maits 2015 unit test – iv aipmt 15.11.14 solution part ... · mentors eduserv: paruslok...

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[ 1 ]MAITS 2015 UNIT TEST – IV AIPMT 15.11.14 SOLUTION

PART - I PHYSICSSOLUTION

1. (3)16 191 (2 3) 10 1.6 10 A 8mA towards right

2. (3)conceptual

3. (4)

1 2 3m : m : m 1:3 :5

m A ; RAl

4. (2)

1 2(2 )A A A

l l l

1 2

2

5. (3)2

2E R t

(R r )

2

2E r t

(r r )

6. (2)

1 2

2Er r R

i ;

1

1 2

2 rV ; V 0(r r R)

1 1 22r r r R

r2r1

V

R

1 2R (r r )

7. (4)

ninr r

Potential difference across any cellir 0

potential difference across any two points = 0



8. (3)Calculate V across capacitor then

21U CV2

9. (3)

0 0 010 v V r V 510 20 30

10. (3)Considering mesh DBAD

–2i2 + 2i1 + i1 = 2 – 1 3i1 – 2i2 = 1

EB

H

GD

i1

Ai2

i + i1 2

2V, 2

1V 1

3V, 3

1V 1

2

C

Considering mesh DCBD,3(i1 + i2) + (i1 + i2) + 2i2 = 3 – 1 ..... (i)

4i1 + 6i2 = 2

Solving (i) and (ii) 1 25 1i A and i A

13 13

i1 + i2 = A136

Pd across BD, VBD = i2R = V132

..... (ii)

Pd across cell G, VCD = 3 – 3 (i1 + i2) = 3 – 18 21 V13 13

Pd across cell H, VCB = 1 – (i1 + i2) = 07 V13

11. (1)Let x cells be connected correctly and y cells be connected wrongly. Then (x + y) = 12 (given)The net emf of the battery = (x – y) E where E is the emf of one cell.Let R be the total resistance of the circuit, which remains constant.When the battery and the cells aid one another, the net emf = (x – y) E + 2EWhen the battery and the cells oppose each other, the net emf = (x – y) E – 2E

Current i = cetanresistotalemfnet

In the first case, 3 = R

E2E)yx(



In the second case, 2 = R

E2E)yx(

Dividing, we get 2yx2yx

23

3x – 3y – 6 = 2x – 2y + 4 (x – y) = 10 ...... (i)and (x + y) = 12 ...... (ii)Solving (i) and (ii) we get x = 11 and y = 1. Therefore Only one cell is wrongly connected,

12. (2)

210R 2r

...... (1)

8r R2

...... (2)

13. (2)If Ammeter carries no current

600 a d X are in series and voltage across

X 2V

acrossx(6V)V 2

600 X

14. (2)

A10 1A10

i A{as V 10}

B B0; V 0i

15. (2)

B DV V ;

So, current will flow from B to D.16. (2)

They will meet at CMSo distance from P’s is

0.3 1 3 0.75 m0.4 4

17. (3)capacitors areopen circuited in steady state



18. (3)

2

X3A

19. (3) 3 is short circuited

So, voltage across (3 ) 0;

20. (1)

ABR R R 2R

21. (2)

3 0.1A30

i

At null point potential difference between A and galvanometer point will be 0.3 volt

ABV 0.1 10 1volt; so, 30cml

22. (2)

812 (40 )

ll

23 (40 )

ll

3 80 2l l

5 80 ; 16cml l

23. (3)Same concept as question number 2

24. (2)using conservation of linear momentum and taking displacement of wedge = X

( cos ) 0cos

( )

m L X MXmLXM m

25. (4)Conceptual

26. (3)

eq(r R )i



27. (2)V Ri

4 100 2Volt

200

28. (4)Conceptual

29. (1)Conceptual

30. (2)

1EV 3R

4R

2EV 6R

7R

3EV 2R

3R

31. (1)

t /0q q 1 e

32. (1)Conceptual

33. (2)

' 1 2 21 1 2

1 2 1 2

m m 2mV v vm m m m

m M 2M12 (10) 12 20 8 m / sm M m M

34. (1)Conceptual

35. (4)

V

m1 m2

1 1 22 1 2

1 2 1 2

2m m mV V Vm m m m

11

1 2

2m Vm m

2 2 22 22 1 1 2

22 1 1 2 1 1 2

1

1 m v m 2m 4m m21 m m m m (m m )m v2



36. (3)using conservation of linear momentum after explosion m1v1 + m2v2 = 0 find v2 and then KE2

37. (3)after elastic collision between equal masses velocities get interchanged, so find velocity and thenuse conservation of mechanical energy for both balls.

38. (3)As per coservation of linear momentum

ˆ ˆO mvi mvj 2mv ' :

v vˆ ˆv ' i j2 2

v| v ' | 22

KE released2

2 21 1 1 vmv mv (2m) 22 2 2 2

39. (4)after elastic collision between equal masses velocities get interchanged

40. (1)after every collision maximum height gets multiplied by e2

41. (2)if we replace removed mass ,C.M. of combined lies at origin,so

re m o v e d re m o v ed re m a in in g re m ain in gm x m x 0

42. (2)Just m1leaves the wall when spring starts elongating so take x = 0 at that instant and use conservation

of mechanical energy to find v2 as v1 will be = 0, and then find vCM using 1 1 2 2 2 2

cm1 2 1 2

m v m v m vvm m m m

where 2 20 2

1 1kx mv2 2

43. (4)mcone ycone + ycylinder mcylender = 0

2 21 hr h . r (2r). r 03 4 9621872863

solving h 2r 644. (4)

fc 0

cmc 0

m 0 m 0.12 16 0.12xm m 12 16

45 (3)if we replace removed mass, C.M. of combined lies at origin,so

re m o v e d re m o v ed re m a in in g re m ain in gm x m x 0



PART- II : CHEMISTRY

SOLUTION46. (1)

1·81

1·8 1·8

2·8 1.8

0.642

moles of glucose = 1 – 0.642 = 0.357

% = 0.357 × 100% = 35.7%

47. (2)

2 22AB g 2AB g B g

1 0 0

1 2x P1 x

2x P1 x

x P1 x

2

P 2

2x xP P1 x 1 xK

1 2x P1 x

48. (4)

5 3 2PCl g PCl g Cl g

3 3 2

3 – x 3 + x 2 + x

3 – x = 1.5

x = 1.5

3PCl 3 1.5 4.5

49. (2)

H ve

low temp and high pressure.

50. (1)

3 3CH B is Lewis alid



51. (3)

pH = 10

4OH 10

10–4 in 1000 ml.

so in 100 ml. 10–5

Total OH– : 6.02 × 1018

52. (1)

20 0.01 2 30 0.01[OH]50

30.1 2 1050

OHP 3 log2

pH = 14 – 2.7

= 11.3

53. (2)

Meq. of CH3COOH = 20 × 0.1

Meq. of CH3COOK = 50 × x

pH = pKa + 50xlog2

x = 0.4

54. (2)

2 2300 10 200 10[OH ]500

31[OH ] 2 10500

OHP 2 7

pH 11 3

55. (2)

12 2 210 [Mg ] [OH ]

12 210 0.01 [OH ]

10[OH ] 10



5OH 10

POH = 5

pH = 9

56. (4)

6

[H ] in pH 2 ; 10

[H ] in pH 6 ; 10

104 times

57. (3)

OH (salt)P pKa log(Bane)

OH 0.2P 4 74 log0.3

58. (3)

At equivalence point complete neutrallisationn takes place ;

1pH 7 (pKa pKb)2

pH 7 (pKa pKa)

59. (1)

(salt)pH pKa log(Acid)

4 110 log

pH = 4

60. (2)

2NaOH HCI NaCI H O

76. (4)

I : R—C—CH3

OLiAIH4 RCHCH3

OH

P/HI RCH CH2 3

RCH CH2 3 (Clemmensen)Zn(Hg)/Conc. HCl

N H /C H ONa2 4 2 5 RCH CH2 3 (Wolff-Kishner)

II :

III :

IV :

Thus II, III, IV.



77. (3)

active HRMgX RH

3 33 4

(CH ) COHCH MgBr CH

Note : 3 3CH COH is source of active H and decomposes RMgX.

78. (3)

3 2 2HIO 5HI 3I 3H O

79. (1)

ILi Cu3 2 3 2 3 2 2(CH ) CHBr (CH ) CHLi [(CH ) CH] CuLi

(CH3) CHCH Br2 2

3 2 2 3 2(CH ) CH CH CH(CH )

80. (1)

Elimination of two halogens from a dihalide with Zn dust is Anti elimination.

81. (1)

Addition of Br2 is selectively anti.

Thus, product formed is a mixture of d-and-l and thus racemic.

82. (3)

Addition of OsO4/H2O2 is syn thus meso isomer is formed. Thus (c) is incorrect.

83. (2)

(a) A H O/H2

+

C—CH CH2 3

CH3

OH

major due to1.2-hydride shift

(b) A Hg(OAc) /NaBH /NaOH2 4 CH—CHCH3

CH3

OH

(B)

(c) 3

2 2

(i) BH(ii) H O / OH

A 2 2CH CH CH

CH3

OH



84. (3)

85. (3)

Addition of 2 2X H O is anti.

86. (2)Conceptual

87. (3)

CHCOONa

CHCOONacurrent

CHCOO

CHCOO

–

–+ 2Na+

At anode CH COO

CH COO

–

–+ 2CO + 2e2

–CH

CH

At cathode 2Na 2e 2Na

2 22Na 2H O 2NaOH H

88. (2)

CH CH + H—OH CH CHOH2

–+

Vinyl alcohol

Otautomerise

CH3 – C – H

89. (2)Conceptual

90. (2)By Li/NH3, there is Anti-addition forming (B).



PART- III : BOTANY91. (3)

In single - celled organisms like bacteria, unicellular algae or Amoeba, the usage of the two terms -growth and reproduction is not very clear as reproduction is synonymous with growth, i.e., increasein number of cells.

92. (1)

Cohn (1872) recognised four basic types of shapes in bacteria - cocci, bacilli, spirilla and vibrio.

93. (2)

Neurospora (pink or red mould) is a genus of Ascomycete fungi. The best known species in thegenus is N. crassa, a common model organism in biology.

94. (2)

Natural system of classification is based on natural affinities among the organisms and considers,not only the external features, but also internal features, like ultrastructure, anatomy, embryologyand phytochemistry. Such a classification for flowering plants was given by George Bentham andJoseph Dalton Hooker.

95. (4)

96. (2)

97. (2)

In Whittaker’s five kingdom system of classification multicellular eukaryotes are distributed amongkingdoms Fungi, Plantae and Animalia.

98. (3)

99. (3)

Fusion between one large, non-motile (static) female gamete and a smaller, motile male gamete istermed oogamous type of sexual reproductin, e.g., Volvox, Fucus, etc.

100. (4)

The plant body of liverworts (Class-Hepaticopsida) is thalloid, i.e., thallus like, dorsiventral, prostrate,horizontally growing and dichotomously branched.

The plant body of mosses (Class-Bryopsida / Musci) consists of two stages - protonema stage(juvenile stage) and foliose or leafy stage (adult stage).

101. (3)

As we go higher from species to kingdom, the number of common characteristics goes on dicreasing.Lower the taxa, more are the characteristics that the members within the taxon share. Higher thecategory, greater is the difficulty of determining the relationship to other taxa at the same level.Hence, the problem of classification becomes more complex.

102. (2)

103. (3)



104. (2)

Dinoflagellates are mostly marine and photosynthetic.

105. (1)

106. (2)

The plant body, in bryophytes, is more differentiated than that of algae.

107. (2)

108. (4)

In gymnosperms, the stems are unbranched (e.g., Cycas, except C. rumphii) or branched (e.g.,Pinus, Cedrus).

109. (3)

Human being is the organisms who is aware of himself, i.e., has self - consciousness.

110. (1)

111. (2)

The cell wall in archaebacteria contains proteins (pseudomurein) and non-cellulosic polysaccharides.It lacks peptidoglycan. The cell membrane has a monolayer branched chain lipids which decreasemembrane fluidity.

112. (4)

The walls of diatons are embedded with silica and thus the walls are indestructible.

113. (2)

In the members of class oomycetes, cellulose is predominant in the cell wall.

114. (2)

115. (2)

116. (3)

Equisetum belongs to Class-Sphenopsida of Division Pteridophyta.

117. (1)

The gymnosperms (GK. Gymnos = naked; sperma = seed) are plants in which the ovules are notenclosed by any ovary wall and remain exposed, both before and after fertilization.

118. (1)

119. (3)

In brown algae (e.g., Ectocarps, Dictyota, Laminaria, Sargassum, Fucus, etc.) food is stored as acomplex carbohydrate, which may be in the form of laminarin or mannitol.

120. (3)

121. (2)



122. (4)

Some red dinoflagellates like Gonyaulax undergo such rapid multiplication that they make the seaappear red (red tides).

123. (3)

Most fungi are saprophytic.

124. (2)

125. (4)

In bryophytes the main plant body is gametophyte haploid. Members of bryophytes are calledamphibians of plant kingdom. The unique feature of bryophytes is that the sporophyte is dependentupon gametophyte both physically and nutritionally.

126. (4)

Heterotrophic bacteria oxidise various organic substances.

127. (2)

128. (2)

Phytophthora infestans causes late blight of pototo. The great Hunger or Potato famine of the 1840sin Ireland and other parts of Europe was due to late blight of potato.

129. (4)

Spore of ferns germinates to produce a heart shaped monoecious gametophyte, known as prothallus.

130. (1)

131. (3)

132. (4)

Unlike bryophytes and pteridophytes, in gymnosperms (e.g., Cycas, Pinus, etc.), the male andfemale gametophytes do not have an independent free living existence. They remain within thesporangia retained on the sporophyte.

133. (2)

The plant body of liverworts (Riccia, Marchantia, etc) is thalloid, i.e., thallus like.

134. (4)

135. (1)

Oospore and sporogonium (the sporophyte of bryophytes) are diploid.



PART- IV : ZOOLOGY136. (2)

137. (1)

Peripatus isconnecting link between annelida and arthropoda.

138. (3)

139. (4)

House fly has sponging type of mouth parts.

140. (2)

House fly lays egg at wet-place but not in aquatic medium.

141. (1)

Class crinoidea of echinoderms are sedentry.

142. (2)

143. (3)144. (1)

Trypanosoma gambience is digenetic.

145. (4)146. (1)147. (3)

Echinodermata (star fish) characterized by absence of excretory organ.

148. (3)149. (3)

Neopilina has internal segmentation.

150. (1)Cuttle fish (sepia) is member of cephalopoda.

151. (1)

152. (4)

153. (4)

154. (1)

155. (2)

Larva of sponges are free swimming but adult are fixed.

156. (1)

157. (3)

158. (2)

Nematoda is group of pseudocoelomate animal.

159. (2)



160. (4)

161. (4)

162. (3)

163. (1)

164. (4)

165. (1)

Star fish is member of echinodermata.

166. (2)

Lateral projection of body wall is called parapodia eg-Nereis.

167. (2)

Scoliodon and whale are chordates.

168. (4)

169. (1)

170. (3)

171. (3)

172. (2)

Latimaria living fossil.

173. (2)

174. (3)

175. (4)

Scoliodon is member of charilaginous fish.

176. (1)

177. (1)

178. (2)

Sea squirt (Herdmania).

179. (3)

180. (1)

maits 2015 unit test – iv aipmt 15.11.14 solution part ... · mentors eduserv: paruslok...

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