solutions_aiats medical-2016 test-09_(code-a & b)
TRANSCRIPT
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8/18/2019 Solutions_AIATS Medical-2016 Test-09_(Code-A & B)
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Test - 9 (Code A) (Answers & Hints) All India Aakash Test Series for Medical-2016
1/9
1. (2)
2. (3)
3. (3)
4. (3)
5. (3)
6. (4)
7. (1)
8. (3)
9. (1)
10. (2)
11. (4)
12. (1)
13. (2)
14. (3)
15. (3)
16. (4)
17. (3)
18. (3)
19. (2)
20. (2)
21. (1)
22. (4)
23. (1)
24. (2)
25. (1)
26. (4)
27. (2)
28. (2)
29. (1)
30. (3)31. (3)
32. (4)
33. (2)
34. (2)
35. (1)
36. (4)
37. (2)38. (2)
39. (4)
40. (1)
41. (4)
42. (1)
43. (3)
44. (4)
45. (1)
46. (2)
47. (1)
48. (3)
49. (4)
50. (1)
51. (3)
52. (4)
53. (3)
54. (2)
55. (2)
56. (2)
57. (4)
58. (2)
59. (1)
60. (1)
61. (2)
62. (4)
63. (3)
64. (3)
65. (2)
66. (3)67. (1)
68. (2)
69. (3)
70. (1)
71. (2)
72. (4)
ANSWERS
TEST - 9 (Code-A)
All India Aakash Test Series for Medical-2016
Test Date : 03-04-2016
73. (2)74. (3)
75. (4)
76. (4)
77. (2)
78. (4)
79. (4)
80. (4)
81. (3)
82. (3)
83. (4)
84. (4)
85. (3)
86. (1)
87. (4)
88. (4)
89. (4)
90. (3)
91. (3)
92. (1)
93. (4)
94. (3)
95. (3)
96. (2)
97. (3)
98. (2)
99. (1)
100. (3)
101. (3)
102. (2)103. (3)
104. (3)
105. (3)
106. (2)
107. (2)
108. (3)
109. (3)110. (4)
111. (2)
112. (1)
113. (2)
114. (1)
115. (4)
116. (2)
117. (3)
118. (4)
119. (4)
120. (2)
121. (4)
122. (2)
123. (4)
124. (2)
125. (4)
126. (3)
127. (3)
128. (2)
129. (1)
130. (3)
131. (4)
132. (2)
133. (3)
134. (3)
135. (2)
136. (1)
137. (3)
138. (3)139. (1)
140. (3)
141. (4)
142. (2)
143. (2)
144. (4)
145. (2)146. (1)
147. (3)
148. (2)
149. (4)
150. (3)
151. (3)
152. (1)
153. (3)
154. (4)
155. (1)
156. (2)
157. (4)
158. (4)
159. (3)
160. (3)
161. (2)
162. (4)
163. (3)
164. (2)
165. (1)
166. (4)
167. (2)
168. (2)
169. (2)
170. (3)
171. (3)
172. (1)
173. (2)
174. (1)175. (1)
176. (3)
177. (3)
178. (2)
179. (1)
180. (1)
click here for code-B solutions
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All India Aakash Test Series for Medical-2016 Test - 9 (Code A) (Answers & Hints)
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Hints to Selected Questions
[ PHYSICS]
1. Answer (2)
2. Answer (3)
R =2
PQ
r
R% = l % + 2r %
=1 0.02
2 100100 1
3. Answer (3)
4. Answer (3)
v x
= 1 m/s v y = 2 m/s, tan =
y
x
v
v
y = tan –2
2
1
2x
x g
v
5. Answer (3)
A E
W
V pw
V pg
V wg
N
S
V wg
pg V
sin = wg V
40 sin = 20
= 30°
6. Answer (4)
P will be the maximum height.
7. Answer (1)
A
T 1
T 2
C
B90°–
mg
30°
1
sin120
T
=
sin(60 )
Mg
(T 1)min
where sin (60° + ) = 1
= 30°
8. Answer (3)
AN
B
mg sin
V
m g c o
s
Between A and B
mg (R – R cos) + 0 = 21
2mv
mg cos – N =2
mv
R , N = 0
9. Answer (1)
qE
qE cos
mg q E
c o s s i n
N N
mg s i n
N = mg cos + q E sin
mg sin = N + q E cos
10. Answer (2)
V 12
is uniform along – y axis.
12V
= 1 1
ˆ ˆcos 37 sin 37u i u j –
2 2
ˆ ˆcos 53 sin 53u i u j
= 1 2
ˆcos 37 cos 53u u i
1 2
ˆsin 37 sin 53u u j
1 2cos 37 cos 53 0u u
1 2
4 3
5 5u u
1 24 3u u
11. Answer (4)
12. Answer (1)
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Test - 9 (Code A) (Answers & Hints) All India Aakash Test Series for Medical-2016
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13. Answer (2)
14. Answer (3)
15. Answer (3)
C = C v –
1
R
u , PV n = constant
P =K
T constant
PV
T
P 2V = K
1
2PV = K
16. Answer (4)
mV 0 = 2m v
21 (2 )
2
m v = 21
2K x
17. Answer (3)
18. Answer (3)
A1V 1 = A
2V 2
19. Answer (2)
' w o
w s
V V V f f
V V V
20. Answer (2)
21. Answer (1)
22. Answer (4)
ˆi
P mv i
ˆ
3i
v P m j mv
i f P P
23. Answer (1)
5 – 10i = 2 + i x = 4
24. Answer (2)
25. Answer (1)
2 6 61
5 12 10 120 12 102
= 2 61
5 12 102
Heat
26. Answer (4)
0
0
640 693 2.3log
10
N
t N
27. Answer (2)
28. Answer (2)
3
h
kTm
29. Answer (1)
30. Answer (3)
E
S
X
W
Y Y
N
M = NiA
M = m x – my
31. Answer (3)
32. Answer (4)
33. Answer (2)
2 2 2
rms
200 100 50
2 2 2
V
34. Answer (2)
0 1.5 1
I f f
u f
1
2 .5
I
4I cm
35. Answer (1)
36. Answer (4)
– eV = 2 21
2m v u
37. Answer (2)
No. of fission =3
3
10
200 1.6 10
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All India Aakash Test Series for Medical-2016 Test - 9 (Code A) (Answers & Hints)
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38. Answer (2)
39. Answer (4)
S 1 + S
2 = 0
( –1) A1 + ( –1) A
2 = 0
40. Answer (1)
41. Answer (4)
42. Answer (1)
43. Answer (3)
– ray is E.M. radiation
44. Answer (4)
f o + f
e= 60
f o + f
e+10 = V
o + f
e
V o
= 65, f o = 55
1 1
o oV u
=1
of
45. Answer (1)
E = B C
46. Answer (2)
Number of gram equivalents of oxygen = number of
gram equivalents of aluminium
⇒
w
8 =
1.35
9
⇒ w = 1.2 g
47. Answer (1)
Basicity of H3PO
2 is one
48. Answer (3)
2
2 2
1 2
1 1 1R ( )
n nZ
2
2 2
1 1 1
R (2)(2) (4)
Divide and multiply by (4)2
2
2 2
1 1 1R (4)
(4) (8)
49. Answer (4)
50. Answer (1)
Cr belongs to 6th group
51. Answer (3)
O sp
2
52. Answer (4)
F–Be–F
53. Answer (3)
Vc = 3 b
[ CHEMISTRY]
54. Answer (2)
1 mole NaOH + 1 mole HF give 68 kJ/mole
55. Answer (2)
A + B B
G =
°pG G R
= 500 [ 100 ( 200)]
= – 500 + 300
= – 200
G = – 2.303 RT Log Ke
– 200000 = – 2.303 8.31 T log 103
T = 3483.4 K
56. Answer (2)
Kp and total pressure remain constant
57. Answer (4)
MnO +5I I + Mn– – 2+
24
+7
–5
2
58. Answer (2)
Fact
59. Answer (1)
60. Answer (1)
Conc. H SO2 4
COOH
COOH CO + CO + H O2 2
61. Answer (2)
62. Answer (4)
Correct according to IUPAC rules
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Test - 9 (Code A) (Answers & Hints) All India Aakash Test Series for Medical-2016
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63. Answer (3)
O
O
will not show tautomerism
64. Answer (3)It is dissymmetric
65. Answer (2)
Acidic character – I
66. Answer (3)
Fact
67. Answer (1)
is aromatic
68. Answer (2)
FeCl3 + NaSCN 3
Blood red
Fe(SCN) +NaCl
69. Answer (3)
CH – CH – CH – CH3 3
CH3
Cl
, the total reactivity.
is 2 3.8 = 7.6 for 2° - carbon atom
70. Answer (1)
CH – C = CH CH – C = O + HCHO3 2 3
CH3
O3
Zn/H O2
CH3
71. Answer (2)
SO2 is reducing agent
72. Answer (4)
C
B
A
D
AB =a
2, AD = 2a ,
3AC a
2
73. Answer (2)
P = 100 m
A
P
P
= B
100 1X
700 7
bT =
bK m
= BbB A
X 1000K
1 X M
=
1
1000710
1 1671
1
7
= 10 K
74. Answer (3)
2Cl
red red 2
P0.059E E log
n Cl
.
75. Answer (4)
For zero order reaction
t1 + t
2 + t
3 = 10 + 5 + 2.5 = 17.5 s.
76. Answer (4)
Argyrol is Ag sol (–ve)
77. Answer (2)
2 2 2 Ag S 4NaCN 2Na[Ag(CN) ] Na S .
78. Answer (4)
Ionic PCl5 is [ PCl
4 ]+ [ PCl
6]–
79. Answer (4)
Fact
80. Answer (4)
All are correct
81. Answer (3)
Fe3+ has d 5 system
82. Answer (3)
Ionization and linkage both isomerism are possible.
83. Answer (4)
C6H5Cl has partial double bond character.
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All India Aakash Test Series for Medical-2016 Test - 9 (Code A) (Answers & Hints)
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84. Answer (4)
Fact
85. Answer (3)
A is CH – C – CH3 3
NOH
Oxime
CH – CH – CH
3 3
NO
2- Nitrosopropane
86. Answer (1)
CH CH3 2
CH CH COOH CH CH CH CH NH2 2 3 2 2 2 2
HN3
CH CH CH3 2 2
CH NH CH – CH CH CH OH2 2 3 2 2 2
HNO
2
Conc.
H SO
443 K
2 4
HCl
87. Answer (4)
Br 2
(aq) forms saccharic acid from glucose but not
from fructose.
88. Answer (4)
CH = C – CH = CH and CH = C – CH = CH2 2 2 2
CH3
Isoprene
Cl
Chloroprene
89. Answer (4)
MgCO3 and Al(OH)
3 are antacid.
90. Answer (3)
OH
CHCl2
OO
OH
+ NaOH + CHCl3
O–H
C = O
O
NaOH
H
In this reaction electrophile is CCl2
91. Answer (3)
X X aaC C
XX aaC
X YaaC
X XYaa
XaYa
X aC
92. Answer (1)
n = 20
Caruncle (2n) = 40
tube nucleus (n) = 20
PEC (3n) = 60
Nucellus (2n) = 40
93. Answer (4)Zea mays – C
4 plant
94. Answer (3)
Collenchyma
95. Answer (3)
Insectivorous plants
[ BIOLOGY
]
96. Answer (2)
TTRR
TR
TtRR
(Tall-Red)
XttRr
tR tr
TtRr
(Tall-Pink)
97. Answer (3)
Oxidative phosphorylation in mitochondria.
98. Answer (2)
Diapause is a stage of suspended development.
99. Answer (1)
Dry to mesic conditions.
100. Answer (3)
23 S rRNA.
101. Answer (3)
Hypogynous and perigynous flowers.
102. Answer (2)
(b) Metaphase I
(d) Anaphase I
(a) Anaphase II
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Test - 9 (Code A) (Answers & Hints) All India Aakash Test Series for Medical-2016
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103. Answer (3)
Largest tiger reserve.
104. Answer (3)
Parietal placentation
105. Answer (3)
106. Answer (2)80 dB
107. Answer (2)
Autosomal dominant trait
108. Answer (3)
Pyruvate act as oxidant in lactic acid fermentation.
109. Answer (3)
ER, golgi body, lysosomes and vacuoles in
endomembrane system.
110. Answer (4)
LDP.
111. Answer (2)
Secondary phloem is developed on outer side of
vascular cambium.
112. Answer (1)
Ethylene
113. Answer (2)
Fucus – Brown algae
Polysiphonia – Red algae
114. Answer (1)
CMP-11, 2015, Paris.
115. Answer (4)
Dorsiventral thalloid gametophyte in liverworts.
116. Answer (2)
Nucleosomes in eukaryotes.
117. Answer (3)
Diatoms.
118. Answer (4)
(a) Gases (b) Bacteria, fungi (c) Public transport
119. Answer (4)
Viroid is naked RNA.
120. Answer (2)
121. Answer (4)
Ulothrix , Spirogyra – cellulosic wall.
122. Answer (2)
Hydrophyte.
123. Answer (4)
Axillary buds.124. Answer (2)
Consciousness.
125. Answer (4)
Functionally similar.
126. Answer (3)
Night and early morning when absorption is high.
127. Answer (3)
Pericarp, placenta and thalamus are edible in guava.
128. Answer (2)
129. Answer (1)
130. Answer (3)
131. Answer (4)
Fungi, bacteria and cyanobacteria in organic farming
132. Answer (2)
PEG and NaNO3
133. Answer (3)
ETC and oxidative phosphorylation.
134. Answer (3)
Albugo candida is parasite on mustard.
135. Answer (2)
Haplo-diplontic.
136. Answer (1)
The first digit in E.C. indicates the class of theenzyme and hence the mode of its activity.
137. Answer (3)
Proton pump inhibition results in reduced HClrelease.
138. Answer (3)
Thymosine is a hormone from thymus gland.
139. Answer (1)
Regression of corpus luteum is triggered by reducedlevel of LH of secretory phase.
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All India Aakash Test Series for Medical-2016 Test - 9 (Code A) (Answers & Hints)
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140. Answer (3)
The first haploid cells in testes are secondary
spermatocytes while in ovaries are secondary
oocyte and first polar body.
141. Answer (4)
Annelids generally have closed circulatory system.
142. Answer (2)
Mitochondrial DNA from Africans populations has
maximum variations through mutations.
143. Answer (2)
Elytra are mesothoracic wings. Two pairs of thoracic
and eight pairs of abdominal spiracles are present in
cockroach. Unjointed anal styles are found only in
male cockroaches.
144. Answer (4)
Urethra has pseudostratified non-ciliated epithelium
while all other parts have transitional epithelium.
145. Answer (2)
Haem is the prosthetic group and it is a part of the
active site of enzyme.
146. Answer (1)
Lens and cornea, both are easily transplantable due
to avascular nature.
147. Answer (3)
Male amphibians have copulatory pads on first digit
of forelimbs.
148. Answer (2)
GIFT – Gametic Intrafallopian Transfer.
149. Answer (4)
Vector mediated gene transfer is indirect method of
gene transfer.
150. Answer (3)
Dystrophy is a genetic disorder, characterized by
lack of dystrophin protein.
151. Answer (3)
HIV reduces number of helper T cells by destroying
them.
152. Answer (1)
Frequency of recessive allele (q) = 0.3
So, frequency of dominant allele (p) = 0.7
Hence frequency of heterozygotes (2 pq) = 0.42
153. Answer (3)
AB positive blood group is a universal recipient due
to absence of agglutinins/antibodies.
154. Answer (4)
Interferons are secreted by virally infected cells and
other cells of the body to protect the non-infected
cells.155. Answer (1)
Autonomic neurons in submucosa (meissner’s
plexus) and muscularis externa (auerbach’s plexus)
control the muscles of gut wall.
156. Answer (2)
Melatonin hormone from pineal gland sets up the
biological clock of the body.
157. Answer (4)
Ca2+ ions trigger the exocytosis of synaptic vesicles
from synaptic knots.
158. Answer (4)
Deglutition is a voluntary process in which uvula
prevents the food from rising up in the nasopharynx
and epiglottis prevents the food entry in glottis
159. Answer (3)
Annelids and arthropods are metamerically
segmented schizocoelomates.
160. Answer (3)
Macroglia are ectodermal in origin.
161. Answer (2)
Malpighian tubules are located at the junction of
midgut and hindgut.
Pharyngeal glands of cockroach have four secretory
parts and two receptacles.
162. Answer (4)
Hemidesmosomes connect the cells to their
basement membranes.
163. Answer (3)164. Answer (2)
Charge and size both play an important role in
separation of DNA fragments.
165. Answer (1)
Synarthrose/fibrous joints are completely immovable
joints.
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Test - 9 (Code A) (Answers & Hints) All India Aakash Test Series for Medical-2016
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166. Answer (4)
167. Answer (2)
Ca2+ ions are involved in blood clotting.
168. Answer (2)
Syphilis is curable in primary stage and its chancres
are painless.
169. Answer (2)
Insertionally inactivated lac z’ gene results in failure
to utilize ‘X–gal’ by recombinant bacteria, which fail
to produce blue color, hence appearing white.
170. Answer (3)
171. Answer (3)
Dissociation of oxygen from (Hb) causes a right
shift in the curve.
172. Answer (1)
Secondary/anamnestic response is always vigorous
due to memory cells.
173. Answer (2)
Earliest autotrophs were chemoautotrophs and were
anoxygenic.
174. Answer (1)
Dehydration triggers on ‘RAAS’ and enhanced tubular
reabsorption of Na+ ions.
175. Answer (1)
Outcrossing is mating of animals of same breed but
not having any common ancestors for at least 4–6
generations, introducing variations but not causing
drastic changes.
176. Answer (3)
Corpora allata produce juvenile hormone.
177. Answer (3)
Ringworms are fungal infections.
178. Answer (2)
179. Answer (1)
180. Answer (1)
Retroviruses are used as ideal cloning vectors for
eukaryotic host cells.
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Test - 9 (Code B) (Answers & Hints) All India Aakash Test Series for Medical-2016
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1. (3)
2. (2)
3. (1)
4. (3)
5. (2)
6. (3)
7. (2)
8. (4)
9. (4)
10. (2)
11. (3)
12. (4)
13. (4)
14. (4)
15. (1)
16. (1)
17. (3)
18. (4)
19. (4)
20. (2)21. (3)
22. (4)
23. (3)
24. (2)
25. (3)
26. (4)
27. (4)
28. (1)
29. (1)
30. (2)31. (1)
32. (1)
33. (4)
34. (3)
35. (2)
36. (4)
37. (3)38. (1)
39. (3)
40. (2)
41. (1)
42. (1)
43. (1)
44. (1)
45. (4)
46. (1)
47. (4)
48. (4)
49. (2)
50. (3)
51. (1)
52. (4)
53. (2)
54. (1)
55. (1)
56. (4)
57. (2)
58. (2)
59. (4)
60. (2)
61. (2)
62. (1)
63. (4)
64. (4)
65. (4)
66. (3)67. (1)
68. (4)
69. (3)
70. (1)
71. (4)
72. (1)
ANSWERS
TEST - 9 (Code-B)
All India Aakash Test Series for Medical-2016
Test Date : 03-04-2016
73. (1)74. (2)
75. (2)
76. (3)
77. (1)
78. (4)
79. (2)
80. (4)
81. (4)
82. (4)
83. (1)
84. (2)
85. (1)
86. (3)
87. (2)
88. (1)
89. (3)
90. (4)
91. (4)
92. (1)
93. (1)
94. (4)
95. (2)
96. (3)
97. (3)
98. (4)
99. (1)
100. (1)
101. (2)
102. (4)103. (2)
104. (4)
105. (2)
106. (4)
107. (2)
108. (2)
109. (1)110. (4)
111. (2)
112. (3)
113. (4)
114. (3)
115. (4)
116. (2)
117. (1)
118. (1)
119. (4)
120. (4)
121. (1)
122. (1)
123. (1)
124. (4)
125. (1)
126. (3)
127. (3)
128. (4)
129. (1)
130. (4)
131. (1)
132. (1)
133. (2)
134. (3)
135. (1)
136. (3)
137. (3)
138. (4)139. (1)
140. (1)
141. (3)
142. (1)
143. (4)
144. (3)
145. (1)146. (1)
147. (2)
148. (4)
149. (4)
150. (2)
151. (3)
152. (4)
153. (1)
154. (2)
155. (2)
156. (1)
157. (1)
158. (2)
159. (2)
160. (4)
161. (3)
162. (2)
163. (3)
164. (3)
165. (1)
166. (1)
167. (2)
168. (4)
169. (1)
170. (1)
171. (4)
172. (2)
173. (4)
174. (4)175. (2)
176. (1)
177. (3)
178. (1)
179. (1)
180. (3)
click here for code-A solutions
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All India Aakash Test Series for Medical-2016 Test - 9 (Code B) (Answers & Hints)
2/9
Hints to Selected Questions
[ PHYSICS]
1. Answer (3)
E = B C
2. Answer (2)f o + f
e= 60
f o + f
e+10 = V
o + f
e
V o
= 65, f o = 55
1 1
o oV u
=1
of
3. Answer (1)
– ray is E.M. radiation
4. Answer (3)
5. Answer (2)
6. Answer (3)
7. Answer (2)
S 1 + S
2 = 0
( –1) A1 + ( –1) A
2 = 0
8. Answer (4)
9. Answer (4)
No. of fission =3
3
10
200 1.6 10
10. Answer (2)
– eV = 2 21
2m v u
11. Answer (3)
12. Answer (4)
0 1.5 1
I f f
u f
1
2 .5
I
4I cm
13. Answer (4)
2 2 2
rms
200 100 50
2 2 2V
14. Answer (4)
15. Answer (1)
16. Answer (1)
E
S
X
W
Y Y
N
M = NiA
M = m x – my
17. Answer (3)
18. Answer (4)
3
h
kTm
19. Answer (4)
20. Answer (2)
0
0
640 693 2.3log
10
N
t N
21. Answer (3)
2 6 61
5 12 10 120 12 102
= 2 61
5 12 102
Heat
22. Answer (4)
23. Answer (3)
5 – 10i = 2 + i x = 4
24. Answer (2)
ˆi
P mv i
ˆ
3i
v P m j mv
i f P P
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25. Answer (3)
26. Answer (4)
27. Answer (4)
' w o
w s
V V V f f
V V V
28. Answer (1)
A1V 1 = A
2V 2
29. Answer (1)
30. Answer (2)
mV 0 = 2m v
21 (2 )2
m v = 21
2K x
31. Answer (1)
C = C v –
1
R
u , PV n = constant
P =K
T constant
PV
T
P 2V = K
1
2PV = K
32. Answer (1)
33. Answer (4)
34. Answer (3)
35. Answer (2)
36. Answer (4)
V 12
is uniform along – y axis.
12V
= 1 1
ˆ ˆcos 37 sin 37u i u j –
2 2ˆ ˆcos 53 sin 53u i u j
= 1 2
ˆcos 37 cos 53u u i
1 2
ˆsin 37 sin 53u u j
1 2cos 37 cos53 0u u
1 2
4 3
5 5u u
1 24 3u u
37. Answer (3)
qE
qEcos
mg q E
c o s s i n
N N
mg s i n
N = mg cos + q E sin
mg sin = N + q E cos
38. Answer (1)
AN
B
mg sin
V
m g c o
s
Between A and B
mg (R – R cos) + 0 = 21
2mv
mg cos – N =2
mv
R , N = 0
39. Answer (3)
A
T 1
T 2
C
B
90°–
mg
30°
1
sin120
T
= sin(60 )
Mg
(T 1)min
where sin (60° + ) = 1
= 30°
40. Answer (2)
P will be the maximum height.
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41. Answer (1)
A E
W
V pw
V pg
V wg
N
S
V wg
pg V
sin = wg V
40 sin = 20
= 30°
42. Answer (1)
v x
= 1 m/s v y = 2 m/s, tan =
y
x
v
v
y = tan –2
2
1
2x
x g
v
43. Answer (1)
44. Answer (1)
R =2
PQr
R% = l % + 2r %
=1 0.02
2 100100 1
45. Answer (4)
46. Answer (1)
OH
CHCl2
OO
OH
+ NaOH + CHCl3
O–H
C = O
O
NaOH
H
In this reaction electrophile is CCl2
47. Answer (4)
MgCO3 and Al(OH)
3 are antacid.
48. Answer (4)
CH = C – CH = CH and CH = C – CH = CH2 2 2 2
CH3
Isoprene
Cl
Chloroprene
49. Answer (2)
Br 2 (aq) forms saccharic acid from glucose but not
from fructose.
50. Answer (3)
CH CH3 2
CH CH COOH CH CH CH CH NH2 2 3 2 2 2 2
HN3
CH CH CH3 2 2
CH NH CH – CH CH CH OH2 2 3 2 2 2
HNO2
Conc.
H SO
443 K
2 4
HCl
[ CHEMISTRY]
51. Answer (1)
A is CH – C – CH3 3
NOH
Oxime
CH – CH – CH
3 3
NO
2- Nitrosopropane
52. Answer (4)
Fact
53. Answer (2)
C6H5Cl has partial double bond character.
54. Answer (1)
Ionization and linkage both isomerism are possible.
55. Answer (1)
Fe3+ has d 5 system
56. Answer (4)
All are correct
57. Answer (2)
Fact
58. Answer (2)
Ionic PCl5 is [ PCl
4 ]+ [ PCl
6]–
59. Answer (4)
2 2 2 Ag S 4NaCN 2Na[Ag(CN) ] Na S .
60. Answer (2)
Argyrol is Ag sol (–ve)
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61. Answer (2)
For zero order reaction
t1 + t
2 + t
3 = 10 + 5 + 2.5 = 17.5 s.
62. Answer (1)
2Cl
red red 2
P0.059E E logn Cl
.
63. Answer (4)
P = 100 m
A
P
P
= B
100 1X
700 7
bT =
bK m
= BbB A
X 1000K
1 X M
=
1
1000710
1 1671
1
7
= 10 K
64. Answer (4)
C
B
A
D
AB =a
2, AD = 2a ,
3AC a
2
65. Answer (4)
SO2 is reducing agent
66. Answer (3)
CH – C = CH CH – C = O + HCHO3 2 3
CH3
O3
Zn/H O2
CH3
67. Answer (1)
CH – CH – CH – CH3 3
CH3
Cl
, the total reactivity.
is 2 3.8 = 7.6 for 2° - carbon atom
68. Answer (4)
FeCl3 + NaSCN 3
Blood red
Fe(SCN) +NaCl
69. Answer (3)
is aromatic
70. Answer (1)
Fact
71. Answer (4)
Acidic character – I
72. Answer (1)
It is dissymmetric
73. Answer (1)
O
O
will not show tautomerism
74. Answer (2)
Correct according to IUPAC rules
75. Answer (2)
76. Answer (3)
Conc. H SO2 4
COOH
COOH CO + CO + H O2 2
77. Answer (1)
78. Answer (4)
Fact
79. Answer (2)
MnO +5I I + Mn– – 2+
24
+7
–5
2
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80. Answer (4)
Kp and total pressure remain constant
81. Answer (4)
A + B B
G =
°pG G R
= 500 [ 100 ( 200)]
= – 500 + 300
= – 200
G = – 2.303 RT Log Ke
– 200000 = – 2.303 8.31 T log 103
T = 3483.4 K
82. Answer (4)
1 mole NaOH + 1 mole HF give 68 kJ/mole
83. Answer (1)
Vc = 3b84. Answer (2)
F–Be–F
85. Answer (1)
O sp
2
86. Answer (3)
Cr belongs to 6th group
87. Answer (2)
88. Answer (1)
2
2 2
1 2
1 1 1R ( )
n nZ
2
2 2
1 1 1R (2)
(2) (4)
Divide and multiply by (4)2
2
2 2
1 1 1R (4)
(4) (8)
89. Answer (3)
Basicity of H3PO
2 is one
90. Answer (4)
Number of gram equivalents of oxygen = number of
gram equivalents of aluminium
⇒ w
8 =
1.35
9
⇒ w = 1.2 g
91. Answer (4)
Haplo-diplontic.92. Answer (1)
Albugo candida is parasite on mustard.
93. Answer (1)
ETC and oxidative phosphorylation.
94. Answer (4)
PEG and NaNO3
95. Answer (2)
Fungi, bacteria and cyanobacteria in organic farming
96. Answer (3)
97. Answer (3)
98. Answer (4)
99. Answer (1)
Pericarp, placenta and thalamus are edible in guava.
[ BIOLOGY
]
100. Answer (1)
Night and early morning when absorption is high.
101. Answer (2)
Functionally similar.
102. Answer (4)
Consciousness.
103. Answer (2)
Axillary buds.
104. Answer (4)
Hydrophyte.
105. Answer (2)
Ulothrix , Spirogyra – cellulosic wall.
106. Answer (4)
107. Answer (2)
Viroid is naked RNA.
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108. Answer (2)
(a) Gases (b) Bacteria, fungi (c) Public transport
109. Answer (1)
Diatoms.
110. Answer (4)
Nucleosomes in eukaryotes.
111. Answer (2)
Dorsiventral thalloid gametophyte in liverworts.
112. Answer (3)
CMP-11, 2015, Paris.
113. Answer (4)
Fucus – Brown algae
Polysiphonia – Red algae
114. Answer (3)
Ethylene
115. Answer (4)
Secondary phloem is developed on outer side of
vascular cambium.
116. Answer (2)
LDP.
117. Answer (1)
ER, golgi body, lysosomes and vacuoles in
endomembrane system.
118. Answer (1)Pyruvate act as oxidant in lactic acid fermentation.
119. Answer (4)
Autosomal dominant trait
120. Answer (4)
80 dB
121. Answer (1)
122. Answer (1)
Parietal placentation
123. Answer (1)
Largest tiger reserve.
124. Answer (4)
(b) Metaphase I
(d) Anaphase I
(a) Anaphase II
125. Answer (1)
Hypogynous and perigynous flowers.
126. Answer (3)
23 S rRNA.
127. Answer (3)
Dry to mesic conditions.
128. Answer (4)
Diapause is a stage of suspended development.
129. Answer (1)
Oxidative phosphorylation in mitochondria.
130. Answer (4)
TTRR
TR
TtRR
(Tall-Red)
XttRr
tR tr
TtRr
(Tall-Pink)
131. Answer (1)
Insectivorous plants
132. Answer (1)
Collenchyma
133. Answer (2)
Zea mays – C4 plant
134. Answer (3)
n = 20
Caruncle (2n) = 40
tube nucleus (n) = 20
PEC (3n) = 60
Nucellus (2n) = 40
135. Answer (1)
X X aaC C
XX aaC
X YaaC
X XYaa
Xa Ya
X aC
136. Answer (3)
Retroviruses are used as ideal cloning vectors for
eukaryotic host cells.
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137. Answer (3)
138. Answer (4)
139. Answer (1)
Ringworms are fungal infections.
140. Answer (1)
Corpora allata produce juvenile hormone.
141. Answer (3)
Outcrossing is mating of animals of same breed but
not having any common ancestors for at least 4–6
generations, introducing variations but not causing
drastic changes.
142. Answer (1)
Dehydration triggers on ‘RAAS’ and enhanced tubular
reabsorption of Na+ ions.
143. Answer (4)
Earliest autotrophs were chemoautotrophs and were
anoxygenic.
144. Answer (3)
Secondary/anamnestic response is always vigorous
due to memory cells.
145. Answer (1)
Dissociation of oxygen from (Hb) causes a right
shift in the curve.
146. Answer (1)
147. Answer (2)
Insertionally inactivated lac z’ gene results in failure
to utilize ‘X–gal’ by recombinant bacteria, which fail
to produce blue color, hence appearing white.
148. Answer (4)
Syphilis is curable in primary stage and its chancres
are painless.
149. Answer (4)
Ca2+ ions are involved in blood clotting.
150. Answer (2)
151. Answer (3)
Synarthrose/fibrous joints are completely immovable
joints.
152. Answer (4)
Charge and size both play an important role in
separation of DNA fragments.
153. Answer (1)
154. Answer (2)
Hemidesmosomes connect the cells to their
basement membranes.
155. Answer (2)
Malpighian tubules are located at the junction of
midgut and hindgut.
Pharyngeal glands of cockroach have four secretory
parts and two receptacles.
156. Answer (1)
Macroglia are ectodermal in origin.
157. Answer (1)
Annelids and arthropods are metamerically
segmented schizocoelomates.
158. Answer (2)Deglutition is a voluntary process in which uvula
prevents the food from rising up in the nasopharynx
and epiglottis prevents the food entry in glottis
159. Answer (2)
Ca2+ ions trigger the exocytosis of synaptic vesicles
from synaptic knots.
160. Answer (4)
Melatonin hormone from pineal gland sets up the
biological clock of the body.
161. Answer (3)
Autonomic neurons in submucosa (meissner’s
plexus) and muscularis externa (auerbach’s plexus)
control the muscles of gut wall.
162. Answer (2)
Interferons are secreted by virally infected cells and
other cells of the body to protect the non-infected
cells.
163. Answer (3)
AB positive blood group is a universal recipient due
to absence of agglutinins/antibodies.
164. Answer (3)
Frequency of recessive allele (q) = 0.3
So, frequency of dominant allele (p) = 0.7
Hence frequency of heterozygotes (2 pq) = 0.42
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165. Answer (1)
HIV reduces number of helper T cells by destroying
them.
166. Answer (1)
Dystrophy is a genetic disorder, characterized by
lack of dystrophin protein.
167. Answer (2)
Vector mediated gene transfer is indirect method of
gene transfer.
168. Answer (4)
GIFT – Gametic Intrafallopian Transfer.
169. Answer (1)
Male amphibians have copulatory pads on first digit
of forelimbs.
170. Answer (1)
Lens and cornea, both are easily transplantable due
to avascular nature.
171. Answer (4)
Haem is the prosthetic group and it is a part of the
active site of enzyme.
172. Answer (2)
Urethra has pseudostratified non-ciliated epithelium
while all other parts have transitional epithelium.
173. Answer (4)
Elytra are mesothoracic wings. Two pairs of thoracic
and eight pairs of abdominal spiracles are present in
cockroach. Unjointed anal styles are found only in
male cockroaches.
174. Answer (4)
Mitochondrial DNA from Africans populations has
maximum variations through mutations.
175. Answer (2)
Annelids generally have closed circulatory system.
176. Answer (1)
The first haploid cells in testes are secondary
spermatocytes while in ovaries are secondary
oocyte and first polar body.
177. Answer (3)
Regression of corpus luteum is triggered by reduced
level of LH of secretory phase.178. Answer (1)
Thymosine is a hormone from thymus gland.
179. Answer (1)
Proton pump inhibition results in reduced HCl
release.
180. Answer (3)
The first digit in E.C. indicates the class of the
enzyme and hence the mode of its activity.