-
7/30/2019 Rt Solutions-04!03!2012 XIII VXY Paper I Final Test Code a Sol
1/19
13th VXY (Date: 04-03-2012) Final Test
PAPER-1
Code-A
ANSWER KEY
PHYSICS
SECTION-2PART-A
Q.1 B
Q.2 B
Q.3 A
Q.4 B
Q.5 C
Q.6 C
Q.7 D
Q.8 A
Q.9 D
Q.10 C
Q.11 A,C
Q.12 C
Q.13 B,C
Q.14 B,D
Q.15 C,D
PART-C
Q.1 0050
Q.2 0006
Q.3 0006
Q.4 0400
Q.5 1000
Q.6 0003
MATHS
SECTION-1PART-A
Q.1 C
Q.2 C
Q.3 D
Q.4 C
Q.5 A
Q.6 B
Q.7 B
Q.8 C
Q.9 A
Q.10 C
Q.11 B,C,D
Q.12 A,C,D
Q.13 B,C
Q.14 A,B,C
Q.15 B,C
PART-C
Q.1 0004
Q.2 0055
Q.3 0025
Q.4 0019
Q.5 0641
Q.6 0020
CHEMISTRY
SECTION-3PART-A
Q.1 B
Q.2 C
Q.3 B
Q.4 B
Q.5 C
Q.6 A
Q.7 CQ.8 B
Q.9 A
Q.10 C
Q.11 A,B,C
Q.12 A,B,C
Q.13 B
Q.14 A,B,C,D
Q.15 A,C
PART-C
Q.1 0050
Q.2 0009
Q.3 Bonus
Instruction for Q.4
Partial Marks will be awarded (+2) for correct
value of 'ab' (+2) for correct value of 'cd' and
(+1) as bonus if both are correct.
Q.4 ab = 20 and cd = 28
Q.5 0006
Q.6 0094
-
7/30/2019 Rt Solutions-04!03!2012 XIII VXY Paper I Final Test Code a Sol
2/19
MATHEMATICS
Code-A Page # 1
PART-A
Q.1
[Sol. I =
1
1
xxxx dx6321nx l
Using King
I =
1
1
xxxxdx6321nx l
2I =
1
1
xxx
xxx
dx6321
6321nx l =
1
1xxx
xxxx
dx1236
63216nx l
(Multiplying Nr and Dr by 6x)
=
1
1
2 dx6nx l
2I = 2 ln 6 1
0
2 dxx =3
2ln 6 I = 3
1ln 6. Ans.]
Q.2
[Sol.
Let g(x) = ex f(x)
g'(x) = ex (f '(x) f(x)) > 0 Rxxx0
gI
g(x) is increasing on R
g(x) > g(x0) 0xx e
x f(x) >
0
0
x)x(fe 0
f(x) > 0 0xx [given f (x0) = 0]
Q.3[Sol. Since f(x) is continuous at x = 0, so at x = 0, both left and right limits must exist and both must be
equal to 3.
Now,2x
5xcos)xsinx1( =
2
2
x
.....x2
)5(
(Using series expansion of sin x and cos x)
If )x(fLim0x
exist, then + + 5 = 0 and )x(fLim0x
= 3 2
= 3
= 1, =
4.
Since, 3x
xx1Lim
x
1
2
3
0x
0
x
xxLim
2
3
0x
Now, )1(x1Lim x1
0x
= e = 3 e2 = 9
Hence, 2 + 2 + 2 + e2 = ( 1)2 + ( 4)2 + (0)2 + (3)2 = 1 + 16 + 0 + 9 = 26. Ans.]
-
7/30/2019 Rt Solutions-04!03!2012 XIII VXY Paper I Final Test Code a Sol
3/19
MATHEMATICS
Code-A Page # 2
Q.4
Sol. Domain is x [1, 1]Given, sin1x = 2tan1x
2
1
x1
xtan =
2
1
x1
x2tan
2x1
x2
= 2
x1
x2
x = 0 or (1 x2)2 = 4 (1 x2) (1 x2) (3 + x2) = 0 x = 1, 0, 1 .......(1)
tan1 )1x(x + cosec1 2xx1 = 2
x(x 1) 0 x x2 0 x(x 1) = 0 x = 0, 1 .......(2)Now, (1) (2) gives x = 0, 1Hence, number of common solution are 2. Ans.]
Q.5
[Sol. Equation of tangent at M
2
1,3 is 3 x + 2y = 4
Also, distance of PQ from (0, 0) =7
4O
x
y
P
QM
x + y = 52 211
y
4
x 22
2
1,3
Now, PQ = 27
165 = 2
7
19
Area of (OPQ) = 719
27
4
2
1
= 7194
. Ans.]
Q.6
[Sol. f(x) = sin1x + cos1x + tan1
x
1+ tan1(x)
x [ 1, 0) f(x) =22
= 0
y
(1, )
(1,0)(1,0)x
(0, /2)
x (0, 1] f(x) =
22Required area (shaded region)
= 2 2
122
1
sq. units. Ans.]
-
7/30/2019 Rt Solutions-04!03!2012 XIII VXY Paper I Final Test Code a Sol
4/19
-
7/30/2019 Rt Solutions-04!03!2012 XIII VXY Paper I Final Test Code a Sol
5/19
MATHEMATICS
Code-A Page # 4
(ii) Acute angle between L1
= 0 and P1
= 0 = P2
(x-axis)
cos =001109
0)1(0013
00,1,:axis-xofsDr'
10,3,:0LofsDr' 1
cos =10
3 cot = 3 = cot1 3 = cot1
= 3 Ans.(iii) Image of the point (5, 0, 3) in the plane P
2(x-y plane) is (5, 0, 3)
Point of intersection of the line L1 = 0 with x-y plane is -(3l + 2, 0 l + 4) l + 4 = 0 l = 4 (14, 0, 0)Now line passing through (5, 0, 3) & (14, 0, 0) is
3
z
0
y
9
14x
1
z
0
y
3
14x
Ans. ]
Q.11
[Sol. As, | cc1 cc
2| = | (r + r
1) (r + r
2) | = constant
where | r1 r
2| < c
1c
2
locus of C is a hyperbola with foci c1 and c2 i.e., (4, 0) and (4, 0).
Also, 2a = | r1 r
2| = 2 a = 1
Now, e =a2
ae2=
2
8= 4
C2C1
(4, 0)(4, 0)
C
So, b2 = 12 (42 1) = 15
Hence, locus of centre of circle is hyperbola, whose equation is15
y
1
x22
= 1.
Now, verify the options. Ans.]
Q.12Sol. Coordinates of B (p, q p2)
Coordinates of C
a
b1,
a
b2
Since mid point of B and C is origin. Hence2
a
bp
= 0 ap + b = 0 (D)
Similarly q p2 +a
b1
2
= 0 (C)
Since a > 0 and abscissa of point C is positive. Hencea
b> 0 b is negative
Hence ab < 0 (Option B is false).
Sum of the roots of equation (x2 + 2px + q) (ax2 + 2bx + 1) = 0 is
a
b2p2 = 2
a
bp =
a
2(ap + b) = 0 (A) is true. ]
-
7/30/2019 Rt Solutions-04!03!2012 XIII VXY Paper I Final Test Code a Sol
6/19
MATHEMATICS
Code-A Page # 5
Q.13
[Sol.514/bin
10
2
2 2x
1x
=20
x
1x
[11th, 02-01-2011,P-2, PQ]
Tr + 1
= 20Crx20 r(1)r
rx
1= 20C
rx20 2r(1)r
20 2r = 0 r = 10So, term independent of x = 20C
10. Now verify alternatives. Ans.]
Q.14
[Sol. f(x) = x2 1
0
1
0
1
0
22dtxdt)t(fx2dt)t(f
f(x) = x2 + A + 2Bx x2 = A + 2Bx
A = 1
0
1
0
22 dtBt2Adt)t(f
B = 1
0
1
0
dtBt2Adt)t(f
f(x) = 3x 3Now verify all the option. Ans.]
Q.15
[Sol.
(A) disjoint mutually exclusiveTwo mutually exclusive events E and F with P(E) and P(F) > 0 cannot be independent
(B) P(E) + )FE(P + )FE(P = P(E) + P(F) P(E F) + )FE(P = 1 Exhaustive
(C) FEP > P(E) or)F(P
)FE(P > P(E) ......(1)
Now,)E(P
)EF(P >
)E(P
)F(P)E(P> P(F) (C) is correct.
(D) Every element of set of positive integer can be of the form 6n, 6n + 1, ......., 6n + 5.
But favourable are 6n + 1 or 6n + 5.
So, required probability =6
2=
3
1. Ans.]
-
7/30/2019 Rt Solutions-04!03!2012 XIII VXY Paper I Final Test Code a Sol
7/19
MATHEMATICS
Code-A Page # 6
PART-C
Q.1
[Sol. Applying cosine rule, 2ab cosC = a2 + b2 c2 22 22 cos 8
= 2 + 2 + 2 c2
using cos8
=
2
22
2
2222
= 4 + 2 c2 c2 = 2 2 c = 22
Hence abc=
22222 = 2 (abc)2
= 4 Ans.]
Q.2
[Sol. A =
ww0b11
ca1
| A | = w(1 + b a c) 0 (for non-singular matrices) 1 + b a + cNow total matrices = 4 4 4 = 64 and 1 + b = a + c
2
2
4
1
i
i
1i
1
i
i
1i
i
i
1
1
i
i
1
1
1
111
cab
Total possibility = 9Total possibility of 1 + b = a + c are 9Hence 64 9 = 55 Ans.]
Q.3
[Sol.22
221
1
at
at2
at
at2 = 1
t1t2
= 4
M
T
90
Q(t )2
P(t )1
at t , a(t + t )1 2 1 2
O
N
x
y
(h,k)
Now T a t1t2, a(t
1+ t
2)
and N a )2tttt( 212221 , a t1t2(t1 + t2)Hence 2h = a((t
1+ t
2)2 + 2)
& 2k = a(t1
+ t2) (1 t
1t2)
2k = 5a(t1
+ t2)
t1
+ t2
=a5
k2
2h =
2
a5
k22
2h =
2
22
a25a50k4
50ah = 4k2 + 50a2
25ah = 2k2 + 25a2
25 2(h 2) = 2k2 or y2 =2
25(x 1)
Hence latus rectum =2
25 2L = 2
2
25= 25. Ans.]
-
7/30/2019 Rt Solutions-04!03!2012 XIII VXY Paper I Final Test Code a Sol
8/19
MATHEMATICS
Code-A Page # 7
Q.4
[Sol. Bowl A
6R
4B
5 ballsPut inBowl B
Let A : One ball drawn from the bowl B found to be blue.
B1
: 1 R + 4 B from bowl A to B.
B2
: 2 R + 3 B from bowl A to B.
B3
: 3 R + 2 B from bowl A to B.
B4 : 4 R + 1 B from bowl A to B.B
5: 5 R + 0 B from bowl A to B.
P (B1) =
510
44
16
C
CC=
510C
6; P (B
2) =
510
34
26
C
CC=
510 C
60; P (B
3) =
510
24
36
C
CC=
510 C
120
P (B4) =
510
14
46
C
CC=
510C
60; P (B
5) =
510
04
56
C
CC=
510 C
6
1BAP = 54
; 2BAP = 53
; 3BAP = 52
; 4BAP = 51
; 5BAP = 0
ABP 2 =
5
1i
ii
22
BAPBP
BAPBP
ABP 2 =124836
5
24
36
=
1435
2
3
=
5
402
3
= 4215
=14
5
n
m
Hence (m + n) = 19. Ans.]
Q.5[Sol. log
2| z 3 | > log
2| z 3 | > 0
log2| z 3i | > log
2| z 3 | (z 3, 3i)
put z = x + iy
| z 3i | > | z 3 |
| x + i (y 3) | > | x 3 + iy |
x2 + (y 3)2 > (x 3)2 + y2
x > y
| amp ( z ( 1 + i)) | 4
(3, 0)
(5, 0)
(0, 2)
(1, 1)
A
| z | = 5
O
B
(0, 5)
x
y
x = y
y > x
x > y
4 amp (z (1 + i))
4
Region enclosed by z satisfying all the three given inequalities is sector AOB in which point (3, 0) and the
points on the line segment OB are not included.
Area of the sector AOB = Area of quater circle =4
25=
16
625
a + b = 625 + 16 = 641. Ans.]
-
7/30/2019 Rt Solutions-04!03!2012 XIII VXY Paper I Final Test Code a Sol
9/19
MATHEMATICS
Code-A Page # 8
Q.6
[Sol. If and are the roots of x2 + t2x 2t = 0, then we have + = t2 and = 2t
so that 2
2
2
22
)(
2)(
)(
=
t
1
4
t2 and
t2
1
t4
11
)(
122
.
Now, I = dx11
x1
x
2
1
22
= dxt2
1
t4
1x
t
1
4
tx
2
1
2
22
= 3
t4
3
8
t32
2
.
Now 2
2
t4
3
8
t3
32
6=
4
23(Using A.M. G.M.)
Hence, I 8
9+ 3
Hence a = 3, b = 9 and c = 8. a + b + c = 20. Ans.]
-
7/30/2019 Rt Solutions-04!03!2012 XIII VXY Paper I Final Test Code a Sol
10/19
PHYSICS
Code-A Page # 1
PART-A
Q.4
[Sol. P1A = P
2A + mg
P1
> P2
P1 P2 ]
Q.5
[Sol. F = qE = 5 106d
9= 1.5 104
d = 4
6
105.1
1045
= 0.30 m ]
Q.6
[Sol.
y
x+ve x-direction. ]
Q.7
[Sol. IA
= I + 4I + 2I42 cos 2
= 5I
IB
= I + 4I + 2I42 cos = I
IA
IB = 4I ]
Q.8
[Sol. n = 4 3 E = 3eV
=3
1242= 414 nm ]
Q.9
[Sol. E = exactly the energy difference ]
Q.10[Sol. E
4 2 = 8eV
E2 1 = 10 eV
< 8eVE
3 2 = 5eV
> 5eV ]
-
7/30/2019 Rt Solutions-04!03!2012 XIII VXY Paper I Final Test Code a Sol
11/19
PHYSICS
Code-A Page # 2
Q.11
[Sol. Current leads the voltagei VR
V0
V VC L
V0
= 2R
2CL
V)VV(
= 25100 = 55 V
P.F. =0
R
V
V=
55
5=
5
1]
Q.12
[Sol. T2 =GM
42
R3
T =GM
2R3/2 ]
Q.13
[Sol. r = 90I
r
II sin I = 1 sin r = cos I = cot I
sin QC
= 1 sin 90
sin QC
= 1
sin QC
=
1= tan I
QC
= sin1 (tan I) = sin1 (cot r) ]
Q.14
[Sol. v = 22
2
AA
=
2
3 v
max
a = 2x =2
A2
=2
amax
]
PART-C
Q.1
[Sol. 6rv = B = 3
4
r3
PL g
=9
2r2
v
gPL
=9
2
7.0
100075.1)9.0(2
= 50 poise ]
-
7/30/2019 Rt Solutions-04!03!2012 XIII VXY Paper I Final Test Code a Sol
12/19
PHYSICS
Code-A Page # 3
Q.2
[Sol. f = mg 8
R3cos 60
30
60
mg
3R/8
30
N1
N2
60
N2
= f
N1
= mg
f =16
mg3
fN1
16
3 32 = 6 ]
Q.3
[Sol. q = q1
11CR
t
e
= q2
22CR
t
e
2
1
q
q=
2211 CR
1
CR
1t
e
R = 6 ]
Q.4
Sol. PL 6 102 g = 600 g
mg + 600 g = PL 1000 g
m = 1000600 = 400 gm
Q.5
[Sol.dt
dE=ATT4
T4
= )1067.5()05.0(4
)1067.5(
82
2
= 1012
[A = 4r2
]
T = 1000 K ]
Q.6
[Sol. L1
= L2
=2
L= 9 105 H
R1
= R2
=2
R= 3
eqL
1
=1L
1
+2L
1
Leq = 2
L1
= 2
9
10
5 H
eqR
1=
1R
1+
2R
1 R
eq=
2
R1
= 1.5
=R
L=
2
3
102
9 5= 3 105 sec. ]
-
7/30/2019 Rt Solutions-04!03!2012 XIII VXY Paper I Final Test Code a Sol
13/19
CHEMISTRY
Code-A Page # 1
PART-A
Q.1
[Sol. (B) Trigonal also called Rhombohedral
= = 90 and a = b = c ]
Q.2
[Sol. (A) A
Hybridisation : sp3d
Number of angles at 90 = 12
2
x x
x x
x
x
90
90
(B) A
Hybridisation : sp3
d2
Number of angles at 90 = 12
x x
x x
x
x
(C) A
Hybridisation : sp2
dNumber of angles at 90 = 4
x x
x x90
(D) A
x
x
xx
Hybridisation : sp3d
Number of angles at 90=6
x 12090
]
Q.3
[Sol. Cl C CH HOCCH
Na OC CONa+ +
Na OCCH OH+
2
Cl CCH
O O
OO
O O
O
Omorereactive
OH
O
OH
NaOH Cannizaro
+
]
Q.4
[Sol. (B) 0.3 M K2Fe[Fe(CN)
6] will give maximum number of ions in the solution. ]
-
7/30/2019 Rt Solutions-04!03!2012 XIII VXY Paper I Final Test Code a Sol
14/19
CHEMISTRY
Code-A Page # 2
Q.5
[Sol.
423 SOHHNO
ArN
S,
ONaCH3
]
Q.6[Sol. (A)
7N15 n/p = 8/7
n/p > 1 so7N15 will be -emitter ]
Q.7
[Sol.
OH OH
H HH
OH H
CHOH2
O
OOH
H OH
H OH
HO
H
CHOH2
Glycosidiclinkage
Due to this it will be reducing sugar the ring can open into aldehydic
form and it can undergo mutarotation.]
Q.8
[Sol.
ClBeCl ClBeOHH O2sp
+slow HCl
HCl
FastBe
B
Cl
Cl
Cl
OH
OH2
OH2
+
+
H O2
Be(OH)2[Be(OH) ]2 n or
white ppt
OH[Be(OH) ]4
2
Beryllate ion is formed in basic medium
]
Q.9
[Sol. BCl3
+ 3H2O H
3BO
3+ 3HCl
H3BO3 + 2H2O H3O+ + 4)OH(B
During the hydrolysis of BCl3the medium becomes acidic due to formation of HCl hence in the presence
HCl feasibility of second reaction ie formation of 4)OH(B is least.]
-
7/30/2019 Rt Solutions-04!03!2012 XIII VXY Paper I Final Test Code a Sol
15/19
CHEMISTRY
Code-A Page # 3
Q.10
[Sol. (i) Under ordinary condition, H2O molecule can not approach to the vaccant antibonding M.O. of
ClCl bond (yet having low energy) due to steric crowding over central carbon atom.
(ii) CCl bond dissociation energy is not very high as its magnitude is less than that of CF. ]
Q.11
[Sol. (A) K4[Fe(CN)
6] 4K+ [Fe(CN0
6]4
Na2HPO
3 2Na+ + HPO
32
higher negative charge over [Fe(CN)6]4so it can cause more coagulation in a basic dye.
(B) Starch aqua-sol is a lyophilic sol so it can be used as protective colloid.
(C) Freundlich adsorption isotherm fails at high pressure ]
Q.12
[Sol. (A) O O O
O O
O O
Si Si(B) O O O
O O
O O
Cr Cr
(C) O O O
O O
O O
S S
(D) or
Si
SiSiO
O
OO
O
OO
O O
]
Q.13
[Sol. H
H
Me
Me
It can show GI It do not posses COSIt posses POS It can show OI (it is meso isomer) ]
-
7/30/2019 Rt Solutions-04!03!2012 XIII VXY Paper I Final Test Code a Sol
16/19
CHEMISTRY
Code-A Page # 4
Q.14
[Sol. (A) Z =C
CC
RT
VP=
8
3
Z < 1
attractive forces are dominating
(B)TC
T1
T2
(L+G)Gas
Liquid
P
V
Liquid and gas can not be distinguished above critical temperature.
(C) At very high pressure , Z > 1
1)ideal(V
)real(V
m
m
Vm
(real) > Vm
(ideal)
(D) For y-Intercept, X = 0
PVmT
P
When PO
Gas shows ideal behaviour
PVm
= RT
RT
PVm
y = R ]
-
7/30/2019 Rt Solutions-04!03!2012 XIII VXY Paper I Final Test Code a Sol
17/19
CHEMISTRY
Code-A Page # 5
Q.15
[Sol. (A) & (C) Mozingo & WolfKishner is applicable
(B) N2H
2donot effect C
||O
(D) Ring expension take place ]
PART-C
Q.1
[Sol. Mole of H2O2 at t = 0 1
Kt
693.0
2/1
t1/2
= 10 sec.
Number of half lives in 20 sec. =10
202
Mole of H2O
2remaining after 20 sec. =
4
1
H2O
2+ 2H+ + 2I I2 + 2H2O
(Reactant)H(Product)HH ofofoReaction= [60 +2 (290)][200 + 2 (60)]
=520 + 320 200 kJ
Energy released when4
1mole H
2O
2reacts
200 4
1
50 kJ Ans. ]
Q.3
[Sol.O
H
O
O
O
87
6
5
4
3
2
1
+HCHO
O
O
O
O
O
H
H
HCHO
+ CH CHO3
8
7
6
5
4
3 21
]
-
7/30/2019 Rt Solutions-04!03!2012 XIII VXY Paper I Final Test Code a Sol
18/19
CHEMISTRY
Code-A Page # 6
Q.4
[Sol. Ag2CrO
4+ AgNO
3
0.5 litre 0.5 litre
saturated solution 2 106 M
Ag2CrO
4 2Ag+ + CrO
42
2S S
When mixed with 0.5 litre AgNO3, total volume1 litre
AgNO3Ag+ + NO
3
2 106 2 106
In the solution:
[Ag+] =2
S21026
= [S + 106]
[CrO42] =
2
S
[NO3] =
2
1026
= 106
Given :2
10
2
S 6
S = 106
KSP
[Ag2CrO
4] = [Ag+]2 [CrO
42]
= [106 + 106]22
106
= 4 10122
106
= 2 1018
ab = 19
18
10
102
= 20
In the solution we have
Ag+, CrO42, NO
3, H+, OH
M
1000K]Ag[
Agm
;
6
Ag
102
1000K50
Ag
K = 107 ;M
1000K]CrO[
24CrO2
4m
2/10
1000K100
6
CrO24
; 0.5 107 = 24CrO
K
6
NO
3m10
1000K]NO[ 3
; 3NO
K = 0.75 107
-
7/30/2019 Rt Solutions-04!03!2012 XIII VXY Paper I Final Test Code a Sol
19/19
CHEMISTRY
7H
m10
1000K]H[
; 350 = 7H
10
1000K
H
K = 0.35 107
7
OHm
10
1000K]OH[
; 7OH
10
1000K200
OHK = 0.2 107
K = 107 + 0.5 107 + 0.75 107 + 0.35 107 + 0.2 107
= 2.8 107
= 28 108
cd = 28 ]
Q.5
[Sol. Complex Number of geometrical isomers [CoCl2Br
2]2
0
[Rh(en)3]3+ 0
[Cr(en)2Br
2]+ 2
[Pt(en)Cl2] 2
[Co(NH3)3(NO
2)3] 2
[Pt(NH3)2(SCN)
2] 2
[Cr(NH3)2
Br2Cl
2] 5 ]
Q.6
[Sol. Isoelectric pH of a aminoacid =2
21
aa pKpK
8.2 =2
7 2apK 2
apK = 16.47 = 9.4 ]