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CLASS-XII_STREAM-SB+2_PAGE # 1
PART-I (1 Mark)MATHEMATICS
1. I = !"
#
$
&
10
01
A = !"
#$&
0i
i0
A2 = !"
#$&
10
01
A3 = !
"
#$&
0i
i0
A4 = ' ( A4n = '
))))' + A + A2 + A3 = 0
=)' + A + A2 + A3 + ............+ A2010
= (' + A + A2 + A3) + A4 (' )+ A + A2 + A3) ..... + A2008 (' )+ A + A2)
= 0 + !"
#$&
0i
i0= !
"
#$&
0i
i0.
ANSWER KEYHINTS & SOLUTIONS (YEAR-2010)
Ques. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Ans. C B D B C B B A C B B C A C D
Ques. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30Ans. D D D C A A A B C D A D A B D
Ques. 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45
Ans. D B C B B A A D A C B C D A B
Ques. 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60
Ans. B D A A A C B C D B A B A C A
Ques. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75
Ans. A A C A C C A A C D A A C B B
Ques. 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90
Ans. D C A A D C B B C B C A B D C
Ques. 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105Ans. D A B B A A B D B C A A D D
Ques. 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120
Ans. C B A B D B C D B B D A C A A
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CLASS-XII_STREAM-SB+2_PAGE # 2
2. Let a, ar, ar2 are sides.
Case 1 : r > 1
a + ar > ar2
( r2 r 1 < 0
( r * ++
,
-
.
.
/
01
2,
2
51
Case 2 : r = 1 equilateral triangle
Case 3 : r < 1
ar2 + ar > a
( r2 + r 1 > 0 ( r * ) ++
,
-
.
.
/
0 23
2
51,
2
51
4 )r lies in the interval
!
!
"
#
.
.
/
0 22
2
52,
2
51.
3. Number of diagonals passing through centre = 6
4 Number of rectangles = 6C2= 15
4. 25 2, 3 + 45 , 3 + 45 2, 5 5 5 2
6 1 3 i , 1 + 32 i , 1 32 i, 6
where 5 =2
i31 2, 5 2 =
2
i31
1 + 32 i , 1 32 i are conjugate of each other..
4 Least possible degree = 5
5. Tangent at (2, 1)
y.2 = 2 (x + 1)
( y = x + 1
( x y + 1 = 0
equation to circle
(x 1)2 + (y 2)2 + 7(x y + 1) = 0
putting x = 14 + (y 1)2 + 7( y) = 0
( y2 4y 7y + 8 = 0
D = 0 ( (7)+ 4)2 = 32
y =4
D)4( 827
4 y =2
024 28 = 2 2
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CLASS-XII_STREAM-SB+2_PAGE # 3
6. 9BKN = 6
9
1
APB
BKN:
9
9 +
,
-./
0:4:
3
1
AB
BN
1
2
BN
AN
( 9 APB =1
9 6
=1
54
4 9 ABC = 108.
7. 2
2
2
2
b
y
a
x2 = 1 , a > b > 0
h =3
x, k =
3
y
2
2
2
2
b
)k3(
a
)h3(2 = 1 which is ellipse
8. 1
6
1
7 sin1cos;
9. (1 + tan 1) (1+ tan 2) .........(1+ tan 45)
(1 + tan A) (1 + tan B) = 2 if A + B = 45
Ans = 223
10. f is differentiable function .
f?(a) f?(b) > 0 ( f is increasing at x = a and b
or f is decreasing at x = a and b
minimum number of roots of f?(a) = 0 in (a, b) is 2,
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CLASS-XII_STREAM-SB+2_PAGE # 4
11. When x is less than 49 then f(x) has negative value.
Which is not possible.
when x > 49
then f?(x) = 49 (x 41)48 + 41 (x 49)40
+ 2009 (x 2009)2008
so sing of f?(x) does not change
f?(x) > 0( non real except one positive root.
12.
f?(x) is given smooth curve hence differentiable i- given domain
By first derivative test
x = a, c is point of local maxima
x = b is point of local minima
13. @ :3
i
4dx)x(f
shaded area = Trapezium Area Area under curve
= 2
)13(2
7
2
5+,
-./
02
4 = 2
14. 'n
= dx.)x(log
e
1
n
@
'n
= dxx.x
)x(logn)x(logx
e
i
1ne
1
n
@
'n
= (e 0) n 'n
1
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CLASS-XII_STREAM-SB+2_PAGE # 5
'n
+ n 'n1
= e
'2011
+ 2011 '2010
= e
4 '100
+ 100 '99
= e
15. x2 + y2< 100 ( Area bounded = 100 >
sin (x + y) > 0Hence common area of region bounded = 50 > as in half the region sin (x + y) > 0 and in other half
sin (x + y) < 0
16. n(S) = 7C3
= 35 n (* ) = 7 + 7 + 7 = 21 P (* ) =5
3
35
21:
17. w).vu(
A , vu
A =230
112
kji
= )6(k)4(j)32(i 22 = 53
greatest when x2 + y2 = 1
fmax.
= 17 Ans. (D)
18.
Number of ways = 4P2 4P
2 10 = 1440
19. 1 + x2 + x4 + .........+ x2010
=2
10062
x1
)x(1
= )x1)(x1(
)x(1 21006
2
=)x1(
)x1(
)x1(
)x1( 10061006
2
2
= (1 + x1006))x1(
)x1(
)x1(
)x1( 503503
2
2
= (1 + x1006) (1 + x + x2 + ...............+ x502) (1 x + x2 x3 + ............ + x502)
4 n 1 = 502 n = 503
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CLASS-XII_STREAM-SB+2_PAGE # 6
20. an
= 3an1
+ 1
= 3(3an2
+ 1) + 1
= 32 an2
+ 3 + 1
= 32 (3an3
+ 1) + 3 + 1
= 33 an3
+ 32 + 3 + 1
aN
= 3n a0
+ 3N1 + 3N2 + .... 1
Method - Ia
2010= (1 + 3 + 32 + 33 + 34) + ....... + 32009
= (1 + 3 + 32 + 33 + 34)+ 35 (1 + 3 + 32 + 33 + 34)+ 310 (1 + 3 + 32 + 34 + 35) +.....+32005 (1+ 3 + 32 + 33 + 34)
each (1 + 3 + 32 + 33 + 34) is divisble by
Method-II
a2010
=2
1)1242(
2
1)243(
2
13
2
13 40240240252010 2:::
A
= 11 k (k * I)
Hence remainder is 0
PHYSICS
21.
F B1mg B
2(m + M)g = M.
m
mg1
F = B1mg + B
2(M + m)g + B
1Mg
= B1g(M + m) + B
2g(m + M)
= (B1g + B
2g) (M + m)
22. T =k
1
mm
mm2
21
21++,
-../
0
2>
we can use concept of reduced mass.
23. 2T cos= = ma ;2
2
x4
xmT2
2
3
= a
If x < < < . f =
mT4
21>
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CLASS-XII_STREAM-SB+2_PAGE # 7
24. AU C = mean distance between Sun and Earth
and Time period = 125 year
T2D r3
(1 yr)2D (1AU)3
(125)2
D
3
a
2
4.0r
!"
#
$%
& 2
4 25 = !"
#$%
& 2
2
4.0ra
ra
= 50 0.4
= 49.6.
25. Zero
26. 3R
KQrE D r inside
2r
KQE D 2r
1outside.
27. There is no net effect of outside charge.
29. tanC =h
r
from snell's law
sinC =B
1
r =1
h
23
30. B2
> B3
B1
sin i1
= B2
sin r
B1
< B2
from diagram2
3
2
1
B
BE
B
B
For condition of TIR
B2
> B3
and B1> B
3
So, B2
> B1
> B3
.
31. A = T/t0
2
AA = 42
120000=
2222
10100120
A
A= 7500 (D).
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CLASS-XII_STREAM-SB+2_PAGE # 8
32. 5 2 A = g
A
g:5
f =>2
1
A
g
= 2105.2
10
2
13
A>= 3.18
33. dQ = 0
dU = dw
++
,
-
.
.
/
023 dV
V
anCdT
2
2
= dVV
an
nbV
nRT2
2
!!"
#
$$%
&3
3
CdT =nbV
nRT3
.dV
@ @ 3:3 nbVdV
T
dT
nR
C
nTnR
C3 = n(V nb) + C
n (V nb) nR
CnT = K
[n(V nb) + n TC/nR] = K
ln(V nb) TC/nR = K
(V nb) TC/nR = constant Answer is (C) option.
35.
v0
= 39.6 kmph = 39.6 185 = 2.2 5 = 11 m/s
t1
=330
d
t2
=330
3011d A3=
330
d
330
3011A
9 t = t1 t
2= 1
time interval = 30 1 = 29 second
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CLASS-XII_STREAM-SB+2_PAGE # 9
36. v =M
RT
4
1
32
2
M
M
v
v
0
H
H
0 :::
37. d = 0.1 mm
D = 1m
7 = 600 nm
given IR
= 75% of maximum = 75% of 4' = 3'0
'R
= 3' = ' + ' + 2 ' cos H
H =3
>( 9x =
D
dy=
>
7
2
3
>=
6
7
y =d6
D7= 3
9
101.06
1060013
3
A
A( y = 1 mm
38.
vcom
=21
01
mm
0vm
2
2( x
i=
21
21
mm
)(m)0(m
2
2
xcm
= xi+ v
comt
xcm
=21
01
21
2
mm
tvm
mm
)(m
22
2
39. According to given information
tan = =r
a
r = a cot
A B Camv
rr
mv2= qvB ( )v =
m
qBr( )v =
m
cotqBa =
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CLASS-XII_STREAM-SB+2_PAGE # 10
CHEMISTRY
41. [Co(dien)Cl3]
dien = NH2 CH
2 CH
2 NH CH
2 CH
2 NH
2
[Co(dien)Cl3] have two geometrical isomers cis and trans.
cis form is optically active but trans form is optically inactive.
42. The ligands which allow back bonding to sufficient extent are called ! acid acceptors.Example CO, CN, NO etc.
43. O22 : Total electron = 18 ; Bond order = 1.
Peroxide (O22) : (I 1s)2 (I *1s)2 (I 2s)2 (I *2s)2 (I 2p
z)2 (>2p2
x= >2p2
y) (>*2p
x2 = >*2p2
y)
Bond order = !"
#$%
&
2
NN AB=
2
810= 1.
44. E =7
hc=
1
1031062.6 834 A3= 1.988 1025 J.
45. t1/2
(half-life) is independent of initial concentration so reaction is first order.
46. ClF3
F
Cl
F
87.5
87.5
F
F
..
..
F nearly 'T' shaped.
47. 3AlCl
O||
ClCR
J K
Friedal craft acylation
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CLASS-XII_STREAM-SB+2_PAGE # 11
48. > > >
Order of stability of conjugate bases
> > >
49. anti conformation is most stable
50. ePaTh0
1
234
91
234
902K
51. Entropy change is positive, but enthalpy does not change.
52. On increasing temperature, change in activation energy is not significant.
53. For FCC unit cell number of atoms per unit cell
Z = 8[corner] 8
1+ 6 [face center]
2
1= 4.
For body center unit cell number of atoms per unit cell
Z = 8[corner] 8
1+ 1 [body center] 1 = 2.
54. CO2
CO +2
1O
2;
1CK = 9.1 1012 .....(i)
H2O H
2+
2
1O
2;
2CK = 7.1 1012 .....(ii)
Target eq.
CO2
+ H2
CO + H2O K
C= ?
Target eq. can be deduce by
[Eq. (i) Eq(ii)] ( KC
=2
1
C
C
K
K
=12
12
101.7
101.93
3
A
A= 1.28.
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CLASS-XII_STREAM-SB+2_PAGE # 12
55. R J K P
t=0 R0
0
t=t (R0 x) x
Rate = dt
]R[d= K[R]1
=dt
dx= K[R
0 x] = @
x
00
dx.)xR(
1= @
t
0
dt.K
= ln ++
,
-
.
.
/
0
xR
R
0
0= Kt
[R] = (R0 x) = R
0eKt
56. PCl3F
2:
57. (I) & (IV) are enantiomeric pair
I IV
58. 2NO2(g) J K 2NO (g) + O
2(g)
NO2is reactant so its concentration is decrease with time while NO and O
2are product so their concentration
increases with time.
Formation of moles of NO is double than O2.So X = NO, Y = O
2and Z = NO
2.
59. Cyclic
Planar
Complete conjugation
(4n + 2) >e
Tropylium carbocation
60. Anti addition takes place.
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CLASS-XII_STREAM-SB+2_PAGE # 13
PART-II (2 Mark)MATHEMATICS
81.
n
4/1
2/1
x2
1x +
,
-./
02 =
n
4/3
2/n
x2
11x +
,
-./
02
= xn/2LM
LN
O
LP
LQ
R+,
-./
022+
,
-./
02+
,
-./
02 .........
x2
1c....
x2
1c
x2
1cc
4/3rn
2
4/32n
4/31n
0n
nc0
, 22
n1
n
2
c,
2
care in AP
( 1,2
n,
8
)1n(nare in APAP
( n = 1 +8
)1n(n
( 8(n 1) n (n 1) = 0
( (n 1) (8 n) = 0 ( n = 1 or n = 8
but n = 1 i.e. not posible
( n = 8
Now
8
4/1
2/1
x2
1x +
,
-./
02 =
r
4/1
8
0r
r82/1r
8
x2
1)x(c +
,
-./
0S
:
=4
r34
r8
0r
r8 x
2
1c +
,
-./
0S
:
required number of terms = 3 (r = 0, 4, 8)
82. Roots of 2x2 + 2x = 1 = 0 are ,2
i183
It will satisfies the (x + 1)n r = 0
(
n
12
i1+,
-./
02
8= r
(
n
12
i1+,
-./
02
8= r, r * R
(
n
4
in
e2
1
++
,
-
.
.
/
0
++,
-../
0>
8
= r
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CLASS-XII_STREAM-SB+2_PAGE # 14
n
4
i
e++
,
-
.
.
/
0 >8
will be real only if n is multiple of 4.
( For n = 4000
( r =
4000
2
1
++,
-
../
0
= 10004
1
83. y = ax + b and 1b
y
a
x 22:2
fig. 1 and 3 incorrect ()a < 0 b the equation of line y = ax + b)
Now for a > 0 and b < 0 fig. 2 is correct
84. Area of cyclic quadratic and is maximum only if area of 9ABC and area of 9ADC is maximum
( Area of 9ABC is maximum if ABC is equilateral 9 .
( AC = 2R sin 60 = 3 R.
maximum Area of 9 )ABC =2)R3(
4
3=
2R4
33
maximum area of 9 ADC =2
1(AC) (DM)
=2
1++
,
-
.
.
/
060cot
2
R3)R3(
=4
R3 2
area of ABCD = 222 R3R4
3R
4
33:2
85.
f(x) is monotonic increasing in [a, T )
( f?(x) > 0 U x * [a,T )
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CLASS-XII_STREAM-SB+2_PAGE # 15
( f?(x) is monotonic decreasing in (T , b] ( f?(x) < 0 U x * (T , b]
Now g?(x) =x
)x(f
g?(x) = 2x
)x(f)x(xf?
h(x) = xf?(x) f(x)
( h?(x) = f?(x) + xfV (x) f?(x)
( h?(x) = xfV < 0 { x > 0 and fV (x) < 0 concavity}
( h(x) is M.D.
So, at x = a it may be M.I. followed by M.D. till x = b
or
M.D. through out a
But cant be M.I. followed by M.D.
So (B) is correct answer.
86.
Let radius of base of given cone is r and length
as h ( V1
=3
1>r2h ........(1)
Let radius of base 1 and R1 and height H. For the cane with apex O
V2 = 31 >R2H ..........(2)
NowHh
h
R
r: ( i
r
R
h
H:
( H = hMNO
PQR
r
R1
V = +,
-./
0>
r
R1hR
3
1 2
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CLASS-XII_STREAM-SB+2_PAGE # 16
++
,
-
.
.
/
0>:
r
R2R2h
3
1
dR
dv 2
= 0 ( R =3
r2
For maximum volume
R 3
r2( V
2= +
,
-.
/
0>
r
R1hR
3
1 2
= +,
-./
0>
3
21
9
r4h
3
1 2
= h81
4 2> .......(2)\ (27
4
1V
V2 :
87. f?(x) = 1 + f(x) (dx
dy= 1 + y
( n (1 + y) = x + c
( y = ex + c 1
( y = 7)ex 1
( f(x) = 7)ex 1 ..... (1)
Now f(x) = x + @x
0
dt)t(f
7ex 1 = x + 7et (t)0x
7ex 1 = x + 7ex x 7
( 7)= 1
( f(x) = ex 1
for f(x) = 0 ( ex 1 = 0
( W x = 0
88.
A =2
1.>)X
2
>+
2
1.>)X
2
>): )
2
2>
89. Let P is origin and p.v. of A, B, C are a, b, c respectively
Given 0PC3PB2PA
:22K
( 0c3b2a :22
...... (i)
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CLASS-XII_STREAM-SB+2_PAGE # 17
Area of 9ABC =2
1|a
b
+ b
c
+ c
a
|
=2
1|a
++,
-../
0
2
ac3
+ ++,
-../
0 3
2
ac3
c
+ c
a
|
=2
1|2
3( c
a
) 0 0 +2
1( c
a
) + c
a
|
=2
3|c
a
| ...... (ii)
Area of 9 APC =2
1|c
a
|
(APC
ABC
9
9=
1
3
90. 6m + 2m+n. 3m + 2n = 332 = 4 83
( 2m2. 3m + 2m+n2 3m + 2n2 = 83 .......(1)
83 is prime number so that this is possible only if
m = 2
( by (1) 32 + 2n 32 + 2n2 = 83
( 2n. 9 + 2n2 = 74
( 2n2 (36 + 1) = 74
( 2n2 = 21
( n = 3Hence. m2 + mn + n2
= 4 + 6 + 9 = 19
PHYSICS
91. h10upto........g
Vr
g
Vr
g
V th20
420
220 3
!!"
#
$$%
&222
=g
V20 [1 + r2 + r4 + .............upto 10th] h
= hr1
)r(1h2
2
102
3!!"
#
$$%
&
3
3
= hr1
r1h2
2
20
3!!"
#
$$%
&
3
3.
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CLASS-XII_STREAM-SB+2_PAGE # 18
92. 2r
GMm=
r
mv2( v =
r
GM
T =V
r2>T =
GM
r2
3
>
T = rGM
r2>
= GM
r
2
3
>
geff
= g 5 2 Re cos2H
f(g 5 2 Re cos260) = (g 5 2Re cos20)
fg g = 5 2R !"
#$%
&3 1
4
f
g[f 1] =4
R25[f 4] ( 2R
Gm(f 1) =
4
R25(f 4)
)4f(
)1f(GM4
2 35
3
= R3
( )4f(
)1f(GMT
2
2
3>
3
= R3
)4f(
)1f(GMT2
2
3>
3= R3 Y =
3R3
4
M
>=
)1f(GT4
)4f(32 3
3>
93. a = 2r 2
32
2
q
Kq2
2
q
Kq2cos30
= r3
kr = 2
2
q
kq2
cos30
k.r.3r2 =04
2
>Z.q2
2
3
r =
3/1
0
2
k12
3q++
,
-
.
.
/
0
>Z
94. Consider the infinite ladder circuit shown below :
6
6
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CLASS-XII_STREAM-SB+2_PAGE # 19
Let the equivalent impedence of circuit be ZSo, Z = wL + Z?
Z = 5 L +C
C
XZ
ZX
2= 5 L +
C
1Z
C
1Z
52
5A
On solving we get,
Z =C2
LC4CLLC 222 3585
For Z to be purely inductive 0LC4CL 222 :35 = 0
5 =LC
2Ans.
95.
13
1
V1=
R
13(
R
1
v1
3: (
1
Rv1
3
B:
)VR2(V
1
1f 33
B3 =
R
1
3
B3
++,
-../
0
3
B3
B3
3
B3:
1
RR2
R
1
V
1
f
)RR2R2(
)1(
R
1
B33
33
3= !
"
#$
%
&
3
B3
3
2
1
R
1
!"
#$%
&
3
B333
)2(
)2(
R
)1(= ++
,
-../
0
3
3+,
-./
0 3
2
2
R
1( v
f=
)1(2
)2(R
3
3
96. V(x) =2
1kx2 V
0cos +
,
-./
0
a
x
G =dx
dV3
= kx + V0 sin +,
-
./
0
a
x
. a
1
Since, x
7/29/2019 kvpy sol-20101
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CLASS-XII_STREAM-SB+2_PAGE # 20
97. 9Q = 9U + 9W
nCPdt = nC
vdt +
T3
3
1
)TT(nR 12
Heat capacity in this process
CP
= Cv+
T31
nR.
98. W = Area of 9ABC
W = +2
1(P
2P
1) (V
2V
1)
P1 BA
C
V2V1
P2
{It is clockwise}Heat released at CA :It is at constant volumeSo, Q = nC
VdT
Q = n2
3R (T
A TT
C)
Q1 = 2
3
(P1 V2
P2V2)
Heat released at A B (Constant pressure)
Q2= nC
PdT =
2
5nR (T
B TT
A)
Q2
=2
5(P
1V
1 P
1V
2)
Heat absorbed at BC9Q = du + 9 W du = 0Q
BC+ Q
1+ Q
2=
9W
QBC
= 9 W + Q1
+ Q2
=21 (P
2 P
1)(V
2V
1)
25 (P
1V
1P
1V
2)
23 (P
1V
2P
2V
2)
=2
1[(P
2P
1)(V
2V
1) 5P
1V
1+ 5P
1V
2 3P
1V
2+ 3P
2V
2]
=2
1[P
2V
2 P
2V
1 P
1V
2+ P
1V
1 5P
1V
1+ 5 P
1V
2 3P
1V
2+ 3P
2V
2]
=2
1[ 4P
1V
1+ 4P
2V
2+ P
1V
2 P
2V
1]
Q = 2 [(P2V
2 P
1V
1) +
2
1(P
1V
2 P
2V
1]
99. V K [ V (
0mv2
12 2 2+
,
-./
0= 0 + 0 +
,
-./
0
d
Kqq
d = ++
,
-
.
.
/
0
>Z 2
2
0 mv
q.
4
1= 2
0
2
mv4
q
>Z
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CLASS-XII_STREAM-SB+2_PAGE # 21
100. N0 [ Initial nucleon
at t = 0 J K7N
0N
Add. at a constant rate (7N C) =dt
dN3
@
k
0
dt
= @+
,
-.
/
0
37
N
N0 CN
dN
t =NN0
)CN(n1
377
3
t = ++
,
-
.
.
/
0
37
37
73
CN
CNn
1
0
( e7t = CN
CN
0 37
37
7N C = e7t (7N0 C)
N =7
27
73 teC(7N
0 C) =
7
C+ N
0e7t
7
C.e(7t)
=7C [1 e(7t)] + N
0edt .
CHEMISTRY
101. H2C
2O
4.2H
2O (GMM = 126)
Molarity (M) =100126
100052.2
A
A= 0.2M
M1V
1= M
2V
2
0.2 10 = M2 500
M2
= 4 103.
valence factor of H2C
2O
4.2H
2O = 2.
Normality = [M] V.F. = 2 [4 10 3] = 8 103 N
Amount of oxalic acid (mg/mL) =3
33
10
10126104 A3
+,
-./
0
mL
mg= 0.504.
Note : Options are not match with solution.
102. J K\H J K
]Br
J KHBr + CH
3OH
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CLASS-XII_STREAM-SB+2_PAGE # 22
103.
104. A J K B
9H = 7.5 kJ/mole ; 9S = 25 J/mole.
9G = 9 H T9S at equilibrium 9 G = 0
0 = 7.5 103 T(25)
T =25
105.7 3A= 300 K
105. [Mg2+] = 102 M
)2)OH(Mg(SPK = [Mg
2+] [OH]2 = 1.0 1012
= [OH]2 =2
12
10
100.13
3A
= 1010
[OH] = 1 105
pOH = 5 ; pH = 9
106. For FCC unit cell
4r = a2
a = ++,
-../
0 A
2
4.1414= 400 pm
Volume of unit cell a3 = (400)3 = 6.4 107 pm3.
107. Formula of cyclic silicate : [SinO
3n]2n ; Cyclic silicates : (SiO
32)
nor (SiO
3)
n2n
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CLASS XII STREAM SB 2 PAGE # 23
108. Target eq., 2B (s) + 3H2(g) K B
2H
6(g)
Given that H2O () K H
2O (g) 9H
10 = 44 kJ .....(i)
2B + 3/2 O2
(g) K B2O
3(s) 9H
20 = 1273 kJ .....(ii)
B2H
6(g) + 3 O
2(g) K B
2O
3(s) + 3 H
2O (g) 9H
30 = 2035 kJ .....(iii)
H2(g) + 1/2 O
2(g) K H
2O () 9H
40 = 286 kJ .....(iv)
Target eq. : Eq.(ii) Eq.(iii) + 3[Eq.(iv)] + 3 [Eq.(i)]
= 1273 [ 2035] + 3[286] + 3[44] = 36 KJ/mole.
109. K3[Fe(CN)
6]
3 + x 6 = 0 x = + 3
26Fe = 3d6 4s2
26Fe3+( d5 with S.L. ( t
2g2,2,1, eg0,0 CFSE = 2.0 9
0+ 2P ~ 2.0 9
0
Number of unpaired electron = 1.
So, B = )2n(n 2 = 3 B.M
110. 9Tf= K
f molality
= 0.925 = 1.85 !"
#$%
&
A
A
92M
10008
M = 173.9 ~ 174Closest answer = 160.
* * * * *