Post on 11-Dec-2021
SAFE HANDS, AKOLA
MAJOR TEST# 07 (NEET) SOLUTIONS
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1. (a)
Sol. Resistance [R] = [ML2T–3A–2]
2. (d)
Sol. d = 2r
m
volume
mass
3. (a)
Sol. If ball is thrown with velocity u, then time of flight g
u
velocity after :sectg
u
t
g
uguv = gt.
So, distance in last 't' sec : 20 .)(2)( 2 hggt
.2
1 2gth
4. (c)
Sol. Given 42 dtctbtay
v = 3420 dtctbdt
dy
Putting ,0t vinitial = b
So initial velocity = b
Now, acceleration (a) 21220 tdcdt
dv
Putting t = 0, ainitial = 2c
5. (c)
Sol. Let the velocities be B & A
,
A = B (given), )given( 3 R
R
1
2
t sec h
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MAJOR TEST# 07 (NEET) SOLUTIONS
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3
Cos AB 2 B A
Cos AB 2 - B A
22
22
Putting A = B we obtain = 120°,
Hence, (C) is correct
6. (b)
Sol. 8.9
30sin8.92sin2 o
g
uT
= 1sec
7. (b)
Sol. Initial velocity smJi /ˆ8ˆ6 (given)
Magnitude of velocity of projection 22yx uuu
22 86 = 10 m/s
Angle of projection 6
8tan
x
y
u
u
3
4
5
4sin and
5
3cos
Now horizontal range g
uR
2sin2
10
5
3
5
42)10( 2
meter6.9
8. (a)
Sol. F = nmV
= (50 × 10–3) (400) = 10 N
9. (b)
Sol. a =
10. (b)
Sol.
m
Fr
60
30
Mass
forceNet
)40048( 2 xxdx
d
dx
dUF
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MAJOR TEST# 07 (NEET) SOLUTIONS
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For the equilibrium condition
.
11. (a)
Sol. K = = [t3 – t2 + t]24 = 46J
12. (b)
Sol. As W = K
Force is along negative x-axis and displacement is along + x-axis
W = negative
Hence
K = negative
13. (a)
Sol. Velocity of centre of mass .
14. (a)
Sol. Distance covered by wheel in 1 rotation =
(Where D= 2r = diameter of wheel)
Distance covered in 2000 rotation
= 2000 = (given)
15. (b)
Sol. Moment of inertia of disc about a diameter = (given)
Now moment of inertia of disc about an axis perpendicular to its plane and passing through a point on its rim
= .
16. (a)
0dx
dUF
0416 x 16/4x mx 25.0
4
2
Pdt
321
332211
mmm
vmvmvmvcm
100
ˆ1050ˆ1030ˆ1020 kji kji ˆ5ˆ3ˆ2
Dr 2
D m3105.9
meterD 5.1
IMR 2
4
1IMR 42
IIMR 6)4(2
3
2
3 2
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MAJOR TEST# 07 (NEET) SOLUTIONS
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Sol.
and the X- axis is given by
Dot product of these two vectors is zero i.e. angular momentum is perpendicular to X-axis.
17. (a)
Sol. Here, angular momentum is conserved, i.e.
L = I constant. At A.
the moment of inertia I is least. So angular speed and therefore the linear speed of planet at A is maximum.
18. (a)
Sol. Here, mass of the particle = M
Mass of the spherical shell = M
Radius of the spherical shell = R
Let O be centre of spherical shell. Gravitational potential at point P due to particle at O is
Gravitational potential at point P due to spherical shell is
19. (d)
Sol. Escape velocity, ….(i)
Where and be the mass and radius or the earth respectively. The orbital velocity of a satellite close
to the earth’s surface is
…. (ii)
From (i) and (ii), we get
20. (c)
Sol. VPn = (V + V)(P + P)n
prL
243
121
ˆˆˆ
kji
kjkji ˆ2ˆˆ2ˆˆ0
kji ˆ0ˆ0
)2/R(
GMV1
21 VVV
R
GM3
R
GM
R
GM2
R
GM
)2/R(
GM
E
Ee
R
GM2v
EM ER
R
GMv E
a
ae v2v
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MAJOR TEST# 07 (NEET) SOLUTIONS
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VPn = VPn
= – n
K = – =
21. (d)
Sol. Elastic potential energy per unit volume is
As both the wires are made up of same material, so their Young’s modulus is same for both the wires.
As both the wires are stretched by the same load, therefore
22. (a)
Sol. 4P
P
2
1
or P1 = 4P2
21 r
s44
r
s4 or
4
1
r
r
2
1
64
1
4
1
r
r
v
v3
32
31
2
1
23. (a)
Sol. Here, ,m102cm2r,Nm06.0S 2
1
1
m105cm5r 2
2
Since bubble has two surfaces initial surface area of the bubble 222
1 )102(42r42
24 m1032
Final surface area of the bubble 222
2 )105(42r42
24 m10200
V
V1
P
Pn1
V
V
P
P
V/V
P
n
P
Y
)stress(
2
1u
2
2)stress(u
222
211
22
21
2
1
)A/F(
)A/F(
)stress(
)stress(
u
u
21 FF
1
16
1
2
D
D
D
D
A
A
u
u44
1
2
2
21
22
2
1
2
2
1
SAFE HANDS, AKOLA
MAJOR TEST# 07 (NEET) SOLUTIONS
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Increase is surface area
= 44 103210200
24 m10168
work done = Surface tension × increase is surface area 41016806.0
mJ12.3J1012.3 3
24. (b)
Sol. AS the velocity of water at the bottom of the river is zero.
11 sm518
518hkm18dv
Also, sPa10poise10,m5dx 32
Force of viscosity, dx
dvAF
Shearing stress, dx
dv
A
F
233
Nm105
510
25. (c)
Sol. 22 yav 22
2ya
a
22
2
4ya
a
2
3 Ay [As
22
max avv
26. (b)
Sol. y = A cos ( )xkt standard equation
y = 3 cos (100 xt ) – given equation
So K = and k
22 cm
27. (d)
Each charge will experience two forces each of magnitude F inclined at an angle of 600. Their resultant is
given by F360cosF2FF2/10222
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28. (b)
Since lines of force are denser at B hence electric field is maximum at B
29. (d)
By using relation cnC .3/1 ccC 2.)8( 3/1
30. (a)
Sol. Given Wheatstone bridge is balanced because S
R
Q
P . Hence the circuit can be redrawn as follows
31. (c)
Sol. Use R = A
2
1
R
R
=
2
22
1
11
A
A
32. (b)
Sol. In steady state, the capacitor arm presents an infinite resistance. So the potential difference across C is
that across r2.
Current through r2 = 21 rr
V
P.D. across r2 = 21
2
rr
Vr
33. (a)
Sol. K.E = QU magnetic moment = i × Area = T
Q× R2
T = qB
m2 R =
qB
mKE2 =
qB
mU2
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MAJOR TEST# 07 (NEET) SOLUTIONS
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Magnetic moment = m2
BQ2
×
QB
Um2 Magnetic moment = QU
34. (b)
Sol. coilwirenet BBBislooptodueB
iswiretodueB
Bnet is more than Bcoil.
35. (c)
Sol. L = 2r and r = 2
L M = IA = I(r2) = I
2
2
L
=
4
IL2
36. (a)
Sol. = sinMB and )cos1( MBW
2)60cos1(
MBMBW o . Hence =
WMB
MB o 32
360sin
37. (b)
Sol. Rate of cooling difference in temperature
dT
dt
Kdt
dT
dT = K.dt
In First Case
dT = 6159=2
= 60 30 = 30
dt = 4 minutes
60
1
4x30
2
dt
dTK
For second case dT 2
50 30 20
.min620x
2
K
dTdt
601
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MAJOR TEST# 07 (NEET) SOLUTIONS
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38. (a)
Sol. Temperature on any scale can be converted into other scale by LFPUFP
LFPx
= constant for all scales
20150
20x
=
100
60 x = 980C
39. (a)
Sol. Root means square velocity smM
RTvrms /1930
3 (given)
gmkgRT
M 210219301930
30031.83
)1930(
3 3
2
i.e. the gas is hydrogen.
40. (a)
Sol. µ = = = = 1.41
41. (d)
Sol. Path difference
2
2211
2 i.e. constructive interference obtained at the same point So, resultant intensity
IIIIIIR 25)49()( 2221 .
42. (c)
Sol. By using Einstein's equation E = W0 + Kmax
max1.26 K eVK 9.3max
Also .9.3max0 V
KV
43. (a)
Sol. For infrared > 700 nm i.e., wavelength is greater than 700 nm
2
Asin
2
Amsin
º30sin
º45sin2
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MAJOR TEST# 07 (NEET) SOLUTIONS
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– 0.56 eV
– 0.85 eV
– 1.51 eV
– 3.4 eV
n =
n = 5
n = 4
n = 3
n = 2
n = 1 – 13.6 eV
1 =
2i
2f
yn
1
n
1R
ni = ; nf = 3
1= 1.097 × 107
0
9
1
= 097.1
109 7
= 820 × 10–9 m
44. (a)
Sol. By using 3/1Ar
10
6
5
8
125
273/13/1
2
1
2
1
A
A
r
r
45. (c)
Sol.
92 90
–
91
–
92 5
82 2
–
84
82
2+
80
78
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MAJOR TEST# 07 (NEET) SOLUTIONS
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46. (b)
Sol. AgBrFCHAgFBrCH 33
This reaction is known as swarts reaction.
47. (c)
Sol. RMgXMgRX thus Grignard reagent (RMgX) is formed by reaction of dry magnesium (Mg)
with alkyl halides (RX) in the presence of ether.
48. (a)
Sol. order of reactivity of different halo compounds towards nucleophilic substitution reactions are:
Allyl chloride > vinyl chloride > chlorobenzene.
49. (b)
Sol. In tertiary alkyl halides steric hindrance does not allow substitution by 2SN mechanism in which the
nucleophile attacks on the carbon atom and the reaction takes place in single step.
50. (b)
Sol. The reactivity of different alkyl halides towards 2SN reaction decreases in the order.
Methyl halides > 1° halides > 2° halides > 3° halides.
halides1
)d(
CHCHHCCHBr
)c(
CHCHCHCHBr
)a(
CHCHCHCHCHBr
32|
2
|
322
32222
3CH
3CH
halides3
)b(
CHCHCCH 32
|
|3
3CH
Br
51. (d)
Sol.
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MAJOR TEST# 07 (NEET) SOLUTIONS
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52. (a)
Sol. 1SN reaction proceeds with racemisation.
53. (b)
Sol. The ease of dehydrohalogenation of alkyl halides with alcoholic KOH is 3° > 2° > 1°,
This order of alkyl halides can be explained on the basis of the stability of the alkene formed after
dehydrohalogenation of haloalkanes. 3° alkyl halides on dehydrohalogenation forms more substituted
alkenes, which is more stable (and formed at faster rate), while primary alkyl halides yield least
substituted alkenes, which is less stable (and formed at slower rate.)
54. (b)
Sol.
%)20(Butene1
CHHCCHCH
CHCHCHCHHCHCCHCH
223
%)80(Butene233
)alc(KOH
etanBromobu23
|
23
Br
In elimination reaction of alkyl halide major product is obtained according to Saytzeff’s rule, which
states that when tow alkenes may be formed, the alkene which is most substituted one predominates.
55. (b)
Sol. when two molecules of the same alkyl halide react with Na in dry ether, an alkane with double the
numbr of carbon atoms is formed. This is known as Wurtz reaction.
56. (a)
Sol. 1SN mechanism
(i) ncarbocatio3
ClC)CH(ClC)CH( 3333
(ii) OHC)CH(C)CH( 33OH33
For 1SN mechanism the reactivity of carbocaiton are 3° > 2° > 1°
57. (a)
Sol. It act as a stronger nucleophile from the carbon end because it will lead to the formation of C-C
bond which is more stable than C –N bond
58. (c)
Sol. Chlorofluorocarbons (CFC’s) or freons are used as refrigerant in refrigerators and air conditioners.
59. (b)
Sol. The reaction is electrophilic substitution reaction
halidesodiumAlkanehalidAlkyl
XNa2RRRXNa2XRetherdry
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MAJOR TEST# 07 (NEET) SOLUTIONS
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60. (a)
Sol. Allylic halides are the compounds in which the halogen atoms is bonded to an 3sp -hybridised
carbon atom next to carbon – carbon double bond (C=C) i.e. to an allylic carbon.
61. (b)
Sol. During chlorination of benzene, anhydrous 3AlCl being a Lewis acid helps in generation of the
elecetrophile Cl by combining with the attacking reagent.
The electrophile Cl attacks the benzene ring in this reaction.
62. (b)
Sol. Gem-dihalides are named as alkylidene halides having halogen atoms on the same carbon atom.
63. (a)
Sol. Benzylic halides show high reactivity towards the 1SN reaction
The carbocation thus formed gets stabilized through resonance as shown in the figure.
64. (a)
Sol. For the same alkyl halide boiling point increases as the mass of halogen increases.
65. (d)
Sol.
)tertiary(anemethylprop2Bromo2
CHCCH
)ondary(secene2bromopent4)primary(ene2Bromobut1
CHHCCHCHCH,BrCHCHCHCH
3
|
|3
3|
323
3CH
Br
Br
66. (d)
Sol.
2, 4, 6-trinitrophenol or picric acid
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MAJOR TEST# 07 (NEET) SOLUTIONS
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67. (a)
Sol.
68. (c)
Sol. – OH group is attached to primary carbon.
69. (b)
Sol.
Cyclohexanol is a secondary alcohol because OH group is linked to o2 carbon.
70. (b)
Sol. OHHC 52 and 33 CHOCH are isomers.
71. (b)
Sol. Hydroboration oxidation (Industrial preparation of alcohol)
BCHCHCHHBCHCHCH 3323ether
Dry6223 )(
2
13
OHCHCHCHBCHCHCHOH
2233323 3)( 22
72. (d)
Sol.
73. (b)
Sol.
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74. (c)
Sol. NaIHOCHCHICONaHC 52525252 75 (b)
Sol.
76. (d)
Sol. Acetone reacts with Grignard’s reagent to give tertiary alcohol.
alcoholbutyl -ter33323 )()( 2 OHCCHMgBrCHOCCH
OH
77. (b)
Sol.
78. (c)
Sol. NaBrCHOCHCHBrCONaCH
ether)al (symmetricetherDimethyl
3333
NaBr
CH
CH
OCHCCHBrCH
CH
CH
NaOCCH
ether)ical (unsymmetretherbutyl -Methyl ter
3
3
3
|
|33
3
3
|
|3
79. (b)
Sol. AgBrHCOHCOAgBrHC 22 5252Dry
252
If we take moist OAg2 then alcohol is formed
AgOHOHOAg 222 AgBrOHHCAgOHBrHC 5252
80. (b)
Sol.
ethoxide sodiumdimethyl 22,
3
3
3
|
|3 CHClNa
CH
CH
OCCH
ether butyl -Methyl
3
3
3
|
|3
t
NaClCH
CH
CH
OCCH
SAFE HANDS, AKOLA
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81. (b)
Sol. 4NaBH and 4LiAlH attacks only carbonyl group and reduce it into alcohol group. They do not attack
on double bond.
4
aldehydecinnamic 56
NaBHCHCHOCHHC
alcoholcinnamic 256 . OHCHCHCHHC
82. (a)
Sol. IOCHHlCICHOHC 36366 5
83. (a)
Sol. YX
COOHCHCHOHCHCHCH 23Oxidation
223
Since on oxidation same no. of carbon atoms are obtained in as therefore alcohol is primary
84. (a)
Sol. HClCHHCClCHHC 356AlCl
anhyd.366
3
,
It is a Friedel-craft’s reaction.
85. (c)
Sol.
86. (d)
Sol. –NO2 group in benzene ring shows – and – M effect, which deactivates the ring towards
electrophilic substitution but activates towards nucleophilic substitution
87. (b)
Sol. Symmetrical alkane with even no. of carbon atoms can be prepared by Wurtz reaction.
88. (b)
Sol. Photochemical chlorination of alkanes is a free radical process and the reaction is initiated by
homolysis of halogens to form free radicals. ••h
2 ClClCl
89. (a)
Sol. Lower alkanes )toCC( 41 are gaseous at room temperature.
90. (b)
Sol. Kerosene is obtained as a fraction during fractional distillation of petroleum. It is a mixture of
aliphatic hydrocarbons.
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MAJOR TEST# 07 (NEET) SOLUTIONS
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91- a
121- b
151- d,
92- b
122- c
152- d
93- c
123- d
153- c
94- b
124- c
154- c
95- b
125- d
155- d
96- d
126- c
156- a
97- c
127- c
157- c
98- b
128- a
158- d
99- c
129- c
159- c
100- b
130- a
160- a
101- d
131- c
161- a,
102- b
132- a
162- b
103- c
133- c
163- b
104- c
134- b
164- c
105- c
135- a
165- b
106- c
136- d
166- a
107- d
137- a
167- c
108- c
138- a
168- a
109- a
139- b
169- a
110- b
140- b
170- a
111- b
141- c
171- c
112- d
142- a
172- a
113- c
143- c
173- c
114- c
144- a
174- c,
115- a
145- c
175- a
116- b
146- b
176- b,
117- b
147- c
177- c
118- c
148- a
178- c
119- a
149- d
179- a
120- c
150- c
180- b