---
title: "Pycnometer data, water content, γsub"
subject: "Soil Mechanics"
chapter: 2
key: "semester_4/soil_mechanics:pycnometer-water-content-submerged"
source: https://www.bctengineer.com/q/semester_4/soil_mechanics/pycnometer-water-content-submerged
site: BCT Engineering
---

# Pycnometer data, water content, γsub

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

- Asked in: 2076 Baisakh, 3 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/soil_mechanics/pycnometer-water-content-submerged
- Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-2

## Question

From the pycnometer test, the specific gravity of soil solid of the soil specimen is found to be 2.65. Also, the dry unit weight of this soil specimen is found to be 15 kN/m³. If one cubic meter of this soil specimen weighs 18 kN/m³, determine (i) water content, (ii) degree of saturation, and (iii) submerged unit weight of this soil specimen.

## Short answer

Water content=20% (from γ/γd−1). Void ratio e=0.733 yields degree of saturation S=wG/e=72.3%. Submerged unit weight=γsat−γw=19.15−9.81=9.34 kN/m³.

## Answer

#### Given
$G = 2.65$, $\gamma_d = 15$ kN/m³, $\gamma = 18$ kN/m³, $\gamma_w = 9.81$ kN/m³.

#### (i) Water content
$$
w = \frac{\gamma}{\gamma_d} - 1 = \frac{18}{15} - 1 = 0.20 = 20\%
$$

#### (ii) Degree of saturation
Void ratio:
$$
e = \frac{G\gamma_w}{\gamma_d} - 1 = \frac{2.65 \times 9.81}{15} - 1 = 0.733
$$
$$
S = \frac{wG}{e} = \frac{0.20 \times 2.65}{0.733} = 0.723 = 72.3\%
$$

#### (iii) Submerged unit weight
Imagine the soil saturated at the same void ratio:
$$
\gamma_{sat} = \frac{(G+e)\gamma_w}{1+e} = \frac{(2.65+0.733) \times 9.81}{1.733} = 19.15\ \text{kN/m}^3
$$
$$
\gamma' = \gamma_{sat} - \gamma_w = 19.15 - 9.81 = 9.34\ \text{kN/m}^3
$$

**Answer: $w = 20\%$, $S = 72.3\%$, $\gamma' = 9.34$ kN/m³.**

## Check yourself

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

1. How is w found from gamma = 18 and gamma_d = 15 kN/m3?
   A. w = gamma_d/gamma - 1 = -16.7%
   B. w = gamma/gamma_d - 1 = 20%
   C. w = gamma - gamma_d = 3%
   D. w = gamma/gamma_d = 120%
   Answer: B. gamma = gamma_d(1+w), so w = 18/15 - 1 = 0.20. (Key point: Water content from unit weights)

2. Which relation gives S once e is known?
   A. S = w/e = 27%
   B. S = e/(wG) = 138%
   C. S = wG/e = 72.3%
   D. S = G/e = 361%
   Answer: C. From Se = wG, with e = 0.733, S = 0.2x2.65/0.733 = 72.3%. (Key point: Degree of saturation)

3. How is the submerged unit weight obtained?
   A. Find gamma_sat at the same void ratio, then subtract gamma_w: 9.34 kN/m3
   B. Use the bulk unit weight 18 minus 9.81 = 8.19 kN/m3
   C. Use the dry unit weight 15 minus 9.81 = 5.19 kN/m3
   D. Multiply gamma_sat by S
   Answer: A. gamma' = gamma_sat - gamma_w with gamma_sat = (G+e)gamma_w/(1+e) = 19.15. (Key point: Submerged unit weight method)

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