Solution 3 (Class XII)

8/8/2019 Solution 3 (Class XII)

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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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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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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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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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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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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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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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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


15/24


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


16/24


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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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


19/24


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


20/24


=)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


21/24


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


22/24


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


23/24


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


24/24


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

Solution 3 (Class XII)

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Transcript of Solution 3 (Class XII)