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AIPMT SAMPLE PAPER- 1(SOLUTIONS)
PHYSICS
Key: 1.c 2.b 3.b 4.c 5.b 6.b 7.a 8.c 9.d 10.a 11.c 12.c 13.d 14.b 15.a
16.b 17.a 18.b 19.d 20.a 21.b 22.b 23.c 24.a 25.b 26.d 27.b 28.a 29.a 30.b
31.a 32.d 33.d 34.d 35.c 36.c 37.a 38.b 39.a 40.c 41.b 42.a 43.c 44.b 45.b
1.Vavg = i i
=
d+dd + d
= kmph
2. at = 3t dvdt = t ⇒∫ dv = ∫ tdt v =
v/t=2 = 6ms-1
ac = R
=
= 18ms-2
3.v = ai + b − ct j The actual velocity is given by, V = u i + (u − gt)j
On comparing the two equations,
ux =a
uy = b
g = c
Range; R = g
=
4. W = ∫F d +∫F d
= k ∫ xdx + ∫ dλ
= k − + −
= (8+12)k
= 20k
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5. F = R−
F1 = R −
⇒ F1=F
6. When heated every linear dimension has to increase.
. It is a y li p o ess ∆ = f o I la , ∆Q = ∆U+∆W ⇒∆Q=∆w
5 = wAB+wBC+wCA
5 = 10(2-1)+0+wCA ⇒wCA= − 5j
8. Total energy is zero in the case of escape velocity.
9. The two capacitors finally arrive at a common potential. They system loses energy but total
charge remains the same
10. Initially
mg=K x0 ----(1)
when one spring is cut
mg - kx0 = ma
mg - g- ma (from (1))
⇒ a= g. This result as can be seen is independent of mass of the block.
11. FE = qeBn FE = qvB(-n
Thus it both E and B are doubled the above conditions are not affected.
12.
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V0=R[ − − ] = 0
13. When x = 0.3m
10= k .
When x = 0.45m
U= × k × .
U= × . × .
= 10
= 22.5 J ∴ Additional energy is
U -10 =22.5-10
= 12.5J
14. Heat gained by the ball = heat lost by water
200×c×(60-20)=200 (1)(80-60)
2c = 1 ⇒ c =
15. On comparing
Kx = π
K = π
⇒ π = π⇒ = λ =30
Distance between a node and the next anti-node is
= = .
16. Applying conservation of momentum
1(21)-2(4) = 1(1)+2v ⇒ 2v = 12
V= 6 ms-1
We have,
E = −−
= −− − = = .
17. Q =cv
dQ= cdv
10-6
= c 10-3 ⇒c = 10
-3F.
18. RP =
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Rα = = R
RP : Rα =1:2
19. WKT in magnitude 2
21 0.01 10
a x
2
5T
20. = %
W.K.T K =
dk= dp dkk = pm × mp dp
= 4
= 4×100%
= 400%
21. Net charge is a dissolve is zero . ∴� =0.
22. Let R=
Case I (series)
H1 =T −TR
=T −TR
Case II (parallel)
H2 =T −TR T − TR ⇒ 2H1= ⇒ H2 = 4 H1
We have,
H1t1 = H2t2
H1(12) = 4H1t2
t2= 3s.
23. We have
BMN=0
BNO=iR
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BOP =iπR
B = BMN +BNO +BOP
= iπR π+
24. IAA=mBl2
IBB= mAl2 = = ⇒mB=3mA
Taking mA at the origin
XCM=++
=
25. for diatomic gns
CV= R, CP= R
r- = R = = .
26.
T1+T2 = mg ∑C =
T1(0)-mg +T2 =
T2 = mg
T2 = g
=
27. Factual information
28. y = 0.2 sin (�� − �
y│ = . , = . = . sin . π − π
=0.2 sin(- . π
= - . si π+ π
= -0.2 (-1)
= 0.2
29. E-field inside a shell is zero. Thus EiE = alwaλs.
30. We have,
2=E +ER+ +
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And 1 =E −ER+ + ⇒ 2(E1-E2) = E + E
E =3E2 EE =
31. Using Angular impulse- Angular momentum theorem
(10 l)=Iw - 0
10 =× w ⇒ w= 15rad s
-1
We have, K = Iw
= × × ×
= = 75 J
32. The fundamental frequency is
Fo =
= ×
= 80 Hz
It will resonate for 80Hz, 240 Hz, 400 Hz,_ _ _ _ _
33. m1 : m2 : m3 =1 : 3: 5
A1l1: A2l2 : A3l3 = 1:3:5
Also l1:l2:l3 = 5:3:1 l : l : l = : : ⇒ : : = : :
“i e they ha e sa e esisti ity s
R1:R2:R3= : :
= 25:3:
= 125:15:1
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34.
u-v =10 vu =
⇒ 3v=10 ⇒v= ⇒ u =
v − u = f − — = f ⇒ f = -4.4 cm
35. E = −
=− − x
= 12x
E│ = =
= 24 Vm-1
36. All there capacitors have same p.d, i.e., they are in parallel.
C1=∈
C2=∈
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C3=∈
Ceff = C1+C2+C3
= ∈ K + K + K
= ∈ + +
= ∈
. ‘ = Ω
XL = πfL
= ×π× × .π
= Ω
Z=√R + X
=√ +
= Ω l =V Z
=
= 2 A
38. ∈ = -M∆∆
15×103=-M×
−. ⇒M = 5H
39. Factual information
40. =
II i = (√I + √I )(√I − √I )
=+− =
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41. We have, n sin 300=1 sin 90
0 ⇒ n = 2
Also V =
= × 8
= 1.5×108 ms
-1
42. factual information
. ‘ α A
There fore, RR = (AA )
=
= 2
44. q=2t2+8t+1 c
I =
= 4t+8
I│ = = +
= 12 A
45. Factual information
CHEMISTRY
KEY: 46.b 47. d 48. b 49. b 50. c 51. b 52. d 53. c 54. b 55. a
56. b 57. a 58. b 59. b 60. d 61. a 62. a 63. b 64. a 65. d
66. a 67. a 68. b 69. d 70. b 71. c 72. c 73. a 74. d 75. b
76. b 77. d 78. d 79. b 80. d 81. d 82. b 83. c 84. c 85. b
86. d 87. a 88. b 89. a 90. b
Solutions:
46. (b)
Sol: Conceptual
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47. (d)
Sol:CH − CH CH = CH − CH + O Z → CH CH CHO+ CH CHO H O
48. (b)
Sol: The substance which on heating converts directly into vapors is purified by sublimation.
49. (b)Sol: n-alkanes on heating to 300°C in presence of AlCl3/HClisomerise to give branched chain
alkanes
(having lower boiling point than normal chain alkanes). This process is known as isomerization.
CH3 – CH2 – CH2 – CH3
n-butane
� � / � °� → CH3
|
CH3 – CH – CH3
Isobutene (low b.p)
50. (c)
Sol: Acidic strength follows the order
CH ≡ CH > CH2 = CH2> CH3 – CH3
Therefore, basic character follows the reverse order
CH3 – CH2> CH2 = CH > CH ≡C
51. (b)
Sol: –NO2 group is meta directing group
52. (d)
Sol: CH3 CH Cl CH2 – CH2 OH + KOHaq→ CH3 – CH – CH2CH2 OH + KCl
|
OH
53. (c)
Sol:EWG increases reactivity towards nucleophilic substitution, EDG decreases reactivity towards
nucleophilic substitution.
54. (b)
Sol:3C2H5OH + PBr3→ 3C2H5Br + H3PO3
(X)
C2H5Br + KOHalc.→ C2H4
(Y)
C2H4 + HHSO4→ C2H5 – HSO4
→ C2H5OH
(Z)
55. (a)
Sol:
56. (b)
Sol:H2O + CCl−→ :CCl2 + Cl–
Electrophile (dichlorocarbene)
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57. (a)
Sol:
58. (b)
Sol:
59. (b)
Sol: R – NC + 4(H) i→ R – NH – CH3
60. (d)
Sol: Rate = − ��� [ ] = + ��� [ ]
��� [ ] = − ��� [ ]
61.(a)
Sol:Half life of the given reaction is independent of initial concentration , hence it is a first order reaction
[ / ∝ � − ] 62. (a)
Sol: For OV,
�� (�. . , ��⊝ �� ���⊝ ) = 0.414
∴ �� = . = .
63. (b)
Sol:In Combustion, energy releases. Hence it is exothermic ∆H = -ve
64. (a)
Sol: The balanced Equation is
2MnO⊝ + C O − + H⨁ ⟶ Mn+ + CO + H O
65. (d)
Sol: No. Of E ui ale ts of Al deposited = . = 0.5
Volume of H2 gas at STP = No. Of equivalents × Equivalent Volume
= 0.5×11.2 = 5.6L
66. (a)
Sol: mole fraction of each will be =1/3
Therefore partial pressure of each =150/3=50mm
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Equilibrium constant=50x50/50=50mm
67. (a)
Sol:PE = -2KE ∴ PE will change from -2x to −
Change in potential energy = −
-(-2x)
= −
+2x =
68. (b)
Sol:m=3, l=3, n=4
fo = , Nu e of a es ill e
69. (d)
Sol:[H+] = 10
-3M, [H
+] = 10
-4M, [H
+]=10
-5M for the given acids
M = V + V + VV ∴ V1=V2=V3
=(10-5
×1)+(10-4
×1)+(10-3
×1)=3.7×10-4
M.
70.(b)
Sol: Most soluble compound is that which have highest KSp value. KSp of MnS (7×10-16
) is highest
71. (c)
Sol:Volume of balloon = πr3
= × × 03 =4190.47m
3
Mass of the air displaced = 4190.47×1.2=5028.56Kg
No of moles of helium in the balloon=VRT
= × . ×. ×
= 170344
Mass of helium = 4×170.344×103 = 681.376Kg
Mass of filled balloon = 681.376+100 =781.376Kg
Pay load = Mass of air displaced – Mass of filled balloon
=5028.56-781.376
=4247.184Kg
72. (c)
Sol: KE= RT
73. Ans: (a)
Sol: Partial pressure of N2 in air (P ) = PTotal× X (in air)
= 5×0.6 P (in air) = K × XN (in H2O)
5×0.6 = 1× × XN (in H2O)
X in 10 moles of water = × .× = 3×10
–5
X = + H
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3×10–5
= + ⇒ nN ×3×10–5
+ 3×10–5×10 = n2
3×10–4
= nN (1–3×10–5
)
nN = 3×10–4
74. (d)
Sol:Tefflon is a polymer of tetra fluoro ethylene but not a mineral .
75. Ans(b)
Sol: Conceptual
76. (b)
Sol: Cr has s s p s p d s configuration in the ground state
Total electrons with m=0 are 2+2+2+2+2+1+1=12
77. (d)
Sol: [�� � ] − is with hybridization and � =
[�� � ]is with hybridization and � = BM
[Cu NH ] +is with hybridization with � = 1.732 BM [� ] − is with hybridization with � = 0 BM
78.(d)
Sol: boiling point of � > � due intermolecular hydrogen bonding
79. (b)
Sol:
4 2 2 3
3 2 3
5 2 3 4
3 2 3 3
SCl H O H SO HCl
NCl H O NH HOCl
PCl H O H PO HCl
AsCl H O H AsO HCl
80. Ans(d)
Sol: Acidic nature of hydrides increases from H O to H Te
81.Ans(d)
Sol: conceptual
82.(b)
Sol: for H O volume strength= normality x 5.6
83.(c)
Sol: it is an application of micro cosmic salt
84. Ans:(c)
Sol: Conceptual
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85. Ans(b)
Sol: acidic nature of oxides increases in a period from left to right
86. (d)
Sol: Greater charge on cation
87. Ans (a)
Sol.Comceptual
88. Ans (b)
Sol: Silver
89. Ans (a)
Sol: A, D and E
90. Ans. (b)
Sol: ORLON
BIOLOGY
Key: 91. b 92. c 93. b 94. c 95. c 96. d 97. a 98. b 99. b 100. a
101. c 102. b 103. b 104. a 105. d 106. c 107. b 108. b 109. a 110. a
111. c 112. c 113. c 114. b 115. a 116. b 117. d 118. a 119. c 120. c
121. c 122. d 123. d 124. b 125. c 126. d 127. b 128. b 129. c 130. a
131. c 132. a 133. b 134. d 135. c 136. a 137. c 138. b 139. b 140. b
141. b 142. d 143. b 144. a 145. c 146. d 147. d 148. b 149. c 150. d
151. d 152. a 153. b 154. c 155. d 156. b 157. b 158. a 159. b 160. b
161. c 162. d 163. b 164. a 165.d 166.d 167. d 168. c 169.b 170. a
171. d 172. a 173. a 174. c 175. c 176. c 177. a 178. c 179. d 180. b