---
title: "Moist soil 20 kg numerical"
subject: "Soil Mechanics"
chapter: 2
key: "semester_4/soil_mechanics:moist-soil-mass-numerical"
source: https://www.bctengineer.com/q/semester_4/soil_mechanics/moist-soil-mass-numerical
site: BCT Engineering
---

# Moist soil 20 kg numerical

Soil Mechanics · Chapter 2: Solids- Water –Air Relations and Index properties of soils

- Asked in: 2078 Chaitra, 4 marks
- Video: [Dry unit weight, Porosity, Void ratio and Degree of Saturation || Soil Mechanics problem solved](https://www.youtube.com/watch?v=OkC1THwBqVQ) — Civil Engineering
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/soil_mechanics/moist-soil-mass-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-2

## Question

A mass of moist soil is 20 kg and its volume is 0.011 m³. After oven drying, the mass reduces to 16.5 kg. Assume G = 2.70. Determine water content, dry density, degree of saturation and porosity.

## Short answer

Water content 21.2%, dry density 1500 kg/m³, degree of saturation 71.6%, porosity 44.4%.

## Answer

#### Given
$M = 20$ kg, $V = 0.011$ m³, dry mass $M_s = 16.5$ kg, $G = 2.70$, $\rho_w = 1000$ kg/m³.

#### Solution
Mass of water: $M_w = 20 - 16.5 = 3.5$ kg.

**Water content**
$$
w = \frac{M_w}{M_s} = \frac{3.5}{16.5} = 0.212 = 21.2\%
$$
**Dry density**
$$
\rho_d = \frac{M_s}{V} = \frac{16.5}{0.011} = 1500\ \text{kg/m}^3
$$
**Void ratio** (from $\rho_d = \dfrac{G\rho_w}{1+e}$):
$$
e = \frac{G\rho_w}{\rho_d} - 1 = \frac{2700}{1500} - 1 = 0.80
$$
**Degree of saturation**
$$
S = \frac{wG}{e} = \frac{0.2121 \times 2.70}{0.80} = 0.716 = 71.6\%
$$
**Porosity**
$$
n = \frac{e}{1+e} = \frac{0.80}{1.80} = 0.444 = 44.4\%
$$

**Answer: $w = 21.2\%$, $\rho_d = 1500$ kg/m³, $S = 71.6\%$, $n = 44.4\%$.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Water content w for 20 kg moist and 16.5 kg dry is:
   A. 21.2%
   B. 17.5%
   C. 3.5%
   D. 82.5%
   Answer: A. w = 3.5/16.5 = 21.2% (divide by dry mass). (Key point: Water content)

2. Dry density is:
   A. 1250 kg/m3
   B. 1818 kg/m3
   C. 1500 kg/m3
   D. 2700 kg/m3
   Answer: C. 16.5/0.011 = 1500 kg/m3. (Key point: Dry density)

3. Void ratio is obtained from:
   A. e = G rho_w/rho_d - 1 = 0.80
   B. e = wG
   C. e = n/(1+n)
   D. e = 1 - n
   Answer: A. 2700/1500 - 1 = 0.80. (Key point: Void ratio route)

4. Degree of saturation S is:
   A. 44.4%
   B. 100%
   C. 71.6%
   D. 80%
   Answer: C. S = wG/e = 0.2121 x 2.7/0.80 = 71.6%. (Key point: Saturation)

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