ANSWERS · All India Aakash Test Series for Medical-2017 Test - 5 (Code E) (Answers & Hints) 2/9 Hints to Selected Questions [PHYSICS] 1. Answer (2) Charge distribution would be unsymmetrical
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Test - 5 (Code E) (Answers & Hints) All India Aakash Test Series for Medical-2017
1/9
1. (2)
2. (4)
3. (2)
4. (4)
5. (1)
6. (4)
7. (2)
8. (2)
9. (1)
10. (1)
11 (4)
12. (1)
13. (2)
14. (3)
15. (3)
16. (4)
17. (4)
18. (2)
19. (2)
20. (2)
21. (3)
22. (1)
23. (4)
24. (1)
25. (4)
26. (3)
27. (2)
28. (1)
29. (2)
30. (1)
31. (3)
32. (3)
33. (1)
34. (3)
35. (3)
36. (3)
ANSWERS
TEST - 5 (Code-E)
All India Aakash Test Series for Medical-2017
Test Date : 19-02-2017
37. (3)
38. (1)
39. (2)
40. (2)
41. (3)
42. (1)
43. (2)
44. (3)
45. (1)
46. (1)
47. (3)
48. (2)
49. (4)
50. (1)
51. (4)
52. (2)
53. (1)
54. (2)
55. (4)
56. (4)
57. (2)
58. (2)
59. (2)
60. (4)
61. (3)
62. (3)
63. (1)
64. (3)
65. (3)
66. (1)
67. (2)
68. (4)
69. (2)
70. (1)
71. (1)
72. (3)
73. (3)
74. (3)
75. (1)
76. (4)
77. (2)
78. (1)
79. (3)
80. (3)
81. (4)
82. (4)
83. (2)
84. (1)
85. (2)
86. (1)
87. (2)
88. (3)
89. (3)
90. (3)
91. (2)
92. (2)
93. (3)
94. (1)
95. (4)
96. (2)
97. (3)
98. (2)
99. (4)
100. (3)
101. (1)
102. (2)
103. (3)
104. (2)
105. (3)
106. (1)
107. (2)
108. (3)
109. (1)
110. (4)
111. (4)
112. (3)
113. (2)
114. (2)
115. (2)
116. (3)
117. (1)
118. (2)
119. (2)
120. (2)
121. (1)
122. (3)
123. (2)
124. (1)
125. (2)
126. (3)
127. (1)
128. (3)
129. (1)
130. (1)
131. (4)
132. (4)
133. (2)
134. (2)
135. (1)
136. (2)
137. (2)
138. (1)
139. (3)
140. (3)
141. (4)
142. (2)
143. (2)
144. (4)
145. (4)
146 (2)
147. (3)
148. (2)
149. (3)
150. (2)
151. (4)
152. (2)
153. (2)
154. (3)
155. (4)
156. (3)
157. (3)
158. (4)
159. (2)
160. (3)
161. (4)
162. (1)
163. (3)
164. (3)
165. (3)
166. (1)
167. (4)
168. (4)
169. (3)
170. (3)
171. (2)
172. (1)
173. (3)
174. (1)
175. (2)
176. (1)
177. (3)
178. (2)
179. (4)
180. (3)
All India Aakash Test Series for Medical-2017 Test - 5 (Code E) (Answers & Hints)
2/9
Hints to Selected Questions
[ PHYSICS]
1. Answer (2)
Charge distribution would be unsymmetrical inside
cavity but symmetry on outer surface of cavity.
Potential will be
0
kq
r.
O
++
+
+
++
+
+
+
+
q
2. Answer (4)
kqV
r
(27 )
k qV
r,
3 3( ) 27r r
3. Answer (2)
enc
0
q
4. Answer (4)
Fq
⇓
E
P
Fdipole
5. Answer (1)
E
Equipotential surface
6. Answer (4)
Work done = PE = qV
Now, VA = V
B = V
C = V
D = V
E
Hence V is zero in all cases
Work done = 0
7. Answer (2)
E�
will change but | |E�
doesn't change also
potential doesn't change.
8. Answer (2)
When key is closed, to make the potential of both
the sphere same, all charge of inner sphere flows to
the outer sphere.
9. Answer (1)
10. Answer (1)
The equivalent circuit will be
CC
C C
C
C CA B
eq
77 F
8
CC
11 Answer (4)
We require 16 such branches in parallel
Hence total number of capacitors required = 32
8 F 8 F 8 F 8 F1000 V
eq
82 F
4C
12. Answer (1)
eq
1 2 3 4–
1 1 1 1
1 1 1 1
1 1 1 1
E
eq
41V
4E
eq
1 1 1 1 1
1 1 1 1r
eq
1
4r
13. Answer (2)
R l
Two parts would be parallel to each other.
1 2eq
1 2
R RR
R R
, eq
8
9R
1
2.3
R
RA
,
2
4.3
R
RA
Test - 5 (Code E) (Answers & Hints) All India Aakash Test Series for Medical-2017
3/9
14. Answer (3)
Under no deflection condition it will be balanced
Wheatstone bridge
80
20 80
R
R = 20
15. Answer (3)
No current will pass through resistance 3 and 6
Hence i1 = i
2
1
2
1i
i
16. Answer (4)
Rt = R
0(1 + t)
1 = R0(1 + 0.00125 × 27)
2 = R0(1 + 0.00125 × t)
1 1 0.00125 27
2 1 0.00125 t
1 + 0.00125 t = 2(1 + 0.00125 × 27)
t = 854°C
1127 K
17. Answer (4)
200 k
100 k
a
b
30 k
Equivalent resistance between a and ground
200 30100 k 126 k
230
⎛ ⎞ ⎜ ⎟⎝ ⎠
Current = 300
mA = 2.38 mA126
Now reading of voltmeter = 200 30 300
230 126
= 62 volt
18. Answer (2)
Average speed is nonzero.
19. Answer (2)
Equilibrium is unstable if Q and q are of opposite
polarity and Q is displaced along the line joining two
q charges.
20. Answer (2)
+Q
Flux of field by symmetry = 0
4
Q
21. Answer (3)
a
a
a
Electric field = 0 by symmetry, all fields will cancel
each other
Distance of charge from centre of cube = 3
2
a
Potential due to each
0
34
2
qV
a
Net potential = 8 V
0
4
3
q
a
22. Answer (1)
The given figure is equivalent to two identical
capacitors in parallel.
Hence, 02 A
Cd
All India Aakash Test Series for Medical-2017 Test - 5 (Code E) (Answers & Hints)
4/9
23. Answer (4)
2 x
VE
x
3 y
VE
y
4 z
VE
z
ˆ ˆ ˆ–2 – 3 4E i j k �
24. Answer (1)
old2
LT
g
new
22 2 2
– –2
L L LT
qE g gg g
m
Ratio, old
new
1
2
T
T ,
new
old
2 :1T
T
25. Answer (4)
.E A � �
Here E A� �
= 0
26. Answer (3)
2.5 101
10 2
⎛ ⎞ ⎜ ⎟⎝ ⎠
L
R L
2.5 101
10 2
⎛ ⎞ ⎜ ⎟⎝ ⎠x
R L
27. Answer (2)
Energy of system =
2
0
64
q
a
Since there will be 6 pair interactions.
28. Answer (1)
2 sin2
k
Er
2
q
r
29. Answer (2)
Consider potential of point P due to complete sphere
be V1 then
2
1
2
3V k R
Now consider potential that would have been created
by removed position
2
29
k RV
Hence potential of point P = V1 – V
2
2 2
22 5–
3 9 9
k R k Rk R
2
0
5
36
R
30. Answer (1)
Let q be the charge on inner shell upon earthing
Vinner
= 0
0
10
4 3
q q
r r
⎡ ⎤ ⎢ ⎥ ⎣ ⎦
–3
qq
i.e., 3
q will flow from inner shell to the earth.
31. Answer (3)
i
3
10 V 4 V
10 – 42 A
3i
Power dissipated across resistor = i2 R = 12 W
32. Answer (3)
From graph,
i = 2t
Heat =
3 3
2 2
0 0
4 . 5 180 Ji Rdt t dt ∫ ∫
Test - 5 (Code E) (Answers & Hints) All India Aakash Test Series for Medical-2017
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33. Answer (1)
q2
q1
q4q
3
rWhen broughtto one point
r
( + + + )q q q q1 2 3 4
1 2 3 4
centre
( )k q q q qV
r
1 2 3 4
centre
( )k q q q qV
r
Hence change in potential = zero
34. Answer (3)
Only option (3) shows correct variation.
35. Answer (3)
The figure can be reduced as
A
B
C3
C
Both capacitors are in parallel.
Hence, eq
4
3 3
C CC C
36. Answer (3)
2
1
(120)P
R
2
2
(110)P
R
% drop =
2 2
1 2
21
– 120 –110100 16%
120
P P
P
37. Answer (3)
Current through both the sections will be same.
Heat energy per unit volume
2i Rt
V
Now, 2
dx
R A
V A dx A
2
2
. .i tH
A
2
1H
A
Now,
2 21 1
2 4
A B
B A
H A
H A
⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠
38. Answer (1)
Q
q
q q
2
3kqQ
a2
2
kq
a2
2
kq
a
For equilibrium,
2
2 2
33
kqQ kq
a a
3
qQ
But Q should be of opposite polarity from q
Hence, –3
qQ
39. Answer (2)
qE
6rVT
6rVT = qE
6T
qEV
r
40. Answer (2)
Across PQ
R/3
R/2
R
P
RQ
5
11PQ
RR
41. Answer (3)
lR
A
l
AR
All India Aakash Test Series for Medical-2017 Test - 5 (Code E) (Answers & Hints)
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[ CHEMISTRY]
42. Answer (1)
.E A � �
ˆ ˆ ˆ(5 2 ). (1 ) 5i j i
43. Answer (2)
18– 2 2 – 6 – 3 5
3A B
V V
VA – V
B = 19 volt
44. Answer (3)
When switch 1 is connected
+q –q
q CE =
When switch 2 is connected and after reaching
steady state
–q +q
Work done by battery = 2qE = 2CE2
Heat generated = 2CE2
45. Answer (1)
In steady state condition no current will pass through
3 ohm resistor and current in the circuit will be 1 A
through 2 .
Potential difference across 10 F is 2 V
Hence charge = 20 C
46. Answer (1)
Cresol is used as froath stabilizer.
47. Answer (3)
Sphalerite is ZnS.
48. Answer (2)
SnO2 has magnetic impurities.
49. Answer (4)
Pb is more reactive than the other three.
50. Answer (1)
When Fe2O
3 impurity is in excess, Bayer's process
is used.
51. Answer (4)
All will decompose because these are carbonates
and hydroxides ore.
52. Answer (2)
2O
2A
1HgS HgO Hg O
2
53. Answer (1)
2Na[Ag(CN)2] + Zn 2Ag + Na
2[Zn(CN)
4]
54. Answer (2)
According to definition.
55. Answer (4)
Al powder can reduce all these.
56. Answer (4)
Fact.
57. Answer (2)
2 2
1Ag O 2Ag O
2
58. Answer (2)
Spongy iron forms in fusion zone and then converted
to pig iron.
59. Answer (2)
Carbon (graphite rods) are used as anode.
60. Answer (4)
Mond’s process.
61. Answer (3)
Hg is not transition elements.
62. Answer (3)
Fact.
63. Answer (1)
64. Answer (3)
65. Answer (3)
Ru shows +8.
66. Answer (1)
Mn2+ has d5 system (half filled d-orbital).
67. Answer (2)
Sc3+ has 18e– [Ne] 3s2 3p6
Test - 5 (Code E) (Answers & Hints) All India Aakash Test Series for Medical-2017
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68. Answer (4)
Mn2O
7 has highest oxidation state.
69. Answer (2)
Zn2+ has d10 system.
70. Answer (1)
2
4 4 2 23MnO 4H 2MnO MnO 2H O
1 mole MnO4
2– = 2
3 mole MnO
4
–, 1
3 mole MnO
2
71. Answer (1)
4 2 4 2 22KMnO K MnO MnO O (g)
72. Answer (3)
HNO3 is good oxidizing and HCl is reducing agent.
73. Answer (3)
o
redE (KMnO
4) is maximum in neutral medium.
o 7 4
redE (Mn , Mn ) 1.69 V
74. Answer (3)
f 0 , f 7 and f 14 are colourless.
75. Answer (1)
76. Answer (4)
COO–
COO–
(ox), CH2
COO–
NH2
(gly)
77. Answer (2)
It has different ligands.
78. Answer (1)
EAN = 28 + 0 + 2 × 4 = 36
79. Answer (3)
(en) has, di, in its name.
80. Answer (3)
In K4[Fe(CN)
6], Fe does not exist as free Fe2+ ion in
anion sphere of [Fe(CN)6]4–.
81. Answer (4)
Fact.
82. Answer (4)
All does not show optical isomerism.
83. Answer (2)
It has d3 system.
84. Answer (1)
For low spin complex, splitting energy is large
(0 > P) compare to pairing energy.
Orbital splitting energy in octahedral coordination
entities is more than tetrahedral entities for same
metal ion and ligand.
85. Answer (2)
According to CFT.
86. Answer (1)
0 t
9
4 is correct
87. Answer (2)
4d has 50% more CFSE than 3d.
88. Answer (3)
Pb
C H2 5
C H2 5
C H2 5 C H
2 5
89. Answer (3)
It has 6 32g gt e configuration and unpaired electron
in eg.
90. Answer (3)
[ BIOLOGY
]
91. Answer (2)
Dioscorea, XO
92. Answer (2)
93. Answer (3)
It is an intersex.
94. Answer (1)
95. Answer (4)
(a) 0%
(b) Two
96. Answer (2)
X-body was discovered by Henking.
97. Answer (3)
All India Aakash Test Series for Medical-2017 Test - 5 (Code E) (Answers & Hints)
8/9
98. Answer (2)
2 2 425%
4 4 16
99. Answer (4)
100. Answer (3)
101. Answer (1)
It can be expressed in females in homozygous
condition.
102. Answer (2)
103. Answer (3)
104. Answer (2)
GAG is replaced by GTG.
105. Answer (3)
106. Answer (1)
107. Answer (2)
108. Answer (3)
Tall plants = 12/16
Plants with long sized starch grains = 4/16
109. Answer (1)
Phenotypes = 4, Genotypes = 8.
110. Answer (4)
111. Answer (4)
112. Answer (3)
113. Answer (2)
Algae show external fertilization.
114. Answer (2)
115. Answer (2)
116. Answer (3)
Agave produces fleshy buds called bulbils.
117. Answer (1)
118. Answer (2)
119. Answer (2)
Rice, bamboo, radish.
120. Answer (2)
121. Answer (1)
122. Answer (3)
Monocliny involves formation of bisexual flowers.
123. Answer (2)
-cellulose bands are present in endothecium.
124. Answer (1)
125. Answer (2)
126. Answer (3)
127. Answer (1)
Castor, maize, rice.
128. Answer (3)
129. Answer (1)
130. Answer (1)
131. Answer (4)
132. Answer (4)
133. Answer (2)
Abiotic pollinator in Zea mays.
134. Answer (2)
135. Answer (1)
136. Answer (2)
137. Answer (2)
Tunica albuginea is situated under tunica vaginalis.
138. Answer (1)
139. Answer (3)
140. Answer (3)
Number of chromosomes in gamete of dog = 39
141. Answer (4)
In haploid organism gametes are formed through
mitosis.
142. Answer (2)
143. Answer (2)
Fraternal twins or dizygotic twins are developed from
two different ova.
144. Answer (4)
Test - 5 (Code E) (Answers & Hints) All India Aakash Test Series for Medical-2017
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145. Answer (4)
Fructose is present in secretion of seminal vesicles.
146 Answer (2)
147. Answer (3)
Ovulation will take place on 40 – 14 = 26th day.
148. Answer (2)
Unejaculated sperms are reabsorbed in epididymis
and vas deferens.
149. Answer (3)
150. Answer (2)
hCG keeps corpus luteum active. Relaxin softens
pubic symphysis. Progesterone maintains cervical
mucus plug.
151. Answer (4)
152. Answer (2)
153. Answer (2)
Human placenta is deciduous, haemochorial,
chorionic and metadiscoidal.
154. Answer (3)
155. Answer (4)
156. Answer (3)
157. Answer (3)
Adrenal medulla and pineal gland are ectodermal.
Lining of urinary bladder is endodermal.
158. Answer (4)
159. Answer (2)
Human Immunodeficiency Virus causes AIDS.
160. Answer (3)
The incidence of STDs are reported to be very high
among persons in the age group of 15-24 years.
Genital warts is caused by Human Papilloma Virus
(HPV).
161. Answer (4)
162. Answer (1)
163. Answer (3)
164. Answer (3)
165. Answer (3)
166. Answer (1)
167. Answer (4)
Lactational amenorrhoea = 6 months.
168. Answer (4)
169. Answer (3)
170. Answer (3)
In vasectomy the two vasa deferentia are interrupted
by giving cuts and the ends tied.
171. Answer (2)
ICSI (Intra cytoplasmic sperm injection) involves in
vitro fertilisation.
172. Answer (1)
Trichomoniasis is caused by Trichomonas vaginalis
a flagellate protozoan.
173. Answer (3)
AIH is artificial insemination husband.
174. Answer (1)
Gossypol inhibits spermatogenesis.
175. Answer (2)
Teratozoospermia – Defective sperm morphology.
176. Answer (1)
VDRL is antibody detection.
177. Answer (3)
178. Answer (2)
179. Answer (4)
180. Answer (3)
Genital herps is caused by a virus.
� � �
Test - 5 (Code F) (Answers & Hints) All India Aakash Test Series for Medical-2017
1/9
1. (3)
2. (1)
3. (4)
4. (3)
5. (1)
6. (4)
7. (4)
8. (3)
9. (1)
10. (1)
11. (1)
12. (1)
13. (3)
14. (1)
15. (1)
16. (3)
17. (3)
18. (4)
19. (4)
20. (1)
21. (2)
22. (3)
23. (2)
24. (3)
25. (1)
26. (4)
27. (4)
28. (4)
29. (2)
30. (2)
31. (1)
32. (1)
33. (4)
34. (3)
35. (2)
36. (3)
ANSWERS
TEST - 5 (Code-F)
All India Aakash Test Series for Medical-2017
Test Date : 19-02-2017
37. (1)
38. (4)
39. (4)
40. (2)
41. (3)
42. (2)
43. (4)
44. (2)
45. (4)
46. (1)
47. (1)
48. (3)
49. (4)
50. (3)
51. (4)
52. (3)
53. (4)
54. (4)
55. (2)
56. (1)
57. (1)
58. (3)
59. (4)
60. (4)
61. (3)
62. (1)
63. (3)
64. (3)
65. (3)
66. (3)
67. (4)
68. (2)
69. (4)
70. (3)
71. (1)
72. (1)
73. (3)
74. (1)
75. (1)
76. (2)
77. (4)
78. (4)
79. (4)
80. (2)
81. (4)
82. (2)
83. (3)
84. (4)
85. (2)
86. (3)
87. (2)
88. (4)
89. (1)
90. (3)
91. (3)
92. (4)
93. (4)
94. (2)
95. (2)
96. (3)
97. (3)
98. (1)
99. (3)
100. (3)
101. (4)
102. (3)
103. (4)
104. (1)
105. (3)
106. (4)
107. (4)
108. (4)
109. (3)
110. (1)
111. (4)
112. (4)
113. (4)
114. (1)
115. (2)
116. (4)
117. (3)
118. (1)
119. (4)
120. (1)
121. (1)
122. (4)
123. (1)
124. (4)
125. (3)
126. (1)
127. (2)
128. (4)
129. (1)
130. (4)
131. (2)
132. (3)
133. (1)
134. (4)
135. (4)
136. (1)
137. (2)
138. (4)
139. (1)
140. (3)
141. (4)
142. (3)
143. (1)
144. (3)
145. (2)
146. (3)
147. (1)
148. (2)
149. (2)
150. (3)
151. (1)
152. (1)
153. (1)
154. (3)
155. (4)
156. (1)
157. (4)
158. (2)
159. (1)
160. (1)
161. (2)
162. (1)
163. (4)
164. (4)
165. (2)
166. (4)
167. (3)
168. (4)
169. (1)
170. (4)
171. (2)
172. (2)
173. (4)
174. (4)
175. (2)
176. (1)
177. (1)
178. (3)
179. (4)
180. (4)
All India Aakash Test Series for Medical-2017 Test - 5 (Code F) (Answers & Hints)
2/9
Hints to Selected Questions
[ PHYSICS]
1. Answer (3)
In steady state condition no current will pass through
3 ohm resistor and current in the circuit will be 1 A
through 2 .
Potential difference across 10 F is 2 V
Hence charge = 20 C
2. Answer (1)
When switch 1 is connected
+q –q
q CE =
When switch 2 is connected and after reaching
steady state
–q +q
Work done by battery = 2qE = 2CE2
Heat generated = 2CE2
3. Answer (4)
18– 2 2 – 6 – 3 5
3A B
V V
VA – V
B = 19 volt
4. Answer (3)
.E A � �
ˆ ˆ ˆ(5 2 ). (1 ) 5i j i
5. Answer (1)
lR
A
l
AR
6. Answer (4)
Across PQ
R/3
R/2
R
P
RQ
5
11PQ
RR
7. Answer (4)
qE
6rVT
6rVT = qE
6T
qEV
r
8. Answer (3)
Q
q
q q
2
3kqQ
a2
2
kq
a2
2
kq
a
For equilibrium
2
2 2
33
kqQ kq
a a
3
qQ
But Q should be of opposite polarity from q
Hence, –3
qQ
Test - 5 (Code F) (Answers & Hints) All India Aakash Test Series for Medical-2017
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9. Answer (1)
Current through both the sections will be same.
Heat energy per unit volume
2i Rt
V
Now, 2
dx
R A
V A dx A
2
2
. .i tH
A
2
1H
A
Now,
2 21 1
2 4
A B
B A
H A
H A
⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠
10. Answer (1)
2
1
(120)P
R
2
2
(110)P
R
% drop =
2 2
1 2
21
– 120 –110100 16%
120
P P
P
11. Answer (1)
The figure can be reduced as
A
B
C3
C
Both capacitors are in parallel.
Hence, eq
4
3 3
C CC C
12. Answer (1)
Only option (1) shows correct variation.
13. Answer (3)
q2
q1
q4q
3
rWhen broughtto one point
r
( + + + )q q q q1 2 3 4
1 2 3 4
centre
( )k q q q qV
r
1 2 3 4
centre
( )k q q q qV
r
Hence change in potential = zero
14. Answer (1)
From graph,
i = 2t
Heat =
3 3
2 2
0 0
4 . 5 180 Ji Rdt t dt ∫ ∫
15. Answer (1)
i
3
10 V 4 V
10 – 42 A
3i
Power dissipated across resistor = i2 R = 12 W
16. Answer (3)
Let q be the charge on inner shell upon earthing
Vinner
= 0
0
10
4 3
q q
r r
⎡ ⎤ ⎢ ⎥ ⎣ ⎦
–3
qq
i.e., 3
q will flow from inner shell to the earth.
17. Answer (3)
2 sin2
k
Er
2
q
r
18. Answer (4)
Consider potential of point P due to complete sphere
be V1 then
2
1
2
3V k R
Now consider potential that would have been created
by removed position
2
29
k RV
All India Aakash Test Series for Medical-2017 Test - 5 (Code F) (Answers & Hints)
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Hence potential of point
P = V1 – V
2
2 2
22 5–
3 9 9
k R k Rk R
2
0
5
36
R
19. Answer (4)
Energy of system =
2
0
64
q
a
Since there will be 6 pair interactions.
20. Answer (1)
2.5 101
10 2
⎛ ⎞ ⎜ ⎟⎝ ⎠
L
R L
2.5 101
10 2
⎛ ⎞ ⎜ ⎟⎝ ⎠x
R L
21. Answer (2)
.E A � �
Here, E A� �
= 0
22. Answer (3)
old2
LT
g
new
22 2 2
– –2
L L LT
qE g gg g
m
Ratio, old
new
1
2
T
T ,
new
old
2 :1T
T
23. Answer (2)
2 x
VE
x
3 y
VE
y
4 z
VE
z
ˆ ˆ ˆ–2 – 3 4E i j k �
24. Answer (3)
The given figure is equivalent to two identical
capacitors in parallel.
Hence, 02 A
Cd
25. Answer (1)
a
a
a
Electric field = 0 by symmetry, all fields will cancel
each other
Distance of charge from centre of cube = 3
2
a
Potential due to each
0
34
2
qV
a
Net potential = 8 V
0
4
3
q
a
26. Answer (4)
+Q
Flux of field by symmetry = 0
4
Q
27. Answer (4)
Equilibrium is unstable if Q and q are of opposite
polarity and Q is displaced along the line joining two
q charges.
28. Answer (4)
Average speed is nonzero.
Test - 5 (Code F) (Answers & Hints) All India Aakash Test Series for Medical-2017
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29. Answer (2)
200 k
100 k
a
b
30 k
Equivalent resistance between a and ground
200 30100 k 126 k
230
⎛ ⎞ ⎜ ⎟⎝ ⎠
Current = 300
mA = 2.38 mA126
Now reading of voltmeter = 200 30 300
230 126
= 62 volt
30. Answer (2)
Rt = R
0(1 + t)
1 = R0(1 + 0.00125 × 27)
2 = R0(1 + 0.00125 × t)
1 1 0.00125 27
2 1 0.00125 t
1 + 0.00125 t = 2(1 + 0.00125 × 27)
t = 854°C
1127 K
31. Answer (1)
No current will pass through resistance 3 and 6
Hence i1 = i
2
1
2
1i
i
32. Answer (1)
Under no deflection condition it will be balanced
Wheatstone bridge
80
20 80
R
R = 20
33. Answer (4)
R l
Two parts would be parallel to each other.
1 2eq
1 2
R RR
R R
, eq
8
9R
1
2.3
R
RA
,
2
4.3
R
RA
34. Answer (3)
eq
1 2 3 4–
1 1 1 1
1 1 1 1
1 1 1 1
E
eq
41V
4E
eq
1 1 1 1 1
1 1 1 1r
eq
1
4r
35. Answer (2)
We require 16 such branches in parallel
Hence total number of capacitors required = 32
8 F 8 F 8 F 8 F1000 V
eq
82 F
4C
36. Answer (3)
The equivalent circuit will be
CC
C C
C
C CA B
eq
77 F
8
CC
37. Answer (1)
38. Answer (4)
When key is closed, to make the potential of both
the sphere same, all charge of inner sphere flows to
the outer sphere.
39. Answer (4)
E�
will change but | |E�
doesn't change also
potential doesn't change.
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[ CHEMISTRY]
46. Answer (1)
47. Answer (1)
It has 6 32g gt e configuration and unpaired electron
in eg.
48. Answer (3)
Pb
C H2 5
C H2 5
C H2 5 C H
2 5
49. Answer (4)
4d has 50% more CFSE than 3d.
50. Answer (3)
0 t
9
4 is correct.
51. Answer (4)
According to CFT.
52. Answer (3)
For low spin complex, splitting energy is large
(0 > P) compare to pairing energy.
Orbital splitting energy in octahedral coordination
entities is more than tetrahedral entities for same
metal ion and ligand.
40. Answer (2)
Work done = PE = qV
Now, VA = V
B = V
C = V
D = V
E
Hence V is zero in all cases
Work done = 0
41. Answer (3)
E
Equipotential surface
42. Answer (2)
Fq
⇓
E
P
Fdipole
43. Answer (4)
enc
0
q
44. Answer (2)
kqV
r
(27 )
k qV
r,
3 3( ) 27r r
45. Answer (4)
Charge distribution would be unsymmetrical inside
cavity but symmetry on outer surface of cavity.
Potential will be 0
kq
r.
O
++
+
+
++
+
+
+
+
q
53. Answer (4)
It has d3 system.
54. Answer (4)
All does not show optical isomerism.
55. Answer (2)
Fact.
56. Answer (1)
In K4[Fe(CN)
6], Fe does not exist as free Fe2+ ion in
anion sphere of [Fe(CN)6]4–.
57. Answer (1)
(en) has, di, in its name.
58. Answer (3)
EAN = 28 + 0 + 2 × 4 = 36
59. Answer (4)
It has different ligands.
60. Answer (4)
COO–
COO–
(ox), CH2
COO–
NH2
(gly)
61. Answer (3)
62. Answer (1)
f 0 , f 7 and f 14 are colourless.
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[ BIOLOGY
]
91. Answer (3)
92. Answer (4)
93. Answer (4)
Abiotic pollinator in Zea mays.
94. Answer (2)
95. Answer (2)
96. Answer (3)
97. Answer (3)
98. Answer (1)
99. Answer (3)
Castor, maize, rice.
100. Answer (3)
63. Answer (3)
o
redE (KMnO
4) is maximum in neutral medium.
o 7 4
redE (Mn , Mn ) 1.69 V
64. Answer (3)
HNO3 is good oxidizing and HCl is reducing agent.
65. Answer (3)
4 2 4 2 22KMnO K MnO MnO O (g)
66. Answer (3)
2
4 4 2 23MnO 4H 2MnO MnO 2H O
1 mole MnO4
2– = 2
3 mole MnO
4
–, 1
3 mole MnO
2
67. Answer (4)
Zn2+ has d10 system.
68. Answer (2)
Mn2O
7 has highest oxidation state.
69. Answer (4)
Sc3+ has 18e– [Ne] 3s2 3p6
70. Answer (3)
Mn2+ has d5 system (half filled d-orbital).
71. Answer (1)
Ru shows +8.
72. Answer (1)
73. Answer (3)
74. Answer (1)
Fact.
75. Answer (1)
Hg is not transition elements.
76. Answer (2)
Mond’s process.
77. Answer (4)
Carbon (graphite rods) are used as anode.
78. Answer (4)
Spongy iron forms in fusion zone and then converted
to pig iron.
79. Answer (4)
2 2
1Ag O 2Ag O
2
80. Answer (2)
Fact.
81. Answer (4)
Al powder can reduce all these.
82. Answer (2)
According to definition.
83. Answer (3)
2Na[Ag(CN)2] + Zn 2Ag + Na
2[Zn(CN)
4]
84. Answer (4)
2O
2A
1HgS HgO Hg O
2
85. Answer (2)
All will decompose because these are carbonates
and hydroxides ore.
86. Answer (3)
When Fe2O
3 impurity is in excess, Bayer's process
is used.
87. Answer (2)
Pb is more reactive than the other three.
88. Answer (4)
SnO2 has magnetic impurities.
89. Answer (1)
Sphalerite is ZnS.
90. Answer (3)
Cresol is used as froath stabilizer.
All India Aakash Test Series for Medical-2017 Test - 5 (Code F) (Answers & Hints)
8/9
101. Answer (4)
102. Answer (3)
103. Answer (4)
-cellulose bands are present in endothecium.
104. Answer (1)
Monocliny involves formation of bisexual flowers.
105. Answer (3)
106. Answer (4)
107. Answer (4)
Rice, bamboo, radish.
108. Answer (4)
109. Answer (3)
110. Answer (1)
Agave produces fleshy buds called bulbils.
111. Answer (4)
112. Answer (4)
113. Answer (4)
Algae show external fertilization.
114. Answer (1)
115. Answer (2)
116. Answer (4)
117. Answer (3)
Phenotypes = 4, Genotypes = 8.
118. Answer (1)
Tall plants = 12/16
Plants with long sized starch grains = 4/16
119. Answer (4)
120. Answer (1)
121. Answer (1)
122. Answer (4)
GAG is replaced by GTG.
123. Answer (1)
124. Answer (4)
125. Answer (3)
It can be expressed in females in homozygous
condition.
126. Answer (1)
127. Answer (2)
128. Answer (4)
2 2 425%
4 4 16
129. Answer (1)
130. Answer (4)
X-body was discovered by Henking.
131. Answer (2)
(a) 0%
(b) Two
132. Answer (3)
133. Answer (1)
It is an intersex.
134. Answer (4)
135. Answer (4)
Dioscorea, XO
136. Answer (1)
Genital herps is caused by a virus.
137. Answer (2)
138. Answer (4)
139. Answer (1)
140. Answer (3)
VDRL is antibody detection.
141. Answer (4)
Teratozoospermia – Defective sperm morphology.
142. Answer (3)
Gossypol inhibits spermatogenesis.
143. Answer (1)
AIH is artificial insemination husband.
Test - 5 (Code F) (Answers & Hints) All India Aakash Test Series for Medical-2017
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144. Answer (3)
Trichomoniasis is caused by Trichomonas vaginalis
a flagellate protozoan.
145. Answer (2)
ICSI (Intra cytoplasmic sperm injection) involves in
vitro fertilisation.
146. Answer (3)
In vasectomy the two vasa deferentia are interrupted
by giving cuts and the ends tied.
147. Answer (1)
148. Answer (2)
149. Answer (2)
Lactational amenorrhoea = 6 months.
150. Answer (3)
151. Answer (1)
152. Answer (1)
153. Answer (1)
154. Answer (3)
155. Answer (4)
156. Answer (1)
The incidence of STDs are reported to be very high
among persons in the age group of 15-24 years.
Genital warts is caused by Human Papilloma Virus
(HPV).
157. Answer (4)
Human Immunodeficiency Virus causes AIDS.
158. Answer (2)
159. Answer (1)
Adrenal medulla and pineal gland are ectodermal.
Lining of urinary bladder is endodermal.
160. Answer (1)
161. Answer (2)
162. Answer (1)
� � �
163. Answer (4)
Human placenta is deciduous, haemochorial,
chorionic and metadiscoidal.
164. Answer (4)
165. Answer (2)
166. Answer (4)
hCG keeps corpus luteum active. Relaxin softens
pubic symphysis. Progesterone maintains cervical
mucus plug.
167. Answer (3)
168. Answer (4)
Unejaculated sperms are reabsorbed in epididymis
and vas deferens.
169. Answer (1)
Ovulation will take place on 40 – 14 = 26th day
170 Answer (4)
171. Answer (2)
Fructose is present in secretion of seminal vesicles.
172. Answer (2)
173. Answer (4)
Fraternal twins or dizygotic twins are developed from
two different ova.
174. Answer (4)
175. Answer (2)
In haploid organism gametes are formed through
mitosis.
176. Answer (1)
Number of chromosomes in gamete of dog = 39.
177. Answer (1)
178. Answer (3)
179. Answer (4)
Tunica albuginea is situated under tunica vaginalis.