ENGINEERS ACADEMY Soil Mechanics Properties of Soils | 1
QUESTION BANK
1. The liquid limit and plastic limit of sample are 5 Assertion (A): If the water table is very near to 65% and 29% respectively. The percentage of the subgrade of the road. It will ultimately cause the soil fraction with grain size finer than 0.002 cracking of the road surface. mm is 24. The activity ratio of the soil sample Reason (R): The consistency of the soil will is change from plastic to liquid state leading to its (a) 0.50 (b) 1.00 volumetric decrease. (c) 1.5 (d) 2.00 (a) Both A and R are true and R is the correct explanation of A 2. The given figure indicate the weights of different pycnometers: (b) Both A and R are true but R is not a correct explanation of A (c) A is true but R is false (d) A is false but R is true 6. The standard plasticity chart to classify fine grained soils is shown in the given figure. Empty Pycnometer Pycnometer Pycnometer Pycnometer +Dry Soil +Soil + Water +Water W W W W 1 2 3 4 PI% The specific gravity of the solids is given X A-Line by
W2 (a) 20 35 50 WW4 2 %LL The area marked X represents WW1 2 (b) (a) silt of low plasticity (WW)(WW)3 4 2 1 (b) clay of high plasticity (c) organic soil of medium plasticity W2 (c) (d) clay of intermediate plasticity WW3 4 7. A soil sample is having a specific gravity of 2.60 and a void ratio of 0.78. The water content in WW2 1 (d) percentage required to fully saturate the soil at that (WW)(WW)2 1 3 4 void ratio would be 3. A soil sample has a shrinkage limit of 10% and (a) 10 (b) 30 specific gravity of soil solids as 2.7. The porosity (c) 50 (d) 70 of the soil at shrinkage limit is 8. A dry soil has mass specific gravity of 1.35.If (a) 21.2% (b) 27% the specific gravity of solids is 2.7, then the void (c) 73% (d) 78.8% ratio will be 4. In a wet soil mass, air occupies one-sixth of its (a) 0.5 (b) 1.0 (c) 1.5 (d) 2.0 volume and water occupies one-third of its 9. A clay sample has a void ratio of 0.50 in dry volume. The void ratio of the soil is state and specific gravity of solids = 2. 70. Its (a) 0.25 (b) 0.5 shrinkage limit will be (a) 12% (b) 13.5% (c) 1.00 (d) 1.50 (c) 18.5% (d) 22% # 100-102, Ram Nagar, Bambala Puliya Email : info @ engineersacademy.org Pratap Nagar, Tonk Road, Jaipur-33 Website : www.engineersacademy.org Ph.: 0141-6540911, +91-8094441777 ENGINEERS ACADEMY 2 | Properties of Soils Civil Engineering 10. A soil has liquid limit of 60% plastic limit of 35% 15. The moisture content of a clayey soil is gradually and shrinkage limit of 20% and it has a natural decreased from a large value. What will be the moisture content of 50%. The liquidity index of correct sequence of the occurrence of the soil is following limits? (a) 1.5 (b) 1.25 1. Shrinkage limit (c) 0.6 (d) 0.4 2. Plastic limit 11. Consider the following statements in relation to 3. Liquid limit the given sketch: Select the correct answer using the codes given below: Volume (cc) Weight (g) (a) 1, 2, 3 (b) 1, 3, 2 0.2 Air 0 0.3 Water 0.3 (c) 3, 2, 1 (d) 3, 1, 2 0.5 Solids 0.1 16. The initial and final void ratios of a clay sample in a consolidation test are 1 and 0.5, respectively. 1. Soil is partially saturated at degree of If initial thickness of the sample is 2.4 cm, then saturation = 60% its final thickness will be 2. Void ratio = 40% (a) 1.3 cm (b) 1.8 cm 3. Water content = 30% (c) 1.9 cm (d) 2.2 cm 4. Saturated unit weight = 1.5 g/cc 17. Given that Plasticity index (PI) of local soil = 15 Which of these statements is/are correct? and PI of sand = zero, for a desired PI of 6, the (a) 1, 2 and 3 (b) 1, 3 and 4 percentage of sand in the mix should be (c) 2, 3 and 4 (d) 1, 2 and 4 (a) 70 (b) 60 12. A soil has a liquid limit of 45% and lies above (c) 40 (d) 30 the A-line when plotted on a plasticity chart. 18. A clayey soil has liquid limit = w ; plastic limit The group symbol of the soil as per IS soil L = w and natural moisture content = w. The Classification is p consistency index of the soil is given by (a) CH (b) CI
(c) CL (d) MI wL w wL w p (a) (b) 13. The dry density of a soil is 1.5 g/cc. If the wLP w wL w saturation water content were 50% then its
saturated density and submerged density would, wP w wL w p respectively, be (c) (d) wLP w wP w (a) 1.5 g/cc and 1.0 g/cc 19. Consider the following statements: (b) 2.0 g/cc and 1.0 g/cc 1. ‘Relative compaction’ is not the same as (c) 2.25 g/cc and 1.25 g/cc ‘relative density’. (d) 2.50 g/cc and 1.50 g/cc 2. Vibrofloatation is not effective in the case 14. A fill having a volume of 1,50.000 cum is to be of highly cohesive soils. constructed at a void ratio of 0.8. The borrow 3. ‘Zero air void line’ and 100% saturation line pit soil has a void ratio of 1.4. The volume of are not identical. soil required (in cubic meters) to be excavated from the borrow pit will be Which of these statements is/are correct? (a) 1,87,500 (b) 2,00,000 (a) 1 and 2 (b) 1 and 3 (c) 2,10,000 (d) 2,25,000 (c) 2 and 3 (d) 3 alone # 100-102, Ram Nagar, Bambala Puliya Email : info @ engineersacademy.org Pratap Nagar, Tonk Road, Jaipur-33 Website : www.engineersacademy.org Ph.: 0141-6540911, +91-8094441777 ENGINEERS ACADEMY Soil Mechanics Properties of Soils | 3 20. A soil has mass unit weight , water content Codes: A B C D ‘w’ (as ratio). The specific gravity of soil solids (a) 4 3 5 1 (b) 5 4 3 1 = G, unit weight of water = w ; S the degree of saturation of the soil is given by (c) 4 1 5 2 (d) 5 1 3 2 1 w S 24. If an unconfined compressive strength of 4 kg/ (a) 1 w (1 w) cm2 in the natural state of clay reduces by four G times in the remoulded state, then its sensitivity w will be (b) S 1 (a) 1 (b) 2 w (1 w) G (c) 4 (d) 8 25. The value of porosity of a soil sample in which (1 w) (c) S the total volume of soil grains is equal to twice 1 w (1 w) the total volume of voids would be G (a) 75% (b) 66.66% w (c) 50% (d) 33.33% (d) S 1 26. A soil has a liquid limit of 40% and plasticity w (1 w) wG index of 20%. The plastic limit of the soil will be 21. The saturated and dry densities of a soil are (a) 20% (b) 30% respectively 2000 kg/m3 and 1500 kg/m3. The (c) 40% (d) 60% water content (in percentage) of the soil in the 27. A sample of saturated sand has a dry unit weight saturated state would be of 18 kN/m3 and a specific gravity of 2.7. If (a) 25 (b) 33.33 density of water is 10 kN/m3, the void ratio of (c) 50 (d) 66.66 the soil sample will be 22. If a soil sample of weight 0.18 kg having a (a) 0.5 (b) 0.6 volume of 10–4 m3 and dry unit weight of 1600 (c) 0.4 (d) 0.9 kg/m3 is mixed with 0.02 kg of water then the Common Data for Questions :28 & 29 water content in the sample will be For constructing an embankment, the soil is transported (a) 30% (b) 25% from a borrow area using a truck which can carry (c) 20% (d) 15% 6m3 of soil at a time. The details are as follows. 23. Match List-I (Terms) with List-II (Formulae) and select the correct answer using the codes Borrow Truck Field Property given below the lists: area (loose) (compacted) List-I List-II Bulk density 1.66 1.15 1.82 V (g / cc) A. Void Ratio 1. V V Water content 8 6 14 (%) WW B. Porosity 2. WS 28. The quantity of soil to be excavated from the borrow pit, in m3 for a compacted earth fill of VW 3 C. Degree of saturation 3. 100 m is VV (a) 104 cum (b) 146 cum W (c) 98 cum (d) 87 cum D. Water content 4. 29. The number of truck loads of soil required to V obtain 100m3 of compacted earth fill
VV (a) 12 nos. (b) 56 nos. 5. VS (c) 25 nos. (d) 33 nos # 100-102, Ram Nagar, Bambala Puliya Email : info @ engineersacademy.org Pratap Nagar, Tonk Road, Jaipur-33 Website : www.engineersacademy.org Ph.: 0141-6540911, +91-8094441777 ENGINEERS ACADEMY 4 | Properties of Soils Civil Engineering 30. When the product of rock weathering is not 39. A borrow pit soil has a dry density of 17 kN/m3. transported as sediment but remains in place, is How many cubic meters of this soil will be called required to construct an embankment of 100 m3 (a) alluvial soil (b) glacial soil volume with a dry density of 16 kN/m3. (c) residual soil (d) aeoline soil (a) 94 m3 (b) 106 m3 31. Aeolian soils are (c) 100m3 (d) 90m3 (a) Residual soils (b) Wind deposits 40. The void ratio and specific gravity of a soil are (c) Gravity deposits (d) Water deposits 0.65 and 2.72 respectively. The degree of 32. If the porosity of a soil sample is 20%, the void saturation (in percent) corresponding to water ratio is content of 20% is (a) 0.20 (b) 0.80 (a) 65.3 (b) 20.9 (c) 1.00 (d) 0.25 (c) 83.7 (d) 54.4 33. Consistency Index for a clayey soil is [{LL= 41. A dry soil sample has equal amounts of solids Liquid Limit, PI = Plasticity Index, w = natural and voids by volume. It void ratio and porosity moisture content] will be LL w w PL Void ratio Porosity (%) (a) (b) PI PI (a) 1.0 100% (c) LL – PL (d) 0.5 w (b) 0.5 50% 34. If soil is dried beyond its shrinkage limit, it will (c) 0.5 100% show (d) 1.0 50% (a) Large volume change 42. The plasticity index and the percentage of grain (b) Moderate volume change size finer than 2 microns of a clay sample are (c) Low volume change 25 and 15, respectively. Its activity ratio is (d) No volume change (a) 2.5 (b) 1.67 35. The toughness index of clayey soils is given by (c) 1.0 (d) 0.6 (a) Plasticity index/Flow index 43. A soil sample having a void ratio of 1.3, water (b) liquid limit /Plastic limit content of 50% and a specific gravity of 2.60, (c) Liquidity index /plastic limit is in a state of (d) Plastic limit/Liquidity index (a) partial saturation (b) full saturation 36. A soil sample in its natural state has mass of (c) over saturation (d) under saturation 2.290 kg and a volume of 1.15 × 10–3 m3. After being oven dried, the mass of the sample is 44. The natural void ratio of a sand sample is 0.6 and its density index is 0.6. If its void ratio in the 2.035 kg. Gs for soil is 2.68. The void ratio of the natural soil is loosest state is 0.9, then the void ratio in the (a) 0.40 (b) 0.45 densest state will be (c) 0.55 (d) 0.53 (a) 0.2 (b) 0.3 37. Principle involved in the relationship between (c) 0.4 (d) 0.5 submerged unit weight and saturated weight of 45. Which one of the following correctly represents a soil is based on the dry unit weight of a soil sample which has (a) Equilibrium of floating bodies a bulk unit weight of t at a moisture content of (b) Archimedes’ principle w%? (c) Stokes’ law wt w (d) Darcy’s law (a) (b) t 1 100 100 38. A soil sample has a void ratio of 0.5 and its porosity will be close to 100 t (100 w) (a) 50% (b) 66% (c) t (d) 100 w 100 (c) 100% (d) 33% # 100-102, Ram Nagar, Bambala Puliya Email : info @ engineersacademy.org Pratap Nagar, Tonk Road, Jaipur-33 Website : www.engineersacademy.org Ph.: 0141-6540911, +91-8094441777 ENGINEERS ACADEMY Soil Mechanics Properties of Soils | 5 46. Given that coefficient of curvature = 1.4, 51. Assertion (A): A soil is at its liquid limit if the consistency index of the soil is equal to zero. D30 = 3 mm, D10 0.6 mm. Based on this information of particle size Reason (R): The consistency index of a soil is distribution for use as subgrade, this soil will be defined as ratio of (liquid limit minus the natural taken to be water content) to (natural water content minus plastic limit). (a) uniformly-graded sand Codes given below : (b) well-graded sand (a) Both and R are true and R is the correct (c) very fine sand explanation of A (d) poorly-graded sand (b) Both and R are true but R is not a correct 47. The following data were obtained from a liquid explanation of A limit test conducted on a soil sample. (c) A is true but R is false (d) A is false but R is true Number of 17 22 25 28 34 blows 52. Consistency as applied to cohesive soils is an Water indicator of its 63.8 63.1 61.9 60.6 60.5 Content (a) density (b) moisture content (c) shear strength (d) porosity The liquid limit of the soil is: (a) 63.1% (b) 62.8% 53. The ratio of saturated unit weight to dry unit (c) 61.9% (d) 60.6% weight of a soil is 1.25. If the specific gravity of 48. The void ratios at the densest, loosest and the solids (Gs) is 2.65, the void ratio of the soil is natural states of a sand deposit are 0.2, 0.6, (a) 0.625 (b) 0.663 and 0.4 respectively. The relative density of (c) 0.944 (d) 1.325 the deposit is 54. In the phase diagrams given the change due to (a) 100% (b) 75% initial state changing into final state is shown (c) 50% (d) 25% due to consolidation. Depth of soil layer undergoing consolidation is H; e is initial void 49. While computing the values of limits of 0 ratio; e is final void ratio; e is change in void consistency and consistency indices, it is found r ratio. that liquidity index, has a negative value. 1. Liquidity index cannot have a negative value e e and should be taken as zero. 0 Water e1 Water 2. Liquidity index can have a negative value. 3. The soil tested is in semisolid state and stiff. 4. The soil tested is in medium soft state. Which of these statements are correct? (a) 1 and 4 (b) 1 and 3 (c) 2 and 4 (d) 2 and 3 50. Which one of the following represents relative Initial State Final State density of saturated sand deposit having moisture Indicate which of the following expressions gives content of 25%, if maximum and minimum void settlement of the layer. ratio of sand are 0.95 and 0.45 respectively and e e specific gravity of sand particles is 2.6? (a) H log10 (b) log10 H 1 e0 1 e0 (a) 40% (b) 50% e e (c) 60% (d) 70% (c) (d) H 1 e0 1 e0 # 100-102, Ram Nagar, Bambala Puliya Email : info @ engineersacademy.org Pratap Nagar, Tonk Road, Jaipur-33 Website : www.engineersacademy.org Ph.: 0141-6540911, +91-8094441777 ENGINEERS ACADEMY 6 | Properties of Soils Civil Engineering
55. match List-I (Unit/Test) with List-II (Purpose) w (G e) and select the correct answer using the codes C. Submerged density 3. (1 e) given below the lists w (G 1) List-I D. Saturated density 4. (1 e) A. Casagrande’s apparatus Codes: A B C D B. Hydrometer (a) 2 1 4 3 C. Plate load test (b) 2 3 4 1 D. Oedometer (c) 4 1 2 3 (d) 4 3 2 1 List-II 58. What are the respective values of void ratio, 3 1. Determination of grain size distribution porosity ratio and saturated density (in kN/m ) 2. Consolidation characteristics for a soil sample which has saturation moisture content of 20% and specific gravity of grains as 3. Determination of consistency limits 2.6? (take density of water as 10 kN/m3) 4. Determination of safe bearing capacity of (a) 0.52, 1.08, 18.07 (b) 0.52, 0.34, 18.07 soil (c) 0.77, 1.08, 16.64 (d) 0.52, 0.3, 20.14 Codes: A B C D 59. Embankment fill is to be compacted at a density (a) 1 3 2 4 of 18 kN/m3. The soil of the borrow area is at (b) 1 3 4 2 a density of 15 kN/m3. What is the estimated 3 (c) 3 1 2 4 number of trips of 6 m capacity truck for hauling 3 (d) 3 1 4 2 the soil required for compacting 100 m fill of the embankment? (Assume that the soil in the 56. Two soil sample A and B have porosities borrow area and that in the embankment are at n = 40% and n = 60% respectively. What is A B the same moisture content) the ratio of void ratios e : e ? A B (a) 14 (b) 18 (a) 2 : 3 (b) 3 : 2 (c) 20 (d) 23 (c) 4 : 9 (d) 9 : 4 60. Well-graded dense saturated sands have high 57. Match List-I (Densities) with List-II shear strength because (Expressions) and select the correct answer using (a) such sands have a better grade (superior the codes given below the lists: type of sand grains resulting in higher
(Symbols G, e, w and S stand for specific gravity strength of soil grains, void ratio, unit weight of water (b) such sands have lower water content, which and degree of saturation respectively) increases shear strength List-I List-II (c) such sands have better interlocking of grains, higher inter-particle contacts and higher inter- w G Se particle frictional resistance resulting in higher A. Dry density 1. (1 e) strength (d) presence of water in such sands induces
wG capillary pressure generating higher inter B. Moist density 2. (1 e) granular stresses, which generate apparent cohesion and hence higher shear strength.
# 100-102, Ram Nagar, Bambala Puliya Email : info @ engineersacademy.org Pratap Nagar, Tonk Road, Jaipur-33 Website : www.engineersacademy.org Ph.: 0141-6540911, +91-8094441777 ENGINEERS ACADEMY Soil Mechanics Properties of Soils | 7 AANSWERSNSWERS ANDAND EEXPLANATIONSXPLANATIONS 1. Ans. (c) Soils above A-line are clays and soils below A- line are silts and organic soils. Plasticity Index Equation of A-line is, PI = 0.73 (w – 20) Activity = % of clay fraction L 7. Ans. (b) 65 29 36 Given, S = 100% 1.5 1.25 24 24 Se 100 0.78 w 30% 2. Ans. (d) G 2.60 Specific Gravity of solids is given by 8. Ans. (b) WW2 1 Gs (WW)(WW)2 1 3 4 Gs Gm G s (1 n) 3. Ans. (a) 1 e G is mass specific gravity e m Shrinkage limit, w 100 s G Gm is specific gravity of solids G 2.7 e 0.1 2.7 0.27 e s 1 e 1 1 Porosity, Gm 1.35
e 0.27 9. Ans. (c) n 100 100 21.2% 1 e 1 0.27 At shrinkage limit, soil is fully saturated.
4. Ans. (c) e 0.5 w 100 100 18.5% V s Void ratio, e= v G 2.5 V s 10. Ans. (c) VV = air void + water filled voids Liquidity Index
1VV 1 V VVV V w w 50 35 15 6 3 2 ; s v 2 P 0.6 wLP w 60 35 25
e V / 2 1.0 Consistency Index = 1 – 0.6 = 0.4 V / 2 11. Ans. (b) 5 Ans. (c) V The volume of soil increases from plastic limit w 100 60% Degree of saturation S = V to liquid limit. The cracking in soil is due to v reduction in bearing capacity and consequent partially saturated. failure and heaving. 0.3 100 60% 6. Ans. (d) 0.2 0.3
% LL (Liquid Limit) Plasticity Vv 0.2 0.3 < 35 Low end Void ratio e 1 Vs 0.5 35 – 50 Intermediate > 50 High 1 100 100%
# 100-102, Ram Nagar, Bambala Puliya Email : info @ engineersacademy.org Pratap Nagar, Tonk Road, Jaipur-33 Website : www.engineersacademy.org Ph.: 0141-6540911, +91-8094441777 ENGINEERS ACADEMY 8 | Properties of Soils Civil Engineering Water content 18. Ans. (a) 19. Ans. (a) W ww 0.3 30% ‘Zero air void line1 and 100% saturation line are W 1.0 s identical. Saturated unit weight 20. Ans. (b) 21. Ans. (b) W 1.0 0.3 0.2 t 1.5 g / cc V 0.2 0.3 0.5 t 1500 2000 In case of fully saturated condition air voids will d 1 w 1 w be filled by water. w = 0.3333 = 33.33% 12. Ans. (b) 22. Ans. (b) Liquid limit 45% lies between 35% to 50% for Dry weight of sample = 1600 × 10 r–4 = 0.1 6 intermediate plasticity. The soil above A line kg Weight of water in soil before mixing should be given symbol Cl. additional quantity = 0.18 – 0.16 = 0.02 kg After 13. Ans. (c) mixing water the total quantity of water = 0.02 sat d (1 w) 1.5 1.5 2.25g/cc + 0.02 = 0.04 kg water
sub sat w 2.25 1.0 1.25g/cc 0.04 Thus, water content = 100 25% 14. Ans. (b) 0.16 Volume of solids will remain same in fill and 23. Ans. (d) barrow pit Degree of saturation, V VV Vs ; VWw w 1 e 1 efill 1 e borrow pit S 100 100 VVv v 150000 V ; V 2,00,000cum Density of water is 1 gm/cc. 1 0.8 1 1.4 24. Ans. (c) 15. Ans. (c) (UCS) 4 Liquid limit > Plastic limit > Shrinkage limit Sensitivity natural 4 (UCS) 16. Ans. (b) remoulded (4 / 4) 25. Ans. (d) H e; H 0.5 1 Given, Vs = 2Vv H 1 e0 2.4 1 1 V ev 0.5 H 0.6 cm; Hf H H Void ratio, Vs = 2.4 – 0.6 = 1.8 cm e 0.5 1 17. Ans. (b) Porosity, n or33.33% Assuming in mix, x part is sand and (1 – x) part 1 e 1.5 3 is soil 26. Ans. (a) Plastic limit, w = W – PI = 40 – 20 – 20% x(PI) (1 x)PI p L PI of mix= sand soil 1 27. Ans. (a) G 6 x 0 (1 x) 15 d1 e w x 1 6 9 0.6 60% 15 15
# 100-102, Ram Nagar, Bambala Puliya Email : info @ engineersacademy.org Pratap Nagar, Tonk Road, Jaipur-33 Website : www.engineersacademy.org Ph.: 0141-6540911, +91-8094441777 ENGINEERS ACADEMY Soil Mechanics Properties of Soils | 9
G 2.7 10 147 e w 1 0.5 = = 24.5 nos ; d 18 6 28. Ans. (a) Say 25 nos. 29. Ans. (c) 30. Ans. (c) 31. Ans. (b) Truck Embankment 32. Ans. (d)
Borrow n 0.20 pit e = 0.25 1 n 1 0.2 33. Ans. (a) 34. Ans. (d) 35. Ans. (a)
1 1.66 gm cc 2 1.15 gm cc 3 1.82gm cc 36. Ans. (d) Given G = 2.68, w1 8% w2 6% w3 14% 3 Take w = 1000 Kg/m d1 d2 d3
V1 V2 V3 W 2.290 = = 1991 Kg/m3 V 1.15 103 Water content, 1 d = 1 1 w1 WW d 2.290 2.035 w = 100 = 100 W 2.035 1.66 d = 1.537gm cc d1 1 0.08 = 12.53%
.15 1991 3 = 1.085gm cc d = = = 1769.3 Kg/m d2 1 0.06 1 W 1 0.1253 G 1000 2.68 1.82 w d = ; 1769.3 = d = 1.596gm cc 1 e 1 e 3 1 0.14 e = 0.52 To find volume of borrow pit (V ) the following 1 37. Ans. (b) equation may be used. 38. Ans. (d) V1 d3 V 1.596 = ; 1 = e 0.5 V d 100 1.537 n = 0.333 = 33.3% 3 1 1 e 1 0.5 3 3 V1 = 103.84 m say 104 m 39. Ans. (a) To find number of trunk load : Borrow pit: V2 d3 V 1.596 3 2 d = 17kN/m ; Volume = V 1 V = ; = 1 3 d2 100 1.085 3 Embankment: V2 = 147 m 3 3 =16kN/m ; V2 = 100 m No. of truck loads d2 Using the relationship, # 100-102, Ram Nagar, Bambala Puliya Email : info @ engineersacademy.org Pratap Nagar, Tonk Road, Jaipur-33 Website : www.engineersacademy.org Ph.: 0141-6540911, +91-8094441777 ENGINEERS ACADEMY 10 | Properties of Soils Civil Engineering V1 d2 V 16 0.9 0.6 1 0.9 – e = V = ; = min 2 d1 100 17 0.6 3 V1 = 94.11 m emin = 0.9 – 0.5 = 0.4 40. Ans. (c) 45. Ans. (c)
wG 0.20 2.72 100 t t e = ; 0.65 = d = Sr Sr w (100 w) 1 100 Sr = 0.837 = 83.7% 41. Ans. (d) 46. Ans. (b) Coefficient of curvature Vv Void ratio, e = 1 Vs 2 2 D30 D30 Cc = D60 = Porosity, DD60 10 CDc 10
e 1 Coefficient of uniformity, n = 100 = 100 50% 1 e 1 1 2 2 D30 D 3 1 60 42. Ans. (b) Cu = 2 = 17.9 CDc 10 D10 0.6 1.4 Activity As 1 < Cc < 3 and Cu > 6 so it is well graded Plasticity Index sand. = Percent of clay particles finer than 2 m 47. Ans. (c) Liquid limit is the water content corresponding 25 = 1.67 to number of blows of 25. 15 48. Ans. (c) As activity is more than 1.25 so it is active soil. Relative density or Density index, 43. Ans. (b) e e Gw = Se max 100 50% ID = emax e min 260 50 S = 100% 49. Ans. (d) 1.3
Therefore soil is fully saturated. wL w Consistency index = wL w p S 100% always w w P Remember w can be more than 100% IL wLP w e can be more than 1.0. w can be grater than wp 50. Ans. (c) 44. Ans. (c) At saturated moisture content void ratio is emax e Density Index = 0.6 Gw 26 25 emax e min e = 0.65 S 100 Void ratio in loosest state, e = 0.9 max Relative density, Void ratio in natural state, e = 0.6
# 100-102, Ram Nagar, Bambala Puliya Email : info @ engineersacademy.org Pratap Nagar, Tonk Road, Jaipur-33 Website : www.engineersacademy.org Ph.: 0141-6540911, +91-8094441777 ENGINEERS ACADEMY Soil Mechanics Properties of Soils | 11 57. Ans. (a) e e 0.95 0.65 max 0.6 Dr = Moist density, emax e min 0.95 0.45 G Se Dr = 60% = w 51. Ans. (c) 1 e
For w = wL For dry density put S = 0 For saturated density put S = 1 wL w Consistency index = 0 Submerged density wLP w = saturated density – density of water 52. Ans. (c) 58. Ans. (d) Consistency of soil refers to the resistance Given w = 20%, G = 2.6, S = 100% offered by it against forces that tend to deform or rupture the soil aggregate. It is related to Gw 2.6 20 e e= 0.52 ; n = 0.34 strength. S 100 1 e 53. Ans. (b) G e 2.6 0.52 = w = 10 sat sat 1 e 1 0.52 Given = 1.25, Gs = 2.65 d = 20.53 kN/m3 3 G e G If density of water is taken as 9.81 kN/m = w s ; = w s sat 1 e d 1 e 3 sat = 20.14 kN/m 59. Ans. (c) sat Gs e 2.65 e = ; 1.25 = In problems of fill and borrow pit, the volume of d Gs 2.65 soil solids shall be equated. e = 0.663 VV 54. Ans. (d) Vs = 1 e G(1 w) w e e H e H 0 f As the moisture content of borrow area and that Settlement, H = 1 e 1 e 0 0 of embankment are same. 55. Ans. (d) V1 1 = V2 2 56. Ans. (c) For 100 m fill at embankment, the volume n 0.4 2 required from borrow pit A eA = 1 nA 0.6 3 100 18 V = 120m3 2 15 n 0.6 3 B eB = 1 n 0.4 2 120 B Number of trips of truck = 20 6 e A 4 60. Ans. (c) e = B 9
# 100-102, Ram Nagar, Bambala Puliya Email : info @ engineersacademy.org Pratap Nagar, Tonk Road, Jaipur-33 Website : www.engineersacademy.org Ph.: 0141-6540911, +91-8094441777