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HINTS AND SOLUTIONS

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1. P. Maheswari worked on test tube fertilization, intraovarion pollination in angiosperms and in vitroculture of immature embryos

2. Nostoc is a photosynthetic cyanobacateria that can fix atmospheric N2

in its specialised heterocystcells.

3. Albugo is a parasitic fungus related to Phycomycetes which infects Brassicaceae members5 A. Prothallus of pteridophytes and gametophytes of spermatophytes are atracheophytic

C. Both zygote and spore mother cells are diploid structures that can undergo meiosis7. Dryopteris is a "Pteropsidan" pteridophyte that possess membranous true indusium of its own that

protect the sorus8. Liliaceae members are hypogynous. In Arachis monadelphous and decandrous condition of stamens

occur.10. Dioecious species exibit only xenogamy as sexes are separate .14. In B&H classification

a) number of coharts in Bicarpellatae are : 4.

b) number of natural orders ( families) in monocots : 34.

c) Series in monochlamydae : 8.d) Subclasses in dicotyledonae : 3.

15. Two male genomes ( 2 male gametes) and three female genomes ( egg and diploid central cell) areinvolved in double fertilization

20. Glycosidic bonds exist between sugar molecules and nitrogen bases .21. Inhibition of succinic dehydrogenease by melanate (which closely resembles the substrate succi-

nate) is due to competitive inhibition .

23. In a cross between YY×yy, F2 generation exibit 1:2:1 ratio

Y Y × y y

Y y

Y y F1 Generation

On selfingYy Yy×

YY Yy

Yy yy

Y

y

+ Y y

YY:Yy:yy1 : 2 : 1

26. Accumulation of ions in roothair increases water potential gradient that helps in intake of water byosmosis.

27. In C4 plants, PEP (3C) get regenerated in mesophyll cells.

29. Majority of viral infections don't result in death. But Rabies and Ebola can cause death.

31.

N15N15

N15 N14N15 N14

Ist generation

N15 N14N14 N14 N15 N14

N14 N14

2nd generation

N15 : N14 = 2 :6 = 1 : 3

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34. Same deficiency symptom can be expressed by many elements Ex: Chlorosis in Plants can be

caused by the deficiency of N, K, Mg, S, Fe or Mn etc.

35. For net production of one hexose sugar , C3 plant consumes 18ATP9 : 1

Net production of ATP due to incomplete oxidation are 2 ATP

37. "Mycobacterium" is the genus that can cause tuberculosis (M. tuberculosis) and

Leprosy ( M. leprae)

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42. Cryopreservation is a technique of Ex - situ conservation

43. Member of a species shares the same ecological niche

44. Simple squamous is other name of pavement epithelium

49. Eyeball of frog is covered by nictitating numbering

50. Type of vertebrae

Caecilians - Amphicoelus

Urodela - Opisthocoelous

Anura - Procoelous

52. Primary host for Plasmodium is female anopheles mosquito

54. Transverse row - synchronous

Longitudinal - Metachronous

55. That leaves from intestine is formed after first moult so it is second stage larva.

That reaches to intestines is formed after third moult so it is fourth stage larva.

56. Arthropod with four pairs of legs is arachnida member - Palamnaeus

58. Spiracles absent in 1st thoracic segment & 9 , 10 abdominal segments.

Ganglia absent in 5th, 8th, 9th, 10th abdominal segments.

65. Urinary bladda contains transistional epithelium which can stretch

69. Cauda equina is formed inside the neural canal of vertebral column

70. Hyper polarisation due to efflux of k+ ions

72. Parathormonl stimulates oseoclasts which are involved bone resorption

74. Free circulating antibodies are produced by plasma cells

77.2 192

0.161200

number of recessive individualsq

total number of individuals� � �

So q=0.4

Then p=0.6

So heterozygous genotypic frequence ( 2 pq) = 2 x 0.6 x 0.4 = 0.48

78. Polygenic trait means trait under the control of many genes

79. XY linked genes are found in homologous portions of X & Y chromosomes

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��� ��81. Linear momentum = m � = MLT–1

Angular momentum = mvr = ML2T–1

82. A = 7i + 6j

A B� = 11i + 9j

B� = 4i + 3j

and B = 2 24 3 5� �

83. Using � 2 – u2 = 2as

2u

2� �� �� � – u2 = 2a (3)

23u

4

�

= 6a (1)

Again � 2– u2 = 2as

O2 – 2

u

2� �� �� � = 2as

2u

4

�

= 2as (2)

(1)

(2) gives 3 = 6a

2a3s = 1m

84. H1 = 2 2u sin

2g

�

H2 =2 2 2 2u sin (90 ) u cos

2g 2g

� � ��

1

2

H

H = Tan2�

12 2 2

H HH

Tan Tan 60� � �

� � = H/3

85. Mg = n(2m )

t

�

10 × 1000 =10 2 5

1

� � � �

� = 1000m/s = 1m/s

86. � =

3d d t

dt dt 3

� �� � �� �

x

= t2

At t = 0, � 1 = 0At t = 2sec, v2 = 22 = 4m/s

W = 2 22 1

1 1m m

2 2� � �

= 2 21

(2) (4 0 ) 16 J2

� �

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87. mg = 1 × 9.8 = 9.8 N

fs = usF = 0.3 × 20 = 6N

fk = ukF = 0.2 × 20 = 4N

Body moves

mg fka

m

�

� � = 9.8 4

1

�

= 5.8 m/s2, down

88. cm� = 1 2 31 2 3

1 2 3

m m m

m m m

� � � � �

� �

= 20 (10i) 30(10 j) 50(10k)

20 30 50

� �

� �

= 2i + 3j + 5k

89. T = 3 mg (1 + cos� ) = 3mg � �12

1� = 3mg/2

90. I = 2mr

4(1)

I1 = 2

2mrmr

2�

= 3

2 mr2 =

3

2 (4I) = 6I

91. g = 2

GM

R

g1 = � �

� �

M12

1 2 2R2

GGM

(R )� = 2g

92. T = 2g

�

l

T� � l

93. e = F

AY

l

e1 : e2 = y2 : y1 = 6 : 1

e

21 1

e em1 6 7

� �� � �� ���

94. F = 22TA 2TA

d v� =

22 70 (40)

0.05

�

= 44.8 × 105 dyne = 44.8 N

95. A1 � 1 = A2 � 2 �r1 = r2 2

1 1 2 2r r� � � r2 = r/4

21

2 12

r16

r

� �� � � � �� �� �

96. l1 � 1 = l1� 2

� l1 : l2 = � 2 = � 1 = 19 : 23

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97. PV = m

RTM

1 1 1 2

2 2 2 1

m P V T

m P V T� � �

= 75 3 320

76 3.2 300� � = 7.5 : 76

98. Isothermal, P1V1 = P2V2

12

PP

2� �

Adiabatic, r r1 1 2 2P V P V�

12 r

PP

2�

pf is more in isothermal

� P� is less in isothermal

99. Conceptual theory

100.1 2 1 2

0Kt 2

� � � � � �� �� � �� �

for I step,

60 40 60 40K 10

7 2

� �� �� �� �

�

20

7 = (40) (1)

for II step, 12

t = K(24) (2)

(1)

(2) gives t = 7 min

101.2

20

� ��

�

�� = 40 cm

2

� = 20 cm

102. Conceptual theory

103.2

hr

u 1�

�

= 169

28

1� = 28 7 cm

104. Theroy based

105. Shift = �

� (u–1)t

2u� = �

� (u–1)t

u� = 1t

�� = 1.72

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106. n = 1 MB

2 I�

1

2

n

n =1

2

B

B

40

20=

6

2

36 10

B

�

�

62B 9 10 T�

� � �

107. V1 + V2 = 0

1 2

0

q q10

4 d� �� �� �� � �x x

0

1 2 60

4 16

�� �� �� �� � �x x

x = 4 cm

E = E1 + E2 = 1 2

20

q q1

4 (d )

� ��� �� � � 2x x

= 9 × 109

9 9

2 2 2 2

2 10 6 10

(4 10 ) (12 10 )

� �

� �

� �� ��� �

� �� � = 15000 N/C

108. Cs = 2uF

C

3 = 2uF � C = 6uF

Cp = 3C = 18 uF

U = 1

2(Cp) (V

2) = 1

2 (18u) (12)2 = 1296 uJ

109. V = E – ir

� 50 = E – 11r (1)

60 = E – r (2)

Solving (1) & (2)

r =1�

110.1 2

1 2

E E 3

E E 1.5

��

�

E1 + E2 = 2E1 – E2

3E2 = E1 ; E1 : E2 = 3.1

111. W = n(u0mi) = (I) � �74 10 (5) (40)�

� � = 58 10 J�

� �

Independent of r

112. Theory question

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113. ig = 15mA

Vg = 750 mv

� G = g

g

V 750m

i 15m� = 50�

n = 3g

i 5 1000

i 315 10�� �

�

� S = G

n 1� = 1000

3

50

1� = 150

997�

114. e = Bv l �

= � �5 518

(3 10 ) (20) 720�

� �

= 0.12V

115. Eph = KEm + �

4 = 2 + �

� W = 2ev

012400 12400

2� � �

� = 6200 A°

116.h h

m 2mE� � �

�

1 2 2

2 1 1

m E

m E

�� �

� =

2m 1

m 8� = 1 : 2

117. Theory question

118. m� = 0.09% of m

E = 2( m) C�

= 3 8 20.09

(1 10 ) (3 10 )100

�

� � � � �� �� � = 108 10 J��

119. Theory question

120. cV 0.5 1

I AR 600 1200

� � �

0.9519

1 1 0.95

�� � � �

�� �

C

B

I

I� �

CB

I 1I A

1200 19� � �

� �

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���� ����121. (2) Use the formula of photoelectric effect

� 0h = �h – kinetic energy

122. (1) Ring chlorination in first step

Side chain chlorination in second step followed by dehydrochlorination

123. (2) � � � � �G H T S

Usually �G and �H values differ

124. (2) Orthoboric acid on strong heating finally gives bonic oxide

125. (2) 0.1 mole gas is initially present

Gas retained : Gas escaped = 570 : (760 – 570) = 3 : 1

126. (2) Availability of oxygen is limiting factor, so it is consumed

127. (3) X = H2SO

4 and Y = H

2S

2O

8 . Only Y has a peroxy bond

128. (1) base HSO4

– + H+ � H2SO

4 acid

129. (1) H+ = � �

� �

1 1 2 2 3 2

1 2 3

V M V M V M

V V V = � � � � �

� �

75 0.2 25 1 300 0

75 25 30 = �

40 1

400 10

130. (2) Li is lighest. Density of K is less than than of Na

131. (4) All covalent bonds are present in solid silicon network structure

132. (2) Eco-friendly with no toxicity of chemicals

133. (3) � � �G nF(emf)

134. (1) For zero order reaction, half-life is proportional to initial concentration

135. (2) Al3+ is more effective than Mg2+ or K+

136. (4) O+ is highest ability of attracting bonded electrons and O– has test

137. (4) �* 2py orbital has unpaired electron

138. (4) Element with higher oxidation potential act as reductant

139. (2) NO + NO2 �

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030 C N2O

3

140. (4) Flame colours are to be remembered

141. (1) ClO2 has unpaired electron

142. (4) Lime stone is basic flux to remove silica impority

143. (4) Sequenece to be remembered

144. (3) Only [Cr (CN)6]3– has unpaired electrons

145. (4) All three types of isomerism

146. (1) C:H:O = 18 3 14

: :12 1 16

= 1.5 : 3 : 1.5 = 1 : 2 : 1

147. (2) 2-Methyl but-2-ene has better hyper conjugation

148. (1) Only 1-yne give aldehyde

149. (2) Bromination at C-1

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150. (1) –OH group on phenyl ring is more favourable

151. (1) More the branching, less is boiling point

152. (2) Rosenmund's partial reduction ; –Cl to –H

153. (1) CH3COOH + HOC

2H

5

�

����2H O CH

3COOC

2H

5

154. (2) Only primary amines give carbylamine test

155. (1) � � � 3R C N CH MgBr ��� � �

3CH|

R C NMgBr �

���� � �

3

3

CH|H O R C O

156. (3) Polyacrilonitrile called orlon

157. (2) Count the number based on base pairs

158. (1) Hemiacetal in cyclic form

159. (3) Acetyl salicilic acid, called aspirin

160. (3) All three molecule has one lone pair each on Xe

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