rt solutions-01!01!2012 xii abcd paper ii code a
TRANSCRIPT
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12th ABCD (Date: 01-01-2012) Final Test
PAPER-2
Code-A
ANSWER KEY
MATHS
SECTION-1
PART-A
Q.1 A
Q.2 B
Q.3 C
Q.4 C
Q.5 D
Q.6 C
Q.7 B
Q.8 C
Q.9 B
Q.10 D
Q.11 C
Q.12 A
Q.13 B
Q.14 C
Q.15 D
Q.16 B
PART-B
Q.1 (A) S
(B) Q
(C) Q
PART-C
Q.1 0140
Q.2 0005
Q.3 0025
Q.4 0005
PHYSICS
SECTION-2
PART-A
Q.1 C
Q.2 B
Q.3 C
Q.4 D
Q.5 C
Q.6 B
Q.7 C
Q.8 D
Q.9 A
Q.10 C
Q.11 C
Q.12 C
Q.13 B
Q.14 D
Q.15 A
Q.16 D
PART-B
Q.1 (A) R
(B) P,Q,S
(C) P,Q,R,S
PART-C
Q.1 0100
Q.2 0020
Q.3 0007
Q.4 0096
CHEMISTRY
SECTION-3
PART-A
Q.1 A
Q.2 B
Q.3 A
Q.4 C
Q.5 A
Q.6 D
Q.7 C
Q.8 D
Q.9 C
Q.10 C
Q.11 B
Q.12 C
Q.13 D
Q.14 B
Q.15 A
Q.16 A
PART-B
Q.1 (A) Q
(B) R,S
(C) P,S
PART-C
Q.1 0005
Q.2 0175
Q.3 0003
Q.4 7268
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MATHEMATICS
Code-A Page # 1
PART-A
Q.1
[Sol. Tangent to the parabola
y2 = 4x is y = mx +m
1....(i)
m2x my + 1 = 0,
As, it touches the circle x2 + y2 = 1, soS(1,0)O
y
x
1mm
1
24
m4 + m21 = 0
m2 = tan2 =2
411 =
4
152
2
15= 2 sin 18 Ans.]
Q.2
[Sol. The equation of the tangent is)i..(..........1
2
3
b
y
2
1
a
x
Auxiliary circle is x2 + y2 = a2.................(ii)
C is the centre.
Combined equation of CL, CM is obtained by homgenising (ii) with (i), i.e.,
x2 + y2a2 0b2
y3
a2
x2
Since LCM = 90
1
4
1+1 0
b4
a32
2
4
7
b4
a32
2
90
M
P( =60)
(0, 0)C
y
x
L
7b2 = 3a2 7 a2 (1e2) = 3a2
Hence e =7
2Ans. ]
Q.3
[Sol. We must have |2 4| < 5 5 < 2 4 < 51 < ( 2)2 < 9 0 ( 2)2 < 9 3 < 2 < 3
D(z)
2 F1F2
1 < < 5 0 < < 5 = 1, 2, 3, 4 4 values. ][Note : = 0 is not possible because, | z 2 | + | z 4| = 5 will represent a circle.]
Q.4
[Sol. From above figure,
P(C (A B)') = 1
5
1
15
1
10
1 BA
S
C=
30
62330 =
30
19Ans.
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MATHEMATICS
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Q.5
[Sol. Equation of line through O(0, 0, 0) and perpendicular to the plane 2x y z = 4, is
1
0z
1
0y
2
0x
= t (let)
Any point on it is (2t, t, t)As above point lies on the plane 3x 5y + 2z = 6, so
6t + 5t 2t = 6 9t = 6 t =
3
2.
Co-ordinates of point of intersection are
3
2,
3
2,
3
4 (x
0, y
0, z
0) [Given]
Hence, (2x03y0 + z0) = 4 Ans.]
Q.6
Sol. Equation of chord of contact with respect to point (4, 2) is
2a
x4 2
b
y2= 1 and with respect to point (2, 1) is 1
b
y
a
x222
.
Now, according to given condition,
2
2
2
2
b
1a
2
b
2a
4
= 1 4
4
a
b=
4
1 2
2
a
b=
2
1
Now, e =2
3
2
11
a
b1
2
2
Ans.
Q.7
Sol. As, z lies on the curve arg(z + i) =4
, which is a ray originating from (i) and lying right side of
imaginary axis making an angle4
with the real axis in anticlockwise sense.
O45
Re(z)
Im(z)
(4+3i)
(43i)
The value of | z(4 + 3i) | + | z (4 3i) | will be minimum when z, 4 + 3i, 4 3i are collinear. Minimum value = distance between (4 + 3i) and (4 3i)
= 22 )33()44( = 3664 = 100 = 10. Ans.]
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Q.8
Sol. Equation of normal at P is
(y 1) = 2(x 1) 2x + y = 3
At x =4
1, y =
2
7
0,
4
1
y,
4
1
4
1x
P(1, 1)
C
y
radius =
22
2
71
4
11
=
4
25
16
25 =
4
55Ans.
Paragraph for question nos. 9 to 11
[Sol. Let dr's of line L be , so equation of line L is
kcjbiatr
......(1)
If L intersects L1
at P, so shortest distance between them is zero.
0 =
kj2ikcjbiakj2ikcjbiakji2
3
4,
3
10,
3
10
Q
(5, 5, 2)P
0O
L : r =2
L : r =1
8i3
3j + k
+ (2i j+ k)
(2i + j k)+ (i 2j + k)
121
cba
kji
kji2
= 0 a + 3b + 5c = 0 .......(2)
Similarly, L intersects L2
at Q, so shortest distance between them is zero.
0 =
kji2kcjbia
kji2kcjbiakj33
i8
112
cba
kji
kj33
i8
= 0
3a + b 5c = 0 .......(3)
On solving (2) and (3), we get2
c
5
b
5
a
.
So, the equation of line L is k2j5i5tr
.
Any point on line L is A' (5t, 5t, 2t). If A' is the point of intersection of L and L1, so A' will also satisfy
L1, we get
1
1t2
2
1t5
1
2t5
2 (5t 2) = 5t 1 10t + 4 = 5t 1 t = 1So, co-ordinates of P are (5, 5, 2).
Similarly, if A' (5t, 5t, 2t) is the point of intersection of L and L2, so A' will also satisfy L2, we get
1
1t2
1
3t5
23
8t5
5t + 3 = 1 2t t =
3
2
So, co-ordinates of Q are
3
4,
3
10,
3
10
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Code-A Page # 4
(i) Given M (1, 2, 3) and N (2 + , 1 2, 1)
MN = P.v of N P.v ot M = k)4(j)21(i)1(
Also, n
= normal vector of plane k3j4ir
= k3j4i .
Now, nMN
= 0 1 ( + 1) + 4 (1 + 2) + 3 ( 4) = 0
12 = 7 =12
7 Option (B) is correct.
(ii) The normal vector of the plane through P (5, 5, 2) and Q
3
4,
3
10,
3
10and perpendicular to the
plane kjir
+ 1 = 0 is parallel to the vector = kjiPQn
=
1113
2
3
5
3
5kji
=
111
255
kji
3
1
= k0j3i3
3
1 = ji
The required equation of plane, is 1 (x 5) + 1 (y + 5) + 0 (z 2) = 0
x + y = 0 or jir
= 0 Option (D) is correct
(iii) Volume of tetrahedron OPAB (where O is origin) = OBOAOP6
= |
502
310
255
|6
1
=6
1[5(5) + 5(6) + 2(2)] =
6
51=
2
17. Option (C) is correct.]
Paragraph for question nos. 12 to 14
Sol. For the given ellipse, 116y
25x
22
, 53
25161e . So, eccentricity of hyperbola = 3
5 .
Let the hyperbola be, 1B
y
A
x2
2
2
2
... (1)
Then, B2 = A2
1
9
25=
9
16A2. Also, foci of ellipse are (3, 0).
As, hyperbola passes through (3, 0). So, 2A
9= 1 A2 = 9, B2 = 16
Equation of hyperbola is 116
y
9
x 22
(i) Vertices of hyperbola are (3, 0) (A) is correct.Focal length of hyperbola = 10 (B) is incorrect.
Equation of directrices of hyperbola are x = 5
9. (C) is incorrect.
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Code-A Page # 5
(ii) Any point of hyperbola is P(3sec, 4tan).Equation of auxiliary circle of ellipse is x2 + y2 = 25.
Equation of chord of contact to the circle x2 + y2 = 25, with respect to P(3 sec, 4 tan), is3x sec + 4y tan = 25 ... (1)
If (h, k) is the mid point of chord of contact, then its equation is
hx + ky 25 = h2 + k225 hx + ky = h2 + k2 ... (2)As, equations (1) and (2) represent the same straight line, so on comparing, we get
22 kh
25
k
tan4
h
sec3
sec =
22 kh
25.
3
h, tan =
4
k
kh
2522
Eliminating , we get,
2
22kh
25
16
k
9
h22
= 1. (As, sec2tan2 = 1)
Locus of (h, k) is
22222
25
yx
16
y
9
x
(iii) Required area of quadrilateral = 2(A2 + B2) = 2(9 + 16) = 50 Ans.]
Q.15
[Sol. Statement-1: We have baxa
0bxa
atbx
, for some scalar t.
atbx
= k)t31(j)t21(i)t2(
Now, xa
= 0 t =2
1
2
k
2
i3x
x
= 4
1
4
9
= 2
5
Statement-1 is false.
Obviously, Statement-2 is true. ]
Q.16
[Sol. Option (B) is true.
S-1: As, f ' (1+) = 0 = f '(11) f is differentiable at x = 1.So, f is differentaible x R.
S-2: As, f(1 + h) < f(1) < f (1 h).
Where h is sufficienty small positive quantity, so f(x) is decreasing at x = 1.
f(x) has neither local maximum nor local minimum at x = 1.
But, S-2 is not explaining S-1. Ans.]
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Code-A Page # 6
PART-B
Q.1
[Sol.
(A) Clearly, ABC is equilateral.
Now, ar. (ABC) =2
zz4
3
=2
z4
33= 348 (Given)
B( z)
A(z)
Re(z)
C( z) 2
2 /3
2 /3
Im(z)
O(0,0)
64z2
8z . Ans.]
(B) Equation of chord of hyperbola 11
y
2
x22
, whose mid-point is (h, k) is
2
hxky =
2
h2
1
k2
(using T = S1)
As, it is tangent to the circle x2 + y2 = 4, so
22
22
k4
h
k2
h
= 2
2
22
22
k4
h4k
2
h
Locus of (h, k) is (x22y2)2 = 4(x2 + 4y2) = 4.
(C) We know that ac = (semi-minor axis)2 = 4
Now,
4
4
dx}x2{ = 2/1
0
dx}x2{16 =
2
18
2
18
dx}x2{ = 2
1
0
dx}x2{16 = 2
1
0
dxx216 = 2
1
0
dxx32
= 322
1
0
2
2
x
= 32
8
1= 4 Ans. ]
PART-C
Q.1
[Sol. p = sinv
=|c|cv
= |c|
cv|c|v222
=3
4)3)(6( =
3
14
B(3,5,2) M Q
A(2,3,1)v=(1, 2, 1)
c=(1, 1, 1)
rr r
r
= v ^ cp
30p2 = (30)
3
14= 140 Ans.]
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Code-A Page # 7
Q.2
[Sol. 2
4
5
2
0
4
1
0
4
5
4
1
dxxcosdxxsindxxcosdx)x(F
=
2
45
4
5
41
4
1
0
xsinxcosxsin
y
x
O 1/4
5/4 2
=
2
1
2
2
2
1=
22
10
10
2
024
10dx)x(F
24
10dx)x(F
24
22= 5. Ans.]
Q.3
[Sol. Equation of normal at P1
(4 cos 1, 3 sin
1) is
7sin
y3cos
x4
11
... (1)
Also, equation of CQ1
is
y =
1
1
cos
sinx ... (2)
Solving (1) and (2), we get
1cos
x4
1sin
3
1
1
cos
sinx = 7
y
xC
(0, 0)
P (4 cos , 3sin )1 1 1 x +y =162 2
K1
1
Q1 1 1(4 cos , 4sin )
1
cos
x
= 7 x = 7 cos 1, y = 7 sin 1
So, K1
= (7 cos 1, 7 sin
1) CK
1= 7
Similarly, CK2
= CK3
= ..... = CKn
= 7
n
1ii
175CK 7n = 175 n =7
175= 25. Ans.]
Q.4
[Sol. Given, y = x2
Now,axdx
dy
=ax
x2 = 2a
Equation of tangent is (y a2) = 2a (x a) BOx
y
A (a, a )2
0,
2
a
Put y = 0, we get
x = a 2
a=
2
a.
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Code-A Page # 8
Also, equation of OA is y = ax
Area = a
0
2dx)xax( = k
2
a
2
a2
a
0
32
3
x
2
ax
=
4
ka3
3
a
2
a33
=4
ka3
6
a3
=4
ka3
k =3
2
q
p(Given) (p + q)
least= 5. Ans.]
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Code-A Page # 1
PART-A
Q.2
[Sol. Temperature quickly rises for monoatomic gas as Q = nCvT so it exert more pressure as compare to
diatomic gas. ]
Q.3
[Sol. f f + 4 f + 8 ....... 2f = f + 4 (561)
Third one f + 8 = 228 Hz. ]
Q.4
[Sol. 2f
f= 2u
u+ 2v
vu =v for optical bench
21
f d(f) =
errorminfor33 du
dv
v
2
u
2u
= 0
du
dv= 3
3
u
v= 2
2
u
v v = u
for u = 2f, error is min. ]
Q.5
[Sol. from left : cm65.1
x
Bx
t
=3/2
from right : cm45.1
xt
so t = 15 cm ]
Q.6
[Sol. =r2 [where is a constant]Charge in the shell (element)
dq = (4r2dr) = (4)r4dr
Charge enclosed in sphere of radius r, q = 4 r
0
4drr =5
r45
By Gauss's theorem,
at r = R/2 (Er = R/2
)
2
2
R4 =
0
5
5
)2/R(4
Er = R/2
=
0
3
40
R
Total charge enclosed, Q =5
R45
Er = 2R
=0
5
5
R4
2)R2(4
1
=
0
3
20
R
2/Rr
R2r
E
E
= 2 ]
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Code-A Page # 2
Q.7
[Sol. Vmin
= lg5 ]
Q.8
[Sol. Com remains at rest. ]m1 m2
L
x1 x2
CM 21
21 mm
Lmx
]
Q.11
[Sol. N.L.
fr=0equili.
mg sin
x0kx0
At equilibrium kx0= mg sin
k(x + x0)mg sin fr = ma
Equi.
fr mg sin
xk(x0 0+x )
a
R
a
for R = (MR2
)
R
a
from here we get a = xm2
k,
m2
k0
If instead solid cylinder is used then2
MRI
2
then xm3
k2a , 0m3
k2'
amplitude does not depend on physical system. ]
Q.14
[Sol. Loop is at rest so only induced electric field. Flux through loop after a time t = t
2/3200
2
220
])tvz(R[2
tbrNR
(z v t)0 0
v t0z=z0
2/3200
2
20
])tvz(R[2
NIRB
]
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PHYSICS
Code-A Page # 3
Q.15
[Sol. Since Einside
conductor = 0
dsE
over guassian surface = 0
q0
dsE
=0
inq
=0
surfaceonin0 qq
qin
on surface =q0
]
Q.16
[Sol. Speed increases & becomes constant. ]
PART-B
Q.1
[Sol.(A) (P) : There is no convectional current = 0(R) : At the position shown E
is downward & later it becomes upward avg = 0 for other locations
also with the same argument.(Q) : Since I = 0 B = 0
(S) : B
& since 0
]
PART-C
Q.1
[Sol. =1000
2501 =
400
Q1 L
QL
= 100 J ]
Q.2
[Sol. 4He21H
1+ 3H
1
m = Hem 42
11 Hm 13 Hm =0.021271 E = mc2
=19.81 MeV
ve sign shows that energy is supplied. ]
Q.3
[Sol. = LBHm = 0.655796
d
=m
dm+
L
dL+
B
dB+
H
dH
d
=
50.3
05.0
00.3
05.0
50.4
05.0
32
1= 0.07045
d
= 7% ]
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Q.4
[Sol. V =3
r3
4 .... (1)
2r4
Q
.... (2)
p0
2/2 +p0
r
s4pp2 0
0
2
.... (3)
pp0
= 0 .... (4)
r
s4
r16
Q
2
142
2
0
s4r34
r16
sn
2
1 332
0
0
n = 96 ]
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PART-A
Q.1
[Sol. Due to maximum hydration of Li+ ]
Q.2
[Sol. Sb2S
3is the negatively charged sol. ]
Q.3
[Sol.
OH
+ HCHO tionPolymerisa
Bakellite ]
Q.4
[Sol. Boron dissolves into concentrated oxidising acids. ]
Q.5
[Sol.O + O
3 2
1 mol
moles of O3
=100
180 = 0.8 mole
O3
O2
+ [ O ]
0.8 0.8 mol
2I + [O] I2
2 0.8 0.8= 1.6 mol ]
Q.6[Sol. Due to Back bonding ]
Q.7
[Sol. Non reducing suger's do not show the phenomena of muta-rotation.]
Q.8
[Sol. Vander Waal's force sizeBoiling point Vander waal's forces ]
Q.9,10,11
[Sol. (9) P = 350 mm of Hg
BoBA
oAT xPxPP
20 10
9400
10
1
= 362
P < PT
ve deviationV
mix< 0 Ans.
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(10) Tb
= Kb m
= 2.7 1009
10001
=9
27= 3
Tb
solution = Tb
sovlvent + Tb
= 300 + 3 = 303 Ans.
(11) 2A A2
1 0
1 /2
= 0.6 = 0.4 0.3 Total = 0.7P
S= P solvent
= 400 7.09
9
PS
=7.9
3600mm of Hg ]
Q.12,13,14
[Sol.
CHO
OO
O
OH O
NH2Br
OH
O3 KOH
KOH
CrOH NNH22 3
HBr NH3H2/Pd
excess excess
EtONaZn/H O2
P1
P5
P2
P3
P4
P6
P7
]
Q.15
[Sol. I /Br + conc H2SO
4 I
2/ Br
2
3H2SO
4+ 2KX 2KHSO
4+ X
2 + SO
2+ 2H
2O ]
PART-B
Q.1
[Sol. (A) For zero order reaction : x = kt
or t1/8
=k8
a
t1/8
8k = a
t1/4
=k4
a
=k4
k8t8/1
t1/4
= 10 2 = 20 min Q
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(B) For first order reaction :
k =xa
alog
t
303.2
t3/4
=a4/1
alog
t
303.2= 4log
k
303.2
t7/8
= 8logk
303.2
a8
1
alog
k
303.2
3
2
t
t
8/7
4/3 3
2
t
20
8/7
8/7t2
60
8/7t = 30 RIt takes time to complete first order reaction.S
(C) P,S(If order > 1 then completion time is infinite) ]
PART-CQ.1
[Sol. If oxidiation state of non metal is greater than highest oxidation state considering oxidation state of
oxygen is (2) then there will be peroxy linkage H3PO
5, H
2SO
5, H
2S
2O
8, H
4P
2O
8, HNO
4]
Q.2
[Sol. A(s) | A(aq) (0.5 M) || B+(aq) (0.5M) | B
A A2+ + 2e2B+2e 2B_________________
A + 2B+ A2+ + 2B .... (i)_________________
0.991 =5.05.0
5.0log
2
06.0Eocell
(log2 = 0.3)
1Eocell
ab = 01
When current stops , Ecell
= 0 and reaction (i) is almost 100 % complete
A + 2B+ A2+ + 2B0.5 0.5
At eq. 0 0.5 + 2
5.0
= 0.75
cd = 100 [A2+]= 75
abcd = 0175 ]
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Code A Page # 4
Q.3
[Sol. This reaction represents chromyl chloride test
A = salt containing ClB = CrO
2Cl
2(d3S) hybridisation of Cr
C = CrO42
D = PbCrO4
]
Q.4
[Sol.
CH=CH2
3AlCl/ClCMe||O
CMe
HClZnHg
CH CH2 3
NBS
CHCH3
Br
Me CO Na3
CH=CH2
]