Conventional Question Practice Programe CE (Test-1...

IES M

ASTER

CE (Test-1), Objective Solutions, 5th March 2016 (1)

1. (a)

2. (b)

3. (b)

4. (d)

5. (d)

6. (a)

7. (d)

8. (d)

9. (d)

10. (c)

11. (b)

12. (b)

13. (d)

14. (b)

15. (d)

16. (b)

17. (d)

18. (b)

19. (d)

20. (a)

21. (a)

22. (a)

23. (d)

24. (b)

25. (d)

26. (c)

27. (d)

28. (c)

29. (d)

30. (d)

31. (c)

32. (b)

33. (a)

34. (b)

35. (c)

36. (d)

37. (b)

38. (a)

39. (d)

40. (d)

41. (c)

42. (b)

43. (c)

44. (a)

45. (c)

46. (d)

47. (d)

48. (c)

49. (b)

50. (a)

51. (d)

52. (b)

53. (a)

54. (c)

55. (b)

56. (c)

57. (c)

58. (a)

59. (c)

60. (a)

61. (b)

62. (c)

63. (b)

64. (d)

65. (b)

66. (a)

67. (b)

68. (b)

69. (b)

70. (c)

71. (b)

72. (c)

73. (a)

74. (d)

75. (a)

76. (b)

77. (c)

78. (a)

79. (b)

80. (a)

81. (d)

82. (d)

83. (b)

84. (c)

85. (c)

86. (b)

87. (b)

88. (c)

89. (b)

90. (c)

91. (d)

92. (d)

93. (d)

94. (d)

95. (a)

96. (d)

97. (d)

98. (c)

99. (c)

100. (d)

101. (a)

102. (d)

103. (a)

104. (d)

105. (d)

106. (c)

107. (a)

108. (c)

109. (a)

110. (b)

111. (a)

112. (b)

113. (a)

114. (b)

115. (d)

116. (c)

117. (c)

118. (a)

119. (d

120. (d)

Conventional Question Practice ProgrameDate: 5th March, 2016

ANSWERS

IES M

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(2) CE (Test-1), Objective Solutions, 5th March 2016

1. (a)

For copperd = 2cm

= 20 mmFor steel

di = 2 cm= 20 mm

d0 = 4 cm= 40 mm

s = 100N/mm2

Es = 2Ec

Copper

Steel

4cm

2cm

Strain in steel = strain in copper

s

sE

= c

cE

c =c

s

100 EE

c = 50 N/mm2

2. (b) Since force in all the three rods will besame stress in B will be 4 times that in A

since area of B is 14 th of that of A.

= 1.2 × 10–5/°CE = 2 × 105 MPa

T = 50°C

20mm 20mm 20mm

R R

YX

AB

C

20mm 10mm 20mm

Since supports are rigid, relativedisplacement of Y w.r.t X = 0

A B cL t L T L T

– CA B

A B C

R LRL RL 0E A E A E A

=

LA = LB = LC = L

A B C

L R R R3L TE A A A

=

3E T = 2 2 2

1 1 1R20 10 20

4 4 4

R = 5 5

2 2 2

3 2 10 1.2 10 504 4 420 10 20

= 18849.56 N

A =2

R 60 MPa20

4

=

B =2

R 240 MPa10

4

=

3. (b)E = 3 × 105 MPaG = 1.2 × 105 MPa

E =9KG

3K G

3 × 105 =

5

5

9 K 1.2 103K 1.2 10

3K + (1.2 × 105) = 3.6 K0.6 K = 1.2 × 105

K = 2 × 105 MPa

4. (d)G = 70 GPaK = 150 GPa

E = 9KG

3K G

= 9 150 703 150 70

= 181.73 GPa

182 GPa

E = 2G 1

182 = 2 70 1

0.3

Option (d) is correct.

5. (d)L = 2 mt1 = 20°t2 = 60°

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CE (Test-1), Objective Solutions, 5th March 2016 (3)Temperature change t = t2 – t1

= 40°CE = 200 × 103 MPa

610 10 / C

Yielding of the wall = 0.2 mm

A

L

1E

A

L

BR1

B1

th

R1

A BB1

B2

ad

BB1 = th t L

B1B2 = Expansion restricted by supports

= BB2 = Yielding of support

B1B2 = thad

LE

(–ve sign shows compressive stress)BB1 = BB2 + B1B2

th ad

tht L EE

th thEL

3

th200 10 L t

2000

= –100[(2000 × 10 × 10–6 × 40) – 0.2]= –60 MPa= 60 MPa compressiveOption (d) is correct.

6. (a)Member 1,Area (A1) = 10 × 10 mm2

= 100 mm2

E1 = EMember 2,E2 = E/2

A2 = ?Given, P1 = P2 : and L1 = L2

and 1 2

1 1 2 2

1 1 2 2

PL P LE A E A

E1A1 = E2A2

E × 100 = 2E A2

A2 = 200If a is the length of the side of squarea2 = 200a = 14.142 mmOption (a) is correct.

7. (d) The time required to reach a specified stagein the consolidation process is

t ~2m.d

K

where, t = the time required to complete somepercentage of the consolidationprocess.

m = the compressibility of the mineralskeleton

d = the thickness of the soil mass (perdrainage surface)

K = the permeability of the soil.Permeability increase with increase in size ofgrain, therefore consolidation time t will reduce.Hence option 3 is incorrect. Consolidation timeis indepedent of applied stress.

8. (d)

t =2

vv

v v w

T d KCC m

where, t = conslidation timeTv = time factor

= 2·U4

for U < 0.6

where U = degree of conslidationd = drainage path

Cv = coefficient of volume changeK = permeability

m v = coefficient of volume change

w = unit weight of water

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(4) CE (Test-1), Objective Solutions, 5th March 2016Squeezing more water means going to higherdegree of consolidation, which will take moretime.Higher stress change, higher compressiblity andhigher volume of soil will all lead to higher volumeof squeezed water.Consolidation creates a denser soil with changedsoil peroperties like lower permeability, highershear strength.

10. (c) Higher the value of group index, porer isthe quality of soil as pavement.

11. (b)

well graded

soil

Uniform soil% finer

than

Particle size on log scale

An uniform soil means most of soil particlesare in a limited range of size. That doesnot mean particle size is small. E.g., a soilhaving particle size range between 2mmto 4mm will be categorized uniform sand.Hence option 1 is incorrect.

Angular particles carry high charge densityat edges. They tend to form honeycombstructure which is of lower density. Henceoption 2 is correct.Well graded soil usually compacts to higherdensities.

The test to determine the maximum densityusually involves some form of vibrations.The test to determine minimum densityusually involves pouring oven-dried soil intoa container. Unfortunately, the details ofthese tests have not been entirelystandardized and values of the maximumdensity and minimum density for a givengranular solid depend on the procedure usedto determine them.

12. (b) The one-dimensional consolidation equationcan be solved numerically by the methodof finite element.

13. (d) Density Index = max

Dmax min

e eI 100%e e

Loosest state

Present State

Densest State

e eminemax

using wd

G1 e

we can derive that

maxd(min)

1(1 e ) and min

d(max)

1(1 e )

Then, ID = d(min) d

d(min) d(max)

1 1

100%1 1

Gs remains constant throughout; hence isadvantageous to use in deriv ingrelationships.

The strength of the dry soil is assessed bybreaking and crumbling between the fingers.

14. (b) Gap graded and step graded are both thesame classification of a soil.

Poorly graded is also known as uniformlygraded.

16.(b) A

BP

C

1m1m 0.75 mm

E = 200 GPa= 200 × 103 MPa

AAB = 1 cm2

= 100 mm2

ABC = 2 cm2

= 200 mm2

Gap = 0.75 mmThere will be no deformation in BC so,

AB

AB

AB

PL0.75EA

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3P 1000 0.75

200 10 100

P = 75 × 200= 15 × 103 N= 15 kNOption (b) is correct.

17. (d) Final position of bolt is at sec A-A whenbolt is tightened the sleeve is compressedwhereas bolt is elongated

Movement of nut = (y – x) =

= compression of sleeve + extension ofbolt

sleeve

Nut

x

y

A B C D

A

BC D

18. (b)Load (P) = 16 kNDiameter (d) = 16 mm% Elongation = 0.04%

f i

i

L L100 0.04L

4f i

i

L L4 10L

Strain (E) = 4 × 10–4

E = StressStrain =

P/A

= 3 2

4

16 10 1644 10

= 3

2 416 10 416 4 10

= 198.94 × 103 MPa= 199 GPa

Poisson’s ratio Lateral StrainLongitudinal Strain

=

f i

i

f i

i

d dd

L LL

=

2

20.01 10

0.04 10= 0.25

E = 2G 1

G = E

2 1

= 199

2 1 0.25

= 79.6 80 GPaStatement (2) and (4) are correct.Option (b) is correct.

19. (d)Toughness is the ability to absorb mechanicalenergy upto failure. Area under load deformationcurve upto fracture is called toughness; whilearea under stress strain curve upto fracture iscalled modulus of toughness.

Note : Toughness is desirable against impactloading. As failure strain is more in ductilematerial, mild steel is more tough than castiron.

20. (a)

45°

45°

= ±400 MPa

x

y

It is a case of pure shear stress

2 1

Centre

On the plane at 45° from x–plane.

1 2 400 MPa

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(6) CE (Test-1), Objective Solutions, 5th March 2016and 0Option (a) is correct.

21. (a)

A

B

x

y

xy 0

x y

xysin2 cos22

=

sin2 02

= 0Option (a) is correct.

22. (a)

45° x

y

1

2

3

6x 1 1000 10

6y 3 600 10

6

45 2 800 10

x y45 x y

12 2

xycos2 sin2

2

800 × 10–6 = (800 × 10–6) + 12 (400 × 10–6)

xycos90 sin90

2

xy 0

It means shear strain along x and y axis arezero so x and y plane are principal plane, so

direction of the major principal strain is alonggauge 1.

Option (a) is correct.23. (d)

d = 100mmV = 10 kN

= 104 N

max = avg43

=4

22

4 10 N3 mm100

4

=16 MPa3

24. (b) The upper and lower limits of the range ofwater content over which the soil exhibitsplastic behaviour are defined as the liquidlimit and the plastic limit respectively.

29. (d) Most of the particles in the silt range andcoarser are approximately equidimensionaland most of those in the clay size are farfrom equidimensional.

34. (b) The degree of compaction of a soil ismeasured in terms of dry density.

36. (d) Void ratio = Volume of voidsVolume of solids

=

1 16 3 1.0

1 116 3

37. (b) For rectangular section,

b =

3M y M h/2

bh /2 =I

b = 26Mbh

If h = 2h

b

h

b = 2 26M 6Mbh b 2h

=

= 21 6M4 bh

b = b

4

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CE (Test-1), Objective Solutions, 5th March 2016 (7)38. (a)

A

3kN

B1m 2m

x

RBCRA

x

x1kN/m

taking moment about point A

(RB × 3) – (3 × 1) – 33 12

= 0

3RB = 3 + 4.5RB = 2.5 kN

RA + RB = 3 + (1 × 3)RA + RB = 6

RA = 3.5 kNMmax = (B.M.)c

= (RA × 1) – 11 12

= 3 kN–mMmax = 3 × 106 N–mm

y =150 75mm

2=

b = maxM yI

= 6

4 43 10 N-mm 75mm

300 10 mm

= 75 MPa

39. (d)

Maximum bendingstress in rectangularbar

Maximumbending stressin circular bar

=

33

M b 32Mdb 2b

12

3 312M 32M8b d

b = 1/33 d

64

= 1/3 1/33 d

4

2

Weight of rectangularbar b 2b LWeight of circular bar

d L4

= 2

28bd

=

2/3 2/38 316

=

2/3

1/33

2


40. (d)Bending stress in a beam is given by

=MyI

=MZ

Where Z is section modulus.

For beams with same cross section, Z is same.It is given that cross sectional area is same.We cannot infer that Z is same

A

B

= 1max 2

2max 1

M ZM Z

Also For Beam A :

w

+ M1max

2w8

For Beam B :

w/4

–M =2 max2w

8

M1max = M2max

A

B

= 2

1

ZZ

Since, Z1 and Z2 depends on shape of cross

section, hence A

B

also depends on shape of

cross section.

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(8) CE (Test-1), Objective Solutions, 5th March 201641. (c)

a

a

AB

Ba/y

A

ByAy

AA

For square beam, shear stress,

= NA

F AyI b

As F, INA and b remains same,

Ay

A

B =

a aa2 4

a a aa4 4 8

A

B =

1 8 42 3 3

42. (b)A2 = 4 A1

22d

4

= 214 d

4

d2 = 2d1

A case of simply supported beam loadedwith a point load P is used.

P

A. Mmax = PL

2

1

MM = 1

B. Deflection( ) = 3 3

4

PL PL3EI 3E D

64

41

D

1

2

=

4412

4 41 1

2DD 16D D

2 = 11

16

C. Bending stress,

b = max

3

M MZ D

32

b 31

D

b 1

b 2

=3 3

2 1

1 1

D 2D 8D D

b 2 = 118

D. Section Modulus (Z) = 4

max

D64

y D/2

I

Z = 3D32

Z D3

1

2

ZZ =

3 31 1

2 1

D D 1D 2D 8

Z2 = 8 Z1

43. (c)N = 250 R.P.M.d = 45 mmP = 50 kW = 50 × 103 W

2u 427 N/mm

P = T·W

T = 3P 50 10

W 2 25060

= 350 10 60 N.m.

500

= 650 10 60 N-mm

500

Maximum shear stress

= 316T

d = 106.27 N/mm2

FOS = 427 4.01

106.27

44. (a)d = 8 cm = 80 mm

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CE (Test-1), Objective Solutions, 5th March 2016 (9)M = 3000 N–mT = 4000 N–mEquivalent twisting moment

Te = 2 2M T

= 5000 N–m= 5 × 106 N–mm

Max. shear stress = e3

16Td

=

6

316 5 10

80

= 156.25 MPa

Equivalent bending moment

Me = 2 21 1M M T2 2

= 63 5 101500 10

2

= 4 × 106 N–mm

Maximum normal stress = e3

32 Md

=

6

332 4 10

80

= 250 MPa

Option (a) is correct.

46. (d)Diameter (d) = 2 cm

Permissible shear stress 2s 10 kN/cm

TJ r

T = Jr

T = 3d

16

= 32 10

16

= 5 kN-cm


47. (d)For solid shaftd = 100 mmL = 1000 mm = 60 N/mm2

T = 3d

16

= 3100 60

16

For hollow shaftd0 = 100 mmdi = 50 mm = 60 N/mm2

TJ

4 40 i

0

d d32Td2

= 4 4

0 i

0

d d16 d

= 4 460 100 50

16 100

= 60 937500

16

3

60 937500T 1516T 16100 60

16

15T T16

Hence T should be reduced by = 15T T16

= T

16Option (d) is correct.

48. (c)

Young’s modulus (E) = Normal stressNormal strain

Modulus of Rigidity (G) = Shear stressShear strain

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Bulk modulus (K) = Hydrostatic stressVolumetric strain

Poisson’s ratio ( ) = Lateral strain

Longitudinal strain

49. (b)

eff =L eff =2Leff

L=2

effL=2

51. (d) If all the air in a soil could be expelled bycompaction, the soil would be in a state offull saturation and the dry density would bethe maximum

55. (b) Poorly graded sands will have particle ofuniform size leading to poor packing of soilstructure. Hence they will have lower frictionangle.Inorganic clays lie above A-lie in plasticitychart.

Unconfined Consistencycompressivestrength (qu)

< 25 very soft25 – 50 Soft50 – 100 medium100 – 200 stiff200 – 400 very stiff> 400 hard

57. (c) Oven drying method, torsion balancemethod and radiation method are used todetermine water content.

58. (a) Shrinkage ratio

=

L d

d

L s

V V 10 6100% 100%V 6(w w ) 100% (50 15)%

SR = 1.905

59. (c) Measuring flask method is used fordetermination of specific gravity of soilparticles.

60. (a)

Soil type dry unit weight (kN/m3)in loosest state

Gravel 16Coarse sand 15Fine sand 14Coarse silt 13

62. (c)

H

h/2

Bulb

Jar

He = effective depth= distance from suspension surface to

the level at which density is beingmeasured

=h VhH2 2A

where,

hH2

= depth suspension

= distance from suspension surface tocentre of bulb

= (also known as) Hydrometer reading.Vh = Volume of hydrometer bulbA = internal area of cross section of jar.

63. (b) The angularity of particles has greatinfluence on the behaviour of coarse-grainedsoils.

64. (d) The relative density of a soil gives a moreclear idea of the denseness than does thevoid ratio.

66. (a) Due to repeated cyclical loading, the yieldstrength of metal reduces. This is termedas fatique.

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CE (Test-1), Objective Solutions, 5th March 2016 (11)Both ducticlity and malleability are due toplasticity.

69. (b)

terminal velocity, v = 90.2 D2 where D is inmm and v is in cm/s.here D = 0.075 mm

v = 90.2 × (0.075)2 = 0.50 cm/s

71. (b) Plasticity of the soil is due to adsorbedwater.

72. (c) In comparison to atterberg limits of normalsoils, expansive soils have

More liquid limit

Less plastic limit

Less shrinkage limit

More volumetric shrinkage.

This essentially means expansive soils havea large plastic region, represented by highervalues of plasticity index.

73. (a) Natural soils generally have a sensitivity >1. However stiff clays having fissures andcracks sometimes exhibit a sensitivity valueless than 1, since weaknesses; i.e., fissuresand cracks present in an undisturbed samplewill be removed upon remoulding.

Plasticity index is never negative. If WL<WP then IP is taken as zero.Plasticity of soil is due to presence ofelectrically charged minerals and water (bi-polar in nature) in the adsorbed layer ofsoil.Highly sensitive soils lose higher proportionsof shear strength on remoulding.Correspondingly when left alone, regain inshear strength (thixotropy) is also higher.

76. (b) The plasticity index increases as the amountof clay fraction increases.

A high value of toughness index indicatesa high percentage of colloidal active clayminerals.

77. (c) Relationship between atterberg limits andengineering properties of soil:

Properties Fix WL and Fix IP andincrease IP increase WL

Strength increases decreasesToughness increases decreasesCompressibility constant increasesPermeability decreases increasesRate of vol. decreases increaseschange

78. (a) Ic = P p

w w w w100 100I w w

l l

l

= 45 30 15100 100 75%45 25 20

79. (b) IP = w wp 45 25 20% l

82. (d) Auger boring is generally used in soils whichcan stay open without casing or drilling mud.Clays, silts and partially saturated sandscan stand unsupported. For soils which cannot stand unsupported especially for sandysoils below water table, a casing is normallyrequired.

84. (c)In case of pure shear,

xy

xy

x

x

x 0

y 0

xy max

x y 2 21,2 x y xy

1 42 2

21,2 max

10 42

1,2 max

1 max

2 max

Option (c) is correct.

85. (c)L = 8 mInitially,

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A A1

Aluminium

A

L

A1Steel

5mm

Final,

A2

A A1

A1Steel A

AluminiumA2 A3

5mm

6s 12 10 / C

6a 23 10 / C

t1 = 30°CTotal elongation of Aluminium – Total

elongation of steel= (AA3) – (A1A2)= (AA1 + A1A2 + A2A3) – (A1A2)= AA1 + A2A3

= 10 mm

al sL t L t 10

3 6 68 10 t 23 10 12 10 10

3 610t

8 10 11 10

= 113.636°Ct2 – 30° = 113.636°Ct2 = 143.636°COption (c) is correct.

86. (b)

1 is acting along longitudinal direction

1

2

2

2

1

2

Strain in lateral direction

2 1 2

1 E E E

Strain in lateral direction when 1 acting alone.

1

2 EGiven,

2

1 2

2 1 2 1

E E E 2E

2 1 11

E E 2E

1

2 12

2 12 1

Option (b) is correct.87. (b)

1m 2m

10 kN

C

ARA

RB

RA + RB = 10Overall deformation,

total AC CB 0

A AR (1) (R 10)(2)

0EA EA

RA + 2RA – 20 = 03RA = 20

RA = 20 kN3

RB = 103 kN

Option (b) is correct.

88. (c)

1 40 MPa

2 20 MPa

Since these are principal planes, so shearstress on this planes is zero.If we consider these two planes, along x andy axis.

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xx 40 MPa

yy 20 MPa

xy 0

30

1

xx yy xx yyx xy30

cos2 sin22 2

= 40 20 40 20 cos60 0sin602 2

= 30 + (10 cos 60°)

= 130 102

= 35 MPaOption (c) is correct.

89. (b)

xx 60 MPa

yy 120 MPa

xy 40 MPa

Radius of the Mohr circle

= 2

xx yy 2xy2

= 2

2120 60 402

= 2 230 40= 50 MPaOption (b) is correct.

90. (c)Centre of Mohr circle always lies on x–axis.

Co–ordinate of centre = 1 2 , 02

where 1 and 2 are the normal stresseson two mutually perpendicular planes.

Option (c) is correct.

91. (d)1. Two dimensional stresses applied to a thin

plate in its own plane represent the planestress condition.

2. Under plane stress condition,

3 0

so, 1 1 21E

2 2 11E

3 1 21E

so, in plane stress condition the strain inthe direction perpendicular to plane is notzero.

3. Normal and shear stresses may occur.Simultaneously on a plane.Statement (1) and (3) are correct.Option (d) is correct.

92. (d)

xy

y

xx

y

xy

In order to completely specify the state of stressin two dimensions in a component, we need toshow normal and shear stresses on twomutually perpendicular planes passing throughthe point.With single plane, one cannot show the stateof stress. Even any two planes passing throughthe point are not enough to show its state ofstress.

93. (d) Soils compacted at a water content morethan the optimum water content usually havea dispersed structure.

3

4

Dry

den

sity

Water content

2

1

94. (d) Soils compacted at a water content morethan the optimum water content usually havea dispersed structure if the compactioninduces large shear strains and aflocculated structure if the shear strain arerelatively small.

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Dry side

Wet side

Shea

r stre

ss

Normal stress ( )

Failure Envelops

The soils compacted dry of the optimumhave a steeper stress-strain curve than thoseon the wet side.

Dry side

Wet side

Dev

iato

r st

ress

Axial strain

98. (c) Tempers and pneumatic-tyred rollers arecalled universal methods.

Tempers are effective for compacting soilsin a confined space of all types.Pneumatic-tyred rollers are extremely usefulfor compacting all types of soils.

99. (c) Compaction methods can not remove allthe air voids and therefore, the soil neverbecomes ful ly saturated. Thus, thetheoretical maximum dry density is onlyhypothetical.

The amount of compaction in the field shouldbe approximately equal to that in thelaboratory. The standard proctor test isadequate to represent the compaction offills behind retaining walls and in highwaysand earth dams when light rollers are used.In such cases, the optimum water contentobtained from the standard proctor test canbe used as control criterion. However, insituations where heavier compaction isrequired for example in modern highwaysand runways, the standard proctor test doesnot represent the equivalent compaction inthe laboratory.

The water content in the field can bedetermined indirectly using a proctor needle.

100.(d) The procedure for conducting the test issimilar to that in the standard proctor testbut the soil is compacted only in 2 layers.It is claimed that the optimum water contentand the dry density obtained in the test arealmost equal to that in the standard proctortest.

The soils compacted wet of the optimumhave relatively flatter stress-strain curve anda corresponding lower value of the modulusof elasticity.

Dry side

Wet side

Dev

iato

r st

ress

Axial strain

101. (a)Plastic deformation is a function of applied

stress temperature and strain rate becauseplastic deormation is accompanied by changesin both the internal (stress and strain rate) andexternal (temperature) state of the material.


102. (d)Assertion :

1

1

2

2

x

y

z

Plane state of stress,

1 2x E E

2 1y E E

1 2z E E

= 1 2E

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CE (Test-1), Objective Solutions, 5th March 2016 (15)So, a plane state of stress does not results

in a plane state of strain.Reason :

x

y

z

Uniaxial state of stress

x E

y E

z E

So, a uniaxial state of stress results in a 3–dimension state of strain.

Assertion is wrong, Reason is correct.Option (d) is correct.

103. (a)Mohr’s circle represents the transformation

of second order tensor. Stresses, strains andarea moment of inertia are second order tensor,so Mohr’s circle construction is possible forstresses, strains and area moment of inertia.


104. (d)

1 = 1 2E

2 = 2 1E

1

2

1

2

105. (d) Ductile materials have permanent strains.Rubber is not ductile as it regains its originalshape after release of stress.

106. (c) In composite shaft,

Applied torque = T = T1 + T2

Angle of twist = 1 2

1 21 2

1P 2 P

TL T LG I G I

108.(c) The removal of the salt in the soil poresresults in a soil that loses much of itsstrength when disturbed.

112.(b) The halloysite mineral has a crystalstructure similar to Kaolinite. Because ofthe ability of halloysite to adsorb waterbetween the sheets soils containing thismineral can exist at a high water contentand a low density.

115.(d) This method for the determination of watercontent is quite suitable for coarse-grainedsoils from which the entrapped air can beeasily removed.

116.(c) A minimum thickness of 20 mm is usuallyrequired to get uniform distribution ofpressure on the sample.

117.(c) The correlation is not perfect. A liberal factorof safety should be provided if the design isbased only on index properties.

118.(a)0.9800.9901.0001.0101.0201.0301.0401.050

–20–10

010203040

50

Hydrometer reading Rh

Density

Hg

Hydrometer

As sedimentation progresses, density ofliquid goes on reducing; hydrometer settlesfurther into liquid and reading reduces.

119.(d) In AASHTO, there is no place for organicsoils. Organic soils are also classified asOL and OH and as peat (Pt) if highly organicin USC system.

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