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

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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%. |

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