Q21GATE 2010MCQ

An open ended steel barrel of 1m height and 1m diameter is filled with saturated fine sand having coefficient of permability of $10^{-2}$ m/s. The barrel stands on a saturated bed of gravel. The time required for the water level in the barrel to drop by 0.75m is

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📖 Explanation

\begin{aligned} \text { Time required } &=\frac{\text { Drop \in water level }}{\text { Coefficient of permeability }} \\ &=\frac{0.75}{10^{-2}}=75 \mathrm{s} \end{aligned}

Q22GATE 2008MCQ

The ground conditions at a site are shown in the figure below. The total stress, pore water pressure and effective stress (kN/ $m^{2}$ ) at the point P are, respectively

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📖 Explanation

Total stress, $\sigma=\gamma_{\text {sat }} \times 5=21 \times 5=105 \mathrm{kN} / \mathrm{m}^{2}$ Pore water pressure, $\mathrm{u}=\gamma_{\mathrm{w}} \times 5=10 \times 5=50 \mathrm{kN} / \mathrm{m}^{2}$ Effective stress, $\bar{\sigma}=\sigma-u=105-50=55 \mathrm{kN} / \mathrm{m}^{2}$

Q23GATE 2008MCQ

The ground conditions at a site are shown in the figure below. The saturated unit weight of the sand (kN/ $m^{3}$ ) is

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📖 Explanation

\begin{aligned} \therefore \quad e&=\frac{G w}{S} \\ \Rightarrow \quad e&=\frac{2.7 \times 0.2}{1} \Rightarrow e=0.54 \\ \therefore \quad \gamma_{\text {sat }}&=\left(\frac{G+e}{1+e}\right) \times \gamma_{w}=\left(\frac{2.7+0.54}{1+0.54}\right) \times 10 \\ &=21.04 \mathrm{kN} / \mathrm{m}^{3} \end{aligned}

Q24GATE 2007MCQ

Match the following groups

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Q25GATE 2007MCQ

Water is flowing through the permeability apparatus shown in the figure. The coefficient of permeability of the soil is k m/s and the porosity of the soil sample is 0.50. What are the discharge velocity and seepage velocity through the soil sample ?

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📖 Explanation

Seepage velocity $\left(v_{s}\right)=\frac{v}{n}=\frac{v}{0.5}=2 v$ [porosity n=0.5] as per Darcy's law, v=ki as $\quad i=1$ $v=k \; m / s$ Discharge velocity =k Seepage velocity =2k

Q26GATE 2006MCQ

Which of the following statement is NOT TRUE in the context of capillary pressure in solids ?

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Q27GATE 2005MCQ

In a constant head permeameter with cross section area of 10 $cm^{2}$ , when the flow was taking place under a hydraulic gradient of 0.5, the amount of water collected in 60 seconds is 600 cc. The permeability of the soil is

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📖 Explanation

\begin{aligned} Q &=k \mathrm{i} A \quad \Rightarrow \quad k=\frac{Q}{i A}=\frac{600 / 60}{0.5 \times 10} \\ &=2.0 \mathrm{cm} / \mathrm{s} \end{aligned}

Q28GATE 2005MCQ

Assuming that a river bed level does not change and the depth of water in river was 10 m, 15 m and 8 m during the months of February, July and December respectively of a particular year. The average bulk density of the soil is 20 kN/ $m^{3}$ . The density of water is 10 kN/ $m^{3}$ . The effective stress at a depth of 10 m below the river bed during these months would be

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📖 Explanation

Since the river bed level does not change and the depth of water is changing only above the river bed, the effective stress below the river bed will not change. Total stress at a depth of 10 m below river bed, $\sigma=20 \times 10=200 \mathrm{kN} / \mathrm{m}^{2}$ Pore water pressure, $u=10 \times 10=100 \mathrm{kN} / \mathrm{m}^{2}$ $\therefore$ Effective stress,

\begin{aligned} \bar{\sigma} &=\sigma-u=200-100 \\ &=100 \mathrm{kN} / \mathrm{m}^{2} \end{aligned}

Thus effective stress will be $100 \mathrm{kN} / \mathrm{m}^{2}$ each in the month of February, July and December

Q29GATE 2004MCQ

A 10 m thick clay layer is underlain by a sand layer of 20 m depth (see figure below). The water table is 5 m below the surface of clay layer. The soil above the water table is capillary saturated. The value of $\gamma _{sat}$ is 19 kN/ $m^{3}$ . The unit weight of water is $\gamma _{w}$ . If now the water table rises to the surface, the effective stress at point P on the interface will

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📖 Explanation

In first case, when the water table is 5m below the surface and the clay over water table is capillary saturated, the effective stress is given by

\begin{aligned} \bar{\sigma}_{1} &=\gamma_{\text {sat }} \times 5+\gamma \times 5 \\ &=\gamma_{\text {sat }} \times 5+\left(\gamma_{\text {sat }}-\gamma_{\mathrm{w}}\right) \times 5 \\ \Rightarrow \quad \bar{\sigma}_{1} &=10 \gamma_{\text {sat }}-5 \gamma_{\mathrm{w}} \; \; \; \ldots(i) \end{aligned}

In the second case, when water table rises to the surface, the effective stress is given by

\begin{aligned} \bar{\sigma}_{2} &=\gamma \times 10=\left(\gamma_{\text {sat }}-\gamma_{\mathrm{w}}\right) \times 10 \\ &=10 \gamma_{\text {sat }}-10 \gamma_{\mathrm{w}} \\ &=\left(10 \gamma_{\mathrm{sat}}-5 \gamma_{\mathrm{w}}\right)-5 \gamma_{\mathrm{w}} \quad[\because \text { from (i) }) \\ \Rightarrow \bar{\sigma}_{2} &=\bar{\sigma}_{1}-5 \gamma_{\mathrm{w}} \end{aligned}

Thus in second case, the effective stress at the interface will decrease by $5 \gamma_{\mathrm{w}}$ .

Q30GATE 2003MCQ

For the soil strata shown in figure, the water tables is lowered by drainage by 2 m and if the top 2 m thick silty sand stratum remains saturated by capillary action even after lowering of water table, the increase in effective vertical pressure in kPa at mid-height of clay layer will be

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📖 Explanation

In the capillary zone, pore pressure is negative. The total stress will not change due to lowering of water table because the stratum is still saturated. Thus, the increase in effective vertical pressure will be only due to pore pressure. Thus increase in effective stress, $=h \gamma_{w}=2 \times 10=20 \mathrm{kN} / \mathrm{m}^{2}$

Q31GATE 2001MCQ

The coefficients of permeability of a soil in horizontal and vertical directions are 3.46 and 1.5 m/day respectively. The base length of a concrete dam resting in this soil is 100 m. When the flow net is developed for this soil with 1:25 scale factor in the vertical direction, the reduced base length of the dam will be

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📖 Explanation

\begin{aligned} K_{x}&=3.46 \mathrm{m} / \mathrm{day} \\ K_{y}&=1.5 \mathrm{m} / \mathrm{day} l_{b}&=100 \mathrm{m} \text{(base length)}\\ \end{aligned}

scale factor in vertical direction =1: 25 scale factor in horizontal direction $=\frac{1}{25} \sqrt{\frac{K_{y}}{K_{x}}}=\frac{1}{37.969}$ $\therefore$ Reduced base length = scale $\times l_{\mathrm{b}}$ $=\frac{100}{37.969}=2.63 \mathrm{m}$