solution 3 (class xii)
TRANSCRIPT
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1Pre-foundation Career Care Programmes (PCCP) Division 1
PART-I (1 Mark)MATHEMATICS
1. Given : a1, a
2, a
3.........AP and a
1, a
2, a
4, a
8......GP.
Let common difference of A.P. = da
2= a
1+ d
a4
= a1
+ 3d
a8
= a1
+ 7d
1
2
a
a=
2
4
a
a=
4
8
a
a= r
1
1
a
da = da
d3a
1
1
= d3a
d7a
1
1
= r
(a1 + d)2 = a1(a1 + 3d)a
12 + d2 + 2 a
1d = a
12 + 3a
1d
d2 = a1
d (d 0)
d = a1
....(i)
Hence,1
2
a
a= r ;
1
1
a
da = r
1
11
a
aa = r (using (i))
r = 2.
ANSWER KEYHINTS & SOLUTIONS (PRACTICE PAPER-3)
Ques. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Ans. A B C B D B D A A C B B D C A
Ques. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30Ans. B C B C D B B D D C C D C A D
Ques. 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45
Ans. D C A B B C A A A B C D C C D
Ques. 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60
Ans. A C B A B A D C D B D A D B B
Ques. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75
Ans. A C C C B B C D A D D D B A B
Ques. 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90
Ans. B B D C A C B C B B D C D B C
Ques. 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105Ans. D A B C D A C C B B B D A A B
Ques. 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120
Ans. C A C D B C B A B A D B A C C
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2Pre-foundation Career Care Programmes (PCCP) Division 2
2.k
101
T
T
1k
k till k = 10
Tk> T
k1
Let k = 11
T11
< T10
T10
is maximum at k = 10.
3. x = 2 + 3 + 6
22x = 263
x2 + 2 2 2 = 9 + 26
x2 7 = 28
(x2 7)2 = 64 2
So, smallest possible value of n is 4.
4. Let the three players are A, B, C.
Now, each player get 0 score after playing 9 games. It happened only when each player wins 3 games
and loss 6.
So,
A win 3 games out of 9 9C3
B win 3 games out of remaining 6 6C3
C win 3 games out of remaining 3 3C3
So, required way = 9C3
6C3 3C
3
=!6123
!6789
!3123
!3456
1
= 1680.
5.
B C
A(2, 3)
(x, y)(4, 0)
(2, z)
O
O is circumcenter
OA = OB = OC = circumradius(2 2)2 + (z 3)2 = (4 2)2 + (0 z)2
z2 + 9 6z = 4 + z2
9 6z = 4
5 =6z
6
5= z
Circumcenter = 22 )22()3z( = | z 3 | = 365
=6
13.
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3Pre-foundation Career Care Programmes (PCCP) Division 3
6.
P P'
(5, 15) (21, 15)L
A
Mid point of PP =
2
1515,
2
215
L = (13, 15)
Point A will be (13, 0)By property PA + PA = 2a
PA = 22 )150()513(
= 22564
= 289 = 17 cm
PA = 22 )150()2113(
= 22564
= 289 = 17 cm
2a = PA + PA2a = 17 + 172a = 34 cm
So, length of major axis = 2a = 34 cm.
7.
B C
P(10, 10)
(a, b)(0, 6) 2x + 3y = 18
PB = PC(10 0)2 + (10 6)2 = (a 10)2 + (b 10)2
100 + 16 = a2 + 100 20a + b2 + 100 20ba2 + b2 20a 20b + 84 = 0 ....(i)
Also (a, b) i.e. on 2x + 3y = 182a + 3b = 18
a = 9 2b3
Using equation (i)
2
2
b39
+ b2 20
2
b39 20b + 84 = 0
81 +4
b9 2 27b + b2 180 + 30b 20b + 84 = 0
4
b13 2 17b 15 = 0
13b
2
68b
60 = 0
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13b2 78b + 10b 60 = 0
13b(b 6) + 10 (b 6) = 0
b = 6 or b =13
10
When b = 6, then a = 9 2
63 = 0
When b =13
10, then a = 9 +
132
103
= 9 +26
30=
13
132
8a + 2b = 8 13
132+ 2
13
10
= 7913
201056
.
8. cosec2( + ) sin2() + sin2(2) = cos2()
cosec2( + ) + sin2(2) =1
)(sin)(cos 22
cosec2( + ) = 1 sin2(2)cosec2( + ) = cos2(2)Minimum value of cosec2( + ) is 1 and maximum value of cos2(2) is 1. They will be equal for the value 1.
+ =2
.....(i)
2 = 0 .....(ii)By adding (i) & (ii)
3 =2
=6
=3
sin() = sin (6
3
) = sin (
6
) =
2
1.
9. sinx + siny =5
7....(1)
cos x + cosy =5
1....(2)
By (1)2 + (2)2 we get
2 + 2sinx siny + 2 cosx cosy = 2
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5Pre-foundation Career Care Programmes (PCCP) Division 5
sinx siny + cosx cosy = 0
cos(x y) = 0
x y = 90By (1) (2) we get
sinx cosx + sinx cosy + siny cosx + siny cosy =25
7
sin(90 + y)cosx + sin(x + y) + sin(x 90) cos y =25
7
cosy cosx + sin(x + y) cosx cosy =25
7
sin(x + y) =25
7.
10.6xy =
0
1
(1, 1)
y = sin x
Clearly, curve meet each other twice in 2 34 56 78 9
10 11 Total 10 Times.
11. f(x) is differentiable on R.So, it will be contincous on R.Continuity at x = 0LHL
0xlim x
xsin 2
Put x = 0 h, then h 0
0hlim h
)h0sin( 2
0hlim h
h
h
hsin
= 0
RHL
0xlim x2 + ax + b
Put x = 0 + h, then h 0
0hlim h
2 + ah + b = b
Value of f(x) at x = 0f(0) = b.
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f(x) is contineous at x = 0
LHL = RHL = f(0)0 = b = bb = 0
Differentiability at x = 0LHD
0h
lim
h
fhf
)0()0(
0hlim
h
bh
sinh2
0hlim 2
2
h
hsin= 1
RHD
0hlim h
)0(f)h0(f
0hlim h
bbahh2
0hlim h
)ah(h = a.
f(x)is differentiable at x = 0, LHD = RHDa = 1.
12. Let point p(x1, y
1) is on the curve y2 = 4x.
y12 = 4x
1 x
1=
4
y2
1
PA = 21
21 )3y()0x(
AP2 = x1
2 + y12 6y
1+ 9
AP2 = x1
2 + y12 6y
1+ 9
Let AP = zz2 = x
12 + y
12 6y
1+ 9
z2 =
22
1
4
y
+ y
12 6y
1+ 9
z2 =
16
y4
1 + y12 6y
1+ 9
Diff. w.r.t. y1
2z1dy
dz=
16
y43
1 + 2y1 6
2z1dy
dz=
4
y3
1 + 2y1 6
=4
24y8y 13
1
2z 1dy
dz
= (y1
2) (y12
+ 2y1 + 12)
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7Pre-foundation Career Care Programmes (PCCP) Division 7
For the critical points
1dy
dz= 0
(y1 2)(y
12 + 2y
1+ 12) = 0
y1
= 2
y12 = 4x
1
(2)2 = 4x1 x
1= 1.
2
2
1dy
dz
+ 2z 2
1
2
dy
zd= (y
12 + 2y
1+ 12) + (y
1 2) ( 2y
1+ 2)
= y1
2 + 2y1
+ 12 + 2y12 4y
1+ 2y
1 4
= 3y1
2 + 8.
when y1
= 2 and1dy
dz= 0
21
2
dy
zd> 0
z is min at (1, 2)
Minimum distance = 22 )32()01( = 11 = 2 .
13. We can find the answer through option as the sum of weight of packet taken from trucks is 1022870 gm
and its unit digit is 0. The truck that have heavier bags have unit digit 0. So, the truck have lighter bags in
which the sum of weight of bags must have unit digit 0.
So, according to option D. i.e. truck no. 2, 8
Track 2 have 21 bags and total weight = 21 999 gm = .......8 gmTruck have 27 bags and total weight = 27 999 = 128 999 gm = ......2 gm
So, the unit digit of the weight contain by truck 2, 8 together is 0.
14. dx)]x2cos([)xcos(1
0
= dx0cos)xcos(2/1
0 + dxcos)xcos(1
2/1
=dx)xcos(
2/1
0
dx)xcos(
1
2/1
=
xsin
2/1
0
xsin
1
2/1
=
1
1
0
=2
.
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15. IN
=
n
x)nxcos( 101
0
+ 1
0
9
n
dxx)nxcos(10
= 0 +
1
0
8
1
0
n
dxx)nxsin(
n
9
n
10
n
x)nxsin( 9
=
1
0
8
2dxx)nxsin(
n
910
=
1
0
10dx)nxsin(
n
!10
= 0 as Denom
16. y = x2 & y = 1 x2
Point of intersections of graphs x2 = 1 x2
2x2 = 1
x = 2
1
Point of intersections =
2
1,
2
1and
2
1,
2
1.
Area under graph :
=
2/1
2/1
22 )x1(x
=
2/1
2/1
2 1x2 =
2/1
2/1
3
x3
x2
= 2 21
262
= 226
62 =
23
4=
3
22.
17. k4i3a
and b
= k12j5
54)3(a 22
and 1312)5(b 22
Therefore, a vector which bisects the angle is 13
k4i3
+ 5 (
k12j5
) =
k8j25i39
.
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9Pre-foundation Career Care Programmes (PCCP) Division 9
19. Let ......5.3.2M 321 xxx , ......5.3.2N 321 yyy xi & yi w
m/d
d=
N/d
d
12
12 1x
13
13 2x
15
15 3x
....... =
13
13
12
12 21 yy
.........
N/d
M/d
d/1
d/1
=
....13/1
13
1
12/1
12
1
....13/1
13
1
12/1
12
1
21
21
yy
xx
21
21
21
21
yy
yyxx
xx
32
)......13)(12(.....32
).....13)(12(= 1
M
N .
20.m
m2
C
Afromelementoneonly
CC nmCnmnmm 210 (1 + n)m
PHYSICS
27. Sphere is hollow so potential inside sphere will be same as that on surface.
28. Heat supplied Q = du + W (at contat pressure)PV = RTPdV = RdT
dT =R
PdV
Q = CVdT + PdV
Q = CV
R
PdV+ PdV
Work done at constant pressue, W
W = PdV
PdV
PdVR
PdVC
W
Q V
1R
CV
W
Q (For diatomic gos, CV = R
2
5)
1R2
R5
W
Q
2
7
W
Q
7
2
Q
W
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10Pre-foundation Career Care Programmes (PCCP) Division 10
29.
seriesbalmerFor3
1
2
1R
1
seriesymanlFor2
1
1
1R
1
22
22
(
36
49
9
1
4
1 )
36/5
4/3
= 5
27
5
9
1
3
27
5
30. q
A
q
B unchanged
charge divides2q and
2q
Than, on touching2
qsphere to q
Charge divides4
q3
2
q2/q
force between4
q3
2
qR
f= 2
2
R8
q3K
f=8
3 F
8
3
R
kq2
2
31. Intially block enters in the magnetic filed rate of change in flux will be constant so costant current will
produce, when it mass in side the magnetic field there is no change in magnetic flux, current I = 0, when
it use the filed the rate of the change in flux will be again constant between in decrecaing order so contant
current will induced on opposite.
32. No change in moment of inertia
34.
EA
O +qq
Electric field at each point of OA obtained to it and opposite to direction of dipole moment.
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11Pre-foundation Career Care Programmes (PCCP) Division 11
38.
m
a
Total force in upward direction m (g + a) because mass m is stationary on inclined plane and whole
system is accelerated with acceleatration a in upward.
39. Force of positve charge = Electric force + Magnetic force
F = (qE + qVB)
This force is in upward direction so no any particle will pass through the hole.
40. Potential energy at H height = Kinetic energy at the lowest point of circular path.
mgH =2
1mv2
To complete the circular motion minmum velocity at lowest point will be V = gR5
mgH =2
1m (5gR)
H =2
5R
CHEMISTRY
41. According to Grahams law
Rate of diffusion massMolar
1
due to highest molar mass of CO2rate of diffusion is slowest.
42. Moles of H2
=2
3, Moles of O
2=
32
4
Kinetic energy of n moles of gas =
2
3nRT
so,oxygenofenergyKinetic
hydrogenofenergyKinetic=
RTn2
3
RTn2
3
2
1
=2
1
n
n
=32/4
2/3
= 12 : 1
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44. ClF3 sp3d hybridisation,
but due to presence of two lp on central atom Cl, according to VSEPR theory shape is T
45. HCO3
+ H+ H2CO
3
Bronsted base
HCO3
H+
+ CO32
Bronsted acid
47. Isoelectronic means same no. of electronsCO has 6 + 8 = 14 electronsCN has 6 + 7 + 1 = 14 electrons
48. CO2, due to sp hybridisation bond angle = 180
49. Diethyl ether, because it is inert towards the Grignard reagent
50. CH3 CH
2 CH
2 CHO + CH
3 CH
2 CH
2 MgBr
CH CH CH C CH CH CH3 2 2 2 2 3
H
OMgBr
H O3+
CH CH CH C CH CH CH3 2 2 2 2 3
H
OH
Achiral Secondary alcohol
51. [Ni (PPh3)
2Cl
2] dsp2 hybridisation, because PPh
3is strong ligand hence pairing of electrons
takes place[NiCl
4]2 sp3 hybridisation, because Cl is weak ligand hence pairing of electrons is not
takes place
52. 16H + 2MnO + 5COO+ 4
COO
53. Suppose equilibrium constant for the following reaction is K1
N2+ 3H
22NH
3; K
1= 3
22
23
]H][N[
]NH[-------- (i)
and equilibrium constant for the following reaction is K2
21 N2 + 2
3 H2 NH3 ; K2 = 2/32
2/12
3]H[]N[
]NH[-------- (ii)
square the both side of equation (ii)
K22 = 3
22
23
]H][N[
]NH[
K22 = k
1[by equation (i)
K2
= 1k
K2
= 41 [ K1 = 41]
K2
= 6.4
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54. Suppose reaction is 2A Productaccording to rate lawRate R = k [A]2
or2
1
R
R= 2
2
21
]A[
]A[
according to question2
1
RR = 2
1
21
2
]A[]A[
2]A[]A[ 12
2
1
R
R= 4
R2
=4
R1
55. HCO3
H+
+ CO32
Conjugate base
NH3 H+ + NH
2
Conjugate base
56. (II) & (IV)Because both have close system of conjugated double bond and follow Huckels (4n+2) e rule.
57. N
H
p of N takes part in resonance with conjugated double bonds, so it is not easily available on N for theprotonation.
N
p is not taken part in resonance so easily available for the protonation.
N
H
O
due to high E.N. of O availability of p on N decreases.
N
H
No extra effect, so availability of p on N increases.
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58. Gauche conformer.
because angle between same groups is 60
59. Suppose initial quantity = No
after 75% completion of the reaction
remaining quantity N = No 100
25
= 4
No
T =K
303.2log
N
No
T =K
303.2log
4/N
N
o
o
T =K
386.1-------- (i)
T1/2
=K
693.0
K =30
693.0-------- (ii)
so by equation (i) and (ii)
T =30/693.0
386.1
T = 60 min.
60. Concentration of H+ ions in H2SO
4solution = 2 0.1 = 0.2 M
So no. of moles of H+ ions in 10 ml H2SO
4
solution =1000
102.0 = 0.002
concentration of OH ions in 0.1 KOH solution
= 1 0.1
= 0.1 M
So no. of moles of OH ions in 10 ml KOH
solution =1000
101.0
= 0.001
after mixing remaining moles of H+ ions = 0.002 0.001
= 0.001
so concentration of H+ ions in mixture of
solutions =
1010
001.0
1000 = 0.05 M
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PART-II (2 Mark)MATHEMATICS
81. p(x) = a0
+ a1x + ........+ a
nxn
p(0) = 7a
0= 7
p(1) = a0 + a1 + a2 + ............+ an = 9p( 1) = a0 a
1+ a
2........... = 1
p(2) = a0
+ 2a1
+ 4a2
+ .......... = 13p( 2) = a
0 2a
1+ 4a
2........... = 15
p(1) + p( 1) = 2[a0
+ a2
+ .........] = 10a
0+ a
2+ a
4= 5 .....(1)
7 + a2
+ a4
= 5a
2+ a
4= 2 .....(2)
p(2) + p( 2) = 13 152(a
0+ 4a
2+ .........) = 2
a0
+ 4a2
+ 16a4
= 14a
2+ 16a
4= 8 .....(2)
p(3) = 25
a0 + 3a1 + 9a2 + ........ = 25a
0+ 3a
1+ 9a
2+ 27a
3+ 81a
4+ 243a
5= 25 .....(3)
From (1) and (2)
4a + 4a = 82 44a + 16a = 82 4
+
a = 0 and a = 24 2
Smallest possible value of n is 3.
82.
abc
2 0)ba( 1
ab
a2
|a b| < c ...... (1) [Triangle inequalities]|b c| < a ...... (2)|c a| < b ...... (3)Squaring and addinga2 + b2 + c2 < 2ab + 2bc + 2ca
2ab
a2
So, b [1,2).
83. y = 4|3x| 5
6 1 3 7
12
Ox
6
x4
2xx
12
5
When, x < 1y = | 3 x 4 | 5y = x 1 5y = x 6When 1 x < 3y = | 3 x 4 | 5= | x 1| 5= x + 1 5= x 4When, 3 x < 7y = |x 7| 5y = 7 x 5
y = 2 x
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16Pre-foundation Career Care Programmes (PCCP) Division 16
When, x 7y = | x 7| 5
= x 7 5
= x 12.
Area bounded region
=
1
6dx)6x( +
3
1dx)4x( +
7
3dx)x2( +
12
7dx)12x(
=
1
6
2
x62
x
+
3
1
2
x42
x
+
7
3
2
2
xx2
+
12
7
2
x122
x
=
6
2
1
36
2
36+
12
2
9
4
2
1+
2
4914
2
96 +
144
2
144
84
2
49
= 2
11
18
2
15
2
9
2
21
2
3
72 + 2
119
=2
130
2
48 90
= 65 24 90
= 49 sq. unit.
84.b
b
a b
a
A B
D C
cos =ab2
aba 222 =
a2
b....(i)
cos(180 ) = 2
222
b2
abb
cos = 2
22
b2
ab2
cos = 2
22
b2
b2a ....(ii)
From (i) & (ii)
2
22
b2
b2a =
a
b
2
3
b
a
2
b
a 1 = 0
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Letb
a= x
x3 2x 1 = 0
(x + 1)(x2 x 1) = 0
x = 1 or x =
2
)1)(1(411
=2
51
x cannot be negative
x =2
1)15(
85. an
=2
a1 1n
a1
=2
a1 0
a1
=2
cos1
a1
=
2
2cos2 2
a1
=2
cos
a2
=2
a1 1
=2
2cos1
=
2
4cos2
2
= cos 22
--
an
= cos n2
nlim 4n(1 a
n)
nlim
4n
n2cos1
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18Pre-foundation Career Care Programmes (PCCP) Division 18
nlim 4n 2sin2 1n2
nlim
1n1n
1n
2n2
22
2sin22
1n1n 22
nlim
2
2
1n1n
1n
2
22
2sin
=2
2.
86. f(x) = (sin x)sinx
f(x) = xsinlogxsine
Minimum value of sinxlog(sinx) is 0.Maximum value of (sin x)sinx is e0 = 1.
Maximum value of sinxlog(sinx) is e
1.
Minimum value of (sin x)sinx is e1
e
.
87. 10
1
dxx
1
= xlog10
1= log10 = 2.303
B = 1 +2
1+ ....+
9
1
= 1 + 0.5 + 0.33 + 0.25 + 0.20 + 0.16 + 0.14 + 0.12 + 0.11 2.81
C =2
1+ ....+
9
1+
10
1 2.81 1 + 0.1 =1.91
So, C < A < B and B A 0.51, A C = 2.303 1.91 0.40. So, B A > A C.
88. r
rr
r
60
60
r
rA B
CD
As we want the distance between two point is at least r. Now when the point A, B are at distance r.Thenthe angle made arc BA is 60.Now as chord AB come closer to centre the length of chord AB is increased that is it is greater than r andthe angle is also increases i.e. from 60 to 180 and now when chord AB move way from centre then the
length of chord AB decreases , when chord AB reach CD the length of AB equal to r and the angle changfrom 180 to 60
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19Pre-foundation Career Care Programmes (PCCP) Division 19
So, the angle required for desired conditions = 2(180 60) = 240Total angle for all around the circle = 360
So, required propability =360
240=
3
2
89. Let aN
be pth
digit no.
So
1
5
)5(4 1P< N
1
5
)15(4 P2 10P1 < a
N
9
8(10P1)
5P1 1 < N 5P1
min
N
Nlog
alog
=
)15ln(
)102ln(P
1P=
5ln510
)15()10(ln10P1P
P1P
=5ln
10ln= 105log
max
N
Nlog
alog
=
)15ln(
)110(9
8ln
1P
P
=
5ln5)110(9
8
)15(10ln109
8
1PP
1PP
= 105log
By sandwitch therom limit is 105log .
90. S = nji1
|ji|
= |11| + |1 2| + -------- |1n| =2
n)1n(
|2 2| + -------- |2 n| =2
)1
n)(2
n .......
S =
N
1r2
)1rn)(rn(
=3C
1n
PHYSICS
91. Mass of sphere of radius r
m = 3
3
R
Mr
3R
3
4
M
Xcm
=mM
)rR(m0M
m
R3/4
Mr3
4
3
3
Xcm =
3
3
3
3
R
MRM)rR(
R
Mr
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20Pre-foundation Career Care Programmes (PCCP) Division 20
=)rrRR)(rR(
)rR(r
R
rRR
)rR(r22
3
3
333
3
X
cm= 22
3
rrRR
r
92. For planeo convex lens
R1)1(
F1L
FL
=1
R
Refraction through less
u
1
R
1
V
1
This v must be centre of mirror
u
1
1r
R
1
R
1
u =
R
93. In cyclic process u = 0u =W = Area of loop= (P
1 P
2) (V
2 V
1)
94. On comparing both the figures
x =R6)xR(
)R6)(xR(
x2 + xR 6R2 = 0
x =2
)R64(Rx 22
x = R22
R5R
95. Torque about point A
RsinmgMRMR5
2 22
f
A
O
= R7sing5
Applying Newtons law,Ma = Mgsin Mgcos
gcosgsin7
sing5 gcos
7
sing2
= tan12
7
Velocity of sound =M
RTT Velocity of sound
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21Pre-foundation Career Care Programmes (PCCP) Division 21
96.
H
F
a/2A
mg
For height greater than HBalancing torque about point A
F H =2
amg .........(1)
For height less than HF = gh .........(2)From (1) and (2)
=H2
a
97.
mV /r2
mg
N
sin
r
mvcosmgsinmgcos
r
MV 22
sin
rVg
2
= cos
rVg
2
tan =rgV
rgV2
2
98.
3 5
4S1
S2
Path difference = 5 4 = 1 mFor constructive interferencen
1= 1m
For destructive interferenceFor n = 1,
1= 1m,
2= 2m
m12
)1n2( 2
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22Pre-foundation Career Care Programmes (PCCP) Division 22
99. Two small blocks slide without losing contact with the surface along two frictionless tracks 1 and 2,starting at the same time with same initial speed v. Track 1 is perfectly horizontal, while track 2 has adip in the middle, as shown.
1
2 V
V
FinishStart
Which block reaches the finish line first ?
[Hint : Use velocity-time graph to solve](A) Block on track 1 reaches the finish line first(B*) Block on track to reaches the finish line first(C) Both blocks reach the finish line at the same time(D) It depends on the length of the dip in the second track, relative to the total length of the tracks.
100. Q = u + PV (1st law of thermodynamics )
mv
= u + 1.01 105 ( 310
1
8.1/1
1 )
u 20.8 105 J kg1
CHEMISTRY
101. Millimoles of NH4OH in 10 ml of 0.1 M NH
4OH solution = 0.1 10 = 1
millimoles of NH4Cl in 10 ml of 1M NH
4Cl solution = 1 10 = 10
pOH = pKb
+ log]Base[
]Salt[
6 = pKb
+ log1010/1
1010/10
[ pOH = 14 pH = 14 8 = 6]
6 = pKb
+ log10pK
b= 6 1
pKb
= 5
102. 2C4H
10+ 13O
2 8CO
2+ 10H
2O H = 2658 kJ/mol
Butane present in cylinder = 11.6 kg= 11600 g
=58
11600mol
Combustion of 1 mol of C4H
10gives = 2658 kJ energy
Combustion of 11600/58 mol of C4H
10gives =
58
116002658 kJ
= 531600 kJ energy
energy consumes in 1 day = 15000 kJ
so 531600 kJ energy will be consumed in =15000
531600 35 days
103. W = d V = 0.879 50 = 43.95 , Kf= 5.12 Kg kg mol1
= 5120K g mol1
Tf=
mW
wK of
5.51 5.03 =95.43m
643.05120
m =096.21
16.3292
m = 156 g mol1
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23Pre-foundation Career Care Programmes (PCCP) Division 23
104. Zn + 2Ag+ Zn2+ + 2Ag
(0.04M) (0.28M)
Ecell
= E n
059.0log
]Zn[]Ag[
]Ag][Zn[2
22
Ecell
= 2.57 2
059.0log
1)04.0(
128.02
2
{ [Ag] = [Zn] = 1}
by solving the equation we get
Ecell
2.50 V
105. Co + Con. HCl2+
CoCl42
HOH(excess)
[Co(H O) ]2 62+
Pink
106. ln k =T
11067+ 31.33
2.303 log k =T
11067+ 31.33
suppose k1and T
1are rate constant and temperature in case-I
and k2
and T2
are rate constant and temperature in case-II
So, 2.303 log k1
= 1T
11067+ 31.33 --------- (1)
2.303 log k2=
2T
11067+ 31.33 --------- (2)
by subracting equation (2) from equation (1)
2.303 log2
1
k
k= 11067
21 T
1
T
1
2.303 log1
1
k2
k= 11067
2T
1
298
1[ k
2= 2k
1]
2.303 (0.3010) =2T
11067
298
11067
by solving the above equation we get
T2 303.7 K
T2 31C
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24Pre-foundation Career Care Programmes (PCCP) Division 24
107. K1
=]Br][CuCl[
]Cl][BrCuCl[2
4
23
------- (i)
K2
=]Br][BrCuCl[
]Cl][BrCuCl[2
3
222
------- (ii)
K3 = ]Br][BrCuCl[
]Cl][CuClBr[2
22
2
3 ------- (iii)
K4
=]Br][CuClBr[
]Cl][CuBr[2
3
24
------- (iv)
equilibrium constant for given equation
K = 324
323
]Br][CuCl[
]Cl][CuClBr[------- (v)
by multiplying the right hand side of equation (i), (ii) and (iii) we get right hand side of equation (v)it means K = K
1K
2K
3
108.
OH
Br in CS2 2
HBr
Br
NaOH
H O2
OH
Br
ONa
MeINaI
Br
OMe
(X)
(Y)
109. eyHexPbTh 0142
20682
23490
By comparing mass no.234 = 206 + 4x + oyx = 7By comparing nuclear charge90 = 82 + 2x 1yy = 82 + 2 7 90y = 6
110.
O
AlCl3 I2/NaOH
OH OH
IIO
MeIIO
Me
IIO
OH
Haloformreaction
fries
rearrangement