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Home›Geotechnical Engineering›Properties Of Soils

GATE CE · Geotechnical Engineering

Properties Of Soils Practice Questions

Generate GATE-level questions on Properties Of Soils in Geotechnical Engineering. Focus on core concepts, previous year patterns, and numerical problem-solving techniques.

39 questions · 20 PYQs · 0 AI practice · GATE CE 2027

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Q1GATE 2025NAT

A clayey soil has a moisture content of $18 \%$ , a specific gravity of soil solids of 2.74 , and a degree of saturation of $65 \%$ . The soil soaks up water during a rain event, and the degree of saturation increases to $85.2 \%$ . The change of the volume during the soaking is negligible. The new moisture content (in %) of the soil will be _________ (rounded off to 2 decimal places).

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📖 Explanation

Given: Initial moisture content, $w_1 = 18% = 0.18$ Specific gravity, $G = 2.74$ Initial degree of saturation, $S_1 = 65% = 0.65$ Final degree of saturation, $S_2 = 85.2% = 0.852$ Assuming negligible change in volume, the void ratio $e$ remains constant. The relation between moisture content, saturation, and void ratio is: $w = \frac{S \cdot e}{G}$ Rearranging to find $e$ : $e = \frac{w_1 \cdot G}{S_1}$ $e = \frac{0.18 \cdot 2.74}{0.65} = 0.75877$ Now calculate the new moisture content $w_2$ : $w_2 = \frac{S_2 \cdot e}{G}$ $w_2 = \frac{0.852 \cdot 0.75877}{2.74} = 0.2361$ $w_2 = 23.61%$

Q2GATE 2025MCQ

For a partially saturated soil deposit at a construction site, water content (w) is 15%, degree of saturation (S) is 67%, void ratio (e) is 0.6 and specific gravity of solids in the soil ( $G_s$ ) is 2.67. Consider unit weight of water as $9.81\ \text{kN/m}^3$ . To fully saturate $5\ \text{m}^3$ of this soil, the required weight of water (in kN) will be ______ (round off to the nearest integer).

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📖 Explanation

Given: $w = 15% = 0.15$ $S = 67% = 0.67$ $e = 0.6$ $G_s = 2.67$ $V = 5\ \text{m}^3$ $\gamma_w = 9.81\ \text{kN/m}^3$ To fully saturate the soil, we need to bring degree of saturation $S$ from 0.67 to 1.00. Volume of voids per unit volume of soil: $V_v = \frac{e}{1 + e} = \frac{0.6}{1 + 0.6} = 0.375$ So in 5 m? of soil, total void volume: $V_{v,\text{total}} = 0.375 \times 5 = 1.875\ \text{m}^3$ Currently 67% of voids are filled with water, so water volume present: $V_{w,\text{present}} = 0.67 \times 1.875 = 1.25625\ \text{m}^3$ To fully saturate: Required water volume = $V_{w,\text{required}} = 1.875 - 1.25625 = 0.61875\ \text{m}^3$ Weight of required water: $W_w = \gamma_w \times V_{w,\text{required}} = 9.81 \times 0.61875 = 6.07\ \text{kN}$ Rounded to nearest integer: $\boxed{6\ \text{kN}}$

Q3GATE 2024MSQ

Which of the following statement(s) is/are CORRECT?

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📖 Explanation

Shrinkage limit is defined as the maximum water content at which further reduction in water content of soil does not lead to reduction in the volume of soil as water is being just replaced by air. Higher is the shrinkage limit lower is the compressibility (volume change). Dilatancy correction is applied to the already corrected N-values for overburden pressure. At large value of shearing strain, both the initially loose and initially dense sands approach a constant value of void ratio called critical void ratio.

Q4GATE 2024NAT

A homogeneous earth dam has a maximum water head difference of $15 \mathrm{~m}$ between the upstream and downstream sides. A flownet was drawn with the number of potential drops as 10 and the average length of the element as $3 \mathrm{~m}$ . Specific gravity of the soil is 2.65 . For a factor of safety of 2.0 against piping failure, void ratio of the soil is ____(rounded off to 2 decimal places).

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📖 Explanation

$(\Delta \mathrm{H})$ total head loss $=15 \mathrm{~m}$ $\left(n_{d}\right)$ number of equipotential drops $=10$ (I) length of flow field $=3 \mathrm{~m}$ (G) specific gravity $=2.65$ (FOS) factor of safety $=2$ (e) void ratio $=$ ? $(\Delta h)$ equipotential drop of head

\begin{aligned} & =\frac{\text { Total head loss }}{\text { no. of equipotential drop }} \\ \Delta \mathrm{h} & =\frac{\Delta \mathrm{H}}{\mathrm{N}_{\mathrm{d}}}=\frac{15}{10}=1.5 \mathrm{~m} \end{aligned}

FOS against quick sand condition $=\frac{i_{c r}}{i}$ (i) hydraulic gradient $=\frac{\Delta \mathrm{h}}{\mathrm{l}}=\frac{1.5}{3}=0.5$ $2=\frac{i_{c r}}{0.5},\left(i_{c r}\right)$ ciritical hydraulic gradient $=1$ Critical hydraulic gradient $\left(\mathrm{i}_{\mathrm{cr}}\right)=\frac{\mathrm{G}-1}{1+\mathrm{e}}=1=\frac{2.65-1}{1+\mathrm{e}}$ (e)void ratio $=0.65$

Q5GATE 2023MCQ

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

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📖 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

Q6GATE 2023NAT

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).

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📖 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}

Q7GATE 2022MCQ

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?

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📖 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 $\uparrow \rightarrow S\uparrow \rightarrow K \uparrow$

Q8GATE 2021NAT

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 ____________

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📖 Explanation

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

Q9GATE 2021MCQ

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

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📖 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}

Q10GATE 2020MCQ

Soil deposit formed due to transportation by wind is termed as

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📖 Explanation

Soil deposited by wind is Aeolian soil.

Q11GATE 2020MCQ

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

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📖 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}

Q12GATE 2020NAT

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 ________.

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📖 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}

Q13GATE 2019NAT

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 _______

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📖 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}

Q14GATE 2019MCQ

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

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📖 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.

Q15GATE 2019MCQ

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

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📖 Explanation

SC $\rightarrow$ Clayey sand

Q16GATE 2019NAT

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_______

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📖 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 $= 50 + 0.9375 = 50.94 kg$

Q17GATE 2019NAT

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 ______

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📖 Explanation

Outside clearance $=\left ( \frac{D_2 -D_4}{D_4} \right ) \times 100$ Outside clearance $=\left ( \frac{100 -90}{90} \right ) \times 100 =11.11\%$

Explanation Figure 2

Q18GATE 2018MCQ

The clay mineral, whose structural units are held together by potassium bond is

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📖 Explanation

Kaolinite $\rightarrow$ Hydrogen bond Illite $\rightarrow$ lonic Bond/Potassium Bond Montmorillonite $\rightarrow$ Water Bond

Q19GATE 2018NAT

A core cutter of 130 mm height has inner and outer diameters of 100 mm and 106 mm, respectively. The area ratio of the core cutter (in %, up to two decimal places) is ______

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📖 Explanation

\begin{aligned} \text { Area ratio } &=\frac{A_{\text {outer }}-A_{\text {inne }} r}{A_{\text {inner }}} \times 100 \\ &=\frac{\frac{\pi}{4} D_{0}^{2}-\frac{\pi}{4} D_{i}^{2}}{\frac{\pi}{4} D_{i}^{2}} \times 100 \\ &=\frac{\frac{\pi}{4}(106)^{2}-\frac{\pi}{4}(100)^{2}}{\frac{\pi}{4}(100)^{2}} \times 100 \\ &=12.36 \% \end{aligned}

Q20GATE 2018NAT

In a shrinkage limit test, the volume and mass of a dry soil pat are found to be 50 $cm^{3}$ and 88 g, respectively. The specific gravity of the soil solids is 2.71 and the density of water is 1 g/cc. The shrinkage limit (in %, up to two decimal places) is ______

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📖 Explanation

Dry soil mass =88 gm Volume of dry soil = 50 cc Dry density of soil mass,

\begin{aligned} \rho_{d}&=\frac{M_{d}}{V_{d}}=\frac{88}{50} \mathrm{gm} / \mathrm{cc}\\ \therefore \text{ Shrinkage limit, } w_{s}&=\frac{1}{R}-\frac{1}{G}=\frac{1}{\left(\rho_{d} / \rho_{w}\right)}-\frac{1}{G} \\ w_{s} &=\frac{\rho_{w}}{\rho_{d}}-\frac{1}{G} \\ &=\frac{1}{1.76}-\frac{1}{2.71}=0.1990 \end{aligned}

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