rt solutions-01!01!2012 xii abcd paper i code a
TRANSCRIPT
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12th ABCD (Date: 01-01-2012) Final Test
PAPER-1
Code-A
ANSWER KEY
MATHS
SECTION-1
PART-A
Q.1 C
Q.2 C
Q.3 C
Q.4 B
Q.5 D
Q.6 C
Q.7 B
Q.8 A
Q.9 D
Q.10 B,C,D
Q.11 A,C,D
Q.12 C,D
Q.13 A,C,D
PART-C
Q.1 0001
Q.2 0004
Q.3 0006
Q.4 0009
Q.5 0007
Q.6 0008
Q.7 0000
Q.8 0003
PHYSICS
SECTION-2
PART-A
Q.1 B
Q.2 C
Q.3 B
Q.4 B
Q.5 A
Q.6 D
Q.7 C
Q.8 D
Q.9 A
Q.10 B,D
Q.11 C,D
Q.12 B,D
Q.13 B,C
PART-C
Q.1 0800
Q.2 0001
Q.3 0200
Q.4 2200
Q.5 0003
Q.6 0015
Q.7 0009
Q.8 0080
CHEMISTRY
SECTION-3
PART-A
Q.1 B
Q.2 D
Q.3 A
Q.4 C
Q.5 B
Q.6 D
Q.7 A
Q.8 C
Q.9 A
Q.10 A,B,C
Q.11 A,B,C
Q.12 A,B
Q.13 B,C
PART-C
Q.1 0200
Q.2 0006
Q.3 0003
Q.4 0003
Q.5 0009
Q.6 0006
Q.7 4000
Q.8 0014
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MATHEMATICS
Code-A Page # 1
PART-A
Q.1
Sol. Given, 3
2 III
7dx)x(''f)x3(
32)x('f)x3( + 3
2
7dx)x('f
0 (f '(2)) + f(3)f(2) + 7 f(3) = f '(2) + f(2) + 7 = 4 + (1) + 7 = 10. Ans.
Q.2
Sol. Given, tan1 x + tan1 y + tan1(xy) =12
11...(1)
At x = 1;4
+ 2 tan1 y =
12
11 2 tan1 y =
3
2 y = 3
Differentiate both sides of equation (1) with respect to x, we get
2x1
1
+ 2
y1
1
y ' + 2
)xy(1
1
(xy' + y) = 0
21 +
41 y' +
41 (y' + 3 ) = 0
21 +
43 +
21 y' = 0 2 + 3 + 2y' = 0
So, y' =12
3. Ans.
Q.3
[Sol. (a + b + c2) (a + b2 + c) = | a + b + c2 |2 = (a2 + b2 + c2abbcca) (ab)2 + (bc)2 + (ca)2 = 2 (a = b, | bc | = 1) or (b = c, | ac | = 1) or (c = a, | ab | = 1)| bc | = 1
(b, c) = {(2, 1), (1, 2), (3, 2), (2, 3), (4, 3), (3, 4), (5, 4), (4, 5), (6, 5), (5, 6)}
Required probability =666
103
=36
5Ans.]
Q.4
Sol. As, f(0+) = f(0) = f(0) = 0, so f(x) is continuous at x = 0.
Further, f(0 + h) > f(0) and f(0h) > f(0) where h is sufficiently small positive quantity.
Hence, f(x) has local minimum at x = 0.
Q.5
[Sol. We have, dyy
dxydyx2
y
xd =dy
y
x=y + C
As, y(1) = 1 C = 2 y
x=y + 2
Now, y() =3 3
= 5 = 15. Ans.]
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MATHEMATICS
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Q.6
[Sol. Given, f (x) =xtan1
9xtan4xtan2
2
=
xtan1
)xtan2(22
+
xtan1
xtan14
2
2
+ 5 = 2 sin 2x + 4 cos 2x + 5
Rf= 205,205
Hence, (M + m) = 10 Ans.
Aliter : f(x) = xtan1
xtan2
2
+ xtan1
xtan42
2
+ xtan1
92 = sin2x + 2 sin 2x + 9 cos2x
= 1 + 4(1 + cos 2x) + 2 sin 2x = 5 + 2 sin 2x + 4 cos 2x. ]
Paragraph for question nos. 7 to 9
[Sol.mb
Given, f(x) = x2 ex f '(x) = x ex (2x)
+
0 2
Sign scheme of f '(x) O
x
y
4
1y
2e
4y,2x
x = 2
Graph of f(x) = x e2 x
As, g(x) =
xe2
0
2dt
t1
)t('f g'(x) =
)e41(
e2)e2('fx2
xx
=
)e41(
)e1(ee8x2
xe2x2 x
.
Now, verify alternatives. ]
Q.10
Sol. S = {1, 2, 3, ... n}
E1
= No. is divisible by 2.
E2 = No. is divisible by 3.If n = 6k say n = 6
S = {1, 2, 3, 4, 5, 6}
P(E1) =
6
3=
2
1; P(E
2) =
6
2=
3
1
P(E1 E
2) =
6
1= P(E
1).P(E
2) (B) is correct.
If n = 6k + 2 say n = 8
S = {1, 2, 3, 4, 5, 6, 7, 8}
P(E1) =
84 =
21 ; P(E
2) =
82 =
41
Here, P(E1E
2) =
8
1= P(E
1) P(E
2) (C) is correct.
Note that : P(E1 E
2) =
10
1 P(E
1) P(E
2)
Not independent dependent. ]
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MATHEMATICS
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Q.11
[Sol. Given, = 2 tan = tan 2
tan =
2tan1
tan2
2x
y
0
0
=
20
2
0
00
y1x
1xy2
2
02
0
0
0 y)1x(
)1x(2
2x
1
3x02y
02 = 3
(1,0) (2,0)x
y
M(x , y )0 0
(0,0)A B
Locus of M is hyperbola 3x2y2 = 3
Now, verify alternatives. Ans.]
Q.12
[Sol. Given, f(x) =
0x,0
1x0,)x2(cos1xx1
Note: f (x) is not defined at x = 1 for 0 ......(1)f(x) is continuous at x = 0 but for continuous at x = 1.
f (1) = f(1)
2 hsinhLim 20h
= 0 20h
hLim
= 0
+ 2 > 0 >2 ......(2)Hence, (1) (2) 0
Note: f (x) is differentiable in (0, 1). Ans.]
Q.13
[Sol. Now, slope of LM =0t
2
=
t (given)
2 = t2 = 2 + t2
Let mid point of LM is (h, k)
(h, k)
y
x
slope = m = t(given)
(0, 0)
M(0, ) L (t, 2)
Now, 2h = t and 2k = + 2 = t2 + 4 On eliminating t, we get
k = 2(h2 + 1)
Locus of (h, k) is y = 2(x2 + 1) x2 =2
1(y2)
Now, verify alternative. Ans.]
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MATHEMATICS
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PART-C
Q.1
[Sol. Clearly, the given curves intersect at x = 0,1k
k2
Required area = dxk
xkxx
1k
k
0
22
2
= 2
k1k6
1
Now, for maximum area,2
k
1k
will be minimum k +k
1= 2 k = 1. Ans.]
Q.2
[Sol. Let c = 14x, a = 14, b = 14 + x
Now, cos B =13
5 (14 + x)2 = (14x)2 + 1422(14x) 14
13
5 x = 1.
So, a = 14, b = 15, c = 13.
Hence, r =s
=
21
84= 4. Ans.]
Q.3
Sol. Sn
= C0C
1+ C
1C
2+ ... + C
n1C
n
Sn
= nC0
.nCn1
+ nC1
.nCn2
+ ... + nCn1
.nC0
= 2nCn1
Sn
= 2nCn1
Now,n
1n
SS =
415
1nn2
n
2n2
CC
=4
15 415
!n2)!1n()!1n(
)!2n(!n)!2n2(
n)2n(
)1n2)(1n(
=8
158(2n2 + 3n + 1) = 15n2 + 30n
16n2 + 24n + 8 = 15n2 + 30n n26n + 8 = 0 (n4) (n2) = 0 n = 2 or 4 Sum of all values of n = 6 ]
Q.4
[Sol. P(HHH) =
33
6
1
3
1
118
5
2
1
18
13
= 8
1
18
5
8
1
18
13
= 8
1
p + q = 9 Ans.]
Q.5[Sol Given, A = (I + B) (IB)1
Now, AT = T1)BI(BI = T1)BI( (I + B)T
= 1T)BI( (I + BT) = (I + B)1 (IB)Also, AAT = (I + B) (IB)1 (I + B)1 (IB)
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MATHEMATICS
Code-A Page # 5
= (I + B) ((I + B) (IB))1(IB) = (I + B) ((IB) (I + B))1 (IB)= (I + B) (I + B)1 (IB)1 (IB) = I I = I2 = I.
| AAT | = | I | = 1 | A |2 = 1As, | A | > 0 | A | = 1.Hence, det.(2A)det. (adj A) = 8 det.( A)(det.( A))2 = 8(1)(1)2 = 7 Ans.
Q.6
[Sol. Given,dx
dyx2y = x4 y2 or y
x
2
dx
dy = x3y2. Now dividing by y2, we get
x
1
y
2
dx
dy
y
12
= x3 ........(i)
This is a Bernouli's differential equation, substituting ty
2
, we get
dx
dt
dx
dy
y
22
. So, equation (i) becomes
xt
dx
dt
2
1
= x3
tx2
dx
dt
= 2x3
IF =2xnxn2
dxx
2
xeee2
ll
So, general solution is given by
x2t =6
x2 6+ C
y
2x2=
3
x6+ C
y
2=
3
x4+ 2
x
C
If x = 1, y =6 C = 0
3
x
y
2 4
y =
4x
6i.e. f(x) =
4x
6
Now,dx
dy= 24x5 = 5x
24. Hence,
5
1
3xdx
dy
=
3
24= 8. Ans.]
Q.7
Sol. Given equation of planes are
P1
: x + y + 1 = 1 ... (1)
P2
: x + 2ay + z = 2 ... (2)
P3
: ax + a2y + z = 3 ... (3)
If 3 planes intersect in a line, then
0
1aa
1a21
111
2
Applying C1C
1C
2and C
2C
2C
3
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MATHEMATICS
Code-A Page # 6
011aaa
11a2a21
100
22
On expanding along C1, we get
(12a) (a21)(aa2)(2a1) = 0 1 + 3a2a2 = 0 a =2
1, 1.
But, for a = 1, planes P1
and P3
are parallel and for a =2
1, planes P
1and P
2are parallel.
Hence, we conclude that 3 planes will never intersect in a line for any real value of a.
Q.8
[Sol. Clearly, d =22
4
3
2
1
2
1
4
3
=16
1
16
1 =
8
1=
22
1
So, | sin | = d22 | sin | = 1
y
x
0,
2
1
2
1,0
4
3
,2
1
2
1,
4
3
y=
x
O
=2
,
2
,
2
3.
Hence, the number of values of are 3.]
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PHYSICS
Code-A Page # 1
PART-A
Q.1
[Sol. F2fmax
= 0
F = 2fmax
= 2mgfmaxfmax
F
F'f 'max
= 0
F' = f 'max
= 2mg = F
f 'max
F'
m
m ]
Q.2
[Sol. Pressure is atmospheric
]
Q.3
[Sol. P =
....P
V
P
V
P
V
V
3
2
2
2
1
2
2
]
Q.4
[Sol. R =g
2sinu2 =
10
11002
= 1000 m
R
R=
u
u2+
2sin
2cos x2
1000
R=
100
12 R = 20 m
980 m < R < 1020 m ]
Q.5
[Sol. I= I' (angular momentum conservation)
' ='I
I = 2
'T
1= 2
T
1
T' = T
2
= 24
2
]Q.6
[Sol.
m(satellite)
v1
v||v
Earth
mvr = constant ]
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Q.7
[Sol.p
h =
mqV2
h
KEm2
h ]
Q.8
[Sol. Fring width ;d
D
If V decreases, then increases & hence increases. ]
Q.10
[Sol. During discharging ; initial current
Iin
=R
Vin
equal & not zero.
(B) correct
When2
VV in
2
Vin
= RC/tin
eV
t = RC ln (2)R same so lesser time for lesser capacitance.
(D) correct ]
Q.11
[Sol. VA2
1P
22
Also
TV =
Area
T
=2
d
4
T
When T 4 times
V 2 times P 2 times (C) correct
Also d
d4
TAd
42
1P
2
222
(D) correct ]
Q.12
[Sol. Stationary magnet generates magnetic field only which does not affect piece of paper. When it moves, itgenerates electric field also which affects pieces of paper.
Torque due to magnetic force
= MB sin
Here B
is to plane parallel to M
= 0 ]
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Q.13
[Sol. For adiabatic process, PV = constant.
ln P =ln V + constant.
|slope| =
1
> 2
f1
< f2
and1V
C 2V
C
[ f = 12
and Cv
=1
R
]
PART-C
Q.1
[Sol. In both cases, Weight = Bouyant force
Initially, bVg =
w gV
3
2
b=
3
2
w
After wards, bVg =
oil g
6V5
w3
2 =
oil
6
5
oil
=5
4
w=
5
4 100 = 800 kg/m3. ]
Q.2
[Sol. K.Eloss
= P.E.gain
mghR
vI
2
1mv
2
12
2
2
2
R
vI
2
1= mg
g4
v32
2
mv2
1
2
mR2
1I =
2)5.0(8
2
1= 1 kg m2 ]
Q.3.
[Sol. eA (T4T04) = mS
dt
dT
dt
dT=
mS
TTeA4
04
S2
d
3
4
)TT(4
de
4
04
2
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On putting values and solving, we get
dt
dT= 200 103C/s ]
Q.4
[Sol. Train to wall :
f1
=
uv
vf v speed of sound, u speed of train
Wall to train / passenger :
f2
=
v
uvf1
=
uv
uvf = 1800
34340
34340= 2200 Hz ]
Q.5
[Sol.
m
Nuetron
m
H-atom
2m
u/2
Maximum possible loss of energy is2
.E.Kin
when collision is perfectly inelastic as shown.
(K.E.loss
)max
=2
5.25= 12.75 eV
This is sufficient to raise the electron upto 3rdshell. While returning first shell threephotons are possible.]
Q.6
[Sol. Initially,)1(2
Rf
= 10 R = 10 cm
In water
=~ +
1strefraction u1
= ; R = + 10 cm
v
1
)3/4(=
10
)3/4(1 v
1=30 cm
refraction at lens u2 = v1 =30 cm
v2
= fu
fu
2
2
= 1030
)10()30(
= 15 cm ]
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PHYSICS
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Q.7
[Sol.
30 30
q
Q Q
2
kqQ2
cos 30 = ma
2
89
)2.0(
101092
2
3= 3 106 a
a = 9 m/s2 ]
Q.8
[Sol. V1 + V2 + V3 = 120
20 (0.5) V
1+ V
2+ V
3= 110 V .... (1)
Also V1
+ V2
= 60 V3
= 11060 = 50V
Now V2
+ V3
= 90 V2
= 40V
R3R2R1
120V, 20
V1 V2 V3
I=0.5A
R2
=5.0
40= 80 ]
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CHEMISTRY
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PART-A
Q.1
[Sol. Radius of 4th orbit = x 42 16x2r = n 2 16 x = 4 = 8x ]
Q.2
[Sol. ZnS + 2HCl ZnCl2 + H2 S ]
Q.3
[Sol. Presence of electron withdrawing group increases the rate of BAC2
reaction while rate electron
releasing group decrease the rate of BAC2
reaction. ]
Q.4
[Sol. Normal boiling point = 100C at 1 atmStandard boiling point at 1 bar 1 bar < 1 atm
as P Boiling point less than 100C ]
Q.5
[Sol. Only Mn & Mg metals liberates H2
gas with dil. HNO3
M(Mg, Mn) + dil. 2HNO3 M(NO
3)2
+ H2
3Cu + 8HNO3 3Cu(NO
3)2
+ 2NO + 4H2O
4Zn + 10HNO3 4Zn(NO
3)2
+ NH4NO
3+ 3H
2O
or
Fe ]
Q.6
[Sol.
Br2
CF3
Cl
FeBr3
CF3
Cl
Br
Nitration
CF3
ClO N2 Br
NH 2 NH2
CF3
NHNH2
O N2 Br]
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Q.7,8,9
[Sol.
Gas colourless (A)=
KMnO + H4+
Exposure/
FeCl 3
AgN
O3
(X) = SO32
(Y)=S O2 32
Violet/Purple(C)= [Fe(S O ) ]2 3 2
White ppt. (E) = Ag S O2 2 3
(D)= Fe2+
+ S O4 62
Green
Black (F)= Ag S + H SO2 42
[B + X Y]
XZn+HCl
(A)(i)
(ii)
H S2
(B) = S
Fe3+
H2 S + Na2 [Fe(CN)5NO]No reaction colourless solutionH
2S + OH S2 ; Na
2S + Na
2[Fe(CN)
5NO] Na
4[Fe(CN)
5NOS] (Violet colour) ]
Q.10
[Sol. (i) Water is more acidic than EtOH(ii) Bond length is not equal since R.S. are not equivalent.
(iii) Ph2
NH
is weaker base than MeNH2, due to involvement of lone pair of electron in resonance.
(iv) If key atom is lone pair bearing it will show +M effect. ]
Q.12
[Sol. Due to high ionisation energy of Be and Mg ]
Q.13
[Sol. (B)
O
O
(C)
OH
OH
(D)
O
(E)
OH
OH
and can be changed in by using conc. H2SO
4and H
3PO
4]
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PART-C
Q.1
[Sol. Edge length = BAr2r2
= 2 60 + 2 100 320 pmMolecular mass of AB = 320
= 3 A
O
a
N
Mz
310
A
)10320(
N
3204
2.3
x
; x =
50
104
= 200 ]
Q.2
[Sol. NH4NO
3 N2O + 2H2O
NH4
NO2
N2
+ 2H2
O
Except Na+ , K+, Rb+, Cs+ all metal nitrate give NO2
on heating
M(NO3)y M2Oy + NO2 + O2 ]
Q.3
[Sol. Lassaigne method
Sodium Extract + FeSO4
+ NaOH, boil and cool + FeCl3
+ conc. HCl
Blue and green colour and how the presence of nitrogen.]
Q.4
[Sol. S =
asp
K
]H[1K
S =
5.310
101105.2
5
413
= 36105.2 13 =30 106 ]
Q.5
[Sol. CH3
CH2
CHOCH CH=CH2 2
OH O
H CH =CH/G.I.3 C
OH C CH3H C3
O
C =CH2H C3
OCH
2= CHOCH
3
O]
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Q.6
[Sol. (a) PH4Cl + NaOH PH3 + H2O + NaCl
(b) P4
+ 3 NaOH + 3H2O PH3 +3NaH2PO2
(c) Ca3P
2+ 6H
2O 2PH3 + 3Ca(OH)2
(d) 4H3PO
3 PH
3 + 3H
3PO
4
(e) PCl5 + H2SO4 SO2Cl2 + 2POCl3 + 2HCl
(f) 2H3PO
2 2PH
3 + H
3PO
4
(g) P2Cl
4+ 4H
2O H4P2O4 + 4HCl
(h) 2AlP + 3H2SO
4 Al2(SO4)3 + 2PH3
(i) P4O
10+ 6H
2O 4H3PO4 ]
Q.7
[Sol. 2CH4
+ 4O2 2CO
2+ 4H
2O
H =900 2 =1800 kJ
G = o Reactantfo
Productsf GG
= (400) 2 + 4 (120)[2(40) + 0]=800480 + 80=1200 kJ / moleG = HTS
T
GHS
osystem
=
300
kJ)1200()1800( = kJ
300
600
=2 kJ =2000 J / KFinal answer 4000 Ans. ]
Q.8
[Sol. (a)
Cl
= 1
Cl
= 4
Cl
= 2Cl
= 4
Cl= 1
Cl= 2
x = 14 Ans. ]