**PART-I (1 Mark)**

**MATHEMATICS**

**1.** Given : a_{1}, a_{2}, a_{3} ...AP and a_{1}, a_{2}, a_{4}, a_{8} ...GP.
Let common difference of A.P. = d

a_{2 }= a_{1} + d
a_{4} = a_{1} + 3d
a_{8} = a_{1} + 7d
1
2
a
a
=
2
4
a
a
=
4
8
a
a
= r
1
1
a
d
a
= a_{a} 3_{d}d
1
1
= _{a}a _{3}7_{d}d
1
1
= r
(a_{1} + d)2_{ = a}
1(a1 + 3d)
a_{1}2_{ + d}2_{ + 2 a}
1d = a1
2_{ + 3a}
1 d
d2_{ = a}
1 d (d 0)
d = a_{1} ....(i)
Hence,
1
2
a
a
= r ;
1
1
a
d
a
= r
1
1
1
a
a
a
= r (using (i))
r = 2.

**ANSW ER KEY**

**HINTS & SOLUTIONS (PRACTICE PAPER-3)**

**Que s.** **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
**Que s.** **16** **17** **18** **19** **20** **21** **22** **23** **24** **25** **26** **27** **28** **29** **30**
**Ans.** B C B C D B B D D C C D C A D
**Que s.** **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
**Que s.** **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
**Que s.** **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
**Que s.** **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
**Que s.** **91** **92** **93** **94** **95** **96** **97** **98** **99** **100** **101** **102** **103** **104** **105**
**Ans.** D A B C D A C C B B B D A A B
**Que s. 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

**2.**
k
101
T
T
1
–
k
k
till k = 10
T_{k} > T_{k–1}
Let k = 11
T_{11} < T_{10} T10 is maximum at k = 10.
**3.** x = 2 + 3+ _{6}

##

x 2##

2 =##

##

2 6 3 x2_{ + 2 }– 2 2 = 9 + 6 2 x2 – 7 = 8 2 (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 9C_{3}

B win 3 games out of remaining 6 6C_{3}

C win 3 games out of remaining 3 3C3

So, required way = 9_{C}
3 ×
6_{C}
3 ×
3_{C}
3
=
!
6
1
2
3
!
6
7
8
9
×
!
3
1
2
3
!
3
4
5
6
×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 = (z3)2(22)2 = | z – 3 | = _{6} 3
5
_{ = }
6
13
.

**6.**
P P'
(5, 15) (21, 15)L
A
Mid point of PP’ =
2
15
15
,
2
21
5
L = (13, 15)
Point A will be (13, 0)
By property PA + PA’ = 2a
PA = _{(}_{13} _{5}_{)}2 _{(}_{0} _{15}_{)}2
= _{64}_{}_{225}
= _{289} = 17 cm
PA’=

### (

### 13

_{}

### 21

### )

2_{}

### (

### 0

_{}

### 15

### )

2 = 64225 = 289 = 17 cm 2a = PA + PA’ 2a = 17 + 17 2a = 34 cmSo, 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 – 20b
a2_{ + b}2
– 20a – 20b + 84 = 0 ....(i)

Also (a, b) i.e. on 2x + 3y = 18 2a + 3b = 18

a = 9 –

2 b 3 Using equation (i)

2
2
b
3
9
+ b2
– 20
2
b
3
9 _{– 20b + 84 = 0}
81 +
4
b
9 2
– 27b + b2– 180 + 30b – 20b + 84 = 0
4
b
13 2
– 17b – 15 = 0
13b2
– 68b – 60 = 0

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 6 3 = 0 When b = 13 10 , then a = 9 + 13 2 10 3 = 9 + 26 30 = 13 132 8a + 2b = 8 × 13 132 + 2 × 13 10 =

### 79

### 13

### 20

### 1056

###

###

.**8.**cosec2

_{(}

+ ) – sin2(– ) + sin2(2– ) = cos2(– )

cosec2_{(}
+ ) + sin2(2– ) =
1
)
(
sin
)
(
cos2 2
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}

sinx siny + cosx cosy = 0 cos(x – y) = 0

x – y = 90º

By (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.**
6
x
y =
0
1
(1, 1)
y = sin x

Clearly, curve meet each other twice in 2– 3

4– 5
6– 7
8– 9
10– 11
Total 10 Times.
**11.** f(x) is differentiable on R.
So, it will be contincous on R.
Continuity at x = 0
LHL
–
0
x
lim
_{x}
x
sin 2
Put x = 0 – h, then h 0
0
hlim _{h}
)
h
0
sin( 2
0
hlim _{h}
h
h
h
sin
= 0
RHL
0
x
lim _{x}2_{ + ax + b}
Put x = 0 + h, then h 0
0
hlim h
2_{ + ah + b = b}
Value of f(x) at x = 0
f(0) = b.

f(x) is contineous at x = 0 LHL = RHL = f(0) 0 = b = b b = 0 Differentiability at x = 0 LHD 0 hlim

_{h}

_{h}

*f*

*h*

*f*

###

###

###

### )

### (

### 0

### )

### 0

### (

0 hlim h b h sinh2 0 hlim 2 2 h h sin = 1 RHD 0 hlim_{h}) 0 ( f ) h 0 ( f 0 hlim

_{h}b – b ah h2 0 hlim

_{h}) a h ( h = a. f(x)is differentiable at x = 0, LHD = RHD a = 1.

**12.** Let point p(x_{1}, y_{1}) is on the curve y2_{ = 4x.}

y1
2_{ = 4x}
1 x1 =
4
y_{1}2
PA = 2
1
2
1 0) (y 3)
x
(
AP2_{ = x}
1
2_{ + y}
1
2
– 6y_{1} + 9
AP2_{ = x}
1
2_{ + y}
1
2
– 6y1 + 9
Let AP = z
z2_{ = x}
1
2_{ + y}
1
2
– 6y1 + 9
z2_{ = }
2
2
1
4
y
+ y_{1}2
– 6y1 + 9
z2_{ = }
16
y_{1}4
+ y_{1}2
– 6y1 + 9
Diff. w.r.t. y_{1}
2z
1
dy
dz
=
16
y
4 _{1}3
+ 2y_{1}– 6
2z
1
dy
dz
=
4
y13 _{ + 2y}
1– 6
=
4
24
y
8
y_{1}3 _{1}
2z
1
dy
dz
= (y_{1}– 2) (y_{1}2 + 2y_{1} + 12)

For the critical points
1
dy
dz
= 0
(y_{1}– 2)(y1
2_{ + 2y}
1 + 12) = 0
y_{1} = 2
y_{1}2 = 4x_{1}
(2)2 = 4x_{1}
x1 = 1.
2
2
1
dy
dz
+ 2z 2
1
2
dy
z
d
= (y_{1}2_{ + 2y}
1 + 12) + (y1– 2) ( 2y1 + 2)
= y_{1}2_{ + 2y}
1 + 12 + 2y1
2
– 4y1 + 2y1– 4
= 3y_{1}2_{ + 8.}
when y_{1} = 2 and
1
dy
dz
= 0
2
1
2
dy
z
d
> 0
z is min at (1, 2)
Minimum distance = (10)2(23)2 = 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 = 2}1

× 999 gm = ...8 gm

Truck have 27_{ bags and total weight = 2}7

× 999 = 128 × 999 gm = ...2 gm

So, the unit digit of the weight contain by truck 2, 8 together is 0.

**14.** cos( x)cos([2x] )dx
1
0

###

= cos( x)cos0dx 2 / 1 0###

+ cos( x)cos dx 1 2 / 1###

= cos( x)dx 2 / 1 0###

_{– }cos( x)dx 1 2 / 1

###

=_{}

###

###

###

###

###

###

###

### x

### sin

1/2 0 –_{}

###

###

###

###

###

###

###

### x

### sin

1 2 / 1 = 1 – 1 0 = 2 .**15.** I_{N} =
n
x
)
nx
cos( 10
1
0
+

_{}

1
0
9
n
dx
x
)
nx
cos(
10
= 0 +
_{}

###

###

###

###

###

###

1 0 8 1 0 n dx x ) nx sin( n 9 n 10### n

### x

### )

### nx

### sin(

### 9

= ###

1 0 8 2 sin(nx)x dx n 9 10 = ###

1 0 10 sin(nx)dx 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
1
and _{}
2
1
,
2
1
.
Area under graph :

=

_{}

2
/
1
2
/
1
2
2 _{(}

_{1}

_{x}

_{)}x =

_{}

2
/
1
2
/
1
2
1
x
2 _{ = }2 / 1 2 / 1 3 x 3 x 2 = 2 × 2 1 2 6 2 = 2 2 6 6 2 = 2 3 4 = 3 2 2 .

**17.**a3i4kand b = 5j12k

###

4 5 ) 3 ( a 2 2 and b (5)###

12 13 2 2 Therefore, a vector which bisects the angle is 13

_{}

3i4k_{}

+ 5 (5j 12k
) = 39i 25j 8k .

**19.** Let M 2x1.3x2.5x3...
, N 2y1.3y2.5y3...
xi & yi w

###

m / d d =###

N / d d _{} 1 – 2 1 – 2x1 1 – 3 1 – 3x2 1 – 5 1 – 5x3 ... =

_{} 1 – 3 1 – 3 1 – 2 1 – 2y1 y2 ...

###

###

N / d M / d d / 1 d / 1 = .... 1 – 3 / 1 1 – 3 1 1 – 2 / 1 1 – 2 1 .... 1 – 3 / 1 1 – 3 1 1 – 2 / 1 1 – 2 1 2 1 2 1 y y x x 2 1 2 1 2 1 2 1 y y y y x x x x 3 2 )... 1 – 3 )( 1 – 2 ( ... 3 2 )... 1 – 3 )( 1 – 2 ( = 1 M N .**20.**C 2 m m A from elementonlyone

C C m n m n mC n m 2 1 0 (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 = RT
PdV = RdT
dT =
R
PdV
Q = C_{V}dT + PdV
Q = C_{V}
R
PdV
+ PdV

Work done at constant pressue, W W = PdV PdV PdV R PdV C W Q V 1 R CV W Q

(For diatomic gos, CV = R

2 5 ) 1 R 2 R 5 W Q 2 7 W Q 7 2 Q W

**29.**
series
balmer
For
3
1
2
1
R
1
series
yman
l
For
2
1
1
1
R
1
2
2
2
2
(
36
4
9
9
1
4
1
)
36
/
5
4
/
3
=
5
27
5
9
1
3
27
5
**30.** q
A
q
B _{unchanged}
charge divides
2
q
and
2
q
Than, on touching
2
q
sphere to q
Charge divides
4
q
3
2
q
2
/
q
force between
4
q
3
2
q
R
f= _{2}
2
R
8
q
3
K
f=
8
3
× F
8
3
R
kq
2
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.**

E A

O +q

–q

**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 1

mv2

To complete the circular motion minmum velocity at lowest point will be V = 5gR

mgH = 2 1 m (5gR) H = 2 5 R

**CHEMISTRY**

**41.** According to Graham’s law

Rate of diffusion _{Molar}_{mass}

1

due to highest molar mass of CO_{2} rate of diffusion is slowest.

**42.** Moles of H_{2} =
2
3
, Moles of O_{2} =
32
4

Kinetic energy of n moles of gas = 2 3 nRT so, oxygen of energy Kinetic hydrogen of energy Kinetic = RT n 2 3 RT n 2 3 2 1 = 2 1 n n = 32 / 4 2 / 3 = 12 : 1

**44.** ClF_{3} sp3d hybridisation,

but due to presence of two lp on central atom Cl, according to VSEPR theory shape is ‘T’

**45.** HCO_{3}–_{ + H}+
H2CO3
Bronsted base
HCO_{3}–_{ }
H+ + CO3
2–
Bronsted acid

**47.** Isoelectronic means same no. of electrons
CO has 6 + 8 = 14 electrons

CN–_{ has 6 + 7 + 1 = 14 electrons}

**48.** CO_{2}, due to sp hybridisation bond angle = 180º

**49.** Diethyl ether, because it is inert towards the Grignard reagent
**50.** CH_{3}– CH2– CH2– CHO + CH3– CH2– CH2– MgBr
CH 3– CH – CH – C – CH – CH – CH2 2 2 2 3
H
OMgBr
H O3
+
CH 3– CH – CH – C – CH – CH – CH2 2 2 2 3
H
OH

Achiral Secondary alcohol

**51.** [Ni (PPh_{3})_{2} Cl_{2}] dsp2 hybridisation, because PPh3 is 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 K_{1}

N_{2 }+ 3H_{2} 2NH_{3} ; K_{1} = 3
2
2
2
3
]
H
][
N
[
]
NH
[
--- (i)
and equilibrium constant for the following reaction is K_{2}

2
1
N_{2} +
2
3
H_{2} NH_{3} ; K_{2} = 3/2
2
2
/
1
2
3
]
H
[
]
N
[
]
NH
[
--- (ii)
square the both side of equation (ii)

K_{2}2_{ = }
3
2
2
2
3
]
H
][
N
[
]
NH
[
K_{2}2_{ = k}

1 [by equation (i)

K_{2} = k1

K_{2} = 41 [ K1 = 41]

**54.** Suppose reaction is 2A Product

according to rate law
Rate R = k [A]2
or
2
1
R
R
= 2
2
2
1
]
A
[
]
A
[
according to question
2
1
R
R
= _{2}
1
2
1
2
]
A
[
]
A
[
2
]
A
[
]
A
[ 1
2
2
1
R
R
= 4
R2 = _{4}
R_{1}
**55.** HCO_{3}–
H+ + CO_{3}2–
Conjugate base
NH_{3} H+ + NH_{2}–
Conjugate base

**56.** (II) & (IV)

Because both have close system of conjugated double bond and follow Huckel’s (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 the

protonation.

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

**58.** Gauche conformer.

because angle between same groups is 60º

**59.** Suppose initial quantity = N_{o}
after 75% completion of the reaction
remaining quantity N = N_{o}×
100
25
=
4
N_{o}
T =
K
303
.
2
log
N
No
T =
K
303
.
2
log
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 H}
2SO4 solution = 2 × 0.1 = 0.2 M

So no. of moles of H+_{ ions in 10 ml H}
2SO4
solution =
1000
10
2
.
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 10 1 . 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 = 10 10 001 . 0 × 1000 = 0.05 M

**PART-II (2 Mark)**

**MATHEMATICS**

**81.**p(x) = a

_{0}+ a

_{1}x + ...+ a

_{n}xn p(0) = 7 a

_{0}= 7 p(1) = a

_{0}+ a

_{1}+ a

_{2}+ ...+ a

_{n}= 9 p(– 1) = a0– a1 + a2 ... = 1 p(2) = a

_{0}+ 2a

_{1}+ 4a

_{2}+ ... = 13 p(– 2) = a0– 2a1 + 4a2 ... = – 15 p(1) + p(– 1) = 2[a0 + a2 + ...] = 10 a

_{0}+ a

_{2}+ a

_{4}= 5 ...(1) 7 + a

_{2}+ a

_{4}= 5 a

_{2}+ a

_{4}= – 2 ...(2) p(2) + p(– 2) = 13 – 15 2(a

_{0}+ 4a

_{2}+ ...) = – 2 a

_{0}+ 4a

_{2}+ 16a

_{4}= – 1 4a

_{2}+ 16a

_{4}= – 8 ...(2) p(3) = 25 a

_{0}+ 3a

_{1}+ 9a

_{2}+ ... = 25

a_{0} + 3a_{1} + 9a_{2} + 27a_{3} + 81a_{4} + 243a_{5} = 25 ...(3)

From (1) and (2)
4a + 4a = 2 4 – 8
4a + 16a = 2 4 – 8
– – +
a = 0 and a = 4 2 – 2
Smallest possible value of n is 3.
**82.**

###

abc 2_{0}) b – a (

_{ab}1 a2

###

###

|a – b| < c ... (1) [Triangle inequalities] |b – c| < a ... (2) |c – a| < b ... (3)Squaring and adding

a2_{ + b}2_{ + c}2_{ < 2ab + 2bc + 2ca}
2
ab
a2
So, b [1,2).
**83.** y = |x3|4 – 5
– 6 – 1 3 7
12
O
–_{ x }
–_{ 6}
x – 4
2 _{–}_{x}
x – 1
2
– 5
When, x < – 1
y = | 3 – x – 4 | – 5
y = – x – 1 – 5
y = – x – 6
When –1 x < 3
y = | 3 – x – 4 | – 5
= | – x – 1| – 5
= x + 1 – 5
= x – 4
When, 3 x < 7
y = |x – 7| – 5
y = 7 – x – 5
y = 2 – x

When, x 7

y = | x – 7| – 5

= x – 7 – 5

= x – 12.

Area bounded region

=

_{}

1
6
dx
)
6
x
(
+ _{}

3
1
dx
)
4
x
(
+ _{}

7
3
dx
)
x
2
( _{ + }

###

12 7 dx ) 12 x ( = 1 6 2 x 6 2 x + 3 1 2 x 4 2 x _{ + }7 3 2 2 x x 2

_{ + }12 7 2 x 12 2 x = 6 2 1 – 36 2 36 + 12 2 9 – 4 2 1 + 2 49 14

_{– } 2 9 6

_{ + } 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 = ab 2 a b a2 2 2 = a 2 b ....(i) cos(180º – ) = 2 2 2 2 b 2 a b b – cos = 2 2 2 b 2 a b 2 cos = 2 2 2 b 2 b 2 a ....(ii) From (i) & (ii)

2 2 2 b 2 b 2 a =

*a*

*b*

### 2

3 b a – 2 b a – 1 = 0Let
b
a
= x
x3
– 2x – 1 = 0
(x + 1)(x2
– x –1) = 0
x = – 1 or x =
2
)
1
)(
1
(
4
1
1
=
2
5
1
x cannot be negative
x =
2
1
)
1
5
(
**85.** a_{n} =

### 2

### a

### 1

###

_{n}

_{1}a

_{1}= 2 a 1

_{0}a

_{1}= 2 cos 1 a

_{1}=

### 2

### 2

### cos

### 2

2###

a_{1}= 2 cos a

_{2}=

### 2

### a

### 1

###

_{1}= 2 2 cos 1 =

### 2

### 4

### cos

### 2

2###

= cos 2 2 -a_{n}= cos n 2 nlim 4 n

_{(1 }– an) nlim 4 n

_{}

_{n}2 cos 1

nlim 4 n × 2sin2 n 1

### 2

###

nlim 1 n 1 n 1 n 2 n 2 2 2 2 sin 2 2 × n1 n 1 2 2 nlim_{2}2 × 1 n 1 n 1 n 2 2 2 2 sin = 2 2 .

**86.**f(x) = (sin x)sinx

f(x) =

### e

sinxlogsinxMinimum 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 e

1

### e

.**87.**

_{}

10
1
dx
x
1
= _{}

### log

### x

_{}

10_{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
r
r
r
60º
60º
r
r
A B
C
D

As we want the distance between two point is at least r. Now when the point A, B are at distance r.Then the angle made arc BA is 60º.

Now as chord AB come closer to centre the length of chord AB is increased that is it is greater than r and the angle is also increases i.e. from 60º to 180º 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 chang from 180º to 60º

So, the angle required for desired conditions = 2(180 – 60) = 240

Total angle for all around the circle = 360º

So, required propability = 360 240

=
3
2
**89.** Let a_{N} be p_{th} digit no.

So
1
–
5
)
5
(
4 P–1
< N
1
–
5
)
1
–
5
(
4 P
2 × 10P–1 < aN _{9}
8
(10P
–1)
5P–1
– 1 < N 5P–1
min
N
N
log
a
log
=
)
1
–
5
ln(
)
10
2
ln(
P
1
P
=
5
ln
5
–
10
)
1
–
5
(
)
10
(ln
10
P
1
–
P
P
1
–
P
=
5
ln
10
ln
= 10
5
log
max
N
N
log
a
log
=
)
1
–
5
ln(
)
1
–
10
(
9
8
ln
1
P
P
= (10 _{–}1)5 ln5
9
8
)
1
–
5
(
10
ln
10
9
8
1
P
P
1
P
P
= log10_{5}

By sandwitch therom limit is log10_{5} .

**90.** S =

###

i j n 1 | j – i | = |1–1| + |1 – 2| + --- |1–n| = 2 n ) 1 – n ( |2 – 2| + --- |2 – n| = 2 ) 1 – n )( 2 – n ... S =_{}

N
1
r 2
)
1
r
–
n
)(
r
–
n
(
=n1C_{3}

**PHYSICS**

**91.** Mass of sphere of radius r

m = _{3}
3
R
Mr
R3
3
4
M
X_{cm }=
m
M
)
r
R
(
m
0
M
m
R
3
/
4
M
r
3
4
3
3
X_{cm }= _{}
_{3}
3
3
3
R
M
R
M
)
r
R
(
R
Mr

=
)
r
rR
R
)(
r
R
(
)
r
R
(
r
R
r
R
R
)
r
R
(
r
2
2
3
3
3
3
3
3
_{}
Xcm = 2 2
3
r
rR
R
r

**92.** For planeo convex lens

R
1
)
1
(
F
1
L
F_{L} =
1
R

Refraction through less

u 1 R 1 V 1

This v must be centre of mirror

–
u
1
1
r
R
1
R
1
u =
R
**93.** In cyclic process u = 0
u =W = Area of loop
= (P_{1}– P_{2}) (V_{2}– V_{1})

**94.** On comparing both the figures

x =
R
6
)
x
R
(
)
R
6
)(
x
R
(
x2_{ + xR}
– 6R2 = 0
x =
2
)
R
6
4
(
R
x 2 2
x = 2R
2
R
5
R

**95.** Torque about point A

R
sin
mg
MR
MR
5
2 _{2} 2
f
A
O
=
R
7
sin
g
5

Applying Newton’s law,

Ma = Mgsin – Mgcos gcosgsin– 7 sin g 5 gcos 7 sin g 2 = tan–1 2 7 Velocity of sound = M RT T Velocity of sound

**96.**

H F

a/2 A

mg

For height greater than H Balancing torque about point A F × H =

2 a

mg ...(1)

For height less than H

F = gh ...(2)
From (1) and (2)
=
H
2
a
**97.**
mV /r2
mg
N
sin
r
mv
cos
mg
sin
mg
cos
r
MV2 2
sin
r
V
g
2
= cos
r
V
g
2
tan =
rg
V
rg
V
2
2
**98.**
3 _{5}
4
S1
S2
Path difference = 5 – 4 = 1 m

For constructive interference n1= 1m

For destructive interference
For n = 1, _{1}= 1m, _{2}= 2m
m
1
2
)
1
n
2
( _{2}

**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 a
dip in the middle, as shown.

1

2 V

V

Finish Start

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 ×10
5_{ (}
3
10
1
8
.
1
/
1
1
)
u 20.8 × 105 J kg–1

**CHEMISTRY**

**101.**Millimoles of NH

_{4}OH in 10 ml of 0.1 M NH

_{4}OH solution = 0.1 × 10 = 1 millimoles of NH

_{4}Cl in 10 ml of 1M NH

_{4}Cl solution = 1 × 10 = 10 pOH = pK

_{b}+ log ] Base [ ] Salt [ 6 = pK

_{b}+ log 10 10 / 1 10 10 / 10 [ pOH = 14 – pH = 14 – 8 = 6] 6 = pK

_{b}+ log10 pK

_{b}= 6 – 1 pK

_{b}= 5

**102.** 2C_{4}H_{10} + 13O_{2} 8CO_{2} + 10H_{2}O H = – 2658 kJ/mol

Butane present in cylinder = 11.6 kg = 11600 g =

58 11600

mol

Combustion of 1 mol of C4H10 gives = 2658 kJ energy
Combustion of 11600/58 mol of C_{4}H_{10} gives =

58 11600 2658

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 mol
–1
= 5120K g mol–1
T_{f} =
mW
w
K_{f} _{o}
5.51 – 5.03 =
95
.
43
m
643
.
0
5120
m =
096
.
21
16
.
3292
m = 156 g mol–1

**104.** Zn + 2Ag+
Zn2+ + 2Ag
(0.04M) (0.28M)
E_{cell} = Eº –
n
059
.
0
log
]
Zn
[
]
Ag
[
]
Ag
][
Zn
[
2
2
2
E_{cell} = 2.57 –
2
059
.
0
log
1
)
04
.
0
(
1
28
.
0
2
2
{ [Ag] = [Zn] = 1}

by solving the equation we get
E_{cell} 2.50 V
**105.** Co + Con. HCl 2+
CoCl4
2–
HOH(excess)
[Co(H O) ]2 6
2+
Pink
**106.** ln k =
T
11067
–
+ 31.33
2.303 log k =
T
11067
–
+ 31.33

suppose k_{1} and T_{1} are rate constant and temperature in case-I
and k_{2} and T_{2} are rate constant and temperature in case-II
So, 2.303 log k_{1} = –
1
T
11067
+ 31.33 --- (1)
2.303 log k_{2 }= –
2
T
11067
+ 31.33 --- (2)

by subracting equation (2) from equation (1)
2.303 log
2
1
k
k
= –11067 _{}
2
1 T
1
–
T
1
2.303 log
1
1
k
2
k
= –11067
2
T
1
–
298
1
[ k2 = 2k1]
2.303 × (–0.3010) =
2
T
11067
298
11067
–

by solving the above equation we get
T_{2} 303.7 K

**107.** K_{1} =
]
Br
][
CuCl
[
]
Cl
][
Br
CuCl
[
–
–
2
4
–
–
2
3
--- (i)
K_{2} =
]
Br
][
Br
CuCl
[
]
Cl
][
Br
CuCl
[
–
–
2
3
–
–
2
2
2
--- (ii)
K_{3} =
]
Br
][
Br
CuCl
[
]
Cl
][
CuClBr
[
–
–
2
2
2
–
–
2
3
--- (iii)
K_{4} =
]
Br
][
CuClBr
[
]
Cl
][
CuBr
[
–
–
2
3
–
–
2
4
--- (iv)
equilibrium constant for given equation

K = 2– – 3 4 3 – – 2 3 ] 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_{1}K_{2}K_{3}
**108.**
OH
Br in CS2 2
–HBr
Br
NaOH
–H O2
OH
Br
ONa
Me–I
–NaI
Br
OMe
(X)
(Y)
**109.** _{90}234Th_{82}206Pbx_{2}4Hey_{–}0_{1}e
By comparing mass no.

234 = 206 + 4x + oy x = 7

By comparing nuclear charge
90 = 82 + 2x 1y
y = 82 + 2 × 7 – 90
y = 6
**110.**
O
AlCl3 I2/NaOH
OH OH
II
O
Me
II
O
Me
II
O
OH
Haloform
reaction
fries
rearrangement