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Test - 13 (Code-A) (Answers) All India Aakash Test Series for Medical-2018
1/7
1. (2)
2. (2)
3. (4)
4. (4)
5. (3)
6. (1)
7. (4)
8. (2)
9. (3)
10. (3)
11. (1)
12. (3)
13. (1)
14. (2)
15. (1)
16. (3)
17. (3)
18. (2)
19. (2)
20. (2)
21. (2)
22. (3)
23. (1)
24. (3)
25. (4)
26. (2)
27. (3)
28. (3)
29. (2)
30. (1)
31. (1)
32. (3)
33. (2)
34. (2)
35. (4)
36. (2)
Test Date : 29/04/2018
TEST - 13 (Code A)
All India Aakash Test Series for Medical - 2018
ANSWERS
37. (2)
38. (4)
39. (2)
40. (1)
41. (2)
42. (3)
43. (1)
44. (2)
45. (4)
46. (3)
47. (2)
48. (2)
49. (1)
50. (3)
51. (2)
52. (3)
53. (3)
54. (2)
55. (1)
56. (2)
57. (2)
58. (1)
59. (2)
60. (1)
61. (2)
62. (3)
63. (2)
64. (2)
65. (2)
66. (3)
67. (3)
68. (1)
69. (1)
70. (2)
71. (1)
72. (1)
73. (2)
74. (2)
75. (4)
76. (1)
77. (4)
78. (1)
79. (4)
80. (2)
81. (1)
82. (1)
83. (3)
84. (2)
85. (3)
86. (4)
87. (2)
88. (2)
89. (4)
90. (4)
91. (2)
92. (4)
93. (4)
94. (1)
95. (2)
96. (2)
97. (3)
98. (3)
99. (3)
100. (4)
101. (3)
102. (1)
103. (2)
104. (4)
105. (3)
106. (4)
107. (1)
108. (2)
109. (4)
110. (3)
111. (2)
112. (3)
113. (2)
114. (2)
115. (3)
116. (2)
117. (4)
118. (3)
119. (4)
120. (1)
121. (4)
122. (1)
123. (2)
124. (4)
125. (2)
126. (2)
127. (4)
128. (3)
129. (2)
130. (4)
131. (2)
132. (4)
133. (4)
134. (4)
135. (2)
136. (4)
137. (1)
138. (2)
139. (3)
140. (2)
141. (1)
142. (2)
143. (2)
144. (2)
145. (3)
146 (2)
147. (2)
148. (3)
149. (4)
150. (3)
151. (1)
152. (1)
153. (1)
154. (3)
155. (3)
156. (4)
157. (4)
158. (1)
159. (3)
160. (3)
161. (3)
162. (2)
163. (2)
164. (4)
165. (4)
166. (4)
167. (2)
168. (4)
169. (2)
170. (1)
171. (2)
172. (4)
173. (3)
174. (3)
175. (1)
176. (4)
177. (1)
178. (2)
179. (2)
180. (2)
All India Aakash Test Series for Medical-2018 Test - 13 (Code-A) (Hints)
2/6
PHYSICS
1. 2 –1dv
a tdt
Acceleration is positive in interval 1
12
t and
velocity is negative
2. ˆ ˆ( ) , ,A v
v ai b ct j v a u b ct �
At t = ta, v
v = 0
ta =
b
c
Time of flight T = 2ta =
2b
c
Range = uHT =
2ab
c
3.
u2 u
1
v2 v
1
H
1 1ˆ ˆ
v u i gtj �
2 2ˆ ˆ
v u i gtj �
20v v⇒
� �
1 2 1 20v v v v ⇒
� � � �
t = 1 2
u u
g
21
2H gt 1 2
2
u uH
g⇒
4. Using equation, 21
2s ut at w.r.t. balloon.
srel
= 0, urel
= v1 + v
2, a
rel = –g
2
1 2
10 ( )
2v v t gt
t = 0 and 1 2
2( )v v
g
5. Block will remain in rest after removing F1.
6. Fnet
= ma
Tcos – f = ma
7. Only in case of uniform circular motion resultant
force is towards the centre. t
a w� �
HINTS
8. P = F.v F v
P v2
9.
30°
x
l/4
l/4
3cos30
4 8
l lx
11. At height, 2
1
gg
h
R
⎛ ⎞⎜ ⎟⎝ ⎠
4
gg
2
1 1
41
h
R
⎛ ⎞⎜ ⎟⎝ ⎠
h = R
12. Energy conservation
2 2 1 2
1 1 2 2
1 10 –
2 2
Gmmmv m v
d
m1v
1 = m
2v
2
Relative velocity = v1 + v
2
OR
2 1 2
red red
1
2
Gm mm v
d
1 2
red
2 ( )G m mv
d
⇒
13. W = B
3 3 34 4( – ) 1
3 3R r g R g
W
Br
R
3 3
3
– 1R r
R
3
1 –11–
r
R
⎛ ⎞ ⎜ ⎟⎝ ⎠
Test - 13 (Code-A) (Hints) All India Aakash Test Series for Medical-2018
3/6
14. Velocity become constant after a particular time
16. Heating of glass surface of bulb is due to radiation.
17.1
2PV = constant
19. dQ = msdT
2
3
1
Q aT dT ∫ = 15
4
a
20. E x2
x E
21.1
2 2
v Tf
L L
22.v
f
apparent
–
⎛ ⎞ ⎜ ⎟⎝ ⎠
v vf f
v v
fapparent
= 2f3
vv
25. Intensity 2
1
r
Photocurrent Intensity
26. Kmax
= E – Electron reaching at collector have maximum energy
= Kmax
+ eV
27.3
1
nf
n
log log 3lognf C n
C is constant.
28.
n = 5
n = 1
Photon
P mv= hP
h
mv h
vm
29. A = A0e–t
4
0
0
1
93
AA A
⎛ ⎞ ⎜ ⎟⎝ ⎠
31.
22 1
1
Rf
u
⎛ ⎞⎜ ⎟
⎝ ⎠
32.
1
2sin
n
a
⎛ ⎞ ⎜ ⎟⎝ ⎠ for maxima
1
1 13 2 6500 Å
2 2
⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
a a
1
56500 4642.8 Å
7
33.tan
tancos
34. E = hx >
m >
R
35. emf induced is zero.
37. Power = rate of heat generation = F.v
Q = Fv
QF
v
38.2 2
x yv v v
2 2
0 02
xv v v
2 2 2
0 04
xv v v
03
xv v
x x xv u a t
03 0
qEv t
m
03mv
tqE
39.2
qM L
m
( )2
qmvr
m
1
2qvr
All India Aakash Test Series for Medical-2018 Test - 13 (Code-A) (Hints)
4/6
40. P = I2R
41. Slope 1
R
R T
42. Qi = 60 C
Qf = kQ
i
180 = k(60) k = 3
43.
E2
E1
d
E1 = E
2
1 2
k dE
R
2
Q
R
1 32
kQdE
R
44.2
kqE
r
( – 1)k n qV
r
2
( – 1)( – 1)
V k n qn r
rE
kq
r
45.
(1, 2)
y x = 2
x
y
Electric field lines are
perpendicular to equipotential surface
m1 × m
2 = –1
CHEMISTRY
46. C –3
[B] [C] 0.15 0.05K 7.5
[A] 1 10
G° = –2.303 RT log KC
= –2.303 × 8.314 × 298 log7.5
= –4993 J mol–1
= –4.993 kJ mol–1
47.2 2 3
N (g) 3H (g) 2NH (g) ���⇀↽���
Initial 1 3 0
Eq. 1– 0.5 3 – 1.5 1
3 3NH NH
PP x P
3
48. Intensive properties are independent of total amount
of substance.
49.
2
2 2 2
1 1 1 n – 4R – R
2 n 4n
⎡ ⎤⎡ ⎤ ⎢ ⎥⎢ ⎥ ⎣ ⎦ ⎣ ⎦2
2
4 n
R n – 4
⎡ ⎤ ⎢ ⎥
⎣ ⎦
2
2
nk
n – 4 so
4k
R
51. Because limiting reagent decides the amount of
product formed.
52. 2 moles of Na2HPO
3 will have equal number of P
atoms as in 1 mole of Ca3(PO
4)2.
53. van der Waals equation at high pressure,
mPV Pb
1RT Rt
55. In N(SiH3)3 p – d back bonds are formed.
56. [Cu(H2O)
4]SO
4.H
2O has ionic, covalent, coordinate
and hydrogen bonds.
57.2 2
1 1 2 2C C
2 20.001
m
0 0
m m
40.1 0.001
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠
0.1 × 16 = 0.001 × 0.001 2
m( )
0.001
m = 40 S cm2 mol–1
58.T A B A B
P P (P – P )X = 100 + (250) X
B
So,A
P 100 torr
B AP – P 250 torr
BP 250 100 350 torr
59. According to Faraday’s first law
W = Zit
120 111.2 5 60 60
x 96500
x = 1.99 2
So tin is present as Sn2+
Test - 13 (Code-A) (Hints) All India Aakash Test Series for Medical-2018
5/6
60. Tb = i × K
b × m
i = 1 + (n–1)63. W = ZQ × Current efficiency
2H+ +
O
O
+ 2e–
OH
OH
64. Slow step is rate determining step
So Rate = Keq
[NOBr2][NO]
2eq
2
[NOBr ]K
[NO][Br ]= (From 1st step)
and 2 eq 2[NOBr ] K [NO][Br ]=
and rate law expression is:
Rate = Keq
K1[NO]2[Br
2]
65. Mn2+ acts as autocatalyst for the reaction.
66. CuFeS2
(copper pyrites)
68. Fe+1 is present in [Fe(H2O)
5NO]SO
4
70.4
226 222288 86
Ra He Rn
71. Structure of diborane
H
H
H
H
B B
H
H
72 As H2SO
5 and H
2S
2O
8 have peroxide linkages
O
S=OO OH
HO
O
S=O
O
OHS – O
O
OH
O
74. As CO2 has carbon in +4 oxidation state.
MnO4
2–
ion disproportionates only when the base
strength is below 0.1 M.
75. 1400 ml obtained by 1 g
22400 ml will be obtained by 16 g
i.e., CH4 so R is –CH
3
76. Because bond length inversely depends on number
of resonating structures.
79. In triplet carbene unpaired es
– are present in pure
p atomic orbitals.
80. CH – C – CH OH 23
CH3
CH3
involves 1, 2 methyl shift
82 Acidic nature of alcohols 1° > 2° > 3° so –OH at C2
acts as base and –OH at C5 as acid when H
2O is
lost (–OH of 3° and H of 2° alcohols are lost)
84.3
2
O
2 4Zn/H OOHC – (CH ) – CHO
85.
O
(as –Cl is better leaving group) then
is formed by Clemmensen reduction
86. As after giving H+ (1) & (3) will become antiaromatic
but (4) here become aromatic.
87.
N
N
H
H+
N
N
H
H
88. NBS is for allylic bromination.
89.2.19 4.25
pI 3.222
90. Chain growth polymerisation can occur through free
radical, cationic or anionic mechanism.
BIOLOGY
92. Saccharomyces is unicellular, no ascocarp formation.
94. Strawberry is an eaterio of achene.
95. Pisum-terminal leaflets as tendrillar.
96. The sporophyte is not free living rhizoids are haploid
and unicellular.
97. Runner-remains prostate, corm = storage.
98. Whorled phyllotaxy – Nerium, Alstonia.
100. (1) & (3) are correct.
101. At anaphase stage of mitosis, each sister chromatid
is called daughter chromosome.
102. Histone protein synthesis, centriole duplication
duplication of organelles.
105. Rhizobium – Unicellular heterotroph and
symbiotically associated to legumes.
All India Aakash Test Series for Medical-2018 Test - 13 (Code-A) (Hints)
6/6
106. CO2 release occurs in mitochondria.
107. Primary CO2 receptor :
PEP in mesophyll cells
RuBP in BSC
110 Ethylene induce female flower formation
111. Chlamydomonas like unicellular form may show
oogamy and only few algae show oogamy
112. Secretory or glandular tapetum secrete
sporopollenin, pollenkitt and compatibility proteins.
114. Turner’s syndrome – 44A + XO
Fused earlobes – Recessive trait
Window’s peak, morphan syndrome – Dominant
traits
116. Autosomal recessive trait.
117. a. In prokaryotes only (yeast is eukaryotic)
b. Do not code for any RNA or protein.
118. Termination sequence is present in eukaryotic
transcription unit.
121. Brandy = 70% Rum = 40%
Vodka = 80% Wine = 9-12%
122. i (inhibitor gene) – Constitutive expression.
123. Statin: Monascus purpureus.
124. Mycorrhizal association is to protect against root
borne pathogens. It absorbs and stores N, P, K and
Ca.
125. Male wasp carry out mating act with flower
Flower resemble female insect (wasp/bee)
126. COP-21 Paris climate summit.
127. FOAM - Friends of arcata marsh
128. Three mile Island of USA – 1979
Chernobyl incident – 1986
Fukushima daiichi – 2011
Nagasaki – 1945
130. Endangered animals : Asiatic Lion, Kashmiri stag,
Tiger, Musk Deer, Red Panda. (5)
131. Snow leopard –Khangchendzonga national park.
133. Water logging salt deposition on roots high soil
salinity
136. Fact
137. Tendons and ligaments are dense regular white
fibrous connective tissue in the form of cord.
138. (1 6) linkage is found only at the branching point
Right end of glycogen is reducing end.
139. Fact
140. Example of a lectin is concanavalin-A.
� � �
141. PEM is seen in marasmus.
142. Cytosine is a nitrogenous base while cytidine is a
nucleoside.
143. Myoglobin attains function at its tertiary structure.
145. BMR increases due to hyper-secretion of thyroxine.
148. Rhodopsin is responsible for scotopic vision. Cones
are present in fovea. Aqueous humor fills space
between cornea and lens.
150. Craniosacral outflow refers to parasympathetic
system and is responsible for lowering heart rate to
normal physiologic value.
152. Columns of Bertini are cortical renal columns
projections in medulla.
153. Roundworms have pseudocoelom.
154. Fertilization is internal.
155. Reptiles and birds are uricotelic in nature.
158. Some air remains in the conducting part of lungs and
does not take part in gaseous exchange. It is called
dead space air.
160. The AV valves remain linked to papillary muscles by
chordae tendineae.
161. Soon after ovulation, as the viability of ovum is ~24
hours after ovulation
162. AMF from Sertoli cells plays a crucial role.
163. d = Luteal phase is always of 14 days and day of
ovulation is always 14 days before the menstrual
phase that is 35 – 14 = 21st day.
164. Parental care and embryo protection are developed
in viviparous animals.
165. According to Hardy weinberg equilibrium
q2 = 16% = 0.16,
then q = 0.4
so p = 0.6 (as p + q = 1)
then 2pq = 2 × 0.4 × 0.6 = 0.48 48%
167. Fact
168. Fact
169. Fact
170. Fact
171. They are applicable to bacterial diseases as it is
cultured in vitro
172. Interferons are secreted from virus infected cells.
173. It is commonly called Bhang, Ganja, marijuana,
charas, hashish.
174. Fact
176. Fertilized eggs recovered usually non-surgically and
Hisardale was developed in Punjab.
Test - 13 (Code-B) (Answers) All India Aakash Test Series for Medical-2018
1/7
1. (4)
2. (2)
3. (1)
4. (3)
5. (2)
6. (1)
7. (2)
8. (4)
9. (2)
10. (2)
11. (4)
12. (2)
13. (2)
14. (3)
15. (1)
16. (1)
17. (2)
18. (3)
19. (3)
20. (2)
21. (4)
22. (3)
23. (1)
24. (3)
25. (2)
26. (2)
27. (2)
28. (2)
29. (3)
30. (3)
31. (1)
32. (2)
33. (1)
34. (3)
35. (1)
36. (3)
Test Date : 29/04/2018
TEST - 13 (Code B)
All India Aakash Test Series for Medical - 2018
ANSWERS
37. (3)
38. (2)
39. (4)
40. (1)
41. (3)
42. (4)
43. (4)
44. (2)
45. (2)
46. (4)
47. (4)
48. (2)
49. (2)
50. (4)
51. (3)
52. (2)
53. (3)
54. (1)
55. (1)
56. (2)
57. (4)
58. (1)
59. (4)
60. (1)
61. (4)
62. (2)
63. (2)
64. (1)
65. (1)
66. (2)
67. (1)
68. (1)
69. (3)
70. (3)
71. (2)
72. (2)
73. (2)
74. (3)
75. (2)
76. (1)
77. (2)
78. (1)
79. (2)
80. (2)
81. (1)
82. (2)
83. (3)
84. (3)
85. (2)
86. (3)
87. (1)
88. (2)
89. (2)
90. (3)
91. (2)
92. (4)
93. (4)
94. (4)
95. (2)
96. (4)
97. (2)
98. (3)
99. (4)
100. (2)
101. (2)
102. (4)
103. (2)
104. (1)
105. (4)
106. (1)
107. (4)
108. (3)
109. (4)
110. (2)
111. (3)
112. (2)
113. (2)
114. (3)
115. (2)
116. (3)
117. (4)
118. (2)
119. (1)
120. (4)
121. (3)
122. (4)
123. (2)
124. (1)
125. (3)
126. (4)
127. (3)
128. (3)
129. (3)
130. (2)
131. (2)
132. (1)
133. (4)
134. (4)
135. (2)
136. (2)
137. (2)
138. (2)
139. (1)
140. (4)
141. (1)
142. (3)
143. (3)
144. (4)
145. (2)
146. (1)
147. (2)
148. (4)
149. (2)
150. (4)
151. (4)
152. (4)
153. (2)
154. (2)
155. (3)
156. (3)
157. (3)
158. (1)
159. (4)
160. (4)
161. (3)
162. (3)
163. (1)
164. (1)
165. (1)
166. (3)
167. (4)
168. (3)
169. (2)
170. (2)
171. (3)
172. (2)
173. (2)
174. (2)
175. (1)
176. (2)
177. (3)
178. (2)
179. (1)
180. (4)
All India Aakash Test Series for Medical-2018 Test - 13 (Code-B) (Hints)
2/6
PHYSICS
1.
(1, 2)
y x = 2
x
y
Electric field lines are
perpendicular to equipotential surface
m1 × m
2 = –1
2.2
kqE
r
( – 1)k n qV
r
2
( – 1)( – 1)
V k n qn r
rE
kq
r
3.
E2
E1
d
E1 = E
2
1 2
k dE
R
2
Q
R
1 32
kQdE
R
4. Qi = 60 C
Qf = kQ
i
180 = k(60) k = 3
5. Slope 1
R
R T
6. P = I2R
7.2
qM L
m
( )2
qmvr
m
1
2qvr
HINTS
8.2 2
x yv v v
2 2
0 02
xv v v
2 2 2
0 04
xv v v
03
xv v
x x xv u a t
03 0
qEv t
m
03mv
tqE
9. Power = rate of heat generation = F.v
Q = Fv
QF
v
11. emf induced is zero.
12. E = h
x >
m >
R
13.tan
tancos
14.
1
2sin
n
a
⎛ ⎞ ⎜ ⎟⎝ ⎠ for maxima
1
1 13 2 6500 Å
2 2
⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
a a
1
56500 4642.8 Å
7
15.
22 1
1
Rf
u
⎛ ⎞⎜ ⎟
⎝ ⎠
17. A = A0e–t
4
0
0
1
93
AA A
⎛ ⎞ ⎜ ⎟⎝ ⎠
18.
n = 5
n = 1
Test - 13 (Code-B) (Hints) All India Aakash Test Series for Medical-2018
3/6
Photon
P mv= hP
h
mv h
vm
19.3
1
nf
n
log log 3lognf C n
C is constant.
20. Kmax
= E – Electron reaching at collector have maximum energy
= Kmax
+ eV
21. Intensity 2
1
r
Photocurrent Intensity
24.v
f
apparent
–
⎛ ⎞ ⎜ ⎟⎝ ⎠
v vf f
v v
fapparent
= 2f3
vv
25.1
2 2
v Tf
L L
26. E x2
x E
27. dQ = msdT
2
3
1
Q aT dT ∫ = 15
4
a
29.1
2PV = constant
30. Heating of glass surface of bulb is due to radiation.
32. Velocity become constant after a particular time
33. W = B
3 3 34 4( – ) 1
3 3R r g R g
W
Br
R
3 3
3
– 1R r
R
3
1 –11–
r
R
⎛ ⎞ ⎜ ⎟⎝ ⎠ 34. Energy conservation
2 2 1 2
1 1 2 2
1 10 –
2 2
Gmmmv m v
d
m1v
1 = m
2v
2
Relative velocity = v1 + v
2
OR
2 1 2
red red
1
2
Gm mm v
d
1 2
red
2 ( )G m mv
d
⇒
35. At height, 2
1
gg
h
R
⎛ ⎞⎜ ⎟⎝ ⎠
4
gg
2
1 1
41
h
R
⎛ ⎞⎜ ⎟⎝ ⎠
h = R
37. 30°
x
l/4
l/4
3cos30
4 8
l lx
38. P = F.v F v
P v2
39. Only in case of uniform circular motion resultant
force is towards the centre. t
a w� �
40. Fnet
= ma
Tcos – f = ma
41. Block will remain in rest after removing F1.
42. Using equation, 21
2s ut at w.r.t. balloon.
srel
= 0, urel
= v1 + v
2, a
rel = –g
All India Aakash Test Series for Medical-2018 Test - 13 (Code-B) (Hints)
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2
1 2
10 ( )
2v v t gt
t = 0 and 1 2
2( )v v
g
43.
u2 u
1
v2 v
1
H
1 1ˆ ˆ
v u i gtj �
2 2ˆ ˆ
v u i gtj �
20v v⇒
� �
1 2 1 20v v v v ⇒
� � � �
t = 1 2
u u
g
21
2H gt 1 2
2
u uH
g⇒
44. ˆ ˆ( ) , ,A v
v ai b ct j v a u b ct �
At t = ta, v
v = 0
ta =
b
c
Time of flight T = 2ta =
2b
c
Range = uHT =
2ab
c
45. 2 –1dv
a tdt
Acceleration is positive in interval 1
12
t and
velocity is negative
CHEMISTRY
46. Chain growth polymerisation can occur through free
radical, cationic or anionic mechanism.
47.2.19 4.25
pI 3.222
48. NBS is for allylic bromination.
49.
N
N
H
H+
N
N
H
H
50. As after giving H+ (1) & (3) will become antiaromatic
but (4) here become aromatic.
51.
O
(as –Cl is better leaving group) then
is formed by Clemmensen reduction
52.3
2
O
2 4Zn/H OOHC – (CH ) – CHO
54 Acidic nature of alcohols 1° > 2° > 3° so –OH at C2
acts as base and –OH at C5 as acid when H
2O is
lost (–OH of 3° and H of 2° alcohols are lost)
56. CH – C – CH OH 23
CH3
CH3
involves 1, 2 methyl shift
57. In triplet carbene unpaired es
– are present in pure
p atomic orbitals.
60. Because bond length inversely depends on number
of resonating structures.
61. 1400 ml obtained by 1 g
22400 ml will be obtained by 16 g
i.e., CH4 so R is –CH
3
62. As CO2 has carbon in +4 oxidation state.
MnO4
2–
ion disproportionates only when the base
strength is below 0.1 M.
64 As H2SO
5 and H
2S
2O
8 have peroxide linkages
O
S=OO OH
HO
O
S=O
O
OHS – O
O
OH
O
65. Structure of diborane
H
H
H
H
B B
H
H
66.4
226 222288 86
Ra He Rn
68. Fe+1 is present in [Fe(H2O)
5NO]SO
4
Test - 13 (Code-B) (Hints) All India Aakash Test Series for Medical-2018
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70. CuFeS2
(copper pyrites)
71. Mn2+ acts as autocatalyst for the reaction.
72. Slow step is rate determining step
So Rate = Keq
[NOBr2][NO]
2eq
2
[NOBr ]K
[NO][Br ]= (From 1st step)
and 2 eq 2[NOBr ] K [NO][Br ]=
and rate law expression is:
Rate = Keq
K1[NO]2[Br
2]
73. W = ZQ × Current efficiency
2H+ +
O
O
+ 2e–
OH
OH
76. Tb = i × K
b × m
i = 1 + (n–1)77. According to Faraday’s first law
W = Zit
120 111.2 5 60 60
x 96500
x = 1.99 2
So tin is present as Sn2+
78.T A B A B
P P (P – P )X = 100 + (250) X
B
So,A
P 100 torr
B AP – P 250 torr
BP 250 100 350 torr
79.2 2
1 1 2 2C C
2 20.001
m
0 0
m m
40.1 0.001
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠
0.1 × 16 = 0.001 × 0.001 2
m( )
0.001
m = 40 S cm2 mol–1
80. [Cu(H2O)
4]SO
4.H
2O has ionic, covalent, coordinate
and hydrogen bonds.
81. In N(SiH3)3 p – d back bonds are formed.
83. van der Waals equation at high pressure,
mPV Pb
1RT Rt
84. 2 moles of Na2HPO
3 will have equal number of P
atoms as in 1 mole of Ca3(PO
4)2.
85. Because limiting reagent decides the amount of
product formed.
87.
2
2 2 2
1 1 1 n – 4R – R
2 n 4n
⎡ ⎤⎡ ⎤ ⎢ ⎥⎢ ⎥ ⎣ ⎦ ⎣ ⎦2
2
4 n
R n – 4
⎡ ⎤ ⎢ ⎥
⎣ ⎦
2
2
nk
n – 4 so
4k
R
88. Intensive properties are independent of total amount
of substance.
89.2 2 3
N (g) 3H (g) 2NH (g) ���⇀↽���
Initial 1 3 0
Eq. 1– 0.5 3 – 1.5 1
3 3NH NH
PP x P
3
90. C –3
[B] [C] 0.15 0.05K 7.5
[A] 1 10
G° = –2.303 RT log KC
= –2.303 × 8.314 × 298 log7.5
= –4993 J mol–1
= –4.993 kJ mol–1
BIOLOGY
93. Water logging salt deposition on roots high soil
salinity
95. Snow leopard –Khangchendzonga national park.
96. Endangered animals : Asiatic Lion, Kashmiri stag,
Tiger, Musk Deer, Red Panda. (5)
98. Three mile Island of USA – 1979
Chernobyl incident – 1986
Fukushima daiichi – 2011
Nagasaki – 1945
99. FOAM - Friends of arcata marsh
100. COP-21 Paris climate summit.
101. Male wasp carry out mating act with flower
Flower resemble female insect (wasp/bee)
102. Mycorrhizal association is to protect against root
borne pathogens. It absorbs and stores N, P, K
and Ca.
All India Aakash Test Series for Medical-2018 Test - 13 (Code-B) (Hints)
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� � �
103. Statin: Monascus purpureus.
104. i (inhibitor gene) – Constitutive expression.
105. Brandy = 70% Rum = 40%
Vodka = 80% Wine = 9-12%
108. Termination sequence is present in eukaryotic
transcription unit.
109. a. In prokaryotes only (yeast is eukaryotic)
b. Do not code for any RNA or protein.
110. Autosomal recessive trait.
112. Turner’s syndrome – 44A + XO
Fused earlobes – Recessive trait
Window’s peak, morphan syndrome – Dominant
traits
114. Secretory or glandular tapetum secrete
sporopollenin, pollenkitt and compatibility proteins.
115. Chlamydomonas like unicellular form may show
oogamy and only few algae show oogamy
116 Ethylene induce female flower formation
119. Primary CO2 receptor :
PEP in mesophyll cells
RuBP in BSC
120. CO2 release occurs in mitochondria.
121. Rhizobium – Unicellular heterotroph and
symbiotically associated to legumes.
124. Histone protein synthesis, centriole duplication
duplication of organelles.
125. At anaphase stage of mitosis, each sister chromatid
is called daughter chromosome.
126. (1) & (3) are correct.
128. Whorled phyllotaxy – Nerium, Alstonia.
129. Runner-remains prostate, corm = storage.
130. The sporophyte is not free living rhizoids are haploid
and unicellular.
131. Pisum-terminal leaflets as tendrillar.
132. Strawberry is an eaterio of achene.
134. Saccharomyces is unicellular, no ascocarp formation.
140. Fertilized eggs recovered usually non-surgically and
Hisardale was developed in Punjab.
142. Fact
143. It is commonly called Bhang, Ganja, marijuana,
charas, hashish.
144. Interferons are secreted from virus infected cells.
145. They are applicable to bacterial diseases as it is
cultured in vitro
146. Fact
147. Fact
148. Fact
149. Fact
151. According to Hardy weinberg equilibrium
q2 = 16% = 0.16,
then q = 0.4
so p = 0.6 (as p + q = 1)
then 2pq = 2 × 0.4 × 0.6 = 0.48 48%
152. Parental care and embryo protection are developed
in viviparous animals.
153. d = Luteal phase is always of 14 days and day of
ovulation is always 14 days before the menstrual
phase that is 35 – 14 = 21st day.
154. AMF from Sertoli cells plays a crucial role.
155. Soon after ovulation, as the viability of ovum is ~24
hours after ovulation
156. The AV valves remain linked to papillary muscles by
chordae tendineae.
158. Some air remains in the conducting part of lungs and
does not take part in gaseous exchange. It is called
dead space air.
161. Reptiles and birds are uricotelic in nature.
162. Fertilization is internal.
163. Roundworms have pseudocoelom.
164. Columns of Bertini are cortical renal columns
projections in medulla.
166. Craniosacral outflow refers to parasympathetic
system and is responsible for lowering heart rate to
normal physiologic value.
168. Rhodopsin is responsible for scotopic vision. Cones
are present in fovea. Aqueous humor fills space
between cornea and lens.
171. BMR increases due to hyper-secretion of thyroxine.
173. Myoglobin attains function at its tertiary structure.
174. Cytosine is a nitrogenous base while cytidine is a
nucleoside.
175. PEM is seen in marasmus.
176. Example of a lectin is concanavalin-A.
177. Fact
178. (1 6) linkage is found only at the branching point
Right end of glycogen is reducing end.
179. Tendons and ligaments are dense regular white
fibrous connective tissue in the form of cord.
180. Fact
