Weight Volume Relationships Questions and Answers

Answer: B

The theoretical maximum dry density is the dry density at the zero air voids condition, when the soil is fully saturated at the given water content:

The other values would come from a wrong water content or from using the formula for a different degree of saturation: 1.605 is lower and 1.832 and 1.902 are higher than the zero air voids value, which is the upper limit of the dry density at 18 percent water content.

Answer: B

The porosity is 0.44 and the void ratio 0.79.

Porosity follows from the two densities, and void ratio from porosity:

\[ \begin{aligned} n &= 1 - \frac{\rho_{b}}{\rho_{s}} = 1 - \frac{1.48}{2.64} \approx 0.44 \\ e &= \frac{n}{1 - n} = \frac{0.44}{0.56} \approx 0.79 \end{aligned} \]

The saturated volumetric moisture content of 36% is not needed. The option 0.79 and 0.44 swaps the two quantities. 0.36 and 0.56 takes the moisture figure as the porosity, and 0.56 and 0.36 takes the solid fraction \(1 - n\) as the porosity.

Answer: A

The porosity is 0.30.

Void ratio \(e\) is void volume over solid volume, while porosity \(n\) is void volume over total volume, so

\[ n = \frac{e}{1 + e} = \frac{0.43}{1.43} = 0.30 \]

0.70 is \(1 - n\), the solid fraction. 0.40 and 0.75 do not follow from the relation; porosity must always be smaller than the void ratio.

Answer: B

The bulk unit weight is 18639.00 N/m³.

For a soil with specific gravity \(G\), void ratio \(e\) and degree of saturation \(S\), the bulk unit weight is

\[ \begin{aligned} \gamma &= \frac{(G + Se)\gamma_{w}}{1 + e} \\ &= \frac{(2.7 + 0.9 \times 0.8) \times 9810}{1.8} \\ &= \frac{3.42 \times 9810}{1.8} = 18639\ \text{N/m}^{3} \end{aligned} \]

18.64 is the same figure in kN/m³, but the question asks for N/m³. 10594.80 is far too low: even the dry unit weight, \(G\gamma_{w}/(1 + e) = 14715\ \text{N/m}^{3}\), is larger, and adding the water in the voids can only raise it. 10.60 is that low value written in kN/m³.

Answer: A

The void ratio is 0.667.

Porosity n and void ratio e are directly related, so only n is needed; the densities and G given are not used.

\[ \begin{aligned} e &= \frac{n}{1 - n} = \frac{0.40}{1 - 0.40} \\ &= \frac{0.40}{0.60} \approx 0.667 \end{aligned} \]

A common slip is to take e = n, or to use the reverse relation \(n = e/(1+e)\), which gives the other options.

Answer: C

The saturated and submerged densities are 2.25 g/cc and 1.25 g/cc.

At saturation the soil holds water equal to 50% of the weight of its solids, so its bulk density is the dry density times \((1 + w)\):

\[ \begin{aligned} \rho_{sat} &= \rho_d (1 + w) = 1.5 \times 1.5 = 2.25 \\ \rho' &= \rho_{sat} - \rho_w = 2.25 - 1 = 1.25 \end{aligned} \]

Both in g/cc. 1.5 and 1.0 repeats the dry density, 2.0 and 1.0 adds the water content as if it were a density, and 2.50 and 1.50 overstates the saturated density.

Answer: D

The saturated density is \(\frac{G+e}{1+e}\rho w\).

Take a soil element with unit volume of solids. Its total volume is \(1 + e\). The solids have mass \(G\rho_w\), and when saturated the voids hold water of mass \(e\rho_w\):

\[ \begin{aligned} \rho_{sat} &= \frac{G\rho_w + e\rho_w}{1 + e} \\ &= \frac{G + e}{1 + e}\rho_w \end{aligned} \]

This is the general form \(\frac{G + Se}{1 + e}\rho_w\) with the degree of saturation S = 1. The option with \(G + eS\) over \(1 + w\) uses the wrong denominator, \(G + S\) drops the void ratio from the water term, and \(G + w\) adds a water content to a specific gravity.

Answer: A

The porosity and void ratio are n = 0.44 and e = 0.8.

The soil displaces 25 cm³ of water, which is the volume of its solids. The rest of the 45 cm³ sample volume is voids:

\[ \begin{aligned} V_v &= 45 - 25 = 20\ \text{cm}^3 \\ n &= \frac{V_v}{V} = \frac{20}{45} = 0.44 \\ e &= \frac{V_v}{V_s} = \frac{20}{25} = 0.8 \end{aligned} \]

Check: \(e = n/(1 - n) = 0.44/0.56 = 0.8\). The pair 0.44 and 0.62 has the right porosity but a void ratio inconsistent with it, and 0.35 or 0.72 for n do not come from these volumes.

Answer: D

The dry density is 20 kN/m³.

Bulk density counts the water along with the solids. Removing the water means dividing by \(1 + w\), with the water content as a fraction:

\[ \begin{aligned} \gamma_d &= \frac{\gamma}{1 + w} \\ &= \frac{22}{1 + 0.10} = 20\ \text{kN/m}^{3} \end{aligned} \]

18.6 kN/m³ does not follow from the relation at all. 22 kN/m³ ignores the water, and 23.2 kN/m³ adds water instead of removing it. Dry density must be less than the bulk value whenever the soil holds any water.

Answer: A

The submerged density is \(\frac{r_{w}(G - 1)}{1 + e}\).

The submerged (buoyant) unit weight is the saturated unit weight minus the unit weight of water:

\[ \begin{aligned} \gamma' &= \frac{(G + e)\gamma_w}{1 + e} - \gamma_w \\ &= \frac{(G - 1)\gamma_w}{1 + e} \end{aligned} \]

The \(G - 1\) is the buoyancy effect: each unit volume of solids loses the weight of an equal volume of water. \(G + 1\) adds the water instead of taking it away, and \(1 - e\) in the denominator has no physical meaning, since total volume per unit volume of solids is \(1 + e\).