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Test - 1 (Code-C) (Answers) All India Aakash Test Series for Medical-2020

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1. (3)

2. (4)

3. (1)

4. (1)

5. (2)

6. (1)

7. (2)

8. (2)

9. (2)

10. (3)

11. (1)

12. (2)

13. (3)

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16. (2)

17. (1)

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20. (2)

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24. (3)

25. (2)

26. (4)

27. (4)

28. (1)

29. (4)

30. (3)

31. (4)

32. (4)

33. (1)

34. (1)

35. (2)

36. (3)

Test Date : 14/10/2018

ANSWERS

TEST - 1 (Code-C)

All India Aakash Test Series for Medical-2020

37. (1)

38. (4)

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All India Aakash Test Series for Medical-2020 Test - 1 (Code-C) (Answers & Hints)

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ANSWERS & HINTS

1. Answer (3)

Hint : The interval in which displacement is zero.

Solution : For the curve, ACB displacement is zero

AB

C

Tt

x

0

0

��

�

av

SV

t

2. Answer (4)

Hint :

Displacement and velocity may be positive or

negative.

2nd

4th

0

+ve

s

v

Solution : Direction of displacement is vertically

downward.

S < 0

Direction of velocity is downward

So, v < 0

3. Answer (1)

Hint : max

2 v S

Solution :

u = 0 vmax v = 0

S2

S2

S1

S1

2

2 max

max 1 12

2

v

v S S ...(i)

[ PHYSICS]

2

2 max

max 2 20 2

2

vv S S

...(ii)

2

max

1 2

1 1

2

vS S S

max

2 2 2 3 100

5

v S

4 15 m/s

4. Answer (1)

Hint : 1 2

1 2

2

av

v vv

v v

Solution : v1 = 36 km/h = 10 ms–1

v2 = 54 km/h = 15 ms–1

1 2

1 2

av

S Sv

t t

1 1 1

210

2 10

S

SS t t

2 2 2

215

2 15

S

SS t t

112 ms

1 1

2 10 15

av

Sv

S

5. Answer (2)

Hint : v

at

Solution : v = t2 + 4t

At t = 1 s, v = 5 ms–1

t = 2 s, v = 12 ms–1

212 57 ms

2 1

a

Test - 1 (Code-C) (Answers & Hints) All India Aakash Test Series for Medical-2020

3/17

6. Answer (1)

Hint : dv

a vdx

Solution:

0 0

1v x

v x

...(i)

0

0

0

v

v x v

x

...(ii)

0

0

vdv

dx x ...(iii)

0 0

0

0 0

v vdva v x v

dx x x

2 2

0 0

200

v v

x

xx

7. Answer (2)

Hint : slope of graphdv

a v tdt

Solution : a = slope of v – t graph

= –ve value (constant)

t

a

0

8. Answer (2)

Hint : Average speed = Total distance travelled

Total time taken

Detailed Solution : From t = 0 to t = 4 s x1

= 4 m

From t = 4 s to t = 6 s x2 = 0

From t = 6 s to t = 10 s x3 = 4 m

Average speed = 14 0 4 8 4

ms10 10 5

9. Answer (2)

Hint : 2

0

1

2S S ut at

Solution:

v

tt0

v0

0

0

0

(constant)

va

t

2

0

1

2x x ut at

20

0

0

10

2

vx x t

t

20

0

0

1

2

vx x t

t

10. Answer (3)

Hint : | Average velocity | = |Total displacement travelled|

Total time taken

Solution:

ABO

| Displacement | = 10 m

Time taken = 5

10 2

s

| Average velocity | = 110 20

ms

2

11. Answer (1)

Hint : Reaction time

Solution:

Distance travelled during reaction time

S1 = 10 × 1 = 10 m ...(i)

Distance travelled during retarded motion

2

2

10050 m

2 2 1

vS

a

...(ii)

Total travelled distance = S1 + S

2 = 60 m

Distance of car from red signal

= 70 – 60 = 10 m

12. Answer (2)

Hint : 2

0

1

2h v t gt

Solution :

7

8

5

4

h

t = 6


4/17

Particle will be at maximum height at t = 6 s.

2110 6 180 m.

2 h

13. Answer (3)

Hint : 2 2

2 1 02v v a l

Detailed solution : 2 2

2 1 02v v a l

2 2

2 1

02

v vl

a

l

2

2 2

1 02

2

lv v a

2 2 2 2

2 2 2 1 1 2

1 0

02 2

v v v vv v a

a

2 2

1 2

2

v vv

14. Answer (2)

Hint : d + x = v0t ...(i)

21

2x at ...(ii)

Solution : 32 + x = 10t ...(i)

211

2x t ...(ii)

32 m

10 ms–1

a = 1 ms–2

x

2132 10 0

2 t t

t2 – 20t + 64 = 0

20 400 256 20 12

2 2t

t = 4, 16

t = 4 s

15. Answer (4)

Hint : Stone will acquire velocity of frame of

reference when it leave frame of reference.

Solution : 2

0

1

2h u t gt

h

u0

2

0

1

2h u t gt

–3.2 = 6t – 5t2

5t2 – 6t – 3.2 = 0

t = 1.6 s

16. Answer (2)

Hint : Average speed = Total travelled distance

Total time taken

Solution : Time taken by cyclist A

362 h

18

Time taken by B cyclist from cycling

1 32 h

2 2

Average speed of B for actual riding

3624 km/h

3

2

17. Answer (1)

Hint : S = S1 + S

2 + S

3

Solution : 2

1

11 4 8 m

2 S

S2 = v

0t2 = (a

1t1)t

2 = 1 × 4 × 10 = 40 m

2

0 2 30 2v a S

3

164 m

2 2

S

S = 40 + 8 + 4 = 52 m

18. Answer (1)

Hint : A A

B B

a u

a u

Detailed solution : tA = t

B = t

0

0 = 20 – aAt0


5/17

0

20A

at

Similarly

0

10

Ba

t

20 2

10 1

A

B

a

a

19. Answer (3)

Hint : Total time (n – 1) T0 time interval

Solution : 21

2h gt

1

2

3

4

5

6

7

Total time = (7 – 1) T0

= 6 T0

21

2h gt

2

0

1{36 }

2h g T

2

018h gT

2

0180 18 10 T

T0 = 1 s

20. Answer (2)

Hint : dv

a vdx

Solution :

v(ms )–1

x(m)3 70

10

5

A

B

5

4

dv

dx

Velocity at t = 5 s

v = 7.5 ms–1

25 37.5(7.5) 9.4 ms

4 4

a

21. Answer (2)

Hint :

2

1

2

1

t

t

t

t

vdt

v

dt

Solution :

52

0

5

0

(30 4 2)t t dt

v

dt

53 2

0

30 42

3 2

(5 0)

t tt

3 2 5

0

1(10 2 2 )

5t t t

11[10 125 50 10] 262 ms

5

22. Answer (3)

Hint : Smaller least count.

Solution : Least count is minimum in = 0.001 m, so

it is most precise.

23. Answer (3)

Hint : In multiplication, product has number of

significant figures equal to least significant figure

present in calculation.

Solution :

V = IR

V = 1.25 × 25.425

= 31.78125

= 31.8 (3 significant figures)

24. Answer (3)

Hint : Limitations of dimensional analysis.

Solution :

1. Two and more physical quantities may have

same dimensional expression.

2. Numerical constant has no dimensions.

3. Method of dimensions can be used only for

product of physical quantities.


6/17

25. Answer (2)

Hint : n1u

1 = n

2u

2

Solution : 1 1 1

2 1

2 2 2

a b c

M L Tn n

M L T

h = 6.6 × 10–34 J-s

[h] = [M1L2T–1]

a = 1, b = 2, c = –1

1 2 1

34

2

kg s6.6 10

10 g 5m 100 s

mn

34 2 216.6 10 10 10

25

= 2.64 × 10–31

26. Answer (4)

Hint : Planck’s constant and angular momentum

have same dimensions.

Solution :

1. Energy density and pressure have same

dimensions.

2. Relative density and plane angle have no

dimension.

1 2 2[M L T ][ ]

[MK]

Q

Cm Q

0 2 2 1[M L T K ]

0 2 2[ ] [M L T ] QL

m

27. Answer (4)

Hint : n1u

1 = n

2u

2

Solution : [E] = [M1L2T–2]

2 kgm2s–2 = [2 kg]1 [2 m]2 [n s]–2

2

2

22 2

n

n2 = 4

n = 2, similarly we can find the relation for

momentum and power.

28. Answer (1)

Hint : Trignometric ratios have no dimensions

Solution : [LHS] = [RHS]

[y] = [Ax2]

1

2[ ] [ ]

yA L

x

[] = [M0L0T0]

[] = [M0L1T0]

[A2] = [M0L1T0]

29. Answer (4)

Hint : [L] = [pavbmc]

Solution : [L] = [M1L2T–1]

[p] = [M1L2T–3]

[v] = [M0L1T–1]

[m] = [M1L0T0]

[M1L2T–1] = k [M1L2T–3]a [M0L1T–1]b [M1L0T–0]c

1 = a + 0 + c

2 = 2a + b + 0

–1 = –3a – b

a = –1, b = 4, c = 2

30. Answer (3)

Hint :

2

21 1

2 2

lk mv m

t

Solution :

2

21 1

2 2

lk mv m

t

100 100 100 2 100 k m l t

k m l t

= 1% + 4% + 2% = 7%

31. Answer (4)

Hint : It has no dimension.

Solution : Solid angle 2

cos dsd

r

0 0 0

2

cos[ ] [M L T ]

dsd

r

32. Answer (4)

Hint : Fundamental forces.

Solution : Gravitational force is the weakest force.

Gravitational forces are central forces, electrostatic

forces are central forces, strong nuclear forces are

not central forces.

33. Answer (1)

34. Answer (1)

Hint : Mass and energy are inter-convertable.

Solution : Conservation laws in nature.

1. Law of conservation of energy

2. Law of conservation of linear momentum

3. Law of conservation of angular momentum

35. Answer (2)

Hint : Raman effect.

Solution : CV Raman won noble prize for scattering

of light by molecules.


7/17

36. Answer (3)

Hint : Non-zero digits are significant.

Solution : Trailing zeros in a number without

decimal are insignificant.

37. Answer (1)

Hint : z = xy

Z X Y

Z X Y

Solution : z = xy

Z X Y

Z X Y

0.4 0.1 1

4 1 5

Z

Z

1

4 5

Z

Z = 0.8 m

Z = (4 ± 0.8) m

38. Answer (4)

Hint : If there is no digit after 5 then last digit should

be even.

Solution : 2.765 rounding off (3 significant)

= 2.76

5.735 = 5.74

39. Answer (4)

Hint : Minimum percentage error.

Solution : Percentage error in 2.000 m

0.001100 0.05%

2.000

40. Answer (1)

Hint : The difference with average value should be

least.

Solution : t1 = |10.00 – 9.95| = 0.05 s

t2 = |10.00 – 10.10| = 0.10 s

41. Answer (3)

Hint : � � �

rel A Ba a a

Solution : � � �

rel A Ba a a

= g – g = 0

42. Answer (1)

Hint : L.C = 1MSD – 1VSD

Solution : L.C = 1MSD – 1VSD

VSDMSD 1

MSD

1 MSD = 1 mm

19 MSD = 20 VSD

1 VSD = 19

VSD20

L.C = 19

1 1 0.05 mm20

43. Answer (4)

44. Answer (2)

Hint : R = R1 + R

2

Solution : Req

= R1 + R

2

1 2

eq 1 2

0.1 0.1

30

R RR

R R R

eq

0.2 2100 100 %

30 3

R

R

45. Answer (3)

Hint : Mean absolute error

1 2| | | | | |

nx x x

n

Solution : xm

= 330 ms–1

x1 = |342 – 330| = 12

x2 = |338 – 330| = 8

x3 = |318 – 330| = 12

x4 = |322 – 330| = 8

112 8 12 810 ms

4

46. Answer (1)

Hint: Orbital angular momentum = hl l 1

2

Solution: For f - orbital, I = 3

Orbital angular momentum = h 3h3 3 1

2

[ CHEMISTRY]

47. Answer (3)

Hint: Chromium has half-filled d-subshell.

Solution: Cr(24) : [Ar]4s13d5

48. Answer (4)

Hint: n-factor for acid is number of replaceable H+

ions

Solution: n-factor = 1

Mol. wt 120Eq. wt. 120

n-factor 1


8/17

49. Answer (4)

Hint: No. of particles = No. of moles × NA

Solution:

Moles of oxygen atom = 0.3 × 14 = 4.2

No. of electrons of oxygen atoms = 4.2 × 8 × NA

= 33.6 NA

50. Answer (2)

Hint: x y 2 2 2

y yC H x O xCO H O

4 2

2

2

Moles of CO obtainedx

y 2 Moles of H O obtained

Solution:

8.8

x 144

5.4y 32

18

Empirical formula = CH3

Molecular wt. 2 30n 4

Empirical formulawt. 15

Molecular formula = n × empirical formula

= 4 (CH3) = C

4H

12

51. Answer (3)

Hint: Only four lines of Balmer series of hydrogen

atom lie in visible region

Solution: For n = 6 (5th excited state) transition to

n = 1

* Maximum possible transitions

= 6 1 5 + 4 + 3 + 2 + 1 =15

* Maximum Paschen transitions

= 3(6 3, 5 3, 4 3)

52. Answer (4)

Hint: For one electron species, energy of subshell

depends only on the value of n.

Solution: For 3rd shell, No. of degenerates orbitals = n2

= 32 = 9.

53. Answer (1)

Hint: Frequency 2

32

z

v znf

r nn

z

Solution:

2

2

2

3

He Li

2

Li He

3

3

T f 2 9

T f 322

1

54. Answer (1)

Hint: Dilution equation, M1V

1 = M

2V

2

Solution: For stock solution,

1

49 1.5 10M 7.5

98

Now, M1V

1 = M

2V

2

7.5 × V1 = 0.1 × 2.5 × 1000

V1 = 33.33 ml

55. Answer (2)

Hint: Normality = Molarity × n-factor

Solution: Molarity of

22

23

3

9.033 10

6.022 10OH 0.3 M

500 10

Molarity of Ca(OH)2

= 0.3

M2

Normality of Ca(OH)2

=

0.32 0.3 M

2

56. Answer (3)

Hint: No. of atoms = No. of molecules × atomicity

Solution: Remaining molecules of CO2

=

323 20440 10

6.022 10 1044

No. of remaining atoms = 59.22 × 1020 × 3

= 1.77 × 1022

57. Answer (4)

Hint: 2 2 3

1 1FeO O Fe O

4 2

Solution:

(72g)

(8g)

2 2 31 mole

0.25 mole

1 1FeO O Fe O

4 2

% increase in wt. 8

100 11.11%72

58. Answer (1)

Hint: h

mv

Solution: H2 has minimum molar mass so have

longest de-Broglie wavelength.

59. Answer (3)

Hint: s, p and d-subshells have 1, 3 and 5 orbitals

respectively.

Solution:

In l subshell, number of orbitals = 2l + 1


9/17

60. Answer (2)

Hint: Electrons in half filled p-subshell have different

orientations.

Solution: Orientation of orbitals is given by magnetic

quantum number.

61. Answer (2)

Hint: 2

2 2

1 2

1 1 1RZ

n n

Solution: For second line of Lyman series :

2

2 2

1 1 1R (1)

x 1 3

1 8R

x 9

9R

8x

For 3rd line of Lyman series : 2

2 2

1 1 1R (1)

1 4

15 9 15 135R

16 8x 16 128x

128x

135

62. Answer (3)

Hint: Ni2+ = [Ar] 3d8

Solution: 29

Cu = 1s2 2s2 2p6 3s2 3p6 4s1 3d10

s-electrons in Cu atom = 7

63. Answer (1)

Hint: 6

n

zv 2.18 10 m/sec

n

Solution: 6

4 He

8

22.18 10v 14

c 2753 10

64. Answer (1)

Hint: For minimum molar mass, one molecule

should contain at least one atom of oxygen

Solution:% of oxygen =wt.of oxygen

100wt.of compound

164 100

Mol.wt.

Minimum mol. wt. = 400 amu

65. Answer (4)

Hint: Molarity (M) = solute

n

V L

Solution: Mass of 1 ml D2O = 1 g

3

1

20M 50 M

1 10

66. Answer (2)

Hint: Law of multiple proportions is illustrated for

compounds which have two same elements.

Solution: CH4 : for 12 g C, wt. of H = 4 g = x

C2H

4 : for 12 g C, wt. of H = 2 g = y

x 4 2

y 2 1 = Simple whole no. ratio

67. Answer (4)

Hint: M1

V1 + M

2 V

2 = M

3 V

3

Solution: 0.5 × 2 + 1 × 1= M3 × 3

3

2M 0.67 M

3

68. Answer (3)

Hint: 2 2 3

32AI O Al O

2

Solution: 2 2 3

32Al O Al O

2

3 2

3KClO KCl O

2

From stoichiometry, 2 mole Al require 3

2 mole O

2

and 3

2 mole O

2 is obtained by 1 mole KCIO

3.

69. Answer (3)

Hint: Divalent metal chloride should be MCl2

Solution: Mol. wt.= 2 × vapour density = 60 × 2 = 120

x + 2 × 35.5 = 120

x = 49

Eq. wt. of metal 49

24.52

70. Answer (1)

Hint: Average atomic mass =

% abundance of isotope atomic mass

100

Solution: 13 x 12 (100 x)

12.011100

x = 1.1%


10/17

71. Answer (1)

Hint: Mass of electron is 9.1 × 10–31 kg

Solution: Positron is the particle having mass equal

to electron but having a unit positive charge

72. Answer (2)

Hint: -particle is He nucleus (He2+) which contains

2 protons and 2 neutrons.

Solution: For neutron, e

0m

For other particles, e

m order is

electron > proton > -particle

73. Answer (4)

Hint: l = 2 means d-subshell

Solution: d-subshell can have maximum 10

electrons.

74. Answer (2)

Hint: -rays, X-rays , UV, visible, IR, microwaves

Wavelength increases

Frequency decreases

Solution: Visible waves have higher frequency than

IR.

75. Answer (1)

Hint:

solvent

solute

solvent

X 1000molality(m) =

X MW

Solution: 0.2 1000

m 13.890.8 18

76. Answer (2)

Hint: Exchange energy of electrons is defined for

degenerate orbitals.

Solution: Exchange energy of electrons is defined

as energy released when an electron exchanges its

position with electron having same spin present in


77. Answer (2)

Hint: 21

KE mv2

, h

mv , v

Solution: 2v v mv 2 1 2

mv KEh h 2 h

27 7 1

34

23.313 10 10 sec

6.626 10

78. Answer (4)

Hint: h

x p4

Solution:h

x m v4

34

3 24

h 6.626 10x

1004 m v4 10 6.626 10

10x 2.5 10 m 2.5Å

79. Answer (4)

Hint: For principal quantum number n, l can be 0, 1,

2, ... (n – 1)

Solution: for n = 2,

l 2, it will have value 0 and 1.

80. Answer (1)

Hint:

2

n

nr 0.53 Å

Z

Solution: n = 2, Z = 4

2

4

2r 0.53 0.53 Å

4

81. Answer (2)

Hint: Ti : [Ar] 4s2 3d2

Solution:

Ti2+ : [Ar] 3d2

82. Answer (1)

Hint: 2

2 2

1 2

1 1E 13.6Z

n n

eV/atom

Solution: For, n1 = 4 and n

2 = 5

9E 13.6 eV/atom

16 25

83. Answer (4)

Hint: Electron density of the p-orbitals lie along the

axes.

Solution: dxy

orbital does not have electron density

along the axis.

84. Answer (3)

Hint: One AVOGRAM =

A

1

N


11/17

85. Answer (2)

Hint: % of an element wt.of element

100wt.of compound

Solution: For YO: % of oxygen =16

100M 16

160040

M 16

M 24

For second oxide: Let YOx

% of oxygen =

16x100

16x 24

1600x20

16x 24

3x

8

x 3 8 3

8

YO YO Y O

86. Answer (2)

Hint: No. of molecules = No. of moles × NA

Solution: H2

: No. of molecules A A

10N 5N

2

O2 : No. of molecules = 10 N

A

CO2

: No. of molecules = 8 N

A

SO2

: No. of molecules = A A

64N N

64 .

87. Answer (4)

Hint: Isoelectronic species have same number of

electrons.

Solution: Species Total electrons

N2

14

CO 14

14

17

2

2O

2O

CN–

14

88. Answer (1)

Hint: Ionisation energy 2

2

Z13.6 eV

n

Solution: Z = 2, n = 1

Ionisation energy

2

2

13.6 254.4eV

1

89. Answer (4)

Hint: Bohr’s model is valid only for one electron

species.

Solution: According to Bohr’s theory, angular

momentum of an electron is quantised.

90. Answer (3)

Hint: nhc

E

Solution: Energy released = 66.26 × 60

J100

34 8

9

60 n 6.626 10 3 1066.26

100 100 10

n = 2 × 1019

[ BIOLOGY]

91. Answer (2)

Hint: Nuclear membrane is found in all eukaryotes.

Solution: Mycoplasma, BGA, Bacillus, purple

photosynthetic bacteria & E.coli are prokaryotes.

92. Answer (3)

Hint: Body of plants and animals are composed of

cells & product of cells was proposed by Theodor

Schwann.

Solution: Theodor Schwann was a British zoologist.

93. Answer (4)

Hint: Chloroplast is the site of photosynthesis and

contain pigments in eukaryotes.

Solution: Chromatophores are membranous

extension in the cytoplasm of a cyanobacterial cell

which have photosynthetic pigments

94. Answer (2)

Hint: This inclusion body provides buoyancy to the

bacteria

Solution: Ribosomes are found in all prokaryotes

and eukaryotes. Cyanophycean granules are found in

cyanobacteria only while gas vacuoles are found in

blue-green algae (BGA) as well as in purple & green

photosynthetic bacteria

95. Answer (3)

Hint: Lipid molecules of plasma membrane have

polar head and non-polar tail.

Solution: Polar head is hydrophilic in nature and it

interacts with water. Rest all the statements

regarding plasma membranes are true.


12/17

96. Answer (4)

Hint: Movement of molecules across the membrane

without help of carrier proteins and ATP is called

simple diffusion

Solution: Simple diffusion occurs only for those

molecules which are neutral or non-polar, while

transport of polar and hydrophilic substances need

carrier proteins.

97. Answer (2)

Hint: Smallest cell organelle is known as organelle

within organelle.

Solution: Ribosome is known as organelle within

organelle and its r-RNA part is synthesized inside the

nucleolus.

98. Answer (3)

Hint: These structures are present in pits.

Solution: Symplast of two adjacent cells are

connected via cytoplasmic strands or

plasmodesmata. They are lined by plasma

membrane.

99. Answer (3)

Hint: The given figure is of golgi bodies

Solution: Enzymatic precursors for lysosomes are

synthesized in ER. Golgi does processing and

packaging of materials for intra as well as extra-cellular

targets.

Rest all the features of golgi are true.

100. Answer (4)

Hint: ER, Golgi, lysosomes & vacuoles function in

a coordinated manner.

Solution: Oxidation of fatty acids, proteins and

carbohydrate occurs inside the mitochondria.

Rest all functions are performed by organelles of

endomembrane system.

101. Answer (3)

Hint: This organelle has hydrolytic enzymes which

become functional at acidic pH.

Solution: Lysosomes have hydrolytic enzymes for

digestion of almost all types of macromolecules

which are functional at acidic pH.

102. Answer (3)

Hint: This organelle is found in almost all eukaryotic

cells and is site of ATP synthesis.

Solution: Mitochondria are sausage shaped, have

their own 70S ribosomes i.e, palade particles and

they are viewed after staining with Janus green.

Usually their number is high in those cells which

have high metabolism.

103. Answer (2)

Hint: ‘S’ is a unit

Solution: ‘S’ stands for Svedberg coefficient or

sedimentation coefficient and it is indirect measure of

density and size of ribosomal sub units.

104. Answer (2)

Hint: These structures are absent in prokaryotes.

Solution: Cytoskeleton are proteinaceous

filamentous structures which provide mechanical

strength & support to the cell.

105. Answer (3)

Hint: Microfilaments are solid unbranched rod like

fibrils.

Solution: Microtubules have diameter of 25 nm.

Intermediate filaments are involved in formation of

scaffolds of chromatin.

106. Answer (4)

Hint: This structure is absent in higher plants.

Solution: Centrioles have 9 peripheral fibrils of

tubulin and these are absent in the centre therefore

the arrangement is 9 + 0.

107. Answer (4)

Hint: In leucoplast granum remains absent.

Solution: Protoplast is a cell without cell wall.

108. Answer (2)

Hint: Both mitochondria and chloroplast are semi

autonomous structures.

Solution: Mitochondria, chloroplast and bacteria all

have ds circular DNA, 70S ribosomes porins on

outer membrane and self duplication ability.

Mitochondria and chloroplast are partially dependent

on nucleus.

109. Answer (4)

Hint: Polyribosomes are formed in cytoplasm

Solution: Polysomes are not formed with the help of

RER

110. Answer (3)

Hint: Nucleolus is found inside the nucleus

Solution: Nucleolus is non-membrane bound

structure found in nucleoplasm of nucleus. They are

larger in cells involved in protein synthesis.

111. Answer (4)

Hint: Chromatin is packed DNA

Solution: Packed DNA has RNA, histones and

some non-histone proteins.


13/17

112. Answer (1)

Hint: Chromatids of chromosomes are held together

at primary constriction.

Solution: Primary constriction is known as

centromere which holds the two halves of a

chromosome.

113. Answer (2)

Hint: Such chromosomes appear L-shaped during

anaphase.

Solution: A chromosome with centromere slightly

away from the centre is submetacentric chromosome.

114. Answer (4)

Hint: Loops of lampbrush chromosomes have hair

like structure and with protein that are known as

informosomes.

Solution: Lampbrush chromosomes form loops

which have hair like structures. These hairs are

bound to protein. Some of which are called

informosome i.e, having mRNA + protein. These

loops participate in transcription and form m-RNA

115. Answer (3)

Hint: These structures help in photorespiration

along with chloroplast and mitochondria.

Solution: Peroxisomes are involved in photorespiration

because they have enzymes for formation as well

as destruction of peroxide.

116. Answer (3)

Hint: These structures are tubular and help bacteria

to attach with rocks also.

Solution: Fimbriae are bristle like structures which

help bacteria to attach with rocks or host cell.

117. Answer (2)

Hint: The cell also has cell wall with plasmodesmata.

Solution: Plant cells have cell wall with

plasmodesmata. Their 90% volume is occupied by

vacuoles.

118. Answer (4)

Hint: Bacterial type ribosomes are also found in

some cell organelles of eukaryotes.

Solution: Both ribosomes and plasma membrane

are similar in prokaryotes and eukaryotes.

119. Answer (2)

Hint: Crossing over occurs in pachytene stage.

Solution: Point at which crossing over occurs forms

the recombination nodule.

120. Answer (3)

Hint: Spindle fibres are not directly attached to the

centromere.

Solution: A disc-shaped structure is found over

centromere through which spindle fibres are attached

called kinetochore.

121. Answer (2)

Hint: Meiosis involves two sequential karyokinesis

and cytokinesis.

Solution: Except DNA duplication and histone

protein synthesis, rest all phenomenon occur twice

in meiosis.

122. Answer (4)

Hint: Tubulin protein synthesis occurs in the stage

where duplication of mitochondria & chloroplast

occurs.

Solution: Most of the cell organelles get duplicated

in G1. But Golgi, chloroplast and mitochondria are

doubled at G2 phase along with tubulin protein

synthesis.

123. Answer (3)

Hint: Interphase is known as the most active stage

of cell cycle.

Solution: Interphase constitute more than 95%

duration of cell cycle of a human cell. Rest all the

features regarding the interphase are true.

124. Answer (3)

Hint: Last phase of karyokinesis in mitosis involves

reappearance of cell organelles and nuclear

membrane.

Solution: At the end of mitosis, the chromosomes

decondense into chromatin which occurs in

telophase.

125. Answer (1)

Hint: Major restriction point inhibits cells to go for

DNA synthesis.

Solution: Major control on cell cycle can be

imposed on 1

G S transition step.

126. Answer (3)

Hint: Number of mitotic divisions to form ‘n’ number

of cells are = n – 1

Solution: Number of generations (n) required to form

‘x’ number of cells are = 2n

For 32 cells

Mitotic divisions = 32 – 1 = 31

Number of generations = 25 = 32, x = 5


14/17

127. Answer (2)

Hint: The given stage is the best stage to study

morphology of chromosomes.

Solution: Given stage is Metaphase I. where,

Bivalent chromosomes align at equator forming

two metaphasic plates

Alignment of bivalents is totally random process.

Spindle fibre attaches to each chromosome of

homologous pair

Separation of chromosomes occur at anaphase I

128. Answer (2)

Hint: Each chromosome has 2 chromatids

Solution: 1 Bivalent = 2 chromosomes

20 Bivalent = 20 × 2 = 40 chromosomes

Since 1 chromosome = 2 chromatids

40 chromosomes = 80 chromatids

129. Answer (2)

Hint: Recombinase enzyme catalyses the process

occurring in pachytene stage.

Solution: Crossing over is enzyme mediated

process and it produces recombinants.

130. Answer (1)

Hint: Dyad of cells appear in last stage of heterotypic

division

Solution: Dyad of cells are formed at the end of

telophase I.

131. Answer (2)

Hint: Cell plate represents the future middle lamella

of adjacent cells.

Solution: Cell plate help in cytokinesis of plant cells

which represent the middle lamella of adjacent cells.

132. Answer (2)

Hint: Treatment of colchicine produces polyploid

condition.

Solution: Colchicine inhibits formation of tubulin and

therefore microtubules. Lack of microtubules results

in inhibition of spindle fibres formation and its affect

will be visible first at metaphase.

133. Answer (4)

Hint: Mitosis does not involve crossing over.

Solution: Mitosis helps in reproduction in unicellular

organisms. It helps in repair and regeneration but

cannot produce recombinants.

134. Answer (2)

Hint: Interkinesis is resting phase between meiosis

I and II.

Solution: Meiosis I and II involves into sequential

karyokinesis. Chromosomes do not decondense upto

DNA level in interkinesis because after telophase I

chromatin recondense in prophase II.

135. Answer (3)

Hint: Meiosis involves crossing over as well as

reduction of chromosome number.

Solution: Four cells produced after telophase II are

genetically dissimilar to each other as well as to

their parents.

136. Answer (2)

Hint : Compounds whose role or function we do not

understand at the moment.

Solution : Primary metabolites are essential for

physiological processes and have identifiable

functions. Secondary metabolities include alkaloids,

antibiotics, rubber, essential oils, spices etc.

137. Answer (3)

Hint : Order of percentage of biomolecules in cells.

Solution : Water > proteins > nucleic acids >

carbohydrates > lipids > ions

70 90% 10 15% 5 7% 3% 2% 1%

A C B D E

138. Answer (4)

Hint : Polymers are found in retentate/acid insoluble

fraction.

Solution : Cysteine, calcium & cytosine are

micromolecules that are obtained in filtrate/acid

soluble fraction. Cellulose is a polymer of glucose

and due to its large size it cannot cross the filtration

membrane.

139. Answer (2)

Hint : Bones are reservoirs of some elements.

Solution : Matrix of bone is largely composed of

hydroxyapatite crystals of calcium phosphate.

Essential fatty acids & essential amino acids are

supplied in diet.

140. Answer (4)

Hint : Identify a molecule which is a monomer.

Solution : Inulin is a polymer of fructose, while

insulin is a polymer of amino acids. Lactose is a

disaccharide. Polymers can be broken into

monomers by addition of water (hydrolysis).

141. Answer (1)

Hint : Dehydration is a condensation reaction

involving loss of water molecule.

Solution : Cystine formation requires removal of H2

and disulfide bonds are formed between two amino

acids. Formation of lecithin, adenosine and collagen

requires formation of ester & glycosidic bonds

respectively.


15/17

142. Answer (3)

Hint : Structural sugar found in RNA.

Solution : Cellulose is unbranched structural

homopolymer of glucose comprising of (1-4)

glycosidic linkages. Ribose is a monomer formed

without glycosidic bond. Cytidylic acid is a

nucleotide found in RNA.

143. Answer (1)

Hint : Reducing sugar.

Solution : Sugars with free aldehyde or ketone

groups are detected by Benedict’s and Fehling’s

solution. Glucose and galactose have free aldehyde

group while fructose has free keto group. Sucrose

has no free keto or aldehyde group.

144. Answer (1)

Hint : Isomerism in hexose sugars.

Solution : Galactose & glucose are aldose sugars

while fructose is ketohexose. Ribose is a pentose

sugar.

145. Answer (1)

Hint : Pentose sugar derived from C5H

10O

5.

Solution : 5 carbon sugar in RNA (ribonucleic acid)

is ribose. DNA has deoxyribose with molecular

formula C5H

10O

4.

146. Answer (3)

Hint : Identify a nucleotide. Ribozyme is RNA acting

as enzyme.

Solution : ATP is adenosine triphosphate.

147. Answer (1)

Hint : Formation of polymers and disaccharides

requires glycosidic bonds.

Solution : Glucose is a monomer of cellulose but

cellulose is unbranched homopolymer glucose.

148. Answer (2)

Hint : Unsaturated fatty acids have low melting

points.

Solution : Gingelly oil/sesame oil and arachidonic

acid are unsaturated fatty acids, hence will have

lower melting points in comparison to other saturated

lipids given.

149. Answer (3)

Hint : Double bonds cannot be introduced beyond 9th

carbon position in most animals usually.

Solution : Linoleic acid has 2 double bonds

(9 & 12th carbon). Plants are able to synthesize

essential fatty acids by introducing double bonds at

12th & 15th carbon. Oleic acid has a single double

bond at 9th position. In most mammals, arachidonic

acid can be formed from linoleic acid.

150. Answer (1)

Hint : Identify a homopolymer.

Solution : Chitin comprises of repeating units of (1

4) N-acetylglucosamine. Nucleic acids (DNA),

proteins i.e. RuBisCO and insulin are

heteropolymers.

151. Answer (4)

Hint : Organic cofactor loosely attached to

apoenzyme.

Solution : Niacin and riboflavin are vitamins requires

for formation of coenzymes such as NAD, FAD &

FMN.

152. Answer (3)

Hint : First alphabet of amino acid is considered as

the basis of single alphabet code.

Solution : Tyrosine is coded by Y, phenylalanine by

‘F’ and glutamic acid by E. This is because

threonine, proline & glycine have already been named

as T, P & G respectively.

153. Answer (1)

Hint : Two or more amino acids can be linked by

peptide bonds.

Solution : Collagen is a heteropolymer of amino

acids. Chitin & cellulose are polysaccharides whose

monomers are linked by glycosidic bonds. Choline is

nitrogenous compound found in lecithin.

154. Answer (4)

Hint : All components given are sugars in raffinose.

Solution : Tripeptide has 3 amino acids &

triglyceride has 3 fatty acids linked to glycerol.

155. Answer (4)

Hint : Active site are formed at tertiary level of

protein organisation. Hydrogen bonds stabilise

predominantly protein structure formed at secondary

level of organisation.

Solution : Ionic bonds, Van der Waals forces,

hydrophobic interactions along with hydrogen bonds

stabilise quaternary structure.

156. Answer (3)

Hint : DNA of a eukaryote comprises of A, T, G, C

as nitrogenous base.

Solution : Cytosine is a nitrogenous base while

cysteine is an amino acid.

157. Answer (2)

Hint : Amino acids are monomers of proteins.

Solution : Amino acids are substituted methanes.

They exist as zwitterions at isoelectric pH.

Trihydroxypropane is glycerol. DNA & RNA are

negatively charged molecules.


16/17

158. Answer (4)

Hint : Identify odd one out w.r.t. macromolecules.

Solution : Lipids are neither polymer nor

macromolecules.

159. Answer (2)

Hint : Thymine is exclusive to DNA.

Solution : Thiamine is vitamin B1 while thymine is

nitrogenous base in DNA.

160. Answer (3)

Hint : Hydrolytic enzymes work by addition of water

to break the bond between biomolecules.

Solution : Hydrolases belong to class III according

to nomenclature given by IUB. Phosphodiesterases

also belong to this class of enzymes.

161. Answer (3)

Hint : While overcoming competitive inhibition Km

value increases.

Solution : Succinate is the substrate for enzyme

succinate dehydrogenase. Malonate is its

competitive inhibitor.

162. Answer (2)

Hint : Positional information of amino acids decides

specificity of enzyme action.

Solution : Apoenzymes are proteinaceous part of

holoenzymes. Enzymes never alter the equilibrium of

the reaction, they act by lowering activation energy

barrier.

163. Answer (3)

Hint : White fat is unilocular and adipocytes are

signet ring shaped cells.

Solution : Brown fat in newly born prevents shivering

in them. White fat acts as a reservior of energy and

the triglycerides are hydrolysed upon need.

164. Answer (4)

Hint : Intercalated discs are interdigitations between

adjacent cells that act as boosters of cardiac

impulse.

Solution : Cardiac fibres are striated and the edges

of sarcomeres are ‘Z’ lines. Connexons are part of

gap junctions present at intercalated discs. Visceral

muscles are uninucleated.

165. Answer (1)

Hint : Muscle fibres associated with limbs and heart

are striped in appearance.

Solution : Striated muscles include both cardiac &

skeletal muscles. The alternate light and dark bands

refer to the horizontal arrangement of actin & myosin

filaments.

166. Answer (2)

Hint : Muscles under the control of our will are

voluntary.

Solution : Smooth and cardiac muscles are

involuntary in nature i.e. they are not under the

control of our will. Smooth muscles contain

unbranched, unstriped, uninucleate fibres.

167. Answer (4)

Hint : Concentric arrangement of calcium salts

surrounding osteocytes.

Solution : The lamellae refer to concentric

arrangement of calcium phosphate salts surrounding

osteocytes in diaphysis of long bones of mammals.

168. Answer (2)

Hint : Exocrine glands release/pour their secretions

through ducts.

Solution : Insulin is a secretion of ductless part of

gland pancreas. Saliva, earwax and milk are

secretions of exocrine glands.

169. Answer (3)

Hint : Endocrine glands lack ducts.

Solution : Endocrine glands do not have ducts and

release their secretion into blood capillaries.

170. Answer (2)

Hint : Multiple layer in epithelial tissue are effective

in protection.

Solution : At surfaces prone to stress (wear & tear)

or abrasion compound epithelium is effective.

171. Answer (2)

Hint : Junctions involved in cementing neighbouring

cells.

Solution : Tight junctions help to stop substances

from leaking across a tissue.

172. Answer (1)

Hint : Stratified i.e. compound epithelium lines

surfaces that face mechanical stress.

Solution : Stratum germinativum comprises of cells

that divide regularly and forms the bottommost layer

of straitified epithelium. Diffusion surfaces and tubular

parts of nephron are lined by single layered simple

epithelium.

173. Answer (4)

Hint : Wine flask shaped cells of alimentary canal.

Solution : Goblet cells are mucus secreting

unicellular glands of alimentary canal. Cerumen and

sweat are secreted by sebaceous & sudorific glands

respectively. Hormones are secretions of endocrine

glands.


17/17

174. Answer (2)

Hint : Neurilemma is a contribution of Schwann

cells.

Solution : Schwann cells form myelin sheath &

neurilemma in neurons of PNS. Oligodendrocytes

form myelin sheath in CNS. Neurilemma is absent in

CNS. Oligodendrocytes are branched cells while

Schwann cells are not branched.

175. Answer (1)

Hint : Nissl's granules are clusters of RNA &

ribosomes.

Solution : Nissl's granules are site of protein

synthesis located in cell body of neuron/soma & in

dendrites (afferent processes). They are absent in

afferent processes also called axons.

176. Answer (4)

Hint : Response to stimulus is a feature of structural

& functional units of neural system.

Solution : Extensibility is a property of muscle fibres

while neurons exhibit both excitability & conductivity.

177. Answer (1)

Hint : I band is composed of actin filaments while A

band comprises of both actin & myosin filaments.

Solution : Length of both actin & myosin filaments

remain constant. Size of A band also remains

contant. Only the extent of overlap between actin &

myosin filaments increases thereby reducing size of

Henson’s zone.

178. Answer (2)

Hint : Brush border membrane (BBM) due to

microvilli increase surface area for reabsorption &

absorption.

Solution : Cilia in trachea push mucus with

entrapped particles in specific (upward) direction.

BBM cuboidal is a feature of tubular part of nephrons

while BBM columnar epithelium dominants in

intestine.

179. Answer (1)

Hint : Keratinised epithelium occurs on water

resistant surfaces in living beings.

Solution : Stratum corneum is the topmost layer of

stratified keratinised epithelium. Pharynx, tongue &

cornea are non-keratinised epithelial surfaces.

Microvilli in alimentary canal support absorption of

nutrients not reabsorption. Transitional epithelium is

distensible/stretchable. Little intercellular matrix is

present in epithelial tissue.

180. Answer (4)

Hint : ‘Osteo’ refers to bone & clasts refers to

destruction/demineralisation.

Solution : Osteocytes & osteoblasts are bone

forming cells while osteoclasts are bone dissolving/

dissolution cells that decrease bone calcium.

��

Test - 1 (Code-D) (Answers) All India Aakash Test Series for Medical-2020

1/17

1. (3)

2. (2)

3. (4)

4. (1)

5. (3)

6. (1)

7. (4)

8. (4)

9. (1)

10. (3)

11. (2)

12. (1)

13. (1)

14. (4)

15. (4)

16. (3)

17. (4)

18. (1)

19. (4)

20. (4)

21. (2)

22. (3)

23. (3)

24. (3)

25. (2)

26. (2)

27. (3)

28. (1)

29. (1)

30. (2)

31. (4)

32. (2)

33. (3)

34. (2)

35. (1)

36. (3)

Test Date : 14/10/2018

ANSWERS

TEST - 1 (Code-D)

All India Aakash Test Series for Medical-2020

37. (2)

38. (2)

39. (2)

40. (1)

41. (2)

42. (1)

43. (1)

44. (4)

45. (3)

46. (3)

47. (4)

48. (1)

49. (4)

50. (2)

51. (2)

52. (3)

53. (4)

54. (1)

55. (2)

56. (1)

57. (4)

58. (4)

59. (2)

60. (2)

61. (1)

62. (2)

63. (4)

64. (2)

65. (1)

66. (1)

67. (3)

68. (3)

69. (4)

70. (2)

71. (4)

72. (1)

73. (1)

74. (3)

75. (2)

76. (2)

77. (3)

78. (1)

79. (4)

80. (3)

81. (2)

82. (1)

83. (1)

84. (4)

85. (3)

86. (2)

87. (4)

88. (4)

89. (3)

90. (1)

91. (3)

92. (2)

93. (4)

94. (2)

95. (2)

96. (1)

97. (2)

98. (2)

99. (2)

100. (3)

101. (1)

102. (3)

103. (3)

104. (4)

105. (2)

106. (3)

107. (2)

108. (4)

109. (2)

110. (3)

111. (3)

112. (4)

113. (2)

114. (1)

115. (4)

116. (3)

117. (4)

118. (2)

119. (4)

120. (4)

121. (3)

122. (2)

123. (2)

124. (3)

125. (3)

126. (4)

127. (3)

128. (3)

129. (2)

130. (4)

131. (3)

132. (2)

133. (4)

134. (3)

135. (2)

136. (4)

137. (1)

138. (2)

139. (1)

140. (4)

141. (1)

142. (2)

143. (4)

144. (1)

145. (2)

146. (2)

147. (3)

148. (2)

149. (4)

150. (2)

151. (1)

152. (4)

153. (3)

154. (2)

155. (3)

156. (3)

157. (2)

158. (4)

159. (2)

160. (3)

161. (4)

162. (4)

163. (1)

164. (3)

165. (4)

166. (1)

167. (3)

168. (2)

169. (1)

170. (3)

171. (1)

172. (1)

173. (1)

174. (3)

175. (1)

176. (4)

177. (2)

178. (4)

179. (3)

180. (2)

All India Aakash Test Series for Medical-2020 Test - 1 (Code-D) (Answers & Hints)

2/17

ANSWERS & HINTS

1. Answer (3)

Hint : Mean absolute error

1 2| | | | | |

nx x x

n

Solution : xm

= 330 ms–1

x1

= |342 – 330| = 12

x2

= |338 – 330| = 8

x3

= |318 – 330| = 12

x4

= |322 – 330| = 8

112 8 12 810 ms

4

2. Answer (2)

Hint : R = R1

+ R2

Solution : Req

= R1

+ R2

1 2

eq 1 2

0.1 0.1

30

R RR

R R R

eq

0.2 2100 100 %

30 3

R

R

3. Answer (4)

4. Answer (1)

Hint : L.C = 1MSD – 1VSD

Solution : L.C = 1MSD – 1VSD

VSDMSD 1

MSD

1 MSD = 1 mm

19 MSD = 20 VSD

1 VSD = 19

VSD20

L.C = 19

1 1 0.05 mm20

5. Answer (3)

Hint : � � �

rel A Ba a a

Solution : � � �

rel A Ba a a

= g – g = 0

6. Answer (1)

Hint : The difference with average value should be

least.

Solution : t1

= |10.00 – 9.95| = 0.05 s

t2

= |10.00 – 10.10| = 0.10 s

[ PHYSICS]

7. Answer (4)

Hint : Minimum percentage error.

Solution : Percentage error in 2.000 m

0.001100 0.05%

2.000

8. Answer (4)

Hint : If there is no digit after 5 then last digit should

be even.

Solution : 2.765 rounding off (3 significant)

= 2.76

5.735 = 5.74

9. Answer (1)

Hint : z = xy

Z X Y

Z X Y

Solution : z = xy

Z X Y

Z X Y

0.4 0.1 1

4 1 5

Z

Z

1

4 5

Z

Z = 0.8 m

Z = (4 ± 0.8) m

10. Answer (3)

Hint : Non-zero digits are significant.

Solution : Trailing zeros in a number without

decimal are insignificant.

11. Answer (2)

Hint : Raman effect.

Solution : CV Raman won noble prize for scattering

of light by molecules.

12. Answer (1)

Hint : Mass and energy are inter-convertable.

Solution : Conservation laws in nature.

1. Law of conservation of energy

2. Law of conservation of linear momentum

3. Law of conservation of angular momentum

13. Answer (1)

Test - 1 (Code-D) (Answers & Hints) All India Aakash Test Series for Medical-2020

3/17

14. Answer (4)

Hint : Fundamental forces.

Solution : Gravitational force is the weakest force.

Gravitational forces are central forces, electrostatic

forces are central forces, strong nuclear forces are

not central forces.

15. Answer (4)

Hint : It has no dimension.

Solution : Solid angle 2

cos dsd

r

0 0 0

2

cos[ ] [M L T ]

dsd

r

16. Answer (3)

Hint :

2

21 1

2 2

lk mv m

t

Solution :

2

21 1

2 2

lk mv m

t

100 100 100 2 100 k m l t

k m l t

= 1% + 4% + 2% = 7%

17. Answer (4)

Hint : [L] = [pav

bm

c]

Solution : [L] = [M1L2T–1]

[p] = [M1L2T–3]

[v] = [M0L1T–1]

[m] = [M1L0T0]

[M1L2T–1] = k [M1L2T–3]a [M0L1T–1]b [M1L0T–0]c

1 = a + 0 + c

2 = 2a + b + 0

–1 = –3a – b

a = –1, b = 4, c = 2

18. Answer (1)

Hint : Trignometric ratios have no dimensions

Solution : [LHS] = [RHS]

[y] = [Ax2]

1

2[ ] [ ]

yA L

x

[] = [M0L0T0]

[] = [M0L1T0]

[A2] = [M0L1T0]

19. Answer (4)

Hint : n1

u1

= n2

u2

Solution : [E] = [M1L2T–2]

2 kgm2s–2 = [2 kg]1 [2 m]2 [n s]–2

2

2

22 2

n

n2 = 4

n = 2, similarly we can find the relation for

momentum and power.

20. Answer (4)

Hint : Planck’s constant and angular momentum

have same dimensions.

Solution :

1. Energy density and pressure have same

dimensions.

2. Relative density and plane angle have no

dimension.

1 2 2[M L T ][ ]

[MK]

Q

Cm Q

0 2 2 1[M L T K ]

0 2 2[ ] [M L T ] QL

m

21. Answer (2)

Hint : n1

u1

= n2

u2

Solution : 1 1 1

2 1

2 2 2

a b c

M L Tn n

M L T

h = 6.6 × 10–34 J-s

[h] = [M1L2T–1]

a = 1, b = 2, c = –1

1 2 1

34

2

kg s6.6 10

10 g 5m 100 s

mn

34 2 216.6 10 10 10

25

= 2.64 × 10–31

22. Answer (3)

Hint : Limitations of dimensional analysis.

Solution :

1. Two and more physical quantities may have

same dimensional expression.

2. Numerical constant has no dimensions.

3. Method of dimensions can be used only for

product of physical quantities.

23. Answer (3)

Hint : In multiplication, product has number of

significant figures equal to least significant figure

present in calculation.


4/17

Solution :

V = IR

V = 1.25 × 25.425

= 31.78125

= 31.8 (3 significant figures)

24. Answer (3)

Hint : Smaller least count.

Solution : Least count is minimum in = 0.001 m, so

it is most precise.

25. Answer (2)

Hint :

2

1

2

1

t

t

t

t

vdt

v

dt

Solution :

52

0

5

0

(30 4 2)t t dt

v

dt

53 2

0

30 42

3 2

(5 0)

t tt

3 2 5

0

1(10 2 2 )

5t t t

11[10 125 50 10] 262 ms

5

26. Answer (2)

Hint : dv

a vdx

Solution :

v(ms )–1

x(m)3 70

10

5

A

B

5

4

dv

dx

Velocity at t = 5 s

v = 7.5 ms–1

25 37.5(7.5) 9.4 ms

4 4

a

27. Answer (3)

Hint : Total time (n – 1) T0

time interval

Solution : 21

2h gt

1

2

3

4

5

6

7

Total time = (7 – 1) T0

= 6 T0

21

2h gt

2

0

1{36 }

2h g T

2

018h gT

2

0180 18 10 T

T0

= 1 s

28. Answer (1)

Hint : A A

B B

a u

a u

Detailed solution : tA = t

B = t

0

0 = 20 – aAt0

0

20

Aa

t

Similarly

0

10B

at

20 2

10 1

A

B

a

a

29. Answer (1)

Hint : S = S1

+ S2

+ S3

Solution : 2

1

11 4 8 m

2 S

S2

= v0

t2

= (a1

t1

)t2

= 1 × 4 × 10 = 40 m

2

0 2 30 2v a S

3

164 m

2 2

S

S = 40 + 8 + 4 = 52 m


5/17

30. Answer (2)

Hint : Average speed = Total travelled distance

Total time taken

Solution : Time taken by cyclist A

362 h

18

Time taken by B cyclist from cycling

1 32 h

2 2

Average speed of B for actual riding

3624 km/h

3

2

31. Answer (4)

Hint : Stone will acquire velocity of frame of

reference when it leave frame of reference.

Solution : 2

0

1

2h u t gt

h

u0

2

0

1

2h u t gt

–3.2 = 6t – 5t2

5t2 – 6t – 3.2 = 0

t = 1.6 s

32. Answer (2)

Hint : d + x = v0

t ...(i)

21

2x at ...(ii)

Solution : 32 + x = 10t ...(i)

211

2x t ...(ii)

32 m

10 ms–1

a = 1 ms–2

x

2132 10 0

2 t t

t2 – 20t + 64 = 0

20 400 256 20 12

2 2t

t = 4, 16

t = 4 s

33. Answer (3)

Hint : 2 2

2 1 02v v a l

Detailed solution : 2 2

2 1 02v v a l

2 2

2 1

02

v vl

a

l

2

2 2

1 02

2

lv v a

2 2 2 2

2 2 2 1 1 2

1 0

02 2

v v v vv v a

a

2 2

1 2

2

v vv

34. Answer (2)

Hint : 2

0

1

2h v t gt

Solution :

7

8

5

4

h

t = 6

Particle will be at maximum height at t = 6 s.

2110 6 180 m.

2 h

35. Answer (1)

Hint : Reaction time

Solution:

Distance travelled during reaction time

S1

= 10 × 1 = 10 m ...(i)

Distance travelled during retarded motion

2

2

10050 m

2 2 1

vS

a

...(ii)


6/17

Total travelled distance = S1

+ S2

= 60 m

Distance of car from red signal

= 70 – 60 = 10 m

36. Answer (3)

Hint : | Average velocity | = |Total displacement travelled|

Total time taken

Solution:

ABO

| Displacement | = 10 m

Time taken = 5

10 2

s

| Average velocity | = 110 20

ms

2

37. Answer (2)

Hint : 2

0

1

2S S ut at

Solution:

v

tt0

v0

0

0

0

(constant)

va

t

2

0

1

2x x ut at

20

0

0

10

2

vx x t

t

20

0

0

1

2

vx x t

t

38. Answer (2)

Hint : Average speed = Total distance travelled

Total time taken

Detailed Solution : From t = 0 to t = 4 s x1

= 4 m

From t = 4 s to t = 6 s x2

= 0

From t = 6 s to t = 10 s x3

= 4 m

Average speed = 14 0 4 8 4

ms10 10 5

39. Answer (2)

Hint : slope of graphdv

a v tdt

Solution : a = slope of v – t graph

= –ve value (constant)

t

a

0

40. Answer (1)

Hint : dv

a vdx

Solution:

0 0

1v x

v x

...(i)

0

0

0

v

v x v

x

...(ii)

0

0

vdv

dx x ...(iii)

0 0

0

0 0

v vdva v x v

dx x x

2 2

0 0

200

v v

x

xx

41. Answer (2)

Hint : v

at

Solution : v = t2 + 4t

At t = 1 s, v = 5 ms–1

t = 2 s, v = 12 ms–1

212 57 ms

2 1

a

42. Answer (1)

Hint : 1 2

1 2

2

av

v vv

v v

Solution : v1

= 36 km/h = 10 ms–1

v2

= 54 km/h = 15 ms–1

1 2

1 2

av

S Sv

t t

1 1 1

210

2 10

S

SS t t


7/17

46. Answer (3)

Hint: nhc

E

Solution: Energy released = 66.26 × 60

J100

34 8

9

60 n 6.626 10 3 1066.26

100 100 10

n = 2 × 1019

47. Answer (4)

Hint: Bohr’s model is valid only for one electron

species.

Solution: According to Bohr’s theory, angular

momentum of an electron is quantised.

[ CHEMISTRY]

2 2 2

215

2 15

S

SS t t

112 ms

1 1

2 10 15

av

Sv

S

43. Answer (1)

Hint : max

2 v S

Solution :

u = 0 vmax v = 0

S2

S2

S1

S1

2

2 max

max 1 12

2

v

v S S ...(i)

2

2 max

max 2 20 2

2

vv S S

...(ii)

2

max

1 2

1 1

2

vS S S

max

2 2 2 3 100

5

v S

4 15 m/s

44. Answer (4)

Hint : Displacement and velocity may be positive or

negative.

2nd

4th

0

+ve

s

v

Solution : Direction of displacement is vertically

downward.

S < 0

Direction of velocity is downward

So, v < 0

45. Answer (3)

Hint : The interval in which displacement is zero.

Solution : For the curve, ACB displacement is zero

AB

C

Tt

x

0

0

��

�

av

SV

t

48. Answer (1)

Hint: Ionisation energy 2

2

Z13.6 eV

n

Solution: Z = 2, n = 1

Ionisation energy

2

2

13.6 254.4eV

1

49. Answer (4)

Hint: Isoelectronic species have same number of

electrons.

Solution: Species Total electrons

N2

14

CO 14

14

17

2

2O

2O

CN–

14


8/17

50. Answer (2)

Hint: No. of molecules = No. of moles × NA

Solution: H2

: No. of molecules A A

10N 5N

2

O2

: No. of molecules = 10 NA

CO2

:

No. of molecules = 8 NA

SO2

: No. of molecules = A A

64N N

64 .

51. Answer (2)

Hint: % of an element wt.of element

100wt.of compound

Solution: For YO: % of oxygen =16

100M 16

160040

M 16

M 24

For second oxide: Let YOx

% of oxygen =

16x100

16x 24

1600x20

16x 24

3x

8

x 3 8 3

8

YO YO Y O

52. Answer (3)

Hint: One AVOGRAM =

A

1

N

53. Answer (4)

Hint: Electron density of the p-orbitals lie along the

axes.

Solution: dxy

orbital does not have electron density

along the axis.

54. Answer (1)

Hint: 2

2 2

1 2

1 1E 13.6Z

n n

eV/atom

Solution: For, n1

= 4 and n2

= 5

9E 13.6 eV/atom

16 25

55. Answer (2)

Hint: Ti : [Ar] 4s2 3d

2

Solution:

Ti2+ : [Ar] 3d2

56. Answer (1)

Hint:

2

n

nr 0.53 Å

Z

Solution: n = 2, Z = 4

2

4

2r 0.53 0.53Å

4

57. Answer (4)

Hint: For principal quantum number n, l can be 0, 1,

2, ... (n – 1)

Solution: for n = 2,

l 2, it will have value 0 and 1.

58. Answer (4)

Hint: h

x p4

Solution:h

x m v4

34

3 24

h 6.626 10x

1004 m v4 10 6.626 10

10x 2.5 10 m 2.5 Å

59. Answer (2)

Hint: 21

KE mv2

, h

mv , v

Solution: 2v v mv 2 1 2

mv KEh h 2 h

27 7 1

34

23.313 10 10 sec

6.626 10

60. Answer (2)

Hint: Exchange energy of electrons is defined for


Solution: Exchange energy of electrons is defined

as energy released when an electron exchanges its

position with electron having same spin present in



9/17

61. Answer (1)

Hint:

solvent

solute

solvent

X 1000molality(m) =

X MW

Solution: 0.2 1000

m 13.890.8 18

62. Answer (2)

Hint: -rays, X-rays , UV, visible, IR, microwaves

Wavelength increases

Frequency decreases

Solution: Visible waves have higher frequency than

IR.

63. Answer (4)

Hint: l = 2 means d-subshell

Solution: d-subshell can have maximum 10

electrons.

64. Answer (2)

Hint: -particle is He nucleus (He2+) which contains

2 protons and 2 neutrons.

Solution: For neutron, e

0m

For other particles, e

m order is

electron > proton > -particle

65. Answer (1)

Hint: Mass of electron is 9.1 × 10–31 kg

Solution: Positron is the particle having mass equal

to electron but having a unit positive charge

66. Answer (1)

Hint: Average atomic mass =

% abundance of isotope atomic mass

100

Solution: 13 x 12 (100 x)

12.011100

x = 1.1%

67. Answer (3)

Hint: Divalent metal chloride should be MCl2

Solution: Mol. wt.= 2 × vapour density = 60 × 2 = 120

x + 2 × 35.5 = 120

x = 49

Eq. wt. of metal 49

24.52

68. Answer (3)

Hint: 2 2 3

32AI O Al O

2

Solution: 2 2 3

32Al O Al O

2

3 2

3KClO KCl O

2

From stoichiometry, 2 mole Al require 3

2 mole O

2

and 3

2 mole O

2

is obtained by 1 mole KCIO3

.

69. Answer (4)

Hint: M1

V1

+ M2

V2

= M3

V3

Solution: 0.5 × 2 + 1 × 1= M3

× 3

3

2M 0.67 M

3

70. Answer (2)

Hint: Law of multiple proportions is illustrated for

compounds which have two same elements.

Solution: CH4

: for 12 g C, wt. of H = 4 g = x

C2

H4

: for 12 g C, wt. of H = 2 g = y

x 4 2

y 2 1 = Simple whole no. ratio

71. Answer (4)

Hint: Molarity (M) = solute

n

V L

Solution: Mass of 1 ml D2

O = 1 g

3

1

20M 50 M

1 10

72. Answer (1)

Hint: For minimum molar mass, one molecule

should contain at least one atom of oxygen

Solution:% of oxygen =wt.of oxygen

100wt.of compound

164 100

Mol.wt.

Minimum mol. wt. = 400 amu


10/17

73. Answer (1)

Hint: 6

n

zv 2.18 10 m/sec

n

Solution: 6

4 He

8

22.18 10v 14

c 2753 10

74. Answer (3)

Hint: Ni2+ = [Ar] 3d8

Solution: 29

Cu = 1s2 2s

2 2p6 3s

2 3p6 4s

1 3d10

s-electrons in Cu atom = 7

75. Answer (2)

Hint: 2

2 2

1 2

1 1 1RZ

n n

Solution: For second line of Lyman series :

2

2 2

1 1 1R (1)

x 1 3

1 8R

x 9

9R

8x

For 3rd line of Lyman series : 2

2 2

1 1 1R (1)

1 4

15 9 15 135R

16 8x 16 128x

128x

135

76. Answer (2)

Hint: Electrons in half filled p-subshell have different

orientations.

Solution: Orientation of orbitals is given by magnetic

quantum number.

77. Answer (3)

Hint: s, p and d-subshells have 1, 3 and 5 orbitals

respectively.

Solution:

In l subshell, number of orbitals = 2l + 1

78. Answer (1)

Hint: h

mv

Solution: H2

has minimum molar mass so have

longest de-Broglie wavelength.

79. Answer (4)

Hint: 2 2 3

1 1FeO O Fe O

4 2

Solution:

(72g)

(8g)

2 2 31 mole

0.25 mole

1 1FeO O Fe O

4 2

% increase in wt. 8

100 11.11%72

80. Answer (3)

Hint: No. of atoms = No. of molecules × atomicity

Solution: Remaining molecules of CO2

=

323 20440 10

6.022 10 1044

No. of remaining atoms = 59.22 × 1020 × 3

= 1.77 × 1022

81. Answer (2)

Hint: Normality = Molarity × n-factor

Solution: Molarity of

22

23

3

9.033 10

6.022 10OH 0.3 M

500 10

Molarity of Ca(OH)2

= 0.3

M2

Normality of Ca(OH)2

=

0.32 0.3 M

2

82. Answer (1)

Hint: Dilution equation, M1

V1

= M2

V2

Solution: For stock solution,

1

49 1.5 10M 7.5

98

Now, M1

V1

= M2

V2

7.5 × V1

= 0.1 × 2.5 × 1000

V1

= 33.33 ml

83. Answer (1)

Hint: Frequency 2

32

z

v znf

r nn

z

Solution:

2

2

2

3

He Li

2

Li He

3

3

T f 2 9

T f 322

1


11/17

[ BIOLOGY

]

91. Answer (3)

Hint: Meiosis involves crossing over as well as

reduction of chromosome number.

Solution: Four cells produced after telophase II are

genetically dissimilar to each other as well as to

their parents.

92. Answer (2)

Hint: Interkinesis is resting phase between meiosis

I and II.

Solution: Meiosis I and II involves into sequential

karyokinesis. Chromosomes do not decondense upto

DNA level in interkinesis because after telophase I

chromatin recondense in prophase II.

93. Answer (4)

Hint: Mitosis does not involve crossing over.

Solution: Mitosis helps in reproduction in unicellular

organisms. It helps in repair and regeneration but

cannot produce recombinants.

84. Answer (4)

Hint: For one electron species, energy of subshell

depends only on the value of n.

Solution: For 3rd shell, No. of degenerates orbitals = n2

= 32 = 9.

85. Answer (3)

Hint: Only four lines of Balmer series of hydrogen

atom lie in visible region

Solution: For n = 6 (5th excited state) transition to

n = 1

* Maximum possible transitions

= 6 1 5 + 4 + 3 + 2 + 1 =15

* Maximum Paschen transitions

= 3(6 3, 5 3, 4 3)

86. Answer (2)

Hint: x y 2 2 2

y yC H x O xCO H O

4 2

2

2

Moles of CO obtainedx

y 2 Moles of H O obtained

Solution:

8.8

x 144

5.4y 32

18

Empirical formula = CH3

Molecular wt. 2 30n 4

Empirical formulawt. 15

Molecular formula = n × empirical formula

= 4 (CH3

) = C4

H12

87. Answer (4)

Hint: No. of particles = No. of moles × NA

Solution:

Moles of oxygen atom = 0.3 × 14 = 4.2

No. of electrons of oxygen atoms = 4.2 × 8 × NA

= 33.6 NA

88. Answer (4)

Hint: n-factor for acid is number of replaceable H+

ions

Solution: n-factor = 1

Mol. wt 120Eq. wt. 120

n-factor 1

89. Answer (3)

Hint: Chromium has half-filled d-subshell.

Solution: Cr(24) : [Ar]4s13d

5

90. Answer (1)

Hint: Orbital angular momentum = hl l 1

2

Solution: For f - orbital, I = 3

Orbital angular momentum = h 3h3 3 1

2

94. Answer (2)

Hint: Treatment of colchicine produces polyploid

condition.

Solution: Colchicine inhibits formation of tubulin and

therefore microtubules. Lack of microtubules results

in inhibition of spindle fibres formation and its affect

will be visible first at metaphase.

95. Answer (2)

Hint: Cell plate represents the future middle lamella

of adjacent cells.

Solution: Cell plate help in cytokinesis of plant cells

which represent the middle lamella of adjacent cells.

96. Answer (1)

Hint: Dyad of cells appear in last stage of heterotypic

division

Solution: Dyad of cells are formed at the end of

telophase I.


12/17

97. Answer (2)

Hint: Recombinase enzyme catalyses the process

occurring in pachytene stage.

Solution: Crossing over is enzyme mediated

process and it produces recombinants.

98. Answer (2)

Hint: Each chromosome has 2 chromatids

Solution: 1 Bivalent = 2 chromosomes

20 Bivalent = 20 × 2 = 40 chromosomes

Since 1 chromosome = 2 chromatids

40 chromosomes = 80 chromatids

99. Answer (2)

Hint: The given stage is the best stage to study

morphology of chromosomes.

Solution: Given stage is Metaphase I. where,

Bivalent chromosomes align at equator forming

two metaphasic plates

Alignment of bivalents is totally random process.

Spindle fibre attaches to each chromosome of

homologous pair

Separation of chromosomes occur at anaphase I

100. Answer (3)

Hint: Number of mitotic divisions to form ‘n’ number

of cells are = n – 1

Solution: Number of generations (n) required to form

‘x’ number of cells are = 2n

For 32 cells

Mitotic divisions = 32 – 1 = 31

Number of generations = 25 = 32, x = 5

101. Answer (1)

Hint: Major restriction point inhibits cells to go for

DNA synthesis.

Solution: Major control on cell cycle can be

imposed on 1

G S transition step.

102. Answer (3)

Hint: Last phase of karyokinesis in mitosis involves

reappearance of cell organelles and nuclear

membrane.

Solution: At the end of mitosis, the chromosomes

decondense into chromatin which occurs in telophase.

103. Answer (3)

Hint: Interphase is known as the most active stage

of cell cycle.

Solution: Interphase constitute more than 95%

duration of cell cycle of a human cell. Rest all the

features regarding the interphase are true.

104. Answer (4)

Hint: Tubulin protein synthesis occurs in the stage

where duplication of mitochondria & chloroplast

occurs.

Solution: Most of the cell organelles get duplicated

in G1

. But Golgi, chloroplast and mitochondria are

doubled at G2

phase along with tubulin protein

synthesis.

105. Answer (2)

Hint: Meiosis involves two sequential karyokinesis

and cytokinesis.

Solution: Except DNA duplication and histone

protein synthesis, rest all phenomenon occur twice

in meiosis.

106. Answer (3)

Hint: Spindle fibres are not directly attached to the

centromere.

Solution: A disc-shaped structure is found over

centromere through which spindle fibres are attached

called kinetochore.

107. Answer (2)

Hint: Crossing over occurs in pachytene stage.

Solution: Point at which crossing over occurs forms

the recombination nodule.

108. Answer (4)

Hint: Bacterial type ribosomes are also found in

some cell organelles of eukaryotes.

Solution: Both ribosomes and plasma membrane

are similar in prokaryotes and eukaryotes.

109. Answer (2)

Hint: The cell also has cell wall with plasmodesmata.

Solution: Plant cells have cell wall with

plasmodesmata. Their 90% volume is occupied by

vacuoles.

110. Answer (3)

Hint: These structures are tubular and help bacteria

to attach with rocks also.

Solution: Fimbriae are bristle like structures which

help bacteria to attach with rocks or host cell.

111. Answer (3)

Hint: These structures help in photorespiration

along with chloroplast and mitochondria.

Solution: Peroxisomes are involved in photorespiration

because they have enzymes for formation as well

as destruction of peroxide.

112. Answer (4)

Hint: Loops of lampbrush chromosomes have hair

like structure and with protein that are known as

informosomes.


13/17

Solution: Lampbrush chromosomes form loops

which have hair like structures. These hairs are

bound to protein. Some of which are called

informosome i.e, having mRNA + protein. These

loops participate in transcription and form m-RNA

113. Answer (2)

Hint: Such chromosomes appear L-shaped during

anaphase.

Solution: A chromosome with centromere slightly

away from the centre is submetacentric chromosome.

114. Answer (1)

Hint: Chromatids of chromosomes are held together

at primary constriction.

Solution: Primary constriction is known as

centromere which holds the two halves of a

chromosome.

115. Answer (4)

Hint: Chromatin is packed DNA

Solution: Packed DNA has RNA, histones and

some non-histone proteins.

116. Answer (3)

Hint: Nucleolus is found inside the nucleus

Solution: Nucleolus is non-membrane bound

structure found in nucleoplasm of nucleus. They are

larger in cells involved in protein synthesis.

117. Answer (4)

Hint: Polyribosomes are formed in cytoplasm

Solution: Polysomes are not formed with the help of

RER

118. Answer (2)

Hint: Both mitochondria and chloroplast are semi

autonomous structures.

Solution: Mitochondria, chloroplast and bacteria all

have ds circular DNA, 70S ribosomes porins on

outer membrane and self duplication ability.

Mitochondria and chloroplast are partially dependent

on nucleus.

119. Answer (4)

Hint: In leucoplast granum remains absent.

Solution: Protoplast is a cell without cell wall.

120. Answer (4)

Hint: This structure is absent in higher plants.

Solution: Centrioles have 9 peripheral fibrils of

tubulin and these are absent in the centre therefore

the arrangement is 9 + 0.

121. Answer (3)

Hint: Microfilaments are solid unbranched rod like

fibrils.

Solution: Microtubules have diameter of 25 nm.

Intermediate filaments are involved in formation of

scaffolds of chromatin.

122. Answer (2)

Hint: These structures are absent in prokaryotes.

Solution: Cytoskeleton are proteinaceous

filamentous structures which provide mechanical

strength & support to the cell.

123. Answer (2)

Hint: ‘S’ is a unit

Solution: ‘S’ stands for Svedberg coefficient or

sedimentation coefficient and it is indirect measure of

density and size of ribosomal sub units.

124. Answer (3)

Hint: This organelle is found in almost all eukaryotic

cells and is site of ATP synthesis.

Solution: Mitochondria are sausage shaped, have

their own 70S ribosomes i.e, palade particles and

they are viewed after staining with Janus green.

Usually their number is high in those cells which

have high metabolism.

125. Answer (3)

Hint: This organelle has hydrolytic enzymes which

become functional at acidic pH.

Solution: Lysosomes have hydrolytic enzymes for

digestion of almost all types of macromolecules

which are functional at acidic pH.

126. Answer (4)

Hint: ER, Golgi, lysosomes & vacuoles function in

a coordinated manner.

Solution: Oxidation of fatty acids, proteins and

carbohydrate occurs inside the mitochondria.

Rest all functions are performed by organelles of

endomembrane system.

127. Answer (3)

Hint: The given figure is of golgi bodies

Solution: Enzymatic precursors for lysosomes are

synthesized in ER. Golgi does processing and

packaging of materials for intra as well as extra-cellular

targets.

Rest all the features of golgi are true.

128. Answer (3)

Hint: These structures are present in pits.

Solution: Symplast of two adjacent cells are

connected via cytoplasmic strands or

plasmodesmata. They are lined by plasma

membrane.


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129. Answer (2)

Hint: Smallest cell organelle is known as organelle

within organelle.

Solution: Ribosome is known as organelle within

organelle and its r-RNA part is synthesized inside the

nucleolus.

130. Answer (4)

Hint: Movement of molecules across the membrane

without help of carrier proteins and ATP is called

simple diffusion

Solution: Simple diffusion occurs only for those

molecules which are neutral or non-polar, while

transport of polar and hydrophilic substances need

carrier proteins.

131. Answer (3)

Hint: Lipid molecules of plasma membrane have

polar head and non-polar tail.

Solution: Polar head is hydrophilic in nature and it

interacts with water. Rest all the statements

regarding plasma membranes are true.

132. Answer (2)

Hint: This inclusion body provides buoyancy to the

bacteria

Solution: Ribosomes are found in all prokaryotes

and eukaryotes. Cyanophycean granules are found in

cyanobacteria only while gas vacuoles are found in

blue-green algae (BGA) as well as in purple & green

photosynthetic bacteria

133. Answer (4)

Hint: Chloroplast is the site of photosynthesis and

contain pigments in eukaryotes.

Solution: Chromatophores are membranous

extension in the cytoplasm of a cyanobacterial cell

which have photosynthetic pigments

134. Answer (3)

Hint: Body of plants and animals are composed of

cells & product of cells was proposed by Theodor

Schwann.

Solution: Theodor Schwann was a British zoologist.

135. Answer (2)

Hint: Nuclear membrane is found in all eukaryotes.

Solution: Mycoplasma, BGA, Bacillus, purple

photosynthetic bacteria & E.coli are prokaryotes.

136. Answer (4)

Hint : ‘Osteo’ refers to bone & clasts refers to

destruction/demineralisation.

Solution : Osteocytes & osteoblasts are bone

forming cells while osteoclasts are bone dissolving/

dissolution cells that decrease bone calcium.

137. Answer (1)

Hint : Keratinised epithelium occurs on water

resistant surfaces in living beings.

Solution : Stratum corneum is the topmost layer of

stratified keratinised epithelium. Pharynx, tongue &

cornea are non-keratinised epithelial surfaces.

Microvilli in alimentary canal support absorption of

nutrients not reabsorption. Transitional epithelium is

distensible/stretchable. Little intercellular matrix is

present in epithelial tissue.

138. Answer (2)

Hint : Brush border membrane (BBM) due to

microvilli increase surface area for reabsorption &

absorption.

Solution : Cilia in trachea push mucus with

entrapped particles in specific (upward) direction.

BBM cuboidal is a feature of tubular part of nephrons

while BBM columnar epithelium dominants in

intestine.

139. Answer (1)

Hint : I band is composed of actin filaments while A

band comprises of both actin & myosin filaments.

Solution : Length of both actin & myosin filaments

remain constant. Size of A band also remains

contant. Only the extent of overlap between actin &

myosin filaments increases thereby reducing size of

Henson’s zone.

140. Answer (4)

Hint : Response to stimulus is a feature of structural

& functional units of neural system.

Solution : Extensibility is a property of muscle fibres

while neurons exhibit both excitability & conductivity.

141. Answer (1)

Hint : Nissl's granules are clusters of RNA &

ribosomes.

Solution : Nissl's granules are site of protein

synthesis located in cell body of neuron/soma & in

dendrites (afferent processes). They are absent in

afferent processes also called axons.

142. Answer (2)

Hint : Neurilemma is a contribution of Schwann

cells.

Solution : Schwann cells form myelin sheath &

neurilemma in neurons of PNS. Oligodendrocytes

form myelin sheath in CNS. Neurilemma is absent in

CNS. Oligodendrocytes are branched cells while

Schwann cells are not branched.


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143. Answer (4)

Hint : Wine flask shaped cells of alimentary canal.

Solution : Goblet cells are mucus secreting

unicellular glands of alimentary canal. Cerumen and

sweat are secreted by sebaceous & sudorific glands

respectively. Hormones are secretions of endocrine

glands.

144. Answer (1)

Hint : Stratified i.e. compound epithelium lines

surfaces that face mechanical stress.

Solution : Stratum germinativum comprises of cells

that divide regularly and forms the bottommost layer

of straitified epithelium. Diffusion surfaces and tubular

parts of nephron are lined by single layered simple

epithelium.

145. Answer (2)

Hint : Junctions involved in cementing neighbouring

cells.

Solution : Tight junctions help to stop substances

from leaking across a tissue.

146. Answer (2)

Hint : Multiple layer in epithelial tissue are effective

in protection.

Solution : At surfaces prone to stress (wear & tear)

or abrasion compound epithelium is effective.

147. Answer (3)

Hint : Endocrine glands lack ducts.

Solution : Endocrine glands do not have ducts and

release their secretion into blood capillaries.

148. Answer (2)

Hint : Exocrine glands release/pour their secretions

through ducts.

Solution : Insulin is a secretion of ductless part of

gland pancreas. Saliva, earwax and milk are

secretions of exocrine glands.

149. Answer (4)

Hint : Concentric arrangement of calcium salts

surrounding osteocytes.

Solution : The lamellae refer to concentric

arrangement of calcium phosphate salts surrounding

osteocytes in diaphysis of long bones of mammals.

150. Answer (2)

Hint : Muscles under the control of our will are

voluntary.

Solution : Smooth and cardiac muscles are

involuntary in nature i.e. they are not under the

control of our will. Smooth muscles contain

unbranched, unstriped, uninucleate fibres.

151. Answer (1)

Hint : Muscle fibres associated with limbs and heart

are striped in appearance.

Solution : Striated muscles include both cardiac &

skeletal muscles. The alternate light and dark bands

refer to the horizontal arrangement of actin & myosin

filaments.

152. Answer (4)

Hint : Intercalated discs are interdigitations between

adjacent cells that act as boosters of cardiac

impulse.

Solution : Cardiac fibres are striated and the edges

of sarcomeres are ‘Z’ lines. Connexons are part of

gap junctions present at intercalated discs. Visceral

muscles are uninucleated.

153. Answer (3)

Hint : White fat is unilocular and adipocytes are

signet ring shaped cells.

Solution : Brown fat in newly born prevents shivering

in them. White fat acts as a reservior of energy and

the triglycerides are hydrolysed upon need.

154. Answer (2)

Hint : Positional information of amino acids decides

specificity of enzyme action.

Solution : Apoenzymes are proteinaceous part of

holoenzymes. Enzymes never alter the equilibrium of

the reaction, they act by lowering activation energy

barrier.

155. Answer (3)

Hint : While overcoming competitive inhibition Km

value increases.

Solution : Succinate is the substrate for enzyme

succinate dehydrogenase. Malonate is its

competitive inhibitor.

156. Answer (3)

Hint : Hydrolytic enzymes work by addition of water

to break the bond between biomolecules.

Solution : Hydrolases belong to class III according

to nomenclature given by IUB. Phosphodiesterases

also belong to this class of enzymes.

157. Answer (2)

Hint : Thymine is exclusive to DNA.

Solution : Thiamine is vitamin B1

while thymine is

nitrogenous base in DNA.

158. Answer (4)

Hint : Identify odd one out w.r.t. macromolecules.

Solution : Lipids are neither polymer nor

macromolecules.


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159. Answer (2)

Hint : Amino acids are monomers of proteins.

Solution : Amino acids are substituted methanes.

They exist as zwitterions at isoelectric pH.

Trihydroxypropane is glycerol. DNA & RNA are

negatively charged molecules.

160. Answer (3)

Hint : DNA of a eukaryote comprises of A, T, G, C

as nitrogenous base.

Solution : Cytosine is a nitrogenous base while

cysteine is an amino acid.

161. Answer (4)

Hint : Active site are formed at tertiary level of

protein organisation. Hydrogen bonds stabilise

predominantly protein structure formed at secondary

level of organisation.

Solution : Ionic bonds, Van der Waals forces,

hydrophobic interactions along with hydrogen bonds

stabilise quaternary structure.

162. Answer (4)

Hint : All components given are sugars in raffinose.

Solution : Tripeptide has 3 amino acids &

triglyceride has 3 fatty acids linked to glycerol.

163. Answer (1)

Hint : Two or more amino acids can be linked by

peptide bonds.

Solution : Collagen is a heteropolymer of amino

acids. Chitin & cellulose are polysaccharides whose

monomers are linked by glycosidic bonds. Choline is

nitrogenous compound found in lecithin.

164. Answer (3)

Hint : First alphabet of amino acid is considered as

the basis of single alphabet code.

Solution : Tyrosine is coded by Y, phenylalanine by

‘F’ and glutamic acid by E. This is because

threonine, proline & glycine have already been named

as T, P & G respectively.

165. Answer (4)

Hint : Organic cofactor loosely attached to

apoenzyme.

Solution : Niacin and riboflavin are vitamins requires

for formation of coenzymes such as NAD, FAD &

FMN.

166. Answer (1)

Hint : Identify a homopolymer.

Solution : Chitin comprises of repeating units of (1

4) N-acetylglucosamine. Nucleic acids (DNA),

proteins i.e. RuBisCO and insulin are

heteropolymers.

167. Answer (3)

Hint : Double bonds cannot be introduced beyond 9th

carbon position in most animals usually.

Solution : Linoleic acid has 2 double bonds

(9 & 12th carbon). Plants are able to synthesize

essential fatty acids by introducing double bonds at

12th & 15th carbon. Oleic acid has a single double

bond at 9th position. In most mammals, arachidonic

acid can be formed from linoleic acid.

168. Answer (2)

Hint : Unsaturated fatty acids have low melting

points.

Solution : Gingelly oil/sesame oil and arachidonic

acid are unsaturated fatty acids, hence will have

lower melting points in comparison to other saturated

lipids given.

169. Answer (1)

Hint : Formation of polymers and disaccharides

requires glycosidic bonds.

Solution : Glucose is a monomer of cellulose but

cellulose is unbranched homopolymer glucose.

170. Answer (3)

Hint : Identify a nucleotide. Ribozyme is RNA acting

as enzyme.

Solution : ATP is adenosine triphosphate.

171. Answer (1)

Hint : Pentose sugar derived from C5

H10

O5

.

Solution : 5 carbon sugar in RNA (ribonucleic acid)

is ribose. DNA has deoxyribose with molecular

formula C5

H10

O4

.

172. Answer (1)

Hint : Isomerism in hexose sugars.

Solution : Galactose & glucose are aldose sugars

while fructose is ketohexose. Ribose is a pentose

sugar.

173. Answer (1)

Hint : Reducing sugar.

Solution : Sugars with free aldehyde or ketone

groups are detected by Benedict’s and Fehling’s

solution. Glucose and galactose have free aldehyde

group while fructose has free keto group. Sucrose

has no free keto or aldehyde group.

174. Answer (3)

Hint : Structural sugar found in RNA.

Solution : Cellulose is unbranched structural

homopolymer of glucose comprising of (1-4)

glycosidic linkages. Ribose is a monomer formed

without glycosidic bond. Cytidylic acid is a

nucleotide found in RNA.


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��

175. Answer (1)

Hint : Dehydration is a condensation reaction

involving loss of water molecule.

Solution : Cystine formation requires removal of H2

and disulfide bonds are formed between two amino

acids. Formation of lecithin, adenosine and collagen

requires formation of ester & glycosidic bonds

respectively.

176. Answer (4)

Hint : Identify a molecule which is a monomer.

Solution : Inulin is a polymer of fructose, while

insulin is a polymer of amino acids. Lactose is a

disaccharide. Polymers can be broken into

monomers by addition of water (hydrolysis).

177. Answer (2)

Hint : Bones are reservoirs of some elements.

Solution : Matrix of bone is largely composed of

hydroxyapatite crystals of calcium phosphate.

Essential fatty acids & essential amino acids are

supplied in diet.

178. Answer (4)

Hint : Polymers are found in retentate/acid insoluble

fraction.

Solution : Cysteine, calcium & cytosine are

micromolecules that are obtained in filtrate/acid

soluble fraction. Cellulose is a polymer of glucose

and due to its large size it cannot cross the filtration

membrane.

179. Answer (3)

Hint : Order of percentage of biomolecules in cells.

Solution : Water > proteins > nucleic acids >

carbohydrates > lipids > ions

70 90% 10 15% 5 7% 3% 2% 1%

A C B D E

180. Answer (2)

Hint : Compounds whose role or function we do not

understand at the moment.

Solution : Primary metabolites are essential for

physiological processes and have identifiable

functions. Secondary metabolities include alkaloids,

antibiotics, rubber, essential oils, spices etc.

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