 cosec x = 3 cosec2x = 9 Ans.]
Q.6
[Sol. Let a, ar, ar2, ........
Now, a + ar = 12 ..........(1)
ar2 + ar3 = 48 ..........(2)
Now,)1(equation
)2(equation
)1r(a
)r1(ar2
= 4
r2 = 4, (As r 1) r =2Also, a =12 (using (1)). Ans.]
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Q.7
[Sol. f(x, y) = sin 2x
x2sinx1
0x
2xxtan1Lim
(1) = el
where l =x2sinx
xxtanLim
20x
= 30x x2
xxtanLim
=6
1.
Hence limit = 61
e Ans.]
Q.8
[Sol. For existence of limit,
)x(fLim)x(fLim
QRx,x
Qx,x
x
y
2
2
23
23
1
1
O
intersection of3 points
Thus, must be root of the equationx32x = x22
x2 (x1)2(x1) = 0
x = 1 or x = 2 or x = 2Sum of all the values of = 1. Ans.]
Paragraph for questions nos. 9 & 10
[Sol. Clearly
4f = c;
4f =
c
1
(i) As )x(fLim
4x
exists, so c =c
1 c = 1 Ans.
(ii) As f (x) is continuous at x =4
, so )x(fLim
4x
=
4f c =
c
1= 1 c = 1 Ans.]
Q.11
[Sol. If , then a 0 P(x) = bx + cNow, P(2) = 2b + c = 9 and P'(3) = b = 5 c =1 P(x) = 5x1
Now,
x
x 5x5
1x5Lim
(1 form) =
x5x5
5x51x5Limxe
= 54
e
Q.12
[Sol. When , then a, b 0 P(x) = c, but P(2) = 9 P(x) = 9
Now,)3x(sin
3)x(PLim
3x
=
)3xsin(
33lim
3x
= 0
0
zerotowardstending
zeroExactSince ]
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Q.13 If = , then the value of
1ex4
tan1
)x(PLimLim
x2x0
is equal to
(A)
16
3(B*)
16
9(C) 2
e4
5(D)
4
e2
[Sol. For = , P(x) = a (x)2
As P(2) = 9 P(2) = a (2)2 = 9 a =2)2(
9
Now,
1e)x(4
tan1
)x()2(
9
Limx2
2
2
x
=
1e)x(4
tan1)x(4
tan1
)x()2(
Limx
2
2x
Putting x = + h, we get
=
1eh4
tan1h4
tan1
h
)2(
9Lim
h
2
20h
)1e(htan1
htan112
h)2(
9
Limh
2
20h
2h
2
20hh
h
)1e(
h
htan2
)htan1(h
2)2(
9Lim
= 2
)2(4
9
=
20 )2(4
9Lim
=
16
9. Ans.]
Q.14
[Sol. Option (C) is correct
For x 0 f (x) = 0 and for x < 0,
f (x) = xsin2 1
range of f (x) is [0, ]
Statement-1 is true but Statement-2 is false ]
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Q.15
[Sol. Put x = tan where
2,
2 2 (, )
g () =
2
2
tan1
tan1= cos 2
Hence g () (1, 0]
Using I.V.T. some c R, g (c) =33
1 ]
PART-C
Q.1
[Sol.
1V
1L
1B
1N
4s'A
required number =!4
!8number of ways when ends in AA
=!3
!
!4
!8 =
!3
!
!3
!
!3
!2
m + n = 10. Ans.]
Q.2
[Sol. Clearly, f (x) must also be continuous at x = 4 and x = 6.
so, f(4) = f (4) = f (4+)
2
a=
2
=
2
b a =1, b = 1
Also, f (6) = f (6) = f(6+)
4
b=
2
+
4
a ba = 2
Hence, a =1, b = 1]
Q.3
[Sol. We have tan xtan y = tan (xy) [1 + tan x tan y]
ytanxtan
y
xy
y
xy
]ytanxtan1[)yx(tanLim
yx
=
)ytanxtan1()yx(
)ytanxtan1()y()yx(tanLim
yx
g(y) =y g(x) =xNow h(x) = Min (x2,x)
x
1 O
x2
x
y
Obviously h(x) is not derivable
at two points.i.e. {1, 0} 2 points.]
Q.4
[Sol. Given sin + sin3 = 1sin2 = cos2 sin2 (1 + sin2)2 = (cos2)2
(1cos2) (2cos2)2 = cos4 (1cos2) (4 + cos44 cos2) = cos4 4 + cos44cos2 + 4cos44cos2cos6 = cos4 cos6 + 4cos48cos2 + 4 = 0Hence, (cos64cos4 + 8cos2) = 4. Ans.]
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Q.5
[Sol. We have f(x) = 1x3 and g5(x) =
)x5(f
1......
)x2(f
1
)x(f
1
5
Now, 35
0x x)x(g1Lim
=
30x x
)x5(f
1......
)x(f
1
51
Lim
=
)x5(f
1
)x4(f
1
)x3(f
1
)x2(f
1
)x(f
1x
1)x5(f
1......1
)x2(f
11
)x(f
1
Lim30x
=
30x x5
)x5(f
)x5(f1
.......)x2(f
)x2(f1
)x(f
)x(f1
Lim
=
5
54321 33333
=5
2
)15(52
= 45. Ans.]
Q.6
[Sol. arc cos
xcosarc
2= arc sin
xsinarc
2
cos1
xsin
2
2 1= sin1
xsin
2 1
cos1
xsin
21
1= sin1
xsin2 1
Let xsin2 1
= where [0, 1] think!
cos1 1 = sin1
21 2sin = sin1 22 = 22 = 2
2 = 22
Hence is either 0 or 1.If = 0 then x = 0if = 1 then x = 1hence sum of all possible value of x is 1 Ans.]
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Q.7
[Sol. Let b1, b
2, ......... b
10be the roots of f (x) = b (b = 5)
b1
+ b2
+ ....... + b10
=9 (Theory of equation)
and c1, c
2, ........ c
10be the roots of f (x) = c (c = 3)
y=c
y=b
C1 C2 C3
B1 B2 B3
P
x
y
c1
+ c2
+ ........ + c10
=9
then cot(B1C
1P) =
)cb(
)cb(11
; cot(B
2C
2P) =
)cb(
)cb(22
, & so on
Hence
10
1nnnPCBcot =
cb)cb(........)cb()cb( 10102211
=cb
)c..........cc()b........bb(2211021
=
35
)9()9(
= 0 Ans.]
Q.8
[Sol. f(x) = sgn )1bx2bx()1axx( 22 For exactly one point of discontinuity 1bx2bx1axx 22 = 0 at exactly one value of x.
Let P(x) = x2
ax + 1 .....(1) and Q(x) = bx2
2bx + 1 ......(2)and D
1and D
2are the discriminant of equation (1) and (2) respectively.
Case-I: D1
= 0 and D2
< 0
a24 = 0 4b24b < 0
a = 2 b (0, 1) rejected because there is no integral value of b.Case-II: D
1< 0 and D
2= 0
a (2, 2) b = 0, 1a =1, 0, 1 b = 0 is rejected
number of ordered pairs are 3.Case-III: D
1= 0 and D
2= 0
a =2, 2 b = 0, 1
For b = 0, a =
2 & 2 and for b = 1, a = 2 number of ordered pairs are 3.
Total number of ordered pairs are 6. Ans.]