Question:
The volume and voids ratio of an undisturbed soil sample are 100 cm\(^3\) and 0.60, respectively. After oven drying, the mass of this sample is reduced from 185 g to 165 g without any shrinkage in the volume. If the specific gravity of this sample is 2.64, the degree of saturation of this soil is _________ %. (Rounded off to 2 decimal places)
The volume and voids ratio of an undisturbed soil sample are 100 cm\(^3\) and 0.60, respectively. After oven drying, the mass of this sample is reduced from 185 g to 165 g without any shrinkage in the volume. If the specific gravity of this sample is 2.64, the degree of saturation of this soil is _________ %. (Rounded off to 2 decimal places)
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To calculate the degree of saturation, you need to find the mass of water, the mass of solids, and use the formula \( S = \frac{W_w}{W_s} \times 100 \).
Updated On: Apr 14, 2025
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Solution and Explanation
The degree of saturation (S) is calculated using the formula:
\[
S = \frac{W_w}{W_s} \times 100
\]
where:
\(W_w\) is the mass of water in the soil, and
\(W_s\) is the mass of the solids in the soil.
Step 1: Calculate the mass of water in the sample
The initial mass of the sample is 185 g, and after oven drying, the mass is reduced to 165 g. Thus, the mass of water lost during drying is: \[ W_w = 185 \, {g} - 165 \, {g} = 20 \, {g} \] Step 2: Calculate the volume of solids
The volume of the soil sample is given as 100 cm\(^3\). The void ratio (\(e\)) is given as 0.60. The void ratio is the ratio of the volume of voids to the volume of solids, i.e., \[ e = \frac{V_v}{V_s} \] where \(V_v\) is the volume of voids and \(V_s\) is the volume of solids. Since the total volume of the sample is 100 cm\(^3\), we can calculate the volume of solids (\(V_s\)): \[ V_s = \frac{V}{1 + e} = \frac{100}{1 + 0.60} = \frac{100}{1.60} = 62.5 \, {cm}^3 \] Step 3: Calculate the mass of solids
The mass of solids can be calculated using the formula: \[ W_s = \rho_s \times V_s \] where \( \rho_s \) is the density of the solids and \(V_s\) is the volume of solids. The density of the solids can be found using the specific gravity (\(G_s\)): \[ \rho_s = G_s \times \rho_w = 2.64 \times 1 \, {g/cm}^3 = 2.64 \, {g/cm}^3 \] Now, calculate the mass of solids: \[ W_s = 2.64 \, {g/cm}^3 \times 62.5 \, {cm}^3 = 165 \, {g} \] Step 4: Calculate the degree of saturation
Finally, we can calculate the degree of saturation: \[ S = \frac{W_w}{W_s} \times 100 = \frac{20 \, {g}}{165 \, {g}} \times 100 = 52.70% \] Thus, the degree of saturation is approximately 52.70%.
\(W_w\) is the mass of water in the soil, and
\(W_s\) is the mass of the solids in the soil.
Step 1: Calculate the mass of water in the sample
The initial mass of the sample is 185 g, and after oven drying, the mass is reduced to 165 g. Thus, the mass of water lost during drying is: \[ W_w = 185 \, {g} - 165 \, {g} = 20 \, {g} \] Step 2: Calculate the volume of solids
The volume of the soil sample is given as 100 cm\(^3\). The void ratio (\(e\)) is given as 0.60. The void ratio is the ratio of the volume of voids to the volume of solids, i.e., \[ e = \frac{V_v}{V_s} \] where \(V_v\) is the volume of voids and \(V_s\) is the volume of solids. Since the total volume of the sample is 100 cm\(^3\), we can calculate the volume of solids (\(V_s\)): \[ V_s = \frac{V}{1 + e} = \frac{100}{1 + 0.60} = \frac{100}{1.60} = 62.5 \, {cm}^3 \] Step 3: Calculate the mass of solids
The mass of solids can be calculated using the formula: \[ W_s = \rho_s \times V_s \] where \( \rho_s \) is the density of the solids and \(V_s\) is the volume of solids. The density of the solids can be found using the specific gravity (\(G_s\)): \[ \rho_s = G_s \times \rho_w = 2.64 \times 1 \, {g/cm}^3 = 2.64 \, {g/cm}^3 \] Now, calculate the mass of solids: \[ W_s = 2.64 \, {g/cm}^3 \times 62.5 \, {cm}^3 = 165 \, {g} \] Step 4: Calculate the degree of saturation
Finally, we can calculate the degree of saturation: \[ S = \frac{W_w}{W_s} \times 100 = \frac{20 \, {g}}{165 \, {g}} \times 100 = 52.70% \] Thus, the degree of saturation is approximately 52.70%.
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