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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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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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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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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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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


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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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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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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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– 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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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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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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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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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


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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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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


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