Question 1 |
A standard penetration test (SPT) was carried out at a location by using a manually operated hammer dropping system with 50 \% efficiency. The recorded SPT value at a particular depth is 28 . If an automatic hammer dropping system with 70 \% efficiency is used at the same location, the recorded SPT value will be
28 | |
20 | |
40 | |
25 |
Question 1 Explanation:
We know that,
Efficiency of blow \alpha \frac{1}{\text { SPT value }}
\Rightarrow \quad \eta_{1} \mathrm{~N}_{1}=\eta_{2} \mathrm{~N}_{2}
Hence,
0.50 \times 28=0.70 \times \mathrm{N}_{2}
or \quad \mathrm{N}_{2}=\frac{0.50 \times 28}{0.7}=20
Option (B) is correct
Efficiency of blow \alpha \frac{1}{\text { SPT value }}
\Rightarrow \quad \eta_{1} \mathrm{~N}_{1}=\eta_{2} \mathrm{~N}_{2}
Hence,
0.50 \times 28=0.70 \times \mathrm{N}_{2}
or \quad \mathrm{N}_{2}=\frac{0.50 \times 28}{0.7}=20
Option (B) is correct
Question 2 |
A soil having the average properties, bulk unit weight =19 \mathrm{kN} / \mathrm{m}^{3}; angle of internal friction =25^{\circ} and cohesion =15 \mathrm{kPa}, is being formed on a rock slope existing at an inclination of 35^{\circ} with the horizontal. The critical height (in \mathrm{m} ) of the soil formation up to which it would be stable without any failure is _____ (round off to one decimal place).
2.1 | |
3.6 | |
5.1 | |
8.2 |
Question 2 Explanation:
Given,
\quad \gamma_{\mathrm{t}}=19 \mathrm{kN} / \mathrm{m}^{3}
\phi=25^{\circ}
C= 15 \mathrm{kPa}
\beta=35^{\circ}
For \mathrm{C}-\phi soil
\mathrm{FOS}=\frac{\mathrm{C}+\gamma_{\mathrm{t}} z \cos ^{2} \beta \tan \phi}{\gamma_{\mathrm{t}} z \cos \beta \sin \beta}
For critical height FOS=1
\begin{aligned} \text { FOS } & =\frac{\mathrm{C}+\gamma_{\mathrm{t}} \mathrm{z} \cos ^{2} \beta \tan \phi}{\gamma_{\mathrm{t}} \mathrm{zcos} \beta \sin \beta}=1 \\ & =19 \times \mathrm{H}_{\mathrm{C}} \times \cos 35^{\circ} \sin 35^{\circ} \\ & =15+19 \times \mathrm{H}_{\mathrm{C}} \times \cos ^{2} 35 \tan 25 \\ \mathrm{H}_{\mathrm{C}} & =5.03 \mathrm{~m} \end{aligned}
\quad \gamma_{\mathrm{t}}=19 \mathrm{kN} / \mathrm{m}^{3}
\phi=25^{\circ}
C= 15 \mathrm{kPa}
\beta=35^{\circ}
For \mathrm{C}-\phi soil
\mathrm{FOS}=\frac{\mathrm{C}+\gamma_{\mathrm{t}} z \cos ^{2} \beta \tan \phi}{\gamma_{\mathrm{t}} z \cos \beta \sin \beta}
For critical height FOS=1
\begin{aligned} \text { FOS } & =\frac{\mathrm{C}+\gamma_{\mathrm{t}} \mathrm{z} \cos ^{2} \beta \tan \phi}{\gamma_{\mathrm{t}} \mathrm{zcos} \beta \sin \beta}=1 \\ & =19 \times \mathrm{H}_{\mathrm{C}} \times \cos 35^{\circ} \sin 35^{\circ} \\ & =15+19 \times \mathrm{H}_{\mathrm{C}} \times \cos ^{2} 35 \tan 25 \\ \mathrm{H}_{\mathrm{C}} & =5.03 \mathrm{~m} \end{aligned}
Question 3 |
Let \psi represent soil suction head and K represent hydraulic conductivity of the
soil. If the soil moisture content \theta increases, which one of the following
statements is TRUE?
\psi decreases and K increases. | |
\psi increases and K decreases. | |
Both \psi and K decrease. | |
Both \psi and K increase. |
Question 3 Explanation:
h_c\propto \frac{1}{R}
K\propto S
Water content \uparrow \rightarrow R\uparrow \rightarrow h_c \downarrow \rightarrow \psi \downarrow
Water content \uparrow \rightarrow S\uparrow \rightarrow K \uparrow
K\propto S
Water content \uparrow \rightarrow R\uparrow \rightarrow h_c \downarrow \rightarrow \psi \downarrow
Water content \uparrow \rightarrow S\uparrow \rightarrow K \uparrow
Question 4 |
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 4 Explanation:
G_{m}=\frac{55+40+5}{\frac{55}{2.55}+\frac{40}{2.65}+\frac{5}{2.70}}=2.596
Question 5 |
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 5 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}
There are 5 questions to complete.
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