KAB MODEL TS EAMCET (E) 2017(Engineering)

KEY

1) 3 2) 2 3) 4 4) 2 5) 2 6) 3 7) 4 8) 4 9) 2 10) 4

11) 3 12) 13) 1 14) 3 15) 3 16) 2 17) 1 18) 1 19) 1 20) 4

21) 1 22) 3 23) 1 24) 3 25) 4 26) 2 27) 3 28) 1 29) 2 30) 1

31) 3 32) 2 33) 2 34) 4 35) 3 36) 4 37) 2 38) 39) 1 40) 4

41) 3 42) 1 43) 3 44) 2 45) 4 46) 1 47) 4 48) 1 49) 3 50) 4

51) 1 52) 4 53) 2 54) 2 55) 4 56) 1 57) 4 58) 2 59) 2 60) 1

61) 2 62) 1 63) 2 64) 3 65) 4 66) 4 67) 1 68) 4 69) 4 70) 3

71) 1 72) 2 73) 2 74) 1 75) 2 76) 3 77) 2 78) 1 79) 2 80) 3

81) 3 82) 1 83) 4 84) 3 85) 1 86) 3 87) 2 88) 4 89) 1 90) 3

91) 1 92) 4 93) 2 94) 2 95) 2 96) 2 97) 1 98) 3 99) 1 100) 3

101) 2 102) 4 103) 2 104) 2 105) 3 106) 1 107) 3 108) 2 109) 2 110) 3

111) 3 112) 1 113) 4 114) 2 115) 1 116) 4 117) 1 118) 3 119) 4 120) 3

121) 4 122) 1 123) 4 124) 1 125) 3 126) 3 127) 1 128) 1 129) 4 130) 3

131) 1 132) 2 133) 1 134) 2 135) 3 136) 3 137) 3 138) 3 139) 4 140) 3

141) 1 142) 2 143) 2 144) 3 145) 3 146) 4 147) 4 148) 2 149) 4 150) 3

151) 2 152) 2 153) 1 154) 4 155) 1 156) 1 157) 3 158) 2 159) 2 160) 1

KAB MODEL TS EAMCET (E) 2017SOLUTIONS

MATHEMATICS1) 3

The domain is given by 3 0x and 7 0x 3 7x

22

3 7 3 7 2 3 7f x x x x x x x

224 2 21 10 4 2 4 5x x x

22max 5 8f f

2 22min 3 7 4f f f

min max2, 2 2f f

Range = 2, 2 2 .

2) 2

1f x y f y x

y y

y y

10 10x

10 10

y y y y

y y y y

10 10 10 10 x 1

10 10 10 10 x 1

2y x 110

x 1

2y x 110

1 x

10

1 x2y log

1 x

1

10

1 1 xf x y log

2 1 x

3) 4

Sum 23n 3n 2

2

23 3

n n 12 2

n n 1 2n 1 n n 13 3n

2 6 2 2

2n2n 3n 1 3n 3 4

4 2n

2n 24

2n2 n 1

4 2n

n 12

4) 2

AB BA

2 2 2A B A B

5) 2

2 1 1 1 1 2 2 1 1A 2 2A

1 1 1 1 2 2 1 1

3 21 1 2 2 4 4A 4A 2 A

1 1 2 2 4 4

BY induction n n 1A 2 A

6) 3

Differentiate w.r.t ‘x’ on both sides and put 0x we get 1a value is : 0

7) 4

43 2 x

5

3 4

x cos x 2

f x tan x 1 sec 2x f x

sin x x 5

f x dx 0

8) 42

2 2 22 3

2 3

1 0

1 1 1.......

x x

x

1 1 4 ........ 9 times

6 9 54

9) 2

1 1 2 3 5

1 2 5 10

3 5 10 11

i i

z i i z

i i

.z is purely real

10) 4

2 2 2a 2b aw 2bw aw 2bw 2 2 2 2 2a 4ab 4b a w 4ab 4b w 2 2 2a w 4ab 4b w

2 2 2 2a 1 w w 4b 1 w w 12ab

212ab 1 w w 0

11) 388

i cos i sinsin i cos8 88 8

sin i cos i cos i sin8 8 8 8

82

81 cos isin cos2 isin 2 1

8 8

12)

13) 11

2is a root of the given equation and

1 1 1, ,

2 3 4are in H.P.

14) 3

15) 3

The No. of choices

16) 2

Let x1 be the number of stations before the first halting station, x2 between first and second,

x3 between second and third, x4 between third and fourth and x5 on the right of 4th stations.

Then 1 5 2 3 40, 0, , , 1x x x x x satisfying 1 2 3 4 5 8x x x x x

The total number of ways is the number of solution of the above equation

Let 2 2 3 3 4 41, 1, 1y x y x y x

Then (i) reduces to 1 2 3 4 5 5,x y y y x where 2 3 4, , 0.y y y

The number of solution of this equation is 5 5 1 95 1 4C C .

52 2 30

6 7 6 75 5 6 4C C C C 161

17) 1

Substitute above in the expansion and simplify.

18) 1

22006 =4.8668

=4[7 + 1]668 leaves remainder 4

2006 7 268 4 leaves remainder 4

20062 2006 leaves remainder = 0

19) 1

Coefficient of

= 990

20) 4

21) 1

22) 3

= 2 2256 in cos 2 68coss x ecx ec x is minimum when and minimum value is

580.23) 1

1 2 3, , ,n n n nr r r rc a c b c c c d

11 11112 3 21 x x x 1 x 1 x

11 11 11 11 11 11 2 11 220 1 11 0 1 11C C x ..... C x C C x ..... C x

11 11 114 11 11 110 2 2 1 4 0x C C C C C C

11 5 11 66 56 5

1 1. . 1C a C a abb b

9060 , 30

30

2x

100 100cos x sin x 1 100 100cos x 1 sin x

100cos x 11001 sin x 1

100sin x 0

sin x 0

x n

24) 3

Domain of is [–1, 1]

Range is 1 1tan 1 tan 12 2

25) 4

cosh sinhf x x x

1 2 3 ....... nf x f x f x f x

26) 2

height of the tower

........... (1)

................... (2)

............................... (3)

1 1 1sin cosx x Tan x

,2 4 2 4

3,

4 4

x x x xe e e e

2 2

x x1

2e e2

31 2 nxx x x1 2 3 nf x x x ...... x e e e .....e

,AB a BC b mts

PQ h mts

cot cota b x

a b x hh

cot 2 cot 2b x

b x hh

cot 3 cot 3x

x hh

cot cot 2 sin 3

cot 2 cot 3 sin

ha a

b h b

1 2cos 2

Q

P ABcx b a

23

27) 3

28) 1

29) 2

1 2

4R1

S a S b r r

2

2

4R1

CS a S b 4R cos2

2

tan c / 2S a S b

2 S a S b

s a S b

a cot A

cos A2R.sin A.

sin A

2R cos A

2R cos A cosB cosC

A B C2R 1 4sin sin sin

2 2 2

4R A B C2R 1 sin sin sin

R 2 2 2

r2R 1

R

R r2R

R

2 R r

2 1

2 3

b c

b c

2 1tan tan tan 2 1

2 2 22 3

B C A B C

045B C

30) 1

or

31) 3

2

3

AC

A

32) 2

Let 1 2 3ˆˆ ˆd d i d j d k . Then 2 2 2 2

1 2 3 51........... 1d d d d

From (1)

33) 2

34) 4

Where

Maximum value of

2 a b cb c c a a bK say

11 12 13 36

a 7K, b 6K,c 5K

2 2 26K 5K 7K 1

cos A2 6K 5K 5

AB AE BC DC ED AB BC AE ED DC

1 2 3 1 3 2

. . . 1 1cos 2 2 4 3

3 5

d a d b d cd d d d d d

d a d b d c

1 2 3 1 2 35 2 0, 4 5 3 0d d d d d d

31 21 2 35 , , 5

5 1 5

dd dd d d

1 5 5d i j k

3a b p q

3 sin ,p q p q

i j k

b c 2 1 1 i 3 0 j 6 1 k 0 1

1 0 3

3i 7j k

a b c a. b c a. 3i 7 j k

a 3i 7 j k cos

a,3i 7 j k

9 49 1cos

59 cos

a b c is 59

35) 3

36) 4

The equation of two lines are

and 0r a b

is parallel to and is parallel to

[For intersection of their equations]

37) 2

E1=Event that the maximum number=6

E2=Event that the minimum number=3

Favourable cases to E1=5C2 among favourable case to E1.

(3, 4, 6) and (3, 5, 6) are favourable to E2.

38)39) 1

40) 4

Error is atleast 15 paise when we round off 15p, 16p, ……49p, 50p, 51p….. 85p.

.

2cos0 sin

3a b c a b c

12 3

0r b a

r b a r a b

1

r b pa

r a qb

p q

2

1 22

1 1

2 1

5 5C

n E EEP

E n E

0 1 215

13 16 16 16

2 C C CP x

16

137.

2

71

100P E

41) 3

First we give two coins to each of three persons. Remaining coins are 7. These seven can

be distributed among three in 3 1

7 3 1C

way i.e., 2

9C ways, which is 36.

42) 1

N(S) = 16, n(E) = 6

43) 3

and variance

44) 2

148, 146, 144, ….. is an A.P. with a = 148, d = –2

nth term

Average 125

45) 4

at because the locus of P is .

46) 1

The transformed equation of is

Comparing this equation with we get k – 1 = 0

6 8 10 24Mean 8

3 3

2 2 226 8 10

83

36 64 100 864

3 3

a n 1 d 148 n 1 2

150 2n

148 146 ..... 150 2n125

n

n296 n 1 2

2 125 n 24n

4,0

7Q

40,

7R

7 7 4 0x y

x X h, y Y k

xy x 2y 6

X h Y k X h 2 Y k 6 0

XY k 1 X h 2 Y hk h 2k 6 0

xy c

h 2 0&c hk h 2k 6 k 1,h 2&c 2 1 2 2 6 4

47) 4

As the line passes through (13, 32), we have

Thus the line is

The equation of line parallel to has slope 4. Thus

Then the equation to line k is

The distance between lines k and c is

48) 1

Hint given curve 22 4 1y x

Vertex , 1,2 , Focus S = (0, 2)

Point of intersection of and the curve is = (0, 4)

reflected ray passes through focus S = (0, 2).

Equation of the line along reflected ray travels is x=0.

49) 3

2 22 5 2 0 2 2 0x xy y x y x y

represents the lines 2 0 1 , 2 0 2x y x y

Clearly the origin O is the point of intersection of (1) and (2)

Let OAB be the triangle such that (1) represents and (2) represents

Since A lies in (1), A=(I, 2I) for some I.

Since B lies in (2), B = (2k, k) for some k

Centroid 2 2 2 2 2 2, , 1,1 1, 1

3 3 3 3 3 3

I k I k I k I k I k I k

2 3,2 3 1, 1, 1,2 2,1I k I k I k A B

Slope of

Equation of is y – 2 = –1 (x – 1)

13 32 32 13 81 1 b 20

5 b b 5 5

x y1,i.e., 4x y 20

5 20

4x y 20 3 3

4, cc 4

4x y 3

2 2

20 3 23

174 1

4y

2 22x 5xy 2y 0

OA

OB

l 2k 3,2l k 3 l 1,k 1, A 1,2 B 2,1

1 2ABis 1

2 1

AB

y 2 x 1 x y 3 0

50) 4

Point of intersection of

Given lines are concurrent 4 / 3,2 / 3 lies on

51) 1

Length of perpendicular from centre < radius.

52) 4

Let PA and PB be the tangents drawn from the point P(h, k) to the given circle with centre

C(–2, 3). So that and

From triangle PCA,

The locus of P(h, k) is therefore x2 + y2 + 4x – 6y + 9 = 0

53) 2

P=(–9,7) C=(3, –2); cp=d the require d point divides PC in the ratio –(d + r):r.

54) 2

Chord of S=0, is

Equation circle 0S L Centre lies on L=0. We get value.

55) 4

Length of the latusrectum

56) 1

Any tangent to the parabola is

It is tangent to 22y x

2 2

2 0, 4.4 4

m mm m m

2 22 2

hg bg gh afx 2xy 35y 4x 44y 12 0 is ,

ab b ab h

1 22 35 2 2 1 1 22 48 24 4 2

, , ,35 1 35 1 36 36 3 3

5x y 8 0 5 4 / 3 2 / 3 8 0 20 2 24 0 2 4 2

2APB APC CPB 090PAC PBC

2 2sin and CA = 4 9 (9sin 13cosCA

CP

2sin 2CP 2 24 4 6 13h k h k

' 0S 1 0S S

123

4

2y x 21

4y mx m

57) 4

2 2 2 26 0 6 5 0 6 6 5r r and r r also r r r r

58) 2

59) 2

Equation of the tangent at to hyperbola

Slope of the tangent and slope of the normal

Equation of the normal at If it passes through ,

then

60) 1

Equation of plane through (3, 2, –1) is

a(x–3) + b(y – 2) + c(z + 1) = 0 ……………(i)

also (3, 4, 2) and (7, 0, 6) lie on Eq. (i) , then

0.a + 2b + 3c = 0 …………… (ii)

And 4a – 2b + 7c = 0 …………… (iii)

Eliminating a,b,c from eqs. (i), (ii), and (iii), we get

We get. 5x+3y–2z=23

2 2144 8116 7.

25 25b b

( , / )ct c t 2xy c is

22cx

cty ct

21/ t 2t2( , / ) / ( )ct c t is y c t t x ct ( ', / ')ct c t

2 3 3' / ( ' ) ( ') '( ' ) ' 1ct c t t ct ct t t t t t t t t

3 2 1

0 2 3 0

4 2 7

x y z

23

61) 2

Let DC’s of shortest distance line are l,m,n which is perpendicular to both the given lines

62) 1

The plane meets the coordinate axes in

Centroid of triangle ABC is

63) 2

64) 3

2

20 2

2

2 tan2 tan

1 tanlim ........simplify.1 tan

11 tan

x xx x

x

xx

7x 11y 13z 3003

A 429,0,0 ,B 0,273,0 ,C 0,0,231

429 / 3,273/ 3,231/ 3 143,91,77

x 0 x 0 x 0

log 1 ax log 1 bx a bf 0 Lt f x Lt Lt a b

x 1 ax 1 bx

65) 42 4 6 3 5

30

1 1 ......... .......2! 4! 6! 3! 5!

lim

x x x x xx a b x

x

2 4

20

1 .......... .......3! 2!

lim

b aa b x x

x

1 0; 16 2

b aa b

66) 4

Hint: 1d dx ddydy dy dydx

67) 1

Put

68) 44 2

22

11

1

x xy x x

x x

2 1dy

x ax bdx

Then 2 1a b 2 2 4 1 5a b

69) 4

For increasing , 1 0,x

' 0f x only when R – {–1, 0}.

70 3

3 2f x x ax bx c

'f x Never zero

No extreme values

x tan

2 2

x sin dfog x f f sin tan x fog x 1

dx1 x 1 sin

2 1

' 0 0.1

x xf x

x x

2' 3 2f x x ax b

2

2

4 12

4 3 0

a b

a b

71) 1

1

3dy y

dx x

Slope at given point is m=1

Equation of tangent is 2

ax y

Perpendicular distance from (0, 0) to 2

ax y is

2

a

72) 2

is not derivable at ‘0’ and 0 1,2

Lagrange’s mean value theorem is not applicable for

73) 2

Resolve in the partial fractions.

74) 1

75) 2

76) 3

Given series can be written as

. Then use definite integration as limit of sum.

xf x x f ' x f

x

f x x

1 1 1sin 2x cos 2x dx cos 2x sin 2x c cos 2x sin 2x c

2 2 2

1 1 12.sin 2x / 4 c sin 2x / 4 sin 2x 5 / 4 5 / 4

2 2 2

2 2

1 27 2 7 7

1 7cos sec 17

sin cos sin sin

x xdx dx dx I I

x x x

2

1 7

sec

sin

xI dx

x

2

1 7 7 8

sec tan tan cos7

sin sin sin

x x x xI dx dx

x x x 1 27

tan

sin

xI I

x

9

1

1 n

xr

rLt

n n

77) 2

78) 1

=

79) 2

Solving we get

The area is

80) 3

Consider the equation at the conics has center at (0, 0). ax2 + 2hxy + by2 + c = 0

a,h,b are parameter.

00

sin 4 sin 4cos 4

4 4

xx t xg x tdt

0

cos 4x

g x t dt

0

sin 4 1 1sin 4 sin 4

4 4 4

xt

x x

0 sin 4 0g

g x g n g

1

1

0

1tan

1 1dx

x x

11

0

1tan

1 1

x xdx

x x

1 1

1 1

0 0

tan tan 1x dx x dx 1

1

0

2 tan x dx

4 24 32 0 2,2x x x

22

22

8

4 4

xS dx

x

231

2

44 2

2 12 3

x xTan

PHYSICS81) 3

Units and Measurements- Dimensional analysis

82) 10.1 0.1 0.1 0.7

10 20 5 20

V l b h

V l b h

;

Hence the percentage error is 0.7100 100 3.5%

20

V

V

83) 4Conceptual

84) 3Motion in a straight line- Kinematic equations for uniformly accelerated motion

85) 1

2 21 1Fx Kx mV ,

2 2 calculate for V

86) 3

ˆ ˆAs V 5 2t i t j

ˆ ˆx ya a i a j

ˆ ˆ5 2i j

ˆ ˆ( )x yF ma i m g a j

2 2| | ( ) 26x yF m a g a N

87) 22

21 mlmgh I , I

2 3

88) 4Apply law of conservation of energy

89) 1Conceptual

90) 3

1 2

2

d d

ay

max

ax

m(g+ay)

91) 1

Suppose, height of liquid in each arm before rising the temperature is l.

With temperature rise height of liquid in each arm increases i.e.l1>l and l2>l

Also 1 2

1 21 1

l ll

t t

1 1 2 2 2 1l l t l l t 1 2

2 1 1 2

l l

l t l t

92) 4Conceptual

93) 2

2 2 2 211 1 1 2 2

2

k1f k 4 mf ;k 4 mf

2 m

p 1 2k k k

94) 2

2Y r eYAeF

95) 2

21

2 1V

m RmV U nC T T

M

96) 3

Let the temperature of junction be , under steady state condition

100 0A BK K

40 C

97) 1

Conceptual

98) 3

t 2g

99) 1

For resonance, the amplitude must be maximum which is possible only when the

denominator of expression is zero i.e. 2 0a b c 2 4

2

b b ac

a

100) 3

Frequency of reflected sound heard by driver O

S

V Vn n

V V

It is given that 2 .n n Hence, 2 car

car

V Vn n

V V

.3car

VV

101) 2

102) 4

1 1 1

75 12x x

103) 2Tan 1

tansin c

1c sin cot

104) 2

By using phase difference 2

( )

For path difference , phase difference 21 and for path difference /4, phase difference

2 = /2. Also by using 204 cos

2I I

21 1

22 2

cos ( / 2)

cos ( / 2)

I

I

2

22

cos (2 / 2) 1/ 2 1/ 2

cos2

K

I

2 .2

KI

105) 3Conceptual

106) 1

1 2 ,q CV 2q CV

Now condenser of capacity C is filled with dielectric K, therefore C2 = KC

As charge is conserved

'1 2 2( 2 ) 'q q C C V

2

3

)2(

3'

K

V

CK

CVV

107) 3

Hence 3

2RR eq .

108) 2

g gV 50 A 100 ,i 50 A

1) 50V

n 100005mV

R G n 1 100 9999 999900

2) 10V

n 20,0005mV

R G n 1 100 19999 200k

109) 2

At 1

Et , i

R

At t = 1 sec, 2E1 e

R

21

22

i e

i e 1

110) 3Magnetic field at any point lying on the current carrying straight conductor is zero. Here H1 = Magnetic field at M due to current in PQ.H2= Magnetic field at M due to QR+ magnetic field at M due to QS+ magnetic field at M due to PQ

11 1

30

2 2

HH H 1

2

2

3

H

H

111) 3

Given LX R

and Z 2Rrms rmsP V i cos

20E

4R

R

R

R

RR

R

R

R R

R

R

R

R

A B

2R/3

2R/32R/3

R BA

2R/3

112) 1

n2 r n

113) 4M e

L 2m

114) 2

1 1

2 2

K h

K h

115) 1

Conceptual116) 4

10 t0

t0

N e1

e N e

117) 1

012400A

E in eV

118) 3

d

iV

neA

119) 4

max min

max min

E Em

E E

120) 3

l4

CHEMISTRY121) 4

4 2

1f

f

T KCl i KCl

T X i x i x

20.5

4ix

For association of 3 molecules 11 1i x

n

1

1 1 0.53

0.75

122) 1

1TV constant 1

2 1

1 2

T V

T V

123) 4Orthoboric acid (H3BO3) is a Lewis acid, it accepts OH– from water

3 3 2 4H BO H O B OH H

It is not a H+ donor.

124) 1‘NO’ is paramagnetic due to odd electrons.

125) 3Soframicine - Anti septicAspirin - Anti pyreticValium - Sedative

126) 3

2

4NiCl is paramagnetic as ' 'Cl being a weak ligand, does not cause pairing-up of

d-electrons.

127) 1In hcp unit cell Effective number of atoms are 6, packing fraction is 74% and coordination number is 12. The arrangement of atoms in layers is ABABAB….

128) 1

3 23 3Co NH NO exists in facial and meridonial forms which are Achiral.

129) 4In, Chlorine atom is at its highest possible Oxidation state therefore cannot undergo

disproportionation

130) 3

NO2 group is strongly deactivating

Cl is weakly deactivating

CH3 is weakly activating

131) 1

Carboxylic acids are stronger acids than phenols

–‘Cl’ increases acidity and –CH3 group decreases acidity due to hyper conjugation.

132) 2

133) 1

i) In Reimer-Tiemann reaction, phenon turns into salicylaldehyde in presence of alk-

KMnO4.

ii) In sandmeyer reaction, benzene diazonium chloride turns into chlorobenzene in

presence of cuprous chloride and HCl.

iii) In Gatterman-Koch reaction, Benzene turns into benzaldehyde in presence of CO, HCl

and AlCl3.

iv) In Etard reaction, Toluene turns into Benzaldehyde in presence of CrO2Cl2.

134) 2Electrometallurgical process is employed to extract highly electropositive elements by taking their fused salts.Anhydrous 2ZnCl can’t be prepared by heating alone the hydrated salt of 2ZnCl since it

undergoes hydrolysis with its own water of crystallization.Bauxite containing 2SiO as impurity is known as White Bauxite, which is purified by

Serpeck’s Process.Bauxite containing 2 3Fe O as impurity is known as Red Bauxite which is purified by

Baeyer’s process.135) 3

Tollen’s reagent oxideses Aldehydes to Carboxylates and HCOOH to CO2.

136) 3

Order of the reaction is equal to the sum of powers of concentration of the reactants in rate

law expression.

For any chemical reaction, A B products; x yRate k A B ; order = x y

137) 3

4 3NH NO 2N O

4 2NH Cl NaNO 2N (Impure)

3 2Ba N 2N (Pure)

3 2NH Cl excess 3NCl HCl

3 2 2 4NH excess Cl N NH Cl

138) 3

meq of acid = meq of KOH

139) 4

22 21 2

1

2

1 1 1;

2

HZ Rn n

n

n

2 2

1 1 14

2R

1 1

HR

140) 3

141) 1

2F oxidizes water to 2O

142) 2

2 2 4

4 4 2

31632

2 3216

r n M

r n M

O O CH

CH CH O

OH

O

143) 2

BuNa – N is a type of rubber an Elastomer.

144) 3

5 2 3 2PCl SO POCl SOCl

4 2 3 2 2 2P 8SOCl 4PCl 4SO 2S Cl

2 5 5 2 5 3C H OH PCl C H Cl POCl HCl

3 2 2CH COOH SOCl CHCOCl SO HCl

Hydrolysis3 3 3PCl H PO

2 2 2 2S Cl CH CH mustardgas

145) 3

Ag ion has strong affinity for X ions, so in presence of aq 3AgNO reaction take place

by 1SN mechanism, via formation of carbocation intermediate.

146) 4

2

31 3

2 2

NHK

N H

2

2

2 2

NOK

N O

23 1

22 2

H OK

H O

3 2 2

52NH O 2NO 3H O

2 ,

2 3

25

22

3 2

NO H OK

NH O

147) 4

For 4 3HClO ,CH COOH, HCl - pH < 7

For NaCl, pH = 7

148) 2

i) Fructose is a reducing Sugar where as Sucrose is non-reducing.

ii) D-Glucose & D-Mannose have same configuration at C-3, C-4 & C-5.

Therefore they give same Osazone.

149) 4

0Au e Au, E 1.69V .......... 1

3 0Au 3e Au,E 1.40V .......... 2

From 2 1 , 3Au 2e Au 0 0 03 2 1G G G

2 F E 3 F 1.4 1 1.69 F

0E 1.255V

3Au /Au

0E 1.255V

150) 3

In NF3 and ClO3-the central atoms N and Cl have one lone pair and 3 bond pairs.

151) 2

Proteins give violet colouration with Biuret reagent.

152) 2

Saturated compounds cannot exhibit ring-chain isomerism

153) 1

154) 4

In the partial hydrolysis of 6Xe F different products formed are 4 2 2Xe O F and XeO F

155) 1

2 2

( ) ( ) ( )

Na H O NaOH H

A C B

2 2 22

( )

Zn NaOH Na ZnO H

D

2 4 4 2Zn H SO ZnSO H

2 2 2 3eq eqn H O n Na SO

25 0.3 20

1000 1000

N

40.3 0.24

5N N

.

5.6. 0.24 5.6 1.344

V SN

V S

156) 1

2 5 3PCCPDC

C H OH CH CHO

3 2 3KOHCH CHO I CHI

157) 3

Cannizaro reaction is given by aldehydes with no alpha hydrogen.

158) 2

3 3 3 4 3 2 2

( ) ( )

KOHCH COOH NH CH COONH CH CONH H O

X Y

159) 2

As2S3 is a negative sol, so it is coagulated by a positive ion Al3+

160) 1

B.O.D. of fairly pure water = 1ppm