---
title: "Dry sand cylinder saturation numerical"
subject: "Soil Mechanics"
chapter: 2
key: "semester_4/soil_mechanics:dry-sand-cylinder-saturation"
source: https://www.bctengineer.com/q/semester_4/soil_mechanics/dry-sand-cylinder-saturation
site: BCT Engineering
---

# Dry sand cylinder saturation numerical

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

- Asked in: 2077 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/dry-sand-cylinder-saturation
- Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-2

## Question

Dry sand is poured into a cylindrical container (internal diameter 0.2 m and height 0.2 m) and just filled up to its top. The weight of the dry sand in the container is found to be 10 kg. By adding water, this dry sand sample is fully saturated with water. Let the void ratio of this sand sample be 0.54, which remains constant throughout the saturation process. Taking the specific gravity of soil solids as 2.65, find: (i) the amount of water needed to fully saturate the dry sand sample and its water content at full saturation; (ii) the amount of water to be added to the dry sand sample to achieve 80% degree of saturation. Mention the assumed condition if any.

## Short answer

(i) 2.04 kg of water needed, water content 20.4%. (ii) 1.63 kg of water for 80% saturation. Assumes given void ratio and dry mass directly define solids and void volumes.

## Answer

#### Given
Container: $D = 0.2$ m, $H = 0.2$ m, so $V = \frac{\pi}{4}(0.2)^2(0.2) = 6.283 \times 10^{-3}$ m³. Dry sand $M_s = 10$ kg, $G = 2.65$, $e = 0.54$ (constant). $\rho_w = 1000$ kg/m³.

**Assumption:** the stated void ratio of 0.54 is used directly with the given mass of sand; the mass of solids fixes $V_s$, and the void volume is $V_v = eV_s$. (Using the container volume with $e = 0.54$ would give a slightly different mass of 10.8 kg, so the data are only approximately consistent.)

#### Volume of solids
$$
V_s = \frac{M_s}{G\rho_w} = \frac{10}{2.65 \times 1000} = 3.774 \times 10^{-3}\ \text{m}^3
$$

#### (i) Water to saturate the sand
$$
V_v = eV_s = 0.54 \times 3.774 \times 10^{-3} = 2.038 \times 10^{-3}\ \text{m}^3
$$
Water needed $= 2.038 \times 10^{-3} \times 1000 = 2.04$ kg (litres).

Water content at saturation:
$$
w = \frac{M_w}{M_s} = \frac{2.038}{10} = 20.4\% \quad (\text{also } w = e/G = 0.54/2.65 = 20.4\%)
$$

#### (ii) Water for 80% saturation
$$
M_w = S\,e\,V_s\rho_w = 0.80 \times 2.038 = 1.63\ \text{kg}
$$
So **1.63 litres** of water must be added to the dry sand.

**Answer: (i) 2.04 kg of water, $w = 20.4\%$; (ii) 1.63 kg (litres) of water.**

## Check yourself

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

1. Volume of solids is found from:
   A. Ms/(G rho_w) = 3.774 x 10^-3 m3
   B. Container volume
   C. Ms x G
   D. Ms/rho_w
   Answer: A. 10/(2.65 x 1000) = 3.774e-3 m3. (Key point: Volume of solids)

2. Water needed to saturate is:
   A. 1.63 kg
   B. 3.77 kg
   C. 2.04 kg
   D. 10 kg
   Answer: C. Vv = 0.54 x 3.774e-3 = 2.038e-3 m3, i.e. 2.04 kg. (Key point: Water to saturate)

3. Water content at saturation is:
   A. 20.4%
   B. 54%
   C. 26.5%
   D. 10%
   Answer: A. w = e/G = 0.54/2.65 = 20.4%. (Key point: Saturated water content)

4. Water for 80% saturation is:
   A. 0.8 kg
   B. 2.04 kg
   C. 1.63 kg
   D. 1.20 kg
   Answer: C. 0.80 x 2.038 = 1.63 kg. (Key point: 80% saturation)

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