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2016

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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


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


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

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