Upload
vishal27042233
View
221
Download
0
Embed Size (px)
Citation preview
7/30/2019 Rt Solutions-IITJEE 2012-05-06 XII a Paper II Code B
1/16
12th (Date: 06-05-2012) Review Test-1
PAPER-2
Code-B
ANSWER KEY
PHYSICS
SECTION-1
PART-A
Q.13
B
Q.24
C
Q.31
C
Q.42
C
Q.5 B
Q.6 A
Q.7 D
Q.8 B
Q.9 A
Q.10 D
Q.1113
B,D
Q.1214
A,C
Q.1315
D
Q.1411
A,B,C,D
Q.1512
D
PART-C
Q.13
0002
Q.21
0002
Q.32
0008
Q.4 0003
Q.5 0007
CHEMISTRY
SECTION-2
PART-A
Q.1 D
Q.2 B
Q.3 B
Q.4 C
Q.5 C
Q.6 C
Q.7 D
Q.8 B
Q.9 A
Q.10 C
Q.11 A,B,C
Q.12 A,B,C,D
Q.13 A,C,D
Q.14 A,B,C,D
Q.15 A,B,C,D
PART-C
Q.1 0004
Q.2 0004
Q.3 0003
Q.4 0007
[Only for (A) Batch]
Q.5 0004
[Only for (AN1 & AN2, AH2)
Batch]
Q.5 2312
MATHS
SECTION-3
PART-A
Q.1 C
Q.2 A
Q.3 B
Q.4 D
Q.5 C
Q.6 A
Q.7 B
Q.8 C
Q.9 D
Q.10 B
Q.11 A,C
Q.12 B,C,D
Q.1362
C,D
Q.1476
A,B,C,D
Q.1566
A,B
PART-C
Q.1 0000
Q.2 0001
Q.3 0003
Q.4 0009
Q.5 0005
7/30/2019 Rt Solutions-IITJEE 2012-05-06 XII a Paper II Code B
2/16
Code-B Page # 1
PHYSICS
PART-A
Q.1
[Sol. Number of images =30
3601 = 111
Last image will coincide irrspective of position of object. ]
Q.2
[Sol. k = l
yA
keff
=21
21
kk
kk
=
2221
21
LyLy
Ayy
]
Q.3
[Sol.t
Q
. l = KA (T
1T
2) = constant graph will be rectangular hyperbola. ]
Q.4
[Sol. Velocity of mirror parallel to mirror will not contribute in image velocity speed of mirror perpendicular to
mirror =2v speed of image = 2
2v = v ]
Q.5
[Sol. As object approaches C, the ray passing through F hits the line AB closer to C. ]
Q.6
[Sol. Image location =1
L+n
b+
n
b+
1
L= 2L +
n
b2]
Q.7
[Sol. |1 sin = n3 sin |q' for both the rays will be same. ]
Q.8
[Sol. t =108
5.1
3
10120003
=
2
12 102 = 60 ms ]
Q.9
[Sol. sin = n0
sin r
sin (90
r) 0c
n
n
cos r 0
c
n
n
1 20
2
n
sin 2
0
2c
n
n
sin2 n02n
c2
sin 2c
20 nn ]
7/30/2019 Rt Solutions-IITJEE 2012-05-06 XII a Paper II Code B
3/16
Code-B Page # 2
PHYSICS
Q.10
[Sol. Closer is the ................ of n02 and n
c2, move will be precision required. ]
Q.11
[Sol. P =areaBase
Weight]
Q.12
[Sol. Both H2
and He have same degrees of translational freedom, but H2
have more number of degrees of
freedom. Also speedM
1]
Q.13
[Sol. Rate of fall of temperature difference of temperature ]
Q.14
[Sol. P1V
1= P
2V
2= nRT
at constant temperature
p V
B< V
A
U = f(T) ]
PART-C
Q.1
[Sol.V
1+
15
1
=
60
2
V
1=
30
12
V = 30 cm
m =15
30
= 2 ]
Q.2
[Sol. i sin i = n sin r
r
90rsin (90r)
n
1
22
2
n
1
n
isin1
n21 sin2 i
nmax
= 11 = 2 ]
7/30/2019 Rt Solutions-IITJEE 2012-05-06 XII a Paper II Code B
4/16
Code-B Page # 3
PHYSICS
Q.3
[Sol. sin =glassn
3
4
nglass
=33
8
t =
33
8
103
1039
8=
8
1039
10
39
8
t =10
1088
= 8 109 s ]
Q.4
[Sol.2
1=
3
z1
1
30
3
z+ 1 = 2
z = 3 m ]
Q.5
[Sol. Let depth of ice formed is x374.4 106 = (7x) (900) (80 103)
x =863
4.374
cm
x =863
37440
cm
x =56
4160cm =
7
520cm
If depth of water converted into ice is y, then
1000 y = 900 7
520
y =7
952cm
so d2d
1=
3
47
952
3.17
520
=
7
2713
7
400=
7
400351= 7 cm ]
7/30/2019 Rt Solutions-IITJEE 2012-05-06 XII a Paper II Code B
5/16
Code-B Page # 1
CHEMISTRY
PART-A
Q.1
[Sol. r = k [CH3COOCH
3][H+]
r = k' [CH3COOCH
3] where k' = k [H+]
Also k' =t
0
VV
VVlog
93.6
303.2
=25
50log
93.6
303.2
k' = 0.1 min1
k =]H[
'k =
1.0
1.0= 1 M1 min1 ]
Q.2
[Sol.
Me(3)
Et(2)
F(1)(4)H
(1) (2) (3) clock wise (R) (S) because 4th group is on Horizontal line.]
Q.3
[Sol. According to the intermediates attraction forces SiO2, NaCl, MgF
2and ZnO having higher boiling point
than H2O.
C = C
HH
H
H
H
H
Cz
x
y
single bond multiple bond
sp3
sp2
Bond angle of sp2 hybridised carbon atom is more than sp3 hybridised Carbon atom, so z is minimum .
Multiple bondSingle bond repulsion > SingleSingle bond so y > x. ]
Q.4
[Sol. Compound which least surface area have minimum vander waal's force attraction. so its boiling point
should be minimum or
B.PBranching
1]
7/30/2019 Rt Solutions-IITJEE 2012-05-06 XII a Paper II Code B
6/16
Code-B Page # 2
CHEMISTRY
Q.5
[Sol. ln
21
a
1
2
T
1
T
1
R
E
k
k
ln
310
1
300
1
R
E)eln(
k
k a
300
310
1310300
10
R
Ea
Ea
=10
R310300
620
1
300
1
R10
R310300
k
kln
300
620
= 16 ]
Q.6
[Sol.
equatorial down
equatorial upIf two groups is on same side (up-up or down-down) it is cis-form and if on opposite side (up-down)
then it is trans form. ]
Q.7
[Sol. (A) According to berry pseudorotation in PF5
molecule all the PF bond lengths seems to likelysame but actually they are different.[Axial PF bond length > Equatorial PF bond length]
(B) H C = S2
F
FCl
Cl
Due to the higher covalent radii of Cl, internuclear distance of SCl bond is more than SFbond
(C)Xe
F
F
F
F
F
F
sp3d3 hybridisation distorted octahedrom / Capped octahedron ]
7/30/2019 Rt Solutions-IITJEE 2012-05-06 XII a Paper II Code B
7/16
Code-B Page # 3
CHEMISTRY
Paragraph for question nos. 8 to 10
[Sol.(i) CH4
+ OH 3CH + H2O
r = k[CH4][ OH ] where k = 4 106 and [ OH ] constant
Reaction is pseudo Ist orderr = k' [CH
4]
k' = k [ OH ]
= 4 106 1 1015 = 4 109
tav
=7
91025
104
1
'k
1
sec.
(ii&iii) tav
of CH4
is 25 107 sec.
and tav
of isoprene = 1511 10102
1
= 4000 sec.
tav
of CH4is much more than isoprene, there for it will travel longer distance as well as remain for longer
time in atmosphere.]
Q.11
[Sol. H = bafa )E()E( H =2 Kcal/mole ba )E( = 8 Kcal/mole
fa )E( =2 + 8 = 6 Kcal/mol
also fraction of molecules crossing energy barrier is given as RT/Eae
Fraction of molecules A crossing energy barrier for forward reaction = 5002/6000e = e6
Fraction of molecules B crossing energy barrier is backward reaction = 5002/8000e = e8
Aslo keq
=b
f
k
k= RT/)Ea(
2
RT/)Ea(1
b
f
eA
eA
But A1
= A2
Given
Keq
=RT/)EaEa( bfe
=
RT/He
= )5002/2000(e = e2 ]
Q.12
[Sol. We can not bisect all the things into two equal halves.]
Q.13
[Sol. (A)N = N N N
H H
H
H H
H
(NN) Bond order 2 1
Bond order lengthbond
1
7/30/2019 Rt Solutions-IITJEE 2012-05-06 XII a Paper II Code B
8/16
Code-B Page # 4
CHEMISTRY
(B) H H H
O+
lone pairbond pair repulsion > bond pairbond pair
H H
H
H
N+ 1
so HOH < HNH
(C)
H H
H
H
N+ 1
x
H H
CH3
H
N+ 2
y
Due to the large size of CH3
steric repulsion is dominant > H. H3CNH is more than 10928'
so2
> 1
and y less than x.
Bond angle HNH in NH4+ > CH
3NH
3
(D)
NN N = N
H
H H
H
H H1 2
2 > 1
SingleSinglebond repulsion
Multiple bondSingle bondrepulsion
]
Q.14
[Sol.
H
Cl
H
ClPOS & COS absent (Chiral molecule)
H
CH3 Me
HPOS & COS absent (Chiral molecule)
COOH
COOH
OH
H
H
HO POS absent and COS absent (Chiral molecule)
OHC C(Me)3
CH CH Me2 2NO2
Rings are perpendicular to each other due to steric repulsion and groups are
differents on every ring so, POS and COS is not present, so it must be chiral
molecule.]
7/30/2019 Rt Solutions-IITJEE 2012-05-06 XII a Paper II Code B
9/16
Code-B Page # 5
CHEMISTRY
Q.15
[Sol. Correct order of bond length
H2+ < H
2 N
2+ < N
2 O
2+ < O
2 NO > NO+ ]
PART-C
Q.1
[Sol.
For nth order (n > 1) reaction
1n]A[
1 vs time graph is a straight line always.
4]A[
1vs time graph is given straight line
n1 = 4n = 5
Reaction is of 5th order
5
A ]A[Kdt
dA
5A ]A[K
dt
dA
t
0
A5
]A[
]A[
dtk]A[
]A[d
0
tK]A[
1
]A[
1
4
1A4
04
tK4]A[
1
]A[
1A4
04
4KA
= tan45 = 1
KA
= 1/4 KR
=2
KA=
8
1
ROR = KR
[A]5 =8
1[2]5 = 0004 Ans.]
Q.2
[Sol. Total stereogenic area is 2 Total geometrical isomerism = 22 = 4]
7/30/2019 Rt Solutions-IITJEE 2012-05-06 XII a Paper II Code B
10/16
Code-B Page # 6
CHEMISTRY
Q.3
[Sol. Number of structures in which bond angle around nitrogen other than the central nitrogen atom, is equal
to 180
N
N
N
N
N
+ +sp sp
N
N
N
N
N
+ +sp sp
N
N
N
N
N
+ +sp sp
Number of structures in which bond angle around nitrogen other than the central nitrogen atom, is not
equal to 180
N
N
N
N
N
+
+
sp2
sp
N
N
N
N
N
+
+sp
2
N
N
N
N
N++
sp2
sp2
]
Q.4
[Sol. (i) PF4I IP
F F
FF
90
90
90 PF3I2 FP
F I
IF
90
90 90
>120
Bond angle between identical (1) (0)
atoms < 120 > 90
(ii)90
>120B
Cl
FF 90
>120B
Br
FF 90
B
I
FF
Bond angle between identical (1) (1) (1)atoms < 120 > 90
(iii)
>90120
(1)
C
O
HH
>120
>90120
>90 single bond - single bond repulsion ]
7/30/2019 Rt Solutions-IITJEE 2012-05-06 XII a Paper II Code B
11/16
Code-B Page # 7
CHEMISTRY
[Only for (A) Batch]
Q.5
[Sol. Only those carbocations will rearrange which stabilis in each step
only a, c, e, g will rearrange. ]
[Only for (AN1 & AN2) Batch]
Q.5
[Sol. (a) Diastereomers (non-superimposable non-mirror images) 2(b) Identical (superimposable mirror images) 3(c) Enantiomers (non-superimposable mirror images) 1(d) Diastereomers (non-superimposable non-mirror images) 2 ]
7/30/2019 Rt Solutions-IITJEE 2012-05-06 XII a Paper II Code B
12/16
MATHEMATICS
Code-B Page # 1
PART-A
1.
1x
k
xn
kLim
1x l= 2
Put x = 1 + h
)h1(nh
)h1(nh(kLim
0h
l
l
= 2
2
2
0h h
.......2
hhhk
Lim
= 2 2
k= 2. k = 4
2. k(x) = xsgn1 =
0x;2
0x;1
0x;0
. So, k(x) is discontinuous at x = 0. Ans.
3. Expression
(cos4
1 + cos4
2 + cos4
3 ++ cos4
179 ) (sin4
1 + sin4
2 + sin4
3 + + sin4
179)= cos 2 + cos 4 + cos 6 + ............. + cos (358)
= cos
1sin
1179sin2
3582
= cos (180) = 1. Ans.
4. f (x) = sin1(sin x) =
2
5x
2
3,2x
23x
2,x
2x
2,x
g (x) = cos1(cos x) =
2x,x2
x0,x
Now, verify alternatives.
5. x + sin y = 101 ......(1)
x + 101 cos y = 100 ......(2)
(1) (2) sin y 101 cos y = 1
sin y = 1 + 101 cos y
y
2,0
cos y [0, 1]
possible only if sin y = 1 y =2
and cos y = 0 y =
2
sin y = 1 x = 100 [x + y] = 101. Ans.
7/30/2019 Rt Solutions-IITJEE 2012-05-06 XII a Paper II Code B
13/16
MATHEMATICS
Code-B Page # 2
6. As g(x) is continuous at x =2
, so [ST-6]
2g = 1)1exact(
]x[sin1
1Lim
x2
2
2x
p2 1 = 1 p2 = 2 p = 2 . Ans.
7.
x
x 1x
x
e
1xLim
Let x =t
1
t1
0t t1
1
e
1
t
1Lim = t1
t1
0t )t1(t
e)t1(Lim
e
1
=
t
eeLim
e
1 t)t1(n
0t2
l
=
1
t)t1(n
1t
)t1(n1eLim
e1
1t
)t1(n
0tl
l
l
= 20t tt)t1(nLim
e1
l =
e21 . Ans.
Paragraph for question nos. 8 to 10
Sol. According to given information, we must have f(x) a polynomial of degree 4 with leading coefficient 3.
So, f(x) = 3 (x 2) (x 3) (x + 1) (x + 6) + (x2 + 1)
8. f(0) = 109. Ans.
9. )6x(3
1x)x(f
Lim
2
6x
= )6x(3
)6x()1x()3x()2x(3
Lim6x
= (
8) (
9) (
5) =
360. Ans.
10. We have g(x) =)x(f1x
12
=)1x()6x()1x()3x()2x(3)1x(
122
=)6x()1x()3x()2x(3
1
, x 6, 1, 2, 3
But x
2
5,
2
15, so points of discontinuous are three i.e., x = 6, 1, 2. Ans.
7/30/2019 Rt Solutions-IITJEE 2012-05-06 XII a Paper II Code B
14/16
MATHEMATICS
Code-B Page # 3
11. Clearly f(x) is continuous everywhere
but non-derivable at 3 points viz.2
e,e,
2
ex
.
y=ex
y=x+ y= x
y=xe
(0, )
(0,e)
( ,0)( ,0) (e,0)x = e 2 O
(0,0)
x
y
Graph of y = f (x)
2
ex
+
=
Now verify alternatives from graph.
Also, Range of f(x) is
2
e,
12.(A) f (x) is discontinuous at x = 1, as f (1) = f (1) = 0, but f (1+) = 4.
(B) g (x) is continuous at x = 0, as g(0) = g (0) = g(0+) = 1.
(C) h (x) is continuous x R as difference of two continuous function is also a continuous function.(D) k (x) is continuous at x = 0 as k(0+) = k(0) = k(0) = 1. ]
13. Given, >x2sin
4
31
1
2 x R
> 4 x R. Ans.]
14.
(A) f(e) = 1 ; f(e+) = 1 ; f(e) = 1
(B) Here, a = 13, l =3
1; so a +
l
1= 13 +
3
1
1= 13 3 = 10.
(C) f(x) = x2 sgn x =
0x,x
0x,0
0x,x
2
2
.
y
x(0,0)
(1,1)
(1,1)
x=1 x=1
(D) As, f is continuous on [1, 1] and f (1) = 4, f (1) = 3
there exists a number r such that | r | < 1 and f (r) = .]
15. Let f(x) = 4x22x + k
1 1or
1 1
Here D 0 k 4
1
........(1)
1A2
B1
1
4
11 ........(2)
f( 1) > 0 k > 6 ........(3)f(1) > 0 k > 2 ........(4)
So, (1) (2) (3) (4)
k
4
1,2 . Ans.
7/30/2019 Rt Solutions-IITJEE 2012-05-06 XII a Paper II Code B
15/16
MATHEMATICS
Code-B Page # 4
PART-C
1. Given, f (x) = 3 2 |x|x | x | 1
=
0x,11xx
0x,11xx1xx3 3
Hence, f (x) = 1 x Ry =1
x
y
So, f (x) is continuous and differentiable x R.
Hence, number of points of non-derivability of f (x) is 0.
2.4
27
xtan
xsin4)bax(Lim
0x
As x 0, Dr 0So, Nr 0 for existence of limit. So
b 2 = 0 b = 2.
Now,x
xsin4)2ax(Lim
0x
=
4
27
0
0
xsin42axx)xsin4()2ax(
Lim2
0x
=
4
27
x
xsinax4xaLim
22
0x= 27 4a 1 = 27 a =
4
28= 7.
Hence, (a 3b) = 7 3(2) = 1. Ans.
3. a =
2n3
3
1n
1n= M =
2n
2
2
)1nn(
)1nn(
1n
1n
=
1nn
1nn.....
21
3
13
7
7
3
1n
1n
2n
n.....
4
6
3
5
2
4
1
32
2
=1nn
3
2
)1n(nLim
2n
=
2
3
2
3a
N =
1n1
21
n21
)n1(=
1n
2
2n
n
n
1n=
1n
2
)2n(n
)1n(
N =
1n1n n
1n
2n
1n
=n...........321
)1n(n.....432
)2n)(1n......(54
)1n......(..........43
3
2
=
2n
)1n(2
= 2 2b
(ab) =2
3 2 = 3 Ans.
7/30/2019 Rt Solutions-IITJEE 2012-05-06 XII a Paper II Code B
16/16
MATHEMATICS
Code-B Page # 5
4. Givenysin
xsin= 3 .......(1)
andycos
xcos=
2
1.....(2)
Now,y2sin
x2sin=
ycos
xcos
ysin
xsin=
2
13 =
2
3
Now,y2cos
x2cos=
1ycos2
1xcos22
2
........(3)
(1)2 9 sin2 y = sin2x ........(4)(2)2 cos2y = 4 cos2x 1 sin2y = 4 4 sin2x sin2y = 4 sin2x 3
Put the value of sin2y in equation (4) 36 sin2x 27 = sin2 x sin2 x =35
27
sin2y =35
3
Hence,y2cos
x2cos=
1ycos2
1xcos22
2
=
ysin21
xsin212
2
=
35
321
35
2721
=29
19
Hence,y2cos
x2cos
y2sin
x2sin =
2
3
29
19=
58
49=
q
p
| p q | = | 49 58 | = | 9 | = 9. Ans.]
5. We have f (x) = 1|x| =
x1;1x
1x0;1x
0x1;1x1x;1x
)x(ff = 1|)x(f| = 11|x|
=
x2;2x
2x1;2x1x0;x
0x1;x
1x2;2x
2x;2x
(0,0)
y=x
y=x2
y=x+2
y=x+2
y=x
y=x
2
x=1 x=2x=1
(1,1) (1,1)
x=2x
y
Graph of y = f f (x)( )
So, )x(ff , is non-derivable at five points viz. x = 2, 1, 0, 1, 2.