APEX ME-CEE MOCK TEST SET-06 SOLUTION

ZOOLOGY SOLUTIONS 23. b) Zygote form in stomach of mosquito about 9 to

1. c) The stratum germinativum layer has specialized 10 days after the blood meal. The process of fusion
columnar cells called basal cells. It can be confused of male and female gametes is called syngamy.
with cuboidal cells. But they are derived from the Syngamy is anisogamous due to dissimilar structure
undifferentiated columnar epithelium. of gametes. Hence, their fusion is called anisogamy.
2. c) When ligaments are damaged, the knee joint may 24. a) Chordae tendinae are strong and elastic chords of
become unstable. Ligament damage often happens fibres connecting atrioventricular valves with the
from a sports injury. wall of ventricles.
3. d) In human, the process of erythropoiesis starts 25. b)
initially in the yolk sac, then switches to the fetal 26. c) When a sperm enters into the egg of frog, second
liver in the second gestation month. After birth, meiotic division occurs.
erythropoeisis occurs in bone marrow. 27. c) Nucleic acids (DNA and RNA) rich in carbon,
4. c) The structures present in neuroplasm hydrogen, oxygen, nitrogen and phosphorous are
characteristic of neurons are Nissl bodies which are only capable of reproduction at their own, therefore,
found to be aggregation of RER rich in RNA and are considered as basic of life.
concerned with protein synthesis. 28. c)
5. d) Vitamin K - Antihaemorrhagic vitamins 29. c) In scientific terms, the evolutionary history and
Note: vitamin E (Tocopherols) - sterility relationship of organism or group of organism is
6. a) called its phylogeny.
7. a) FRC is the sum of residual volume and expiratory 30. a)
reserve volume (1200+ 1200) = 2400 mL 31. c) Amnion is present only in mammals, birds and
8. d) Serum differs from blood in lacking clotting reptiles.
factors. Serum do not contain any proteins required 32. c) Only reptile in which heart is completely 4
for the clotting of blood.. chambered is crocodile.
9. b) 33. a)
10. d) Humans have 2, 3, 3, 3, 3 digital formula for both 34. d) Sea pentagon - Echinodermata
hands and feet. Sea butterfly - Mollusca
11. c) The receptors sites for neurotransmitters are Sea mouse- Annelida
present on post synaptic membrane. 35. a) Arachnida: Tick, mite, spider, scorpion
12. a) Sympathetic nerves are made up of efferent fibres Insecta: cockroach, silver fish
which are derive from the thoracic and lumbar Myriapoda: millipede, centipede
regions of the spinal cord. Therefore, symphathetic 36. c) Porifera : cell aggregate body plan
nerves in mammals arise from the thoracolumbar Coel + platyhelminthes blind sac body plan
region that is T1-T12 and L1 to L2 of spinal cord. 37. b)
13. a) 38. c) Gastro-vascular cavity is for digestion and
14. a) Leydig cells also known as interstitial cells of circulation.
Leydig, are found adjacent to the seminiferous 39. b) Rheotaxis- water and air current
tubules in the testicle. They produce testosterone in 40. a) Reduction of digit number and lengthening of
the presence of LH. limbs for faster.
15. d) Ovulation: This phase happens when the ovary Movement - cursorial adaptation
releases that mature egg down the fallopian tube on
its way to fertilization. This is the shortest phase of Botany
the cycle, lasting just 24 hours. 41. c) DNA can be stained with basic
16. c) A benign tumour is typically surrounded by an Fuschin, Feulgen and Rossenbeck.
outer fibrous sheath of connective tissue or the 42. c) Pili or fimbriae of bacteria help in
epithelium. attachment on host. Sex pili also help on
17. b) Isoniazid drug used against TB.
transfer of genetic material during
18. b)
bacterial conjugation.
19. c) Pheretima posthuma - no larval stage (direct
development). 43. c) Two types of viral life cycles are
20. c) Spermatheca- diverticulum → storage, present: killing host is lytic life cycle and
Ampulla→ nourishment leaving host cell is lysogenic life cycle.
21. d) Earthworm is protandrous hermaphrodite. 44. b) Algae and fungi have unicellular non-
22. b) Male gamete from microgametocyte jacketed sex organs.
Female gamete from macrogametocyte

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 APEX ME-CEE MOCK TEST SET-06 SOLUTION
45. b) Functionally, chloroplast is associated kinetochore. It shows Fuelgen negative
with photosynthesis and prepares food due to least amount of DNA.
materials like starch. In algae, storage 58. d) The process of aggregation of
food materials are stored in pyrenoids. homologous chromosomes during
46. d) The Penicillium, a fungus produces meiosis is called synapsis. It forms
penicillin (antibiotic), which can stop complex group of two homologous
growth or even can kill bacteria like chromosome and called synaptonemal
Salmonella. Hence, the relationship is complex.
ammensalism or antibiosis. 59. c) Colour blindness is X linked sex
47. d) The elaters are produced from diploid disease. Every X-linked character is
sporophytic structure while antherozoids, expressed in male due to hemizygous
perichaetium and scales are structures of condition (for X) but may be expressed
gametophyte so they are haploid. (homozygous, XcXc) or suppressed in
48. b) Ferns have line of spore dispersal female (heterozygous, XcX).
called stomium.Calyptra, elaters and 60. d) Structure associated with ribosomal
peristomes are absent in ferns. factory is nucleolus.
49. b) Megasporophylls aggregate around 61. b) Every nucleoside consists of nitrogen
central cone axis and form female cone. base and pentose sugar. The
50. b) Cyathium is specialized condensed phosphorylated nucleoside is nucleotide.
cymose inflorescence in which naked 62. c) Sickle cell anaemia shows
flowers are present. There is single pleiotropism, i.e., multiple effects of
female flower covered by 5 stamens or single gene and not related to sex
male flowers. chromosomes.
51. a) Epigeal germination is common 63. a) The invention of monoclonal
character of dicots and hypogeal antibodies (MABs) through genetic
germination is more common in engineering is accredited to Georges
monocots. Kohler, Cesar Milstein and Niels Kaj
52. b) Lianas and epiphytes are common Jerne (1975); who shared the Nobel
character of tropical regions.Epiphyte Prize in Medicine (1984) for the
gets water and minerals from atmosphere discovery.
and show parasitic character for space or 64. a) Pollen grain has haploid number of
shelter. chromosomes, so, haploid plant can be
53. b) Primary productivity is concerned with produced from germinating pollen
autotrophic produce and secondary grains.
productivity is related with consumers. 65. c) Lysine, histidine and arginine are basic
54. c) Linderman proposed 10% energy side chains at neutral pH. Tyrosine and
transfer law (2nd law of valine are neutral and Glutamine is
thermodynamics) and states that 10% acidic amino acid.
energy is transferred one trophic level to 66. b) Mice coat colour shows recessive
another epistasis with genotype 9:3:4. In mice
55. c) Elephas(elephant), Panthera(tiger), albinism (white coat) is produced by a
Gavialis(gharial) and Manis(pangolin) all recessive gene aa. There is a different
are protected animals of Nepal. Among gene B which in the dominant state (BB
them Gavialis is reptile and rest are and Bb) produces grey coat colour called
mammals. agouti, and when recessive (bb) leads to
56. d) Acid rain consists of H2SO4 an HNO3 black coat colour.
formed due to presence of oxides of 67. c) Emasculation the process of removal
nitrogen and sulphur in atmosphere. of stamen from bisexual flower before
57. c) Chromosome has narrow constriction anthesis to check self-pollination during
having least DNA and high tubulin plant breeding.
protein and called centromere or 68. a) Heredity is the transmission of traits
from one parent to offspring.
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 APEX ME-CEE MOCK TEST SET-06 SOLUTION
69. c) Periderm is secondary tissue, produced
from secondary cork cambium
(phellogen).It includes secondary cortex
(phelloderm), cork cells (phellem) and 82. d) Let formula of metal chloride is MClx
cork cambium (phellogen). Molecular mass of MClx = (E×x+x ×
70. c) In fully turgid cell, turgor pressure is 35.5) = (3x + 35.5x)
maximum with minimum DPD or suction or (35.5+3)x = 2 × V.D. = 2 × 70 = 140
pressure. or x =
71. a) For determining essentiality of carbon
dioxide for photosynthesis, Moll’s half ≈4
leaf experiment is needed. In this ∴Atomic mass = E × x = 3 × 4 = 12.
experiment, half part of leaf is inserted in 83. d) s is obtained from spectral evidence.
vessel containing KOH to absorb 84. b) Splitting of spectral lines under the
atmospheric CO2. influence of magnetic field is called
72. b) During photorespiration, CO2 is Zeeman effect.
produced in mitochondria as respiration. 85. 85. b) sp, sp2 and sp3 respectively.
73. d) All prokaryotes like bacteria, 86. a) H2O (acid) + NH3(base) → NH4OH
cyanobacteria, mycoplasm produce 38 87. b) In H3PO3, O.N. of P = +3
ATP from complete oxidation of glucose 88. c) Electricity required to deposit 27 g of
molecule during aerobic respiration. Al = 3 F So, electricity required to
74. b) During conversion of fumaric acid deposit 9 g of Al= 3F27×9 = 1 F. ( ∵
from succinic acid, FAD is reduced to Al3+ + 3e¯ → Al)
FADH2 in presence of enzyme succinate 89. c)
dehydrogenase.
75. c) Dormancy of seed is caused by
abscisic acid (ABA) and seed
germination is promoted by gibberellin.
76. c) The edible part of pomegranate is
fleshy seed testa which is not produced 90. a) PCl5(g) ⇋ PCl3(g) + Cl2(g)
without fertilization (parthenocarpy)
77. b) Endothecium and nucellus are nutritive
tissues of megasporangium.
78. b) Fertilizers have major or macro-
elements, which are required in large
amount for metabolism. 91. b) CaCl2; it is a salt of strong acid and
79. d) DNA helicase is important enzyme for weak base.
uncoiling of DNA helix during its 92. a)
replication.
80. c) Quarantine is process of controlling
and checking plant diseases for certain
time or in certain places.

Chemistry
81. b) Mass of oxygen which gets displaced
from metal oxide=0.4g
Now, 0.4 g of oxygen combines
with metal = 3.2g 93. d)
8 g of oxygen combines with
metal =
94. b) The existence of a substance in more
than one solid modifications is known as
∴GEW of metal = 64 g

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 APEX ME-CEE MOCK TEST SET-06 SOLUTION
polymorphism as that of allotropes of 111. a) Lead pencils do not contain
element. lead. Lead pencils contain graphite.
95. c) 112. 112. c) Picric acid has three
electron-withdrawing –NO2 groups, so
its phenolic group exhibits strong acidic
character like that of –COOH.
113. 113. a) CH3−CH2–Br + alc. KOH
→ CH2=CH2+ KBr + H2O.
96. c) N1V1 + N2V2 + N3V3 = N4V4 The reaction is called
0.2×50 + 0.1×50 + 0.2×100 = N4×200 ⇒ dehydrohalogenation.
N4 = 35/200 = 0.175 N. 114. b) 3 ether metamers are possible
97. b) Burning of ½ mol of S8 produce heat for C4H10O.
= 1590 kJ. 115. a) Nitro group is electron
Burning of 1 mol of S8 produce heat = 2 withdrawing group, so it deactivates the
×1590 = 3180 kJ. ring towards electrophilic substitution.
Since reaction is exothermic, the value of 116. d) 2, 2, 3, 3 - tetramethyl pentane,
ΔH = −3180 kJ mol−1. 2, 2 - dimethyl pentane and 2, 2, 3 -
98. a) Here 0.2 mole of NaOH is limiting trimethyl pentane do not contain an
reagent isopropyl group. 2 - Methyl pentane
∴Neutralisation of 1 mole of NaOH contains one isopropyl group.
release heat = 57.0 kJ
Neutralisation of 0.2 mole of NaOH
release heat = 57.0×0.2 = 11.4 kJ.
99. c) The transition metals and their
compounds are used as catalysts. Because
of the variable oxidation states they may
form intermediate compound with one of
the reactants. These intermediate
provides a new path with lower activation
energy.
100. a) It is due to hydrogen bonding
when H2O forms a cage like structure in
solid ice and density is reduced. 117. c) Carbene is the intermediate in
101. a) Li and Mg are diagonally Reimer - Tiemann reaction.
related elements. 118. d) Shortest C – C distance (1.20
102. a) Be(OH)2 is amphoteric while Å) is in acetylene. As acetylene has sp
Ca(OH)2, Sr(OH)2 and Ba(OH)2 are all hybridisation, the bond length increases
basic. in the order
103. a) first increases then decreases. 119. b) Kjeldahl’s method is used to
104. d) Purification of bauxite involves estimate nitrogen.
Baeyer’s process, Hall’s process and 120. a) Electrolysis of concentrated
Serperk process. aqueous solution of an alkali (Na or K)
105. d) The ions responsible for hard salt of saturated monocarboxylic acid
water are soluble in water. gives alkane while that of dicarboxylic
106. b) oscillation of loose electrons acid gives alkenes.
107. b) Rest all are the uses of boric 121. d) CH3CH2COOH + NaHCO3→
acid. CH3CH2COONa + H2O+ CO2
108. a) AlCl3 anhydrous is covalent 122. b) Vinyl chloride shows
but AlCl3.6H2O is ionic. resonance.
109. a) Green vitrol is FeSO4.7H2O 123. c) The reaction is called Etard
110. a) Iron obtained from blast reaction.
furnace is known as pig Iron

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 APEX ME-CEE MOCK TEST SET-06 SOLUTION
124. d) All aldehydes and methyl 3𝑅𝑇
Vrms = √ ∝ √𝑇
ketones gives this reaction. 𝑀
125. b) Calcium salt of formic acid on 142. c) Cubical expansivity of material of container
dry distillation gives formaldehyde. = 3 × α/3 = α which is equal to that of liquid. So
126. b) C6H5N2Cl + CH3OH → level of liquid remains almost stationary.
40 𝑡−5
C6H5OCH3 + N2 + HCl 143. c) 100 = 95−5
𝑅𝑒𝑑𝑢𝑐𝑡𝑖𝑜𝑛
127. c) 𝐶𝐻3 𝑁𝐶 → 𝐶𝐻3 𝑁𝐻𝐶𝐻3 t = 410C
128. a) 2C6H5I Cu→C6H5 - C6H5. 144. c) ∆𝑄 = 𝑚𝑠∆𝜃
∆𝑄
129. d) pepsin s = 𝑚∆𝜃 when ∆𝜃 = 0, s = ∞
130. c) Natural rubber is found to be a 3𝑅𝑇
polymer of cis-isoprene. (CH2= 145. a) Crms = √ ∝ √𝑇
𝑀
C(CH3).CH=CH2) 𝐶2 𝑇 927+273
= √𝑇2 = √ 27+273 = 2
𝐶1 1
PHYSICS SOLUTION
𝐺𝑚1 𝑚2 C2 = 2C1
131. b) F= 1
𝑑2
𝐹𝑑2
146. a) λ∝ 𝑇
G=𝑚 𝑇𝐴 𝜆 4800
1 𝑚2 = 𝜆𝐵 = 3600 = 4: 3
[𝑀𝐿𝑇 −2 ][𝐿2 ] 𝑇𝐵 𝐴
∴[G]= [𝑀2 ]
= [𝑀−1 𝐿3 𝑇 −2 ] 𝐶
147. a) µ =𝑉
132. b) Because the vertical components of 𝐶 3 ×108
velocities of both the bullets are same and equal to V= = = 1.25 X 108 m/s
𝜇 2.4
2ℎ 148. a) Focal length of the combination can be
zero and t= √ 𝑔
calculated as
133. a) For jumping he presses the spring platform, 1 1 1 1 1
= 𝑓 + 𝑓 = +40 + −25
𝐹
so the reading of spring balance increases first and 1
200
2

finally it becomes zero. F=− cm

3
134. c) n = 50 100 100
𝑛2 502
∴𝑃= 200 = = −1.5 𝐷
𝐹 −
%∆K= (2𝑛 + )% = (2 × 50 + )% = 3
100 100 𝐴 + 𝐷𝑚𝑖𝑛
sin
125% 149. a) µ = 𝐴
2
sin
135. a) ve = √2𝑔𝑅 g = constant sin(600 + 𝐷𝑚 )
2

ve ∝ √𝑅 √2 = 2
𝐴
sin
𝑣𝑒1 𝑅1 𝑅 1 2
= √𝑅 = √2𝑅 = Dmin = 300
𝑣𝑒2 2 √2
𝑥𝑑 40 ×25 200 2
136. d) In the state of weightlessness or in gravity 150. d) f = 𝑥−𝑑= 40−25 = 𝑐𝑚 = 3m
3
free space, water will rise to the upper end of the 1 3
P = 𝑓 = 2 𝐷 = 1.5𝐷
tube of any length.
(𝑠𝑡𝑟𝑒𝑠𝑠)2 2
𝐼 9
137. a) Energy per unit volume = 2𝑌 √𝐼 1 + 1 √ +1
𝐼𝑚𝑎𝑥 2 1 4
138. a) Since the system is free from any external 151. c) = [ ] =[ ]=
𝐼𝑚𝑖𝑛 𝐼 9 1
√𝐼1 −1 √ −1
2 1
force.
acm = 0 and initially they are at rest vcm = 0 152. b) µ = tan ip = tan 600 = √3
1 𝜆 𝜆 𝜋 𝜆 𝑉
139. c) Ep = 4 total energy 153. b) ∆𝑥 = 2𝜋 × 𝜙 = 2𝜋 × = 6 = 6𝑓 =
3
1 1 1 360
𝑚𝑤 2 𝑦 2 = 4 × 2 𝑚𝑤 2 𝐴2 6 ×500
= 0.12 m = 12 cm
2
𝐴2 𝑉
y2 = 154. b) For closed pipe f = 4𝑙
4
y = ± 𝐴/2 V= 4lf= 4 × 0.2 × 250 = 200 m/s
140. b) AV = constant 155. d) The sounds of different source are said to
1 1 differ in quality. The number of overtones and their
v ∝ 𝐴 ∝ 𝑟2 relative intensities determines the quality of any
𝑣1 𝑟 2 2 2
= (𝑟2 ) = (3) = 4: 9 musical sound.
𝑣2 𝜎
1 156. c) For plane infinite sheet of charge E= 2𝜀 =
141. c) In isothermal expansion, temperature 0

remains unchanged. independent of distance

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 APEX ME-CEE MOCK TEST SET-06 SOLUTION
0
E∝r 178. c) The colour of star depends on wavelength
157. c) W= q(∆V) = q(0) = 0 of maximum intensity radiation. According to
𝜀 𝜀 𝐴 1
158. c) C = 0 𝑟 Wein’s displacement law λm ∝ . Hence colour of
𝑑 𝑇
C∝A star depends on its surface temperature.
𝑖𝑆 50 ×12 179. b) The particle that is unstable in free space is
159. c) ig = 𝑆+𝐺 i.e. 10 = 12+𝐺
neutron.
G = 48Ω 180. d) Mass of proton = mass of anti-proton = 1
160. d) Resistivity depends only on the material of amu = 931 MeV
the conductor So minimum energy of ϒ-photon = 2 x 931 =
𝑡 +𝑡
161. c) tn = 𝑖 2 𝑐 1862 MeV
𝑡𝑖 + 15 MAT
280 = 2 181. c)
ti = 5450C 182. b)
162. c) Time ts = t1+ t2 = 35 min 183. c)
163. c) 1 Tesla = 104 Gauss 184. c)
164. d) No magnetic lines of force passes through
185. c)
the steel box.
165. a) Transformer works on ac only. 186. b)
166. c) Two coils are placed close to each other.
The mutual inductance of the pair of coils depends
upon relative position and orientation of the two
coils.
167. c) Because power = i2R, if R=0, then P=0
168. a) Gain in kinetic energy = qV= (1e) x 1v =
1eV
ℎ𝐶 187. d)
169. c) Work function φ = 𝜆
ℎ𝐶 6.626 ×10−34 ×3 ×108 188. c)
λ= = = 5.65 x 10-7 m = 189. a)
𝜙 2.2 ×1.6 ×10−19
565nm 190. a)
170. c) Stopping potential for photoelectrons 191. d) 32 × 2 + x = 68
depends on both the frequency of the incident light 192. d)
and nature of the cathode material. 193. a)
171. c) A direct X-ray photograph of the intestines 194. d)
is not generally taken by the radiologists because the 195. b)
X-rays will pass through the intestines without
causing a good shadow for any useful diagnosis.
172. c) Final energy of electron = -13.6+ 12.1 = -
1.51eV, which is corresponds to third Let son's age = x & father 's age = 3x ⇒
level i.e. n = 3. Hence number of spectral lines 3x – 5 = 4 (x – 5) ⇒ x = 15
𝑛(𝑛−1) 3(3−1)
emitted = = =3 196. b)
2 2
𝑡 197. b) Bat is a mammal
173. b) n = 𝑇 10 /5 = 2
1
2
198. b)
𝑁 1 1
Undecayed part 𝑁 = 2𝑛 = 4 = 25%
0
Decay = 100 – 25 = 75%
174. d) The average value of output direct current 199. d) 2365 + 1111 ⇒ 3476
2𝐼
in a full wave rectifier is 𝜋0 . 200. b)
𝛽 100
175. c) α = 1+𝛽 = 1+100 = 0.99 
176. a) In an NPN transistor, the emitter current
is slightly more than the collector current.
IE = IB + IC
IE > IC
177. b) The output of OR gate is Y = A+B.

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