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Test - 1

Description: Test - 1
Number of Questions: 20
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Tags: Test - 1 Soil Mechanics

The results for sieve analysis carried out for three types of sand, P, Q and R, are given in the adjoining figure. If the fineness modulus values of the three sands are given as FMP, FMQ and FMR, it can be stated that

  1. $FM_Q = \sqrt{FM_P \times FM_R}$

  2. FMQ = 0.5 (FMP + FMR)

  3. FMP > FMQ > FMR

  4. FMP < FMQ < FMR

Answer: 1
Explanation:

 

Likelihood of general shear failure for an isolated footing in sand decreases with

  1. decreasing footing depth

  2. decreasing intergranular packing of the sand

  3. increasing footing width

  4. decreasing soil grain compressibility

Answer: 2
Explanation:

 Likehood of general shear failure for an isolated footing in sand decreases with decreasing intergranular packing of sand.

For a saturated sand deposit, the void ratio and the specific gravity of solids are 0.70 and 2.67, respectively. The critical (upward) hydraulic gradient for the deposit would be

  1. 0.54

  2. 0.98

  3. 1.02

  4. 1.87

Answer: 2
Explanation:

 $e = 0.70, g = 2.67\\
\text{Critical hydraulic gradient causes effective stress to zero and}\\
i_c = \frac{G - 1}{1 + e} = \frac{2.67 - 1}{1 + 0.7} = 0.98$

Two geometrically identical isolated footings, X (linear elastic) and Y (rigid), are loaded identically (shown alongside). The soil reactions will

  1. be uniformly distributed for Y but not for X

  2. be uniformly distributed for X but not for Y

  3. not be uniformly distributed for both X and Y

  4. both 2 and 3

Answer: 1
Explanation:

 

Quicksand condition occurs when

  1. the void ratio of the soil becomes 1.0

  2. the upward seepage pressure in soil becomes zero

  3. the upward seepage pressure in soil becomes equal to the saturated unit weight of the soil

  4. the upward seepage pressure in soil becomes equal to the submerged unit weight of the soil

Answer: 4
Explanation:

 Quick sand condition occurs when the upward seepage pressure in soil becomes equal to submerged unit weight of the soil. So, effective stress is equal to zero.

A soil is composed of solid spherical grains of identical specific gravity and diameter between 0.075 mm and 0.0075 mm. If the terminal velocity of the largest particle falling through water without flocculation is 0.5 mm/s, that for the smallest particle would be

  1. 0.005 mm/s

  2. 0.05 mm/s

  3. 5 mm/s

  4. 50 mm/s

Answer: 1
Explanation:

 

The e-log p curve shown in the figure is representative of

  1. normally consolidated clay

  2. over consolidated clay

  3. under consolidated clay

  4. normally consolidated clayey sand

Answer: 2
Explanation:

 e - log P curve or e Vs P curve on semi log graph paper is for over consolidated clay as p increases and 'e' decreases.

Deposit with flocculated structure is formed when

  1. clay particles settle on seabed

  2. clay particles settle on fresh water lake bed

  3. sand particles settle on river bed

  4. sand particles settle on seabed

Answer: 2
Explanation:

 Flocculated structure is formed when clay particles settle on fresh water lake bed.

Dilatancy correction is required when a strata is

  1. cohesive and saturated and also has N value of SPT > 15

  2. saturated silt/fine sand and N value of SPT < 10 after the overburden correction

  3. saturated silt/fine sand and N value of SPT >15 after the overburden correction

  4. coarse sand under dry condition and N value of SPT < 10 after the overburden correction

Answer: 3
Explanation:

 Dilatancy correction is also known as correction due to submergence of the soil and applied for the silt and fine sand is given as
$$N_c = 15 + \frac{1}{5}(N_0 - 15)$$

Where, $N_0 = $ value of SPT after correction applied due to overburden pressure.

A precast concrete pile is driven with a 50 kN hammer falling through a height of 1.0 m with an efficiency of 0.6. The set value observed is 4 mm per blow and the combined temporary compression of the pile, cushion and the ground is 6 mm. As per Modified Hiley Formula, the ultimate resistance of the pile is

  1. 3000 kN

  2. 4285.7 kN

  3. 8.333 kN

  4. 11905 kN

Answer: 2
Explanation:

 

In a compaction test, G, w, S and e represent the specific gravity, water content, degree of saturation and void ratio of the soil sample respectively. If yw represents the unit weight of water and yd represents the dry unit weight of the soil, the equation for zero air voids line is

  1. $y_d \frac{Gy_w}{1 + Se}$

  2. $y_d \frac{Gy_w}{1 + G_W}$

  3. $y_d y\frac{Gy_w}{e + y_wS}$

  4. $y_d \frac{G_w}{1 + Se}$

Answer: 1
Explanation:

 

Group symbols assigned to silty sand and clayey sand are respectively

  1. SS and CS

  2. SM and CS

  3. SM and SC

  4. MS and CS

Answer: 3
Explanation:

 $\text{SM - Silty Sand} \\
\text{SC - Clayey Sand}$

When a retaining wall moves away from the back-fill, the pressure exerted on the wall is termed as

  1. passive earth pressure

  2. swelling pressure

  3. pore pressure

  4. active earth pressure

Answer: 4
Explanation:

 Active earth pressure - When there is a horizontal strain and retaining wall moves away from bachfill. Passive earth pressure - When retaining wall moves toward the backfill. Pore pressure - Neutral pressure and constant in all directions.

A fine grained soil has liquid limit of 60 and plastic limit of 20. As per the plasticity chart, according to IS classification, the soil is represented by the letter symbols

  1. CL

  2. CI

  3. CH

  4. CL-ML

Answer: 3
Explanation:

 

A clay soil sample is tested in triaxial apparatus in consolidated-drained conditions at a cell pressure of 100 kN/m2. What will be the pore water pressure at a deviator stress of 40 kN/m2?

  1. 0 kN/m2

  2. 20 kN/m2

  3. 40 kN/m2

  4. 60 kN/m2

Answer: 2
Explanation:

 

The relationship among specific yield (Sy), specific retention (Sr) and porosity ($\eta$) of an aquifer is

  1. Sy = Sr + $\eta$

  2. Sy = Sr – $\eta$

  3. Sy = $\eta$– Sr

  4. Sy = Sr + 2$\eta$

Answer: 3
Explanation:

 $\eta = \text{porosity of an aquifer}\\
\hspace{0.2cm} = \text{specific yield + specific retention}\\
\hspace{0.2cm} = S_y + S_r\\
S_y = \eta - S_r$

Compaction by vibratory roller is the best method of compaction in case of

  1. moist silty sand

  2. well graded dry sand

  3. clay of medium compressibility

  4. silt of high compressibility

Answer: 2
Explanation:

 Compaction by vibratory roller - well graded dry sand.

The number of blows observed in a Standard Penetration Test (SPT) for different penetration depths are given as follows:

 
Penetration of sampler Number of blows
0-150 mm 6
150-300 mm 8
300-450 mm 10

The observed N value is

  1. 8

  2. 14

  3. 18

  4. 24

Answer: 3
Explanation:

 In standard penetration test (SPT), the number of blows required for first 15 cm penetration is no counted and treated as sheeting apparatus inside the soil and total number of blows for next two 15 cm blows is taken into account for the calculation of standard penetration test. So, N = 8 + 10 = 18

If $\sigma_h, \sigma_w, \sigma_n$ and $\sigma_w$ represent the total horizontal stress, total vertical stress, effective horizontal stress and effective vertical stress on a soil element respectively, the coefficient of earth pressure at rest is given by

  1. $\frac{\sigma_h}{\sigma_w}$

  2. $\frac{\sigma_h}{\sigma_w}$

  3. $\frac{\sigma_w}{\sigma_h}$

  4. $\frac{\sigma_w}{\sigma_h}$

Answer: 1
Explanation:

 

For a sample of dry, cohesionless soil with friction angle $\phi$, the failure plane will be inclined to the major principal plane by an angle equal to

  1. $\phi$

  2. 45°

  3. 45° - $\phi$/2

  4. 45° + $\phi$/2

Answer: 4
Explanation:

 

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