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Test - 5 (Code A) (Answers & Hints) All India Aakash Test Series for Medical-2017
1/10
1. (3)
2. (4)
3. (1)
4. (3)
5. (2)
6. (1)
7. (2)
8. (3)
9. (3)
10. (2)
11 (1)
12. (4)
13. (2)
14. (3)
15. (3)
16. (2)
17. (4)
18. (3)
19. (2)
20. (2)
21. (4)
22. (1)
23. (3)
24. (3)
25. (2)
26. (2)
27. (3)
28. (1)
29. (3)
30. (3)
31. (4)
32. (1)
33. (2)
34. (4)
35. (2)
36. (2)
ANSWERS
TEST - 5 (Code-A)
All India Aakash Test Series for Medical-2017
Test Date : 29-01-2017
37. (3)
38. (1)
39. (1)
40. (2)
41. (4)
42. (1)
43. (1)
44. (4)
45. (1)
46. (2)
47. (2)
48. (3)
49. (3)
50. (1)
51. (2)
52. (3)
53. (4)
54. (2)
55. (4)
56. (4)
57. (4)
58. (4)
59. (4)
60. (2)
61. (4)
62. (3)
63. (1)
64. (3)
65. (3)
66. (2)
67. (2)
68. (2)
69. (3)
70. (4)
71. (3)
72. (1)
73. (4)
74. (3)
75. (2)
76. (2)
77. (1)
78. (2)
79. (4)
80. (3)
81. (3)
82. (1)
83. (4)
84. (2)
85. (1)
86. (1)
87. (1)
88. (4)
89. (4)
90. (2)
91. (4)
92. (2)
93. (1)
94. (1)
95. (3)
96. (3)
97. (3)
98. (2)
99. (1)
100. (3)
101. (2)
102. (4)
103. (4)
104. (4)
105. (4)
106. (1)
107. (4)
108. (2)
109. (2)
110. (3)
111. (2)
112. (3)
113. (1)
114. (1)
115. (2)
116. (1)
117. (1)
118. (3)
119. (2)
120. (4)
121. (2)
122. (1)
123. (4)
124. (3)
125. (1)
126. (4)
127. (3)
128. (3)
129. (4)
130. (2)
131. (4)
132. (3)
133. (1)
134. (1)
135. (4)
136. (1)
137. (4)
138. (4)
139. (2)
140. (3)
141. (4)
142. (1)
143. (4)
144. (3)
145. (2)
146 (3)
147. (3)
148. (3)
149. (4)
150. (2)
151. (2)
152. (4)
153. (2)
154. (2)
155. (3)
156. (1)
157. (1)
158. (3)
159. (4)
160. (3)
161. (2)
162. (2)
163. (4)
164. (2)
165. (4)
166. (3)
167. (4)
168. (4)
169. (1)
170. (2)
171. (2)
172. (3)
173. (4)
174. (3)
175. (2)
176. (1)
177. (4)
178. (1)
179. (1)
180. (2)
All India Aakash Test Series for Medical-2017 Test - 5 (Code A) (Answers & Hints)
2/10
Hints to Selected Questions
[ PHYSICS]
1. Answer (3)
qv0
q
r
v
By angular momentum conservations
024mv mvr ...(i)
By energy conservations
22 2
0
1 1
2 2
kqmv mv
r
2
2 2
0 0 20
1 1 24. .
2 2 4
qmv m v
rr
2 2
2 0 0 0
0 20
12 41
2 4
mv mv
mv
rr
2
1 12 1
2 rr
r2 – 2r – 24 = 0
r = 6 m
2. Answer (4)
++
+
+
+
+
–
–
–
–
c q2R
Vc = V
q + V
shell
Vshell
= 0
As total charge on shell = 0
3. Answer (1)
21
2U CV
10 20 10dU dV
CVdt dt
= 2000 J/s = 2 mJ/s
4. Answer (3)
High resistance would ensure more heat production
in short circuit condition, low melting point would
melt the wire fairly easy and thus breaking circuit.
5. Answer (2)
Let charge/volume =
3
0
a
3
0
4
3a
Dividing, 3
4
4
3
6. Answer (1)
2
1 1
1
EH R t
R r
⎡ ⎤ ⎢ ⎥⎣ ⎦
2
2 2
2
EH R t
R r
⎡ ⎤ ⎢ ⎥⎣ ⎦
Equating H1 = H
2
1 2r R R
7. Answer (2)
2
1– –
x
VE
x x
2
1– –
y
VE
y y
2
1– –
z
VE
z z
2 2 2
1 1 1ˆ ˆ ˆ– – –E i j k
x y z
�
at (1, 1, 1)
ˆ ˆ ˆ– – –E i j k
�
| | 3 N/CE �
Test - 5 (Code A) (Answers & Hints) All India Aakash Test Series for Medical-2017
3/10
8. Answer (3)
dqi
dt = Slope of charge-time graph
i = –5 A
H = i2 Rt
H t
9. Answer (3)
I = neAVd
1 2
2 (2 )d d
neAV ne A V
1
2
1:1d
d
V
V
10. Answer (2)
50 V
R30 V
10
20
I
I
1 2
Loop-1
30 – 50 + 20 I = 0
I = 1 ampere
Loop-2
50 – IR – 20I = 0
50 – R – 20 = 0
R = 30 11. Answer (1)
C1
C2
A
B
120°
Resistance per unit central arc angle
36 18
/ rad2
Resistance of � �
1 2
18 212
3
AC B AC B
eq
1 1 1 1 1
24 24 12 12R
Req
= 4
12. Answer (4)
A B
2 F
1 F
1 F
VC
VA
VC
VA
1 F
VCV
B
VAV
B
AB
2 F
1 F2 FC
⇒
2 1 82 F
3 3
AB
C
13. Answer (2)
t
q
dq
dt decreases with time
14. Answer (3)
2 2H i Rdt R i dt ∫ ∫H = R × Area under i2-t graph
Area = 1
(6 2) 4 162
H = 160 J
15. Answer (3)
Uniformly accelerated motion can have parabolic or
straight line trajectory .
16. Answer (2)
2QQ
q
Final potential difference between spheres will be E
(2 – ) ( )–
K Q q K Q qE
R R
0– 2
2
Qq ER
All India Aakash Test Series for Medical-2017 Test - 5 (Code A) (Answers & Hints)
4/10
17. Answer (4)
qE
T
mg
T cos = mg
T sin = qE
18. Answer (3)
Final charge on capacitor would be CE.
19. Answer (2)
( 3 3)3 3
2 3 3AB
R RR R
R
Solving R = 3
20. Answer (2)
10 V
2 k 3 k
6 k
Req
of 3 k and 6 k= 2
Potential drop across 2 kwill be 5 V and
across combinations of 3 kand 6 kwould also
be 5 V. Hence, reading = 5 V
21. Answer (4)
Value of E will determine direction of flow of current.
22. Answer (1)
–
g
VR G
I
–3
10– 20 9980
10R
23. Answer (3)
enc
0
q
, initially qenc
increases till l
tu
then
becomes constant and then decreases when rod
comes completely out of cube.
24. Answer (3)
All three capacitors are in parallel hence Ceq
= 9 C
25. Answer (2)
Potential first decreases to R, then remains constant
inside shell and then decreases to zero at infinity
26. Answer (2)
E
curved
= base
box
= ER2
27. Answer (3)
Charge distribution before closing the key
4 V
2 F 6 F
8 C 8 C+6 C +6 C
4 F 4 F+ +
Charge distribution after closing the switch
4 V
2 F 6 F
+10 C
+5 C +9 C
4 F 4 F
–5 C –9 C
–10 C
4 C
+6 C –6 C
Charge flown through the switch = 4 C
28. Answer (1)
No current will pass through the grounded resistor
Hence reading of ammeter will be zero.
29. Answer (3)
+10 0 0 2 –2 10
Test - 5 (Code A) (Answers & Hints) All India Aakash Test Series for Medical-2017
5/10
Final charge distribution on the plates will be
x = 10
y = 0
z = –2
x + y + z = 8 C
30. Answer (3)
r
q
kqV
r = 5 volt
when drops are merged, let radius be R
3 34 48
3 3r R
R = 2r
Hence, its potential will be
(8 ) (8 ) 4
2
k q k q kq
R r r = 20 V
Hence, difference = 15 V
31. Answer (4)
If charges are both positive or both negative PE
decreases if they are of opposite sign PE increases
32. Answer (1)
P
E
When PE is chosen to be zero when P E� �
PE = – PEcos
So at 3
–2
PEPE
and PE when dipole is parallel to the field is –PE.
Now when reference potential is chosen to be of
3
its PE becomes 0.
ie PE increases by 2
PE
hence PE in parallel position will be
– –2 2
PE PEPE
33. Answer (2)
10
20
i
1 2 4 t
Charge flown = area under i-t graph
= 35
34. Answer (4)
Terminal voltage = E – iR
Given E – 10 × R = 50
E – 1 × R = 60
Solving,10
9R
550V
9E
35. Answer (2)
Potential difference across 100 resistance should
be 5 V as voltmeter and 100 resistance are in
parallel
req
of voltmeter and 100 should be 50
r of voltmeter = 100
36. Answer (2)
R1
R2
3 V 2 V
In series current are same through both
3 = IR1
2 = IR2
All India Aakash Test Series for Medical-2017 Test - 5 (Code A) (Answers & Hints)
6/10
1
1 1 1 2
22 2 1
2
3
2
l
R A l A
lR l A
A
2
2
3 6.
2 1
B
A
r
r
2
2
12 :1
4 ⇒ B A
BA
r r
rr
37. Answer (3)
Let charge induced on sphere at any moment = q
Potential of sphere at any time = 0
0kq kq
x R
–
qRq
x
2
dq qR dxi
dt dtx
⎛ ⎞ ⎜ ⎟⎝ ⎠
2
qRVi
x
38. Answer (1)
Just after closing the switch, branch containing only
capacitor will make the circuit short, hence E
iR
39. Answer (1)
Time period of SHM is not affected by constant force
hence will remain unchanged.
40. Answer (2)
l
Work done by electric force = PE
qEl sin = mgl(1 – cos)
2
2sin cos 2sin.
2 2 2qEl mgl
tan2
qE
mg
⎛ ⎞ ⎜ ⎟⎝ ⎠
1
2tanqE
mg
⎡ ⎤ ⎢ ⎥⎣ ⎦
41. Answer (4)
1
CR
42. Answer (1)
Only charge and electric field remain unchanged
43. Answer (1)
r
dr
di = J. 2r dr
0
0
. 2
R
di J r r dr ∫ ∫
3
02
3
J Ri
44. Answer (4)
45. Answer (1)
qi
t
i = 2, t = 1
q = 2 C
Charge of 1e– = +1.602 × 10–19C in magnitude
In 1 C, number of electron
– 18
–19
16.25 10
1.602 10e
2 C charge will have 12.5 × 1018 = 1.25 × 1019
Test - 5 (Code A) (Answers & Hints) All India Aakash Test Series for Medical-2017
7/10
[ CHEMISTRY]
46. Answer (2)
Siderite is FeCO3
47. Answer (2)
Fact
48. Answer (3)
ZnS + 4NaCN Na2S + Na
2[Zn(CN)
4]
49. Answer (3)
When Fe2O
3 is in excess, Bayer's process is
followed.
50. Answer (1)
Complex formed is Na[Ag(CN)2]
51. Answer (2)
In cynide process NaCN is used.
52. Answer (3)
Dolomite is CaCO3. MgCO
3
53. Answer (4)
All three are correct.
54. Answer (2)
Aluminium is highly reactive.
55. Answer (4)
Fact
56. Answer (4)
Fe2O
3, MnO
2 and P
4O
10 are reduced to Fe, Mn and
P respectively.
57. Answer (4)
In pig iron carbon is 4 g, in cast iron it is 3 g
(%) decrease = 4 – 3
100 25%4
58. Answer (4)
Fact
59. Answer (4)
60. Answer (2)
It is semiconductor
61. Answer (4)
Ag2+ has d9 and Ag3+ has d8 system.
62. Answer (3)
63. Answer (1)
Ni according to the trend
64. Answer (3)
Sc2+, Zn3+ are not possible.
65. Answer (3)
+7 is the highest oxidation number of Mn (Group
VII)
66. Answer (2)
2Cu /Cu
E 0.34 V
67. Answer (2)
SO2 + V
2O
5 SO
3 + V
2O
4
2 4 2 2 5
1V O O V O
2
68. Answer (2)
Ni2+ and Pt4+ are stable
69. Answer (3)
Fact
70. Answer (4)
pH 42– 2–2 7 4Cr O CrO
�����⇀↽�����
71. Answer (3)
Fact
72. Answer (1)
In CeO2, cerium has noble gas configuration
73. Answer (4)
They do not react with CaCO3
74. Answer (3)
75. Answer (2)
Fact
76. Answer (2)
ICH NH
2 2
CH2
ICH
2
CH NH2 2
NH (dien)
77. Answer (1)
According to the rules
All India Aakash Test Series for Medical-2017 Test - 5 (Code A) (Answers & Hints)
8/10
78. Answer (2)
EAN = 28 – 2 + 2 × 4 = 26 + 8 = 34
79. Answer (4)
Fact
80. Answer (3)
All surrounding atoms are same
81. Answer (3)
Fe3+ has an unpair electron
82. Answer (1)
Cr3+ : xx xx
3d
xx
4s
xx xx xx
4p
83. Answer (4)
[PtCl4]2– is
Cl
Cl
PtCl
Cl
2–
dsp2
84. Answer (2)
Only (a) and (b) are correct
85. Answer (1)
Fact
86. Answer (1)
Fact
87. Answer (1)
Cl– is weak ligand
88. Answer (4)
All behave as acid ligand
89. Answer (4)
Mn+ is electron deficient hence strength of Mn–C
bond will be less
90. Answer (2)
It has 6 42g gt e configuration.
[ BIOLOGY
]
91. Answer (4)
Conidia is a feature of ascomycetes and
deutomycetes.
92. Answer (2)
93. Answer (1)
Orange and mango which are polycarpic.
94. Answer (1)
95. Answer (3)
Embryo stage is absent in algae.
96. Answer (3)
97. Answer (3)
98. Answer (2)
Foliar buds are present on leaves.
99. Answer (1)
100. Answer (3)
Zoidogamous in Funaria, Chara, Marchantia and
Adiantum.
101. Answer (2)
Embryo sac represents female gametophyte.
102. Answer (4)
A – It is the epidermis which is protective in function.
103. Answer (4)
All are correct.
104. Answer (4)
105. Answer (4)
106. Answer (1)
107. Answer (4)
Has an unwettable stigma.
108. Answer (2)
109. Answer (2)
110. Answer (3)
Chemotropic growth of pollen tube.
111. Answer (2)
112. Answer (3)
113. Answer (1)
114. Answer (1)
It is also called perisperm.
115. Answer (2)
A – Cock (ZZ)
B – Hen (ZW)
116. Answer (1)
117. Answer (1)
Test - 5 (Code A) (Answers & Hints) All India Aakash Test Series for Medical-2017
9/10
118. Answer (3)
119. Answer (2)
120. Answer (4)
Ratio is (1 : 1)3
121. Answer (2)
122. Answer (1)
123. Answer (4)
124. Answer (3)
It is a physical mutagen, others are chemical
mutagen.
125. Answer (1)
126. Answer (4)
127. Answer (3)
0.6%
128. Answer (3)
129. Answer (4)
The ratio is 9 : 7
130. Answer (2)
131. Answer (4)
Both (1) and (2)
132. Answer (3)
Thalassemia, Cystic fibrosis and sickle cell anaemia.
133. Answer (1)
134. Answer (1)
or Dizygotic twins
135. Answer (4)
136. Answer (1)
Genetic similarity between parent and young one
occurs in asexual reproduction.
137. Answer (4)
A(Horse) = 60 years
B(Cow) = 25 years
C(Crow) = 15 years
138. Answer (4)
Longitudinal binary fission - Euglena & Vorticella
Plasmotomy – Pelomyxa
139. Answer (2)
Sharks have internal fertilisation.
140. Answer (3)
Haploid males are produced through parthenogensis
in honey bee.
141. Answer (4)
Sertoli cells secrete ABP which concentrates
testosterone in seminiferous tubules.
142. Answer (1)
Fraternal twins are dizygotic twins.
143. Answer (4)
144. Answer (3)
145. Answer (2)
Ectodermal – Spinal cord, Pituitary, Iris muscles
Mesodermal – Notochord, spleen, adrenal cortex,
blood vessels
Endodermal – Pancreas, thymus, liver, lining of
urinary bladder and thyroid
146 Answer (3)
Expulsion of baby occurs in expulsion stage.
147. Answer (3)
FSH stimulates sertoli cells to convert spermatids
into sperms.
148. Answer (3)
Castration – Removal of primary sex organs
Oopherectomy – Removal of ovaries
Orchidectomy – Removal of testes
149. Answer (4)
First movement of foetus and appearance of hair on
head is observed in fifth month.
150. Answer (2)
151. Answer (2)
152. Answer (4)
During cleavage nuclear cytoplasmic ratio increases.
153. Answer (2)
Fast block to check polyspermy involves influx of
Na+ causing depolarisation of membrane of ovum.
154. Answer (2)
A preovulatory surge of LH is necessary for ovulation
to take place.
155. Answer (3)
156. Answer (1)
157. Answer (1)
158. Answer (3)
All India Aakash Test Series for Medical-2017 Test - 5 (Code A) (Answers & Hints)
10/10
159. Answer (4)
FSH and LH increase to high levels.
160. Answer (3)
Spermiogenesis involves formation of sperm from
spermatid.
161. Answer (2)
Endoderm is first germinal layer to be developed
during embryonic development.
162. Answer (2)
Gonorrhoea – Neisseria gonorrhoeae
Chancroid – Haemophilus ducreyi
Genital Herpes – Herpes simplex virus
163. Answer (4)
164. Answer (2)
IUI-Intra uterine insemination
165. Answer (4)
166. Answer (3)
Contraceptive method Average failure rate
Coitus interruptus 23%
Barrier methods 10-15%
Oral contraceptives 2-2.5%
Rhythm method 20-30%
167. Answer (4)
168. Answer (4)
Norplant is a subcutaneous implant.
169. Answer (1)
170. Answer (2)
171. Answer (2)
172. Answer (3)
Circumcision is surgical removal of prepuce.
173. Answer (4)
174. Answer (3)
Demographic transition occurs when birth rate
equals death rate.
175. Answer (2)
176. Answer (1)
177. Answer (4)
Norgynon, femsheild and DMPA (Depot medroxy
progesterone acetate) are female contraceptives.
178. Answer (1)
ICSI – Intra cytoplasmic sperm injection
179. Answer (1)
180. Answer (2)
AIDS is a viral STD.
� � �
Test - 5 (Code B) (Answers & Hints) All India Aakash Test Series for Medical-2017
1/10
1. (3)
2. (2)
3. (3)
4. (3)
5. (2)
6. (4)
7. (3)
8. (3)
9. (1)
10. (4)
11. (4)
12. (2)
13. (4)
14. (3)
15. (2)
16. (1)
17. (1)
18. (3)
19. (1)
20. (4)
21. (4)
22. (1)
23. (1)
24. (3)
25. (2)
26. (4)
27. (4)
28. (1)
29. (2)
30. (4)
31. (3)
32. (1)
33. (4)
34. (2)
35. (3)
36. (4)
ANSWERS
TEST - 5 (Code-B)
All India Aakash Test Series for Medical-2017
Test Date : 29-01-2017
37. (1)
38. (1)
39. (4)
40. (3)
41. (4)
42. (1)
43. (3)
44. (2)
45. (1)
46. (4)
47. (2)
48. (4)
49. (3)
50. (3)
51. (3)
52. (4)
53. (2)
54. (3)
55. (1)
56. (1)
57. (2)
58. (4)
59. (3)
60. (4)
61. (4)
62. (1)
63. (2)
64. (3)
65. (1)
66. (2)
67. (3)
68. (4)
69. (2)
70. (4)
71. (1)
72. (1)
73. (3)
74. (1)
75. (4)
76. (4)
77. (4)
78. (4)
79. (2)
80. (2)
81. (4)
82. (4)
83. (2)
84. (1)
85. (4)
86. (3)
87. (1)
88. (1)
89. (4)
90. (4)
91. (4)
92. (3)
93. (1)
94. (1)
95. (4)
96. (4)
97. (2)
98. (1)
99. (1)
100. (2)
101. (3)
102. (1)
103. (4)
104. (3)
105. (4)
106. (2)
107. (4)
108. (1)
109. (3)
110. (3)
111. (4)
112. (1)
113. (3)
114. (1)
115. (2)
116. (1)
117. (4)
118. (2)
119. (2)
120. (1)
121. (4)
122. (2)
123. (4)
124. (2)
125. (4)
126. (1)
127. (3)
128. (4)
129. (1)
130. (1)
131. (3)
132. (1)
133. (3)
134. (4)
135. (2)
136. (2)
137. (3)
138. (3)
139. (2)
140. (3)
141. (2)
142. (1)
143. (2)
144. (1)
145. (4)
146. (4)
147. (1)
148. (2)
149. (2)
150. (1)
151. (2)
152. (4)
153. (2)
154. (4)
155. (4)
156. (1)
157. (2)
158. (1)
159. (3)
160. (3)
161. (1)
162. (4)
163. (4)
164. (2)
165. (4)
166. (4)
167. (2)
168. (1)
169. (1)
170. (1)
171. (4)
172. (1)
173. (2)
174. (3)
175. (2)
176. (1)
177. (4)
178. (2)
179. (2)
180. (3)
All India Aakash Test Series for Medical-2017 Test - 5 (Code B) (Answers & Hints)
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Hints to Selected Questions
[ PHYSICS]
1. Answer (3)
qi
t
i = 2, t = 1
q = 2 C
Charge of 1e– = +1.602 × 10–19C in magnitude
In 1 C, number of electron
– 18
–19
16.25 10
1.602 10e
2 C charge will have 12.5 × 1018 = 1.25 × 1019
2. Answer (2)
3. Answer (3)
r
dr
di = J. 2r dr
0
0
. 2
R
di J r r dr ∫ ∫
3
02
3
J Ri
4. Answer (3)
Only charge and electric field remain unchanged.
5. Answer (2)
1
CR
6. Answer (4)
l
Work done by electric force = PE
qEl sin = mgl(1 – cos)
2
2sin cos 2sin.
2 2 2qEl mgl
tan2
qE
mg
⎛ ⎞ ⎜ ⎟⎝ ⎠
12tan
qE
mg
⎡ ⎤ ⎢ ⎥⎣ ⎦
7. Answer (3)
Time period of SHM is not affected by constant force
hence will remain uncharged.
8. Answer (3)
Just after closing the switch, branch containing only
capacitor will make the circuit short, hence E
iR
9. Answer (1)
Let charge induced on sphere at any moment = q
Potential of sphere at any time = 0
0kq kq
x R
–
qRq
x
2
dq qR dxi
dt dtx
⎛ ⎞ ⎜ ⎟⎝ ⎠
2
qRVi
x
10. Answer (4)
R1
R2
3 V 2 V
In series current are same through both
3 = IR1
2 = IR2
Test - 5 (Code B) (Answers & Hints) All India Aakash Test Series for Medical-2017
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1
1 1 1 2
22 2 1
2
3
2
l
R A l A
lR l A
A
2
2
3 6.
2 1
B
A
r
r
2
2
12 :1
4 ⇒ B A
BA
r r
rr
11. Answer (4)
Potential difference across 100 resistance should
be 5 V as voltmeter and 100 resistance are in
parallel
req
of voltmeter and 100 should be 50
r of voltmeter = 100
12. Answer (2)
Terminal voltage = E – iR
Given E – 10 × R = 50
E – 1 × R = 60
Solving,10
9R
550V
9E
13. Answer (4)
10
20
i
1 2 4 t
Charge flown = area under i-t graph
= 35
14. Answer (3)
P
E
When PE is chosen to be zero when P E� �
PE = – PEcos
So at 3
–2
PEPE
and PE when dipole is parallel to the field is –PE.
Now when reference potential is chosen to be of
3
its PE becomes 0.
i.e., PE increases by 2
PE
hence PE in parallel position will be
– –2 2
PE PEPE
15. Answer (2)
If charges are both positive or both negative PE
decreases if they are of opposite sign PE increases.
16. Answer (1)
r
q
kqV
r = 5 volt
when drops are merged, let radius be R
3 34 48
3 3r R
R = 2r
Hence, its potential will be
(8 ) (8 ) 4
2
k q k q kq
R r r = 20 V
Hence, difference = 15 V
17. Answer (1)
+10 0 0 2 –2 10
All India Aakash Test Series for Medical-2017 Test - 5 (Code B) (Answers & Hints)
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Final charge distribution on the plates will be
x = 10
y = 0
z = –2
x + y + z = 8 C
18. Answer (3)
No current will pass through the grounded resistor
Hence reading of ammeter will be zero.
19. Answer (1)
Charge distribution before closing the key
4 V
2 F 6 F
8 C 8 C+6 C +6 C
4 F 4 F+ +
Charge distribution after closing the switch
4 V
2 F 6 F
+10 C
+5 C +9 C
4 F 4 F
–5 C –9 C
–10 C
4 C
+6 C –6 C
Charge flown through the switch = 4 C
20. Answer (4)
E
curved
= base
box
= ER2
21. Answer (4)
Potential first decreases to R, then remains constant
inside shell and then decreases to zero at infinity.
22. Answer (1)
All three capacitors are in parallel hence Ceq
= 9 C
23. Answer (1)
enc
0
q
, initially qenc
increases till l
tu
then
becomes constant and then decreases when rod
comes completely out of cube.
24. Answer (3)
–
g
VR G
I
–3
10– 20 9980
10R
25. Answer (2)
Value of E will determine direction of flow of current.
26. Answer (4)
10 V
2 k 3 k
6 k
Req
of 3 k and 6 k= 2
Potential drop across 2 kwill be 5V and across
combinations of 3 kand 6 kwould also be 5 V.
Hence, reading = 5 V
27. Answer (4)
( 3 3)3 3
2 3 3AB
R RR R
R
Solving R = 3
28. Answer (1)
Final charge on capacitor would be CE.
29. Answer (2)
qE
T
mg
T cos = mg
T sin = qE
Test - 5 (Code B) (Answers & Hints) All India Aakash Test Series for Medical-2017
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30. Answer (4)
2QQ
q
Final potential difference between spheres will be E
(2 – ) ( )–
K Q q K Q qE
R R
0– 2
2
Qq ER
31. Answer (3)
Uniformly accelerated motion can have parabolic or
straight line trajectory .
32. Answer (1)
2 2H i Rdt R i dt ∫ ∫H = R × Area under i2-t graph
Area = 1
(6 2) 4 162
H = 160 J
33. Answer (4)
t
q
dq
dt decreases with time
34. Answer (2)
A B
2 F
1 F
1 F
VC
VA
VC
VA
1 F
VCV
B
VAV
B
AB
2 F
1 F2 FC
⇒
2 1 82 F
3 3
AB
C
35. Answer (3)
C1
C2
A
B
120°
Resistance per unit central arc angle
36 18
/ rad2
Resistance of � �
1 2
18 212
3
AC B AC B
eq
1 1 1 1 1
24 24 12 12R
Req
= 4
36. Answer (4)
50 V
R30 V
10
20
I
I
1 2
Loop-1
30 – 50 + 20 I = 0
I = 1 ampere
Loop-2
50 – IR – 20I = 0
50 – R – 20 = 0
R = 30
37. Answer (1)
I = neAVd
1 2
2 (2 )d d
neAV ne A V
1
2
1:1d
d
V
V
38. Answer (1)
dqi
dt = Slope of charge-time graph
i = –5 A
H = i2 Rt
H t
All India Aakash Test Series for Medical-2017 Test - 5 (Code B) (Answers & Hints)
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39. Answer (4)
2
1– –
x
VE
x x
2
1– –
y
VE
y y
2
1– –
z
VE
z z
2 2 2
1 1 1ˆ ˆ ˆ– – –E i j k
x y z
�
at (1, 1, 1)
ˆ ˆ ˆ– – –E i j k
�
| | 3 N/CE �
40. Answer (3)
2
1 1
1
EH R t
R r
⎡ ⎤ ⎢ ⎥⎣ ⎦
2
2 2
2
EH R t
R r
⎡ ⎤ ⎢ ⎥⎣ ⎦
Equating H1
= H2
1 2r R R
41. Answer (4)
Let charge/volume =
3
0
a
3
0
4
3a
Dividing, 3
4
4
3
42. Answer (1)
High resistance would ensure more heat production
in short circuit condition, low melting point would
melt the wire fairly easy and thus breaking circuit.
43. Answer (3)
21
2U CV
10 20 10dU dV
CVdt dt
= 2000 J/s = 2 mJ/s
44. Answer (2)
++
+
+
+
+
–
–
–
–
c q2R
Vc = V
q + V
shell
Vshell
= 0
As total charge on shell = 0
45. Answer (1)
qv0
q
r
v
By angular momentum conservations
024mv mvr ...(i)
By energy conservations
22 2
0
1 1
2 2
kqmv mv
r
2
2 2
0 0 20
1 1 24. .
2 2 4
qmv m v
rr
2 2
2 0 0 0
0 20
12 41
2 4
mv mv
mv
rr
2
1 12 1
2 rr
r2 – 2r – 24 = 0
r = 6 m
Test - 5 (Code B) (Answers & Hints) All India Aakash Test Series for Medical-2017
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[ CHEMISTRY]
46. Answer (4)
It has 6 42g gt e configuration.
47. Answer (2)
Mn+ is electron deficient hence strength of Mn–C
bond will be less
48. Answer (4)
All behave as acid ligand
49. Answer (3)
Cl– is weak ligand
50. Answer (3)
Fact
51. Answer (3)
Fact
52. Answer (4)
Only (a) and (b) are correct
53. Answer (2)
[PtCl4
]2– is
Cl
Cl
PtCl
Cl
2–
dsp2
54. Answer (3)
Cr3+ : xx xx
3d
xx
4s
xx xx xx
4p
55. Answer (1)
Fe3+ has an unpair electron
56. Answer (1)
All surrounding atoms are same
57. Answer (2)
Fact
58. Answer (4)
EAN = 28 – 2 + 2 × 4 = 26 + 8 = 34
59. Answer (3)
According to the rules
60. Answer (4)
ICH NH
2 2
CH2
ICH
2
CH NH2 2
NH (dien)
61. Answer (4)
Fact
62. Answer (1)
63. Answer (2)
They do not react with CaCO3
64. Answer (3)
In CeO2
, cerium has noble gas configuration
65. Answer (1)
Fact
66. Answer (2)
pH 42– 2–2 7 4Cr O CrO
�����⇀↽�����
67. Answer (3)
Fact
68. Answer (4)
Ni2+ and Pt4+ are stable
69. Answer (2)
SO2
+ V2
O5
SO3
+ V2
O4
2 4 2 2 5
1V O O V O
2
70. Answer (4)
2Cu /Cu
E 0.34 V
71. Answer (1)
+7 is the highest oxidation number of Mn (Group
VII)
72. Answer (1)
Sc2+, Zn3+ are not possible.
73. Answer (3)
Ni according to the trend
74. Answer (1)
75. Answer (4)
Ag2+ has d9 and Ag3+ has d8 system.
76. Answer (4)
It is semiconductor
77. Answer (4)
All India Aakash Test Series for Medical-2017 Test - 5 (Code B) (Answers & Hints)
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[ BIOLOGY
]
91. Answer (4)
92. Answer (3)
or Dizygotic twins
93. Answer (1)
94. Answer (1)
Thalassemia, Cystic fibrosis and sickle cell anaemia.
95. Answer (4)
Both (1) and (2)
96. Answer (4)
97. Answer (2)
The ratio is 9 : 7
98. Answer (1)
99. Answer (1)
0.6%
100. Answer (2)
101. Answer (3)
102. Answer (1)
It is a physical mutagen, others are chemical
mutagen.
103. Answer (4)
104. Answer (3)
105. Answer (4)
106. Answer (2)
Ratio is (1 : 1)3
107. Answer (4)
108. Answer (1)
109. Answer (3)
110. Answer (3)
111. Answer (4)
A – Cock (ZZ)
B – Hen (ZW)
112. Answer (1)
It is also called perisperm.
113. Answer (3)
114. Answer (1)
115. Answer (2)
116. Answer (1)
Chemotropic growth of pollen tube.
78. Answer (4)
Fact
79. Answer (2)
In pig iron carbon is 4 g, in cast iron it is 3 g
(%) decrease = 4 – 3
100 25%4
80. Answer (2)
Fe2
O3
, MnO2
and P4
O10
are reduced to Fe, Mn and
P respectively.
81. Answer (4)
Fact
82. Answer (4)
Aluminium is highly reactive.
83. Answer (2)
All three are correct.
84. Answer (1)
Dolomite is CaCO3
. MgCO3
85. Answer (4)
In cynide process NaCN is used.
86. Answer (3)
Complex formed is Na[Ag(CN)2
]
87. Answer (1)
When Fe2
O3
is in excess, Bayer's process is
followed.
88. Answer (1)
ZnS + 4NaCN Na2
S + Na2
[Zn(CN)4
]
89. Answer (4)
Fact
90. Answer (4)
Siderite is FeCO3
Test - 5 (Code B) (Answers & Hints) All India Aakash Test Series for Medical-2017
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117. Answer (4)
118. Answer (2)
119. Answer (2)
Has an unwettable stigma.
120. Answer (1)
121. Answer (4)
122. Answer (2)
123. Answer (4)
All are correct.
124. Answer (2)
A – It is the epidermis which is protective in function.
125. Answer (4)
Embryo sac represents female gametophyte.
126. Answer (1)
Zoidogamous in Funaria, Chara, Marchantia and
Adiantum.
127. Answer (3)
128. Answer (4)
Foliar buds are present on leaves.
129. Answer (1)
130. Answer (1)
131. Answer (3)
Embryo stage is absent in algae.
132. Answer (1)
133. Answer (3)
Orange and mango which are polycarpic.
134. Answer (4)
135. Answer (2)
Conidia is a feature of ascomycetes and
deutomycetes.
136. Answer (2)
AIDS is a viral STD.
137. Answer (3)
138. Answer (3)
ICSI – Intra cytoplasmic sperm injection
139. Answer (2)
Norgynon, femsheild and DMPA (Depot medroxy
progesterone acetate) are female contraceptives.
140. Answer (3)
141. Answer (2)
142. Answer (1)
Demographic transition occurs when birth rate
equals death rate.
143. Answer (2)
144. Answer (1)
Circumcision is surgical removal of prepuce.
145. Answer (4)
146. Answer (4)
147. Answer (1)
148. Answer (2)
Norplant is a subcutaneous implant.
149. Answer (2)
150. Answer (1)
Contraceptive method Average failure rate
Coitus interruptus 23%
Barrier methods 10-15%
Oral contraceptives 2-2.5%
Rhythm method 20-30%
151. Answer (2)
152. Answer (4)
IUI-Intra uterine insemination
153. Answer (2)
154. Answer (4)
Gonorrhoea – Neisseria gonorrhoeae
Chancroid – Haemophilus ducreyi
Genital Herpes – Herpes simplex virus
155. Answer (4)
Endoderm is first germinal layer to be developed
during embryonic development.
156. Answer (1)
Spermiogenesis involves formation of sperm from
spermatid.
All India Aakash Test Series for Medical-2017 Test - 5 (Code B) (Answers & Hints)
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157. Answer (2)
FSH and LH increase to high levels.
158. Answer (1)
159. Answer (3)
160. Answer (3)
161. Answer (1)
162. Answer (4)
A preovulatory surge of LH is necessary for ovulation
to take place.
163. Answer (4)
Fast block to check polyspermy involves influx of
Na+ causing depolarisation of membrane of ovum.
164. Answer (2)
During cleavage nuclear cytoplasmic ratio increases.
165. Answer (4)
166. Answer (4)
167. Answer (2)
First movement of foetus and appearance of hair on
head is observed in fifth month.
168. Answer (1)
Castration – Removal of primary sex organs
Oopherectomy – Removal of ovaries
Orchidectomy – Removal of testes
169. Answer (1)
FSH stimulates sertoli cells to convert spermatids
into sperms.
170 Answer (1)
Expulsion of baby occurs in expulsion stage.
� � �
171. Answer (4)
Ectodermal – Spinal cord, Pituitary, Iris muscles
Mesodermal – Notochord, spleen, adrenal cortex,
blood vessels
Endodermal – Pancreas, thymus, liver, lining of
urinary bladder and thyroid
172. Answer (1)
173. Answer (2)
174. Answer (3)
Fraternal twins are dizygotic twins.
175. Answer (2)
Sertoli cells secrete ABP which concentrates
testosterone in seminiferous tubules.
176. Answer (1)
Haploid males are produced through parthenogensis
in honey bee.
177. Answer (4)
Sharks have internal fertilisation.
178. Answer (2)
Longitudinal binary fission - Euglena & Vorticella
Plasmotomy – Pelomyxa
179. Answer (2)
A(Horse) = 60 years
B(Cow) = 25 years
C(Crow) = 15 years
180. Answer (3)
Genetic similarity between parent and young one
occurs in asexual reproduction.