Test - 1 (Answers & Hints )
All India Aakash Test Series for Medical-2013
T ES ES T – 1
ANSWERS 1.
(2)
41.
(3)
81.
(4)
121. (2)
161. (1)
2. 3.
(1) (3)
42. 43.
(4) (4)
82. 83.
(2) (3)
122. (4)
162. (3) 163. (4)
4. 5. 6.
(3) (2) (1)
44. 45. 46.
(1) (4) (3)
84. 85. 86.
(4) (3) (2)
124. (4)
7. 8.
(1) (1)
47. 48.
(1) (2)
87. 88.
(4) (2)
127. (1) 128. (3)
167. (4) 168. (1)
9.
(1)
49.
(2)
89.
(2)
129. (3)
169. (4)
10. 11.
(1) (3)
50. 51.
(4) (2)
90. 91.
(3) (4)
130. (4)
170. (3) 171. (2)
12. 13.
(3) (1)
52. 53.
(2) (3)
92. 93.
(3) (4)
14. 15.
(1) (3)
54. 55.
(4) (2)
94. 95.
(3) (4)
134. (4)
16. 17. 18.
(2) (3) (4)
56. 57. 58.
(2) (4) (3)
96. 97. 98.
(2) (2) (1)
136. (3)
19. 20.
(4) (4)
59. 60.
(1) (3)
99. (3) 100. (3)
139. (3)
21. 22.
(2) (2)
61. 62.
(2) (2)
101. (3) 102. (2)
141. (4)
23.
(2)
63.
(3)
103. (3)
24. 25.
(1) (4)
64. 65.
(4) (4)
104. (4)
143. (2) 144. (3)
105. (3)
145. (3)
26. 27. 28.
(1) (4) (1)
66. 67. 68.
(1) (3) (3)
106. (1)
146 (1) 147. (3)
29. 30.
(3) (2)
69. 70.
(3) (1)
109. (4)
31. 32.
(1) (1)
71. 72.
(2) (2)
111. (3)
33. 34.
(4) (2)
73. 74.
(4) (1)
113. (4) 114. (4)
35.
(4)
75.
(3)
115. (3)
36.
(4)
76.
(2)
116. (1)
37.
(4)
77.
(2)
117. (3)
38. 39. 40.
(4) (4) (2)
78. 79. 80.
(4) (2) (4)
118. (4)
107. (3) 108. (2) 110. (4) 112. (3)
119. (4) 120. (3)
123. (2) 125. (4) 126. (3)
131. (2) 132. (4) 133. (3) 135. (1) 137. (3) 138. (4) 140. (4) 142. (3)
148. (1) 149. (4) 150. (4) 151. (4)
164. (1) 165. (1) 166. (2)
172. (1) 173. (4) 174. (3) 175. (3) 176. (4) 177. (3) 178. (1) 179. (4) 180. (2) 181. (3) 182. (4) 183. (3) 184. (1) 185. (2) 186. (3) 187. (4) 188. (3) 189. (1) 190. (4) 191. (2)
152. (3) 153. (1)
192. (3) 193. (4)
154. (1)
194. (2)
155. (4) 156. (4)
195. (2)
157. (1) 158. (3) 159. (2) 160. (4)
196. (3) 197. (1) 198. (2) 199. (1) 200. (1) 1/9
All India Aakash Test Series for Medical-2013
Test - 1 (Answers & Hints )
Hints to Selected Questions [PHYSICS] 1.
Answer (2) Entropy =
2.
7. Q M1L2 T −2 = = [M1L2 T −2K −1 ] T K
ΔKE 8.
(5.00 ± 0.05) R.D = (5.00 ± 0.05) − (4.00 ± 0.05)
⇒
9.
ΔRD 0.05 0.1 = + 5.00
10.
Answer (3)
−1
3
⎛ 1 g ⎞ ⎛ 1 cm ⎞ ⎛ 1 s ⎞ Now n 2 = 6.67 × 10 ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 1 kg ⎠ ⎝ 1 m ⎠ ⎝ 1 s ⎠
11.
2
L
a 1 F2 F1 / ez F1 M = = = 2 12 2 2 a2 a 1ez a1e z e z
Answer (3)
−2
−8
1 2 S '− S 1 × 100 where S = at 2 and S ' = at ' 2 S 2
ΔS % =
and t ' = 1.3t
= 6.67×10 –11. Answer (2)
= L−1
F 1
Answer (3) Dimensional formula of G is = M –1L3T –2
L
Answer (1)
M 2 =
i.e. convert 1 J/s into cgs.
12.
Answer (3)
2 1 1 = + v av v1 v 2
Δm
−1 −1 = v1−1 + v 2−1, d [2v av 2v av ] = d [v1−1 ] + d [v2− 1 ]
Δm 0.1 0.1 = +
2
6.
=
M 1 =
60 J 60 J per minute = 60sec = 1 J/s
5.
Answer (1) dx
1 1 = + 5.00 100 10
4.
Answer (1)
∫ x 2
1.00
ΔRD 3.
Δv av 2 v av
=
Δv1 v12
m
+
1
Δv 2 v 22
=
ΔF F 1
+
Here, Δv1 = 0.5, Δv 2 = 0.1,v 1 = 15,v 2 = 30,v av = 20
14.
Answer (1)
Answer (1)
15.
Answer (3)
d θ π cos d θ × 100 is minimum = = × 100 π sin θ tan θ
when tan θ is maximum
⇒ θ = 90º 2/9
a
Δm = 0.2 Answer (1)
dm × 100 is minimum m
Δa
1
13.
=
2ΔR × 100 = 4% R
L d 2 x x = 2 dimensionally = 2 2 T dt t
5.00 ± 0.05 1.00 ± 0.1
RD
× 100 =
KE
Answer (1)
R.D =
Answer (1)
y 5y 0 4y 0
y 0 x 0 2x 0
5x 0
x
Test - 1 (Answers & Hints )
16.
All India Aakash Test Series for Medical-2013
Answer (2)
22.
Answer (2)
L.C. = 1MSD –1VSD
23.
Answer (2)
where 20 VSD = 19 MSD
1
⎡1 kg ⎤ n 2 = 12 ⎢ ⎥ ⎣5 g⎦
19 MSD 20
1 VSD =
⎡ 1m ⎤ ⎢ 10 cm ⎥ ⎣ ⎦ 1
17.
2
2
19 1 MSD = MSD ⇒ L.C.= 1 MSD – 20 20
⎡1000 g⎤ ⎡100cm ⎤ = 12 ⎢ ⎥ ⎢ ⎥ ⎣ 5 g ⎦ ⎣ 10cm ⎦
Answer (3)
= 12 × 200 × 100 = 24 ×10 4 24.
4
4
Area =
4
∫3 ydx = ∫ 3
⎛ x 3 ⎞ x dx = ⎜ ⎟ ⎝ 3 ⎠3
Answer (1)
2
K ∝ V a Ab F c solve for a, b, c
= 18.
3
3
4 3 64 27 37 − = − = 3 3 3 3 3
25.
Answer (4)
26.
Answer (1)
Answer (4)
dy = 6 x dx
y = x 2 Inx dy dx 2 2 d In In x x x = + dx dx dx
= 6 × 3 = 18 27.
⎡ 1⎤ = x 2 ⎢ ⎥ + 2 x Inx ⎣ x ⎦
Answer (4) 3 3 3(1 − 0.002 × 3) = = 3 3 (3 + 0.006) [3(1 + 0.002)] 33
= x [1 +2 Inx ] 19.
Answer (4)
=
(4.9 × 105 ) − (0.35 × 105 ) = 4.55 × 105 20.
Answer (4)
[D] =
28.
Answer (1)
29.
Answer (3) Compare given equation with
1 = T −1 T
( x − x1 )2
30.
[D] T −1 L 1 −1 = −1 = = LT [B] L T
Δ
π
=
0.1 s
31.
= 0.2 cm
Now,
Δg g
× 100 =
Δ
× 100 + 2
ΔT
0.2 cm 0.1 = × 100 + 2 × × 100 200 cm 50 =
1 + 0.4 = 0.5 10
Answer (2) Mean value
Answer (2) Δt =
( y − y1 ) 2
+ =1 a2 b 2 Where x 1 and y1 represent centre.
1 [B] = = L−1 L
21.
1 − 0.006 9
T
∫ 0 sinθd θ = − [cos θ]0π = − [cos π − cosθ] . π−0
π−0
π−0
Answer (1) y = mx + C
× 100
Here intercept C is negative, and slope is positive. 32.
Answer (1) d dy (ωt − kx ) = cos(ωt − kx ) dx dx dy d = cos(ωt − kx ) (ωt − kx ) dt dt
3/9
All India Aakash Test Series for Medical-2013
Test - 1 (Answers & Hints )
33.
Answer (4)
43.
Answer (4)
34.
Answer (2)
44.
Answer (1)
35.
Answer (4)
We cannot add physical quantites of different
log1018 = log10 2 × 3 × 3 = log102 + 2log103 36.
Answer (4) 45.
m ∝ paF bt c
−1
2
P
1
ΔP
=1 37.
P
Answer (4) 1
⎡ 1 ⎤ = 100 ⎢ ⎥ ⎣1000 ⎦
1
1
⎡1 cm ⎤ ⎡ 1 s ⎤ ⎢ 1 m ⎥ ⎢ 60 s ⎥ ⎣ ⎦ ⎣ ⎦
−2
46.
P
1
⎡ 1 ⎤ ⎢100 ⎥ [ 60 × 60 ] ⎣ ⎦
39.
Answer (4)
x
+
× 100 =
Δy y
Δx x
× 100 +
Δy y
× 100
× 100 +
ΔQ × 100 Q
= 1+2 = 3%
3600 = 3.6 1000
Answer (4)
Δx
Answer (3)
ΔP
47.
38.
=
= 2+1 = 3%
⎡ 1g ⎤ n 2 = 100 ⎢ ⎥ ⎣1 kg ⎦
=
Answer (4)
ΔP
by solving a = –1, b = 2, c = 1
⎡ 1W ⎤ ⎡ 1 N ⎤ ⎡ 1 s ⎤ ⇒ m 2 = 1⎢ ⎥ ⎢ ⎥ ⎢ −3 ⎥ ⎣1000 W ⎦ ⎣ 1000N ⎦ ⎣ 10 s ⎦
x × y = z is not possible x + y
dimensions
Answer (1) Mean absolute value =
=
1 m3 = (100 cm)3 = 106 cm3
1.008 + 1.010 + 1.012 3
3.030 = 1.010 3
40.
Answer (2)
Absolute error in 1st measurement = 1.010–1.008
41.
Answer (3)
= 0.002 Absolute error in 2nd measurement = 1.010–1.010
a = M0L0 T0 t
=0 Absolute error in 3rd measurement = 1.010–1.012
a = M0L0T1
= –0.02
a F = b
Mean absolute error
1 1 −2
⇒
MLT
⇒
b = =
M0L0 T1 = b
=
M0L0 T1
% error=
1 1 −2
MLT
=
Answer (4) Most accurate is the least percentage error
0.1 0.1 0.1 0.1 × 100, × 100, × 100, × 100 23.1 25.1 27.1 29.1 4/9
mean absolute error × 100 mean absolute value
M –1L –1T3
ab = [ M 0 L 0 T 1 ][M –1 L –1 T 3 ] = M –1 L –1 T 4 42.
0.002 + 0 + .002 0.004 = = 0.00133 3 3
0.00133 × 100 1.010
48.
Answer (2)
49.
Answer (2)
50.
Answer (4)
Test - 1 (Answers & Hints )
All India Aakash Test Series for Medical-2013
[CHEMISTRY] 51. Answer (2)
65. Answer (4)
1 mol CH4 = 1 mole C atom = 4 mol of H atom 52. Answer (2) Fact
2H2 + O2 ⎯⎯→ 2H2O(l) 2 vol 1 vol Given 26 ml 24 ml Here H2 is the limiting reagent.
53. Answer (3)
Volume of O2 left = 24 – 13 = 11 ml
Fact
66. Answer (1)
54. Answer (4)
67. Answer (3)
C60H122 = 842 g 6 × 1023 molecules 1 molecule
⎯→
⎯→
Mol. wt. = 2 × V.D. = 2 × 130 = 260
842 g
842 6 × 10
23
68. Answer (3)
= 1.4 × 10−21g
55. Answer (2)
Initial moles of O2 = 5 No. of moles of O2 reacted = 5 – 2 = 3 moles Here C3H8 is limiting reagent
Fact
5 mole O2 reacts with 1 mole C3H8
56. Answer (2) 78.9 × 1× 100 ≈ 2.6 × 104 Mol. mass = 0.30 57. Answer (4) Equal volume contains equal number of molecules
1
∴ 3 mole O2 reacts with 5 × 3 mole C3H8 =
3 mole 5
69. Answer (3)
58. Answer (3) 18 g H2O
⎯→
72 g H2O
⎯→
16 g oxygen At. mass = 16 × 72 = 64 g 18
70. Answer (1)
59. Answer (1) Al2 (SO4 )3
3+
2Al ⎯⎯→
3SO + 24−
60. Answer (3)
Eq. mass =
%=
61. Answer (2)
1 mol 2 mol
25 O2 ⎯⎯ → 8CO2 + 9H2O 2 8 mol 16 mol
62. Answer (2)
15 × 100 = 20% 15 + 60
72. Answer (2) M=
49 1 1 × = = 0.25 M 98 2 4
Fact
73. Answer (4)
63. Answer (3)
Fact
d=
At. mass 56 = = 18.66 valency 3
71. Answer (2)
Fact
C8H18 +
6.4 6.4 = = 206 Sp. heat 0.031
m 26 g = = 1.08 g/ml V 24 ml
74. Answer (1)
2NaHCO3 + 2HCl ⎯⎯→ 2NaCl + 2CO 2 + 2H 2O
64. Answer (4)
5 C2H2 + O2 ⎯⎯→ 2CO2 + H2O 2 1V
2.5 V
No. of moles of HCl reacted = 0.2 ×
20 4 = 1000 1000
= 4 × 10 –3 mol 5/9
All India Aakash Test Series for Medical-2013
2 mol HCl evolves 4×
10 –3
mol
⎯→
⎯→
= 2 × 22.4 L CO 2
2 × 22.4 × 4 × 10 −3 L = 2 = 89.6 mL
1 mol NaHCO3 = 3 mol O atoms = 48 g
86. Answer (2) Fact Fact Chalk is a compound so ultimate particle will be molecule
Fact 77. Answer (2)
89. Answer (2)
Eq. mass of acid =
Mol. mass Basicity
28.9 =
Basicity of H3PO3 = 2, H3BO3 = 1 78. Answer (4) Mass No. of moles = No. of g atoms = At. mass
79. Answer (2)
14 × N × 100 194
No. of nitrogen atoms =
1 O2 2
2 g H2 = 1 mole = 6 × 1023 molecules
91. Answer (4)
Fact
0.5 × 10 =
1 × V2 1000
93. Answer (4)
V2 = 0.5 × 10 × 1000 = 5000
Vol. of water added = 5000 – 10 = 4990
4 NH3 + 5O2
⎯⎯→ 4NO + 6H2O
4 × 17 g NH3
⎯⎯→ 5 mol O2
6.8 g NH3
81. Answer (4) MgO CO ⎯⎯→ + 2
Mass of pure MgCO3 = 1000 × 84 g MgCO3
⎯→
750 g MgCO3
⎯→
Ozone = O3 75 = 750 g 100
95. Answer (4) N=
22.4 L 22.4 × 750 L = 200 L 84
4.9 1000 × 1.5 × = 1.5 49 100
96. Answer (2) Fact
82. Answer (2)
97. Answer (2)
Fact
Fact
83. Answer (3)
98. Answer (1)
1 11.2 L = mol = 3 × 1023 molecules 2 84. Answer (4)
6/9
5 6.8 × 4 × 17
⎯⎯→
94. Answer (3)
22.4 L
84 g
1 mol
Mol. mass 2
92. Answer (3)
N1V1 = N2 V2
Cl – =
2 = 0.025 mol 80
Vapour density =
80. Answer (4)
28.9 × 194 =4 14 × 100
90. Answer (3)
2 g SO3 =
F2 + 2KOH ⎯→ 2KF + H2O +
MgCO3
Fact
88. Answer (2)
76. Answer (2)
⇒
85. Answer (3)
87. Answer (4)
75. Answer (3)
⇒
Test - 1 (Answers & Hints )
Fact 99. Answer (3) Fact 100. Answer (3)
8 mol valence electrons
Fact
=0.5 mol
Test - 1 (Answers & Hints )
All India Aakash Test Series for Medical-2013
[BIOLOGY] 101.
Answer (3) Protista– Amoeba , Paramoecium Euglena , Gonyaulax
120.
Rhizopus –Zygomycetes gametangial copulation. 121.
Monera – Nostos , Anabaena, Plantae – Spirogyra 102.
Answer (2)
103.
Answer (3) Fungi – Albugo , Rhizopus .
104.
Answer (4) Present in gut of ruminants.
105.
Answer (3) Chlorophyll a
106.
Answer (1)
107.
Answer (3)
Answer (2) Mycoplasma are pathogenic both to plants and animals, flagella absent.
109.
122.
Answer (4)
123.
Answer (2) Produce basidiospore. Diakaryophase long lived.
124.
Answer (4)
125.
Answer (4)
126.
Answer (3) Plant viruses
127.
Answer (1)
128.
Answer (3) Viroids, infectious to plants only.
129.
110.
Answer (4)
111.
Answer (3)
112.
Answer (3)
130.
Answer (4)
114.
Answer (4)
131.
115.
Answer (3)
116.
Answer (1) Euglena –Sexual reproduction absent.
117.
Answer (3)
132.
Answer (4) Sac and club fungi
119.
Answer (4) Penicillium
Answer (4) Both (1) & (3) Pair of contrasting characters-couplet. Each statement-lead.
133.
Answer (3) At higher level number of common characters decreases.
134.
Answer (4) Species
135.
Answer (1)
136.
Answer (3) Descending or ascending arrangement of categories is called hierarchy.
Oospore–Zygote (Oomycetes) 118.
Answer (2) Manual–Useful for identification of name of species. Monograph one taxon.
Short-Smooth Long-Tinsel
Answer (4) Bacteriophage, B-collar
Dispersed by air and formed under unfavourable conditions. 113.
Answer (3) Endomycorrhiza
Answer (4) Ginkgo, Alnus
Answer (2) Drosophila of plant kingdom
Binary fission A–cell wall, B-cell membrane 108.
Answer (3)
137.
Answer (3)
138.
Answer (4) Classification taxonomy is based on easily observable characters. 7/9
All India Aakash Test Series for Medical-2013
139.
Answer (3)
140.
Answer (4)
141.
Answer (4)
155. Answer (4)
Name of the author appears after specific epithet. 142.
Answer (2)
Answer (3) Order–Poales
145.
In class ophiuroidea arms are present but ambulacral groove is absent. In crinoidea arms have ambulacral groove. 158. Answer (3)
Answer (3) Preserved plants and animals. Insects–Insect boxes
In phylum platyhelminthes the class turbellaria includes free living flat worms like Planaria . Only tapeworms absorb nutrients of the host directly through the body surface. 159. Answer (2)
146 Answer (1) For sessile advantageous, from all sides. animals on the
Crustaceans have two pairs of antennae. In class arachnida antennae are absent and they have four pairs of walking legs arising from cephalothorax. 157. Answer (1)
Growth and reproduction involves cell divisions. 144.
Coelomocytes are amoeboid corpuscles that help in phagocytosis. They also engulf the excretory waste and carry them to the skin gills for removal in echinoderms. 156. Answer (4)
Answer (3) Cellular organisation is unexceptionally present in all living organisms.
143.
Test - 1 (Answers & Hints )
animals, radial symmetry is as it allows the food to be gathered Bilateral symmetry arose when the ocean floor became mobile.
147. Answer (3) Contraction of the circular muscles increases the pressure of the coelomic fluid in the segments. This increased pressure elongates the animal and pushes the anterior end forwards. 148. Answer (1) 149. Answer (4) In molluscs external skeleton is shell made up of calcium carbonate. 150. Answer (4) In phylum arthropoda circulatory system is open. 151. Answer (4) Hormiphora , Ctenoplana , Beroe belongs to phylum ctenophora but Porpita belongs to phylum coelenterata. 152. Answer (3) Arthropods are unisexual i.e., dioecious.
Trypanosoma gambiense is endoparasite, blood parasite, but it is extracellular parasite. It is present in C.S.F. of brain. 160. Answer (4) 161. Answer (1) Spongilla is fresh water sponge but Euspongia , Chalina , Proterion all are present in marine water. 162. Answer (3) 163. Answer (4) The animal illustrated in the figure D is scorpion. Scorpions and spiders respire by book lungs. 164. Answer (1) Land snails do not have gills, they respire by pulmonary sac. Clam is a bivalve so radula is absent. 165. Answer (1) In class arachnida antennae are absent. 166. Answer (2) The animal illustrated in the figure 2 is Aurelia . In Aurelia metagenesis is absent. 167. Answer (4) 168. Answer (1)
153. Answer (1)
169. Answer (4)
154. Answer (1)
170. Answer (3)
Free swimming dipleura larva is common ancestral larva of phylum, echinodermata, hemichordata and chordata. 8/9
171. Answer (2) Digestion is sponges is intracellular, it occur in choanocytes and phagocytes.
Test - 1 (Answers & Hints )
All India Aakash Test Series for Medical-2013
172. Answer (1)
186. Answer (3) In Aschelminthes, only longitudinal muscles are present.
173. Answer (4) 174. Answer (3)
In Aschelminthes the cuticle has keratin not chitin.
Schistostoma is the blood fluke.
187. Answer (4)
175. Answer (3)
188. Answer (3) In hemichordates, sexes are separate
Shell → is absent in Octopus . Radula is absent in bivalves
189. Answer (1) In earthworms, the circulatory system is closed.
176. Answer (4)
190. Answer (4)
In the life of Plasmodium , the formation of gametes occurs in the intestine,stomach of mosquito
Annelids have bilateral symmetry
177. Answer (3) 178. Answer (1)
191.
Answer (2)
192.
Answer (3) Phylum is second highest category of animals.
In platyhelminthes there is pseudometameric segmentation.
193.
Answer (4) Conidio present septat, branchedmycelian.
179. Answer (4)
194.
180. Answer (2)
Answer (2) Producer decomposes protists.
(a) is Pila and (b) is Octopus . In Octopus shell is absent. Octopus has closed circulatory system.
195.
Answer (2) Simple structure, complex behaviour.
181. Answer (3)
196. Answer (3)
182. Answer (4)
Eggs of Ascaris require more oxygen, more moisture low temperature to start the development. These conditions are provided outside the body of host in soil.
183. Answer (3) 184. Answer (1)
197. Answer (1)
Cestum is a ctenophore
198. Answer (2)
185. Answer (2)
199. Answer (1)
The larva of Taenia is cysticercus larva or bladder worm.
200. Answer (1)
9/9
