Question 1 |

In an aggregate mix, the proportions of coarse aggregate, fine aggregate and mineral filler are 55%, 40% and 5%, respectively. The values of bulk specific gravity of the coarse aggregate, fine aggregate and mineral filler are 2.55, 2.65 and 2.70, respectively. The bulk specific gravity of the aggregate mix (round off to two decimal places) is ____________

2.6 | |

3.21 | |

5.64 | |

6.25 |

Question 1 Explanation:

G_{m}=\frac{55+40+5}{\frac{55}{2.55}+\frac{40}{2.65}+\frac{5}{2.70}}=2.596

Question 2 |

A partially-saturated soil sample has natural moisture content of 25% and bulk unit weight of 18.5 \mathrm{kN} / \mathrm{m}^{3}. The specific gravity of soil solids is 2.65 and unit weight of water is 9.81 \mathrm{kN} / \mathrm{m}^{3}. The unit weight of the soil sample on full saturation is

21.12 \mathrm{kN} / \mathrm{m}^{3} | |

19.03 \mathrm{kN} / \mathrm{m}^{3} | |

20.12 \mathrm{kN} / \mathrm{m}^{3} | |

18.50 \mathrm{kN} / \mathrm{m}^{3} |

Question 2 Explanation:

\begin{aligned} \mathrm{w} &=0.25, \gamma_{\mathrm{t}}=18.5 \mathrm{kN} / \mathrm{m}^{3} \\ \mathrm{G}_{\mathrm{s}} &=2.65, \gamma_{\mathrm{w}}=9.81 \\ \gamma_{\mathrm{t}} &=\frac{G_{\mathrm{S}} \gamma_{W}(1+w)}{1+e} \\ \Rightarrow \qquad \qquad \qquad \qquad \mathrm{e} &=\frac{2.65 \times 9.81 \times 1.25}{18.5}-1\\ \Rightarrow \qquad \qquad \qquad \qquad e&=0.756\\ \text{At full saturation}, \quad \mathrm{S}&=1\\ \Rightarrow \qquad \qquad \qquad \quad \gamma_{\mathrm{sat}}&=\frac{\left(G_{\mathrm{S}}+e\right) \gamma_{\mathrm{W}}}{1+e}\\ \gamma_{\text {sat }} &=\frac{(2.65+0.756) \times 9.81}{1.756} \\ &=19.03 \mathrm{kN} / \mathrm{m}^{3} \end{aligned}

Question 3 |

A soil has dry weight of 15.5 kN/m^3, specific gravity of 2.65 and degree of saturation
of 72%. Considering the unit weight of water as 10 kN/m^3, the water content of the soil
(in %, round off to two decimal places) is ________.

19.28 | |

0.19 | |

0.28 | |

28.12 |

Question 3 Explanation:

\begin{aligned} \gamma _d&=15.5 kN/m^3\\ G&=2.65\\ S&=72\% \\ \gamma _d&=\frac{G\gamma _w}{1+e}\\ &=\frac{2.65 \times 10}{1+e}=15.5\\ e&=0.7096\\ w&=\frac{Se}{G}=\frac{0.72 \times 0.7096}{2.65}=0.1928\\ w&=19.28\% \end{aligned}

Question 4 |

A sample of 500 g dry sand, when poured into a 2 litre capacity cylinder which is partially
filled with water, displaces 188 cm^3 of water. The density of water is 1 g/cm^3. The specific
gravity of the sand is

2.72 | |

2.66 | |

2.55 | |

2.52 |

Question 4 Explanation:

\begin{aligned} W_s&=500gm \\ V_s&=188cc \\ \gamma _s&=\frac{W_s}{V_s} \\ G_s&=\frac{\gamma _s}{\gamma _w}\\&=\frac{500/188}{1}=2.66 \end{aligned}

Question 5 |

Soil deposit formed due to transportation by wind is termed as

aeolian deposit | |

alluvial deposit | |

estuarine deposit | |

lacustrine deposit |

Question 5 Explanation:

Soil deposited by wind is Aeolian soil.

Question 6 |

The dimensions of a soil sampler are given in the table.

For this sampler, the outside clearance ratio (in percent, round off to 2 decimal places) is ______

For this sampler, the outside clearance ratio (in percent, round off to 2 decimal places) is ______

8.96 | |

11.11 | |

16.2 | |

18.6 |

Question 6 Explanation:

Outside clearance =\left ( \frac{D_2 -D_4}{D_4} \right ) \times 100

Outside clearance =\left ( \frac{100 -90}{90} \right ) \times 100 =11.11\%

Question 7 |

The notation "SC" as per Indian Standard Soil Classification System refers to

Sandy clay | |

Silty clay | |

Clayey silt | |

Clayey sand |

Question 7 Explanation:

SC\rightarrowClayey sand

Question 8 |

If the fineness modulus of a sample of fine aggregates is 4.3, the mean size of the particles in the sample is between

150\mu m \; and \; 300\mu m | |

300\mu m \; and \; 600\mu m | |

1.18\mu m \; and \; 2.36\mu m | |

2.36\mu m \; and \; 4.75\mu m |

Question 8 Explanation:

F.M of mean size 2.36 = 5

F.M of mean size 1.18 = 4

Hence for F.M of 4.3, the mean size of aggregate will be between 1.18 mm and 2.36 mm.

F.M of mean size 1.18 = 4

Hence for F.M of 4.3, the mean size of aggregate will be between 1.18 mm and 2.36 mm.

Question 9 |

A box measuring 50 cmx50cmx50cm is filled to the top with dry coarse aggregate of mass 187.5 kg. The water absorption and specific gravity of the aggregate are 0.5% and 2.5, respectively. The maximum quantity of water (in kg, round off to 2 decimal places) required to fill the box completely is_______

40.56 | |

50.94 | |

80.50 | |

90.50 |

Question 9 Explanation:

Total volume of Box =0.5 \times 0.5\times 0.5=0.125m^3

wt of aggregates = 187.5 kg

sp. gravity = 2.5

Unit weight =2.5 \times 9.81=24.525kN/m^3

wt of water absorbed =\frac{0.5}{100}\times 187.5=0.9375kg

vol of aggregate =\frac{187.5 \times 9.81}{1000 \times 24.525}=0.075m^3

volume of voids =0.125-0.075=0.05m^3

wt of water in voids =0.05 \times 1000=50kg

wt of water required to fill the box = 50 + 0.9375 = 50.94 kg

wt of aggregates = 187.5 kg

sp. gravity = 2.5

Unit weight =2.5 \times 9.81=24.525kN/m^3

wt of water absorbed =\frac{0.5}{100}\times 187.5=0.9375kg

vol of aggregate =\frac{187.5 \times 9.81}{1000 \times 24.525}=0.075m^3

volume of voids =0.125-0.075=0.05m^3

wt of water in voids =0.05 \times 1000=50kg

wt of water required to fill the box = 50 + 0.9375 = 50.94 kg

Question 10 |

A soil has specific gravity of its solids equal to 2.65. The mass density of water is 1000 kg/m^3. Considering zero air voids and 10% moisture content of the soil sample, the dry density (in kg/m^3, round off to 1 decimal place) would be _______

2094.9 | |

1845.6 | |

2547.4 | |

1258.9 |

Question 10 Explanation:

G_s=2.65,

\gamma _w=1000kg/m^3

Degree of saturation, S = 100% (Zero air voids)

w = 10%

The dry density corresponding to zero air voids or 100% saturation,

\begin{aligned} \gamma _d&= \frac{\gamma _wG_s}{1+wG_s}\\ &= \frac{1000\times 2.65}{1+0.1 \times 2.65}\\ &= 2094.9kg/m^3 \end{aligned}

\gamma _w=1000kg/m^3

Degree of saturation, S = 100% (Zero air voids)

w = 10%

The dry density corresponding to zero air voids or 100% saturation,

\begin{aligned} \gamma _d&= \frac{\gamma _wG_s}{1+wG_s}\\ &= \frac{1000\times 2.65}{1+0.1 \times 2.65}\\ &= 2094.9kg/m^3 \end{aligned}

There are 10 questions to complete.