Question 1 |

An uncompacted heap of soil has a volume of 10000 m^3 and void ratio of 1. If the soil is compacted to a volume of 7500 m^3, then the corresponding void ratio of the compacted soil is __________. (round off to one decimal place)

0.2 | |

0.8 | |

0.5 | |

0.1 |

Question 1 Explanation:

\begin{aligned}
(V_S)_{\text{heap}}&=(V_S)_{\text{compacted soil}}\\
\left ( \frac{V_T}{1+e} \right )_{\text{heap}}&=\left ( \frac{V_T}{1+e} \right )_{\text{compacted soil}}\\
\frac{10000}{1+1}&=\frac{7500}{1+e}\\
e&=0.5
\end{aligned}

Question 2 |

OMC-SP and MDD-SP denote the optimum moisture content and maximum dry density obtained from standard Proctor compaction test, respectively. OMC-MP and MDD-MP denote the optimum moisture content and maximum dry density obtained from the modified Proctor compaction test, respectively. Which one of the following is correct?

OMC-SP \lt OMC-MP and MDD-SP \lt MDD-MP | |

OMC-SP \gt OMC-MP and MDD-SP \lt MDD-MP | |

OMC-SP \lt OMC-MP and MDD-SP \gt MDD-MP | |

OMC-SP \gt OMC-MP and MDD-SP \gt MDD-MP |

Question 2 Explanation:

\begin{array}{r} \mathrm{OMC}-\mathrm{SP} \gt \mathrm{OMC}-\mathrm{MP} \\ \mathrm{MDD}-\mathrm{SP} \lt \mathrm{MDD}-\mathrm{MP} \end{array}

Question 3 |

Following statements are made on compacted soils, wherein DS stands for the soils compacted on
dry side of optimum moisture content and WS stands for the soils compacted on wet side of
optimum moisture content. Identify the incorrect statement.

Soil structure is flocculated on DS and dispersed on WS. | |

Construction pore water pressure is low on DS and high on WS. | |

On drying, shrinkage is high on DS and low on WS. | |

On access to water, swelling is high on DS and low on WS. |

Question 4 |

Two series of compaction tests were performed in the laboratory on an inorganic clayey soil
employing two different levels of compaction energy per unit volume of soil. With regard to the
above tests, the following two statements are made.

I The optimum moisture content is expected to be more for the tests with higher energy.

II The maximum dry density is expected to be more for the tests with higher energy.

The CORRECT option evaluating the above statements is

I The optimum moisture content is expected to be more for the tests with higher energy.

II The maximum dry density is expected to be more for the tests with higher energy.

The CORRECT option evaluating the above statements is

Only I is TRUE | |

Only II is TRUE | |

Both I and II are TRUE | |

Neither I nor II is TRUE |

Question 4 Explanation:

Optimum moisture content of the soil decreases with increase compactive efforts.

Question 5 |

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 \gamma _{w} represents the unit weight of water and \gamma _{d} represents the dry unit weight of the soil, the equation of zero air voids line is

\gamma _{d}=\frac{G\gamma _{w}}{1+Se} | |

\gamma _{d}=\frac{G\gamma _{w}}{1+Gw} | |

\gamma _{d}=\frac{Gw}{e+\gamma _{w}S} | |

\gamma _{d}=\frac{Gw}{1+Se} |

Question 5 Explanation:

The bulk unit weight of soil is given by

\gamma=\frac{(G+S e) \gamma_{w}}{1+e}

For dry unit weight, degree of saturation,

S=0 \quad \therefore \gamma_{d}=\frac{G \gamma_{w}}{1+e}

But we know that

G w=\text { Se } \quad \therefore \gamma_{d}=\frac{G \gamma_{w}}{1+\frac{G w}{S}}

For zero air voids, degree of saturation,

S=100 \% \quad \therefore \gamma_{d}=\frac{G \gamma_{w}}{1+G w}

\gamma=\frac{(G+S e) \gamma_{w}}{1+e}

For dry unit weight, degree of saturation,

S=0 \quad \therefore \gamma_{d}=\frac{G \gamma_{w}}{1+e}

But we know that

G w=\text { Se } \quad \therefore \gamma_{d}=\frac{G \gamma_{w}}{1+\frac{G w}{S}}

For zero air voids, degree of saturation,

S=100 \% \quad \therefore \gamma_{d}=\frac{G \gamma_{w}}{1+G w}

Question 6 |

Deposit with flocculated structure is formed when

clay particles settle on sea bed | |

clay particles settle on fresh water lake bed | |

sand particles settle on river bed | |

sand particles on sea bed |

Question 6 Explanation:

Flocculated structure is formed when clay particles settle on sea bed, since sea bed is polar in nature.

Question 7 |

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

moist silty sand | |

well graded dry sand | |

clay of medium compressibility | |

silt of high compressibility |

Question 8 |

In a standard proctor test, 1.8 kg of moist soil was filling the mould (volume
= 94.4cc) after compaction. A soil sample weighing 23 g was taken from the
mould and ovendried for 24 hours at a temperature of 110^{\circ}C. Weight of the
dry sample was found to be 20 g. Specific gravity of soil solids is G = 2.7. The
theoretical maximum value of the dry unit weight of the soil at that water content
is equal to

4.67kN/m^{3} | |

11.5kN/m^{3} | |

16.26kN/m^{3} | |

18.85kN/m^{3} |

Question 8 Explanation:

Bulk density of the soil

=\frac{1.8 \times 1000}{944}=1.907 \mathrm{gm} / \mathrm{cc}

Water content of soil

=\left(\frac{23-20}{20}\right) \times 100=15 \%

For, theoretical maximum dry unit weight, degree of saturation should be 100 %.

\begin{aligned} S e &=G w \\ \Rightarrow \quad e &=\frac{2.7 \times 0.15}{1} \Rightarrow e=0.405 \end{aligned}

The theoretical maximum dry unit weight may be given as

\gamma_{d, \max }=\frac{G \gamma_{w}}{1+e}=\frac{2.7 \times 9.81}{1+0.405}=18.85 \mathrm{kN} / \mathrm{m}^{3}

=\frac{1.8 \times 1000}{944}=1.907 \mathrm{gm} / \mathrm{cc}

Water content of soil

=\left(\frac{23-20}{20}\right) \times 100=15 \%

For, theoretical maximum dry unit weight, degree of saturation should be 100 %.

\begin{aligned} S e &=G w \\ \Rightarrow \quad e &=\frac{2.7 \times 0.15}{1} \Rightarrow e=0.405 \end{aligned}

The theoretical maximum dry unit weight may be given as

\gamma_{d, \max }=\frac{G \gamma_{w}}{1+e}=\frac{2.7 \times 9.81}{1+0.405}=18.85 \mathrm{kN} / \mathrm{m}^{3}

Question 9 |

A clayey soil has a maximum dry density of 16 kN/m^{3} and optimum moisture content of 12%. A contractor during the construction of core of an earth dam
obtained the dry density 15.2 kN/m^{3} and water content 11%. This construction is acceptable because

The density is less than the maximum dry density and water content is on
dry side of optimum | |

The compaction density is very low and water content is less than 12% | |

The compaction is done on the dry side of the optimum | |

Both the dry density and water content of the compacted soil are within the
desirable limits |

Question 10 |

Compaction of an embankment is carried out in 500 mm thick layers. The rammer used for compaction has a foot area of 0.05 m^{2} and the energy imparted in every drop of rammer is 400 Nm. Assuming 50% more energy in each pass over the compacted area due to overlap, the number of passes required to develop compactive energy equivalent of Indian Standard light compaction for each layer would be

10 | |

16 | |

20 | |

26 |

Question 10 Explanation:

As per Indian Standard light compaction test, a
hammer of 2.6 kg is allowed to fall from a height of
310 mm and 3 layers are tamped 25 times in a
mould of volume 1000 cc .

\therefore Energy imparted

\begin{array}{l} =2.6 \times 25 \times 3 \times 9.81 \\ \times \frac{310}{1000}=593.01 \mathrm{N}-\mathrm{m} \end{array}

For embankment compaction, the volume of soil covered by rammer

\begin{aligned} &=0.05 \times 10^{4} \times \frac{500}{10} \\ &=25000 \mathrm{cc} \end{aligned}

\therefore Energy imparted in every drop of hammer

\begin{aligned} &=\left(1+\frac{50}{100}\right) \times 400\\ &=600 \mathrm{N}-\mathrm{m} \end{aligned}

If 'n' number of passes are required to develop equivalent energy to Indian Standard light compaction test, then

n \times \frac{600}{25000}=\frac{593.01}{1000}

\Rightarrow \quad n=24.71 \simeq 25\text{( approx.)}

However, If the data of Standard Proctor Test are taken i.e. weight of hammer =2.495 kg; Volume of mould =944cc, then the value of 'n' will be close to 26.

\therefore Energy imparted

\begin{array}{l} =2.6 \times 25 \times 3 \times 9.81 \\ \times \frac{310}{1000}=593.01 \mathrm{N}-\mathrm{m} \end{array}

For embankment compaction, the volume of soil covered by rammer

\begin{aligned} &=0.05 \times 10^{4} \times \frac{500}{10} \\ &=25000 \mathrm{cc} \end{aligned}

\therefore Energy imparted in every drop of hammer

\begin{aligned} &=\left(1+\frac{50}{100}\right) \times 400\\ &=600 \mathrm{N}-\mathrm{m} \end{aligned}

If 'n' number of passes are required to develop equivalent energy to Indian Standard light compaction test, then

n \times \frac{600}{25000}=\frac{593.01}{1000}

\Rightarrow \quad n=24.71 \simeq 25\text{( approx.)}

However, If the data of Standard Proctor Test are taken i.e. weight of hammer =2.495 kg; Volume of mould =944cc, then the value of 'n' will be close to 26.

There are 10 questions to complete.