Solution 2

HINTS & SOLUTIONS (PRACTICE PAPER-2) ANSWER KEY Que s.

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

Ans.

D 17

A 18

D 19

C 20

C 21

B 22

C 23

C 24

B 25

B

C

A

D

B

Que s.

A 16

26

27

28

29

30

Ans.

C

A

A

D

A

D

C

B

D

C

D

Que s.

31

32

33

34

35

36

37

38

39

40

41

A 42

C 43

B 44

B 45

Ans.

C 46

A 47

B 48

D 49

B 50

D 51

C 52

D 53

A 54

D 55

D 56

B 57

B 58

B 59

A 60

B 61

A 62

A 63

C 64

A 65

D 66

C 67

B 68

D 69

A 70

B 71

A 72

D 73

B 74

B 75

A 76

C 77

A 78

A 79

D 80

B 81

C 82

B 83

C 84

B 85

A 86

C 87

B 88

B 89

C 90

B Que s. 91 B Ans. Que s. 106 B Ans.

A 92 A

B 93 C

C 94 B

B 95 A

B 96 C

A 97 D

A 98 B

D 99 D

D 100 D

D 101 D

C 102 C

C 103 B

C 104 A

A 105 A

107 A

108 A

109 D

110 A

111 D

112 D

113 B

114 A

115 C

116 A

117 B

118 D

119 A

120 C

Que s. Ans. Que s. Ans. Que s. Ans.

PART-I (1 Mark) MATHEMATICS 1.

Sol.

The equation z2 = z , where z is a complex number, has (A*) 4 solution (B) 2 solution (C) no solution (D) infinitely many solutions z = a + ib

z = a – ib It is given : z2 = z (a + ib)2 = a – ib a2 – b2 + i 2ab = a – ib  a2 – b2 = a and 2ab = – b 2ab + b = 0 b(2a + 1) = 0 b = 0 or a = When b = 0,

So,

1 2

a2 – b2 = a a2 – 0 = a a (a – 1) = 0 a = 0, or a = 1 2

1   1  1 When a = ,   – b2 = 2 2  2 

Pre-foundation Career Care Programmes (PCCP) Division

11

1 1 – b2 = 4 2 1 1 3 + = 4 2 4

b2 =



3 2 z = 0 + 0i or 1 + 0i



z=

b=±

1 1 3 3 +i. or –i 2 2 2 2 There will be four solutions possible.



2.

x 1 x 1     2/3  1/ 3  x  1 x  x 1/ 2  x

10

 ( x1/ 3  1)( x 2 / 3  x1 / 3  1) ( x1/ 2  1)( x1/ 2  1)     ( x 2 / 3  x1/ 3  1) x1/ 2 ( x1/ 2  1)  

(x x

1/ 3

 1)  (1  x 1 / 2 )

10

10



10

1/ 3

 x 1 / 2 General Term



Tr + 1

= (– 1)r 10 Cr (x1/3)10 – r (x-1/2)r = (– 1)r 10 Cr x

For the independent term

10 r r  3 2

10  r r  =0 3 2

10  r r  3 2 20 – 2r = 3r 5r = 20 r = 4.

10! So, independent term = (– 1)4 10 C4 = 4!6! = 210.

3.

a a2

a3  1

b b2

b3  1

c c2

c3 1

a a2

a3

a a2

1

b b

2

b

3

b b

2

1

c c

2

c

3

c c

2

1

+

=0

1 a a2 abc

1 b b2 1 c c2

=0

a a2 1 –1

b b2 1 c c2 1

Pre-foundation Career Care Programmes (PCCP) Division

=0

22

1 a a2 abc

1 b b

a 1 a2

2

1 c c2

+1

1 a a2 abc

1 b b

c 1 c2

=0

1 a a2

2

1 c c2

b 1 b2

–1

1 b b2 1 c c2

=0

1 a a2 1 b b2 1 c c2  4.

(abc – 1) = 0

abc = 1.

ax2 – 6xy + y2 = 0 m1 + m2 =

2h b

m1 + m2 =

6 1

m 1m 2 =

a b

a =a 1 Given : m2 = m12 m 12 + m 1 – 6 = 0 (m1 + 3) (m1 – 2) = 0 m1 = – 3 and m1 = 2 and m 1m 2 = a m1 × m12 = a m 13 = a a is positive, so a = (2)3 = 8. m 1m 2 =

5.

x2 + 4y2 =1 ....(i) 2 2 4x + y = 4 y2 = 4 – 4x2 put in (i) x2 + 4(4 – 4x2) = 1 x2 + 16 – 16x2 = 1 16 – 1 = 15x2 15 = 15x2 x2 = 1 x=±1 y2 = 4 – 4x2 = 4 – 4(1) y2 = 0 y=0  (1, 0) and (– 1, 0) are two common point.

Pre-foundation Career Care Programmes (PCCP) Division

33

6.

Diameter = major axis = 2a radius = a Area of circle = a2

(0, b)

Let the equation of the ellipse be

y=

x2 a2



y2 b2

(– a, 0)

 1.

2

(0, – b) a

1 2 a = 4 × 3

Area of ellipse =

2

a –x

(a, 0)

(0, 0)

b a2  x 2 a

b y= a



0

b a 2  x 2 dx a a

1 x b x a 2  x 2  a 2 sin 1  =4×  2 a 0 a 2 2 1 2 b  a    a = 4 × 3 a  4 

b 1  a 3

7.

2

 1 1   3

2

e=

b 1   a

PD =

PA  PB 18  2 = =9 PC 4

=

=

8 2 2 = . 9 3 B

N O 18 4 P 9 C 2 M A

1 BN = AB = 10 2 1 13 CD = 2 2 PM = MC – PC

D

MC =

13 5 –4= 2 2 5 ON = PM = 2 In ONB =

OB =

5 10 2    2

2

=

100 

25 = 4

425 5 17 = 4 2

5 17 × 2 = 5 17 cm. 2 cos 175 = cos (180 – 5) = – cos 5 cos 185 = cos (180 + 5) = – cos 5 cos 355 = cos(360 – 5) = cos 5  cos 5 + cos 10 + cos 15 + ......+ cos 355 = cos5 + cos 10 + .....+ cos 85 + cos 90º + cos(180º – 85) + cos (180º – 80º) + .......+ cos (180º – 5) + cos 180º + cos (180º + 5) + cos(180º+ 10) + .......+ cos (180º + 85) + cos 270 + cos(360 – 85) + ......+ cos(360 – 5) = cos 5 + cos 10 + .......+ cos 85 + 0 – cos 85 – cos 80 – ..........– cos 5 – 1 – cos 5 – cos 10 – cos 85 + 0 + cos 85 + cos 80 + ...........+ cos 5 =–1

Diameter = 2 × OB = 8.

Pre-foundation Career Care Programmes (PCCP) Division

44

9.

By AA similarity QCP ~ PBA

y D

Q a– y

C a–x

3

ax ay 3 = = a x 4

5

P 4

ax 3 = a 4

A

x B

a

a 4

x=

ay 3 = x 4 ay 3 = a/4 4

y=

13a 16

In PAB 42 = x2 + a2 2

a 16 =   + a2 4

16 =

17a 2 16

a2 =

16  16 256 = 17 17

Area of square ABCD = a2 =

10.

256 . 17

By Angle bisector theorem

A

cos B =

11.

s= s=

7 3 2  4 2  22 9  16  4 21 = = = . 8 23 4 24 24

4

x 2 = x=4 2 1

x=

2

B

2

D 1

C

18 x  6 = 9x + 3 2 (9 x  3)( 2)(3 x  1)( 6 x )

= 6 (3 x  1)( 3 x  1)( x ) = 6(3x + 1) x x is a perfect square, so in 1 to 20, 4 perfect square i.e. possible values are {1, 4, 9, 16}.

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55

12.

f(1) = 0 as lim

h0

f (1  h) exists. h

f (1  h)  f (1) f ‘(1) exists. h 0 h So, f is differentiable at 1.  lim

13.

14.

Statement I is true that the derivative of an odd differentiable function is always even and statement II is also true that If f(x) is differentiable at a point x0 and g(x) is not differentiable at x0, then f(x) g(x) is not differentiable at x0.

x   

0,

0  x 1

 1,  1  x  0

 sin( 1)   sin1,  1  x  0 f(x) =   1  0, 0  x 1

lim f ( x )  sin1

x 0 

f (x)  0 and xlim 0   lim f ( x ) does not exists. x 0

15.

x–

x > 0 for all x > 1.

and –

x + 4 x > 0 for all x < 0 So, range of f is [0, ). x

16.

f(x) =

t

 e (t  1)(t  2) dt 0

f’(x) = ex(x – 1)(x – 2) For decreasing f’(x) < 0 ex(x – 1)(x – 2) < 0 (x – 1)(x – 2) < 0 x  (1, 2). 1

17.



x | x |3 / 2 dx

1 0

=



1 0

=

[ ex > 0]



1

x(  x )3 / 2 dx +

1



x( x )3 / 2 dx

0

x 5 / 2 ( 1)3 / 2 dx +

1



x 5 / 2 dx

0

0

= (– 1)

3/2

1

 x7 / 2   x7 / 2    +    7 / 2  –1  7 / 2  0

= (– 1)3/2

2 2 [0 – (– 1)7/2] + [1 – 0] 7 7

= – (– 1)5

2 2 2 2 4 + = + = . 7 7 7 7 7

Pre-foundation Career Care Programmes (PCCP) Division

66

18.

 x f ( x)f (x)......f 1

10 ( x )

2

1 dx

Let f11(x) = t

dx x f1( x )f2 ( x )......f10 ( x ) = dt

 1.dt

=t+C = f11(x) + C.

19.

Applying AM and GM condition : x1  2x 2  3 x 3  6 x1x 2 x 3 1 / 3 3 4  6 x1x 2 x 3 1/ 3 3

Cubing both sides

64  6 x1x 2 x 3 27 x1x 2 x 3 

32 81

Now also

x1  x 2  x 3 2 2 2  x1 x 2 x 3 3

2

2

2

2



2

2

x1  x 2  x 3  32    3  81   210 x1 + x2 + x3  3  38  2

2

2

2

2

2

1/ 3



2/3

1/ 3

   

x1 + x2 + x3 

38  2   9 9

x 12 + x 22 + x 32 

24  0 .6 9

1/ 3

x 12 + x 22 + x 32  1 . 6 So, according options least value is 2. 20.

12 3 4 5 6 7 at each position two posibility i.e. either H or T So total posible outcomes = 27 As we want three tails is sequence and it happened at 7th turn. So at position 5, 6, 7 we have tails and position 4 must have head otherwise our requirement is fullfilled at position 6. So, the position 1, 2, 3 can have anything i.e. H/T = 23 To get tail at 1, 2,3 position we have only one probability. So, total favourable cases = 23 – 1 = 7 required probability =

7 27

=

Pre-foundation Career Care Programmes (PCCP) Division

7 . 128

77

PHYSICS 22.

PV = nRT PdV + VdP = 0

1 dv 1   V dP P  = P–1 –

23.

(Answer is C)

1 for point charge r2 Force solid sphere

E

Inside the sphere E  r and out side the sphere E 

24.

=

r2

L r 4

So,  = 25.

1

L r 4

Angular momentum,  = mvr Magnetic moment,  = IA = nqr2 =

 qr 2 2

v vqr 2  = 2r qr  2 or

26.

vqr q     2mvr 2m

kT c m

(Answer is C)

2  2 1/ 2

ML T  = 2T M 1

1/ 2

= [M0L0T0]

Which dimensionally ratisfies

1 kT f = c m f 27.

(Answer is D)

Average power, =  Av2

28.

Wave length,  =

29.

 = 5 × 10–5 cm  = 500 nm (B)

30.

1 vA2w2 2 (Answer is A)

d 0.2  6  103  = 0.5 × 10–4 m D 24 (Answer is B)

By angular momentum conservation m1v1r1 = m2v2r2 ( m1 = m2 )

So, v2 =

5.6  10 4  9  1010 5.6  1012

Pre-foundation Career Care Programmes (PCCP) Division

= 900 m/s (Answer is B)

88

31.

y = y1 + y2 = 2A cos (2Kx) cos(2cost) x = 0, y = 0 (Ans. C)

  20m

32.

A  m (some as x)  

1 sec–1 t

1   sec–1 t  = angle(dimensionless) 33.

34.

(Answer is A)

KE = 0.01864 × 931 = 17.4 Mev KEf – KEi = 17.4 Mev As proton is at rest (B) option is correct

mv 2 kq1q2  2 r r kq1q2 mr

v=

v = 1.6 × 10–19

9  10 9 9.11 10 31  0.53  10 10

v = 2.24 × 106 m/s

(Answer is D)

I

35.

F

F=

F=

F

0 1 2  2d 4  10 7  4000  4000  60  10 2 2  1.5  10 2

F = 128N repulsion

(Answer is B)

37.

(C)

38.

Phase different between VR and VC is 90º so,

VR2  VC2 = 220

(Answer is D)

39.

Distance between to consecutive destructive interference is 1.7cm, so total number of points of destructive interference on the line PQ is 4. (Answer is A)

40.

(D)

Pre-foundation Career Care Programmes (PCCP) Division

99

CHEMISTRY 41.

The ionization energy or enthalpy of Na is greater than that of Li but the hydration enthalpy of Li is higher than that of Na. Difference between ionization enthalpy of both is less than difference between their hydration enthalpy so overall oxidation potential of Li is greater than Na. Therefore Li metal is a better reducing agent than Na metal.

42.

Compressibility K =  Kx – Ky =

1 P

1 1 – 1 2

= 0.5 atm–1 OH

43. CO2H

44.

2CH3

CHO

conc. NaOH

CH3

CH2OH + CH3

COOH

Cannizzarro reaction 45.

On heating solution becomes unsaturated  solubility and conductance both are increase.

46.

According to graph, segment BC represents isobaric process. It means pressure is constant so according to Charle’s law V  T [on constant pressure] From B to C volume decreases so temperature will also decreases.

H 6

H 3

4

1

47. 7

5

OH

2

when same groups present in same side of the double bond then geometrical isomer is known as Z – isomer. Due to back donation electron defficiency of [B] almost neutral.

+

+ Cl

Cl

Cl 49.

en

Co

en

en

Cl trans

Pre-foundation Career Care Programmes (PCCP) Division

Co en cis/d

/////////////////////////

48.

+

Cl Cl Co

en

en

10 10

50.

xH2S + 2NaNO3 +2HCl  yS + zNO + k NaCl + 4H2O Put the following value of variables in above equation x = 3, y = 3, z = 2, k = 2 3H2S + 2NaNO3 + 2HCl  3S + 2NO + 2NaCl + 4H2O

51.

 R = k [X]1/3 [Y]2/3 order of reaction =

1 2 + =1 3 3

52.

Because Zn has higher oxidation potential than Cu. so Zn loses its electrons at an anode and convert in Zn2+ ions and come into the solution. These electrons flow externally from zinc to Cu by wire.

53.

Because ideal mixture of benzene and toluene follows Raoult’s law according to Raoult’s law Pmix = PA + PB Pmix = XAPAº + XBPBº Pmix = XAPAº + (1 – XA) PBº Pmix = XA(PAº – PBº) + PBº This is a linear equation, by comparing with y = mx + c y = Pmix x = XA m = PAº – PBº c = PBº according to above data graph will be like following -

PBº Pmix PAº XA = 0 XA = 1 XA molefraction (benzene )

54.

sp3d2 hybridization explains the bonding of complex which has C.N. (co-ordination number) 6 so answer may be either [Fe(CN)6]3– or [Fe(H2O)6]2+. But H2O is a weak ligand and CN– is a strong ligand. Because weak ligand form an outer orbital complex with metal ion and this is possible when sp3d2 hybridization takes place.

55.

2NaCl + H2SO4  Na2SO4 + 2HCl MnO2 + 4HCl  MnCl2 + 2H2O + Cl2

56.

Bond order =

1 [N – Na] 2 b

Pre-foundation Career Care Programmes (PCCP) Division

11 11

1 [10 – 5] = 2.5 2

 B.O. of O2+ =

Nb = no. of electron present in BMO

1 [10 – 7] = 1.5 Na = no. of electron present in ABMO 2 1 B.O. of O22– = [10 – 8] = 1 2 1  Inter atomic distance  B.O  order of inter atomic distance ; O22– > O2– > O2+ B.O. of O2– =

57.

amylose [it is a part of starch]

58

t1/ 2 = t/12 =

59.

60.

0.693 K 0.693

6.93  10 – 3

t1/2 = 100 s Because unit cell has six faces and every facial atom is a part of two unit cell. It means only half part of one facial atom belongs to one unit cell. 1 so, total no. of facial atoms in fcc unit cell = 6 × =3 2 =



W T1 – T2 = Q1 T1

W 800 – 200  800 100

W = 75 J

PART-II (2 Mark) MATHEMATICS 81.

a=A– d b=A c=A+d a + b + c = 3A = A=



3 2

1 =b 2

a+b+c= a+c= b2 =

3 2

3 3 1 –b= – =1 2 2 2

....(i)

a 2c 2 2

 1   = ± ac 2 When

ac = c=

1 4

1 4a

...(ii)

Pre-foundation Career Care Programmes (PCCP) Division

12 12

From (i) & (ii) 1 =1 4a 4a2 – 4a + 1 = 0 (2a – 1)2 = 0 2a – 1 = 0

a+

1 2 By this we get a=b=c But a < b < c, so we take a=

When

– ac =

1 4

1 4a

c=

....(iii)

from (i) & (iii) a–

1 =1 4a

4a2 – 4a – 1 = 0 a=

=

4  16  16 8

44 2 1 2 = 8 2

as b > a 

a=

1 2 2

82.

g’(x) is changing its slope from positive to negative as it passes through g(2). So, g(2) is largest.

83.

92  × 180º = 140º B =   9 

1 = 2 =

D

x

180 º 140 º = 20º 2

z 3

C

3 = 140 – 2 = 120º In ACD

y 2  x 2  z2 cos 120º = 2xy

2

y A

x

1 x

B

y 2  x 2  z2 1 = 2xy 2 – xy = y2 + x2 – z2 z2 = x 2 + y2 + xy

Pre-foundation Career Care Programmes (PCCP) Division

13 13

84.

0 * *  *    * 0 *  * * * 0  * T AA =           * * *  0 As the diagonal element of resulting matrix are zero.  multiplication R1 of A and C1 of AT (i.e. = R1 of A) = 0 It is possible only when all element in R1 of A is zero In the same way we can say that element of all 10 rown of A is zero so for the above condition we can formm only 1 matrix i.e. A = null matrix.

(0, b) C E

85.

(– a, 0) A

(0, 0) O

(–ae, 0)

F

B

D (0, – b)

(i)

As line DE and AC are  so product of there slope = – 1.

b  0 b × =–1 0  ae a b b × =–1 ae a 2

b e=   a (ii) and line CF and AD are also  b0 b  0 × =–1 0  ae 0a



b b × =–1 ae a 2

b e=   a (iii) and line AO is  to CD 00 2b × –1 0a 0



2

b From (i) & (ii) case e =   a But in (iii) case ‘e’ cannot be determined. So, we can not determined uniquely. 

86.



0

[ x ]e  x dx =

= 0 +  e x



1

 0e 0

2 1

x

2

dx +



1

(1) e  x dx +

3

 2e 2

x

dx + ......

x 3 2

 +  2e  + .........

= – (e–2 – e–1) – 2(e–3 – e–2) – 3(e–4 – e–3) ......... = e–1 + e– 2 + e–3 ............

=

1 e1

+

1 e2

+

1 e3

+ ..........

Pre-foundation Career Care Programmes (PCCP) Division



S =

1 e 1 1 e

=

1 . e 1

14 14

87.

  

/2



0

=

  cos1003 xdx    

lim g( x )

x





lim

lim x

= 89.

=

 2008

e f ( t ) dt

e

g' ( x )

=

 2  1004

0 f(x)

x

=

=

e f ( t ) f ( x ) dt x

=

 cos1004 xdx  

0

g( x )

=

0

(1002  1000  998 .....2) (1003  1001  .....1)   × 1003  1001  ....... 1 1004  1002  ....... 2 2

x

88.

/2



lim 0 

x 

lim 0 

x

x



0

e f ( t )  g( x )e f ( x )

f ' ( x )e f ( x )

g( x ) f ' (x) 3 x 3  ..... 16 x 3  ....

3 . 16

dy = sin(x + y) + cos(x + y) dx let x + y = u Differentiating wrt x

1+

dy du = dx dx

dy du = –1 dx dx



dy = sin(x + y) + cos(x + y) dx

du – 1 = sinu + cosu dx du = sinu + cosu + 1 dx du  dx sin u  cos u  1 Integrating both sides du

 sin u  cos u  1   dx

Pre-foundation Career Care Programmes (PCCP) Division

15 15

du



2 tan u / 2 2

1  tan u / 2



1  tan 2 u / 2 2

1  tan u / 2

 dx 1



sec 2 (u / 2)du

 2 tan u / 2  1  tan

2

u / 2  1  tan 2 u / 2



sec 2 (u / 2)du  dx 2 tan u / 2  2



sec 2 (u / 2)du  dx 2(tan u / 2  1)

 dx







Put 1+ tanu/2 = t (1/2)sec2u/2du = dt



dt  dx t



log |t| = x + c log |1+ tanu/2| = x + c log 1  tan

(x  y) =x+c 2

It passes through origin.  log1 = 0 + c c = 0. So, log 1  tan 90.

(x  y) = x. 2

Total mappings of bijection is 6 !. for self inverse

6! 2! 2! 2! 3! +  2 pairing

P=

4! 6 C 4 2! 2! 2!   

2 going to their own value

+

6 C2  1    

4 going to their own value

= 76

76 19 = . 720 180

Pre-foundation Career Care Programmes (PCCP) Division

16 16

PHYSICS d 91.

M2

M1 r1

r2

M1r1 = M2r2 and r1 + r2 = d

M2 d so r1 = M  M 1 2 2r1 T = v 1

........(i)

........(ii)

Here

M1v12 = r1

GM1M2 d2

GM2r1

or v1 =

d

From (ii) T=

T=

T=

2r1d GM2r1

2d r1 GM2 2d

T=

GM2

M2d M1  M2

2d3 / 2 G(M1  M2 )

92.

a = r = 20 × 0.5 = 10 m/s2

93.

For min

(Answer is B)

hc  min = K.E. min =

hc K.E.

6.6  10 34  3  10 8 min = 30  10 3  1.6  10 19 = 4.14 × 10–111 m

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17 17

94.

95.

 V  V0   340  9  App. frequency, N =  V  V  N   340  9   90   S   N = 94.9 kHz

Mass of water, m =

kg 10 3 m =  × 18×10–2 × (3.96)2 × 10–4 × 0.92 ×103 mLf =

t = 96.

  18  (3.96 )2  0.92

mL f  t = KA (   ) 1 2

KA (1  2 ) t 

  18(3.96)2  0.92  333  103  10 3

t = 9.2 s

103  400    (3.96)2  15

(Answer is A)

Range, R = 2 (H  h) h (i) h = H/4 HH  X1 = 2  H   44 

3H/4 H/2 2

3 X1 = H 2 (ii) h = H/2 X2 = H (iii) h = 3H/4 X3 = H

3

H/4

1 X2

3 2

X1= x3

X2 > X1 = X3 (Answer is C)

CHEMISTRY 101.

Xe contains 8 electrons in its outermost shell or valence shell. In XeF2, Xe uses 2 electrons for  bonds so it contains 3 lp. In XeF4., Xe uses 4 electrons for  bonds, so it contains 2 lp.

N2 + 3H2

102. initial moles

4

2NH3

16

0

at equilibrium

[according to question ammonia gas is produced = 4 mol]  2 = 4 put the value of  = 2 4–2

16 – 3 × 2

2

10

4

concentration 2 at equilibrium V

10 V

4 V

moles at equilibrium

Pre-foundation Career Care Programmes (PCCP) Division

2×2

18 18

KC

=

[NH3 ]2 [N2 ][H2 ]3 2

4   V =  2  10 3     V  V  = =

42 2  10 3 8

× V2

× (10)2

10 3

(V = 10L]

= 0.8 mol–2 lit2 103.

Kw 0  Depression in freezing point TF = m W 0 In given two cases m0, W and K are constants so TF  w0 ( TF )1 ( w 0 )1 = ( TF )2 ( w 0 )2 2.9 5.5 – 4 = 5.5 – 2.5 2.9  x 1.5 2.9 = 3 2.9  x x = 2.9 g I

105.

OH

II '

2

2R and 5R III

III ' 5

II

OH I '

106.

Zn + 2Ag+ (0.0001M)  Zn2+ (0.1M) + 2Ag by nernst equation Ecell = Eº –

[ Zn2  ][ Ag]2 0.059 log10 [ Ag ]2 [ Zn] n

 [Ag] = [Zn] = 1  Ecell = Eº –

[ Zn2  ] 0.059 log10 [ Ag ]2 2

 Ecell = 1.56 –

(0.1) 0.059 log10 ( 0 . 0001)2 2

 Ecell = 1.56 – 0.2065  Ecell  1.35 V 107.

Cu(Y)

(CH3)2SiCl2 2CH3Cl + Si  (X)

SiCl4 + 2CH3MgBr  (CH3)2SiCl2 + 2Mg(Br)Cl (Z) Pre-foundation Career Care Programmes (PCCP) Division

19 19

108.

according to Arrhenius theory K = Ae

–

E RT

according to question K is same in both cases, so

E1 E2 RT1 = RT2 E1 and T1 are activation energy and temp. in absence of catalyst

400 500 = T2 625

E2 and T2 are activation energy and temp. in presence of catalyst

T2 = 600K 109.

28

Ni  [Ar]3d84s24p0

Ni2+  [Ar]3d8 4s0 4p0

[Ar] because CN– is a strong ligand w.r.t. Ni2+ ion, so pairing of electrons takes place.

[Ar] dsp2 hybridization

[Ar] 2

dsp hybridized Unhybridized orbitals orbitals

In dsp2 hybridization geometry of complex is square planar. 110.

238 206  82 Pb 92 U 

 x 24He  y –01e

by balancing of mass no. 238 = 206 + 4x + 0y x=8 by balancing of nuclear charge 92 = 82 + 2x – 1y 92 = 82 + 2 × 8 – y y=6

Pre-foundation Career Care Programmes (PCCP) Division

20 20