India’s Premier Coaching Institute Dr. DHOTE’s ACADEMY · 2 Final Round Test Series NEET - 2020 Dr.DHOTE’S ACADEMY, Latur Office : (02382) 247222, 8766475788 Let’s BEAT The

Final Round Test Series NEET - 20201

Dr.DHOTE’S ACADEMY, Latur Office : (02382) 247222, 8766475788

Let’s BEAT The NEET

Final Round Test Series NEET - 2020Class : XII + Repeater Date : 24.03.2020 Test No. : 09

India’s Premier Coaching Institute

Dr. DHOTE’s ACADEMY

HINTS AND SOLUTIONS

01. (1) L = 4m

Y = 9 × 1010

A

F = YY

F = AY

= (2 × 10–3)2 × 9 × 109 × 100

1

= × 4 × 10–6 × 9 × 107

= 360 N.

02. (2) y1 y2 k1 k2

Keq

= K1 + K

2

A2Y =

Ay1+

Ay2

y = 2

yy 21

03. (3) F = h

xA = 0.4 × 1011 × 1 × .005 ×

1

1002. 2

= 4×104 N

04. (3)V

V=

B

p= 11

5

1025.1

101

= 8×10–7

05. (4) V = 1/2 K(2)2

V1 = 1/2 K(10)2

then V1 = 5V

06. (3) vt =

)(r

9

2 2

07. (3) mg – 6 rv = ma

08. (4) No medium is present in vacuum

09. (3)A

F = YY

If Y &

are constant

F = AY

F A F’ = 4F

10. (4)2

1

p

p=

22

11

vm

vm, m r3, v r2




p r5 then 2

1

p

p=

32

1

11. (1) F = h

x

410164

500

= 2

6

104

x102

x = m32

105 2 = 0.156 cm

12. (3) The gravitational force remains constant. The viscous

force increases with increase in velocity. The net force

decreases and finally becomes zero when terminal

velocity is reached.

13. (2)2

1

r

r= b

2

1

= a

2

1

y

y = c

m

2m

r y1 1 1

r y2 2 2

steel

brass

l1 =

11

1

yA

)mg3(

l2 =

22

2

yA

)mg2(

2

1

= 22

112

1 yAyA2

3

= 2

3

cb

a2 =

cb2

a32

14. (4) depth = 200 m

310100

1.0

V

V

density = 1 × 103

g = 10

b = v/v

p

= v/v

hg

b = 200 × 10 × 103 × 1000 = 2 × 109

15. (4)2

1

r

r =

2

1

PE (per unit volume) =

2

A

F

y2

1

PE 1/A2

2

1

PE

PE = 2

1

2

2

A

A = 16 : 1

16. (1) mg – FV – F

up = ma = 0

mg = FV + F

up... (i)

Now

F– mg – FV – F

up = 2mg – mg – F

V – F

up

= mg – FV – F

up = 0

Acceleration in second case = 0

ball will move upward with constant speed = 10 cm/s

17. (4) Y =0r

k= 10103

7

= 2.33 ×1010 N / m2

18. (4) Y = Strain

Stress= Constant

It depends only on nature of material.

19. (4)

20. (2) Energy stored = 2

1 stress × strain × volume

u = volume

energy= e

5

1

= Y

S

2

1

Y

SS

2

1 2

21. (2) Tension in wire remains same

22. (2) u = Y2

S

Y

)stess(

2

1 22

23. (1) =

AL2

F

LA

F

2

1




24. (3) W =L

YA 2

= 2

3610

10502

2)10(10102

= 2 × 10–2 J

25. (2) Angle of shere = L

r =

100

104 1× 30° = 0.12°

26. (2) r = L 10–2 0.8 = 2 = 0.004

27. (4)

28. (4) geffective

= 0

29. (3) U = 2

1 Y (strain)2 Volume

U = 0.075 J

30. (3)

31. (3) U = 2

1 Y (strain)2 Volume

32. (4)

33. (3)

34. (3)

35. (1) Work done = 2

1 mgh =

2

1 × 5 ×10 × 3

36. (3) A liquid has no length and no shape, but it has only

definite volume and so, it possesses only bulk modulus.

37. (1) Breaking stress for a wire depends only on material.

38. (2) Young’s modulus of a substance is independent of

dimension s of wire.

39. (3) F = –hAdx

dv

h = dv

dx

A

F

Writing the dimensions

[h] = ]LT[

]L[

]L[

]MLT[12

2

= [ML–1T–1]

40. (4) Elastic energy stored in the wire is

U = 2

1× stress × strain × volume

= ALL

L

A

F

2

1

= LF2

1

= 2

1 × 200 × 1 × 10–3 = 0.1 J

41. (1) Force requied to increase the length of rod :

F = L

YA

= 100

1.0104109 610

= 360 N

42. (1) Volume elasticity coefficient :

B = 100/1.0

gh

V/V

p

= 1000/1

8.910200 3

= 19.6 × 108

43. (3) Tensile force on each surface = Y

F

Lateral strain due to other two forces acting on

perpendicular surfaces = Y

F2

Total increase in length = Y

F)21(

44. (4) F = Kx

F = L

YA

2D

L

A

L

1001

10021

502

20022




233

300

3

100

200

4

1

504

Here 4 correct.

45. (4) Elastic energy stored in the wire is

U = 2

1 stress × strain × volume

= 2

1

LA

F l ×AL

= 2

1FDl

= 2

1× 200 ×1 ×10–3 = 0.1

46. (3)

Amorphous solids neither have ordered

arrangement (i.e. no definite shape) nor have sharp

melting point like crystals, but when heated, they become

pliable until they assume the properties usually related

to liquids. It is therefore they are regarded as super-

cooled liquids.

47. (3)

Silicon due to its catenation property form network

solid.

48. (3)

Orthorhombic geometry has cba and

090 . The shape of match box obey this

geometry.

49. (4)

In a triclinic crystal has no notation of symmetry.

50. (1)

In NH3 molecule, the original appearance is r e pe a t ed

as a result of rotation through 120o. Such as axis is said

to be an axis of three-fold symmetry or a triad axis.

51. (1)

Na2O has antifluorite (A

2B) type structure.

52. (2)

Cationic radius increases down the group

and decreases along the period.

53. (3)

Distance between centres of cation and anion

pmd

2542

508

2

pmrorrorpmrr aaac 144254110254

54. (2)

3023330

0

3 10)1002.6()400(

1002

10

Na

Mn

55. (3)

o90γβα,a cb is the parameter for

crystal triclinic

56. (1)

Cubelongs to cubic crystal which has dimensions a

= b = c and o90γβα

57. (1)

Diamond and graphite are polymorphous because both

have similar chemical composition but different

arrangement of constituent particles i.e., carbon

58. (1)

Two dimensional lattices are = 5 [Square, Rectangle,

Rhombus, Parallelogram, Hexagonal]

Three dimensional lattices are = 14 [Bravais lattices]

59. (1)

A corners of cube Total 8 atoms at corner, which

have contribution at corner 8

1 So 1

8

18

number of atoms of A per unit cell




Bcentre of body at centre of body one atom

contributed completely. So, 1 × 1 = 1

i.e., A1B

1 = AB

60. (2)

ABC3

At corners number of atom 18

18 (contribution at

corner) A

At body centre = 1 × 1 = 1

At face centre 3

2

16

face)ation(Contribut

face)(Total

ABC3

61. (1)

Apresent at body centre in given figure A

So, its number = 1 × 1 = 1 (contributed completely)

Bpresent at corners 1

8

18

ion)(Contribut

atom) (Total

B

Cpresent at two opposite face

C12

12

face)eachation(Contribut

face)atatoms2(Total

So, formula ABC

62. (3)

32

16faceeachofcentresatPresent

18

18cornersatPresent

ccpCu

Cu4Total

34

112centreedgeatAg

ion)(Contribut

edge)atatoms(Total

Au at body centre contributed completely

111 Cu

4Ag

3Au

1

63. (3)

1

8

18A

ion)(Contribut

atoms)(Total

32

16centreFaceB

ion)(contribut

atoms)(Total

Now, one atom is missing from one corner

718atoms)(total

A8

7

8

17So,

corner)ation(contribut

2473

8

7 BABA

64. (1)

1A18

18cornersA

3B34

112edgesB

1C111centrebodyC

CAB3

65. (2)

Network covalent solids have highest melting point due

to network structure in which bonds are tightly held like

SiO2




Layered structure so, it require very high temperature

to melt it.

(Close packed structures, calculation of packing

efficiency, close packing in ionic compouds)

66. (1)

Accp total atoms

42

16

8

184

Bat tetrahedral voids doubl of number of atoms

2 × 4 = 8 given B 8N.

So, number of A atoms is half the number of (B) atoms,

i.e., 42

Nor4

2

8

67. (3)

BA Rock salt type f.c.c.number of atoms

= 4

842

.2 atomsofnovoidlTetrahedra

B constituting the lattice i.e., i.e.,4

A 25%of tetrahedral void

28100

25

242 ABorBA

68. (4)

For tetrahedral void the radius ratio

0.4140.225r

r

69. (2)

2

16

8

184fccccpTi

givencarbonat octahedral holes = equal to no.

atoms 4 So, Tii4C

4TiC

Givenhydrogenat tetrahedral holes double of

number of atoms 842

284 TiHHTi

70. (2)

In given figure (X) is present at centre of edge which

contributed

4

1

71. (3)

332211 VMVMVM

]V[VM300.3200.5 213

0.3850

19M30],[20M910 32

72. (2)

BMM1000d

M1000m

73. (2)

(x)solubilityKp H

...(i)0.05K1 H

...(ii)xK3 H




x

0.05

3

1

mol/litre0.1530.05x

74. (1)

332211 VMVMVM

]V[VM5201.24801.5 213

1000

624720

520480

5201.24801.5M2

M1.3441000

1344

75. (2)

V.P

1B.P.

If B.P. is low means liquid can easily boil so, liquid forms

vapours easily - vapours increases = vapour pressure

increases, among all compounds boiling point of CH3OH

is min. B.P. = 65oC so it will have maximum vapour

pressure.

76. (3)

Boiling pont of the liquid is the temperature at which

vapour pressure of liquid equals to atomspheric pressure.

77. (3)

Solute-Solvent interactions > Solute-solute or solvent-

solvent interaction

interactions are high

So, bonds cannot easily break thats why vapours

decreases So. V.P. decreases negative deviation.

78. (2)

In positive deviation solution

Solute-solvent interactions < Solute-solute or solvent -

solvent interaction

Bond can break easily Vapurs increases V.P..

increases

V.P.High that’s why B.P. = Low min. boiling

azeotrope.

0ΔHmix

0ΔVmix

79. (3)

o

B

o

A PP Vapour pressure of (A) is morre than (B)

so mole fraction of (A) is more in vapour phase than

liquid phase. Because vapours are more formed

A(v)A(l) χχ

80. (2)

Positive deviation shows

minimum B.P. azeotrope

V.P. = HighB.P. = minimum

Sol. (2)

4.5562.619

2.619

pp

py

EtOHMeOH

MeOH

(Vapour)MeOH

0.3657.175

2.619

4.5562.619

4.556

pp

py

MeOHEtOH

EtOHEtOH

0.6357.175

4.556

83. (3)

Pressure cooker reduces cooking time for food because

B.P. of water involved in cooking is increased due to

which more heat is transferred to food and food can

easily cook.

83. (2)




Osmotic pressure, CRTπ

atm0.932980.082150342

10001.95

84. (1)

AAo

o

x750

750760(solute),x

p

pp

Ax750

10

0.013xA 85. (1)

o

bbb TTT

f

o

ff TTT For same solvent (given)

o

bT = constant

o

fT = constant

So mTΔT bb

(given)samemmTΔT ff So T

b & T

f also becomes same

86. (2)

In osmosis solvent molecules pass from solvent to solution

through semipermeable membrane.

87. (3)

30027273Ti(CST)π

ClNaNaCl

2)(i

atm4.922300)0.0821(0.1 88. (1)

volume

wi.e.,g/L8.6Urea

4214CONHNHmassMolar 22

601612

]π[πisotonic 21

STCSTC 21

21 CC

VM

W

VM

W

ww

100M

10005

60

8.6

w

g/mol348.88.6

10605Mw

Errata : 5% solution of unknown solute but it must be

written 5% weight/volume means 5 g of solute present

in 100 mL of solution

89. (4)

In relative lowering of vapour pressure elevation in B.P.

depression in freezing point get minium for high molecular

masses but osmotic pressure can’t lowered as much

for high mol. masses

90. (2)

58.5NaClM

)Cl2(Nai(NaClm)i(kΔT

w

bb

f(solvent)

o

bbkwM

1000w0.512TT

10058.5

1000w0.512099.07100

g5.3100.512

54.405

100.512

10w0.5120.93

Documents

India’s Premier Coaching Institute Dr. DHOTE’s ACADEMY · 2 Final Round Test Series NEET - 2020 Dr.DHOTE’S ACADEMY, Latur Office : (02382) 247222, 8766475788 Let’s BEAT The