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Final Round Test Series NEET - 20201
Dr.DHOTE’S ACADEMY, Latur Office : (02382) 247222, 8766475788
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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
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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
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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
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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
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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
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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
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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)
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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