# Civil Engineering - Soil Mechanics and Foundation Engineering - Discussion

### Discussion :: Soil Mechanics and Foundation Engineering - Section 1 (Q.No.20)

20.

A partially saturated sample of soil has a unit weight of 2.0 g/cm3 and specific gravity of soil particles is 2.6. If the moisture content in the soil is 20%, the degree of saturation is

 [A]. 20% [B]. 77% [C]. 92% [D]. none of these.

Explanation:

No answer description available for this question.

 Iiit Basar said: (Aug 29, 2013) Unit weight of soil = 2.0g/cm3. Dry unit weight = unit weight/(1+w). Here w = 0.2. Hence dry unit weight = 2/(1+0.2). = 1.66. Void ratio = ((specific gravity * unit weight of water )/dry unit weight)-1. = 0.566. Degree of saturation = w * specific gravity /void ratio. = 0.9182. = 91.82%.

 Ramu Vemula said: (Jul 18, 2014) Unit weight of soil = 2g/cm3. Dry unit weight = unit weight/(1+w). w = 0.2 & G = 2.6. Dry unit weight = 2/(1+0.2) = 1.81. e = G*1/dry unit weight-1 = 2.6*1/1.81-1 = 0.566. Deg of saturation = (G*w)/e = (2.6*0.2)/0.566 = 0.9187. Deg of saturation = 91.87%.

 Priyanka said: (Mar 2, 2015) Unit weight of soil = 2g/cm3. Dry unit weight = Unit weight/(1+w). W = 0.2 & G = 2.6 unit weight of water = 9.81. Dry unit weight = 2/(1+0.2) = 1.66. E = g*9.81/dry unit weight-1 = (2.6*9.81/1.66)-1 = 14.36. S, degree of saturation = (g*w)/e = (2.6*0.2)/14.30 = 0.036. S = 3.6%.

 Prashant said: (Dec 29, 2015) Y = Yw*(G+e*Sr)/(1+e). Put e*Sr = W*G. Y = Yw*(G+G*W)/(1+e). Put values: e = 0.56. Sr = W*G/e. Sr = 92%

 Nawaz said: (Feb 28, 2016) Here very simple given data, Y = 2g/cm^3, G = 2.6, w (water content) = 0.2. Sr = ?, We know-Dry unit weight (Yd) = Y/1+w. So, 2/1 + 0.2 = 1.66. Again we know, 1 + e = GYw/Yd, So, 1+e = 2.6*1/1.66 = 1.56 - 1 = 0.56. So, Sr = WG/e. So, Sr = 0.2*2.6/0.56 = 0.92. So, 92% answer.

 Pratyush said: (Oct 21, 2016) Yes @Priyanka. I'm getting the same answer can anyone justify why Yw =1 not 9.81?

 Anonymous said: (Dec 31, 2016) Look at the units.Unit weights are given in g/cm^3 and unit weight of water is 1 g/cm^3 in CGS system.

 Souvik said: (Apr 10, 2017) Why don't we consider the saturated unit of soil because in question it is mentioned that the soil sample is partially saturated? Why don't we ignore the formula Ysat = (G+e) * Yw/(1+e)? Can anyone clarify the answer?

 Meghana said: (May 4, 2017) @Souvik. If we use that formula, we get e = 0.6. By using Gw=Sr.e. We get Sr = 86.6%.

 Sub said: (Oct 23, 2017) Unit weight of soil = 2.0g/cm3. Dry unit weight = unit weight/(1+w). Here w = 0.2. Hence dry unit weight = 2/(1+0.2). = 1.66. Void ratio = ((specific gravity * unit weight of water )/dry unit weight)-1. = 0.566. Degree of saturation = w * specific gravity /void ratio. = 0.9182. = 91.82%.

 Himanshu Gupta said: (Dec 2, 2017) @Souvik. Partially saturated means the soil have a degree of saturation less than 100%. If it is completely saturated then there is no need to find a degree of saturation.

 Kiruthika said: (Apr 15, 2018) @Himanshu Gupta. The question is to find the degree saturation; I am getting the answer as S=87%, By using for:eS=wG, e=.62.

 Anil said: (Jul 31, 2018) 77% is correct I think.

 Jitendra Gkp said: (Jan 10, 2019) b=G*w(1+w)/(1+e). e=.56. Sr=wG/e=92%.

 Ujjval said: (Jan 17, 2019) W=.2, bulk density = 2g/cc. Dry density = 2/1 +.2 = 1.667g/cc. e = [(2.6 * 1)/1.67-1] = .56. S = (2.6 * 0.1)/.56 = .9265. OR S = 92.6%.